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Solutions_Manual_Continuum_Mechanics_Lai

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Worked solutions to end-of-chapter problems from Lai, Rubin and Krempl, Introduction to Continuum Mechanics (Elsevier, 2010). The opening section covers Chapter 2 on indicial notation: summation convention, Kronecker delta, permutation symbol identities, matrix and index forms, and symmetric and antisymmetric tensors. Later chapters follow. It is a published reference kept in a folder of downloaded physics books, not Phil's own work.

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Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-1 CHAPTER 2, PART A 2.1 Given []102 1 012 a n d 2 303 3ij iSa⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤==⎣⎦⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ Evaluate (a) iiS, (b) ij ijSS , (c) ji jiSS , (d) jk kjSS (e)mmaa , (f) mn m nSa a , (g) nm m nSa a ------------------------------------------------------------------------------- Ans. (a) 11 22 33 113 5iiSS S S=++= + + = . (b) 22222 2 22 2 11 12 13 21 22 23 31 32 33 ij ijSS S S S S S S S S S=++++++++= 104014909 2 8++++++++= . (c) ji jiSS =ij ijSS =28. (d) 11 2 2 3 3 j kk j kk kk kkSS SS SS SS=++ 11 11 12 21 13 31 21 12 22 22 23 32 31 13 32 23 33 33SS SS SS SS SS SS SS SS SS=++++++++ () () () () () () () () ()()()()()()()()()() 1 1 00 23 00 1 1 20 32 02 33 2 3= ++++ ++++= . (e) 222 12 3 1491 4mma a aaa= + + =++= . (f) 11 2 2 3 3 mn m n n n n n n nSa a S a a S a a S a a=++= 11 1 1 12 1 2 13 1 3 21 2 1 22 2 2 23 2 3 31 3 1 32 3 2 33 3 3S a aS a a S a aS a aS a a S a aS a aS a a S a a++++++++ () () () () () () () () () ()()()()()()()()()()()( ) () () () () () ()111 012 213 021 122 223 331 032 333 106041 2902 7 5 9 .=+ + + + + + + + =+++++ +++ = (g) nm m nSa a =mn m nSa a =59. __________________________________________________________________ 2.2 Determine which of these equations have an identical meaning with 'j ii jaQ a= . (a)'m pp maQ a= , (b) 'q pq paQ a= , (c) 'n mm naa Q= . ------------------------------------------------------------------------------- Ans. (a) and (c) __________________________________________________________________ 2.3 Given the following matrices []12 3 0 0, 0 5 1 20 2 1ii jaB⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤==⎣⎦⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ Demonstrate the equivalence of the subscripted equations and corresponding matrix equations in the following two problems. (a) [] [] [] and ii j jbB a b B a== , (b) [][] []T and ij i j sBa a s a B a== ------------------------------------------------------------------------------- Ans. (a) ()()()()()() 1 1 11 1 12 2 13 3 21 30 02 2ii j j j jbB a bB a B aB a B a=→ ==++= + + = 2 2 21 1 22 2 23 3 3 3 31 1 32 2 33 3 2, 2jj jj bB aB a B aB a bB aB a B aB a== ++=== ++= . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-2 [] [] []2301 2 0510 2 0212 2bB a⎡ ⎤ ⎡⎤ ⎡⎤ ⎢ ⎥ ⎢⎥ ⎢⎥== =⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥⎣ ⎦ ⎣⎦ ⎣⎦. Thus, [] [] [] gives the same results as ii j jbB a b B a== (b) () () () () () () () () ()11 1 1 12 1 2 13 1 3 21 2 1 22 2 2 23 2 3 31 3 1 32 3 2 33 3 3 2( 1 ) ( 1 ) 3( 1 ) ( 0 ) 0( 1 ) ( 2 ) 0( 0 ) ( 1 ) 5 (0)(0) 1 (0)(2) 0 (2)(1) 2 (2)(0) 1 (2)(2) 2 4 6.ij i j sB aa B aa B aa B aa B a a B a a B a a Ba a Ba a Ba a==+++ ++ + ++ + = + + + +++++= + = and [][] [] [ ] [ ]T2301 2 102051 0 1022 24 6 0212 2saB a⎡ ⎤ ⎡⎤ ⎡⎤ ⎢ ⎥ ⎢⎥ ⎢⎥== = = + =⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥⎣ ⎦ ⎣⎦ ⎣⎦. __________________________________________________________________ 2.4 Write in indicial notation the matrix equation (a) [][][] AB C= , (b) [][][]TD BC= and (c) [][][] []TE BC F= . ------------------------------------------------------------------------------- Ans. (a) [][] [] ij i m m j AB C A B C=→ = , (b)[][][]T ij mi mj DB C A B C=→ = . (c) [][][] []T ij mi mk kj E BC F E B C F=→ = . __________________________________________________________________ 2.5 Write in indicial notation the equation (a) 222 12 3 sAAA=++ and (b) 222 222 12 30 xxxφφφ∂∂∂++= ∂∂∂. ------------------------------------------------------------------------------- Ans. (a) 222 12 3 ii sAAAA A=++= . (b) 222 2 222 12 300 iixx xxxφφφ φ∂∂∂ ∂++= → =∂∂ ∂∂∂. __________________________________________________________________ 2.6 Given that =ij i jSa a and =ij i jSa a′′ ′ , where =im i maQ a′ and =j nj n aQ a′ , and ik jk ijQQδ= . Show that =ii iiSS′ . ------------------------------------------------------------------------------- Ans. == = = =i j m im n jn m i n jmn i i m i n imn m nmn mm m m i iS Q aQ a Q Q aa S Q Qaa aa aa S S δ ′′ →= = . __________________________________________________________________ 2.7 Write ii ij jvvavtx∂∂=+∂∂ in long form. ------------------------------------------------------------------------------- Ans. 11 1 111 11 2 3 1231j jvv v vvvia v vv vtx t xx x∂∂ ∂∂∂∂=→ = + = + + +∂∂ ∂∂∂ ∂. 22 2 222 21 2 3 1232j jvv v vvvia v vvvtx txxx∂∂ ∂∂∂∂= → =+ =+ + +∂∂ ∂∂∂∂. 33 3 333 31 2 3 1233j jvv v vvvia v vv vtx txxx∂∂ ∂∂∂∂=→ = + = + + +∂∂ ∂∂∂∂. __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-3 2.8 Given that 2ij ij kk ijTE Eμλδ=+ , show that (a) ()22ij ij ij ij kkTE EE Eμλ=+ and (b) ()2 224( 4 3 )ij ij ij ij kkTT E E Eμ μλ λ =+ + ------------------------------------------------------------------------------- Ans. (a) 2(2 ) 2 2 2 ( )ij ij ij kk ij ij ij ij kk ij ij ij ij kk ii ij ij kkT E E E EE E E EE E E EE E Eμλ δ μ λ δ μ λ μ λ= + =+ =+ =+ (b) () () ()2 22 22 2 2 22(2 )(2 ) 4 2 2 42 2 4( 4 3 ) .ij ij ij kk ij ij kk ij ij ij ij kk ij kk ij ij kk ij ij ij ij ii kk kk ii kk ii ij ij kkTT E E E E E E E E E E EE E E E E E E EE Eμλ δ μλ δ μ μ λ δμ λ δ λ δ δ μ μλ μλ λ δ μμ λ λ=+ + = + + += + + + =+ + __________________________________________________________________ 2.9 Given that =ii j jaT b , and =ii j jaT b′′′, where =ii m maQ a′and =ij im jn mnTQ Q T ′. (a) Show that im mn n im jn mn jQT b QQT b′′ ′= and (b) if =ik im kmQQδ, then ( ) 0kn n jn jTb Q b′′−=. ------------------------------------------------------------------------------- Ans. (a) Since =ii m maQ a′ and =ij im jn mnTQ Q T ′, therefore, =ii j jaT b→. im m im jn mn jQa QQT b′′= (1), Now, = =ii j j mm j j m n naT b a T b Tb′′′ ′ ′ ′ ′ ′→= , therefore, Eq. (1) becomes im mn n im jn mn jQT b QQT b′′ ′= . (2) (b) To remove imQfrom Eq. (2), we make use of =ik im kmQQδby multiplying the above equation, Eq.(2) withikQ. That is, i k i m m n n i k i m j n m nj k m m n n k m j n m nj k n n j n k njQ Q Tb Q Q Q Tb Tb Q Tb Tb Q Tb δ δ ′′ ′ ′′ ′ ′′ ′=→ = → = () 0kn n j n jTb Q b′′→−= . __________________________________________________________________ 2.10 Given [] []10 2 a n d 2 03iiab⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥==⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ Evaluate [ ]id, if ki j k i jda bε= and show that this result is the same as ()k kd=×⋅abe . ------------------------------------------------------------------------------- Ans. ki j k i jda bε=→ 1 1 2 3 123 3 2 132 23 32 2 2 312 3 1 132 1 3 3 1 1 3 3 3 123 1 2 213 2 1 1 2 2 1(2)(3) (0)(2) 6 (0)(0) (1)(3) 3 (1)(2) (2)(0) 2ij i j ij i j ij i jda b a b a b a b a b da b a ba b a b a bda b a b a b a b a bεεε εεε εεε= =+= − = − = ==+= − = − = − = = + =−= − = Next, () ( ) () 12 23 1 2322 3 6 3 2== − + ab e + e e + e e e e×× . () () () 11 2 2 33 6, 3, 2dd d=⋅ = =⋅ = −=⋅ =abe abe abe×× × . __________________________________________________________________ 2.11 (a) If 0ijk ijTε=, show that ij jiTT= , and (b) show that ij ijkδε=0 ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-4 Ans. (a) 1 231 23 321 32 23 32 23 32 1, 0 0ij ij for k T T T T T T T ε εε == → + = → − → = . 2 312 31 132 13 31 13 31 13 2, 0 0ij ij for k T T T T T T T ε εε == → + = → − → = . 3 123 12 213 21 12 21 12 21 3, 0 0ij ij for k T T T T T T T ε εε == → + = → − → = . (b) ()()()()()() 11 11 22 22 33 33 10 10 10 0ij ijk k k kδε δε δ ε δε=++= + + = . __________________________________________________________________ 2.12 Verify the following equation:ijm klm ik jl il jkεεδ δ δ δ=− . (Hint): there are 6 cases to be considered (i) ij=, (2) ik=, (3) il=, (4) jk=, (5) jl=, and (6) kl=. ------------------------------------------------------------------------------- Ans. There are 4 free indices in the equation. Theref ore, there are the following 6 cases to consider: (i) ij=, (2) ik=, (3) il=, (4) jk=, (5) jl=, and (6) kl=. We consider each case below where we use LS for left side, RS for right side and repeated indices with parenthesis are not sum: (1) For () () () () () () , L S= 0 , 0.i im k l m ik il il ik ij R S εεδ δ δ δ == = − = (2) For ik=, () 1 ()1 () 2 ()2 () 3 ()3 () () () () LS= , ij i l ij i l ij i l ii j l i l j i RS εεε εε ε δ δ δ δ++ = − 0 if LS=RS = 0 if 1 if jl jli jli≠ ⎧ ⎪==⎨ ⎪=≠⎩. (3) For il=, () () () () () () LS= , i j m kim ik ji i i j k RS εεδ δ δ δ = − 0 if LS=RS = 0 if 1 if jk jki jki≠ ⎧ ⎪==⎨ ⎪−= ≠⎩ (4) For jk=, () () () () () () LS= , ij m j l m ij j l i l j j RS εεδ δ δ δ = − 0 if LS=RS = 0 if 1 if il il j il j≠ ⎧ ⎪==⎨ ⎪−= ≠⎩ (5) For jl=, () () () () () () LS= , ij mkj m i k j j ij j k RS εεδ δ δ δ = − 0 if LS=RS = 0 if 1 if ik ik j ik j≠ ⎧ ⎪==⎨ ⎪=≠⎩ (6) For kl=, () () () () () () LS= =0, 0i j m k km ik jk ik jk RS εεδ δ δ δ = −= __________________________________________________________________ 2.13 Use the identity ijm klm ik jl il jkεεδ δ δ δ=− as a short cut to obtain the following results: () 2ilm jlm i jaεεδ= and (b) 6ijk ijkεε=. ------------------------------------------------------------------------------- Ans. (a) 3 2i l m j l m ij l l i l l j ij ij ijεεδ δ δ δδ δδ=−= − = . (b) (3)(3) 9 3 6ijk ijk ii jj i j ji iiεεδ δ δ δ δ=−=− = − = . __________________________________________________________________ 2.14 Use the identity ijm klm ik jl il jkεεδ δ δ δ=− to show that ( ) ( ) ( ) ××⋅ − ⋅ ab c = a c ba b c . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-5 ------------------------------------------------------------------------------- Ans. ( ) ( ) = ( )mm i j k jki i j k m jk m iab c a b cε ε ab c =e e e e×× × × =( ) =ijk m j k nmi n ijk nmi m j k n jki nmi m j k nab c ab c ab cε εε ε ε ε = ee e ()jn km jm kn m j k n jn km m j k n jm kn m j k n ab c ab c ab c δδδ δ δ δ δ δ=− = − eee ()( )knkn j jnnabc abc=−= ⋅ − ⋅ ee a c b a b c . __________________________________________________________________ 2.15 (a) Show that if ij jiTT=− , 0ij i jTa a= and (b) if ij jiTT=−, and ij jiSS= , then 0ij ijTS= ------------------------------------------------------------------------------- Ans. Since ij i j ji j iTa a T aa= (switching the original dummy index to ij and the original index to ji ), therefore 20 0ij i j ji j i ij j i ij i j ij i j ij i jTa a T aa Taa Ta a Ta a Ta a== − = − → = →= . (b) ij ij ji jiTS T S= (switching the original dummy index to ij and the original index to ji ), therefore, 2 0 0ij ij ji ji ij ji ij ij ij ij ij ijTS T S TS TS TS TS== − = − →= → = . __________________________________________________________________ 2.16 Let () () /2 and / 2ij ij ji ij ij jiTS S RS S=+ =− , show that , ij ji ij jiTTR R==− , and Rij ij ijST=+ . ------------------------------------------------------------------------------- Ans. () () /2 /2ij ij ji ji ji ij ijTS S T S S T=+ →=+ = . () () () /2 /2 /2ij ij ji ji ji ij ij ji ijR SS R S S SS R=− →=− = −− = − . () () += / 2 + / 2ij ij ij ji ij ji ijTR S S S S S +− = . __________________________________________________________________ 2.17 Let 123(, , )fxx x be a function of 12 3,, a n d xxx and 123(, , )ivxx x be three functions of 12 3,, a n d xxx . Express the total differential and i df dv in indicial notation. ------------------------------------------------------------------------------- Ans. 123 12 3i iff ffdf dx dx dx dxxx x x∂∂∂∂=++=∂∂∂∂. 123 12 3ii i i im mvv v vdv dx dx dx dxxx x x∂∂∂∂=++=∂∂ ∂∂. __________________________________________________________________ 2.18 Let ijAdenote that determinant of the matrix ijA⎡⎤⎣⎦. Show that 123 ij ijk i j kAA A Aε= ------------------------------------------------------------------------------- Ans. 123 1 1 123 2 2 123 3 3 123 ijk i j k jk j k jk j k jk j kA AA A AA AAA AAAε εεε=++ 123 11 22 33 132 11 32 23 231 21 32 13 213 21 12 33 312 31 12 23 321 31 22 13 11 22 33 11 32 23 21 32 13 21 12 33 31 12 23 31 22 13 11 12 13 21 22 23 31 32 33A A AA A AA A AA A AA A AA A A AAA AAA AAA AAA AAA AAA AAA AAA AAAεεεεεε=+++++ =−+−+− = __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-6 CHAPTER 2, PART B 2.19 A transformation Toperate on any vector ato give Ta = a / a , where a is the magnitude of a. Show that Tis not a linear transformation. ------------------------------------------------------------------------------- Ans. Since aTa =afor any a, therefore ()a+bTa + b =a+b. Now =+abTa + Tbab therefore ( ) ≠ Ta + b T a + T b and Tis not a linear transformation. _________________________________________________________________ 2.20 (a) A tensor Ttransforms every vector ainto a vector Ta = m a× where m is a specified vector. Show that Tis a linear transformation and (b) If 12+ m=e e , find the matrix of the tensor T. ------------------------------------------------------------------------------- Ans. (a) ( ) ( )αβα β α β α β α β== T a +b m a +b m a + m b =ma +mb =T a +T b . ×× × × × Thus, the given Tis a linear transformation. (b) 11 1 2 1 3 ()=+ = − Te = m e e e e e×× , 22 1 2 2 3 ()=+= Te = m e e e e e× × , 33 1 2 3 2 1 ()=+ = − + Te = m e e e e e e×× . Thus, []001 00 1 11 0⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥−⎣⎦T . _________________________________________________________________ 2.21 A tensor Ttransforms the base vectors 12and ee such that 112 212 and − Te = e + e Te = e e . If 12 122 3 and 3 2a= e + e b= e + e , use the linear property of Tto find (a) Ta,(b) Tb, and (c) ()Ta + b . ------------------------------------------------------------------------------ Ans. ()() 1 2 1 2 12 12 12 (a) (2 3 ) 2 3 2 3 5 == − = − T a = T e+e T e+T e e+ e + e e e e . ()() 12 1 21 2 1 2 1 2 (b) (3 2 ) 3 2 =3 2 =5 =− Tb = T e + e Te + Te e + e + e e e + e . () () 12 12 1 (c) ( ) = 5 5 10 −+ = Ta + b T a + T b= e e e + e e . _________________________________________________________________ 2.22 Obtain the matrix for the tensor Twhich transforms the base vectors as follows: 11 3 2 23 3 1223 , 3 − Te = e + e , Te = e + e Te = e + e . ------------------------------------------------------------------------------ Ans. []20 1 01 3 13 0−⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T . _________________________________________________________________ 2.23 Find the matrix of the tensor Twhich transforms any vector ainto a vector ( ) b=ma n⋅ where ()() ()() 12 13 2/2 a n d 2/2 − m= e +e n= e +e . ------------------------------------------------------------------------------ Ans. () ()()() () 11 1 1 2 1 2 2/2 2/2 /2 n ⎡⎤ =− = −⎣⎦Te = m e n m = e + e e + e ⋅ . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-7 () 22 2 0 n= Te = m e n m = m = 0 ⋅ . () ()()() () 33 3 1 2 1 2 2/2 2/2 /2 n ⎡⎤ ==⎣⎦Te = m e n m = e + e e + e ⋅ . Thus, []1/2 0 1/2 1/2 0 1/2 00 0−⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T . _________________________________________________________________ 2.24 (a) A tensor Ttransforms every vector into its mirror image with respect to the plane whose normal is2e. Find the matrix of T. (b) Do part (a) if the plane has a normal in the 3e direction. ------------------------------------------------------------------------------ Ans. (a) 11 2 2 3 3 , , == − =Te e Te e Te e , thus, []100 01 0 001⎡ ⎤ ⎢ ⎥=−⎢ ⎥ ⎢ ⎥⎣ ⎦T . (b) 11 2 2 3 3 , , == = −Te e Te e Te e , thus, []10 0 01 0 00 1⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥−⎣ ⎦T . _________________________________________________________________ 2.25 (a) Let Rcorrespond to a right-hand rotation of angle θ about the 1x-axis. Find the matrix of R. (b) do part (a) if the rotation is about the 2x-axis. The coordinates are right-handed. ------------------------------------------------------------------------------ Ans.(a) 11 2 1 2 3 3 1 2 3 , 0 c o s s in , 0 s in c os θθθ θ == = −Re e Re e + e + e Re e e + e . Thus, []10 0 0c o s s i n 0s i n c o sθθ θθ⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦R . (b) 13 1 2 2 3 3 1 sin cos , , cos sinθθθ θ =− = =R e e+ e R e e R e e+ e . Thus, []cos 0 sin 01 0 sin 0 cosθ θ θ θ⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦R . _________________________________________________________________ 2.26 Consider a plane of reflection which passes through the origin. Let nbe a unit normal vector to the plane and let rbe the position vector for a point in space. (a) Show that the reflected vector for ris given by 2( ) =−Tr r r n n ⋅ , where Tis the transformation that corresponds to the reflection. (b) Let 123() / 3n= e +e +e , find the matrix of T. (c) Use this linear transformation to find the mirror image of the vector 12323 a=e + e + e . ------------------------------------------------------------------------------ Ans. (a) Let the vector r be decomposed into two vectors and ntrr , where nr is in the direction of n and tris in a direction perpendicular to n. That is, nris normal to the plane of reflection and tris on the plane of reflection and tn=+rr r . In the reflection given by T, we have, and nn t t=− =Tr r Tr r , so that () 22 ( )tn t n n n n+ =−=− −= − = − Tr = Tr Tr r r r r r r r r r n n ⋅ . (b) 123 1 2 3() / 3 1 / 3 → n= e +e +e e n=e n=e n= ⋅⋅⋅ . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-8 () () 11 1 1 123 1 2 3 2( ) 2 1/ 3 ( ) / 3 2 2 / 3 ⎡⎤ =− =− = − −⎣⎦Te e e n n e e + e + e e e e ⋅ . () () 22 2 2 1 23 1 2 3 2( ) 2 1/ 3 ( ) / 3 2 2 / 3 ⎡⎤ =− − = −+−⎣⎦Te e e n n = e e + e + e e e e ⋅ . () () 33 3 3 123 1 23 2( ) 2 1/ 3 ( ) / 3 2 2 / 3 ⎡⎤ =− − = −− +⎣⎦Te e e n n = e e + e + e e e e ⋅ . []12 2 121 2322 1−−⎡⎤ ⎢⎥=− −⎢⎥ ⎢⎥−−⎣⎦T . (c) [] [] ()12 312 2 1 3 121 2 2 2 3 2322 1 3 1−− −⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− − = −→ − + +⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥−− −⎣⎦ ⎣ ⎦ ⎣ ⎦Ta T a = e e e . _________________________________________________________________ 2.27 Knowing that the reflected vector for ris given by 2( ) =− Tr r r n n ⋅ (see the previous problem), where Tis the transformation that corresponds to the reflection and nis the normal to the mirror, show that in dyadic notation, the reflection tensor is given by 2 − T=I n n and find the matrix of Tif the normal of the mirror is given by 123() / 3n= e +e +e , ------------------------------------------------------------------------------ Ans. From the definition of dyadic product, we have , 2 ( ) 2 () ( 2 () ) ( 2) 2=− − − = − → −Tr r r n n = r nn r = Ir nn r I nn r T = I nn ⋅ . For [] 12311 1 1 22() / 3 [ 2 ] 1 1 1 1 1 1 13311 1 1⎡⎤⎡ ⎤ ⎢⎥⎢ ⎥→= =⎢⎥⎢ ⎥ ⎢⎥⎢ ⎥⎣⎦⎣ ⎦n= e +e +e n n . 12 2 1[] [ ][ 2 ] 2 1 2322 1−−⎡⎤ ⎢⎥→=− = − −⎢⎥ ⎢⎥−−⎣⎦TI n n . _________________________________________________________________ 2.28 A rotation tensor Ris defined by the relation 12 23 31 , , = == Re e Re e Re e (a) Find the matrix of Rand verify that TRR = I and det 1R= and (b) find a unit vector in the direction of the axis of rotation that could have been us ed to effect this particular rotation. ------------------------------------------------------------------------------ Ans. (a) [] [][]T001 010001 100 100 001100 010 010 100010 001⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥→= =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦R= R R , []001 det 1 0 0 1 010= R= . (b) Let the axis of rotation be 11 2 2 33ααα++ n= e e e , then [] [ ] [ ]→− → Rn = n R I n = 01 21 3 1 2 2 3 310 1 0 1 1 0 0 0, 0, =0 01 1 0α αα α α α α α α− ⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥−= → − + = − = −⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣⎦. Thus, 123ααα== , so that a unit vector in the dir ection of the axis of rotation is 123() / 3++ n= e e e . _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-9 2.29 A rigid body undergoes a right hand rotation of angle θabout an axis which is in the direction of the unit vector m. Let the origin of the coordinates be on the axis of rotation and rbe the position vector for a typical point in the body. (a) show that the rotated vector of ris given by: () ( ) () = 1 cos + cos sinθθ θ−+ Rr m r m r m r ⋅× , where Ris the rotation tensor. (b) Let 123() / 3m= e +e +e , find the matrix for R. ------------------------------------------------------------------------------ Ans. (a) Let the vector r be decomposed into two vectors and mprr , where mr is in the direction of mand pris in a direction perpendicular to m, that is, p m=+ rr r . Let /p p≡pr r be the unit vector in the direction of pr, and let ≡qm p×. Then, ( m,,p q ) forms an orthonormal set of vectors which rotates an angle of θabout the unit vector m. Thus, mm= Rr r and ( ) cos sinpp θθ =+ Rr r p q . From p m=+ rr r , we have, () () { } () {} () () () {} () () ()cos sin cos sin cos sin cos sin cos 1 cos sin cos 1 cos sinp mp m p p m pp m m m m mm mθθ θ θ θθ θ θ θθ θ θθ θ=+= + + = + + =+ + =− + − + =+ −+ − =+ −+Rr Rr Rr r p q r r p r m p r rm r r r r m r r r rr m r r rr m r × ×× ×× We note that ()m=rr m m⋅ , so that () cos ( ) 1 cos sinθθ θ=+ −+Rr r r m m m r ⋅ ×. (b) Use the result of (a), that is, () cos ( 1 cos sinθθ θ=+ −+Rr r r m) m r ⋅ ×, we have, () 11 1 1 cos ( ) 1 cos sinθθ θ=+ ⋅− +Re e e m m m e ×, () 22 2 2 cos ( ) 1 cos sinθθ θ=+ ⋅ − + Re e e m m m e ×, () 33 3 3 cos ( ) 1 cos sinθθ θ=+ ⋅− +Re e e m m m e ×. Now, 123() / 3m= e +e +e , therefore, 123 =1 / 3= me= me me⋅⋅⋅ () () () 13 2 2 3 1 32 11/ 3 ( ), 1/ 3 ( ), 1/ 3 ( )−− − me = e + e me= e e me = e + e×× × . Thus, () () ( ) () () { } () ( ) () {} () ( ) () {}11 1 1 11 2 3 3 2 12 3cos ( ) 1 cos sin cos 1/ 3 ( ) 1 cos sin 1/ 3 ( ) 1/3 1 2cos 1/3 1 cos sin 1/ 3 1/3 1 cos sin 1/ 3θθ θ θθ θ θθ θ θ θ=+ ⋅− + +− + − =+ + − + + − −Re e e m m m e =e e +e +e e +e ee e× () () ( ) () () ( ) () {} () ( ) () ( ) () {}22 2 2 21 2 3 3 1 12 3cos ( ) 1 cos sin cos 1/ 3 ( ) 1 cos sin 1/ 3 ( ) 1/3 1 c o s 1/ 3 s i n 1/3 1 2c o s 1/3 1 c o s s i n 1/ 3θθ θ θθ θ θθ θ θ θ=+ ⋅ − + =+ − + =− − + + +− +Re e e m m m e e e +e +e e -e ee e× () () ( ) () () ( ) () {} () ( ) () {} () ( )33 3 3 31 2 3 2 1 12 3cos ( ) 1 cos sin cos 1/ 3 ( ) 1 cos sin 1/ 3 ( ) 1/3 1 cos 1/ 3 sin 1/3 1 cos sin 1/ 3 1/3 1 2cosθθ θ θθ θ θθ θ θ θ=+ ⋅− + =+ − + =− + +− − + +Re e e m m m e e e +e +e - e +e ee e× Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-10 []() () () () ( ) () () ()1 2cos 1 cos 3 sin 1 cos 3 sin 11 cos 3 sin 1 2cos 1 cos 3 sin3 1c o s 3 s i n 1c o s 3 s i n 12 c o sθ θθθθ θ θθθ θ θθθθ θ⎡⎤+− − − + ⎢⎥ =− + + − −⎢⎥ ⎢⎥−− −+ +⎢⎥⎣⎦T . _________________________________________________________________ 2.30 For the rotation about an arbitrary axis mby an angle θ, (a) show that the rotation tensor is given by (1 cos )( ) cos sin θ θθ −+ R= m m I+ E , where mmdenotes that dyadic product of m and Eis the antisymmetric tensor whose dual vector (or axial vector) is m, (b) find the AR, the antisymmetric part of Rand (c) show that the dual vector for ARis given by (sin ) θm. Hint, () ( ) () = 1 cos + cos sinθθ θ−+ Rr m r m r m r ⋅× (see previous problem). ------------------------------------------------------------------------------ Ans. (a) We have, from the previous problem, ()()( ) = 1 cos + cos sinθθ θ−+ Rr m r m r m r ⋅ ×. Now, by the definition of dyadic product, we have () () mrm =m m r⋅ , and by the definition of dual vector we have, mr = E r× , thus () =1 c o s ( ) + c o s s i nθ θθ −+ Rr mm r r Er (){ } 1c o s ( ) + c o s s i nθθ θ−+ =m m I E r , from which, () 1c o s ( ) + c o s s i nθ θθ −+ R= m m I E . (b) AT() / 2=− →RR R (){ }(){ }AT T2 1 cos ( ) + cos sin 1 cos ( ) + cos sin θθ θθ θ θ−+ − − + R = mm I E mm I E . Now [] []T ij j imm m m⎡⎤ ⎡⎤===⎣⎦ ⎣⎦mm mm , and the tensor E, being antisymmetric, T−E= E , therefore, A22 s i n θ R= E , that is, Asinθ R= E . (c) dual vector of A(sin )(dual vector of ) sinθ θ ==RE m . _________________________________________________________________ 2.31 (a) Given a mirror whose normal is in the direction of 2e. Find the matrix of the tensor S which first transforms every vector into its mirror image and then transforms them by a o45 right- hand rotation about the 1e-axis. (b) Find the matrix of the tensor T which first transforms every vector by a o45 right-hand rotation about the 1e-axis, and then transforms them by a reflection with respect to the mirror (whose normal is 2e). (c) Consider the vector 123(23 )a= e + e + e , find the transformed vector by using the transformation S. (d) For the same vector 123(23 )a= e + e + e , find the transformed vector by using the transformation T. ------------------------------------------------------------------------------ Ans. Let 12 and TT correspond to the reflection and the rotation respectively. We have [] 11 1 12 2 13 3 1100 =, , 0 1 0 001⎡ ⎤ ⎢ ⎥=− = → = −⎢ ⎥ ⎢ ⎥⎣ ⎦Te e Te e Te e T . () () [] 21 1 22 2 3 23 2 3 210 0 11=, , 0 1 /2 1 /2 22 01 /2 1 /2⎡ ⎤ ⎢ ⎥== − → = − ⎢ ⎥ ⎢ ⎥⎣ ⎦Te e Te e +e Te e +e T . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-11 (a) [] [ ] [ ] 2110 0 1 0 0 100 01 /2 1 /2 0 10 0 1 /2 1 /2 001 0 1/ 2 1/ 2 0 1/ 2 1/ 2⎡⎤ ⎡ ⎤ ⎡⎤⎢⎥ ⎢ ⎥ ⎢⎥== − − = − − ⎢⎥ ⎢ ⎥ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ − ⎣⎦ ⎣⎦ ⎣ ⎦ST T . (b) [][] [ ] 1210 0 1 0 0 100 0 1001 /2 1 /2 0 1 /21 /2001 0 1/ 2 1/ 2 0 1/ 2 1/ 2⎡⎤ ⎡⎤ ⎡⎤⎢⎥ ⎢⎥ ⎢⎥== − − = − ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦TT T . (c) [] [] []10 0 1 1 01 / 21 / 2 2 5 / 2 3 0 1 /2 1 /2 1 /2⎡⎤ ⎡ ⎤ ⎡⎤⎢⎥ ⎢ ⎥ ⎢⎥== − − = − ⎢⎥ ⎢ ⎥ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ − ⎣⎦ ⎣⎦ ⎣ ⎦bS a . (d) [] [] []10 0 1 1 01 / 2 1 / 2 2 1 / 2 3 0 1 /2 1 /2 5 /2⎡⎤ ⎡ ⎤ ⎡⎤⎢⎥ ⎢ ⎥ ⎢⎥== − = ⎢⎥ ⎢ ⎥ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣ ⎦cT a . _________________________________________________________________ 2.32 Let Rcorrespond to a right-hand rotation of angle θ about the 3x-axis (a) find the matrix of 2R. (b) Show that2Rcorresponds to a rotation of angle 2θabout the same axis (c) Find the matrix of nR for any integer n. ------------------------------------------------------------------------------- Ans. (a) []cos sin 0 sin cos 0 00 1θθ θθ−⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦R . 22 2 22cos sin 2sin cos 0 cos sin 0 cos sin 0 sin cos 0 sin cos 0 2sin cos cos sin 0 00 1 00 1 0 0 1θθ θ θ θθ θθ θθ θθ θ θ θθ⎡ ⎤ −− −−⎡⎤ ⎡⎤⎢ ⎥⎢⎥ ⎢⎥⎡⎤→= = − ⎢ ⎥⎢⎥ ⎢⎥⎣⎦⎢ ⎥⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎢ ⎥⎣ ⎦R . (b) 22 22 2cos sin 2sin cos 0 cos 2 sin 2 0 2sin cos cos sin 0 sin 2 cos 2 0 00 1 0 0 1θθ θ θ θθ θθ θ θ θ θ⎡⎤−− −⎡ ⎤⎢⎥⎢ ⎥⎡⎤=− =⎢⎥⎢ ⎥ ⎣⎦⎢⎥⎢ ⎥⎣ ⎦ ⎢⎥⎣⎦R . Thus, 2Rcorresponds to a rotation of angle 2θabout the same axis (c) cos sin 0 sin cos 0 00 1nnn nnθθ θθ−⎡⎤ ⎢⎥⎡⎤=⎢⎥⎣⎦ ⎢⎥⎣⎦R . _________________________________________________________________ 2.33 Rigid body rotations that are small can be described by an orthogonal transformation*ε=+RI R where 0ε→as the rotation angle approaches zero. Consider two successive small rotations 1Rand 2R, show that the final result does not depend on the order of rotations. ------------------------------------------------------------------------------ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-12 Ans. () () ()** * * 2 * * * * 2 * * 21 2 1 2 1 21 2 1 21 εε ε ε ε ε ε=+ + = + + + = + + + R R IRIR IR R R RI RR R R . As 0ε→, ()** 21 2 1 12ε≈+ + = RR I R R R R . _________________________________________________________________ 2.34 Let TandSbe any two tensors. Show that (a) TTis a tensor, (b) TT T(= T+ S T + S ) and (c) TT T()= TS S T . ------------------------------------------------------------------------------- Ans. Let a,b,c be three arbitrary vectors and αβbe any two scalars, then (a) TT T() ()αβ αβ α β α β== aT b + c b + c T a bT a + cT a = aTb + aTc⋅⋅ ⋅ ⋅ ⋅ ⋅ () ( )TTT TT() = αβ α β αβ → =a T b+ T c T b+ c T b+ T c⋅ . Thus, TTis a linear transformation, i.e., tensor. (b) TT T() () = a T+S b b T+S a=b T a+b S a=a T b+a S b⋅⋅ ⋅ ⋅ ⋅ ⋅ TT T TT() ( ) →= =a T +S b T+S T +S⋅ . (c) TT T T T T T() () ( ) ( ) ()=→ = a T S b b T S a=b T S a = S a T b=a S T b T S S T⋅⋅ ⋅ ⋅ ⋅ . _________________________________________________________________ 2.35 For arbitrary tensors Tand S, without relying on the component form, prove that (a) 1T T 1()( )−−= TT and (b) 11 1()−− −= TS S T ------------------------------------------------------------------------------- Ans. (a) 11 T 1 T T 1 T T 1( ) () ()( )−− − − −=→ =→ =→ = T TIT T ITT IT T . (b) 11 1 1 1() ( ) ( )−− − − −== = TS S T T SS T TT I , thus, 11 1()−−−= TS S T . _________________________________________________________________ 2.36 Let {}{} andii′ ee be two Rectangular Cartesian base vectors. (a) Show that if im i mQ′=ee , then ii m mQ′=ee and (b) verify mi mj ij im jmQQ QQδ== . ------------------------------------------------------------------------------- Ans. (a) i mi m i j mi m j mi mj ji j jm m i im mQQ Q Q Q Q δ ′′ ′′=→ = = = → =→ =ee e e e e e e ee ⋅⋅ . (b) We have, ij i j ijδ′′== ee ee⋅⋅ , thus, ij i j mi m nj n mi nj m n mi nj mn mi mj Q Q QQ QQ QQ δ δ ′′= === =ee e e e e⋅⋅ ⋅ . And mn ij i j im m jn n im jn m n im jn im jm Q Q QQ QQ QQ δ δ == = = =ee e e e e⋅⋅ ⋅ . _________________________________________________________________ 2.37 The basis {}i′eis obtained by a o30 counterclockwise rotation of the {}iebasis about the 3e axis. (a) Find the transformation matrix []Qrelating the two sets of basis, (b) by using the vector transformation law, find the components of 123+ a= e e in the primed basis, i.e., find ia′ and (c) do part (b) geometrically. ------------------------------------------------------------------------------ Ans. (a) oo o o 11 2 2 12 3 3cos30 sin30 , sin30 cos30 , ′′ ′=+ = −+ =ee e e ee e e . Thus, [] ioo oocos30 sin 30 0 sin 30 cos30 0 00 1⎡⎤ −⎢⎥ =⎢⎥ ⎢⎥⎢⎥⎣⎦eQ . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-13 (b) [] [ ][]''1 T'21 '33/2 1 /2 0 32 1/2 3 /2 0 1 0 2 00 1 0 0iia a a⎡⎤⎡⎤ ⎡⎤⎡⎤⎢⎥⎢⎥ ⎢⎥⎢⎥′ =→ = − = → ⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎣⎦ ⎢⎥⎢⎥ ⎣⎦ ⎣⎦⎣⎦eeaQ a a = e (c) Clearly 123+ a= e e is a vector in the same direction as 1′eand has a length of 2. See figure below _________________________________________________________________ 2.38 Do the previous problem with the {}i′ebasis obtained by a o30clockwise rotation of the {}iebasis about the 3eaxis. ------------------------------------------------------------------------------- Ans. (a) oo o o 11 2 212 3 3cos30 sin30 , sin 30 cos30 , ′′′=− =+ =ee e eee e e . Thus, [] ioo oocos30 sin 30 0 sin 30 cos30 0 00 1⎡⎤ ⎢⎥ =−⎢⎥ ⎢⎥⎢⎥⎣⎦eQ . (b) [] [ ][]''1 T'21 2 '33/2 1 /2 0 1 3 1/2 3/2 0 1 3 3 00 1 0 0iia a a⎡⎤⎡⎤− ⎡⎤⎡⎤⎢⎥⎢⎥ ⎢⎥⎢⎥′′ =→ = = → + ⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦ ⎣⎦⎣⎦eeaQ a a = ee (c) See figure below _________________________________________________________________ 2.39 The matrix of a tensor Twith respect to the basis {}ieis Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-14 []15 5 500 50 1−⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦T Find 11 12 31, a n d TT T′′ ′ with respect to a right-handed basis {}i′ewhere 1′e is in the direction of 232−+ee and 2′e is in the direction of 1e. ------------------------------------------------------------------------------ Ans. The basis {}i′eis given by: 12 3 2 1 3 1 22 3( 2 )/ 5 , , ( 2 )/ 5′′ ′ ′ ′=− + = = = +ee e e e e e ee e × . 11 1 10 15 5 0 1 / 5 2 / 5 500 1 / 5 4 / 5 50 1 2/ 5T⎡⎤−⎡⎤⎢⎥ ⎢⎥⎡⎤ ′′′== − −= ⎢⎥ ⎣⎦ ⎢ ⎥⎢⎥ ⎢⎥−⎣⎦ ⎣⎦eT e⋅ . 12 1 215 5 1 01 / 5 2 / 5 5 0 0 0 1 5 / 5 50 1 0T−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤ ′′′== − = −⎣⎦ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥−⎣⎦ ⎣ ⎦eT e⋅ . 31 3 10 15 5 0 2/ 5 1/ 5 5 0 0 1/ 5 2/5 50 1 2/ 5T⎡⎤−⎡⎤⎢⎥ ⎢⎥⎡⎤ ′′′== − = ⎢⎥ ⎣⎦ ⎢ ⎥⎢⎥ ⎢⎥−⎣⎦ ⎣⎦eT e⋅ . _________________________________________________________________ 2.40 (a) For the tensor of the previous problem, find ijT⎡⎤′⎣⎦, i.e., []'ieTif {}i′eis obtained by a o90 right hand rotation about the 3e axis and (b) obtain iiT′ and the determinant ijT′and compare them with iiTand ijT. ------------------------------------------------------------------------------ Ans. (a) [] 122 1 3301 0 , , 1 0 0 001−⎡ ⎤ ⎢ ⎥′′ ′== −= → =⎢ ⎥ ⎢ ⎥⎣ ⎦eee e ee Q . [] [][] []T01 015 5 0 1 0 0 5 0 100 5 0 0 1 0 0 5 1 5 00 1 5 01001 0 51ijT−− − ⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ′ ⎡⎤′= = =− =−⎣⎦ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦TQ T Q (b) 11 22 33 0112 , 2 5ii ijTT T T T′′′′ ′=++= + + = = − . 11 22 33 1012 , 2 5ii ijTT T T T=++= + + = = − . _________________________________________________________________ 2.41 The dot product of two vectors iiaa= e and iibb= e is equal to iiab. Show that the dot product is a scalar invariant with respect to orthogonal transformations of coordinates. ------------------------------------------------------------------------------- Ans. From im i maQ a′= and 'im i mbQ b= , we have, 'i i mi m ni n mi ni m n mn m n m m i ia b Q aQb Q Qab ab ab a b δ ′=== = = . __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-15 2.42 If ijTare the components of a tensor (a) show that ij ijTT is a scalar invariant with respect to orthogonal transformations of coordinates, (b) evaluate ij ijTT with respect to the basis {}iefor []100 125 123 i⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦ eT , (c) find []′T, if ii′=eQ e , where []001 100 010 i⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦eQ and (d) verify for the above [] []and′TT that ij ij ij ijTT TT′′= . ------------------------------------------------------------------------------ Ans. (a) SinceijTare the components of a tensor, ij mi nj mnTQ Q T′= . Thus, () ( ) ( )ij ij mi nj mn pi qj pq mi pi nj qj mn pq mp nq mn pq mn mnT T QQ T QQ T QQ Q Q TT TT TT δδ ′′== = = (b) 22222 2 22 2 11 12 13 21 22 23 31 32 33 114 2 514 9 4 5ij ijTT T T T T T T T T T=++++++++= + + ++ + + = . (c) [] [][] []T010100001 010001 251 001125100 001251 231 100123010 100231 001⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ′== = =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦TQ T Q (d) 4 25 1 4 9 1 1 45ij ijTT′′=+ +++++= . _________________________________________________________________ 2.43 Let [] []and′TT be two matrices of the same tensor T, show that [] []det =det′TT . ------------------------------------------------------------------------------ Ans. TTdet det det det ( 1)( 1)det det ⎡⎤ ⎡⎤⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤⎣⎦ ⎣⎦⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦′′=→ = = ± ± =TQ T Q T Q Q T T T . _________________________________________________________________ 2.44 (a) If the components of a third order tensor are ijkR, show that iikRare components of a vector, (b) if the components of a fourth order tensor are ijklR, show that iiklRare components of a second order tensor and (c) what are components of ...iikR , if ...ijkR are components of a tensor of thnorder? ------------------------------------------------------------------------------- Ans. (a) Since ijkRare components of a third order tensor, therefore, ijk mi nj pk mnp iik mi ni pk mnp mn pk mnp pk nnpR QQ Q R R QQ Q R Q R Q R δ ′′=→ == = , therefore, iikRare components of a vector. (b) Consider a 4thorder tensor ijklR, we have, ijkl mi nj pk ql mnpq iikl mi ni pk ql mnpq m n pk ql m n pqp k q l n n p q R QQQ QR R QQQ QR Q QR Q QR δ ′′=→ == = , therefore, iiklRare components of a second order tensor. (c) ...iikR are components of a tensor of the ( 2)thn− order. _________________________________________________________________ 2.45 The components of an arbitrary vector aand an arbitrary second tensor Tare related by a triply subscripted quantityijkR in the manner i ijk jkaR T= for any rectangular Cartesian basis {}ie. Prove that ijkRare the components of a third-order tensor. ------------------------------------------------------------------------------ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-16 Ans. Since i ijk jkaR T= is true for any basis, therefore, i ijk jkaR T′′′= ; Since ais a vector, therefore, im i maQ a′= and since Tis a second order tensor, therefore, ij mi nj mnTQ Q T′= . Thus, ()i mi m ijk jk mi mjk jkaQ a R T QR T′′ ′=→= . Multiply the last equation with siQand noting that sim i s mQQδ= , we have, ()'' '' ''si ijk jk si mi mjk jk si ijk jk sm mjk jk si ijk jk sjk jkQR T QQ R T QR T R T QR T R T δ =→ = → = ''si ijk mj nk mn sjk jk si ijk mj nk mn smn mnQR Q Q T R T QR Q Q T R T→= →= . Thus, () 0'smn si mj nk ijk mnRQ Q Q R T−= . Since this last equation is to be true for all mnT, therefore, smn si mj nk ijkR QQ Q R ′ = , which is the transformation law for components of a third order tensor. _________________________________________________________________ 2.46 For any vector aand any tensor T, show that (a) A0 aTa =⋅ and (b) SaT a = aT a⋅⋅ , where ASand TT are antisymmetric and symmetric part of Trespectively. ------------------------------------------------------------------------------ Ans. (a) ATis antisymmetric, therefore, AT A()=− TT , thus, AA T A A A() 2 0 0 −→ → a T a=a T a= a T a a T a= a T a=⋅⋅ ⋅ ⋅ ⋅ . (b) Since SAT=T +T , therefore, SA S A S() aT a = a T+ T a = aT a + aTa = aT a⋅⋅ ⋅ ⋅ ⋅ . __________________________________________________________________ 2.47 Any tensor can be decomposed into a symmetric part and an antisymmetric part, that is SAT=T +T . Prove that the decomposition is unique. (Hin t, assume that it is not true and show contradiction). ------------------------------------------------------------------------------- Ans. Suppose that the decomposition is not unique, then ,we have, SAS A S S A A() ( ) =→ − − = T=T +T S +S T S + T S 0 . Let abe any arbitrary vector, we have, SS AA S S A A() ( ) 0 0−− = → − − = aT Sa + aT S a a T aa S a + a T aa S a⋅⋅ ⋅ ⋅ ⋅ ⋅ . But AA0 == aTa aSa⋅⋅ (see the previous problem). Therefore, S S SS SS SS0( ) 0 0 −= →− = → − = → = aT a aS a a T S a T S T S⋅⋅ ⋅ . It also follows from SS AA() ( )−− =TS + T S 0 that AA=TS . Thus , the decomposition is unique. _________________________________________________________________ 2.48 Given that a tensor Thas the matrix []123 456 789⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T , (a) find the symmetric part and the anti-symmetric part of Tand (b) find the dual vector (or axial vector) of the anti-symmetric part of T. ------------------------------------------------------------------------------- Ans. (a) [][]{}T S123 147 2 61 0 135 11 1456 258 61 01 4 35722 27 8 9 3 6 9 10 14 18 5 7 9⎧⎫⎡⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎪⎪⎢⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎡⎤ += + = = ⎨⎬⎢⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣⎦⎪⎪⎢⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎩⎭T= T T . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-17 [][]{}T A123 147 0 2 4 0 1 2 11 1456 258 2 0 2 1 0 122 2789 369 4 2 1 8 2 1 0⎧⎫ −−− − ⎡⎤ ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎪⎪⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎡⎤ −= − = − = − ⎨⎬⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎦⎪⎪⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣⎦ ⎣ ⎦ ⎣ ⎦⎩⎭T= T T . (b) AA A A 23 1 31 2 12 3 1 2 3 1 2 3() ( 1 2 1 ) 2TT T=− + + =−− + − = − +te e e e e e e e e . _________________________________________________________________ 2.49 Prove that the only possible real eigenvalues of an orthogonal tensor Qare 1λ=± . Explain the direction of the eigenvectors corresponding to them for a proper orthogonal(rotation) tensor and for an improper orthogonal (reflection) tensor. ------------------------------------------------------------------------------ Ans. Since Qis orthogonal, therefore, for any vector n, we have, Qn Qn = n n⋅⋅. Let nbe an eigenvector, then λ Qn = n , so that → Qn Qn = n n⋅⋅ 22 2() () ( 1 ) () 0 1 0 1λλ λ λ=→ − = → − = → = ± nn nn nn⋅⋅ ⋅ . The eigenvalue 1λ= (Qn = n ) corresponds to an eigenvector parallel to the axis of rotation for a proper orthogonal tensor (rotation te nsor); Or, it corresponds to an eigenvector parallel to the plane of reflection for an improper orthogonal te nsor (reflection tensor). The eigenvalue 1λ=− , (− Qn = n ) corresponds to an eigenvector perpendicular to the axis of rotation for an o180 rotation; or, it corresponds to an eigenvector perpendicular to the plane of reflection. _________________________________________________________________ 2.50 Given the improper orthogonal tensor []12 2 121 2322 1−− ⎡ ⎤ ⎢ ⎥=−−⎢ ⎥ ⎢ ⎥ −−⎣ ⎦Q . (a) Verify that []det 1=− Q . (b) Verify that the eigenvalues are 1 and 1 λ=− (c) Find the normal to the plane of reflection (i.e., eigenvectors corresponding to 1 λ=− ) and (d) find the eigenvectors corresponding 1 λ=(vectors parallel to the plane of reflection). ------------------------------------------------------------------------------- Ans. (a) []()3det 1/ 3 (1 8 8 4 4 4) ( 27) / 27 1= −−−−− =− = −Q . (b) () { }2 12 33 / 3 1, I 1 / 3 (1 4) (1 4) (1 4) 1, I 1I= = = − + − + − =− =− → 32 210 ( 1 ) ( 1 )0 1 , 1 ,1 λλλ λ λ λ−− + = →− − = → =− (c) For 1λ=− , 123 1 23 12 312 2 2 1 2 2 2 110 , 1 0 , 1 033 3 3 3 3 3 3 3ααα α αα αα α⎛⎞ ⎛⎞ ⎛⎞+−−= −+ +−= −−+ +=⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠. That is, 123 1 23 12 320 , 2 0 , 2 0ααα α αα αα α−−= −+ −= −−+ = , thus, 123ααα==, therefore, 123() / 3=±n e +e +e , this is the normal to the plane of reflection. (d) For 1λ=, 123 1 23 12 312 22 1 22 2 110 , 1 0 , 1 033 33 3 33 3 3ααα α αα αα α⎛⎞ ⎛⎞ ⎛⎞− − − = − +− − = − − +− =⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ All three equations lead to 123 3 12 0 ααα α αα++= →= − − . Thus, 11 2 2 1 2 3222 12 31[( ) ]αα α α ααα=− + ++ne + e e , e.g., 12 31(2 ) 6=−ne + e e etc. these vectors are all perpendicular to 123() / 3=±n e +e +e and thus parallel to the plane of reflection. _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-18 2.51 Given that tensors and R S have the same eigenvector nand corresponding eigenvalue 11 and rs respectively. Find an eigenvalue and the corresponding eigenvector for the tensor =TR S . ------------------------------------------------------------------------------ Ans. We have, 11andrs==Rn n Sn n , thus, 11 1 1s sr s = = Tn RSn = R n = Rn n . Thus, an eigenvalue for =TR S is 11rswith eigenvector n. _________________________________________________________________ 2.52 Show that if nis a real eigenvector of an antisymmetric tensor T, then the corresponding eigenvalue vanishes. ------------------------------------------------------------------------------ Ans. ()λ λ → Tn = n n Tn = n n ⋅⋅ . Now, from the definition of transpose, we have TnT n = nTn⋅⋅ . But, since Tis antisymmetric, i.e.,T=− TT , therefore, T− nTn = nT n⋅ ⋅. Thus, 20 0 −→ → nT n = nT n nT n = nT n =⋅⋅ ⋅ ⋅ . Thus, ( ) 0 0λ λ→= nn=⋅ . _________________________________________________________________ 2.53 (a) Show that ais an eigenvector for the dyadic product abof vectors and a b with eigenvalue ab⋅, (b) find the first principal scalar invariant of the dyadic product aband (c) show that the second and the third principal scalar invariants of the dyadic product abvanish, and that zero is a double eigenvalue of ab. ------------------------------------------------------------------------------ Ans. (a) From the definition of dyadic product, we have, ( ) ( ) = ab a a b a ⋅, thus ais an eigenvector for the dyadic product abwith eigenvalue ab⋅. (b) Let ≡Ta b , then ij i jTa b= and the first scalar invariant of abis ii i iTa b==ab⋅. (c) 22 23 1 1 13 11 12 2 32 33 31 33 21 220000ab ab a b a b ab abIab ab ab ab ab ab=++= + + = . 11 12 13 1 2 3 32 1 2 2 2 31 2 3 1 2 3 31 32 33 1 2 30ab ab ab b b b Ia b a b a ba a a b b b ab ab ab b b b=== . Thus, the characteristic equation is 32 2 11 1 1 2 3 0( ) 0 , 0 II Iλλ λλ λ λ λ−= → − = → = = = . _________________________________________________________________ 2.54 For any rotation tensor, a set of basis {}i′emay be chosen with 3′e along the axis of rotation so that 11 2 2 12 3 3c os sin , sin c os , θθθ θ ′′′ ′ ′′ ′ ′=+ = −+ =Re e e Re e e Re e , where θ is the angle of right hand rotation. (a) Find the antisymmetric part of R with respect to the basis {}i′e, i.e., find A[] i′eR . (b) Show that the dual vector of AR is given by A 3 sinθ′=te and (c) show that the first scalar invariant of R is given by 1 2cos θ+ . That is, for any given rotation tensor R, its axis of rotation and the angle of rotation can be obtained from the dual vector of ARand the first scalar invariant of R. ------------------------------------------------------------------------------ Ans. (a) From 11 2 2 12 3 3co s sin , sin co s , θθθ θ ′′′ ′ ′′ ′ ′=+ = −+ =Re e e Re e e Re e , we have, []Acos sin 0 0 sin 0 sin cos 0 sin 0 0 00 1 00 0'' ii ''iiθθ θ θθ θ−−⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥⎡⎤ =→ =⎢⎥ ⎢⎥ ⎣⎦ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ee eeRR (b) the dual vector (or axial vector) of ARis given by Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-19 A 23 1 31 2 12 3 1 2 3 3() ( 0 0 s i n ) s i nTTT θθ ′′ ′′ ′′ ′ ′ ′ ′=− + + =− + − =te e e e e e e . (c) The first scalar invariant of R is 1cos cos 1 1 2cosIθθθ++=+ = . __________________________________________________________________ 2.55 The rotation of a rigid body is described by 12 23 31 , , = == Re e Re e Re e . Find the axis of rotation and the angle of rotation. Use the result of the previous problem. ------------------------------------------------------------------------------ Ans From the result of the previous problem, we have, the dual vector of ARis given by A '3 sinθ=te , where '3eis in the direction of axis of rotation and θis the angle of rotation. Thus, we can obtain the direction of axis of rotation and the angle of rotation θ by obtaining the dual vector of AR. From 12 23 31 , , == =Re e Re e Re e , we have, [] ()AA 123001 0 1 1 11100 1 0 122010 1 1 0− ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤ =→ = − → = + +⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦RR t e e e . Thus , ()123 A '333 22 3++==ee ete , where ()'31 2 31 3=+ +ee e e is in the direction of the axis of rotation and the angle of rotation is given by sin 3 / 2θ= , which gives oo60 or 120θ= . On the other hand, the first scalar invariant of R is 0. Thus, from the result in (c) of the previous problem, we have, 112 c o s 0I θ+== , so that cos 1/ 2θ=− which gives oo120 or 240θ= . We therefore conclude that o120θ= . _________________________________________________________________ 2.56 Given the tensor []100 01 0 00 1−⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦Q . (a) Show that the given tensor is a rotation tensor. (b) Verify that the eigenvalues are 1 and 1λ=− . (c) Find the direction for the axis of rotation (i.e., eigenvectors corresponding to 1λ=). (d) Find the eigenvectors corresponding 1λ=− and (e) obtain the angle of rotation using the formula 112 c o sI θ=+ (see Prob. 2.54), where 1Iis the first scalar invariant of the rotation tensor. ------------------------------------------------------------------------------- Ans. (a) []det 1=+Q , and [] [] [ ]I= QQ I therefore it is a rotation tensor. (b) The principal scalar invariants are: 123 1, 1, 1 II I=−= −= → characteristic equation is ()()32 2111 0 λλλ λ λ+− − =+ − = → the eigenvalues are: 1,1,1 λ=− . (c) For 1λ=, clearly, the eigenvector are: 3±n= e , which gives the axis of rotation. (d) For 1λ=− , with eigenvector 11 2 2 3 3ααα++ n= e e e , we have 12 30 0 , 0 0 , 2 0ααα=== . Thus, 12 3arbitrary, arbitrary, 0α αα = == . The eigenvectors are: 22 11 2 2 1 2 , 1 αα α α++ = n= e e . That is, all vectors perpendicu lar to the axis of rotation are eigenvectors. (e) The first scalar invariant of Q is 1 1 I=−. Thus, 1 2 c o s1 c o s1θ θθ π +=− → =− → = . ( We note that for this problem, the antisymmetric part of Q=0 , so that Asin θ=t0 = n , of which θπ= is a solution). _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-20 2.57 Let Fbe an arbitrary tensor. (a) Show that TFF and TFF are both symmetric tensors. (b) If F=Q U=V Q , where Qis orthogonal and and U V are symmetric, show that 2TU= F F and 2TV= F F (c) If λand nare eigenvalue and the corresponding eigenvector for U, find the eigenvalue and eigenvector for V. [note corrections for text] ------------------------------------------------------------------------------- Ans. (a) TT TT T T() ( )== FF F F FF , thus TFF is symmetric. Also TT TT T T() ( )== FF F F FF , therefore, TFF is also symmetric. (b) TT T T T T T T 2→→ → F = Q U F = UQ FF = UQQ U = UU FF = U . T T TT T T TT 2→→ → F = VQ F = Q V FF = VQQ V = VV FF = V . (c) Since F=Q U=V Q , and λ Un = n , therefore, ( ) ( ) ( ) λ λ =→ VQn = QUn Q n V Qn = Qn , therefore, Qnis an eigenvector for Vwith the eigenvalue λ. _________________________________________________________________ 2.58 Verify that the second principal scalar invariant of a tensor Tcan be written: () 2 /2ii jj ij ji IT T T T=− . ------------------------------------------------------------------------------ Ans. 222 2 11 22 33 11 22 33 11 22 22 33 33 11() 2 2 2ii jjTT T T T T T T T T T T T T=++ = +++ + + . 22 2 1 1 2 2 3 3 11 12 21 13 31 21 12 22 23 32 31 13 32 23 33 ij ji j j j j j jT TT TT T T T TT TT TT TTT TT TT TT=++= ++++ ++++ . Thus, 22 2 11 22 33 11 22 22 33 33 11(2 2 2 )ii jj ij jiT T T T T T TT TT TT T−= + + + + + 22 2 11 22 33 12 21 13 31 23 32(2 2 2 )TT T T T T T T T−+++ + +11 22 12 21 22 33 23 32 33 11 13 312( )TT TT T T TT TT TT= −+−+− . Thus, () /2ii jj ij jiTT TT−11 22 12 21 22 33 23 32 33 11 13 31()TT TT T T TT TT TT=− +− + − 22 23 11 13 11 12 2 32 33 31 33 21 22TT TT TTITT TT TT=++= . _________________________________________________________________ 2.59 A tensor has a matrix []T given below. (a) Write the characteristic equation and find the principal values and their corresponding princi pal directions. (b) Find the principal scalar invariants. (c) If 123,,nn n are the principal directions, write [] inT. (d) Could the following matrix []Srepresent the same tensor Twith respect to some basis. []540 41 0 003⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T , []72 0 21 0 00 1⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦S . ------------------------------------------------------------------------------ Ans. (a) The characteristic equation is: []254 0 4 1 0 0 (3 ) (5 )( 1 ) 16 (3 )( 4 21) (3 )( 3)( 7) 0 00 3λ λλ λ λ λ λ λ λ λ λ λ− −− = → − − −− − = − − − = − + − = − Thus, 12 33, 3, 7λλλ== −= . For 13,λ= clearly, 13=±ne . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-21 For 2 3λ=− 222 12 1 2 3 1 2 3 12 12 3 2 1 3 2 1 2( 5 3 ) 4 0 , 4 ( 1 3 ) 0 , (3 3 ) 0 , 1 8 4 0, 4 2 0, 6 0 2 , 0. ( 2 ) / 5.αα α α α α α α αα αα α α α α++= + − += += + + = →+= += = → = − = → = ± − ne e For 37λ= 222 12 1 2 3 1 2 3 12 12 3 12 3 3 1 2(5 7) 4 0, 4 ( 1 7) 0, (3 7) 0, 1 24 0 , 48 0 , 4 0 2 , 0 . ( 2 ) / 5 .αα α α α α α α αα αα α ααα−+= + − −= −= + + = →− + = − = − = → = = → =± + ne e (b) The principal scalar invariants are: 12 35 1 3 7, ( 5 16) ( 3 0) (15 0) 9, 15 48 63II I=−+= = −− + −− + − = − = − − = − . We note that () () ()32 2 279 6 3 0 7 97 0 7 (9 ) 0λλλ λλ λ λ λ−− + = → − −− = → − − = , same as obtained in (a) (c) [] {}123,,300 03 0 007i⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦n nn nT . (d) 72 0 det 2 1 0 3 63 00 1⎡⎤ ⎢⎥=−≠ −⎢⎥ ⎢⎥−⎣⎦, therefore, the answer is NO. Or, clearly one of the eigenvalue for []S is 1−, which is not an eigenvalue for []T, therefore the answer is NO. _________________________________________________________________ 2.60 Do the previous problem for the following matrix: []300 004 040⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T ------------------------------------------------------------------------------ Ans. (a) The characteristic equation is: 230 0 0 0 4 0 (3 )( 16) (3 )( 4)( 4) 0 04 0λ λλ λ λ λ λ λ− −= → − − = − − + = − Thus, 12 33, 4, 4λλλ=== − . For 13,λ= clearly, 11=±ne , because 113=Te e . For 24λ= 222 12 3 2 3 1 2 3 12 3 2 3 1 2 3 2 2 3(3 4) 0, (0 4) 4 0, 4 (0 4) 0, 1 0, 4 4 0, 4 4 0 0, , ( ) / 2.αα α α α α α α αα α α α α α α−= −+= + −= + + = −= − + = − = →= = →= ± + ne e For 3 4λ=− 222 12 3 2 3 1 2 3 12 3 2 3 1 2 3 3 2 3(3 4) 0, (0 4) 4 0, 4 (0 4) 0, 1 7 0, 4 4 0, 4 4 0 0, , ( ) / 2αα α α α α α α αα αα αα α α+= ++= + += + + = →= += += → = = − → = ± − ne e (b) 12 33, (0 0) (0 16) (0 0) 16, 48II I== − + − + − = −= − . () () ()32 2 23 16 48 0 3 16 3 0 3 ( 16) 0λλ λ λλ λ λ λ−−+ = → − − − = → − − = , same as in (a) . (c)[] {}123,,30 0 04 0 00 4i⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦n nn nT Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-22 (d)[]72 0 det det 2 1 0 (7 4) 3 48 00 1⎡⎤ ⎢⎥== − − = − ≠ −⎢⎥ ⎢⎥−⎣⎦S , therefore, the answer is NO. Or, clearly one of the eigenvalue for []S is 1−, which is not an eigenvalue for []T, therefore the answer is NO. _________________________________________________________________ 2.61 A tensor Thas a matrix given below. Find the principal values and three mutually perpendicular principal directions: []110 110 002⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T ------------------------------------------------------------------------------ Ans. The characteristic equation is: () () ()2 2211 0 11 0 0 2 ( 1) 1 2 ( 2 ) 2 0 00 2λ λλ λ λ λ λ λ λ λ− ⎡⎤ −= → − − − = − − + = − − =⎣⎦ −. Thus, 12 30, 2λλλ== = . That is, there is a double root 23 2λλ==. For 10λ=, 222 12 1 2 3 1 2 3 12 3 1 2 3 1 1 2(1 0) 0, (1 0) 0, (2 0) 0, 1 0, 2 0 , 0, ( ) / 2.αα α α α ααα αα α α α α−+ = + −= −= + + = →+= = →= − = →= ±− ne e For 23 2λλ== , one eigenvector is clearly 3n. There are infinitely many others all lie on the plane whose normal is 11 2 () / 2=± −ne e . In fact, we have, 222 12 1 2 3 1 2 3 2 12 3 12 3 1 23 3(1 2) 0, (1 2) 0, (2 2) 0, 1 0, 0 0 , 1 2 ( ),αα α α α ααα αα α αααα α α α α−+ = + −= −= + + = →− + = = → = = = − → =± + + ne e e which include the case where 33 0, 1αα== ± → = ± ne . _________________________________________________________________ CHAPTER 2, PART C 2.62 Prove the identity ()dd d dt dt dt+=TSTS + , using the definition of derivative of a tensor. ------------------------------------------------------------------------------ Ans. () 00 00{ ( ) ( )} { ( ) ( )} { ( ) ( )} { ( ) ( )}lim lim {( ) ( ) } { {( ) ( ) }lim lim .tt ttdt t t t t t t t t t t t dt t t tt t tt t d d t t dt dtΔ→ Δ→ Δ→ Δ→+ Δ+ + Δ − + + Δ− + + Δ−+= =ΔΔ +Δ − +Δ −+= +ΔΔTS T S TT SSTS TT SS T S_ _________________________________________________________________ 2.63 Prove the identity ()dd d dt dt dt=STTS T + S using the definition of derivative of a tensor. ------------------------------------------------------------------------------ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-23 Ans. () 0( ) ( ) () ()lim tdt t t t t t dt t Δ→+Δ +Δ −=ΔTS T STS 0() () () ( ) () ( ) ( ) ( )lim ttt tt tt t tt t t t t Δ→+Δ +Δ − +Δ + +Δ −=ΔTS TS TS T S 00 00( ){ ( ) ( )} { ( ) ( )} ( )lim lim { ( ) ( )} { ( ) ( )}()l i m l i m ()tt tttttt t tt tt tt tt t tt t ddtttt d t d tΔ→ Δ→ Δ→ Δ→+Δ +Δ − +Δ −=+ΔΔ +Δ − +Δ −=+= +ΔΔTS S T T S SS TT S TTS T S . _________________________________________________________________ 2.64 Prove that T Tdd dt dt⎛⎞=⎜⎟⎝⎠TTby differentiating the definition TaT b = bTa⋅⋅ , where and a b are constant arbitrary vectors. ------------------------------------------------------------------------------ Ans. TT(/ ) ( / )dd t d d t→ aT b = bTa a T b = b T a⋅⋅ ⋅ ⋅ . Now, the definition of transpose also gives ()T(/ ) /dd t dd taT b = bT a⋅⋅ . Thus, ()T T/( / )dd t d d tbT a = bT a⋅⋅ . Since and a b arbitrary vectors, therefore, T Tdd dt dt⎛⎞⎛⎞=⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠TT. _________________________________________________________________ 2.65 Consider the scalar field 2 11 2332x xx xφ=+ + . (a) Find the unit vector normal to the surface of constant φat the origin (0,0,0) and at (1,0,1) . (b) wh at is the maximum value of the directional derivative of φat the origin? At (1,0,1)? (c) Evaluate / dd rφ at the origin if 13() dd s=+re e . ------------------------------------------------------------------------------ Ans. (a) 12 11 23(2 3 ) 3 2 , xx xφ∇= + + + ee e ( ) 33 1 2 3 1 2 3 at (0,0,0), 2 , at (1,0,1), 2 3 2 , 2 3 2 / 17 φφ∇= → ∇= + + → + + en = e eee n = eee . (b) Atmax (0,0,0), ( / ) 2 dd rφφ =∇ = in the direction of 3=ne . At (1,0,1) , max (/ ) 1 7dd rφφ =∇ = . (c) At o3 1 3 (0,0,0), / ( ) / 2 ( ) / 2 2 dd r d d rφφ=∇ = + = re e e⋅⋅ . _________________________________________________________________ 2.66 Consider the ellipsoidal surface defined by the equation22 22 22// / 1xa yb zc++= . Find the unit vector normal to the surface at a given point ( , , ) xyz. ------------------------------------------------------------------------------ Ans. Let 22 2 222(, ,) 1xyzfx y z abc=++− , then 123 222 2 2 22222 2 2, , f xfyfz x y zfxyzabca b c∂∂∂=== → ∇ = + +∂∂∂eee , thus, 1/222 2 123 222 2 2 2222 2 2 2=fx y z x y z f abc a b c−⎡⎤∇⎛⎞⎛⎞⎛⎞ ⎛ ⎞=+ + + +⎢⎥⎜⎟⎜⎟⎜⎟ ⎜ ⎟∇⎝⎠⎝⎠⎝⎠ ⎝ ⎠⎢⎥⎣⎦ne e e . _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-24 2.67 Consider the temperature field given by: 123xxΘ= . (a) Find the heat flux at the point (1,1,1)A , if k−∇ Θq= . (b) Find the heat flux at the same point if −∇Θ q= K , where []00 02 0 003k k k⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦K= ------------------------------------------------------------------------------ Ans. 12 21 12 1 233 ( ) ( ) 3 ( )A xx x xΘ= →∇Θ= + → ∇Θ = + ee e e . (a) 123( ) kk−∇ Θ = − +q= e e . (b) [] [ ] 1200 3 3 02 0 3 6 ( 3 6 ) 003 0 0kk kk k k k⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥−∇ Θ = − = − → − +⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦q= K q = e e . _________________________________________________________________ 2.68 Let 123(, , )xxxφ and 123(, , )xxxψ be scalar fields, and let 123(, , )xxx v and 123(, , )xxx w be vector fields. By writing the subscripted co mponents form, verify the following identities. (a) ( )φψφ ψ∇+ = ∇ + ∇ , sample solution: ()[( ) ]i ii ixx xφψφ ψφψφ ψ∂+∂ ∂∇+= = += ∇ + ∇∂∂ ∂. (b) div( ) div div+= +vw v w , (c) div( ) ( ) (div )φφφ=∇+ vv v⋅ and (d) div(curl ) 0 =v . ------------------------------------------------------------------------------ Ans. (b) ()div( ) div divii i i ii ivw v w xx x∂+ ∂ ∂+= =+ = +∂∂ ∂vw v w . (c) ()div( ) (div ) ( )ii i ii ivvvxx xφ φφφφ φ∂∂ ∂== + = + ∇∂∂ ∂vv v ⋅. (d) curl div(curl )j kk k ijk i ijk i ijk ijk kj i j i jv vv v x xx x x xεε ε ε∂ ∂∂ ∂ ∂∂−=→ = =∂∂ ∂ ∂ ∂ ∂v= e e v . By changing the dummy indices, ( , ij j i→→ ) we have, kk ijk jik ij jivv xxx xεε∂ ∂ ∂ ∂=∂∂∂ ∂. Thus, kk ijk ijk ij jivv xxx xεε∂∂∂∂=−∂∂ ∂∂20 0kk ijk ijk ij i jvv xx x xεε⎛⎞ ∂∂∂∂→= → = ⎜⎟⎜⎟ ∂∂ ∂ ∂⎝⎠. Thus, div(curl ) 0 =v . _________________________________________________________________ 2.69 Consider the vector field 22 2 11 32 23x xx++ v= e e e . For the point (1,1,0) , find (a) ∇v, (b) ()∇vv, (c) div and curl vv and (d) the differential dvfor 123() / 3 dd sr= e +e +e . ------------------------------------------------------------------------------ Ans.(a) [] []()1 3 1,1,0 22 0 0 200 00 2 0 0 0 02 0 0 2 0x x x⎡⎤⎡ ⎤ ⎢⎥⎢ ⎥∇→ ∇⎢⎥⎢ ⎥ ⎢⎥⎢ ⎥⎣⎦⎣ ⎦v= v = . (b) [] 12001 2 () 0 0 0 0 0 () 2 0201 0⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥∇= =→ ∇⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦vv vv = e . (c) 11 div 2 0 0 2 at (1,1,0), div 2 xx++= → v= v= . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-25 ()33 21 2 1 12 3 2 3 1 23 31 12curl 2vv vv v vxxxx xx xx⎛⎞ ⎛⎞ ⎛⎞ ∂∂∂∂ ∂ ∂−+ −+ −= −⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂∂ ∂∂ ⎝⎠ ⎝⎠ ⎝⎠v= e e e e . At (1,1,0) , () 11 curl 2 1 0 2 −= v= e e . (d) () ( ) [][] []/3 200 2 /3 At 1,1, 0 , 0 0 0 / 3 0 020 /3 2 /3ds ds dd d d d s ds ds⎡⎤ ⎡⎤ ⎡⎤⎢⎥ ⎢⎥ ⎢⎥∇→= ∇ = = ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎢⎥ ⎣⎦ ⎣⎦v= v r v v r . 13 2( ) /3 dd s→v= e +e _________________________________________________________________ CHAPTER 2, PART D 2.70 Calculate div ufor the following vector field in cylindrical coordinates: (a) 20, rzuu u A B rθ== = + . (b) sin / , = 0rzur u uθθ= =, and (c) 22sin / 2, = cos / 2, 0rzur u r uθθθ== . ------------------------------------------------------------------------------ Ans.(a) 2 10 , d i v 00000rr z rzu uu uuu u A B rrr r zθ θθ∂∂ ∂== = + → + ++ = + + + =∂∂ ∂u= . (b)22 1s i n/ , = 0 d i v s i n/ 0s i n/ 00rr z rzu uu uur u u r rrr r zθ θθθ θθ∂∂ ∂== → + + + = − + + + =∂∂ ∂u= (c) 22sin / 2, = cos / 2, 0rzur u r uθθθ== 1div sin sin / 2 sin / 2 0 sinrr z u uu urr r rrr r zθθθθ θθ∂∂∂→+ + + = − + + =∂∂ ∂u= . _________________________________________________________________ 2.71 Calculate ∇ufor the following vector field in cylindrical coordinate: / , , 0rzuA r uB r uθ == = . ------------------------------------------------------------------------------ Ans. []2 21 0 10 00 0 1rr r r zz zuu u Au Brr zr uu u AuBrr z r uu u rr zθ θθ θθ θ θ⎡⎤∂∂ ∂⎛⎞ ⎡ ⎤− −− ⎜⎟⎢⎥ ⎢ ⎥ ∂∂ ∂⎝⎠⎢⎥ ⎢ ⎥⎢⎥∂∂ ∂⎛⎞ ⎢ ⎥=+ =⎢⎥⎜⎟ ⎢ ⎥∂∂ ∂⎝⎠⎢⎥ ⎢ ⎥⎢⎥∂∂ ∂ ⎢ ⎥⎢⎥ ⎢ ⎥⎣ ⎦ ∂∂ ∂⎢⎥⎣⎦u∇ . _________________________________________________________________ 2.72 Calculate div ufor the following vector field in spherical coordinates 2/, 0ruA r B r uuθφ =+ == ------------------------------------------------------------------------------ Ans. 2 2(s i n ) ()11 1div sin sinru u ru rr r rφ θθ θθθ φ∂ ∂ ∂→+ +∂∂ ∂u= Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-26 ()23 22 2113.BrA r A r B Arrrr r⎧⎫∂∂⎛⎞=+ = + =⎨⎬⎜⎟∂∂⎝⎠⎩⎭ _________________________________________________________________ 2.73 Calculate ∇ufor the following vector field in spherical coordinates: 2/, 0ruA r B r uuθφ =+ == . ------------------------------------------------------------------------------ Ans. []11 sin cot 11 sin cot 11 sinrr r r ru u uu u rr r r r u uu u u rr r r r uu u u u rr r r rφ θ φ θθ θ φφ φ θθθ φ θ θθ φ θ θθ φ⎡⎤ ⎛⎞ ∂∂ ∂⎛⎞−− ⎢⎥ ⎜⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝⎠ ⎢⎥ ⎢⎥⎛⎞ ∂∂ ∂⎛⎞⎢⎥∇= + − ⎜⎟ ⎜⎟∂∂ ∂⎢⎥⎝⎠ ⎝⎠⎢⎥∂∂ ∂ ⎛⎞ ⎢⎥++ ⎜⎟ ⎢⎥∂∂ ∂ ⎝⎠ ⎣⎦u 3 3 3/0 0 2 / 0 0 0/ 0 0 / 0 00 / 0 0 /r r rur AB r ur A B r ur A B r⎡⎤ ∂∂ −⎡⎤ ⎢⎥ ⎢⎥== +⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ + ⎣⎦ ⎣⎦. _________________________________________________________________ 2.74 From the definition of the Laplacian of a vector, ()2div curlcurl ∇∇ −v= v v , derive the following results in cylindrical coordinates: ()22 2 2 22 2 22 212 1rr r r r rv vv v v v rr rr zr rθ θ θ⎛⎞ ∂ ∂∂ ∂ ∂∇= + + − + −⎜⎟⎜⎟ ∂∂ ∂∂ ∂⎝⎠v and ()22 2 2 22 2 2 2 211 2r vv v v v v rr rr z r rθ θθ θ θ θ θ θ∂∂ ∂ ∂ ∂∇= + + + + −∂∂ ∂∂ ∂v . --------------------------------------------------------------------------- Ans. Let ()rv be a vector field. The Laplacian of vis ()2div curlcurl ∇∇ −v= v v . Now, 1divrr z v vv v rr r zθ θ∂∂∂=+ + +∂∂ ∂v , so that () r 2 22 zr 22 2 2 2 22 211 1div 11 1 1 11 1rr z rr z rr z r r r z rvv vv v vv v rr r r z r r r r z vv v vv v v v v v zrr r z r r r r r z rr r v v rr rrθθ θ θθ θ θθθ θ θθ θ θ θ∂∂∂∂ ∂∂∂∂⎛⎞ ⎛⎞∇ = ++ + + ++ +⎜⎟ ⎜⎟∂ ∂ ∂∂∂ ∂ ∂∂⎝⎠ ⎝⎠ ⎛⎞ ∂∂ ∂∂∂ ∂ ∂ ∂∂⎛⎞++ + += +− + − + ⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ∂ ∂∂+++∂∂ ∂ve e ee 2 22 2 z 211 1.rz r r z v vv v v v rz z r r z r z zθ θθθ θ⎛⎞ ⎛ ⎞ ∂ ∂∂ ∂ ∂++ + + + ⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟ ∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ⎝⎠ ⎝ ⎠ee Next, r θ z11curlzr z r vv v vv v v rz z r r r rθθ θ θθ∂∂∂∂ ∂ ∂⎛⎞ ⎛ ⎞ ⎛⎞−+ −+ + − ⎜⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎝ ⎠v= e e e , so that Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-27 () 2 22 2 22 2211curlcurl 11 1,rr z r rr zvv vv v rr r r z z r vv vv v rr z r rr zθθ θθθθ θθ θ∂ ∂∂ ∂ ∂∂⎛⎞ ⎛⎞+− − − ⎜⎟ ⎜⎟∂∂ ∂ ∂∂ ∂ ⎝⎠ ⎝⎠ ⎛⎞ ⎛⎞ ∂∂ ∂∂ ∂+− − −⎜⎟ ⎜⎟⎜⎟ ⎜⎟∂∂ ∂ ∂∂ ∂∂⎝⎠ ⎝⎠v= = ()11curlcurlzr vv v vv zr z r r r rθθ θ θθ θ∂∂∂∂∂∂⎛⎞ ⎛ ⎞=− − + −⎜⎟ ⎜ ⎟∂∂∂∂ ∂ ∂⎝⎠ ⎝ ⎠v ()11 1curlcurlrz rz z zv vv vv v rz r rz r r r zθ θθ∂ ∂∂ ∂∂ ∂∂∂ ⎛⎞ ⎛⎞ ⎛⎞=− + − − −⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ ∂⎝⎠ ⎝⎠ ⎝⎠v Thus, ()2 22 2 22 211 1rr r z rvv vv v v rr r r r z rr rθθ θθ⎛⎞ ∂∂∂∂ ∂∇= + − + − +⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂∂∂⎝⎠v 2 22 22 2 2 22 22 2 2 2 2 2 211 1 1 2 1rr zr r r r r vv v vv vv v v v v rr z r r r rr z r r z r rθθ θ θθ θ θθ⎛⎞ ⎛⎞ ⎛ ⎞ ∂∂ ∂ ∂∂ ∂∂∂ ∂ ∂−+ − + − = ++ − + −⎜⎟ ⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂∂ ∂⎝⎠ ⎝ ⎠ ⎝⎠. ()2 22 2 22 211 1 1rr z v vv v rr rz rrθ θθθ θ θ⎛⎞ ∂ ∂∂ ∂∇= + + +⎜⎟⎜⎟∂∂ ∂ ∂∂ ∂ ⎝⎠v 2 22 22 211 1 1 1 1zr r r z vv v v vv v v v zr z r r r r r r r z rrθθ θ θ θθ θ θ θ θ⎛⎞ ∂∂ ∂∂∂ ∂ ∂ ∂∂∂⎛⎞ ⎛ ⎞−− + + − = + + + ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ∂∂∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝ ⎠ ⎝⎠ 22 22 22 2 211 1 1zr r vv v v vv v rz r r rr zr r rθθ θ θ θ θθ⎛⎞ ⎛ ⎞ ∂∂∂∂∂ ∂+− + + + − − +⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟∂∂ ∂ ∂∂ ∂∂∂ ⎝⎠ ⎝ ⎠ 22 2 22 2 2 2 211 2r vv v v v v rr rr z r rθ θθ θ θ θ θ∂∂ ∂ ∂ ∂=+ +++ −∂∂ ∂∂ ∂. _________________________________________________________________ 2.75 From the definition of the Laplacian of a vector, ()2div curlcurl ∇∇ −v= v v , derive the following result in spherical coordinates: ()22 2 2 2 2 22 3 2 2 2 2 2 2 2 2sin 12 1 c o t 1 2 2 sin sin sinrr r r r rv v rv rv v v v r rr r r r r r rφ θθ θ θ θφ θθ φ θ θ⎛ ⎞∂ ∂ ∂∂ ∂∂ ∂∇= − + + + − −⎜ ⎟⎜ ⎟ ∂∂ ∂ ∂ ∂∂ ∂⎝ ⎠v ------------------------------------------------------------------------------ Ans. From () ()2 2sin 11 1 sin sinrrv v v rr r rφ θθ θθθ φ∂ ∂ ∂++∂∂ ∂ , we have, ()2 2 r2 2 2 2sin sin 11 1 1 11 1div sin sin sin sin sin 11 1 1 sin sin sin,t ha t i sr r rvv vv rv rv r r rr r r rrrr v v rv rr r r rφ φ θ θ θ φ θ φθθ θθ θ φ θ θθ θ φ θ θφ θ θ θφ∂∂ ∂∂ ∂∂∂∂∇= + + + + + ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂++ + + ∂∂ ∂ ∂⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎛⎞ ⎜⎟ ⎝⎠ve e e ()()()22 2 2 2 r22 3 2 2sin sin 12 1 1 11div sin sin sin sinrrrv rv vv vv rr r r r rr r r rφφ θθθθ θθ θ θ φ φ θθ∂∂ ∂∂ ∂∂ ∇= − + − + − ∂∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠v e Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-28 () ()2222 32 2 2 2 2 2sin sin sin 11 11 c o s 1 1 sin sin sin sinrrv v vv v rr r rr r rφ θθ θ θθθ θ θ θθ θ θ φ θθ θ θ θ∂ ∂ ∂∂ ∂ ∂++ + − + ∂∂ ∂ ∂ ∂ ∂∂⎛ ⎞⎛⎞⎜ ⎟ ⎜⎟⎜ ⎟ ⎝⎠⎝ ⎠e () ()2 2 2 32 2 2 2 2sin 11 1 sin sin sinrrv v v r rr rφ θ φθ φφ θθθ θ φ⎛⎞ ∂ ∂ ∂ ∂⎜⎟++ +⎜⎟∂∂ ∂ ∂ ∂⎝⎠e. Also, r θsin 111 1 1 1curlsin sin sinrrvr v vr v vv rrr r r r r rφφ θθ φθ θθ θ φ θ φ θ∂∂⎛⎞ ⎛ ⎞ ∂∂ ∂∂ ⎛⎞−− − ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎝ ⎠v= e + e + e so that r θ11 1 11curlcurl sinsin sin sin 11 1 1 1 1 sin sin sin 11 1 sinrr r rrv rv vv rr r r rr r v vr v vrrr r r r r r r rv vrrr r r rφ θ φ θθ φθθθ θ φ θ φ θ θφ θ θ θφ θ θφ⎧⎫ ∂ ⎛⎞ ∂∂∂ ∂∂⎛⎞⎪⎪−− − ⎨⎬ ⎜⎟ ⎜⎟∂∂∂ ∂ ∂ ∂⎝⎠⎪⎪ ⎝⎠ ⎩⎭ ⎧⎫ ∂⎛⎞ ∂∂ ∂ ∂∂ ⎛⎞ ⎪⎪−− − ⎨⎬ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎪⎪ ⎝⎠ ⎩⎭ ∂ ⎛∂ ∂−∂∂ ∂v= e +e + sin 11 1 sin sinv v rr rφ θ φθ θθ θ θ φ⎧⎫ ∂ ⎞⎛ ⎞ ∂ ∂ ⎪⎪−− ⎨⎬⎜⎟ ⎜ ⎟∂∂ ∂ ⎪⎪⎝⎠ ⎝ ⎠⎩⎭e i.e., 2 2 22 r 22 2 2 2 2curlcurl 1c o t 1 1 1 1 sin sinrr rrv rv rv vv v rr r r r r rr rφ θθ θ θθ φθθ φ θ⎧⎫ ⎛⎞ ⎛⎞ ∂ ∂∂∂∂ ∂ ⎛⎞ ⎪⎪⎜⎟ −+ −− −⎜⎟⎨⎬ ⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂∂∂∂ ⎝⎠ ⎪⎪⎝⎠ ⎝⎠ ⎩⎭v= e 2 22 2 θ 22 2 2 2 2 2sin 11 1 1 1 1 sinrr rv vr v r v r v vv v rr r r r rr r r rφ θθ θ θθ φθ θ θ θθφ⎧ ⎫ ⎛⎞ ⎛⎞ ∂ ∂∂ ∂ ∂ ∂∂ ∂⎛⎞ ⎪ ⎪⎜⎟ −− − − − − + ⎜⎟ ⎨ ⎬ ⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂∂ ∂ ∂∂ ⎝⎠ ⎪ ⎪ ⎝⎠ ⎝⎠ ⎩ ⎭+ e 2 22 2 2 2 2 22 2 211 1 1 1 1 1 sin sin sin sin 11 c o ssin sinsin sinrr rrv rv rv vv v rr r r r rr r r vv vvv rrφφ φ φ φφ θθ φθ φ θφ θφ θ θθθθθ φ θ φ θθ⎧⎫⎛⎞ ∂∂ ∂ ⎛⎞ ∂∂ ∂⎪⎪⎜⎟ −− + + − ⎜⎟⎜⎟∂∂ ∂ ∂ ∂ ∂⎪⎪ ∂ ⎝⎠ ⎪⎝ ⎠ ⎪⎨⎬⎛⎞ ∂∂ ⎪⎪ ⎛⎞ ∂∂⎜⎟−− + − + − ⎜⎟ ⎪⎪⎜⎟ ∂∂ ∂ ∂∂ ⎝⎠ ⎪⎪ ⎝⎠ ⎩⎭+e Thus, ()2div curlcurl ∇= ∇ −vv v gives: ()()() () ()22 2 2 2 2 22 3 2 2 2 2 22 22 2 2 2 2sin sin 12 1 1 11 sin sin sin sin 1c o t 1 1 1 1 sin sinrr r rr rrv rv vv vv rr r r r rr r r r rv rv rv vv v rr r r r r rr rφ φ θθ φ θθθθ θθθ θ φ φ θθ θ θθ φθθ φ θ⎛ ⎞ ∂∂ ∂∂ ∂∂⎜ ⎟∇= − + − + −⎜ ⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎧ ⎛⎞ ⎛⎞ ∂ ∂∂∂∂ ∂ ⎛⎞⎜⎟ −− + − − −⎜⎟⎨ ⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂∂∂∂ ⎝⎠ ⎝⎠ ⎝⎠v ⎫⎪⎪⎬ ⎪⎪⎩⎭ i.e., ()22 2 2 2 2 22 3 2 2 2 2 2 2 2 2sin 12 1 c o t 1 2 2 sin sin sinrr r r r rv v rv rv v v v r rr r r r r r rφ θθ θ θ θφ θθ φ θ θ⎛ ⎞∂ ∂ ∂∂ ∂∂ ∂∇= − + + + − −⎜ ⎟⎜ ⎟ ∂∂ ∂ ∂ ∂∂ ∂⎝ ⎠v _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-29 2.76 From the equation TT(div ) div( ) tr( ) −∇ Ta = T a T a⋅ [See Eq. 2.29.3)] verify that in polar coordinates, the θ-component of the vector (div ) T is:1(div )rr rTT T T rr rθ θθ θ θ θθ∂∂+=+ +∂∂T . ------------------------------------------------------------------------------ Ans. TT T T θθ θ (div ) div( ) tr( ) (div ) div( ) tr( ) −∇ → −∇ Ta = T a T a Te = T e T e⋅⋅ Now, TT rr θθ r θ rθθ r θθ θ θ rr r r rTT TT T Tθ θθ θ θ θ θ =+ =+→ = = = =Te e e , Te e e e T e e Te , e T e e Te ⋅⋅ ⋅⋅ TT θ r θθ1.. , d i v ( )rr rTT Tie T Trr rθ θθ θ θθ θθ∂∂=+→ =+ +∂∂Te e e Te . Also []T θ rθθ θ0/ 01 / 01 /010/ 00 00rr r rr rrTT T r rr TT T rθ θθ θ θ− −− ⎡⎤⎡ ⎤ ⎡⎤ ⎡⎤⎡⎤ =+→ ∇= →∇= = ⎢⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎣⎦ − ⎣⎦ ⎣⎦ ⎣⎦⎣ ⎦ee e e T e Thus, TT θθ11(div ) div( ) tr( ) (0 / )rr rr r rTT T TT T TTrrr r rr rθ θθ θ θ θθ θ θ θθθθ∂∂∂ ∂ +−∇ =+ + − − =+ +∂∂ ∂∂T= T e Te . _________________________________________________________________ 2.77 Calculate div Tfor the following tensor field in cylindrical coordinates; 22, , c onsta nt, 0rr zz r r rz zr z zBBTA T A T TTT TTT rrθθ θ θ θ θ =+ =− = = = = = = = ------------------------------------------------------------------------------ Ans. 3312 2(div ) 0rr r rr rz rTT T TT BB rr r z rrθθ θ θ∂−∂∂++ + = − + =∂∂ ∂T= . 1(div ) 0rr r zTT T T T rr r zθθ θ θ θ θ θθ∂∂ + ∂++ + =∂∂ ∂T= . 1(div ) 0z zr zz zr zT TT T rr z rθ θ∂∂∂++ + =∂∂ ∂T= . _________________________________________________________________ 2.78 Calculate div Tfor the following tensor field in cylindrical coordinates; 23 2 35 3 3 5 35 22233 3, , , 0, .rr zz rz zr rr z zAz Br z Az Az Bz Ar BrzTT T T T RR R R R RR TTTT R r zθθ θθ θ θ⎛⎞ ⎛ ⎞=− = = − + == − + ⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ==== = + ------------------------------------------------------------------------------ Ans. 22 35 5 3513 3 3(div )rr r rr rz rTT T TT Az Br z Brz Ar Brz rr r z r z RR R RRθθ θ θ⎛⎞⎛⎞ ∂−∂∂ ∂∂++ + = − − −+ ⎜⎟⎜⎟⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂⎝⎠⎝⎠T= 22 2 2 35 5 5 3 5 511 3 3 1 3 133Bz Brz BrAz Br z r Ar z Brzrr r z z zR RR R RR R∂∂ ∂ ∂ ∂ ∂⎛⎞ ⎛⎞=− −− −++⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂⎝⎠ ⎝⎠ 22 46 5 5 4 5 631 5 6 3 36 1 5Az R Br z R Brz Brz Ar R Bzr Brz R rr z z RR R R R R R⎛⎞ ⎛⎞∂∂ ∂ ∂=− + − − −− + −⎜⎟ ⎜⎟⎜⎟ ⎜⎟∂∂ ∂ ∂⎝⎠ ⎝⎠ 33 57555 5731 5 6 3 361 5Arz Br z Brz Brz Arz Bzr Brz RRRRR RR⎛⎞ ⎛ ⎞=− + − − + − +⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-30 ()33 22 757 7 5 5 515 15 15 15 15 15 150r z rz rz rz rz rz rzBB r z RRR R R R R⎛⎞ ⎛⎞ ⎛ ⎞= − + = + −=−=⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝ ⎠ ⎝⎠. 1( d i v ) 00000rr r zTT T T T rr r zθθ θ θ θ θ θθ∂∂ + ∂++ + = + + + =∂∂ ∂T= 1(div )z zr zz zr zT TT T rr z rθ θ∂∂∂++ +∂∂ ∂T=23 2 35 3 5 3 533 3 A Brz Az Bz A Bz rzRR R R R R⎛⎞ ⎛ ⎞ ⎛ ⎞∂∂=− + − + − +⎜⎟ ⎜ ⎟ ⎜ ⎟⎜⎟ ⎜ ⎟ ⎜ ⎟∂∂⎝⎠ ⎝ ⎠ ⎝ ⎠ 22 2 2 22 4 2 35 5 7 35 5 7 353 3 15 3 9 15 3AA r B z B r z AA z B z B z AB z RR R R RR R R RR⎛⎞ ⎛ ⎞ ⎛ ⎞= − −+− − −+− − +⎜⎟ ⎜ ⎟ ⎜ ⎟⎜⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝ ⎠ () ()22 22 22 22 35 5 7 33 5 53 3 15 15 3 3 15 150A A Bz Bz A A Bz Bzrz rz RR R R RR R R⎛⎞ ⎛ ⎞=− + + − + + = =− + − + =⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. _________________________________________________________________ 2.79 Calculate div Tfor the following tensor field in spherical coordinates; 332, , 0rr r r r rBBTA TTA TTTTTT rrθθ φφ θ θ θφ φθ φ φ = − == + ====== ------------------------------------------------------------------------------ Ans. () ()2 2sin 11 1(div ) = - sin sinrr r r rrT TTT T rr r r rφθθ φφ θθ θθ θ φ∂ ∂+ ∂++∂∂ ∂T ()2 2 22 4 22 4 4 411 2= 2 12 2 222 20 .rrrT TT BA BArrr r r r r r r BA BA BA BArr r rr r r rrθθ φφ∂ + ∂⎛⎞ ⎛ ⎞−= − − + ⎜⎟ ⎜ ⎟∂∂ ⎝⎠ ⎝ ⎠ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞=+ − + = + − + =⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ () ()3 3cot sin 11 1(div ) = + sin sin cot cot=+ 0 .r rrrT T TTT T rrr r r T T rrθ θφ θ θ φφ θθ θ φφ θθθ θ θθ θ φ θ θ∂ ∂− − ∂++∂∂ ∂ −=T () ()3 3sin cot 11 1(div ) = + = 0sin sinr rrrT T TT T T rrr r rφ φθ φφ φ φ θφ φθ θ θθ θ φ∂ ∂ ∂− +++∂∂ ∂T . _________________________________________________________________ 2.80 From the equation TT(div ) div( ) tr( ) −∇ Ta = T a T a⋅ [See Eq. 2.29.3)] verify that in spherical coordinates, the θ-component of the vector (div ) T is: 3 3cot () ( s i n )11 1(div )sin sinrr rT TTT rT T rr r r rθφ θ θ φφ θθ θ θθ θ θθ θ φ∂− − ∂∂=+ + ++∂∂∂T . ------------------------------------------------------------------------------ Ans. TT T T θθ θ (div ) div( ) tr( ) (div ) div( ) tr( ) −∇ → −∇ Ta = T a T a Te = T e T e⋅⋅ . Now, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 2-31 T θ r θ rTT Tθθ θ θ φ φ=++→ Te e e e2 T θ 2() ( s i n )11 1div( )sin sinrT rT T rr r rθφ θθ θ θ θθθ φ∂ ∂∂=+ +∂ ∂∂Te . Also, [] θ rθθ01 / 0 010 00 0 00c o t /r rφ θ−⎡⎤ ⎢⎥=++→ ∇=⎢⎥ ⎢⎥⎣⎦ee ee e T θ0/ c o t / 01 / 0 00 0 0 / c o t /00c o t / 0/c o t /rr r r rr r rr rrTTT T r T r r TTT T r T r r TTT T r T rθφ φ θθ θφ θ θ φ θ φθ φφ φ φ φ φθ θ θ θ⎡⎤ ⎡ ⎤ − −⎡⎤⎢⎥ ⎢ ⎥⎢⎥⎡⎤→∇= = − ⎢⎥ ⎢ ⎥⎢⎥ ⎣⎦⎢⎥ ⎢ ⎥⎢⎥ − ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦Te T θcottrrT T rrφφθθ⎡⎤→∇ = − +⎣⎦Te . Thus, TT θθ (div ) div( ) tr( )θ −∇ T= T e Te 2 2cot ( ) ( sin )11 1 sin sinrrTT rT T T rr r r r rθφ φφ θθ θ θθ θ θθ θ φ∂ ∂∂=+ ++ −∂∂ ∂ 3 3cot () ( s i n )11 1 sin sinrr rTT rT T T T rr r r r r rθφ φφ θθ θ θ θθ θ θθ θ φ∂ ∂∂=− + ++ −∂∂ ∂. _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-1 CHAPTER 3 3.1 Consider the motion: 11 o 2 2 3 3(1 ) / (1 ), , x kt X kt x X x X=+ + = = . (a) Show that reference time is ott=. (b) Find the velocity field in spatial coordinates. (c) Show that the velocity field is identical to that of the following motion: () 11 2 2 3 31, , x kt X x X x X=+ = = . ----------------------------------------------------------------------------------- Ans. (a) At ott=, 11 2 2 3 3 , , xXx X x X=== . Thus, ott=is the reference time. (b) In material description, 11 o 2 3 /( 1 ) , 0 vk X k t vv=+ = = . Now, from 11 o(1 ) / (1 )x kt X kt=++ , 1o 1(1 ) / (1 ) X kt x kt →= + + , therefore, 11 1 2 3 /( 1 ) , 0 vk Xk x k tvv→= = + == . (c) For() 11 2 2 3 31, , x kt X x X x X=+ = = , 11 2 3 , 0 vk Xvv→= == 11 2 3 /( 1 ) , 0 vk x k tvv→= + == , which are the same as the velocity components in (b). _________________________________________________________________ 3.2 Consider the motion: 11 2 2 3 3 , , x tX x X x Xα=+ = = , where the material coordinates iX designate the position of a particle at 0t=. (a) Determine the velocity and acceleration of a particle in both a material and a spatial descrip tion. (b) If the temperature field in spatical description is given by 1Axθ= , what is its material description? Find the material derivative of θ, using both descriptions of the temperature. (c) Do part (b) if the temperature field is 2Bxθ= ----------------------------------------------------------------------------------- Ans. (a) Material description: () 11 1 2 3 fixed// , 0 iXvD x D t x t v v α−== ∂ ∂= = = , () 11 1 2 3 fixed// 0 , 0 iXaD v D t v t aa−== ∂ ∂= = = . Spatial description: The same as above 12 31 2 3 , 0 , 0 vv v a a aα=== === .. (b) The material description of θ is ()1 AtXθα=+ . Using the material description: () () 11 /( / ) AtX D D t tAtX Aθαθ αα ⎡⎤ =+ → = ∂ ∂ +=⎣⎦. Using the spatical description: 1Axθ=→ 12 3 1230 ( )( 0 ) ( 0 )( 0 ) ( 0 )Dvv v A ADt t x x xθ θθθθα α∂∂∂∂=+ + + = + + + =∂∂ ∂ ∂. (c) Using the material description: 22 /( / ) ( ) 0 BX D Dt t BXθ θ =→= ∂ ∂= . Using the spatical description: 2Bxθ=→ () () () () 12 3 1230( 0 ) 0 0 0 0Dvv v BDt t x x xθθθθθα∂∂∂∂=+ + + = + + + =∂∂ ∂ ∂. _______________________________________________________________________ 3.3 Consider the motion 22 11 2 1 2 3 3 , , xXx X t X x Xβ ==+= , whereiXare the material coordinates. (a) at 0t=, the corners of a unit square are at (0,0,0), (0,1,0), (1,1,0) and (1,0,0) AB C D . Determine the position of ABCD at 1t=and sketch the new shape of the square. (b) Find the velocity v and the acceleration in a material description a nd (c) Find the spatial velocity field. ----------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-2Ans. For the material line ( )() 123 2, , , 0 , ,0AB X X X X = ; at 1t=, ()() 123 2,, 0 , , 0xx x X= For the material line ( )() 123 1, , , , 1 ,0BC X X X X = ; at 1t=, ()( )2 123 1 1,, , 1 , 0xx x X X β=+ For the material line ( )() 123 1, , , ,0 ,0AD X X X X = ; at 1t=, ()()2 123 1 1,, , , 0xx x X X β= For the material line ( )() 123 2, , , 1 , ,0CD X X X X = ; at 1t=, ()( ) 123 2,, 1 , , 0xx x X β=+ The shape of the material square at 1 t= is shown in the figure. (b) fixed fixed, iiii ii XXxvvatt−−∂∂⎛⎞ ⎛⎞== →⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠22 1 3 21 1 3 21 0, 2 ; 0, 2 vv v X t aa a X ββ == = == = (c) Since 11xX= , in spatial descrip. 22 13 2 113 2 1 0, 2 ; 0, 2 vv v x t aa a x ββ == = == = __________________________________________________________________ 3.4 Consider the motion: 22 121 2 2 2 3 3 , , x Xt X x k Xt X x Xβ=+= += (a) At 0t=, the corners of a unit square are at (0,0,0), (0,1,0), (1,1,0) and (1,0,0) AB C D . Sketch the deformed shape of the square at 2t=. (b) Obtain the spatial description of the velocity field. (c) Obtain the spatial description of the acceleration field. --------------------------------------------------------------------------------------------- Ans. (a) For material line ( )() 123 2,, , 0 , , 0AB X X X X = ; at 2t=, ()( )2 123 2 2 2,, 4 , 2 , 0xx x X k X X β=+ . For material line ( )() 123 1, , , , 1 ,0BC X X X X = ; at 2t=, ()( ) 123 1,, 4 , 2 1 , 0xx x X k β=+ + . For material line ( )() 123 1, , , ,0 ,0AD X X X X = ; at 2t=, ()() 123 1,, , 0 , 0xx x X= . For mat. line ( )() 123 2, , , 1 , ,0CD X X X X = ; at 2t=, ()( )2 123 2 2 2,, 4 1 , 2 , 0xx x X k X X β=++ . The shape of the material square at 2 t= is shown in the figure. x x12 ABC DC’ 2k4 B’ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-3b) fixed fixed, iiii ii XXxvva tt−−∂∂== → ∂∂⎛⎞ ⎛⎞⎜⎟ ⎜⎟⎝⎠ ⎝⎠, 22 12 2 2 3 12 2 32, , 0 ;2 , 0vX t v k X vaX a aββ= == = = = . (c) 22(1 )xk tX=+ → , () () ()22 22 2 12 3 1 2 3 2222, , 0 ; , 01 11xt k x xvv v a a akt kt ktββ== = = = =+ ++. _________________________________________________________________ 3.5 Consider the motion: () 11 1 2 2 3 3 , , xks X t X x X x X=+ + = = . (a) For this motion, repeat part (a) of the prev ious problem. (b) Find the velocity and acceleration as a function of time of a particle that is in itially at the orgin. (c) Find the velocity and acceleration as a function of time of the particles that are passing through the origin. ----------------------------------------------------------------------------------- Ans. a) For material line ( )() 123 2,, , 0 , , 0AB X X X X = ; at 2t=, ()() 123 2,, 2 , , 0xx x k s X= . For material line ( )() 123 1, , , , 1 ,0BC X X X X = ; at 2t=, ()( ) 123 1 1,, 2 2 , 1 , 0xx x k s k X X=++ . For material line ( )() 123 1, , , ,0 ,0AD X X X X = ; at 2t=, ()( ) 123 1 1,, 2 2 , 0 , 0xx x k s k X X=++ . For material line ( )() 123 2, , , 1 , ,0CD X X X X = ; at 2t=, ()( ) 123 2,, 2 2 1 , , 0xx x k s k X=+ + . The shape of the material square at 2 t= is shown in the figure. AB C DA’B’C’ D’2k2k(s+1) 1 x1x2 s (b) fixed fixedand iiii ii XXxvva tt−−∂∂== → ∂∂⎛⎞ ⎛⎞⎜⎟ ⎜⎟⎝⎠ ⎝⎠, () 11 2 3 1 2 3 , 0 , 0 ; 0 vk s X v v aa a=+= = = = = . Thus, for the particle ( )() 123,, 0 , 0 , 0XX X = , 12 3 1 2 3 , 0, 0 and 0, 0, 0 vk s v v a a a=== === (c) ()11 1xk s X t X=+ + → ()11 1 1 1 () / ( 1 ) xk s tk t X X xkst kt =++ → =− + , thus, in spatial descriptions, ()() ()1 1 12 3 1 2 3 , 0, 0 and 0, 0, 011ks x xk s tvk s v v a a akt kt⎧⎫ + −⎪⎪=+ = = = = = =⎨⎬++⎪⎪⎩⎭. At the position ()() 123, , 0,0,0xx x= , 12 3 1 2 3 / (1 ), 0, 0 and 0, 0, 0 vk s k tv v a a a=+= = = = = . _________________________________________________________________ 3.6 The position at time tof a particle initially at ()123,,XXX is given by 22 11 2 2 2 3 3 3 2, , x XX t x X k X t x Xβ=− =− = , where 1 and 1 kβ==. (a) Sketch the deformed shape, at time 1t=of the material line OAwhich was a straight line at 0t=with the point Oat () 0,0,0 and the point Aat () 0,1,0 . (b) Find the velocity at 2t=, of the particle which was at (1,3,1) at 0t=. (c) Find the velocity of the particle which is at (1,3,1) at 2t=. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-4--------------------------------------------------------------------------------------------- Ans. With 1 and 1 kβ== , 22 11 2 2 23 3 3 2, , x XX t x X X t x X=− =− = For the material line OA:123 2(, , ) ( 0 , , 0 )XX X X = : at 1t=, 2 12 2 2 3 2, , 0 xX x X x=−= = . Thus, the deformed shape of the material line at 1 t= is a parabola given in the figure shown. A A’ Ox1x2 (b) 2 11 2 2 2 3 3 3 /4 , / , / 0 v Dx Dt X t v Dx Dt X v Dx Dt== − == − == For the particle 123(, , ) ( 1 , 3 , 1 )XX X = , at 2t=, 12 324 72, 1, 0(3) (2) . vv v=−= − = − = (c) The particle, which is at 123(, , ) ( 1 , 3 , 1 )xx x= at 2t=, has the material coordinates given by the following equations: 2 12 2 3 31 8 , 3 2 , 1XX X X X=− =− = →12 3201, 5, 1 XX X= == 22 12 2 3 3 4 4(5) (2) 200, 1, 0. vX t v X v→= − = − = − = −= − = _________________________________________________________________ 3.7 The position at time tof a particle initially at 123(, , )XXX is given by: () ()11 12 2 2 12 33 , , x Xk XX t xXk XX t xX=+ + =+ + = , (a) Find the velocity at 2 t=, of the particle which was at (1,1,0) at the reference time 0 t=. (b) Find the velocity of the particle which is at (1,1,0) at 2 t=. ----------------------------------------------------------------------------------------- Ans. (a) () ()1 1 12 2 2 12 3 3 /, /, / 0 . v D xD tk X X v D x D tk X X v D xD t== +== +== For the particle 123(, , ) ( 1 , 1 , 0 )XX X = , at 2t=, 123 (1 1) 2 , (1 1) 2 , 0 vk k v k k v=+= = += = (b) The particle, which is at ()() 123, , 1,1, 0xx x= at 2t=, has the material coordinates given by the following equations: () ()11 2 21 2 3 12 , 1 2 , 0X kX X X kX X X=+ + =+ + = . 12311, , 0 14 14XX X kk→= = = ++, 12 1 2 32() , 0 14kvv k X X v k→== + = = + _________________________________________________________________ 3.8 The position at time t of a particle initially at ( ) 123,,XXX is given by 22 11 2 2 2 2 3 3 , , xXX t x X k X t x Xβ=+ =+ = , where 1 and 1 kβ==. (a) for the particle which was initially at (1,1,0), what are its positions in the following instant of time: 0, 1, 2tt t=== . (b) Find the initial position for a particle which is at (1,3,2) at 2t=. (c) Find the acceleration at 2 t= of the particle which was initially at (1,3,2) and (d) find the acceleration of a particle which is at (1,3,2) at 2 t=. -------------------------------------------------------------------------------------------- Ans. With 1 and 1 kβ== , 22 112 2 22 3 3 , , xXX t xX X t xX=+ =+ = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-5(a) 123 1 2 3 0 ( ,,)( , , )( 1 , 1 , 0 )tx x x X X X=→ = = , 2 123 1 2 2 2 3 1 ( ,,)( , , )( 2 , 2 , 0 )tx x x X X X X X=→ = + + = 2 123 1 2 2 2 3 2 ( ,,)( 4 , 2 , )( 5 , 3 , 0 )tx x x X X X X X=→ = + + = (b) 22 112 2 22 3 3 , xXX t xX X t xX=+ =+ = , at 2 12 2 3 21 4, 3 3, 2tX X X X=→= + = = 12 3 3, 1, 2 XX X→= − = = . (c) 22 2 112 2 22 3 31 22 2 3 , 2 , , 0 xXX t x X X t xX v X t v Xv=+ =+ =→ = = = . 2 12 2 32, 0 , 0aX a a→= = = . For ( )() 123,, 1 , 3 , 2XX X = ,()2 12 323 1 8 , 0aa a→= = == at any time. (d) The initial position of this particle was obtained in (b), i.e., 12 3 3, 1, 2 XX X→= − = = . Thus, 22 12 2 322 ( 1 ) 2 , 0 , 0aX a a→= = = = = . _________________________________________________________________ 3.9 (a) Show that the velocity field /( 1 )iivk x k t=+ corresponds to the motion () 1iixXk t=+ and (b) find the acceleration of this motion in material description. ----------------------------------------------------------------------------------------- Ans. (a) From () () () 1 a n d / 1 = / 1ii i i i iixX kt X x kt v kX kx kt=+ =+ → = + . (b) 0iiivk X a=→ = , or ()()()()()()22 2011 1 11 1iji j ii i i i ij j xf i x e dkx k v v kx kx k x kavt x kt kt kt kt kt ktδ −∂∂⎛⎞=+ = − + = − + =⎜⎟∂∂ + + + + +⎝⎠ +. _________________________________________________________________ 3.10 Given the two dimensional velocity field: 2 , 2xyvy v x=−= . (a) Obtain the acceleration field and (b) obtain the pathline equation. ----------------------------------------------------------------------------------------- Ans. (a) () ( ) ( )02 ( 0 ) 224xxx xx yvvvav v y x xtxy∂∂∂=+ + = + − + − = −∂∂∂, () ( ) ( )02 ( 2 ) 2 04yyy yx yvvv av v y x ytxy∂∂∂ =+ + = + − + = −∂∂∂, i.e., 4 4x y x y−− a= e e (b) 2 a n d 2 0dx dy dy xyx x d x y d ydt dt dx y=− = → =− → + = , 22 22constant= , x yX Y→+= + Or, ()22 2 a n d 2 2 2 2 4 0dx dy d x dy d xyx x xdt dt dt dt dt=− = → =− =− → + = sin 2 cos 2 and cos 2 sin 2 xAt B t y A t Bt→= + = − + , where , AY B X=−= . _________________________________________________________________ 3.11 Given the two dimensional velocity field: , xyvk x v k y==− . (a) Obtain the acceleration field and (b) obtain the pathline equation. ----------------------------------------------------------------------------------------- Ans. (a) () ( ) ( )20( ) 0xxx xx yvvvav v k x k k y k xtxy∂∂∂=+ + = + + − =∂∂∂ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-6() ( ) ()20( 0 )yyy yx yvvv av v k x k y k k ytxy∂∂∂ =+ + = + + −− =∂∂∂, That is, ()2 x y kx y+ a= e e (b) 0 ln ln lnxt Xdx dx xkx kdt x X kt ktdt x X=→ = →− = → =∫∫ktxXe→= . Similarly derivation gives ktyY e−→= . Or, xyX Y= where (),XYare material coordinates. _________________________________________________________________ 3.12 Given the two dimensional velocity field: 22() , 2xyvk xy v k x y=− = − . Obtain the acceleration field. ----------------------------------------------------------------------------------------- Ans. ()() ( )220( 2 ) 2 2xxx xx yvvva v v k x y kx kxy kytxy∂∂∂=+ + = + − + − −∂∂∂22 22( )xkx y=+ . ()() ()2202 2 2yyy yx yvvv av v k x y k y k x y k xtxy∂∂∂ =+ + = + −−− −∂∂∂22 22( )yk x y=+ . That is, ()()22 2 xy 2kx y x y++ a= e e _________________________________________________________________ 3.13 In a spatial description, the equation to evaluate the acceleration ()D Dtt∂=+ ∇∂vvvvis nonlinear. That is, if we consider two velocity fields ABand vv , then AB A + B+ ≠ aaa , where ABand aa denote respectively the acceleration fields corresponding to the velocity fields ABand vv each existing alone, A+Ba denotes the acceleration field corresponding to the combined velocity field AB+vv . Verify this inequality for the velocity fields: AB 21 12 21 1222 , 22x xx x =− + = −ve e v e e -------------------------------------------------------------------------------------------- Ans. From ()D Dt t∂=+ ∇∂vvvv 21 2 1 AB 12 1 224 2 4 00 2 0 0 2=,24 2 4 02 0 0 2 0x xx x x xx x−− −−⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤⎡⎤ ⎡⎤ +== += ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣⎦ −− − − ⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦aa AB 1 1 22 1 1 2244 , 44x xx x →= − − = − −ae e ae e AB 11 2 288x x →+= − −aa e e . On the other hand, AB+vv = 0 , so that A+B0= a. Thus, AB A + B+ ≠ aaa _________________________________________________________________ 3.14 Consider the motion: ()() 11 2 2 13 3 , sin sin , xXx X t X x X ππ == + = (a) At 0 t=, a material filament coincides with the straight line that extends from () 0,0,0 to () 1,0,0 . Sketch the deformed shape of this filament at 1/ 2, 1 and 3 / 2 tt t=== . (b) Find the velocity and acceleration in a material and a spatial description. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-7----------------------------------------------------------------------------------------- Ans. (a) Since 11 3 3 and xXx X== , therefore there is no motion of the particles in the 13 and x xdirections . Every particle moves only up and down in the 2x direction. When 22 1 22 22 1 1/ 2 sin , 1 , 3 / 2 sintx X X t x X tx X X π π =→ =+ = → = =→ =− The deformed shapes of the material at thr ee different times are shown in the figure. xy t=0, t=1t=1/2 t=3/2(1,0) (b) ()() 12 1 30, cos sin , 0vv t X v ππ π== = , () ( )2 12 1 30, sin sin , 0aa t X a πππ ==− = Since 11xX= , the spatial descriptions are of the same form as above except that 1Xis replaced with 1x. _________________________________________________________________ 3.15 Consider the following velocity and temperature fields: 22 22 1 1 2 2 12 12() / ( ) , = ( )x xx x k x xα++ Θ + v= e e (a) Write the above fields in polar coordinates and discuss the general nature of the given velocity field and temperature field (e.g.,what do the flow and the isotherms look like?) (b) At the point () 1,1, 0A , determine the acceleration and the material derivative of the temperature field. ----------------------------------------------------------------------------------------- Ans. (a) In polar coordinates, 11 2 2 r x xr+=ee e , where 222 12 rxx=+ and re is the unit vector in the rdirection, so that 2, =r krrαΘ v= e . Thus, the given velocity field is that of a two dimensional source flow from the origin, the flow is purely radial with radial velocity inversely proportional to the radial distance from the origin. With 2=krΘ , the isotherms are circles. (b) From and 0rvvrθα== , and Eq. (3.4.12) 2 2 2300 0rr r rrvv vv vavtr r r r rrθθ ααα θ∂∂∂ ⎛⎞ ⎛ ⎞=+ + − = + −+ + = − ⎜⎟ ⎜ ⎟∂∂∂ ⎝⎠ ⎝ ⎠. 0r rvv v v v vavtr r rθθ θ θ θ θθ∂∂∂=+ + + =∂∂∂. That is, 23/rrα−a= e . At the point (1,1, 0), 2Ar = , 23 2/( 2) 2/4rr αα−= −a= e e . () 02 2rv Dvk r kDt t r r rθ ααθΘ∂ Θ ∂ Θ ∂ Θ ⎛⎞=+ + = + = ⎜⎟∂∂∂ ⎝⎠. _________________________________________________________________ 3.16 Do the previous problem for the following velocity and temperature fields: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-8 ( )()21 12 22 12 22 12, =xxkx x xxα−+Θ+ +eev= ------------------------------------------------------------------------------------------ Ans. With 222 12 1 2 cos , sin and xrx r x x rθθ== + = , we have ( )( ) 21 12 1 2 2 22 2 12sin cos and =xx rkrr xx rθαα θ θ α +− +== Θ +-e e e ev= e Particles move in concentric circles wi th their speed inversely proportional to r. Isotherms are circles. (b) With 0, rvvrθα== , we have, from Eq.(3.4.12). 2 2 2 31rr r rrvv vv vavtr r r r r rθθ α α θ∂∂∂ ⎛⎞= + + − =− =− ⎜⎟∂∂∂ ⎝⎠, 0r rvv v v v vavtr r rθθ θ θ θ θθ∂∂∂=+++ =∂∂∂ i.e., 23 r/rα−a= e . At the point , 2Ar= , therefore, 22/ 4rα−a= e θ 02 0rDkrDt t rα Θ∂ Θ ⎛⎞=+ ⋅ ∇ Θ+ ⋅ = ⎜⎟∂ ⎝⎠v= e e . _________________________________________________________________ 3.17 Consider : 11Xk x=X+ e . let ()()()1 11 2 =/ 2 + dd SXe e &()()()2 21 2 =/ 2+dd S − Xe e be differential material elements in the undeformed configuration. (a) Find the deformed elements () ()12and ddxx . (b) Evaluate the stretches of these elements 11 2 2/ and /ds dS ds dS and the change in the angle between them. (c) Do part (b) for 21 and 10kk−== and (d) compare the results of part (c) to that predicted by the small strain tensor E. ------------------------------------------------------------------------------------------- Ans. (a) 11 1 2 2 3 3 , , xXk Xx XxX=+ = =→ []10 0 01 0 , 00 1k dd+⎡⎤ ⎢⎥= →⎢⎥ ⎢⎥⎣⎦Fx = F X () ()()11 11 1210 0 1 01 0 1 1 2200 1 0k dS dSdd k+⎡⎤ ⎡ ⎤ ⎛⎞ ⎛⎞⎢⎥ ⎢ ⎥ ⎡⎤⎡ ⎤ =→ = + +⎜⎟ ⎜⎟ ⎣ ⎦ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦⎝⎠ ⎝⎠⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦xx e e . () ()()22 22 1210 0 1 01 0 1 1 2200 1 0k dS dSdd k+−⎡⎤ ⎡ ⎤ ⎛⎞ ⎛⎞⎢⎥ ⎢ ⎥ ⎡⎤⎡ ⎤ =→ = − + +⎜⎟ ⎜⎟ ⎣ ⎦ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦⎝⎠ ⎝⎠⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦xx e e . (b) ()2 12 12111 2ds dskdS dS⎛⎞== + +⎜⎟⎝⎠. Let γ be the decrease in angle (from o90 ), then ()/2πγ−is the angle between the two deformed differential elements. Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-9() () ()() ()12 2 12 2 12 1211 1cos 1 12 22 11k dS dS ddkds ds ds ds kπγ−+ + ⋅ ⎛⎞ ⎛⎞ ⎛⎞ ⎡⎤ −= = − + += → ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥⎣⎦ ⎝⎠ ⎝⎠ ⎝⎠ ++2xx () ()211sin 11k kγ−+ += ++2 . (c) For 1k=, 12 1253, s in 25ds ds dS dSγ == = − . For 210k−= , ()2 12 12111 1 2 2 1 1.01 1.005 22ds dskk kdS dS⎛⎞ ⎛⎞== + + ≈ + = + = =⎜⎟ ⎜⎟⎝⎠ ⎝⎠. () ()211 20 . 0 1sin 0.0099 22 1 1 . 0 1 11k kkradiankk kγγ−+ + −− −=≈ = = → = −++ ++2 (− sign indicates increase in angle). (d) 11 1 1 2 3 , 0 kX u kX u u−→ = = = u=x X= e , [] [ ]00 00 000 000 000 000kk⎡⎤⎡⎤ ⎢⎥⎢⎥→∇ = → =⎢⎥⎢⎥ ⎢⎥⎢⎥⎣⎦⎣⎦uE , () [] []' 11 2 1 1001 11 1110 0 0 0 1 110 022 2 20000 0kk kE⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥′=+ → = = =⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦ee e , 11 1 1.00522ds dS k ds kEdS dS−′== → = + = , same as the result of part (c). Also with () 21 21 2′=e- e + e [] [] 12 1200 1 11110 0 0 0 1 110 0 222 2000 0 0kk kE Ek−− ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥′′→= = = − → = −⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦ Thus, the decrease in angle = k−, or the increase in angle is 0.01 0.0099 ≈ . _________________________________________________________________ 3.18 Consider the motion: x=X+A X , where Ais a small constant tensor (i.e., whose components are small in magnitude and independent of iX). Show that the infinitesimal strain tensor is given by T() / 2E= A+A . ----------------------------------------------------------------------------------------- Ans. () −→ ∇ ∇ u=x X=A X u= A X . Since Ais a constant, therefore, ()() ∇∇ =∇u= A X A X . Now, [] [] /ij i jXX δ ⎡⎤ ⎡ ⎤∇∂ ∂ ==⎣⎦ ⎣ ⎦X= I →∇u=AT() / 2→E= A+A _______________________________________________________________________ 3.19 At time t, the position of a particle, initially at ( ) 123,,XXX is defined by: 5 11 3 2 2 2 3 3 , , , 10 xXk Xx Xk XxX k−=+ =+ = = . (a) Find the components of the strain tensor and (b) find the unit elongation of an element initially in the direction of 12+ee . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-10----------------------------------------------------------------------------------------- Ans. (a) 11 1 32 2 2 23 3 3 , , 0 uxXk Xu xX k XuxX=− = =− = =− = [] [ ][] []T 00 0 0 / 2 00 0 02000 / 20 0kk kk k⎡⎤ ⎡ ⎤∇+ ∇ ⎢⎥ ⎢ ⎥→∇ = → = =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦uuuE (b) Let () 11 2 1 1 1 11 2E ′′ ′ ′=→ = ⋅ee + e e E e []5 1100/ 2 1 11 0110 0 0 122 2/2 0 0 0k kEk k−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥′→= = =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ _________________________________________________________________ 3.20 Consider the displacements: 22 4 11 1 2 2 2 3 (2 ), , 0, 10 uk X X X u k X u k−=+ = = = . (a) Find the unit elongations and the change of a ngles for two material elements () ()12 11 2 2 and dd X d d X==Xe X e that emanate from a particle designated by 12 X=e +e . (b) Sketch deformed positions of these two elements. ----------------------------------------------------------------------------------------- Ans. (a) []12 1 240 02 0 00 0kX kX kX kX+⎡⎤ ⎢⎥∇=⎢⎥ ⎢⎥⎣⎦u , At () ()[] [ ] 12350 5 / 2 0 , , 1,1, 0 , 0 2 0 / 2 2 0 00 0 0 00kk kk XX X k k k⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=∇ = → =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦uE . Unit elong. for ()1 11 dd X=Xe is 4 1155 1 0 Ek−== × , unit elong. for ()2 22 dd X= Xe is 4 2222 1 0 Ek−== × . Decrease in angle between them is 4 1221 0E k radian−== . (b) For ()1 11 dd X=Xe , () ()()()()11 1 11 11 11 51 5 dd d d X k d X k d X=+ ∇ =+ = +xX u X e e e , For ()2 22 dd X= Xe , () ()()() 22 2 22 21 22 21 22 (2 ) ( 1 2 ) dd d d X k d X k d X k d X k d X=+ ∇ =+ + = + +xX u X e e e e e The deformed positions of these two elements are shown below: dX1dX1dX2 PP’ u1=3k u=2kdX2k (1+5k)dX (1+k)2 2 Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-11_________________________________________________________________ 3.21 Given displacement field: 4 11 2 3 , 0 , 10 uk Xuu k−== = = . Determine the increase in length for the diagonal element OA of the unit cube (see figure below) in the direction of 123e+ e + e (a) by using the strain tensor and (b) by geometry. ----------------------------------------------------------------------------------------- Ans. (a) [] []00 000 000k⎡⎤ ⎢⎥==⎢⎥ ⎢⎥⎣⎦uE . Let () 11 2 31 3′=ee + e + e , then the unit elongation in the '1e- direction is []4 11 1 1001 11 01110 0 0 133 30001k kE−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥′′′=⋅ = ==⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦eE e . (b) From the given displacement field, we s ee that the unit cube becomes longer in the 1xdirection by an amount of k, while the other two sides remain the same. The diagonal OAbecomes ' OA, (see Figure), where 3 OA= and 22 2' ( 1 ) 1 1 3 2 3 ( 1 2 /3 /3 )OA k k k k k=++ + =+ += + + 21 / 2'3 ( 1 2 / 3 / 3 ) 3OA OA k k→−= + + − . Using binomial theorem, ()1/221 2 /3 /3 1 ( 1/2 ) ( 2 /3 ) . . . 1 /3kk k k++ = + + ≈ + Thus, '3 ( 1 / 3 ) 3 3 / 3 ( ' ) / / 3OA OA k k OA OA OA k−= + −= → − = , same as that obtained in part (a). _________________________________________________________________ 3.22 With reference to a rectangular Cartesian coordina te system, the state of strain at a point is given by the matrix []453 0 34 11 0 01 2−⎡⎤ ⎢⎥=− ×⎢⎥ ⎢⎥−⎣⎦E . (a) What is the unit elongation in the direction of 12 322+ee + e ? (b) What is the change in angle be tween two perpendicular lines (in the undeformed state) emanating from the point and in the directions of 12 322+ee + e and 1336−ee ? ----------------------------------------------------------------------------------------- Ans. Let 11 2 3(2 2 ) / 3′=+ee e + e , the unit elongation in this direction is: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-12[]44 11 1 153 02 15 8221 3 4 12 1 0 1 09901 2 1E− −⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥′′′=⋅ = − × = ×⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥−⎣⎦ ⎣ ⎦eE e . Let () 21 3136 45′=−ee e , then the decrease in angle between the two elements is: []44 12 1 253 0 3 23 222 2 2 1 3 4 1 0 1 0 1 0. 34 5 4 501 2 6E rad−−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥′′ ′=⋅ = − × = ×⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥−−⎣⎦ ⎣ ⎦eE e _________________________________________________________________ 3.23 For the strain tensor given in the previous problem, (a) find the unit elongation in the direction of1234−ee and (b) find the change in angle between two elements in the dir. of 1334−ee and 1343+ee . ----------------------------------------------------------------------------------------- Ans. (a) Let () 11 21345′=−ee e , the unit elongation in this direction is: []2 44 4 11 1 153 0 3 13 734 0 3 4 14 1 0 1 0 1 . 4 8 1 052 501 20E− −−⎡⎤ ⎡ ⎤ ⎛⎞ ⎢⎥ ⎢ ⎥′′′= ⋅ = − − − ×= ×=× ⎜⎟ ⎢⎥ ⎢ ⎥⎝⎠⎢⎥ ⎢ ⎥−⎣⎦ ⎣ ⎦eE e (b) Let () ()'' ''11 32 1 31134 a n d 4355=− =+ee ee e e , then the decrease in angle between these two elements is: []2 ''' 4 4 4 '' ''12 1 253 04 17 22 2 2 3 0 4 3 4 1 0 10 10 2.88 10 .52 501 2 3E rad−− −⎡⎤ ⎡ ⎤ ⎛⎞ ⎢⎥ ⎢ ⎥=⋅ = − − × =× = × ⎜⎟ ⎢⎥ ⎢ ⎥⎝⎠⎢⎥ ⎢ ⎥−⎣⎦ ⎣ ⎦eE e _________________________________________________________________ 3.24 (a) Determine the principal scalar invariants fo r the strain tensor given below at the left and (b) show that the matrix given below at the right can not represent the same state of strain. []453 0 34 11 0 01 2−⎡⎤ ⎢⎥=− ×⎢⎥ ⎢⎥−⎣⎦E , 4300 060 1 0 002−⎡⎤ ⎢⎥×⎢⎥ ⎢⎥⎣⎦ ----------------------------------------------------------------------------------------- Ans. (a) ()44 1542 1 0 1 1 1 0 I−−=+ +× =× , 88 8 8 253 4 1 5010 10 10 28 1034 1 2 02I−− − −−=× + × +× = ×− 12 12 353 0 34 1 1 0 1 7 1 0 01 2I−−=− × = × − Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-13(b) For 4300 060 1 0 002−⎡⎤ ⎢⎥×⎢⎥ ⎢⎥⎣⎦, 12 336 10I−=× , which is different from the 3I in (a), therefore, the two matrices can not represent the same tensor. _________________________________________________________________ 3.25 Calculate the principal scalar invariants fo r the following two tensors. What can you say about the results? () ()1200 0 0 00 a n d 0 0 000 0 0 0ττ ττ− ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤ ⎡⎤== −⎢⎥ ⎢ ⎥⎢⎥ ⎢⎥⎣⎦ ⎣⎦⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦TT . ----------------------------------------------------------------------------------------- Ans. For ()1 {}00 00 000 iτ τ⎡⎤ ⎢⎥⎡⎤=⎢⎥⎢⎥⎣⎦⎢⎥⎣⎦ eT , 2 12 3 0, , 0II I τ ==− = . For ()2 {}00 00 00 0 iτ τ−⎡⎤ ⎢⎥⎡⎤=−⎢⎥⎢⎥⎣⎦⎢⎥⎣⎦ eT2 12 3 0, , 0II I τ == −= We see that these two tensors have the same princi pal scalar invariants. This result demonstrates that two different tensors can have the same thr ee principal scalar invariants and therefore the same eigenvalues (in fact, 12 3,, 0λτλ τλ== −= ). However, corresponding to the same eigenvalue τ, the eigenvector for ()1Tis 12() / 2+ee , whereas the eigenvector for()2Tis 12() / 2−ee . We see from this example that having the same principal scalar invariants is a necessary but not sufficient condition for the two tensors to be the same. _________________________________________________________________ 3.26 For the displacement field: ( )22 6 11 2 2 3 3 1 3 1 , , 2 , 10 uk X u k X Xuk X X X k−== =+ = , find the maximum unit elongation for an element that is initially at () 1, 0, 0 . ----------------------------------------------------------------------------------------- Ans. [] ()1 32 31 120 0 0 22 0 2kX kX kX kX X k X⎡⎤ ⎢⎥∇=⎢⎥ ⎢⎥+⎣⎦u , thus, for ( )() 123,, 1 , 0 , 0XX X = , [] [ ]20 0 20 000 000 20 2 0 2kk k kk k k⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥∇= → =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦uE , the characteristic equation for this tensor is: () ( )2 2 12 320 00 0 0 2 0 0 , 3 , . 02kk kk k k kkλ λλ λ λ λ λ λ− ⎡⎤ −= → − − − = → = = =⎢⎥⎣⎦ − Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-14Thus, the maximum unit elongation at () 1, 0, 0 is 6 233 1 0kλ−== × . _________________________________________________________________ 3.27 Given the matrix of an infinitesimal strain tensor as []12 22 2200 00 00kX kX kX⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥ − ⎣⎦E . (a) Find the location of the particle that does not undergo any volume change. (b) What should the relation between 12and kk be so that no element changes its volume? ----------------------------------------------------------------------------------------- Ans. (a)()() 11 22 33 1 2 2 20dVEE E k k XdVΔ=++=− = . Thus, the particles which were on the plane 20 X=do not suffer any change of volume. (b) If () 12 1220 , . , 2kk i e kk−= = , then no element changes its volume. _________________________________________________________________ 3.28 The displacement components for a body are: 22 4 11 2 2 3 1 3 ( ), (4 ), 0, =10 uk X X u k X X u k−=+ = − = . (a) Find the strain tensor. (b) Find the change of length per unit length for an element which was at ()1,2,1 and in the direction of 12e+ e . (c) What is the maximum unit elongation at the same point ()1,2,1 ? (d) What is the change of volume for th e unit cube with a corner at the origin and with three of its edges along the positive coordinate axes? ----------------------------------------------------------------------------------------- Ans. (a) [] [ ]11 33 320 2 0 0 08 0 0 4 000 04 0kX k kX kk X k X kX⎡ ⎤ ⎡⎤ ⎢ ⎥ ⎢⎥∇=− → =⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦uE (b) At ()1, 2,1 , []20 0 00 4 04 0k k k⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦E , for () []' '' '11 2 1 1 1 120 0 1 11, 1 1 0 0 0 4 12 204 00k E kk k⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=⋅ = =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦e= e+ e e E e (c) The characteristic equation is ()( )2 220 0 04 0 2 4 0 04k kk k kλ λλ λ λ− ⎡⎤ −=→ − − =⎢⎥⎣⎦ − 1232, 4, 4kk kλλλ →= = = − . The maximum elongation is 4k. (d) Change of volume per unit volume 12iiE kX== , which is a function of 1X. Thus, 112 11 1 122 ( 1 ) ooVk X d Vk X d X k XkΔ= = = =∫∫. _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-153.29 For any motion, the mass of a particle (materia l volume) remains a constant (conservation of mass principle). Consider the mass to be the pr oduct of its volume and its mass density and show that (a) for infinitesimal deformation o (1 )kkEρ ρ+= where oρdenote the initial density and ρ, the current density. (b) Use the smallness of kkEto show that the current density is given by o(1 )kkE ρρ=− . ----------------------------------------------------------------------------------------- Ans. (a) o oo o oo o1dV dV dV dVdV dVdV dV dVρρρρ ρ ρ⎛⎞ +Δ Δ=→ == = + ⎜⎟ ⎝⎠, For small deformation, okkdVEdVΔ=→ () o 1kkE ρρ=+ . (b) From bionomial theorem, for small ()1, 1 + 1kk kk kkE EE−≈− , thus, ()()1 oo11kk kkE E ρρ ρ−=+ =− . _________________________________________________________________ 3.30 True or false: At any point in a body, there al ways exist two mutually perpendicular material elements which do not suffer any change of angle in an arbitrary small deformation of the body. Give reason(s). ----------------------------------------------------------------------------------------- Ans. True. The strain tensor Eis a real symmetric tensor, for which there always exists three principal directions, with respect to which, the matrix of Eis diagonal. That is, the non-diagonal elements, which give one-half of the change of angle between the elements which were along the principal directions, are zero. _________________________________________________________________ 3.31 Given the following strain components at a point in a continuum: 6 11 12 22 33 13 23 , 3 , 0, 10 EEEk E k EE k−=== = == = Does there exist a material element at the point which decreases in length under the deformation? Explain your answer. ----------------------------------------------------------------------------------------- Ans. [] () ( ) ()()2 2 2 12 300 00 0 , 3 0 003 0 0 3 32 0 3 , 0 , 2 .kk k k kk k k k k k kk kk k kλ λλ λ λ λλ λ λ λ λ− ⎡⎤ ⎢⎥ ⎡⎤=→ − = → − − − =⎢⎥ ⎢ ⎥ ⎣⎦⎢⎥ − ⎣⎦ →−−+= → = = =E Thus, the minimum unit elongation is 0. Therefore, there does not exist any element at the point which has a negative unit elongation (i.e., decreases in length). _________________________________________________________________ 3.32 The unit elongation at a certain point on the surface of a body are measured experimentally by means of strain gages that are arranged o45apart (called the o45strain rosette) in the direction of () 11 221,a n d 2ee + ee . If these unit elongation are designated by , , abc respectively, what are the strain components 11 22 12, and EEE ? Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-16----------------------------------------------------------------------------------------- Ans. With () 11 21 2′=ee + e , we have, [] ()11 12 13 11 1 1 21 22 23 11 12 21 22 31 32 331 11110 1220EEE E EEE EEEE EEE⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥′′′=⋅ = = + + +⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦eE e , with 12 21E E= , ()() 11 22 11 11 12 22 12 111222EEEE E E E E+′′=+ + → = − . Thus, the strain components are: () 11 22 12 , , 2acEa Ec Eb+=== − . _________________________________________________________________ 3.33 (a) Do the previous problem, if the measured strains are 6200 10−× , 650 10−× and 6100 10−× in the direction 11 2, a n d ′ee e respectively. (b) Find the principal directions, assuming 31 32 33 0 EEE=== . (c) How will the result of part b be altered if 330 E≠. ----------------------------------------------------------------------------------------- Ans. (a) With 6 11200 10 E−=× , 6 11 50 10E−′=× and 6 22100 10 E−=× , we have, from the results of the previous problem, 66 11 22 12 11200 10050 10 100 1022EEEE− − + + ⎛⎞′=− =− × = −× ⎜⎟⎝⎠ (b) 11 12 2 12 22 11 22 120 00 ( ) ( ) 0 00EE EE E E Eλ λλ λ λ λ− ⎡⎤ −= →−− − =⎣⎦ − ()()22 11 22 11 22 12 0 EE E EE λλλ⎡⎤→+− + + − =⎣⎦, () ()2 2 11 22 11 22 12 1,2 34 , 02EE EE E λλ+± − + == , thus, () () ( )22 6 6 1,2 36200 100 200 100 4 100 261.8 1010 , 02 38.2 10λλ− − −⎡⎤+± − + − ×⎢⎥=× = =⎢⎥×⎢⎥⎣⎦ The principal direction for 3λ is 3e. The principal directions corresponding to the other two eigenvalues lie on the plane of 12 and ee . Let () 11 2 2 1 2 1 1 1 1 2 2 cos sin , then E 0 E αα θ θ λ α α+≡ + −+ = n= e e e e , Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-17()11 2 11 2EtanEλ αθα−→= = , For 6 o 11 1 1 12261.8 200 61.8261.8 10 , tan = 0.618 31.7100 100E Eλλθ θ− − −=× = == −→ = −−−, Or, 12 0.851 0.525=−ne e For 6o 21 1 2 1238.2 20038.2 10 , tan = 1.618 58.3100E Eλλθ θ− − −=× = = → =− Or, 12 0.525 0.851=+ne e . (c) If 330 E≠, then the principal strain corresponding to the direction 3e is 33Einstead of zero. Nothing else changes. _________________________________________________________________ 3.34 Repeat the previous problem with 6 11 11 22 1000 10 EEE−′=== × . ----------------------------------------------------------------------------------------- Ans. (a) From the results of Problem 3.32, 6 11 22 12 1120001000 10 022EEEE− +⎛⎞′=−= − × = ⎜⎟⎝⎠, (b) and (c) []3 3 3310 0 0 01 0 0 00 E− −⎡⎤ ⎢⎥⎢⎥= ⎢⎥ ⎢⎥⎣⎦E , the principal strains are 310− in any directions lying on the plane of 12 and ee and the principal strain33E is in 3e direction. _________________________________________________________________ 3.35 The unit elongation at a certain point on the surface of a body are measured experimentally by means of strain gages that are arranged o60apart (called the o60strain rosette) in the direction of () () 11 2 1 211,a n d 22− ee + 3 e e + 3 e . If these unit elongation are designated by , , abc respectively, what are the strain components 11 22 12, and EEE ? ----------------------------------------------------------------------------------------- Ans. With '' '11 2 1 1 2() / 2 , ( ) / 2== −ee + 3 e e e + 3 e , we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-18()11 12 13 ' ''11 1 1 21 22 23 11 12 22 31 32 331 1113 0 3 2 3 3440EEE EE E E E E E EEE⎡⎤ ⎡⎤⎢⎥ ⎢⎥⎡⎤ =⋅ = = + + ⎢⎥ ⎢⎥ ⎣⎦⎢⎥ ⎢⎥⎣⎦ ⎣⎦eE e (i) ()11 12 13 "' " " 11 1 1 21 22 23 11 12 22 31 32 331 1113 0 3 2 3 3440EEE EE E E E E E EEE−⎡⎤ ⎡⎤⎢⎥ ⎢⎥⎡⎤ =⋅ = − = − + ⎢⎥ ⎢⎥ ⎣⎦⎢⎥ ⎢⎥⎣⎦ ⎣⎦eE e (ii) (i) & (ii), []'' ' ' 22 11 11 111122 2 233E EE E b c a⎡⎤ →= + − = + −⎣⎦, '' ' 11 11 1233EE bcE−−== , 11E a=. _________________________________________________________________ 3.36 If the o60 strain rosette measurements give 66 62 1 0, b1 1 0, c1 . 5 1 0a− −−=× =× = × , obtain 11 12 22, and EEE . Use the formulas obtained in the previous problem. ----------------------------------------------------------------------------------------- Ans. Using the formulas drived in the previous problem, we have, [] () ()() ( )66 221122 2 1 2 1 . 5 2 1 0 1 1 033Eb c a− −⎡⎤ =+ − = + − × = ×⎣⎦, 6 121 10 32 3bcE− −== − × , 6 1121 0 E−=× . _________________________________________________________________ 3.37 Repeat the previous problem for the case 6b= c 2000 10 a−==× . ----------------------------------------------------------------------------------------- Ans. [] ( ) () ( ) ()63 22112 2 2 2000 2 2000 2000 10 2 1033Eb c a− −⎡⎤ =+ − = + − × = ×⎣⎦, 12 0 3bcE−== , 3 1121 0 E−=× _______________________________________________________________________ 3.38 For the velocity field: 2 21kxv= e , (a) find the rate of deforma tion and spin tensors. (b) Find the rate of extension of a material element dd sx= n where () 12 /2 n= e +e at 1253+ x= e e . ----------------------------------------------------------------------------------------- Ans. 2 12 2 3 , 0 vk xvv== = , [] [ ] [] [] []22 2 SA 2202 0 0 0 0 0 0 0 0 00 , 00 000 0 0 0 0 0 0kx kx kx kx kx⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥→∇ = → =∇ = =∇ =−⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦vD v W v (b) At the position 1253+ x= e e , [] [ ]03 0 0 3 0 30 0 , 30 0 00 0 0 0 0kk kk⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥== −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦DW Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-19For the element dd sx= n with 12() / 2n= e +e , the rate of extension is: () () []03 0 1 11103 0 0 1 3200 0 0nnk D kk⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=⋅ =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦nD n = . _________________________________________________________________ 3.39 For the velocity field: 1 11tk xα⎛⎞+ ⎜⎟+⎝⎠v= e , find the rates of extension for the following material elements: ()1 11 dd s=xe and ()()()2 21 22 dd s=+x/ e e at the origin at time 1 t=. ----------------------------------------------------------------------------------------- Ans. 12 3 1, 01tkvv vxα⎛⎞+== =⎜⎟+⎝⎠[]() ( ) []2 1 /1 0 0 00 0 00 0tk xα⎡⎤−+ +⎢⎥ →∇ = =⎢⎥ ⎢⎥ ⎢⎥⎣⎦vD . At ()() 123 1 and at , , 0,0,0tx x x== , []() 10 0 00 0 00 0kα⎡ ⎤ −+ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦D . Rate of extension for ()1 11 dd s=xe is () 11 1 D kα=−+ ; for ()()()2 21 22 dd s=+x/ e e , it is: []() ()' 1110 0 1 11110 0 0 0 1 12200 0 0k D kα α⎡⎤−+ ⎡⎤ ⎢⎥ ⎢⎥== − +⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ _________________________________________________________________ 3.40 For the velocity field ()()12 cos sintxπ v= e (a) find the rate of deformation and spin tensors, and (b) find the rate of extension at 0t=for the following elements at the origin: () () ()()()12 3 11 2 2 3 1 2, a nd / 2 dd s d d s d d s== =xe x ex e + e . ----------------------------------------------------------------------------------------- Ans. (a) With ()() 12 1 30, cos sin , 0vv t x v π == = , [] [ ]( ) ()1 1100 0 0 c o s c o s / 2 0 cos cos 0 0 cos cos / 2 0 0 00 0 0 0 0tx tx txππ ππ ππ⎡ ⎤ ⎡⎤ ⎢ ⎥ ⎢⎥∇= → =⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦vD , []() ()1 10c o s c o s / 2 0 cos cos / 2 0 0 00 0tx txππ ππ⎡⎤ − ⎢⎥=⎢⎥ ⎢⎥⎣⎦W . (b) At 0t= and () () 123, , 0,0,0xx x= , []0/ 2 0 /2 0 0 00 0π π⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦D . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-20For ()1 11 dd s=xe , rate of extension is 11D=0, for ()2 22 dd s=xe , 220 D= and for ()()()3 31 2/2 dd s=xe + e , []' 110/ 2 0 1 11 1 0/ 200 12200 0 0Dπ ππ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥= =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ _________________________________________________________________ 3.41 Show that the following velocity components correspond to a rigid body motion. 123 2 13 312 , , vxxv xxvxx=− = − + =− ---------------------------------------------------------------------------------------- Ans. [] [ ]01 1 0 0 0 1 0 1 000 11 0 0 0 0− ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥∇= − → =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥−⎣⎦ ⎣ ⎦vD Therefore, the velocity field is a rigid body motion.. _________________________________________________________________ 3.42 Given the velocity field r1 rv= e , (a) find the rate of deformati on tensor and the spin tensor and (b) find the rate of extension of a radial material line element. ----------------------------------------------------------------------------------------- Ans. With 1, 0rzvv vrθ== = , we have, using Eq. (2.34.5) [] [] [ ][]2 21 1v 00 1100 , 00 0 1rr r r zz zvv v rr zr vv vvrr z r vv v rr zθ θθ θθ θ θ⎡⎤∂∂ ∂⎛⎞ ⎡⎤− − ⎢⎥⎜⎟⎢⎥ ∂∂ ∂⎝⎠⎢⎥⎢⎥⎢⎥∂∂ ∂⎛⎞ ⎢⎥=+ = = =⎢⎥⎜⎟ ⎢⎥∂∂ ∂⎝⎠⎢⎥⎢⎥⎢⎥∂∂ ∂ ⎢⎥⎢⎥⎢⎥⎣⎦ ∂∂ ∂⎢⎥⎣⎦vD W 0∇ . (b) The rate of extension for a radial element is 21 rrD r=− . _________________________________________________________________ 3.43 Given the two-dimensional velocity field in polar coordinates: 40, 2rvv rrθ==+ (a) Find the acceleration at 2 r= and (b) find the rate of deformation tensor at 2 r=. ----------------------------------------------------------------------------------------- Ans. (a) Using Eq. (3.4.12), ()2 2142rr r rrv v vv vav v rtr r r r rθ θ θθ∂∂ ∂ ⎛⎞ ⎛⎞= + + − =− =− + ⎜⎟ ⎜⎟∂∂ ∂ ⎝⎠ ⎝⎠, 0rrvv v vav vtr rθθ θ θ θθ∂∂ ∂ ⎛⎞=+ + +=⎜⎟∂∂ ∂ ⎝⎠. At 2 r=, 2(6) / 2 18ra=−= − , 0 aθ=. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-21(b) Eq. (2.34.5) []2 21 4 vv 02 0 4 10 20rr rvv rr r r v vvvr rr rθθ θ θθθ θ⎡⎤∂∂ ⎡ ⎤ ⎛⎞ ⎛⎞ ⎡⎤− −+ − ⎢⎥⎜⎟ ⎜⎟ ⎢ ⎥ ⎢⎥ ∂∂⎝⎠ ⎝⎠ ⎢⎥ ⎢ ⎥ →= = = ⎢⎥⎢⎥ ⎢ ⎥ ∂ ∂∂⎛⎞ ⎛⎞ ⎢⎥− + ⎢⎥ ⎢ ⎥ ⎜⎟ ⎜⎟ ⎢⎥∂⎣⎦ ∂∂ ⎝⎠ ⎝⎠ ⎣ ⎦ ⎣⎦v∇ . [][ ]2 S 204 / 4/ 0r r⎡⎤−=∇ = ⎢⎥ ⎢⎥−⎣⎦Dv , at 2 r=, []01 10−⎡⎤=⎢⎥−⎣⎦D . _________________________________________________________________ 3.44 Given the velocity field in spherical coordinates: 20, 0, sinrBvvv A r rθφ θ⎛⎞=== + ⎜⎟⎝⎠ (a) Determine the acceration field and (b) find the rate of deformation tensor. ----------------------------------------------------------------------------------------- Ans. (a) From Eq. (3.4.16), 2 2 2 21sin sin sinrr r r rrvv v vv v v Bav v v A r tr r r r r rφφ θ θφ θ θ θθ φ∂∂ ∂ ∂= + + − + − =− =− + ∂∂ ∂ ∂⎛⎞ ⎛⎞ ⎛⎞⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ 2 2 2cos sincos cot sinrrvv vv v v v Bav v v A r tr r r r r rφφ θθ θ θ θ θφθθθθ θθ φ∂∂ ∂ ∂=+ + ++ − = − = − + ∂∂ ∂ ∂⎛⎞ ⎛⎞ ⎛⎞⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ sin cos 0 sinrrvv v v v vav v v tr rrφφ φ φ φ θ φθ θθ θθ φ∂∂ ∂ ∂ =+ + + + + = ∂∂∂ ∂⎛⎞ ⎜⎟⎝⎠ (b) Eq. (2.35.25) → []11 sin cot 11 sin cot 11 sinrr r r rv v vv v rr r r r v vv v v rr r r r vv v v v rr r r rφ θ φ θθ θ φφ φ θθθ φ θ θθ φ θ θθ φ⎡⎤ ⎛⎞ ∂∂ ∂⎛⎞−− ⎢⎥ ⎜⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝⎠ ⎢⎥ ⎢⎥⎛⎞ ∂∂ ∂⎛⎞⎢⎥=+ − ⎜⎟ ⎜⎟⎢⎥∂∂ ∂⎝⎠ ⎝⎠⎢⎥ ∂∂ ∂⎢⎥ ⎛⎞++ ⎜⎟ ⎢⎥∂∂ ∂⎢⎥ ⎝⎠ ⎣⎦v∇00 cot00 1,0v r v r vv rrφ φ φφθ θ⎡ ⎤− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=−⎢ ⎥ ⎢ ⎥ ∂∂⎢ ⎥ ⎢ ⎥∂∂⎣ ⎦, thus the nonzero components of rate of deformation tensor are: 313sin2 2rvv BDrr rφφ φ θ∂⎛⎞=−+ = −⎜⎟∂⎝⎠, 33cot 11 1cos 022vv BBDA Arr rrφφ θφθθθ∂ ⎛⎞ ⎡⎤⎛⎞ ⎛⎞=− + =−+ ++ =⎜⎟ ⎜⎟ ⎜⎟⎢⎥∂ ⎝⎠ ⎝⎠⎣⎦ ⎝⎠. _________________________________________________________________ 3.45 A motion is said to be irrotational if the sp in tensor vanishes. Show that the following velocity field is irrotational: 222 22 12 12 2, xxrxx r−+=+eev= ----------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-22Ans. 222 21 12 1 2 22, , xxvv r x x rr=− = = + , →[]11 2 2 32 3 12 1 2 22 1 1 23 3 12 1 222 1 221vv x x rr x xx x rr r vv x x rr xx x x rr r∂∂ ∂ ∂ ⎡ ⎤⎡ ⎤−+⎢ ⎥⎢ ⎥∂∂ ∂ ∂⎢ ⎥⎢ ⎥ ∇= =⎢ ⎥⎢ ⎥∂∂ ∂∂−− ⎢ ⎥⎢ ⎥∂∂ ∂ ∂⎣ ⎦⎣ ⎦v , 222 12 12 1 11 222 , a l s o ,x x rr rrxx r xx xr x r∂∂ ∂=+→ = → = =∂∂ ∂, [] [] []22 12 2 1 44S 22 21 1 2 442 =0 . 2xx x x rr xx x x rr⎡⎤ −⎢⎥ ⎢⎥∇= ∇ → =⎢⎥−⎢⎥ − ⎣⎦vv W _________________________________________________________________ 3.46 Let () ()12 12and dd s d d s==xn x m be two material elements that emanate from a particle Pwhich at present has a rate of deformation D. (a) Consdier ()()( )12/( )DD t d d ⋅xx to show that ()() 12 1211cos sin 2Dd s Dd s D ds Dt ds Dt Dtθθθ⎡⎤+− = ⋅ ⎢⎥ ⎣⎦mD n where θ is the angle between and m n . (b) Consider the case of () ()12dd=xx , what does the above formula reduce to? (c) Consider the case where 2πθ=, i.e., ()1dxand ()2dxare perpendicular to each other, where does the above formula reduces to? ----------------------------------------------------------------------------------------- Ans. () ()()() () ()() ()() () ()()()2 12 1 2 1 12 1 2 DD D ddd d d d dd d dDt Dt Dt⎛⎞⎛⎞⎜⎟ ⋅= + = ∇ +∇⎜⎟⎜⎟ ⎝⎠⎝⎠xxx x x x v xx x v x ⋅⋅ ⋅ ⋅ ()()() ()()() ()()(){}() () () TT 12 1 2 1 2 1 22 dd d d d d d d=∇ +∇= ∇ + ∇ =⋅xv xxv xx v v x x D x⋅⋅ ⋅ . With () ()12 12and dd s d d s==xn x m , the above formula give, () ( ) () ( ) 12 12 12 12 2c o s 2DDds ds ds ds ds ds ds dsDt Dtθ ⋅= ⋅ → = ⋅nm nD m nD m . Thus, () () () ()12 2 1 12 12coscos cos 2Dds Dds Dds ds ds ds ds dsDtD t D tθθθ++= ⋅ nD m , ()()() ()12 1211cos sin 2 2Dd s Dd s D ds Dt ds Dt Dtθθθ⎧⎫⎪⎪→+ − = ⋅ = ⋅⎨⎬⎪⎪⎩⎭nD m mD n . (b) For , () ()12dd d s==xx n the above formula ()()() ()1 nnDd sDds Dt⎧⎫⎪⎪→= ⋅ =⎨⎬⎪⎪⎩⎭nD n , no sum on n. (c) For ()1dxperpendicular to ()2dx, o90θ= , we have, ()22nmDDDtθ−=⋅ = nD m . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-23_________________________________________________________________ 3.47 Let 123,,eee and 123,,DD D be the principal directions and corresponding principal values of a rate of deformation tensor D. Further, let () ()( ) 12 3 11 2 2 3 3, and dd s d d s dd s== =xe x ex e be three material elements. Consider the material derivative ()() () (){ }12 3/DD t d d d xx x⋅× and show that () 1231Dd VDD DdV Dt=++ , where 123 dV ds ds ds= . ----------------------------------------------------------------------------------------- Ans. Since the principal directions are (or can alwa ys be chosen to be) mutually perpendicular, therefore, () () ()12 3 123 dd d d s d s d s d V⋅×= =xx x . ()() () () () 123 3 12 23 13 12Dd s d sd s Dd s Dd V Dd s Dd sds ds ds ds ds dsDt Dt Dt Dt Dt→= = + + , ()()()()3 12 11 22 33 12 3111 1 Dd s Dd V Dd s Dd sD DDdV Dt ds Dt ds Dt ds Dt→=++= + + . _________________________________________________________________ 3.48 Consider a material element dd sx= n (a) Show that () () /DD t −⋅ n = Dn + Wn n Dn n , where Dis rate of deformation tensor and W is the spin tensor. (b) Show that if nis an eigenvector of D, then, D Dt=nWn = nω× ----------------------------------------------------------------------------------------- Ans. (a)()1()D D Dds D Dds Dds ds ds dsDt Dt Dt Dt ds Dt Dt⎛⎞ ⎛⎞= + =+ =+ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠nn nnn n n D n n ⋅ . [see Eq.(3.13.12) ]. We also have, ()() () ()DDds d d dsDtD t== ∇ ∇nx v x = v n , therefore, () () ( ) () ( ) ()DD Dt Dt⎛⎞∇+→ = ∇ − =−⎜⎟⎝⎠nnvn = nnD n vn nnD n D + Wn nnD n ⋅⋅ ⋅ . (b) If nis an eigenvector ofD, then λ Dn = n , therefore, () ( )D Dtλλ =− ⋅ = − =nD+W n n n D n n+W n n W n . That is, D Dt=nWn. Since Wis antisymmetric →Wn = nω×, where ωis the dual vector for W. Thus D Dt=nWn = nω×. That is, the principal axes of D rotates with an angular velocity given by the dual vector of the spin tensor. _________________________________________________________________ 3.49 Given the following velocity field: ()2 12 3 2 1 2 3 1 3 2, , vk x xv x xvk x x=− = − = for an incompressible fluid, determine the value of k, such that the equation of mass conservation is satisfied. ----------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-24Ans. 3 12 11 12300 0 1v vvxk x kxx x∂ ∂∂++= → − + = → =∂∂∂ _________________________________________________________________ 3.50 Given the velocity field in cylindrical coordinates: ( , ), 0rzvf r v vθθ= == . For an incompressible material, from the conservation of mass principle, obtain the most general form of the function ( , ) frθ. ----------------------------------------------------------------------------------------- Ans. The equation of continuity for an incompressible material is [see Eq.(3.15.11)]: () ( )1100 0 ,rr z v vv v fffr fr grr r z rr r rθθθ∂∂∂ ∂∂++ + = → + = → = → =∂∂ ∂ ∂ ∂. Therefore, ( ) / fgrθ= , where ()gθis an arbitrary function of θ. _________________________________________________________________ 3.51 An incompressible fluid undergoes a two-dimensional motion with cos /rvk rθ= . From the consideration of the principle of conservation of mass, find vθ, subject to the condition that 0 at 0 vθθ== . ----------------------------------------------------------------------------------------- Ans. ()3/2cos 1 1cos2r rv kvkr r rθθ∂ ⎛⎞=→ = − ⎜⎟∂ ⎝⎠, 3/2(c o s)rv k r rθ= 3/21( c o s) 2rrvv k rr rθ ∂ ⎛⎞→+ = ⎜⎟∂ ⎝⎠. The equation of continuity for an incompressible fluid is [see Eq.(3.15.11)]:10rr z v vv v rr r zθ θ∂∂∂++ + =∂∂ ∂. Thus, ()cos sin.22v kkvf r rrθ θθθ θ∂⎛⎞ ⎛⎞=− → =− +⎜⎟ ⎜⎟∂⎝⎠ ⎝⎠ Since 0 at 0 vθθ==, Therefore, () 0fr=. Thus, sin 2kv rθθ⎛⎞=−⎜⎟⎝⎠. _________________________________________________________________ 3.52 Are the following two velocity fields isochoric (i.e., no change of volume)? (i) 222 11 2 2 12 2, xxrxx r+=+eev= and (ii) 222 21 12 12 2, xxrxx r−+=+eev= ----------------------------------------------------------------------------------------- Ans. (i) With 22 11 2 2 /, /, vx rv x r== 222 12 rxx=+ , 2 222 11 1 12 1 2 23 2 4 11 12 22 2 22 2 1 21 2 23 24 24 42 2 22 1 22211. 2 2 , 2 2 . 22 2 211 2 2 2, 0.vx x rr rrxx r x r xxx xx rr rr vx x v vx x r xx x x rr rr rr rr r⎛⎞ ∂ ∂∂ ∂=− =− =+→ = = ⎜⎟∂∂ ∂∂ ⎝⎠ ∂∂ ∂ ∂=− =− + =− − =−=∂∂ ∂ ∂ (ii) 22 12 2 1 /, /, vx r v x r=− =222 12 rxx=+ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-25222 12 2 1 12 1 34 11 1 2 1 12 1 2 21 12 34 4 4 22 1 222= 2 2 22 2 2, 0.vx x x rrrxx r xxx x rr v x xx v v x x xx r xx x x rr r r⎛⎞ ∂ ∂∂= =+→ = ⎜⎟∂∂ ∂ ⎝⎠ ∂∂ ∂ ∂= − = − + =−=∂∂ ∂ ∂ _________________________________________________________________ 3.53 Given that an incompressible and inhomogene ous fluid has a density field given by 2kxρ= . From the consideration of the principle of conservation of mass, find the permissible form of velocity field for a two dimensional flow () 30v=. ----------------------------------------------------------------------------------------- Ans. Since the fluid is incompressible, therefore, () 12 1 2 2 1200 0 0 0 0 . Dvv v v k vDt t x xρ ρρ ρ∂∂∂=→ + + =→+ + =→ =∂∂ ∂ The conservation of mass equation of an incomp ressible fluid in two dimensional flow is ()12 1 12 2 12 100 , 0 . vv vvf xvxx x∂∂ ∂+= →= → = =∂∂ ∂ _________________________________________________________________ 3.54 Consider the velocity field: 1 11x ktα +v= e . From the consideration of the principle of conservation of mass, (a) Find the density if it depends only on time t, i.e., ( ) tρρ= , with () o 0ρρ= . (b) Find the density if it depends only on 1x, i.e., 1ˆ()xρρ= , with oˆ() *xρρ= . ----------------------------------------------------------------------------------------- Ans.(a) Equation of conservation of mass is 3 12 12 3 123 1 2 30v vvvv vtx x x x x xρρρρρ⎛⎞ ∂ ∂∂ ∂∂∂ ∂++++ + += ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠. With 1 12 3 , 01xvv vktα= ==+, () () o/ oo 00l n l n 1 111t k dd d tkt ktdt kt kt kρ α ρρα ρ ρ α ρραρρ ρ−→+ = → = − → = − +→= +++∫∫. (b) with ()1xρρ= and 1 12 3 , 01xvv vktα== =+ 1 o3 12 1 12 3 123 1 2 3 1 o 11 1 11 o 1 *0011 0, ln ln **x xv vv x dvv vtx x x x x x k t d x k t x dx x ddxdx x x xρ ρα ρρρρ ρ αρρ ρρ ρρρ ρρρ⎛⎞ ∂ ∂∂ ∂∂ ∂∂++ ++ + += → + = ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ + + ⎝⎠ →+ = →= −→= − → = ∫∫ where o ρis the density at1oxx=. _________________________________________________________________ 3.55 Given the velocity field: ( ) 11 2 2xtx tα+ v= e e . From the consideration of the principle of conservation of mass, determine how the fluid density varies with time, if in a spatial description, it is a function of time only. ----------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-26Ans. Equation of conservation of mass is 3 12 12 3 123 1 2 30v vvvv vtx x x x x xρρρρρ⎛⎞ ∂ ∂∂ ∂∂∂ ∂++++ + += ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠. With 11 2 2 3 , , 0 vx t v x t vα α = == , ()2 o2 o o 002 l nt t ddtt t d t t edtρ α ρρρ ρρα α α α ρ ρρρ−++ = → = − →= − → = ∫∫. _________________________________________________________________ 3.56 Show that im ik km kmiWE E X XX∂∂ ∂=−∂∂∂, where 1 2im im miuuEX X⎛⎞∂∂=+⎜⎟∂∂⎝⎠is the strain tensor and 1 2im im miuuWX X⎛⎞∂∂=−⎜⎟∂∂⎝⎠is the rotation tensor. ------------------------------------------------------------------------------------------ Ans. 22 222 211 22 1 2 1 2im i m i m kk m i m k i k ik k m mk mi mi ik ik km i kk m mk i im k m iWu u u u XX XX X X X X uuu u XX XX XX X X uu u u EE X XX X X X x x⎛⎞ ⎛⎞ ∂∂ ∂ ∂ ∂∂=− = − = ⎜⎟ ⎜⎟⎜⎟ ∂∂∂∂ ∂ ∂∂ ∂ ⎝⎠ ⎝⎠ ⎛⎞∂∂∂∂+−− = ⎜⎟⎜⎟∂∂ ∂∂ ∂∂ ∂ ∂⎝⎠ ⎛⎞⎛⎞ ⎛ ⎞∂∂ ∂∂ ∂∂∂∂+− + =− ⎜⎟⎜⎟ ⎜ ⎟⎜⎟∂∂∂ ∂ ∂ ∂ ∂ ∂⎝⎠ ⎝ ⎠⎝⎠ _________________________________________________________________ 3.57 Check whether or not the following distributi on of the state of strain satisfies the compatibility conditions: []12 1 2 4 12 3 3 23 1 3, , 10XX X X kX X X X k XX X X−+⎡⎤ ⎢⎥=+ =⎢⎥ ⎢⎥ + ⎣⎦E ----------------------------------------------------------------------------------------- Ans. Yes. We note that the given ijEare linear in 12 3, and XXX and the terms in the compatibility conditions all involve s econd derivatives with respect to iX, therefore these conditions are obviously satisfied by the given strain components. _________________________________________________________________ 3.58 Check whether or not the following distributi on of the state of strain satisfies the compatibility conditions: []22 2 12 3 1 3 22 4 23 1 2 13 1 20, 1 0XX X X X kX X X k XX X X−⎡⎤ +⎢⎥ ⎢⎥=+ = ⎢⎥ ⎢⎥⎣⎦E ----------------------------------------------------------------------------------------- Ans. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-2722 2 11 22 12 22 12 21 22 2 33 23 22 22 23 3220 0 0 , O K 2 0 2 0, not satisfiedEE E XX XX EE EkXX XX∂∂ ∂+= → + =∂ ∂∂ ∂∂∂+= → + ≠∂ ∂∂ The given strain components are not compatible. _________________________________________________________________ 3.59 Does the displacement field: 3 11 2 1 2 3 3sin , , cosuX u X X u X=== correspond to a compatible strain field? ----------------------------------------------------------------------------------------- Ans. Yes. The displacement field obviously exists. In fact, the displacement field is given . There is no need to check the compatibility conditions. Wh enever a displacement field is given, there is never any problem of compatibility of strain components. _________________________________________________________________ 3.60 Given the strain field: 4 12 21 1 2 , 1 0 EEk X X k−== = and all other 0ijE=. (a) Check the equations of compatibility for this st rain field and (b) by attempting to integrate the strain field, show that there does not exist a con tinuous displacement field for this strain field. ----------------------------------------------------------------------------------------- Ans. (a)22 2 11 22 12 22 12 2120 0 2EE EkXX XX∂∂ ∂+= → + ≠∂ ∂∂. This compatibility condition is not satisfied. (b) () ()12 11 1 1 2 3 22 2 2 1 3 120 0 , . Also, 0 0 ,uuE uu X X E u u X XXX∂∂=→ =→ = =→ =→ =∂∂. Now, ()()() ()12 3 21 3 12 12 1 2 2 3 1 3 21 2 1,,22 , ,uXX u XX uuE kX X f X X g X XXX X X∂∂ ∂∂=+→ = + = +∂∂ ∂ ∂, That is, ()() 12 2 3 1 32, ,kX X f X X g X X=+ . Clearly, there is no way this equation can be satisfied, because the right side can not have terms of the form of 12XX. _________________________________________________________________ 3.61 Given the following strain components: () () 11 2 3 22 33 2 3 12 13 231,, ,, 0 Ef X X E E f X X E E Eν α α== = −= = = . Show that for the strains to be compatible, () 23, fXX must be linear in 23and X X. ----------------------------------------------------------------------------------------- Ans () ()22 22 22 2 2 23 23 33 13 11 22 12 11 22 2 22 2 12 13 21 2 31 3,, 1120 , 20fX X fX X EE EE E E XX XX XX X XX X αα∂∂ ∂∂ ∂∂ ∂ ∂+ = →= + = →=∂∂ ∂∂ ∂ ∂∂ ∂ , ()2 2 23 23 31 11 12 23 1 1 2 3 23, 10fX X EE EE XX X X X X XX α∂ ⎛⎞∂∂ ∂∂ ∂=−++ → =⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂∂ ⎝⎠, Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-28()()()222 23 23 23 22 23 23,,,0, 0, 0fX X fX X fX X XX XX∂∂∂===∂∂ ∂∂. () 23, fXX→ is a linear function of 23 and X X. We note also 22 2 22 33 23 22 22 2 2 13 32 3 202E E E ff XX XX X Xν α⎛⎞ ∂∂∂ ∂∂+= − += = ⎜⎟⎜⎟ ∂ ∂∂ ∂ ∂ ⎝⎠, 2 2 31 23 22 12 31 31 2 2 3 10EE EE f XX XX X X X Xν α⎛⎞∂∂ ∂∂ ∂∂=− = = − + + ⎜⎟∂∂ ∂∂ ∂ ∂ ∂ ∂ ⎝⎠, 2 2 33 23 31 12 12 12 3 3 1 20EE E E f XX XX X X X Xν α⎛⎞ ∂∂ ∂ ∂ ∂∂=− = = − + + ⎜⎟∂∂ ∂∂ ∂ ∂ ∂ ∂ ⎝⎠. Thus, if() 23, fXX is a linear function of 23and X X, then all compatibility equations are satisfied. _________________________________________________________________ 3.62 In cylindrical coordinates (),,rzθ , consider a differential volume bounded by the three pairs of faces: and ; = and = ; and . rr rrd r d zz zzd z θθθ θ θ == + + == + The rate at which mass is flowing into the volume across the face rr=is given by ()() rvr d d zρθ and similar expressions for the other faces. By demanding that the net rate of inflow of mass must be equal to the rate of increase of mass inside the differential volume, obtain the equation of conservation of mass in cylindrical coordinates. Check your answer with Eq. (3.15.7 ). ----------------------------------------------------------------------------------------- Ans. Mass flux across the face rr=into the differential volume dVis ()() rvr d d zρθ . That across the face rrd r=+ out of the volume is ()() rrrd rvr d r d d zρ θ=++ . Thus , the net mass flux into dVthrough the pair of faces and rr rrd r==+ is ()()()()()() rr r rrr rrd r rr rrd rv rd dz v r dr d dz v v rd dzρ θρ θρ ρ θ== + = = +⎡⎤ −+ = −⎣⎦ () rrrd rvd r d d zρ θ=+− . Now, () ()()()r rrrr rrd rvv v rd dz dr rd dzrρρρ θ θ== +⎡⎤∂⎡⎤ −= − ⎢⎥ ⎣⎦∂⎣⎦ and () ()() () rr r rrrd rv drd dz v d v drd dz v drd dzρ θρρ θ ρ θ=+⎡⎤ −= − + = −⎣⎦, where we have dropped the higher order term involving ()r dv d r d d zρθ ⎡⎤⎣⎦which approaches zero in the limit compared to the terms involving only three differentials. Thus, the net mass flux into dVthrough the pair of faces and rr rrd r== + is ()r rvrv d r d d zrρρθ⎧⎫∂⎛⎞−−⎨⎬⎜⎟∂⎝⎠⎩⎭. Similarly, the net mass flux into dVthrough the pair of faces and d θθθ θ θ= =+ is ()vdd r d zθρθθ∂⎛⎞−⎜⎟∂⎝⎠, and the net mass flux into dVthrough the pair of faces and z zz zd z==+ is Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-29()zvdz dr rdzρθ∂⎛⎞⎡⎤ −⎜⎟ ⎣⎦∂⎝⎠ Thus, the total influx of mass through these three pairs of faces is: ()()1 r rzv v vvdr rd dzrr r zθρ ρ ρρθθ⎧⎫ ∂ ∂∂ ⎛⎞ ⎛⎞ ⎛⎞⎪⎪−+ + +⎨⎬⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝⎠ ⎝⎠ ⎪⎪⎩⎭ On the other hand, the rate of increase of mass inside dVis ()rd drdz rd drdzttρρθ θ∂∂=∂∂. Therefore, the conservation of mass principle gives, ()()1 r rzv v vvdr rd dz rd drdzrr r z tθρ ρ ρρ ρθθθ⎧⎫ ∂ ∂∂ ∂ ⎛⎞ ⎛⎞ ⎛⎞⎪⎪−+ + + =⎨⎬⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂∂⎝⎠ ⎝⎠ ⎝⎠ ⎪⎪⎩⎭, That is: 10rr z v vv v trr r zθρ ρρ ρρ θ∂ ∂∂∂ ⎛⎞ ⎛⎞++ + + = ⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎝⎠, Or, 10rr z rzvv vv vvvtr r z r r r zθθ ρρ ρ ρρθθ⎧⎫ ∂ ∂∂ ∂∂ ∂ ∂ ⎛⎞ ⎛⎞++ + + + ++=⎨⎬ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎩⎭. This is the same as Eq.(3.15.7). _________________________________________________________________ 3.63 Given the following deformation in r ectangular Cartesian coordinates: 13 2 1 3 23, , 2x Xx Xx X= =− =− Determine (a) the deformation gradient F, (b) the right Cauchy-Green tensor Cand the right stretch tensor U, (c) the left Cauchy-Green tensor B, (d) the rotation tensor R, (e) the Lagrangean strain tensor *E(f) the Euler strain tensor *e, (g) ratio of deformed volume to initial volume, (h) the deformed area (magnitude and its normal) for the area whose normal was in the direction of 2eand whose magnitude was unity for the undeformed area. ----------------------------------------------------------------------------------------- Ans. (a) []00 3 100 02 0⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥−⎣⎦F , (b) [][][]T01 000 3 1 0 0 00 2 100 0 4 0 30 0 0 2 0 0 0 9−⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥== − − =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦ ⎣ ⎦CF F , [][]1/2100 020 003⎡⎤ ⎢⎥==⎢⎥ ⎢⎥⎣⎦UC . (The only positive definite root). (c) [][] []T00 3 01 0 9 0 0 100 00 2 0 1 0 02 0 3 0 0 0 0 4− ⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥== − − =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥−⎣⎦ ⎣⎦ ⎣ ⎦BF F . (d) [][] []100 3 10 0 00 1 100 0 1 / 2 0 100 02 0 0 0 1 / 3 01 0−⎡⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢⎥== − = −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥−−⎣⎦ ⎣ ⎦ ⎣⎦RF U . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-30(e) []000 0 0 0 11030 03 / 2022008 0 0 4⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤== =⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦*EC - I , (f) 14/9 0 0 1000200 3 / 8−⎡ ⎤ ⎢ ⎥⎡⎤ ⎡ ⎤=− =⎢ ⎥ ⎣⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦*eI B . (g) () () () odet 9 1 4 6V VΔ== =ΔB , (h) ()()T1 oodet dd A−A= F F n , []1 o06 0 11, det 6, 0 0 3620 0dA−−⎡ ⎤ ⎢ ⎥= =−⎢ ⎥ ⎢ ⎥⎣ ⎦F= F , o2=ne , [] ()() () ()T1 oo 300 2 0 0 1det 1 6 6 0 0 1 0 3603 0 0 3dd A d−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− = → −⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦⎢⎥ ⎢ ⎥ ⎢ ⎥−−⎣⎦ ⎣ ⎦ ⎣ ⎦A= F F n A = e . _________________________________________________________________ 3.64 Do the previous problem for the following deformation: 12 23 3 12 , 3 , x Xx XxX=== . ----------------------------------------------------------------------------------------- Ans. (a) []020 003 100⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦F . (b) [][][]T001020 100 200003 040 030100 009⎡⎤⎡ ⎤ ⎡ ⎤ ⎢⎥⎢ ⎥ ⎢ ⎥== =⎢⎥⎢ ⎥ ⎢ ⎥ ⎢⎥⎢ ⎥ ⎢ ⎥⎣⎦⎣ ⎦ ⎣ ⎦CF F . [][]1/2100 020 003⎡⎤ ⎢⎥==⎢⎥ ⎢⎥⎣⎦UC . (The only positive definite root). (c) [][] []T020001 400 003200 090 100030 001⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥== =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦BF F . (d) [][] []10201 0 0 010 00301 / 2 0 001 1000 0 1 / 3 100−⎡⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢⎥== =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦ ⎣⎦RF U . (e) []000 0 0 0 11030 03 / 2022008 0 0 4⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤=− = =⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦*EC I , (f) 13/8 0 0 104 / 9 0200 0−⎡⎤ ⎢⎥⎡⎤ ⎡ ⎤=− =⎢⎥ ⎣⎦ ⎣ ⎦ ⎢⎥⎣⎦*eI B . (g) () () () odet 4 9 1 6V VΔ== =ΔB . (h) ()()T1 oodet dd A−A= F F n , []1 oo 2006 11, det 6, 3 0 0 ,6020dA−⎡⎤ ⎢⎥= ==⎢⎥ ⎢⎥⎣⎦F= F n e Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-31[] ()() () ()T1 oo 10300 3 1det 1 6 0 0 2 1 0 366000 0dd A d−⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎢ ⎥ ⎢⎥ ⎢⎥== →⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥⎣⎦⎢ ⎥ ⎢⎥ ⎢⎥⎣ ⎦ ⎣⎦ ⎣⎦A= F F n A =e _________________________________________________________________ 3.65 Do Prob. 3.63 for the following deformation: 11 2 3 3 2 , 3 , 2 xXx Xx X== = − ----------------------------------------------------------------------------------------- Ans. (a) []100 003 02 0⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦F . (b) [][][]T10 0 1 0 0 100 00 20 0 3 040 03 0 0 20 009⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥==− =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣⎦ ⎣ ⎦CF F , The only positive definite root is[][]1/2100 020 003⎡⎤ ⎢⎥==⎢⎥ ⎢⎥⎣⎦UC . (c) [][] []T100 1 00 1 0 0 003 0 0 2 0 9 0 02 0 0 3 0 0 0 4⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥== − =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥−⎣⎦ ⎣⎦ ⎣ ⎦BF F . (d) [][] []1100 1 0 0 100 003 0 1 / 2 0 001 02 0 00 1 / 3 01 0−⎡⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢⎥== =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥−−⎣⎦ ⎣ ⎦ ⎣⎦RF U , (e) []000 103 / 202004⎡⎤ ⎢⎥⎡⎤=− =⎢⎥ ⎣⎦ ⎢⎥⎣⎦*EC I , (f) 100 0 104 / 9 02003 / 8−⎡ ⎤ ⎢ ⎥⎡⎤ ⎡ ⎤=− =⎢ ⎥ ⎣⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦*eI B . (g) () () () odet 1 9 4 6V VΔ== =ΔB . (h) ()()T1 oodet dd A−A= F F n , []1 oo 260 0 11, det 6, 0 0 3 ,602 0dA−⎡⎤ ⎢⎥= =− =⎢⎥ ⎢⎥⎣⎦F= F n e [] ()() () ()T1 oo 3600 0 0 1det 1 6 0 0 2 1 0 3603 0 0 3dd A d−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== → −⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦⎢⎥ ⎢ ⎥ ⎢ ⎥−−⎣⎦ ⎣ ⎦ ⎣ ⎦A= F F n A = e _________________________________________________________________ 3.66 Do Prob. 3.63 for the following deformation: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-3212 21 332, , 3x Xx X x X== − = ----------------------------------------------------------------------------------------- Ans. (a) []02 0 100 00 3⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦F . (b) [][][]T01 0 0 2 0 1 0 0 200 1 0 0 0 4 0 00300 3 0 0 9−⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥== − =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦CF F , [][]1/2100 020 003⎡⎤ ⎢⎥==⎢⎥ ⎢⎥⎣⎦UC . (The only positive definite root). (c) [][] []T02 0 0 1 0 4 0 0 1 0 0 200 0 1 0 00 3 003 0 0 9− ⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥== − =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣⎦ ⎣ ⎦BF F . (d) [][] []102 0 1 0 0 01 0 100 01 / 2 0 100 00 3 0 0 1 / 3 00 1−⎡⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢⎥== − = −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦ ⎣⎦RF U . (e) []000 103 / 202004⎡⎤ ⎢⎥⎡⎤=− =⎢⎥ ⎣⎦ ⎢⎥⎣⎦*EC I , (f) 13/8 0 0 1000200 4 / 9−⎡ ⎤ ⎢ ⎥⎡⎤ ⎡ ⎤=− =⎢ ⎥ ⎣⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦*eI B (g) () () () od e t 419 6V VΔ== =ΔB . (h) ()()T1 oodet dd A−A= F F n , []1 oo 206 0 11, det 6, 3 0 0 ,6002dA−−⎡⎤ ⎢⎥= ==⎢⎥ ⎢⎥⎣⎦F= F n e 1T oo 103 0 0 3 1[] ( d e t ) [ ] [ ] = ( 1 ) ( 6 ) 6 0 0 1 0 3600 2 0 0dd A d−⎡ ⎤ ⎡⎤ ⎡⎤ ⎢ ⎥ ⎢⎥ ⎢⎥−= →⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥⎣ ⎦ ⎣⎦ ⎣⎦A= FF n A =e _________________________________________________________________ 3.67 Given 11 2 2 2 3 3 3, , xXX x X x X=+ = = . Obtain (a) the deformation gradient and F the right Cauchy-Green tensor C, (b) The eigenvalues and eigenvector of C, (c) the matrix of the stretch tensor1and −U U with respect to the ie-basis and (d) the rotation tensor Rwith respect to the ie-basis. ----------------------------------------------------------------------------------------- Ans. (a) [] [][][]T130 100130 1 3 0 010 , 310010 31 00 001 001001 0 0 1⎡⎤ ⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥== = =⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣ ⎦FC F F . (b) the characteristic equation is Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-33()()2 1 , 2 12313 0 31 0 0 0 1 1 1 1 0 , 00 1 11 121 410.908326, =0.0916735, 12λ λλ λ λ λ λλ λ λ− −= → − − + = − ±−=→ = = For 110.908326λ= , () () ()11 2 2 11 1 11 2 1 21 3 0 1 / 3 3.302775 , 13.302775 0.289785 0.957093 .3.450843λα α α λα α−+ = → = − − = =+= +ne ee e For 20.0916735λ= , () () ()21 2 2 21 1 21 2 1 21 3 0 1 / 3 0.3027755 , 10.3027755 0.957093 0.289784 .1.044832λα α α λα α−+ = → = − − = − =− = −ne e e e For 33 31,λ==ne , (c) The matrices with respect to the principal axes are as follows []10.9083 0 0 0 0.0916735 0 00 1i⎡⎤ ⎢⎥=→⎢⎥ ⎢⎥⎣⎦nC []3.30277 0 0 0 0.302774 0 00 1i⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦nU . 10.302774 0 0 0 3.302772 0 00 1i−⎡⎤ ⎢⎥⎡⎤=⎢⎥ ⎣⎦ ⎢⎥⎣⎦nU . The matrices with respect to the ie-basis are given by the formula [] [][] []T {} { }ii=enUQ U Q : []0.289785 0.957093 0 3.30277 0 0 0.289785 0.957093 0 0.957093 0.289785 0 0 0.302774 0 0.957093 0.289785 0 00 1 0 0 1 00 1i⎡⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢⎥=− −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦ ⎣⎦eU =0.554704 0.832057 0 0.832057 3.05087 0 00 1⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦. 10.289785 0.957093 0 0.302774 0 0 0.289785 0.957093 0 0.957093 0.289785 0 0 3.302772 0 0.957093 0.289785 0 00 1 0 0 1 00 1i−⎡⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢⎥⎡⎤=− −⎢⎥ ⎢ ⎥ ⎢⎥ ⎣⎦ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦ ⎣⎦eU =3.050852 0.832052 0 0.832052 0.554701 0 00 1− ⎡⎤ ⎢⎥−⎢⎥ ⎢⎥⎣⎦. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-34(d) [] []11 3 0 3.050852 0.832052 0 0.55470 0.83205 0 0 1 0 0.832052 0.554701 0 0.83205 0.55470 0 001 0 0 1 0 0 1i−− ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎡⎤== − = −⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦eRF U . _________________________________________________________________ 3.68 Verify that with respect to rectangular Cart esian base vectors, the right stretch tensor Uand the rotation tensor Rfor the simple shear deformation , 11 2 2 2 3 3 , , xXk Xx XxX=+= = , are given by: With 21 / 2(1 / 4)fk−=+ , [] () []2/2 0/2 0 /2 1 /2 0 , /2 0 00 100 1fk ffk f kf k f kf f⎡⎤⎡ ⎤⎢⎥⎢ ⎥=+ = −⎢⎥⎢ ⎥⎢⎥⎢ ⎥⎣ ⎦ ⎢⎥⎣⎦UR . ----------------------------------------------------------------------------------------- Ans. [] ()2/2 0/2 0 /2 0 /2 1 /2 0 00 100 1fk ffk f kf f kf k f⎡⎤⎡⎤⎢⎥⎢⎥=− + ⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦ ⎢⎥⎣⎦RU () ()() () () ( )() ( ) () ()22 2/2 /2 /2 /2 1 /2 0 /2 /2 /2 /2 1 /2 0 00 1fk f k f f k f k f k f kf f f kf kf kf f k f⎡⎤++ +⎢⎥ ⎢⎥=− + − + +⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ()() () []22 22 221/ 4 1/ 4 010 0 1 / 4 0 0 1 0 the given 00100 1fk k fkk fk⎡⎤++⎢⎥ ⎡⎤ ⎢⎥ ⎢⎥=+ = =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦⎢⎥⎣⎦F Since the decomposition of Fis unique, therefore, the given Rand Uare the rotation and the stretch tensor respectively. _________________________________________________________________ 3.69 Let () () () () 11 2 2 12 , dd S d d S==XN X N be two material elements at a point P. Show that if θ denotes the angle between their respective deformed elements ()( )12 12=a n d dd s d d s = xm x n , then, (2) (1) 12cosCNNαβ α βθλλ= , where () ()12 (1) (2) 12 12 12, , a n d ds dsNNdS dSαα αα λλ == = =Ne N e . ----------------------------------------------------------------------------------------- Ans. () () () () () ()()() 12 1 2 1 2 1 2 Tdd d d d d d d⋅= ⋅ =⋅ =⋅x x FX FX X FFX X CX , () () () () ()()()()2 12 1 1 2 12 12 2 1 2 1 12 12 1 2cos ( ) ( ), cos .ds ds dS dS dS dS N N CNN dS dSNNds dsααβ β αβ α β αα ββθ θλλ→= ⋅ = →= ⋅=NC N eC e eC e⋅ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-35_________________________________________________________________ 3.70 Given the following right Cauchy-Green deformation tensor at a point []90 0 04 0 000 . 3 6⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦C (a) Find the stretch for the material elements which were in the direction of 12 3, ,and ee e . (b) Find the stretch for the material element which was in the direction of 12+ee . (c) Find cos θ, where θ is the angle between () () 12 and ddxx where ()1 11 dd S=Xe and ()2 21 dd S= Xe deform to () ()12 12 and dd s d d s==xm x n . ----------------------------------------------------------------------------------------- Ans. (a) For the elements which were in 12 3, ,and ee e direction, the stretches are 11 22 33,,CC C , that is, 3, 2 and 0.6 respectively. (b) Let () [] []' 11 2 1 190 0 1 9 11 1 1 3'1 1 0 0 4 0 1 1 1 0 422 2 2000 . 3 60 0C⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=+ → = = =⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦ee e . That is, the stretch for 1' dd SX= e is ()' 11 /1 3 / 2ds dS C== . (c) o 120c o s 0 9 0 C θθ=→ =→= . There is no change in angle. (note, 123,,}{e e e are principal axes for C. _________________________________________________________________ 3.71 Given the following large shear deformation: 112 2 2 3 3 , , xXXxXxX=+ = = . (a) Find the stretch tensor U(Hint: use the formula given in problem 3.68) and verify that =2UC , the right Cauchy-Green deformation tensor. (b) What is the stretch for the element which was in the direction 2e? (c) Find the stretch for an element which was in the direction of 12+ee . (d) What is the angle between the deformed elements of 11 2 2and dS dSe e ?. ----------------------------------------------------------------------------------------- Ans. (a) For 11 2 2 2 3 3 , , xXk Xx XxX=+ = = , from Prob. 3.68, we have [] ()2/2 0 /2 1 /2 0 00 1fk f kf k f⎡⎤ ⎢⎥ =+⎢⎥ ⎢⎥ ⎢⎥⎣⎦U where1 2 2 14kf−⎛⎞ =+⎜⎟⎜⎟⎝⎠. Thus, with 1 k=, 2/ 5f= [] ()/2 0 1 1/2 0 1 1/2 0 2/2 3/2 0 1/2 3/2 0 1/2 3/2 0 500 1 0 0 1 / 00 5 / 2ff ff f f⎡ ⎤ ⎡⎤ ⎡ ⎤⎢ ⎥ ⎢⎥ ⎢ ⎥=== ⎢ ⎥ ⎢⎥ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦U . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-36[] [] () () []11 / 2 0 11 / 2 0 1 1 0 41/2 3/2 0 1/2 3/2 0 1 2 05001 00 5 / 2 00 5 / 2⎡⎤ ⎡⎤ ⎡⎤⎢⎥ ⎢⎥ ⎢⎥== =⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦UU C . (b) The stretch for the element which was in the direction 2eis 22 2 C= . (c) Let () 11 2'/ 2=ee + e , [] []' 111101 2 11 51101 2 0 1 1103 5 / 222 20010 0dsCdS⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== = → =⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦. (d) ()()o 12 12 121cos 1 2 cos 1 cos 45 2ds dsCdS dSθθ θ θ⎛⎞ ⎛⎞=→ = → =→ = ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠. _________________________________________________________________ 3.72 Given the following large shear deformation: 11 2 2 2 3 3 2, , xXX x X x X=+ = = (a) Find the stretch tensor U(Hint: use the formula given in problem 3.68) and verify that =2UC , the right Cauchy-Green deformation tensor. (b) What is the stretch for the el ement which was in the direction 2e. (c) Find the stretch for an element which was in the direction of 12+ee . (d) What is the angle between the deformed elements of 11 2 2and dS dSe e . ----------------------------------------------------------------------------------------- Ans. For 11 2 2 2 3 3 , , xXk Xx XxX=+ = = , from Prob. 3.68, we have [] ()2/2 0 /2 1 /2 0 00 1fk f kf k f⎡⎤ ⎢⎥ =+⎢⎥ ⎢⎥ ⎢⎥⎣⎦U where1 2 2 14kf−⎛⎞ =+⎜⎟⎜⎟⎝⎠. Thus, with 2 k=, 1/ 2f= [] ()2/2 011 0 1/2 1 /2 0 1 3 0 2 00 2 00 1fk f kf k f⎡⎤ ⎡ ⎤⎢⎥ ⎢ ⎥=+ =⎢⎥ ⎢ ⎥⎢⎥ ⎢ ⎥⎢⎥ ⎣ ⎦ ⎣⎦U . []/2 0 1 1 0 1/2 0 1 1 0 200 1 00 2fk f kf f⎡⎤ ⎡⎤⎢⎥ ⎢⎥=− = − ⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎣⎦ ⎣⎦R . [] [] []211 0 11 0 120 113 0 13 0 250 2001 00 2 00 2⎡⎤ ⎡⎤ ⎡⎤⎢⎥ ⎢⎥⎛⎞ ⎢⎥== =⎜⎟⎢⎥ ⎢⎥ ⎢⎥⎝⎠⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦UU C . (b) The stretch for the element which was in the direction 2eis 22 5 C= . (c) Let () 11 2 /2 ′=ee + e , Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-37 [] [] 111201 3 111 102501 1 107 5 5 2 . 2 3 6220010 0dsCdS⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥′== = → = =⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦. (d) ()()12 12 122cos 1 5 cos 2 cos 5ds dsCdS dSθθ θ⎛⎞ ⎛⎞=→ = → = ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠. _________________________________________________________________ 3.73 Show that for any tensor () ()()1 123,, , d e t d e tjn njmmA XX XX X−∂ ∂ ∂∂AA = A A ----------------------------------------------------------------------------------------- Ans 13 11 12 11 12 13 11 12 13 11 12 13 23 21 22 21 22 23 21 22 23 21 22 23 31 32 33 31 32 33 31 32 33 31 32 33mmm mm m m mmmA AA AAAXXX AAA A A A A AAAAA A A A A A AX XXXAAA A A A A A A AAAX XX∂ ∂∂ ∂∂∂∂ ∂ ∂∂=→ = + +∂∂ ∂ ∂∂∂∂ ∂∂∂AA . Let c ijA denote the cofactor of ijA, i.e., 22 23 21 23 11 12 32 33 31 33,e t c .ccAA AAAAAA AA== − Then, 13 11 12 21 22 11 12 13 21 22 ...ccccc mm m m m mA AA AAAAAAAXX X X X X∂ ∂ ∂∂ ∂∂=+++++∂∂ ∂ ∂ ∂ ∂A That is, ij c ij mmA AXX∂∂=∂∂A. On the other hand, () ()11detdetc ji c jiij ijA A−−=→ = AA AA. Thus, () ()11det detij nj ji jnmm mAA X XX−−∂∂ ∂==∂∂∂AAA AA . _________________________________________________________________ 3.74 Show that if TU = 0 , where the eigenvalues of Uare all positive (nonzero), then T=0 . ----------------------------------------------------------------------------------------- Ans. Using the eigenvectors of Uas basis, we have, [] [ ] [ ]11 12 13 1 1 11 2 12 3 13 21 22 23 2 1 21 2 22 3 23 31 32 33 3 1 31 2 32 3 3300 00 00TTT T T T TTT T T T TTT T T Tλλ λ λ λλ λ λ λλλ λ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥==⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦TU = T U Thus, TU = 0 gives, all 0ijT=, that is, T=0 . _________________________________________________________________ 3.75 Derive Eq. (3.29.21), that is, 22 2 θθ oo o o+rrrBrr zθ θθ θ⎛⎞ ⎛ ⎞ ⎛⎞∂∂∂=+⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝ ⎠ ⎝⎠ ----------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-38Ans. T θθ θ θ θ θB=⋅ =⋅eB e eF F e . From Eq. (3.29.15). we have, To o o θ r θ z oo o o+rrr rr zθθθ θ∂∂∂=+∂∂ ∂Fe e e e , thus, oo o o o o θθ θ r θ z θ r θθ θ z oo o o o o o o++rrr r r rBrr z r r zθθθθ θ θ θθ⎛⎞∂∂∂∂ ∂ ∂=⋅ + = ⋅ ⋅ + ⋅⎜⎟∂∂ ∂∂ ∂ ∂⎝⎠eF e e e eF e eF e eF e . Since, oo o θ r θθ θ z oo o o , rr r rr zθ θθ θ∂∂ ∂⋅⋅ ⋅ =∂∂∂e Fe = , e Fe = e Fe , [See Eq. (3.29.10)], therefore, 22 2 θθ oo o o+rrrBrr zθ θθ θ⎛⎞ ⎛ ⎞ ⎛⎞∂∂∂=+⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝ ⎠ ⎝⎠. _________________________________________________________________ 3.76 Derive Eq. (3.29.23), i.e., rz o o oo oo o o+rz r z rzBrr r r zz θθ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞∂∂∂ ∂∂ ∂=+⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂∂⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ----------------------------------------------------------------------------------------- Ans. T rz r z r zB=⋅ =⋅eB e eF F e , from Eq. (3.29.16), we have, Too o zr θ z oo o ozzz rr zθ∂∂∂=+ +∂∂ ∂Fe e e e , thus, oo o o o o rz r r θ zr r r θ rz oo o o o o o o,+zzz z z zBrr z r r zθθ⎛⎞∂∂∂ ∂ ∂ ∂=⋅ + + = ⋅ ⋅ + ⋅⎜⎟∂∂ ∂∂ ∂ ∂⎝⎠e F e e e e Fe e Fe e Fe . From Eq. (3.29.9), oo o rr r θθ z oo o o , rr r rr z θ∂∂ ∂⋅⋅ ⋅ =∂∂∂eF e = , eF e = eF e , thus, rz oo o o o o o o+rz r z rzBrr r r zz θθ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞∂∂ ∂ ∂ ∂∂=+⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂∂⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠. _________________________________________________________________ 3.77 From ()()() oo o o o o ,,, , ,,, , ,,, rr rz t rz tz z rz tθθ θ θ θ=== ,derive the components of 1−B with respect to the basis at x. ----------------------------------------------------------------------------------------- Ans. From 1dd−X=F x , where oo o r θ zo r o o θ oz and dd r r d d z d d r r d d z θθ++ + + x= e e e X= e e e , we have, ( )oo o 1 or o o θ oz r θ z dr r d dz dr rd dz θθ−++ = + +ee e F e e e () () ()o1 o1 o1 or r r θ rz dr dr rd dz θ−− −→= ⋅ + ⋅ + ⋅ eF e eF e eF e ()()()o1 o1 o1 oo o rr r θ rzrr rdr d dz dr rd dzrzθθθ−− − ∂∂∂→++=⋅+ ⋅+⋅∂∂ ∂eF e eF e eF e o1 o1 o1 ooo rr r θ rz , , rrr rr z θ−− −∂∂∂→⋅ = ⋅ = ⋅ =∂∂∂eF e eF e eF e . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-39o1 o1 o1 oo oo oo r θ z o1 o1 o1 ooo r θ zSimilarly, , , . , , .zz zrrr rr z zzz rr zθθθθ θθ θ θ−−− −− −∂∂∂⋅= ⋅= ⋅=∂∂∂ ∂∂∂⋅= ⋅= ⋅=∂∂∂eF e eF e eF e eF e eF e eF e Thus, 1o o o 1 o o o oo o o o o o o rr θ r 1o o o oo o o zr, .zz zrr z r r z rrr r r r rr z zzzθθ θθ θ θθθ θ−− −∂∂ ∂ ∂∂ ∂=+ + = + +∂∂∂ ∂∂∂ ∂∂ ∂=+ +∂∂∂Fe e e e Fe e e e Fe e e e Also, we have, () () () ()TT1oo 1 1oo 1 oo rr r r θ rr θ T1oo 1 o zr r z, , .rr rr r zθ−− −− −−∂∂⋅⋅ = ⋅⋅ =∂∂ ∂⋅= ⋅ =∂eF e = e F e eF e = e F e eF eeFe Thus, () () ()TT1o 1o ooo o o o o o o rr θ zr θ z T1o ooo zr θ z, .rrr r r r rr z r r z zzz rr zθθ θθ θθ θ−− −∂∂∂ ∂ ∂ ∂=+ + = + +∂∂∂ ∂ ∂ ∂ ∂∂∂=++∂∂∂Fe e e eFe e e e Fe e e e The components of1−Bwith respect to the basis at xare: () ()() () () ()1T11 T 1 1 rr r r r r 22 2TT T1o 1o 1o oo o o o o o o rrr r .rr zB rr z r r z rrr r r rθθθ−−− − − −− −=⋅ =⋅ =⋅ ∂∂ ∂ ∂ ∂ ∂⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞=⋅ + ⋅ + ⋅ =+ +⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂∂∂ ∂ ∂ ∂⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠e B eeF F eeF F e eF e eF e eF e () ()() () () ()1T11 T 1 1 θθ θ θ θ θ 22 2TT T1o 1o 1o oo o o o o o o θ r θθ .zB rr z r r z r r r rrrθθ θθθ θθθθ θ θ−−− − − −− −=⋅ =⋅ =⋅ ∂∂ ∂∂ ∂ ∂ ⎛⎞ ⎛ ⎞ ⎛⎞=⋅ + ⋅ +⋅ = + +⎜⎟ ⎜ ⎟ ⎜⎟∂ ∂ ∂ ∂∂∂ ⎝⎠ ⎝ ⎠ ⎝⎠eB e e F F e e F F e eF e eF e eF e 22 2 1 oo o o zzrr zBzzzθ −∂∂ ∂⎛⎞⎛ ⎞⎛⎞=+ +⎜⎟⎜ ⎟⎜⎟∂∂∂⎝⎠⎝ ⎠⎝⎠. ()()() () () ()TT T1o 1o 1o o oo o o o oo oo o o rrr r1T11 T 1 1 r θ r θ r θ zr rr z r r r r z z rrr r r r r r zB θθ θθ θ θθθθ θ θ−− −−−− − − ∂∂ ∂∂ ∂ ∂ ∂ ∂ ∂=⋅ + ⋅ +⋅ = + + ∂∂∂ ∂ ∂ ∂ ∂ ∂ ∂=⋅ =⋅ =⋅ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠eF e eF e eF ee B ee F F ee F F e () ()() () () ()1T11 T 1 1 rz r z r z TT T1o 1o 1o o oo o o o oo oo o o rrr rrz zB rr z r r r r z z zzzr z r z r zθθθ θ−−− − − −− −=⋅ =⋅ =⋅ ∂∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞=⋅ + ⋅ +⋅ = + +⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂∂∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠eB e e F F e e F F e eF e eF e eF e 1 o o oo oo o o zrr r r zzBrz r z rzθθθ θθ θ−∂∂ ∂ ∂ ∂∂⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞=+ +⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂∂⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-40_________________________________________________________________ 3.78 Derive Eq. (3.29.47), that is, oo222 θθ oo oo oorr zCrrrθ θθθ⎛⎞ ⎛⎞ ⎛⎞∂∂∂=++⎜⎟ ⎜⎟ ⎜⎟∂∂∂⎝⎠ ⎝⎠ ⎝⎠ ----------------------------------------------------------------------------------------- Ans. oooo o T o θθ θ θ θ θC=⋅ =⋅eC e eF F e . Now, o θ r θ z oo oo oo[Eq.3.29.3]rr z rrrθ θθθ∂ ∂∂=++∂∂∂Fe e e e , therefore, oooT oT oT θθ θ r θ z θ r θθ oo oo oo oo oorr z r rCrrr r rθθ θθθ θ θ⎛⎞∂∂∂ ∂ ∂=⋅ + + = ⋅ + ⋅⎜⎟∂∂∂ ∂ ∂⎝⎠eF e e e eF e eF e oT θ z ooz rθ∂+⋅∂eF e . Now, from Eqs. (3.29.14) (3.29.15) and (3.29.16), oT oT oT θ r θθ θ z oo oo oo,,rr z rrrθ θ θθ∂∂∂⋅= ⋅= ⋅=∂∂∂eF e eF e eF e , thus. oo222 θθ oo oo oorr zCrrrθ θθθ⎛⎞ ⎛⎞ ⎛⎞∂∂∂=++⎜⎟ ⎜⎟ ⎜⎟∂∂∂⎝⎠ ⎝⎠ ⎝⎠ _________________________________________________________________ 3.79 Derive Eq. (3.29.49), ooθ oo o oo o oo orrrr r zzCrr r r rrθθ θθ θ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞∂∂∂ ∂∂ ∂=+ +⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ------------------------------------------------------------------------------------------ Ans. oooo o T o rθ rθ r θ C=⋅ =⋅eC e eF F e . Now, o θ r θ z oo oo oo[Eq.(3.29.3)]rr z rrrθ θθθ∂ ∂∂=++∂∂∂Fe e e e , oooT rθ rr θ z oo oo oorr zCrrrθ θθθ⎛⎞∂∂∂=⋅ + + =⎜⎟∂∂∂⎝⎠eF e e eoT oT rr r θ oo oorr rrθ θθ∂∂⋅+ ⋅∂∂eF e eF e oT rz ooz rθ∂+⋅∂eF e . From Eqs. (3.29.14), (3.29.15) and (3.29.16) oT oT oT rr r θ rz ooo,,rr z rrrθ ∂∂∂⋅= ⋅= ⋅=∂∂∂eF e eF e eF e , Thus, ooθ oo o oo o oo orrrr r zzCrr r r rrθθ θθ θ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞∂∂ ∂∂ ∂∂=+ +⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠. _________________________________________________________________ 3.80 Derive the components of1−Cwith respect to the bases at X. ----------------------------------------------------------------------------------------- Ans. () ()oo1T1 oT o o1 1 o o1 o1 o1 ooo rr r r r r r θ rz ==rrrrrCrr z θ−−− − − − − ∂ ∂∂⋅⋅ = ⋅ + ⋅ + ⋅∂∂∂e F F eeF F e eF e eF e eF e Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-41oo o o oorr r r rr rr r r zz θθ∂∂ ∂ ∂ ∂∂=+ +∂∂ ∂ ∂ ∂∂. [See Eqs.(3.29.30), (3.29.31) and (3.29.32)]. () ( )oo1T1 oT o o1 1 o o1 o1 o1 oo oo oo r θ r θ rr r θ rz oo o oo o oo o== .rrrrCrr z rr r r rr rrr r zzθθ θθ θ θθ θ θθ−−− − − − − ∂ ∂∂⋅⋅ = = ⋅ + ⋅ + ⋅∂∂∂ ∂∂ ∂∂ ∂∂⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞=+ +⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠e F F eeF F e eF e eF e eF e The other components can be similarly derived. _________________________________________________________________ 3.81 Derive components of Bwith respect to the basis {} rθz ,,eee at xfor the pathline equations given by ( , , , ), ( , , , ), z=z( , , , ) r rXYZt XYZt XYZt θθ == . --------------------------------------------------------------------------------------------- Ans. From r θ zX Y Z and d dr rd dz d dX dY dZ θ+++ + x= e e e X= e e e and (,,, ) , (,,, ) , z = z (,,, ) r rXYZt XYZt XYZt θθ == , we have, r θ zX Y Z r θ zX YZddd d r r d d z d X d Y d Z rr r r r rdX dY dZ dX dY dZXY Z X Y Z zz zdX dY dZ dX dY dZXY Zθ θθθ→+ + + + ∂∂∂ ∂ ∂∂⎛⎞ ⎛ ⎞→+ + + ++⎜⎟ ⎜ ⎟∂∂∂ ∂ ∂∂⎝⎠ ⎝ ⎠ ∂∂∂⎛⎞++ + = + +⎜⎟∂∂∂⎝⎠x = F X x = e e e = Fe Fe Fe ee eF eF eF e Xr θ zY r θ z , , rr z rr z XXX YYYθθ∂∂∂ ∂∂∂→= + + = + +∂∂∂ ∂∂∂Fe e e e Fe e e e Zr θ zrr z Z ZZθ∂∂∂=+ +∂∂∂Fe e e e , and TT Xr r X Yr r Y ,, .rretcXY∂∂⋅= ⋅= ⋅= ⋅=∂∂eF e e F e eF e e F e TT r XYZ X Y Z , , rrr r r r XYZ X Y Zθθθθ ∂∂∂ ∂ ∂ ∂=++ = + +∂∂∂ ∂ ∂ ∂Fe e e e Fe e e e T XYZ zzzz X YZ∂∂∂=++∂∂∂Fe e e e . The components of Bare: 222 T rr r Xr Yr Z rrrrr r r rBX YZ X Y Z∂∂∂ ∂ ∂ ∂ ⎛⎞⎛⎞⎛⎞=⋅ = ⋅ + ⋅ + ⋅ = + + ⎜⎟⎜⎟⎜⎟∂∂∂ ∂ ∂ ∂ ⎝⎠⎝⎠⎝⎠eF F e eF e eF e eF e . T r θ rX rY rZ .rrrrBXYZ rr rr rr XX YY ZZθθθθ θθθ∂∂∂= ⋅=⋅ +⋅ +⋅∂∂∂ ∂∂ ∂∂ ∂∂⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞=++⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠eF F e eF e eF e eF e _________________________________________________________________ 3.82 Derive the components of 1−Bwith respect to the basis {} rθz ,,eee at xfor the pathline equations given by ( , , , ), ( , , , ), = ( , , , ) XXr z t Y Yr z t Z Zr z tθ θθ == . ----------------------------------------------------------------------------------------- Ans. From r θ zX Y Z and d dr rd dz d dX dY dZ θ+++ + x= e e e X= e e e and Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-42(,,, ) , (,,, ) , =(,,, ) XXr z t Y Yr z t Z Zr z tθ θθ == , we have, 11 1 1 XYZ r θ z XY 11 1 Zr θ z.d d dX dY dZ dr rd dz XX X YY Ydr d dz dr d dzrz r z ZZ Zdr d dz dr rd dzrzθ θθθθ θθθ−− − − −− −→++ + + ∂∂∂ ∂∂∂⎛⎞ ⎛ ⎞→+ + ++ +⎜⎟ ⎜ ⎟∂∂ ∂ ∂∂ ∂⎝⎠ ⎝ ⎠ ∂∂∂⎛⎞+++ = + +⎜⎟∂∂ ∂⎝⎠X = F x e e e= Fe Fe Fe ee eF e F e F e Thus, 11 rX Y Z θ XYZ 1 z XYZ, .XYZ X Y Z rrr r r r XYZ zzzθθθ−− −∂∂∂ ∂ ∂ ∂=++ = + +∂∂∂ ∂ ∂ ∂ ∂∂∂=++∂∂∂F e e e e F e eee Fe e e e and () () ()TT11 11 rX X r rY Y r T11 rZ Z r, , , etc. that is, X Y rr Z r−− −− −−∂∂⋅⋅ = ⋅⋅ =∂∂ ∂⋅⋅ =∂eF e = eF e eF e = eF e eF e = e F e () () ()TT11 Xr θ zY r θ z T1 Zr θ z, .X XX YYY rr z rr z ZZZ rr zθθ θ−− −∂∂∂ ∂∂∂=+ + =+ +∂∂∂ ∂∂∂ ∂∂∂=+ +∂∂∂Fe e e eFe e e e Fe e e e Thus, () ( ) () ()1T T T1T 1 1 1 1 r r r r rX rY rrXYBrr−−− − − − ∂ ∂=⋅ =⋅ = ⋅ + ⋅∂ ∂eF F eeF F e eF e eF e ()222T1 rZZ XYZ rr r r−∂∂ ∂ ∂ ⎛⎞⎛⎞⎛⎞+⋅ = + + ⎜⎟⎜⎟⎜⎟∂∂ ∂ ∂ ⎝⎠⎝⎠⎝⎠eF e . () () ()TT T11 1 1 1 r θ rX rY rXYBrrθθθ−− − − − ∂∂=⋅ = ⋅ + ⋅∂∂eF F e eF e eF e ()T1 rZZ XX YY ZZ rr r r r r rθ θθθ−∂∂ ∂ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞+⋅ = + + ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠eF e ., etc. _________________________________________________________________ 3.83 Verify that (a) the components of Bwith respect to {} rθz ,,eee can be obtained from T⎡⎤⎣⎦FF and (b) the component of C, with respect to {}ooo rθz ,,eee can be obtained from T⎡⎤⎣⎦FF , where []F is the matrix of the two points deformation gradient tensor given in Eq. (3.29.12). ----------------------------------------------------------------------------------------- Ans. (a) Eq. (3.29.12) → Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-43o o o o ooo T o oo o oo oo oo o o o o ooorrrr r z r r z rrr rrr r r z r r z rrr zzzr r z r r z zzzθ θ θθθ θ θ θθθ θ θ⎡⎤ ⎡ ⎤∂∂∂∂ ∂∂ ⎢⎥ ⎢ ⎥∂∂ ∂∂∂∂⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂∂∂∂ ∂∂⎡⎤=⎢⎥ ⎢ ⎥⎣⎦ ∂∂ ∂∂∂∂⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂∂∂∂ ∂∂⎢⎥ ⎢ ⎥∂∂ ∂∂∂∂⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦FF → 22 2 o oo o o o oo oo o o, rr rrrr r r r r r rBBrr z r r r r z zθθ θθ θθ θ⎛⎞⎛ ⎞⎛⎞∂∂∂ ∂ ∂ ∂ ∂ ∂ ∂=+ + = + +⎜⎟⎜ ⎟⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝⎠⎝ ⎠⎝⎠ etc. (b) ooo o o o o T oo oo oo o oo o ooo o o o or r zrrr rrr r r z r r z rrr rrr r r z r r zzzz zzz r r zθ θ θ θθθ θθθ θ θ θ⎡⎤ ⎡ ⎤∂ ∂∂∂∂∂ ⎢⎥ ⎢ ⎥∂∂∂∂∂ ∂⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂∂∂∂ ∂ ∂⎡⎤=⎢⎥ ⎢ ⎥⎣⎦ ∂∂∂ ∂∂ ∂⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂ ∂∂∂∂∂⎢⎥ ⎢ ⎥∂∂∂∂∂ ∂⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦FF → oo o o22 2 o o o o oo o oo o oo, rr rrr z r rrr z zCCrrr r r r r r rθθθ θ θ θθ⎛⎞⎛ ⎞⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞∂∂∂ ∂ ∂∂ ∂ ∂ ∂=+ + = + +⎜⎟⎜ ⎟⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂∂ ∂ ∂∂ ∂ ∂ ∂⎝⎠⎝ ⎠⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠. _________________________________________________________________ 3.84 Given oo o o, , rr k z zzθθ== + = . (a) Obtain the components of the Left Cauchy-Green tensor B, with respect to the basis at the current configuration (),,rzθ . (b) Obtain the components of the right Cauchy-Green tensor Cwith respect to the basis at the reference configuration. ----------------------------------------------------------------------------------------- Ans. (a), Using Eqs (3.29.19) to (3.29.24). we obtain 22 2 oo o o1rrrrrBrr z θ⎛⎞⎛ ⎞⎛⎞∂∂∂=+ +=⎜⎟⎜ ⎟⎜⎟∂∂ ∂⎝⎠⎝ ⎠⎝⎠, () ()22 2 2 22 oo o oo1rrrrB kr krrr zrθθθθθ θ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞∂∂∂=+ += + = +⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠, 22 2 oo o o1zzzzzBrr z θ⎛⎞ ⎛ ⎞ ⎛⎞∂∂∂=+ +=⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝ ⎠ ⎝⎠, oo o o o o oo0rrr r r rrBrr r r zzθθθ θ θθ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞∂∂ ∂ ∂ ∂∂==⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂∂⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠++ , o o oo oo o o0rzrz r z rzBrr r r zz θθ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞∂∂ ∂ ∂ ∂∂=+ =⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟∂∂ ∂ ∂ ∂∂⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠+ , oo o o o o oozzr z r zrB rkrr r r zzθθθ θ θθ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞∂∂ ∂ ∂ ∂∂=+ =⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂∂⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠+ . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-44. Thus, [] ()210 0 01 01rk rk rk⎡⎤ ⎢⎥ =+⎢⎥ ⎢⎥ ⎢⎥⎣⎦B . (b) Using Eqs.(3. 29.43) to (3. 29.51), we have, oo o o2 2 2 222 o o o o oo oo o1, 1rrrr z r r zCCrrr r r rθθθθ θθθ⎛⎞⎛ ⎞⎛⎞ ⎛ ⎞⎛ ⎞⎛ ⎞∂∂∂ ∂ ∂ ∂=+ += = + + =⎜⎟⎜ ⎟⎜⎟ ⎜ ⎟⎜ ⎟⎜ ⎟∂∂∂ ∂ ∂ ∂⎝⎠⎝ ⎠⎝⎠ ⎝ ⎠⎝ ⎠⎝ ⎠, ()oo22 2 2 zz ooo1zzzrr zCr kθ ⎛⎞ ⎛⎞ ⎛⎞∂∂∂=++= +⎜⎟ ⎜⎟ ⎜⎟∂∂∂⎝⎠ ⎝⎠ ⎝⎠, oo oooo o oo o oo o oo o o oo0, 0,r rzrrr r zzCrr r r rr rr r r zzCrz r z rzθθθ θθ θ θθ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞∂∂ ∂∂ ∂∂=+ +=⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛⎞∂∂ ∂∂ ∂∂=+ +=⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ooθ oo o o o o oo ozrr rr zzCr kzr zr zrθθ θθθ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞∂∂ ∂∂ ∂∂=++=⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠. Thus, [] ()210 0 01 01rk rk rk⎡⎤ ⎢⎥ =⎢⎥ ⎢⎥ + ⎢⎥⎣⎦C . _________________________________________________________________ 3.85 Given ()1/22, / , ra X b Y a z Z θ =+ = = , where (),,rzθ are cylindrical coordinates for the cuurent configuration and () ,,XYZ are rectangular coordinates for the reference configuration. (a) Obtain the components of []Bwith respect to the basis at the current configuration and (b) calculate the change of volume. ----------------------------------------------------------------------------------------- Ans. (a) Using Eqs.(3.29.59) to (3.29.64), we have, 222 2 rrrrr aBX YZ r∂∂∂⎛⎞⎛⎞⎛⎞⎛ ⎞=++=⎜⎟⎜⎟⎜⎟⎜ ⎟∂∂∂⎝⎠⎝⎠⎝⎠⎝ ⎠, 222 2rrrrBX YZ aθθθθθ∂∂∂⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞=++=⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟∂∂∂⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ 222 1zzzzzBXYZ∂∂∂⎛⎞⎛⎞⎛⎞=++=⎜⎟⎜⎟⎜⎟∂∂∂⎝⎠⎝⎠⎝⎠, 0rrr rr rrBXX YY ZZθθθθ∂∂ ∂∂ ∂∂⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞=++=⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠, 0rzrz rz rzBXX YY ZZ∂∂ ∂∂ ∂∂⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=++=⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠, 0zrz rz rzBXX YY ZZθθθθ∂∂ ∂∂ ∂∂⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞=++=⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 3-45Thus, []() ()2 2/0 0 0/ 0 00 1ar ra⎡⎤ ⎢⎥ ⎢⎥=⎢⎥ ⎢⎥ ⎣⎦B . (b) det 1B= =1, thus, there is no change of volume. _________________________________________________________________ 3.86 Given ( ), ( ), ( ) rr X g Y zh Z θ== = , where ()() , , and , ,rz X Y Zθ are cylindrical and rectangular Cartesian coordinate with respect to the current and the reference configuration respectively. Obtain the components of the right Cauchy-Green Tensor Cwith respect to the basis at the reference configuration. ----------------------------------------------------------------------------------------- Ans. Using Eqs.(3.29.68) etc. we have, () ( ) ()22 2 22 2 XX YY (), ( ), ( )ZZrr zCr X C r g Y C h ZXXXθ ∂∂∂⎛⎞⎛ ⎞⎛⎞′′ ′ =+ += = =⎜⎟⎜ ⎟⎜⎟∂∂∂⎝⎠⎝ ⎠⎝⎠ XY YZ XZ 0, 0, 0rr r r zzCC CXY X Y XYθθ ∂∂ ∂ ∂ ∂∂⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞=+ += = =⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟∂∂ ∂ ∂ ∂∂⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠. []() () ()2 2 2() 0 0 0( ) 000 ( )rX gY hZ⎡⎤′⎢⎥ ⎢⎥ ′ =⎢⎥ ⎢⎥ ′⎣⎦C , where () / , . ,r X dr dX etc′≡ _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-1CHARTER 4 4.1 The state of stress at a certain point in a body is given by :[]123 245 . 350 iMPa⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦ eT . On each of the coordinate planes (with normal in 123,,ee e directions), (a) what is the normal stress and (b) what is the total shearing stress ------------------------------------------------------------------------------ Ans. (a) The normal stress on the 1eplane (i.e., the plane whose normal is in the direction1e) is 1 .MPa, on the 2e plane is 4 .MPa, and on the 3e plane is 0 .MPa (b) The total shearing stress on the 1eplane is 222 3 13 =3.61 . MPa += On the 2eplane is 222 5 29 =5.39 . MPa += , and on the3eplane is 2235+ 34 5.83 MPa == . _________________________________________________________________ 4.2 The state of stress at a certain point in a body is given by : []21 3 14 0 . 30 1 iMPa−⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥ − ⎣⎦ eT (a) Find the stress vector at a point on the plane whose normal is in the direction of 12 322e+ e + e . (b) Determine the magnitude of the norm al and shearing stresses on this plane. ------------------------------------------------------------------------------- Ans. (a) The stress vector on the plane is t=T n , where 12 3 =( 2 2 )/3++ nee e . Thus [] 12321 3 2 5 1114 0 2 6 , ( 5 6 5 ) / 33330 1 1 5−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥−= → + +⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦t= t = e e e MPa . (b) Normal stress ( )( ) 12 3 123 =( 1/9 ) 2 2 5 6 5 3 .nTM P a=⋅ + + + + =nt e e e e e e ⋅ Magnitude of shearing stress 2 286 / 9 9 0.745 .snTT M P a=− = − =t Or, ( )()( )() 123 12 3 13 = 5 6 5 /3 3 1/3 2 2 2 /3 5/3 0 . 7 4 5sn s TT=+ + − + + = − + → = =Tt - n e e e e ee e e __________________________________________________________________ 4.3 Do the previous problem for a plane passing through the point and parallel to the plane 123234xxx−+= . ------------------------------------------------------------------------------- Ans. (a) The normal to the plane is 123(23 ) / 1 4−+ n= e e e . [][][]→ t=T n []21 3 1 1 3 1114 0 2 9 14 1430 1 3 0−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥→− − = −⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦t= , 12 1 2 (1 / 14 )(13 9 ) 3.47 2.41→− = −t= e e e e MPa . (b) Normal stress []13 112 393 1 / 1 4 2 . 2 1 .140nTM P a⎡⎤ ⎢⎥=⋅ − −= =⎢⎥ ⎢⎥⎣⎦nt = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-2Magnitude of shearing stress ()2 2 2250 /14 2.21 3.60 .snTT M P a=− = − =t Or, 12 1 2 3 123 (3.47 2.41 ) (2.21/ 14)( 2 3 ) 2.88 1.23 1.77sn T=− − − − + = − −Tt n = e e e e e e e e , 3.60 .sTM P a→= _________________________________________________________________ 4.4 The stress distribution in a certain body is given by []12 1 20 100 100 100 0 0 100 0 0x x x x− ⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦T MPa. Find the stress vector acting on a plane which passes through the point ( ) 1 / 2, 3 / 2, 3 and is tangent to the circular cylindrical surface 22 12 1 xx+=at that point. ------------------------------------------------------------------------------ Ans. Let 22 12 fxx=+ , then the unit normal to the circle 1 f= at a point () 12,xxis given by 11 2 2 11 2 222 1222 44xx fx xfxx+ ∇== +∇+een= e e . At the point ( ) 1 / 2, 3 / 2, 3 , ()12132+ n= e e . and []05 0 5 0 3 50 0 0 50 3 0 0⎡⎤ − ⎢⎥=⎢⎥ ⎢⎥−⎣⎦T , thus, [] 12 31/2 05 0 5 0 3 2 5 3 50 0 0 3 / 2 25 25 3 25 25 3 0 50 3 0 0 25 3⎡⎤ ⎡ ⎤ ⎡ ⎤ − ⎢⎥ ⎢ ⎥ ⎢ ⎥=→ + − ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥−−⎣⎦ ⎣ ⎦ ⎣ ⎦t= t = e e e MPa. _________________________________________________________________ 4.5 Given 11 221 ., 1 . TM P a T M P a== − , and all other 0ijT=at a point in a continuum. (a) Show that the only plane on which the stress v ector is zero is the plane with normal in the 3edirection. (b) Give three planes on which there is no normal stress acting. ------------------------------------------------------------------------------ Ans. (a) []11 22 1 1 2 2 3100 01 0 000 0nn nn n n n⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥−= − → −⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣⎦t= t = e e . 12 3 0 nn→== → t=0 t=e . (b) 22 12 nTn n=⋅ −nt = . Thus, the plane with 22 12 0 nn−=has no normal stress. These include 31 2 1 2, ( )/ 2, =( )/ 2== −nen e + e n e e etc. _________________________________________________________________ 4.6 For the following state of stress []10 50 50 50 0 0 . 50 0 0MPa− ⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦T , find 11 13 and TT′′ where 1′e is in the direction of 12323++eee and 2′e is in the direction of 123+− eee . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-3------------------------------------------------------------------------------ Ans. 11 2 3 21 2 3(23 ) / 1 4 , ( ) / 3′′=+ + =+−eee e e e e e , thus [] []' 1110 50 50 1 40 111 2 3 50 0 0 2 1 2 3 50 90 /14 6.43 .14 1450 0 0 3 50TM P a−− ⎡⎤ ⎡ ⎤⎡ ⎤ ⎢⎥ ⎢ ⎥⎢ ⎥= = =− =−⎢⎥ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢ ⎥⎢ ⎥−−⎣⎦ ⎣ ⎦⎣ ⎦ 31 2 1 2 3'' ' ( 54 ) / 4 2=× = −+ −ee e e e e , therefore, []' 1310 50 50 5 11 2 3 50 0 0 4 450 / 588 =18.6 58850 0 0 1TM P a−− ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥==⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥−−⎣⎦ ⎣ ⎦. _________________________________________________________________ 4.7 Consider the following stress distribution []2 0 00 00 0xαβ β⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦T where and αβare constants. (a) Determine and sketch the distribution of the stress vector acting on the square in the 10x=plane with vertices located at ()()()() 0,1,1 , 0, 1,1 , 0,1, 1 and 0, 1, 1− −− − . (b) Find the total resultant force and moment about the origin of the stress vectors acting on the square of part (a). ------------------------------------------------------------------------------ Ans. (a) The normal to the plane 10x=is 1e, thus, 1 21 2xαβ=+ete e . On the plane, there is a constant shearing stress β in the 2edirection and a linear distribution of normal stress 2xα, (see figure). (b) ()11 R2 1 2 2 3 1 2 1104 dA x dx dxαββ −−== =∫∫ ∫Ft e + e e + e . () () () ()11 o2 2 3 3 2 1 2 2 3 11 111 3 2 2 3 1 2 3 2 2 3 2 312 3 3 11 1400 233dA x x x dx dx xx x x x dx dxαβ αβα α α−− −− −=× = ⎡⎤ =− − = + − = − ⎢⎥ ⎢⎥⎣⎦∫∫ ∫ ∫∫Mx t e + e e + e e+ e e e e e e× _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-44.8 Do the previous problem if the stress distribution is given by 2 11 2Txα= and all other 0ijT=. ------------------------------------------------------------------------------ Ans. (a) The normal to the plane 10x=is1e, thus, 12 21xα=ete . On the plane, there is a parabolic distribution of normal stress 2 2xα, (see figure). x1x3 x2(0,1,-1) (0,1,1)(0,-1,-1) (0,-1,1) αα (b) ()11 2 R2 1 2 3 1 114 3dA x dx dxαα −−== =∫∫ ∫Ft e e . () () ()11 2 o2 2 3 3 2 1 2 3 11 11 23 232 23 2 3 2 3 1100dA x x x dx dx xx x d xd xα αα−− −−== == +∫∫ ∫ ∫∫Mx t e + e e e- e e e×× _________________________________________________________________ 4.9 Do problem 4.7 for the stress distribution: 11 12 21 3 , TT T xα α === and all other 0ijT=. ------------------------------------------------------------------------------ Ans. (a) The normal to the plane 10x=is1e, thus, 1 13 2 xαα=+ete e . On the plane, there is a constant normal stress of αand a linear distribution of shearing stress 32xαe, (see figure). x1x3 x2(0,1,-1) (0,1,1)(0,-1,-1) (0,-1,1)α α (b) ()11 R1 3 2 2 3 1 114 dA x dx dxααα −−== + =∫∫ ∫Ft e e e . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-5() ( ) ()11 o2 2 3 3 1 3 2 2 3 11 11 2 31 32 23 2 3 1 114 3dA x x x dx dx xx x d x d xαα αααα−− −−=× = + =− + = −∫∫ ∫ ∫∫Mx t e + e e e ee - e e× _________________________________________________________________ 4.10 Consider the following stress distribution for a circular cylindrical bar: []32 3 20 00 00x x x xαα α α−⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T (a) What is the distribution of the stress vect or on the surfaces defined by (i) the lateral surface22 23 4 xx+= , (ii) the end face10x=, and (iii) the end face 1x=l? (b) Find the total resultant force and moment on the end face1x=l . ------------------------------------------------------------------------------ Ans. (a) The outward unit normal vector to the lateral surface 22 23 4 xx+=is given by 22 33 12x x+=een . The outward unit normal vector to 10x=is 21=− ne and that to 1x=l is 31=ne . Thus, 32 32 2300 0 100 0 0200 0xx xx xxαα α α− ⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥⎡⎤=− = →=⎣⎦ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣⎦1 1 nntt . () 13 2 2 3 3 2 2 3 x xx x αα αα =− =− − + = − 2ntT e e e e e () 1 3 22 3 3 22 3 x xx x αα αα== − + = − + 3ntT e e e e e (b) On the end face 1x=l, 13 2 2 3 x x αα == − + 3ntT e e e ( )()() R3 2 2 3 2 3 3 2 0 dA x x dA x dA x dAαα α α == − + = − + =∫∫ ∫ ∫ 3n Ft e e e e . [note: the axes are axes of symmetry, the integrals are clearly zero]. ( )( ) ()o2 2 3 3 3 2 2 3 22 22 2 3 231 1 1 1 0022 8dA x x x x dA xx d A r r d r r d rαα ααα π π απ α=× = × − + =+ = = =∫∫ ∫∫ ∫Mx t e + e e e ee e e _________________________________________________________________ 4.11 An elliptical bar with lateral surface defined by 22 2321 xx+= has the following stress distribution: []32 3 202 20 0 00xx x x−⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T . (a) Show that the stress vector at any point () 123,,xxx on the lateral surface is zero. (b) Find the resultant force, and resultant moment, about the origin O, of the stress vector on the left end face 10x=. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-6Note: 22 23 and 42 82xd A xd Aπ π==∫∫. ------------------------------------------------------------------------------ Ans. (a) The outward unit normal vector to the lateral surface 22 2321 xx+= is given by: 22 33 122 232 4x x x x+= +een , thus, [] [ ] [ ]32 13 222 232302 0 0 120 0 0 400 2 0xx xx xxxx− ⎡⎤ ⎡ ⎤⎡ ⎤ ⎢⎥ ⎢ ⎥⎢ ⎥== − =⎢⎥ ⎢ ⎥⎢ ⎥+⎢⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦⎣⎦tT n (b) On the left end face 10x=, 1−n= e , the stress vector is 32 232x x− t= e e , ( )()() R3 2 2 3 2 3 3 2 22 0 dA x x dA x dA x dA== − = − =∫∫ ∫ ∫Ft e e e e . [note: the axes are axes of symmetr y, the integrals are clearly zero]] () ( ) ()o2 2 3 3 3 2 2 3 22 23 1 1 12 22. 42 82 22dA x x x x dA xx d Aππ π=× = × − ⎧⎫=− + =− + =− ⎨⎬ ⎩⎭∫∫ ∫Mx t e + e e e ee e _________________________________________________________________ 4.12 For any stress state T, we define the deviatoric stress Sto be ()/3kkT− S=T I , where kkTis the first invariant of the stress tensor T. (a) Show that the first invariant of the deviatoric stress vanishes. (b) Given the stress tensor []65 2 100 5 3 4 . 24 9kPa−⎡⎤ ⎢⎥=⎢⎥ ⎢⎥−⎣⎦T , evaluate S. (c) Show that the principal directions of the stress tensor coin cide with those of the deviatoric stress tensor. ------------------------------------------------------------------------------ Ans. (a) From ()/3kkT− S=T I, we have, ()()() tr tr / 3 tr / 3 3 0kk kk kkTT T− −= S= T I= . (b) [] ()[]6 5 2 0 500 200 100 5 3 4 1800 / 3 500 300 400 . 2 4 9 200 400 300kPa−−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=− = −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥−−⎣⎦ ⎣ ⎦SI (c) Let nbe an eigenvector of T, then λ Tn = n . Now ()() /3 /3kk kkTT λ −− Sn = Tn In = n n , that is λ′ Sn = n where ()/3kkTλλ′=− . Thus, nis also an eigenvector of Swith eigenvalue ()/3kkTλ− . _________________________________________________________________ 4.13 An octahedral stress plane is one whose nor mal makes equal angles with each of the principal axes of stress. (a) How many independent octahedral planes are there at each point? (b) Show that the normal stress on an octahedral plane is given by one-third the first stress invariant. (c) Show that the shearing stress on the octahedral plane is given by () () ()1/222 2 12 23 311 3sTT T T T T T⎡⎤=− + − + −⎢⎥⎣⎦, where 123,,TT T are principal values of the stress tensor. ------------------------------------------------------------------------------ Ans. (a) There are four independent octahedral planes. They are given by the following unit normal vectors: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-7123 123 123 123 1234 ,, 3333−− +− −====e+ e + e e+ e e e e e e e ennn , n We note that 123 3−e+ e + egives the same plane as 1n, etc. (b) Using the principal directions as the orthonormal basis, the matrix of Tis diagonal, i.e., []1 2 300 00 00T T T⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T . The normal to an octahedral plane is 123 3±± ee e, thus, []1 2 3001 1111 0 0 1300 1nT TT T⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥=⋅ ± ± ±⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥±⎣⎦ ⎣⎦nT n = where in this equation, the row matrix and column matrix of nhave the same elements, that is if the row matrix is [] 11 1− then the column matrix is 1 1 1⎡⎤ ⎢⎥−⎢⎥ ⎢⎥⎣⎦. Thus, [] ()1 21 2 3 3001 11111 0 0 13300 1nT TT T T T T⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥=± ± ± =+ +⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥±⎣⎦ ⎣⎦. (c) () ( )2 22 2 2 3 2 2 3 12 3 12 3 1 2 1 3 2 31122239snT T TTT TTT T T T T T T=− = + + − + + + + +nt ( )() () ()22 2 223 12 3 1 2 1 3 2 3 1 2 2 3 3 121 99T T T TT TT T T T T T T T T⎡ ⎤=+ + − − −=− + − + −⎢ ⎥ ⎣ ⎦ That is, () () ()1/222 2 12 23 311 3sTT T T T T T⎡⎤=− + − + −⎢⎥⎣⎦ _________________________________________________________________ 4.14 (a) Let and m n be two unit vectors that define two planes and M N that pass through a point P. For an arbitrary state of stress defined at the point P, show that the component of the stress vector mtin the n-direction is equal to the component of the stress vector ntin the mdirection. (b) If 12and m=e n=e , what does the results of (a) reduce to? ------------------------------------------------------------------------------ Ans. (a) The component of the stress vector mtin the n-direction is ⋅=⋅mnt nT m and the component of the stress vector ntin the mdirection is T⋅=⋅ ⋅nmt mT n = nTm . Since Tis symmetric, therefore, T⋅⋅nTm = nT m , therefore, ⋅=⋅mnnt mt . (b) If 12 and m=e n=e , then 21 1 2 12 21 TT ⋅=⋅→=eeet e t . _________________________________________________________________ 4.15 Let mbe a unit vector that defines a plane Mpassing through a point P. Show that the stress vector on any plane that contains the stress traction mt, lies in the M plane. ------------------------------------------------------------------------------ Ans. Referring to the figure below, where mis perpendicular to the plane M, andmt is the stress vector for the plane. Let Nbe any plane which contains the vector mtand let nbe the unit vector perpendicular to the plane N. Then nt= T n . We wish to show that ntis perpendicular to m. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-8 Now, T0, ⋅⋅ ⋅ ⋅⋅ =nmt m=T n m=n T m=n T m=n t because mtis on the Nplane. Thus, 0,⋅=ntm so that ntlies on the Mplane. _________________________________________________________________ 4.16 Let mtand ntbe stress vectors on planes defined by the unit vector mand nrespectively and pass through the point P. Show that if kis a unit vector that determines a plane that contains mtand nt, thenktis perpendicular to mand n. ------------------------------------------------------------------------------ Ans. Since kis a unit vector that determines a plane that contains mt a n d nt, therefore, × ×mn mnttk=tt. Since , , and == =km ntT k t T m tT n , therefore, T0⋅×⋅=⋅ ⋅ ⋅ ⋅= =×mm n km mntt tmt mT k = kTm = kT m = k ttt, similarly, T0⋅×⋅= ⋅ ⋅ ⋅ ⋅= =×nm n kn mntt tnt nT k = kTn = kT n = k ttt. _________________________________________________________________ 4.17 Given the function22(, ) 4fxy x y=− − , find the maximum value of fsubjected to the constraint that 2 xy+= . ------------------------------------------------------------------------------ Ans. Let 22(, ) 4 ( 2 )gxy x y x y λ =− − + +− , then we have the following three equations to solve for , and xyλ: 20gxxλ∂=− + =∂, 20gyyλ∂=− + =∂ and 2 xy+=. Thus, 2 0 2 , 2 0 2 x xy y λλλ λ −+= →= −+= →= , therefore, xy=→ 22 2 1 xyx x y+=→ =→== . That is, maxfoccurs at 1 xy==. That is, 22 max 4( 1 ) ( 1 ) 2 f=− − = _________________________________________________________________ 4.18 True or false: (i) Symmetry of stress tensor is not valid if the body has an angular acceleration. (ii) On the plane of maximum normal stress, the shearing stress is always zero. ------------------------------------------------------------------------------ Ans. (i) False. (ii) True. _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-94.19 True or false: (i) On the plane of maximum shearing stress, the normal stress is always zero. (ii) A plane with its normal in the direction of 12322− e+ e e has a stress vector 12350 100 100 . MPa − t= e + e e It is a principal plane. ------------------------------------------------------------------------------ Ans. (i) Not true in general. Maybe true in some special cases. (ii) True. We note that ( ) 1231 2 350 100 100 50 2 2 −= − t = e+ e e e+ e e . Therefore, t is normal to the plane, so that there is no shearing stress on the plane. That is, it is a principal plane. _________________________________________________________________ 4.20 Why can the following two matrices not represent the same stress tensor? 100 200 40 40 100 60 200 0 0 ., 100 100 0 . 40 0 50 60 0 20MPa MPa⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦. ------------------------------------------------------------------------------ Ans. The first scalar invariant for the first matrix is 50 .MPa The first scalar invariant for the second matrix is 160 .MPa They are not the same, therefore, they can not represent the same stress tensor. _________________________________________________________________ 4.21 Given []01 0 0 0 100 0 0 . 00 0MPa⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T (a) Find the magnitude of shearing stress on the plane whose normal is in the direction of 12e+ e . (b) Find the maximum and minimum normal stresses and the planes on which they act. (c) Find the maximum shearing stress and the plane on which it acts. ------------------------------------------------------------------------------ Ans. (a) Let ()121 2+ n= e e . Then 01 0 0 0 1 1 11 0 01 0 000 1 1 2200 0 0 0⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥==⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦nt i.e., 100 shearing stress 0sT =→ =ntn . (b) The characteristic equation is ()2201 0 0 0 100 0 0 0 100 0 00λ λλ λ λ− −=→ − − = − 12 3100 ., 100 ., 0MPa MPaλ λλ →= = − = . The maximum normal stress is 100 .MPaand the minimum normal stress is 100 MPa− . For 11 2100 ., 100 100 0,MPaλ αα =− + = so that 12αα= , 11 2() / 2=+ne e . For 2 100 .,MPa λ=−12 1 2 100 100 0ααα α+= → = − , 21 2() / 2=−ne e . (c) ()()() ()max min max100 100100 .22nn sTT TM P a− −−== = The maximum shearing stress acts on the planes 12() / 2± n= n n , i.e., on the planes 12and ee . _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-104.22 Show the equation for the normal stress on the plane of maximum shearing stress is ()()max min 2nn nTT T+ = . ------------------------------------------------------------------------------ Ans. Let {} 123,,nn n be the principal axes of the stress tensor with principal values 123TT T>> , then []1 2 300 00 00T T T⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T . On the plane () 13 /2± n= n n , the shearing stress is a maximum. On this plane, the normal stress is: []() ()1 13 m a x m i n 2 3001 110 10 0 022 200 1nn nnTTT TTTT T T⎡⎤ ⎡⎤+ + ⎢⎥ ⎢⎥⋅→= ± = =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥±⎣⎦ ⎣⎦=n T n _________________________________________________________________ 4.23 The stress components at a point are given by: 11 22100 ., 300 ., T MPa T MPa= = 33400 TM P a= . 12 13 23 0 TTT=== . (a) Find the maximum shearing stress and the planes on which they act. (b) Find the normal stress on these planes. (c) Are there any plane/planes on which the normal stress is 500 . MPa? ------------------------------------------------------------------------------ Ans. (a) The maximum normal stress is clearly 33400 TM P a= ., acting on the 3eplane and the minimum normal stress is clearly 11100 . TM P a= , acting on the 1eplane. Thus, the maximum shearing stress is ()max400 100150 .2sTM P a−== , acting on the plane ()131 2± n= e e . (b) [] []100 0 0 1 100 111 0 1 0 300 0 0 1 0 1 0 250 .220 0 400 1 400nTM P a⎡⎤ ⎡ ⎤⎡ ⎤ ⎢⎥ ⎢ ⎥⎢ ⎥=± =±=⎢⎥ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢ ⎥⎢ ⎥ ±± ⎣⎦ ⎣ ⎦⎣ ⎦ Note: We can also use the result of Prob. 4.22 to obtain max min 400 100250 .22nTTTM P a+ +== = (c) No, because max 400 TM P a= _________________________________________________________________ 4.24 The principal values of a stress tensor Tare 1210 ., 10 .TM P a T M P a= =− and 330 .TM P a= If the matrix of the stress is given by: []11 3300 012 1 0 . 02T MPa T⎡⎤ ⎢⎥=×⎢⎥ ⎢⎥⎣⎦T , find the values of 11 33 and TT . ------------------------------------------------------------------------------ Ans. () ( ) ()() () ()11 1 3 3 1 1 3 3 1 1 3 3 3 31 1 3 3 1 1 1 1 3 3 1 110 10 30 10( 1 ) 2 ( ) =2 (i) 10 10 30 10 4 3 4 (ii)IT T T T T T IT T T T T T=−+= + + → = + → − =− = −→ − = − ()()2 33 33 33 33 (i) and (ii) 3 2 4 6 5 0. TT T T →− = − − → − + = Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-1133[6 36 20] / 2 3 2. T=± − = ± Thus, 33T is either 5 or 1. To determine which is the correct value, we check () ( ) ( ) () ( ) ( ) ()2 23 3 1 1 3 3 1 1 10 30 10 10 10 30 10 4 IT T T T ⎡ ⎤ =− + − + = − + +⎣ ⎦ 33 11 33 11 30 TT TT→+ +− = . (iii) Try 331 T=first, from (i), 111 T=, so that (iii) is clearly satisfied. Next try 335 T=, eq (i) gives1125 3 T=−= − , then left side of (iii) becomes 5 ( 3)(5) 3 3 16 0 +−− − = − ≠ . Thus, 33 111 and 1 TT== . _________________________________________________________________ 4.25 If the state of stress at a point is: []300 0 0 02 0 0 0 . 00 4 0 0kPa⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T , find (a) the magnitude of the shearing stress on the plane whose normal is in the direction of ( ) 12 322++ee e and (b) the maximum shearing stress. ------------------------------------------------------------------------------ Ans. (a) Let ()12 3122 ,3++ n= e e e then [] ()123300 0 0 2 600 11 1 0 00 200 0 2 400 6 4 433 30 0 400 1 400⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− = − → =− +⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦nntt e e e () 442 22 4100 80012 8 4 88.89 .99 10 68 10 6410 6.76 260 .98 1n sn sTk P a TT T k P a=⋅ = −+ = = ××=− = − = ×→ =n nnt t (b) ()() max400 200300 .2sTk P a−−== _________________________________________________________________ 4.26 Given []140 410 . 001MPa⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T (a) Find the stress vector on the plane whose normal is in the direction of 12+ee . (b) Find the normal stress on the same plane. (c) Find the magnitude of the shearing stress on the same plane. (d) Find the maximum shearing stress and the planes on which this maximum shearing stress acts. ------------------------------------------------------------------------------ Ans. (a) Let ()121 2+ n= e e , then [] ()121401 5 11 54101 5 22 20010 0⎡ ⎤ ⎡⎤ ⎡⎤ ⎢ ⎥ ⎢⎥ ⎢⎥== → = +⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥⎣ ⎦ ⎣⎦ ⎣⎦nntt e e . (b) ()155 5 .2nTM P a=⋅ = + =nnt (c) 2 2225 25 0snTT=−=−=nt Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-12(d) The characteristic equation is () ()2 2 12 3 11 4 0 5 , 3 , 1λλ λλ λ⎡⎤−−−= → = = − =⎢⎥⎣⎦. Thus, () ()max min5 . and 3 .nnT MPa T MPa== − For () () ()() 12 1 2 1 2 max5 . 1 5 4 0 1/ 2nTM P a αα α α =− + = → = → = +1ne e . For () () ()() 12 1 2 1 2 min3 . 1+3 4 0 1/ 2nTM P a αα αα =− + = → =− → = −2ne e . Thus, ()() max534 .2sTM P a−−== , acting on the plane whose normal is ()()12 1 2 1/ 2 and ±→ n= n n n=e n=e . _________________________________________________________________ 4.27 The stress state in which the only non-vanishing stress components are a single pair of shearing stresses is called simple shear. Take 12 21TT τ== and all other 0ijT=. (a) Find the principal values and principal directions of this stress state. (b) Find the maximum shearing stress and planes on which it acts. ------------------------------------------------------------------------------ Ans. (a) With []00 00 000τ τ⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T, the characteristic equation is ()22 12 3 0, , 0 . λλ τ λ τλ τλ−= → = = − = For () ()() 11 2 1 2 1 2 , 0 0 1/ 2λτ τ ατ α αα=−+ = → = → = +1ne e . For () ()() 21 2 1 2 1 2 0+ 0 1/ 2λτ τ α τ α αα=− → + = → =− → = −2ne e . For 33 30λ=→ = ne . (b) ()() max2sTτττ−−== , acting on the plane whose normal is ()()12 1 2 1/ 2 and ±→ n= n n n=e n=e . _________________________________________________________________ 4.28 The stress state in which only the three normal stress components do not vanish is called a tri-axial state of stress. Take 11 1 22 2 33 3 , , TT Tσσσ === with 123σσσ>> and all other 0ijT=. Find the maximum shearing stress and the plane on which it acts. ------------------------------------------------------------------------------ Ans. () ()13 13 max1, 2 2sTσσ−=± n= e e . _________________________________________________________________ 4.29 Show that the symmetry of the stress tensor is not valid if there are body moments per unit volume, as in the case of a polarized anisotropic dielectric solid. ------------------------------------------------------------------------------ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-13Ans. Let ** * 11 22 33MM M=++*Me ee be the body moments per unit volume. Then referring to the figure shown below, the total moments of all the surface forces and the body force and body moment about the axis which passes through the center point Aand parallel to the 3x axis is : ()()()()()() ()() ( ) ()() () () () ()()21 2 3 1 21 21 2 3 1 3 * 12 1 3 2 12 12 1 3 2 3 1 2 3 22 123 1 2 3/2 /2 /2 /2 1/12 (density)cMT x x x TTx x x T xx x T T xx x M xx x xx x x x α=Δ ΔΔ ++ ΔΔ ΔΔ − ΔΔ Δ − + Δ ΔΔ Δ + ΔΔΔ ⎡⎤ =Δ Δ Δ Δ + Δ⎢⎥⎣⎦∑ where 3αis the 3xcomponents of the angular acceleration of the element. We now let 1230, 0, 0 xx xΔ→Δ→Δ→ and drop all terms of small quantities of higher order than () 123xxxΔΔΔ , we obtain, ()()()** 2 1 123 1 2 123 3 123 1 2 2 1 3 0 T x xx T x xx M x xx T T MΔΔΔ − ΔΔΔ + ΔΔΔ =→ − = , Similarly, one can show that ** 13 31 2 23 32 1 and TT M TT M−= −= . _________________________________________________________________ 4.30 Given the following stress distribution: []() ()12 1 2 1 2 12 1 2 1 2 2,0 ,2 0 00xx Tx x Tx x x x x⎡ ⎤ + ⎢ ⎥=−⎢ ⎥ ⎢ ⎥⎣ ⎦T , find 12T so that the stress distribution is in equilibrium w ith zero body force and so that the stress vector on the plane 11x=is given by ()() 21 2215xx++ − t= e e . ------------------------------------------------------------------------------ Ans. The equations of equilibrium are 0ij i jT Bxρ∂ +=∂. Now with 0iB=, we have, 13 11 12 12 12 2 1 123 210 ( )T TT TTx f xxxx x∂ ∂∂ ∂++= += → = − +∂∂∂ ∂. 23 21 22 12 12 1 2 123120 2 ()T TT TTx g xxxx x∂ ∂∂ ∂++=− = → = +∂∂∂∂, thus, 12 1 22 Tx x C=−+ . 31 32 33 12300 0 .TTT xxx∂∂∂++= → =∂∂∂ To determine C, we have, the stress vector on the plane 11x=is ()( ) 111 1 12 1 23 1 3 12 1 12 212xTT T x x x x C=⎡⎤ =++=+ +− +⎣⎦t=T e e e e e e . Thus, ()()()() 21 2 2 21 22 1 2 1 212 15 3 2 3xx C xx C T x x+ +−+ =+ +− →= → = −+ ee e e . _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-144.31 Consider the following stress tensor: []23 32 23 30 0 0xx x x xTα−⎡ ⎤ ⎢ ⎥=− −⎢ ⎥ ⎢ ⎥−⎣ ⎦T . Find an expression for33T such that the stress tensor satisfies the equa tions of equilibrium in the presence of body force vector 3g−B= e , where gis a constant. ------------------------------------------------------------------------------ Ans. The equations of equilibrium are 0ij i jT Bxρ∂ +=∂. With 12 3 0, B BB g=== − , we have, 13 11 12 1 12300000 ,T TTBxxxρ∂ ∂∂+++= + + + =∂∂∂23 21 22 2 12300000 ,T TTBxxxρ∂ ∂∂+++= + + + =∂∂∂ 31 32 33 33 33 3 123 3 3 33 3 1 210 1 1( , )TTT T T gBgxxx x x gTx f x xρρα ρα ρ α⎛⎞ ∂∂∂ ∂ ∂+++= − + − = →= + ⎜⎟∂∂∂ ∂ ∂ ⎝⎠ ⎛⎞→= + +⎜⎟⎝⎠ _________________________________________________________________ 4.32 In the absence of body forces, the equilibri um stress distribution for a certain body is () 11 2 12 21 1 22 1 2 33 11 22 , , , /2 , a l l ot he r 0ij T A xT T x T B x C xT T T T== = = += + = . Also, the boundary plane 12 0 xx−= for the body is free of stress. (a) Find the value of C and (b) determine the value of and AB . ------------------------------------------------------------------------------ Ans. (a) The equations of equilibrium are 0ij i jT Bxρ∂ +=∂. With 0, iB= , we have, 13 11 12 1 12300000 ,T TTBxxxρ∂ ∂∂+++= + + + =∂∂∂ 23 21 22 2 12310 0 0 1T TTBC Cxxxρ∂ ∂∂+++= + + + = → = −∂∂∂, 31 32 33 3 12300000TTTBxxxρ∂∂∂+ + + =+++=∂∂∂. (b) The unit normal to the boundary plane 12 0 xx−=is () 12 /2− n= e e . Thus, on this plane (note 12xx=), we have, [] ()11 1 1 11 1 1 1 1 11 2201 0 1101 0 2200 / 2 0 0 0Ax x Ax x xB x x x B x x TT⎡⎤ − ⎡⎤⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢⎥⎢ ⎥ ⎢ ⎥=− − = − + =⎢⎥ ⎢⎥⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢⎥⎢ ⎥ ⎢ ⎥ + ⎣⎦⎣ ⎦ ⎣ ⎦ ⎣⎦t , thus, 11 1 11 0 1 and 0 2 Ax x A x Bx x B−= →= − += →= . _________________________________________________________________ 4.33 In the absence of body forces, do the following stress components satisfy the equations of equilibrium: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-15 () ()()22 2 22 2 2 2 11 2 1 2 22 1 2 1 33 1 2 12 21 1 2 13 31 23 32, , , 2, 0 , 0 .Tx x x T x x x T x x TT x x TT TTαν αν α ν αν⎡⎤ ⎡⎤=+− =+− = +⎣⎦ ⎣⎦ == − == == ------------------------------------------------------------------------------ Ans. The equations of equilibrium are 0ij i jT Bxρ∂ +=∂. With 0, iB= , we have, 13 11 12 111 123220 0 0 ,T TTBxxxxxρα να ν∂ ∂∂+++= − + + =∂∂∂ 23 21 22 22 2 123220 0 0T TTBx xxxxρα ν α ν∂ ∂∂+++= − + + + =∂∂∂ 31 32 33 3 12300000TTTBxxxρ∂∂∂+ + + =+++=∂∂∂. Yes, the equations of equilibrium are all satisfied. _________________________________________________________________ 4.34 Repeat the previous problem for the stress distribution []12 12 121 2 120 23 0 00xx xx xx x x xα+−⎡⎤ ⎢⎥=−−⎢⎥ ⎢⎥⎣⎦T ------------------------------------------------------------------------------ Ans. The equations of equilibrium are 0ij i jT Bxρ∂ +=∂. With 0, iB= , we have, ()13 11 12 1 12301 1 0 0 0 0 ,T TTBxxxρα∂ ∂∂+++= → − + = → =∂∂∂ 23 21 22 2 123(2 3 0) 0T TTBxxxρα∂ ∂∂+++= − + ≠∂∂∂ No, the second equation of equilibrium is not satisfied. _________________________________________________________________ 4.35 Suppose that the stress distribution has th e form (called a plane stress state) []()() ()()11 1 2 12 1 2 12 1 2 22 1 2,, 0 ,, 0 00 0Tx x Tx x Tx x Tx x⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T (a) If the state of stress is in equilibri um, can the body forces be dependent on3x? (b) If we introduce a function () 12,xxϕ such that 22 2 11 22 12 22 12 21, and TT Txx xxϕ ϕϕ ∂∂ ∂== = −∂∂ ∂∂, What should be the function () 12,xxϕ for the equilibrium equations to be satisfied in the absence of body forces? ------------------------------------------------------------------------------ Ans. (a) ()() 11 1 2 12 1 2 13 11 12 11 123 1 2,,0Tx x Tx x T TTBBxxx x xρρ∂∂ ∂ ∂∂+++= + +=∂∂∂ ∂ ∂. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-16()() 21 1 2 22 1 2 23 21 22 22 123 1 2,,0Tx x Tx x T TTBBxxx x xρρ∂∂ ∂ ∂∂+++= + +=∂∂∂ ∂ ∂. Thus, 12 and BB must be independent of 3x. (b) 22 2 2 13 11 12 1 22 2 123 1 2 1 2 1 1 22 200 0T TTBxxx x x x x x x xx xϕϕ ϕ ϕρ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ∂ ∂∂ ∂∂ ∂ ∂ ∂∂ ∂∂+++ → − + + = − = ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ∂∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠. 22 2 2 23 21 22 2 22 2 123 1 1 22 2 2 11 100 0T TTBxxx x x x x x x xx xϕϕ ϕ ϕρ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ∂ ∂∂ ∂ ∂ ∂∂ ∂∂ ∂∂+++= − + + + = − + = ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ∂∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠. Thus, the equations of equilibrium are satisfied for any function () 12,xxϕ which is continuous up to the third derivatives. _________________________________________________________________ 4.36 In cylindrical coordinates (),,rzθ , consider a differential vol ume of material bounded by the three pairs of faces : and ; = and = ; and . rr rrd r d zz zzd z θθθ θ θ == + + == + Derive the and rθequations of motion in cylindrical coordina tes and compare the equations with those given in Section 4.8. ------------------------------------------------------------------------------ Ans. From the free body diagram above, we have, ()()()() () cos / 2rr r r rr r rF T rd dz T dT r dr d dz T drdz dθ θθ θ=− + + + −∑ ()() ()()() cos / 2 sin / 2 sin / 2rrT dT drdz d T drdz d T dT drdz dθ θ θθ θθ θθ θθ θ ++ − −+ ()()()()() rz rz rz r rT rd dr T dT rd dr B rd drdz rd drdz aθθ ρ θ ρ θ ⎡ ⎤ −+ + + =⎣ ⎦. Now, 2311cos 1 ...and sin ...22 ! 2 2 2 3 ! 2dd d d dθθ θ θ θ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=+ + = − +⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠and keeping only terms involving products of three differe ntials (i.e., terms involving prod uct of 4 or more differentials drop out in the limit when these diffe rentials approach zero), we have, () () () () () ()2/ 2rr rr r rz r rT drd dz dT rd dz dT drdz T drdz d dT rd dr B rd drdz rd drdz aθθ θ θθ θ θρθ ρ θ++ − ⎡⎤ ++=⎣⎦ Dividing the equation by rd drdzθ , we get, r rr rr rz rrTT TT TB arr r drzθθ θρρθ∂∂∂++− ++=∂∂. This is Eq. (4.8.1) Next, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-17()()()() () sin / 2rr r r F T rd dz T dT r dr d dz T drdz dθθ θ θ θ θθ θ=− + + + +∑ ()() ()()() sin / 2 cos / 2 cos / 2rrT dT drdz d T drdz d T dT drdz dθ θ θθ θθ θθ θθ θ ++ − ++ ()()()()() zz zT rd dr T dT rd dr B rd drdz rd drdz aθ θθ θ θθθ ρ θ ρ θ ⎡ ⎤ −+ + + =⎣ ⎦. Again, 2311cos 1 ...and sin ...22 ! 2 2 2 3 ! 2dd d d dθθ θ θ θ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=+ + = − +⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ and keeping only terms involving products of thr ee differentials, we have, ()() ()()() () ()2/ 2rr r zT drd dz dT rd dz T drdz d dT drdz dT rd dr Br d d r d z r d d r d z aθθ θ θ θ θ θθθ θθ θ ρθ ρ θ++ + + ⎡⎤ +=⎣⎦ Dividing the equation by rd drdzθ , we get, 1rr r zTT T TTB arr r r zθθ θ θ θ θ θθρρθ∂∂ ∂++ + ++=∂∂ ∂, this is Eq. (4.8.2). _________________________________________________________________ 4.37 Verify that the following stress field satisfies the z-equation of equilibrium in the absence of body forces: 23 2 35 3 3 5 3533 3, , , , 0rr zz rz r zzr z A z zz rr zTA T T A T A TT RR R R R RRθθ θ θ⎛⎞ ⎛ ⎞⎛⎞ =− = = −+ = −+ = =⎜⎟ ⎜ ⎟⎜⎟⎜⎟ ⎜ ⎟⎜⎟⎝⎠ ⎝ ⎠⎝⎠ 222R rz=+ ------------------------------------------------------------------------------ Ans. The zequation of equilibrium in cylindrical coordinate is: 10z zr zr zz zT TT TBrr r zθρθ∂∂∂++ + + =∂∂ ∂. Now , R rR z rR zR∂ ∂= =∂ ∂ so that 22 2 35 3 45 6 22 2 2 22 2 2 3 5 57 3 5 5731 3 3 1 5 13 3 1 5 13 3 1 5, 0zr zT rr z r R z r z RAArr r r r RR R RR R T rz r z rz r zAA RR R R RR R Rθ θ⎛⎞ ⎛ ⎞ ∂ ∂∂ ∂ ∂=− + =− − + −⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟∂∂ ∂ ∂ ∂⎝⎠ ⎝ ⎠ ⎛⎞ ⎛⎞ ∂=− − + − =− − + − =⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ∂⎝⎠ ⎝⎠ 2 3513 rzT zAr RR⎛⎞ =− +⎜⎟⎜⎟⎝⎠, 23 2 24 34 5 6 3 5 5 713 9 1 5 13 9 1 5 zzT zR z z R z z zAAzz z RR R R R R R R⎛⎞ ⎛ ⎞ ∂ ∂∂=− − + − =− − + −⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟∂∂ ∂⎝⎠ ⎝ ⎠ Thus, 22222 2 2 4 35 5 5 5 5 7 73 333391 5 1 5 zr rz zzTTT rzzzz r z zArrz RR R R R R R R⎛⎞ ∂∂++= − −−+++− − ⎜⎟⎜⎟ ∂∂⎝⎠ 22222 2 2 4 35 5 5 5 5 7 73 333391 5 1 5 rzzzz r z zA RR R R R R R R⎛⎞ = − −−+++− −⎜⎟⎜⎟⎝⎠ () ()22 2 2222 2 2 2 35 5 7 3 5 5 731 531 5 3 3 1 5 1 50rz z rzzR z z RAA RR R R R R R R⎛⎞ ++ ⎛⎞⎜⎟=− − + − =− − + − = ⎜⎟⎜⎟ ⎜⎟⎜⎟ ⎝⎠⎝⎠. _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-184.38 Given the following stress field in cylindrical coordinates: 232 222 55533 3, , , 0 , 222rr zz rz r zPzr Pz Pz rTT TT T T R r z R RRθθ θ θπ ππ=− =− =− = = = = + Verify that the state of stress satisfies the equati ons of equilibrium in the absence of body forces. ------------------------------------------------------------------------------ Ans. 222 55 531 3 3 22 2rr r rr rz TT T TT Pzr Pzr Pz r rr d r z r r z R RRθθ θ θ ππ π⎛⎞ ⎛⎞ ⎛⎞ ∂−∂∂ ∂∂++ += − + − + − ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂⎝⎠ ⎝⎠ ⎝⎠ ()22 2 22 55 5 5 533 1 1 3 3 3 1 22 22 2Pz Pzr Pzr Pr Pz rrzrr r z zR RR R Rππ ππ π⎛⎞ ⎛⎞ ⎛⎞∂∂ ∂ ∂ ⎛⎞ ⎛⎞ ⎛⎞= − +− + − +− +− ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ 22 56 5 5 631 5 3 3 P1 5 22 2Pzr Pzr R Pzr zr Pz r R rz RR R R Rππ π π π⎛⎞ ⎛⎞ ∂∂⎛⎞ ⎛ ⎞=− + − + − + ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ()2233 57 7 5 71515 15 15 150 22 2 2 2Pzr r zPzr Pzr Pz r Pzr RR R R Rππ π π π⎛⎞ + ⎛⎞ ⎛⎞⎜⎟=− + + =− + = ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎝⎠ ⎝⎠⎝⎠. 2 100000rr zTT TT rr r zθθθ θ θ θ∂∂ ∂+ + + =+++=∂∂ ∂ 22 3 55 513 1 3 3 22 2z zr zr zz zT TT T Pz r Pz r PzBrr r z r r z R RRθρθ ππ π⎛⎞ ⎛⎞ ⎛⎞ ∂∂∂ ∂∂++ + + = −+ −+ − ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂⎝⎠ ⎝⎠ ⎝⎠ 22 2 2 3 55 5 5 533 1 3 9 3 1 22 22 2Pz Pz r Pz Pz Pz rz R RR R Rππ ππ π⎛⎞ ⎛⎞ ⎛ ⎞ ∂∂⎛⎞ ⎛⎞=− +− − − −⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ∂∂⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ 22 2 2 3 56 5 5 631 5 3 9 1 5 22 2 2 2Pz Pz r R Pz Pz Pz R rz RR R R Rππ π π π⎛⎞ ⎛⎞ ⎛ ⎞ ∂∂= − + −−+⎜⎟ ⎜⎟ ⎜ ⎟⎜⎟ ⎜⎟ ⎜ ⎟ ∂∂⎝⎠ ⎝⎠ ⎝ ⎠ 22 2 2 2 4 2 2 2 2 57 5 5 7 5 731 5 3 9 1 5 1 515 0 22 2 2 2 2 2Pz Pz r Pz Pz Pz Pz r zPz RR R R R R Rππ π π π π π⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞⎛⎞ +=− + − − + =− + =⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟⎜⎟⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟⎜⎟⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠⎝⎠ _________________________________________________________________ 4.39 For the stress field given in Example 4.9.1, determine the constants and AB if the inner cylindrical wall is subjected to a uniform pressure ipand the outer cylindrical wall is subjected to a uniform pressure op. ------------------------------------------------------------------------------ Ans. The given stress field is: 22, , c onstant a nd 0rr zz r rz zBBTA T A T TT T rrθθ θ θ =+ =− = = = = . On the outer wall, orr=, and o rrTp=− , and on the inner wall, irr=, and i rrTp=− , therefore, we have, o 2 oBpA r−=+ (i) and i 2 iBpA r−=+ (ii). () ()() ()2222ii oooi i o oi 22 22 22 io oi oi , prp r pp r r BBpp B A rr rr rr− −→−=−→= = −−. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-19_________________________________________________________________ 4.40 Verify that Eq. (4.8.4) to (4.8.6) are satisfied by the stress field given in Example 4.9.2 in the absence of body forces. ------------------------------------------------------------------------------ Ans. The given stress field is: 332, , 0rr r rBBTA TT A TTT rrθθ φφ θ φ θφ = − == + === . () ()2 2 2 23 3 3sin 11 1- sin sin 12 2 2 200 00 0 .rr r rrT TTT T rr r r r BB B BrA A A Arr r r rrr r rφθ θφ φ θθ θθ θ φ∂ ∂+ ∂++∂∂ ∂ ∂⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞= − ++− + = + ++− + =⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟∂⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ () ()3 3cot sin 11 1+ sin sin cot cot000 + 0 .r rrrT T TTT T rr r r r T T rrθ θφ θ θ φφ θθ φφ θθθ θ θθ θ φ θ θ∂ ∂− − ∂++∂∂ ∂ =++ − = ()()3 3sin cot 11 1+ 0 0 0+ 0 0sin sinr rrrT T TT T T rr r r rφ φθ φφ φ φ θφθ θ θθ θ φ∂ ∂ ∂− +++ = + + =∂∂ ∂. _________________________________________________________________ 4.41 In Example 4.9.2, if the spherical shell is subjected to an inner pressure ipand an outer pressure op, determine the constant and AB . ------------------------------------------------------------------------------ Ans. From the example, we have, 32 rrBTA r=− , thus, o 33 oi22 and iB BpA p A rr−=− −=− ()() ()33 33 oo i oo i 33 33 oi oi and 2i ip pr r pr p rAB rr rr− −→= − = − −−. _________________________________________________________________ 4.42 The equilibrium configuration of a body is described by: 11 2 2 3 31116 , , 44x Xx X x X== −= − . If the Cauchy stress tensor is given by: 111000 .,and all other 0ij TM P a T== , (a) calculate the first Piola –Kirchhoff stress tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is 1eplane and (b) calculate the second Piola-Kirchhoff tensor a nd the corresponding pseudo stress vector for the same plane. ------------------------------------------------------------------------------ Ans. From 11 2 2 3 31116 , , 44x Xx X x X== −= − , we obtain the deformation gradient Fand its inverse as: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-20[]116 0 0 1/16 0 0 0 1/ 4 0 , 0 4 0 and det 1 00 1 / 4 00 4−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤ =− = −⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥ −− ⎣⎦ ⎣ ⎦FF F = . (a) The first Piola-Kirchhoff stress tensor is, from ()()T1 odet−=TF T F : []()[]()()T1 o1000 0 0 1/16 0 0 1000 /16 0 0 det 1 0 0 0 0 4 0 0 0 0 . 00 00 0 4 0 0 0MPa−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== − =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦TF T F For a unit area in the deformed state in the 1e direction, its undeformed area is () ()T oo oo 116 0 0 1 16 110 1 / 4 0 0 0 1 6det00 1 / 4 00dA dA⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− = → =⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦nF n = n eF. That is, its undeformed plane is also 1eplane. The corresponding pseudo stress vector is given by =oo otT n , where o1=ne . Thus [] () 11000 /16 0 0 1 0 0 0 0 1000 /16 00 0 0⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=→ = → =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦oo o o otT n t t e We note that the pseudo stress vector is in the sa me direction as the Cauchy stress vector and the intensity of the pseudo stress vector is 1/16 of the Cauchy stress vector simply because the undeformed area is 16 times the deformed area and both areas have the same normal direction. (b)The second Piola-Kirchhoff stress tensor is, from ()()T11det−−=TF F T F% () []()()T111/16 0 0 1000 0 0 1/16 0 0 det 1 0 4 0 0 0 0 0 4 0 00 4 0 0 0 00 4 1/16 0 0 1000 /16 0 0 1000 / 256 0 0 04 0 0 0 0 0 0 0 . 00 4 0 0 0 0 0 0MPa−−⎡ ⎤⎡ ⎤⎡ ⎤ ⎡⎤ ⎢ ⎥⎢ ⎥⎢ ⎥⎡⎤ ⎡⎤== − −⎢⎥ ⎣⎦ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣⎦⎣⎦⎢ ⎥⎢ ⎥⎢ ⎥− − ⎣ ⎦⎣ ⎦⎣ ⎦ ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥−=⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦TF F T F% The corresponding pseudo stress vector is given =o tT n%% , where o1=ne . Thus, ()1 1000 / 256 =→ =o tT n t e%%% . _________________________________________________________________ 4.43 Can the following equations re present a physically acceptable deformation of a body? Give reason. 11 2 3 3 211, , 422x Xx Xx X=− = =− . ------------------------------------------------------------------------------ Ans. []1/2 0 0 00 1 / 2 d e t 1 04 0−⎡⎤ ⎢⎥=→ −⎢⎥ ⎢⎥−⎣⎦FF = . The given equations are not acceptable as a physically acceptable deformation because it give s a negative ratio of deformed volume to the undeformed volume. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-21_________________________________________________________________ 4.44 The deformation of a body is described by: () () 11 2 2 3 34, 1 / 4 , 1 / 4x Xx X x X== − = − . (a) For a unit cube with sides along the coordinate axes what is its deformed volume? What is the deformed area of the 1eface of the cube? (b) If the Cauchy stress tensor is given by: 11100 .,and all other 0ij T MPa T= =, calculate the first Piola –Kirchhoff stress tensor and the corres ponding pseudo stress vector for the plane whose undeformed plane is 1eplane. (c) Calculate the second Piola-Kirchhoff tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is 1eplane. Also, calculate the pseudo differential force for the same plane. ------------------------------------------------------------------------------ Ans. From () () 11 2 2 3 34, 1 / 4 , 1 / 4x Xx X x X== − = − , we have (a) []40 0 01 / 4 0 d e t 1 / 4 00 1 / 4⎡⎤ ⎢⎥=− →⎢⎥ ⎢⎥ − ⎣⎦FF = , thus () () o d e t 1/4 ( 1 ) 1/4 dV dV dV=→ = = F . ()() () ( ) ( ) ( )T1 oo 11/4 0 0 1 1/4 det 1 1/ 4 0 4 0 0 1/ 4 0 1/16 004 0 0dd A d−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− = →⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦A= F F n A= e That is, the deformed volume is 1/4 of its original volume and the 1eface of unit area deformed into an area 1/16 of it original area and remain in the same direction. These results are quite obvious from the geometry of the deformation. (b) The first PK stress tensor is: []()[]()T1 o100 0 0 1/ 4 0 0 100 /16 0 0 1d e t 0 00 0 4 0 0 00 .400 0 0 0 4 0 0 0MPa−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎛⎞ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== − = ⎜⎟ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎝⎠⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦TF T F The corresponding pseudo stress vector for 1e-plane in the deformed state, whose undeformed plane is also 1e-plane, is given by oo o=tT n , whereo1=ne , that is ()1 100 /16 . MPa =ote The Cauchy stress vector on the 1eface in the deformed state is 1100 . MPa=te Clearly the Cauchy stress vector has a larger magnitude because the area in the deformed state is 1/16 of the undeformed area. (c) The second PK stress tensor is: []1 o1/ 4 0 0 100 /16 0 0 100 / 64 0 0 04 0 0 0 0 0 0 0 . 004 0 0 0 0 0 0MPa−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎡⎤⎡⎤== − =⎣⎦ ⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦TF T% The corresponding pseudo stress vector for the 1e-plane in the deformed state, whose undeformed plane is also 1e-plane, is given by =o tT n%% , where o1n= e . Thus, ()1 100 / 64 . MPa =te% The Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-22pseudo force df%is related to the force 1 (= 100 / 16 for 1)dd A d A = ft = eoo o by the formula -1ddf= F f% , Thus, []-1 11/ 4 0 0 100 /16 10004 0 064004 0dd d⎡⎤ ⎡ ⎤ ⎛⎞ ⎢⎥ ⎢ ⎥⎡⎤⎡⎤ =− → ⎜⎟ ⎣⎦ ⎢⎥ ⎢ ⎥ ⎣⎦⎝⎠⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦f= F f f = e%% ______________________________________________________________ 4.45 The deformation of a body is described by: 11 2 2 2 3 3 , , xX k Xx Xx X=+ = = . (a) For a unit cube with sides al ong the coordinate axes what is its deformed volume? What is the deformed area of the 1eface of the cube? (b) If the Cauchy stress tensor is given by: 12 21 100 .,and all other 0ij TT M P a T== = , calculate the first Piola – Kirchhoff stress tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is 1e plane and compare it with the Cauchy stress vector in the deformed state. (c) Calculate the second Piola-Kirchhoff tensor a nd the corresponding pseudo stress vector for the plane whose undeformed plane is 1e plane. Also, calculate the ps eudo differential force for the same plane. ------------------------------------------------------------------------------ Ans. From 11 2 2 2 3 3 , , xX k Xx Xx X=+ = = , we have (a) []10 010 d e t 1 001k⎡⎤ ⎢⎥=→⎢⎥ ⎢⎥⎣⎦FF = , thus () oo det 1 dV dV dV dV= →= = F . ()() () ()T1 oo 1 210 0 1 1 det 1 1 1 0 0 00 1 0 0dd A k k d k−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− = − → −⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦A= F F n A=e e That is, the deformed volume is the same as its original volume and the 1eface of unit area deformed into an area 21k+ of it original area and whose normal is in the direction of 12k−ee . These results are quite obvious from the geometry of the deformation. (b) The first PK stress tensor is: []()[]()T1 o0 100 0 1 0 0 100 100 0 det 100 0 0 1 0 100 0 0 . 00 0 0 0 1 0 0 0k kM P a−− ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== − =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦TF T F The corresponding pseudo stress vector for the 12() k−ee plane, whose undeformed plane is the 1eplane, is given by oo o=tT n , whereo1=ne . Thus, () 12 100 . kM P a=− +ote e The Cauchy stress vector on the 12() k−ee face in the deformed configuration is [] [ ] [] () 122201 0 0 01 11 0 0100 0 0 1100 0 0kk kk⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥== − → − +⎢⎥ ⎢ ⎥++⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦tT n t = e e The Cauchy stress vector has a smaller magnitude because the deformed area is 21k+ times the undeformed area. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-23(c) The second PK stress tensor is: []1 o1 0 100 100 0 200 100 0 0 1 0 100 0 0 100 0 0 . 001 0 0 0 0 0 0kk k MPa−−− −⎡ ⎤ ⎡⎤ ⎡⎤ ⎢ ⎥ ⎢⎥ ⎢⎥⎡⎤⎡⎤== =⎣⎦ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎣⎦ ⎢ ⎥ ⎢⎥ ⎢⎥⎣ ⎦ ⎣⎦ ⎣⎦TF T% The corresponding pseudo stress vector for the 12k−ee plane, whose undeformed plane is the 1eplane, is given by =o tT n%% , where o1=ne . Thus, () 12 100 2 . kM P a=−+te e% The pseudo force df%is related to the force () ( ) 12 100 for 1 dd A k d A =−+ = f= t e eoo o by the formula -1ddf=F f% , Thus, [] ()-1 1210 100 0 1 0 1 100 2 0010kk dd d k−−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤⎡⎤ =→ −⎣⎦ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦f= F f f = e + e%% . _________________________________________________________________ 4.46 The deformation of a body is described by: 11 2 2 3 32, 2 , 2x Xx X x X=== . (a) For a unit cube with sides al ong the coordinate axes, what is its deformed volume? What is the deformed area of the 1eface of the cube? (b) If the Cauchy stress tensor is given by: 100 0 0 01 0 00 . 00 1 0 0Mpa⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦, calculate the first Piola –Kirchhoff stress tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is the 1eplane and compare it with the Cauchy stress vector on its deformed plane, (c) calculate the second Piola-Kirchhoff tensor and the corresponding pseudo stress vector for the plane whose undeformed plane is the 1eplane. Also, calculate the pseudo diffe rential force for the same plane. ------------------------------------------------------------------------------ Ans. From, 11 2 2 3 32, 2 , 2x Xx X x X=== we have (a) []200 020 d e t 8 002⎡⎤ ⎢⎥=→⎢⎥ ⎢⎥⎣⎦FF = , thus () oo det 8 8 dV dV dV dV= →= = F . ()() () () ()T1 oo 11/2 0 0 1 1/2 det 1 8 0 1/ 2 0 0 8 0 4 00 1 / 2 0 0dd A d−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== →⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦A= F F n A= e (b) The first PK stress tensor is: []()[]()()T1 o100 0 0 1/ 2 0 0 400 0 0 det 8 0 100 0 0 1/ 2 0 0 400 0 . 0 0 100 0 0 1/ 2 0 0 400MPa−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥== =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦TF T F The corresponding pseudo stress vector for the 1e plane in the deformed state, whose undeformed plane is also 1e plane, is 1 400 . MPa =ote The Cauchy stress vector on the 1e plane is 1 100 . MPa=te The Cauchy stress vector has a smaller magnitude because the area is four times larger. (c) The second PK stress tensor is: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 4-24[]1 o1/ 2 0 0 400 0 0 200 0 0 0 1/ 2 0 0 400 0 0 200 0 . 0 0 1/ 2 0 0 400 0 0 200MPa−⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥⎡⎤⎡⎤== =⎣⎦ ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦TF T% The corresponding pseudo stress vector for the 1e plane in the deformed state, whose undeformed plane is also 1e plane, is 1 200 . MPa=te% The pseudo force df%is related to the force ( ) 1 400 for 1 dd A d A == f= t eoo o by the formula -1ddf=F f% . Thus, []-1 11/2 0 0 1 400 0 1/ 2 0 0 200 00 1 / 2 0dd d⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎡⎤⎡⎤ =→⎣⎦ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦f= F f f = e%% . _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-1 CHAPTER 5, PART A 5.1 Show that the null vector is the only isotropic vector. (Hint: Assume that ais an isotropic vector, and use a simple change of basis to equate the primed and unprimed components). -------------------------------------------------------------------------------- Ans. For an isotropic a, by definition, [][] ii′=eeaa , where {}{} andii′ ee are any two orthonormal bases. That is [] [ ][] [] [ ][]TT ii i ii i′=→ =ee e ee eaQ a aQ a for all [] ieQ . Method I. Choose []10 0 01 0 00 1−⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥ − ⎣⎦Q , then 11 22 3310 0 01 0 00 1aa aa aa−⎡⎤⎡ ⎤⎡⎤ ⎢⎥⎢ ⎥⎢⎥=−⎢⎥⎢ ⎥⎢⎥ ⎢⎥⎢ ⎥⎢⎥ − ⎣⎦⎣⎦⎣ ⎦gives 11 2 2 3 3 0, 0, 0 aa a a aa=− = =− = =− = . In other words, the only isotropic vector is the null vector. Method II. The matrix equation [] [ ][]T ii i=ee eaQ a , with the same basis for each matrix, is equivalent to the equation T=aQ a . That is, a is an eigenvector forTQfor any orthogonal tensor Q. But clearly, there is no non-zero vector which is an eigenvector for all orthogonal tensors. ________________________________________________________________________________________________________________________________________________ 5.2 Show that the most general isotropic s econd-order tensor is of the form of αI, where αis a scalar and Iis the identity tensor. ------------------------------------------------------------------------------- Ans. For an isotropic T, by definition, [][] ii′=eeTT , where {}{} andii′ ee are any two orthonormal bases. Choose []100 01 0 00 1−⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦Q , then [] [][][]T ii i i=ee e eTQ T Q gives 11 12 13 11 12 13 11 12 13 21 22 23 21 22 23 21 22 23 31 32 33 31 32 33 31 32 33 11 12 13 21 22 23100 100 100 01 0 01 0 01 0 00 1 00 1 00 1TTT TTT TTT TTT TTT TTT TTT TTT TTT TT T TT T T−− − − ⎡⎤⎡⎤ ⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥== −⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ − ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦⎣⎦ ⎣ ⎦ −− =− −12 12 21 21 13 13 31 31 31 32 330, 0, 0 0. TT TT TT TT TT⎡⎤ ⎢⎥→ =− = =− = =− = =− =⎢⎥ ⎢⎥⎣⎦ Next, the choice of []100 01 0 001⎡ ⎤ ⎢ ⎥=−⎢ ⎥ ⎢ ⎥⎣ ⎦Q gives, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-2 11 11 11 22 23 22 23 22 23 32 33 32 33 32 330 0 100 0 0 100 100 0 0 0 0 10 0 0 10 0 10 0 0 001 0 001 001 0TT T TT TT TT TT TT TT⎡⎤⎡⎤ ⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥=− − =− −⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ − ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦⎣⎦ ⎣ ⎦ 11 22 23 23 23 32 32 32 3300 00 , 0 . 0T TTT T T T TT⎡⎤ ⎢⎥= −→= − = = − =⎢⎥ ⎢⎥−⎣⎦ Next, the choice of []010 100 001⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦Q gives, 11 11 22 22 22 11 11 22 33 33 3300 0 1 0 00 0 1 0 00 00 1 0 0 00 1 0 0 00 00 0 0 1 00 0 0 1 00TTT TT T T T TTT⎡⎤⎡⎤⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥⎢⎥⎢⎥ ⎢⎥ ⎢⎥== → =⎢⎥⎢⎥⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦⎣⎦⎣⎦ Finally, the choice of []100 001 010⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦Q gives 11 11 11 33 11 33 33 1100 00 00 00 . 00 00TT TT T T TT⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥=→ =⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ Thus, [] [] 11 22 33 12 21 13 31 23 32 = and 0 TT T TTTTTT αα == ====== →= TI ________________________________________________________________________________________________________________________________________________ 5.3 For an isotropic linearly elastic body, (a) verify the (),YEμμ λ= as given in Table 5.1. (b) Obtain the value of μas / 0YEλ→ ------------------------------------------------------------------------------- Ans. From Table 5.1, ()()2 223 03Y YY YEEEEμμλμ λ μ λμ−= →+− −=− () ()233 8 4YY YEE Eλ λλ μ−−+ − + →= . (b) () () ( ) ( )233 1 8 / / 3 / 4YY Y YEE E Eλ λλ λ μ−−+− + − = As / 0YEλ→, () ( ) ()218 / / 3 / 14 / / 9YY YEE Eλλ λ+− → + , where we have used the binomial theorem. Thus, { ( 3 ) ( 3 ) [ 1 4 ( / )/9 ] }/4 ( 3 ) ( / )/9 ( 3 / ) ( )/9YY Y Y Y Y YEE E E E E E μλλ λ λ λ λ→− − + − + = − = − Thus, as /0YEλ→, /3YEμ→ . ________________________________________________________________________________________________________________________________________________ 5.4 From () ( )11 2YEνλνν=+− ,()2 12μνλν=−and () 1 3kλν ν+= obtain (),YEμμν= and (,) kkμν= Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-3 ------------------------------------------------------------------------------- Ans. () ( ) () ()2 11 2 1 2 2 1YYEEν μνλμννν ν== → =+− − +. ()() ()21 23 12 1 3 12kkμν μν νλννν+== → =−+ −. ________________________________________________________________________________________________________________________________________________ 5.5 Show that for an incompressible material ( 1/2ν→ ) that (a) /3 , , , b u t 2 /3YEk kμ λλ μ=→ ∞ → ∞ − = (b) 2 ( / 3)kkTμ T= E+ I where kkT is constitutively indeterminate. ------------------------------------------------------------------------------ Ans. (a) From Table 5.1, we have 22,, . 2(1 ) 2(1 1/ 2) 3 (1 )(1 2 ) 3 3YY Y YEE E Ekkνμ λλ μ λ μνν ν== == → ∞ + = → − =++ + − (b) In general, 2 eλμ T= I+ E . Now, from Eq.(5.4.2), we have (2 3 )kkTeμλ=+. As 1/2ν→ , λ→∞ , and 3kkTeλ→ so that 23kkTμ T= I+ E . We note that because of incompressibility, kkTwill be constitutively indeterminate. It becomes determinate when the boundary conditi on(s) is (are) taken into account. ________________________________________________________________________________________________________________________________________________ 5.6 Given ijkl ij klAδδ= andijkl ik jlBδδ= . (a) Obtain 11jkA and11jkB . (b) Identity those 11jkA that are different from11jkB . ------------------------------------------------------------------------------- Ans. (a) 11 11 11 1 1 ,kl kl kl kl k lABδδδ δ δ=== . (b) 1111 1122 1133 11 1111 11 1, all other 0, 1, all other 0kl kl AAA A B B=== = = = . 1122 1122 1133 1133 , AB AB≠ ≠ . ________________________________________________________________________________________________________________________________________________ 5.7 Show that for an anisotropic linear elastic ma terial, the principal directions of stress and strain are in general not coincident. ------------------------------------------------------------------------------- Ans. We have, ij ijkl klTC E= . Let iebe the principal basis for E, then [] ieEis diagonal. Thus, 12 12 1211 11 1222 22 1233 33 kl kl TC EC EC EC E==++ . This equation shows that in general, 120 T≠. Similarly, in general 13 230 and 0 TT≠≠ . Thus the matrix of Tis not diagonal with respect to the principal basis of E. ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-4 5.8 If the Lamé Constants for a material are: 66119.2 (17.3 10 ), 79.2 (11.5 10 ) GPa psi GPa psiλμ=× = × Find Young's modulus, Poisson's ratio and the bulk modulus. ------------------------------------------------------------------------------- Ans. From Table 5.1, we have, () ()() 79.2 32 119.2 79.23 119.2 2 79.2 206 YE GPaμλ μ λμ+ ++⎡⎤+⎣⎦== = ()630 10 psi× 119.20.32( ) 2(119.2 79.2)λνλμ== =++, () 2/ 31 1 9 . 22 7 9 . 2 / 31 7 2 kG P aλμ=+ = + = 6(25 10 ) psi× . ________________________________________________________________________________________________________________________________________________ 5.9 Given Young's modulus 103 YE GPa= and Poisson's ratio 0.34ν= . Find the Lamé constants and λμ. Also find the bulk modulus. ------------------------------------------------------------------------------- Ans. () ( )() () ()()6 0.34 10381.7 11.8 10 1 1 2 1.34 0.32YEGPa psiνλνν== = ×+− ()( )6 10338.4 5.56 10 21 2 1 . 3 4YEGPa psi μν=== ×+× () ( )62 / 3 81.7 2 38.4 / 3 107.3 15.6 10 kG P a p s iλμ=+ = + = × ________________________________________________________________________________________________________________________________________________ 5.10 Given Young's modulus 193 YE GPa= ., shear modulus 76 GPaμ= . Find Poisson's ratioν, Lamé constant λand the bulk modulus k. ------------------------------------------------------------------------------- Ans. ()()()()6 27 6 0 . 2 7 193 21 1 0.27, 89.1 12.9 1022 7 6 1 2 1 0 . 5 4YEGPa psiμννλμν=− = − = = = = ×−− () ( )62 / 3 89.1 2 76 / 3 140 20.3 10 kG P a p s iλμ=+ = + = × ________________________________________________________________________________________________________________________________________________ 5.11 The components of strain at a point of structural steel are: 666 11 22 33 66 12 23 1336 10 , 40 10 , 25 10 12 10 , 0, 30 10EE E EE E− −− −−=× =× =× =× = =× Find the stress components. ( )( )66119.2 17.3 10 , 79.2 11.5 10 GPa psi GPa psiλμ=× = × ------------------------------------------------------------------------------- Ans. From eλμ T= I+2 E , we have, with ()6636 40 25 10 101 10e− −=+ +× =× Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-5 []() ( ) ( )661 0 0 36 12 30 17.7 1.9 4.75 119.2 101 0 1 0 10 2 79.2 12 40 0 10 1.9 18.4 0 . 0 0 1 30 0 25 4.75 0 16.0MPa−−⎡⎤ ⎡ ⎤⎡ ⎤ ⎢⎥ ⎢ ⎥⎢ ⎥××⎢⎥ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢ ⎥⎢ ⎥⎣⎦ ⎣ ⎦⎣ ⎦T= + = ________________________________________________________________________________________________________________________________________________ 5.12 Do the previous problem if the strain components are: 66 6 11 22 33 6 12 23 13100 10 , 200 10 , 100 10 100 10 , 0, 0EE E EE E− −− −=× = −× =× =− × = = ------------------------------------------------------------------------------- Ans. From eλμ T= I+2 E , we have, with ()6100 200 100 10 0e−=−+× = []()6100 100 0 15.8 15.8 0 2 79.2 100 200 0 10 15.8 31.7 0 0 0 100 0 0 15.8MPa−−−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥−− × = −−⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦T= . ________________________________________________________________________________________________________________________________________________ 5.13 An isotropic elastic body ( ) 207 , 79.2YE GPa GPaμ == has a uniform state of stress given by:[]100 40 60 40 200 0 60 0 200MPa⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T . (a) What are the strain components? (b) What is the total change of volume fo r a five centimeter cube of the material? ------------------------------------------------------------------------------- Ans. (a) We would like to use the equation1(1 ) ( )ij ij kk ij YET TEννδ ⎡ ⎤ =+ −⎣ ⎦, therefore, we first obtain ()2071 1 0.30622 7 9 . 2YEνμ=− = − = , then obtain 11 22 33 100kkTT TT M P a=++= . Thus, [] () 3 3100 40 60 1 0 0 11.31 40 200 0 0.306 (100) 0 1 0 60 0 200 0 0 1 100 52.4 78.6 0.483 0.253 0.380 152.4 292 0 0.253 1.41 0 10 207 1078.6 0 231 0.380 0 1.12YE −⎧⎫⎡⎤ ⎡ ⎤ ⎪⎪⎢⎥ ⎢ ⎥=− −⎨⎬⎢⎥ ⎢ ⎥⎪⎪⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦⎩⎭ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=− = − ×⎢⎥ ⎢ ⎥×⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦E (b) Dilatation ()330.483 1.41 1.12 10 0.193 10kk eE− −== −+ ×= × . Total change of volume = () ()()()3335 0.193 10 24.1 10 VV e− −Δ= = × = ×3cm. ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-6 5.14 An isotropic elastic sphere ( ) 207 , 79.2YE GPa GPaμ == of 5 cm radius is under the uniform stress field []620 23 0 000MPa⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥⎣⎦T Find the change of volume for the sphere. ------------------------------------------------------------------------------ Ans. ()2071 1 0.3062 2 79.2YEνμ=− = − = , 1(1 ) ( )ij ij kk ij YET TEννδ ⎡ ⎤ =+ −⎣ ⎦ gives [] ()56 2 0 1 0 0 3.35 1.26 0 11.31 2 3 0 0.306 (3) 0 1 0 1.26 2.34 0 10 000 0 0 1 0 0 0 . 4 4 3YE−⎧⎫⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎪⎪⎢⎥ ⎢ ⎥ ⎢ ⎥=− − = −×⎨⎬⎢⎥ ⎢ ⎥ ⎢ ⎥⎪⎪⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦⎩⎭E Thus, ()3 55 3 450.567 10 0.567 10 2.96 103eVπ − −−⎛⎞ =× → Δ = × = × ⎜⎟⎜⎟⎝⎠ ________________________________________________________________________________________________________________________________________________ 5.15 Given a motion () () 11 12 2 2 12 , xXk XX x X k XX=+ + =+ − , show that for a function ( , ) fab a b= (a) 12 1 2(, ) ( , )O ( )fxx fX X k=+ , ()() 12 1 2 11,,O( )fxx fXXkxX∂∂=+∂∂ , where O( ) 0 as 0 kk→→ ------------------------------------------------------------------------------- Ans. (a) () ()() 12 1 2 1 1 2 2 1 2,fxx x x X kX X X kX X ⎡ ⎤⎡ ⎤ = = ++ +−⎣ ⎦⎣ ⎦ () (){ }() ()2 1 2 1 12 2 12 1212XXk X XX X XX k XXXX=+ − + + + + − . That is, () () 12 12,Ofxx X X k=+ , where O( ) 0 as 0 kk→→ , i.e. ()() ()() 12 1 2 12 1 2,, O ( ) ,,fxx fX X k fxx fX X=+ → ≈ as 0k→. (b)() () 12 1 2 2 2 1 2 2 1,( ) Offxx x x x X k X X X kx∂=→ = = + −= +∂, and ()12 1 2 2 1,ffXX X X XX∂=→ =∂. Thus, 11f f x X∂∂≈∂∂ as 0k→. ________________________________________________________________________________________________________________________________________________ 5.16 Do the previous problem for 22(,)fab a b=+ ------------------------------------------------------------------------------- Ans. (a)() ( ) ( )22 22 12 1 2 1 1 2 2 1 2,fxx x x X kX X X kX X ⎡ ⎤⎡ ⎤ =+= + + + + −⎣ ⎦⎣ ⎦ () (){} () (){ }22 22 2 12 1 1 2 2 1 2 1 2 1 2 2 XX k X X X X X X k X X X X=++ + + − + + +− . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-7 That is, () ( )22 12 1 2,Ofxx X X k=++ , where O( ) 0 as 0 kk→→ , i.e. ()() ()() 12 1 2 12 12,, O ( ) ,fxx fX X k fxx fX X=+ → ≈ as 0 k→. (b) () () ( )22 12 1 2 1 1 1 2 1 1,2 2 2 + = 2 Offxx x x x X kX X X kx∂=+→ = = + +∂ and ()22 12 1 2 1 1,2ffXX X X XX∂=+→ =∂. Thus, 11f f x X∂∂≈∂∂ as 0k→. ________________________________________________________________________________________________________________________________________________ 5.17 Given the following displacement field in an isotropic linearly elastic solid: ()22 4 13 2 23 1 3 1 2 , , , 1 0 uk X Xu k X X uk X X k−=== − = (a) Find the stress components and (b) in the abse nce of body forces, is the state of stress a possible equilibrium stress field? -------------------------------------------------------------------------------- Ans. (a) [] [ ]() () () ()32 3 1 2 31 3 1 2 12 1 2 1 200 2 2 02 0 2222 0 2 2 0kX kX X X X kkX kX X X X kX kX X X X X⎡ ⎤+ ⎡⎤ ⎢ ⎥ ⎢⎥∇= → = −⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥−+ −⎣⎦ ⎣ ⎦uE Thus, [] []() () () ()31 2 31 2 12 1 202 2 02 2 0 2 22 0kkXX X Ek X X X XX X Xμμ⎡ ⎤+ ⎢ ⎥=→ = = −⎢ ⎥ ⎢ ⎥ +−⎣ ⎦TE Since the displacement components are small (of the order of k), therefore, iixX≈ , so that () () 11 22 33 12 21 3 13 31 1 2 23 32 1 2 0, 2 , 2 , 2 TT T TT k xTT k x x TT k x x μμ μ === == == + == − . (b) Substituting the above stress components into the equations of equilibrium, we have, 13 11 12 12300 0 0 0T TT xxx∂ ∂∂++= → + + =∂∂∂, 23 21 22 1230 0000T TT xxx∂ ∂∂++= → + + =∂∂∂and 31 32 33 12302 2 0 0TTTkkxxxμμ∂∂∂++= →−+ =∂∂∂. Thus, all equations of equilibrium are satisfied. Since the stress field is obtained from a given displace ment field, therefore, the state of stress is a possible equilibrium stress field. ________________________________________________________________________________________________________________________________________________ 5.18 Given the following displacement field in an isotropic linearly elastic solid: 4 12 3 21 3 31 2 , , , 10 u k X X u k XX u k XX k−=== = (a) Find the stress components and (b) in the abse nce of body forces, is the state of stress a possible equilibrium stress field? ------------------------------------------------------------------------------- Ans. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-8 (a) [] [ ]32 31 210 0 0kX kX kX kX kX kX⎡⎤ ⎢⎥∇= =⎢⎥ ⎢⎥⎣⎦uE , Thus, [] []32 31 210 02 2 0 0kkX X E kX X XXμμ⎡⎤ ⎢⎥=→ = =⎢⎥ ⎢⎥⎣⎦TE Since the displacement components are small (of the order of k), therefore, iixX≈ , so that 11 22 33 12 21 3 13 31 2 23 32 1 0, 2 , 2 , 2 TT T TT k xTT k xTT k x μ μμ === == == == . (b) Substituting the above stress components into the equations of equilibrium, we have, 13 11 12 12300 0 0 0T TT xxx∂ ∂∂++= → + + =∂∂∂, 23 21 22 1230 0000T TT xxx∂ ∂∂++= → + + =∂∂∂and 31 32 33 1230 0000TTT xxx∂∂∂++= → + + =∂∂∂. Thus, all equations of equilibrium are satisfied. Since the stress field is obtained from a given displacement fi eld, therefore, the state of stress is a possible equilibrium stress field. ________________________________________________________________________________________________________________________________________________ 5.19 Given the following displacement field in an isotropic linearly elastic solid: ()24 12 3 21 3 3 1 2 3 , , , 1 0 u k X X u k XX u k XX X k−=== + = (a) Find the stress components and (b) in the abse nce of body forces, is the state of stress a possible equilibrium stress field? -------------------------------------------------------------------------------- Ans. (a) [] [ ]32 31 21 30 0 2kX kX kX kX kX kX kX⎡⎤ ⎢⎥∇= =⎢⎥ ⎢⎥⎣⎦uE , Thus, []()[] 3322 2kkEk X k X λμ =→ = TI + E Since the displacement components are small (of the order of k), therefore, iixX≈ , so that [] ()33 2 33 121 32 2xx x kx x x x xxλμ μ μλ μ μμλ μ⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥+⎣ ⎦T . (b) Substituting the above stress components into the equations of equilibrium, we have, 13 11 12 12300 0 0 0T TT xxx∂ ∂∂++= → + + =∂∂∂, 23 21 22 1230 0000T TT xxx∂ ∂∂++= → + + =∂∂∂and ()31 32 33 12300 0 2 0TTT xxxλμ∂∂∂++= → + + + ≠∂∂∂. Thus, the stress field is not an equilibrium stress field in the absence of body forces. The given st ate of stress is not a possible equilibrium stress field. ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-9 5.20 Show that for any function ( ) fs, the displacement 1 () uf s= where 1 L sxc t=± satisfies the wave equation 122 2 11 22 Luuc tx∂∂= ∂∂ ------------------------------------------------------------------------------- Ans. () ()2 22 2 11 22 2 2 22 11 22 2 11 1 1, .LL Luu df s df d f s d fcc ctd s t d s t td s d s uu df s df d f s d f xd s xd s x xd s d s∂∂∂∂== ± → = ± =∂∂ ∂ ∂ ∂∂∂∂== → = =∂∂ ∂ ∂ Thus, 22 2 22 11 222 1LLuu dfcc td s x∂∂== ∂ ∂ ________________________________________________________________________________________________________________________________________________ 5.21 Calculate the ratio of the phase velocities /L T cc for Poisson 's ratio equal 1/ 3, 0.49 and 0.499 . ----------------------------------------------------------------------------------------------------------- Ans. From Table 5.1, we have () ()21 21 2212 12 2 2 1μν μνμ νλλ μν νλ μ ν− −=→ + = → =− −+ −, Thus, 22 ( 1 ) 12L Tc cλμν μ ν+ −==−. Thus, for ()()() () ()1/3 , / 2 ( 2/3 )/ 1 2/3 4/3 / 1/3 2 . 0.49, / 2(0.51) / 1 0.98 1.02 / 0.02 7.14.0.499, / 2(0.501) / 1 0.998 1.002 / 0.002 22.4.LT LT LTcc cc ccν ν ν==− = = ==− ====− = = __________________________________________________________________ 5.22 Assume a displacement which depends only 2and x t, i.e., ()2,, 1 , 2 , 3iiuu x t i== . Obtain the differential equations which ()2,iuxt must satisfy in order to be a possible motion in the absence of body forces. ------------------------------------------------------------------------------- Ans. From the Navier equations, we have, 22 22 1 2 22 3// 0 , / / , / 0 . eu x e x e x u x e x=∂ ∂ →∂ ∂ = ∂ ∂ =∂ ∂ ∂ ∂ = Thus, ()22 2 2 22 2 2 2 o1 1 2 1 1 2 2 2 22 22 o 2 22 22 22 2 2 22 2 2 2 o2 2 2 2 2 2 22 2 2 22 2 2 2 o3 3 2 3 3 2(/ )( /) (/ ) ( /) , ( / ) ( )( / ) ( / ), (/ ) ( 2 ) ( /) (/ ) / (/ )( /) (/ ) ( /T L Tut u x ut c u x ut ux ux ut ux ut c ux ut ux ut c uxρμ ρλ μμ ρλ μ ρμ∂∂ = ∂∂→ ∂∂ =∂∂ ∂∂ = +∂ ∂+ ∂ ∂ →∂∂= + ∂ ∂→ ∂∂=∂ ∂ ∂∂ = ∂ ∂→ ∂∂ =∂ ∂ ). Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-10 _________________________________________________________________ 5.23 Consider a linear elastic medium. Assume the following form for the displacement field: ()() 13 3 2 3 sin sin , 0 ux c t x c t u uεβ α β⎡⎤=− + + = =⎣⎦ (a) What is the nature of this elastic wave (l ongitudinal, transverse, direction of propagation?) (b) Find the strains, stresses and determine unde r what condition(s), the equations of motion are satisfied in the absence of body forces. (c)Suppose that there is boundary at 30x=that is traction free. Under what condition(s) will the above motion satisfy this b oundary condition for all time. (d) Suppose that there is boundary at 3x=lthat is also traction free. What further conditions will be imposed on the above motion to satisfy this boundary condition for all time. ------------------------------------------------------------------------------- Ans. (a) Transverse wave, propagating in the 3edirection. (b)The only nonzero strain components are: ()()()()() 13 31 1 3 3 3 1/2 / /2 cos cos E Eu x x c t x c t εβ β α β⎡ ⎤ == ∂ ∂ = − + +⎣ ⎦. The only nonzero stress components are: ()()()() 13 31 1 3 3 3 /c o sc o s TT ux x c t x c t με μ β β α β ⎡ ⎤ ==∂∂= −+ +⎣ ⎦, 1xequation of motion is: ()00 022 2 2 2 2 11 3 3 1 1// /ut T x c u u cρ ρβ βμ μ ρ ∂∂ = ∂∂ → − = − → = . The other two equations are 0=0. (c) The boundary condition on 30x=is: ()()[ ] 31 300 , 0 c o s c o s 0 1Tt c t c t βαβ α −= → = → + = → = −Te . (d) The boundary condition on 3x=lis, ()() ()() 31 30 , 0 cos cos 0,[note 1]. 2sin sin 0 sin 0 / , 1,2,3...Tt c t c t ct n nββ α ββ β β π⎡⎤ =→ =→ − − + = = −⎣⎦ →= → = → = =Te ll l ll l _________________________________________________________________ 5.24 Do the previous problem (Prob. 5.23) if the boundary 30x=is fixed (no motion) and 3x=lis still traction free. ------------------------------------------------------------------------------- Ans. (a) and (b) are the same as in the previous problem. (c) The boundary condition on 30x= is: () [ ] 110, 0 sin sin 0 1ut u c t c t εβ α β α =→=− + = →= . (d) The boundary condition on 3x=lis: ()() ()() ()31 30, 0 c o s c o s 0 2cos cos 0 cos 0 / 2 , 1 ,3,5...T t ct ct ct n nββ ββ β β π⎡⎤ =→ =→ − + + =⎣⎦ →= → = → = =Te ll l ll l _________________________________________________________________ 5.25 Do Problem 5.23 if the boundary 30x=and 3x=l are both rigidly fixed (no motion) -------------------------------------------------------------------------------- Ans. (a) and (b) are the same as in the previous problem 5.23. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-11 (c) The boundary condition on 30x=is: () [ ] 110, 0 sin sin 0 1ut u c t c t εβ α β α =→=− + = →= (d) The boundary condition on 3x=lis: () ()() 11,0 s i n s i n 0 sin cos 0 / , 1,2,3...ut u c t c t ct n nεβ β ββ β π⎡⎤ =→ = − + + =⎣⎦ →→ → = =ll l ll __________________________________________________________________ 5.26 Do Problem 5.23, if the assumed displacement field is of the form: ()() 33 3 1 2 sin sin , 0 ux c t x c t u uεβ α β⎡⎤=− + + = =⎣⎦ -------------------------------------------------------------------------------- Ans. (a) Longitudinal, propagating in the 3edirection. (b)The only nonzero strain components are: () ()() 33 3 3 3 3 /c o s c o s E ux x c t x c t εβ β α β⎡ ⎤ =∂ ∂ = − + +⎣ ⎦. The nonzero stress components are: ()()()()() 1 1 2 2 333 3 33 33 33 /, / 2 / 2 / TT uxT ux ux ux λλ μ λ μ== ∂∂ = ∂∂+∂∂=+ ∂∂ , where ()()()() 33 3 3/c o s c o sux x c t x c t εβ β α β⎡ ⎤ ∂∂= −+ +⎣ ⎦. 3xequation of motion is: () () ()00 022 2 2 2 2 33 3 3 3 3// 2 2 /ut T x c u u cρ ρβ β λ μ λ μ ρ ∂∂ = ∂∂ → − = − + → = + . The other two equations are 0=0. (c) The boundary condition on 30x=is: ()()()()()() 33 300 , 0 2 c o s c o s 0 1T t ct ct λμ ε β β αβ α ⎡⎤ −= → = →+ + = → = −⎣⎦Te . (d) The boundary condition on 3x=lis: ()() ()() 33 30, 0 c o s c o s 0 . [ N o t e 1 ] , 2sin sin 0 sin 0 / , 1,2,3...Tt c t c t ct n nββ α ββ β β π⎡⎤ =→ =→ − − + = = −⎣⎦ →= → = → = =Te ll l ll l __________________________________________________________________ 5.27 Do the previous problem, Problem 5.26, if the boundary 30x=is fixed (no motion) and 3x=lis traction free ( t=0 ). ------------------------------------------------------------------------------- Ans. (a) and (b) are the same as the previous problem, problem 5.26. (c) The boundary condition on 30x=is: () [ ] 330, 0 sin sin 0 1ut u c t c t εβ α β α =→=− + = →= . (d) The boundary condition on 3x=l is, with 1α=, ()() ()() ()33 30, 0 c o s c o s 0 2cos cos 0 cos 0 / 2 , 1 ,3,5...Tt c t c t ct n nββ ββ β β π⎡⎤ =→ =→ − + + =⎣⎦ →= → = → = =Te ll l ll l __________________________________________________________________ 5.28 Do Problem 5.26, if the boundary 30x= and boundary 3x=l are both rigidly fixed. ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-12 Ans. (a) (b) and (c) are the same as in Prob. 5.27, with 1 α=. (d) The boundary condition on 3x=lis () ()() 33,0 s i n s i n 0 sin cos 0 sin 0 / , 1,2,3...ut u c t c t ct n nεβ β ββ β β π⎡⎤ =→ = − + + =⎣⎦ →= → = → = =ll l ll l __________________________________________________________________ 5.29 Consider the displacement field: () 123,,,iiuu x x x t= . In the absence of body forces, (a) obtain the governing equation for iufor the case where the motion is equivoluminal and (b) obtain the governing equation for the dilatation efor the case where the motion is irrotational ( ) //ij j iux u x∂∂= ∂ ∂ . ------------------------------------------------------------------------------- Ans. From the Navier equations of motion, Eq. (5.6.4) ()22 oo 2ii i ij juu eBx xx tρρ λ μ μ∂∂ ∂=+ + +∂∂∂ ∂, we have (a) with 0 and =0i eB= , 22 o 2ii jjuu xx tρμ∂∂=∂∂ ∂. (b) For irrotational motion 2jj j ii ji j jj ii jiuu u uu e x xx x x x x x x∂∂ ∂∂∂ ∂ ∂∂=→ = = =∂∂∂ ∂∂ ∂∂ ∂∂. Thus, () () () ()2 22 oo 22 22 2 o22 2.ii ii i i i i iiuu ee e e x xx xx x tt ee xx tρλ μ μ λ μ ρ λ μ λμ ρ∂∂ ∂∂ ∂ ∂ ∂=+ + =+ → =+∂∂ ∂ ∂ ∂ ∂ ∂∂ +∂∂→=∂∂ ∂ __________________________________________________________________ 5.30 (a) Write a displacement field for an infin ite train of longitudinal waves propagating in the direction of 1234+ee . (b) Write a displacement field for an infinite train of transverse waves propagating in the direction of1234+ee and polarized in the 12xxplane. ------------------------------------------------------------------------------- Ans. Let ()() n1 21/5 3 4=+ee e , then ()() n1 21/5 3 4 x x ⋅= +xe . Also, ()() t1 2 1/5 4 3=± −ee e (a) Equation 5.10.8 of Example 5.10.3 gives nn2sinLctπεη⎛⎞=⋅ − −⎜⎟⎝⎠ux e el. Thus, 12 12 12 334 34 32 42sin , sin , 055 5 55 5LLxx xxuc t u c t uεπ επηη⎡⎤ ⎡⎤⎛⎞⎛⎞= + −− = + −− =⎢⎥ ⎢⎥⎜⎟⎜⎟⎝⎠⎝⎠⎣⎦ ⎣⎦ll (b) Equation 5.10.10 of Example 5.10.3 gives nt2sinTctπεη⎛⎞=⋅ − −⎜⎟⎝⎠ux e el. Thus, 12 12 12334 34 42 32sin , sin , 055 5 55 5TTxx xxuc t uc t uεπ επηη⎡⎤ ⎡⎤⎛⎞ ⎛⎞= ± +− − = +− − =⎢⎥ ⎢⎥⎜⎟ ⎜⎟⎝⎠ ⎝⎠⎣⎦ ⎣⎦mll Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-13 __________________________________________________________________ 5.31 Solve for 23 and εε in terms of 1εfrom the following two algebra equations: 21 3 3 1 1(cos 2 ) (sin 2 ) cos 2 n εαε αε α+= (i) 21 3 1 111sin 2 (cos 2 ) sin 2nεαε α ε α−= − , (ii) ------------------------------------------------------------------------------- Ans. ()132 2 13 1 11cos 2 (sin 2 )1cos 2 (sin 2 )sin 2 1sin 2 (cos 2 )n nnnαα α αααα⎡ ⎤ Δ= =− +⎢ ⎥ ⎣ ⎦ − Thus, ()2 11 13 3 11 11 22 1 13 1 11cos 2 (cos 2 ) (sin 2 ) 1 sin 2sin 2 cos2 (cos 2 ) (sin 2 ) sin 2 sin 4nn n n nε εα αα ε εααα εαα α α⎡⎤−− ⎡⎤ ⎡ ⎤⎢⎥=⎢⎥ ⎢ ⎥⎢⎥ − Δ ⎣ ⎦ ⎣⎦−⎢⎥⎣⎦ ⎡⎤ −=−⎢⎥Δ⎢⎥⎣⎦ That is, () () ()22 13 1 1 21 31 22 22 13 1 13 1(cos 2 ) (sin 2 ) sin 2 sin 4, cos 2 (sin 2 )sin 2 cos 2 (sin 2 )sin 2n n nnαα α αεε εε α αα α αα−== ++ __________________________________________________________________ 5.32 A transverse elastic wave of amplitude 1ε incidents on a traction free plane boundary. If the Poisson's ratio 1/ 3 ν= , determine the amplitudes and angles of reflection of the reflected waves for the following two incident angles (a) 10α= and (b)o 115α= . -------------------------------------------------------------------------------- Ans. From Eq. (5.11.14), we have, for 1/3ν= () ( ) () () () () / 1 2 /2 1 1 2/3 /2 1 1/3 1/3 /2 2/3 1/2TLnc c νν == − − = − − = = . Thus, () 31 1/2 sin sinαα= . Using this equation, and Equations () () ()22 13 1 1 21 31 22 22 13 1 13 1(cos 2 ) (sin 2 ) sin 2 sin 4, cos 2 (sin 2 )sin 2 cos 2 (sin 2 )sin 2n n nnαα α αεε εε α αα α αα−== ++ we have, (a) () 12 3 130 0 and 1/ 2 sin sin 0 0.αα α αα=→ = = =→ = Also, the above equations give 21εε=, and 30ε=,. That is, there is no reflected longitudinal wave. There is only a reflected transverse wave of the same amplitude which completely cancels out the incident transverse wave. (b) oo 1215 15αα=→ = and oo 33 sin 2sin15 0.5176 31.17 ,αα== → = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-14 () () () () () () () ()o2 o o 21 1 2oo o 0 31 1 2oo o(cos30 ) 1/ 4 (sin 62.34 ) sin 30 0.742 cos30 1/ 4 (sin 62.34 ) sin 30 sin 60 / 20.503 cos30 1/ 4 (sin 62.34 ) sin 30εεε εεε− == + == + That, the reflected transverse wave has an amplitude 210.742ε ε= , with a reflected angle of o 21 15αα== . The reflected longitudinal wave has an amplitude 310.503ε ε= ,with a reflected angle of o 331.17α= . __________________________________________________________________ 5.33 Referring to Figure 5.11.1 (Section 5.11), consid er a transverse elastic wave incident on a traction-free plane surface () 20 x=with an angle of incident1αwith the2xaxis and polarized normal to12xx, the plane of incidence. Show that the boundary condition at 20x=can be satisfied with only a reflected transver se wave that is similarly polarized. What is the relation of the amplitudes, wavelengths, and direction of propagation of the incident and reflected wave? -------------------------------------------------------------------------------- Ans. Let the plane of incidence be 12xxplane with the angle of incidence of the transverse wave be 1α. That is, 1n1 1 1 2sin cosαα =−ee e . The waves are polarized normal to the plane of incidence, therefore, 12 0 uu==, and 31 12 2 sin sin uεϕε ϕ=+ , with 11 1 2 1 1 21 2 2 2 2 1222( sin cos ), ( sin cos )TT xx c t x x c tπ πϕ αα η ϕ ααη =− − − =+ − −ll The nonzero stress components are: () () () ( )13 31 3 1 1 1 1 1 2 2 2 2 23 32 3 2 1 1 1 1 2 2 2 2/ 2 / cos sin + / cos sin , / =2 / cos cos + / cos cos .TT ux TT uxμπ μ ε ϕ α ε ϕ α μπ μ ε ϕ α ε ϕ α⎡ ⎤ == ∂∂ =⎣ ⎦ ⎡ ⎤ == ∂∂ −⎣ ⎦ll ll The 3xequation of motion ()22 33 1 1 3 2 2// /o ut T x T xρ∂∂= ∂ ∂ + ∂ ∂ gives: ()() () ()()()22 22 2 2 11 1 2 2 2 11 1 2 2 2 222 / sin / sin 2 / sin + / sin /.oT oT T oc ccρπε ϕ ε ϕπ μ ε ϕ ε ϕ ρμ μ ρ⎡⎤ ⎡⎤ +=⎣⎦ ⎣⎦ →= → =ll ll The traction free boundary at 20 x= requires that 12 22 32 0 TTT=== on the surface, thus, () () 211 1 1 2 2 2 20/ cos cos + / cos cos 0xεϕ α εϕ α=⎡⎤−=⎣⎦ll , where 11 1 1 21 2 2 1222( sin ), ( sin )TT xc t x c tππϕ αη ϕ αη =− − =− −ll. Thus, the boundary condition is satisfied if 1 2 12 12 12 , , , ααε ε ηη====ll . __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-15 5.34 Do the problem of Section 5.11.(Reflection of Plane Elastic Waves, Figure 5.11-1) for the case where the boundary 20 x= is fixed. ------------------------------------------------------------------------------- Ans. As in Section 5.11, we assume ()()() () () ()11 1 12 2 23 3 3 21 1 12 2 2 3 3 3 3cos sin cos sin sin sin sin sin sin sin cos sin , 0u uuαε ϕ αε ϕ αε ϕ αε ϕ αε ϕ αε ϕ=+ + =− + = where 11 1 2 1 1 21 2 2 2 2 12 31 3 2 3 3 322( sin cos ), ( sin cos ) 2(s i n c o s )TT Lx x ct x x ct xx c tπ πϕ αα η ϕ ααη πϕα α η= − −− = + −− =+ − −ll l The equations of motion are satisfied with () ( ) ()22 00 2/, /LTccλμρ μ ρ=+ = Now, at 20 x=, ()()() () () ()2 211 1 2 2 2 3 3 30 11 1 2 2 2 33 30cos sin cos sin sin sin 0 sin sin sin sin cos sin 0x xαε ϕ αε ϕ αε ϕ αε ϕ αε ϕ αε ϕ= =⎡⎤ ++ =⎣⎦ ⎡⎤ −+ =⎣⎦ Thus, at 20 x=, 123sin sin sinϕϕϕ== , so that () ()1 1 111 221 33 12 3 22 2 33 222 2(s i n ) (s i n ) (s i n ) , ,TT L xc t x c t xc t pqπ ππϕα η α η α η ηη ηη′ ′ =− − =− − =− − ′′=− ± =− ±ll l ll Thus, as in Section 5.11, we have, with /TLnc c= , 21 31 , n==ll ll , 12αα= , 31 sin sinnαα= , 21 31 , nηηη η′ ′= =. However, the relations between the am plitudes are different. In fact, from ()()() () () ()12 33 11 12 33 11cos sin cos , sin cos sin .αεα ε α ε αε αε αε+= − −= we can obtain, ()()()() ()()()()31 3 1 2 13 1 3sin sin cos cos sin sin cos cosααα αεααα α−=+, ()()()()1 3 13 1 3sin 2 sin sin cos cosαεααα α−=+. ________________________________________________________________ 5.35 A longitudinal elastic wave is incident on a fixed boundary 20 x= with an incident angle of 1αwith the 2xaxis (similar to Fig. 5.11.1 of Section 5.11). (a) Show that in general, there are two reflected waves, one longitudinal a nd the other transverse (also polarized in the incident plane12xx). (b)Find the amplitude ratio of reflected to incident elastic waves. ------------------------------------------------------------------------ Ans. (a) Let ()()() () ( ) ()11 1 12 2 2 3 3 3 21 1 1 2 2 2 3 3 3 3sin sin sin sin cos sin cos sin cos sin sin sin , 0,whereu uuαε ϕ αε ϕ αε ϕ αε ϕ αε ϕ αε ϕ=+ + =− + − = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-16 11 1 2 1 1 21 2 2 2 2 12 31 3 2 3 3 322( sin cos ), ( sin cos ) 2(s i n c o s )LL Txx c t x x c t xx c tπ πϕ αα η ϕ ααη πϕα α η=− − − =+ − − =+ − −ll l The equations of motion are satisfied with () ( ) ()22 00 2/, /LTccλμρ μ ρ=+ = . Now, at 20x=, () () () () ( ) ()2 211 1 2 2 2 3 3 30 11 1 2 2 2 3 3 30sin sin sin sin cos sin 0 cos sin cos sin sin sin 0x xαε ϕ αε ϕ αε ϕ αε ϕ αε ϕ αε ϕ= =⎡⎤ ++ =⎣⎦ ⎡⎤−+ − =⎣⎦ Thus, at20x=, 123sin sin sinϕϕϕ== , so that () ()11 1 1 1 2 1 22 3 1 33 22 2 33 3(2 / )( sin ) (2 / )( sin ) (2 / )( sin ), ,.LLT xc t x c t xc t pqϕπα η πα η πα η ηη ηη′ ′ =− − =− − =− − ′′=− ± =− ±ll l ll Thus, we have, () ()11 1 1 1 2 1 22 3 1 33 22 2 33 2(2 / )( sin ) (2 / )( sin ) (2 / )( sin ), ,.LLT xc t x c t xc t pqϕπα η πα η πα η ηη ηη′ ′ =− − =− − =− − ′′=− ± =− ±ll l ll Thus, 33 12 1 2 123 1 2 3 1 2 3sin sin sin, , LLTcccα η ααη η ′′= = == ==lll l l l l l l So that 1 2 21 3 1 3 1 21 3 1 , , , sin sin , , nn n ααα α η η η η ′′ == = = ==ll l l where /TLnc c= . We note that unlike the problem in Sect. 5.11, here 31 3 1 , sin sinnnαα = = ll ( instead of 31 3 1 , sin sin nn αα ==ll ). With 123sin sin sinϕϕϕ== , we have ()()()()()() 12 33 11 12 33 11sin cos sin , cos sin cosαεα ε α εα εα εα ε+= − −= Thus, () ()() 31 1 1 3 21 1 3 1 3/ sin 2 / cos , / cos / cosεε α αα εε αα αα=− − = + − __________________________________________________________________ 5.36 Do the previous problem (Prob. 5.35) for the case where 20 x= is a traction free boundary ------------------------------------------------------------------------------- Ans. Let ()()() () ( ) ()11 1 12 2 2 3 3 3 21 1 1 2 2 2 3 3 3 3sin sin sin sin cos sin cos sin cos sin sin sin , 0,whereu uuαε ϕ αε ϕ αε ϕ αε ϕ αε ϕ αε ϕ=+ + =− + − = 11 1 2 1 1 21 2 2 2 2 12 31 3 2 3 3 322( sin cos ), ( sin cos ) 2(s i n c o s )LL Txx c t x x c t xx c tπ πϕ αα η ϕ ααη πϕα α η=− − − =+ − − =+ − −ll l The equations of motion are satisfied with () ( ) ()22 00 2/, /LTccλμρ μ ρ=+ = . At 20 x=, 21 22 23 0 TTT=== . Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-17 ()()() 12 21 22 11// 0 a n d + 2 / / 0u x ux ux u x λμ λ ∂∂ + ∂∂ = ∂∂+∂∂= . That is, ()()()()()() 11 1 1 2 2 2 2 33 3 3/ cos sin 2 / cos sin 2 / cos cos2 0εϕ α ε ϕ α εϕ α −+ + = ll l (i) and () () ( ) () ()()()22 11 1 1 2 2 2 2 33 3 3 3/ 2 cos cos / 2 cos cos /2 s i n c o sc o s0ελμ α ϕε λμ α ϕ εμα α ϕ⎡⎤ ⎡⎤++ +⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎡⎤ +− =⎣⎦ll l (ii) In order for the above two equa tions to be satisfied for all 1and x t, we must have, at 20 x= 123 cos cos cosϕϕϕ== , which gives 33 3 12 1 2 2 123 1 2 3 1 2 3sin sin sin, , LLT q ccc pα η ααη η== = = = =ml ml lll l l l ll l Thus, 12 2 1 3 1 3 1 2 2 1 3 3 1 , , , s in si n , , nn p q n ααα α η η η η== = = = =ll l l m l m l , where /TLnc c= . (i) and (ii) now gives ()()()()()() 11 1 2 2 1 33 3/s i n 2 / s i n 2 / c o s 2 0εα ε α ε α−+ + =ll l (iii) () ()22 11 1 21 1 33 3 3 22 2 11 1 22 22 21 3 3 3 1 1( / )( 2 cos ) ( / )( 2 cos ) ( / )(2 )sin cos 0 22 cos 2 2 sin 2sin ( ) 1 2 sin ( /)(2 )sin cos 1 2 sin nnnnε λ μαε λ μαε μ αα λμλμ αλμμ αμ αμ μμεα ε μ α α ε α++ +− = ⎛⎞++= + −=− = ⎜⎟⎝⎠ −− = − −lll (iv) Since ()22 2 2 2 11 1 1 222 cos 2 2 sin 2sin 1 2 sin n nλμ μλμα λ μ μα μ α αμ⎛⎞++= + −=− = − ⎜⎟⎝⎠, and 31n=ll , therefore, (iii) and (iv) become ()() 12 33 1 1 sin 2 cos2 sin 2nnαεα ε ε α+= (v) () ( )22 22 12 3 3 3 11 1 2 sin 2 sin cos 1 2 sinnn nαεα α ε α ε −− = − − (vi) (v) and (vi) give 22 3 11 22 2 1 13 1 32s i n 2 ( 1 2 s i n ) sin 2 sin 2 (1 2 sin )cos2nn nnε αα ε ααα α−= +−, 22 2 13 1 3 2 22 2 1 13 1 3s i n2 s i n2 ( 1 2 s i n )c o s2 s i n2 s i n2 ( 1 2 s i n )c o s2nn nnααα α ε ε ααα α−−= +− __________________________________________________________________ 5.37 Verify that the thickness stretch vibration given by Eq.(5.12.3), i.e., 11 1( cos sin )( cos sin )LL uA k x B k x C c k t D c k t=+ + does satisfy the longitudinal wave equation ()()2 22 2 2 11 1//L ut c ux∂∂= ∂ ∂ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-18 --------------------------------------------------------------------------------- Ans. 11 1( cos sin )( cos sin )LL uA k x B k x C c k t D c k t=+ + , 22 2 11 1 1 22 2 11 1/( c o s c o s ) ( c o s s i n ) /( ) ( c o sc o s ) ( c o s s i n )LL LL Lux k Ak x Bk x Cc k t Dc k t u t kc A kx B kx C c kt D c kt∂∂ = − + + ∂∂ = − + + that is, ()2 22 2 2 2 11 1 1 1/ a n d /L ux k u ut c k u∂∂ = − ∂∂ = − . Thus, 22 2 2 2 11 1//Lcux ut∂∂= ∂ ∂ __________________________________________________________________ 5.38 (a) Find the thickness-stretch vibration of a plate, where the left face (10x=) is subjected to a forced displacement 1tαω u=( c o s ) e and the right face1x=l is free. (b)Determine the values of ωthat give resonance. --------------------------------------------------------------------------------- Ans. Let (a) 11 1( cos sin )( cos sin )LL uA k x B k x C c k t D c k t=+ + . Using the boundary condition1 (0, ) ( cos )ttαω u= e , we have , 1 cos (0, ) cos sinL L tu t A C c k tA D c k tαω==+ Thus, 11 1 , / , 0 ( cos sin )cosL AC k c D u kx BC kx tαωα ω== = → = + . At 1x=l, 11 12 13 0 TTT=== . Now, ()() 11 1 1 2/ Tu xλμ=+∂ ∂ , thus () 111/0xux=∂∂=l, i.e., ( sin cos )cos 0 tankk B C kt B Ckα ωα −+ = → =ll l , 11 1 [ c o s (/ ) t a n ( / ) s i n (/ ) ] c o sLL L ux c c x c tαωω ω ω →= + l . (b) Resonance occurs at : / / 2, 1,3,5...Lcn nω π= = l __________________________________________________________________ 5.39 (a) Find the thickness stretch vibration if the 10x=face is being forced by a traction ()1 cos tβωt= e and the right hand face 1x=l is fixed. (b) Find th e resonance frequencies. -------------------------------------------------------------------------------- Ans. (a) 11 1( cos sin )( cos sin )LL uA k x B k x C c k t D c k t=+ + . At 1 1 1 11 1 21 2 31 3 1 11 21 310, , ( ) cos cos , 0xT T T t T t T T βωβ ω == −= −= − + + = → = − = =ne tT e e e e e Since()() 11 1 1 2/ Tu xλμ=+ ∂∂ , therefore, the boundary condition at 10x=is: ()() 111 02/ c o sxux tλμβ ω=+∂ ∂ − = , ()2( ) ( c o s s i n ) c o sLL kB C ck t D ck t tλμβ ω→+ + − = , ()0, ,2L Dc k B Ckβωλμ→= = = −+, ()11 1(c o s s i n ) c o s2uA C k x k x tkβωλμ→= −+ At 1x=l, () () () () ()1 11 1( cos sin )cos 0 cos sin 022 tan tan cos sin cos22 2LL LL LuA C k k t A C k kkk cx c xAC k u tkc c cβ βωλμ λμ βω β ω βωωλμ ω λμ ω λμ=− = → − = →++ ⎡⎤ =→ = − ⎢⎥++ + ⎢⎥⎣⎦ll ll ll (b) Resonance occurs at Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-19 () / 2 , 1,3,5...Lnc nωπ== l __________________________________________________________________ 5.40 (a) Find the thickness-shear vibr ation if the left hand face 10x=has a forced displacement 3 cos tαω u=( ) e and the right-hand face 1x=lis fixed. (b) Find the resonance frequencies. -------------------------------------------------------------------------------- Ans. (a) Let 31 1 1 2( cos sin )( cos sin ), 0TT u A k x B k x C ck t D ck t u u=+ + = = In the absence of body forces, the 3xNavier equation of motion (5.6.7) gives: 1122 2 222 33 3 3 22 2 2 2 2 3 23 1()oouu u eux tx x x t xρλ μ μ ρ μ⎛⎞∂∂ ∂ ∂ ∂∂∂⎜⎟ =+ + + + → =⎜⎟∂ ∂∂ ∂ ∂ ∂ ∂⎝⎠ leads to () ( )22 2 33 /oT T ock u ku cρ μμ ρ −= − → = . The boundary condition at 10x=, () 30, cosut t αω→ =( )3( ) ( co s sin ) c os 0 , , /TT T u A C c kt D c kt t D AC k c αωα ω =+ = → = = = . 31 1(c o s s i n ) c o s uk x B C k x tα ω →= + . The boundary condition at 1x=l, 3(,) 0ut=→l3( cos sin )cos 0 cot u k BC k t BC kα ωα =+ = → = − ll l . [] 31 1 cos( / ) cot( / )sin( / ) cosTT T ux c c x c tαωω ω ω →= − l . (b) Resonance occurs at / , 1,2,3...Tnc nωπ= =l __________________________________________________________________ 5.41 (a) Find the thickness-shear vibr ation if the left hand face 10x=has a forced displacement 23 cos sin ttαωω+ u= ( e e ) and the right-hand face 1x=lis fixed. (b) Find the resonance frequencies. ------------------------------------------------------------------------------- Ans. (a) If the left hand face 10x=has a forced displacement 2 cos tαω u= e and the right-hand face 1x=lis fixed, it is clear from the result of the previous problem, ()()() 21 1 1 3 cos / cot / sin / cos , 0TT T ux c c x c t u uαω ω ω ω⎡⎤=− = =⎣⎦l . If the left hand face 10x=has a forced displacement 3 sin tαω u= e and the right-hand face 1x=lis fixed, the displacement field is clearly given by ()()() 31 1 1 2 cos / cot / sin / sin , 0TT T ux c c x c t u uαω ω ω ω⎡⎤=− = =⎣⎦l Thus, the solution to the present problem can be obtained by superposition to be 10u=, ()()() 21 1 cos / cot / sin / cosTT T ux c c x c tαωω ω ω⎡⎤=−⎣⎦l , ()()() 31 1 cos / cot / sin / sinTT T ux c c x c tαωω ω ω⎡⎤=−⎣⎦l . (b) Resonance occurs at / , 1,2,3...Tnc nωπ= =l Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-20 __________________________________________________________________ 5.42 A cast iron bar, 200 cm long and 4 cm in diam eter, is pulled by equal and opposite axial force P at its ends. (a) Find the maximum normal and shearing stresses if P=90,000N. (b) Find the total elongation and lateral contraction. ( ) 103 ., 0.3YEG P a ν == -------------------------------------------------------------------------------- Ans. 22 4 2(4 10 ) / 4 12.6 10 Amπ−−=× =× . () () () ()46 6 max max( ) / 90,000 / 12.6 10 71.4 10 , / 2 35.7 10nsaT PA N T P A N−== ×= × = = × . 69 3 62 9 5( ) ( / )( / ) (71.4 10 ) 2 / (103 10 ) 1.39 10 , ( / )( / ) (0.3)(71.4 10 ) (4 10 ) / (103 10 ) 0.832 10 .Y dYbP A E m PA dE mδ δν− −−== × × × = × =− = × × × × =− ×l l ____________________________________________________________________ 5.43 A composite bar, formed by welding two sle nder bars of equal le ngth and equal cross- sectional area, is loaded by an axial load Pas shown in Figure below. If Young's moduli of the two portions are(1) (2)and YYE E, find how the applied force is distributed between the two halves. ------------------------------------------------------------------------------- Ans. Taking the whole bar as a free body, let 1P be the compressive reactional force from the right wall to the bar and 2P be the compressive reactional force from the left wall to the bar, then the equation of static equilibrium requires 12 PPP=− . (i) There is no net elongation of the composite bar, therefore, 12 (1) (2)0 YYPP AE AE+=ll (ii) Combining Eq. (i) and (ii), we obtain 12 ( 2 )( 1 ) ( 1 )( 2 ), 1( / ) 1( / )YY Y YPPPP EE E E−== ++. (iii) _________________________________________________________________ 5.44 A bar of cross-sectional area Ais stretched by a tensile force Pat each end. (a) Determine the normal and shearing stresses on a plane with a normal vector which makes an angleαwith the axis of the bar. (b) For what value of αare the normal and shearing stresses equal? (c) If the load carrying capacity of the bar is based on the shearing stress on the plane defined by oαα= to be less than oτwhat is the maximum allowable load P? ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-21 Ans. []00 000 000σ⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦T , 12 cos sinαα+ n= e e , (a) For the plane with a normal given by 12 cos sinαα+ n= e e , we have, [] [ ] []2 100c o s c o s 000s i n 0 c o s c o s , 000 0 0nTσα σ α α σα σ α⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=→ → = ⋅⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦t=T n= t = e tn = ()2 2 2 22 24 22 2 22 2cos cos cos 1 cos cos sinsnTT σασ ασ α α σ α α =−= − = − =t sin 2 / 2sTσα →= . (b) ()22 sin 2cos cos sin cos cos sin cos 02σασα α α α α αα=→ = → − = . Thus, (i) cos 0 / 2 0, and (ii) sin cos / 4 / 2sn snTT TT ααπ α α απ σ= →= →== = →= →== . (c) o osin 2 2 2s i n 2o oσαττσα≤→≤ . Max allowable o2 sin 2oPAτ α≤ _________________________________________________________________ 5.45 A cylindrical bar, whose lateral surface is constrained so that there can be no lateral expansion, is then loaded w ith an axial compressive stress 11Tσ=−. (a) Find 22 33 and TT in terms of σ and the Poisson's ratio ν, (b) show that the effective Young's modulus () 11 11/YeffE TE≡ is given by ()2(1 ) / (1 2 )YeffE ννν =− −− . [note misprint in text]. ------------------------------------------------------------------------------- Ans. (a) () 22 22 33 1100 ET T T ν =→ − + = , () 33 33 11 2200 ET T T ν =→− + = . Thus, 22 33TTννσ−= − and 33 22TTννσ−=− . From these two equations, we have, 22 33 /( 1 ) TT νσν==− − . (b) ()22 11 11 22 3311 2 1 22111 1YY Y YET T TEE E Eνσσν σ ν ννσ ννν ν⎡⎤⎛⎞ ⎛ ⎞ ⎡⎤ −− − − ⎛⎞⎡⎤=− + = − + = − = ⎢⎥⎜⎟ ⎜ ⎟ ⎜⎟⎢⎥ ⎣⎦ ⎜⎟ ⎜ ⎟ −− −⎝⎠⎣⎦ ⎢ ⎥ ⎝⎠ ⎝ ⎠⎣⎦ Thus, () ()()() () ( )11 2 11 1111 12 1 12YY YYeff effEE TEEEEνν σ νν νν−− −≡→ == =−+ −−. __________________________________________________________________ 5.46 Let the state of stress in a tension specimen be given by 11 and all other =0ij TTσ= . (a) Find the components of the deviatoric stress defined by ()o1/3kkT =−TT I . (b) Find the principal scalar invariants of oT ------------------------------------------------------------------------------- Ans. (a)o 11 22 33 11 11 /3 /3 2 /3kk kkTT TT TT T σσ σ σ =++= →=− = − = . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-22 oo o o o 22 22 33 12 13 23 /3 /3 , 0kk TTT T TTT σ =− = − = === . (b) () () () () () () () () ()oo o 11 12 23 3 oo o o oo 2 21 1 2 22 2 3 31 1 3 3 oo o 3 31 1 2 2 3 32 / 3/ 3/ 3 0 . 2 /3 /3 /3 /3 2 /3 /3 /3 . 2 / 3 / 3 / 3 2 / 27.IT T T IT T T T T T IT T Tσσσ σσ σσ σσ σ σσσ σ=++= − − = =++= − + −− + − = − ==− − = _________________________________________________________________ 5.47 A circular cylindrical bar of length lhangs vertically under gravity force from the ceiling. Let 1xaxis coincides with the axis of th e bar and points downward and let the point() () 123, , 0,0,0xx x= be fixed at the ceiling. (a) Verify that the following stress field satisfies the equations of equilibrium in the presence of the gravity force: () 11 1Tg xρ=−l , all other 0ijT=and (b) verify that the boundary conditions of zero surface traction on the lateral face and the lower end face are satisfied and (c) obtained the resultant force of the surface traction at the upper face. ------------------------------------------------------------------------------- Ans. (a) The body force per unit volume is given by1gρρB= e . Thus, with () 11 1Tg xρ=−l , we have, 13 11 12 12300 0T TTgg gxxxρρ ρ∂ ∂∂+++ = − + + + =∂∂∂ and the other two equations are trivially satisfied. (b) On the bottom end face 1x=l, () 111 1 1 1 1, xTg ρ== =− = n=e t=T e e e 0lll . On the lateral face, []11 22 33 2 300 0 0 , 0 0 0 0 00 0 0T nn n n⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥+= →⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣⎦n= e e t = t=0 . (c) On the top face at 10x=, () 111 1 1 1 1 1 0, 0xTg g ρρ=− − =− =− − =−n= e t= T e e e e ll Let the area of the face be A, then the resultant force is 11 Ag AWρ−=− t= e e l where Wg Aρ=lis the weight of the bar and the minus sign indicates that the resultant force at the ceiling is upward which balances the weight of the bar. __________________________________________________________________ 5.48 A circular steel shaft is subjected to twisting couples of 2700 Nm. The allowable tensile stress is 0.124 GPa . If the allowable shearing stress is 0. 6 times the allowable tensile stress, what is the minimum allowable diameter? ------------------------------------------------------------------------------- Ans. () () ()4 max max 32// 2tt ns t pMaMTT M a aI aπ π== = = . Thus Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-23 ()()()() ()()93 6 3 3 9 22 2700 2 27000.6 0.124 10 23.1 10 0.6 0.124 10 2.85 10 2.85 5.7 .am a am c m d c mπ π− −≤× → ≥ = × × →≥ × = →≥ _________________________________________________________________ 5.49 In Figure 5P.2, a twisting torque tMis applied to the rigid disc A. Find the twisting moments transmitted to the circular shafts on either side of the disc. Figure 5P.2 ------------------------------------------------------------------------------- Ans. Let1Mand2Mbe the twisting moments transmitted to the left and the right shaft respectively. Then equilibrium of the disc demands that 12 t M MM+= (i) In addition, the disc is rigid, therefore, the angle of twist of the left shaft at the disc relative to the left wall must equal the angle of twist of the right shaft at the disc relative to the right wall, i.e., 11 2 2 11 2 2 ppMMMMIIμμ=→ =llll (ii) Thus, 21 12 12 12, tt M MM M⎛⎞ ⎛⎞==⎜⎟ ⎜⎟++⎝⎠ ⎝⎠ll ll ll (iii) for 12=ll , 12 /2 .t MM M== _________________________________________________________________ 5.50 What needs to be changed in the solution fo r torsion of a solid circular bar obtained in Section 5.14 for it to be valid for torsion of a hollow circular bar with inner radius aand outer radius b? ------------------------------------------------------------------------------- Ans. The hollow circular bar differs from the solid ci rcular bar in that there is an inner lateral surface which is also traction free. However, th e normal to the inner lateral surface differs from that to the outer surface only by a sign so that the zero surface traction in the inner surface is also satisfied since that for the outer surface is satis fied. However, in calculating the resultant force and resultant moment due to the surface traction on the end faces, the integrals are now to be integrated over the circular ring area between by and ra rb==rather than the whole solid Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-24 circular area of radius b. Thus, the only change that needs to be made is that the polar area second moment pIis now given by ()44 2pIb aπ=− . __________________________________________________________________ 5.51 A circular bar of radius or is under the action of an axial tensile load Pand a twisting couple of tM. (a) Determine the stress throughout th e bar. (b) Find the maximum normal and shearing stress ------------------------------------------------------------------------------- Ans. Superpose the solutions for tension and for torsion, we have, with , t pM P AIσβ=≡ (a) 11 12 21 3 13 31 2 , , , all other 0ij TT Tx T T x Tσ ββ == = − = = = . (b) The characteristic equation is ()() ()32 2 2 23 2 2 2 222 32 3 2 3 200 0 0 , w h e r e 0xx x xx r r x x xσλ β β βλ σ λ λ λ βλ β λ λ σ λ β βλ−− −− = → − + + = → − −= = + − Thus, 22 2 1,2 34, 02r σσ βλλ±+= =. Thus ()22 2 22 2 max4 1, 422nsrTT rσσ βσβ++== + . __________________________________________________________________ 5.52 Compare the twisting torque which can be tran smitted by a shaft with an elliptical cross- section having a major diameter equal to twice the minor diameter with a shaft of circular cross-section having a diameter equal to the major diameter of the elliptical shaft. Both shafts are of the same material. Also compare the un it twist (i.e., twist angle per unit length) under the same twisting moment. Assume that the maximum twisting moment which can be transmitted is controlled by the maximum shearing stress. ---------------------------------------------------------------------------------- Ans. (a) For an elliptical shaft with major diameter 2band minor diameter 2a (i.e., ba>), ()() max 22tell sM T abπ= . For a circular shaft with radius b, ()() max 32tcir sM T bπ= , thus ()()() () ()22 max 2 322 1, 24tt tell cir ell s tcirMM M aaTMb a ab bππ⎛⎞ ⎛ ⎞== → = = = ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. (b) 22 33 42' , 'ell t cir tabM M ab bαα μπ μπ⎛⎞ ⎛⎞ +==⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠, thus, ()()()22 2 3325'= =5' 22ell cirba b a a aaα α+ = . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-25 __________________________________________________________________ 5.53 Repeat the previous problem except that the circular shaft has a diameter equal to the minor diameter of the elliptical shaft. -------------------------------------------------------------------------------- Ans. (a) For an elliptical shaft with major diameter 2band minor diameter 2a (i.e., ba>), ()() max 22tell sM T abπ= . For a circular shaft with radius a, ()() max 32tcir sM T aπ= , thus, ()()() () ()max 2 322 2, 2tt tell cir ell s tcirMM M baTMa a ab aππ⎛⎞⎛ ⎞== → = = = ⎜⎟⎜ ⎟⎝⎠⎝ ⎠. (b) 22 2 4 33 4 6' 25 5' , ' '2 1 6 8ell ell t cir t cirab a aMM ab a aαααα μπ μπ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ +== → = =⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠. __________________________________________________________________ 5.54 Consider torsion of a cylindrical bar with an equilateral triangular cross-section as shown in Fig. P.5.3. (a) Show that a warping function ()23 23 33Cx x xϕ=− generate an equilibrium stress field. (b) Determine the constant C, so as to satisfy the traction free boundary condition on the lateral surface 2x a=. With Cso obtained, verify that the other two lateral surfaces are also traction free. (c) Evaluate the shear stress at the corners and along the line 30x=.(d) Along the line30x=where does the greatest shear stress occur? -------------------------------------------------------------------------------. Ans. (a) 22 2 2 33 22 2 2 23 2 36, 6, t h u s 0Cx Cx xx x xϕϕ ϕ ϕ∂∂ ∂ ∂== − + = ∂∂ ∂ ∂, so that equations of equilibrium are satisfied. 2x3x (2, 0 )a− (, 0 )a (b) For the lateral surface 2x a=, () 222 1 2 1 3 2 1=, ' / 0xaTx xμα μϕ=⎡⎤==− + ∂ ∂ =⎣⎦ne t = T e e e [] 232 3 3 3'6 ' 6 ' / 6xax Cx x x Cax C aαα α=→= →= → = . On the lateral surface ()() ()() 32 3 2 2 31 / 3 2 321 / 2 3 xx a x x a=+ → − = → − + n= e e ()() ()( ) 2 3 12 13 1 1/2 3 1/2 3 TT −+ = − + t=T n= T e T e e . Now, for ()()23 23 3 '/ 6 3ax xxϕα=− Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-26 12 3 2 3 2 3 22 13 2 3 2 2 3'(/ ) '( ' / ) ( ) , '(/ ) '( ' / 2 ) ( ) .Tx xx a x x Tx xx a x xμα μϕ μα μ α μαμϕ μαμ α=− + ∂ ∂ =− + =+ ∂ ∂ =+ −, Therefore, ()22 12 13 3 2 3 2 2 3'32 2 2 3 32TT a x x x a x x xaμα⎡⎤ −+ = − + + −⎣⎦. With 23 32 x xa=− , () () () () () () () ()22 12 13 3 3 3 3 3 3 22 2 2 2 33 3 3 3 3 3 22 22 2 333 3 3 3 3'32 2 3 2 2 3 3 2 3 3 22 '2 2 3 4 64 3 3 3 4 342 '246 1 24 3 4 3 2 3 3 3 3 0 .2T T a x xa x a xa xaxa ax x ax ax a x ax a xa ax ax ax ax a a x x xaμα μα μα⎡⎤ ⎛⎞−+ = − − + − + − − ⎜⎟ ⎢⎥⎝⎠ ⎣⎦ ⎡⎤=+ − + + − + − + −⎣⎦ ⎡⎤=+ + − − + + − + − =⎣⎦ That is, on ()() 321/ 3 2 , xx a=+ t=0 . Clearly, for the lateral surface ()() 32 1/ 3 2 x xa =− + , t=0 . (c) at the corner () 2, 0a− , ()() 12 3 2 3 '' / 0 Tx a x xμα μ α=−+ = and ()()() ( ) ()22 2 13 2 2 3 '' / 2 2 '' / 2 4 0 Tx a x x a a aμ α μα μ α μα=+ − = − + = . At the corners (),3aa± , ()() ()() 12 3 2 '/ ' 3 0 Tx a x a a aμα μα= −= ± −= and () ()22 2 2 13 2 2 3 '/ 2 ' 3 / 2 0 Tx x x a a a a aμα μα⎡⎤ ⎡⎤=+ − = + − =⎣⎦ ⎣⎦. That is, the shear stress at all three corners are zero. Along 30x=, ()() 12 3 2 3 '' / 0 Tx a x xμα μ α=− + = , ()()()22 2 13 2 2 3 2 2 '/ 2 ' / 2 2 Tx x x a a a x xμα μα⎡⎤=+ − = +⎣⎦. (d) () ( ) () 213 2 2 2 13/ ' /2 2 2 0 ' /2xadT dx a a x x a T a μα μα=−=+ = → = − → = . But at ()() 23,, 0xxa= , () ()()22 2 13 2 2 3 '/ 2 ' / 2 3 / 2 ' Tx x x a a a a aμαμ α μ α⎡⎤ ⎡ ⎤=+ − = + =⎣⎦ ⎣ ⎦. Thus, along 30x=, the greatest shear stress occurs at ()() 23,, 0xxa= with () 3/ 2 'sTa μα = . __________________________________________________________________ 5.55 Show from the compatibility equations that the Prandtl's stress function () 23,xxψ for torsion problem must satisfy the equation22 32 32constant xxψψ∂∂+= ∂∂ ------------------------------------------------------------------------------- Ans. With 12 13 32, TTx xψ ψ ∂∂== −∂∂, and all other 0ijT=, we have, the nonzero strain components are:12 13 3211, , all other 022ij EE Exxψψ μμ∂∂== − =∂∂. All equations of compatibility are identically satisfied except the following two: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-27 2 31 23 22 12 31 2 2 3 1EE EE xxx x x x⎛⎞∂∂ ∂∂∂=−++⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎝⎠, 2 33 23 31 12 12 3 3 1 2EE E E xxx x x x⎛⎞ ∂∂ ∂ ∂∂=−++⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎝⎠ which leads to 23 23 2 230x xxψψ⎛⎞∂∂ ∂+=⎜⎟⎜⎟∂∂∂⎝⎠, 22 32 3 320x xxψψ⎛⎞∂∂ ∂+= ⎜⎟⎜⎟∂∂∂⎝⎠→22 32 32constant xxψψ∂∂+= ∂∂ ________________________________________________________________________________________________________________________________________________ 5.56 Given that the Prandtl' stress function for a rectangular bar in torsion is given by ()() () ()21/ 2 3 2 33 1,3,5cosh / 2 32 ' 111 c o scosh / 2 2n nnx a nx a nb a a nπ π μαψπ π∞− =⎛⎞ ⎧ ⎫ ⎪⎪=− −⎜⎟ ⎨⎬⎜⎟⎪⎪ ⎝⎠ ⎩ ⎭∑ The cross section is defined by 23 and ax a bx b−≤≤ − ≤≤ . Assume ba>, (a) Find the maximum shearing stress. (b) Find the maximum normal stress and the plane it acts. -------------------------------------------------------------------------------- Ans. We know that when a rectangular membrane, fixed on its side, is subjected to a uniform pressure on one side of the membrane, the deformed surface has a maximum slope at the mid point of the longer side. Thus, based on the membrane analogy discussed in Example 5.17.3, on any plane 1constantx= , the maximum shearing stress occurs on the mid point of the longer side. That is at the point 2x a= and 30x=. From the given function () 23,xxψ , we obtain the stress components as ()() () () ()() () ()21/ 2 3 2 12 33 3 1,3,5 21/ 2 3 2 13 33 2 1,3,5sinh / 2 32 ' 1 1c o s .2c o s h / 2 2 cosh / 2 32 ' 1 11 s i n .2c o s h / 2 2n n n nnx a nx anTx an b aa n nx a nx anTx an b a a nπ π ψμ α π π π π π ψμ α π π π∞− = ∞− =⎛⎞ ⎧ ⎫∂ ⎪⎪⎛⎞== − − ⎜⎟ ⎨⎬⎜⎟ ⎜⎟∂ ⎝⎠⎪⎪ ⎝⎠ ⎩ ⎭ ⎛⎞ ⎧ ⎫∂ ⎪⎪=− = − −⎜⎟ ⎨⎬⎜⎟∂ ⎪⎪ ⎝⎠ ⎩ ⎭∑ ∑ () ( )()( ) ()3/ 2 23 13 12 22 1,3,5 , 0, note sin / 2 1 , 1 ,3,5.. 16 ' 1 11, 0cosh / 2n nAt x a x n n aTTnb a nπ μα π π+ ∞ === = − = ⎧⎫⎪⎪ ⎛⎞=−= ⎨⎬ ⎜⎟⎝⎠ ⎪⎪⎩⎭∑ That is ()()max 2 2 1,3,516 ' 1 11cosh / 2s naTnb a nμα π π∞ =⎧ ⎫ ⎪ ⎪ ⎛⎞=− ⎨ ⎬ ⎜⎟⎝⎠ ⎪ ⎪ ⎩⎭∑ . Or, since 2 2 1,3,51 8n nπ∞ ==∑ , therefore ()()max 2 2 1,3,516 ' 1 12'cosh / 2s naTanb a nμαμαπ π∞ =⎧⎫⎪⎪ ⎛⎞=− ⎨⎬ ⎜⎟⎝⎠ ⎪⎪⎩⎭∑ Since at this point, the only nonzero stress components are () 13 31 13 and TT T = , therefore, the characteristic equation is 32 130 Tλλ−+= so that the maximum normal stress is () () 13 max maxnsTT T== , which acts on plane whose normal is in the direction ()() 13 1/ 2 ±ee . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-28 _________________________________________________________________ 5.57 Obtain the relationship between the twisting moment tMand the twist angle per unit length 'αfor a rectangular bar under torsion. Note: 4 44111 ...96 35π+++ = . ------------------------------------------------------------------------------- Ans. We have, [see Eq. (5.18.10)], ()() () ()21/ 2 3 2 33 1,3,5cosh / 2 32 ' 111 c o scosh / 2 2n nnx a nx a nb a a nπ π μαψπ π∞− =⎛⎞ ⎧ ⎫ ⎪⎪=− −⎜⎟ ⎨⎬⎜⎟⎪⎪ ⎝⎠ ⎩ ⎭∑ . Thus, if let 2 332 ' aAμα π⎛⎞ =⎜⎟⎜⎟⎝⎠ and () ()3 3cosh / 2()cosh / 2nx aFxnb aπ π= , then, we have, ()()() ()() 1/ 2 1/ 2 22 23 3 2 33 1,3,5 1,3,52 1121 2 c o s 21 c o s ( )22 .t aa b nn aa bnnMd A nx nxAb d x A F x d x d xaa nn MNψ ππ∞∞−− −− −==== ⎛⎞ ⎛ ⎞⎡⎤−− − ⎜⎟ ⎜ ⎟⎢⎥ ⎜⎟ ⎜ ⎟ ⎣⎦⎝⎠ ⎝ ⎠ ≡−∫ ∑∑ ∫∫ ∫ Now, ()()3/ 2 2 224cos 2sin 1 , 1,3,522a n anx anadx nan nπ π ππ+ −⎛⎞ ⎛⎞== − =⎜⎟ ⎜⎟⎝⎠ ⎝⎠∫, therefore, ()()() () () ()1/ 2 3 2 2 34 4 4 1,3,5 1,3,5 1,3,518 1 3 2 ' 121 2 c o s 2 2 22a n ann nnx aMA b d x A b aba nn nπ μα π π∞∞ ∞− −== =⎛⎞ ⎛⎞=− = = ⎜⎟ ⎜⎟⎝⎠ ⎝⎠∑∑ ∑ ∫ . Next, () ()() ()3 33 3cosh / 2 2sinh / 224() t a n hcosh / 2 cosh / 2 2bb bbnx a nb a aa n bF x dx dxnb a n nb a n aπ π π ππ π π−−⎛⎞ ⎛⎞=== ⎜⎟ ⎜⎟⎝⎠ ⎝⎠∫∫, so that ()() ()4 1/ 2 2 33 2 35 5 1,3,5 1,3,564 ' 2 112 1 cos ( ) tanh22ab n abn na nx nbN A F x dx dxaa nnμα π π π∞ ∞− −−= =⎛⎞⎡⎤ ⎜⎟ =− =⎢⎥ ⎜⎟ ⎣⎦⎝⎠∑∑ ∫∫ Thus, () ()() () ()()4 3 44 5 5 1,3,5 1,3,5 443 45 5 1,3,564 ' 2 32 ' 1 122 t a n h2 64 ' 2 32 ' 122 t a n h .96 2t nn na nbMM N a ba nn a nbaba nμα μαπ ππ μα μα π π ππ∞∞ == ∞ =⎛⎞⎛⎞⎜⎟ ≡−= − ⎜⎟⎜⎟ ⎝⎠⎝⎠ ⎛⎞⎛⎞ ⎛⎞⎜⎟ =− ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎝⎠ ⎝⎠⎝⎠∑∑ ∑ Or, () ()3 55 1,3,5' 192 122 1 t a n h32t nan bMa bba nμαπ π∞ =⎡ ⎤⎛⎞ ⎛ ⎞=− ⎢ ⎥ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦∑ _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-29 5.58 In pure bending of a bar, let 22 33 L R MM=+= − Me e M , where 23 and ee are not along the principal axes, show that the flexural stress 11T is given by () ()22 3 32 2 23 3 32 3 11 2 322 33 22 23 33 22 23MI MI MI MITx x II I II I+ +=− + −− ------------------------------------------------------------------------------- Ans. Refer to Section 5.19, we had [s ee Eq.(5.19.4)(5.19.6) and (5.19.7)] 11 2 3Tx xβγ=+ , where 22 3 2 2 3 3 3 2 3 , M II M IIβγβ γ=+= − − Solving the above two equations for and βγ, in terms of 23and M M, we obtain ()22 3 32 2 2 33 22 23MIM I II Iβ+=− − and ()23 3 32 3 2 33 22 23MIM I II Iγ+= −. Thus, () ()2 2 33 2 2 2 3 33 2 3 11 2 322 33 22 23 33 22 23MI MI MI MITx x II I II I+ +=− + −−. _________________________________________________________________ 5.59 From the strain components for pure bending 23 23 11 22 33 12 13 23 22 22, , 0 YYMx MxEE E E E EIE IEν== = −= = = Obtain the displacement field ------------------------------------------------------------------------------- Ans. Integration of 2 11 3 2 2 3 33 3 22/ , / , / , where YMu x Ax u x Ax u x Ax AIEνν ∂∂ = ∂∂ = − ∂∂ = − ≡ gives () () ()2 13 1 1 2 3 2 3 2 2 1 3 3 3 3 1 2 ,, ,, / 2 , uA x xf x x u A x x f x x u A x f x x νν =+ = − + = − + (i) where ()()() 12 3 21 3 31 2,, ,a n d , fxx f x x f x x are integration functions. Substituting (i) into 12 21 13 31 23 32// 0 , // 0 a n d // 0ux u x ux ux u x ux∂ ∂+ ∂ ∂= ∂ ∂+ ∂ ∂= ∂ ∂+ ∂ ∂= , we obtain () () () () () ()12 3 2 21 3 1 1 3 12 3 3 31 2 1 22 21 3 3 31 2 2 3 1,/ ,/ ( ) ,/ ,/ ( ) ,/ ,/ ( )fxx x f x x x gx fxx x f x x x gx fxx x f xx x g x∂∂ = − ∂∂ = ∂∂ = − ∂∂ = ∂∂ = − ∂∂ = (ii) where ()()() 123,, gxg x g x are integration functions. Integrations of (ii) give, () 1 132 43 1 223 6 2 () () a n d () fgxx g x f g xx g x=+ =+ (iii) () 21 3 15 3 23 1 38 1 ( ) ( ), and ( ) fgxx gx f gxx g x−= + = + (iv) () () 32 2 1 7 2 33 1 29 1 () a n d ( ) fgxx g x f gx x gx−= + −= + (v) From (iii), () 1 3 1 3 12 2 1 224 3 2 326 2 1 22( ) , ( ) , ( ) , gxa x b g xa x b g xb x c g xb x c=+ =+ =+ =+ (vi) From (iv) and (vi), () () 31 1 1 3 81 1 1 3 53 3 3 3 , , ( ) gxa x b g xb x cg x b x c=− + =− + − = + (vii) From (v) (vi),(vii), () () 19 1 2 1 4 7 2 3 2 40, , ag x b x c g x b x c== + = + (viii) Thus, 11 22 32 23 3 1 13 3 2 13 24 , , fbx b x c f bx bx c f b x bx c=++ =−+ = −−− (ix) Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-30 So that, 22 13 1 1 2 2 3 2 2 3 2 3 3 1 1 3 22 22 2 2 33 2 1 3 2 4 22,, .2YY YMMu x x b xb xc u x xb xb xcIE IE Mux b x b x cIEν ν=+ + + = −+ − + =− − − − (x) _________________________________________________________________ 5.60 In pure bending of a bar, let 22 33 L R MM=+= − Me e M , where 23and ee are along the principal axes, show that the neutral axis, (t hat is, the axis on the cross section where the flexural stress 11T is zero) is, in general, not parallel to the couple vectors. -------------------------------------------------------------------------------- Ans. From Eq.(5.19.10), we have, 23 32 11 22 33MxM xTII=− , thus the neutral axis is given by: 23 32 22 330Mx Mx II−= . That is, the neutral axis is given by 33 22 23 3 2x M I x IM⎛⎞=⎜⎟ ⎝⎠. Thus, only when 22 33II= is the neutral axis parallel to the couple vector22 33 L R MM=+= − Me e M . __________________________________________________________________ 5.61 For plane strain problem, derive the bi-h armonic equation for the Airy stress function ------------------------------------------------------------------------------- Ans. We have [Eq.(5.20.7)} () () () ()22 22 22 11 22 22 22 21 1 2 2 12 13 23 33 121111 , 11 , 1(1 ) , 0.YY YEE EE xx xx EE E E Ex xϕ ϕϕ ϕνν ν νν ν ϕν∂∂ ∂∂= − −+ = − −+ ∂∂ ∂∂ ∂=− + = = = ∂∂⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦ () () () ()22 44 44 22 11 22 24 2 2 24 2 2 22 2 1 11 1 211 , 11YYEEEE x xx x x xx xϕϕ ϕϕνν ν νν ν⎡⎤ ⎡⎤∂∂ ∂∂ ∂∂=− −+ =− −+⎢⎥ ⎢⎥∂∂ ∂ ∂ ∂∂ ∂ ∂⎢⎥ ⎢⎥⎣⎦ ⎣⎦, 2 2 12 22 22 21 1 222 ( 1 )YEE xxx xϕν∂ ∂=− + ∂∂∂ ∂. Thus, the compatibility equation 22 2 11 22 12 22 2 2 22 2 120EE E xx x x⎛⎞∂∂ ∂+−= → ⎜⎟⎜⎟∂∂∂ ∂⎝⎠ () ()(){} ()44 4 2 44 2 2 21 2 1 44 4 44 4 2 44 2 2 44 2 2 21 2 1 21 2 112 1 2 1 0 , 12 0 2 0xx x x xx x x xx x xϕϕ ϕνν ν ν ϕϕ ϕ ϕϕ ϕν⎡⎤⎧⎫∂∂ ∂⎪⎪→− + + +− + =⎢⎥⎨⎬∂∂ ∂ ∂ ⎢⎥⎪⎪⎩⎭⎣⎦ ⎛⎞ ⎛⎞∂∂ ∂ ∂∂ ∂→ − ++ = → ++ = ⎜⎟ ⎜⎟⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂∂ ∂ ∂⎝⎠ ⎝⎠ __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-31 5.62 For plane stress problem, derive the bi-harm onic equation for the Airy stress function ------------------------------------------------------------------------------- Ans. From 22 22 2 11 22 12 22 22 12 21 1 211 ( 1 ), , YY YEE EE EE x x xx xxϕϕϕ ϕ ν ϕνν⎛⎞⎛⎞∂∂ ∂∂ + ∂=− =− = −⎜⎟⎜⎟⎜⎟⎜⎟ ∂∂ ∂∂ ∂∂⎝⎠⎝⎠, [Eq. (5.22.3)] we get 22 2 44 4 2 2 11 22 12 2 2 4 4 22 22 22 12 2 1 1 2 12 12 122 , 2 2 2 YYEE EEExx x xx xx x x xx xϕϕϕ ϕ ϕνν⎛⎞∂∂ ∂ ∂∂ ∂ ∂ ∂+= + − = −−⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂∂ ∂∂⎝⎠ The compatibility equation 22 2 11 22 12 22 12 212E EE xx xx∂∂ ∂+=∂∂ ∂∂ then gives 44 4 2 2 4 4 22 22 22 1 2 12 12 1222 2 x x xx xx xxϕϕ ϕ ϕ ϕνν∂∂ ∂ ∂ ∂→+− = − − → ∂ ∂ ∂∂ ∂∂ ∂∂Thus, 44 2 44 2 2 12 1 220 xx x xϕϕ ϕ∂∂ ∂++= ∂∂ ∂ ∂ __________________________________________________________________ 5.63 Consider the Airy stress function 22 11 212 3 2x xx x ϕα α α=+ + . (a) Verify that it satisfies the bi-harmonic equation. (b) Determine the in-plane stresses 11 12 22, TT a n d T . (c) Determine and sketch the tractions on the four rectangular boundaries 112 20, , 0,x xb x x c==== .(d) As a plane strain solution, determine 13 23 33, , and all the strain componentsTTT . (e) As a plane stress solution, determine 13 23 33, , TTT , and all the strain components . ------------------------------------------------------------------------------- Ans. (a) 44 44 42 2 12 1 2/0 , /0 , / 0xxx xϕϕϕ∂ ∂= ∂ ∂= ∂ ∂ ∂= , thus 22 11 212 3 2x xx x ϕα α α=+ + satisfies the bi-harmonic equation. (b) 22 2 22 11 2 3, 12 1 2 2, 22 1 1 /2 = / /2 Tx T x x T xϕ αϕ α ϕ α =∂ ∂ = −∂ ∂ ∂ =− =∂ ∂ = (c) ( ) ()1 1 1 1 12 1 2 3 1 2 2 1 11 1 12 1 2 3 1 2 2 2 2 1 2 12 2 2 2 1 1 2 1 21 2 12 2 2 2 1 1 2On 0, 2 , on , 2 , on 0, 2 , on , 2 .xT T x b T T xT T x c T Tαα αα αα αα=− = −+= −+ = = +=− = − =− + = − = = + =− +t = Te e e e e t = Te e e e e t = Te e e e e t = Te e e e e Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-32 α α 2 2 α2α2 α2α2 3 3α21 α21bc (d) As a plane strain solution, ()() () ( ) () ( ) () ( ) () ( ) () ( ) () ( )33 11 22 3 1 13 23 13 23 33 22 11 11 22 3 1 22 22 22 11 1 3 12 12 2,2 , 0, 0, 1/ ( 1 ) 1 2 1/ ( 1 ) 1 , 1/ ( 1 ) 1 2 1/ ( 1 ) 1 , 1/ 1 1/ 1YY YY YYTT T T T E E E EE T T E EE T T E EE T Eνν α α νν ν ν α ν ν α νν ν ν α ν ν α νν α=+ = + = = = = = ⎡⎤ ⎡ ⎤= − −+ = − −+⎣⎦ ⎣ ⎦ ⎡⎤ ⎡ ⎤= − −+ = − −+⎣⎦ ⎣ ⎦ =+ = −+ (e) As a plane stress solution, ()()()() () ( ) () () () ( ) ( ) () ( )() ()33 13 23 13 23 11 11 22 3 1 22 22 11 1 3 12 12 2 2 33 11 22 3 10, 0, 1 / 2 1 / , 1/ 2 1/ , 1/ 1 1 / /2 . 1/ 2 / .YY YY Y Y YYTTT EE E E T T E EE T T E EE T E EE T T Eνα ν α ν αν α ν α ν α μ νν α α=== = = = − = − =− = −= + = − + = − ⎡⎤=− + = − +⎣⎦ Note, for this problem, since 11 22TT+ is a linear function of 12and x x, in fact, a constant, therefore, all the compatibility equations are satisfied so that 33E is meaningful and 3udoes exist. _________________________________________________________________ 5.64 Consider the Airy stress function 2 12xxϕα= . (a) Verify that it satisfies the bi-harmonic equation. (b) Determine the in-plane stresses 11 12 22, TT a n d T . (c) Determine and sketch the tractions on the four rectangular boundaries 112 20, , 0,x xb x x c==== . (d) As a plane strain solution, determine 13 23 33, , TTT and all the strain components. (e) As a plane stress solution, determine 13 23 33, , TTT . and all the strain components . ------------------------------------------------------------------------------- Ans. (a) 44 44 42 2 12 1 2/0 , /0 , / 0xxx xϕϕϕ∂∂ =∂∂ =∂∂ ∂ = , thus 2 12xxϕα= satisfies the bi-harmonic equation. (b) 22 2 22 11 2 12 1 2 1 22 1 2 /0 , = / 2 , /2 Tx T x x x T x xϕ ϕα ϕ α =∂ ∂ = −∂ ∂ ∂ =− =∂ ∂ = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-33 ( ) ()1 1 11 1 21 2 1 2 1 1 11 1 21 2 2 2 2 1 21 2 22 1 1 22 1 1 2 2 1 21 2 22 1 1 22 1 1 2(c) On 0, 2 0, On , 2 . On 0, 2 2 2 . O n , 22 22 .x TT x x b TT b xT T x x x xc T T x x x cαα αα α αα αα= − =− + = = = = + =− =− = −+= − = == + = − + = − +t= T e e e e t=T e e e e t= T e e e e e e t=T e e e e e e e α2 bb cα2c α2x1x1α2x1 ox2 (d) As a plane strain solution, () () ( ) ( ) () ( ) () ( ) ( )33 11 22 2 13 23 13 23 33 2 11 11 22 2 22 22 22 11 2 12 12 12, 0 . 0 , 1/ ( 1 ) 1 2 1 / , 1/ ( 1 ) 1 2 ( 1 )/ , 1/ 1 2 1 / .YY YY YYTT T x T T E E E E ET T E x EE T T E x EE T E xνν α νν ν α ν ν νν ν α ν να ν=+ = = = = = = ⎡⎤ ⎡ ⎤ =− − + = − +⎣ ⎦ ⎣⎦ ⎡⎤ ⎡ ⎤=− − + = −⎣⎦ ⎣ ⎦ ⎡⎤ =+ = − +⎣⎦ (d) As a plane stress solution, ()()() ( ) () () ( ) () ( ) () ( ) ( )33 13 23 13 23 33 11 22 2 11 11 22 2 22 22 11 2 12 12 1 10, 0, 1 / 2 / . 1/ 2 / , 1/ 2 / . 1/ 1 2 1 / / .YY YY Y Y YYTTT EE E E T T E x E ET T E xE ET T E x EE T E x xνα ν να ν ν α να ν α μ⎡⎤ === = = = − + = −⎣⎦ =− = − =− = ⎡⎤ =+ = − + = −⎣⎦ Note, for this problem, since 11 22TT+ is a linear function of 12and x x, therefore, all the compatibility equations are satisfied so that 33E is meaningful and 3udoes exist. _________________________________________________________________ 5.65 Consider the Airy stress function ()44 12x x ϕα=− . (a) Verify that it satisfies the bi- harmonic equation. (b) Determ ine the in-plane stresses 11 12 22, T T and T . (c) Determine and sketch the tractions on the four rectangular boundaries 112 20, , 0,x xb x x c==== . (d) As a plane strain solution, determine 13 23 33, , and all the strain componentsTTT . (e) As a plane stress solution, determine 13 23 33, , and all the strain componentsTTT . -------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-34 Ans. (a) 44 44 42 2 12 1 2/ 2 4, / 2 4, / 0xx x xϕα ϕ α ϕ∂∂ = ∂∂ = − ∂∂ ∂ = , thus ()44 12x x ϕα=− satisfies the bi-harmonic equation. (b) 22 2 22 2 2 11 2 2 22 1 1 12 1 2 / 1 2, / 1 2, / 0 Tx x T x x T x xϕα ϕ α ϕ=∂ ∂ =− =∂ ∂ = =−∂ ∂ ∂ = . () ()22 1 1 1 1 12 1 2 2 1 1 11 1 12 1 2 2 1 22 2 2 1 2 12 2 2 1 2 2 21 2 12 2 2 1 2(c) On 0, 12 , On , 12 . On 0, 12 , On , 12 .xT T x x b T T x xT T x x c T T xαα αα=− = −+= = = += − =− = −+= − = = +=t= T e e e e t=T e e e e t= T e e e e t=T e e e e bc oα21x22x2 xα21x22 α2x112α2x112 1 (d) As a plane strain solution, () () () ( ) () ( ) () ( ) () ( ) () ( )22 33 11 22 1 2 13 23 13 23 33 12 12 22 2 2 11 11 22 2 1 22 2 2 22 22 11 1 212 , 0, 0, 1 / 1 0. 1 / (1 ) 1 12 1 / (1 ) 1 . 1 / (1 ) 1 12 1 / (1 ) 1 .Y YY YYTT T x x T T E E E E E T EE T T E x x EE T T E x xνα ν ν νν ν α ν ν ν νν ν α ν ν ν=+ = − = = = = = = += ⎡⎤ ⎡ ⎤=− − + = − − + +⎣⎦ ⎣ ⎦ ⎡⎤ ⎡ ⎤=− − + = − + +⎣⎦ ⎣ ⎦ (d) As a plane stress solution, () ( ) () () () ( ) () () () ( ) () ( ) () ()22 33 13 23 13 23 11 11 22 2 1 22 22 22 11 1 2 12 12 22 33 11 22 2 10, 0, 1 / 12 1 / , 1/ 12 1/ , 1/ 1 0. .1 / 1 2 1 / .YY YY Y YYTTT EE E E T T E x x EE T T E x x EE T EE T T E x xνα ν να ν ν να ν=== = = = − = − + =− = += + = ⎡⎤=− + = −⎣⎦ Since 11 22TT+ is not a linear function of 12and x x, 33Eis meaningless, because 3udoes not exist. __________________________________________________________________ 5.66 Consider the Airy's stress function 23 12 12xxx xϕα=+ . (a) Verify that it satisfies the bi- harmonic equation. (b) Determ ine the in-plane stresses 11 12 22, TT a n d T . (c) Determine the condition necessary for the traction at 2x c=to vanish and (d) determine the tractions on the remaining boundaries 11 20, and 0xx b x=== . ------------------------------------------------------------------------------- Ans. (a) 44 44 42 2 12 1 2/0 , /0 , / 0xxx xϕϕϕ∂ ∂= ∂ ∂= ∂ ∂ ∂= , thus 23 12 12xxx xϕα=+ satisfies the bi- harmonic equation. (b) 22 22 2 2 11 2 1 1 2 22 1 12 1 2 2 2 /2 6 , /0 , / 23 Tx x x x T x T x x x xϕα ϕ ϕ α=∂ ∂ = + =∂ ∂ = =−∂ ∂ ∂ =− − . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-35 22 2 22 1 2 1 2 2 2 1 (c)On , ( 2 3 ) , 2 3 0 2 3 3 / 2 x cT T c c c cc cc αα α α == + = − − → − − = → = − → = −t=T e e e e . () ()() () ( ) ()2 11 1 1 1 2 1 2 2 2 2 2 2 2 11 1 1 1 2 1 2 2 1 2 2 2 22 1 2 1 2 2 2(d) On 0, 2 3 3 . On , 3 2 3 . On 0, 0.xT T x x x x c xb T T b xc x xc xT Tα =− = −+=+ = − == + = − − − =− = −+=t= T e e e e e t=T e e e e e t= T e e e __________________________________________________________________ 5.67 Obtain the in-plane displacement compon ents for the plane stress solution for the cantilever beam from the following stra in strain-displacement relations. 2 2 11 2 2 1 2 11 22 12 2 12, , 44YYuP x x u P x x PhE EE xxE I x E I Iν μ⎛⎞⎛⎞ ∂∂== == − = − ⎜⎟⎜⎟⎜⎟ ∂∂ ⎝⎠⎝⎠. ------------------------------------------------------------------------------- Ans. () ()22 1 12 1 2 2 12 12 11 2 2 2 1 12 22 22 22 12 1 1 22 22 21 2 1 2 12 1 1, ,22 2,44 2 24 2YY Y Y YY Yu P xx P x x u P xx P xxuf x u f xx EI EI x EI EI u u Px df Px df Ph Phxxxx I E I d x E I d x I Px df df E I dx dxνν ν μμ∂∂=→ =+ = − → = − +∂∂ ⎛⎞ ⎛⎞⎛⎞ ⎛⎞ ∂∂+= −→ +− += − ⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ∂∂ ⎝⎠ ⎝⎠⎝⎠ ⎝⎠ →+ = −2 2 2 2 2 2.22 4YPx PP hxEI I Iν μμ⎛⎞ ⎛⎞ ⎛⎞+− + ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ In the above equation, the left side is a function of 1xonly, right side is a function of 2xonly, thus both sides must equal to the same constant, say 1c. That is, 22 3 12 2 1 1 11 2 1 1 2 11 23 3 22 2 12 2 2 21 1 2 1 2 3 2.22 6 22 4 6 3 22 4YY Y YYPx df df Px Pxcc f c x cEI d x d x EI EI df Px Px x PP h PP hx cf x c x cd x EI I I EI I Iνν μμ μμ+= →= − → =− + ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=− + − → =− + − + ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ Thus, 2 23 3 12 2 2 12 1 2 3 23 12 1 21 1 2,26 6 2 2 .26YY YYPx x Px Px Phux c x cEI EI I I Px x Pxuc x cEI EIν μμ ν⎛⎞⎛⎞⎛⎞=+ −+ − + ⎜⎟⎜⎟⎜⎟ ⎜⎟⎝⎠⎝⎠⎝⎠ =− − + + _________________________________________________________________ 5.68 (a) Let the Airy stress function be of the form 1 2() c o smxfxπϕ=l. Show that the most general form of Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-36 2()fxis()21 2 2 2 3 2 2 4 2 2 cosh sinh cosh sinhmm m m fxC x C x C x x C x x λλ λ λ=++ + . (b) Is the answer the same if 1 2() s i nmxfxπϕ=l? ------------------------------------------------------------------------------- Ans. (a) The function 12(, )xxϕ must satisfy the bi-harmonic equation. Now, 24 24 11 22 24 11 2 42 4 4 11 22 2 4 4 12 2 2 2() c o s , () c o s , cos , cos ,mx mx mmfx fx xx mx mx md f d f xx d x x d xππ ϕπ ϕ π ππ ϕπ ϕ∂∂⎛⎞ ⎛⎞=− =⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠ ∂∂⎛⎞=− =⎜⎟∂∂ ∂ ⎝⎠ll ll ll l Thus, 42 44 4 2 4 4 1 2 42 2 4 2 4 11 2 2 2 22c o s ( ) 2 0mx m m df dffx xx x x d x d xπ ϕϕ ϕ π πϕ⎡⎤ ∂∂ ∂ ⎛⎞ ⎛⎞∇= + + = − + = ⎢⎥⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎝⎠⎢⎥⎣⎦ll l. Therefore, 42 24 42 222 0, where mm mdf df mf dx dxπλλ λ−+ = ≡l. The characteristic equation for the above ODE is 42 2 420mm DDλλ−+= . The roots of this equation consists of two sets of double roots. They are: , , ,m mmm Dλλλλ=−− . Thus, ()21 2 2 2 3 2 2 4 2 2 cosh sinh cosh sinhmm m m fxC x C x C x x C x x λλ λ λ=++ + . (b) Yes, the same __________________________________________________________________ 5.69 Consider a rectangular bar defined by 123, , x cx c bx b −≤≤ − ≤≤ − ≤≤ll , where /blis very small. At the boundaries 2x c=±, the bar is acted on by equal and opposite cosine normal stress 1 cos , where /mm mAx mλ λπ=l (per unit length in3xdirection). (a) Obtain the in-plane stresses inside th e bar. (b) Find the surface tractions at 1x=±l. Under what conditions can these surface tractions be removed without affecting 22 12 and TT (except near 1x=±l)? How would 11Tbe affected by the removal. Hint: Assume ()21cos , where /mm fx x mϕλ λ π== land use the results of the previous problem ------------------------------------------------------------------------------- Ans. (a) Boundary conditions are () 212 0xcT=±=, () 222 1 cosmm xcTA x λ=±= Let ()21cos , where /mm fx x mϕλ λ π== l. Then (see previous problem) , ()21 2 2 2 3 2 2 4 2 2 cosh sinh cosh sinhmm m m fxC x C x C x x C x x λλ λ λ=++ + . The in-plane stresses are: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-37 ()()22 22 2 1 2 1cosmm Tf x x xϕλλ∂== − ∂, ()2 22 11 2 1 2 2/c o sm Td f d x x xϕλ∂== ∂, () 12 2 1 12/s i nmm Td f d x xxxϕλ λ∂∂=− =∂∂. Now, applying the boundary condition: () ()() () () 222 22 1 1 1 cos cos cos /mm m m mm m m xcTA x f c x A x f c A λλ λ λ λ=±=→ − ± =→ ± = − . From () ()2/mm fc A λ ±= − , ()() fcf c+=− , so that 23 0 CC== and ()2 14cosh sinh /mm m m fcC c C c c A λ λλ ±= + = − (i) Applying the other boundary condition: () () 2212 2 0/ 0xc xcTd f d x=± =±=→= → ( ) 14sinh sinh cosh 0mm m m mCc C c c cλλ λ λ λ ++ = (ii) (i) and (ii) give () 14 22cosh sinh 22 s i n h, sinh 2 2 sinh 2 2mm m mm m m mm mm mmcc c AA cCCcc ccλλλ λλ λλ λλ λλ⎡⎤ + ⎡ ⎤=− =⎢⎥ ⎢ ⎥++ ⎣ ⎦ ⎣⎦. With ()21 2 4 2 2 cosh sinhmm fxC x C x x λλ=+ , we have , ()() () { } (){} {}22 22 2 1 1 2 4 2 2 1 22 2 1cos cosh sinh cos cosh sinh cosh sinh sinh cos 2 .sinh 2 2mm m m m m m mm mm mmm m m mmTf x x C x C x x x cc cx xxc x Accλλ λ λ λ λ λ λλ λλ λλλ λ λλ=− =− + ⎡⎤ +− ⎢⎥=+ ⎢⎥⎣⎦ () () (){} ()2 12 2 1 1 2 4 2 2 2 1 22 2 1/ sin sinh sinh cosh sin cosh sinh sinh cosh 2s i n .sinh 2 2mm m m m m m m m mm m m m m m m mmTd f d xx C x C x x x x ccx c x x Axccλλ λ λ λ λ λ λ λ λλλ λ λλ λλλ⎡⎤ == + +⎣⎦ ⎡⎤−+ ⎢⎥=+ ⎢⎥⎣⎦ () () ( )2 22 11 2 1 2 2 22 2 2 1/c o s cosh cosh sinh sinh cosh2c o s .sinh 2 2m mm m m mm m m m mmTd f d x x x cc x c xx xAxccϕλ λλ λ λ λλ λλλλ∂== = ∂ ⎡⎤−++ ⎢⎥+⎣⎦ (b) Surface tractions at 1x=±lare: () 12 2 ,[ ] s i n 0 Tx m π ±==l . ()() ( ) 22 2 2 11 2cosh cosh sinh sinh cosh,2 c o ssinh 2 2mm m m mm m m mmcc x c xx xTx A mccλλ λ λ λλ λπλλ⎡⎤−++±= ⎢⎥+⎣⎦l At 1x=±l,11Tis an even function of 2x, which gives rise to equal a nd opposite resultant force of magnitudeRF at the two ends. Removal of thes e resultants will have little effects on 12 22 and TT , Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-38 if /clis very large. However, 11Twill need to be modified by subtracting the normal stress (RF /Area) caused by the resultant forces. _________________________________________________________________ 5.70 Verify that the equations of equilibrium in polar coordinates are satisfied by 22 22 211 1, , rr rTT Trr rr rrθθ θϕϕϕ ϕ θ θ∂∂ ∂ ∂ ∂ ⎛⎞=+ = = − ⎜⎟∂∂ ∂∂∂ ⎝⎠. -------------------------------------------------------------------------------- Ans. 223 2 2 22 2 2 3 2 2 23 2 22 2 3()111 1 11 1, 111 1 1 1 1 1rr rT rT rr r r r r r r r rr r r r T rr r r r r r rr r rθθ θϕϕ ϕ ϕ ϕ ϕ θθ θ ϕ ϕϕ ϕ ϕ θθ θ θθθ θ⎛⎞ ⎛ ⎞ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂=+ = +− − = −⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟ ∂∂ ∂ ∂∂ ∂ ∂ ∂ ∂⎝⎠ ⎝ ⎠ ⎡⎤⎛⎞ ⎛ ⎞ ⎡⎤ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂⎛⎞=− =− − =− − ⎢⎥⎜⎟ ⎜ ⎟ ⎜⎟⎢⎥ ⎜⎟ ⎜ ⎟ ∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂∂ ⎝⎠⎣⎦ ⎢⎥⎝⎠ ⎝ ⎠⎣⎦2θ⎡ ⎤ ⎢ ⎥ ∂ ⎢ ⎥ ⎣ ⎦ Thus, [See Eq.(4.8.1), 232 3 2 2 22 2 3 2 2 2 3 2 2()11 11 1 1 1 10r rr TT rT rr r r rrrr r r rr r rθθ θ θ ϕϕϕ ϕ ϕ ϕ θθ θ θ∂ ∂+−∂∂ ⎡⎤ ⎛⎞ ⎛ ⎞∂∂∂ ∂ ∂ ∂=+ − − − − = ⎢⎥ ⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂ ∂ ∂∂ ⎢⎥ ⎝⎠ ⎝ ⎠ ⎣⎦ Similarly, 2 23 2 22 2 2 23 22()11 1 1 1 11 1,rrTrrrr r r r r r rr r r T rr r rrθ θθϕ ϕϕ ϕ θθ θ θ ϕϕ θθ θ⎡⎤⎛⎞ ⎛ ⎞ ⎡⎤ ∂ ∂∂ ∂ ∂∂ ∂ ∂⎛⎞=− =− − =− ⎢⎥⎜⎟ ⎜ ⎟ ⎜⎟⎢⎥ ⎜⎟ ⎜ ⎟ ∂∂ ∂ ∂∂ ∂ ∂ ∂ ∂∂ ⎝⎠⎣⎦ ⎢⎥⎝⎠ ⎝ ⎠⎣⎦ ∂∂∂ ∂==∂∂ ∂∂ ∂ Thus, [See Eq.(4.8.2)]2 2()110.rrT T rr rθθ θ θ∂∂+=∂∂ __________________________________________________________________ 5.71 From the transformation law : 11 12 21 22cos sin cos sin sin cos sin cosrr r rTT TT TT TTθ θθ θθθθ θ θθθ θ− ⎡⎤ ⎡⎤ ⎡ ⎤⎡ ⎤= ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦⎣ ⎦⎣⎦ ⎣⎦ and 22 2 11 22 12 22 12 21, and TT Txx xxϕ ϕϕ ∂∂ ∂== = −∂∂ ∂∂, obtain 2 2211 rrTrr rϕ ϕ θ∂ ∂⎛⎞=+⎜⎟∂ ∂⎝⎠ ------------------------------------------------------------------------------- Ans. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-39 () () () ()11 12 21 22 22 2 2 11 12 22 22 11 12 22 2 2 22 11 12 11 22 12cos sin cos sin sin cos sin cos cos 2 sin cos sin sin cos cos sin cos sin cos sin sin cos 2 sinrr r rTT TT TT TT TT T T T T TT T T T Tθ θθ θθθ θ θ θθ θ θ θθ θθ θ θ θ θ θθ θ θ θ θ θ− ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤= ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥−⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ++ −+ − = −+ − + − (), cosθ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 222 12 1 1 22 122 1 12 22 22 1 12 12 222 2 11 2 2// c o s , / / s i n , sin costan / , / , 1sin cos , 1sin cos sinrxx r x x r r xx r xx xxxxr r xx xx r xr x xr r Trr r xθθ θθθθ θ ϕϕ ϕ θϕ ϕθθθθ ϕϕ ϕθθ θθ−= + → ∂∂= = ∂∂= = −=→ ∂ ∂ == − ∂ ∂ == ++ ∂∂ ∂∂ ∂∂ ∂=+= +∂∂ ∂∂ ∂∂ ∂ ∂∂ ∂ ∂⎛⎞== + + ⎜⎟∂∂ ∂∂⎝⎠ 22 2 2 2 22 2 22 2 2 2 2 22 21c o ssin cos 11 1 1 c o ssin cos sin sin cos cos sin cos cos 2cos sin 2cos sisinrr r rr r r r r r rr rr r r rrϕϕθθθθθ ϕ ϕϕϕ ϕ ϕ ϕ θθθ θ θ θ θ θθθ θ θ θ ϕθ ϕ θ ϕ θ θ ϕ θθθ θ∂∂ ∂⎛⎞+⎜⎟∂∂ ∂⎝⎠ ⎛⎞ ⎛ ⎞ ∂∂ ∂ ∂ ∂ ∂ ∂=+ − + + + − ⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟∂∂ ∂ ∂∂ ∂ ∂ ∂∂ ⎝⎠ ⎝ ⎠ ∂∂ ∂∂=++ + −∂∂ ∂ ∂∂2n, rθϕ θ∂ ∂ 111 2 22 2 1 22 2 2 2 2 22 2 2sincos , sin sin sincos cos cos sin sin 2sin cos 2sin coscosr xr x xr r Trr r r r r x rr r r rr rϕϕ ϕ θϕ ϕθθθθ ϕϕ ϕ θ ϕ ϕ θ θθθ θθθ θ ϕ θϕ θ ϕ θ θ ϕ θ θϕθθ θ θ∂∂ ∂∂ ∂∂ ∂=+= −∂∂ ∂∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂⎛⎞ ⎛⎞== − − − ⎜⎟ ⎜⎟∂∂ ∂ ∂∂ ∂∂⎝⎠ ⎝⎠ ⎛⎞∂∂ ∂∂∂=+ + − +⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂∂⎝⎠. 12 21 22 2 2 2 2 2 2 22 2 2 2sin sin coscos sin cos sin cos sin cos cos sin sin coscos sin .Txx r r r r r r rr r r r r rr r rϕϕ ϕ θ ϕ ϕ θ θθθ θθθ θ ϕ θθϕ θ ϕ θ ϕ θϕ θϕ θθ ϕθθθθθ θ θ∂∂ ∂∂ ∂ ∂ ∂ ∂ ⎛⎞ ⎛⎞−= = − + − ⎜⎟ ⎜⎟∂∂ ∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ∂∂ ∂ ∂ ∂ ∂ ∂=−+ − + − −∂ ∂∂ ∂ ∂ ∂ ∂ ∂∂ Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-40 22 11 12 22 22 2 2 2 2 2 22 2 2 22 2 2 2 2 22 2 2 2 22cos 2 sin cos sin cos cos 2cos sin 2cos sinsin cos sin cos sin cos coscos sin 2 sinrrTT T T rr r r rr r rr rr r r rθθ θθ ϕθ ϕ θ ϕ θ θ ϕ θ θ ϕθθθθ θ ϕ θ θ ϕ θϕ θϕ θ ϕθθθθθ θ θϕ=+ += ⎛⎞∂∂ ∂∂∂++ + − ⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂∂⎝⎠ ∂∂ ∂ ∂ ∂−+ − +∂∂∂ ∂ ∂∂− ∂−∂ 22 2 2 2 22 22 2 2sin cos sin cos sin sin 2sin cos 2sin cossin cos .rr r rr r r rr rθθ θθ ϕ θ ϕθ ϕ θ ϕ θ θ ϕ θ θ ϕθθθθ θ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟∂⎜⎟ −∂∂ ⎝⎠ ⎛⎞∂∂ ∂∂∂++ + −+⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂∂⎝⎠ () () () ()22 2 22 4 4 22 22 2 2 44 2 2 3 3 3 3 22 2 2 33 3 31sin cos 2 cos sin 2sin cos 12cos sin 2sin cos sin cos sin cos cos sin sin cos 2sin cos cos sin sin cos sin cosrrTrr rr r rr rrϕϕ ϕ ϕθθ θ θ θθ ϕϕθθθ θ θ θ θ θ θ θ θ θθθ ϕθ θ θθ θθ θθ⎛⎞∂∂ ∂ ∂=− + + + + ⎜⎟⎜⎟ ∂ ∂∂ ∂⎝⎠ ∂∂++ + − − + −∂∂ ∂−+−∂∂.θ That is, 2 2211 rrTrr rϕ ϕ θ∂ ∂⎛⎞=+⎜⎟∂ ∂⎝⎠. __________________________________________________________________ 5.72 Obtain the displacement field for the plane strain solution of the axis-symmetric stress distribution from that for the plane stress solution obtained in Section 5.28. ------------------------------------------------------------------------------- Ans. From Section 5. 29, we have, for plane stress solution, [See Eq.(5.29.15) and .(5.29.16) and note () 21YEμν=+] ()() () ()1sin cos 2112 1 l n ( 1 ) 2 1r HGAuB r r B r C rrθθ μννν ν ν =+ + +⎡⎤−+ + − − + + −⎢⎥⎣⎦, 2=c o s s i n(1 )BruH G F rθθθθμν+− ++. To obtain the corresponding displacement field for the plane strain solution, we replace the Poisson ratio νwith / (1 )νν−in the above equation [see Section. 5.26]. That is, () ()11 211 , 1111 1 1ν ννννννν ν− ⎛⎞ ⎛⎞+→+ = −→− =⎜⎟ ⎜⎟−− −−⎝⎠ ⎝⎠. Thus, for plane strain: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-41 () ()() () ()() () () () ()() ()111 2 1 2 112l n 2 s i n c o s 21 1 1 1 sin cos 21 2 l n 21 2 s i n c o s .121 2 l n 21 22 YrAuB r r B r C r H G r HG ABr r B r Cr H G ErABr r B r Crr ννν νθθ μνν ν ν θθ νν θ θννμ +−− −=− + − + + + −− − − =+ + =− + − − + − + +⎡⎤ ⎢⎥ ⎣⎦ ⎡⎤−+ − − + −⎢⎥⎣⎦ ⎡⎤ ⎢⎥⎣⎦ and 2( 1 ) 4( 1 ) ( 1 )= cos sin cos sin YBr Bru H GF r H GF rEθθνθ ν νθθ θθμ−− ++− + = +− + . _________________________________________________________________ 5.73 Let the Airy stress function be ( )sin frnϕ θ = , find the differential equation for ( ) fr. Is this the same ODE for ( ) fr if ( )cos frnϕ θ = ? --------------------------------------------------------------------------------- Ans. 22 2 22( )sin 'sin ''sin ; cos sinfrn fn f n n fn n f nr rϕϕ ϕϕϕ θθ θθ θθ θ∂∂∂∂=→ = → = = → = −∂∂ ∂ ∂. Thus, ()22 22 2 22 2 22 2 211 1 1 ' 1() s i n ' 's i n s i nffrn n f f n g rnrr rr r rr rr rϕϕ ϕθ θθ θθ⎛⎞ ∂∂ ∂ ∂∂ ∂ ⎛⎞++ = ++ = −+ ≡ ⎜⎟ ⎜⎟ ⎜⎟ ∂∂∂∂ ∂∂ ⎝⎠ ⎝⎠ , where ()22 2 22 2'1 1''fd d ngrn f f frr d rrd r r⎛⎞ ⎛⎞≡− += +− ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠. Now, () () ()22 22 22 22 2 22 2 22 2 22 22 22 22 22 2211 1 1 11sin 11 1sin sin 0.gr nrr r r rr rr rr rr dd n dd n dd ngr n f r nrd r rd r rd r dr r dr r dr rϕϕ ϕθ θθ θ θθ⎛⎞ ⎛ ⎞ ⎛⎞∂∂ ∂∂∂ ∂ ∂∂ ∂++ ++ = ++⎜⎟ ⎜ ⎟ ⎜⎟⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂∂∂ ∂∂ ∂∂ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎛⎞ ⎛⎞ ⎛⎞ =+− =+− +− =⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ Therefore, ()22 22 22 22110dd n dd nfrrd r rd r dr r dr r⎛⎞ ⎛⎞ +− +− =⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠. The same equation will be obtained if ( )cos frnϕ θ = _________________________________________________________________ 5.74 Obtain the four independent solutions for the following equation 22 2 2 22110dd n d f d f nfrd r r rd r r dr dr⎛⎞ ⎛ ⎞ +− +− =⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-42 --------------------------------------------------------------------------------- Ans. Let mfr= . () () ()() () () ()()22 222 22 22 22 2 2 22 2 4 22 2 2 2 22 211 1132 2 0 20 .mm mdf d f nfm m m n r m n rrd r dr r dd n d f d f nfm n m m m n rrd r rd r dr r dr r mn m n−− −⎛⎞⎡⎤ +− = − + − = −⎜⎟⎣⎦ ⎜⎟⎝⎠ ⎛⎞ ⎛ ⎞⎡⎤ +− +− =− −− + − − =⎜⎟ ⎜ ⎟⎣⎦ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎡⎤ →− −−=⎢⎥⎣⎦ Thus, 123 4 , , 2 , 2 mn mn m n m n=+ =− = + = − . For 0n≠and 1n≠, the four independent solutions for f are: 22,, a n dnnn nrrr r+−+ + − +. For 0n=, 12 34 0, 2 mm mm== == . Two independent solutions for fare given by 2 and Cr . Additional solutions are given by ()00ln lnnn nndrr r rdn =→⎛⎞== ⎜⎟⎝⎠, and 22 2 00ln lnnn nndrr r r rdn++ →→⎛⎞⎡⎤== ⎜⎟ ⎣⎦⎝⎠. The four independent solutions are: 22, ,ln and lnCr r r r . For 1n=, 14 1 mm== , in addition to 13,,rr r−, we have, ()11ln lnnn nndrr r r rdn =→⎛⎞== ⎜⎟⎝⎠ Thus, the four independent solutions are: 13,, a n d l nrr r r r−. __________________________________________________________________ 5.75 Evaluate () () 00cos , sinnn nnddrn r ndn dnθθ = =⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦, () ()2 11cos cosnn nnddr n and r ndn dnθθ−+ = =⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ -------------------------------------------------------------------------------- Ans. () ( ) () ( )00 00cos ln cos sin ln . sin ln sin cos .nn n nn nn n nndrn r r n r n rdn drn r r n r ndnθθ θ θ θ θθθ θ== ==⎡⎤ ⎡⎤=− =⎢⎥ ⎣⎦ ⎣⎦ ⎡⎤ ⎡⎤=+=⎢⎥ ⎣⎦ ⎣⎦ () ()22 2 11 11c o s l nc o s s i n l nc o s s i n cos ln cos sin ln cos sinnn n nn nn n nndrn r r n r n r r rdn drn r r n r n r r rdnθ θθ θ θ θ θ θθ θ θ θ θ θ−+ −+ −+ == ==⎡⎤⎡⎤=− − = − −⎢⎥ ⎣⎦⎣⎦ ⎡⎤⎡⎤=−= −⎢⎥ ⎣⎦⎣⎦ ________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-43 5.76 In the Flamont Problem (Sect. 5.37) , if the concentrated line load F, acting at the origin on the surface of a 2D half-space (defined by / 2 / 2 πθπ−≤≤ ), is tangent to the surface and in the direction of o90θ= , show that: 2s i n, 0rr rFTT Trθθ θθ π⎛⎞=−= =⎜⎟⎝⎠ . ------------------------------------------------------------------------------- Ans. The boundary conditions are: 0 at / 2, 0r TT rθθ θ θπ === ± ≠ . (i), () ()/2 /2 /2 /2cos sin 0 (ii), sin cosrr r rr rT T rd T T rd Fππ θθ ππθθ θ θθ θ −−−= += −∫∫. (iii) From the stress field obtained in Sect. 5.37, ( )1 552 cos 2 sin , 0, 0rr rTr B B T Tθθ θ θθ−=− = = , (iv) we obtain, from Eqs.(ii) and (iv) : ()/2 /2 22 55 5 5 /2 /22c o s s i n 2 0 2 c o s 0 0BB dB d Bππ ππθθ θ θ θ −−−= → = → =∫∫ . From Eqs.(iii) and (iv) ()/2 /2 22 55 5 5 5 /2 /22sin 2 2 sin 2 sin 2 22FBB d F Bd F B F Bππ πππθθ θ θ θπ−−− = −→ − = −→ =→ =∫∫ Thus 5sin 2 sin2 , 0, 0rr rFTB T Trrθθ θθθ π⎛⎞ ⎛⎞=− =− = =⎜⎟ ⎜⎟⎝⎠ ⎝⎠. (v) _________________________________________________________________ 5.77 Verify that the displacement field for the Flamont Problem under a normal force P is given by (){} () (){} 1s i n 2 l n c o s , 1 s i n 2 l n s i n 1c o sr YYPPur u rEEθ νθθ θ ν θ θ ν θθππ=− − + = + + − − , The 2D half space is defined by /2 /2πθπ−≤≤ . ------------------------------------------------------------------------------- Ans. From the given displacement field, we have, () () (){} (){} () (){}2c o s/, . . , . 1 cos 2ln cos 1 cos 1 sin 2c o s1 sin 2ln cos 1 cos 1 cos .rr rr r rr YY r Y YY YT PEu r i e EEr E u PurE PP PrEE Eθθ π νθ θ νθ ν θ θθπ νθνθ θ θ ν θ ν θππ π⎛⎞=∂ ∂ =− = ⎜⎟⎝⎠ ∂+= + + − − +−∂ −− + =+− − = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-44 That is, 12 c o s,. . , .rr r YYu T PEu i e ErE r Eθ θθ θθν νθ θπ∂⎛⎞=+ = = −⎜⎟∂⎝⎠ Next, ()() { } 1 sin 1 sin 12 s i n2 2s i n 2s i n0, . .,2 0 .r r YY rr YYP u u PEurr E rE r PPie E TEr Erθ θθ θθνθ νθ θ θπ π θθμππ−+ + ∂ ∂⎛⎞=− + = − +⎜⎟∂∂⎝⎠ =− + = = = _________________________________________________________________ 5.78 Show that Eq. (5.38.6), i.e., ()1 4(1 )ν=−∇ ⋅ + Φ−uxΨΨ can also be written as: ()() 24 1μ νφ=− − +∇ ⋅ +ux ψψ where () () 21 21, ν νφμμ− −=− Φ=−Ψψ ------------------------------------------------------------------------------- Ans. With () () 21 21, ν νφμμ−−=− Φ=−Ψψ , we have, () 21ν μ−⋅−⋅ → xxΨ= ψ . ()()()21 11 4(1 ) 2νφνμ μ−→= − ∇⋅+ Φ = − + ∇⋅+−ux xΨΨ ψψ . That is, ()() 24 1μ νφ=− − +∇ ⋅ +ux ψψ . ________________________________________________________________________________________________________________________________________________ 5.79 Show that with ()1 4(1 )ii n n iuxxν∂=Ψ − Ψ +Φ−∂, the Navier Equations become : ()()2 2 214 021 2n ni i iixBxxμνν⎛⎞∂∇ Ψ ∂∇ Φ−− − ∇ Ψ + + = ⎜⎟⎜⎟−∂ ∂⎝⎠ ------------------------------------------------------------------------------- Ans. ()11 4 ( 1) 4 ( 1)n ii n n i n i ii iux xx xx νν⎛⎞∂Ψ ∂ ∂Φ=Ψ − Ψ +Φ =Ψ − +Ψ + ⎜⎟−∂ − ∂ ∂ ⎝⎠ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-45 2 2124(1 )im n m n im m m m m muexx xx x x x x ν⎛⎞ ∂∂ Ψ ∂ Ψ ∂ Ψ ∂Φ→= = − + + ⎜⎟⎜⎟ ∂∂ − ∂ ∂ ∂∂ ∂⎝⎠ 2 2 2 22124(1 ) (1 2 ) 1 2(1 ) 4(1 )mn m n ii m i m m i m i m m mn ni im i m m iexx xx x x x xx x x x xxx x x x xν ν νν⎛⎞⎧⎫ ∂Ψ ∂ Ψ ∂Ψ∂∂ ∂ ∂ ∂ ∂ Φ ⎪⎪⎜⎟ =− + + ⎨⎬⎜⎟ ∂∂ ∂ −∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎪⎪⎩⎭⎝⎠ ⎛⎞ ∂Ψ ∂ Ψ∂− ∂ ∂=− + ∇ Ψ + ∇ Φ ⎜⎟⎜⎟ ∂∂ − − ∂∂∂ ∂⎝⎠ ()2 22 12 2 ( 1 ) 4 12 ( 1 )mn ni ii m i m miexxx x x x x xμμ μ νν ν ν⎛⎞ ∂Ψ ∂ Ψ ∂∂ ∂ ∂=− + ∇ Ψ + ∇ Φ ⎜⎟⎜⎟ −∂ −∂ ∂ − − ∂ ∂ ∂ ∂⎝⎠ Also, []2 22 2 2 2 22 224(1 ) 14 ( 1 )2 ( 1) 4 ( 1)j i in n i jj i j i i j nn i ji i iuxxx x x x x xxx x xμμμν μμννν⎛⎞ ∂Ψ ∂ ∂∂ ∂= ∇Ψ− ∇Ψ+ + ∇Ψ+ ∇Φ ⎜⎟⎜⎟ ∂∂ − ∂ ∂ ∂ ∂⎝⎠ ∂Ψ ⎛⎞∂∂=− − ∇Ψ +∇Ψ − − + ∇Φ ⎜⎟−∂ ∂ − ∂ ∂ ⎝⎠ Thus, ()() () ( ) () ()()2 22 2 22 222 2 14 1 2212 4 ( 1 ) 12 1421 2i nn i jj i i i nn i iiu exxx x x x xxxμμμν ν ν ννν ν μνν⎛⎞ ∂ ∂∂ ∂+= − −∇ Ψ − − − ∇ Ψ + − ∇ Φ ⎜⎟∂∂ − ∂ − − ∂ ∂ ⎝⎠ ⎛⎞∂∂=− ∇Ψ − − ∇Ψ+ ∇Φ ⎜⎟−∂ ∂⎝⎠ i.e., ()()2 22 21412 2 12i nn i jj i i iu exxx x x xμμμννν⎛⎞ ∂ ∂∂ ∂+= − ∇ Ψ − − ∇ Ψ + ∇ Φ ⎜⎟∂∂ − ∂ − ∂ ∂ ⎝⎠, so that the Navier Equations become: ()()2 2 214 021 2n ni i iixBxxμνν⎛⎞∂∇ Ψ ∂∇ Φ−− − ∇ Ψ + + = ⎜⎟⎜⎟−∂ ∂⎝⎠. _________________________________________________________________ 5.80 Consider the potential function given in Eq. (5.38.32) [See Example 5.38.5], i.e., () () R, R R ψφ φ = e=ψ , where 2 2 220dd Rd R dRφφφ∇= + = and 2 22220dd Rd R dR Rψψ ψ⎛⎞ +−= ⎜⎟⎜⎟⎝⎠. Show that these functions generate the following displacements, dilatation and stresses as given in Eq.(5.38.5) to (5.38.38): Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-46 (a) Displacements: 2( 3 4 ) , 0RdduR u udR dRθβψφμν ψ⎛⎞=+ − + + = =⎜⎟⎝⎠ (b) Dilation: () 12 2 dedR Rνψψ μ−⎧⎫=− + ⎨⎬⎩⎭ (c) Stresses: () ()2 224 2 4RRddTdR R dRψ ψφνν=− + − + , ()3121ddTTdR R R dRββ θθψψφν⎧ ⎫== − − +−⎨ ⎬⎩⎭ ------------------------------------------------------------------------------- Ans. With RR=xe , we have, Rψ⋅xψ= , thus ()() 24 1μν φ=−− + ∇ ⋅ +→ux ψψ () () () RR R R 24 1 3 4dd dRRdR dR dRψφμν ψ ψ φ ν ψ⎛⎞=− − + + = − + + + ⎜⎟⎝⎠ue e e e , i.e., () 23 4RdduRdR dRψφμν ψ=−+ + + , (b) The non zero strain components are: () ()22 2 2 22 2 222 3 4 2 4R RRu dd d d d d dER RRd Rd R d R dR dR dR dRψ ψψ φ ψ ψ φμμ ν ν⎛⎞ ∂== − +++ + = − ++ + ⎜⎟⎜⎟ ∂⎝⎠. But 22 22 2 222 2 20dd d d RdR R dR dR R dR Rψψψ ψψ ψ+− = → = − () ()22 2 22 2222 4 4 1RRdd d dERdR R R dR dR dRψψφ ψ ψφμν ν∂→= − +++ = − − + +∂. ()2 122 3 4Ru ddEER Rd RR d Rββ θθμ ψψφμμ ν=== − + + + . Therefore, () () () () ()2 2 2 22122 4 1 2 3 4 2 2 22 224 24 2 12 .dd d deedR R R dR R dR dR dd d d dR R R dR dR R dRψψφ ψψ φμμ ν ν ψψ φ φ ψ ψνν ν⎧ ⎫=→ =− − + + + − + + + ⎨ ⎬⎩⎭ ⎧⎫⎪⎪ ⎛⎞=−+ +−+ + + = − − + ⎨⎬ ⎜⎟⎝⎠ ⎪⎪⎩⎭ () 12 2 dedR Rνψψ μ−⎛⎞→= − + ⎜⎟⎝⎠. (c) the stresses are: () () ()2 2 2 222 222 4 112 24 2 4 .RR RRdd dTe EdR R dR R dR dd dR R dRμνψ ψ ψ ψ φμν νν ψψ φνν⎛⎞=+= −+ − − + + ⎜⎟− ⎝⎠ =− + − + () ()22 122 3 412 3112 .dd dTT e EdR R R dR R dR dd dR R R dRββ θθ θθμνψ ψ ψ ψ φμν νν ψψ φν⎛⎞== + = − + + − + ++ ⎜⎟− ⎝⎠ =− − + ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-47 5.81 Consider the following potential functio ns for axis-symmetric problems: ()()ˆ , , , rz Rφφφ β== 0=ψ , 22 ˆ0 φφ∇=∇ = , where ( , , ) rzθ and ( , , ) Rθβ are cylindrical and spherical coordinates respectively with z as the axis of symmetry, θthe longitudinal angle and βthe angle between z axis and Re(the azimuthal angle). Shows that these functions generate the following displacements, dilatation and stresses: Cylindrical coordinates (a) Displacements: 2, 0 , 2 =rzuu urzθφ φμμ∂∂==∂∂ (b) Dilation: 0 e= (c) 2 222 12, 212 12rr rrTe ET e Err rθθ θθμνφ μ ν φμμνν∂ ∂=+ = =+=−− ∂ ∂ 22 222 , 0, 0, 212zz zz r z rz rzTe EE E T Erz zθθμνφ φμμν∂ ∂=+ = = == =−∂ ∂ ∂ Spherical coordinates: (d) Displacements: ˆˆ 12, 0 , 2 =Ruu uRRθβφ φμμβ∂∂==∂∂ (e) Dilation: 0e= (f) Stresses: 22 22 2ˆˆ ˆ 22 1 12, 212 12RR RRTe E Te ERR RRββ ββμνφ μ ν φ φμμνν β∂ ∂∂=+= =+= +−− ∂ ∂∂ 2ˆˆ 21 c o t2 , 0, 012R Te E T TRR Rθθ θθ θ θβμν φ β φμνβ∂∂=+ = + ==−∂ ∂ 2 2112RRTERR Rββφφμβ β∂ ∂== −∂∂∂ ------------------------------------------------------------------------------- Ans. With rz RrzR=+=xe e e , ()()ˆ , , , rz Rφφφ β== 0=ψ we have,[see Eqs.(2.34.4) and (2.35.15)] () ( ) rzRˆˆ 124 1 2rzR Rβφφφ φμν φ μβ∂∂∂ ∂=− − +∇ ⋅ + → = + = +∂∂∂ ∂ux u e e e e ψψ That is, in cylindrical coordinates 2, 0 , 2 =rzuu urzθφ φμμ∂ ∂==∂ ∂ and in spherical coordinates; ˆˆ 12, 0 , 2 =Ruu uRRθβφ φμμβ∂ ∂==∂ ∂ (b) The non zero strain components are: In cylindrical coordinates: [See Eqs.(3.7.20)] Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-48 22 22 22 122 , 2 , 22 0, 0, 2rr z rr zz rz rz r zuu uEEErr r r z rz uuEE Ezr r zθθ θθμ φ φφμμ μ μμ φμμ∂∂∂∂ ∂== = = ==∂∂ ∂∂∂ ∂∂ ∂ ⎛⎞== = += ⎜⎟∂∂ ∂ ∂⎝⎠ ()22 2 22122 0rr zz eE E Err rzθθφφ φμμ φ∂∂ ∂=+ + = + + = ∇ =∂∂ ∂ In spherical coordinates: [see Eqs.(3.7.21)] 22 22 2ˆˆ ˆ 2 21 122 , 2 +RR RRu uuEER RR R R RRβ ββμ φ μφ φμμ μβ β∂ ∂∂∂ ∂== = =+∂∂∂∂∂ 2ˆˆ 2c o t 2 1c o t2 , 0, 0R Ru uEE ERR R R Rβ θθ θ θβμβμ φβ φμβ∂∂=+ =+ = =∂∂ 22 22 22 22ˆˆˆˆ 21 1 1 1 1 1222 ˆˆ ˆˆ 12 2 1 1 2R Ruu uERR RR R R R RR RR RR RRββ βμ φφφφμβ ββ β β φφ φφ ββ ββ⎛⎞∂ ⎛⎞∂ ∂ ∂∂ ∂=− + =− + − ⎜⎟ ⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂⎝⎠ ⎝⎠ ⎛⎞ ⎛⎞∂∂∂∂=− = −⎜⎟ ⎜⎟⎜⎟ ⎜⎟∂∂ ∂ ∂∂ ∂⎝⎠ ⎝⎠ ()22 22 2 2ˆˆ ˆ ˆ ˆ11 1 c o t22RR e EEERR RR RR Rββ θθφ φφφβ φμμβ β∂ ∂∂ ∂ ∂= + + = + +++∂∂∂ ∂∂ 22 2 22 2 2ˆˆ ˆ ˆ12 c o t0RR RR Rφφ φ β φφβ β∂∂ ∂ ∂++ + = ∇ =∂∂ ∂∂ [see Eq.2.35.37)] (c) the stresses are: In cylindrical coordinates: 2 222 12, 212 12rr rrTe ET e Err rθθ θθμνφ μ ν φμμνν∂ ∂=+ = =+ =−− ∂ ∂ 22 222 , 0, 0, 212zz zz r z rz rzTe EE E T Erz zθθμνφ φμμν∂ ∂=+ = = == =−∂ ∂ ∂ In spherical coordinates: 22 22 2ˆˆ ˆ 22 1 12, 212 12RR RRTe E Te ERR RRββ ββμνφ μ ν φ φμμνν β∂ ∂∂=+= =+= +−− ∂ ∂∂ 2ˆˆ 21 c o t2 , 0, 012R Te E T TRR Rθθ θθ θ θβμν φ β φμνβ∂∂=+ = + ==−∂ ∂ 2 2112RRTERR Rββφφμβ β∂ ∂== −∂∂∂ These are the formulas given in Example 5.38.6. __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-49 5.82 For the potential functions given in Eq.(5.38.46), [see Example 5.38.7)], i.e., : ()z ,, 0Rψβ φ = e =ψ , where 20ψ∇=, shows that these functions generate the following displacements iu, dilatation eand the stresses ijT (in spherical coordinates) as give n in Eq.(5.38.47) to (5.38.50):: (a) Displacements: 2 ( 34 ) R c o s, 2 ( 34 )s i n c o s , 0Ruu uRβθψψμν ψβ μ ν ψ β ββ⎧⎫ ∂∂ ⎧⎫=− − − = − + =⎨⎬ ⎨ ⎬∂∂ ⎩⎭ ⎩⎭. (b) Dilation: ()sin22 4 c o seRRψβψμν ββ⎛⎞∂∂=− − − ⎜⎟∂∂⎝⎠. (c) Stresses 2 22s i n2(1 )cos cosRRTRRR Rψ ψνβ ψνβ ββ⎛⎞ ∂∂ ∂=− − − −⎜⎟⎜⎟ ∂∂ ∂ ⎝⎠, ()2 2sin cos21 c o s ( 2 2 ) TRR Rββψ βψ β ψνβ νβ β⎛⎞ ∂∂ ∂=− − − − −⎜⎟⎜⎟ ∂∂ ∂ ⎝⎠, () ()121 c o s 21 s i nsinTRRθθψ ψνβ ν βββ⎛⎞ ⎛⎞ ∂∂=− − − − +⎜⎟ ⎜⎟∂ ∂ ⎝⎠ ⎝⎠, 212(1 ) cos cos sin (1 2 )RTR RRβψ ψψνβ β β νββ⎡⎤ ∂∂ ∂=− − − − −⎢⎥∂ ∂∂ ∂ ⎢⎥⎣⎦. 0RTTθθ β== ------------------------------------------------------------------------------- Ans (a) with RR zR, c o s β = = xe e e and Rz z R R R cos , cos sinβ ψψβ β β ⋅⋅ = = −xe e e e e=ψ , ()()()() z 24 1 4 1 R c o sμν φν ψ ψ β=− − +∇ ⋅ + =− − +∇ →ux e ψψ ()() () () RR12 4 1 cos sin R cos R cosRRβ β μν ψ β βψ β ψ ββ∂∂=− − − + +∂∂ue ee e ()() RR R 4 1 cos sin cos R cos cos sinRβ ββψψνψ β β ψ β β β ψ ββ⎡ ⎤ ∂∂=− − − + + + − ⎢ ⎥∂∂ ⎣ ⎦ee e e e e () () R cos 3 4 R 3 4 sin cosRβψψβν ψ ν ψ βββ⎡ ⎤ ∂∂ ⎡⎤=− − − + − + ⎢ ⎥ ⎢⎥∂∂ ⎣⎦ ⎣ ⎦ee . (b) The strain components are: 22 222 2 ( 3 4) R c o s ( 2 4) R c o sR RRuERR R R RRψψ ψ ψψμμν β ν β⎧⎫ ⎧ ⎫∂ ∂∂ ∂ ∂∂⎪⎪ ⎪ ⎪== − −− − = − −− ⎨⎬ ⎨ ⎬∂∂ ∂ ∂ ∂∂ ⎪⎪ ⎪ ⎪⎩⎭ ⎩ ⎭ 11 12 2 + (3 4 ) sin cos (3 4 ) R cosRu uERR R R Rβ ββψψμμν ψ β β ν ψ βββ β∂⎛⎞ ⎧⎫∂∂ ∂ ⎧⎫== − + − − − ⎨⎬ ⎨ ⎬ ⎜⎟∂∂ ∂ ∂ ⎩⎭ ⎩⎭ ⎝⎠ 2 2sin cos 1(2 4 ) (3 4 ) cos (3 4 ) R cosRR R R Rβψ ψ β ψ ψν νβ ν ψ ββ β⎧⎫ ∂∂ ∂ ⎪⎪ ⎧⎫=− + − + − − −⎨⎬ ⎨ ⎬∂∂ ∂ ⎩⎭ ⎪⎪⎩⎭ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-50 2 2sin cos(2 4 ) cosR RRβψβ ψ ψνββ β∂ ∂∂=− + +∂ ∂ ∂. cot 1c o t2 2 (3 4 ) R cos (3 4 ) sin cosRu uERR R R Rβ θθβ ψ βψμμ ν ψ β ν ψ β ββ⎛⎞ ⎧ ⎫ ∂∂ ⎧⎫=+ = − − − + − + ⎨ ⎬⎨ ⎬ ⎜⎟∂∂ ⎩⎭ ⎩⎭ ⎝⎠ cotcosRRψβ ψββ⎛⎞∂∂=+⎜⎟∂∂⎝⎠. ()21c o s22 2 ( 3 4 ) c o s ( 3 4 ) s i nR Ruu uERR R R R R Rββ ββψ ψ ψ ψμ μν β ν βββ β⎧⎫∂ ⎛⎞∂ ∂∂ ∂ ⎪⎪ ⎧⎫=− + = − − − + − − ⎨⎬ ⎨ ⎬ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎩⎭ ⎪⎪ ⎝⎠ ⎩⎭ 2sin cos(3 4 ) (3 4 )sin cosRR R Rββψ ψ ψνψ ν β ββ β⎧ ⎫ ⎧⎫ ∂∂ ∂⎪ ⎪−− + +− +⎨⎬ ⎨ ⎬∂ ∂∂ ∂ ⎩⎭ ⎪ ⎪ ⎩⎭ 2cos4(1 ) 2cos 2(1 2 )sinR RRβψψ ψνβ ν βββ∂ ∂∂=− − + + −∂ ∂∂ ∂. i.e., 2cos22 ( 1 ) c o s ( 1 2 ) s i nRER RRββψψ ψμν βν βββ∂ ∂∂=− − + + −∂ ∂∂ ∂. ()2 2cot2( 2 4 ) R c o s c o sRREE ERR R Rθθ ββψψ ψβ ψμ νβ ββ⎧⎫ ⎛⎞ ∂∂ ∂ ∂⎪⎪++ = −− − + + ⎨⎬ ⎜⎟∂∂ ∂∂ ⎝⎠ ⎪⎪⎩⎭ 2 2sin cos(2 4 ) cosR RRβψβ ψ ψνββ β∂ ∂∂+− + +∂ ∂ ∂ () ()22 22 22 22 22 2 2cos 14R c o s c o s ( 2 4 ) s i nsin 1c o t s i nR cos 4 cos (2 4 ) 2s i n s iRc o s 4 c o s ( 2 4 ) 2 4 c o sRR R R RR RR R RR R R Rψψ β ψ ψνβ β ν βββ β ψψ β ψ ψ β ψβν β νββ β ψψβ ψ ψβν β ν ν ββ⎧⎫∂∂ ∂ ∂⎪⎪=+ + ++ −⎨⎬∂∂∂∂ ⎪⎪⎩⎭ ⎛⎞∂∂ ∂ ∂ ∂=+ + + + −⎜⎟⎜⎟ ∂∂ ∂ ∂∂⎝⎠ ∂∂ ∂ ∂⎛⎞=− + + − = − − −⎜⎟∂∂ ∂ ∂⎝⎠n.Rβψ β⎛⎞ ∂ ⎜⎟∂ ⎝⎠ where we have used, the relation: 22 22 2 212 c o t0RR RR Rψψ ψ β ψ β β⎛⎞ ∂∂ ∂∂+ ++ =⎜⎟⎜⎟ ∂∂ ∂∂⎝⎠. Thus, ()sin22 4 c o seRRψβψμν ββ⎛⎞∂∂=− − − ⎜⎟∂∂⎝⎠. (c) The stresses are: 2s i nwith 2 cos12eRRμνψ β ψνβν β⎛⎞∂∂=− −⎜⎟−∂ ∂ ⎝⎠, we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-51 2 2 2 22s i n2 2 cos (2 4 ) R cos12 2s i n( 2 2 )cos R cos .RR RRTe ERR R R RR Rμν ψ β ψ ψ ψμ νβ ν βνβ ψνβ ψ ψνβ ββ⎧⎫ ⎛⎞∂∂ ∂ ∂ ⎪⎪=+= − − − −− ⎨⎬ ⎜⎟−∂ ∂ ∂ ∂ ⎝⎠ ⎪⎪ ⎩⎭ ∂∂ ∂=−+ + +∂∂ ∂ Similarly, ()2 22s i n c o s22 1 c o s ( 2 2 )12Te ERR Rββ ββμνψ β ψ β ψμν β ννβ β∂ ∂∂=+= − − + − +−∂ ∂ ∂. () ()2122 1 c o s 2 1 s i n12 s i nTe ERRθθ θθμνψ ψμν β ν βν ββ⎛⎞ ∂ ∂=+ = − − + −+ ⎜⎟−∂ ∂ ⎝⎠. 2cos22 ( 1 ) c o s s i n ( 1 2 )RRTER RRβββψψ ψμν β β νββ∂ ∂∂== − − + + −∂ ∂∂ ∂. 0RTTθθ β== _________________________________________________________________ 5.83 Show that ()1/Ris a harmonic function (i.e., it satisfies the Laplace Equation ()21/ 0R∇=), where R is the radial distance from the origin. ------------------------------------------------------------------------------- Ans (a) 2222 3 12 12 3 12 3, therefore, , , x xx RR RRxxxx RxRx R∂ ∂∂=++ = = =∂∂∂ so that 2 2 11 1 1 23 2 3 4 3 5 11 133 11 11 1 and x xx x R xR x R R R Rx R R R R∂∂ ∂ ⎛⎞ ⎛⎞ ⎛⎞=− =− =− + =− +⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂ ⎝⎠ ⎝⎠ ⎝⎠. Similarly, 2 2 22 3 2 23 5 23 5 233 3 11 11,x x RRx RR x RR∂∂⎛⎞ ⎛⎞=− + =− +⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠ Thus, 2 22 22 2 2 3 12 222 3 5 3 5 3 5 12 3 222 12 3 35 3 33 33 11 1 1 3( ) 33 30.jjx xx xx R R x x x R RR RR R xxx RR R R⎛⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ∂∂ ∂ ∂⎛⎞ ⎛⎞= + + = −+ + −+ + −+⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ∂∂ ∂∂∂⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ++=− + =− + =1 ________________________________________________________________________________________________________________________________________________ 5.84 In Kelvin's problem, we used the potential functionzψeΨ= where in cylindrical coordinates: 222, AR rzRψ== + .Using the results in Example 5.38.6, obtain the stresses. ------------------------------------------------------------------------------- Ans. 22 32 5 313 1, , zz Rz R zR Rψψψ∂∂== − = −∂ ∂, 22 2 32 5 3 3 531 3, rr r r z rz r z R rR R R Rψψ ψ∂∂ ∂ ∂ ⎛⎞=− → = − = − = ⎜⎟∂∂ ∂ ∂ ∂ ⎝⎠, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-52 ()22 2 23 5 3 3 53322 1 2rrzr z z z r zTzz rR R R R Rψψνν ν⎛⎞ ∂∂ ⎛⎞=− + =− − + − =− − + ⎜⎟⎜⎟⎜⎟ ∂∂ ⎝⎠⎝⎠. ()33 322 1 2zz z r zTzr r r R RRθθψψνν ν∂∂ ⎛⎞ ⎛⎞ ⎛ ⎞=− + =− − + − =− − ⎜⎟ ⎜⎟ ⎜ ⎟∂∂ ⎝⎠ ⎝⎠ ⎝ ⎠. () () ()22 35 3 53312 12 12rzrr z r r zTz zrr z RR R Rψψνν ν⎛⎞ ∂∂ ⎛⎞ ⎛ ⎞ ⎛ ⎞=− − + =− − − + = − + ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ∂∂ ∂ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠. 22 3 25 3 3 5 331 32(1 ) 2(1 ) (1 2 )zzzz z zTz zz zR R R R Rψψνν ν⎛⎞ ∂∂=− −=− + −= + − ⎜⎟⎜⎟∂ ∂ ⎝⎠. _________________________________________________________________ 5.85 Show that for 222ln( ), CR z R r zϕ=+= + , 2 231 ()zCRRz rRϕ⎧ ⎫ ∂=−⎨ ⎬+ ∂⎩⎭. ------------------------------------------------------------------------------- Ans. 1, () ()Cr C rR z R r r R R zϕ ϕ ∂∂==∂+ ∂ +. ()22 2 2 2 22 2 2 2 22 22 2 2 22111() () ()() ()() () () ()() ()rr C r rCR r Rz Rz r R R Rz R Rz rR RR z Cr r C R r z R r RR z RR z RR z RR z RR z RR Cz R z CR z z R RR z R RR z RRϕ ⎧ ⎫ ⎧⎫∂∂ ∂ ⎪ ⎪=+ = − − +⎨⎬ ⎨ ⎬∂+ + ∂ + + ∂ ⎪ ⎪ ⎩⎭ ⎩⎭ ⎧⎫ ⎧ ⎫ + −+ ⎪⎪ ⎪ ⎪=− − += − ⎨⎬ ⎨ ⎬++ + ++⎪⎪ ⎪ ⎪⎩⎭ ⎩ ⎭ ⎧⎫ ⎧−+ −⎪⎪ ⎪=− = ⎨⎬ ⎨++⎪⎪⎩⎭31.()zCRRz R⎫⎧ ⎫ ⎪=−⎬⎨ ⎬+ ⎪⎪ ⎩⎭ ⎩⎭ ____________________________________________________________________________________________________________________________________________________________________________ 5.86 Given the following potential functions: z (/ ) , ( 1) zϕ φν ϕ ∂∂ = − e =ψ , where 222ln( ), CR z R r zϕ=+= + . From the results of Example 5.38.4, and Eqs (i), (ii) (iii) of Section 5.40, obtain { }253/ ( 1 2 ) / [ ( ) ]rrTC r z R R R z ν =− − + , (){ }312 / 1 / [ ( ) ] TC z R R R zθθ ν=−− + + . 35(3 / )zzTC z R= , 25(3 ) /rzTC r zR= . -------------------------------------------------------------------------------- Ans. 32 22 22 53 3 512 31 2 1 3 1 2,() () ()rrTzrr rz r rz r zCz CRRz R Rz R Rz RR R Rϕϕ ϕν νν∂∂ ∂⎛⎞=+ + ⎜⎟∂ ∂∂ ∂ ⎝⎠ ⎧⎫ ⎧ ⎫⎛⎞ ⎛ ⎞ ⎛⎞ − ⎪⎪ ⎪ ⎪=− + + − = −⎜⎟ ⎜ ⎟⎨⎬ ⎨ ⎬ ⎜⎟ ⎜⎟ ⎜ ⎟ ++ +⎝⎠ ⎪⎪ ⎪ ⎪⎝⎠ ⎝ ⎠⎩⎭ ⎩ ⎭ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-53 () ()22 2 33 3 312( 1 2 ) 2 ( 12 ) 12 ( 12 )(( 112()zTrr z rr z Cz z Cr C Cz C rR R z R R z RR R zCRR z Rθθϕϕ ϕνν νν ν ν ν⎧⎫∂∂ ∂⎪⎪=− + + −⎨⎬∂∂ ∂∂⎪⎪⎩⎭ ⎛⎞ ⎛⎞=− − + − + − =− − + −⎜⎟ ⎜⎟+ + ⎝⎠ ⎝⎠ ⎧⎫=− − + ⎨⎬+ ⎩⎭, 32 3 3 32 3 5 3 533 zzzz z zTz C C zz R RR Rϕϕ ⎧⎫⎛⎞ ∂∂ ⎪⎪=− = − −+ = ⎜⎟⎨⎬⎜⎟∂∂ ⎪⎪⎝⎠⎩⎭, 32 253 rzrzTz C rz Rϕ⎛⎞∂==⎜⎟⎜⎟∂∂⎝⎠, 0rzTTθθ==. _________________________________________________________________ 5.87 The stresses in Boussinesq problem in cylindrical coordinate are given by: () ()2 5312 12 31, 2( ) 2 ( )z z rrF F rz zTTRRz R Rz RRθθνν ππ⎧⎫ −− ⎧ ⎫ ⎪⎪=− − =− − +⎨⎬ ⎨⎬++ ⎪⎪ ⎩⎭ ⎩⎭, 3 53 2z zzFzT Rπ=− , 2 53 2z rzFrzT Rπ=− , 0rzTTθθ==. Obtain the stresses in recta ngular Cartesian coordinates. -------------------------------------------------------------------------------- Ans. cos sin 0 0 cos sin 0 sin cos 0 0 0 sin cos 0 00 1 0 0 0 1xx xy xz rr rz yx yy yz zr zz zx yz zzTTT TT TTT T TT TTTθθθθ θ θ θθ θθ⎡⎤ −⎡⎤ ⎡ ⎤ ⎡⎤⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥=− ⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥⎢⎥⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦ ⎣⎦ ⎢⎥⎣⎦ () ()22 22cos sin sin cos cos sin cos sin cos sin cos sin 1rr rr rz rr rr rz rz rzTT T T T TT T T T TTθθ θθ θθ θθθ θθ θ θ θθθθ θ θθ⎡⎤ +−⎢⎥ ⎢⎥−+ ⎢⎥ ⎢⎥⎣⎦. Thus, () ()22 2 22 53cos sin 12 12 31cos sin2( ) 2 ( )xx rr z zTT T F F rz z RR z RR z RRθθθθ ννθ θππ=+ ⎧⎫ −− ⎧⎫ ⎪⎪=− − − − +⎨⎬ ⎨⎬++ ⎪⎪ ⎩⎭ ⎩⎭ ()()()2 22 2 5312 12 12 3cos sin sin2( ) ( )zz F xz RR z RR z RRνν νθ θθπ⎡⎤ −− −=− − − +⎢⎥++ ⎢⎥⎣⎦ ()()2 22 53 212 12 3( )cos 2cos 12( )zz F xz zRzRR z RR Rννθθπ⎡⎤ −− ⎧ ⎫=− − + + − +⎢⎥ ⎨ ⎬+⎩⎭ ⎢⎥⎣⎦ () ()()2 22 53 212 12 12 3() c o s2 c o s2( ) ( )zz F xz zRzRR z RR z RR Rνν νθθπ⎡⎤ −− − ⎧ ⎫=− − + + + −⎢⎥ ⎨ ⎬++ ⎩⎭ ⎢⎥⎣⎦ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-54 ()()()22 2 2 53 212 12 12 3( )cos2( ) ( )zz F xz Rz R z R RR z RR z RR Rνν νθπ⎡⎤ ⎧⎫ −− − −+−⎪⎪=− − + +⎢⎥ ⎨⎬++ ⎪⎪ ⎢⎥ ⎩⎭ ⎣⎦ ()()()22 2 2 53 212 12 12 3( )cos2( ) ( )zz F xz Rz R z R RR z RR z RR Rνν νθπ⎡⎤ ⎧⎫ −− − −+−⎪⎪=− − + +⎢⎥ ⎨⎬++ ⎪⎪ ⎢⎥ ⎩⎭ ⎣⎦ Now, () ()22 22 2 2 22cos cos / , , cosxxrx x r r z RRzRz Rzθθ θ=→ = + = = =+− − Therefore, ()()() () () () ()() ()22 2 2 53 2 22 2 53 212 12 12 3( ) 2( ) ( ) 12 12 12 31.2( ) ( )z xx zz F xz Rz R z R xTRRz R Rz RzRz RR R z F xz x x RR z RR z R R z RR Rνν ν π νν ν π⎡⎤ ⎧⎫ −− − −+−⎪⎪=− − + +⎢⎥ ⎨⎬++ + − ⎪⎪ ⎢⎥ ⎩⎭ ⎣⎦ ⎡⎤ ⎧⎫ −− − ⎪⎪=− − + − +⎢⎥ ⎨⎬+++ ⎪⎪ ⎢⎥ ⎩⎭ ⎣⎦ _________________________________________________________________ 5.88 Obtain the variation of zzTalong the z axis for the case where the normal load on the surface of an elastic half-space is uniform with intensity oq, and the loaded area is a circle of radiusor with its center at the origin. -------------------------------------------------------------------------------- Ans. Using Eq.(5.41.3), we have, o3 3 o o 55 o2 332 ''r zzrqr d rzr d rTz q R Rπ π′=′′ ′′=− =−∫∫, where o 22 2 3 o 43R zzzdRRrz R d R r d rT z q R′′′ ′ ′ ′ ′=+→ = →= − ′∫ o 33 3 oo oo o 332 2 3 / 2 ' oo13 3' ( )R zz Rzqz qzTz q q q RR r z=⎡⎤→= − − = −= −⎢⎥+ ⎣⎦. __________________________________________________________________ 5.89 For the potential function 2 1z cos DRψ β−= e, where ( , , ) Rβθare the spherical coordinates with βas the azimuthal angle. Find and RRRTTβ. ------------------------------------------------------------------------------- Ans. 2 23 4 2 11 1 1 2cos 2 cos 6 cos , sin DR DR DR DRR Rψψ ψψ βββ ββ−− − − ∂∂ ∂= → =− → = =−∂∂ ∂. From Example 5.38.7, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-55 () ( ) () ()2 2 34 2 1 23 12s i n2(1 )cos cos 2s i n2(1 )cos 2 cos cos 6 cos sin 25 c o s 2 .RRTRRR R DR R R RR DRψψ ν β ψνβ ββ νβνββ β β β νβ ν−− − −∂∂ ∂=− − + +∂∂ ∂ ⎡⎤=− − − + + −⎢⎥⎣⎦ ⎡⎤=− −⎣⎦ () () ( ) []2 23 3 1 333 1 3 112(1 )cos cos sin (1 2 ) 2(1 )cos sin cos 2 sin sin (1 2 ) 2 cos 2(1 ) cos sin 2 sin cos 2 cos sin (1 2 ) cos sin 2(1 ) 2 2(1 2 ) 2RTRR R DR R RR DR R R DR DRβνψ ψ ψββ β νββ νβ ββ ββ ν β νβ β β β β β ν ββ ν ν−− − −−− −−∂ ∂ ∂=− + + −∂∂ ∂ ∂ −⎡⎤=− − + + −−⎢⎥⎣⎦ ⎡⎤=− + − −⎣⎦ =− + − − =3(1 ) cos sin .νβ β−+ __________________________________________________________________ 5.90 For the potential function, () ()32 1 12 ,[ 3 c o s 1 / 2 ]R CR C R φφ β β− −== − +%% , where ( , , ) Rβθare the spherical coordinates with βas the azimuthal angle, obtain and RRRTTβ. -------------------------------------------------------------------------------- Ans. With () ( ) ()32 1 12 ,/ 2 3 c o s 1R CR C R φβ β−−=− +%()() ( ) ()()42 2 12/2 3 3 c o s 1CR C RRφβ−− ∂→= − − + −∂% ()2 52 3 12 263 c o s 1 2CR C R Rφβ− − ∂→= − + ∂% 34 113s i n c o s 9s i n c o s and C R C RRφφβββ βββ−− ∂∂ ∂=− → =∂∂ ∂%% . From Example 5.38.6, we have ()2 52 3 12 2ˆ 63 c o s 1 2RRTC R C R Rφβ− − ∂== − + ∂. ()2 43 5 11 1ˆˆ 11 1 19 sin cos 3 sin cos 12 sin cosRTC R C R C RRR R R Rβφφβββ β β βββ−− −⎛⎞∂∂ ⎡⎤=− = − − =⎜⎟ ⎢⎥ ⎜⎟∂∂ ∂ ⎣⎦ ⎝⎠ . 0RTTθθ β==. _________________________________________________________________ CHARTER 5, PART B 5.91 Demonstrate that if only 23and E Eare nonzero, then Eq.(5.46.4) becomes Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-56 []22 23 2 23 32 33 32CCEUEECCE⎡ ⎤⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦. ------------------------------------------------------------------------------- Ans. Eq. .(5.46.4) gives [] () ()12 2 13 3 22 2 23 3 23 2 33 3 23 2 2 2 22 3 3 3 2 3 23 3 3 24 2 34 3 25 2 35 3 26 2 36 320 0 0 0CE CE CE CE CE CEUE E E C E C E E C E C ECE CE CE CE CE CE+⎡⎤ ⎢⎥+⎢⎥ ⎢⎥+⎡ ⎤ == + + + ⎢⎥ ⎣ ⎦+⎢⎥ ⎢⎥+⎢⎥+⎢⎥⎣⎦. This is the same as [] () ()22 23 2 23 2 2 2 22 3 3 3 2 3 23 3 3 23 33 3CCEEE E C EC E E C EC ECCE⎡⎤ ⎡ ⎤⎡ ⎤ =+ + + ⎢⎥ ⎢ ⎥ ⎣ ⎦⎣⎦ ⎣ ⎦. _________________________________________________________________ 5.92 Demonstrate that if only 13and E Eare nonzero, then Eq.(5.46.4) becomes []11 13 1 13 31 33 32CC EUE ECC E⎡ ⎤⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ -------------------------------------------------------------------------------- Ans. [] () ()11 12 13 14 15 16 1 12 22 23 24 25 26 13 23 33 34 35 36 3 13 14 24 34 44 45 46 15 25 35 45 55 56 16 26 36 46 56 66 11 1 1 1 3 3 31 3 1 3 3 30 20 0 0 00 00CCCCCC E CCCCCC CCCCCC EUE ECCCCCC CCCCCC CCCCCC ECE CE ECE CE⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ = ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦ ⎡⎤=+ + +⎣⎦ This is the same as [] () ()11 13 1 1 3 1 1 11 13 3 3 13 1 33 3 13 33 3122CC EE EE E C C E E C E C E UCC E⎡⎤ ⎡ ⎤⎡⎤=+ + + = ⎢⎥ ⎢ ⎥ ⎣⎦⎣⎦ ⎣ ⎦ __________________________________________________________________ 5.93 Write stress strain laws for a monoclinic el astic solid in contracted notation, whose plane of symmetry is the 12xxplane. -------------------------------------------------------------------------------- Ans. All 0ijklC= where the indices ijklcontain an odd number of 3. Therefore, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-57 11 12 13 16 12 22 23 26 13 23 33 36 44 45 45 55 16 26 36 6600 00 00[]000 0 000 0 00CCC C CCC C CCC CCCC CC CCC C⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ =⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎣⎦. ________________________________________________________________________________________________________________________________________________ 5.94 Write stress strain laws for a monoclinic el astic solid in contracted notation, whose plane of symmetry is the 13xxplane. -------------------------------------------------------------------------------- Ans. All 0ijklC= where the indices ijklcontain an odd number of 2. Therefore, 11 12 13 15 12 22 23 25 13 23 33 35 44 46 15 25 35 55 46 6600 0000[]000 0 00 000 0CCC C CCC C CCC CCCC CCC C CC⎡⎤ ⎢⎥ ⎢⎥⎢⎥ =⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ________________________________________________________________________________________________________________________________________________ 5.95 For transversely isotropic solid with 3e as the axis of transversely isotropy, show from the transformation law ijkl mi nj rk sl mnrsCQ Q Q Q C′= that 1113 0 C′= (See Sect.5.50) ------------------------------------------------------------------------------- Ans. Since 33 13 23 31 321, 0 QQ Q Q Q===== , therefore, 1 1 1 3 1113 111 3 3 3 111 3 1 111 1 3 2 111 2 3 11 11 1 11 3 11 21 1 12 3 21 11 1 21 3 21 21 1 22 3 .m n r s mnrs m n r mnr m n r mnr n r nr n r nr rr rr rr r rC Q QQQC Q QQQC Q QQC QQQC QQQC QQQC QQQC QQQC QQQC′=== = + =+++ Now, all ijklCwith odd number of either 1 or 2 are zero because 1eplane and 2e-plane are planes of material symmetry. Thus, 1 1 13 1 1 23 1 2 13 1 2 23 0rr rr rr r rQC QC QC QC= === . Thus, 1113 0 C′= __________________________________________________________________________________________________________________________________________________ 5.96 Show that for a transversely isotropic elastic material with 3e as the axis of transverse isotropy, 1133 2233CC= (see Sect.5.50) . --------------------------------------------------------------------------------- Ans. 11 2 2 12 3 3c os sin , sin cos , βββ β ′′ ′=+ = −+ =ee e e ee e e 11 12 21 22 33 31 13 23 32cos , sin , sin , cos , 1, 0 QQ Q Q Q Q Q Q Qβ βββ = = − = = = ==== . Thus, 1 2 3 31 2 3 3 1 2 3 3 3 3 3 31 2 3 3 1 1 2 1 3 3 2 1 2 2 3 3 11 12 1133 11 21 1233 21 12 2133 21 22 2233 .m n r s mnrs m n mn m n mn n n n n C QQQQC QQQQC QQC Q QC QQC QQC QQC QQC QQ C′=== = + =+++ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-58 Now 1233 2133 0 CC== because 1e plane (as well as 2e-plane) is a plane of material symmetry. Thus, ()1233 11 12 1133 21 22 2233 1133 2233 1133 2233cos sin sin cos cos sin .CQ Q CQ Q C C C CCβββ β ββ′=+= − + =− + Again, 1233 0 C′=, because 1′e is also a plane of symmetry. Thus 1133 2233CC= . ________________________________________________________________________________________________________________________________________________ 5.97 Show that for a transversely isotropic elastic material with 3e as the axis of transverse isotropy(see Sect.5.50) 22 2 2 2 2 1111 1122 1212 2222 (sin ) (cos ) (sin ) 2 (cos ) (sin ) (cos ) 0 CC C Cββ β β ββ ⎡⎤ ⎡⎤+− + − − =⎣⎦ ⎣⎦. -------------------------------------------------------------------------------- Ans. Since 13 31 23 32 0 QQQQ==== and 0ijklC=when the indices ijklcontain an odd number of either 1 or 2, therefore, 1 2 2 2 1222 1 1222 1 2 1222 2 11 12 2 2 11 11 22 2 2 12 21 12 2 2 21 21 22 2 2 22 11 12 12 12 1111 11 12 22 22 1122 11 22 12 22 1212 11 22 22 12m n r s mnrs n r s nrs n r s nrs rs r s rs r s rs r s rs r sC QQQQC Q QQQC QQQQC Q QQQC Q QQQC QQQQC QQQQCQ QQQC Q QQQC Q QQQC Q QQQC′== + =+++=+++ 1221 21 12 12 22 2112 21 12 22 12 2121 21 22 12 12 2211 21 22 22 22 2222 . QQQQC QQQQC QQQQC QQQQC++++ Thus, 33 3 3 1222 1111 1122 1212 1221 3333 2112 2121 2211 2222 22 2 2 2 1111 1122 12c o s( s i n) s i n( c o s) ( c o s) s i n ( c o s) s i n (sin ) cos (sin ) cos (sin ) cos (cos ) sin cos sin sin (cos sin ) 2(cos sin )C CCCC CCCC CC Cββ ββ β β β β ββ ββ ββ ββ ββ β β β β β′=− − − − ++++ =− + − + −2 12 2222cos . Cβ ⎡⎤ −⎣⎦ where we have used , and ijkl jikl ijkl jilk ijkl klijCC CC CC= == . Now, 1222 0 C′= because 1′e is also a plane of symmetry, therefore, 22 2 2 22 1111 1122 1212 2222 sin (cos sin ) 2(cos sin ) cos 0 CC C Cββ β β ββ+− + − − = . ________________________________________________________________________________________________________________________________________________ 5.98 In Section 5.50, we obtained the reduction in the elastic coefficients for a transversely isotropic elastic solid by demanding that each Sβplane is a plane of material symmetry. We can also obtain the same reduction by demanding the ijklC′be the same for all β. Use this procedure to obtain the result: 1133 2233CC= . ------------------------------------------------------------------------------- Ans. Since 31 13 32 23 33 0, 1 QQQQ Q==== = , therefore, 1133 1 1 3 3 1 1 33 33 33 1 1 33 mnr s m n r s mn m n mnm n C Q QQQC Q QQQC Q QC′=== . Now, 0ijklC=when the indices contain an odd number of either 1 or 2, therefore, 22 1133 11 11 1133 21 21 2233 1133 2233 cos sin C QQC QQC C C ββ ′=+=+ . Now, 1133 1133CC′= for all β, therefore, 22 2 2 1133 1133 2233 1133 2233 cos sin sin sin CC C C C ββ β β=+→= . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-59 Thus, 1133 2233CC= . ________________________________________________________________________________________________________________________________________________ 5.99 Invert the compliance matrix for a transverse ly isotropic elastic solid to obtain the relationship between ijCand the engineering constants. That is, verify Eq. (5.53.2) and (5.53.3) by inverting the following matrix: []12 1 1 3 1 3 21 1 1 31 3 13 1 13 1 31/ / / /1 / / // 1 /E EE AE E E EE Eνν νν νν−− ⎡⎤ ⎢⎥=− −⎢⎥ ⎢⎥−−⎣⎦ ------------------------------------------------------------------------------- Ans. []12 1 1 3 1 3 21 1 1 31 3 13 1 13 1 31/ / / /1 / / // 1 /E EE AE E E EE Eνν νν νν−− ⎡ ⎤ ⎢ ⎥=− −⎢ ⎥ ⎢ ⎥ −−⎣ ⎦ [] () ()() 21 13 31 13 31 21 21 21 21 13 31 22 13 1311det 1 2 2 1 1 2A EE EEννν νν νν ν ν νν Δ= = − − − = + − − . Now, 31 3 13 1//EEνν=→ () () ( )()2 21 21 31 1 3 21 22 13 131111 2/ 1 EED EE EEνν ν ν Δ= + − − = + where () ( )2 21 31 1 312/ D EE νν=− − . ()()()2231 1 313 1 3 13 1 11 31 13 13 1 3 21 1 3 211( / ) 1/ /111/1 / 11EE EE EE ECEE DE E Dν νννν νν⎡ ⎤ − − ⎣ ⎦== − =−Δ+ + 22 11CC= ()()()2 21 3 112 1 1 2 1 33 21 3 1 21 1 3 13 2 11( / ) 1/ /111//1 / 1EE EE ECE EEE EE Dν ννν ν⎡ ⎤ − − ⎣ ⎦== − =−ΔΔ + ()( ) ()2 12 1 3 1 1 321 1 31 3 12 21 31 13 13 1 3 13 2 1/ // 11 /1 / 1E EE EECEE EE Dνν νννν νν ν+ −−=− = + =−ΔΔ + ()21 1 31 3 31 1 13 21 31 31 13 1 3 13// 11 1/ /EE ECEE EEDνν ννν νν−−== + =− ΔΔ ()13 1 3 31 1 23 31 21 31 21 1 31 3 131/ /11 //EE ECEE EEDν ννν ννν−=− = + =−−ΔΔ ________________________________________________________________________________________________________________________________________________ 5.100 Obtain Eq.(5.53.6) from Eq. (5.53.2) and (5.53.3). ------------------------------------------------------------------------------- Ans. From Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-60 ()2 31 1 31 11 211( / ) 1EEECDν ν⎡⎤−⎣⎦=+ and ( ) ()2 12 1 3 1 1 3 12 21/ 1E EE CDνν ν+ =+ ()( ) ()2 2 12 1 3 1 1 3 13 1 1 3 11 12 21 21/ 1( / ) 11E EE EE E CCDDνν ν νν⎡⎤ + −⎣⎦−= −++ (){ }(){}()2 11 1 21 31 1 3 21 21 2112 ( / )11 1EE EEE DDDνννν ν=− − = =++ + Thus, [see Eq.5.53.5], ()1 12 2121EGν=+. ________________________________________________________________________________________________________________________________________________ 5.101 Invert the compliance matrix for an orthot ropic elastic solid to obtain the relationship between ijCand the engineering constants. ------------------------------------------------------------------------------- Ans. Let []1A−=1 31 21 12 3 11 12 13 32 12 12 22 23 12 3 13 23 33 13 23 12 31 1 1EE ECCC CCCEE ECCC EE Eνν ν ν νν−⎡⎤−−⎢⎥ ⎢⎥⎡ ⎤⎢⎥⎢ ⎥−− =⎢⎥⎢ ⎥⎢⎥⎢ ⎥⎣ ⎦ ⎢⎥ ⎢⎥−− ⎢⎥⎣⎦ [][] 12 23 31 13 21 32 13 31 23 32 21 12 1231detAEE Eννν ννν νν νν νν−−− − −≡Δ= . Since 23 2 31 3 12 1 12 23 31 21 32 13 231EE E EEEνννννν ννν⎛⎞⎛⎞ ⎛⎞== ⎜⎟⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠⎝⎠, therefore, [] 13 21 32 13 31 23 32 21 12 12312 EE Eννν νν νν νν−− − −Δ= . Next ()32 23 32 23 11 32 23 23 23 3 2 2 3 231 11 1 1111EECEE E E E E EEν ννννν− ⎛⎞== − = − ⎜⎟ΔΔ Δ ⎝⎠−, etc., () ()31 31 21 21 23 23 12 21 31 23 13 31 21 32 23 32 23 23 23 2 311 11, 11EE EECCEE EE EE E Eνννν νν ν νν ννν−− −− =− = + = = +ΔΔ ΔΔ−− Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-61 ()31 13 23 32 31 12 32 12 13 131 11EECEE EEν ννννν− =− = +ΔΔ−− _________________________________________________________________ 5.102 Obtain the restriction given in Eq.(5.54.8) for engineering constants for an orthotropic elastic solid -------------------------------------------------------------------------------- Ans. ()31 21 12 3 21 12 32 12 21 12 21 12 12 12 3 1 2 12 13 23 12 31 1 11det 1 1 01 1EE E EE EE E E E EE EE Eνν ν ν ννν ννν νν⎡⎤−−⎢⎥⎡⎤⎢⎥ −⎢⎥⎢⎥⎢⎥ −− → = − → − >⎢⎥⎢⎥⎢⎥−⎢⎥⎢⎥⎣⎦⎢⎥−− ⎢⎥⎣⎦, But, 2 12 21 21 21 1 2 21 12 21 12 2 111 0EE EE E Eννν νννν =→ − = − > →> . Also, 2 2 12 2 1 21 12 12 1211 0EE EEνννν−= −> → > Next, ()32 23 32 23 32 23 23 23 231 1det 1 1 01EE EE EEν νν ννν⎡⎤−⎢⎥ ⎢⎥ =−→ − >⎢⎥−⎢⎥ ⎣⎦. But, 22 32 23 3 2 32 23 23 32 32 2 311 1E E E EE Eνννν ν ν =→ − = − = − , 22 3 2 32 23 23 32 321 0 and E E EEννν ν−> → > > . Also, ()31 13 22 3 1 31 13 13 31 13 13 13 1 13 3 131 11 1det 1 1 11EE E E EEE E E E E E EEν νν ν νν⎡⎤−⎢⎥⎛⎞ ⎛⎞⎢⎥ =−=−=− ⎜⎟ ⎜⎟⎢⎥⎝⎠ ⎝⎠−⎢⎥ ⎣⎦ ()22 3 1 31 13 13 31 3110 a n d E E E Eνν ν ν−> → < < . ________________________________________________________________________________________________________________________________________________ 5.103 Write down all the restrictions for the engi neering constants for a monoclinic solid in determinant form (no need to expand the determinant). Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-62 -------------------------------------------------------------------------------- Ans. 12 1 2 3 1 3 4 1 4 11 11 12 1 2 32 3 42 4 22 22 13 1 23 2 3 43 4 33 3 14 1 24 2 34 3 4 23 56 5 6 31 56 5 6 121/ / / / 0 0 /1 / / / 0 0 // 1 / / 0 0 /// 1 /0 0 2 000 0 1 / / 2 000 0/ 1 / 2EE E G E T EE E G E T EE E G E T EEE G E GG E GG Eνν η νν η νν η ηηη μ μ−− ⎡⎤⎡⎤ ⎢⎥⎢⎥−−⎢⎥⎢⎥ ⎢⎥⎢⎥ −−=⎢⎥⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎢⎥⎣⎦⎣⎦3 23 31 12T T T⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎣⎦ (i) 1234560, 0, 0, 0, 0, 0 EE EGGG>>>>>> 12 1 2 23 2 3 3 4 3 4 12 1 23 2 23 2 3 34 3 4 56 5 6 1 3 1 3 24 2 4 1 4 1 4 56 2 6 13 1 3 24 2 4 14 1 41/ / 1/ / 1/ /( ) 0, 0, 0// / 1 / / 1 / 1/ / 1/ / 1/ / 1/ /0, 0, 0, 0/1 / /1 / /1 / / 1 /EE E E E GiiEE E E E G GG E E EG E G EG EE EG E Gνν η νν ν η μν ηη μν ηη−−>> >−− − −>> > >− (iii) 12 1 2 3 1 3 23 2 3 4 2 4 12 1 2 32 3 23 2 3 43 4 13 1 23 2 3 24 2 34 3 4 13 1 3 4 1 4 1 2 1 2 4 1 4 13 3 43 4 12 1 2 42 4 14 1 34 3 4 141/ / / 1/ / / /1 / / 0 , / 1 / / 0 // 1 / // 1 / 1/ / / 1/ / / /1 / /0 , / 1 / / // 1 / /EE E E E G EE E E E G EE E EE G EE GE E G EE G E E G EE G Eνν ν η νν ν η νν ηη νη νη νη ν η ηη η−− − −− > − > −− −− −> − 12 4 2 40 /1 /EGη> (iv) 12 1 2 3 1 3 4 1 4 12 1 2 32 3 42 4 13 1 23 2 3 43 4 14 1 24 2 34 3 41/ / / / /1 / / /0// 1 / / /// 1 /EE E G EE E G EE E G EEE Gνν η νν η νν η ηηη−− −−>−− _________________________________________________________________ CHAPRTR 5, PART C 5.104 Show that if a tensor is objective, then its inverse is also objective . ------------------------------------------------------------------------------- Ans. Let Tbe an objective tensor, then in a change of frame: o *( ) ( ) ( ) tt=+− xc Qx x T() ()tt *T= Q T Q . Taking the inverse of this equation, we get, since 1T−=QQ . ()11T 1 T() () () ()tt t t−−−= *T= Q T Q Q T Q . Thus, 1−Tis objective. ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-63 5.105 Show that the rate of deformation tensor ()T[] / 2=∇+∇Dvv is objective. [See Example 5.56.2)]. ------------------------------------------------------------------------------- Ans. From Eq.(5.56.13), we have ()TT * () ()tt=+v* Q v Q QQ & ∇∇ . Thus, () ()T T TT * () ()tt=+ v*) Q v Q Q Q & (∇ ∇ . Now, ()TT T T T/( )dd t →+→= − QQ = 0 Q Q QQ Q Q QQ && & & . Thus, () () () ()TT TT T T T T ** () () () () () ()tt t t t t ⎡⎤ =+ + − = +⎢⎥⎣⎦v*+ v * ) Q v Q Q Q Q v Q Q Q Q v v Q && ∇( ∇ ∇ ∇ ∇ ∇ T() ()tt →=D* Q DQ . ________________________________________________________________________________________________________________________________________________ 5.106 Show that in a change of frame, the spin tensor ()T[] / 2=∇−∇Wvv transforms in accordance with the equation TT() ()tt=+W* Q WQ Q Q & . [See Example 5.56.2)]. ------------------------------------------------------------------------------- Ans. From Eq.(5.56.13), we have ()TT * () ()tt=+v* Q v Q QQ & ∇∇ . Thus () ()T T TT * () ()tt=+ v*) Q v Q Q Q & (∇ ∇ . Now, ()()TT T T T/dd t +→= − QQ = 0 = Q Q QQ QQ Q Q && & & , ()T TT T * () ()tt →= − v*) Q v Q Q Q & (∇ ∇ . Thus, ( ) () ()T TT T ** () [ ] () 2tt −= − +v * v*) Q v v Q QQ & ∇( ∇ ∇ ∇ TT() ()tt →= +W* Q WQ Q Q &. ____________________________________________________________ 5.107 Show that in a change of frame, the mate rial derivative of an objective tensor Ttransforms in accordance with the equationTTT() () () ()tt tt=+ +*T QTQ Q TQ Q TQ&&&& , where a super-dot indicates material derivative. Thus the material derivative of an objective tensor T is not objective. ------------------------------------------------------------------------------- Ans. Since Tis objective, therefore, in a change of frame, T() ()tt=*TQ T Q . Taking the material derivative of this equation and noting that *tt=, we have, TTT=++*T QTQ QTQ QTQ&&&& . Since T() ()tt≠*TQ T Q&& , therefore, D DtTis non-objective. ________________________________________________________________________________________________________________________________________________ 5.108 The second Rivlin-Ericksen tensor is defined by: () ()T 21 1 1 1 1 , wher e D Dt=+∇ + ∇ ≡AA Av v A A A&& , where ()T 12== ∇∇AD v + v . Show that 2Ais objective. [See Prob. Error! Reference source not found. and Error! Reference source not found. ]. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-64 ------------------------------------------------------------------------------- Ans. From Prob. 5.105, we had, TT 11 () () () ()tt t t=→ =**DQ D Q A Q A Q . TTT 1 111 () () () () () ()tt tt tt →= + + *A Q AQ Q AQ Q AQ&& & &. (i) We also have, from Eq.(5.56.13), ()TT * () ()tt=+v* Q v Q QQ & ∇∇ . Thus ()T 11****Av * + v * A∇∇ () ()TT T T T T T 11 ]]++ + = QA Q [Q v Q QQ [Q v Q QQ QA Q && ∇∇ () ()()T TT T T T T 11 1 1 ]] +++ = [QA v Q QA Q QQ [Q v A Q QQ QA Q && ∇∇ . Since TT TT T(/ ) 0 0DD t =→ + =→ = − QQ QQ QQ QQ QQ && & & , therefore, ()T 11****Av * + v * A∇∇ = () ()()T TT T T T T 11 1 1 ]] −+− [QA v Q QA Q QQ [Q v A Q QQ QA Q && ∇∇ i.e., ()T 11****Av * + v * A∇∇ () ()T TT T T 11 1 1 ] +− − = [ Q A v Q Q v AQ Q AQ Q AQ && ∇∇ (ii) (i) and (ii) give ()T 11 1** → ** *A+ A v * + v *A& ∇∇ () ()T TTT T T TT 111 1 1 11 ] ++ + −− =Q AQ Q AQ Q AQ + [ Q A v Q Q v AQ Q AQ Q AQ&& & & & ∇∇ () () () ()TT TT T T 11 1 1 1 1⎡ ⎤ ++⎢ ⎥ ⎣ ⎦=Q AQ +Q A v Q Q v AQ =Q A +A v v A Q&& ∇∇ ∇ ∇ . Thus, () ()T 11 1 + A+ A v v A&∇∇ is objective. _________________________________________________________________ 5.109 The Jaumann Derivative of a second order objective tensor Tis :+−TT WW T& , where Wis the spin tensor. Show that the Jaumann derivative of Tis objective. [See Prob. 5.106 and Prob. 5.107] -------------------------------------------------------------------------------- Ans. We have, since Tis objective, therefore, in a change of frame, T=*TQ T Q . In Prob.5.106, we had TT() ()tt=+W* Q WQ Q Q & and in Prob. 5.107, we had TTT() () () ()tt tt=+ +*T QTQ Q TQ Q TQ&&&& . Also, ()TT T T T/DD t =→ + → = − QQ 0 QQ QQ QQ QQ && & & Thus, ()TT T T T T TT TT T T, .=+ − = − =*T W* QTQ QWQ QTQ QQ QTWQ QT Q * W * T = QWQ QTQ + Q Q QTQ QWTQ + Q TQ&& && ()TTT→− − −− **T W * W * T = Q T W W T QQ T QQ T Q && . Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-65 ( ) ()TTT T T T TT.−=+++− −− =+−** *T +T W * W * T QTQ QTQ QTQ Q TW WT Q QT Q Q TQ QTQ Q TW WT Q&& & & && & That is, ( )T−= + − ** *T + T W* W*T Q T T W W T Q&& . Therefore, the Jaumann derivative of T, that is , ( ) +−TT WW T& is objective. ________________________________________________________________________________________________________________________________________________ 5.110 The second Piola Kirchhoff stress tensor T%is related to the first Piola-Kirchhoff stress tensor oTby the formula1 o−=TF T% , or to the Cauchy stress tensor T by 11 T(det ) ( )−−=TF F T F% Show that, in a change of frame, *T= T%% . [See Example 5.56.3 and Example 5.57.1] -------------------------------------------------------------------------------- Ans. In Example 5.56.3 and Example 5.57.1, we obtained that in a change of frame, *( ) t=FQ F and oo*=TQ T . Thus, 11 1 1 1 oo o o*(( ) )t−− − − −== = **T = F T Q F QT F Q QT F T% . That is, *T= T%% . ________________________________________________________________________________________________________________________________________________ 5.111 Starting from the constitutive assumption that ( ) =TH F and ()=**TH F , where Tis Cauchy stress and Fis deformation gradient, show that in order that the assumption be independent of observers, ( ) HF must transform in accordance with the equation T()= QTQ H QF . (b) Choose TQ=R to obtain T()=TR H U R , where Ris the rotation tensor associated with Fand Uis the right stretch tensor. (c) Show that ( ) =Th U% , where 11(det ) ( )−−h= U U H U U . Since 2=CU , therefore, we may write ( ) =Tf C . ------------------------------------------------------------------------------- Ans. (a) In a change of frame, T=*TQ T Q and *=FQ F , therefore, T() ( )=→**TH F Q T Q = H Q F . (b) From T()= QTQ H QF , with TT T() → Q=R R T R=H R F . But TT()= RF = R R U U where Uis the right stretch tensor. Therefore, TT T() ( )→ RT R = H RF T = R H U R . (c) 1T1 T−−→→ F=R U R=F U R =U F , thus, T1 1 T 1 T 1 1 1() () ( ) ()− −− − −−→→ T=R H U R T=F U H U U F F T F =U H U U . Now since det detJ= F= U , we can write 1T 1 1 1() ( d e t ) ( ) J− −− −FT F = U UH U U . The left side of the above equation is the second Piola-Kirchhoff stress tensor T%and the right side is a function of the right stretch tensor U. Thus, ( ) T=h U% , or since 2U= C , one can write () T=h C% . ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-66 5.112 From ()1/22 , = , , =1/rX c Y z Z cαβ θ α=+ = , obtain the right Cauchy-Green deformation tensor B. ------------------------------------------------------------------------------- Ans., we have, with ()1/22 , = , rX c Y z Zαβ θ=+= , () ( ) ()1/2 1/2 122 2 , 02rr rXXXr Y Zααβ α α αβ−− ∂∂ ∂=+ =+ == =∂∂ ∂ 0, , 0; 0, 0, 1zz zcXY Z XY Zθθθ∂∂∂ ∂∂∂=== ===∂∂∂ ∂∂∂ Thus, Using Eq.3.29.59 to 3.29.64, 222 2 rrrrrBX YZ rα ∂∂∂⎛⎞⎛⎞⎛⎞⎛ ⎞=++=⎜⎟⎜⎟⎜⎟⎜ ⎟∂∂∂⎝⎠⎝⎠⎝⎠⎝ ⎠, ()222 2 rrrB rcXYZθθθθθ∂∂∂⎛⎞ ⎛⎞ ⎛⎞=++=⎜⎟ ⎜⎟ ⎜⎟∂∂∂⎝⎠ ⎝⎠ ⎝⎠ 222 1zzzzzBXYZ∂∂∂⎛⎞⎛⎞⎛⎞=++=⎜⎟⎜⎟⎜⎟∂∂∂⎝⎠⎝⎠⎝⎠, 0rrr rr rrBXX YY ZZθθθθ∂∂ ∂∂ ∂∂⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞= ++= ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ 0rzrz rz rzBXX YY ZZ∂∂ ∂∂ ∂∂⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=++=⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠, 0zrz rz rzBXX YY ZZθθθθ∂∂ ∂∂ ∂∂⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞=++=⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ________________________________________________________________________________________________________________________________________________ 5.113 From 2 13 1 3, , , 1 rR K Z zZλθ λλ λ== Θ + = = , obtain the right Cauchy-Green deformation tensor B. -------------------------------------------------------------------------------- Ans. With 2 13 1 3, , , 1 rR K Z zZλθ λλ λ== Θ + = = , we have, 2 13 1 3 1 3, , , 1 , , 0, 0, 0, 1, , 0, 0, .rR K Z zZ rr rKRZ R Z zzz RZλθ λλ λ θθθλ λ== Θ + = = ∂∂ ∂ ∂ ∂ ∂== = = = =∂∂ Θ ∂ ∂ ∂ Θ ∂ ∂∂∂===∂∂ Θ ∂ Using Eq. (3.29.19) to Eq. (3.29.24) and noting that oo o , , rR zZθ≡≡Θ ≡ , 22 2 2 1()rrrrrBRR Zλ∂∂∂⎛⎞⎛ ⎞⎛⎞=+ +=⎜⎟⎜ ⎟⎜⎟∂∂ Θ ∂⎝⎠⎝ ⎠⎝⎠. 0r rrr r r rrB BRR R R ZZθ θθθ θ∂∂ ∂ ∂ ∂∂⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞== =⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟∂∂ ∂ Θ∂ Θ∂∂⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠++ . ()( )()22 2 2 22 2 1rrrrB rK rKRR ZRθθθθθλ∂∂∂⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞=+ += + = +⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟∂∂ Θ ∂⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠. 22 2 2 3()zzzzzBRR Zλ∂∂∂⎛⎞⎛ ⎞⎛⎞=+ +=⎜⎟⎜ ⎟⎜⎟∂∂ Θ ∂⎝⎠⎝ ⎠⎝⎠. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 5-67 0rz zrrz r z rzB BRR R R ZZ∂∂ ∂ ∂ ∂∂⎛⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛⎞=+ = =⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜⎟∂∂ ∂ Θ∂ Θ∂∂⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝⎠+ . 3 z zzr z r zr zrB rK BRR R R ZZ ZZθ θθθ θ θλ∂∂ ∂ ∂ ∂∂ ∂∂⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞ ⎛⎞=+ == =⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟ ⎜⎟∂∂ ∂ Θ∂ Θ∂∂ ∂∂⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ ⎝⎠+ . ________________________________________________________________________________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-1 CHAPTER 6 6.1 In Figure P 6-1, the gate AB is rectangular with width 60bc m= and length L= 4 m. The gate is hinged at the upper edge A. Neglect the weight of the gate, find the reactional force at B. Take the specific weight of water to be 98003/Nm and neglect frictions. hinge30o 4m3m A B sso Figure P 6-1 ------------------------------------------------------------------------------- Ans. Take the gate AB as a free body. With osmeasured from the water surface along the inclined plane to point A, smeasured from point A along the length of the plate (AB) and 30α=oas shown in the figure, we have, () o sin ( ), thus dF pdA g s s bds ρα⎡⎤== +⎣⎦, ()23 oo 00s i n s i n23L AB ALLMR L s d F b g s s s d s b g s ρα ρα⎧ ⎫ ⎪ ⎪=→ = = + = + ⎨ ⎬ ⎪ ⎪ ⎩⎭∑ ∫∫. Therefore, () ( ) ( )()4 o40 . 5 sin sin 30.6 4 9800 5.1 10 .23 2 3Bs LR bL g Nααρ⎧⎫ ⎧⎫ ⎪⎪=+ = + = ×⎨⎬ ⎨ ⎬⎩⎭ ⎪⎪⎩⎭ _________________________________________________________________ 6.2 The gate AB in Figure P 6-2 is 5 m long and 3 m wide. Neglect the weight of the gate, compute the water level h for which the gate will start to fall. Take the specific weight of water to be 98003/Nm. h20,000 N A B5m 60o∇ h20,000 N A B5m 60o∇ FW Figure P 6-2 -------------------------------------------------------------------------------- Ans. Consider the gate plus the triangular re gion of water above the gate as the free body diagram. Then, Horizontal force from water to gate: 2(/ 2 ) ( ) / 2 Fg hb h g b hρρ== acting at 1/3 from base. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-2Weight of water on gate: o2W= (1/ 2)[ ( tan 30 )] / (2 3) gb h h gbhρρ = . 0( 1 / 3 ) ( / 3 )/ 3 ( )BMW h F h P A B=→ + = →∑ () () ( )3 20000 (5) 9( ) 915.31 2.48 .2 2 9800 3PA Bhh mgbρ== = → = __________________________________________________________________ 6.3 The liquids in the U-tube shown in Figure P 6-3 is in equilibrium. Find 2h as a function of 123 1 3, , , and hhρρρ . The liquids are immiscible. ρ3 ρ1 ρ21h 2h3h 12 Figure P 6-3 -------------------------------------------------------------------------------- Ans. 11 1 2 3 32 2 , p gh p gh ghρ ρρ == + , 12 1 13 32 2 2 1 1 3 32 () / p p gh gh gh h h h ρρρ ρ ρ ρ =→ = + →= − . _________________________________________________________________ 6.4 In Figure P 6-4, ,the weightRW is supported by the weightLW, via the liquids in the container. The area under RWis twice that under LW. Find RW in terms of 12,, ,L L WAρρ , and h ( ) 21 and assume no mixingρρ< . h ρρ 12WR WL LA RA1 234 Figure P 6-4 ------------------------------------------------------------------------------- Ans. 321 1p pp g hρ ==+ , () 432 1 1 2p pg h p g hρρ ρ=− =+− () 41 1 2 R RR R W p A p A ghA ρρ ==+ − , i.e., () () 11 2 1 222 2 2R LL LL W p A ghA W ghA ρρ ρρ =+ − = + − _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-36.5 Referring to Figure P 6-5, the radius and length of the cylinder are and rL respectively, The specific weight of the liquid is γ. (a) Find the buoyancy force on the cylinder and (b) Find the resultant force on the cylindrical surfa ce due to the water pressure. The centroid of a semi-circular area is 4/ 3rπfrom the diameter. Figure P 6-5 ------------------------------------------------------------------------------- Ans. (a) Buoyancy force is the net upward force due to the water pressure on the left half of the boundary of the cylinder which is submerged in the water. It is therefore equal to the weight of the water displaced by this left half. That is, Buoyancy force = 2(/ 2 )rLγπ . (b) Horizontal water force: ()()2(2 / 2) 2 2xF rr L r Lγγ== . The line of action of xFis 2/ 3r above the ground. The line of action of yF(the buoyancy force) passes through the centroid of the semi-circular area, i.e., 4 / 3 rπleft of the diameter. _________________________________________________________________ 6.6 A glass of water moves vertically upward with a constant acceleration a. Find the pressure at a point whose depth from the surface of the water is h. Take the atmospheric pressure to beap. ------------------------------------------------------------------------------- Ans. Let z axis be appointing vertically upward, then () ()dp dpg ag a p g a z Cdz dzρρ ρ ρ−−=→ −= +→ = − + + . At the instant of interest, let the origin be at the free surface, then a Cp= , the atmospheric pressure. Thus, () a ppg a zρ−= − + . At a point which is at , zh=− ()a ppg a hρ−= + . __________________________________________________________________ 6.7 A glass of water moves with a constant acceleration ain the direction shown in Figure P 6-6. (a) Show that the free surface is a plane and find its angle of inclination and (b) find the pressure at the point A. Take the atmospheric pressure to beap. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-4gy a Aroθx h Figure P 6-6 ------------------------------------------------------------------------------- Ans. (a) With respect to the coordinate s shown, the governing equations are: (i) cos , (ii) sin , (iii) 0pp pag axy zρθ ρ ρθ∂∂ ∂−= −−= −=∂∂ ∂, thus () (iii) ( , ), (i) cos ( )pdfpp x y p a xf yyd yρθ∂→= →= − + → =∂. ()() ( ) () (ii) sin sin cos sindfgaf g a y C p a x g a y Cdyρθρθ ρ θ ρθ→ = −+ → = −+ + → = − −+ + . At the instant of interest, let the origin be at the center of the surface, then a Cp= .and ()() cos sina p ax g ay pρθρ θ=− − + + . On every point on the free surface, a pp= , therefore, ()() cos sin 0ax g ayρθρ θ−− + = . Thus, the free surface is a plane. The angle of inclinations is given by costansindy a dx g aθβθ== −+. (b) At the point A, o, x ry h=−= − . Thus, ()() o cos sin a p arg a h pρθ ρ θ=+ + + _________________________________________________________________ 6.8 The slender U-tube shown in Figure P 6-7 moves horizontally to the right with an acceleration a. Determine the relation between , and ahl . h za Figure P 6-7 ------------------------------------------------------------------------ Ans. The slope of the free surface is given by a g−. Thus ah ahg g−= −→=l l. _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-56.9 A liquid in a container rotates w ith a constant angular velocity ω about a vertical axis. Show that the free surface is a paraboloid given by 22/( 2 ) zr gω= where the origin is on the axis of rotation and z is measured upward from the lowest point of the free surface. ------------------------------------------------------------------------------- Ans. Let z be pointing vertically upward with the origin at the lowest point of the free surface. We have, ()2(i) and (ii) 0pprgrzρω ρ∂∂−=− −−=∂∂ ()pdfpg z f rrd rρ∂→= − + → =∂. 22 22 2(i) , 22pd f r rrf C p g z Crd rρω ρωρω ρ∂→== → = +→ = −+ +∂. At ( , ) (0,0)rz= , a pp= , therefore, 22 2arp gz pρωρ=−+ + . The free surface is characterized bya pp= , therefore, the equation of the surface is: 22 2rzgω= . _________________________________________________________________ 6.10 The slender U-tube rotates with an angular velocity ω about the vertical axis shown in Figure P 6-8. Find the relation between 12 1 2() , , a n d hhh r rδ ω≡− . r1r2ω o1h 2h Figure P 6-8 ------------------------------------------------------------------------------- Ans. The equation for the free surface is given by (see the previous problem) 22/( 2 ) zr gω= , where the origin is on the axis of rotation and z is measured upward from the lowest point of the free surface. Thus, we have, ()22 22 2 22 12 12 1 2 1 2 and -22 2rrzz z z r rg ggωω ω== → − = , 1212 but,zz hh−=− 22 2 12 1 2 (-) / ( 2 ) hh r r g ω →−= . ____________________________________________________________ 6.11 For minor altitude differences, the atmosphe re can be assumed to have constant temperature. Find the pressure and density distribution for this case. The pressure p, density ρ and absolute temperature Θ are related by the ideal gas law p Rρ=Θ. -------------------------------------------------------------------------------- Ans.: Let gravity be in the negative 3xdirection, then we have 123/ 0 , / 0 , /pxp x p x g ρ ∂∂= ∂∂= ∂∂= − (i) Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-6Thus, pdepends only on3x. Let opdenote the pressure at 0x=, then, we have () 3 / 33 o o 3ln lngRx dp dp g ggd x p x p p p edx p R Rρ−Θ=− → =− → =− + → =ΘΘ (ii) If oρis the density at 30x=, then ()3 / ogRxeρρ−Θ= . ____________________________________________________________ 6.12 In astrophysical applications, an atmosp here having the relation between the density ρ and the pressure pgiven by () oo//nppρρ= , where opand oρare some reference pressure and density, is known as a polytropic atmosphere. Find the distribution of pressure and density in a polytropic atmosphere. ------------------------------------------------------------------------ Ans. Let z axis point upward, then / dp dz g ρ=− . From () oo//nppρρ= , we have, 1/ 1/ oo , where nnCp C pρρ−== . Thus, 1/ 1//nndp dz Cp g p dp Cgdz−=− → =− oo1/pzn pzp dp Cgdz−→= −∫∫. Thus, (A) for 1n≠, () () ()() () ()() () ()oo1/ 1/ 1/ oo 1/ 1/ 1/ 1/ oo o o o o o o[ / ( 1)] [( 1) / ] [( 1) / ] {( 1) / }pz nn nn nn zp nn nn nnnn p C g z p p n n C g zz p pn n p g z z p p n n g z z ρρ−− − −− −−−= − → − = − − − → ⎡ ⎤ =− − − = − − −⎣ ⎦ ()()() /1 1/ 1 oo o o1nn n npp p g z znρ− −− −⎡⎤=− −⎢⎥⎣⎦. (B) for 1 n=, () () ()oooo o o 1/ oo o oln( / ) exp exppz pz ndpCgdz p p Cg z z p p Cg z zp pp p g z z ρ−⎡ ⎤ = − → = −− → = −−⎣ ⎦ ⎡⎤ →= − −⎣⎦∫∫ __________________________________________________________________ 6.13 Given the following velocity field for a Newtonian liquid with viscosity μ=0.982 .mPa s ()522.05 10 lb×s/ft−× : () ()1 11 2 2 2 1 3 , , 0 , 1 vc x xv c x x v c s−=− + = − = = For a plane whose normal is in the 1e direction, (a) find the excess of the total normal compressive stress over the pressure p, and (b) find the magnitude of the shearing stress. ------------------------------------------------------------------------------ Ans. (a) () 11 11 11 11 22 Tp D T p D μμ=− + → − − =− , where 1 1 11 11 svDcx− ∂== − = −∂, Thus, ()()() 11 20 . 9 8 2 1 1 . 9 6 . Tp m P a−− = − − = (b) 12 12 12 212 2 1.96 .vvTD c m P axxμμ μ⎛⎞∂∂==+ = − = − ⎜⎟∂∂⎝⎠ 3 1 13 13 3120v vTDxxμμ⎛⎞∂∂==+ =⎜⎟∂∂⎝⎠. Thus, the magnitude of shearing stress = 1.96 mPa . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-7__________________________________________________________________ 6.14 For a steady parallel flow of an incompressible linearly viscous fluid, if we take the flow direction to be 3e, (a) show that the velocity field is of the form 120, 0vv== and 31 2 (, ) vv x x= (b) If 12 2(, )vx x k x= , find the normal and shear stresses on the plane whose normal is in the direction of 23+ee in terms of viscosity μand pressure p. (c) On what planes are the total normal stresses given by p. ------------------------------------------------------------------------------- Ans. (a) From the equation of continuity 3 12 1230v vv xx x∂ ∂∂++=∂∂∂, we get, 3 30v x∂=∂, thus 3vis independent of 3xi.e., 31 2 (, ) vv x x= . (b) with120, 0vv== and 32vk x= , we have, [] [ ] [ ] [ ] [ ]000 0 0 0 0 0 000 0 0 / 2 2 0 00 0 / 2 0 0p kp p k kk k pμ μ μ− ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥∇= → = → = − + = −⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦vD T I D On the plane with23() / 2+ n= e e , [] [ ] []00 0 0 1101 2201np p kk p T k p kp k pμμ μ μμ−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− = − → = ⋅ −⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ −− ⎣⎦ ⎣ ⎦ ⎣ ⎦t=T n tn = . ()() () ()2 2 22 2 2 102snT T kp kp kp μμ μ⎡⎤ =− = − + − − − =⎢⎥⎣⎦t . (c) []11 22 3 32 300 0 0pn p n p kn p n k n kp n k n p nμμ μμ−− ⎡⎤⎡ ⎤ ⎡⎤ ⎢⎥⎢ ⎥ ⎢⎥−= − +⎢⎥⎢ ⎥ ⎢⎥ ⎢⎥⎢ ⎥ ⎢⎥ −− ⎣⎦ ⎣⎦⎣ ⎦t= , 222 12 3 1 nnn++= () ( )22 2 1 2 32 23 3 32 2nTp n p n k n n k n n p n p k n n μμ μ → = ⋅ − + − ++− = − + tn = . Thus, 32 2 3 2 0 and/or 0 pk n np n nμ−+ = −→ = = That is, on any plane ()() 13 1 2,0, and , ,0nn n n , where 222 12 3 1 nnn++= , the normal component of stress is p−, these include the three coordinate planes ()()() 1,0,0 , 0,1,0 and 0,0,1 . __________________________________________________________________ 6.15 Given the following velocity field for a Newt onian incompressible fluid with a viscosity 0.96 . mPa sμ= : ()22 1 1 11 2 2 1 2 3 , 2 , 0, 1 vk x x v k x x v k s m−−=− = − == . At the point (1,2,1) m and on the plane whose normal is in the direction of 1e, (a) find the excess of the total normal compressive stress over the pressure p and (b) find the magnitude of the shearing stress. ------------------------------------------------------------------------------ Ans. (a) Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-8[] [ ][ ]12 1 2 21 2 122 0 4 4 0 22 0, 4 4 0 . 00 0 0 0kx kx p kx kx kx kx kx p kx pμμ μμ−− + −⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥∇= − − = = − −−⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦vD T At []48 0 (1,2,1) and for 1, 8 4 0 00p kp pμμ μμ−+ −⎡⎤ ⎢⎥== − − − −⎢⎥ ⎢⎥ − ⎣⎦T . On 1e-plane ()11 43 . 8 4 . Tp m P a μ −− = − = − (b) on the same plane, 87 . 6 8 .sTm P aμ== _________________________________________________________________ 6.16 Do Problem 6.15 except that the plane has a normal in the direction 1234+ee . ------------------------------------------------------------------------------- Ans. [] [ ][ ]12 1 2 21 2 122 0 4 4 0 22 0 4 4 0 00 0 0 0kx kx p kx kx kx kx kx p kx pμμ μμ−− + −⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥∇= − − = → = − −−⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦vD T . At []48 0 (1,2,1) and for 1, 8 4 0 00p kp pμμ μμ−+ −⎡⎤ ⎢⎥== − − − −⎢⎥ ⎢⎥ − ⎣⎦T , [] [ ] []48 0 3 / 5 3 2 0 184 0 4 / 5 4 4 0500 0 0pp pp pμμμ μ μμ−+ − − −⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥=− − − =−−⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥ − ⎣⎦ ⎣ ⎦ ⎣ ⎦t=T n . () ()14 432 0 3 44 0 425 5nTp p pμμμ ⎡⎤ =⋅= − − +− − = −−⎣⎦tn . 44()5nTpμ−− = . (b) () () ( )22 2 22 113 20 4 40 25 440 200025 25pp p pμ μμ μ ⎡⎤=− − + − − = + +⎢⎥⎣⎦t ()2 2288 1936 52 5npTpμμ =+ + . () ()22 2 2 64 8 25 5sn sTT Tμμ =− = → =t . __________________________________________________________________ 6.17 Use the results of Sect. 2.34., chapter 2 and the constitutive equations for the Newtonian viscous fluid, verify the Navier Stokes Equation in the r-direction for cylindrical coordinates, i.e., Eq. (6.8.1). ------------------------------------------------------------------------------- Ans. For a Newtonian fluid, the stress tensor in cylindrical coordinates is given by: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-9[] 21 31 32v 12 112 2rr r z rz zv vv v vprr r r z r vv vvTprr z r vTT pzθθ θθμμ μθ μμθ θ μ⎡⎤ ∂ ∂∂ ∂ ∂⎛⎞ ⎛⎞−+ − + +⎢⎥ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎢⎥ ⎢⎥ ∂∂ ∂ ⎛⎞ ⎛ ⎞=− + + +⎢⎥ ⎜⎟ ⎜ ⎟∂∂ ∂⎝⎠ ⎝ ⎠ ⎢⎥ ⎢⎥ ∂−+ ⎢⎥∂ ⎢⎥⎣⎦T The Equations of motion in terms of the stress comp onents in the r-direction is [see Eq.(4.8.1)]: 1rr r rr rz rrTT T TTB arr r zθθ θρρθ∂−∂∂++ + + =∂∂ ∂ We also have the equation of continuity [see Eq.(2.34.6) or Eq.(6.8.4)]: 10rr z v vv v rr r zθ θ∂∂∂⎛⎞++ + =⎜⎟∂∂ ∂⎝⎠ Now, 2 222rr r r rrvT v pTprrr rμμ∂∂ ∂ ∂=− + → =− +∂∂∂ ∂ 2 2 22 2vv 11 1 1 1r rr rvT v vvTrr r r r r rrθθθ θ θ θμμθ θθ θ θ⎛⎞ ∂∂ ∂ ∂ ∂∂⎛⎞=− + → = − + ⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ 22112rr rr TT v vv rr r rrθθ θμθ−∂ ∂⎛⎞=− −⎜⎟∂∂⎝⎠ 22 2rz r z r z rzvv T v vTzr z r z zμμ⎛⎞ ∂∂ ∂ ∂ ∂⎛⎞=+ → = + ⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ Thus, 2 2 2 2 22 22 2 2 2 21(div ) = 2 v 11 1 1 12rr r rr rz r r rr r r zTT T TT v p rr r z r r vv vv v v v rr r r r z rr r r zθθ θ θθ θμθ μμ μθθ θ θ∂−∂∂ ∂ ∂++ + = − +∂∂ ∂ ∂ ∂ ⎛⎞ ⎛⎞ ∂∂ ∂ ∂∂ ∂ ∂ ⎛⎞+− + + − − + +⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ⎝⎠ ⎝⎠ ⎝⎠T 22 2 22 2 2 2 2 2 22 22 211 2 v11 1rr r r r rr r zv vv v v v p rr r rr z r r v vv v v rr r r r z rr rθ θθμθ θ μθθ⎡⎤ ∂ ∂∂ ∂ ∂∂=− + + + + − − ⎢⎥∂∂ ∂∂∂ ∂⎢⎥⎣⎦ ⎛⎞ ∂∂ ∂∂ ∂+−+− + +⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂⎝⎠ But using the equation of continuity, we have, 2 22 22 2vv11 1 10rr r z r r z v vv v v v v v rr r r r z r r r r z rr rθθ θ θθ θ⎛⎞ ∂∂ ∂ ∂∂ ∂ ∂ ∂ ∂⎛⎞−+− + + = + + + =⎜⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ⎝⎠ ⎝⎠ Thus, 22 2 22 2 2 2 2(div ) + = 11 2 =rr r rr r r r rrBa v vv v v v pB arr r rr z r rθρρ μ ρρθ θ→ ⎡⎤ ∂ ∂∂ ∂ ∂∂−+ + + + − − +⎢⎥∂∂ ∂∂∂ ∂⎢⎥⎣⎦T __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-106.18 Use the results of Sect. 2.35, chapter 2 and the constitutive equations for the Newtonian viscous fluid, verify Navier-Stokes Equation in th e r-direction in spherical coordinates, i.e., Eqs. (6.8.5). -------------------------------------------------------------------------------- Ans. For a Newtonian fluid, the stress tensor in spherical coordinates is given by: []112sin cot 11 12sin cot 12sinrr r r r rvv vv vv vprrr r r r r vv vv vTprr r r r v v vTT prr rφφ θθ φφ θθ θφ φ θ φφ θμμ μθθ φ θμμθθ φ θ θμθφ⎡⎤ ⎡⎤ ∂ ⎛⎞ ⎡⎤ ∂ ∂∂ ∂⎛⎞−+ − + − +⎢⎥ ⎢⎥⎜⎟ ⎢⎥⎜⎟∂∂ ∂ ∂∂⎝⎠ ⎢⎢ ⎥ ⎥ ⎣⎦ ⎝⎠⎣⎦⎢⎡ ⎤∂ ⎛⎞ ∂∂⎛⎞ ⎢=− + + − + ⎢ ⎥ ⎜⎟ ⎜⎟ ⎢∂∂ ∂⎝⎠ ⎢ ⎥ ⎝⎠⎣ ⎦ ⎢ ⎢∂⎛⎞⎢ −+ + +⎜⎟∂ ⎢ ⎝⎠ ⎣⎦T⎥ ⎥ ⎥ ⎥ ⎥⎥ ⎥ The equation of motion in the r-direction is given by [see Eq.(4.8.4)] () ()2 2sin 11 1- sin sinrr r r rrrT TTT TB arr r r rφθ θφ φ θθρρθθ θ φ∂ ∂+ ∂++ + =∂∂ ∂. We also have the equation of contin uity [see Eq.(2.35.26) or Eq.(6.8.8)] cot 2 110sinrrv vv vv rr r r rφ θθ θ θθ φ∂⎛⎞ ∂∂⎛⎞++ + + = ⎜⎟ ⎜⎟∂∂ ∂⎝⎠ ⎝⎠. Now, ()22 2212 22rrrrrTvv pp rr r r r rrμ∂ ⎡⎤∂∂ ∂=− − + + ⎢⎥∂∂ ∂ ∂⎢⎥⎣⎦, () 2 2 22 2 2 2sin 1 sin 11 1 1 1cot .r rrT r vv v v vv rr r r rr r rθ θθ θ θθ θθ μθ μθ θθ θ∂ ∂ ⎡ ⎤ ⎛⎞ ⎡⎤ ∂∂ ∂ ∂∂⎛⎞=− + + − + ⎢ ⎥ ⎜⎟ ⎢⎥⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎢ ⎥ ⎣⎦ ⎝⎠⎣ ⎦ 2 2 22 2 211 1 1 sin sin sin sinr rTv v v rr r rrφ φφμθφφ θ φ θφ θ⎡⎤⎛⎞∂∂ ∂ ∂⎢⎥=− +⎜⎟⎜⎟∂∂ ∂ ∂ ∂ ⎢⎥⎝⎠⎣⎦ 22 2 2cot 2 21 122 sinrTT v vv v p rr rr r rθθ φφ φ θθ θμμθφ θ+∂ ⎛⎞ ∂⎛⎞−= + − + − + ⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠. Thus. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-11() ()2 2 2 22 2 2 22 22 2 2 2 2 2sin 11 1 sin sin 22 1 12c o t 11 1 1 1 sin sinrr r r rr r rrrT TT T T rr r r r vv vv v pp rr r r r r rr r vv vv rr rr r rφθ θ φ φ θ θθ θθθ θθ θ φ μμ θθ μμθθ θθ φ θ∂ ∂+ ∂++ − =∂∂ ∂ ⎡⎤ ⎡⎤ ∂ ∂∂ ∂ ∂ ⎛⎞−−+ + + − + ⎢⎥ ⎢⎥⎜⎟∂∂ ∂ ∂ ∂ ⎝⎠ ⎢⎥ ⎣⎦ ⎣⎦ ⎡⎤⎛⎞ ∂ ∂∂ ∂∂+− + + −⎢⎥⎜⎟⎜⎟ ∂∂ ∂ ∂∂⎢⎥⎝⎠⎣⎦2 22 2 21 sin cot 2 21 122 sinrvv rr v vv v p r rr r rφ φ φ θθφ θφ θμμθφ θ⎡ ⎤ ⎛⎞ ∂⎢ ⎥ + ⎜⎟⎜⎟ ∂ ∂∂ ⎢ ⎥ ⎝⎠⎣ ⎦ ∂⎛⎞ ∂⎛⎞+− + − + ⎜⎟ ⎜⎟∂∂⎝⎠ ⎝⎠ 22 2 22 2 2 22 2 2 22 2 2 2 22 2 2 22 21 1 1cot sin 2c o t 22 sin 2 21 1 1 1 1cot cotsin sinrr r r r r rr rvv v v v v p rr r rr r rr v vv rr r v vv v v vv v rr r r rr r rr r r rφ θθ φ θθ θ θμθθ θθ φ θμφθθ μθ θθθ φ θ⎛⎞∂∂ ∂ ∂ ∂∂=− + + − + + + ⎜⎟⎜⎟∂∂ ∂∂∂ ∂⎝⎠ ∂ ⎛⎞ ∂+− − −⎜⎟∂∂ ⎝⎠ ∂ ∂∂ ∂ ∂∂++ − +−+ −− +∂∂ ∂ ∂ ∂ ∂∂2v rφ θφ⎡ ⎤∂⎢ ⎥∂∂ ⎢ ⎥⎣ ⎦ Now, differentiate the equation of cont inuity with respect to r, we have, cot 2 110sinrrv vv vv rr r r r rφ θθ θ θθ φ∂ ⎛⎞ ∂ ∂∂++ + + =⎜⎟∂∂ ∂ ∂⎝⎠, that is, 2 2 2 22 2 2 22 21 1 1 1 sin sin cot cot0rr rvv vv vv v rr r r r r rr r r vv rr rφφ θθ θθθθθ φ φ θ θ θ∂∂ ∂∂ ∂∂+− + − + − +∂∂ ∂ ∂ ∂ ∂∂∂ ∂−=∂ Thus, () ()2 2sin 11 1 sin sinrr r rrT TT T T rr r r rφθθ φφ θθ θθ θ φ∂ ∂+ ∂++ −∂∂ ∂ 22 2 22 2 2 22 2 2 22 22 21 1 1cot sin 2c o t 22 sinrr r r r rvv v v v v p rr r rr r rr v vv rr rφ θθμθθ θ θφ θμφθθ⎛⎞∂∂ ∂ ∂ ∂∂=− + + − + + + ⎜⎟⎜⎟∂∂ ∂∂∂ ∂⎝⎠ ∂ ⎛⎞ ∂+− − −⎜⎟∂∂ ⎝⎠ () ()2 2 22 2 2 2 2211 1sin sin sin 22sin sin sinrr rvv prvrr r rr r vv rrφ θμθθθθ θφ μθφθθθ⎡⎤ ∂∂ ∂∂ ∂ ∂ ⎛⎞=− + + + ⎢⎥ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎢⎥⎣⎦ ∂ ⎡⎤ ∂+− −⎢⎥∂∂ ⎣⎦ Finally, the Navier-Stokes equation in r direction is: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-12() ()2 2 22 2 2 2 2211 1sin sin sin 22sin sin sinrr r rrvv prvrr r rr r vvB a rrφ θμθθθθ θφ μθ ρ ρφθθθ⎡⎤ ∂∂ ∂∂ ∂ ∂ ⎛⎞=− + + + ⎢⎥ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎢⎥⎣⎦ ∂ ⎡⎤ ∂+− − + =⎢⎥∂∂ ⎣⎦ __________________________________________________________________ 6.19 Show that for a steady flow, the streamline containing a point Pcoincides with the pathline for a particle which passes through the point P at some time t. ------------------------------------------------------------------------------- Ans. For a steady flow, the velocity at every poin t on a streamline does not change with time. Therefore, any particle, which is at a point P on the streamline at a given time t, will move along the streamline at all time. That is, its pathline coincides with the streamline containing the point P. We can also demonstrate this mathematically as follows: For steady flow, the velocity field is independent of time, that is, ( v=vx ) . Let ( ) t x=x be the pathline, then, the differential system for the pathline is: (){} ()o , subjected to the condition dttdt==oxvx x x (1) Let ( ) s x=x be the parametric equation fo r the streamline passing thoughox, then the differential system for the streamline is: (){} ()o , subjected to the condition dssds==oxvx x x (2) The two differential systems are identi cal. They determine the same curve. _________________________________________________________________ 6.20 Given the two dimensional velocity field 12 12 2, 01kx xvvkx t= =+ (a) Find the streamline at time t, which passes through the spatial point () 12,αα and, (b) find the pathline equation ( ) t x=x for a particle which is at () 12,XX at time ot ------------------------------------------------------------------------------- Ans. (a) Since the flow is in 1edirection only, therefore, both the streamline and the pathline are straight line in the 1e direction. The streamline equation wh ich passes through the spatial point () 12,αα is simply22xα= . (b) The pathline for a particle which is at () 12,XX at time otis simply22x X= . To find the time history of the particle along the pathline, i.e., to find ( ) t x=x with o()t X=x ,we have, 1 1o11 2 1 2 1 2 21 2 1 2 o1ln ln11 1x t Xtdx kx X dx kX x kX tdtdt kX t x kX t X kX t+=→ = → =++ +∫∫, 2 11 2 2 2o1 and 1kX tx Xx XkX t+→= =+. _________________________________________________________________ 6.21 Given the two dimensional flow: 12 2 , 0 vk xv== Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-13(a) Obtain the streamline passing through the point () 12,αα. (b) Obtain the pathline for the particle which is at () 12,XXat 0t=, including the time history of the particle along the pathline ------------------------------------------------------------------------------ Ans. (a) The streamline is clearly22xα= . (b) The pathline for the particle which is at () 12,XX at time 0 is simply22x X= . To find the time history of the particle along the pathline, i.e., to find ( ) t x=x with (0) X=x , we have, 21 22 2 1 1 2 0, , 0dx dxx X kX x X kX t tdt dt=→ = → = → = + ≤≤ ∞ . _________________________________________________________________ 6.22 Do Prob. 6.21 for the following velocity field : 12 2 1 , vx v xω ω = =− . ------------------------------------------------------------------------------- Ans. (a) From 12 1 2 21 1 1 2 2 21, 0dx dx dx xxx x d x x d xds ds dx xωω== − → = − → + = , 22 22 2 2 12 121 2xxCxx αα →+=→+=+ . The streamline is a circle. (b) Since the flow is steady, clearly, th e pathline is also a circle given by 22 2 2 12 1 2x xXX+= + . To find the time history of the particle along the pathline, i.e., to find ()t x=x with (0) X=x , we have, 22 22 12 1 2 1 21 1 1 22, 0dx dx d x dx d xxx x xdt dt dt dt dtωω ω ω ω== − → = = − → + = , 1 121sin cos , cos sindxxAt B t x A t Btdtω ωω ωω→= + →= = − . 11 2 2 12 1 2 2 1 at 0, , , thus, sin cos , cos sint x X x X x Xt Xt x X t X t ω ωω ω == = = + → = − _________________________________________________________________ 6.23 Given the following velocity field in polar coordinates (),rθ: ,02rQvvrθπ= =. (a) Obtain the streamline passing through the point () oo,rθ, (b) Obtain the pathline for the particle which is at (),RΘ at 0t=, including the time history of the particle along the pathline ------------------------------------------------------------------------------- Ans. Both the streamline and the path lines are radial lines with θ=constant. (a) the streamline passing through the point oo, rrθθ== is oθθ=. (b) the pathline for the particle which is at (),RΘ at 0t= is θ=Θ. To find the time history of the particle, we have, 22 00,22rt Rdd r Q Q Qrdr dt r R tdt dt rθθπ ππ=→= Θ = → = → = + ∫∫. _________________________________________________________________ 6.24 Do Prob. 6.23 for the following velocity field in polar coordinates (),rθ: 0, /rvv C rθ== . ------------------------------------------------------------------------ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-14Ans. Both the streamline and the path lines are circles constant r= . (a) the streamline passing through the point o ,rrθθ== is orr= (b) the pathline for the particle which is at (),RΘ at 0t=is rR=. To find the time history of the particle, we have, 220dr d C d C CrR R tdt dt R dt R Rθθθ =→=→ = → = →= + Θ . _________________________________________________________________ 6.25 From the Navier-stokes equations, obtain Eq. (6 .11.2) for the velocity distribution of the plane Couette flow. ------------------------------------------------------------------------------- Ans. With 2xaxis pointing vertically upward, we have () 12 2 3 1 2 3 , 0, 0 and 0 vv xv v aa a== = = = = , thus, with ()2 ppx= , the Navier-Stoke’s equation in the 1xdirection become 2 12 2 2 2 2 20 . At 0, 0, 0. dvvC x C x v C dxμ=→ = + = = → =oo 2o 1 2At , .vvx dv v C v xdd== → = → = _________________________________________________________________ 6.26 For the plane Couette flow, if in addition to the movement of the upper plate, there is also an applied negative pressure gradient 1/px∂∂, obtain the velocity distribution. Also obtain the volume flow rate per unit width. ------------------------------------------------------------------------------- Ans. With 2xaxis pointing vertically upward, we have () 12 2 3 1 2 3 , 0, 0 and 0 vv xv v aa a== = = = = , thus, the Navier-Stoke’s Equations become, 2 2 11 1 200pd v p xx x dxμ⎛⎞ ∂∂ ∂=− + → = ⎜⎟∂∂ ∂ ⎝⎠ 21 2 2 1 31 3 3 10 0 0 0pp p xx x x x pp p xx x x x⎛⎞ ⎛⎞ ∂∂ ∂∂ ∂=− → = = ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎛⎞ ⎛⎞ ∂∂ ∂ ∂ ∂=− → = = ⎜⎟ ⎜⎟∂∂ ∂ ∂ ∂ ⎝⎠ ⎝⎠ Thus 1constantp xα∂=≡ −∂ 2 2 2 12 2 2 22x dvvC x C dxαα μμ⎛⎞=− → =− + + ⎜⎟⎝⎠, 22 2 oAt 0, 0 0. At ,xv C x d v v==→ = = = → 2 o o1 122v ddvC d Cdαα μμ⎛⎞ ⎛⎞=− + → = + →⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ()2 2 oo 2 22 2 222 2vv x dvx x d x xddαα α μμ μ⎧⎫⎛⎞ ⎛⎞ ⎛ ⎞⎪⎪=− + + = − + ⎨⎬⎜⎟ ⎜⎟ ⎜ ⎟⎪⎪⎝⎠ ⎝⎠ ⎝ ⎠⎩⎭. The volume flow rate per unit width is given by Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-15()2 o 2 22 2 2 00 32 3 2 oo22 62 2 1 2 2ddv xdQv x d x x d xd vv dd d d d ddαα μμ αα α μμ μ⎡⎤ ⎧⎫⎛⎞ ⎛⎞== − + + ⎢⎥ ⎨⎬⎜⎟ ⎜⎟⎢⎥⎝⎠ ⎝⎠⎩⎭ ⎣⎦ ⎧⎫ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎧⎫=− + + = + ⎨⎬ ⎨ ⎬ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎩⎭ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎩⎭∫∫ _________________________________________________________________ 6.27 Obtain the steady uni-directional flow of an incompressible viscous fluid layer of uniform depth dflowing down an inclined plane which makes an angle θ with the horizontal. ------------------------------------------------------------------------------- Ans. With 2x axis normal to the flow and pointing away from the fluid and 1x axis in the flow direction, we are looking for the velocity field in the following form: () 12 2 3 ,0 ,0 vv xv v= == , which clearly satisfies the continuity equation. Now the N-S equations give 2 2 11 1 20s i n 0pd v pgxx x dxρθ μ⎛⎞ ∂∂ ∂=− + + → = ⎜⎟∂∂ ∂ ⎝⎠. 21 2 2 10c o s 0pp pgxx x x xρθ∂∂ ∂ ∂ ∂=− − → = =∂∂ ∂ ∂ ∂ 31 3 3 100 0pp p xx x x x∂∂ ∂ ∂ ∂=− = → = =∂∂ ∂ ∂ ∂. Thus 1pCx∂=∂. The constant Ccan be determined from the pressure condition on the free surface (2x d=), where pressure a pp= , the atmospheric pressure which is independent of1x, thus 1/0 0px C∂∂= →= so that 1/0px∂∂= for the whole flow field. Thus, 22 21 22 2 220s i n s i n s i ndv dv d vg gg x Cdx dx dxρθ μ μ ρθ μ ρθ=+ → = −→ = − + 2 2 12 2 sin2xvg C x Cμρθ→= − + + . At 220, 0 (non slip condition) 0 xv C==→ = . At 21 2 2 1 , shear stress 0 / =0 C = gdsin xd T d v d x μ ρθ == → → . 2 2 gsin d2xvxμρθ⎛⎞→= − ⎜⎟⎝⎠. __________________________________________________________________ 6.28 A layer of water (362.4 /gl b f tρ= ) flows down an inclined plane (o30θ= ) with a uniform thickness of 0.1 ft. Assuming the flow to be laminar, what is the pressure at any point on the inclined plane. Take the atmospheric pressure to be zero. ------------------------------------------------------------------------------ Ans. With flow in the 1xdirection, the N-S equation in the 2xdirection (pointing away from the inclined plane) gives, [note: pis independent of 13and x x, see Prob. 6.27] () 22/c o s 0 c o spxg p g x Cρθ ρθ −∂ ∂ − = → =− + . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-16() ()() 2 2 At , 0 cos , cosa x dp p g d C p g d x ρθ ρθ == = → = → = − . At () ( ) () ()o2 20 cos 62.4cos30 0.1 5.40 / x pg d l b f tρθ =→= = = . We can also obtain the same result by us ing the fact that the piezometric head /( )p gzρ+=constant for any points on the same plane perpendicular to the direction of the flow (see example 6.7.2), therefore () cosab ab b a bppzz p g z z g dggρ ρθρρ+= +→= − = . __________________________________________________________________ 6.29 Two layers of liquids with viscosities12and μμ, densities 12and ρρrespectively, and with equal depths b, flow steadily between two fixed horizon tal parallel plates. Find the velocity distribution for this steady uni-directional flow. Neglect body forces. ------------------------------------------------------------------------------- Ans. We are looking for velocity fields in the tw o layers in the following form corresponding to the uni-directional steady laminar flows: For the top layer: ()()()()() 2 12 3 , 0tt t tvv xvv== = . For the bottom layer: ()()()()() 2 12 3 , 0bb b bvv x vv== = . From the N-S equations for the top layer, we have () () () () () () () () ()22 11 2 1 1 21 22 1 31 33 10/ / ( / ) ( / ) 0 0 / ( / )( / ) ( / )( / ) 00 / (/ ) ( / ) (/ ) ( / ) 0tt t ttt tttpxd v d x x px px x px x pxpx x px x pxμ =−∂ ∂ + → ∂ ∂ ∂ ∂ ==−∂ ∂ → ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ==−∂ ∂ → ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ = Thus, () 11/ (a constant)tpx α ∂∂ = − . Now, () () () 22 2 1 2 1 1 2 1 21 1 1 2 1 21 // ( / 2 ) .tt td v dx dv dx x A v x A x Bμα μ α μ α =− → =− + → =− + + Similarly from the N-S equations for the bottom layer, we have, () 12/ (a constant)bpx α ∂∂ = − . () () 2 2 2 22 2 2 22 22 2 /( / 2 )bbdv dx x A v x A x Bμα μ α =− + → =− + + . The constants 112 2,, ,ABA B will be determined from the bounda ry and the interface conditions: At 2xb= (the top plate), ()0tv=, 2 11 1 0( / 2 ) bA b Bα=−+ + (1) At 2x b=− (the bottom plate), ()0bv=, 2 22 2 0( / 2 ) bA b Bα=−− + (2) At 20x= (the interface), there is no slip be tween the two layers of flow, i.e., () ()tbvv=→ 11 2 2//BBμμ= (3) Also, according to Newton’s 3rd law, the action and reaction at the interface between the two fluid must be equal and opposite, that is, both the shear stress and the normal stress must by continuous at 20x=. Since () ()()() ()() 22 2212 1 22 2 12 1200 00/, /tb tt xx xxT d vd x AT d vd x A μμ == ==== == Therefore, 12AA= , (4) and () () () () () () 222 211 22 22 22 22000 0(, 0 ) , (, 0 ) , tbt b tb xxx xTp x Tp x T T === ==− =− = → () () () () 1 1 11 11 1 2 1(, 0 ) (, 0 ) ( / ) (, 0 ) ( / ) (, 0 ) / .tb t bpxp x p x x p x x p x αα =→ ∂ ∂ = ∂ ∂ → = ≡ ∂ ∂ Now, Eqs.(1)(2)(3)(4) determine the four constants 112 2,, ,ABA B as a function of 1/px α=−∂ ∂ : Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-17()22 21 1 2 12 1 2 12 12 12, ,2AA b Bb Bbμμ μ μαα αμμμ μ μ μ⎧⎫ ⎛⎞ ⎛⎞ −⎪⎪== = =⎨⎬ ⎜⎟ ⎜⎟++ +⎪⎪ ⎝⎠ ⎝⎠ ⎩⎭. Thus, () ()2 2 22 1 1 12 12 12 2 2 22 1 2 22 12 12,22 .22t bx bvx b x bvx bμμ μμαμμ μμ μμ μμαμμ μμ⎡⎤⎛⎞ ⎛⎞−=− − −⎢⎥⎜⎟ ⎜⎟++ ⎢⎥⎝⎠ ⎝⎠⎣⎦ ⎡⎤⎛⎞ ⎛⎞−=− − −⎢⎥⎜⎟ ⎜⎟++ ⎢⎥⎝⎠ ⎝⎠⎣⎦ __________________________________________________________________ 6.30 For the Couette flow of Section 6.15, (a) obtain the shear stress at any point inside the fluid (b) obtain the shear stress on the outer and inner cylinder (c) obtain the torque which must be applied to the cylinders to maintain the flow. ------------------------------------------------------------------------------- Ans. (a) Eq.(6.15.4) and (6.15.7), give ()22 2 2 12 1 2 2 1 /, w h e r e / ( ) vA r B r B r r rrθ=+ = Ω − Ω − . Thus, ()22 12 1 2 22 2 2 212 212rr rrr vdBTT D rdr r rr r rθ θθ θμμμμΩ−Ω ⎛⎞== = = − = −⎜⎟− ⎝⎠ (b) On the outer wall:2rr=, the shear stress is 22 2 12 1 212( ) / ( )rrTT r r rθθ μ== Ω − Ω − , On the inner wall, 1rr= , the shear stress is ()22 2 22 1 212/ ( )rrTT r r rθθ μ== Ω − Ω − . (c) On the outer wall, per unit height, the torque is given by ()()()()()() 2 222 2 121 1 221 2 22 2 22 22 21 21242( 1 ) 2r r rrrr rTr r r rr rrθ θθ θμπ μππ=Ω− Ω Ω− Ω⎡⎤===⎢⎥⎣⎦ −−ee e M The torque on the inner wall is equal a nd opposite to that on the outer wall. __________________________________________________________________ 6.31 Verify the equation2/2βρω μ= for the oscillating plane problem of Section 6.16. -------------------------------------------------------------------------------- Ans. With 22 cos( )xve txβα ωβ ε−=− + , 22 /s i n ( )xvt e t xβωαω β ε−∂∂= − − + , 2222 2/c o s ( ) s i n ( )xxvx e t x e t xβββαω β ε β αω β ε−−∂∂= − − ++ − + and 22 22 222 2 2 22 2 22 22 2 2/c o s ( ) s i n ( ) sin( ) cos( ) 2s i n ( )xx xx xvx e t x e t x et x e t x et xββ ββ ββαω β ε β αω β ε βα ω β ε βα ω β ε βα ω β ε−− −− −∂∂ = − + − − + −− + − − +=− − + Thus 22 2 //vt vxρμ∂∂ = ∂ ∂→ 22 2 22 22sin( ) 2 sin( ) 2/ ( 2 ) .xxet x et xββρω α ωβ ε μ β α ωβ ε ρω μ β β ρω μ−−−− + = − − + →= →= __________________________________________________________________ 6.32 Consider the flow of an incompressible viscous fluid through the annular space between two concentric horizontal cylinders. The radii are and ab . (a) Find the flow field if there is no variation of pressure in the axial direction and if the inner and the outer cylinders have axial Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-18velocities and vabv respectively and (b) find the flow field if there is a pressure gradient in the axial direction and both cylinders are fixed. Take body forces to be zero. -------------------------------------------------------------------------------- Ans. (a) We look for the following form of velocity field in cylindrical coordinates: 0, 0, ( ) and / 0rzvvv v r p zθ=== ∂ ∂ = . The N-S equations give, in the absence of body forces 2 2110, 0 , 0p pd v d v rr r d r drμθ⎛⎞ ∂∂=− =− = + ⎜⎟⎜⎟ ∂∂⎝⎠. The first two equations together with / 0 pz∂∂= give, constant p= . 2 21100 l ndv d v d d v d v d v Crr C v C r Dr dr r dr dr dr dr r drμ⎛⎞ ⎛⎞=+ → = → = → = → = +⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ At , and at , ab ra vv rb vv== == , thus, ln and ln ln( / )ab a bvC a D vC b DvvC a b=+ =+ → − = . () ()ln ln andln / ln /ab a bvv vb vaCDab ba−−→= = . So that, () ()ln lnlnln / ln /ab a bvv vb vavrab ba−−=+ . (b) 0 / and 0 (1/ ) / ( ) prr p p p z θ =−∂ ∂ =− ∂ ∂ → = , 22 2210 0 constantdp d v dv d p dp dz r dr dz dr dzμ α⎛⎞ =− + + → = → = ≡− ⎜⎟⎜⎟⎝⎠, 2 21 dv d v d d v d d v rrrr dr r dr dr dr dr drμαμα αμ⎛⎞ ⎛⎞ ⎛⎞→ + =− → =− → =− →⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ 22 ln22 4dv r dv r C rrC v C r Ddr dr rαα α μμμ→= −+ → = −+ → = −+ + . The boundary conditions are: ( ) ( ) 0 va vb== → () ()() ()() ()22 22 22 2 20l n a n d 0l n44 ln ln11,4 l n/ 4 l n/ 4 l n/abCaD CbD ab ab ab badp dpCDab d z ab d z baαα μμ α μμ μ=− + + =− + + → −− − == − = − () ()() ()22 2 2 2ln ln 1ln4l n / l n /ab b a a b dpvr rd z ba baμ⎡⎤−− ⎢⎥=+ + ⎢⎥⎣⎦. __________________________________________________________________ 6.33 Show that for the velocity field : ( , ), 0xy zvv y z vv= == , the Navier-Stokes equations, with ρB=0, reduces to 22 221constantvvd p dx yzβμ∂∂+= = = ∂∂. ------------------------------------------------------------------------------- Ans. With ( , ), 0xy zvv y z vv== = , we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-1900000 , a n d 0xxxx xx y z y zvvvvav v v a atxyz∂∂∂∂=+ + + = + + + = = =∂∂∂∂, thus, the N-S equations in the absence of body forces are: 22 220, 0 , 0xxvv p pp x yz yzμ⎛⎞∂∂∂∂ ∂=− + + =− =− ⎜⎟⎜⎟∂∂ ∂∂∂⎝⎠. Thus, ( ) ppx= , and 22 22 22 22 22 2 20xx xxvv vv d p dp dp dx x yz yzd x d xμμ⎛⎞ ⎛⎞∂∂ ∂∂ ∂+= → += → = →⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ∂ ∂∂ ∂∂⎝⎠ ⎝⎠, 22 221constantxxvv dp dp dx y z dxβμ∂∂=→ + = ≡∂∂. __________________________________________________________________ 6.34 Given the velocity field in the form of ()22 22//xvv A y az b B== + + , 0yzvv==. Find Aand Bfor the steady laminar flow of a Newtonian fluid in a pipe having an elliptical cross section given by22 22// 1ya zb+=. Assume no body forces and use the governing equation obtained in the previous problem. ------------------------------------------------------------------------------- Ans. The governing equation is [see the previous problem] : 22 221xxvv dp dx yzβμ∂∂+=≡ ∂∂. Now, ( )22 22//xvv A y az b B==+ + → 22 22 22 2 2 2 2222xxvv abAA yz a b a b⎛⎞ ∂∂ + ⎛⎞+= + = ⎜⎟ ⎜⎟ ⎜⎟∂∂ ⎝⎠ ⎝⎠ ()22 2 2 22 222 2ab a bAA ab abββ⎛⎞+→= → =⎜⎟⎜⎟+ ⎝⎠ On the boundary 22 22// 1ya zb+= , no slip condition requires that 0xv=, therefore, ()10AB B A+=→= − , thus, ()22 2 2 22 22 22 2211 2xyz a b yzvA ab ab abβ ⎡⎤⎡⎤⎛⎞ ⎛⎞ =+ − = + −⎢⎥⎢⎥⎜⎟ ⎜⎟⎜⎟ ⎜⎟+ ⎢⎥⎢⎥⎝⎠ ⎝⎠⎣⎦⎣⎦. __________________________________________________________________ 6.35 Given the velocity field in the form of 33 23 3 3xbb bvA z z y z y B⎛⎞ ⎛ ⎞ ⎛ ⎞=+ +− −−+⎜⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝ ⎠,0yzvv== Find Aand Bfor the steady laminar flow of a Newtonian fluid in a pipe having an equilateral triangular cross-section defined by the planes: 0 , 30 , 30 23 3 3bb bzz y z y+=+− =−− = . Assume no body forces and use the governing equation obtained in Prob. 6.33. ------------------------------------------------------------------------------- Ans. The governing equation is 22 221xxvv dp dx yzβμ∂∂+=≡ ∂∂. [See problem 6.33] With () () ( ) /( 2 3 ) 3 / 3 3 / 3xvA z b z y b z y b B=+ +− −− + , Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-20() () ( ) { } / / ( 2 3 ) 3 3 /3 3 3 /3xvy A z b z y b z y b∂∂ = + − − − + − () () /( 2 3 ) 6 Az b y=+ − → ()22/6/ ( 2 3 )xvy A z b∂∂ = − + . Let () ( )( ) () / ( 2 3 ), (,) 3 / 3, (,) 3 / 3f zz b g y zz y b h y zz y b=+ =+ − =− − , then () (,)(,)xvA f z g y z h y z B=+ { } / (,)(,) () (,) (,) xvz A g y z h y z A f z h y z g y z→∂ ∂ = + + Now, () ( )() (,) (,) 3 / 3 3 / 3 2 / 3hyz gyz z y b z y b z b+= + −+ − −= − {} ()()( )22() (,) (,) 2 / 3 / ( 2 3 ) 2 / 3 / 3fz h y z gy z z b z b z z b b += − + = − − ()222 2 2( , ) ( , ) /3 3 2 /3 / 33gyzhyz z b y z b z b y =− − =− + − . Thus, {} () () ()22 2 2 2 2 2(,)(,) () (,) (,) 2/ 3 / 3 3 2 / 3 / 3 3 3/ 3 3xvAg y z h y z Af z h y z g y zz Az b z b y A z z b b A z b z y∂=+ +∂ =− +−+ − − = − − () ()22/6 3 / 3 6 / ( 2 3 )xvzA z b A z b∂ ∂ =− =− . Thus, () () ()22 2216/ ( 2 3 ) 6/ ( 2 3 ) 6 / 3xxvv dpAz b Az b Abdx yzβμ∂∂+= −+ +− = − =≡ ∂∂, from which, /( 2 3 ) Abβ=− . The non-slip condition on the boundary requires 0B=. _________________________________________________________________ 6.36 For the steady-state, time dependent parallel flow of water ( density 3310 /Kg mρ= , viscosity,3210 / Nsm μ−= ) near an oscillating plate, calculate the wave length for 2 cpsω= . ------------------------------------------------------------------------------- Ans. ()22 cosxva e t xβωβε−=− + , the wave length is given by 2 / πβ, where 2ρωβμ= . Here we have, 3310 /Kg mρ= , 4 /rad sωπ= , 3210 / Nsm μ−= , thus ()() ()3 31 3 3 310 422 =10 2 wave length= 2.51 102 10 21 0mmπρω π πβπμβ−− −== → == × _________________________________________________________________ 6.37 The space between two concentric spherical shells is filled with an incompressible Newtonian fluid. The inner shell (radius ir ) is fixed; the outer shell (radius or ) rotates with an angular velocity Ωabout a diameter. Find the velocity distribution. Assume the flow to be laminar without secondary flow. ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-21Ans. We look for solution in the form of () 0, 0, sinrvvv f rθφ θ === . This velocity field clearly satisfies the continuity equa tion [see Section 6.8, Eq.(6.8.8)]: cot 2 110sinrrv vv vv rr r r rφ θθθ θθ φ∂ ∂ ∂++ + + =∂∂ ∂. The N-S equations in spherical c oordinates give[see Section 6.8]: 21 v p rrφ ρ∂−= − ∂ (1), 2cot 1 v p rrφθ ρθ∂− ∂−= (2), ()2 221 sin11 10s i nsinp rvrvrrrrφ φρθ φμθρθ θ θ∂=− ∂⎡⎤ ∂⎛⎞∂∂ ∂ ⎛⎞++⎢⎥⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎢⎥⎝⎠⎣⎦ (3). 22 2 Eq.(1) 0, eq.(2) eq.(3) 0 and 0pp p rφφ θ φ φ→= → →∂∂ ∂==∂∂ ∂∂ ∂∂. Thus, / constant pφ∂∂ = . The constant must be zero, otherwise p will not be single-valued. Eq. (3) now becomes, with / 0 pφ∂∂ = and ()sin vf rφ θ= , () 2 222 sin0s i nfr dd frdr drrrθθ =⎛⎞−⎜⎟⎝⎠. That is, 2 22 220 + 2 20dd f d fd frf r r fdr dr dr dr⎛⎞−=→ − = ⎜⎟⎝⎠. The general solution of this equation is: 2/ fAr B r=+ . Thus, ()2/s i n vA r B rφ θ =+ . The inner shell (radius ir ) is fixed; therefore, at , 0irrvφ==→ ()20/iiArB r=+ (4) The outer shell (radius or ) rotates with an angular velocity Ω, therefore, at () () ()2 oo o o o, sin sin / sin rr v r r A r B rφθθ θ == Ω → Ω = + → ()2 oo o / rA r B rΩ= + (5) Equations (4) and (5) are two equations for the two unknowns Aand B: ()3 o 33 o irA rrΩ= −, ()33 o 33 oi irrB rr=− Ω −, and 2sinBvA r rφ θ⎛⎞=+⎜⎟⎝⎠. _________________________________________________________________ 6.38 Consider the following velocity field in cylindri cal coordinates for an incompressible fluid: () , 0 , 0rzvv r v vθ= == (a) Show that rAvr=where Ais a constant so that the equa tion of conservation of mass is satisfied. (b) If the rate of mass flow through the circular cylindrical surface of radius rand unit length (in z direction) is mQ, determine the constant Ain terms of mQ. ------------------------------------------------------------------------------- Ans. ( a) The equation of continuity is ()110z rv vrvrr r zθ θ∂∂ ∂++ =∂∂ ∂. Thus, ()10rr rdArv rv A vrd r r=→= → = . (b) () ( )21222mm m rm rQQ Q Avr Qv Arr rππ ππ= → =→ =→ = . _________________________________________________________________ 6.39 Given the following velocity field in cylindrical coordinates for an incompressible fluid: (, ) , 0 , 0rzvv r v vθθ= == Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-22(a) Show that ()/rvf rθ= , where ()fθis any function of θ. (b) In the absence of body forces, show that 22 240df ffk dρ μ θ++ + = , 222 2fkp C rrμμ=++ , where kand Care constants. ------------------------------------------------------------------------------- Ans. (a). The equation of continuity is ()110z rv vrvrr r zθ θ∂∂ ∂++ =∂∂ ∂. Thus, ()10( ) ( ) /rr rrv rv f v f rrrθθ∂=→ = → =∂ (b).The N-S equation in the r direction gives 1vrr r r rz rv vv v v pvv Btr r z rθ θθρ∂∂ ∂ ∂ ∂ ⎛⎞++ − += − + ⎜⎟∂∂ ∂ ∂ ∂ ⎝⎠ 22 2 22 2 2 2 211 2rr r r r v vv v v v rr rr z r rθ μ ρθ θ⎡⎤ ∂ ∂∂ ∂ ∂+++ + − − →⎢⎥∂∂ ∂∂ ∂⎢⎥⎣⎦22 33 211 f pd f r rr dμ ρρ θ⎛⎞∂−= − + ⎜⎟⎜⎟∂⎝⎠ (1) The N-S equation in the z direction gives / 0 pz−∂∂ = → p is independent of z. The N-S equation in the θ direction gives 23 212 1 2 20rv p pd f p d f rr d d rr rμμμ ρθρ θ ρθρ θ θ θ∂ ∂∂ ∂⎛⎞ ⎛⎞ ⎛⎞=− + →− + = → = ⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎝⎠ ⎝⎠ 2324()f pf d gpg rrd r rrμμ ∂→= + → = − +∂. Thus, Eq.(1) gives: 23 2 24fr d g d ffdr dρ μμ θ⎛⎞−= − ++ ⎜⎟⎝⎠22 3 24df f r d gfdr dρ μμ θ⎛⎞ ⎛⎞→+ + =⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠. The left side of the above equation is a function of θ, the right side is a function of r, therefore, they must be equal to a constant, say, k−, i.e., 22 3 224 2df f r d g kf kg Cdr drρμ μμ θ⎛⎞ ⎛⎞++ = = − → = +⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠. Therefore, 22 22 2240 , 2df f f kf kp C dr rρμ μ μ θ⎛⎞ ++ + = = + +⎜⎟⎜⎟⎝⎠. _________________________________________________________________ 6.40 Consider the steady two dimensional channel fl ow of an incompressible Newtonian fluid under the action of an applied negative pressure gradient1/px∂∂, as well as the movement of the top plate with velocity ovin its own plane.[See Prob. 6.26]. Determine the temperature distribution for this flow due to viscous dissipation when both plates are maintained at the same fixed temperatureoθ. Assume constant physical properties. ------------------------------------------------------------------------------- Ans. From the result of Prob. 6.26, we have ()()2 1o 2 2 2 23 // 2 , 0 . vv x d x d x v v αμ =+− = = Let the temperature distribution be denoted by ()2xΘ=Θ . From Eq. (6.18.3), we have, 2 inc jjDcDtx xρκΘ∂ Θ=Φ +∂∂, where ( )22 2 2 2 2 11 22 33 12 13 23 22 2 2inc DD D D D DμΦ= + + + + + , represents the heat generated through viscous forces. For this problem, only 12Dis nonzero, thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-23() () ()22 2 2 oo 12 2 2144 2 222 2incvvDd x d xddααμμ μμμ⎡⎤ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞Φ= = + − = + − ⎢⎥ ⎢⎥⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠⎣⎦ ⎣⎦. () ()2322 oo 22 2 2 22223 2vvdx dx Cdx d xμα μ α κμ κ αμ⎡⎤ ⎡⎤⎛⎞ ⎛⎞ ∂Θ ∂ Θ=− + − → = + − + →⎢⎥ ⎢⎥⎜⎟ ⎜⎟∂ ∂ ⎝⎠ ⎝⎠⎣⎦ ⎣⎦ ()43 o 22 222 12vdx C x Ddμα μ κα⎡⎤⎛⎞Θ=− + − + + ⎢⎥⎜⎟⎝⎠⎣⎦. 4 3 o 2o o 2 4 3 o 2o o 2At 0, .2 12 At , .2 12v dxDd v dx dC d Ddμαθθμ κα μαθθμ κα⎛⎞=Θ = → = − + + ⎜⎟⎝⎠ ⎛⎞=Θ = → = − − + + ⎜⎟⎝⎠ Thus, 4 3 o o 22 12v dDdμαθμ κα⎛⎞=+ + ⎜⎟⎝⎠and 44 3 oo 2022 12vv ddCdddμα α μμ κα⎧⎫⎛⎞ ⎛⎞⎪⎪=+ − − +⎨⎬⎜⎟ ⎜⎟⎝⎠ ⎝⎠⎪⎪⎩⎭ 44 3 oo 222 12vv ddCdd dμα α μμ κα⎧⎫⎛⎞ ⎛⎞⎪⎪→= − + − − ⎨⎬⎜⎟ ⎜⎟⎝⎠ ⎝⎠⎪⎪⎩⎭. _________________________________________________________________ 6.41 Determine the temperature distribution in the plane Poiseuille flow where the bottom plate is kept at a constant temperature 1Θand the top plate2Θ. Include the heat generated by viscous dissipation. ------------------------------------------------------------------------------- Ans. For the plane Poiseuille flow [see Eq.(6.12.9)], ()()22 12 2 3 1 /2 , 0 , / 0 vb x v v p xαμ α=− = = ≡ − ∂ ∂ > Let the temperature distribution be denoted by ()2xΘ=Θ . From Eq. (6.18.3), we have, 2 inc jjDcDtx xρκΘ∂ Θ=Φ +∂∂, where ( )22 2 2 2 2 11 22 33 12 13 23 22 2 2inc DD D D D DμΦ= + + + + + , represents the heat generated through viscous forces. For this problem, only 12Dis nonzero, 2 2 22 1 12 2 12 2 2 211 14422 2incvD xDx xxαα αμμμμ μ⎛⎞ ∂== −→ Φ = = − = ⎜⎟∂ ⎝⎠. Thus, 22 2 2 2 2 20inc jjDdcxDt x x dxαρκ κμΘ∂ Θ Θ=Φ + → = + →∂∂22 2 2 2 1 21 dpxx dxκμ⎛⎞Θ∂=−⎜⎟∂⎝⎠ 23 2 211 3x dpCdx xκμ⎛⎞Θ∂→= − + → ⎜⎟∂⎝⎠24 2 2 11 12x pCx Dxκμ⎛⎞∂Θ=− + +⎜⎟∂⎝⎠. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-2424 22 2 11At ,12pbx bC b Dxκμ⎛⎞∂=+ Θ=Θ →Θ =− + + ⎜⎟∂⎝⎠ 24 21 1 11At ,12pbx bC b Dxκμ⎛⎞∂=− Θ=Θ →Θ =− − + ⎜⎟∂⎝⎠. Thus, 2 4 12 11 21 2pD bxκμ⎛⎞ Θ+ Θ ∂=+ ⎜⎟∂⎝⎠. 21 2CbΘ−Θ= . __________________________________________________________________ 6.42 Determine the temperature distribution in th e steady laminar flow between two coaxial cylinders (Couette flow) if the temperatures at th e inner and the outer cylinders are kept at the same fixed temperatureoθ. ------------------------------------------------------------------------------- Ans. For Couette flow, we have, 0, , 0rzBvv A rvrθ==+ = , where ()22 22 12 1 2 22 1 1 22 22 21 21, rr rrAB rr rrΩ−Ω Ω− Ω== −−. The only nonzero rate of deformation is 2v 11 1v22r rvv v BDrr r r rθθ θ θθθ⎧⎫ ∂∂ ∂ ⎧⎫ ⎛⎞=− + = − + = −⎨⎬ ⎨ ⎬⎜⎟∂∂ ∂⎝⎠ ⎩⎭ ⎩⎭. ()2 2 2 24422 4inc rB BD rrθμμμ⎛⎞Φ= = − = ⎜⎟⎝⎠. 2 inc jjDcDt x xρκΘ∂ Θ=Φ+ →∂∂ 2 4410Bd drrd r d r rμκΘ⎛⎞→= + → ⎜⎟⎝⎠22 3342 dd B d BCrdr dr dr r rrμμ κκΘΘ⎛⎞=−→ =+⎜⎟⎝⎠ 2 2lnBCrD rμ κ→Θ=− + + . 2 oo 2 2 oo o 2At , ln . At , ln .ii i o oBrr Cr D r Brr Cr D rμ κ μ κ=Θ = Θ → Θ = −+ + =Θ = Θ → Θ = − + + Thus, 22 2 22/l noi i o iorr r BCr rrμ κ⎛⎞⎛⎞⎛⎞ −=⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠, {}2 22 o 22ln ln ln / lnoo oo i i ii o irr BDr r r rrr r rμ κ⎡⎤=Θ + −⎢⎥ ⎣⎦. _________________________________________________________________ 6.43 Show that the dissipation function for a compressible fluid can be written as that given in Eq. (6.17.10). ------------------------------------------------------------------------------- Ans. From () ()2 22 2 2 2 2 11 22 33 11 22 33 12 13 23 22 2 2 DD D DD D D D DλμΦ= + + + + + + + + , we get, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-25()() () ( )22 2 222 11 22 33 11 22 11 33 22 33 12 13 23 22 4 DD D D D D DD D DDDλμ λ μΦ= + + + + + + + + + We now verify that this is the same as () () () () () ()22 2 2 11 22 33 11 22 11 33 22 33 222 12 13 2322/ 33 4DD D DD DD DD DDDλμ μ μ⎡ ⎤Φ= + + + + − + − + −⎢ ⎥ ⎣ ⎦ ++ + Expanding the above equation, we have, ()( ) ()() () ()22 2 11 22 33 11 22 11 33 22 33 22 2 222 11 22 33 11 22 11 33 22 33 12 13 232/ 3 2 2 2 2/ 32 2 2 2 4 .DD D D D D D D D DD D D D D D D D DDDλμ μμΦ= + + + + + + ⎡⎤++ + − + + + + +⎣⎦ i.e., ()() () ()() ()() () () () () () ()22 2 11 22 33 11 22 11 33 22 33 22 2 222 11 22 33 11 22 11 33 22 33 12 13 23 22 2 11 22 33 11 22 11 33 22 33 222 12 13 2322/ 3 2 2 23 4/ 3 2/ 32 2 2 4 2 /3 4 /3 2 2 /3 2 /3 4.DD D D D D D D D DD D D D D D D D DDD DD D D D D DD D DDDλμ λμ μμ μ λμ μ λμ μ μ⎛⎞Φ= + + + + + + + ⎜⎟⎝⎠ ++ + − + + + + + =++ + + + + + + −++ + That is () () () ( )22 2 222 11 22 33 11 22 11 33 22 33 12 13 23 22 4 DD D D D D DD D DDD λμ λ μ Φ= + + + + + + + + + Thus, Φ=Φ __________________________________________________________________ 6.44 Given the velocity field of a linearly viscous fluid 11 2 2 3 , , 0 vk xv k xv==− = (a) Show that the velocity field is irrotational. (b) Find the stress tensor. (c) Find the acceleration field. (d) Show that the velocity field satisfies the Navier-Stokes equations by finding the pressure distribution directly from the equations. Neglect body forces. Take o pp= at the origin. (e) Use the Bernoulli equation to find the pressure dist ribution. (f) Find the rate of dissipation of mechanical energy into heat. (g) If 20 x=is a fixed boundary, what condition is not satisfied by the velocity field. ------------------------------------------------------------------------------- Ans. (a) [] [ ] []00 00 0 0 0 00 00 , 0 0 0 000 000 0 0 0kk kk⎡⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢⎥ ⎢ ⎥∇= − → = − =⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣⎦ ⎣ ⎦vD W therefore, the flow is irrotational. (b) 11 22 33 12 13 23 2, 2, , 0 Tp k T p k Tp T T T μ μ =− + =− − =− = = = . (c) 2 111 2 22 2 300 00 000 0 0kx ak k x ak k x k x a⎡⎤⎡⎤⎡⎤ ⎡ ⎤⎢⎥⎢⎥⎢⎥ ⎢ ⎥⎢⎥ =− −=⎢⎥⎢⎥ ⎢ ⎥⎢⎥⎢⎥⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦⎣⎦ ⎢⎥⎣⎦. (d) Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-26()222 22 111 11 222 11 12 3vvv p pkx kxx x xxxρμ ρ⎛⎞∂∂∂∂∂=− + + + → =− ⎜⎟⎜⎟∂∂∂∂∂⎝⎠ ()222 22 222 22 222 22 12 3vvv p pkx kxx x xxxρμ ρ⎛⎞∂∂∂∂∂=− + + + → =− ⎜⎟⎜⎟∂∂∂∂∂⎝⎠ and 3 30 is independent of pp xx∂=− →∂. Thus, ()22 2 2 12 2 12 2 12 2 2 22 2 22 2 12 1 2() ()()2 . At 0,22ookx d fx d fx ppkx p fx kxxx d x d x kx kf Cp xx C x x p p C pρρρ ρ ρ∂∂=− → =− + → = →− =∂∂ →= − +→= − + + = = = →= That is, ()() ()()22 2 2 2 12 12 /2 /2oo p k x xpp v vpρρ=− + + → =− + + . (e) From the Bernoulli Equation, we have ()22 22 2 12 12 at origin 22 2 12constant 022 2 /2 .o op vv vv pv p p pp kx xρρ ρ ρ ρ⎡⎤+++= → + =+ =+ ⎢⎥ ⎢⎥⎣⎦ →= − + (f) () () ()2 22 2 2 2 2 2 2 2 11 22 33 11 22 33 12 13 23 22 2 2 2 4 D DD D DD D D D k k kλ μμ μ Φ= + + + + + + + + = + = (h) if 20 x= is a fixed boundary, then v must be zero there. But ( ) 11 2 2 11 0 kx x k x−=≠ v= e e e 2at 0x=, therefore the non slip boundary condition at 20x= is not satisfied for a viscous fluid. _________________________________________________________________ 6.45 Do Problem 6.44 for the following velocity field: ()22 11 2 2 1 2 3 , 2 , 0 vk x x v k x xv=−= − = . ------------------------------------------------------------------------------- Ans. (a) [] [ ] []12 1 2 21 2 122 0 0 0 0 0 22 0 2 0 , 0 0 0 00 0 0 0 0 0 0 0kx kx x x kx kx k x x−−⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢ ⎥∇= − − → = − − =⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦vD W therefore, the flow is irrotational (b) 11 1 22 1 33 12 13 23 4, 4, , 0 Tp k x T p k x Tp T T T μ μ =− + =− − =− = = = (c) () () ()22 22211 21211 2 22 2 22 1 1 2 2 1 2 3222 0 22 0 2 2 00 000kx x xkx xak x k x ak x k x k x xk x x x a⎡ ⎤+ ⎡⎤−⎢ ⎥ − ⎡⎤⎡⎤ ⎢⎥⎢ ⎥ ⎢⎥⎢⎥ ⎢⎥ =− − − = +⎢ ⎥ ⎢⎥⎢⎥ ⎢⎥⎢ ⎥ ⎢⎥⎢⎥ ⎢⎥ ⎣⎦⎣⎦⎢ ⎥ ⎣⎦⎣ ⎦ (d) Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-27() () () ()222 22 2 22 2 111 11 2 11 2 222 11 12 3 222 22 2 22 2 222 21 2 21 2 222 22 12 3 322 22 0 is independent of vvv p pkx x x kx x xxx xxx vvv p pkx x x kx x xx x xxx ppxxρμ ρ ρμ ρ⎛⎞∂∂∂∂∂⎡⎤ ⎡⎤ += − + ++ → += − ⎜⎟⎜⎟ ⎣⎦ ⎣⎦ ∂∂∂∂∂⎝⎠ ⎛⎞∂∂∂∂∂⎡⎤ ⎡⎤ += − + ++ → += − ⎜⎟⎜⎟ ⎣⎦ ⎣⎦ ∂∂∂∂∂⎝⎠ ∂=− →∂3 Thus, ()4 22 2 2 2 2 2 2 1 11 2 1 2 2 1 2 12 22( ) 22x p pd fkx x x p k xx fx kxxx xd xρρ ρ⎛⎞ ∂∂⎡⎤ += − → = − + + → = − + ⎜⎟⎜⎟ ⎣⎦ ∂∂⎝⎠ and () ()22 2 22 2 2 2 21 2 21 2 1 2 22 24 23 2 2 222 2 2.2p dfkx x x kx x x kxxx dx kx dfkx f Cdxρρ ρ ρρ∂⎡⎤ ⎡⎤ += − → − += − +⎣⎦ ⎣⎦ ∂ →− = → =− + Since o pp= at origin, therefore, o Cp= , () ()2224 2 24 22 11 2 2o 1 2 o222kkp x x xx p p xx pρρ→ = −+ + + → = −+ + . Or, since () () ()2222 22222 2 2 2 22 11 2 2 1 2 1 2 1 2 1 2 1 2 , 2 , 4 vk x x v k x x v v k x x x x k x x⎡⎤=− = − + = −+ = +⎢⎥⎣⎦ ()22 12 o /2 p vv pρ→= − + + . (e) From the Bernoulli Equation, we have ()22 22 2 12 12 at origin 222 2 12constant 022 2 /2 .o op vv vv pv p p pp kx xρρ ρ ρ ρ⎡⎤+++= → + =+ =+ ⎢⎥ ⎢⎥⎣⎦ →= − + (f) () () ( )( )2 22 2 2 22 2 2 2 2 11 22 33 11 22 12 1 2 1 2 22 0 2 8 8 1 6 D DD D D D k x k x k x xλμ μ μΦ= + + + + + = + + = + (h) if 20 x= is a fixed boundary, then vmust be zero there. But2 11 2 0 at 0 kx x≠= v= e , therefore the non slip boundary condition at 20x=is not satisfied for a viscous fluid. _________________________________________________________________ 6.46 Obtain the vorticity vector for the plane Poiseuille flow. ------------------------------------------------------------------------------- Ans. With ()( ) ()22 12 2 /2 vv x b x αμ == − , where 1/px α=−∂ ∂ and 23 0 vv==, the spin tensor is [][]12 120/ 0 1/0 0200 0Avx vx∂∂ ⎡⎤ ⎢⎥=∇ = − ∂ ∂⎢⎥ ⎢⎥⎣⎦Wv Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-28The vorticity tensor is 2 W and the vorticity vector is twice the axial vector ()11 2 32 1 13 2 21 3 3 3 3 2 3 22 1122vv x pWWW xxx xα μμ⎛⎞∂∂ ∂22 + + = − = − = = − ⎜⎟∂∂ ∂⎝⎠1ee e e e e e ς= ω= . _________________________________________________________________ 6.47 Obtain the vorticity vector for the Hagen-Poiseuille flow. ------------------------------------------------------------------------------- Ans. With 2 2, = 0 , 44zrdpvr v vzθααμ⎛⎞ ∂=− = = −⎜⎟⎜⎟ ∂⎝⎠, the spin tensor is [][] 21 31 32v 11 10 100 222 1100 0 02 100 02rr z z A z zv vv vvrr r z rr v vWzr vWWrθθ θθ θ⎡⎤ ∂ ∂∂ ∂⎛⎞ ⎛⎞−− − ⎡ ⎤∂⎛⎞ ⎢⎥ ⎜⎟ ⎜⎟− ∂∂∂ ∂ ⎝⎠ ⎝⎠ ⎢ ⎥⎜⎟ ⎢⎥∂⎝⎠⎢ ⎥ ⎢⎥ ∂∂⎛⎞⎢ ⎥ == − = ⎢⎥ ⎜⎟∂∂ ⎢ ⎥ ⎝⎠ ⎢⎥∂⎛⎞⎢ ⎥ ⎢⎥⎜⎟⎢ ⎥ ⎢⎥ ∂⎝⎠⎣ ⎦⎢⎥⎣⎦Wv∇ The vorticity tensor is 2W and the vorticity vector is twice the axial vector () rz1 22z zr z r r zv rpWWW W rrzθ θθ θ θ θ θα μμ∂ ∂ ⎛⎞2 2 + + =2 =− = =− ⎜⎟∂∂⎝⎠ee e e e e e ς= ω= . _________________________________________________________________ 6.48 For a two-dimensional flow of an incompre ssible fluid, we can express the velocity components in terms of a scalar function ψ (known as the Lagrange stream function) by the relations , xyvvyxψ ψ ∂∂== −∂∂. (a) Show that the equation of conservation of mass is automatically satisfied for any (),xyψ which has continuous s econd partial derivatives. (b) Show that for two-dimensional flow of an incompressible fluid, ψ=constants are streamlines. (c) If the velocity field is irrotational, then i ivxϕ∂=−∂ where ϕ is known as the velocity potential. Show that the curves of cons tant velocity potential constant ϕ= and the streamline ψ=constant are orthogonal to each other. (d) Obtain the only nonzero vorticity component in terms of ψ. ------------------------------------------------------------------------------- Ans. (a) With and xyvvyxψ ψ ∂∂== −∂∂, we have, 0.y xv v xy x y y xψψ ∂∂ ∂∂ ∂∂+=−=∂∂∂ ∂∂ ∂ (b) From ( , ) xyCψ=, we have, constant constant/0/y xv dy x dydd x d yx yd x y d x vψψψψ ψψψ==∂∂ ∂ ∂ ⎛⎞ ⎛⎞=+= → = − → = ⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎝⎠ Thus, ( , ) xyCψ=are streamlines. (c) From (, )xy Cϕ=→ constant constant/0/x yv dy x dydd x d yx yd x y d x vϕϕϕϕ ϕϕϕ==∂∂ ∂ ∂ ⎛⎞ ⎛⎞=+= → = − → = − ⎜⎟ ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎝⎠ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-29Thus, constant constant1dy dy dx dxϕψ==⎛⎞ ⎛⎞→= −⎜⎟ ⎜⎟⎝⎠ ⎝⎠. (d) ()22 xyz z z 22y x zy xz yxv vWWWxy yxψψ⎛⎞ ∂⎛⎞∂ ∂∂22 + + = − = − + ⎜⎟ ⎜⎟⎜⎟ ∂∂ ∂∂ ⎝⎠ ⎝⎠eee e e ς= ω= . _________________________________________________________________ 6.49 Show that 2 o 221aVy x yψ⎛⎞ =−⎜⎟⎜⎟+⎝⎠represents a two-dimensional irrotational flow of an inviscid fluid. ------------------------------------------------------------------------------- Ans. With 2 o 221aVy x yψ⎛⎞ =−⎜⎟⎜⎟+⎝⎠, we have, () ()() () () () ()() ()()22 22 2 oo 22 2 322 22 22 2 22 2 2 2 o o 22 3 2 322 22 22 2222 22 2 14 82oxa xa aVy Vyx xxy xy xy ay xa y xVy a V xxy xy xy xyψψ ψ⎛⎞ ⎛ ⎞− ∂∂⎜⎟ ⎜ ⎟=→ = +⎜⎟ ⎜ ⎟∂ ∂⎜⎟ ⎜ ⎟++ +⎝⎠ ⎝ ⎠ ⎛⎞ ⎛⎞ ∂ ⎜⎟ ⎜⎟=− = −⎜⎟ ⎜⎟∂⎜⎟ ⎜⎟++ ++⎝⎠ ⎝⎠ ()22 2 oo 22 22221ay aVVy xyxyψ⎛⎞⎛⎞∂ ⎜⎟=− +⎜⎟ ⎜⎟ ⎜⎟∂ +⎝⎠ ⎜⎟+⎝⎠ ()() ()() ( ) ()22 2 22 oo 22 2 322 22 2222 2 2 2 2 ya ya y yaVV yxy xy xyψ⎛⎞ ⎛ ⎞ ∂⎜⎟ ⎜ ⎟=+ −⎜⎟ ⎜ ⎟∂⎜⎟ ⎜ ⎟++ +⎝⎠ ⎝ ⎠ () ()()22 2 3 2 oo 22 2 322 22 2224 8ya ya y aVV yxy xy xyψ⎛⎞ ⎛ ⎞ ∂⎜⎟ ⎜ ⎟=+ −⎜⎟ ⎜ ⎟∂⎜⎟ ⎜ ⎟++ +⎝⎠ ⎝ ⎠. Thus, ()() () ()() () ()() ()2 22 2 2 2 2 3 2 o oo 22 2 3 2 2 322 22 22 22 22 2222 2 3 2 2 oo 23 3222 22 22 222 82 4 8 888 8 8o ooay ay x y a y a yaVV V yxxy xy xy xy xy ayxay ay x ya ayVV VV xy xy xy xyψψ⎛⎞ ⎛ ⎞ ⎛⎞ ∂∂ ⎜⎟ ⎜ ⎟ ⎜⎟+= − + + −⎜⎟ ⎜ ⎟ ⎜⎟∂∂⎜⎟ ⎜ ⎟ ⎜⎟++ + ++⎝⎠ ⎝ ⎠ ⎝⎠ ⎛⎞ ⎛ ⎞ ⎛⎞+⎜⎟ ⎜ ⎟ ⎜⎟−+ = −⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟++ ++⎝⎠ ⎝ ⎠ ⎝⎠() ()2 3220y xy⎛⎞ ⎜⎟=⎜⎟ ⎜⎟+⎝⎠ Therefore, the given stream function ψrepresents a two-dimensional irrotational flow of an inviscid fluid. __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-306.50 Referring to Figure P 6-9, compute the maximum possible flow of water. Take the atmospheric pressure to be 93.1 .kPa , the specific weight of water 98103/Nm, and the vapor pressure 17.2 . kPa Assume the fluid to be inviscid. Find the length l for this rate of discharge. Δ 5m3m10cm dia Figure P 6-9 ------------------------------------------------------------------------------ Ans. The Bernoulli equation gives, with point 1 at the reservoir top and point 2 at the highest point inside the tube, we have, 22 11 2 2// 2 ( 0 ) // 2 ( 3 )p vg pvgρρ++ = ++ . Thus, assuming 1vto be very small and negligible, we have, with 32 1293,100 ., 17,200 ., 1000 / and 9.81 /p Pa p Pa kg m g m s ρ == = = . () ( ) () ()()2 21 2 2 3 2m a x 2/ 2 / 3 93,100 17,200 /1000 3 9.81 46.47 9.64 / . 9.64 0.1 / 4 0.0757 / .vp p g vm s Q v A m sρ π=− −= − − = →= = = = With point 3 at the exit, we have, 22 22 33// 2 ( 0 ) // 2 ( )pvg pvgρρ++= + + − l now, 23 2 3 , (the vapor pressure), (atm.pressure)va vv p p pp== = ()( ) / ( ) 93,100 17,200 / 9810 7.74 .avp pg mρ =− = − =l _________________________________________________________________ 6.51 Water flows upward through a vertical pipe line which tapers from cross sectional area 1A to area 2Ain a distance of h. If the pressure at the beginning and end of the constriction are 1pand 2prespectively. Determine the flow rate Qin terms of 1212,, ,, a n d AA pp hρ . Assume the fluid to be inviscid. ------------------------------------------------------------------------------- Ans. Let the lower point be denoted as point 1, a nd the upper point denoted as point 2, we have () ()22 2 2 11 2 2 1 2 2 1// 2 ( 0 ) // 2 ( ) / ( ) / 2pvg p vg h p p g h v vρρ ρ++ = ++ → − − = − Let Qbe the flow rate, then 11 2 2 QA v A v== and () ()22 22 12 2 1 2 12 2221 122 1()2ppg h A A pp QQgh QAA AAρ ρ ρ⎡⎤ ⎡⎤−− ⎛⎞⎛⎞ − ⎣⎦⎢⎥ −= − → = ⎜⎟⎜⎟⎢⎥ − ⎝⎠⎝⎠⎣⎦ () ()12 1222 122( )ppg h QA A AAρ ρ⎡⎤−−⎣⎦→= − _________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-316.52 Verify that the equation of conservation of mass is automatically satisfied if the velocity components in cylindrical coordinates are given by 11, , 0rzvv vrz rrθψ ψ ρρ∂∂=− = =∂∂ where the density ρis a constant and ψis any function of rand z having continuous second partial derivatives. ------------------------------------------------------------------------------- Ans. The equation of continuity is ()110z rv vrvrr r zθ θ∂∂ ∂++ =∂∂ ∂. With 11, , 0rzvv vrz rrθψψ ρρ∂∂=−==∂∂, we have, ()2211 1 1 1 1,z rvrv rrr rr r z r r z z z r r r z rψ ψψ ψ ρρ ρ ρ⎛⎞ ⎛⎞ ⎛⎞ ⎛ ⎞ ∂ ∂∂ ∂∂ ∂ ∂ ∂=− =− = = ⎜⎟ ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎝⎠ ⎝ ⎠ ⎝⎠ ⎝⎠ Thus, the equation of continuity is au tomatically satisfied for any function (, )xyψ . _________________________________________________________________ 6.53 From the constitutive equation for a compressible fluid (2 / 3) 2 , /ij ij ij ij ij j jTp D k v xδμδμ δ =− − Δ + + Δ Δ=∂ ∂ , derive the equation 2 3jj ii i ii j j j i jvv Dv v pBkDt x xx x x xxμρρ μ⎛⎞ ⎛⎞∂∂ ∂ ∂∂ ∂=−+ + + ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ∂∂ ∂ ∂ ∂∂ ∂⎝⎠ ⎝⎠ ------------------------------------------------------------------------------ Ans. 221232 2 3ij j i ij ij ij jj j j j i j i ii j j i iTv v pkxx x x x x x v pkxx x x x xδμδμ δ μμ⎛⎞ ∂∂ ∂ ∂∂ Δ ∂ ∂ Δ=− − + + + ⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂ ∂∂⎝⎠ ⎛⎞∂ ∂∂ Δ ∂ Δ ∂ Δ=− − + + + ⎜⎟⎜⎟ ∂∂ ∂ ∂ ∂ ∂⎝⎠ That is, 2 3ij i ji i j j iT v pkx xx x x xμμ∂ ∂ ∂∂ Δ ∂ Δ=− + + +∂∂∂∂ ∂∂. Thus, ij i i jT DvBDt xρρ∂ =+→∂ 2 3ii i iij j iDv v pBkDt x x x x xμρρ μ∂ ∂∂ Δ ∂ Δ=−+ + +∂∂∂ ∂∂ _________________________________________________________________ 6.54 Show that for a one-dimensional, steady, adia batic flow of an ideal gas, the ratio of temperature 12/ΘΘ at sections 1 and 2 is given by () ()2 11 2221112 1112M Mγ γ+−Θ=Θ+− where γis the ratio of specific heat, 1Mand 2Mare local Mach number at section 1 and section 2 respectively. ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-32Ans. 2 constant12pvγ ργ⎛⎞+=⎜⎟−⎝⎠, we have 22 11 2 2 12 12 12p vp vγγ ργ ργ⎛⎞ ⎛⎞+= +⎜⎟ ⎜⎟−−⎝⎠ ⎝⎠. In terms of the Mach numbers 11 1 / M vc= and 22 2 / M vc= , we have, 22 22 11 1 2 2 2 12 12 12p cM p cMγγ ργ ργ⎛⎞ ⎛⎞+= +⎜⎟ ⎜⎟−−⎝⎠ ⎝⎠. For an ideal gas, p Rρ=Θ , and 2 pcRγγρ==Θ , therefore, () ()22 11 1 1 2 2 2 2 12 22 22 22 1 1 12 1 1 1 2 2 212 12 111111 2 2 2 2RR M R R M MMMMργ ργγγ ργ ργ γγ γγγγγγ⎛⎞ ⎛⎞ΘΘ ΘΘ+= +→⎜⎟ ⎜⎟−−⎝⎠ ⎝⎠ ⎛⎞⎛⎞ ΘΘΘ− Θ= − → Θ+ − Θ = Θ+ − Θ →⎜⎟⎜⎟−−⎝⎠⎝⎠ () ()() ()2 2 22 1 11 2 2 2 2 111 / 2 1111 1122 11 / 2MMM Mγγγ γ+−Θ ⎡⎤ ⎡⎤Θ+ − = Θ+ − → =⎢⎥ ⎢⎥Θ+− ⎣⎦ ⎣⎦. _________________________________________________________________ 6.55 Show that for a compressible fluid in isothermal flow with no external work, 2 22dM dv v M= , where Mis the Mach number. (Assume perfect gas). ------------------------------------------------------------------------------- Ans. Since 22 2 2/ a n d M vc c R γ ≡= Θ for ideal gas, therefore, 22/( ) M vRγ≡Θ For isothermal flow, Θ=constant , therefore, 22 22 2222 2 v vdv dM vdv R dvMd MR RR v Mvγ γγγΘ≡→=→= =ΘΘΘ. _________________________________________________________________ 6.56 Show that for a perfect gas flowing through a duct of constant cross sectional area at constant temperature 2 21 2dp dM p M=− . [Use the results of the last problem]. ------------------------------------------------------------------------------- Ans. We have, from ()() constant, 0 / / Av d v dv d dv vρρ ρ ρ ρ=+ = → = − Since constantΘ= , therefore, dp R d dpR d p R dpRρρρρρρΘ=Θ →= Θ→= =Θ Thus, / / dp p dv v=− . From the results of last problem, we have, 22/2 / dM M dv v = , therefore, 22/( 1 / 2 ) ( / )dp p dM M=− . _________________________________________________________________ 6.57 For the flow of a compressible inviscid fluid around a thin body in a uniform stream of speed mVin the 1xdirection, we let the velocity potential be () o1 1Vxϕ ϕ =− + , where 1ϕ is Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-33assumed to be very small. Show that for steady flow the equation governing1ϕ is, with oo o / M Vc= , ()222 2 111 o 222 12 310M xxxϕϕϕ∂∂∂−+ + = ∂∂∂. ------------------------------------------------------------------------ Ans. For steady flow, the equation of continuity is 0i i iivvxxρρ∂∂+=∂∂, in terms of the potential functionϕ, we have, 2 0 ii i ixx x xϕρ ϕρ∂∂ ∂→− − =∂∂ ∂ ∂. (i). The equation of motion is: 22 11 1, note : local sound speedi j ji i iv pp pvc cxx x xρρ ρρ ρρ ρ∂⎡ ⎤ ∂∂ ∂ ∂ ∂=− =− =− = = ⎢ ⎥∂∂∂ ∂ ∂ ∂ ⎣ ⎦ which becomes, 22 2 21 jj i i i jj icxxx x x x xx cϕϕρ ρ ρ ϕ ϕ ρ∂∂ ∂ ∂ ∂∂=− → =−∂∂ ∂ ∂ ∂ ∂∂ ∂ (ii) (ii) into (i) 22 20 ij j i i icxx x x x xϕϕ ϕ ϕ⎛⎞∂∂∂ ∂→− =⎜⎟⎜⎟∂∂ ∂ ∂ ∂ ∂⎝⎠ (iii). Now, with () o1 1Vxϕ ϕ =− + , we have, 1 o1 i iiVx xϕ ϕδ⎛⎞∂ ∂=− +⎜⎟∂∂⎝⎠, 2 2 11 1 o1 o 1 o ij i j ji i j jiVV Vx xx x x x x xϕϕ ϕ ϕϕ ϕδδ⎡ ⎤ ⎛⎞ ⎛ ⎞ ⎛⎞∂∂ ∂ ∂∂∂⎢ ⎥ →= − + +⎜⎟ ⎜ ⎟ ⎜⎟⎜⎟ ⎜ ⎟∂ ∂ ∂∂ ∂ ∂ ∂∂ ⎢ ⎥ ⎝⎠ ⎝⎠ ⎝ ⎠ ⎣ ⎦ 22 2 33 3 11 1 1 11 1 1 11oi j o i j o ij i j iVV Vx xx xx x xϕ ϕϕ ϕδδ δ δ⎡⎤ ⎛⎞∂∂ ∂ ∂≈− + ≈− =− ⎢⎥ ⎜⎟∂∂ ∂ ∂ ∂ ∂ ∂⎢⎥ ⎝⎠ ⎣⎦. Thus, Eq.(iii) 222 2 32 111 2 222 11 2 30ooVc V xx x xϕϕϕ ϕ⎛⎞∂∂∂ ∂→− + + + = ⎜⎟⎜⎟∂ ∂∂∂ ⎝⎠, ()22 2 2 11 1 o 22 2 12 310M xx xϕϕ ϕ⎛⎞∂∂ ∂→→ − + + = ⎜⎟⎜⎟∂∂ ∂⎝⎠. _________________________________________________________________ 6.58 For a one dimensional steady flow of a co mpressible fluid through a convergent channel, obtain (a) the critical pressure and (b) the co rresponding velocity. That is, verify equation (6.30.7) and Eq. (6.30.8) -------------------------------------------------------------------------------- Ans. (a) From Eq. (6.30.6), 1 21 2 22 21 1 112 1pp dmApdt p pγ γγ γργ+ ⎡⎤ ⎧⎫ ⎛⎞⎛⎞ ⎢⎥ ⎪⎪=− ⎨⎬⎜⎟⎜⎟ ⎢⎥−⎝⎠⎝⎠⎪⎪ ⎢⎥⎩⎭ ⎣⎦, (i) we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 6-34() ()21 1 222 11 21 1 1/ 12 2 1 /2 1dd md t pp dmpdp p d t p pγ γ γ γγργγ γ− −⎧ ⎫ ⎛⎞ ⎛⎞ + ⎪ ⎪ ⎛⎞=− ⎨ ⎬ ⎜⎟ ⎜⎟ ⎜⎟− ⎝⎠ ⎝⎠ ⎝⎠⎪ ⎪⎩⎭. (ii) Thus, () ()21/0/dd md t dp p= gives, 21 1 1 22 2 2 11 1 121 2 10, or 0pp p p pp p pγγ γγγ γγγ γγ γγ−− ⎡⎤ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ++ ⎢⎥−= − =⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎢⎥⎣⎦. (iii) That is, 1 2 121p pγ γγ γ γ− ⎛⎞ +=⎜⎟ ⎝⎠, therefore 12 12 1criticalp pγ γ γ− ⎛⎞ ⎛⎞= ⎜⎟ ⎜⎟+⎝⎠ ⎝⎠. (iv) (b) Substituting this critical pressure into Eq. (6.30.4) for the velocity, we get 1 1 2 12 1 1 1 2 11 1 1 122 1 2 1 21111 2 1 1 1critpp p p pvpγ γ γ γγ γ γ γ γρ γρ γρ γ ρ γ− −⎛⎞⎛⎞ ⎛⎞ ⎛⎞⎛⎞ +− ⎜⎟ ⎛⎞⎜⎟ =−=−= = ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟⎜⎟ −−− + + ⎝⎠ ⎝⎠⎝⎠ ⎝⎠⎜⎟ ⎝⎠⎝⎠. From 1 12 1 1 2 2 1 2 2 21 1 2 2 1 1 1 2 p pp p ppγγ γρρ ρ ρρ ρ ρρ ρ ρ ρ ρ−− −⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞⎛⎞=→ =→ =⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎝⎠. But 11 11 22 2 2 12 2 11 1 1 11 2p pp p p pp pγγγγγ γ ρρ ρ ρρ ρ−−−⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞⎛⎞=→ = → =⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎝⎠. Now, at 121 2 11 211,we hav e, 22criticalp pp pγ γγγ ρ ρ− ⎛⎞ ⎛⎞ ⎛⎞ ++⎛⎞== ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠. Thus, 2 2 2 2pvγρ⎛⎞=⎜⎟ ⎝⎠=speed of sound at section (2) . __________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-1 CHAPTER 7 7.1 Verify the divergence theorem divSVdS dV⋅=∫∫vn v for the vector field 122x z+ v= e e , by considering the region bounded by 0, 2,xx== . 0, 2,yy== 0, 2zz==. --------------------------------------------------------------------------------------- Ans. With122x z+ v= e e , we have For the face 0,x=1=−ne , 2x = 0, 0 dS ⋅= − ⋅ = ∫vn vn . For the face 2x=, 1=+ne , 2x = 4, 4 4(4) 16 dS A ⋅⋅ = = = ∫vn = + vn . For the face 0, y= 2=−ne , 2 22 0 0z, (2 ) 2 / 2 4 dS z dz z ⎡⎤ ⋅− ⋅= − = − = −⎣⎦ ∫∫vn = vn . For the face 2, y= 2=+ne , 2 0z, (2 ) 4 dS z dz ⋅⋅ = = +∫∫vn = vn . For the face 0, z=3=−ne , 0, 0 dS ⋅⋅ =∫vn = vn . For the face 2 z=, 3=+ne , 0, 0 dS ⋅⋅ =∫vn = vn . Thus, 16 4 4 16SdS⋅= − + =∫vn and () () div 2 2 2 2 2 16. dV dV== × × =∫∫v So, divSVdS dV⋅=∫∫vn v . ________________________________________________________________________ 7.2 Verify the divergence theorem divSVdS dV⋅=∫∫vn v for the vector field, which in cylindrical coordinates, is rz2rz+ v= e e , by considering the region bounded by 2 r=, 0z= and 4z=. --------------------------------------------------------------------------------------- Ans. For the cylindrical surface 2 r=, () r, 2 2(2) 4, 4 4 4 2 2 (4) 64 rd S d S S π π ⎡⎤ =→⋅ = = →⋅ = == =⎣⎦ ∫∫ne v n = v n . For the end face 0 z=, z0, 0 , 0zzd S= =− → ⋅ =− = → ⋅ = ∫ne v n v n . For the end face 4 z=, 2 z4, 4 , 4 4 21 6zzd S S ππ= =→⋅ = →⋅ == = ∫ne v n = v n . Therefore, 64 16 80SdSπππ ⋅=+=∫vn . ()2 2div 2 2 1 5rr zr vv v rz rr z r r z∂ ∂∂ ∂++ = ++= + + =∂∂ ∂ ∂v= . ()()2div =5 2 4 80VdVπ π= ∫v , thus, divSVdS dV⋅=∫∫vn v . ________________________________________________________________________ 7.3 Verify the divergence theorem divSVdS dV⋅=∫∫vn v for the vector field, which in spherical coordinates isr2r v= e , by considering the region bounded by the spherical surface 2r=. --------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-2Ans. For the spherical surface at 2 r=, ()2 r, 2 2(2) 4 , 4 4 4 4 2 64 rd S d S S π π →⋅ = =→ ⋅ = == = ∫∫n=e v n= v n On the other hand [see Eq.(2.35.26)], ()()233 22211 4 2div 6, div 6 643r drv d r dVdr dr rrππ⎛⎞ == = = ⎜⎟⎜⎟⎝⎠∫v= v . ________________________________________________________________________ 7.4 Show that 3SdS V⋅=∫xn . where x is the position vector and Vis the volume enclosed by the boundary S. --------------------------------------------------------------------------------------- Ans. 3 12 11 2 2 3 3 123, div 1 1 1 3x xxxx xxx x∂ ∂∂++ + + = + + =∂∂∂x= e e e x= . Thus div 3SVdS dV V⋅= =∫∫xn x . ________________________________________________________________________ 7.5 (a) Consider the vector field ϕv= a , where ϕ is a given scalar field and ais an arbitrary constant vector (independent of position). Using the divergence theorem, prove that VSdV dSϕϕ∇=∫∫n . (b) Show that for any closed surface Sthat 0SdS=∫n . --------------------------------------------------------------------------------------- Ans. (a) With ϕv= a , dS dS ϕϕ⋅⋅ → ⋅ = ⋅ ∫∫vn = an vn a n , () div divi i iiaaxxϕ ϕϕ ϕ∂ ∂=== ⋅ ∇∂∂v= a a . Thus, divSSdS dV⋅= →∫∫vn v .SVdS dVϕϕ⋅= ⋅ ∇∫∫an a Since ais arbitrary, therefore, SVdS dVϕϕ=∇∫∫n . (b) Take 1 ϕ=in the results of part (a), we have 0SdS=∫n . ________________________________________________________________________ 7.6 A stress field Tis in equilibrium with a body force ρB. Using the divergence theorem, show that for any volume Vand boundary surface S, that 0SVdS dVρ+ = ∫∫tB . where t is the stress vector. That is, the total resultant force is equipollent to zero. --------------------------------------------------------------------------------------- Ans. The stress vector tis related to the stress tensor Tby t=T n , therefore, divSS VdS dS dV==∫∫ ∫tT n T , thus, ( ) divSV VdS dV dVρρ+=∫∫ ∫tB T + B . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-3But in equilibrium, ( ) div 0ρ= T+ B , therefore, 0SVdS dVρ+ = ∫∫tB . ________________________________________________________________________ 7.7 Let u*define an infinitesimal strain field ()T 1 2⎡ ⎤ ∇∇⎢ ⎥ ⎣ ⎦*E= u * + u * and let **T be the symmetric stress tensor in static equilibrium with a body force ρ**B and a surface traction **t. Using the divergence theorem, verify the following identity (theory of virtual work). ()** * ij ijSV VdS dV T E dVρ+= ∫∫ ∫** **tu * B u *⋅⋅ . --------------------------------------------------------------------------------------- Ans. ( ) () ()TTdiv div( )SS S V VdS dS dS dV dV⎡ ⎤ =⋅ = ⋅ =⎢ ⎥ ⎣ ⎦ =∫∫ ∫ ∫ ∫** ** ** **tu * T n u * n T u * T u * **Tu *⋅ . Now, ()** * *() div divij j ij j j ji j i j ii i iTu T u u uT Tx xx x∂∂ ∂ ∂ == + = ⋅∂∂ ∂ ∂** ** ** ** ** **Tu * T u * + , therefore, ()**div / /ij j i ij j iSVdS dV T u x dV T u x dVρρ ⎡⎤ ⋅+ ⋅= ⋅ ∂ ∂= ∂ ∂⎣⎦ ∫∫ ∫ ∫** ** ** ** ** **tu * Bu * T + B u * + . Now, since ** ** ij jiTT= , therefore, ** * ** * * * * * * * ** ** ** * * ** ** ** **11 1 1 1 22 2 2 2 .jj j ii i ij ij ij ij ij ij ji j i ji ji j ii ij ji ij jjiuu u uu uTE T T T T Tx xx x x x u uuTTTxxx⎛⎞∂∂ ∂∂∂ ∂⎜⎟= + =+=+⎜⎟∂ ∂ ∂∂∂∂⎝⎠ ∂ ∂∂===∂∂∂ Thus, ()** * ij ijSV VdS dV T E dVρ ⋅+ ⋅=∫∫ ∫** *tu * B u * ________________________________________________________________________ 7.8 Using the equations of motion and the divergen ce theorem, verify the following rate of work identity. Assume the stress tensor to be symmetric. 2 2ij ijSV V VDvdS dV dV T D dVDtρρ⎛⎞ ⋅+ ⋅= + ⎜⎟⎜⎟⎝⎠∫∫ ∫ ∫tv Bv --------------------------------------------------------------------------------------- Ans. TTdiv( ) div( )SS S V VdS dS dS dV dV⋅= ⋅=⋅ = =∫∫ ∫ ∫ ∫t v Tn v n T v T v Tv Now, div( ) divij j ij j j ji j i j ii i iTv T v v vT Tx xx x∂∂ ∂ ∂ ==+= ⋅∂∂ ∂ ∂Tv T v + , therefore, ( ) [d i v / ] ( ) /ij j i ij j iSV V VdS dV T v x dV D Dt dV T v xρρ ρ ⋅+ = ⋅ ∂ ∂ = ⋅+ ∂ ∂∫∫ ∫ ∫tv Bv T + B v + v / v ⋅ . Now, ()21=22D D Dv Dt Dt Dtρρ ρ⎛⎞ ⋅ ⎛⎞⋅= ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠vv vv and 1 2jj ii ij ij ij ij ij ji j ivv vvTD T T Tx xx x⎛⎞∂ ∂ ∂∂=+ = =⎜⎟⎜⎟∂∂∂ ∂⎝⎠. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-4Therefore, 2 2ij ijSV V VDvdS dV dV T D dVDtρρ⎛⎞ ⋅+ ⋅= + ⎜⎟⎜⎟⎝⎠∫∫ ∫ ∫tv Bv . ________________________________________________________________________ 7.9 Consider the velocity and density fields1 ()o 11 o=, ttxeααρ ρ−−= ve (a) Check the equation of mass conservation. (b) Compute the mass and rate of increase of ma ss in the cylindrical control volume of cross- section A and bounded by 10x=and13x=. (c) Compute the net mass inflow into the control volume of part (b). Does the net mass inflow equal the rate of mass increase inside the control volume? --------------------------------------------------------------------------------------- Ans. (a). () ()()oo oo div 0tt tt DeeDtαα ρρα ρ ρ α−− −−+= − + = v . That is, the conservation of mass equation is satisfied. (b) Inside the control volume, () () () oo o3 o1 o o03, a n d / 3tt tt ttmd V e A d x e A d m d t e Aαα αρρ ρ α ρ−− −− −−== = = −∫∫. That is, the mass inside the volume is decreasing at the rate of ()o o3tteAααρ−−. (c). Rate of inflow from the face 10x=is zero because at 10x=, 10.v= Rate of outflow from the face 13x= is given by()o 11o 33tt xvA e Aαρα ρ−− == . There is no flow across the cylindrical surface b ecause flow is only in the 1xdirection. Thus, the rate of outflow exactly equals the rate of decrease of mass inside the volume. ________________________________________________________________________ 7.10 (a) Check that the motion ()o 11 2 2 3 3 , , ttxXe x X x Xα−== = corresponds to the velocity field 11=xαve . (b) For a density field() o otteαρρ−−= , verify that the mass contained in the material volume that was coincident with the control volume of Prob. 7.9 at timeot, remains a constant at all times, as it should (conservation of mass). (c) Compute the total linear momentum for the material volume of part (b). (d) Compute the force acting on the material volume --------------------------------------------------------------------------------------- Ans. (a) () o 3 12 111 2 3 , 0 , 0tt x xxvX e x v vtt tααα− ∂ ∂∂== = == ==∂∂ ∂, i.e., 11=xαve . (b) The particles which are at 10x=at time othave the material coordinates10 X=. These particles remain at 10x=at all time. The particles which are at 13x=at time othave the material 1 It should be remarked that, for a real fluid, to achieve the given velocity and density fields in this and some other problems may require body force distributions and/or a pressure density relationship that are not realistic . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-5coordinates13 X=. These particles move in such a way that()o 13ttxeα−= . Thus, to find the mass inside this material volume as a function of time, we have ()() () ()o oo o3 o1 o o 033tte tt tt ttM eA d x eA e Aα αα αρ ρρ− −− −− − ⎡⎤== =⎢⎥⎣⎦ ∫. (c) Linear momentum in the material volume ()() ()() ()() ()()()oo oo oo oo o33 o 1 11 o 1 11 00 2 3 o1 1 1 o 1 o 1 099.22tt tt ttee tt tt tt e tt tt ttev A d x e x A d x eeAx d x eA A eαα ααα α αα αρρ α ρα ρα ρα−− −−− −− − −− −− −== ⎡⎤ ⎢⎥ == = ⎢⎥⎣⎦∫∫ ∫Pe e ee e (d) Force acting on the material volume () o 2 o19 2tt dAedtαρα−==PFe . We see that both the linear momentum and the fo rce increase exponentially with time. This is due to the given data of density and velocity fields , which describe the space occupied by the fixed material increases exponentially with time,o() 103ttxeα− ⎡ ⎤ ≤≤⎣ ⎦,while the density decreases exponentially () o otteαρ−−⎡⎤ ⎢⎥⎣⎦to conserve the mass. We note also that at ott=, the materials occupy the space between 10x=and13x=, and2 o19 2Aρα F= e . ________________________________________________________________________ 7.11 Do Problem 7.9 for the velocity field 11xαv= e and the density fieldo1/kxρρ= and for the cylindrical control volume bounded by 11x=and13x=. --------------------------------------------------------------------------------------- Ans. (a). oo 1 11 2 11 1 1div ( ) 0v Dvx k kDt x x x xρρ ρρρρ α α⎛⎞ ∂∂++ = − + = ⎜⎟⎜⎟ ∂∂⎝⎠v= . That is, the conservation of mass equation is satisfied. (b) Inside the control volume,3o 1o11ln 3, and 0dmmd Vk A d x k Axd tρρρ== = =∫∫. (c). Rate of inflow from the face 11x=is [] 1 1o 11 o 1 1x xvA k xA k Axρρ αα ρ =⎡⎤==⎢⎥ ⎣⎦. Rate of outflow from the face 13x= is given by [] 1 1o 11 o 3 1 3x xvA k xA k Axρρ αα ρ= =⎡⎤==⎢⎥ ⎣⎦. There is no flow across the cylindrical surface because flow is only in the 1xdirection. The net mass inflow is 0 , which is equal to the rate of increase of mass inside the control volume. ________________________________________________________________________ 7.12 The center of mass .cmxof a material volume is defined by the equation . mcmVmd V ρ=∫xx , where mVmd Vρ=∫ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-6Demonstrate that the linear momentum principle may be written in the form .cmSVdS dV mρ+=∫∫tB a where .cmais the acceleration of the mass center. --------------------------------------------------------------------------------------- Ans. We have from the principle of linear momentum: m SV VDdS dV dVDtρρ+=∫∫ ∫tB v Now, since () 0DdVDtρ=, therefore, () () ..cm cmDD DdV dV dV m mDtD t D tρρ ρ=== =∫∫ ∫vx x x v . Therefore, ..cm cm SVDdS dV m mDtρ+= =∫∫tB v a . ________________________________________________________________________ 7.13 Consider the following velocity field and density field o 1 1, 11x ttρ αρα α=+ +v= e (a) Compute the total linear momentum and rate of increase of linear momentum in a cylindrical control volume of cross-sectional area Aand bounded by the planes 11x=and13x=. (b) Compute the net rate of outflow of lin ear momentum from the control volume of (a) (c) Compute the total force on the material in the control volume. (d) Compute the total kinetic energy and rate of in crease of kinetic energy for the control volume of (a). (e) Compute the net rate of outflow of kinetic energy from the control volume. --------------------------------------------------------------------------------------- Ans. (a) Linear momentum is ()33 oo 1 11 1 11 2 1111 1A xdV Adx x dxtt tρρ ααραα α== =++ +∫∫ ∫Pv e e () ()oo 11224 91 22 11AA ttρα ρα αα⎛⎞=− =⎜⎟⎝⎠ ++ee . Rate of increase of linear momentum inside the control volume is ()2 o 138 1A d dt tρα α=− +Pe. (b) Net rate of outflow of linear momentum in 1edirection =()()() () ()112 22 22 oo 11 22 3318 9 1 11 1xxAAAv Avt tt tρα ρ ααρρααα α==⎛⎞ ⎜⎟ −= − =⎜⎟+++ +⎝⎠. (c) Total force = Rate of inc. of P inside control volume + net outflux of P = ()2 o 138 1A tρα α− +e+ ()2 o 138 1A tαρ α+e=0. (d) Total kinetic energy inside the control volume () ()2 33 22 22 oo o 1 11 1 33 1111 1 1 3..22 1 1 2 3 11AA xK E vd V A d x xd xtt ttρ ρα ρα αραα αα⎛⎞== = = ⎜⎟++⎝⎠ ++∫∫ ∫ ()3 o 413. 1A dKEdt tρα α=− +. (e) Net rate of outflow of kinetic energy from the control volume= Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-7() () ()113 33 33 oo 11 33 4 3113 11 1 2 7 22 2 1 11 1 xxAAAv Avt tt tρ αρ ααρρααα α ==⎛⎞⎛⎞ ⎛⎞⎜⎟ −= − = ⎜⎟ ⎜⎟⎜⎟+ ⎝⎠ ⎝⎠ ++ +⎝⎠. ________________________________________________________________________ 7.14 Consider the velocity and density fields ()o 11 o=, ttxeααρ ρ−−= ve For an arbitrary time t, consider the material contained in th e cylindrical control volume of cross- sectional area A , bounded by 10x=and13x=. (a) Determine the linear momentum and rate of increase of linear momentum in this control volume. (b) Determine the outflux of linear momentum. (c) Determine the net resultant force that is acting on the material contained in the control volume. --------------------------------------------------------------------------------------- Ans. (a) Linear momentum inside the control volume: () () () () ()oo oo o33 o 1 11 o 1 11 00 3 o1 1 1 o 1 o 1 099.22tt tt tt tt ttdV e v Adx e x Adx eA x d x eA A eαα αα αρρ ρα ρα ρα ρα−− −− −− −− −−== = ⎡⎤== =⎢⎥⎣⎦∫∫ ∫ ∫Pv e e ee e Rate of increase of linear mome ntum inside the control volume () o 2 o19 2tt dAedtααρ−−=−Pe. (b) Out flux of linear momentum from the control volume in the1edirection: ()()( ) ()()o 11 1 12 2 22 22 2 11 1 1 o30 3 09tt xx x xA v A v Ax Ax A eαρ ρ ρα ρα ρ α−− == = =−= − = . (c) The total force = rate of increase of lin ear momentum inside the control volume + net momentum outflux from the control volume. Thus, () () () oo o 22 2 o 1 o1 o199922tt tt ttA e Ae Aeαα αα ρ ρα ρα−− −− −−−+=F= e e e We see that the force exerted on the material w ithin the control volume decreases exponentially with time. This is due to the give n data of density field and velocity field, which states that within the fixed space defined by103x≤≤ , the density decreases exponentially with time while speed at each spatial point is independent of time. We also note that at ott=,2 o19 2Aρα F= e , the same results was obtained in Problem 7.10. ________________________________________________________________________ 7.15 Do Problem 7.14 for the same velocity field, 11=xαve but with o 1kxρρ= and the cylindrical control volume bounded by 11x=and13x=. --------------------------------------------------------------------------------------- Ans. (a) Linear momentum inside the control volume: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-8() () ()333 o1 1 1 1 o1 1 1 1 o 1 1 o 111 1// 2 dV k x v Adx k x x Adx k A dx k Aρ ρ ρ α ρα ρα== = = =∫∫ ∫ ∫Pv e e e e Rate of increase of linear mome ntum inside the control volume .d dt=P0 (b) Out flux of linear momentum from the control volume in the1edirection: ()() ()() ()( ) 11 1 122 2 2 2 2 2 11 o 1 1o 1 1 o31 3 1// 2 xx x xAv Av k x A x k x A x k Aρ ρρ αρ α ρ α == = =−= − = . (c) The total force = rate of increase of lin ear momentum inside the control volume + net momentum outflux from the control volume 22 o1 o1 02 2 kA kAρα ρα+= F= e e ________________________________________________________________________ 7.16 Consider the flow field () 12 =kx y− ve e with ρ=constant. For a control volume defined by 0, 2, 0, 2, 0, 2xxyyzz====== , determine the net resultant force and moment about the origin (note misprint in text) that are acting on the material contained in this volume. --------------------------------------------------------------------------------------- Ans. Since the flow is steady, the resultant for ce = net linear momentum outflux through the three pairs of faces: (i) through 0 x=and 2 x=, ()() () () () () () ( )22 1 1 12 1220 20 22 2 22 12 12 00 0 22 12 1242 2 42 28 4 1 6 8 .x x xx yz yv dA v dA k x x y dydz k x x y dydz k y dz dy k y dy kkρρ ρ ρ ρρ ρρ= = == == =⎡ ⎤⎡ ⎤ −=−−−⎣ ⎦⎣ ⎦ ⎡⎤ ⎡⎤ ⎢⎥ =− = − ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ =− =−∫ ∫ ∫∫ ∫∫ ∫∫ ∫v v ee ee ee ee ee eei (ii) through 0 y=and 2 y=, ()() ()() ()() () () () ( )22 2 2 12 1220 20 22 222 12 1200 0 22 2 2 12 1 2 1 2224 2 24 24 2 4 8 8 1 6 .yy yy xz x xv dA v dA k y x y dxdz k y x y dxdz k x dz dx k x dx kx x k kρρ ρ ρ ρρ ρρ ρ= = == == = =⎡ ⎤⎡ ⎤ −= − −− − −⎣ ⎦⎣ ⎦ ⎡⎤=− + = − +⎢⎥⎣⎦ ⎡⎤=− + =− + = − +⎣⎦∫ ∫ ∫∫ ∫∫ ∫∫ ∫vv e e e e ee ee ee e e e e (iii) through 0 z=and 2 z=, ()() 33 3200, ( =0) zzvd A vd A vρρ ==−= ∫∫vv . Thus, the total net force () ( ) ( )22 2 12 1 2 1216 8 8 16 8 8 kk kρρ ρ−+− + = + F= e e e e e e . The flow is steady, the resultant moment about a point = net moment of momentum outflux about the same point. Take the point to be the origin, then Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-9(i) through 0 x=and 2 x=, ()( ) ()( ) () () () () () ()2 11 1 2 3 1 22 20 2222 2 2 12 3 1 2 300 2 222 12 3 1 2 3022 4 8 48 1 6 8 1 6 3 2 .x xx yz x yvd A vd A k x x y z x y d y d z k xyz x z x y dydz k yz z y dz dy ky d y kρρ ρ ρρ ρρ= == == = =⎡ ⎤ ×− ×= + + × −⎣ ⎦ ⎡⎤ ⎡⎤=+ = +⎢⎥ ⎣⎦ ⎣⎦ =+ = +∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫rv rv e e e e e ee - e e e - e ee - e e e - e(i) through 0 y=and 2 y=, ()( ) ()( ) ()( )() () () () ( )2 22 1 2 3 1 22 20 2222 2 2 12 3 12 3002 222 12 3 1 23024 2 8 8 4 16 16 8 32 .y yy xzy yvd A vd A k y x y z x y d x d z k y z xyz xy dxdz k z xz x dz dx kx x d x kρρ ρ ρρ ρρ= == === =⎡ ⎤ ×− ×=− + + × −⎣ ⎦ ⎡⎤ ⎡⎤=− − = − −⎢⎥ ⎣⎦ ⎣⎦ =− − = − −∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫rv rv e e e e e ee + e ee + e ee + e e e + e (iii) through 0 z=and 2 z=, ()( ) ()( ) 33200 zzvd A vd Aρρ ==×− ×=∫∫rv rv Thus, ( )( )()22 o1 2 3 1 2 3 1 2 81 6 3 2 1 68 3 2 88 kk kρρ ρ+− +−− = − + M= e e e e e+ e e e . ________________________________________________________________________ 7.17 For Hagen-Poiseuille flow in a pipe, ()22 o1Cr r=−ve . Calculate the momentum flux across a cross-section. For the same flow rate, if the velocity is assumed to be uniform, what is the momentum flux across a cross section? Compare the two results. --------------------------------------------------------------------------------------- Ans. Momentum flux across a cross section = ()() ()222 2 2 2 6 11 o 1 1o2/ 3or o vd A C r r rd r C rρρ π ρ π=− =∫∫ee e Volume flow rate is ()()()22 4 1oo2/ 2or o Q v dA C r r r dr C r ππ == − =∫∫. The uniform flow which has the same flow rate Q is given by : 22 o1 o 1 (/ ) ( / 2 )Qr C rπ= v= e e . The momentum flux across a cross section for this uniform flow is given by 26 11 1 (/ 4 )o Qv C rρπ ρ=ee . Thus, the momentum flux for the Hagen-Poiseuille flow is 4 3 that of the uniform flow. ________________________________________________________________________ 7.18 Consider a steady flow of an inco mpressible viscous fluid of density ρ, flowing up a vertical pipe of radius R. At the lower section of the pipe, the flow is uniform with a speed lv and a pressurelp .After flowing upward through a distance l, the flow becomes fully developed with a parabolic velocity distribution at the upper section, where the pressure is up. Obtain an expression for the fluid pressure drop lupp− between the two sections in terms of ρ,R and the frictional force fF, exerted on the fluid column from the wall though viscosity. --------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-10Ans. Let the control volume encloses the fluid be tween the two sections. The linear momentum theorem states that: For steady flow, Force on the fluid= Momentum outflux – momentum influx. The force on the fluid in the control volume is given by: ()() lu fppA g A Fρ −−−l . The momentum influx through the lower section = ()22 lvRρπ . The momentum outflux through the upper section ()22uvr d rρπ∫, where 22()uvC Rr=− . The constant C can be obtain as follows ()22 4 2 2 0022 () / 2 2 /RR ul l Q v rdr C R r rdr CR v R C v R ππ π π==− = = → =∫∫. Thus, 22 22()l uvvR r R=− . Momentum outflux () () ()222 2 2 2 oo22 2 / ( )RR ulvr d r v R R r r d rρπ ρ π =− ∫∫ () ()()()24 22 2 24 6 2 2 o8/ ( ) 8/ / 6 4 / 3R ll lvR R r r d r vR R v Rρπ ρπ ρπ=− = = ∫. Thus, ()()2 2 22 224/ 3 ( ) ( ) / 3lu f l l lpp A g A F v R v R v R ρρ π ρ π ρ π −− − = − = l () ()22 2 2() () / 3 ()lu l fpp R v R F g Rπρ π ρ π →− = ++ l. That is, ()22/3 /( )lu l fp pv F Rgρ πρ −= + + l. ________________________________________________________________________ 7.19 A pile of chain on a table falls through a hole from the table under the action of gravity. Derive the differential equati on governing the hanging length x. [Assume the pile is large compared with the hanging portion] --------------------------------------------------------------------------------------- Ans. Using a control volume ()2Vc [see Fig. 7.6-1 in Section7.6] enclosing the hanging down portion x of the chain, we can obtain the same equation as that given in Eq. (iv) of Section 7.6, i.e., with μdenoting / ml, mass per unit length: 22/ gx T xd x dtμμ−= (1) where Tis the tension on the chain at the hole. Next, using a control volume enclosing the pile above the table, then, since the particles of the ch ain pile stay essentially at rest at any given instant (except those near the hole), we can assume that the rate of change of momentum inside the control volume is zero (quasi-static approxima tion). Further, we assume that the net force acting at the pile is the tension Tat the hole (the reaction of the supporting table exactly balances the weight of the pile). Then, the momentum principle gives: ()2/ Td x d tμ= (2). Equations (1) and (2) give 2 2 2dx d xgx xdt dt⎛⎞=+ ⎜⎟⎝⎠ (3) We note that this equation is a good approximation wh en the length of the pile is large compared with the hanging portion x. Eventually, when the pile reduces to essentially a flat straight segment on the table, Eq. (vi) of Section 7.6 becomes a better approximation. ________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-117.20 A water jet of 5 . cmdiameter moves at 12 / sec m , impinges on a curved vane which deflects it o60from its original direction. Neglect the weight, obtain the force exerted by the liquid on the vane. (see Fig. 7.6-2 of Example 7.6.2). --------------------------------------------------------------------------------------- Ans. Referring to Fig. 7.6-2, we have, o o12 / , =60 , volume flow rate vm s θ = () () ( )22 2 4 3 4 oo( / 4) 12 (5 10 ) / 4 235.6 10 , 998 (235.6 10 ) 12 282 Qv d m Q v Nππ ρ−− −== × = × = × = Thus, force on the jet = () () ( )oo o1 o 21 2 1 21 cos60 sin 60 282 0.5 282 0.866 141 244 Qv Qv Nρρ−− + = − + = − + ee e e e e . Force on the vane from the jet is 12141 244 N−ee . ________________________________________________________________________ 7.21 A horizontal pipeline of 10 .cmdiameter bends through o90 , and while bending, changes its diameter to 5 . cm The pressure in the 10 . cmpipe is 140 . kPa Estimate the resultant force on the bends when 0.005 3/s e cm .of water is flowing in the pipeline. --------------------------------------------------------------------------------------- Ans . Let () ,,uuuvpA and ( ) ,,dddvpA denote upstream and downstream (speed, pressure and cross-sectional area) respectively and Q the volume discharge. We have, 30.005 / , Qm s= ()() ()()220.005 / 0.1 / 4 0.6366 / , 0.005 / 0.05 / 4 2.546 /udvm s v m s ππ == = = Upstream pressure 140,000 up Pa = . Down stream pressure can be obtained from Bernoulli Equation: 22 22uu ddp vp v ρρ+=+ . Thus, () ( )22 2 2 998140,000 0.6366 2.546 137,00022du u dpp v vρ=+ − = + − = . Let 1ebe the direction of the incoming flow and 2ebe the direction after the o90 bend, then, we have, Momentum outflux = ()()()22 998 .005 2.546 12.7dQρ== ee v . Momentum influx = ()()() 1 998 .005 0.6366 3.18uQvρ== e. Momentum principle gives: 12 21 uu dd d upAp A Q v Q v ρ ρ − += −w ee F e e . ( )( ) () ( ) () ( )12 22 12 1 2Force on water 3.177 140,000( 0.1 / 4) 12.7 137000 0.05 / 4 1100 282 .uu u dd dQv p A Qv p A Nρρ ππ=− + + + =− + + + =− +wFe e ee e e Thus, the force from water to the bend is 12 1100 282 . N −=−wFe e ________________________________________________________________________ 7.22 Figure P7.1 shows a steady water jet of area A impinging onto the flat wall. Find the force exerted on the wall. Neglect weight and viscosity of water. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-12vo vovo --------------------------------------------------------------------------------------- Ans. Let the control volume be coincident with th e outline of the flow shown in the figure. Force on the liquid LF= momentum outflux-momentum influx = o1Qvρ−0e Force on the wall =2 o1 o 1Qv Avρρ=ee . ________________________________________________________________________ 7.23 Frequently in open channel flow, a high speed flow “jumps” to a low speed flow with an abrupt rise in the water surface. This is known as a hydraulic jump . Referring to Fig. p7.2, if the flow rate is Q per unit width, show that when the jump occurs, the relation between 1yand2y.is given by 2 11 21 18 1122yvyygy=− + + Assume the flow before and after the jump is unifo rm and the pressure distribution is hydrostatic. y1y2v1 --------------------------------------------------------------------------------------- Ans. Use a control volume enclosing the water with an upstream section before the jump and a downstream section after the jump. According to the momentum principle, the force on the fluid per unit width is given by (neglect friction from the ground and air) 22 12 2 1/2 /2xF gy gy Qv Qvρρρ ρ=−= − , thus , ()()()22 12 2 11 1 2 1 (/ 2 )g yy Q v v v y v v−= − = − . Conservation of mass gives: () 1 12 2 211 1211 1 2 2 // vy v y v v vy y v v y y y=→ − = − = − . Therefore, we have, () ()22 2 21 2 1 1 1 2(/ 2 )gyy y v y y y −= − . The above equation shows that 12yy= is a root for the equation. This solution corresponds to a flow without a jump. To look for the jump solution, we eliminate the factor () 12yy− and obtain () () ()22 2 21 2 1 1 2 1 2 1 1 22 2 21 1 1 1 1 1 1 1/2 ( 2 / ) 0 11/2 ( 8 / ) /2 1 8 /2gy y y v y y y y v g y yy y v y g y y v g y+= → + − = ⎡⎤=− + + = − ++⎢⎥⎣⎦ ________________________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-137.24 If the curved vane of Example 7.6.2 moves with a velocity ovv<in the same direction as the oncoming jet, find the resultant fo rce exerted on the vane by the jet. --------------------------------------------------------------------------------------- Ans. Fig. 7.6-2 of Example 7.6.2 is reproduced below. vovo AB e1e2 Let the control volume surrounding the jet moves with the vane, then the flow is steady with respect to the moving control volume. Momentum outflux relative to the control volume = () ()2 o1 2 o 1 2() c o ss i n () c o ss i nQv v Av vρθ θ ρ θ θ−+ = − + ee ee Momentum influx relative to the control volume is 2 o1 o 1() ()Qv v Av vρρ−= −ee Thus, since the control volume moves with a constant speed, there is no extra term to be added to the momentum equation for the fixed control vo lume case. Thus, force acting on the jet is () ( )22 2 jet o 1 2 o 1 o 1 2 () c o ss i n () () c o s 1 s i nAv v Av v Av vρθ θ ρρ θ θ ⎡ ⎤ =− + −− =− − +⎣ ⎦Fe e e e e and the force on the vane is ()2 vane o 1 2 () 1 c o s s i nAv vρθ θ ⎡⎤ =−− −⎣⎦Fe e . ________________________________________________________________________ 7.25 For the half-arm sprinkler shown in Fig. P7.3, find the angular speed if 30.566 / sec.Qm= Neglect friction. 1.83 md=2.54 c m --------------------------------------------------------------------------------------- Ans. Let the control volume cVrotate with the arm. Then, rela tive to the control volume, the outflux of moment of momentum about an axis passing through Oand perpendicular to the plane of the paper is ()o3 / QQ Arρ e, where oris the length of the arm. There is no influx of moment of momentum about the same axis since the inflow is parallel to it. Since the control volume is rotating with an angular velocity ωabout the same axis, we need to add terms to the left hand side of Eq. (7.9.8) , the moment of momentum principle. The terms that need to be added are given in Eq. (7.9.9). With 3ω=eω and1xx= e , we have, 2xω×x= eω , ()2 1 xω ××= − xeωω so that ()0 ×× × =xxωω . We also have,o0 and 0ω= = a & , therefore, the only non-zero term is () () ()13 1 22 / dm x Q A Adx ωρ −× × = − × ×∫∫xv e e eωo 2 3o 302rQx d x Q rρω ρω=− =−∫ee . Adding this term to Eq. (7.9.8), whose left ha nd side is zero (because frictional torque is neglected) and whose right hand side is the ne t moment of momentum outflux, we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-142 o3 o 3 o (/) / ( ) Qr Q Q A r Q Arρω ρ ω−= → = − ee . Now, 22 4 2(2.54 10 ) / 4 5.067 10 Amπ−−=× = × , therefore, 4 o/ ( ) 0.566 / [(5.067 10 )(1.83)] 610.4 /QA r r a d sω−=− =− × =− . The minus sign means the rotation is clockwise looking from the top. ________________________________________________________________________ 7.26 The tank car shown in fig. P7.4 contains wate r and compressed air which is regulated to force a water jet out of the nozzle at a constant rate of 3/s e c . Qm The diameter of the jet is .dc m , the initial total mass of the tank car isoM. Neglecting frictional forces, find the velocity of the car as a function of time. d --------------------------------------------------------------------------------------- Ans. Let the control volume cV encloses the whole tank car and moves with the car. Then relative to the control volume, the momentum outflux is ()22 2 11 4/ 4 / ( )QQ d Q dρπ ρ π−= − ee . There is no momentum influx. Since the contro l volume moves with the car which has an acceleration o1ae, therefore, the momentum principle in the 1edirection takes the form [see Eq.(7.8.20}: (with al frictional/resistance force neglected): ( )22 o (/) 4 / ( ) M Qt dv dt Q dρρ π−− = − . ( )22 o /[ 4/ ( ) ] /dv dt Q d M Qt ρπ ρ →= − . Integrating, we have, ( )2 o [4 / ( )]ln vQ d M Q t C πρ =− − + . If the initial velocity is zero then we have ( )2 oo [4 / ( )] ln lnvQd M Q t M πρ⎡ ⎤ =− − +⎣ ⎦. ________________________________________________________________________ 7.27 For the one dimensional problem discussed in Section 7.10, (a) from the continuity equation 11 2 2vvρρ= and the momentum equation 22 12 2 21 1p pv vρρ −= − , obtain 22 2 11 1111vp vp Mγ⎛⎞=− − ⎜⎟ ⎝⎠ (b) From the energy equation 22 12 12 1211 12 1 2ppvvγγ γρ γρ+= +−−, obtain 2 122 2 12 1 2 22 2 1 11 111 221p pvv v v v aa vγγ−−+=+⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠ (c) From the results of (a) and (b), obtain ()2 22 22 11 1122 11011 2ppMMppγγγγγ⎛⎞ ⎛⎞ −⎛⎞−+ − −=⎜⎟ ⎜⎟ ⎜⎟++ ⎝⎠ ⎝⎠ ⎝⎠. --------------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-15Ans. (a) ()22 11 2 2 1 2 2 2 11 1 2 11 2 1 Using , v v pp v v pp v vvρρ ρ ρ ρ=− = − → − = − → () () ()22 1 1 21 21 21 1 1 1 2 2 1 2 2 11 1 11 1 1 11 1 21 2 21 2 2 22 2 2 11 1 11 1 11 12 11/11 1 / 1/ 1/ 111 . T h a t i s , 1 . //1vv v v v p vv v p p v p pp v p v v vp pp v pp v p vv p vp v a Mv vρ ρρ ρ γρ γ γ γ⎛⎞ ⎛⎞−− ⎛⎞ −−= = = − → + = ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎛⎞ −−→+ = → + = − ⎜⎟ ⎝⎠=− (b) 22 2 2 12 1 2 1 1 12 1 2 12 2 11 111 1 1 12 1 2 2 21pp pvv v vp ppρρ ρ γγ γ γ γρ γρ γ ρ γ−−+= + + = + −−→ 22 11 122 2 2 2 11 2 1 2 1 2 22 2 2 2 21 11 1 1 1 2 11 1 1 2 211 1 1 22 2 2, note / and = .11pp ppvv v v v v v aa a a v pa v vγγ γγ ρρ ρ γρ ρ−− −−+= + +=+⎛⎞ ⎛⎞ ⎛⎞→→ ⎜⎟ ⎜⎟ ⎜⎟⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎡⎤ =⎣⎦→ . (c) 22 2 12 2 1 2 22 2 11 11 122 211 1111 22Using ,the equation 111vp v v v pv aa vvp vp Mγγ γ−−+=+⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠=− − → 2 12 22 22 11 2211 11111 221111 11p pppMMpp MMγγ γγ−−++⎡⎤ ⎡⎤ ⎛⎞ ⎛⎞ ⎢⎥ ⎢⎥ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎢⎥ ⎢⎥ ⎝⎠ ⎝⎠ ⎣⎦ ⎣⎦=− − − − 2 2 2 1 12 2 2 22 2 2 21 22 2 411 1 1 1 11 1 2242 2111 221 1111.11 2 111 12 1 11 111222p p Mpp p p pMpp p p p MM M pMMMp MM γγ γγ γ γ γγγγγγ γ γγ⎡⎤⎛⎞ ⎡ ⎤⎛ ⎞ ⎛⎞ ⎛⎞⎢⎥⎜⎟ ⎢ ⎥⎜ ⎟ ⎜⎟ ⎜⎟⎜ ⎟ ⎜⎟ ⎜⎟ ⎢⎥⎜⎟ ⎢ ⎥⎝ ⎠ ⎝⎠ ⎝⎠⎢⎥⎝⎠ ⎣ ⎦⎣⎦ ⎡⎤ ⎛⎞⎡⎤⎡ ⎤ ++ ⎢⎥ ⎜⎟⎢⎥⎜⎟ ⎣ ⎦⎢⎥ ⎣⎦⎝⎠ ⎣⎦−=− + + + − − + − + −−=− + + + Thus, ()2 2 24 2 1 1 11 22 2 11 12 22 111 11 1 1 11 21221 2MMM M MM Mpp ppγ γγγγγγ γγ γγ γ⎡⎤ ⎡⎤ ⎛⎞ + −−⎡ ⎤ ++ + + + ⎢⎥ ⎢⎥ ⎜⎟⎜⎟ ⎣ ⎦⎢⎥ ⎢⎥⎝⎠ ⎣⎦ ⎣⎦−+= . Rearranging, () ()2 2 1 1 22 11 22 22 1 11 2 12 22 111 11 1 2 2 11 12122 11 2MM MM MMM Mpp ppγ γ γ γγ γ γγγγγγγ γ⎡⎤ ⎡ ⎤ ⎛⎞ + −+ ⎢⎥ ⎢ ⎥ ⎜⎟⎜⎟⎢ ⎥ ⎢⎥⎝⎠ ⎣ ⎦ ⎣⎦ ⎡⎤ −−++ + − ⎢⎥− ⎢⎥⎣⎦−+= () ()22 11 22 2 11 12 22 1112 11 1 1121102MM MM Mpp ppγγ γ γγ γ γγ γγ γ⎡⎤ ⎡ ⎤⎡ ⎤ ⎛⎞ +−++ − ⎢⎥ ⎢ ⎥⎢ ⎥ ⎜⎟⎜⎟− ⎢ ⎥⎢ ⎥ ⎢⎥⎝⎠ ⎣ ⎦⎣ ⎦ ⎣⎦−+→= That is, () () ()()2 12 12 22 1121 1 2011 2M Mpp ppγ γγγγ+ ⎛⎞ ⎛⎞−−− − =⎜⎟ ⎜⎟⎜⎟ ⎜⎟++⎝⎠ ⎝⎠. The above equation has two solutions: Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 7-16() ()21 22 21 1 1 2 1 1(1) 11(2) 2 1 , 2 111pp p vp o r p M pργ γ γγγ= ⎡⎤ ⎡ ⎤=− −=− −⎣⎦ ⎣ ⎦++ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-1 CHAPTER 8 8.1 Show that for an incompressible Newtonian fluid in Couette flow, the pressure at the outer cylinder ()o rR= is always larger than that at the inner cylinder. That is, obtain () () ()o 2 o iR rr rr iRTR TR r r d r ρω ⎡⎤ ⎡⎤−− −=⎣⎦ ⎣⎦ ∫ ------------------------------------------------------------------------------- Ans. In Couette flow, 0 and rzBvv v A rrθ == =+ , [see Eq. (6.15.4) and (6.15.7)]. Thus, rr zzTT T pθθ=== − , 0rz zTTθ== , and = function of onlyrdv vTrdr rθθ θμ⎛⎞=−⎜⎟⎝⎠. Thus, the r-equation of motion 2 1rr r rr rz TT T TTrrr r zθθ θρωθ∂ − ∂∂++ + = −∂∂ ∂becomes: 2 rrTrrρω∂=−∂. Now, ()oo 2 iiRRrr RRTdr r r drrρω∂=−∂∫∫. Thus,. () () ()o 2 o iR rr rr i RTR TR r r d r ρω ⎡⎤ ⎡⎤−− −=⎣⎦ ⎣⎦ ∫. The right hand side of this last equation is always positive. ____________________________________________________________ 8.2 Show that the constitutive equation 123 , with / 2 1,2,3nn n n tnλ μ ++ +∂ ∂ = = D, τ=τ τ τ τ τ is equivalent to 22 33 2 12 3 o 12// / / / at a t a t b bt b t+∂∂ +∂ ∂+∂ ∂ = + ∂∂ +∂ ∂2DD D τττ τ where ( )( ) () ( ) ( ) () ()11 2 3 21 2 2 3 3 1 3 1 2 3 o1 2 3 11 2 3 2 1 3 3 2 1 2 123 213 312,, 2, 2 2aa a bb bλλλ λ λλ λλ λ λ λ λ μμμ μ λλ μ λλ μ λλ μλλ μλλ μλλ=++ = + + = ⎡⎤ =+ + = + + + + +⎣⎦ =+ + ------------------------------------------------------------------------------- Ans. ()33 3 3 3 3 3 11 1 1 1 1 1 33 3 3 3 3 11 1 1 1 1 33 112 2j j jj i jii j i i ii j i j i j iii i i ii j i i j ji ji ii i iitt t t tt tλλ λ λ λλμ λ μλ== = = = = = == == = = ≠≠ ==∂ ∂ ∂∂= ∂∂ ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂⎛⎞ ⎛ ⎞ ⎛⎞⎛⎞ ⎛⎞⎜⎟ ⎜ ⎟== ⎜⎟⎜⎟ ⎜⎟⎜⎟ ⎜ ⎟⎝⎠⎝⎠ ⎝⎠ ⎝⎠ ⎝ ⎠ =+ =− + ⎛⎞=− +⎜⎟⎜⎟⎝⎠∑∑ ∑ ∑ ∑ ∑ ∑ ∑∑ ∑∑ ∑ ∑ ∑∑D Dτ τ τ ττ τ ττ τ τ33 3 33 11 1 112j ii ij i ij ji jittμλ == = == ≠≠∂ + ∂=−∑∑ ∑ ∑∑ Dτ τ That is, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-233 211 233 11 22 33 112ii iitt t t t t tλμ λ λ λ λ λ λ ==∂∂∂∂∂ ∂ ∂+++ ∂∂ ∂ ∂ ∂ ∂ ∂⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=−+ + +⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠∑∑ Dτττ ττ τ ττ (i) Next, we have, ( ) ( ) ( ) () ( ) ( ) ()22 2 2 2 2 12 23 31 12 23 31 1 12 23 31 2 2 2 22 22 2 2 12 23 31 3 21 31 1 23 1 12 23 2 22 22 22 31 2 23 31 3 12 3/// // / / // / .ttt tt t t tt tλλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ λλ++∂ ∂ = ++∂ ∂ + ++∂∂ +++∂ ∂ =+∂ ∂ + ∂ ∂ ++∂ ∂ + ∂ ∂+ + ∂ ∂+ ∂ ∂τττ ττ ττ ττ τ i.e., ( ) ()()() () () ()22 12 23 31 1 2 3 2 1 3 3 2 1 22 2 2 231 132 213 2 3 1 3 1 2 22 12 3/ 222 / /// / / /.tt ttt t t tλλ λλ λλ μ λ λ μ λ λ μ λ λ λλ λλ λλ λ λ λ λ λλ⎡⎤ ++∂ ∂ = + + + + +∂ ∂⎣⎦ −+∂∂ −+∂∂ −+∂∂ + ∂∂ +∂ ∂ +∂∂D τ τττ τ τ τ (ii) Finally, we have, 3 33 3 3 12 123 123 123 123 33 3 3 3 12 23 1 13 2 12 3 22 22 22 1 231 13 2 12 3 23 222 222 2 222tt t t tt tt tt ttt tλλλ λλλ λλλ λλλ λλ μ λλ μ λλ μ λλμ λλμ λλμ λλ λ∂ ∂∂ ∂=++ ∂∂∂∂ ⎛⎞ ⎛⎞ ⎛⎞ ∂ ∂∂∂∂∂=− +− +− ⎜⎟ ⎜⎟ ⎜⎟⎜⎟ ⎜⎟ ⎜⎟∂∂ ∂∂ ∂∂⎝⎠ ⎝⎠ ⎝⎠ ∂ ∂∂∂=++− − ∂∂∂ ∂2 22 22 2 2 222DD D DDDτ ττ τ τ ττ τ3 2 13 12 22ttλλ λ∂ ∂− ∂ ∂2 2τ τ that is, 3 3 12 123 231 13 2 12 3 23 13 12 3222 2 2 2222 tttt t t tλλλ λλμ λλμ λλμ λλ λλ λλ∂ ∂∂ ∂∂∂∂=++− − − ∂∂∂∂ ∂ ∂ ∂2 22 222DDD τ ττ τ(iii) Thus, (i) + (ii)+ (iii) gives ( )( ) () ( ) ( ) () ()22 33 123 1 22 33 1 1 2 3 123 1 23 2 1 3 3 2 1 2 123 213 312// / 22 / 2/ .tt t t tλλλ λ λλ λλ λ λ λ λ μμμ μ λλ μ λλ μ λλ μλλ μλλ μλλ+++∂∂ + + + ∂∂ + ∂∂ ⎡⎤ =+ + + + ++ ++ ∂ ∂⎣⎦ ++ + ∂ ∂2DD Dτ ττ τ That is, 23 12 3 o 1 2 23 2taa a b bbt tt t∂+ ∂∂∂ ∂ ∂++= + +∂ ∂∂ ∂2DDDττττ. where ( )( ) () ( ) ( ) () ()11 2 3 21 2 2 3 3 1 3 1 2 3 o1 2 3 11 2 3 2 1 3 3 2 1 2 123 213 312,, 2, 2 2aa a bb bλλλ λ λλ λλ λ λ λ λ μμμ μ λλ μ λλ μ λλ μλλ μλλ μλλ=++ = + + = ⎡⎤ =+ + = + + + + +⎣⎦ =+ + ____________________________________________________________ 8.3 Obtain the force-displacement relationship for the Kelvin-Voigt solid, which consists of a dashpot (with damping coefficient η) and a spring (with spring constant G) connected in parallel. Also, obtain its relaxation function. ------------------------------------------------------------------------------- Ans. Since the spring and the dashpot are connected in parallel, therefore, the total force is given by: sp dash SS S=+ and the total displacement ε is given by spd a s hεεε== . Now, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-3 and =sp dashdSG Sdtεεη= , therefore, dSGdtεεη=+ . To find the relaxation function, we let o()Htεε= , where ( ) Ht is the Heaviside function. Then oo() () SGH t tεηε δ = + . Thus, the relaxation function is o/( ) ( )SG H ttε ηδ =+ . ____________________________________________________________ 8.4 (a) Obtain the force-displacement relationship for a dashpot (damping coefficient oη) and a Kelvin-Voigt solid (damping coefficient ηand spring constant G, see the previous problem) connected in series. (b) Obtain its relaxation function. ------------------------------------------------------------------------------- Ans. (a) Let and kv dSS be the force transmitted by the Kelvin-Voight element and the dashpot respectively and let and kv dεεbe the elongation of the Kelvin-Voight element and the dashpot respectively. Then we have, the total force is given by dk v SS S== (i) and the total displacement is given by dk vεεε=+ (ii), where odk v dk v k vddSS S Gdt dtε εηε η== = = + (iii). From (ii) and (iii) we have, () () oo1dk v kv ddd dSS S GSGdt dt dtεε εε εεηη ηη η=+= +− = + −− (iv). Thus, 2 o 2 odd dd S G d G dt dt dt dtηηεεε ηη η η⎛⎞+=− +⎜⎟ ⎝⎠, or, ()2 o o o 2dd S dSGG d t d tdtηη ηηεεη+= −+ . Thus, the force-displacement relationship is given by: ()2 o o o 2dS d dSGd t d tG dtηη ηηε εη++= + . (v) (b) Let o()Htεε= , where ( ) Ht is Heaviside function. Then Eq. (v) gives () () ()oo o o oo o()G dS G dStdt dtεηε η η δδηη ηη ηη+= +++ +, (vi) where ()tδ is the Dirac function. The integration factor for this ODE is ()o exp / Gtηη⎡⎤+⎣⎦. Thus, () ()oo o oo o o oo()G t Gt Gt G ddSe e t edt dtηη ηη ηηεη εη η δδηη ηη++ +⎡⎤ ⎢⎥=+⎢⎥ ++⎣⎦ and () ()oo o oo o o oo()Gt Gt Gt tt tG dSe e t dt e dtdtηη ηη ηηεη εη η δδηη ηη++ + =−∞ −∞=+++∫∫ ()()oo /( ) /( ) oo o o oo o()t tGt Gt G Get e d tηη ηη εη εη ηδδηη ηη ηη++ −∞ −∞⎧⎫⎡⎤ =+ − ⎨⎬⎣⎦ ++ + ⎩⎭∫. That is, ()() () ()oo o2 oo o o o o o 2 oo o oo() ()Gt Gt Gt o GG GS e et etηη ηη ηηεη εη η εη εη ηδδηη ηη ηη ηη ηη++ +⎧⎫ ⎧⎫⎪⎪ ⎪⎪=+ −= + ⎨⎬ ⎨⎬++ + + + ⎪⎪ ⎪⎪⎩⎭ ⎩⎭. Thus, the relaxation function is () ()o2 oo 2 ooo()Gt G Setηηηη ηδεη ηηη− +=++ +. ____________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-48.5 A linear Maxwell fluid, defined by Eq. (8.1.2 ), is between two parallel plates which are one unit apart. Staring from rest, at time 0t=, the top plate is given a displacement o uv t= while the bottom plate remains fixed. Neglect inertia effects, obtain the shear stress history. ------------------------------------------------------------------------ Ans. The velocity field for the fluid in this motion is given by (inertia neglected) 1o 223 () , 0 vv H t xv v== = , where ()Ht is the Heaviside Function. The only non-zero rate of deformation component is () 12 o /2 Dv H t= . Thus, from the constitutive equation for the linear Maxwell fluid, we obtain, ()12 12 odSSv H tdtλμ+= . Thus, ()// o 12tt v dSe e Htdtλλμ λ⎡⎤=⎣⎦. That is, () ()// / / / oo o 12 o0 01t tttt t t t vv vSe e Ht d t e d t e v eλλ λ λ λμ μμλμλ λλ−∞⎡⎤ == = = −⎣⎦ ∫∫. Thus, the shear stress history is: ()/ 12 o 1tSv eλμ−=− . ____________________________________________________________ 8.6 Obtain Eq. (8.3.1) i.e., ()/ '' ' 2( ) ( ) , w h e r e ( ) /tttt t d t t eλφφ μ λ− −∞−=∫S= D , by solving the linear non-homogeneous ordinary differential equation 2d dtλμ=SS+ D . ------------------------------------------------------------------------- Ans. The integration factor for this ODE is [] exp /tλ. That is the equation can be written as; ()// 2 tt deedtλ λμ λ= SD . Thus , () ( )//2/tttee t d tλλμλ−∞= →∫SD () ( )//2/tttee t d tλλμλ′ − −∞′′ →= ∫S D . That is, ()() ( )()'/'' ' '' 22tt ttet d t t t t d tλμφλ−− −∞ −∞=≡ −∫∫SD D . ____________________________________________________________ 8.7 Show that for the linear Maxwell fluid, defined by Eq. (8.1.2), ()'' ' ()ttt J t d t tφ−∞−=∫, where ()tφ is the relaxation function and ()Jtis the creep compliance function. ------------------------------------------------------------------------------- Ans. Let () 12 oSS H t= be applied to the top plate of a channel of unit depth in which is the linear Maxwell fluid. [ ()Ht is the unit step function, i.e., Heavis ide function]. Neglecting inertia, the velocity field is ()2o 2vx vx= , where ovis the velocity of the top plate. Then from the constitutive equation / 2 dd tλ μ= S+ S D, we obtain () () oo 1 2 o o 22 / 2 / SS t D v d u d tλδ μ μ μ== = +, where ()out is the displacement of the top plate. From ()oo odu SStdtλδμμ= +, we obtain ()oo o o oo ooo o()tt tdu S S Sdt dt S t u t S tdtλλδ λμμ μ μ μ=→ = + = +∫∫∫+. Thus, the creep compliance function is: oo () / ( ) /Jt u S t λμ == + . Since the relaxation function is /() ( / )tteλφμ λ−= , therefore, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-5() ( )() () () ()'/ / //(/) ( ) / ( 1 /) ( ) (1 / ) ( ) .tt tt t tt tt tt ttt t J t dt e t dt e t dt ed t e t d tλ λ λλφμ λ λ μ λ λ λ−− ′−− −∞ −∞ −∞ ′′−− −− −∞ −∞′′ ′ ′ ′ ′ ′−= + = + ′′ ′ =+∫∫ ∫ ∫∫ Now () ()'/ /[]tt t tt t t ed t eλ λλ λ−− ′−− ′=−∞−∞′== ∫ and ()()() () () // / / 2 ' () ( )ttt t tt tt tt tt te t dt de t dt e t e dt tλλ λ λλ λλ λ λ′′ ′ ′−− −− −− −− −∞ −∞ −∞ ′=−∞⎡⎤ ′′ ′ ′ ′== − = −⎢⎥⎣⎦ ∫∫ ∫. Thus, ()()()()2 '' '1 ttt J t d t t t tφλ λ λ λ λλ−∞−= + − = + − =∫. ____________________________________________________________ 8.8 Obtain the storage modulus and loss modulus for the linear Maxwell fluid with a continuous relaxation spectrum defined by Eq. (8.4.1), i.e., ()() / ot Hte dλλφ λλ∞ −=∫. ------------------------------------------------------------------------------- Ans. Let the shear strain be: 12 oiteωγγ= . For this strain history, the rate of deformation history is given by 12 12 o2eit dDidtω γωγ== . Thus, from the constitutive equation, () ( ) 2ttt t d tφ−∞′′ ′−∫S= D , we have () ( ) () 12 12 o2ettitSt t D t d t i t t d tωφω γ φ′ −∞ −∞′′′ ′ ′−=−∫∫= . With () ( )/ o/ttH e dλφ λλ λ∞−⎡⎤=⎣⎦∫, we have , ()() () '/ / / 12 o oo= ottt tt it t i t tHH eSi e e d d t i e e d t dλ λωλ ω λλλωγλ ω γ λλλ−∞∞ ′−− ′′ ′ −∞ =−∞′ ′ ∫∫ ∫ ∫== Now, () ()() 1/ 1 / / 1tt it it ti t tteed t e d t eiλωλ λ ωλ λω λ λω′++ ′′ ′′=−∞ =−∞′′==+∫∫. Thus, () ()* 12 o =o1it it H Sie d G eiω ω λλωλγλ ω∞ ≡+∫= , where () ()* =o1HGi diλλω λλω∞ +∫= is the complex modulus. Now, () ()()() () ()* =o =o1 11 1Hi HGi d i dii iλλλλ ω λω λω λλω λω λω∞∞ −=++ −∫∫= 22 22 22 22 =o =o =o( ) () () () (1 ) (1 ) (1 )iH H Hdd i dλλ λωλ ω λ λ ω λ ω λλ λλ λω λω λω∞∞ ∞+== + ++ +∫∫ ∫ Thus, 22 22 22 =o =o() (), (1 ) (1 )HHGd G dλλλω λ λ ω λλ λ λλ ω λλ ω∞∞′′ ′== ++∫∫ ____________________________________________________________ 8.9 Show that the viscosity μof a linear Maxwell fluid, define by () ( ) 2ttt t d tφ−∞′′ ′−∫S= D , is related to the relaxation function ()tφ and the memory function ()fsby the relation () ()oosds sf s ds μφ∞∞== −∫∫. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-6------------------------------------------------------------------------ Ans. ( ) () () ( ) () ( )0 12 12 12 12=022 2t ts sS t t D t d t s D t sd s s D t sd s φφ φ∞ ′−∞ =∞ =′′ ′ =− = − − = −∫∫ ∫. For simple shearing flow, 12 2 3 , 0 vk xvv== = , 122Dk=, so that () () 12 1200/ssSk s d s S k s d s φμ φ∞∞ ===→ = =∫∫. Now, the memory function ()fsis related to the relaxation function ()sφ by the relation ( ) / ( ) dsd s f sφ= . Thus, () () ()0 00 0()dssds s s s ds sf s dsdsφμφ φ∞∞ ∞ ∞⎡⎤ ==− = −⎣⎦∫∫ ∫. ____________________________________________________________ 8.10 Show that the relaxation function for the Jeffrey model [Eq. (8.2.7)] with 20 a= is given by [note: Reference to Eq.(8.2.7) is miss ing in the problem statement in the text] 1/ o 12 1 1 o1 o 1 o() 1 () , () D i r a c F u n c t i o n2ta b Sb bte t tab a bφδ δγ−⎡⎤⎛⎞== − + = ⎢⎥⎜⎟ ⎢⎥⎝⎠⎣⎦. ------------------------------------------------------------------------ Ans. Let the shear strain 12γ be given by 12 ()Ht γγ=o . Then 12 12 o2/ ( )Dd d t tγγδ = = , where ()tδ is Dirac function. From the constitutive equation, we have, ()11oo o o 12 12 1 12 1 o 1 12 11 1 // o 1 12 o 111 22 2 2ta tab SS bSa b b Stt t a a a t b bSe eta a tγγ γ δ δδδ δγδ⎛⎞ ∂∂ ∂∂+=+ → + = + ⎜⎟∂∂ ∂ ∂ ⎝⎠ ⎛⎞ ∂∂→= + ⎜⎟∂∂ ⎝⎠ 11 1 11 1// / o 12 1 o1 1 // / oo 11 1 1 2 11 1 1 11 1()2( ) 1() () ()ttta ta ta t tta ta tab Sb dtee t d t e d taa d t bb bb b bet t e d t etaa a a aa aδδγ δδ δ−∞ −∞ −∞ −∞→= + ⎡⎤=+ − =+ −⎣⎦∫∫ ∫ Thus, the relaxation function is: () () ()11// oo 12 1 1 1 1 o1 1 o 1 1 1 o o1122 2ta ta bb Sb b b bte t e taa b a a a b bφδ δγ−−⎡ ⎤ ⎛⎞ ⎛⎞≡= − + = − + ⎢ ⎥ ⎜⎟ ⎜⎟⎢ ⎥ ⎝⎠ ⎝⎠ ⎣ ⎦ ____________________________________________________________ 8.11 Given the following velocity field: () 12 1 30, , 0vv v x v=== . Obtain (a) the particle pathline equations using the current time as th e reference time, (b) the relative right Cauchy- Green deformation tensor and (c) the Ri vlin-Ericksen tensors using the equation () ()2 12 -- / 2 . .t ttττ=+ +CI + A A (d) the Rivlin-Ericksen tensor 2Ausing the recursive equation, [] [ ] [] [] [] []T 21 1 1 / DD t=+ ∇ + ∇AA A v v A etc. ------------------------------------------------------------------------------- Ans. (a) Let iix′′=xe be the position at time τ of the particle which is at iixx= e at time t. Then () 123,,,iixx x x x τ ′′= gives the pathline equation. Thus, ()3 12 12 10 (i) , (ii) , 0 (iii)dx dx dxvv v xdd dττ τ′ ′′′ == == = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-7with the initial conditions: () 123,,,iixxxxxt′= . Eq (i) gives () 11 2 3 1 ,, x fxxx x′= =, Eq. (iii) gives () 31 2 3 3 ,, x gxx x x′== . Eq. (ii) becomes, ()2 1dxvxdτ′=→()() 21 1 2 3 ,, x vx hx x xτ′=+ , ()()()() ()()'2 21 1 2 3 1 2 3 21 21 ,, ,, x v xt h xx x h xx x x v xt x x v x t τ →= + → =− → →=+ − . Thus, ()() 11 2 2 1 3 3 , , , xxxxv x t xx τ ′′ ′== +−= (b) [][ ] ( ) () () 1110 0 1 0 0 /1 0 1 0 , / 00 1 0 0 1tt dv dx t k t k dv dxττ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥′=∇ = − = − ≡⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦Fx [][] []() ()()() () ()()2 2 T 2 210 1 0 0 1 0 010 1 0 10 00100 1 0 01 100 0 0 2 00 010 00 0 002001 000 0 00tt tkt k tkt kt kt kktktττ τ ττ ττ⎡ ⎤ ⎡⎤−+ − −⎡⎤⎢ ⎥ ⎢⎥ ⎢⎥== −= − ⎢ ⎥ ⎢⎥ ⎢⎥⎢ ⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎢ ⎥ ⎣ ⎦ ⎡⎤ ⎡⎤ ⎡⎤− ⎢⎥ ⎢⎥ ⎢⎥=+ − + ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦CF F (c) [] []2 12 100 2 0 0 0 0 , 0 0 0 , 000 0 00kk dvkkdx⎡⎤ ⎡⎤⎢⎥ ⎢⎥== = ⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎣⎦ ⎣⎦AA (d) [] [] [] [] []T 1 21 1D Dt⎡⎤=+ ∇ + ∇⎢⎥⎣⎦AAA v v A , where []100 00 000k k⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦A . Thus, [] [ ]()111 1 1 1 000ij k k ij ijDDvDt t Dt t x∂∂∂ ⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤=+ ∇→ = + = + =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂∂ ∂ ⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦AAA A AAv . [] [] [] []22 T 1100 0 0 0 0 0 0 0 0 0 0 0 00 0 , 00 0 000000 0 00 0 00kk k kk⎡ ⎤⎡ ⎤ ⎡⎤ ⎡⎤⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥∇= = ∇ = ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣ ⎦⎣ ⎦Av v A . Thus, [] [] [] [] []2 T 21 120 0 00 000 0k ⎡ ⎤ ⎢ ⎥=∇ + ∇ = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦AA vv A . ____________________________________________________________ 8.12 Given the following velocity field: 11 2 2 3 , , 0 vk x v k x v=−= = . Obtain (a) the particle pathline equations using the current time as the reference time, (b) the relative right Cauchy- Green deformation tensor and (c) the Ri vlin-Ericksen tensors using the equation () ()2 12 -- / 2 . .t ttττ=+ +CI + A A (d) the Rivlin-Ericksen tensor 2Aand 3Ausing the recursive equation, [] [ ] [] [] [] []T 21 1 1 / DD t=+ ∇ + ∇AA A v v A etc. ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-8Ans. (a) (a) Let iix′′=xe be the position at time τ of the particle which is at iixx= e at time t. Then () 123,,,iixx x x x τ ′′= gives the pathline equation. Thus, 3 12 11 2 2 (i) , (ii) , 0 (iii)dx dx dxvk x v k xdd dτττ′ ′′′′== − == = with the initial conditions: () 123,,,iixxxxxt′= . Now, () () ()()1 '1 11 2 3 1 2 3 1 11 1 1 1 1ln , , , , ln ln ln ln ln .ktdxkx x k g x x x g x x x x ktd xk x k t x xktx x eτττ ττ−−′′=− → =− + → = + ′′ ′→= − ++ →−= − − → = Similarly, () 2 22 2kt dxkx x x edτ τ− ′′′=→ = and () 31 2 3 3 ,, x fxxx x′== Thus, () () 11 2 2 3 3 , , ,kt ktxxe x x e x xττ−− −′′′== = (b) [][ ]() ()[][] []() ()2 T 200 0 0 00 , 0 0 00 1 0 0 1kt kt kt kt tt t t tee eeττ ττ−− − − −−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥⎢⎥ ⎢ ⎥′=∇ = = =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎣ ⎦Fx C F F Since ()() () ()2322 2 4812 . . .23 !kt kkek t t tτττ τ−=− + − − +mmm , therefore, [][]()() ()23 23 2340 0 80 0 20 0 0 2 0 0 4 0 0 8 0 ...23 !00 0 0 0 0 0 0 0tkk ktttk k kτττ⎡⎤ ⎡ ⎤ − −⎡⎤⎢⎥ ⎢ ⎥ −− ⎢⎥⎢⎥ ⎢ ⎥ =+− + + +⎢⎥⎢⎥ ⎢ ⎥⎢⎥⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦CI (c) [] [] []23 23 12 340 0 80 0 20 0 0 2 0, 0 4 0, 0 8 000 0 0 0 0 0 0 0kk k kk k⎡ ⎤⎡ ⎤ − −⎡⎤⎢ ⎥⎢ ⎥⎢⎥⎢ ⎥⎢ ⎥ == =⎢⎥⎢ ⎥⎢ ⎥⎢⎥⎣⎦ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦AA A (d) with 11 2 2 3 , , 0 vk x v k x v=− = = , [] [] 100 2 0 0 00 0 2 0 00 0 0 00kk kk−−⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥∇= → =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦vA [] [ ]()111 1 1 1 000ij k k ij ijDDvDt t Dt t x∂∂∂ ⎡ ⎤ ⎡⎤ ⎡ ⎤⎡⎤=+ ∇→ = + = + =⎢ ⎥ ⎢⎥ ⎢ ⎥⎢⎥∂∂ ∂ ⎣ ⎦ ⎣⎦ ⎣ ⎦⎣⎦AAA A AAv [] [] [] [] []T 21 1 2 240 0 20 0 0 0 0 0 20 0 020 0 0 0 0020 04 0 00 0 0 0 0 0 0 0 00 0 0 0 0k kk kk kk k k k=∇ + ∇ ⎡ ⎤−−− −⎡ ⎤ ⎡⎤ ⎡⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥ =+=⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣ ⎦ ⎣⎦ ⎣⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦AA vv A Next, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-9[] []23 23 240 0 40 0 00 04 0 0 0 0 4 0 00 0 0 0 0 0 0 0kk k kk k⎡⎤ ⎡ ⎤ − −⎡⎤⎢⎥ ⎢ ⎥⎢⎥⎢⎥ ⎢ ⎥∇= =⎢⎥⎢⎥ ⎢ ⎥⎢⎥⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦Av . Thus, [][ ] [] [] [] []3 T 3 32 280 0 00 8 0 00 0k k⎡ ⎤ −⎢ ⎥ ⎢ ⎥ =+ ∇ + ∇ = ⎢ ⎥ ⎢ ⎥⎣ ⎦AA v v A . ____________________________________________________________ 8.13 Given the following velocity field: 11 2 2 3 3 , , 2 vk x v k x v k x= == − . Obtain (a) the particle pathline equations using the current tim e as the reference time, (b) the relative right Cauchy-Green deformation tensor and (c) th e Rivlin-Ericksen tensors using the equation () ()2 12 -- / 2 . .t ttττ=+ +CI + A A (d) the Rivlin-Ericksen tensor 2Aand 3Ausing the recursive equation, [] [ ] [] [] [] []T 21 1 1 / DD t=+ ∇ + ∇AA A v v A etc. ------------------------------------------------------------------------------- Ans. (a) Let iix′′=xe be the position at time τ of the particle which is at iixx= e at time t. Then () 123,,,ixx xx τ ′ gives the pathline equation. Thus, 3 12 11 2 2 3 (i) , (ii) , 2 (iii)dx dx dxv kx v kx kxdd dτ ττ′ ′′′′′== == = − with the initial conditions: () 123,,,iixxxxxt′= . Now, () () ()()1 11 1 2 3 1 2 31 11 1 1 1 1l n ,, ,, l n ln ln ln ln .ktdxkx x k g x x x g x x x x ktd xk xk t x xk t xx eτττ ττ−′′′=→ =+ → = − ′′ ′→= +− →−=− → = Similarly, () 2 22 2kt dxkx x x edτ τ− ′′′=→ = and () 2 33ktxx eτ−−′= . Thus, () () () 2 11 2 2 3 3 , , kt kt ktxx e x x e xx eττ τ−− − −′′′=== (b) []() () ()[][] []() () ()2 T 2' 2400 0 0 00 , 0 000 0 0kt kt kt kt t tt t t kt ktee ee eeττ ττ ττ−− −− −− −−⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎡⎤=∇ = = =⎣⎦⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎣ ⎦Fx C F F Since ()() () () ()() () ()2323 2 2323 4481 2 ...,23 ! 16 641 4 ...,23 !kt ktkkek t tt kkek t t tτ τττ τ ττ τ− −−=+ − + − + − + =− − + − − − + therefore, [][] t=CI Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-10()() ()23 23 23 2340 0 800 200 0 2 0 0 4 0 0 8 0 ...23 !00 4 00 1 6 006 4kk ktttk k k k kkτττ⎡⎤ ⎡ ⎤⎡⎤⎢⎥ ⎢ ⎥ −− ⎢⎥⎢⎥ ⎢ ⎥ +− + + +⎢⎥⎢⎥ ⎢ ⎥⎢⎥ − − ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ (c) [] [] []23 23 12 3 2340 0 800 200 0 2 0 , 0 4 0 , 0 8 000 4 0 0 16 0 0 64kk k kk k k kk⎡ ⎤⎡ ⎤⎡⎤⎢ ⎥⎢ ⎥⎢⎥⎢ ⎥⎢ ⎥ == =⎢⎥⎢ ⎥⎢ ⎥⎢⎥ − − ⎣⎦ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦AA A (d) with 11 2 2 3 3 , , 2 vk x v k x v k x=== − , [] [] 100 2 0 0 00 0 2 0 00 2 0 0 4kk kk kk⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥∇= → =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥−−⎣ ⎦⎣ ⎦vA [] [ ]()111 1 1 1 000ij k k ij ijDDvDt t Dt t x∂∂∂ ⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤=+ ∇→ = + = + =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂∂ ∂ ⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦AAA A AAv [] [] [] [] []T 21 1 2 2 240 0 200 0 0 0 0 200 02 0 0 0 0 0 02 0 0 4 0. 00 40 02 0 02 00 4 00 1 6k kk kk kk k k k kk k k k=∇ + ∇ ⎡ ⎤⎡ ⎤ ⎡⎤ ⎡⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥ =+=⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ −− − − ⎣ ⎦ ⎣⎦ ⎣⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦AA vv A Next, [] []23 23 2 2340 0 400 00 04 0 0 0 04 0 00 2 0 0 16 0 0 32kk k kk k k kk⎡⎤ ⎡ ⎤⎡⎤⎢⎥ ⎢ ⎥⎢⎥⎢⎥ ⎢ ⎥∇= =⎢⎥⎢⎥ ⎢ ⎥⎢⎥− − ⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦Av . Thus, [][ ] [] [] [] []3 T 3 32 2 3800 00 8 0 006 4k k k⎡ ⎤ ⎢ ⎥ ⎢ ⎥ =+ ∇ + ∇ = ⎢ ⎥− ⎢ ⎥⎣ ⎦AA v v A Etc. ____________________________________________________________ 8.14 Given the following velocity field: 12 21 3 , , 0 vk x v k x v= == . Obtain (a) the particle pathline equations using the current time as th e reference time, (b) the relative right Cauchy- Green deformation tensor and (c) the Ri vlin-Ericksen tensors using the equation () ()2 12 -- / 2 . .t ttττ=+ +CI + A A (d) the Rivlin-Ericksen tensor 2Aand 3Ausing the recursive equation, [] [ ] [] [] [] []T 21 1 1 / DD t=+ ∇ + ∇AA A v v A etc. ------------------------------------------------------------------------------- Ans. (a) Let iix′′=xe be the position at time τ of the particle which is at iixx= e at time t. Then () 123,,,iixx x x x τ ′′= gives the pathline equation. Thus, 3 12 12 21 (i) , (ii) , 0 (iii)dx dx dxvk x v k xdd dττ τ′ ′′′′== == = Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-11with the initial conditions: () 123,,,iixxxxxt′= . Now, 22 22 11 2 1 21 1 22 11(i) 0 sinh cosh sinh cosh (iv)dx d x dx d xkx k k x k xdd dd xA k B k xA k t B k tττ ττ ττ′′ ′ ′′′ ′ →=→ = = → − = → ′=+→ =+ 1 221(ii) cosh sinh cosh sinh (v)dxxA k B k x A k t B k tkdτττ′′→= = + →= + (iv) and (v) gives 12 1 2sinh cosh , cosh sinh A x kt x kt B x kt x kt=− + = − ( )( ) 11 2 cosh cosh sinh sinh cosh sinh sinh cosh xx kt k kt k x kt k kt k τ ττ τ ′=− +− ( )( ) 21 2 cosh sinh sinh cosh cosh cosh sinh sinh x x kt k kt k x kt k kt k τ ττ τ ′=− + − That is, () () 11 2 cosh sinh xxk t xk tττ ′=− +− , () () 21 2 sinh cosh x xk t x k tττ ′= −+ − , 33x x′= (b) [][ ]()() () () [][] [] ()22 T 22cosh sinh 0 sinh cosh 0 , 00 1 cosh sinh 2cosh sinh 0 2cosh sinh sinh cosh 0 , 00 1tt tt tkt kt kt kt xx x x x xx x x k tττ ττ τ⎡⎤−− ⎢⎥′=∇ = − −⎢⎥ ⎢⎥⎣⎦ ⎡⎤+⎢⎥ ⎢⎥ == + ≡ − ⎢⎥⎢⎥⎣⎦Fx CF F Since 23 45 3 22 4 2 2 4 5c osh 1 O( ) , sinh O( )26 2cosh 1 O( ), sinh O( ), sinh cosh O( )3xxxx x x x xx x x xx x x xx x=+ + =+ + =+ + = + =+ + [] [] () () ()23 32 23 23 2341 2 .. 2 ... 0302 0 42 ... 1 2 ... 0 2 0 0300 000 1 40 0 0 8 0 0 4 0 8 0 0 ...2600 0 00 0txx x k x xx k t kk ttkkτ ττ⎡⎤++ ++⎢⎥ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥=++ ++ = + −⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦⎢⎥ ⎢⎥⎣⎦ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ −−⎢⎥ ⎢⎥+++ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥⎣⎦ ⎣⎦CI Thus, (c) [] [] []23 23 12 340 0 0 8 0 02 0 2 0 0, 0 4 0, 8 0 0 . 00 0 0 00 0 00kk k kk k e t c⎡⎤ ⎡⎤⎡⎤⎢⎥ ⎢⎥⎢⎥⎢⎥ ⎢⎥ == =⎢⎥⎢⎥ ⎢⎥⎢⎥⎣⎦ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦AA A Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-12(d) with 12 21 3 , , 0 vk x v k x v=== , [] [] 100 0 2 0 00 2 0 0 000 0 0 0kk kk⎡⎤⎡ ⎤ ⎢⎥⎢ ⎥∇= → =⎢⎥⎢ ⎥ ⎢⎥⎢ ⎥⎣⎦⎣ ⎦vA . [] [ ]()111 1 1 1 000ij k k ij ijDDvDt t Dt t x∂∂∂ ⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤=+ ∇→ = + = + =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥∂∂ ∂ ⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦AAA A AAv . [] [] [] [] [] [] [] []T 21 1 22 22 1220 0 40 0 02 0 0 0 20 0 0 0 0 2 0 0 4 0 . 00 0 0 0 0 0 00 0 00kk kk kk k k=∇ + ∇ ⎡ ⎤⎡ ⎤⎡⎤ ⎡ ⎤⎢ ⎥⎢ ⎥⎢⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ∇= = → =⎢⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦AA vv A Av A Next, [] []23 23 240 0 0 4 0 00 04 0 0 0 4 00 00 0 0 0 0 00 0kk k kk k⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥∇= =⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦ ⎢⎥⎢⎥⎣⎦⎣⎦Av . Thus, [][ ] [] [] [] []3 T 3 32 208 0 08 0 0 00 0k k⎡ ⎤ ⎢ ⎥ ⎢ ⎥ =+ ∇ + ∇ = ⎢ ⎥ ⎢ ⎥⎣ ⎦AA v v A . ____________________________________________________________ 8.15 Given the velocity field in cylindrical coordinates: () 0, 0, rzvvv v rθ=== , obtain the second Rivlin-Ericksen tensors , 2,3,...NN= A using the recursive formula. ------------------------------------------------------------------------ Ans. []000 000 , 00dvkdrk⎡⎤ ⎢⎥==⎢⎥ ⎢⎥⎣⎦v∇ , [] [] []T 100 000 00k k⎡⎤ ⎢⎥=+ =⎢⎥ ⎢⎥⎣⎦Av v∇∇ Since ()1ijA =constant, independent of time and space, therefore () []11 1 0k ijk ij ijDvDt t⎡⎤ ⎡ ⎤ ∂ ⎛⎞⎛ ⎞=+ ∇= ⎢⎥ ⎢ ⎥⎜⎟⎜ ⎟∂ ⎝⎠⎝ ⎠⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦AAA . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-13[]() ()T 21 1 200 000 00 00 2 00 000000 0 00000 0 00 . 00 00 000 00 0 00kk k k kk k⎡⎤=∇ + ∇⎢⎥⎣⎦ ⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥=+= ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣ ⎦AA vv A []() ()T 32 2 222 00000 00 2 00 000 0 00000 000 0 00 000 .00 0 0 0 0 0 000 0 0 0 0kk k k⎡⎤=∇ + ∇⎢⎥⎣⎦ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥=+=⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦AA v v A Thus, 0, 3,4...N N==A ____________________________________________________________ 8.16 Using the equations given in Appendix 8.1 for cylindrical coordinates, verify that the rrθcomponent of the third order tensor ∇Tis given by: ()1rr rr rrTT T rrθθ θθ+ ∂⎡ ⎤∇= −⎢ ⎥∂⎣ ⎦T ------------------------------------------------------------------------------ Ans. From the equations () = no sum on , sum on , and 1, , 1; 1, 1 , all other 0ij mq j q m i i q q m j ijm m rz r r i j kT Th T T m qx hh r hθθ θ θ θ∂ ∇+ Γ + Γ∂ === Γ = Γ = − Γ = we have, () = =rr rr qr q r rq q r r r r r rrTTTh T T T Tθ θθ θ θ θ θ θ θ θθθ∂∂∇ +Γ+Γ +Γ+Γ∂∂, thus, () () ()( )1 =1 1 =rr rr rr rr rr rrTT TTTr T T Trrθθ θθ θθθ θ+ ∂∂∇+ − + − → ∇−∂∂. ____________________________________________________________ 8.17 Using the equations given in Appendix 8.1 for cylindrical coordinates, verify that the rθθcomponent of the third order tensor ∇Tis given by: ()1rr r rTT T rrθ θθ θθθ∂ −∇= +∂T ------------------------------------------------------------------------------ Ans. From the equations () = no sum on , sum on and 1, , 1; 1, 1 , all other 0.ij mq j q m i i q q m j ijm m rz r r i j kT Th T T m qx hh r hθθ θ θ θ∂ ∇+ Γ + Γ∂ === Γ = Γ = − Γ = we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-14() () () () ( )( ) = = 1 =1 1 = .rr qq r r q q r r r r rr rr r r rr rrTTTh T T Th T T TT T TTr T T Trrθθ θ θθ θ θ θ θ θ θ θ θ θ θθ θθ θθ θ θ θθ θθ θθθθ θθ∂∂∇+ Γ + Γ → ∇+ Γ + Γ∂∂ ∂∂ −→∇ + − + →∇ +∂∂ ____________________________________________________________ 8.18 Using the equations given in Appendix 8.1 fo r spherical coordinates, verify that the rrφcomponent of the third order tensor ∇Tis given by: ()() 1 sinrrrr rrTT T rrφφ φθφ+ ∂∇= −∂T ------------------------------------------------------------------------ Ans. From the equations () = no sum on , sum on and 1, , sin ; 1, sin , sin , cos , 1, cos all other 0ij mq j q m i i q q m j ijm m rr r r r ijkT Th T T m qx hh r h rθφ θ θ φ φ φφ φφθ θθ θφφθθ θθ θ∂ ∇+ Γ + Γ∂ === Γ = Γ = Γ= − Γ= − Γ= − Γ= Γ = we have, () () () ( ) () () () = = sin = sin sin 1 =.sinrr rr qr q r rq q r r r r r rr rr rr rr rr rr rr rrTTTh T T Th T T TTr T T TT TTrrφ φφ φ φ φ φ φ φ φ φφ φφ φ φφ φφφ θθ θφ θφ∂∂∇+ Γ + Γ → ∇+ Γ + Γ∂∂ ∂→∇ + − + − →∂ + ∂→∇ −∂ ____________________________________________________________ 8.19 Using the equations given in Appendix 8.1 fo r spherical coordinates, verify that the φφφcomponent of the third order tensor ∇Tis given by: () () cot 1 sinrrTT TT T rr rφφ θ φφ θ φφθ θφ++ ∂++∂ ------------------------------------------------------------------------------- Ans. From the equations () = no sum on , sum on 1, , sin ; 1, sin , sin , cos , 1, cos all other 0ij mq j q m i i q q m j ijm m rr r rr i j kT Th T T m qx hh r h rθφ θ θ φ φ φφ φφθ θθ θφφθθ θθ θ∂ ∇+ Γ + Γ∂ === Γ = Γ = Γ= − Γ= − Γ= − Γ= Γ = we have, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-15() () () () () ()()() ()() () = = = sin cos cot 1 =sinqq q q rr r r rr r rr rrTTh T T TT h TTTT TTh T T T T TTT TT TT TT TTrr rφφ φφ φ φ φ φ φφφφ φφ φ φ φφ θφ θφφ φ φφ φθ θφφ φφφ φφ φφ φ φ φ θ φ φ θ θ φ φφφφ φφ φφ θ φφ θ φφ θ φφ θ φφ φφφφ φ φ θθφ θ θφ∂∇+ Γ + Γ∂ ∂→∇ + Γ + Γ + Γ + Γ∂ ∂→∇ + + Γ + + Γ∂ ∂=+ + + +∂ ++ ∂→∇ + +∂ ____________________________________________________________ 8.20 Given the velocity field in cylindrical coordinates: () 0, , 0rzvv v r vθ=== , obtain (a) the first Rivlin-Ericksen tensor 1A (b) 1∇A (c) the second Rivlin-Ericksen tensors 2A, using the recursive formula.. ------------------------------------------------------------------------ Ans. []1()v00 100 00 0 1rr r r zz zvv vvr rr zr vv v dvvrr z d r vv v rr zθ θθ θθ θ θ⎡⎤∂∂ ∂⎛⎞⎡ ⎤ − ⎢⎥⎜⎟ −⎢ ⎥ ∂∂ ∂⎝⎠⎢⎥⎢ ⎥⎢⎥∂∂ ∂⎛⎞ ⎢ ⎥=+ =⎢⎥⎜⎟ ⎢ ⎥ ∂∂ ∂⎝⎠⎢⎥⎢ ⎥⎢⎥∂∂ ∂ ⎢ ⎥⎢⎥⎢ ⎥ ⎣ ⎦ ∂∂ ∂⎢⎥⎣⎦v∇ [] [] []() ()T 100 ()00 , 00 0kr dv v rkr kdr r⎡⎤ ⎢⎥ ⎛ ⎞=+ = =− ⎜⎟ ⎢⎥⎝⎠⎢⎥⎣⎦Av v∇∇ () ()() () () () () () () () ()11 1 11 1 11 1 1 11 1k ijk ij ij rr r rz k ijk r z zr z zzDvDt t vvv vv v v vvvθ θθ θθ θ θ θ θθ θ θ θθθ θ θ θ θθ θθ θ θ⎡⎤ ⎡ ⎤ ∂ ⎛⎞⎛ ⎞=+ ∇ ⎢⎥ ⎢ ⎥⎜⎟⎜ ⎟∂ ⎝⎠⎝ ⎠⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎡⎤∇∇∇ ⎢⎥⎡⎤=∇ =∇ ∇ ∇ ⎢⎥⎣⎦⎢⎥∇∇∇⎢⎥⎣⎦AAA AAA AA A A AAA The components of the third order tensor ()1∇A can be obtained from Appendix 8.1 as: () 112rr rr rrAA A k rr rθθ θθ+ ∂⎡⎤∇= − = −⎢⎥∂⎣⎦A , () 110rr r rAA A rrθθ θ θθθ∂ −∇=+ =∂A () 110z rz rzA A rrθ θθ∂∇=− =∂A () 110rr r rAA A rrθθ θ θθθ∂−∇=+ =∂A , () 12k rθθθ∇=A , () 110z rz zA A rrθ θθθ∂∇=+ =∂A Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-16() 110z zr zrA A rrθ θθ∂∇=− =∂A , () 110z zr zA A rrθ θθθ∂∇=+ =∂A , () 110zz zzA rθθ∂∇==∂A Thus, 12/ 0 0 2 / 0 0 02 / 0 0 2 / 0 00 0 0 0 0ijkr k vr Dvk r k v rDtθ−−⎡⎤ ⎡ ⎤⎡⎤⎛⎞ ⎢⎥ ⎢ ⎥== ⎢⎥⎜⎟ ⎢⎥ ⎢ ⎥⎝⎠⎢⎥⎣⎦ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦A, [] []() () 100 0 ( ) / 0 / 0 0 00 / 0 0 0 /0 00 0 0 00 0 0 0k r v r r kdv dr k r dv dr kv r⎡⎤ −⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥== −⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦Av∇ , [] []() ()T 10/ 0 0 0/ 0 0 () / 0 0 0 0 0 / 0 00 0 0 0 0 0 0 0dv dr k r kdv dr vr r k r k v r⎡⎤ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥=− = −⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦vA∇ , [] [] [] []()2 T 1 21 1// 0 0 2 0 0 20 0 0 0 0 0 00 0 0 0 0kd v d r v r k D Dt⎡ ⎤ ⎡⎤ −⎢ ⎥ ⎡⎤ ⎢⎥=+ + = = ⎢ ⎥ ⎢⎥ ⎢⎥⎣⎦ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦AAA v v A ∇∇ . ____________________________________________________________ 8.21 Derive Eq. (8.11.3), i.e., () ()T 1N NN ND Dt+=+ ∇ + ∇AAA v v A . ------------------------------------------------------------------------ Ans. We had {see Eq. (8.11.7)}, () ()1 22 1NN N NN N NND D DD d D dds d d ds d d d dDtD t D t DtD t+ +=⋅ → = ⋅ ⋅ ⋅A xxxA x A x + x x + xA . That is, ()() ()1 2 1N N NN ND Dds d d d d d dDt Dt+ +== ∇ ⋅ ⋅ ⋅ ∇Avx A x + x x + x A vx () ()T N NNDdd d d d dDt⋅∇ ⋅ ∇ ⋅A=x v A x +x A v x +x x () ()T 1N NN NDdd d dDt+⎡⎤⋅∇ ∇ = ⋅⎢⎥⎣⎦A=x v A + A v + x x A x . Thus, () ()T 1N NN ND Dt+=+ ∇ + ∇AAA v v A ____________________________________________________________ 8.22 Let /DD t≡+ −ST T W W T , where Tis an objective tensor and Wis the spin tensor, show that Sis objective, i.e., () ()Ttt=*SQ S Q . ------------------------------------------------------------------------- Ans. Since Tis objective, therefore () ()Ttt=*TQT Q and from Eq. (8.13.13), ( ) () () ()TTdd t t t t=+*WQ / Q Q W Q , therefore, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-17() ()T TT T T T T TT T T.D Dt dD d d dt Dt dt dt dttdt≡+ ⎛⎞⎡⎤⎡ ⎤ =+ ++ + ⎜⎟ ⎣⎦⎣ ⎦⎝⎠ ⎡⎤ ⎡ ⎤−−⎣⎦ ⎣ ⎦* ** * * * TST W - W T QT Q QTQ Q Q QT QTQ Q QTQ QWQ QQ QTQ QWQ Q TQ Now, () () () ()T TT ddtt t tdt dt=→ = −QQQQ I Q Q , therefore, the above equation becomes ()TT TT T TT.dD d d dt Dt dt dt dtdt⎡⎤ ⎛⎞⎡ ⎤ =+ +− + ⎢⎥ ⎜⎟ ⎣ ⎦⎝⎠ ⎢⎥⎣⎦ ⎡⎤⎡⎤ −−⎢⎥ ⎣⎦⎣⎦* QT Q QS T QQ QQ T Q T Q T W Q QTQ QWTQ That is, () () () ()TT Dtt t tDt⎛⎞=+ − =⎜⎟⎝⎠* TSQ T W W T Q Q S Q . ____________________________________________________________ 8.23 Obtain the viscosity function and the two normal stress function for the nonlinear viscoelastic fluid defined by 1 20()t fst s d s∞−⎡ ⎤ −−⎣ ⎦ ∫S= I C ( ) ------------------------------------------------------------------------------ Ans. For 12 2 3 , 0 vk xv v== = , we have [see Section 8.9, Eq.(8.9.12)] []() ()()2 210 10 00 1() ,tk kkt ttτ ττ τ=− ⎡⎤ ⎢⎥ −− +⎢⎥ ⎢⎥⎢⎥⎣⎦C () () ()2 2 110 10 00 1()tkk ktt tττ ττ−+− =−⎡ ⎤ −−⎢ ⎥⎡⎤ −⎢ ⎥⎣⎦⎢ ⎥ ⎢ ⎥ ⎣ ⎦C 22 110 10 00 1()tks k s ks ts−+ =⎡⎤ ⎢⎥⎡⎤−⎢⎥ ⎣⎦⎢⎥⎣⎦C . Thus, 22 10 00 00 0()tks k s ks ts−−− =−⎡ ⎤ ⎢ ⎥⎡⎤−− ⎢ ⎥ ⎣⎦⎢ ⎥⎣ ⎦IC () ()12 12 2 2 00 ,.SSk s f s d s s f s d skμ∞∞ =− → ≡ =−∫∫ ()22 11 2 22 33 0, 0 , 0 Sk s f s d s S S∞ =− = =∫, ()22 11 12 2 2 2 2 23 3 0, 0. SS k s f s d s S Sσσ∞ =−= − =−= ∫ ____________________________________________________________ 8.24 Derive the following transformation laws [Eqs .(8.13.8) and Eq. (8.13.12)] under a change of frame. () () () ()*T * T and tt tt t ττ τ==VQV Q RQR Q ------------------------------------------------------------------------------- Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-18Ans. Since ** * and tt t t t tF= V R F = VR , therefore, from () () () ()*T tt t ττ τ= FQ F Q , we get () () () () () ()** T T T tt t t t t tt ττ τ τ ⎡⎤ ⎡⎤ ==⎣⎦ ⎣⎦V RQV R Q QV Q QR Q , where () ()T tττ ⎡ ⎤⎣ ⎦QV Q is a symmetry tensor and () ()T t tτ⎡⎤⎣⎦QR Q is an orthogonal tensor. Therefore, the uniqueness of the polar decomposition leads to () () () ()*T * T and tt tt t ττ τ==VQV Q RQR Q . ____________________________________________________________ 8.25 From ()L tD Dττ τ=⎡⎤≡⎢⎥ ⎣⎦JT( and () t tD Dττ τ=⎡⎤=∇ ⎢⎥ ⎣⎦Fv , show that =+oTT T D + D T( . [note misprint in the problem in text] ------------------------------------------------------------------------------ Ans. From () ()() ()T L ttττττ= JF T F , we have, ()()() () ()()() ()()()T L TT tt tt ttDD D D DD D Dτ τττττ τ τ τττ τττ=++JF T FTF F F F T Thus, ()()( ) ()T L tD DttDD tττ τ=⎡⎤=∇ + + ∇ ⎢⎥ ⎣⎦J TvT T v [Note () ()T tttt== FF I ] Now, ∇v=D+W , therefore, ()()() () ()T L T tD DD DD t D t D Dtττ τ=⎡⎤=+ + =++ ⎢⎥ ⎣⎦ =++ − = +oJ TTD + W T T D + W DT TD + W T + TW TDT TD + TW WT T + DT TD ____________________________________________________________ 8.26 Consider () ()() ()11 T U ttτττ τ−−= JF T F . Show that (a) () U /tDDτττ=⎡ ⎤⎣ ⎦J is objective and (b) () ()() ( )T U /tDDDDττττ=⎡⎤ =− ∇ − ∇ −⎣⎦o TJT v v T = T T D + D T . ------------------------------------------------------------------------------- Ans. (a) Given() ()() ()()T11 U ttτττ τ−−= JF T F , and () () () ()()T11 U ttτ ττ τ−−=** ** JF T F . In a change of frame (see Section 8.13. Eq.(8.13.6), () () () ()T tt t ττ τ= *FQ F Q , so that () () () () ()() ()()() ()TT11 T 1 1 T and tt t t ttττ τ τ τ τ−− − −== **FQ F Q F Q F Q . Also, T() () () ()tt t t=*TQ T Q . Thus () () () () () ()() ()T1T T 1 T U () () ()ttttττ τ τ τ τ τ τ−−=*JQ F Q Q T Q Q FQ () () ()() ()T11 T() .ttttττ τ−−=QF T F Q That is , () () () ()T U ttττ=* U JQ J Q and () () () ( )()T U //NN N ND Dt D Dtττ ττ =* U JQ J Q . Thus, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-19()()()()U TN N NN tD Dtt DDτ ττ τ ττ=⎡⎤⎛⎞ ⎡⎤⎢⎥⎜⎟= ⎢⎥⎜⎟⎢⎥ ⎢⎥⎣⎦ ⎝⎠⎣⎦* U =tJ JQQ . (b) () () [] ()()T11 U // / /tt t ttDD D Dt D D t D Dτ τ ττττ τ τ τ−− = ==⎡ ⎤⎡⎤ ⎡⎤ =+ +⎣⎦ ⎢ ⎥ ⎣⎦⎣ ⎦ =tJF T T T F . Now, () () [] () ()11 1// 0tt t t t t DD D D ττ τττ τ−− − ⎡⎤ =→ + =→⎣⎦FF I F F F F () () [] ()[]() ( ) ()11 1/// . Thus,tt t t t t t t tDD t D D t D Dττ τ ττ τττ τ−− − = = =⎡⎤ ⎡ ⎤ =− =− =− ∇⎣ ⎦ ⎣⎦ =−∇=tFF F F F v F v () ()() () ()T U /tDDDDDtD tτττ=⎡⎤ =− ∇ − ∇=−−−⎣⎦TTJT v v T T D W D + W T . That is, the upper convected derivative of Tcan be written: ()() ()U ˆ tD D DD tττ τ=⎡⎤≡= − − + = − +⎢⎥ ⎣⎦o J TT + TW WT TD DT T TD DT . ____________________________________________________________ 8.27 Given the velocity field of a plane Couette flow: 12 10, vv k x== . (a) For a Newtonian fluid, find the stress field []T and the co-rotational stress rate ⎡⎤⎣⎦oT. (b) Consider a change of frame (change of observer) described by: []* 11 *22cos sin cos sin, sin cos sin cosxx tt tt x tt tt xωωω ω ωωω ω⎡⎤ −−⎡⎤ ⎡⎤ ⎡⎤==⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦Q Find *,, a n d⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤∇⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦** * *vv DW . (c) Find the co-rotational stress rate for the starred frame (d) Verify that the two stress rates are rela ted by the objective tensorial relation. ------------------------------------------------------------------------------ Ans. (a) [] [ ] []00 0 / 2 0 / 2, , 0/ 2 0 / 2 0kk kk k− ⎡⎤ ⎡ ⎤ ⎡ ⎤∇= = =⎢⎥ ⎢ ⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦vD W . Thus, stress tensor is []00 / 220/ 2 0p kp k p kk pμμμ−−⎡⎤ ⎡ ⎤ ⎡⎤=+ =⎢⎥ ⎢ ⎥ ⎢⎥−−⎣⎦ ⎣ ⎦ ⎣⎦T . [] []2 20/ 2 0/ 2 0 /2 0 /2 0 0pk k k pk k kp k k kp kμμ μ μμ μ⎡ ⎤ −− − −⎡ ⎤ ⎡⎤ ⎡⎤ ⎡ ⎤− =−= ⎢ ⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥−− ⎢ ⎥− ⎣ ⎦ ⎣⎦ ⎣⎦ ⎣ ⎦ ⎣ ⎦TW WT Co-rotational stress rate is: 22 2200 00kk D Dt kkμμ μμ⎡ ⎤⎡ ⎤⎡⎤⎡⎤=+ = ⎢ ⎥⎢ ⎥⎢⎥⎣⎦⎣⎦⎢ ⎥⎢ ⎥−−⎣ ⎦⎣ ⎦o TT . (b) From Eq. (5.56.12) of Chapter 5, we have, [] [ ] () []()T*( / /vd d t d d t ⎡ ⎤ ⎡⎤ =⎣⎦ ⎣ ⎦x*) = Qv + Q x Qv + Q Q x * . Thus, * * 1 1 * *22 20 cos sin sin cos cos sin sin cos cos sin sin cosvx tt t t t t v tt t t t t vxωω ωωω ωωωω ω ω ω ω⎡⎤ ⎡⎤ −− −⎡⎤ ⎡⎤ ⎡ ⎤ ⎡⎤=+⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥−− ⎢⎥ ⎢⎥⎣⎦ ⎣ ⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦. Since, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-20()* 11 ** ** 112 2 12*22cos sincos sin cos sinsin cosxx ttx tx tx v k tx txx tt xωωωω ωωωω⎡⎤ ⎡⎤⎡⎤=→ = + → = + ⎢⎥ ⎢⎥⎢⎥− ⎢⎥ ⎣⎦⎣⎦ ⎣⎦ Therefore, ( ) ()*2 *** *1212 1 2 ** * 2* *222 1 12cos sin sinsin 01 cos 10cos sin costt x t xvv t x xkvt vx x tx t txωω ωωωωωωω ω⎡⎤−+ ⎡⎤ ⎡⎤ ⎡ ⎤−− − ⎡⎤ ⎡⎤ ⎢⎥=+ = +⎢⎥ ⎢⎥ ⎢ ⎥⎢⎥ ⎢⎥ ⎢⎥⎢⎥ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎣⎦ + ⎣⎦ ⎣⎦ ⎣ ⎦ ⎢⎥⎣⎦ from which, we get, ()() ()2 2sin 2 / 2 sin 01*10 cos sin 2 / 2ttk ttωωω ωω⎡⎤−− −⎡⎤⎡⎤ ⎢⎥ ∇= + ⎢⎥ ⎣⎦⎢⎥ ⎣⎦ ⎣⎦v* , []()() () ()s i n2 /2 c o s2 /2 c o s2 /2 s i n2 /2ttkttωω ωω⎡⎤−=⎢⎥ ⎣⎦D* , []01 / 2 0 1 1/2 0 1 0k ω− − ⎡ ⎤⎡ ⎤=+⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦W* . (c) For the Newtonian fluid, the stre ss field in the starred-frame is: []() () ()sin 2 cos 2 cos 2 sin 2pk t k t ktp k tμωμω μω μω⎡⎤−−=⎢⎥−+⎣⎦*T , where the indeterminate pressure pis time independent. Thus, () () () ()cos 2 sin 22sin 2 cos 2tt Dktt Dtωωμωωω⎡⎤−− ⎡⎤=⎢⎥ ⎢⎥− ⎣⎦ ⎣⎦T*, and ()cos 2 sin 2 cos 2 sin 2 sin 2 cos 2 sin 2 cos 2 2kt p k t kt p k t k pk t k t p ktk tμω μω μωμ ωωμωμω μωμω+ + ⎡⎤ ⎡ ⎤⎡⎤=+⎢⎥ ⎢ ⎥ ⎣⎦ −+ − −+ −⎣ ⎦ ⎣⎦**TW cos 2 sin 2 cos 2 sin 2 sin 2 cos2 sin 2 cos 2 2kt p k t kt p k t k p kt k t p kt k tμωμ ω μ ωμ ωωμω μω μω μω−− −−⎡⎤ ⎡⎤⎡⎤=+⎢⎥ ⎢⎥ ⎣⎦ −− −−⎣⎦ ⎣⎦**WT Thus, cos 2 sin 2 cos 2 sin 22sin 2 cos 2 sin 2 cos 2kt k t t tkkkt k t t tμω μω ω ωμωμωμ ω ω ω⎡⎤ ⎡ ⎤⎡⎤ ⎡⎤−= + ⎢⎥ ⎢ ⎥ ⎣⎦ ⎣⎦ −− ⎣⎦ ⎣ ⎦** * *TW WT . Thus, 2cos 2 sin 2/sin 2 cos2ttDD t kttωωμω ω⎡ ⎤⎡⎤ ⎡ ⎤=+ − = ⎢ ⎥ ⎣⎦ ⎣ ⎦ − ⎣ ⎦o* ** * * *TT T W W T (d) [] []T⎡⎤⎣⎦oQT Q 2 2 2cos sin 0 cos sin cos2 sin 2 sin cos sin cos sin 2 cos2 0tt k t t t tktt t t t t kωω μ ω ω ωωμωωω ω ω ω μ⎡⎤−⎡⎤ ⎡⎤ ⎡ ⎤== ⎢⎥ ⎢⎥ ⎢⎥ ⎢ ⎥−− ⎢⎥− ⎣⎦ ⎣⎦ ⎣ ⎦ ⎣⎦ Thus, we have [] []*T⎡⎤ ⎡ ⎤=⎣⎦ ⎣ ⎦o oTQ T Q . ____________________________________________________________ 8.28 Given the velocity field: 11 2 2 3 , , 0 vk x v k x v=−= = . Obtain (a) the stress field for a second-order fluid (b) the co-rotational derivative of the stress tensor ------------------------------------------------------------------------ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-21Ans. (a) [] [ ] [] [ ]00 00 , 0 00 0k k−⎡⎤ ⎢⎥∇= = =⎢⎥ ⎢⎥⎣⎦vD W , [] [] []2 2 2 1140 0 20 0 20 0 20 0 2 020 020020 04 0 00 0 00 0 00 0 0 0 0k kk k kk k k⎡ ⎤−− −⎡⎤ ⎡⎤ ⎡⎤⎢ ⎥⎢⎥ ⎢⎥ ⎢⎥⎢ ⎥ == → = =⎢⎥ ⎢⎥ ⎢⎥⎢ ⎥⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎢ ⎥⎣ ⎦AD A [] () () () ()TT 2 1 11 11 2 2/ 40 0 20 0 0 0 0 0 20 0 020 0 0 0 0020 04 000 0 0 0 0 0 0 0 00 0 0 0 0DD t k kk kk kk k k k⎡⎤ ⎡ ⎤=+ ∇ + ∇ = ∇ + ∇⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎡ ⎤−−− −⎡ ⎤ ⎡⎤ ⎡⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥ =+=⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢ ⎥⎣ ⎦ ⎣⎦ ⎣⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦A A Av v A Av v A The second-order fluid is defined by Eq.(8.18.6): [] []2 11 21 3 2 22 22 12 340 0 40 0 20 0 020 04 0 04 0 00 0 0 0 0 0 0 0p kk k pk k kμμμ μμ μ=− + + + → ⎡ ⎤⎡ ⎤−⎡⎤⎢ ⎥⎢ ⎥⎢⎥⎢ ⎥⎢ ⎥ =− + + +⎢⎥⎢ ⎥⎢ ⎥⎢⎥⎣⎦ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦TI A A A TI 22 11 1 2 3 22 1 2 3 33 24 ( ) , 24 ( ) , T p kk T p kk T p μμ μ μμ μ=− − + + =− + + + =− . To obtain the pressure p, we first calculate the acceleration: [] [ ] [ ] []2 11 2 2200 00 00 0 0 0kx kk x tk k x k x⎡⎤−−⎡⎤ ⎡ ⎤⎢⎥⎢⎥ ⎢ ⎥⎢⎥ =∂ ∂ +∇ = =⎢⎥ ⎢ ⎥⎢⎥⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎢⎥⎣⎦av / v v Equations of motion i jTijaxρ∂=∂ then give 22 12 12 2, , 0 ,pp pkx kxxx xρρ∂∂∂−= −= −=∂∂∂ thus, 22 2 12() / 2 pk x xρ=− + +C . (b) The co-rotational derivative of T: / DD t=+oTT T W - W T . Since W=0 , () 12 1 2 12 1 2100 010 001ij ij ijijTT Dp pvv v vDt x x x x⎡⎤⎡⎤ ∂∂⎡⎤ ⎛⎞∂∂ ⎛⎞ ⎡⎤ ⎢⎥== + = − −⎢⎥ ⎜⎟ ⎢⎥ ⎜⎟ ⎢⎥ ⎢⎥∂∂ ∂∂ ⎣⎦ ⎝⎠⎢⎥ ⎝⎠ ⎣⎦ ⎣⎦ ⎢⎥⎣⎦o TT () ( ) []22 22 2 2 12 1 2 2 1 12100 100 010 010 001 001ppk x x k kx kx kv vxxρρ⎡⎤ ⎡⎤⎛⎞∂∂ ⎢⎥ ⎢⎥=− = − + = −⎜⎟ ⎢⎥ ⎢⎥∂∂⎝⎠⎢⎥ ⎢⎥⎣⎦ ⎣⎦I ____________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-228.29 Show that the Lower Convected derivative of 1A is 2A, i.e., 12=AA( . ------------------------------------------------------------------------- Ans. From Eq.(8.19.22), () () () ( ) ( )o 1 11 1 1 1 1 1 1 T 11 1 11 1 2/ // .DD t DD t DD t=++ = + − ++ =+ + + − =+ ∇ + ∇AA A D D A A A W W A A D D A AA W D D W A AA v v A = A( ____________________________________________________________ 8.30 The Reiner-Rivlin fluid is defined by the constitutive equation: ()() 12 3 22 3 , , , pI I I I φφ −2T= I+S S= D+ D where iIare the scalar invariants of D. Obtain the stress components for this fluid in a simple shearing flow. ------------------------------------------------------------------------------- Ans. In a simple shearing flow, 12 2 3 , 0 vk xv v== = , []2 2 2 23/4 0 0 0/ 2 0 /2 0 0 , 0 /4 0 , , 0400 0 0 0 0k k kkk I I⎡⎤⎡⎤⎢⎥⎢⎥⎡⎤⎢⎥== − =⎢⎥ ⎣⎦⎢⎥⎢⎥⎣⎦ ⎢⎥⎣⎦2D= D [] [] () ()2 22 2 12/4 0 0 0/ 2 0 /4 , 0 /2 0 0 /4 , 0 0 /4 0 00 0 0 0 0k k pk k k kφφ⎡ ⎤⎡⎤⎢ ⎥⎢⎥⎢ ⎥ =− + +⎢⎥⎢ ⎥⎢⎥⎣⎦ ⎢ ⎥⎣ ⎦TI . ____________________________________________________________ 8.31 The exponential of a tensor A is defined as: [] 11exp!N n n=∑ AI + A . If A is an objective tensor, is []exp A also objective? ------------------------------------------------------------------------------- Ans. Yes. Because () () ()() () () () () () ()() ()2TT T T TNNtt tt tt t t tt=→ = = →=** 2 *A Q AQ A Q AQ Q AQ Q A Q AQ A Q That is, ()N*A is objective for all N. As a consequence, [] exp Ais objective. ____________________________________________________________ 8.32 Why is it that the following cons titutive equation is not acceptable: () , p α −∇T= I+S S= v , where v is velocity and αis a constant ------------------------------------------------------------------------- Ans. Because ∇vis not objective. ____________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-238.33 Let daand dAdenote the differential area vectors at time τand time t respectively. For an incompressible fluid, show that 21//NN NN tNttDda D d D D d d d ττττ− ==⎡⎤ ⎡⎤ =⋅ ≡ −⋅⎣⎦ ⎣⎦AC A A M A where dais the magnitude of daand the tensors NMare known as the White-Metzner tensors. -------------------------------------------------------------------------------- Ans. From Eq. (3.27.12), we have, [note here dAis the reference area and dais the area at the running time τ], ()()T1det dd−FF A a= . For an incompressible fluid, () det 1=F, so that ()T1dd−FA a= , ()()()()( )TT T 111 1 1 Tdd d d d d d d−−− − −⋅⋅ FA FA = A F FA = A F F A aa = ⋅⋅ . That is, 21 t da d d−⋅AC A= . Thus, 11 2 ,w h e r eNN N tt NN NN N t ttDD Dd add d d DD Dτ τ τττ τ−− = ==⎡⎤ ⎡⎤ ⎡⎤ =⋅ ≡ −⋅ = −⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦ ⎣⎦CCAA A M A M . ____________________________________________________________ 8.34 (a) Verify that Oldroyd's lower convect ed derivatives of the identity tensor Iare the Rivlin-Ericksen tensor NA. (b) Verify that Oldroyd upper derivatives of the identity tensor are the negative White-Metzner tensors [see Prob. 8. 33 for the definition of White-Metzner tensor]. ------------------------------------------------------------------------------- Ans. (a) The Nth lower convected derivative of Tis given by () ()() ()T/ , where NN LL t ttDD ττ ττ τ τ =⎡⎤ =⎣⎦JJ F T F . For T=I , () () () ()T Lt t tτττ τ== JF F C . Thus, ()N N t L N NN t tD D DDτ ττ ττ= =⎡⎤ ⎡⎤ ==⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦C JA (b) The Nth upper convected derivative of Tis given by () ()() ()()T11/ , where NN UU t ttDD ττ ττ τ τ−− =⎡⎤ =⎣⎦JJ F T F . For T=I , () ()()() ()T11 1 Utt tτττ τ−− −== JF F C . Thus, ()1 N N t U N NN t tD D DDτ ττ ττ− = =⎡⎤ ⎡⎤ == −⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦C JM. ____________________________________________________________ 8.35 Obtain the equation ()T/DD t=+ ∇ ∇TT T v + v T( , where T( is the lower convected derivative of T. ------------------------------------------------------------------------------- Ans. By definition, the lower convected derivative is ()/LtDDτττ=⎡ ⎤⎣ ⎦J , where () ()() ()T Lt tττττ= JF T F . Thus, () () ()T//Lt tt tD DD D t tτ τττ τ= =⎡⎤ ⎡⎤ =⎣⎦ ⎣⎦JF T F ()[] () () ()TT T //tt t tttD D t t tD Dττττ==⎡ ⎤ ++⎣ ⎦FT F F T F . Now, () () ()T T TT// /tt ttDD D D t D D t τττ =⎡⎤ == = ∇⎣⎦FF F v [see Eq.(8.12.3)] and ()tt= FI , therefore, Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-24()()T L tD D DD tττ τ=⎡⎤=+ ∇∇ ⎢⎥ ⎣⎦J TT= T v+ v T( . ____________________________________________________________ 8.36 Consider the following constitutive equation: () ()()**/2 w h e r e /DD t DD tλμ α =≡ +oS+ S D , S S D S+S D and oS is co-rotational derivative of S. Obtain the shear stress function and the two normal stress functions for this fluid. ------------------------------------------------------------------------------- Ans. With 12 2 3 , 0 vk x vv== = , the rate of deformation te nsor and spin tensor are: [] [ ]0 / 20 0 / 20 / 200 , / 200 00 0 0 0 0kk kk⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥== −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦DW . Since the flow is steady, 0t∂=∂S. The co-rotational derivative is , for symmetric S: ()T=− +oS SW WS = SW SW . Now , [] []12 11 12 22 32 T 22 21 11 21 31 32 310 0, 2200 0 0SS S S S kkSS S S S SS−− − −⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥=− =⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ − ⎣ ⎦ ⎣⎦SW SW , 12 11 22 32 11 22 21 31 32 312 220SS SS kSS S S SS−− −⎡⎤ ⎢⎥⎡⎤=−⎢⎥⎣⎦ ⎢⎥−⎣⎦oS , [] []12 11 22 32 11 22 21 31 32 312 220SS S S kSS S S SS+ ⎡ ⎤ ⎢ ⎥+= +⎢ ⎥ ⎢ ⎥⎣ ⎦SD DS . Thus, () */DD t α≡+oSS D S + S D gives () ()()() ()() () () () ()12 11 22 32 * 11 22 21 31 32 3121 1 1 1 11 2 1 1211 0SS S S D kSS S SDtSSαα α α αα α α αα⎡⎤ −+ + − − ⎡⎤ ⎢⎥=+ + − + +⎢⎥ ⎢⎥⎣⎦⎢⎥ −+⎣⎦S Therefore , *2D Dtλμ=→SS+ D () 11 12 10 ( i ) Sk Sλα+−= , ()() 12 11 22 11 ( i i )2kSS S kλαα μ⎛⎞⎡⎤++ + − =⎜⎟⎣⎦⎝⎠, () () 13 23 22 12 1 0 (iii), 1 0 (iv),2kSS S k Sλαλ α+− = + += () 23 13 33 1 0 (v), S =0 (vi)2kSSλα++ = . Now, (iii) , (v) and (vi) give 13 23 33 0 SSS=== . Eq. (i) gives () 11 12 1 Sk Sλα=− , Eq.(iv) gives () 22 12 1 SkSλα=− + , thus, with ()()2 2() 1 1Ak k αλ≡+ − , we have, 12 /( ) Sk A kμ= , ()2 11 1/ ( ) Sk A kλμ α=− , ()2 22 1/ ( ) Sk A kλμ α=− + The shear stress function is 12 /( ) Sk A kμ= . The normal stress functions are: ()22 11 12 2 2 2 23 3 2/ ( ) , 1 / ( ) SS k A k S S k A kσλ μ σ λ μ α≡−= =−= − + . Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-25____________________________________________________________ 8.37 Obtain the apparent viscosity and the normal stress functions for the Oldroyd 3-constant fluid [see (C) of Section 8.20]. ------------------------------------------------------------------------------- Ans. For the simple shearing flow, [] [ ]0 / 20 0 / 20 / 200 , / 200 00 0 0 0 0kk kk⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥== −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦DW . [] [ ][ ] ()12 11 22 32 11 22 21 31 32 312 /2 2 0SS SS kS S S S SS−− −⎡⎤ ⎢⎥=+ − = −⎢⎥ ⎢⎥−⎣⎦oS0 S W W S , [] [] ()12 11 22 23 11 22 12 13 23 132 /2 2 0SS S S kS S S S SS+ ⎡ ⎤ ⎢ ⎥+⎢ ⎥ ⎢ ⎥⎣ ⎦SD + DS = , () ( )12 22 23 22 23422 ˆ /2 2 0 0 20 0SSS kS S−−−⎡⎤ ⎢⎥−= −⎢⎥ ⎢⎥−⎣⎦oS=S S D+D S , [] [ ][ ]2 2/ 200 0/ 2 0 00 0k k⎡⎤−⎢⎥ ⎢⎥ = +−= ⎢⎥ ⎢⎥⎣⎦oD0 D W W D , 2100 0104000k⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦DD = 2 200 ˆ 20 0 0 00 0k⎡⎤−⎢⎥−=⎢⎥ ⎢⎥ ⎣⎦oD=D D , ()220 ˆ 20 0 00 0kk kλμ μ μλ μ⎡ ⎤ −⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎣ ⎦2 2 D+ D 11 1 12 12 1 22 13 1 23 1 12 1 22 22 23 13 1 23 23 332 ˆSk S S k S S k S Sk S S S Sk S S Sλλλ λλ λ−− −⎡⎤ ⎢⎥=−⎢⎥ ⎢⎥−⎣⎦S+ S () 1 2 11 1 12 12 1 22 13 1 23 12 1 22 22 23 13 1 23 23 33ˆ ˆ 2 22 0 00 00 0Sk S S k S S k S kk Sk S S S k Sk S S Sλμλ λλλ λ μ μ λμ λ=→ ⎡ ⎤ −− − −⎡⎤⎢ ⎥ ⎢⎥−= ⎢ ⎥ ⎢⎥⎢ ⎥ ⎢⎥−⎣⎦ ⎣ ⎦2 2S+ S D+ D Thus, 2 22 23 33 13 12 11 1 12 0, , 2 2 SSSS S k S k S k μ λλ μ ==== = − = −2 , so that, we have, ()2 12 11 1 , 2 Sk S kμ μλ λ ==−2, all other 0ijS=. The apparent viscosity is () ( )2 12 1 11 22 1 1 22 33/, = 2 , = 0 kSk TT k T Tημ σμ λ λ σ== − = − − =2 . ____________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-268.38 Obtain the apparent viscosity and the normal stress functions for the Oldroyd 4-constant fluid [see (D) of Section 8.20] ------------------------------------------------------------------------------- Ans. For the simple shearing flow [] [ ]0 / 20 0 / 20 / 200 , / 200 00 0 0 0 0kk kk⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥== −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦DW ,12 11 22 23 11 22 12 13 32 312 220SS SS kSS S S SS−− −⎡ ⎤ ⎛⎞⎢ ⎥=−⎜⎟⎢ ⎥⎝⎠⎢ ⎥ −⎣ ⎦oS , () ( )12 22 23 22 23422 ˆ /2 2 0 0 20 0SSS kS S−−−⎡⎤ ⎢⎥−= −⎢⎥ ⎢⎥−⎣⎦oS=S S D+D S , [] [ ][ ]2 2/ 200 0/ 2 0 00 0k k⎡⎤−⎢⎥ ⎢⎥ = +−= ⎢⎥ ⎢⎥⎣⎦oD0 D W W D , 2100 0104000k⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦DD = , 2 22 2100 100 00 ˆ 20 1 0 0 1 0 0 0 02200 0 0 0 0 0 0 0k kk⎡ ⎤ −−⎡⎤ ⎡ ⎤⎢ ⎥ ⎢⎥ ⎢ ⎥−= − = ⎢ ⎥ ⎢⎥ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣ ⎦oD=D D , ()[]() o o 11 22 330/ 2 0 / 200 00 0k tr S S S kμμ⎡⎤ ⎢⎥=+ +⎢⎥ ⎢⎥⎣⎦SD . ()() 1oˆ ˆ 2 trλμ μ λ+= →2 S+ S S D D+ D ( ) ()11 1 12 12 1 22 o 11 22 33 13 1 23 12 1 22 o 11 22 33 22 23 13 1 23 23 33 22/ 2 /2 20 00 . 00 0Sk S S k S k S S S S k S Sk S k SSS S S Sk S S S kk kλλ μ λ λμ λ λμ μ μ⎡ ⎤ −− + + + − ⎢ ⎥−+ + +⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎡⎤−⎢⎥=⎢⎥ ⎢⎥ ⎣⎦2T hus, 2 22 23 33 13 11 1 12 12 o 11 0, 2 = 2 , / 2 SSSS S k S k S k S k λ λμ μ μ ==== − − + =2 From which, we get, with 2 1o () ( 1 )Bkkλμ≡+ , () ()22 11 1 12 o =2 / ( ), 1 / ( )Sk B k Sk k B kμλ λ μ λ μ−= +22 . Thus, the apparent viscosity is: ()2 12 o/( 1 ) / ( ) kSk k B kημ λ μ== +2 . Normal stress functions are: 2 11 12 2 1 2 2 2 3 3 =2 ( ) / ( ) , =0TT k B k T Tσμ λ λ σ−=− − =2 . ____________________________________________________________ Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-278.39 Given [] {}100 01 0 00 1 i−⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦nQ and [] {}010 000 000 i⎡⎤ ⎢⎥=⎢⎥ ⎢⎥⎣⎦nN and ()T2 T 12 and 2 kk==AN + N A N N . (a) Verify that TT 11 2 2 and =− = QA Q A QA Q A . (b) From () 12, p−T= I+f A A and () ( )TT T 12 1 2,,= Qf A A Q f QA Q QA Q , show that () ( )Tkk=− QT Q T and (c) From the results of part (b), show th at the viscometric functions have the properties: ()()()()()() 11 2 2, , kk k k kk σσ σσ =− − = − = − SS . ------------------------------------------------------------------------------- Ans. (a) []()T2 2 12000 00 0 0 0 0 1 0 =0 0 a n d 2 1 0 0 0 0 0 0 20 0 0 0 0 0 0 0 0 0 000k kk k k⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎡⎤== = ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦AN + N A []T 1 1100 0 0 100 0 0 01 0 0 001 0 0 0 0 01000 0 01 000kk kk−−⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎡⎤ == − = −⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎢⎥ ⎢ ⎥ ⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎣⎦ ⎣ ⎦QA Q A []T2 2 2 2000 000 100 100 01 0 0 2 001 0 0 2 0 00 1 0 0 000 1 0 0 0kk⎡⎤ ⎡⎤ −−⎡⎤ ⎡⎤⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎡⎤ == = ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦QA Q A (b) () () () () () ()()TT T 12 12 TT 12 1 2,, ,, .kp p pp k k⎡⎤=− = −⎣⎦ =− =− −Q T Q Q I+f A A Q I+Q f A A Q I+f Q AQ Q A Q I+f A A Now, ()()()() 112 2 and kk k k−= − −=AAA A , thus,. ( ) () () ( )T 12 , kp k k=− − − QT Q I + f A A and ( ) () () ( )T 12 , kp k k−= − QT Q I + f A A . That is, () ( )Tkk−= QT Q T and () ( )Tkk=− QT Q T (c) []()11 12 13 11 12 13 T 21 22 23 21 22 23 31 32 33 31 32 33100 100 01 0 01 0 00 1 00 1TTT T T T kT T T T T T TTT T T T−− − − ⎡ ⎤⎡ ⎤ ⎡⎤ ⎡⎤ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥⎡⎤⎡⎤ == −⎣⎦ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥⎣⎦ ⎢ ⎥⎢ ⎥ ⎢⎥ ⎢⎥ − ⎣⎦ ⎣⎦ ⎣ ⎦⎣ ⎦QT Q () ( ) () ( ) () ( ) () ( )T 11 11 22 22 33 33 , , kk T k T k T k T k T k T k−=→ − = − = − = QT Q T ()()()()()() 12 12 13 13 23 23 12, , Tk T k Tk T k Tk T k−−= −− = −− = . Thus, ()()()()()() 11 2 2 , kk k k S k S kσσ σσ=− =− = − − , . [Note, in viscometric flow, 13 23 0 TT== ]. ____________________________________________________________ 8.40 For the velocity field given in example 8.21.2, i.e., () 0, 0, rzvvv v rθ=== , (a) obtain the stress components in terms of the shear stress function ()Sk and the normal stress functions ()() 12 and kkσσ , where / kd v d r= , (b) obtain the following velocity distribution for the Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-28Poiseuille flow under a pressure gradient of ( f−): () ( ) /2R rvr f r d rγ=∫, where γis the inverse shear stress function, and (c)obtain the relation () ()32 3/2 1/( ) /RfR f f Q fγπ ⎡⎤= ∂∂⎣⎦. ------------------------------------------------------------------------------- Ans. (a) In example 8.21.2, we see that the velocity field () 0, 0, rzvvv v rθ=== describes a viscometric flow with the nonzero Rivlin-Ericksen tensors given by []() () [] () i i2 12000 00 00 , 0 2 0 00 0 0 0 0kr kr k r⎡ ⎤ ⎡⎤ ⎢ ⎥ ⎢⎥== ⎢ ⎥ ⎢⎥ ⎢ ⎥ ⎢⎥⎣⎦ ⎣ ⎦ n nAA , where () 1z 2 3 , , and /r k r dv drθ === =nen ene (see Example 8.10.2, but, note the differences in the order of bases). Thus the stress components with respect to the basis {}inare given by (See section 8.22): () () 12 ( ) , , , 0zr zz rr rr z rSk S S k S S k S Sθθ θ θ τσ σ=− = − = = = . (b) With ijS depending only on r, the equations of motion become: ()10 ( ), 0 (ii) , 0 ( )rr rr rzSS S pp pir S i i irrr r r zθθ θ− ∂ ∂∂ ∂ ∂+− = = − =∂∂ ∂ ∂ ∂ Eq. (i) gives 0p rz∂∂⎛⎞=⎜⎟∂∂⎝⎠, Eq. (ii) gives 0p zθ∂∂⎛⎞=⎜⎟∂∂⎝⎠ and Eq. (iii) gives 0p zz∂∂⎛⎞=⎜⎟∂∂⎝⎠ Thus, / a constant pzf∂∂ = ≡ − . Eq. (iii) becomes () ()1 2rz rz rzfrCrS f rS fr Srr r r∂∂=− → =− → =− +∂∂. Since rzS must be finite at 0r=, thus, 0C= and /2rzSf r=− . Now, () w h e r e ()rzSk kτ τ = is the shear stress function and / kd v d r= . Thus, () /2 kf rτ=− . Inverting this equation, we have, () ()1/2 /2 kf r f rτγ−=− ≡− . Since ()kτ is an odd function of k, therefore, γ is also an odd function of k, so that () () () /2 / /2 /2 k f rd v d r f rd v f r d rγγ γ=− → =− → =− . Thus, () () ( ) /2R rvR vr f r d r γ −= −∫. Since ()0 vR=, therefore, () ( ) /2R rvr f r d rγ=∫. (c) The volume discharge is given by ()02RQv r r d r π =∫. Therefore, () () ()22 2 2 2 00 0 0 0/2R RR R R dv dvQ vrd r vrr r d r r d r r f r d rdr drππ ππ γ⎧⎫⎡⎤ == − = − = ⎨⎬⎣⎦⎩⎭∫∫ ∫ ∫ Thus, ()2 0// 2RQr f r d rπγ=∫. Let 22 2/ 2 2 / and 4 /fr s dr ds f r s f≡→ ≡ = , then , () ()() ()/2 /222 3 3 2 00 0// 2 8 / / 8RR f R f rs sQ r f r d r s f sd s f Q s sd sπγ γ π γ== === → =∫∫ ∫. Differentiating the last equation with respect to f, we obtain ()322 32 1882222 22 2fQRf Rf Rf R Rf Rf RfRfffγγ γπ∂ ⎧⎫∂⎪⎪⎛⎞ ⎛⎞ ⎛⎞⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞== = ⎨⎬⎜⎟ ⎜⎟ ⎜⎟⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟∂∂ ⎝⎠ ⎝⎠ ⎝⎠⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠⎪⎪⎩⎭. Thus, ()3 321 2fQRf f Rfγ π∂⎛⎞=⎜⎟∂ ⎝⎠. Lai et al, Introduction to Continuum Mechanics Copyright 2010, Elsevier Inc 8-29____________________________________________________________