piaras Part II

11 Differential Equations for Solid Mechanics Simple problems involving homogeneous stress states have been considered so far, wherein the stress is the same throughout the component under study. An exception to this was the varying stress field in the loaded beam, but there a simplified set of elasticity equations was used. Here the question of vary ing stress and strain fi elds in materials is considered. In order to solve such problems, a differential formulation is required. In this Chapter, a number of differential equations will be derived, relating the stresses and body forces ( equations of motion ), the strains and displacements ( strain-displacement relations ) and the strains with each other ( compatibility relations ). These equations are derived from physical principles and so apply to any type of material, although the latter two are derived under the assu mption of small strain. 2 Section 1.1 Solid Mechanics Part II Kelly 31.1 The Equations of Motion In Part I, balance of forces and moments ac ting on any component was enforced in order to ensure that the component was in equilibr ium. Here, allowance is made for stresses which vary continuously throughout a material , and force equilibrium of any portion of material is enforced. One-Dimensional Equation Consider a one-dimensional differential element of length xΔ and cross sectional area A, Fig. 1.1.1. Let the average body force per unit volume acting on the element be b and the average acceleration and density of the element be a and ρ. Stresses σ act on the element. Figure 1.1.1: a differential element under th e action of surface and body forces The net surface force acting is Ax Ax x )( ) ( σ σ −Δ+ . If the element is small, then the body force and velocity can be assumed to va ry linearly over the element and the average will act at the centre of the element. Th en the body force acting on the element is xAbΔ and the inertial force is xaAΔρ . Applying Newton’s second law leads to a bxx x xxAa xAbAx Ax x ρσ σρ σ σ =+Δ−Δ+→Δ=Δ+−Δ+ )( ) ()( ) ( (1.1.1) so that, by the definition of the derivative, in the limit as 0→Δx , a bdxdρσ=+ 1-d Equation of Motion (1.1.2) which is the one-dimensional equation of motion . Note that this equation was derived on the basis of a physical law and must therefor e be satisfied for all materials, whatever they be composed of. The derivative dx d/σ is the stress gradient – physically, it is a m easure of how rapidly the stresses are changing. Example Consider a bar of length l which hangs from a ceiling, as shown in Fig. 1.1.2. A xΔ)(xσ) ( x xΔ+σ x x xΔ+ab, Section 1.1 Solid Mechanics Part II Kelly 4 Figure 1.1.2: a hanging bar The gravitational force is mgF= downward and the body force per unit volume is thus g bρ= . There are no accelerating mate rial particles. Taking the z axis positive down, an integration of the equation of motion gives c gz gdzd+−=→=+ ρσ ρσ0 (1.1.3) where c is an arbitrary constant. The lower end of the bar is free and so the stress there is zero, and so ()zlg−=ρσ (1.1.4) ■ Two-Dimensional Equations Consider now a two dimensional infin itesimal element of width and height xΔ and yΔ and unit depth (into the page). Looking at the normal stress components acting in the x−direction, and allowing for variations in stress over the element surfaces, the stresses are as shown in Fig. 1.1.3. Figure 1.1.3: varying stresses acti ng on a differential element Using a (two dimensional) Taylor series and dropping higher order terms then leads to the linearly varying stresses illustra ted in Fig. 1.1.4. (where ()yxxx xx ,σσ≡ and the partial derivatives are evaluated at ()yx,), which is a reasonable approximation when the element is small. yΔ xΔ ),(yxxxσ) ,( y yxxxΔ+σ ), ( yx xxxΔ+σ) , ( y yx xxx Δ+Δ+σlz Section 1.1 Solid Mechanics Part II Kelly 5 Figure 1.1.4: linearly varyin g stresses acting on a differential element The effect (resultant force) of this linear variation of stress on the plane can be replicated by a constant stress acting over the whole plane, the size of which is the average stress. For the left and right side s, one has, respectively, yyxx xx∂∂Δ+σσ21, yyxxxx xx xx∂∂Δ+∂∂Δ+σ σσ21 (1.1.5) One can take away the stress y yxx∂∂Δ / )2/1(σ from both sides without affecting the net force acting on the element so one finally ha s the representation shown in Fig. 1.1.5. Figure 1.1.5: net stresses acting on a differential element Carrying out the same procedure for the shear stresses contributing to a force in the x−direction leads to the stre sses shown in Fig. 1.1.6. Figure 1.1.6: normal and shear stresses acting on a differential element Take x xba, to be the average acceleration and body force, and ρ to be the average density. Newton’s law then yields xxσyyxx xx∂∂Δ+σσ xxxx xx∂∂Δ+σσyyxxxx xx xx∂∂Δ+∂∂Δ+σ σσ ),(yxxxσxxxx xx∂∂Δ+σσ 2xΔ 1xΔ ),(yxxyσxxxx xx Δ∂∂+σσ1 1,vb ),(yxxxσyyxy xyΔ∂∂+σσ Section 1.1 Solid Mechanics Part II Kelly 6yxa yxbyyy x yxx yx xxy xy xyxx xx xx ΔΔ=ΔΔ+Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂Δ++Δ−Δ⎟ ⎠⎞⎜ ⎝⎛ ∂∂Δ++Δ− ρσσσσσσ (1.1.6) which, dividing through by yxΔΔ and taking the limit, gives x xxy xxa by xρσσ=+∂∂+∂∂ (1.1.7) A similar analysis for force components in the y−direction yields another equation and one then has the two-dimensional equations of motion: y yyy xyx xxy xx a by xa by x ρσσρσσ =+∂∂+∂∂=+∂∂+∂∂ 2-D Equations of Motion (1.1.8) Three-Dimensional Equations Similarly, one can consider a three-di mensional element, and one finds that z zzz zy zxy yyz yy yxx xxz xy xx a bz y xa bz y xa bz y x ρσσσρσσσρσσσ =+∂∂+∂∂+∂∂=+∂∂+∂∂+∂∂=+∂∂+∂∂+∂∂ 3-D Equations of Motion (1.1.9) These three equations express forc e-balance in, respectively, the zyx,, directions. Section 1.1 Solid Mechanics Part II Kelly 7 Figure 1.1.7: from Cauchy’s Exerc ices de Mathematiques (1829) The Equations of Equlibrium If the material is not movi ng (or is moving at constant velocity) and is in static equilibrium, then the equations of motion reduce to the equations of equilibrium , 000 =+∂∂+∂∂+∂∂=+∂∂+∂∂+∂∂=+∂∂+∂∂+∂∂ zzz zy zxyyz yy yxxxz xy xx bz y xbz y xbz y x σσσσσσσσσ 3-D Equations of Equilibrium (1.1.10) These equations express the force balance be tween surface forces and body forces in a material. The equations of equilibrium may also be used as a good approximation in the analysis of materials which have relatively small accelerations. 1.1.2 Problems 1. What does the one-dimensional equation of mo tion say about the st resses in a bar in the absence of any body fo rce or acceleration? 2. Does equilibrium exist for the following tw o dimensional stress distribution in the absence of body forces? 03 22 62/8 4 3 2 22 22 2 =====++=−−==−+= yz zy xz zx zzyyyx xyxx y xy xy xy xy xy x σσσσσσσσσ Section 1.1 Solid Mechanics Part II Kelly 83. The elementary beam theory predicts that the stresses in a circular beam due to bending are )4/ ( 3/) ( ,/4 2 2R I I y RV I Myyx xy xx π σσ σ = −== = and all the other stress components are zero. Do these equations satisfy the equations of equilibrium? 4. With respect to axes xyz0 the stress state is given in terms of the coordinates by the matrix [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = xz zz yz yy xyzz zy zxyz yy yxxz xy xx ij 22 22 00 σσσσσσσσσ σ Determine the body force acting on th e material if it is at rest. 5. What is the accelerati on of a material particle of density -3kgm3.0=ρ , subjected to the stress [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− = 4 24 24 2 2 2 22 2 22 2 2 z z yz xzyz y y xyxz xy x x ijσ and gravity (the z axis is directed vertically upwards from the ground). 6. A fluid at rest is subjecte d to a hydrostatic pressure p and the force of gravity only. (a) Write out the equations of motion for this case. (b) A very basic formula of hydrostatics, to be found in any elementary book on fluid mechanics, is that giving the pressure variation in a static fluid, gh pρ=Δ where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the vertical distance between the two point s in the fluid (the relative depth). Show that this formula is but a special case of the equations of motion. Section 1.2 Solid Mechanics Part II Kelly 91.2 The Strain-Displacement Relations The strain was introduced in Part I: §3.6. E xpressions which relate the displacements of material particles to the strain s for a continuously varying stra in field are derived in what follows. 1.2.1 The Strain-Displacement Relations Normal Strain Consider a line element of length xΔ emanating from position ),(yx and lying in the x- direction, denoted by AB in Fig. 1.2.1. After deformation the line element occupies BA′′, having undergone a translati on, extension and rotation. Figure 1.2.1: deformation of a line element The particle that was originally at x has undergone a displacement ) ,(yxux and the other end of the line element ha s undergone a displacement ) , ( yx xuxΔ+ . By the definition of (small) normal strain, xyxuyx xu ABAB BAx x xxΔ−Δ+=−′=),( ), (* ε (1.2.1) In the limit 0→Δx one has xux xx∂∂=ε (1.2.2) This partial derivative is a displacement gradient , a measure of how rapid the displacement changes through the material, and is the strain at ),(yx. Physically, it represents the (approximate) unit change in length of a line element, as indicated in Fig. 1.2.2. x•• A•• BA′B′ ),(yxux), ( yx xuxΔ+ x x xΔ+y *B Section 1.2 Solid Mechanics Part II Kelly 10 Figure 1.2.2: unit change in length of a line element Similarly, by considering a line element initially lying in the y direction, the strain in the y direction can be expressed as yuy yy∂∂=ε (1.2.3) Shear Strain The particles A and B in Fig. 1.2.1 also unde rgo displacements in the y direction and this is shown in Fig. 1.2.3. In this case, one has xxuBByΔ∂∂=′* (1.2.4) Figure 1.2.3: deformation of a line element A similar relation can be derived by consider ing a line element initially lying in the y direction. A summary is given in Fig. 1.2.4. Form the figure, xu x ux uy xy ∂∂≈∂∂+∂∂=≈/ 1/tanθθ provided that (i) θ is small and (ii) the displacement gradient x ux∂∂/ is small. A similar expression for the angle λ can be derived, and hence the shear strain can be written in terms of displacement gradients. x•• A•• BA′B′ ),(yxuy x x xΔ+y *B), ( yx xuyΔ+A BA′*B xΔxΔ xxuxΔ∂∂B′ Section 1.2 Solid Mechanics Part II Kelly 11 Figure 1.2.4: strains in terms of displacement gradients The Small-Strain Stress-Strain Relations In summary, one has ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂=∂∂= xu yuyuxu y x xyy yyx xx 21εεε 2-D Strain-Displacement relations (1.2.5) 1.2.2 Geometrical Interpretation of Small Strain A geometric interpretation of the strain was given in Part I: §3.6.4. This interpretation is repeated here, only now in terms of displacement gradients. Positive Normal Strain Fig. 1.2.5a, 021,0 ,0 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂==∂∂=>∂∂=xy yu yu xux xyy yyx xx ε ε ε (1.2.6) Negative Normal Strain Fig 1.2.5b, 021,0 ,0 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂==∂∂=<∂∂=xu yu yu xu y x xyy yyx xx ε ε ε (1.2.7) yΔ yΔ xΔyΔ θλ xΔyΔ xΔ xΔxux ∂∂xuy ∂∂yux ∂∂ yuy ∂∂ Section 1.2 Solid Mechanics Part II Kelly 12 Figure 1.2.5: some simple deformations; (a ) positive normal strain, (b) negative normal strain, (c) simple shear Simple Shear Fig. 1.2.5c, yu xu yu yu xux y x xyy yyx xx∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂==∂∂==∂∂=21 21,0 ,0 ε ε ε (1.2.8) Pure Shear Fig 1.2.6a, xu yu xu yu yu xu y x y x xyy yyx xx∂∂=∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂==∂∂==∂∂=21,0 ,0 ε ε ε (1.2.9) Pure Rotation Fig 1.2.6b, 021=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=xu yu y x xyε (1.2.10) Figure 1.2.6: (a) pure shear, (c) pure rotation )a() b()(xuy)(yux− )(xuy)(yuxx) ( x xuxΔ+ y)(xux )a() b( )c()(yux Section 1.2 Solid Mechanics Part II Kelly 131.2.3 The Rotation Form Fig. 1.2.6b and Eqn. 1.2.10, a rigid body rotation of an element occurs when xu yu y x ∂∂−=∂∂ (1.2.11) This leads one to define the rotation of a material particle, zω, the “ z” signifying the axis about which the element is rotating: ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=yu xux y z21ω (1.2.12) The rotation will in gene ral vary throughout a material. When the rotation is everywhere zero, the material is said to be irrotational . Note that any shear strain can be decom posed into a pure shear and a rotation, as illustrated in Fig. 1.2.7. Figure 1.2.7: decomposition of a shear stra in into a pure shear and a rotation 1.2.4 Fixing Displacements The strains give information a bout the deformation of material particles but, since they do not encompass translations and rotations, they do not give informa tion about the precise location in space of particles. To determine this, one must specify three displacement components (in two-dimensional problems). Ma thematically, this is equivalent to saying arbitrary shear strain xy xuy ∂∂=θyux ∂∂=λ ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=+=xu yu y x xy21 21θλε ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=−=yu xux y z21 21λθωzω +rotation (no strain) pure shear ()xyy x xu yuεθλ=+=∂∂=∂∂ 21zω xyε xyε Section 1.2 Solid Mechanics Part II Kelly 14that one cannot uniquely determine the di splacements from the strain-displacement relations 1.2.5. Example Consider the strain field 0 ,01.0 == =xy yy xx εε ε . The displacements can be obtained by integrating the strain-displacement relations: )()( 01.0 xg dy uyfx dx u yy yxx x ==+== ∫∫ εε (1.2.13) where f and g are unknown functions of y and x respectively. Substituting the displacement expressions into the shear strain relation gives )( )( xg yf ′−=′ . (1.2.14) Any expression of the form )( )( yGxF= which holds for all x and y implies that F and G are constant1. Since gf′′, are constant, one can integrate to get CxB xg DyA yf += += )(, )( . From 1.2.14, D C−= , and CxB uCyAx u yx+=−+=01.0 (1.2.15) There are three arbitrary constants of integr ation, which can be determined by specifying three displacement components. For ex ample, suppose that it is known that b a u u ux y x = = = ),0(,0)0,0(,0)0,0( . (1.2.16) In that case, ab C B A / ,0 ,0 −=== , and, finally, xab uyabx u yx)/()/( 01.0 −=+= ( 1 . 2 . 1 7 ) which corresponds to Fig. 1.2.8, with ) /(ab being the (tan of the small) angle by which the element has rotated. 1 since, if this was not so, a change in x would change the left hand side of this expression but would not change the right hand side and so the equality cannot hold Section 1.2 Solid Mechanics Part II Kelly 15 Figure 1.2.8: an element undergoing a normal strain and a rotation ■ In general, the displacement field will be of the form CxB uCyA u yx++ =−+ = LLLL (1.2.18) and indeed Eqn. 1.2.15 is of this form. Physically, A, B and C represent the possible rigid body motions of the material as a whole , since they are the same fo r all material particles. A corresponds to a translation in the x direction, B corresponds to a translation in the x direction, and C corresponds to a positive (counterclockwise) rotation. 1.2.5 Three Dimensional Strain The three-dimensional stress-strain re lations analogous to Eqns. 1.2.5 are ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂=∂∂=∂∂= yu zu xu zu xu yuzu yu xu z y yzz x xzy x xyz zzy yyx xx 21,21,21, , ε ε εε ε ε 3-D Stress-Strain relations (1.2.19) The rotations are ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=zu yu xu zu yu xuy z xz x yx y z21,21,21ω ω ω (1.2.20) 1.2.6 Problems 1. The displacement field in a material is given by ()2, 3 Axy u yxA uy x =−= where A is a small constant. xy ab Section 1.2 Solid Mechanics Part II Kelly 16(a) Evaluate the strains. What is the rotation zω? Sketch the deformation and any rigid body motions of a differe ntial element at the point )1,1( (b) Sketch the deformation and ri gid body motions at the point )2,0( , by using a pure shear strain superimposed on the rotation. 2. The strains in a material are given by αεεαε == =xy yy xx x ,0 , Evaluate the displacements in terms of three arbitrary constants of integration, in the form of Eqn. 1.2.15, CxB uCyA u yx++ =−+ = LLLL What is the rotation? 3. The strains in a material are given by Ax Ay Axyxy yy xx = = = ε ε ε , ,2 where A is a small constant. Evaluate the displacements in terms of three arbitrary constants of integration. What is the rotation? 4. Show that, in a state of plane strain ( 0=zzε ) with zero body force, 22 22 2yu xu y xex x z ∂∂+∂∂=∂∂−∂∂ω where e is the volumetric strain or dilatation , the sum of the normal strains: zz yy xx e εεε++= . Section 1.3 Solid Mechanics Part II Kelly 171.3 Compatibility of Strain As seen in the previous section, the displa cements can be determined from the strains through integration, to within a rigid body motion. In the two- dimensional case, there are three strain-displacement relations but only two displacement components. This implies that the strains are not independent but are related in some way. The relations between the strains are called compatibility conditions . 1.3.1 The Compatibility Relations Differentiating the first of 1.2.5 twice with respect to y, the second twice with respect to x and the third once each with respect to x and y yields ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂+∂∂∂=∂∂∂ ∂∂∂=∂∂ ∂∂∂=∂∂ yxu yxu yx yxu x yxu yy x xy y yy x xx 23 23 2 23 22 23 22 21, ,ε ε ε It follows that yx x yxy yy xx ∂∂∂=∂∂+∂∂ εεε2 22 22 2 2-D Compatibility Equation (1.3.1) This compatibility condition is an equation wh ich must be satisfied by the strains at all material particles. Physical Meaning of th e Compatibility Condition When all material particles in a component defo rm, translate and rotate, they need to meet up again very much like the pieces of a jigsaw puzzle must fit together. Fig. 1.3.1 illustrates possible deformations and rigid body motions for three line elements in a material. Compatibility en sures that they stay toge ther after the deformation. Figure 1.3.1: Deformatio n and Compatibility undeformed deformed - compatibility ensured deformed - compatibility not satisfied Section 1.3 Solid Mechanics Part II Kelly 18The Three Dimensional Case There are six compatibility relations to be satisfied in the three dimensional case : ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+∂∂+∂∂=∂∂∂ ∂∂∂=∂∂+∂∂⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂−∂∂+∂∂=∂∂∂ ∂∂∂=∂∂+∂∂⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂−∂∂=∂∂∂ ∂∂∂=∂∂+∂∂ z y x z yx yx x yz y x y xz xz z xz y x x zy zy y z xy zx yz zz xy yy xxxy zx yz yy zx xx zzxy zx yz xx yz zz yy εεε εεεεεεε εεεεεεε εεεε 2 2 22 222 2 22 222 2 22 22 , 2, 2, 2 (1.3.2) By inspection, it will be seen that these are satisfied by Eqns. 1.2.19. 1.3.2 Problems 1. The displacement field in a material is given by 2, Ay u Axy uy x = = , where A is a small constant. Determine (a) the components of small strain (b) the rotation (c) the principal strains (d) whether the compatibility condition is satisfied 192 One-Dimensional Elasticity There are two types of one-dimensional problems, the elastostatic problem and the elastodynamic problem. The elastostatic prob lem gives rise to a second order differential equation in displacement wh ich may be solved using elementary integration. The elastodynamic problem gi ves rise to the one-dimensional wave equation, whose solution predicts the propa gation of stress waves and vibrations of material particles 20 Section 2.1 Solid Mechanics Part II Kelly 212.1 One-dimensional Elastostatics Consider a bar or rod made of linearly elastic material subjected to some load. Static problems will be considered here, by which is meant it is not necessary to know how the load was applied, or how the material particle s moved to reach the st ressed state; it is necessary only that the load was applied slow ly enough so that the a ccelerations are zero, or that it was applied sufficiently long ago that any vibrations have died away and movement has ceased. The equations governing the stat ic response of the rod are: 0=+bdxdσ Equation of Equilibrium (2.1.1a) dxdu=ε Strain-Displacement Relation (2.1.1b) εσE= Constitutive Equation (2.1.1c) where E is the Young’s modulus, ρ is the density and b is a body force (per unit volume). The unknowns of the problem are the stress σ, strain ε and displacement u. These equations can be combined to give a second order differential equation in u, called Navier’s Equation : 022 =+Eb dxud 1-D Navier’s Equation (2.1.2) One requires two boundary conditions to obtain a solution. Let the length of the rod be L and the x axis be positioned as in Fig. 2.1.1. The possible boundary conditions are then 1. displacement specified at both ends (“fixed-fixed”) LuLu u u = = )(, )0(0 2. stress specified at both ends (“free-free”) L Lσσσσ = = )(, )0(0 3. displacement specified at left-end, stress specified at right-e nd (“fixed-free”): L L u u σσ= = )(, )0(0 4. stress specified at left-e nd, displacement specified at right-end (“free-fixed”): LuLu= = )(, )0(0σσ Figure 2.1.1: an elastic rod x L Section 2.1 Solid Mechanics Part II Kelly 22Note that, from 2.1.1b-c, a stress boundary c ondition is a condition on the first derivative of u. Example Consider a rod in the absence of any body forces subjected to an applied stress oσ, Fig. 2.1.2. Figure 2.1.2: an elastic ro d subjected to stress The equation to solve is 022 =dxud ( 2 . 1 . 3 ) subject to the boundary conditions E dxu E dxu Lx x0 0 0,σ σ=∂=∂ = = (2.1.4) Integrating twice and applying th e conditions gives the solution BxEu+=0σ (2.1.5) The stress is thus a constant 0σ and the strain is Eo/σ . There is still an arbitrary constant B and this physically represents a possibl e rigid body translation of the rod. To remove this arbitrariness, one must specify th e displacement at some point in the rod. For example, if 0)2/(= Lu , the complete solution is oo o ELxEu σσσεσ= =⎟ ⎠⎞⎜ ⎝⎛−= , ,2 (2.1.6) 2.1.1 Problems 1. What are the displacements of material particles in an elastic bar of length L and density ρ which hangs from a ceiling (see Fig. 1.1.2). 2. Consider a steel rod ( GPa 210=E , 3g/cm85.7=ρ ) of length cm 30 , fixed at one end and subjected to a displacement mm1=u at the other. Solve for the stress, strain and displacement for the case of gravity acting along the rod. What is the solution in the absence of gravity. How significant is the effect of gravity on the stress? oσoσ Section 2.2 Solid Mechanics Part II Kelly 232.2 One-dimensional Elastodynamics In rigid body dynamics, it is assumed that when a force is applied to one point of an object, every other point in the object is se t in motion simultaneously. On the other hand, in static elasticity, it is assumed that the object is at re st and is in e quilibrium under the action of the applied forces; the material may well have undergone considerable changes in deformation when first struck, but one is only concerned with the final static equilibrium state of the object. Elastostatics and rigid body dynamics are su fficiently accurate for many problems but when one is considering the eff ects of forces which are applied rapidly , or for very short periods of time, the effects must be consid ered in terms of th e propagation of stress waves. 2.2.1 The Wave Equation Consider now the dynamic problem. In this case one considers the equation of motion: a bdxdρσ=+ Equation of Motion (2.2.1a) dxdu=ε Strain-Displacement Relation (2.2.1b) εσE= Constitutive Equation (2.2.1c) where a is the acceleration. Expressing the acceler ation in terms of the displacement, one then obtains the dynamic vers ion of Navier’s equation, 22 22 tubxuE∂∂=+ ∂∂ρ 1-D Navier’s Equation (2.2.2) In most situations, the body forces will be negligible, and so consider the partial differential equation 22 2 221 tu c xu ∂∂= ∂∂ 1-D Wave Equation (2.2.3) where ρEc= (2.2.4) Equation 2.2.3 is the standard one-dimensional wave equation with wave speed c; note from 2.2.4 that c has dimensions of velocity. The solution to 2.2.3 (see below) shows that a stress wave travels at speed c through the material from the point of disturbance, e.g. a pplied load. When the stress wave reaches a Section 2.2 Solid Mechanics Part II Kelly 24given material particle, the particle vibr ates about an equilib rium position, Fig. 2.2.1. Since the material is elastic, no energy is lost , and the solution predicts that the particles vibrate indefinitely, without damping or decay. Figure 2.2.1: stress wave travelling at speed c through an elastic rod This type of wave, where the disturbance (parti cle vibration) is in the same direction as the direction of wave propagation, is called a longitudinal wave . 2.2.2 Particle Velocities and Wave Speed Before examining the wave equation 2.2.3 directly, first re-express it as tv x∂∂=∂∂ρσ (2.2.5) where v is the velocity. Consider an element of material wh ich has just been reached by the stress wave, Fig. 2.2.2. The length of mate rial passed by the stress wave in a time interval tΔ is tcΔ. During this time interval, the stresse d material at the left-hand side of the element moves at (average) velocity v and so moves an amount tvΔ. The strain of the element is then the change in leng th divided by the original length: cv=ε (2.2.6) Under the small strain assumption, this implies that c v<<1. Let the stress acting on the element be σΔ; the stress on the free side of the element is zero. Then 2.2.5 leads to tv tcΔ=ΔΔρσ (2.2.7) and so cvρσ=Δ (2.2.8) This is the discontinuity in stress across the wave front. 1 note also that the density of the element will change as it is compressed, but again this change in density is small and can be neglected in the linear elastic theory stress free •vibration of stressed particle stress wave at speed c Section 2.2 Solid Mechanics Part II Kelly 25 Figure 2.2.2: stress wave passing through a material element Since εσE=Δ , one has ρ/E c= , as in 2.2.4. The wave speeds for some materials are given in Table 2.2.1. As can be seen, the wave speeds for typical engineering materials are of the order km/s and so particle velocities will be in the range m/s500− . Material ()3kg/mρ ()GPaE ()m/sc Aluminium Alloy 2700 70 5092 Brass 8300 95 3383 Copper 8500 114 3662 Lead 11300 17.5 1244 Steel 7800 210 5189 Glass 1870 55 5300 Granite 2700 3120 Limestone 2600 4920 Perspex 2260 Table 2.2.1: Elastic Wave Speeds for Several Materials 2.2.3 Solution of the Wave Equation The one-dimensional wave equation 2.2.3 has the general solution ()()ctxg ctxf txu ++−=),( (2.2.9) where f and g are any functions2; for example, one solution is ()ctx g efctx+==−sin ,, which can be verified by substitution a nd carrying out the diffe rentiation. The actual forms of the functions f and g can be determined from the initial conditions of the problem. 2 provided they possess second derivatives wave front at time t tcΔ σΔtvΔ wave front at time t tΔ+ Section 2.2 Solid Mechanics Part II Kelly 26Waves due to Initial Displacement Consider the initial conditions 0)0,()( )0,( == xvxU xu Then ()() () ()[] 0)( )0,( )0,(≡′−′−=∂∂≡+= xgxfctuxUxgxf xu x so that, from the second condition, )( )( xg xf= and, from the first, these must equal 2/)(xU . It follows that 2 /) ( ) ( ctxU ctxf −=− and 2 /) ( ) ( ctxU ctxg +=+ , so that the solution is [ ]) ( ) ( ),(21ctxU ctxU txu ++−= Suppose for example that the in itial displacement profile wa s triangular, with maximum displacement uu= at 0=x , extending to L x±= , Fig. 2.2.3. Figure 2.2.3: an initial tr iangular displacement Then [ ]ctx ctx txu+ −Δ+Δ=21),( where ctx±Δ means “a triangular displacement with centre at ctx± and length L2. At time zero, the displacement at 0=x is 0 )0,0(Δ= u as required. At time cL/2 , however, [ ]Lx Lx txu2 2 21),(+ −Δ+Δ= x 0=u 0=u)0,(xux Section 2.2 Solid Mechanics Part II Kelly 27which corresponds to two triangular displacemen t profiles of half the magnitude of the original profile; one is to the left and the othe r is to the right of the original profile, Fig. 2.2.4. Figure 2.2.4: displacements at time 2 L/c As the wave passes, particles displace from their equilibrium point, up to the maximum position and then back again. It can be seen that the solution corre sponds to a wave of disturbed material propagating th rough the material from the s ource, half in one direction and half in the other. 2.2.4 Vibration Analysis Consider now an alternative solution to th e wave equation (see the Appendix to this section, §2.2.6, for details) () ( )∑∞ =+ + = 1sin cos sin cos ),( nn n n n n n n n tc Dtc Cx Bx A txu λ λ λ λ , (2.2.10) The constants DCBA ,,, and eigenvalues3 λ can be obtained from the initial and boundary conditions (see later). The terms xnλcos and xnλsin are called modes or mode shapes . At any given time t, the displacements are a linear combination of these modes. Example modes are shown in Fig. 2.2.5. Some modes will dominate over ot hers, for example perhaps only the first few modes (terms in the series 2.2.10) ar e significant and need be considered. A vibration analysis is one in which the eigenv alues (or, equi valently, the natural frequencies λωc= ) and modes are evaluated without regard to which of them might be important in an application. The boundary conditions alone determine the modes and natural frequencies. Thus a vi bration analysis is carried out without regard to how the vibration is initiated . The exact combination of the m odes for a particular problem is determined from the initial conditions; the in itial conditions will determine the constants DCBA ,,, in the above equations and hence the actual amplitude of vibration. 3 note that some authors use the term “eigenvalue” to mean the quantity ()ncλ in this expression x 0=u0=u)0 (0 == ux L x2=L x 2−= Section 2.2 Solid Mechanics Part II Kelly 28The vibration is termed free if the load is zero or constant; forced vibration occurs when the load itself oscillates. Figure 2.2.5: mode shapes for a vibrating elastic rod Even though a vibration analys is does not completely solve the problem of a material model loaded in a certain way, for example solving for the propagation paths of stress waves, the amplitudes of vibration, and so on, the natural frequencie s and modes are very useful information in themselves, for design and other purposes. Dynamic response analysis or transient response analysis is the calculation of the complete response to any arbitrary boundary and initi al conditions. This is more difficult than the vibration analysis, sin ce it is a time-depe ndent problem. Natural Frequencies The natural frequencies depend on the boundary conditions. There are four possible cases, the same as for the static elasticity problem: 1. fixed-fixed - 0 ),0(=tu , 0 ),(=tLu 2. free-free - 0 /),0(=∂∂txu , 0 /),(=∂∂tLxu 3. fixed-free - 0),0(=tu , 0 /),(=∂∂tLxu (2.2.11) 4. free-fixed - 0 /),0(=∂∂txu , 0 ),(=tLu The natural frequencies and modes for each of these boundary conditions are solved for and given in the Appendix to this section, §2.2 .6 (in the boxes). For example, considering the “fixed-fixed” case, the solution is []∑∞ =+ = 01) sin() sin( ) cos( ),( nn n n n n x ct B ct A txu λλ λ (2.2.12) with Frequencies: K,1,0 ,= == nLcncn nπλω -1-0.500.51 0.2 0.4 0.6 0.8 11stmode 2ndmode3rdmode x Section 2.2 Solid Mechanics Part II Kelly 29Modes: () K,1,0 , sin =n xnλ (2.2.13) One can plot these sine functions over ],0[L to see the displacement profile of each mode (the first three are those plotte d in Fig. 2.2.5 – it can be seen that the higher the mode, the higher the frequency). The boundary conditions in 2.2.11 are all homogeneous (i.e. 0=). In practice, the boundary conditions will not be homogeneous, but one only needs homogeneous boundary conditions to obtain the natural frequencies (see below). Non-Homogeneous Boundary Conditions Consider the following non-homogeneous boundary conditions: BC’s: utu ˆ),0(=, 0),(=tLu (2.2.14) Since the wave equation is linear, the solution can be written as th e superposition of two separate solutions, ),( ),( ),( txutxutxuh p+= (2.2.15) The hu is the homogeneous solution, and is chosen to satisfy the wave equation with homogeneous boundary conditions; pu is some particular solution and accounts for the non-homogeneous boundary condition: BC’s: 0 ),0(=t uh , 0 ),(=tLuh ut up ˆ),0(=, 0 ),(=tLup (2.2.16) Substituting 2.2.15 into the wave equation 2.2.3 gives ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂−=∂∂−∂∂ 22 2 22 22 2 221 1 tu c xu tu c xu p p h h (2.2.17) The left hand side is zero. The right hand side can be made zero by choosing pu to be any particular solution of the wave equation. For a simple constant displacement boundary condition, one can choose the linear function ⎟ ⎠⎞⎜ ⎝⎛−=Lxuxup 1ˆ)( (2.2.18) which can be seen to satisfy 2.2.16b. The complete solution u is illustrated in Fig. 2.2.6. Section 2.2 Solid Mechanics Part II Kelly 30 Figure 2.2.6: displacements as a superp osition of two separate solutions Suppose now that the in itial conditions are IC’s: )( )0,()( )0,( xv xvxu xu == (2.2.19) The initial conditions can be split between hu and pu according to IC’s: )( )0,( ),( )( )0,()( )0,( ),( )( )0,( xv xv xvxv xvxu xu xuxu xu p p p hp p p h = −== −= (2.2.20) Thus, the complete solution is obtained by adding together: (i) the function hu which satisfies the wave equation with homogeneous boundary conditions on displacement, and initial conditions IC’s: )( )( )0,()( )( )0,( xvxv xvxuxu xu p hp h −=−= (ii) the function ⎟ ⎠⎞⎜ ⎝⎛−=Lxuxup 1ˆ)( Thus, using the “fixed-fixed” homogen eous solution from the Appendix, []∑∞ =+ +⎟ ⎠⎞⎜ ⎝⎛−= 0) sin() sin( ) cos( 1ˆ),( nn n n n n x ct B ct ALxutxu λλ λ (2.2.21) and the natural frequencies are given by 2.2.13. The constants n nBA, can be obtained from the initial conditions, as outlined in the Appendix. An important point to be ma de here is that the modes and natural frequencies are determined from (i), i.e. the problem involving the homogene ous boundary conditions, and so, as stated earlier, the non-homogeneous boundary condition does not affect the modes and natural frequencies. Forced Vibration Suppose now that the boundary conditions and initial conditions are given by uˆ 0 L)(xup)0,(xu Section 2.2 Solid Mechanics Part II Kelly 31 BC’s: () 0),(cos ),0( =Ω= tLut tuα, IC’s: () 0)0,(2/ cosˆ)0,( == xvL x u xu π (2.2.22) Again, let ),( ),( ),( txutxutxuh p+= and substitute into the wave equation. In this case, the particular solution will be of the general form 2.2.12, () ( )tc Dtc Cx Bx A up λ λ λλ sin cos sin cos + + = (2.2.23) Applying the boundary conditi ons, one finds that { ▲Problem 1} ()tcx cL cxtxup Ω ⎭⎬⎫ ⎩⎨⎧⎟ ⎠⎞⎜ ⎝⎛Ω⎟ ⎠⎞⎜ ⎝⎛Ω−⎟ ⎠⎞⎜ ⎝⎛Ω= cos sin cot cos ),(α (2.2.24) As with the constant non-homogeneous bounda ry condition, the initial conditions can now be split appropriately between the homogene ous and particular solutions. Again, the complete solution is obtained by adding together: (i) the function hu which satisfies the wave equation with homogeneous boundary conditions on displacement, and initial conditions IC’s: 0)0,(sin cot cos2cosˆ)0,( =⎭⎬⎫ ⎩⎨⎧⎟ ⎠⎞⎜ ⎝⎛Ω⎟ ⎠⎞⎜ ⎝⎛Ω−⎟ ⎠⎞⎜ ⎝⎛Ω−⎟ ⎠⎞⎜ ⎝⎛= xvcx cL cx Lxu xu hh απ (ii) the function 2.2.24 The complete solution is () []∑∞ =+ +Ω ⎭⎬⎫ ⎩⎨⎧⎟ ⎠⎞⎜ ⎝⎛Ω⎟ ⎠⎞⎜ ⎝⎛Ω−⎟ ⎠⎞⎜ ⎝⎛Ω= 0) sin() sin( ) cos(cos sin cot cos ),( nn n n n n x ct B ct Atcx cl cxtxu λλ λα (2.2.25) Resonance occurs when the displacements become “infinite”, which from 2.2.24 occurs when LcncL π=Ω→=Ω0 sin . These are precisely the natural frequencies of the system, i.e. the natural frequencies of (i). Thus the problem of re sonance becomes more prominent when the forcing frequency Ω approaches any of the natural frequencies nλ. Section 2.2 Solid Mechanics Part II Kelly 322.2.5 Problems 1. Consider the case of forced vibration. Use the boundary conditions 2.2.22 to evaluate the constants in the particular so lution 2.2.23 and hence derive the particular solution 2.2.24. 2. Consider a fixed-free problem, with the end 0=x subjected to a forced displacement t uΩ=sinα and the end Lx= free. (a) Find the vibration of the material. What are the natural frequencies? (b) When does resonance occur? [note: the appropriate homogeneous solution and natura l frequencies are given in the Appendix to this section, §2.2.6] 3. Consider a vibrating bar with an oscillatory stress applied to one end, () tΩ=cos 0ασ . The end Lx= is fixed, 0 )(=Lu . (a) Find the vibration of the material. What are the natural frequencies? (b) When does resonance occur? [note: the appropriate homogeneous solution and natura l frequencies are given in the Appendix to this section, §2.2.6] 2.2.6 Appendix to Section 2.2 Method of Separation of Variabl es Solution to the Wave Equation Assuming a separable solution, write )()( ),( tTxXtxu= so that )()( /2 2tTxX tu &&=∂∂ and )()( /2 2tTxX xu ′′=∂∂ . Inserting these into the wave equation gives dXXd X dtTd TcTdXXdc dtTdX 2 22 22 2 22 1 11= →= (2.2.26) This relation states that a function of t equals a function of x and it must hold for all t and x. It follows that both sides of this e xpression must be equal to a constant, say k (if the left hand side were not constant it would change in value as t is changed, but then the equality would no longer hold because the right hand side does not change when t is changed – it is a function of x only). Thus there are two second order ordinary differential equations: 0 ,02 22 22 =− =− kTcdtTdkXdxXd (2.2.27) which have solutions tkc tkc xk xkDe CeT Be Ae X += =+=−,0 (2.2.28) Section 2.2 Solid Mechanics Part II Kelly 33 Modes and Natural Frequencies fo r Homogeneous Boundary Conditions Suppose first that k is positive. Consider homogeneous boundary conditions, that is, 0=u and/or 0 /=∂∂ xu at the end points L x ,0= . Suppose first that 0),0(=tu . Then 0)0( 0)()0( ),0( =→= = X tT Xtu and so 0=+BA . If also 0),(=tLu , then 0=+− Lk LkBe Ae which implies that 0==BA , and 0),(=txu . Similarly, if one uses the conditions 0),0(/=∂∂ txu or 0),(/=∂∂ tLxu , or a combination of zero u and first derivative, one arrives at the same conclusion: a trivial zero solution. Therefore, to obtain a non-zero solution, one must have k negative, and ) sin( ) cos( )( x Bx A xX λ λ+ = , 2λ−=k (2.2.29) The solution for )(tT must then be ) sin( ) cos( )( ct D ct CtT λ λ+ = (2.2.30) and the full solution is ()() ( )()() ( )ct D ct Cx Bx A txu λ λ λ λ sin cos sin cos ),( + + = (2.2.31) There are four possible combin ations of boundary conditions. 1. Fixed-Fixed Here, 0),( ),0( == tLutu . Thus 0 )0(==A X and 0) sin( )( = = L B LX λ . For non-zero B one must have K,1,0 ,/ 0) sin( = ±=→= nLn L πλ λ . Thus one has the infinite number of solutions ) sin( )( x B xXn n n λ= , and the complete general solution is ( DBBCBA == ,)4 []∑∞ =+ = 1) sin() sin( ) cos( ),( nn n n n n x ct B ct A txu λλ λ (2.2.32) with Frequencies: K,2,1 ,= == nLcncn nπλω Modes: () K,2,1 , sin =n xnλ (2.2.33) It can be proved that the seri es 2.2.32 converges and that it is indeed a solution of the wave equation, provided some fairly weak conditions are fulfilled (see a text on Advanced Calculus). 4 the solutions corresponding to negative values of n, i.e. K,2,1 ,/= −= nL nπλ , can be subsumed into 2.2.32 through the constants n nBA,; the solution for 0=n is zero Section 2.2 Solid Mechanics Part II Kelly 34The first three modes are plotted in Fig. 2.2.7. Figure 2.2.7: first three mode shapes for fixed-fixed Case 2. Free-Free Here, 0),(/ ),0(/ =∂∂=∂∂ tLxu txu . Thus 0 )0(==′ B Xλ and 0) sin( )( = −=′ L A LX λλ . Thus the general solution is ( DABCAA == ,) []∑∞ =+ += 10 ) cos() sin( ) cos( ),( nn n n n n x ct B ct A Atxu λ λ λ (2.2.34) with the nλ as for fixed-fixed. Frequencies: K,2,1 ,= == nLcncn nπλω Modes: () K,2,1 , cos =n xnλ (2.2.35) The displacement profiles of the first three modes are shown in Fig. 2.2.8. Figure 2.2.8: first three mode shapes for free-free -1-0.500.51 0.2 0.4 0.6 0.8 1-1-0.500.51 0.2 0.4 0.6 0.8 1 Section 2.2 Solid Mechanics Part II Kelly 35 Case 3. Fixed-Free Here, 0),(/ ),0( =∂∂= tLxu tu . Thus 0 )0(==A X and 0) cos( )( = = L B LX λλ . For non-zero B one must have K K ,2,1,0,1,2 ,2/)1 2( 0) cos( −−= −=→= nL n L π λ λ . The solution is again given by 2.2.32, which is repeated here, []∑∞ =+ = 1) sin() sin( ) cos( ),( nn n n n n x ct B ct A txu λλ λ (2.2.36) only now Frequencies: K,2,1 ,2)1 2(=−== nLc ncn nπλω Modes: () K,2,1 , sin =n xnλ (2.2.37) The displacement profiles of the first three modes are shown in Fig. 2.2.9. Figure 2.2.9: first three mode shapes for fixed-free Case 4. Free-Fixed Here, 0),( ),0(/ ==∂∂ tLutxu . Thus 0 )0(==′ B Xλ and 0) cos( )( = = L A LX λ . For non-zero A one must have 0) cos(=Lλ so the general solution is as for free-free, Eqn. 2.2.34, but with 00=A : []∑∞ =+ = 1) cos() sin( ) cos( ),( nn n n n n x ct B ct A txu λ λ λ (2.2.38) with the nλ as for fixed-free. Frequencies: K,2,1 ,2)1 2(=−== nLc ncn nπλω Modes: () K,2,1 , cos =n xnλ (2.2.39) -1-0.500.51 0.2 0.4 0.6 0.8 1 Section 2.2 Solid Mechanics Part II Kelly 36 The displacement profiles of the first three modes are shown in Fig. 2.2.10. Figure 2.2.10: first three mode shapes for free-fixed Full Solution (incorporating Initial Conditions) (a) Initial Condition on Displacement The initial condition on displacement is )( )0,( 0xu xu= (2.2.40) which give, from 2.2.32, 2.2.34, 2.2.36, 2.2.38, )( ) sin( )0,(0 1xux A xu nn n = =∑∞ =λ fixed-fixed/fixed-free )( ) cos( )0,(0 10 xux A A xu nn n = +=∑∞ =λ free-free (2.2.41) )( ) cos( )0,(0 1xux A xu nn n = =∑∞ =λ free-fixed These can be solved by using the orthogonality condition of the trigonometric functions: ⎩⎨⎧ =≠= =∫ ∫nm Lnmdxx x dxx xL m nL m n,2/,0) cos() cos( ) sin() sin( 0 0λλ λλ (2.2.42) for either of L n Lnn 2/)12(,/ π πλ − = . Thus multiplying both sides of 2.2.41a by ) sin( xmλ and 2.2.41b-c by ) cos( xmλ and integrating over []L,0 gives dxx xuLAnL n ) sin()(2 00λ∫= fixed-fixed/fixed-free K,2,1 ,) cos()(2,)(1 00 00 0 = = = ∫ ∫ndxx xuLA dxxuLAnL nL λ free-free (2.2.43) -1-0.500.51 0.2 0.4 0.6 0.8 1x Section 2.2 Solid Mechanics Part II Kelly 37dxx xuLAnL n ) cos()(2 00λ∫= f r e e - f i x e d (b) Initial Condition on Velocity The initial condition on velocity, () )( 0,0xv xu=& , gives )( ) sin( )0,(0 1xvx cB xu nn n n = =∑∞ =λλ & fixed-fixed/fixed-free )( ) cos( )0,(0 1xv x cB xu nn n n = =∑∞ =λ λ & free-fixed/free-free (2.2.42) Using the orthogonality conditions again gives dxx xvLcBnL nn ) sin()(2 00λλ∫= fixed-fixed/fixed-free dxx xvLcBnL nn ) cos()(2 00λλ∫= free-fixed/free-free (2.2.43) Example Consider the fixed-free case with initial conditions Lx xv xu /2)(,0)(0 0 = = . Thus 0=nA and cL nnL Lc ndxxLnxLc nB nn L n 3 312 21 2 0 )12()1(32)12()1(4 )12(8 2)12(sin)12(8 ππππ π −−=−− −=⎟ ⎠⎞⎜ ⎝⎛− −= ++ ∫ so that K,2,1 ,2)12(), sin() sin()12()1( 32),( 131 3=−==−−=∑∞ =+ nLc nc ct xn cLtxun n n nnnπλω λλπ The period for the first (dominant) mode is cL c T /4 /21 1 ==λπ . The solution is plotted in Fig. 2.2.11 for m/s 5000=c , m1.0=L , for the five times 40 ,16/1 K=i iT (up to the quarter-period). Thereafter, the solution decreases back to zero, down through negative displacements, back to zero and then repeats. Section 2.2 Solid Mechanics Part II Kelly 38 Figure 2.2.11: displacements for fixed-free example Example Consider the free-free case with initial conditions 0)(, )(0 0 = = xvxuxu . Thus 0 =nB and ()() K,2,1 ,1 12cos22 22 000 =−−=⎟ ⎠⎞⎜ ⎝⎛=== ∫∫ nnLudxLxnxLuALudxxLuA nL nL ππ so that ()() Lnx ctnLutxun nn nnπλλλπ=⎥ ⎦⎤ ⎢ ⎣⎡ −−+=∑∞ =,) cos() cos(1 1 2 21),( 12 2 (2.2.34) The period for the first (dominant) mode is cL c T /2 /21 1 = =λπ . The solution is plotted in Fig. 2.2.12 again for m/s 5000=c , m1.0=L , for the nine times 80 ,16/1 K=i iT (up to the half-period). Thereafter, the solution returns back to the initial position and then repeats. 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.100.511.522.5x 10-5 xucLt /= cLt 2/= 0=t Section 2.2 Solid Mechanics Part II Kelly 39 Figure 2.2.12: displacement s for free-free example 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.100.010.020.030.040.050.060.070.080.090.1 xucLt 4/= cLt /=0=t cLt 2/= cLt 4/3= 413 2D Elastostatic Problems in Cartesian Coordinates Two dimensional elastostatic problems are di scussed in this Chapter, that is, static problems of either plane stress or plane stra in. Cartesian coordina tes are used, which are appropriate for geometries which are have straight boundaries. The two-dimensional Navier equations are derived and the Airy st ress function technique is used to solved exactly some important problems. 42 Section 3.1 Solid Mechanics Part II Kelly 433.1 Plane Problems What follows is to be applicable to any tw o dimensional problem, so it is taken that 0==xz yzσσ , which is true of both plane stress and plane strain. 3.1.1 Governing Equations for Plane Problems To recall, the equations governing the elastostatic problem are the elastic stress-strain law (Part I, Eqns. 4.2.11-14), the strain-displacement relations (Eqns. 1.2.5) and the equations of equilibrium (1.1.10) [] [] ()yy xx zzxy xy xx yy yy yy xx xx EE E E σσνεσνενσσενσσε +−=+= −= −=1,1,1 Plane Stress (3.1.1a) [] [ ] ()yy xx zzxy xy yy xx yy yy xx xxE E E σσνσσνεσν νσνενσσννε +=+= −+−+= −−+=1, ) 1(1, ) 1(1 Plane Strain (3.1.1b) ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂=∂∂= xu yuyuxu y x xyy yyx xx 21εεε Strain-displacement relations (3.1.2) 000 =+∂∂=+∂∂+∂∂=+∂∂+∂∂ zzzyyy yxxxy xx bzby xby x σσσσσ Equations of Equilibrium (3.1.3) One way of solving these equations is to re-write the stresses in 3.1.3 in terms of strains by using 3.1.1, and then using 3.1.2 to re-w rite the resulting e quations in terms of displacements only. For example in th e case of plane strain one arrives at Section 3.1 Solid Mechanics Part II Kelly 44() ( )() () () ( )() () 0 21 1221 120 21 1221 12 22 2 2222 2 22 =+ ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ∂∂−+∂∂∂+∂∂−−+=+ ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ∂∂−+∂∂∂+∂∂−−+ yy x yxx y x bxu yxu yu Ebyu yxu xu E ν νννν ννν (3.1.4) These are the 2D Navier’s equati ons, analogous to the 1D version, Eqn. 2.1.2. This set of partial differential equations can be so lved subject to boundary conditions on the displacement. Obviously, in the absence of body forces, any linear displacement field satisfies 3.1.4, for example the field CyByEu CyAxEuo yo x ++−= −+=νσ σ, (3.1.5) with CBA ,, representing the possibl e rigid body motions; this corresponds to a simple tension o xxσσ= . Solving Eqns. 3.1.4 directly for more complex cases is not an easy task. An alternative solution strategy for the plane elastostatic problem is the Airy stress function method described in the next section. 3.1.2 Problems 1. Derive the plane stress Navier equations analogous to 3.1.4. 2. Show that the displacement field 2Ax ux= , ()ν+−= 1/ 4Axy uy , in the absence of body forces, satisfies the plane stress govern ing equations derived in Problem 1 (this solution does not satisfy 3.1.4). Determin e the corresponding stress field and verify that it satisfies the equilibrium equations. 3. Consider the thin plate shown below subjecte d to a uniform pressure p on the top and its own body weight. The plate is perfectly bonded to the base plate. (a) Does the stress distribution 0 ), ( ),( == −+−=xy xx yy hyg p yx σσ ρ σ satisfy the equations of equilibrium? (b) Does it satisfy the boundary conditions at the upper surface, and at the two free surfaces? (c) Suppose now that the plate was made out of elastic material. Show that, in that case, the stresses given above are actually not a correct solution to the problem. Section 3.1 Solid Mechanics Part II Kelly 45 xyp h Section 3.2 Solid Mechanics Part II Kelly 463.2 The Stress Function Method An effective way of dealing with many two dimensional problems is to introduce a new “unknown”, the Airy stress function φ, an idea brought to us by George Airy in 1862. The stresses are written in terms of this new function and a new differential equation is obtained, one which can be solved mo re easily than Navier’s equations. 3.2.1 The Airy Stress Function The stress components are written in the form yxxy xyyyxx ∂∂∂−=∂∂=∂∂= φσφσφσ 22222 (3.2.1) Note that, unlike stress and displacement, th e Airy stress function has no obvious physical meaning. The reason for writing the stresse s in the form 3.2.1 is that, provided the body forces are zero, the equilibrium equations are automatically satisfied, which can be seen by substituting Eqns. 3.2.1 into Eqns. 2.2.3 { ▲Problem 1}. On this point, the body forces, for example gravitational forces, are generally very small compared to the effect of typical surface forces in elas tic materials and may be safely ignored (see Problem 2 of §2.1). When body forces are significant, E qns. 3.2.1 can be amended and a solution obtained using the Airy stress function, but th is approach will not be followed here. A number of examples including non-zero body forces are examined later on, using a different solution method. 3.2.2 The Biharmonic Equation The Compatability Condition and Stress-Strain Law In the previous section, it was shown how one needs to solve the equilibrium equations, the stress-strain constitutive law, and the strain-displacement relations, resulting in the differential equation for displacements, Eqn. 3.1.4. An alternative appr oach is to ignore the displacements and attempt to solve for the stresses and strains only . In other words, the strain-displacement equations 3.1.2 are ignored. However, if one is solving for the strains but not the displacements, one must en sure that the compatib ility equation 1.3.1 is satisfied. Eqns. 3.2.1 already ensures that the equilibrium equations are satisfied, so combine now the two dimensional compatibility relation and the stress-strain relations 3.1.1 to get {▲Problem 2} Section 3.2 Solid Mechanics Part II Kelly 47 () 0 1 2 : strain plane0 2 : stress plane 44 2 24 4444 2 24 44 =−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂∂+∂∂ νφφφφφφ y yx xy yx x (3.2.2) Thus one has what is known as the biharmonic equation: 0 244 2 24 44 =∂∂+∂∂∂+∂∂ y yx xφφφ The biharmonic equation (3.2.3) The biharmonic equation is often writt en using the short-hand notation 04=∇φ . By using the Airy stress func tion representation, the problem of determining the stresses in an elastic body is reduced to that of finding a soluti on to the biharmonic partial differential equation 3.2.3 whose derivati ves satisfy certain boundary conditions. Note that the biharmonic equation is indepe ndent of elastic constants, Young’s modulus E and Poisson’s ratio ν. Thus for bodies in a state of plan e stress or plane strain, the stress field is independent of the material pr operties, provided the boundary conditions are expressed in terms of tractions (stress) 1; boundary conditions on displacement will bring the elastic constants in through the stress-strai n law. Further, the plane stress and plane strain stress fiel ds are identical. 3.2.3 Some Simple Solutions Clearly, any polynomial of degr ee 3 or less will satisfy the biharmonic equation. Here follow some elementary examples. (i) 2Ay=φ one has 0 ,222 == = ∂∂=xy yy xx Ayσσφσ , a state if uniaxial tension (ii) Bxy=φ here, Bxy yy xx −= == σ σσ ,0 , a state of pure shear (iii) Bxy Ay+=2φ here, B Axy yy xx −= = = σ σ σ ,0 ,2 , a superposition of (i) and (ii) 1 technically speaking, this is true only in simply connected bodies, i.e. ones without any “holes”, since problems involving bodies with holes have an implied displacement condition (see, for example, Barber (1992), §2.2). Section 3.2 Solid Mechanics Part II Kelly 483.2.4 Pure Bending of a Beam Consider the bending of a re ctangular beam by a moment 0M, as shown in Fig. 3.2.1. The elementary beam theory predicts that the stress xxσ varies linearly with y, Fig. 3.2.1, with the 0=y axis along the beam-centre, so a good pl ace to start would be to choose, or guess, as a stress function 3Cy=φ , where C is some constant to be determined. Then 0 ,0 ,6 == =xy yy xx Cy σσ σ , and the boundary conditions along the top and bottom of the beam are clearly satisfied. Figure 3.2.1: a beam in pure bending The moment and stress distribution are related through 3 2 0 4 6 Cb dyyC ydy Mb bb bxx = = = ∫∫+ −+ −σ (3.2.4) and so 3 04/b MC= and 3 02/ 3 byMxx=σ . The fact that this la st expression agrees with the elementary beam theory ( I My/−=σ with 3/ 23hb I= , where h is the depth “into the page”) shows that that the beam theory is exact in this simple loading case. Assume now plane strain conditions. In th at case, there is an other non-zero stress component, acting “perpendicular to the page”, 32/ 3) ( byMyy xx zz ν σσνσ =+= . Using Eqns. 3.1.1b, [] [] say , 23) 1() 1(1say , 23 1) 1(1 332 y y bM Ev Ey y bM E E yy xx yyyy xx xx βνσν νσνεαννσσννε =⎟ ⎠⎞⎜ ⎝⎛+−=−+−+==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−=−−+= (3.2.5) and the other four strains are zero. As in §1.2.4, once the strains have been found, the displacements can be found by integrating the strain-displacement relations. Thus 0M 0Mb Section 3.2 Solid Mechanics Part II Kelly 49() )( )(0)( )(21 21)()( 2 21 yf x xgxgyfxxu yuxg y uyyuyf xy uyxu y x xyyy yyxx xx ′−=+′→≡′+′+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=+=→=∂∂=+=→=∂∂= αα εββ εαα ε (3.2.6) Therefore ) (yf′ must be some constant, C− say, so A Cy yf +−=)( , and B x Cxxg +−=2 21)( α . Finally, B Cx y x uA Cyxy u yx +++−=+−= 2 21 2 21βαα (3.2.7) which are of the form 1.2.18. For the case wh en the mid-point of the beam is fixed, so has no translation, 0 )0,0( )0,0( ==y x u u , and if it has no rotation there, 0)0,0(=zω , then the three arbitrary constants are zero, and 2 21 2 21y x uxy u yx βαα +−== (3.2.8) 3.2.5 A Cantilevered Beam Consider now the cantilevered beam shown in Fig. 3.2.2. The beam is subjected to a uniform shear stress τσ=xy over its free end, Fig. 3.2.2a. The boundary conditions are 0),( ),( ,),0( ,0),0( =±=± = = bx bx y yxy yy xy xx σ στ σ σ (3.2.9) It is difficult, if not impossi ble, to obtain concise expressi ons for stress and strain for problems even as simple as this2. However, a concise solution can be obtained by relaxing one of the above conditions. To this end, consider the similar problem of Fig. 3.2.2b – this beam is subjected to a shear force F, the resultant of the shear stresses. The applied force of Fig. 3.2.2b is equiva lent to that in Fig. 3.2.2a if F dyyb bxy=∫+ −),0(σ (3.2.10) 2 an exact solution will usually require an infi nite series of terms for the stress and strain Section 3.2 Solid Mechanics Part II Kelly 50This is known as a weak boundary condition , since the stress is not specified in a point- wise sense along the boundary – only the result ant is. However, from Saint-Venant’s principle (Part I, §3.3.2), the stress field in both beams will be the same except for in a region close to the applied load. Figure 3.2.2: A cantilevered beam subjected to ; (a) a uniform distribution of shear stresses along its free end, (b) a shear force along its free end The elementary beam theory predicts a stress I FxyI Myxx / /=−=σ . Thus a good place to start is to choose the stress function 3xyαφ= , where α is a constant to be determined. The stresses are then 23 ,0 , 6 y xyxy yy xx ασ σασ −== = (3.2.11) However, it can be seen that 0 3 ),(2≠−=± b bxxy α σ . To offset this, one can superimpose a constant shear stress 23bα, in other words amend the stress function to xyb xy2 33ααφ−= ( 3 . 2 . 1 2 ) The boundary conditions are now satisfied and, from Eqn. 3.2.10, 34bF=α (3.2.13) and so ()2 2 3 343,0 ,23y bbFxybF xy yy xx −== = σ σ σ (3.2.14) 3.2.6 Problems 1. Verify that the relations 3.2.1 satis fy the equilibrium equations 2.2.3. 2. Derive Eqn. 3.2.2. 3. A large thin plate is subject ed to certain boundary conditi ons on its thin edges (with its large faces free of stress), leading to the stress function τ xy F xy )a() b(b Section 3.2 Solid Mechanics Part II Kelly 515 23Bx yAx−=φ (i) use the biharmonic equation to express A in terms of B (ii) calculate all stress components (iii) calculate all strain components (in terms of B, E, ν) (iv) derive an expression for the vol umetric strain, in terms of B, E, ν, x and y. (v) check that the compatibility equation is satisfied (vi) check that the equilibrium equations are satisfied 4. A very thick component has the same boundary conditions on any given cross-section, leading to the following stress function: 5 32 44 y yx yx −+=φ (i) is this a valid stress function, i.e. does it satisfy the biharmonic equation? (ii) calculate all stress components (with 4/1=ν ) (iii) calculate all strain components (iv) find the displacements (v) specify any three displacement components which will render the arbitrary constant displacements of (iv) zero 5. For the cantilevered beam discussed in §3.2. 5, evaluate the resultant shear force and moment on an arbitrary cross-section xx=. Are they as you expect? (You will find that the beam is in equilibrium, as expected, since the equilibrium equations have been satisfied.) 6. For the cantilevered beam discussed in §3.2.5, evaluate the strains and displacements, assuming plane stress conditions. Note: to evaluate the three arbitrary constant s of integration, one would be tempted to apply the obvious 0==y xu u all along the built-in end. However, since only weak boundary conditions were imposed, one cannot enforce these strong conditions (try it). Instead, apply the follow ing weaker conditions: (i) the displacement at the built-in end at 0=y is zero ( 0 ==y xu u ), (ii) the slope there, x uy∂∂/ , is zero. 7. Show that the stress function ()[]23 32 22 5 2 2 3 35 2 15 4 2020xh yh yxh y x Lyhp−+ −−− −=φ satisfies the boundary conditions for the simply supported beam subjected to a uniform pressure p shown below. Check the boundary conditions in the weak (Saint- Venant) sense on the shorter left and right hand sides (for both normal and shear stress). Since the normal stress xxσ is not zero at the ends , but only its resultant, check also that the moment is zero at each end. p xy Lh Lp Lp L Section 3.2 Solid Mechanics Part II Kelly 52Note that the elementary beam theory pred icts an approximate fl exural stress but an exact shear stress: ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− = − −=22 32 2 346,6yhxhpx Lyhp xy xx σ σ 8. Consider the dam shown in the figure below. Assume first a general cubic stress function 3 42 32 23 161 21 21 61yC xyC yxC xC + + +=φ Apply the boundary conditions to determine the constants and hence the stresses in the dam, in terms of ρ, the density of water. (Use the stress transformation equations for the sloped boundary and ignor e the weight of the dam.) [Just consider the effect of the water; to these must be added the stresses resulting from the weight of the dam itself, which are given by 0 ,tan1,0 =⎥⎦⎤ ⎢⎣⎡− = =xy s yy xx yx g σβρσσ where sρ is the density of the dam material.] x yρβ 534 2D Elastostatic Problems in Polar Coordinates Many problems are most conveniently cast in term s of polar coordinates. To this end, first the governing differential e quations discussed in Chapter 1 are expressed in terms of polar coordinates. Then a number of impor tant problems involving polar coordinates are solved. 54 Section 4.1 Solid Mechanics Part II Kelly 554.1 Cylindrical and Polar Coordinates 4.1.1 Geometrical Axisymmetry A large number of practical engineering pr oblems involve geometrical features which have a natural axis of symmetry , such as the solid cylinder, shown in Fig. 4.1.1. The axis of symmetry is an axis of revolution ; the feature which possesses axisymmetry (axial symmetry) can be generated by re volving a surface (or line) about this axis. Figure 4.1.1: a cylinder Some other axisymmetric geometries are illu strated Fig. 4.1.2; a frustum, a disk on a shaft and a sphere. Figure 4.1.2: axisymmetric geometries Some features are not only axisymmetric – th ey can be represented by a plane, which is similar to other planes right through the axis of symmetry. The hollow cylinder shown in Fig. 4.1.3 is an example of this plane axisymmetry . axis of symmetr ycreate cylinder by revolving a surface about the axis of symmetry Section 4.1 Solid Mechanics Part II Kelly 56 Figure 4.1.3: a plane axis ymmetric geometries Axially Non-Symmetric Geometries Axially non-symmetric geometries are ones which have a natural axis associated with them, but which are not completely symmetric. Some examples of th is type of feature, the curved beam and the half-space, are s hown in Fig. 4.1.4; the half-space extends to “infinity” in the axial directi on and in the radial direction “below” the surface – it can be thought of as a solid half-cylinder of infinite radius. One can also have plane axially non- symmetric features; in fact, bot h of these are examples of such features; a slice through the objects perpendicular to the axis of sy mmetry will be representative of the whole object. Figure 4.1.4: a plane axis ymmetric geometries 4.1.2 Cylindrical and Polar Coordinates The above features are best described using cylindrical coordinates , and the plane versions can be described using polar coordinates . These coordinates systems are described next. Stresses and Strains in Cylindrical Coordinates Using cylindrical coordinates, any po int on a feature will have specific ),,( zrθ coordinates, Fig. 4.1.5: axisymmetric plane representative of feature Section 4.1 Solid Mechanics Part II Kelly 57r – the radial direction (“out” from the axis) θ – the circumferential or tangential direction (“around” the axis – counterclockwise when viewed from the positive z side of the 0=z plane) z – the axial direction (“along” the axis) Figure 4.1.5: cylindrical coordinates The displacement of a materi al point can be described by the three components in the radial, tangential and axial direct ions. These are often denoted by θuvuur≡≡, and zuw≡ respectively; they are shown in Fig. 4.1.6. Note that the displacement v is positive in the positive θ direction, i.e. the direction of increasing θ. Figure 4.1.6: displacements in cylindrical coordinates The stresses acting on a small element of materi al in the cylindrical coordinate system are as shown in Fig. 4.1.7 (the normal stresses on the left, the shear stresses on the right). plane0=zrz θ rz uw v Section 4.1 Solid Mechanics Part II Kelly 58 Figure 4.1.7: stresses in cy lindrical coordinates The normal strains θθεε,rr and zzε are a measure of the elongation/shortening of material, per unit length, in th e radial, tangential and axial directions respectively; the shear strains z rθθεε, and zrε represent (half) the change in the right angles between line elements along the coordinate directions. The physical meaning of these strains is illustrated in Fig. 4.1.8. Figure 4.1.8: strains in cy lindrical coordinates Plane Problems and Polar Coordinates The stresses in any particular plane of an axisymmetric body can be described using the two-dimensional polar coordinates ()θ,r shown in Fig. 4.1.9. strain at point o rrε = unit elongation of oA θθε = unit elongation of oB zzε = unit elongation of oC θεr = ½ change in angle AoB∠ zθε = ½ change in angle BoC∠ zrε = ½ change in angle AoC∠ rθz ABC orrσrrσzzσ zzσθθσθθσ θσrzθσzrσ Section 4.1 Solid Mechanics Part II Kelly 59 Figure 4.1.9: polar coordinates There are three stress components acting in the plane 0=z : the radial stress rrσ, the circumferential (tangential) stress θθσ and the shear stress θσr, as shown in Fig. 4.1.10. Note the direction of the (positive) shear stress – it is conventional to take the z axis out of the page and so the θ direction is counterclockwise. The three stress components which do not act in this plane, but which act on this plane (z zzθσσ, and zrσ), may or may not be zero, depending on the particular problem (see later). Figure 4.1.10: stresses in polar coordinates rrσrrσθθσ θσr θθσθσrθ r Section 4.2 Solid Mechanics Part II Kelly 604.2 Differential Equations in Polar Coordinates Here, the two-dimensional Cartesian relati ons of Chapter 1 are re-cast in polar coordinates. 4.2.1 Equilibrium equations in Polar Coordinates One way of expressing the equations of equilib rium in polar coordinates is to apply a change of coordinates directly to the 2D Cartesian version, Eqns. 1.1.8, as outlined in the Appendix to this section, §4.2.6. Alternatively, the equations can be derived from first principles by considering an element of material subjected to stresses θθσσ,rr and θσr, as shown in Fig. 4.2.1. The dimensions of the element are rΔ in the radial direction, and θΔr (inner surface) and ()θΔΔ+r r (outer surface) in the tangential direction. Figure 4.2.1: an element of material Summing the forces in the radial direction leads to () () () 02cos2cos2sin2sin ≡ΔΔ−Δ⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+Δ+ΔΔ−Δ⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+Δ−Δ−ΔΔ+⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+=∑ r rr rr r rrrF rr rrrrr rr r θθ θθθθθ θθ σθθθσσθσθθθσσθθσθσσ (4.2.1) For a small element, 1 cos, sin ≈ ≈θθθ and so, dividing through by θΔΔr, () 02≡∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂Δ−−+Δ+∂∂ θσ θσθσσσθ θθ θθr rrrrr rr (4.2.2) A similar calculation can be carried out for forces in the tangential direction { ▲Problem 1}. In the limit as 0 ,→ΔΔθr , one then has the two-dimens ional equilibrium equations in polar coordinates: rrσrrrr rrΔ∂∂+σσ θθσθσr θΔrθθσσθθ θθΔ∂∂+rrr rΔ∂∂+θ θσσθθσσθ θΔ∂∂+r r Section 4.2 Solid Mechanics Part II Kelly 61() 02 101 1 =+∂∂+∂∂=−+∂∂+∂∂ r r rr r r r rrrr rr θ θθ θθθθ σ θσσσσθσσ Equilibrium Equations (4.2.3) 4.2.2 Strain Displacement Relations and Hooke’s Law The two-dimensional strain-displacement relati ons can be derived from first principles by considering line elements initially lying in the r and θ directions. Alternatively, as detailed in the Appendix to this section, §4. 2.6, they can be derived directly from the Cartesian version, Eqns. 1.2.5, ⎟ ⎠⎞⎜ ⎝⎛−∂∂+∂∂=+∂∂=∂∂= ru ru u rru u rru r rrr rr θθ θθ θθ θεθεε 1 211 2-D Strain-Displacement Expressions (4.2.4) The stress-strain relations in polar coordi nates are completely analogous to those in Cartesian coordinates – the ax es through a small material element are simply labelled with different letters. Thus Hooke’s law is now [] [] ()θθθ θ θθ θθ θθ σσνεσνενσσενσσε +−=+= −= −= rr zzr r rr rr rr EE E E1,1,1 Hooke’s Law (Plane Stress) (4.2.5a) [] []θ θ θθ θθ θθ σνεσν νσνενσσννεr r rr rr rrE E E+= −+−+= −−+=1, ) 1(1, ) 1(1 Hooke’s Law (Plane Strain) (4.2.5b) 4.2.3 Stress Function Relations In order to solve problems in polar coordi nates using the stress function method, Eqns. 3.2.1 relating the stress compone nts to the Airy stress function can be transformed using the relations in the Appendix to this section, §4.2.6: θφ θφ θφσφσ θφφσθ θθ∂∂∂−∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂−= ∂∂= ∂∂+∂∂=rr r rr r rrrr rr2 2 22 22 21 1 1, ,1 1 (4.2.6) It can be verified that these equations au tomatically satisfy the equilibrium equations 4.2.3 {▲Problem 2}. Section 4.2 Solid Mechanics Part II Kelly 62The biharmonic equation 3.2.3 becomes 01 12 22 2 22 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂φθ rrr r (4.2.7) 4.2.4 The Compatibility Relation The compatibility relation expressed in polar coordinates is (see the Appendix to this section, §4.2.6) 02 2 1 2 1 22 22 22 2=∂∂−∂∂+∂∂−∂∂∂−∂∂+∂∂ θε εε θε ε θεθ θθ θ θθ r rr r rr r rr rr rr r r (4.2.8) 4.2.5 Problems 1. Derive the equilibrium equation 4.2.3b 2. Verify that the stress functi on relations 4.2.6 satisfy the equilibrium equations 4.2.3. 3. Verify that the strains as given by 4.2.4 satisfy the compatibility relations 4.2.8. 4.2.6 Appendix to §4.2 From Cartesian Coordinates to Polar Coordinates To transform equations from Cartesian to pol ar coordinates, first note the relations )/ arctan( ,sin , cos 2 2xy y x rry rx =+== = θθ θ (4.2.9) Then the Cartesian partial derivatives become θθθθθθθθθθ ∂∂+∂∂=∂∂ ∂∂+∂∂ ∂∂=∂∂∂∂−∂∂=∂∂ ∂∂+∂∂ ∂∂=∂∂ r r y ryr yr r x rxr x cossinsincos (4.2.10) The second partial de rivatives are then Section 4.2 Solid Mechanics Part II Kelly 63⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂+⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂−⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂−⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂=∂∂ θθθθθ θθθ θθθθθ θθθ θθθθθθθθ rr r rrr rr r r r r r r rr r r r x 2 2 22 22 22 222 1 12sin1 1sin cossin sincossin sincos cos cossincossincos (4.2.11) Similarly, ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂− −=∂∂∂⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂=∂∂ θθθθθθθθθθθ θ rr r rrr r yxrr r rrr r y 2 2 22 2 22 22 2 22 22 22 2 22 1 12cos1 1cos sin1 12sin1 1cos sin (4.2.12) Equilibrium Equations The Cartesian stress components can be expres sed in terms of polar components using the stress transformation formulae, Part I, Eqns. 3.4.7. Using a negative rotation (see Fig. 4.2.2), one has () θσσσθθσθσθσθσσθσθσθσσ θ θθθ θθθ θθ 2cos cos sin2sin cos sin2sin sin cos 2 22 2 r rr xyr rr yyr rr xx +− =+ + =− + = (4.2.13) Applying these and 4.2.10 to the 2D Cartesian equilibrium equations 3.1.3a-b lead to () () 02 1cos1 1sin02 1sin1 1cos =⎥⎦⎤ ⎢⎣⎡+∂∂+∂∂+⎥⎦⎤ ⎢⎣⎡−+∂∂+∂∂=⎥⎦⎤ ⎢⎣⎡+∂∂+∂∂−⎥⎦⎤ ⎢⎣⎡−+∂∂+∂∂ r r r r r rr r r r r r r r rrr rrr r rrr rr θ θθ θ θθθθ θθ θ θθθ σ θσσθ σσθσσθσ θσσθ σσθσσθ (4.2.14) which then give Eqns. 4.2.3. Figure 4.2.2: rotation of axes xry θ θ Section 4.2 Solid Mechanics Part II Kelly 64 The Strain-Displacement Relations Noting that θθθθ θθ cos sinsin cos u u uu u u r yr x +=−= , (4.2.15) the strains in polar coordinates can be obtained directly from Eqns. 1.2.5: () ⎟ ⎠⎞⎜ ⎝⎛−∂∂+∂∂−⎟ ⎠⎞⎜ ⎝⎛+∂∂+∂∂=−⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂=∂∂= ru ru u r ru u r ruu ur rxu r r rrx xx θθ θθ θθθθ θθθθθθε 1 212sin1sin cossin cossincos 2 2 (4.2.16) One obtains similar expressions for the strains yyε and xyε. Substituting the results into the strain transformation equa tions Part I, Eqns. 3.8.1, () θεεεθθεθεθεθεεθεθεθεε θθθ 2cos cos sin2sin cos sin2sin sin cos 2 22 2 xy xx yy rxy yy xxxy yy xx rr +− =− + =+ + = (4.2.17) then leads to the equations given above, Eqns. 4.2.4. The Stress – Stress Function Relations The stresses in polar coordina tes are related to the stresses in Cartesian coordinates through the stress transformation equations (t his time a positive rotation; compare with Eqns. 4.2.13 and Fig. 4.2.2) () θσσσθθσθσθσθσσθσθσθσσ θθθ 2cos cos sin2sin cos sin2sin sin cos 2 22 2 xy xx yy rxy yy xxxy yy xx rr +− =− + =+ + = (4.2.18) Using the Cartesian stress – stress function relations 3.2.1, one has θφθφθφσ 2sin sin cos2 2 22 2 22 yx x yrr∂∂∂−∂∂+∂∂= (4.2.19) and similarly for θθθσσr, . Using 4.2.11-12 then leads to 4.2.6. Section 4.2 Solid Mechanics Part II Kelly 65 The Compatibility Relation Beginning with the Cartesian relation 1.3.1, each term can be transformed using 4.2.11-12 and the strain transformation relations, for example () θεθεθεθθθθθ θε θ θθ 2sin sin cos1 12sin1 1sin cos 2 22 2 22 22 22 2 22 r rrxx rr r rrr r x − +×⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂=∂∂ (4.2.20) After some lengthy calculati ons, one arrives at 4.2.8. Section 4.3 Solid Mechanics Part II Kelly 664.3 Plane Axisymmetric Problems In this section are considered plane axisymmetric problems . These are problems in which both the geometry and loading are axisymmetric. 4.3.1 Plane Axisymmetric Problems Some three dimensional (not necessarily plane) examples of axisymmetric problems would be the thick-walled (hollow) cylinder und er internal pressure, a disk rotating about its axis 1, and the two examples shown in Fig. 4.3.1; the first is a complex component loaded in a complex way, but exhibits axisym metry in both geometry and loading; the second is a sphere loaded by con centrated forces along a diameter. Figure 4.3.1: axisymmetric problems A two-dimensional (plane) example would be one plane of the thick-walled cylinder under internal pressure, illustrated in Fig. 4.3.22. Figure 4.3.2: a cross section of an internally pressurised cylinder It should be noted that many problems i nvolve axisymmetric geometries but non- axisymmetric loadings, and vice versa . These problems are not axisymmetric. An example is shown in Fig. 4.3.3 (the problem involves a plane axisymmetric geometry). 1 the rotation induces a stress in the disk 2 the rest of the cylinder is coming out of, and into, the page Section 4.3 Solid Mechanics Part II Kelly 67 Figure 4.3.3: An axially symme tric geometry but with a non-axisymmetric loading The important characteristic of these axisymmetr ic problems is that all quantities, be they stress, displacement, strain, or anythi ng else associated with the problem, must be independent of the circumferential variable θ. As a consequence, any term in the differential equations of §4.2 involving the derivatives 2 2/,/ θ θ ∂∂∂∂ , etc. can be immediately set to zero. 4.3.2 Governing Equations for Plane Axisymmetric Problems The two-dimensional strain-displacement relations are given by Eqns. 4.2.4 and these simplify in the axisymmetric case to ⎟ ⎠⎞⎜ ⎝⎛−∂∂==∂∂= ru rururu rrr rr θ θ θθθ εεε 21 (4.3.1) Here, it will be assumed that the displacement 0 =θu . Cases where 0 ≠θu but where the stresses and strains ar e still independent of θ are termed quasi-axisymmetric problems ; these will be examined in a later section. Then 4.3.1 reduces to 0 , , = =∂∂=θ θθ ε ε εrr r rrru ru (4.3.2) It follows from Hooke’s law that 0 =θσr . The non-zero stresses are illustrated in Fig. 4.3.4. axisymmetric plane representative of feature Section 4.3 Solid Mechanics Part II Kelly 68 Figure 4.3.4: stress components in plane axisymmetric problems 4.3.3 Plane Stress and Plane Strain Two cases arise with plane axisymmetric problems: in the plane stress problem, the feature is very thin and unloaded on its larg er free-surfaces, for example a thin disk under external pressure, as shown in Fig. 4.3. 5. Only two stress components remain, and Hooke’s law 4.2.5a reads [] []rrrr rr EE νσ σ ενσ σ ε θθ θθθθ − =− = 11 or [] []rrrr rr EE νε ενσνε ενσ θθ θθθθ +−=+−= 22 11 (4.3.3) with () 0 , = = +−=θ θθ ε ε σ σνεz zr rr zzE and 0=zzσ . Figure 4.3.5: plane stress axisymmetric problem In the plane strain case, the strains θεεz zz, and zrε are zero. This will occur, for example, in a hollow cylinder under internal pressure, with the ends fixed between immovable platens, Fig. 4.3.6. Figure 4.3.6: plane strain axisymmetric problem rrσ rrσθθσ θθσ Section 4.3 Solid Mechanics Part II Kelly 69Hooke’s law 4.2.5b reads [] []rrrr rr EE νσ σννενσ σννε θθ θθθθ − −+=− −+= ) 1(1) 1(1 or () ( )() [] () ( )() []θθ θθθθ εν νεν νσεν νεν νσ −+− +=−+− += 121 1121 1 rrrr rr EE (4.3.4) with ()θθσ σν σ + =rr zz . Shown in Fig. 4.3.7 are the stresses acti ng in the axisymmetr ic plane body (with zzσ zero in the plane stress case). Figure 4.3.7: stress components in plane axisymmetric problems 4.3.4 Solution of Plane Axisymmetric Problems The equations governing the plane axisym metric problem are the equations of equilibrium 4.2.3 which reduce to the single equation () 01= − +∂∂ θθσ σσ rrrr r r, (4.3.5) the strain-displacement relations 4.3.2 and the stress-strain law 4.3.3-4. Taking the plane stress case, substituting 4. 3.2 into the second of 4.3.3 and then substituting the result into 4.3.5 leads to (with a similar result for plane strain) 01 1 2 22 = − + ur drdu r drud (4.3.6) This is Navier’s equation for plane axisymmetry. It is an “Euler-type” ordinary differential equation which can be solved exac tly to get (see Appendi x to this section, §4.3.8) rCrCu1 2 1+ = (4.3.7) rrσrrσzzσ zzσθθσθθσ Section 4.3 Solid Mechanics Part II Kelly 70With the displacement known, the stresses a nd strains can be evaluated, and the full solution is 2 2 1 2 2 12 2 1 2 2 12 1 1 1 1,1 1 11,11 rCECE rCECErC C rC CrCrCu rrrr ν νσν νσε ε θθθθ ++−=+−−=+ = − =+ = (4.3.8) For problems involving stress boundary conditions, it is best to have simpler expressions for the stress so, introducing new constants ( )ν+ −= 1/2EC A and ()ν− = 12/1ECC , the solution can be re-written as () () () () () ()rEC r EAuEC EC r EA EC r EACrA CrA zz rrrr ν ννεν νεν νεσ σ θθθθ −++−=−=−++−=−+++=+ −= + += 121 14,121 1,121 121,21 2 22 2 (4.3.9) Plane stress axisymmetric solution Similarly, the plane strain solution turns out to be again 4.3.8a-b only the stresses are now {▲Problem 1} () ( )()() ( )()⎥⎦⎤ ⎢⎣⎡+ −+− +=⎥⎦⎤ ⎢⎣⎡+ −−− +=1 2 2 1 2 212121 1,12121 1CrCECrCE rr νν νσ νν νσθθ (4.3.10) Then, with ()ν+ −= 1/2EC A and ( )( )ν ν 21 12/1 − + =ECC , the solution can be written as () () ()⎥⎦⎤ ⎢⎣⎡− + −+=⎥⎦⎤ ⎢⎣⎡− + −+=⎥⎦⎤ ⎢⎣⎡− + ++== + −= + += CrrAEuCrAECrAEC CrA CrA rrzz rr ννννε ννεν σ σ σ θθθθ 2121 12121 1, 2121 14 ,21,21 2 22 2 (4.3.11) Plane strain axisymmetric solution The solutions 4.3.9, 4.3.11 involve two consta nts. When there is a solid body with one boundary, A must be zero in order to ensure finite-valued stresses and strains; C can be determined from the boundary condition. When there are two boundaries, both A and C are determined from the boundary conditions. Section 4.3 Solid Mechanics Part II Kelly 714.3.5 Example: Expansion of a thick circular cylinder under internal pressure Consider the problem of Fig. 4.3.8. The two unknown constants A and C are obtained from the boundary conditions 0)()( =−= bp a rrrr σσ (4.3.12) which lead to 0 2 )( , 2 )(2 2= + = −= + = CbAb p CaAarr rr σ σ (4.3.13) so that ()θθ θθ σ σν σ σ σ + =−++=−−−=rr zz rrabrbpabrbp ,1 /1 /,1 /1 / 2 22 2 2 22 2 (4.3.14) Cylinder under Internal Pressure Figure 4.3.8: an internally pressurised cylinder The stresses through the thickness of the cylin der walls are shown in Fig. 4.3.9a. The maximum principal stress is the θθσ stress and this attains a ma ximum at the inner face. For this reason, internally pressurized vessels often fail there first, with microcracks perpendicular to the inner edge been driven by the tangential stress, as illustrated in Fig. 4.3.9b. Note that by setting tab += and taking the wall thickness to be very small, a tt <<2,, and letting ra=, the solution 4.3.14 reduces to: trptrp pzz rr ν σ σ σθθ = += −= , , (4.3.15) which is equivalent to the thin-walled pressure-vessel solution, Part I, §4.5.2 (if 2/1=ν , i.e. incompressible). r • ba Section 4.3 Solid Mechanics Part II Kelly 72 Figure 4.3.9: (a) stresses in the thick-walled cylinder, (b) microcracks driven by tangential stress Generalised Plane Strain Solution Another useful solution is that for a cylinder wh ich is free to expand in the axial direction. In this case, zzε is not forced to zero as in plane st rain, but allowed to be a constant along the length of the cylinder. The zzσ stress is zero, as in plane stress. This situation is called generalized plane strain . Returning to the full three-dimensional stress -strain equations (Part I, Eqns. 4.2.9), set zz zzε ε= , a constant, and 0= =yz xzε ε . Re-labelling zyx,, with zr,,θ, and again with 0=θσr , one has () [] () [] () [] 0 ) 1()21)( 1() 1()21)( 1() 1()21)( 1( = + + −− +=+ + −− +=+ + −− += θθθθ θθθθ ε εν ενν νσε εν ενν νσε εν ενν νσ rr zz zzzz rrzz rr rr EEE (4.3.16) Substituting the strain-displacement relations 4.3.2 into 4.3.16a-b leads to () ()zzr rzzr r rr ru ruru ru εβαν α β σεβαν β α σ θθ + + +∂∂=+ + +∂∂= (4.3.17) where rrσr br= ar=zzσθθσ p−1 /2 2 2−abp1 /1 / 2 22 2 −+ ababp )a() b(θθσ Section 4.3 Solid Mechanics Part II Kelly 73)21)( 1(,)21)( 1() 1( ν ννβν ννα− +=− +−=E E (4.3.18) Substituting 4.3.16 into the axisymmetric equi librium equation 4.3.5 again leads to the differential equation 4.3.6. The solution for displacement and strain is thus again 4.3.8a- b. The constant axial strain is obtained from 4.3.16c and is ()ν ν ε − −= 1/ 21Czz . The axial displacement is then zz zz u ε= (to within a constant). The stresses are again given by the plane strain relations 4.3. 10 only with the additional term ( )( )ν ν ν 21 1/ 22 12− − − CE . As for plane strain, let ( )ν+ −= 1/2EC A and () ( )ν ν 21 12/1 − + =ECC , and the full solution is () () ()⎥⎦⎤ ⎢⎣⎡− + −+=⎥⎦⎤ ⎢⎣⎡− + −+=⎥⎦⎤ ⎢⎣⎡− + ++=−− + −=−− + += CrrAEuC rAEC rAEC C rA C C rA rrrr ννννε ννεννσννσ θθθθ 2121 12121 1, 2121 11421,1421 2 22 22 2 (4.3.19) Generalised plane strain axisymmetric solution A Transversely isotropic Cylinder Consider now a transversely isotropic cylinder. The strain-displacement relations 4.3.2 and the equilibrium equation 4.3.5 are applicab le to any type of ma terial. The stress- strain law can be expressed as (see Part I, Eqn. 6.2.14) zz rr zzzz rrzz rr rr C C CC C CC C C ε ε ε σε ε ε σε ε ε σ θθθθ θθθθ 33 13 1313 11 1213 12 11 + + =+ + =+ + = (4.3.20) Here, take zz zzε ε= , a constant. Then, using the strain-displacement relations and the equilibrium equation, one again arrives at the differential equation 4.3.6 so the solution for displacement and strain is again 4.3.8a-b. With ( )11 12 2/ C C CA − = and ()12 11 12/ C C CC + = , the stresses can be expressed as zz zzzzzz rr CC CCCCCrACCrA ε σε σε σ θθ 33 12 111313 213 2 42121 ++=+ + −=+ + += (4.3.21) The plane strain solution then follows from 0=zzε and the generalized plane strain solution from 0=zzσ . These solutions reduce to 4.3.11, 4.3.19 in the isotropic case. Section 4.3 Solid Mechanics Part II Kelly 74 4.3.6 Stress Function Solution An alternative solution procedure for axis ymmetric problems is the stress function approach. To this end, first specialise equations 4.2.6 to the axisymmetric case: 0 , ,1 22 =∂∂=∂∂=θ θθ σφσφσr rrr rr (4.3.22) One can check that these equations satisfy the axisymmetric equilibrium equation 4.3.4. The biharmonic equation in polar coordinates is given by Eqn. 4.2.7. Specialising this to the axisymmetric case, that is, setting 0 /=∂∂θ , leads to 01 1 1 22 222 22 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ rr r rr r rr rφ φφ (4.3.23) or 01 1 2 3 22 2 33 44 = + − +drd r drd r drd r drd φ φ φ φ (4.3.24) Alternatively, one could have started with the compatibility relation 4.2.8, specialised that to the axisymmetric case: 02 1 22 =∂∂+∂∂−∂∂ rr rr rrr θθ θθ ε ε ε (4.3.25) and then combine with Hooke’s law 4.3.3 or 4.3.4, and 4.3.22, to again get 4.3.24. Eqn. 4.3.24 is an Euler-type ODE and has solu tion (see Appendix to this section, §4.3.8) D Crr BrrA + + + = 2 2ln ln φ (4.3.26) The stresses then follow from 4.3.22: () () C r BrAC r BrA rr 2 ln232 ln21 22 + + + −=+ + + += θθσσ (4.3.27) The strains are obtained from the stress-strain relations. For plane strain, one has, from 4.3.4, Section 4.3 Solid Mechanics Part II Kelly 75() [] () () [] () ⎭⎬⎫ ⎩⎨⎧− + − + − + −+=⎭⎬⎫ ⎩⎨⎧− + − + − + ++= ν ν ννεν ν ννε θθ 212 ln212 431212 ln212 411 22 C r BrA EC r BrA Err (4.3.28) Comparing these with the strain-displacement relations 4.3.2, and integrating rrε, one has () [] () () [] () ⎭⎬⎫ ⎩⎨⎧− + − ++ +−+= =+ ⎭⎬⎫ ⎩⎨⎧− + − +− +−+= =∫ r C r BrrA Er uF r C r BrrA Edr u rrr r ν ννεν ννε θθ 212 ln21211212 ln21211 (4.3.27) To ensure that one has a unique displacement ru, one must have 0=B and the constant of integration 0=F , and so one again has the solution 4.3.113. 4.3.7 Problems 1. Derive the solution equations 4.3.11 for axisymmetric plane strain. 2. A cylindrical rock specimen is subjected to a pressure p over its cylindrical face and is constrained in the axial direc tion. What are the stresses, including the axial stress, in the specimen? What are the displacements? 3. A long hollow tube is subject ed to internal pressure ip and external pressures op and constrained in the axial direction. What is the stress state in the walls of the tube? What if p p po i = = ? 4. A long mine tunnel of radius a is cut in deep rock. Before the mine is constructed the rock is under a uniform pressure p. Considering the rock to be an infinite, homogeneous elastic medium with elastic constants E and ν, determine the radial displacement at the surface of the tunnel due to the excav ation. What radial stress P arr −=)(σ should be applied to the wall of the tunnel to prevent any such displacement? 5. A long hollow elastic tube is fitted to an i nner rigid (immovable) shaft. The tube is perfectly bonded to the shaft. An external pressure p is applied to the tube. What are the stresses and stra ins in the tube? 3 the biharmonic equation was derived using the expression for compatibility of strains (4.3.23 being the axisymmetric version). In simply connected domains, i.e. bodies without holes, compatibility is assured (and indeed A and B must be zero in 4.3.26 to ensure finite strains). In multiply connected domains, however, for example the hollow cylinder, the compatibility conditio n is necessary but not sufficient to ensure compatible strains (see, fo r example, Shames and Cozzarelli (1997)), and th is is why compatibility of strains must be explicitly enforced as in 4.3.25 Section 4.3 Solid Mechanics Part II Kelly 766. Repeat Problem 3 for the case when the tube is free to expand in the axial direction. How much does the tube expand in the axial direction (take 0=zu at 0=z )? 4.3.8 Appendix Solution to Eqn. 4.3.6 The differential equation 4.3.6 can be solved by a change of variable ter=, so that drdt rtr ert= = =1, log, (4.3.28) and, using the chain rule, dtdu r dtud r drtd dtdu drdt drdt dtud drdt dtdu drd druddtdu r drdt dtdu drdu 2 22 2 22 22 221 11 − = + =⎟ ⎠⎞⎜ ⎝⎛== = (4.3.29) The differential equation becomes 022 =−udtud (4.3.30) which is an ordinary differential equati on with constant coefficients. With teuλ= , one has the characteristic equation 0 12=−λ and hence the solution rCrCeC eCut t 1 2 12 1 + =+ =− + (4.3.31) Solution to Eqn. 4.3.24 The solution procedure for 4.3.24 is similar to that given above for 4.3.6. Using the substitution ter= leads to the differential equation with constant coefficients 0 4 422 33 44 = + −dtd dtd dtd φ φ φ (4.3.32) which, with teλφ= , has the characteristic equation ( )0 22 2= −λλ . This gives the repeated roots solution D Ce Bte Att t+ + + =2 2φ (4.3.33) and hence 4.3.24. Section 4.4 Solid Mechanics Part II Kelly 774.4 Rotating Discs 4.4.1 The Rotating Disc Consider a thin disc rotating w ith constant angular velocity ω, Fig. 4.4.1. Material particles are subjected to a centripetal acceleration 2ωr ar−= . The subscript r indicates an acceleration in the radial di rection and the minus sign indi cates that the particles are accelerating towards the centre of the disc. Figure 4.4.1: the rotating disc The accelerations lead to an in ertial force (per unit volume) 2ωρr Fa−= which in turn leads to stresses in the disc. Th e inertial force is an axisymmetric “loading” and so this is an axisymmetric problem. The axisymmetric equation of equilibrium is given by 4.3.5. Adding in the acceleration term gives the corresponding equation of motion: ()2 1ωρσσσ θθ rr rrrrr−=−+∂∂, (4.4.1) This equation can be expressed as () 01=+−+∂∂ r rrrrbr rθθσσσ, (4.4.2) where 2ωρr br= . Thus the dynamic rotating disc pr oblem has been converted into an equivalent static problem of a disc subject ed to a known body force. Note that, in a general dynamic problem, and unlike here, one does not know what the accelerations are – they have to be found as pa rt of the solution procedure. Using the strain-displacement relations 4.3.2 and the plane stress Hooke’s law 4.3.3 then leads to the differential equation 22 2 221 1 1ωρνrEur drdu r drud −−=−+ (4.4.3) This is Eqn. 4.3.6 with a non-homogeneous term . The solution is derived in the Appendix to this section, §4.4.3: ω•2ωr Section 4.4 Solid Mechanics Part II Kelly 782 32 2 11 811ωρνrE rCrCu−−+= (4.4.4) As in §4.3.4, let ()ν+−= 1/2EC A and ()ν− = 12/1ECC , and the full general solution is, using 4.3.2 and 4.3.3, { ▲Problem 1} () () () ()() () ()() () ()()⎥⎦⎤ ⎢⎣⎡−−−++−=⎥⎦⎤ ⎢⎣⎡−−−++−=⎥⎦⎤ ⎢⎣⎡−−−+++=+−+−=+−++= 32 222 2 222 2 222 222 2 18112 1118112 1118312 1131812138121 r CrrA Eur C rA Er C rA Er CrAr C rA rrrr ρων ν νρων ν ν ερων ν ν ερων σρων σ θθθθ (4.4.5) which reduce to 4.3.9 when 0=ω . A Solid Disc For a solid disc, A in 4.4.5 must be zero to ensure finite stresses and strains at 0=r . C is then obtained from the boundary condition 0)(=brrσ , where b is the disc radius: ()223161,0 b C A ρων+= = (4.4.6) The stresses and displacements are [] ⎥⎦⎤ ⎢⎣⎡ ++−−+=⎥⎦⎤ ⎢⎣⎡ ++−+=−+= 2 2 22 2 22 2 2 31 1 83)(331 83)(83)( r brErur b rr b rrr νν νρωνννρωνσρωνσ θθ (4.4.7) Note that the displacement is zero at the disc centre, as it must be, but the strains (and hence stresses) do not have to be, and are not, zero there. Dimensionless stress and displacement are plotted in Fig. 4.4.2 for the case of 3.0=ν . The maximum stress occurs at 0=r , where 22 83)0( )0( brr ρωνσσθθ+== (4.4.8) Section 4.4 Solid Mechanics Part II Kelly 79The disc expands by an amount 32 41)( bEbu ρων−= (4.4.9) Figure 4.4.2: stresses and displacements in the solid rotating disc A Hollow Disc The boundary conditions for the hollow disc are 0)( ,0)( = = b arr rr σ σ (4.4.10) where a and b are the inner and outer radii respectiv ely. It follows from 4.4.5 that () ()()2 2 2 2223161, 381b a C ba A + += +−= ρων ρων (4.4.11) and the stresses and displacement are ⎥⎦⎤ ⎢⎣⎡ −++++−+−+=⎥⎦⎤ ⎢⎣⎡+++−++=⎥⎦⎤ ⎢⎣⎡−−++= 222 2 2 2 2222 2 2 2 2222 2 2 2 2 11 31 1 83)(331 83)(83)( rbar b arErurbar b a rrbar b a rrr νν νν νρωνννρωνσρωνσ θθ (4.4.12) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.10.20.30.40.50.60.70.80.91 br/()σρων2238 b+ () ()ubE 321 38 ρωνν−+ uθθσ rrσ Section 4.4 Solid Mechanics Part II Kelly 80 which reduce to 4.4.7 when 0=a . Dimensionless stress and displacement are plotted in Fig. 4.4.3 for the case of 3.0=ν and 2.0 /=ba . The maximum stress occurs at the inner surface, where ()⎥⎦⎤ ⎢⎣⎡ +−++=2 22/31143)0( ba bννρωνσθθ (4.4.13) which is approximately twice the solid-disc maximum stress. Figure 4.4.3: stresses and displacements in the hollow rotating disc 4.4.2 Problems 1. Derive the full solution equations 4.4.5 for the thin rotating disc, from the displacement solution 4.4.4. 4.4.3 Appendix: Solution to Eqn. 4.4.3 As in §4.3.8, transform Eqn. 4.4.3 using ter= into 2 32 221ωρνteEudtud −−=− (4.4.14) 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 2.5 br/()σρων2238 b+ () ()ubE 321 38 ρωνν−+ u rrσθθσ Section 4.4 Solid Mechanics Part II Kelly 81The homogeneous solution is given by 4.3.31. Assume a particular solution of the form t pAe u3= which, from 4.4.14, gives t p eEu3221 81ρων−−= (4.4.15) Adding together the homogeneous and particular solutions and transforming back to r’s then gives 4.4.4. Section 6.1 Solid Mechanics Part II Kelly 1206.1 Plate Theory 6.1.1 Plates A plate is a flat structural element for which the thickness is small compared with the surface dimensions. The thickness is usually constant but may be variable and is measured normal to the middle surface of the plate, Fig. 6.1.1 Fig. 6.1.1: A plate 6.1.2 Plate Theory Plates subjected only to in-p lane loading can be solved using two-dimensional plane stress theory 1. On the other hand, plate theory is concerned mainly with lateral loading . One of the differences between plane stress and pl ate theory is that in the plate theory the stress components are allowed to vary through the thickness of the plate, so that there can be bending moments, Fig. 6.1.2. Fig. 6.1.2: Stress distribution through the th ickness of a plate and resultant bending moment Plate Theory and Beam Theory Plate theory is an approximate theory; assumptions are made and the general three dimensional equations of elasticity are reduced. It is very like the beam theory – only 1 although if the in-plane loads are compressive an d sufficiently large, they can buckle (see §6.7) middle surface of plate lateral load M Section 6.1 Solid Mechanics Part II Kelly 121with an extra dimension. It turns out to be an accurate theory provided the plate is relatively thin (as in the beam theory) but also that the deflections are small relative to the thickness . This last point will be discussed further in §6.10. Things are more complicated for plates than for the beams. For one, the plate not only bends, but torsion may occur (it can twist), as shown in Fig. 6.1.3 Fig. 6.1.3: torsion of a plate Assumptions of Plate Theory Let the plate mid-surface lie in the yx− plane and the z – axis be along the thickness direction, forming a right handed set, Fig. 6.1.4. Fig. 6.1.4: Cartesian axes The stress components acting on a typical element of the plate are shown in Fig. 6.1.5. Fig. 6.1.5: stresses acting on a material element y xxxσyyσ xyσzxσyzσz zzσxyz Section 6.1 Solid Mechanics Part II Kelly 122 The following assumptions are made: (i) The mid-plane is a “neutral plane” The middle plane of the plate remains free of in -plane stress/strain. Bending of the plate will cause material above and below this mid-plane to deform in-plane. The mid-plane plays the same role in plate theory as the neutral axis does in the beam theory. (ii) Line elements remain normal to the mid-plane Line elements lying perpendicular to the mi ddle surface of the plate remain perpendicular to the middle surface during deformation, Fig. 6. 1.6; this is similar the “plane sections remain plane” assumption of the beam theory. Fig. 6.1.6: deformed line elements remain perpendicular to the mid-plane (iii) Vertical strain is ignored Line elements lying perpendicular to th e mid-surface do not change length during deformation, so that 0=zzε throughout the plate. Again, this is similar to an assumption of the beam theory. These three assumptions are the basis of the Classical Plate Theory or the Kirchhoff Plate Theory . The second assumption can be relaxed to develop a more exact theory (see §6.10). 6.1.3 Notation and Stress Resultants The stress resultants are obtained by integrating the stresses through the thickness of the plate. In general there will be moments M: 2 bending moments and 1 twisting moment out-of-plane forces V: 2 shearing forces in-plane forces N: 2 normal forces and 1 shear force undeformed line element remains perpendicular to mid-surface Section 6.1 Solid Mechanics Part II Kelly 123They are defined as follows: In-plane normal forces a nd bending moments, Fig. 6.1.7: ∫ ∫∫ ∫ + −+ −+ −+ − −= −== = 2/ 2/2/ 2/2/ 2/2/ 2/ ,, h hyy yh hxx xh hyy yh hxx x dzz M dzz Mdz N dz N σ σσ σ (6.1.1) Fig. 6.1.7: in-plane normal forces and bending moments In-plane shear force and twisting moment, Fig. 6.1.8: ∫ ∫+ −+ −= =2/ 2/2/ 2/,h hxy xyh hxy xy dzz M dz N σ σ (6.1.2) Fig. 6.1.8: in-plane shear force and twisting moment Out-of-plane shearing forces, Fig. 6.1.9: ∫ ∫+ −+ −−= −=2/ 2/2/ 2/,h hyz yh hzx x dz V dz V σ σ (6.1.3) y xxNyN xMyM xxσyyσ y xxyNxyM xyσ xyσ xyNxyM Section 6.1 Solid Mechanics Part II Kelly 124 Fig. 6.1.9: out of plane shearing forces Note that the above “forces” and “mom ents” are actually forces and moments per unit length . This allows one to have moments varyin g across any section – unlike in the beam theory, where the moments are for the complete beam cross-section. If one considers an element with dimensions xΔ and yΔ, the actual moments acting on the element are y Mx Mx My Mxy xy y x ΔΔΔΔ , , , (6.1.4) and the forces acting on the element are y Nx NxNyNxVyVxy xy y x y x ΔΔΔΔΔΔ , , , , , (6.1.5) The in-plane forces, which are analogous to the axial forces of the beam theory, do not play a role in most of what follows. They ar e useful in the analysis of buckling of plates and it is necessary to consider them in more exact theories of plate bending (see later). y xyVzxσyzσ yzσ zxσ xV Section 6.2 Solid Mechanics Part II Kelly 1256.2 The Moment-Curvature Equations 6.2.1 From Beam Theory to Plate Theory In the beam theory, based on the assumptions of plane sections remaining plane and that one can neglect the transverse strain, the strain varies linea rly through the thickness. In the notation of the beam, with y positive up, Ry xx /−=ε , where R is the radius of curvature , R positive when the beam bends “up” (see Part I, Eqn. 4.6.16). In terms of the curvature R xv /1 /2 2=∂∂ , where v is the deflection (see Part I, Eqn. 4.6.35), one has 22 xvyxx∂∂−=ε (6.2.1) The beam theory assumptions are essentially the same for the plate, leading to strains which are proportional to distance from the neutral (mid-plane) surface, z, and expressions similar to 6.2.1. This leads again to linearly varying stresses xxσ and yyσ (zzσ is also taken to be zero, as in the beam theory). 6.2.2 Curvature and Twist The plate is initially undeformed and fl at with the mid-surface lying in the yx− plane. When deformed, the mid-su rface occupies the surface ) ,(yxww= and w is the elevation above the yx− plane, Fig. 6.2.1. Fig. 6.2.1: Deformed Plate The slopes of the plate along the x and y directions are xw∂∂/ and yw∂∂/. Curvature Recall from Part I, §4.6.10, that the curvature in the x direction, xκ, is the rate of change of the slope angle ψ with respect to arc length s, Fig. 6.2.2, ds dx /ψκ= . One finds that xy• •initial position w Section 6.2 Solid Mechanics Part II Kelly 126 ()[]2/322 2 / 1/ xx x ∂∂+∂∂= ωωκ (6.2.2) Also, the radius of curvature xR, Fig. 6.2.2, is the reci procal of the curvature, x xRκ/1= . Fig. 6.2.2: Angle and arc-length used in the definition of curvature As with the beam, when the slope is small , one can take xw∂∂=≈ / tanψψ and x ds d ∂∂≈ / /ψψ and Eqn. 6.2.2 reduces to (and si milarly for the curvature in the y direction) 22 221,1 yw R xw Ryy xx∂∂==∂∂== κ κ (6.2.3) This important assumption of small curvature, or equivalently of assuming that the slope 1 /<<∂∂ xω , means that the theory to be developed will be valid when the deflections are small compared to the overall dimensions of the plate. The curvatures 6.2.3 can be interpreted as in Fig. 6.2.3, as the unit increase in slope along the x and y directions. Figure 6.2.3: Physical mean ing of the curvatures yw x yAB C D ABy∂∂ωyy yΔ∂∂+∂∂ 22ωω Ax∂∂ωxx xΔ∂∂+∂∂ 22ωω xCwwyΔxΔψ xws •xR Section 6.2 Solid Mechanics Part II Kelly 127Twist Not only does a plate curve up or down, it can also twist (see Fig. 6.1.3). The twist is defined analogously to the curvature and is denoted by xyT/1: yxw Txy∂∂∂=21 (6.2.4) The physical meaning of the twist is illustrated in Fig. 6.2.4. Figure 6.2.4: Physical me aning of the twist Principal Curvatures Consider the two Cartesian coordinate sy stems shown in Fig. 6.2.5, the second ( nt−) obtained from the first ( yx−) by a positive rotation θ. The partial derivatives arising in the curvature expressions can be expressed in terms of derivatives with respect to t and n as follows: with ()yxww ,= , an increment in w is yywxxww Δ∂∂+Δ∂∂=Δ (6.2.5) Also, referring to Fig. 6.2.5, with 0=Δn , θ θ sin , cos t y t x Δ=Δ Δ=Δ (6.2.6) Thus θ θ sin cosyw xw tw ∂∂+∂∂=∂∂ (6.2.7) yw xAB C D xyx yΔ∂∂∂+∂∂ωω2yyx xΔ∂∂∂+∂∂ωω2 yAB Ax∂∂ω xD C y∂∂ωCBD ww Section 6.2 Solid Mechanics Part II Kelly 128Similarly, for an increment nΔ, one finds that θ θ cos sinyw xw nw ∂∂+∂∂−=∂∂ (6.2.8) Equations 6.2.7-8 can be inverted to get the inverse relations θ θθ θ cos sinsin cos nw tw ywnw tw xw ∂∂+∂∂=∂∂∂∂−∂∂=∂∂ (6.2.9) Figure 6.2.5: Two different Cart esian coordinate systems The relationship between second derivatives can be found in the same way. For example, ntw nw twnw tw n t xw ∂∂∂−∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂−∂∂=∂∂ 2 22 2 22 222 2sin sin cossin cos sin cos θ θ θθ θ θθ (6.2.10) In summary, one has nt t n yxwnt n t ywnt n t xw ∂∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂−=∂∂∂∂∂∂+∂∂+∂∂=∂∂∂∂∂−∂∂+∂∂=∂∂ ωθωωθθωθωθωθωθωθωθ 2 22 22 22 22 2 22 2 222 22 2 22 2 22 2cos cos sin2sin cos sin2sin sin cos (6.2.11) and the inverse relations xn y t θ xΔyΔtΔ o Section 6.2 Solid Mechanics Part II Kelly 129yx x y ntwyx y x nwyx y x tw ∂∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂− ∂∂=∂∂∂∂∂∂− ∂∂+ ∂∂= ∂∂∂∂∂+ ∂∂+ ∂∂= ∂∂ ωθωωθθωθωθωθωθωθωθ 2 22 22 22 22 2 22 2 222 22 2 22 2 22 2cos cos sin2sin cos sin2sin sin cos (6.2.12) or1 xy x y tnxy y x nxy y x t T R R TT R R RT R R R 12cos1 1cos sin112sin1cos1sin112sin1sin1cos1 2 22 2 θ θθθ θ θθ θ θ +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ − =− + =+ + = (6.2.13) These are the same as the transformation equatio ns for stress and stra in and there will be some angle θ for which the twist is zero; at this an gle, one of the curvatures will be the minimum and one will be the maximum at that point in the plate. These are called the principal curvatures . Similarly, just as the sum of the normal stresses is an invariant, the sum of the curvatures is an invariant2: n t y x R R R R1 1 1 1+=+ (6.2.14) If the principal curvatures are equal, the curvat ures are the same at all angles, the twist is always zero and so the plate deforms locally into the surface of a sphere. 6.2.3 Strains in a Plate The strains arising in a plate are next examine d. Consider a line element parallel to the y axis, of length xΔ. Let the element displace as shown in Fig. 6.2.6. Whereas w was used in the previous section on curvatures to denote displacement of the mid-surface, here, for the moment, let ),,( zyxw be the general vertical displacemen t of any particle in the plate. Let u and v be the corresponding displacements in the x and y directions. Denote the original and deformed length of the element by dS and ds respectively. The unit change in length of the el ement is, using Pythagoras’ theorem3, 1 these equations are valid for any co ntinuous surface; Eqns. 6.2.12 are restricted to nearly-flat surfaces. 2 this is known as Euler’s theorem for curvatures 3 multiplying this expression by () dS dS ds 2/+ gives the Green-Lagrange strain xxE Section 6.2 Solid Mechanics Part II Kelly 1301 12 2 2 −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+=−′′ =−=yw yv yu pqpq qp dSdSds xxε (6.2.15) Figure 6.2.6: deformation of a material fibre in the x direction In the plate theory, it will be a ssumed that the displacement gradients zw yv xv yu xu ∂∂ ∂∂ ∂∂ ∂∂ ∂∂, , , , are small, of order 1ε<< say, so that squares and pr oducts of these terms may be neglected. However, the squares and products of the slopes, yw xw yw xw ∂∂ ∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂⎟ ⎠⎞⎜ ⎝⎛ ∂∂, ,2 2 might be significant (of the same order as the displacement gradients, ε) if there are moderate rotations of the plate. Eqn. 6.2.15 now reduces to 1 212 −⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂+=xz xu xxε (6.2.16) With 2/ 1 1 x x+≈+ for 1<<x , one has (and similarly for the other normal strains) •p′• xxuxuΔ∂∂+)(xxwΔ∂∂ xΔ• •p q ()yuq′ •q′′ ()ywxxwxwΔ∂∂+)( mid-surface Section 6.2 Solid Mechanics Part II Kelly 131zwyw yvxw xu zzyyxx ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂= εεε 22 2121 (6.2.17) Consider next the angle change for line elemen ts initially lying parallel to the axes, Fig. 6.2.7. Let θ be the angle qpr′′′∠ , so that θπγ−= 2/ is the change in the initial right angle rpq∠ . Figure 6.2.7: the deformation of Fig. 6.2.6, showing shear strains Taking the dot product of the of the vector elements qp′′ and rp′′: rpqprpqq rrqq rrqp ′′′′′′′′′′′+′′′′′′′′′′+′′′′′′ =θcos (6.2.18) With displacement gradients of order 1ε<<, take z rp rp rpx qp qp qp Δ=′′′′=′′′=′′Δ=′′′=′′′=′′ so zv xv xw zu zxzxxwzzvxxvzzux ∂∂ ∂∂+∂∂+∂∂=ΔΔΔ⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+⎟ ⎠⎞⎜ ⎝⎛Δ∂∂⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+⎟ ⎠⎞⎜ ⎝⎛Δ∂∂Δ =θcos (6.2.19) p′ ••p qq′••r s xΔ••• xxuxΔ∂∂+ΔxxwΔ∂∂zzvΔ∂∂ r′ zΔ•••zzuΔ∂∂ xxvΔ∂∂r′′ q′′ q′′′r′′′zzwzΔ∂∂+Δ θ Section 6.2 Solid Mechanics Part II Kelly 132For small γ, θγγ cos sin=≈ , so (and similarly for the other shear strains) ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂+∂∂= yw zvxw zuyw xw xv yu yzxzxy 212121 εεε (6.2.20) The normal strains 6.2.17 and the shear strain s 6.2.20 are non-linear. They are the starting point for the various different plate theories. Von Kármán Strains Introduce now the assumptions of the classica l plate theory. The assumption that line elements normal to the mid-plane re main inextensible implies that 0=∂∂=zw zzε (6.2.21) This implies that ()yxww ,= so that all particles at a given ()yx, through the thickness of the plate experinece the same vertical displa cement. The assumption that line elements perpendicular to the mid-plane remain norma l to the mid-plane after deformation then implies that 0 ==yz xzεε . The strains now read 002102121 22 ==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂+∂∂==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂= yzxzxyzzyyxx yw xw xv yuyw yvxw xu εεεεεε (6.2.22) These are known as the Von Kármán strains . Membrane Strains and Bending Strains Since 0=xzε and ),(yxww= , one has from 6.2.20, Section 6.2 Solid Mechanics Part II Kelly 133),( ),,(0yxuxwz zyxuxw zu+∂∂−= →∂∂−=∂∂ (6.2.23) It can be seen that the function ) ,(0yxu is the displacement in the mid-plane. In terms of the mid-surface displacements 0 0 0,,wvu , then, 00 00 0 , , wwywz vvxwz uu =∂∂−=∂∂−= (6.2.24) and the strains 6.2.22 may be expressed as yxwzyw xw xv yuywzyw yvxwzxw xu xyyyxx ∂∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂−⎟ ⎠⎞⎜ ⎝⎛ ∂∂+∂∂= 02 0 0 0 02022 0 0202 2 0 0 21 212121 εεε (6.2.25) The first terms are the usual small-strains, for the mid-surface. The second terms, involving squares of displacement gradients, are non-linear, and need to be considered when the plate bending is fairly large. Th ese first two terms together are called the membrane strains . The last terms, involving second derivatives, are the flexural (bending ) strains . They involve the curvatures. When the bending is not t oo large, one has (droppi ng the subscript “0” from w) yxwzxv yuywzyvxwzxu xyyyxx ∂∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂−∂∂=∂∂−∂∂= 2 0 022 022 0 21εεε (6.2.26) Some of these strains are illustrated in Fi gs. 6.2.8 and 6.2.9; the physical meaning of xxε is shown in Fig. 6.2.8 and some terms from xyε are shown in Fig. 6.2.9. Finally, when the mid-surface strains are negl ected, according to the final assumption of the classical plate theory, one has yxwzywzxwzxy yy xx∂∂∂−=∂∂−=∂∂−=2 22 22 , , ε ε ε (6.2.27) Section 6.2 Solid Mechanics Part II Kelly 134 Figure 6.2.8: deformation of material fibres in the x direction Figure 6.2.9: the deformation of 6.2.8 viewed “from above”; a′, b′ are the deformed positions of the mid-surface points a, b Compatibility The strain field arising in th e plate is two-dimensional, xy yy xxεεε ,, , and so the 2D compatibility relation 1.3.1 must be satisfied: yx x yxy yy xx ∂∂∂= ∂∂+ ∂∂ εεε2 22 22 2 (6.2.28) It can be seen that Eqns. 6.2.26 (6.2 .27) indeed satisfy this condition. p′ pa, qb,q′b′ a′xyxwzywz Δ∂∂∂+∂∂2 •• • ••• θxxvvΔ∂∂+0 0 ywzv∂∂−0•c0v••p′ a′xw ∂∂ w•• xxwwΔ∂∂+xxw xwΔ∂∂+∂∂ 22 xΔ•• •• ap bqxwzu∂∂−0 0u xxuuΔ∂∂+0 0xxwzxu xwzu Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+∂∂−22 0 0 q′ b′•q′′ mid-surface zxxε Section 6.2 Solid Mechanics Part II Kelly 135 6.2.4 The Moment-Curvature equations Now that the strains have been related to th e curvatures, the moment -curvature relations, which play a central role in plate theory, can be derived. Stresses and the Curvatures/Twist From Hooke’s law, taking 0=zzσ , xy xy xx yy yy yy xx xxE E E E Eσνεσνσεσνσε+= −= −=1,1,1 (6.2.29) so, from 6.2.27, and solving 6.2.29a-b for the normal stresses, yxwzExw ywzEyw xwzE xyyyxx ∂∂∂ +−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ −−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ −−= 222 22 222 22 2 111 νσννσννσ (6.2.30) The Moment-Curvature Equations Substituting Eqns. 6.2.30 into the definitions of the moments, Eqns. 6.1.1, 6.1.2, and integrating, one has ()yxwD Mxw ywD Myw xwD M xyyx ∂∂∂−−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= 222 2222 22 1ννν (6.2.31) where ()23 112ν−=EhD (6.2.32) Equations 6.2.31 are the moment-curvature equations for a plate. The moment- curvature equations are analogous to the beam moment-deflection equation EIM xv / /2 2=∂∂ . The factor D is called the plate stiffness or flexural rigidity and plays the same role in the plate theory as does the flexural rigidity term EI in the beam theory. Section 6.2 Solid Mechanics Part II Kelly 136 The signs of the moments, radii of curvature a nd curvatures are illustrated in Fig. 6.2.11. Note that the deflection w may or may not be of the same sign as the curvature. Note also that when 0 / ,02 2>∂∂> x Mxω , when 0 / ,02 2>∂∂> y Myω but, with the sign convention being used, when 0 / ,02<∂∂∂> yx Mxyω . Figure 6.2.11: sign convention for curvatures and moments Stresses and Moments From 6.30-6.31, the stresses and moments are related through 12/,12/,12/3 3 3hzM hzM hzM xy xyy yyx xx += −= −= σ σ σ (6.2.33) Note the similarity of these relations to the beam formula I My/−=σ with 12/3hI= times the width of the beam. 6.2.5 Principal Moments It was seen how the curvatures in different di rections are related, through Eqns. 6.2.11-12. It comes as no surprise, examining 6.2.31, that the moments are related in the same way. Consider a small differential element of a pl ate, Fig. 6.2.12a, subjected to stresses xxσ, yyσ, xyσ, and corresponding moments xy y x M M M , , given by 6.1.1-2. On any perpendicular planes ro tated from the orginal yx− axes by an angle θ, one can find the 0>R 0 /2 2>∂∂ xω 0<R0 /2 2<∂∂ xω 0<M0>M z z Section 6.2 Solid Mechanics Part II Kelly 137new stresses ttσ, nnσ, tnσ, Fig. 6.2.12b (see Fig. 6.2.5), through the stress transformatrion equations (Par t I, Eqns. 3.4.7). Then [][][] xy y xxy yy xx tt t M M Mdzz dzz dzz dzz M θ θ θσθ σθ σθ σ 2sin sin cos2sin sin cos 2 22 2 − + =−+ −+ −=−= ∫ ∫ ∫ ∫ (6.2.34) and similarly for the other moments, leading to ()xy x y tnxy y x nxy y x t M M M MM M M MM M M M θ θθθ θ θθ θ θ 2cos sin cos2sin cos sin2sin sin cos 2 22 2 +− −=+ + =− + = (6.2.35) Also, there exist principal planes, upon which the shear stress is zero (right through the thickness). The moments acting on these planes, 1M and 2M, are called the principal moments , and are the greatest and le ast bending moments which occur at the element. On these planes, the twisting moment is zero. Figure 6.2.12: Plate Element; (a) stresses ac ting on element, (b) rotated element Moments in Different Coordinate Systems From the moment-curvature equations 6.2.31, { ▲Problem 1} ()ntD Mt nD Mn tD M xyyx ∂∂∂−−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+ ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+ ∂∂= ωνωνωωνω 222 2222 22 1 (6.2.36) xxσxyσ xyσnnσ ttσ θtnσ Section 6.2 Solid Mechanics Part II Kelly 138showing that the moment-curvature relations 6.2.31 hold in all Cartesian coordinate systems. 6.2.6 Problems 1. Use the curvature transformation relations 6.2.11 and the moment transformation relations 6.2.35 to derive the mo ment-curvature relations 6.2.36. Section 6.3 Solid Mechanics Part II Kelly 1396.3 Plates subjected to Pure Bending and Twisting 6.3.1 Pure Bending of an Elastic Plate Consider a plate subjec ted to bending moments 1M Mx= and 2M My= , with no other loading, as shown in Fig. 6.3.1. Figure 6.3.1: A plate under Pure Bending From equilibrium considerations, these moments act at all points within the plate – they are constant throughout the plat e. Thus, from the moment -curvature equations 6.2.31, one has the set of coupled part ial differential equations yxw xw yw DM yw xw DM ∂∂∂=∂∂+∂∂=∂∂+∂∂=2 22 22 2 22 22 10, , ν ν (6.3.1) Solving for the derivatives, 0 , ) 1(, ) 1(2 21 2 22 22 1 22 =∂∂∂ −−=∂∂ −−=∂∂ yxw DM M yw DM M xw νν νν (6.3.2) Integrating the first two equations twice gives1 )( )( ) 1(21),( )( ) 1(21 2 12 21 2 2 12 22 1xgyxgy DM Mwyfxyfx DM Mw ++ −−= ++ −−= νν νν (6.3.3) and integrating the third shows that two of these four unknown functions are constants: BxgAyf yFywxGxw= = → =∂∂=∂∂)(, )( )( ),(1 1 (6.3.4) 1 this analysis is similar to that used to evaluate displacements in plane el astostatic problems, §1.2.4 y x1M1M 2M2M Section 6.3 Solid Mechanics Part II Kelly 140Equating both expressions for w in 6.3.3 gives )( ) 1(21)( ) 1(21 22 21 2 22 22 1yf Byy DM MxgAxx DM M−+ −−=−+ −− νν νν (6.3.5) For this to hold, both sides here must be a constant, C− say. It follows that CByAxy DM Mx DM Mw +++ −−+ −−=2 21 2 2 22 1 ) 1(21 ) 1(21 νν νν (6.3.6) The three unknown constants represent an arbi trary rigid body motion. To obtain values for these one must fix three degrees of freedom in the plate. If one supposes that the deflection w and slopes ywxw ∂∂∂∂ /,/ are zero at the origin 0==yx (so the origin of the axes are at the plate-centre), then 0=== CBA ; all deformation will be measured relative to this reference. It follows that ()[] ()[] [ ]2 2 12 2 1 22/ 1 / ) 1(2yMM x MM DMw ν ν ν−+− −= (6.3.7) Once the deflection w is known, all other quantities in the plate can be evaluated – the strain from 6.2.20, the stress from Hooke’s la w or directly from 6.2.30, and moments and forces from 6.1.1-3. In the special case of equal bending moments, with oM M M==2 1 say, one has ()2 2 )1(2yxDMwo++=ν (6.3.8) This is the equation of a sphere. In fact, fr om the relationship between the curvatures and the radius of curvature R, constant) 1( ) 1(22 22 =+=→+=∂∂=∂∂ oo MDRDM yw xw ν ν (6.3.9) and so the mid-surface of the plate in this ca se deforms into the surface of a sphere with radius given by 6.3.9, as illustrated in Fig. 6.3.2. Figure 6.3.2: Deformed plate under Pu re Bending with equal moments Section 6.3 Solid Mechanics Part II Kelly 141The character of the deformed plate is plotted in Fig. 6.3.3 for various ratios 1 2/MM (for 3.0=ν ). Figure 6.3.3: Bending of a Plate When the curvatures 2 2/xw∂∂ and 2 2/yw∂∂ are of the same sign2, the deformation is called synclastic . When the curvatures are of opposite sign, as in the lower plots of Fig. 6.3.3, the deformation is said to be anticlastic . Note that when there is only one moment, 0=yM say, there is still curvature in both directions. In this case, one can solve the moment-curvature equations to get () ()()2 2 2 22 22 2 22 12, ,1y xDMwyw yw DM xwx xν νν ν−−=∂∂−=∂∂ −=∂∂ (6.3.10) which is an anticlastic deformation. In order to get a pure cylindrical deformation, )(xfw= say, one needs to apply moments xM and x yM Mν= , in which case, from 6.3.6, 2 or principal curvatures in the cas e of a more complex general loading 5.1 /1 2=MM 3 /1 2=MM 5.1 /1 2−=MM 3 /1 2−=MM Section 6.3 Solid Mechanics Part II Kelly 1422 2xDMwx= (6.3.11) The deformation for 3 /1 2=MM in Fig. 6.3.3 is very clos e to cylindrical, since there y xM Mν≈ . 6.3.2 Pure Torsion of an Elastic Plate In pure torsion, one has the twisting moment M Mxy= with no other loading, Fig. 6.3.4. From the moment-curvature equations, yxw DM xw yw yw xw ∂∂∂=−−∂∂+∂∂=∂∂+∂∂=2 22 22 22 22 ) 1(, 0, 0νν ν (6.3.12) so that ) 1(,0 ,02 22 22 ν−−=∂∂∂=∂∂=∂∂ DM yxw yw xw (6.3.13) Figure 6.3.4: Twisting of a Plate Using the same arguments as before, integrating these equations leads to xyDMw) 1(ν−−= (6.3.14) The middle surface is deformed as shown in Fig. 6.3.5, for a negative xyM. Note that there is no deflection along the lines 0=x or 0=y . The principal curvatures will occur at 45o to the axes (see Eqns. 6.2.12): y xMM Section 6.3 Solid Mechanics Part II Kelly 143) 1(1,) 1(1 2 1 ν ν −−=−+=DM R DM R (6.3.15) Figure 6.3.5: Deformation for a (negative) twisting moment y x 2R Section 6.4 Solid Mechanics Part II Kelly 1446.4 Equilibrium and Lateral Loading In this section, lateral loads are considered and these lead to shearing forces y xVV,, in the plate. 6.4.1 The Governing Differential Equation for Lateral Loads In general, a plate will at any lo cation be subjected to a lateral pressure q, bending moments xy y x MMM , , and out-of-plane shear forces xV and yV; q is the normal pressure on the upper surface of the plate: ⎩⎨⎧ += −−==2/ ),,(2/ ,0),(h z yxqh zyxzzσ (6.4.1) These quantities are rela ted to each other through force equilibrium. Force Equilibrium Consider a differential plate element with one corner at )0,0(),(=yx , Fig. 6.4.1, subjected to moments, pressure and shear force. Taking forc e equilibrium in the vertical direction (neglecting a po ssible small variation in q, since this will only introduce higher order terms): 0=ΔΔ−Δ⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+−Δ+Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂+−Δ+=∑ yxqyxxVVyVxyyVVxV Fx x xy y y z (6.4.2) Fig. 6.4.1: a plate element subjected to moments, pressure and shear forces Eqn. 6.4.2 gives the vert ical equilibr ium equation qyV xV y x−=∂∂+∂∂ (6.4.3) y x() yyMyΔ+q ()yyVyΔ+() yy MxyΔ+ () xxMxΔ+()xxVxΔ+() xxMxyΔ+ Section 6.4 Solid Mechanics Part II Kelly 145 Next, taking moments about the x axis: () () () () 0 2/ 2/2/ =ΔΔΔ+−Δ⎟ ⎠⎞⎜ ⎝⎛Δ∂∂+ Δ+−ΔΔ++Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂+Δ+−Δ−Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂++Δ−Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂+−Δ=∑ yxqy yyxxVV y yyVy yxyyVVy yx yVxyyMMx MyxxMMy M M x xxy y yy y yxy xy xy x (6.4.4) Using 6.4.3, this reduces to (and similarly for moments about the x-axis), yM xMVyM xMV y xy yxy x x ∂∂+∂∂−=∂∂−∂∂+= (6.4.5) These are analogous to the beam equation dx dMV /= . Relations directly from the Equations of Equilibrium The equilibrium relations 6.4.3, 6.4.5 can also be derived directly from the equations of equilibrium, Eqns. 1.1.9, whic h encompass the force balances: 000 =∂∂+∂∂+∂∂=∂∂+∂∂+∂∂=∂∂+∂∂+∂∂ z y xz y xz y x zz yz xzzy yy xyzx yx xx σσσσσσσσσ (6.4.6) Taking the first of these (which ensu res equilibrium of forces in the x direction), multiplying by z and integrating over th e plate thickness, gives [] 00 2/ 2/2/ 2/2/ 2/2/ 2/2/ 2/2/ 2/2/ 2/ = − +⎥ ⎦⎤ ⎢ ⎣⎡ ∂∂+⎥ ⎦⎤ ⎢ ⎣⎡ ∂∂→=∂∂+∂∂+∂∂ ∫ ∫ ∫∫∫∫ + −+ −+ −+ −+ −+ −+ − dz z dzzydzzxdzzz dzyz dzxz h hzxh h zxh hyxh hxxh hzxh hyxh hxx σ σ σ σσ σ σ (6.4.7) and, since the shear stress zxσ must be zero over the top and bottom surfaces, one has Eqn. 6.4.5a. Applying a similar procedure to the second equilibrium equation gives Eqn. 6.4.5b. Finally, integrating directly the third equilibrium equation without multiplying across by z, one arrives at Eqn. 6.4.3. Section 6.4 Solid Mechanics Part II Kelly 146 Eliminating the shear forces from 6.4.3, 6.4.5 leads to the differential equation q yM yxM xM y xy x−= ∂∂+∂∂∂− ∂∂ 22 2 22 2 (6.4.8) This equation is analogous to the equation p x M=∂∂2 2/ in the beam theory. Finally, substituting in the moment-curva ture equations 6.2.31 leads to1 Dq yw yxw xw−=∂∂+∂∂∂+∂∂ 44 2 24 44 2 (6.4.9) This is sometimes called the equation of Sophie Germain after the French investigator who first obtained it in 18152. This partial differential equation is solved subject to the boundary conditions of the problem, i.e. the fi xing conditions of the plate (see below). Again, when once an expression for ),(yxω is obtained, the strains, stresses, forces and moments follow. Note that the differentia l equation 6.4.7 with 0 =q is trivially satisfied in the simple pure bending and torsion problems considered earlier. Eqn. 6.4.9 can be succinctly expressed as Dqw−=∇2 (6.4.10) where 2∇ is the Laplacian , or “del” operator: 22 22 2 y x∂∂+∂∂=∇ (6.4.11) Note that the Laplacian operator (on w) gives the sum of the curvatures in two perpendicular directions and so it is independent of the directions chosen (see Eqn. 6.2.14). Shear Forces in terms of Deflection From 6.4.5 and the moment-curvature e quations, one has the useful relations 1 note that the moment curvature relations were derived for the case of pure bending; here, as in the beam theory, the possible effect of the sh earing forces on the curvature is ne glected. This is a valid assumption provided the thickness of the plate is small in compar ison with its other dimensions. A more exact theory taking into account the effect of the shear forces on deflection can be developed 2 Germain submitted her work to the French Academy, which was awarding a prize for anyone who could solve the problem of the vibration of plates; Lagrange was on the Academy awarding committee and corrected some of her work, deriving Eqn. 6.4.9 in its final form Section 6.4 Solid Mechanics Part II Kelly 147⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ ∂∂=22 22 22 22 ,yw xw yD Vyw xw xD Vy x (6.4.12) 6.4.2 Stresses in the Plate The normal and in-plane shear stresses have been expressed in terms of the moments, Eqns. 6.2.33. Note that these stresses are ze ro over the mid-surface and attain a maximum at the outer surfaces. Expressions for the remaining stress component s can be obtained from the equations of equilibrium as follows: the first of Eqns. 6.4.6 leads, with 6.4.5a, to zVhzz yM xM hzzMhz yMhz xz y x zx xzx xy xzx xy xzx xy xx ∂∂+−=∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂−=∂∂+⎥⎦⎤ ⎢⎣⎡ ∂∂+⎥⎦⎤ ⎢⎣⎡−∂∂=∂∂+∂∂+∂∂= σσσσσσ 333 3 121212 120 (6.4.13) Integrating now gives (note that xV is independent of z) CzVhx zx + =2 36σ (6.4.14) This shear stress must be zero at th e upper and lower (free -) surfaces, at 2/h z±= . This condition can be used to dete rmine the arbitrary constant C and one finds that (see Fig. 6.1.9) ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−−=2 2/123 hz hVx zxσ (6.4.15) The other shear stress, zyσ, can be evaluated in a similar manner: { ▲Problem 1} ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−−=2 2/123 hz hVy zyσ (6.4.16) In some analyses, these shear stresses are taken to be zero, although they can be quite significant. The only remaining stress component is zzσ. This will never exceed the intensity of the external load on the plate; the lateral load itself, however, is negligibly small in Section 6.4 Solid Mechanics Part II Kelly 148comparison with the in-plane stresses set up by the bending of the plate, and for this reason it is acceptable to disregard zzσ, as has been done, in the plate theory. 6.4.3 Problems 1. Derive the expression fo r shear stress 6.4.16. Section 6.5 Solid Mechanics Part II Kelly 1496.5 Plate Problems in Rectangular Coordinates In this section, a number of important plate problems will be examined using Cartesian coordinates. 6.5.1 Uniform Pressure produc ing Bending in One Direction Consider first the case of a plate which bends in one direction only. From 6.3.11 the deflection and moments are 22 22 )( , )( ),(dxwdD xMdxwdD xM xf wy x ν−= = = (6.5.1) The differential equation 6.4.9 reads Dxq dxwd )( 44 −= (6.5.2) The corresponding equation for a beam is EIxp dxwd /)( /4 4= . If )( /)( xq bxp−= , with b the depth of the beam, with 12 /3bhI= , the plate will respond more stiffly than the beam by a factor of ) 1/(12ν− , a factor of about 10% for 3.0=ν , since () bEI EhD2 23 11 112 νν−=−= (6.5.3) The extra stiffness is due to the constraining effect of yM, which is not present in the beam. 6.5.2 Deflection of a Circular Plate by a Uniform Lateral Load A solution for a circular plate problem is pres ented next. This problem will be examined again in the section which follows usi ng the more natural polar coordinates. Consider a circular plate with boundary 2 2 2a y x=+ , (6.5.4) clamped at its edges and subjected to a uniform lateral load q, Fig. 6.5.1. Section 6.5 Solid Mechanics Part II Kelly 150 Figure 6.5.1: a clamped circular plate s ubjected to a unifo rm lateral load The differential equation for the problem is given by 6.4.9. The boundary conditions are that the slope and deflecti on are zero at the boundary: 2 2 2along 0 ,0 ,0 a y xyw xww =+ =∂∂=∂∂= (6.5.5) It will be shown that the deflection 22 2 2) ( a y xcw −+= (6.5.6) is a solution to the problem. First, this function certainly satisfies 6.5.5. Further, letting 2 2 2),( a y x yxf −+= , (6.5.7) the relevant partial derivatives are () () cywcyxwcxwcyywcxyxwcyyxwcxxwf ycywcxyyxwf xcxwcyfywcxfxw 24 ,8 ,2424 ,8 ,8 , 2424 ,8 , 244 ,4 44 2 24 4433 23 23 332 22 2 2 22 =∂∂=∂∂∂=∂∂=∂∂=∂∂∂=∂∂∂=∂∂+=∂∂=∂∂∂+=∂∂=∂∂=∂∂ (6.5.8) Substituting these into the differential equation now yields Dqc64−= (6.5.9) so the deflection is 22 2 2) (64a y xDqw −+ −= (6.5.10) xyaq Section 6.5 Solid Mechanics Part II Kelly 151This is plotted in Fig. 6.5.2. The maximu m deflection occurs at the plate centre, where Dqaw644 max−= . (6.5.11) Figure 6.5.2: mid-plane deflection of the clamped circular plate The curvature 2 2/xw∂∂ along a radial line 0 =y is displayed in Fig. 6.5.3. The curvature is positive toward the centre of th e plate (the plate curves upward) and is negative towards the edge of the pl ate (the plate curves downward). Figure 6.5.3: curvature in th e clamped circular plate The moments occurring in the plate are, from the moment-curvature equations 6.2.31 and 6.5.8, [] [] xyqMa y xqMa y xqM xyyx ) 1(8) 1( ) 3( )13(16) 1( )13( ) 3(16 2 2 22 2 2 νν ν νν νν −+=+−+++−=+−+++−= (6.5.12) The moment xM along a radial line 0 =y is of the same character as the curvature displayed in Fig. 6.5.3. 22 xw ∂∂ )0,0(),(=yx )0,(),( a yx=w Section 6.5 Solid Mechanics Part II Kelly 152 The out-of-plane shear forces are, from 6.4.5, 2,2qyVqxVy x −= −= (6.5.13) At the plate centre, the expressions become 0 ,) 1(162=== +==y x xy y x V V M aqM M ν (6.5.14) Stresses in the Plate From 6.5.12-13 and 6.2.33, 6.4.15-16, the stresses in the plate are [ ] [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−=⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−=−=+−+++ =+−+++= 2232 2 2 32 2 2 3 2/1432/143) 1(23) 1( ) 3( )13(43) 1( )13( ) 3(43 hz hqyhz hqxxyhqza y xhqza y xhqz zyzxxyyyxx σσν σν ν ν σν νν σ (6.5.15) Converting to polar coordinates ) ,(θr through θ θ sin , cos ry rx = = (6.5.16) and using a stress transformation, ()xy xx yy rxy yy xxxy yy xx rr θσ σσθθσθσ θσ θσ σθσ θσ θσ σ θθθ 2cos sin cos2sin sin cos2sin sin cos 2 22 2 +− =− + =+ + = (6.5.17) leads to the axisymmetric stress field { ▲Problem 1} [ ] [] 0) 1( )13( 43) 1( ) 3( 43 2 2 32 2 3 =+−+ =+−+= θθθ σν ν σν ν σ rrr a r hqza r hqz (6.5.18) Section 6.5 Solid Mechanics Part II Kelly 153At the plate centre, ()ν σσθθ + −== 143 32 hqza rr (6.5.19) At the plate edge ar=, ν σ σθθ 32 32 23,23 hqza hqza rr = = (6.5.20) For the shear stress, the traction acting on a surface parallel to the yx− plane can be expressed as (see Fig. 6.5.4) () ()θ θθθ θθσθθσσσσσ e e e ee ee e t cos sin sin cos + +− =+=+= r zy r zxy zy x zxz rzr (6.5.21) where ie is a unit vector in the direction i. Thus 0 cos sin2/143sin cos2 = +−=⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−= + = zy zx zzy zx zrhz hqr θσ θσ σθσ θσ σ θ (6.5.22) Figure 6.5.4: stress components acting on a surface Note that the maximum st ress in the plate is ()2 max432/, ⎟ ⎠⎞⎜ ⎝⎛= =haqharrσσ (6.5.23) The maximum shear stress, on the other hand, is ()ha q azr / 4/3)0,( ×=σ . Thus the shear stress is of an order ah/ smaller than the normal stress. θzy yσ,eθθσz,e zr rσ,e zx xσ,e Section 6.5 Solid Mechanics Part II Kelly 1546.5.3 An Infinite Plate with Sinusoidal Deflection Consider next the classic plate problem addre ssed by Navier in 1820. It consists of an infinite plate with an undulating “up/dow n” sinusoidal deflection, Fig. 6.5.5, by axw yxwππsin sin ),(0= (6.5.24) Figure 6.5.5: A plate with sinusoidal deflection Differentiation of the deflection leads to the curvatures by ax abwxywby ax bwywby ax awxw πππππππππ cos cossin sinsin sin 2 0222 0 2222 0 22 =∂∂−=∂∂−=∂∂ (6.5.25) and hence the pressure Dyxqyxw b a yw xyw xw ),(),(1 122 2 24 22 2 44 −≡⎟ ⎠⎞⎜ ⎝⎛+= ∂∂+∂∂+ ∂∂π (6.5.26) The pressure thus varies like the deflection. There is no need for supports for the plate since the “up” loads balance the “down” loads. From the moment-curvature relations, xy ab Section 6.5 Solid Mechanics Part II Kelly 155()by ax abDw Mby ax b aDw Mby ax b aDw M xyyx πππνππ νπππνπ cos cos 1sin sin1sin sin1 2 02 22 02 22 0 −−=⎟ ⎠⎞⎜ ⎝⎛+ −=⎟ ⎠⎞⎜ ⎝⎛+ −= (6.5.27) and, from 6.4.12, the shear forces are by ax b abDw Vby ax b aaDw V yx ππππππ cos sin1 11sin cos1 11 2 23 02 23 0 ⎟ ⎠⎞⎜ ⎝⎛+ −=⎟ ⎠⎞⎜ ⎝⎛+ −= (6.5.28) Note that both wq/ and y xM M/ are constant throughout the plate. 6.5.4 A Simply Supported Plate with Sinusoidal Deflection Following on from the previous example, consider now a finite plate of dimensions a and b with the same sinusoidal deflection 6.5.24, simply supported along the edges 0=x , ax=, 0=y , by=. In what follows, take 0w in 6.5.24 to be negative, so that the plate is pushed down towards the centre. According to 6.5.24 and 6.5.27, the deflec tion and slope is ze ro along the supported edges, as required. The vert ical reactions at the supports are given by 6.5.28. However, according to Eqn. 6.5.27c, there are varying non-zero twisting moments over the ends of the plate. Thus the solution given by 6.5. 24-28 is not quite the solution to the simply supported finite-plate problem, unless one ca n somehow apply the ex act required twisting moments over the edges of the plate. It turns out, however, that the solution 6.5.24-28 is a correct solution, except in a region close to the edges of the plate. Th is is explained in what follows. Twisting Moments over “Free” Surfaces Consider an element of material of width dy, Fig. 6.5.6. The element is subjected to a twisting moment dyMxy, Fig. 6.5.6a. This twisting moment is due to shear stresses acting parallel to the plate surface (see Fig. 6. 1.8). This system of horizontal forces can be replaced by the statically equivalent system of vertical forces shown in Fig. 6.5.6b – two forces of magnitude xyM separate by a distance dy. Recalling Saint-Venant’s principle, the difference between the statically equivalent system s of forces of Fig. 6.5.6a and 6.5.6b will lead to differences in the stre ss field within the plate only in a small region very close to the plate-edges. Section 6.5 Solid Mechanics Part II Kelly 156 Figure 6.5.6: Equivalent systems of forces le ading to the same twisting moment; (a) horizontal forces, (b) vertical forces Consider next a distribution of twisting moment along the plate edge, Fig. 6.5.7. As can be seen, this distribution is e quivalent to a distribution of sh earing forces (per unit length) of magnitude yMyVxy x∂∂−=)( (6.5.29) Figure 6.5.7: A distribution of tw isting moments along a plate edge The total vertical react ion along the edges can now be taken to be yMVxy x∂∂− (6.5.30) (and x M Vxy y∂∂− / along the other edges) and this gives a correct solution to the problem. From 6.5.27-28, these reactions are dyMxydyxyM yMMxy xy∂∂+ dyyMMxy xy ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+dyMxydy xyMxyM (a) (b) Section 6.5 Solid Mechanics Part II Kelly 157ax a b abDwxMV Fax a b abDwxMV Fby b b aaDwyMV Fby b b aaDwyMV F bxxy y ybxxy y yyaxy x xayxy x x πνππνππνππνπ sin1 1 1 1sin1 1 1 1sin1 1 1 1sin1 1 1 1 2 2 23 0 ),(2 2 23 0 )0,(02 2 23 0 ),(2 2 23 0 ),0(0 ⎥⎦⎤ ⎢⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ +=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−=⎥⎦⎤ ⎢⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−=⎥⎦⎤ ⎢⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ +=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−=⎥⎦⎤ ⎢⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−= (6.5.31) Corner Forces Integrating 6.5.31 over the f our edges, the resultant upward forces on the four edges (with 00<w , they are all four upward) are ⎥⎦⎤ ⎢⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ −=−=+⎥⎦⎤ ⎢⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ −=−=+ 2 2 22 0 02 2 22 0 x0 1 1 121 1 12 a b abaDw F Fb b aabDw F F yb yxa νπνπ (6.5.32) and the resultant of these may be expressed as () ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −+⎟ ⎠⎞⎜ ⎝⎛+ −=222 2 22 0 up12 1 14ba b aabDw Fνπ (6.5.33) The resultant downward force is, using 6.5.26, 2 2 22 0002 2 24 0 00down 1 14sin sin1 1),( ⎟ ⎠⎞⎜ ⎝⎛+ −=⎟ ⎠⎞⎜ ⎝⎛+ −= = ∫∫ ∫∫ b aabDwdydxby ax b aDw dydxyxq Fba ba ππππ (6.5.34) The difference between upF and downF is due to the re-distributed twisting moment, and is explained a follows: consider again Fig. 6. 5.7, where the edge twisting moments have been replaced with a statically equivalent distribution of shear forces. It can be seen that there results shear forces at the ends of th e plate-edge (the “corners”), where the shear forces xyM have no neighbouring shear force of opposite sign with which to “cancel out”. There are concentrated forces (per unit le ngth) at the plate-co rners of magnitude xyM. Examining Fig. 6.5.7, which shows the edge ax=, the force ) 0,(a Mxy is positive up whereas the force ) ,(ba Mxy is positive down. There are also contributions to the corner Section 6.5 Solid Mechanics Part II Kelly 158forces at ) 0,(a and ) ,(ba from the adjacent edges, shown in Fig. 6.5.8. One finds that the downward concentrated forces at the corner are () () () ()abDw b M PabDw ba M PabDw a M PabDw M P xy bxy abxy axy 2 0 02 02 0 02 0 00 1 2 ),0( 21 2 ),( 21 2 )0,( 21 2 )0,0( 2 πνπνπνπν −−= −=−−= +=−−= −=−−= += (6.5.35) Adding these to downF of Eqn. 6.5.34 now gives the upF of Eqn. 6.5.33. Physically, if one applies a pressure to a simp ly supported plate, the plate will tend to rise at the four corners, in a twisting action. Th e corner forces 6.5.35 are necessary to keep the corners down and so pr oduce the deflection 6.5.24. Figure 6.5.8: corner forces in the simply supported plate The ratio of the resultant downward corner fo rce to the downward for ce due to the applied pressure, downF , is ()()22 222 12 b aba +−ν (6.5.36) For a square plate, this is 2/) 1(ν− ; with 3.0=ν , this is 35%. 6.5.5 A Rectangular Plate Simply Supported at the Edges The above solution can be used to solve the problem of a simply supported plate loaded by any arbitrary pressure distribution, through the use of Fourier series. y x)0,(axyM ),(baxyM),0(bxyM)0,0(xyM Section 6.5 Solid Mechanics Part II Kelly 159Consider again this plate, whose displacement boundary conditions are ()()()() () () () ()0 ,00 , 0, , ,0 ,22 0,22 ,22 ,022 =∂∂=∂∂=∂∂=∂∂==== bx x ya yyw yw xw xwbxw xwyawy w (6.5.37) Assume the deflection to be of the form ∑∑∞ =∞ == 11sin sin ),( mnmnbyn axmA yxwππ (6.5.38) with mnA coefficients to be determined. It can be seen that this function satisfies the boundary conditions. Taking the derivatives of this function, ∑∑∞ =∞ =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−=∂∂ 11222 22 sin sin mnmnbyn axmAam xw ππ π (6.5.39) etc., and substituting into the differential equation 6.4.9, gives ∑∑∞ =∞ =−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ 112 22 22 4),( sin sin mnmn yxqbyn axm bn amA Dπππ (6.5.40) This can be written co mpactly in the form ∑∑∞ =∞ =−= 11),( sin sin mnmn yxqbyn axmCππ ( 6 . 5 . 4 1 ) where 2 22 22 4 ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =bn amDA Cmn mnπ (6.5.42) It remains to choose the coefficients of the se ries so as to satisfy the equation identically over the whole area of the plate. One can evaluate the coefficients as one doe s for ordinary Fourier series, although here one has a double series and so one proceeds as follows: first, multiply both sides of (6.5.40) by () byk/ sinπ where k is an integer, and integrate over y between the limits ],0[b, so that ∑∑ ∫ ∫∞ =∞ =−= 11 0 0sin),( sin sin sin mnb b mn dybykyxq dybyk byn axmCπ πππ (6.5.43) Using the orthogonality condition Section 6.5 Solid Mechanics Part II Kelly 160 ⎩⎨⎧ =≠= ∫kn bkndybyk bynb ,2/,0sin sin 0ππ, (6.5.44) leads to ∑ ∫∞ =−= 1 0sin),( sin2mb mk dybykyxqaxmCb π π (6.5.45) Now there are functions of x only so, multiplying both sides by ) / sin( axjπ and following the same procedure, one has ∫∫ ⎥ ⎦⎤ ⎢ ⎣⎡−=ab jk dxaxjdybykyxq Cba 00sin sin),(22π π (6.5.46) and hence the coefficients mnC are (replacing the dummy subscripts kj, with nm,) ∫∫−=ab mn dxdybyn axmyxqabC 00sin sin),(4 ππ (6.5.47) Thus the coefficients mnA of the original expression for the deflection ) ,(yxw , 6.5.38, are ∫∫− ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −=ab mn dxdybyn axmyxqbn am abDA 002 22 22 4sin sin),(41 ππ π (6.5.48) It is now possible to solve for th e coefficients given any loading ),(yxq over the plate, and hence evaluate the deflection, moments and stresses in the plate, by taking the derivatives of the infinite series for w. This solution is due to Navier and is called Navier’s solution to the rectangular plate problem. A similar solution method has been used by Lévy to solve a more general problem – that of a rectangul ar plate simply supported on tw o opposite sides, and any one of the conditions free, simply-supported, or clamped, along the other two opposite sides. For example, considering a square plate, th is involves using a trial function for the deflection of the form (compare with 6.5.38) ∑∞ == 1sin)( ),( nnaxnyF yxwπ (6.5.49) and then attempting to determine the functions ) (yFn . A Uniform Load In the case of a uniform load q yxq=),( , one has Section 6.5 Solid Mechanics Part II Kelly 161 () () () ()⎪⎩⎪⎨⎧ ==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −=⎥⎦⎤ ⎢⎣⎡−⎥⎦⎤ ⎢⎣⎡−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −= −−− ∫∫ LL ,4,2,0 , 0,5,3,1 ,16) cos(1 ) cos(14sin sin4 2 22 22 62 22 22 40 02 22 22 4 nmnmbn am Dmnqnnbmma bn am Dabqdybyndxaxm bn am DabqAb a mn πππππ ππ π π (6.5.50) The resulting series in 6.5.50 converges rapidly. The deflection at the centre of the plate is then () ∑∑∑∑ ∞ =∞ =−+−∞ =∞ = −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −== 5,3,15 ,3,112/) (2 2 22 645,3,15 ,3,1 1)/(1 162sin2sin mnnmmnmn nbam mn Dqbn mA w πππ (6.5.51) For a square plate, ()() 0040624.011 16 45,3,15 ,3,112/) ( 22 2 64 ×−=−+ −=∑∑∞ =∞ =−+ − Dqan mmn Dqaw mnnm π (6.5.52) Denoting the area 2a by A, this is D qA/ 0041.02−=ω . This can be compared with the clamped circular plate; denoting the area there, 2aπ, by A, the maximum deflection, Eqn. 6.5.11, gives D qA/ 0016.02−=ω . Corner Forces The twisting moment is () ()∑∑∞ =∞ =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−=∂∂∂−−= 112 2 cos cos 1 1 ),( mnmn xybyn axmAabmnv Dyxv D yx Mππ π ω (6.5.53) and the four corner forces requi red to hold the plate down are now Section 6.5 Solid Mechanics Part II Kelly 162() () () ()∑∑∑∑∑∑∑∑ ∞ =∞ =∞ =∞ =∞ =∞ =∞ =∞ = ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−= +=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−+= −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−+= −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−= += 112112 0112 0112 00 cos cos 12 ),( 2cos 12 ),0( 2cos 12 )0,( 212 )0,0( 2 mnmn xy abmnmn xy bmnmn xy amnmn xy n m Aabmnv D ba M Pn Aabmnv D b M Pm Aabmnv D a M PAabmnv D M P ππππππππ (6.5.54) For a uniform load over a square plate, using 6.5.50, the corner forces reduce to () () () () ()() () 0421122 2425,3,15 ,3,122 242 26.02825.0132412 1 21 13241 1324 4 Fav qn mav qn mav qP mnmn ≈×−≈−+−−=+−= ∑∑∑∑ ∞ =∞ =∞ =∞ = πππ (6.5.55) (for 3.0=ν ) where 2 0qa F= is the resultant applied force. 6.5.6 Problems 1. Derive the expressions for the stress co mponents in polar form, for the clamped circular plate under uniform lateral load, Eqn. 6.5.18. Section 6.6 Solid Mechanics Part II Kelly 1636.6 Plate Problems in Polar Coordinates 6.6.1 Plate Equations in Polar Coordinates To examine directly plate problems in polar coordinates, one can first transform the Cartesian plate equations considered in the pr evious sections into ones in terms of polar coordinates. First, the definitions of the moments and forces are now ∫ ∫ ∫+ −+ −+ −= −= −=2/ 2/2/ 2/2/ 2/, ,h hr rh hh hrr r dzz Mdzz Mdzz Mθ θ θθ θ σ σ σ (6.6.1) and ∫ ∫+ −+ −−= −=2/ 2/2/ 2/,h hzh hzr r dz V dz Vθ θσ σ (6.6.2) The strain-curvature relations , Eqns. 6.2.27, can be transformed to polar coordinates using the transformations from Cartesian to polar c oordinates detailed in §4.2 (in particular, §4.2.6). One finds that { ▲Problem 1} ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂+∂∂−−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂−=∂∂−= θθεθεε θθθ rw rw rzw rrw rzrwz rrr 2 222 222 1 11 1 (6.6.3) The moment-curvature relations 6.2.31 become { ▲Problem 2} ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂+∂∂−−−=⎥ ⎦⎤ ⎢ ⎣⎡ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂= θθννθθν θθ rw rw rD Mrw w rrw rD Mw rrw r rwD M rr 2 222 22 222 2 22 1 111 11 1 (6.6.4) The governing differential equation 6.4.9 now reads Dqwrrrr−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂2 22 2 221 1 θ (6.6.5) Section 6.6 Solid Mechanics Part II Kelly 164 The shear forces in terms of de flection, Eqn 6.4.12, now read { ▲Problem 3} ⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂ ∂∂=⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂ ∂∂=22 2 22 22 2 221 1 1,1 1 θ θ θθw rrw r rw rD Vw rrw r rw rD Vr (6.6.6) Finally, the stresses are { ▲Problem 4} θ θ θ θθ σ σ σr r r rr MhzMhzMhz 3 3 312,12,12= −= −= (6.6.7) and ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−−= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−−=2 2 2/123,2/123 hz hV hz hV zr zrθ θσ σ (6.6.8) The differential equation 6.6.5 can be solved using a method similar to the Airy stress function method for problems in polar coordi nates (the Mitchell solution), that is, a solution is sought in the form of a Fourier series. Here , however, only axisymmetric problems will be considered in detail. 6.6.2 Plate Equations for Axisymmetric Problems When the loading and geometry of the plate are axisymmetric, the plate equations given above reduce to 011 2222 =⎥⎦⎤ ⎢⎣⎡+ =⎥⎦⎤ ⎢⎣⎡+ = θθ νν rr Mdrwd drdw rD Mdrdw r drwdD M (6.6.9) Drq drdwrdrd rdrdrdrd rwdrd r drd )( 1 1 12 22 −= ⎭⎬⎫ ⎩⎨⎧ ⎥⎦⎤ ⎢⎣⎡⎟ ⎠⎞⎜ ⎝⎛=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ (6.6.10) 0 ,1 22 =⎥⎦⎤ ⎢⎣⎡+ =θVdrdw r drwd drdD Vr (6.6.11) 0 ,12,12 3 3= −= −=θ θ θθ σ σ σr r rr MhzMhz (6.6.12) and Section 6.6 Solid Mechanics Part II Kelly 1650 ,2/1232 = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−−=θσ σzr zrhz hV (6.6.13) Note that there is no twisting moment, so th e problem of dealing with non-zero twisting moments on free boundaries seen with r ectangular plate does not arise here. 6.6.3 Axisymmetric Plate Problems For uniform q, direct integration of 6.6.10 leads to () DrCrB r rADqrw +++− +−= ln411 ln41 642 24 (6.6.14) with () () 3 3322 223 121 21 831 211 ln241 1631 211 ln241 16 rCrADqr drwdrCB r ADqr drwdrCrB r rADqr drdw ++−=−++ +−=++− +−= (6.6.15) and DrAqr drdw r drwd r drwdD Vr +−=⎥⎦⎤ ⎢⎣⎡−+ =21 1 2 22 33 (6.6.16) There are two classes of problem to consider, plates with a central hole and plates with no hole. For a plate with no hole in it, the conditi on that the stresses rema in finite at the plate centre requires that 2 2/drwd remains finite, so 0==CA . Thus immediately one has 2/qr Vr−= . The boundary conditions at the outer edge ar= give B and D. 1. Solid Plate – Uniform Bending The simplest case is pure bending of a plate, 0M Mr= , with no transverse pressure, 0=q . The plate is solid so 0==CA and one has D rBw += 4/2. The applied moment is []νωνω+=⎥⎦⎤ ⎢⎣⎡+ = 121 1 22 0 DBdrd r drdD M (6.6.17) so ()ν+ = 1/ 20DM B . Taking the deflection to be zero at the plate-centre, the solution is Section 6.6 Solid Mechanics Part II Kelly 166()2 0 12rDMwν+= (6.6.18) 2. Solid Plate Clamped – Uniform Load Consider next the case of clamped plate under uniform loading. The boundary conditions are that 0 /== drdww at ar=, leading to DqaDDqaB64,84 2 −= = (6.6.19) and hence ()22 2 64arDqw − −= (6.6.20) which is the same as 6.5.10. The reaction force at the outer rim is 2/ )( qa aVr−= . This is a force per unit length; the force acting on an element of the outer rim is ()2/θΔ−aqa and the total reaction force around the outer rim is π2qa− , which balances the same applied force. 3. Solid Plate Simply Supported – Uniform Load For a simply supported plate, 0=w and 0=rM at ar=. Using 6.6.9a, one then has {▲Problem 5} DqaDDqaB64 15,8 134 2 νν νν ++−=++= (6.6.21) and hence ()2 2 2 2 15 64raraDqw −⎟ ⎠⎞⎜ ⎝⎛−++−=νν (6.6.22) The deflection for the clamped and simply suppor ted cases are plotted in Fig. 6.6.1 (for 3.0=ν ). Section 6.6 Solid Mechanics Part II Kelly 167 Figure 6.6.1: deflection for a circular plate under uniform loading 4. Solid Plate with a Central Concentrated Force Consider now the case of a plate subjected to a single concentrated force F at 0=r . The resultant shear force acting on any cylindric al portion of the plate with radius r about the plate-centre is )( 2 rrVrπ . As 0→r , one must have an infinite rV so that this resultant is finite and equal to the applied force F. An infinite shear force implies infinite stresses. It is possible for the stresses at the centre of th e plate to be infinite. However, although the stresses and strain might be infinite, the displacements, which are obtained from the strains through integrat ion, can remain, and should remain , finite. Although the solution will be “unreal” at the plate-centre, one can again use Saint-Venant’s principle to argue that the solution obtained will be valid ever ywhere except in a small region near where the force is applied. Thus, seek a solution which has finite displ acement in which case, by symmetry, the slope at 0=r will be zero. From the general axisymmetric solution 6.6.15a, 0 01 = == r r rCdrdw (6.6.23) so 0=C . From 6.6.16 FDA rVrr ≡==π π 2 20 (6.6.24) Thus D FAπ2/= and the moments and shear force become infinite at the plate-centre. The other two constants can be obtained fr om the boundary conditions. For a clamped plate, 0 /== drdww , and one finds that { ▲Problem 7} -4-3-2-100.2 0.4 0.6 0.8 1x ωqD64ar/ clamped simply supported Section 6.6 Solid Mechanics Part II Kelly 168 ()() [ ]ar r raDFw /ln2162 2 2+− =π (6.6.25) This solution results in ) /ln( ar terms in the expressions for moments, giving logarithmically infinite in-plane stresses at the plate-centre. 5. Plate with a Hole For a plate with a hole in it, there will be f our boundary conditions to determine the four constants in Eqn. 6.6.14. For example, fo r a plate which is simply supported around the outer edge br= and free on the inner surface ar=, one has 0)( ,0)(0)( ,0)( = == = bM bwaF aM rr r (6.6.26) 6.6.4 Problems 1. Use the expressions 4.2.11-12, which relate second partial derivatives in the Cartesian and polar coordinate systems, together with the strain transformation relations 4.2.17, to derive the strain-curvature relations in polar coordinates, Eqn. 6.6.3. 2. Use the definitions of the moments, 6.6.1, and again relations 4.2. 11-12, together with the stress transformation relations 4.2.18, to derive the moment-curvature relations in polar coordinates, Eqn. 6.6.4. 3. Derive Eqns. 6.6.6. 4. Use 6.2.33, 6.4.15-16 to derive the stresse s in terms of moments and shear forces, Eqns. 6.6.7-8. 5. Solve the simply supported solid plate proble m and hence derive the constants 6.6.21. 6. Show that the solution for a simply supported plate (with no hole), Eqn. 6.6.22, can be considered a superposition of the clamped solution, Eqn. 6.6.20, and a pure bending, by taking an appropriate defl ection at the plate-centre in the pure bending case. 7. Solve for the deflection in the case of a cl amped solid circular pl ate loaded by a single concentrated force, Eqn. 6.6.25. Section 6.7 Solid Mechanics Part II Kelly 1696.7 In-Plane Forces and Plate Buckling In the previous sections, only bending a nd twisting moments and out-of-plane shear forces were considered. In th is section, in-plane forces are considered also. The in-plane forces will give rise to in-plane membrane st rains, but here it is assumed that these are uncoupled from the bending strains. In other words, the membrane strains can be found from a separate plane stress analysis of th e mid-surface and the bending of the plate does not affect these membrane strains. The po ssible effect of the in-plane forces on the bending strains is the main concern here. 6.7.1 Equilibrium for In-plane Forces Start again with the equations of equilibrium , Eqns. 6.4.6. Integrating the first and second through the thickness of the plate (this time without multiplying first by z), and using the definitions of the in-plane forces 6.1.1-6.1.2, leads to 00 =∂∂+∂∂=∂∂+∂∂ yN xNyN xN y xyxy x (6.7.1) 6.7.2 The Governing Differential Equation Consider an element of the deflected pl ate, Fig. 6.7.1. Only a deflection in the y direction, y∂∂/ω , is considered for clarity. Resolving the components of the in-plane forces into horizontal and vertical components: yxywNx ywNyywNxxywNy ywNxywN FyxxNNyNxyyNNxN F xy xy xyy y y Vxy xy xyy y y H Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂+Δ∂∂−Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+∂∂+Δ∂∂−=Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂++Δ−Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂++Δ−= ∑∑ (6.7.2) These reduce to Section 6.7 Solid Mechanics Part II Kelly 170yxyxwNywNyw xN yNyxywNx ywNyFxyxN yNF xy yxy yxy y Vxy y H ΔΔ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂+∂∂+∂∂ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=ΔΔ⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂=ΔΔ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= ∑∑ 2 22 (6.7.3) Using 6.7.1, one has ∑ ∑ ΔΔ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂+∂∂= = yxyxwNywN F Fxy y V H2 22 ,0 (6.7.4) Considering also a deflection x∂∂/ω , one has for the resultant vertical force : ∑ ΔΔ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂∂+∂∂= yxywNyxwNxwN Fy xy x V 22 2 22 2 (6.7.5) Figure 6.7.1: In-plane forces acting on a plate element When the in-plane forces were neglected, th e vertical stress resisted by bending and shear force was qzz−=σ . Here, one has an additional stress given by 6.7.5, and so the governing differential equation 6.4.7 becomes ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂∂+∂∂+−=∂∂+∂∂∂+∂∂ 22 2 22 44 2 24 44 212ywNyxwNxwNqD y yx xy xy xωωω (6.7.6) ω y xxyyNNy y Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂+ xNyΔ ωωΔ+yxxNNxy xy Δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛Δ∂∂+ Section 6.7 Solid Mechanics Part II Kelly 1716.7.3 Buckling of Plates When compressive in-plane forces are applied to a plate, the plate will at first remain flat and simply be compressed. However, when the in-plane forces reach a critical level, the plate will bend and the deflection will be gi ven by the solution to 6.7.6. For example, consider the case of a simply supported plate subjected to a uniform in-plane compression xN only, Fig. 6.7.2, in which case 6.7.6 reduces to 22 44 2 24 44 2xw DN y yx xx ∂∂=∂∂+∂∂∂+∂∂ ωωω (6.7.7) Following Navier’s method from §6.5.5, assume a buckled shape ∑∑∞ =∞ == 11sin sin ),( mnmnbyn axmA yxwππ (6.7.8) so that 6.7.7 becomes ∑∑∞ =∞ == ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ 112222 22 22 40 sin sin mnx mnbyn axm am DN bn amAππππ (6.7.9) Disregarding the trivial 0 =mnA , this can be satisfied by taking 2 22 22 222 ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −=bn am mDaNxπ (6.7.10) Figure 6.7.2: In-plane compression of a plate The lowest in-plane force xN which will deflect the plate is sought. Clearly, the smallest value on the right hand side of 6.7.10 will be when 1=n . This means that the buckling modes as given by 6.7.8 will be of the form by axmππsin sin (6.7.11) so that the plate will only ever buckle with one half-wave in the direction perpendicular to loading. a Section 6.7 Solid Mechanics Part II Kelly 172 When ba≤, the smallest value occurs when 1=m , in which case the critical in-plane force is ()2 22 cr ⎟ ⎠⎞⎜ ⎝⎛+ −=ba ab bDNxπ (6.7.12) When ba/ is very small, the pl ate is loaded along the re latively long edges and the critical load is much higher than for a square plate. The deflection (buckling mode) corres ponding to this critical load is by axAyxwππsin sin ),(11= (6.7.13) Note that the amplitude 11A cannot be determined from the analysis1. As ba/ increases above unity, the value of m at which the applied load is a minimum increases. When ba/ reaches just over 2, the critical buckling load occurs for 2=m , for which ()2 22 22⎟ ⎠⎞⎜ ⎝⎛+ −=ba ab bDNcrxπ (6.7.14) and corresponding buckling more by axAyxwππsin2sin ),(21= (6.7.15) The plate now buckles in two ha lf-waves, as if the centre-li ne were simply supported and there were two smaller separate plates buckling similarly. As ba/ increases further, so too does m. For a very long, thin, plate, bam /≈ , and so the plate subdivides approximately into squares, each bucklin g in a half-wave. 1 this is a consequence of assuming small deflections; it can be determined when the deflections are not assumed to be small Section 6.8 Solid Mechanics Part II Kelly 1736.8 Plate Vibrations In this section, the problem of a vibrating circular plate will be considered. Vibrating plates will be re-examined again in the next section, using a strain energy formulation. 6.8.1 Vibrations of a Cl amped Circular Plate When a plate vibrates with velocity t∂∂/ω , the third equation of equilibrium, Eqn. 6.6.2c becomes the equation of motion 22 tw z y xzz yz xz ∂∂=∂∂+∂∂+∂∂ρσσσ (6.8.1) With this adjustment, the term q is replaced with 2 2/twh q ∂∂+ρ in the relevant equations; the acceleration term is treated as a transverse load of intensity 2 2/twh∂∂ρ . Regarding the circular plate, one has from the axisymmetric governing equation 6.6.10 (with 0=q ), 222 221 tw Dhwdrd r drd ∂∂−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ρ (6.8.2) Assume a solution of the form ()()φω+ = t rWtrw cos ),( (6.8.3) Substituting into 6.8.2 gives 0142 22 = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ Wkdrd r drd (6.8.4) where Dhkρω=2 (6.8.5) Eqn. 6.8.4 gives the two differential equations 01,012 22 2 22 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−+ =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛++ Wkdrd r drdWkdrd r drd (6.8.6) The solution to these equations are Section 6.8 Solid Mechanics Part II Kelly 174()() ()()krKC krIC W krYC krJC W0 4 03 02 01 , + = + = (6.8.7) where 0J and 0Y are, respectively, the Bessel functions of order zero of the first kind and of the second kind; 0I and 0K are, respectively, the Modified Bessel functions of order zero of the first kind and of the second kind1. These functions are plotted in Fig. 6.8.1 below. For a solid plate with no hole at 0=r , one requires that 04 2==C C , since 0Y and 0K become unbounded as 0→r . The general solution is thus ()()krIB krJA rW0 0 )( + = (6.8.8) Figure 6.8.1: Bessel Functions For a clamped plate, the boundary conditions give ()() ()() 00 )( 0 00 0 =′+′== + = =kaIB kaJAdrdWkaIB kaJA aW ar (6.8.9) where the dash means dxx dJ xJ /)( )(0 0=′ and dxxdI xI /)( )(0 0=′ . Using the relations )( )( ),( )(1 0 1 0 xI xI xJ xJ +=′ −=′ (6.8.10) where 1 1,IJ are Bessel functions of order one, one has () ()() ()kaIkaJ kaIkaJ 11 00−= (6.8.11) 1 by definition , these Bessel functions are the solution of the differential equations 6.8.6. -3-2-1012345 0.5 1 1.5 2 2.5 3z0K 0I 0Y0J kr Section 6.8 Solid Mechanics Part II Kelly 175The roots ka give the frequencies of vibration of the plate. The function ()()()()kaJkaI kaIkaJ1 0 1 0 + (6.8.12) is plotted in Fig. 6.8.2 below. The smallest root is found to be 3.1962. Eqn. 6.8.5 then gives for the frequency, hD aραω21= (6.8.13) where 2158.10=α . Figure 6.8.2: The Function 6.8.12 Further roots ka of 6.8.12 are given in Table 6.8.1. For each of these roots there is a corresponding frequency ω given by Eqn. 6.8.13, for which the value of α is also tabulated. ka α nodal circle 1 3.1962 10.2158 2 6.3064 39.7711 0.3790 3 9.4395 89.1041 0.2548, 0.5833 Table 6.8.1: Roots of Eqn. 6.8.11, frequency factors and nodal circle roots From 6.8.3, 6.8.8-9, the solution for the deflection is ()() ()() ( ) φω+⎥⎦⎤ ⎢⎣⎡− = t krIkaIkaJkrJAtrw cos ),(0 00 0 (6.8.14) -4-3-2-101 0.5 1 1.5 2 2.5 3 3.5 4xka Section 6.8 Solid Mechanics Part II Kelly 176These are an infinite number of deflections, each one corresponding to a root ka. The actual deflection will be a superposi tion of these individual solutions. The term inside the square brackets gives the mode shape of the plate during the vibration. The first three (normalized) mode shapes, corresponding to the first three roots, are shown in Fig. 6.8.3. Figure 6.8.3: Mode shapes for the Clamped Circular Plate The point ar/ where these mode-shapes change si gn are the positions of the so-called nodal circles . These roots of the mode shapes are gi ven in the last column of Table 6.8.1 The General Problem For circular plates not constrained to an axisymmetric response, one must use the more general differential equation 6.6.5 222 22 2 221 1 tw Dhwrrr r ∂∂−=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+∂∂ ρ θ (6.8.15) This time, instead of 6.8.3, assume a solution of the form ()()()φωθ θ + =∑ t n rW trwn sin cos ),,( (6.8.16) Then 6.8.4-6 become 0142 22 22 = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−∂+∂∂Wkrn rd r r (6.8.17) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.500.51 ar/ Section 6.8 Solid Mechanics Part II Kelly 177where k is again given by 6.8.5, and 6.8.6 becomes 01,012 22 22 2 22 22 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−∂∂+ ∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−∂∂+ ∂∂Wk rn rr rWk rn rr r (6.8.18) The solution to these equations are ()() ()()krKC krIC W krYC krJC Wn n n n 4 3 2 1 , + = + = (6.8.19) where one now has Bessel functions of order n. Proceeding as before, one now needs to find roots of the equation ()()()()01 1 = ++ + ka JkaI kaIkaJn n n n (6.8.20) and the deflection is ()() ()() () ( ) φωθ+ ⎥⎦⎤ ⎢⎣⎡− =∑ t n krIkaIkaJkrJA trwn nn n sin cos ),( (6.8.21) The solution for 0=n has been given already. For other values of n, there are n so-called nodal diameters . For example, for 1=n there is one nodal diameter along 2/πθ±= , along which the deflection is zero. The roots of 6.8.20 for this case are given in Table 6.8.2, together with the nodal circle locations. ka α nodal circle 1 4.6109 21.2604 2 7.7993 60.8287 0.4897 3 10.9581 120.0792 0.3497, 0.6390 Table 6.8.2: Roots of Eqn. 6.8.20 ( n=1), frequency factors and nodal circle roots The mode shapes for half the plate for this case of one nodal diameter are shown in Fig. 6.8.4, corresponding to the first two roots in Table 6.8.2. The frequencies corresponding to these solutions are again given by 6.8.13 with the frequency factor α given in the table. Section 6.8 Solid Mechanics Part II Kelly 178 Figure 6.8.4: Mode shapes for the case of one nodal diameter Section 6.9 Solid Mechanics Part II Kelly 1796.9 Strain Energy in Plates 6.9.1 Strain Energy due to Plate Bending and Torsion Here, the elastic strain energy due to pl ate bending and twisting is considered. Consider a plate element bending in the x direction, Fig. 6.9.1. The ra dius of curvature is 2 2/xw R∂∂= . The strain energy due to bending through an angle θΔ by a moment yMxΔ is () xxwyM UxΔ ∂∂Δ=Δ22 21 (6.9.1) Considering also contributions from yM and xyM, one has yxywMyxwMxwM Uy xy x ΔΔ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂∂−∂∂=Δ22 2 22 221 (6.9.2) Figure 6.9.1: a bending plate element Using the moment-curvature relations, one has () () yxyxw yw xw yw xw Dyx yw yxw yw xw xw DU ΔΔ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂ ∂∂−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=ΔΔ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−+ ∂∂ ∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂=Δ 22 22 222 22 222 2222 22 222 22 12212 22 νν ν (6.9.3) This can now be integrated over the complete pl ate surface to obtain th e total elastic strain energy. xθΔ R Section 6.9 Solid Mechanics Part II Kelly 180 6.9.2 The Principle of Minimum Potential Energy Plate problems can be solved using the princi ple of minimum potential energy (see Part I, §5.6). Let extW V−= be the potential energy of the loads, equivalent to the negative of the work done by those loads, and so the potential energy of the system is () () () wVwUw +=Π . The solution is then th e deflection which minimizes ()wΠ . When the load is a uniform lateral pressure q, one has ()xxyxwq W Vext ΔΔ+=Δ−=Δ , (6.9.4) and () yx qwyxw yw xw yw xw DΔΔ ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ + ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂ ∂∂−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=ΔΠ22 22 222 22 22 122ν (6.9.5) As an example, consider again the simply supported rectangular plate subjected to a uniform load q. Use the same trial function 6. 5.21 which satisfies the boundary conditions: ∑∑∞ =∞ == 11sin sin ),( mnmnbyn axmA yxwππ (6.9.6) Substituting into 6.9.5 and integrating over the plate gives () dxdybyn axmA qbyn axm byn axm banmbyn axm bn amAD mnmnmnmnba ⎭⎬⎫+⎥ ⎦⎤ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛− −−⎪⎩⎪⎨⎧ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =Π ∑∑∑∑∫∫ ∞ =∞ =∞ =∞ = ππππ ππ πνπππ sin sincos cos sin sin 12sin sin2 112 2 2 2 224222 22 22 22 4 112 00 (6.9.7) Carrying out the integration leads to 2 5,3,15 ,3,12 22 22 4 112 4 41 2 ππmnabA q abbn amAD mnmn mnmn × + ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ×⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =Π ∑∑ ∑∑∞ =∞ =∞ =∞ = (6.9.8) To minimize the total potential energy, one sets Section 6.9 Solid Mechanics Part II Kelly 1812 22 22 622 22 22 4 1604 4 − ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −=→= +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =∂Π∂ bn am DmnqAmnabq bn am abDAA mnmn mn πππ (6.9.9) which is the same result as 6.5.50. 6.9.3 Strain Energy in Polar Coordinates For circular plates, one can transform the strain energy expression 6.9.3 into polar coordinates, giving { ▲Problem 1} () yxrw rw rw rrw rrww rrw r rw DU ΔΔ ⎪⎭⎪⎬⎫ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂∂−∂∂−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ ∂∂⎪⎩⎪⎨⎧ ×−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂+ ∂∂=Δ 22 2 22 2 222 22 2 22 1 1 1 1121 1 2 θθ θν θ (6.9.10) For an axisymmetric problem, the strain energy is () yxrw rrw rw r rw DU ΔΔ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=Δ1121 2222 22 ν (6.9.11) 6.9.4 Vibration of Plates For vibrating plates, one needs to include th e kinetic energy of the plate. The kinetic energy of a plate element of dimensions yx ΔΔ, and moving with velocity t∂∂/ω is yxtwh K ΔΔ⎟ ⎠⎞⎜ ⎝⎛ ∂∂=Δ2 21ρ (6.9.12) According to Hamilton’s principle, then , the quantity to be minimized is now () () )(wKwVwU −+ . Consider again the problem of a circular pl ate undergoing axisymmetr ic vibrations. The potential energy function is () rdrtwh rdrrw rrw rw r rwDa a ∫ ∫ ⎟ ⎠⎞⎜ ⎝⎛ ∂∂− ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛ ∂∂ ∂∂−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂ 02 0222 221121πρ ν π (6.9.13) Assume a solution of the form Section 6.9 Solid Mechanics Part II Kelly 182 ()φω+ = t rWtrw cos)( ),( (6.9.14) Substituting this into 6.9.13 leads to () rdrWh rdrdrdW r drWd drdW r drWdDa a ∫ ∫− ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ 02 0222 221121ωπρ ν π (6.9.15) Examining the clamped plate, assume a solu tion, an assumption based on the known static solution 6.6.20, of the form ()22 2)( raArW −= (6.9.16) Substituting this into 6.9.15 leads to () ()() [] () drrar hArdrr ra a r ra a AD aa ∫∫ − −+−−−+− 042 2 204 22 4 4 22 4 23 4 1 4 4 2 32 πρων π (6.9.17) Evaluating the integrals leads to ⎟ ⎠⎞⎜ ⎝⎛−10 6 2 101 332ah Da A ωρ π (6.9.18) Minimising this function, setting {}0 /=∂∂A , then gives 328.103320,1 2≈= = αραωhD a (6.9.19) This simple one-term solution is very close to the exact result given in Table 6.8.1, 10.2158. The result 6.9.15 is of course greater than the actual frequency. 6.9.5 Problems 1. Derive the strain energy expression in polar coordinates, Eqn. 6.9.10. Section 6.10 Solid Mechanics Part II Kelly 1836.10 Limitations of Classical Plate Theory The validity of the classical plate th eory depends on a number of factors: 1. the curvatures are small 2. the in-plane plate dimensions are large compared to the thickness 3. membrane strains can be neglected The second and third of these points ar e discussed briefly in what follows. 6.10.1 Moderately Thick Plates As with beam theory, and as mentioned alrea dy, it turns out that the solutions based on the classical theory agree well with the full elasticity solutions (away from the edges of the plate), provided the plate thickness is sm all relative to its othe r linear dimensions. When the plate is relatively thick, one is ad vised to use a more exact theory, for example one of the shear deformation theories: Shear deformation Theories The Mindlin plate theory ( or moderately thick plate theory or shear deformation theory ) was developed in the early-to-mid 1900s to allow for possibl e transverse shear strains. In this theory, ther e is the added complication that vertical line elements before deformation do not have to remain perpendicular to the mid-surface after deformation, although they do remain straight. Thus shear strains yzε and zxε are generated, constant through the thickness of the plate. The classical plate theory is inconsistent in the sense that elements are assumed to remain perpendicular to the mid-plane, yet equ ilibrium requires that stress components yz xzσσ, still arise (which would cause th ese elements to deform). Th e theory of thick plates is more consistent, but it still makes the assumption that 0=zzσ . Note that both are approximations of the exact three-di mensional equations of elasticity. As an indication of the error involved in using the classica l plate theory, consider the problem of a simply supported square plate subj ected to a uniform pre ssure. According to Eqn. 6.5.52, the central deflection is 062 .4) 1000/ /( 4=D qaw . The shear deformation theory predicts 4.060 (for 100 /=ha ), 4.070 (for 50 /=ha ), 4.111 (for 20 /=ha ) and 4.259 (for 10 /=ha ). This trend holds in general; the classical theory is good for thin plates but under-predicts de flections (and over-predicts buckling loads and natural frequencies) in relatively thick plates. An important difference between the thin plat e and thick plate theories is that in the former the moments are related to the curvatures through (using xM for illustration) ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=22 22 yw xwD Mx ν (6.10.1) Section 6.10 Solid Mechanics Part II Kelly 184This is only an approximate relation (although it turns out to be exact in the case of pure bending). The thick plate theory predicts th at, in the case of a uniform lateral load q, the relationship is given by ) 1(2082 2 22 22 νννν−+++⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= qhyw xwD Mx (6.10.2) The thin plate expression will be approximate ly equal to the thick plate expression when the thickness h is very small, since in that case 02→h , or when the ratio of load to stiffness, Dq/ , is small. Some solutions for circular plates using th e various theories are presented next for comparison. Comparison of Solutions for Circular Plates 1. Uniform load q, clamped: Both thin and thick plate theories give ()22 2 64raDq− −=ω (6.10.3) 2. Uniform load q, simply supported: () ⎪⎪ ⎩⎪⎪ ⎨⎧ −+++=⎟ ⎠⎞⎜ ⎝⎛+−++− −= thick18thin 0 ,15 64 2 22 582 2 2 2 hra raDq νννηηννω (6.10.4) 3. Concentrated central load, clamped (Eqn. 6.6.25): ()( ) ⎪⎪ ⎩⎪⎪ ⎨⎧ −−−=+ +− −= exact )/ln(12thick 0thin 0)/ log(216 2 542 2 2 ar har r raDF ννηηπω (6.10.5) 4. Concentrated central load Q, simply supported: Section 6.10 Solid Mechanics Part II Kelly 185() () ()⎪⎪⎪ ⎩⎪⎪⎪ ⎨⎧ −+−−−−−+++ =⎟ ⎠⎞⎜ ⎝⎛+ +−++−= exact12)/ln(12thick18thin 0)/ log(213 16 2 2 22 51 2 542 2 22 522 2 2 raahar hra ahar r raDF νν ννννηηνν πω (6.10.6) Higher-order deformation Theories A further theory known as third-order plate theory has also been developed. This allows for the displacements to vary not only linearly (the previously described Mindlin theory is also called the first order shear deformation theory ), but as cubic functions; the in-plane strains are cubic (3~z) and the shear strains are quadratic. This allows the line elements normal to the mid-surface not onl y to rotate, but also to deform and not necessarily remain straight. 6.10.2 Large Deflections Consider now the assumption that the memb rane stresses may be neglected. To investigate the validity of this, consider an initially circular plate of diameter d, clamped at the edges, and deformed into a sp herically shaped surface, Fig. 6.10.1. Figure 6.10.1: a deform ed circular plate Considering a beam, the length of the neutral ax is before and after deformation can safely be taken to be equal, even when the beam de forms as in Fig. 6.10.1, i.e. one can take dd′=. The reason for this is that the “sup ports” are assumed to move slightly to accommodate any small deflection; thus the neut ral axis of a beam re mains strain-free and hence stress-free. d′Rθ2 dw Section 6.10 Solid Mechanics Part II Kelly 186Consider next the plate. Suppose that the supports could move slightly to accommodate the deformation of the plate so that the curved length in Fig. 6.10.1 was equal in length to the original diameter. One then sees that a compressive circumferential strain is set up in the plate mid-surface, of magnitude ()ddd /′− . To quantify this, note that Rd2/=θ and R d2/ sin′=θ . Thus ()()L− + −=′ !5/ 2/ !3/ 2/ 2/ 2/5 3Rd RdRdRd . Then L+⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛=2 2 0 19201 241 Rd Rd θθε (6.10.7) With Rw/ 1 cos−=θ and L+−= !2/ 1 cos2θθ , one also has L+ −=24 21 3841 81 Rd w wdR (6.10.8) so that, approximately, 22 0 38 dw=θθε (6.10.9) The maximum bending strain occurs at 2/hz= , where )/1)(2/.( R hrr=ε , so 24 dhw rr=ε (6.10.10) One can conclude from this rough analysis that , in order that the membrane strains can be safely ignored, the deflection w must be small when compared to the thickness h of the plate1. The corollary of this is that when there are large deflections, the middle surface will strain and take up the load as in a stretching membrane. For the bending of circular plates, one usually requires that h w 5.0< in order that the membrane strains can be safely ignored without introducing considerable e rror. For example, a uniformly loaded clamped plate deflected to hw= experiences a maximum membrane stress of approximately 20% of the maximum bending stress. When the deflections are large, the membrane strains need to be considered. This means that the von Kármán strains, Eqns. 6.2.22, 6.2.25, must be used in the analysis. Further, the in-plane forces, for example in the be nding Eqn. 6.7.6, are now an unknown of the problem. Some approximate solutions of th e resulting equations have been worked out, for example for uniformly loaded circular and rectangular plates. 1 except in some special cases, for example when a plate deforms into the surface of a cylinder 73DElasticity 188 Section 7.1 Solid Mechanics Part II Kelly 1897.1 Vectors, Tensors and the Index Notation The equations governing three dimensional mechanics problems can be quite lengthy. For this reason, it is essential to use a short-hand notation called the index notation1. Consider first the notation used for vectors. 7.1.1 Vectors Vectors are used to describe physical quantities which have both a magnitude and a direction associated with them. Geometrically, a vector is represented by an arrow; the arrow defines the direction of the vector and the magnitude of the vector is represented by the length of the arrow. Analytically, in what follows, vectors will be represented by lowercase bold-face Latin letters, e.g. a, b. The dot product of two vectors a and b is denoted by ba⋅ and is a scalar defined by θcosbaba=⋅ . (7.1.1) θ here is the angle between the vectors when their initial points coincide and is restricted to the range πθ≤≤0 . Cartesian Coordinate System So far the short discussion has been in symbolic notation2, that is, no reference to ‘axes’ or ‘components’ or ‘coordinates’ is made , implied or required. Vectors exist independently of any coordinate system. The symbolic notation is very useful, but there are many circumstances in which use of the component forms of vectors is more helpful – or essential. To this end, introduce the vectors 3 2 1,, eee having the properties 01 3 3 2 2 1 =⋅=⋅=⋅ ee ee ee , (7.1.2) so that they are mutually perpendicular, and 1 3 3 2 2 1 1 =⋅=⋅=⋅ ee ee ee , (7.1.3) so that they are unit vectors. Such a set of orthogonal unit vectors is called an orthonormal set, Fig. 7.1.1. This set of vectors forms a basis, by which is meant that any other vector can be written as a linear combination of these vectors, i.e. in the form 33 22 11 e e e a a a a ++= (7.1.4) where 2 1,aa and 3a are scalars, called the Cartesian components or coordinates of a along the given three directions . The unit vectors are called base vectors when used for 1 or indicial or subscript or suffix notation 2 or absolute or invariant or direct or vector notation Section 7.1 Solid Mechanics Part II Kelly 190this purpose. The components 2 1,aa and 3a are measured along lines called the 2 1,xx and 3x axes, drawn through the base vectors. Figure 7.1.1: an orthonormal set of ba se vectors and Cartesian coordinates Note further that this orthonormal system {}3 2 1,,eee is right-handed , by which is meant 3 2 1 e ee=× (or 1 3 2 e ee=× or 2 1 3 e ee=× ). In the index notation, the expression for the vector a in terms of the components 3 2 1,,aaa and the corresponding basis vectors 3 2 1,,eee is written as ∑ ==++=3 133 22 11 iiia a a a e e e e a (7.1.5) This can be simplified further by using Einstein’s summation convention , whereby the summation sign is dropped and it is understood that for a repeated index ( i in this case) a summation over the range of the index (3 in this case3) is implied. Thus one writes iiae a= . This can be further shortened to, simply, ia. The dot product of two vectors u and v, referred to this coordinate system, is () ( ) () () () () () () () () () 33 22 113 3 33 2 3 23 1 3 133 2 32 2 2 22 1 2 123 1 31 2 1 21 1 1 1133 22 11 33 22 11 vuvuvuvu vu vuvu vu vuvu vu vuv v v u u u ++=⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅+⋅=++⋅++=⋅ ee ee eeee ee eeee ee eee e e e e e vu (7.1.6) The dot product of two vectors written in the index notation reads iivu=⋅vu Dot Product (7.1.7) 3 2 in the case of a two-dimensional space/analysis 1e2e3e 2a1a3aa Section 7.1 Solid Mechanics Part II Kelly 191The repeated index i is called a dummy index , because it can be replaced with any other letter and the sum is the same; for example, this could equally well be written as jjvu=⋅vu or kkvu. Introduce next the Kronecker delta symbol ijδ, defined by ⎩⎨⎧ =≠=jiji ij,1,0δ (7.1.8) Note that 111=δ but, using the index notation, 3=iiδ . The Kronecker delta allows one to write the expressions defining the orthonormal basis vectors (7.1.2, 7.1.3) in the compact form ij j iδ=⋅ee Orthonormal Basis Rule (7.1.9) Example Recall the equations of motion, Eqns. 1.1.9, which in full read 3 3 333 232 1312 2 323 222 1211 1 313 212 111 a bx x xa bx x xa bx x x ρσσσρσσσρσσσ =+∂∂+∂∂+∂∂=+∂∂+∂∂+∂∂=+∂∂+∂∂+∂∂ (7.1.10) The index notation for these equations is i i jija bxρσ=+∂∂ (7.1.11) Note the dummy index j. The index i is called a free index ; if one term has a fee index i, then, to be consistent, all terms must have it. One free index, as here, indicates three separate equations. 7.1.2 Matrix Notation The symbolic notation v and index notation iive (or simply iv) can be used to denote a vector. Another notation is the matrix notation : the vector v can be represented by a 13× matrix (a column vector ): Section 7.1 Solid Mechanics Part II Kelly 192⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 321 vvv Matrices will be denoted by square brackets, so a shorthand notation for this matrix/vector would be []v. The elements of the matrix []v can be written in the index notation iv. Note the distinction between a vector and a 13× matrix: the former is a mathematical object independent of any coordinate system, the latter is a representation of the vector in a particular coordinate system – matrix notation, as with the index notation, relies on a particular coordinate system. As an example, the dot product can be written in the matrix notation as Here, the notation []Tu denotes the 31× matrix (the row vector ). The result is a 11× matrix, iivu. The matrix notation for the Kronecker delta ijδ is the identity matrix [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 100010001 I Then, for example, in both index and matrix notation: [] [] [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = = 321 321 100010001 uuu uuu u ui jij u uI δ (7.1.12) Matrix – Matrix Multiplication When discussing vector transformation equations further below, it will be necessary to multiply various matrices with each other (of sizes 13×, 31× and 33×). It will be helpful to write these matrix multiplications in the short-hand notation. “short” matrix notation “full” matrix notation[][] [ ] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 321 3 2 1T vvv u uu vu Section 7.1 Solid Mechanics Part II Kelly 193First, it has been seen that the dot pr oduct of two vectors can be represented by [][]vuT or iivu. Similarly, the matrix multiplication [][]Tvu gives a 33× matrix with element form jivu or, in full, ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 33 23 1332 22 1231 21 11 vuvuvuvuvuvuvu vuvu This operation is called the tensor product of two vectors, written in symbolic notation as vu⊗ (or simply uv). Next, the matrix multiplication [] [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ≡ 321 33 32 3123 22 2113 12 11 uuu Q Q QQ Q QQ Q Q uQ is a 13× matrix with elements [][]()jij i uQ≡uQ . The elements of [][]uQ are the same as those of [][]T TQu , which can be expressed as [][]()ijj i Qu≡T TQu . The expression [][ ]Qu is meaningless, but [][]QuT {▲Problem 4} is a 31× matrix with elements [][]()jij i Qu≡ QuT. This leads to the following rule: 1. if a vector pre-multiplies a matrix []Q → the vector is the transpose []Tu 2. if a matrix []Q pre-multiplies the vector → the vector is []u 3. if summed indices are “beside each other”, as the j in jijQu or jijuQ → the matrix is []Q 4. if summed indices are not beside each other, as the j in ijjQu → the matrix is the transpose, []TQ Finally, consider the multiplication of 33× matrices. Again, this follows the “beside each other” rule for the summed index. For example, [][]BA gives the 33× matrix {▲Problem 8} [] []()kj ik ij BA= BA , and the multiplication [][]BAT is written as [][]()kj ki ij BA= BAT. There is also the important identity [][]()[][]T T TAB BA= (7.1.13) Note also the following: Section 7.1 Solid Mechanics Part II Kelly 194(i) if there is no free index, as in iivu, there is one element (ii) if there is one free index, as in jijQu , it is a 13× (or 31×) matrix (iii) if there are two free indices, as in kj kiBA , it is a 33× matrix 7.1.3 Vector Transformation Rule Introduce two Cartesian coordinate systems with base vectors ie and ie′ and common origin o, Fig. 7.1.2. The vector u can then be expressed in two ways: ii ii u u e e u ′′== (7.1.14) Figure 7.1.2: a vector represented us ing two different coordinate systems Note that the ix′ coordinate system is obtained from the ix system by a rotation of the base vectors. Fig. 7.1.2 shows a rotation θ about the 3x axis (the sign convention for rotations is positive counterclockwise). Concentrating for the moment on the two dimensions 2 1xx−, from trigonometry (refer to Fig. 7.1.3), [] [] [] []2 2 1 1 2 12 122 11 cos sin sin cos e ee ee e u u u u uCP BD AB OBu u ′+′+′−′=++−=+= θθ θθ (7.1.15) and so 2 1 22 1 1 cos sinsin cos u u uu u u ′+′=′−′= θθθθ (7.1.16) 2x′2x 1x1x′ 1u2u′1u′ 2u θθ o1e′ 2e′u vector components in second coordinate system vector components in first coordinate system Section 7.1 Solid Mechanics Part II Kelly 195 Figure 7.1.3: geometry of the 2D coordinate transformation In matrix form, these transformation equations can be written as ⎥⎦⎤ ⎢⎣⎡ ′′ ⎥⎦⎤ ⎢⎣⎡−=⎥⎦⎤ ⎢⎣⎡ 21 21 cos sinsin cos uu uu θθθθ (7.1.17) The 22× matrix is called the transformation matrix or rotation matrix []Q. By pre- multiplying both sides of these equations by the inverse of []Q, []1−Q , one obtains the transformation equations transforming from []T 2 1uu to []T 2 1uu′′ : ⎥⎦⎤ ⎢⎣⎡ ⎥⎦⎤ ⎢⎣⎡ −=⎥⎦⎤ ⎢⎣⎡ ′′ 21 21 cos sinsin cos uu uu θθθθ (7.1.18) It can be seen that the components of []Q are the directions cosines , i.e. the cosines of the angles between the coordinate directions: ()j i j i ij xx Q ee′⋅=′ = , cos (7.1.19) It is straight forward to show that, in the full three dimensions, Fig. 7.1.4, the components in the two coordinate systems are also related through [][][] [][][]uQ uuQ u T=′ =′′= ′= KK jji ijij i uQ uuQ u Vector Transformation Rule (7.1.20) 2x′2x 1x1x′ 1u2u′1u′ 2u θθθ A BP D oC Section 7.1 Solid Mechanics Part II Kelly 196 Figure 7.1.4: two different coor dinate systems in a 3D space Orthogonality of the Transformation Matrix []Q From 7.1.20, it follows that [][][] [][][]uQQuQ u T= =′= ′= KK kkj ijjij i uQQuQ u (7.1.21) and so [][][]I QQ= =TKik kj ijQQδ (7.1.22) A matrix such as this for which [][]1 T −=Q Q is called an orthogonal matrix . Example Consider a Cartesian coordinate system with base vectors ie. A coordinate transformation is carried out with the new basis given by 3)3( 3 2)3( 2 1)3( 1 33)2( 3 2)2( 2 1)2( 1 23)1( 3 2)1( 2 1)1( 1 1 e e e ee e e ee e e e a a aa a aa a a ++=′++=′++=′ What is the transformation matrix? Solution The transformation matrix consists of the direction cosines j i j i ij xx Q ee′⋅=′ = ), cos( , so 1x2x 1x′2x′ 3x3x′u Section 7.1 Solid Mechanics Part II Kelly 197[] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =)3( 3)2( 3)1( 3)3( 2)2( 2)1( 2)3( 1)2( 1)1( 1 a a aa a aa a a Q ■ 7.1.4 Tensors The concept of the tensor is discussed in detail in Part III, where it is indispensable for the description of large-strain deformatio ns. For small deformations, it is not so necessary; the main purpose for introducing the tensor here (in a rather non-rigorous way) is that it helps to deepen one’s understanding of the concept of stress. A second-order tensor4 A may be defined as an operator that acts on a vector u generating another vector v, so that v uT=)( , or v Tu= Second-order Tensor (7.1.23) The second-order tensor T is a linear operator , by which is meant () Tb Ta baT +=+ … distributive ()()Ta aTαα= … associative for scalar α. In a Cartesian coordinate system, the tensor T has nine components and can be represented in the matrix form [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 33 32 3123 22 2113 12 11 T T TT T TT T T T The rule 7.1.23, which is expressed in symbolic notation, can be expressed in the index and matrix notation when T is referred to particular axes: [] [] [] vT u= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =321 33 32 3123 22 2113 12 11 321 vvv T T TT T TT T T uuu vTujij i (7.1.24) Again, one should be careful to distinguish between a tensor such as T and particular matrix representations of that tensor. The relation 7.1.23 is a tensor relation , relating vectors and a tensor and is valid in all coordinate systems; the matrix representation of this tensor relation, Eqn. 7.1.24, is to be sure valid in all coordinate systems, but the entries in the matrices of 7.1.24 depend on the coordinate system chosen. 4 to be called simply a tensor in what follows Section 7.1 Solid Mechanics Part II Kelly 198Note also that the transformation formulae for vectors, Eqn. 7.1.20, is not a tensor relation; although 7.1.20 looks similar to the tensor relation 7.1.24, the former relates the components of a vector to the components of the same vector in different coordinate systems, whereas (by definition of a tensor) the relation 7.1.24 relates the components of a vector to those of a different vector in the same coordinate system. For these reasons, the notation uQuij i′= in Eqn. 7.1.20 is more formally called element form , the ijQ being elements of a matrix rather than components of a tensor. This distinction between element form and index notation should be noted, but the term “index notation” is used for both tensor and matrix-specific manipulations in these notes. Example Recall the strain-displacement relations, Eqns. 1.2.19, which in full read ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂=∂∂=∂∂=∂∂= 23 32 23 13 31 13 12 21 1233 33 22 22 11 11 21,21,21, , xu xu xu xu xu xuxu xu xu ε ε εε ε ε (7.1.25) The index notation for these equations is ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= ij ji ijxu xu 21ε (7.1.26) This expression has two free indices and as such indicates nine separate equations. Further, with its two subscripts, ijε, the strain, is a tensor. It can be expressed in the matrix notation []( )( ) () () () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ∂∂ ∂∂+∂∂∂∂+∂∂∂∂+∂∂ ∂∂ ∂∂+∂∂∂∂+∂∂∂∂+∂∂ ∂∂ = 3 3 3 2 2 3 21 3 1 1 3 212 3 3 2 21 2 2 2 1 1 2 211 3 3 1 21 1 2 2 1 21 1 1 / / / / // / / / // / / / / x u x u x u x u x ux u x u x u x u x ux u x u x u x u x u ε 7.1.5 Tensor Transformation Rule Consider now the tensor definition 7.1.23 expressed in two different coordinate systems: [][][]{} [][] [] {}i jij ii jij i x vTux vTu ′ ′′=′′′=′= = inin vT uvT u (7.1.27) From the vector transformation rule 7.1.20, Section 7.1 Solid Mechanics Part II Kelly 199[][][] [][][]vQ vuQ u TT =′ =′=′ =′ jji ijji i vQvuQ u (7.1.28) Combining 7.1.27-28, [][][][][]vQT uQT T′= ′=kkj ij jji vQT uQ (7.1.29) and so [][][][][]vQTQ uT′= ′=kkj ij mi jji mi vQTQ uQQ (7.1.30) (Note that m j mj jji mi u u uQQ ==δ .) Comparing with 7.1.24, it follows that [][][][] [][][] []QTQ TQTQ T TT =′ =′′= ′= KK pq qj pi ijpq jq ip ij TQQ TTQQ T Tensor Transformation Rule (7.1.31) 7.1.6 Problems 1. Write the following in index notation: v, 1ev⋅, kev⋅. 2. Show that jiijbaδ is equivalent to ba⋅. 3. Evaluate or simplify the following expressions: (a) kkδ (b) ijijδδ (c) jk ijδδ 4. Show that [][]QuT is a 31× matrix with elements jijQu (write the matrices out in full) 5. Show that [] []()[][]T T TQu uQ= 6. Are the three elements of [][]uQ the same as those of [][]QuT? 7. What is the index notation for ()cba⋅? 8. Write out the 33× matrices []A and []B in full, i.e. in terms of ,,12 11AA etc. and verify that []kj ik ij BA= AB for 1 ,2== j i . 9. What is the index notation for (a) [][]TBA (b) [][] []vAvT (there is no ambiguity here, since [][]()[][][] []()vAv vAvT T= ) (c) [][] []BABT 10. The angles between the axes in two coordinate systems are given in the table below. 1x 2x 3x 1x′ o135 o60 o120 2x′ o90 o45 o45 3x′ o45 o60 o120 Construct the corresponding transformation matrix []Q and verify that it is orthogonal. Section 7.1 Solid Mechanics Part II Kelly 200 11. Consider a two-dimensional problem . If the components of a vector u in one coordinate system are ⎥⎦⎤ ⎢⎣⎡ 32 what are they in a second coordinate system, obtained from the first by a positive rotation of 30o? Sketch the two coordinate systems and the vector to see if your answer makes sense. 12. Consider again a two-dimensional problem wi th the same change in coordinates as in Problem 11. The components of a 2D tensor in the first system are ⎥⎦⎤ ⎢⎣⎡− 231 1 What are they in the second coordinate system? Section 7.2 Solid Mechanics Part II Kelly 2017.2 Analysis of Three Dime nsional Stress and Strain The concept of traction and stress was introduced and discussed in Part I, §3.1-3.5. For the most part, the discussion was confined to two-dimensional states of stress. Here, the fully three dimensional stress state is examined. There will be some repetition of the earlier analyses. 7.2.1 The Traction Vector and Stress Components Consider a traction vector t acting on a surface element, Fig. 7.2.1. Introduce a Cartesian coordinate system with base vectors ie so that one of the base vectors is a normal to the surface and the origin of the coordinate syst em is positioned at the point at which the traction acts. For example, in Fig. 7.1.1, the 3e direction is taken to be normal to the plane, and a superscript on t denotes this normal: 33 22 11)(3e e e tet t t ++= (7.2.1) Each of these components it is represented by ijσ where the first subscript denotes the direction of the normal and the second denotes the direction of the component to the plane. Thus the three components of the traction vector shown in Fig. 7.2.1 are 33 32 31 , ,σσσ : 3 33 2 32 131)(3e e e teσσσ ++= (7.2.2) The first two stresses, the components acting tangential to the surface, are shear stresses whereas 33σ, acting normal to the plane, is a normal stress. Figure 7.2.1: components of the traction vector Consider the three traction vectors )()()(3 2 1,,e e ettt acting on the surface elements whose outward normals are aligned with the three base vectors je, Fig. 7.2.2a. The three (or six) surfaces can be amalgamated into one diagram as in Fig. 7.2.2b. In terms of stresses, the traction vectors are )(3et 2x 1x3x )ˆ(nt 1e2e3e 32σ31σ33σ Section 7.2 Solid Mechanics Part II Kelly 202() () () 3 33 2 32 313 23 2 22 213 13 2 12 11 32 e e e te e e te e e t 1e1e1e1 σσσσσσσσσ ++=++=++= or () jijie teσ= (7.2.3) Figure 7.2.2: the three traction vectors acting at a point; (a) on mutually orthogonal planes, (b) the traction vectors illustrated on a box element The components of the three traction vectors, i.e. the stress components, can now be displayed on a box element as in Fig. 7.2.3. Note that the stress components will vary slightly over the surfaces of an elemental box of finite size. However, it is assumed that the element in Fig. 7.2.3 is small enough that th e stresses can be treated as constant, so that they are the stresses acting at the origin. Figure 7.2.3: the nine stress components with respect to a Cartesian coordinate system The nine stresses can be conveniently displayed in 33× matrix form: 21σ11σ31σ 12σ22σ32σ 23σ33σ 13σ3x 2x 1x1e()1et 01=x 02=x 03=x2e3e()3et()2et 1x2x3x 1x2x3x 1x2x3x 3x 2x2e3e 1e()1et()2et()3et 1x)a( )b( Section 7.2 Solid Mechanics Part II Kelly 203[] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = 33 32 3123 22 2113 12 11 σσσσσσσσσ σij (7.2.4) It is important to realise that, if one were to take an element at some different orientation to the element in Fig. 7.2.3, but at the same material particle , for example aligned with the axes 3 2 1,,xxx′′′ shown in Fig. 7.2.4, one would then have different tractions acting and the nine stresses would be different also. Th e stresses acting in this new orientation can be represented by a new matrix: [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ′′′′′′′′′ =′ 33 32 3123 22 2113 12 11 σσσσσσσσσ σij (7.2.5) Figure 7.2.4: the stress components with respect to a Cartesian coordinate system different to that in Fig. 7.2.3 7.2.2 Cauchy’s Law Cauchy’s Law , which will be proved below, states that the normal to a surface, iine n= , is related to the traction vector iite tn=)( acting on that surface, according to jji i n tσ= (7.2.6) Writing the traction and normal in vector form and the stress in 33× matrix form, [] [] [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 321 33 32 3123 22 2113 12 11 )( 3)( 2)( 1 )(, , nnn n ttt ti ij i σσσσσσσσσ σ nnn n (7.2.7) and Cauchy’s law in matrix notation reads 1x′2x′ 3x′11σ′12σ′ 13σ′ 31σ′ 33σ′32σ′22σ′ 21σ′ 23σ′ Section 7.2 Solid Mechanics Part II Kelly 204⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 321 33 23 1332 22 1231 21 11 )( 3)( 2)( 1 nnn ttt σσσσσσσσσ nnn (7.2.8) Note that it is the transpose stress matrix which is used in Cauchy’s law. Since the stress matrix is symmetric, one can express Cauchy’s law in the form jij i n tσ= Cauchy’s Law (7.2.9) Cauchy’s law is illustrated in Fig. 7.2.5; in this figure, positive stresses ijσ are shown. Figure 7.2.5: Cauchy’s Law; given the stresses and the normal to a plane, the traction vector acting on the plane can be determined Normal and Shear Stress It is useful to be able to evaluate the normal stress Nσ and shear stress Sσ acting on any plane, Fig. 7.2.6. For this purpose, note that the stress acting normal to a plane is the projection of )(nt in the direction of n, )(ntn⋅=Nσ (7.2.10) The magnitude of the shear stress acting on the surface can then be obtained from 22)( N S σ σ −=nt (7.2.11) 3x 2x 1xn()nt 23σ13σ 33σ12σ 22σ 32σ31σ21σ11σ()n 3t ()n 1t()n 2t Section 7.2 Solid Mechanics Part II Kelly 205 Figure 7.2.6: the normal and shear stress acting on an arbitrary plane through a point Example The state of stress at a point with respect to a Cartesian coordinates system 3210 xxx is given by: [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 12 32 213 12 ijσ Determine: (a) the traction vector acting on a plane through the point whose unit normal is 3 2 1 )3/2( )3/2( )3/1( e e e n − += (b) the component of this traction acting perpendicular to the plane (c) the shear component of traction on the plane Solution (a) From Cauchy’s law, ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 392 31 221 12 32 213 12 31 321 33 23 1332 22 1231 21 11 )( 3)( 2)( 1 nnn ttt σσσσσσσσσ nnn so that 3 2 1)(ˆ 3 )3/2( ee e tn−+−= . (b) The component normal to the plane is .4.29/22)3/2()3/2(3)3/1)(3/2()(≈=++ −=⋅= n tn Nσ (c) The shearing component of traction is ()() ( )[ ]()[] { } 1.2 1 32/12 9222 2 2 32 22)(≈ −−++−=−=N S σ σnt ■ 3x 2x 1xn()ntNσ Sσ Section 7.2 Solid Mechanics Part II Kelly 206 Proof of Cauchy’s Law Cauchy’s law can be proved using force equilibr ium of material elements. First, consider a tetrahedral free-body, with vertex at the origin, Fig. 7.2.7. It is required to determine the traction t in terms of the nine stress components (which are all shown positive in the diagram). Figure 7.2.7: proof of Cauchy’s Law The components of the unit normal, in, are the direction cosines of the normal vector, i.e. the cosines of the angles between the norma l and each of the coordinate directions: ()i i i n=⋅=en en, cos (7.2.12) Let the area of the base of the tetrahedran, with normal n, be SΔ. The area 1SΔ is then αcosSΔ , where α is the angle between the planes, as shown to the right of Fig. 7.2.7; this angle is the same as that between the vectors n and 1e, so Sn SΔ=Δ1 1 , and similarly for the other surfaces: Sn Si iΔ=Δ (7.2.13) The resultant surface force on the body, acting in the ix direction, is then Sn St S St Fjji i j ji i i Δ−Δ=Δ−Δ=∑ σ σ (7.2.14) For equilibrium, this expression must be zero, and one arrives at Cauchy’s law. Note: As proved in Part III, this result holds also in the general case of accelerating material elements in the presences of body forces. 3x 2x 1xn()nt 23σ13σ 33σ12σ 22σ 32σ31σ21σ11σ1SΔ 3SΔ••n 2SΔα 1e Section 7.2 Solid Mechanics Part II Kelly 2077.2.3 The Stress Tensor Cauchy’s law 7.2.9 is of the same form as 7.1.24 and so by definition the stress is a tensor. Denote the stress tensor in symbolic notation by σ. Cauchy’s law in symbolic form then reads nσt= (7.2.15) Further, the transformation rule for stress follows the general tensor transformation rule 7.1.31: [][][][] [][][] []QσQσQσQσ TT =′ =′′= ′ = KK pq qj pi ijpq jq ip ij QQQQ σ σσ σ Stress Transformation Rule (7.2.16) As with the normal and traction vectors, the components and hence matrix representation of the stress changes with coordinate system, as with the two different matrix representations 7.2.4 and 7.2.5. Howe ver, there is only one stress tensor σ at a point. Another way of looking at this is to note that an infinite number of planes pass through a point, and on each of these planes acts a traction vector, and each of these traction vectors has three (stress) components. All of these traction vectors taken together define the complete state of stress at a point. Example The state of stress at a point with respect to an 3210 xxx coordinate system is given by [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −− = 12 02 310 12 ijσ (a) What are the stress components with respect to axes 3210 xxx′′′ which are obtained from the first by a o45 rotation (positive counterclockwise) about the 2x axis, Fig. 7.2.8? (b) Use Cauchy’s law to evaluate the normal and shear stress on a plane with normal ()()3 1 2/1 2/1 e e n + = and relate your result with that from (a) Section 7.2 Solid Mechanics Part II Kelly 208 Figure 7.2.8: two different coordinate systems at a point Solution (a) The transformation matrix is []() ()() ()()() ()()() ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′ ′ ′′ ′ ′′ ′ ′ = 21 2121 21 3 3 2 3 1 33 2 2 2 1 23 1 2 1 1 1 001 00 , cos , cos , cos, cos , cos , cos, cos , cos , cos xx xx xxxx xx xxxx xx xx Qij and I QQ=T as expected. The rotated stress components are therefore ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− =⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ′′′′′′′′′ 23 21 2121 2321 23 2321 2121 21 21 2121 21 33 32 3123 22 2113 12 11 3001 00 12 02 310 12 00 100 σσσσσσσσσ and the new stress matrix is symmetric as expected. (b) From Cauchy’s law, the traction vector is ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −= ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 2121 2121 )( 3)( 2)( 1 2 0 12 02 310 12 nnn ttt so that ()()()3 2 1)(ˆ2/1 2/1 2 e e e tn+ −= . The normal and shear stress on the plane are 2/3)(=⋅= n tn Nσ and 2/3 )2/3(32 22)(=−=−=N S σ σnt The normal to the plane is equal to 3e′ and so Nσ should be the same as 33σ′ and it is. The stress Sσ should be equal to ()()2 322 31σσ ′+′ and it is. The results are 1x1e 2 2ee′=3e 3e′ 2 2xx′=3x 1e′ 1x′3x′ o45 o45 Section 7.2 Solid Mechanics Part II Kelly 209displayed in Fig. 7.2.9, in which the traction is represented in different ways, with components ())( 3)( 2)( 1 ,,n n nttt and ( )33 32 31 ,,σσσ ′′′ . Figure 7.2.9: traction and stresses acting on a plane Isotropic State of Stress Suppose the state of stress in a body is [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = = 000 0 0 00 00 0 σσσ δσσ σij ij (7.2.17) One finds that the application of the stress tensor transformation rule yields the very same components no matter what the new coordinate system{ ▲Problem 3}. In other words, no shear stresses act, no matter what the orienta tion of the plane through the point. This is termed an isotropic state of stress , or a spherical state of stress . One example of isotropic stress is the stress arising in a flui d at rest, which cannot support shear stress, in which case [][]Iσ p−= (7.2.18) where the scalar p is the fluid hydrostatic pressure . For this reason, an isotropic state of stress is also referred to as a hydrostatic state of stress . 7.2.4 Principal Stresses For certain planes through a material particle , there are traction vectors which act normal to the plane, as in Fig. 7.2.10. In this case the traction can be expressed as a scalar multiple of the normal vector, n tnσ=)(. 3en′= 2 2xx′= 1x′23 33=′=σσN )(nt 23 32−=′σ21 31=′σ 2)( 1=nt21 )( 2−=nt21 )( 3=ntSσ 1x Section 7.2 Solid Mechanics Part II Kelly 210 Figure 7.2.10: a purely normal traction vector From Cauchy’s law then, for these planes, ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = = 321 321 33 32 3123 22 2113 12 11 , , nnn nnn n ni jij σ σσσσσσσσσ σσσn nσ (7.2.19) This is a standard eigenvalue problem from Linear Algebra: given a matrix []ijσ, find the eigenvalues σ and associated eigenvectors n such that Eqn. 7.2.19 holds. To solve the problem, first re-write the equation in the form () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ =−=− 000 100010001 ,0 ,321 33 32 3123 22 2113 12 11 nnn nj ij ij σ σσσσσσσσσ σδσ σ 0nIσ (7.2.20) or ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−− 000 321 33 32 3123 22 2113 12 11 nnn σσσσσσσσσσσσ (7.2.21) This is a set of three homogeneous eq uations in three unknowns (if one treats σ as known). From basic linear algebra, this system has a solution (apart from 0 =in ) if and only if the determinant of the coefficient matrix is zero, i.e. if 0 det) det( 33 32 3123 22 2113 12 11 = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−− =− σσσσσσσσσσσσ σIσ (7.2.22) Evaluating the determinant, one has the following cubic characteristic equation of the stress tensor σ, 03 22 13=−+− I I Iσσσ Characteristic Equation (7.2.23) and the principal scalar invariants of the stress tensor are no shear stress – only a normal component to the traction nn tnσ=)( Section 7.2 Solid Mechanics Part II Kelly 211 31 23 122 12 332 31 222 23 11 33 22 11 32 312 232 12 11 33 33 22 22 11 233 22 11 1 2σσσσσσσσσσσσσσσσσσσσσσσσ +−−− =−−−++=++= III (7.2.24) (3I is the determinant of the stress matrix.) The characteristic equation 7.2.23 can now be solved for the eigenvalues σ and then Eqn. 7.2.21 can be used to solve for the eigenvectors n. Now another theorem of linear algebra states th at the eigenvalues of a real (that is, the components are real), symmetric matrix (such as the stress matrix) are all real and further that the associated eigenvectors are mutually orthogonal. This means that the three roots of the characteristic equation are real and that the three associated eigenvectors form a mutually orthogonal system. This is illustrate d in Fig. 7.2.11; the eigenvalues are called principal stresses and are labelled 3 2 1 ,,σσσ and the three corresponding eigenvectors are called principal directions , the directions in which the principal stresses act. The planes on which the principal stresses act (to which the principal directions are normal) are called the principal planes . Figure 7.2.11: the three principal stresses acting at a point and the three associated principal directions 1, 2 and 3 Once the principal stresses are found, as mentioned, the principal directions can be found by solving Eqn. 7.2.21, which can be expressed as 0 ) (0 ) (0 ) ( 3 33 2 32 1313 23 2 22 1213 13 2 12 1 11 =−++=+−+=++− n n nn n nn n n σσσσσσσσσσσσ (7.2.25) Each principal stress value in this equation gives rise to the three components of the associated principal direction vector, 3 2 1,,nnn . The solution also requires that the magnitude of the normal be specified: for a unit vector, 1=⋅nn . The directions of the normals are also chosen so that they form a right-handed set. 11σ 3σ2σ 32 Section 7.2 Solid Mechanics Part II Kelly 212Example The stress at a point is given with respect to the axes 321xxOx by the values [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−−= 1 12 012 6 00 0 5 ijσ . Determine (a) the principal values, (b) the principal directions (and sketch them). Solution: (a) The principal values are the solution to the characteristic equation 0) 15)( 5)( 10( 1 12 012 6 00 0 5 =+−+−= −−−−−− σσσ σσσ which yields the three principal values 15 ,5 ,103 2 1 −=== σσσ . (b) The eigenvectors are now obtained from Eqn. 7.2.25. First, for 101=σ , 0 9 12 00 12 16 00 0 0 5 3 2 13 2 13 2 1 =−−=−−=++− n n nn n nn n n and using also the equation 12 32 22 1 =++ n n n leads to 3 2 1 )5/4( )5/3( e e n +−= . Similarly, for 52=σ and 153−=σ , one has, respectively, 0 4 12 00 12 11 00 0 0 0 3 2 13 2 13 2 1 =−−=−−=++ n n nn n nn n n and 0 16 12 00 12 9 00 0 0 20 3 2 13 2 13 2 1 =+−=−+=++ n n nn n nn n n which yield 1 2e n= and 3 2 3 )5/3( )5/4( e e n + = . The principal directions are sketched in Fig. 7.2.12. Note that the three components of each principal direction, 3 2 1,,nnn , are the direction cosines: the cosines of the angles between that principal direction and the three coordinate axes. For example, for 1σ with 5 /4 ,5/3 ,03 2 1 =−== n n n , the angles made with the coordinate axes 3 2 1,,xxx are, respectively, 0, 127o and 37o. Figure 7.2.12: principal directions ■ 3x 1x2x3ˆn1ˆn 2ˆno37 Section 7.2 Solid Mechanics Part II Kelly 213Invariants The principal stresses 3 2 1 ,,σσσ are independent of any coordinate system; the 3210 xxx axes to which the stress matrix in Eqn. 7.2.19 is referred can have any orientation – the same principal stresses will be found from the eigenvalue analysis. This is expressed by using the symbolic notation for the problem: n nσσ= , which is independent of any coordinate system. Thus the principal stresses are intrinsic properties of the stress state at a point. It follows that the functions 3 2 1,,III in the characteristic equation Eqn. 7.2.23 are also independent of any coordinate system, and hence the name principal scalar invariants (or simply invariants ) of the stress. The stress invariants can also be written neatly in terms of the principal stresses: 321 313 32 21 23 2 1 1 σσσσσσσσσσσσ =++=++= III (7.2.26) Also, if one chooses a coordinate system to coincide with the principal directions, Fig. 7.2.12, the stress matrix takes the simple form [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = 321 0 00 00 0 σσσ σij (7.2.27) Note that when two of the principal stresses are equal, one of the principal directions will be unique, but the other two will be arbitrary – one can choose any two principal directions in the plane perpendicular to the uniquely determined direction, so that the three form an orthonormal set. This stress state is called axi-symmetric . When all three principal stresses are equal, one has an isotropic state of stress, and all directions are principal directions – the stress matrix has the form 7.2.27 no matter what orientation the planes through the point. Example The two stress matrices from the Example of §7.2.3, describing the stress state at a point with respect to different coordinate systems, are [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− = 12 02 310 12 ijσ , [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− =′ 2/3 2/1 2/12/1 3 2/32/1 2/3 2/3 ijσ The first invariant is the sum of the normal stresses, the diagonal terms, and is the same for both as expected: 6 3 13223 23 1 =++=++=I The other invariants can also be obtained from either matrix, and are 3 ,63 2 −== I I ■ Section 7.2 Solid Mechanics Part II Kelly 2147.2.5 Maximum and Min imum Stress Values Normal Stresses The three principal stresses include the maximum and minimum normal stress components acting at a point. To prove this, first let 3 2 1,,eee be unit vectors in the principal directions . Consider next an arbitrary unit normal vector iine n= . From Cauchy’s law (see Fig. 7.2.13 – the stress matrix in Cauchy’s law is now with respect to the principal directions 1, 2 and 3), the no rmal stress acting on the plane with normal n is ()ijij N N nnσσ σ = ⋅=⋅= ,)(nnσ n tn (7.2.28) Figure 7.2.13: normal stress acting on a plane defined by the unit normal n Thus 2 332 222 11 321 321 321 0 00 00 0 n n n nnn nnn N σσσ σσσ σ ++= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = (7.2.29) Since 12 32 22 1 =++ n n n and, without loss of generality, taking 3 2 1σσσ≥≥ , one has ( )N n n n n n n σσσσ σσ =++≥++=2 332 222 112 32 22 1 1 1 (7.2.30) Similarly, ( )32 32 22 1 32 332 222 11 σ σσσσσ ≥++≥++= n n n n n nN (7.2.31) Thus the maximum normal stress acting at a point is the maximum principal stress and the minimum normal stress acting at a point is the minimum principal stress. 3 2 1n()ntNσ principal directions Section 7.2 Solid Mechanics Part II Kelly 215Shear Stresses Next, it will be shown that the maximum shearing stresses at a point act on planes oriented at 45 o to the principal planes and that they have magnitude equal to half the difference between the principal stresses. First, again, let 3 2 1,,eee be unit vectors in the principal directions and consider an arbitrary unit normal vector iine n= . The normal stress is given by Eqn. 7.2.29, 2 332 222 11 n n nN σσσσ ++= (7.2.32) Cauchy’s law gives the components of the traction vector as ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 332211 321 321 )( 1)( 1)( 1 0 00 00 0 nnn nnn ttt σσσ σσσ nnn (7.2.33) and so the shear stress on the plane is, from Eqn. 7.2.11, () ( )22 332 222 112 32 32 22 22 12 12n n n n n nS σσσσσσσ ++−++= (7.2.34) Using the condition 12 32 22 1 =++ n n n to eliminate 3n leads to () () () ()[ ]2 32 2 3 22 1 3 12 32 22 32 22 12 32 12σσσσσσσσσσσ +−+−−+−+−= n n n nS (7.2.35) The stationary points are now obtained by equating the partial derivatives with respect to the two variables 1n and 2n to zero: ()() () () []{} ()() () () []{} 0 20 2 2 2 3 22 1 3 1 3 2 3 2 2 222 2 3 22 1 3 1 3 1 3 1 1 12 =−+−−−−=∂∂=−+−−−−=∂∂ n n nnn n nn SS σσσσσσσσσσσσσσσσσσ (7.2.36) One sees immediately that 02 1==n n (so that 13±=n ) is a solution; this is the principal direction 3e and the shear stress is by definition zero on the plane with this normal. In this calculation, the component 3n was eliminated and 2 Sσ was treated as a function of the variables ),(2 1nn . Similarly, 1n can be eliminated with ) ,(3 2nn treated as the variables, leading to the solution 1en=, and 2n can be eliminated with ) ,(3 1nn treated as the variables, leading to the solution 2en=. Thus these solutions lead to the minimum shear stress value 02=Sσ . A second solution to Eqn. 7.2.36 can be seen to be 2/1 ,02 1 ±==n n (so that 2/13±=n ) with corresponding shear stress values ()2 3 2 41 2σσσ −=S . Two other Section 7.2 Solid Mechanics Part II Kelly 216solutions can be obtained as described earlier, by eliminating 1n and by eliminating 2n. The full solution is listed below, and these are evidently the maximum (absolute value of the) shear stresses acting at a point: 2 11 33 2 21,0, 21, 2121, 21,0, 2121, 21, 21,0 σσσσσσσσσ −=⎟ ⎠⎞⎜ ⎝⎛±±=−=⎟ ⎠⎞⎜ ⎝⎛±±=−=⎟ ⎠⎞⎜ ⎝⎛±±= SSS nnn (7.2.37) Taking 3 2 1σσσ≥≥ , the maximum shear stress at a point is ()3 1 max21σστ −= (7.2.38) and acts on a plane with normal oriented at 45o to the 1 and 3 principal directions. This is illustrated in Fig. 7.2.14. Figure 7.2.14: maximum shear stress at a point Example Consider the stress state examined in the Example of §7.2.4: [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−−= 1 12 012 6 00 0 5 ijσ The principal stresses were found to be 15 ,5 ,103 2 1 −=== σσσ and so the maximum shear stress is ()225 21 3 1 max =−=σστ One of the planes upon which they act is shown in Fig. 7.2.15 (see Fig. 7.2.12) 13 maxτ maxτprincipal directions Section 7.2 Solid Mechanics Part II Kelly 217 Figure 7.2.15: maximum shear stress ■ 7.2.6 Mohr’s Circles of Stress The Mohr’s circle for 2D stress states was discussed in Part I, §3.5.4. For the 3D case, following on from section 7.2.5, one has the conditions 12 32 22 12 32 32 22 22 12 12 22 332 222 11 =++++=+++= n n nn n nn n n N SN σσσσσσσσσ (7.2.39) Solving these equations gives ()() () () () () () () () () () ()2 3 1 32 2 1 2 31 2 3 22 1 3 2 23 1 2 12 3 2 2 1 σσσσσσσσσσσσσσσσσσσσσσσσσσσ −−+−−=−−+−−=−−+−−= S N NS N NS N N nnn (7.2.40) Taking 3 2 1σσσ≥≥ , and noting that the squares of the normal components must be positive, one has that ()() () () () () 000 2 2 12 1 32 3 2 ≥+−−≤+−−≥+−− S N NS N NS N N σσσσσσσσσσσσσσσ (7.2.41) and these can be re-written as () [ ]()[] ()[] ()[] ()[] ()[]2 2 1 212 2 1 21 22 3 1 212 3 1 21 22 3 2 212 3 2 21 2 σσ σσσσσσ σσσσσσ σσσσ −≥+−+−≤+−+−≥+−+ N SN SN S (7.2.42) 3x 1x2x3ˆn1ˆn 2ˆno37maxτ Section 7.2 Solid Mechanics Part II Kelly 218If one takes coordinates ()S Nσσ, , the equality signs here represent circles in ()S Nσσ, stress space, Fig. 7.2.16. Each point ()S Nσσ, in this stress space represents the stress on a particular plane through the material particle in question. Admissible ()S Nσσ, pairs are given by the conditions Eqns. 7.2.42; they must lie inside a circle of centre ()() 0,3 1 21σσ+ and radius ()3 1 21σσ− . This is the large circle in Fig. 7.2.16. The points must lie outside the circle with centre ()() 0,3 2 21σσ+ and radius ()3 2 21σσ− and also outside the circle with centre ()() 0,2 1 21σσ+ and radius ()2 1 21σσ− ; these are the two smaller circles in the figure. Thus the admissible points in stress space lie in the shaded region of Fig. 7.2.16. Figure 7.2.16: admissible points in stress space 7.2.7 Three Dimens ional Strain The strain ijε, in symbolic form ε, is a tensor and as such it follows the same rules as for the stress tensor. In particular, it follows the general tensor transformation rule 7.2.16; it has principal values ε which satisfy the characteristic equation 7.2.23 and these include the maximum and minimum normal strain at a point. There are three principal strain invariants given by 7.2.24 or 7.2.26 and the maximum shear strain occurs on planes oriented at 45o to the principal directions. 7.2.8 Problems 1. The state of stress at a point with respect to a 3210 xxx coordinate system is given by [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−− = 2 1 21 012 12 ijσ Use Cauchy’s law to determine the tract ion vector acting on a plane trough this point whose unit normal is 3/) (3 2 1 e ee n ++= . What is the normal stress acting on the plane? What is the shear stress acting on the plane? 2. The state of stress at a point with respect to a 3210 xxx coordinate system is given by [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −= 2 02013231 ijσ NσSσ • •• 3σ2σ1σ Section 7.2 Solid Mechanics Part II Kelly 219What are the stress components with respect to axes 3210 xxx′′′ which are obtained from the first by a o45 rotation (positive counterclockwise) about the 3x axis 3. Show, in both the index and matrix notation, that the components of an isotropic stress state remain unchanged under a coordinate transformation. 4. Consider a two-dimensional problem. Th e stress transformation formulae are then, in full, ⎥⎦⎤ ⎢⎣⎡− ⎥⎦⎤ ⎢⎣⎡ ⎥⎦⎤ ⎢⎣⎡ −=⎥⎦⎤ ⎢⎣⎡ ′′′′ θθθθ σσσσ θθθθ σσσσ cos sinsin cos cos sinsin cos 22 2112 11 22 2112 11 Multiply the right hand side out and use the fact that the stress tensor is symmetric (21 12σσ= - not true for all tensors). What do you get? Look familiar? 5. The state of stress at a point with respect to a 3210 xxx coordinate system is given by [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −− = 1 0 002/52/102/1 2/5 ijσ Evaluate the principal stresses and the principal directions. What is the maximum shear stress acting at the point? Section 7.3 Solid Mechanics Part II Kelly 2207.3 Governing Equations of Three Dimensional Elasticity 7.3.1 Hooke’s Law and Lamé’s Constants Linear elasticity was introduced in Part I, §4.2. The three-dimensional Hooke’s law for isotropic linear elastic solids (Part I, Eqns. 4.2.9) can be expressed in index notation as ij kkij ij μεελδσ 2+= (7.3.1) where (see also Part I, Eqns. 6.2.21) () ( ) ()νμνννλ+=−+=12,21 1E E (7.3.2) are the Lamé constants ( μ is the Shear Modulus). Eqns. 7.3.1 can be inverted to obtain {▲Problem 1} ()kk ij ij ij σδμλμλσμε2 32 21 +−= (7.3.3) 7.3.2 Navier’s Equations The governing equations of el asticity are Hooke’s law (E qn. 7.3.1), the equations of motion, Eqn. 1.1.9 (see Eqns. 7.1.10-11), i i jija bxρσ=+∂∂ (7.3.4) and the strain-displacement re lations, Eqn. 1.2.19 (see Eqns. 7.1.25-26), ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= ij ji ijxu xu 21ε (7.3.5) Substituting 7.3.5 into 7.3.1 and then into 7.3.4 leads to the 3D Navier’s equations {▲Problem 2} ()i i j ji i jja bxxu xxuρ μ μλ =+∂∂∂+∂∂∂+2 2 Navier’s Equations (7.3.6) These reduce to the 2D plane strain Navier’s equations, Eqns. 3.1.4, by setting 03=u and 0 /3=∂∂x . They do not reduce to the plane stress equations since the latter are only an Section 7.3 Solid Mechanics Part II Kelly 221approximate solution to the equations of elasti city which are valid only in the limit as the thickness of the thin plate of plane stress tends to zero. 7.3.3 Problems 1. Invert Eqns. 7.3.1 to get 7.3.3. 2. Derive the 3D Navier’s equations from 7.3.6 from 7.3.1, 7.3.4 and 7.3.5 Section 7.4 Solid Mechanics Part II Kelly 2227.4 Elastodynamics 7.4.1 Propagation of Waves in Elastic Solids When a stress wave travels through a material, it causes material par ticles to displace by u. It can be shown that any vector u can be written in the form1 a u curl+∇=φ (7.4.1) where φ is a scalar potential and a is a vector. These two terms in the displacement field can be examined separately. The most general displacement field can be obtained by adding both solutions together. Irrotational Waves First looking at the scalar potential term, suppose that the displacement is given by φ∇=u . If one can find a scalar φ such that φ∇=u , then it follows that 0u= curl , or 0 e e ee e e u =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂=∂∂∂∂∂∂= 3 21 12 2 13 31 1 32 233 2 13 2 13 2 1 / / / curl xu xu xu xu xu xuu u ux x x (7.4.2) Thus each of the terms inside the brackets is zero. But these terms represent rotations of material particles (see Eqns. 1.1.20). For example, as illustrated in Fig. 7.4.1, ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−∂∂= 21 12 321 xu xuω (7.4.3) Figure 7.4.1: a rotation 1 from the Helmholtz theory 1x2x 2u1u− 3ω3ω Section 7.4 Solid Mechanics Part II Kelly 223 Thus 0u= curl can be interpreted as no rotation of material particles. A small element of material can still undergo normal and shear strain, but the element will not rotate as a rigid body in space. Taking the displacement field φ∇=u , writing it in index notation, j j x u∂∂= /φ , and substituting into Navier’s equations, leads to the three-dimensional wave equation: () () ( ) ννρν ρμλ 21 11 2,1 22 22 −+−=+=∂∂=∂∂∂ Ectu c xxu Li L k ki (7.4.4) This displacement field thus corresponds to stress waves travelling at speed Lc, causing material to strain but not to rotate. These irrotational waves are also called waves of dilatation . Equivoluminal Waves Consider now the displacement field a ucurl= . If one can find a vector a such that a ucurl= , then it follows that 0=⋅∇u , or VVxu xu xu Δ=++=∂∂+∂∂+∂∂=⋅∇ 33 22 1133 22 11 εεεu (7.4.5) Thus the condition that the di splacement field be divergence -free implies that there is no volume change . There can be normal strains only so long as their sum is zero. Taking k k kk x u∂∂= /ε and substituting into Navier’s equations then leads immediately to 22 2 tu xxui k ki ∂∂=∂∂∂ρ μ (7.4.6) or the three-dimensional wave equation: ()νρρμ +== ∂∂=∂∂∂ 12,1 22 22Ec tu c xxu Ti T k ki (7.4.7) This displacement field thus corresponds to stress waves travelling at speed Tc, causing material to shear. These equivoluminal waves are also called shear waves or waves of distortion . In summary, when an event such as an explosion occurs, two different types of wave emerge, irrotational waves which result in irrotational displacement fields, and equivoluminal waves which result in equivoluminal displacements. These waves travel at different speeds. Section 7.4 Solid Mechanics Part II Kelly 224 7.4.2 Plane Waves At a sufficient distance from any initial disturbance, a stress wave will travel in a plane. It can be assumed that all material particles will displace either parallel to the direction of wave propagation ( longitudinal waves ) or perpendicular to this direction ( transverse waves ). Let the wave travel in the 1x direction. Irrotational (p / longitudinal) Plane Waves Consider particles which displace in the direction of wave propagation according to 1 1 1 ),( e u txu= . This is an irrotational wave since 0u= curl , and the stress wave is governed by the one-dimensional wave equation 212 2 2 1121 tu c xu L∂∂=∂∂ (7.4.8) These longitudinal plane waves are also called p-waves2. Figure 7.4.2: a longitudinal wave Equivoluminal (s / transverse / shear) Plane Waves Consider particles which displace according to 2 1 2 ),( e u txu= . This is an equivoluminal wave since 0=⋅∇u , and the stress wave is governed by the one-dimensional wave equation 222 2 2 1221 tu c xu T∂∂=∂∂ (7.4.9) 2 p stands for “primary” 1x wavefront compression / rarefaction Section 7.4 Solid Mechanics Part II Kelly 225These transverse/shear waves are also called s-waves3. Figure 7.4.3: a transverse wave 7.4.3 Vibration Analysis A vibration analysis can be carried out in exactly the same way as in Chapter 2, only the wave speeds in the 1D wave equations 7.4.8 and 7.4.9 are now different from the 1D speed ρ/E . The particular solutions, forced vibration and resonance theory of Chapter 2 can again be applied here. The analysis here is appropriate for thin plates “infinitely wide” in the 3 2,xx directions, Fig. 7.4.4. The figure shows longitudinal vibration, but one can also have transverse vibration where the particles displace perpendicular to the 1x axis. Figure 7.4.4: stretch vibration of a plate 7.4.4 Waves at Boundaries Plane waves exist in unbounded elastic continua. In a finite body, a plane wave will be reflected when it hits a free surface. In this case, one needs to solve Navier’s equations 3 s stands for “secondary” l1x1x00 == SNσσ Section 7.4 Solid Mechanics Part II Kelly 226subject to the boundary conditions of zero normal and shear stress at the free surface. Waves of both types will in general be reflected for any single type of incident wave. Similarly, when a wave meets an interface between two different materials, there will be reflection and refraction. The boundary c onditions are that the displacements are continuous and the normal and shear stresses are continuous, Fig. 7.4.5 Figure 7.4.5: reflection and refraction of a wave at an interface 7.4.5 Waves at Boundaries The waves discussed thus far are body waves . When a free surface exists, for example the surface of the earth, another type of wave motion is possible; these are the Rayleigh waves and travel along the surface very much like water waves. It can be shown that the speed of Rayleigh waves is between 90% and 95% of Tc, depending on the value of Poisson’s ratio. Similar types of waves can propagate along the interface between two different materials. 7.4.6 Problems 1. Consider the motion () 0 ,0 ,2sin3 2 1 1 ==− = u uctxlu uπ, What are the strains in the material? What are the corresponding stresses? What is the volume change in the material? What is the name (or names) given to the type of wave which causes this kind of motion? 2. Consider the motion () 0 ,2sin ,03 1 2 1 =− == uctxlu u uπ, What are the strains in the material? What are the corresponding stresses? What is the volume change in the material? What is the name (or names) given to the type of wave which causes this kind of motion? 3. Derive an expression for the ratio T Lcc/ in terms of the material’s Poisson’s ratio only. Which is the faster, the longitudinal or transverse wave? )1( )1( )1(, ,ρν E )2( )2( )2(, ,ρν E)2( )1( )2( )1(,y y x x u u u u = = )2( )1( )2( )1(,S S N N σσσσ = = Section 7.4 Solid Mechanics Part II Kelly 227 4. Show that the motion () () ctxlpx u u u u − ===1 2 3 2 12cos cos ,0 ,0π, is equivoluminal. 5. Consider the motion ()() [] 0 ,0 , sin sin3 2 3 3 1 ==+ +− = u uctx ctx u u βα β (i) what kind of elastic stress wave does this involve ? (Sketch the plane of the wave and its direction of propagation.) (ii) what are the strains and stresses. (iii) use the equations of motion to determine the wave speed. Is it what you expected? (iv) Suppose that the plane 03=x is a free surface. Determine α. (v) Suppose also that h x=3 is a free surface. Determine β. 6. Consider a plate with left face ()01=x subjected to a forced displacement 1 sin e u tΩ=α and the right face ()lx=1 free. (i) find the “thickness-stretch” vibration of the plate. What are the natural frequencies? (ii) When does resonance occur? 7. Consider a plate with left face ()01=x subjected to a traction 2 cos e t tΩ−=α and the right face () lx=1 fixed, as shown in the figure below. (i) find the “thickness-shear” vibration of the plate. What are the natural frequencies? (ii) When does resonance occur? [note: assume a displacement 2 1 2 ),( e u txu= ; as with transverse waves, this will satisfy the 1-d wave equation with c being the transverse wave speed. Use the traction to obtain an expression for the shear stress 12σ over the left hand face. When applying the stress boundary condition, you will need the strain-displacement expression and stress-strain law, fixed 1x2x 12σ Section 7.4 Solid Mechanics Part II Kelly 228⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂+∂∂= 12 21 1221 xu xuε , 12 1221σμε= 8. Consider the case of ( )3 2sin cos e e u t tΩ+Ω=α over the left face () 01=x with the right face () lx=1 fixed. Derive an expression for the particular solution and show that it represents circular motion of the particles in the 3 2x x− plane. [hint: evaluate the particular solutions for 2u and 3u separately and then show that 2 2 32 2 r u u=+ for some r (independent of time)] Section 8.1 Solid Mechanics Part II Kelly 2418.1 Introduction to Plasticity 8.1.1 Introduction The theory of linear elasticity is useful for modelling materials which undergo small deformations and which return to their original configuration upon removal of load. Almost all real materials will undergo some permanent deformation, which remains after removal of load. With metals, significant permanent deformations will usually occur when the stress reaches some critical value, called the yield stress , a material property. Elastic deformations are termed reversible; the energy expended in deformation is stored as elastic strain energy and is completely recovered upon load removal. Permanent deformations involve the dissipation of energy; such processes are termed irreversible , in the sense that the original state can be achieved only by the expenditure of more energy. The classical theory of plasticity grew out of the study of metals in the late nineteenth century. It is concerned with materials wh ich initially deform elastically, but which deform plastically upon reaching a yield stress. In meta ls and other crystalline materials the occurrence of plastic deformations at the micro-scale level is due to the motion of dislocations and the migration of grain boundaries on the micro-level. In sands and other granular materials plastic flow is due both to the irreversible rearrangement of individual particles and to the irreversible crushing of individual particles. Similarly, compression of bone to high stress levels will lead to particle crushing. The deformation of micro- voids and the development of micro-cracks is also an important cause of plastic deformations in materials such as rocks. A good part of the discussion in what follows is concerned with the plasticity of metals; this is the ‘simplest’ type of plasticity and it serves as a good background and introduction to the modelling of plasticity in other material-types. There are two broad groups of metal plasticity problem which are of interest to the engineer and analyst. The first involves relatively small plastic strains, of ten of the same order as the elastic strains which occur. Analysis of problems involving small plastic strains allows one to design structures optimally, so that they will not fail when in service, but at the same time are not stronger than they really need to be. In th is sense, plasticity is seen as a material failure 1. The second type of problem involves very large strains and deformations, so large that the elastic strains can be disregarded. These problems occur in the analysis of metals manufacturing and forming processes, which can involve extrusion, drawing, forging, rolling and so on. In these latter-type problems, a simplified model known as perfect plasticity is usually employed (see below), and use is made of special limit theorems which hold for such models. Plastic deformations are normally rate independent, that is, the stresses induced are independent of the rate of deformation (or rate of loading). This is in marked 1 two other types of failure, brittle fracture , due to dynamic crack growth, and the buckling of some structural components, can be modelled reasonably accurately using elasticity theory (see, for example, Part I, §6.1, Part II, §5.3) Section 8.1 Solid Mechanics Part II Kelly 242contrast to classical Newtonian fluids for example, where the stress levels are governed by the rate of deformation through the viscosity of the fluid. Materials commonly known as “plastics” are not plastic in the sense described here. They, like other polymeric materials, exhibit viscoelastic behaviour where, as the name suggests, the material response has both elastic and viscous components. Due to their viscosity, their response is, unlike the plastic materials, rate-dependent . Further, although the viscoelastic material s can suffer irrecoverable deformation, they do not have any critical yield or thre shold stress, which is the characteristic property of plastic behaviour. When a material undergoes plastic deformations, i.e. irrecoverable and at a critical yield stress, and these effects are rate dependent, the material is referred to as being viscoplastic . Plasticity theory began with Tresca in 1864, when he undertook an experimental program into the extrusion of metals and published hi s famous yield criterion discussed later on. Further advances with yield criteria and plas tic flow rules were made in the years which followed by Saint-Venant, Levy, Von Mises, Hencky and Prandtl. The 1940s saw the advent of the classical theory; Prager, Hill, Drucker and Koiter amongst others brought together many fundamental aspects of the theory into a single framework. The arrival of powerful computers in the 1980s and 1990s provided the impetus to develop the theory further, giving it a more rigorous foundation based on thermodynamics principles, and brought with it the need to consider many numerical and computational aspects to the plasticity problem. 8.1.2 Observations fr om Standard Tests In this section, a number of phenomena observed in the material testing of metals will be noted. Some of these phenomena are simplified or ignored in some of the standard plasticity models discussed later on. At issue here is the fact that any model of a component with complex geometry, loaded in a complex way and undergoing plastic deformation, must involve material parameters which can be obtained in a straight forward manner from simple laboratory tests, such as the tension test described next. The Tension Test Consider the following key experiment, the tensile test , in which a small, usually cylindrical, specimen is gripped and stretched, usually at some given rate of stretching. The force required to hold the specimen at a given stretch is recorded, Fig. 8.1.1. If the material is a metal, the deformation remains elastic up to a certain force level, the yield point of the material. Beyond this point, permanent plastic deformations are induced. On unloading only the elastic deformation is recovered and the specimen will have undergone a permanent elongation (and consequent lateral contraction). In the elastic range the force-displacement behaviour for most engineering materials (metals, rocks, plastics, but not soils) is linear. After passing the elastic limit (point A in Fig. 8.1.1), further increases in load are usually required to maintain an increase in displacement; this phenomenon is known as work-hardening or strain-hardening . In Section 8.1 Solid Mechanics Part II Kelly 243some cases the force-displacement curve decreases, as in some soils; the material is said to be softening . If the specimen is unloaded from a plastic state ( B) it will return along the path BC shown, parallel to the original elastic line. This is elastic recovery . What remains is the permanent plastic deformation. If the material is now loaded again, the force-displacement curve will re-trace the unloading path CB until it again reaches the plastic state. Further increases in stress will cause the curve to follow BD. Two important observations concerning the above tension test are the following: (1) after the onset of plastic deformation, the ma terial will be seen to undergo negligible volume change, that is, it is incompressible . (2) the force-displacement curve is more or less the same regardless of the rate at which the specimen is stretched (at least at moderate temperatures). Figure 8.1.1: force/displacement curve for the tension test Nominal and True Stress and Strain There are two different ways of describing the force F which acts in a tension test. First, normalising with respect to the original cross sectional area of the tension test specimen 0A, one has the nominal stress or engineering stress , 0AF n=σ (8.1.1) Alternatively, one can normalise with respect to the current cross-sectional area A, leading to the true stress , AF=σ (8.1.2) elastic loading hardening 0ABD Cunloadload plastic deformation elastic deformation force displacement Yield point Section 8.1 Solid Mechanics Part II Kelly 244in which F and A are both changing with time. For very small elongations, within the elastic range say, the cross-sectional area of the material undergoes negligible change and both definitions of stress are more or less equivalent. Similarly, one can describe the deformatio n in two alternative ways. Denoting the original specimen length by 0l and the current length by l, one has the engineering strain 00 lll−=ε (8.1.3) Alternatively, the true strain accounts for the fact that the “original length” is continually changing; a small change in length dl leads to a strain increment ldl d /=ε and the total strain is defined as the accumulation of these increments: ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛==∫ 0ln 0ll ldll ltε (8.1.4) The true strain is also called the logarithmic strain or Hencky strain . Again, at small deformations, the difference between these two strain measures is negligible. The true strain and engineering strain are related through ()ε ε+=1lnt (8.1.5) Using the assumption of constant volume for plastic deformation and ignoring the very small elastic volume changes, one has also { ▲Problem 3} 0ll nσσ= . (8.1.6) The stress-strain diagram for a tension test can now be described using the true stress/strain or nominal stress/strain definitions, as in Fig. 8.1.2. The shape of the nominal stress/strain diagram, Fig. 8.1.2a, is of course the same as the graph of force versus displacement (change in length) in Fig. 8.1.1. A here denotes the point at which the maximum force the specimen can withstand has been reached. The nominal stress at A is called the Ultimate Tensile Strength (UTS) of the material. After this point, the specimen “necks”, with a very rapid reduction in cross-sectional area somewhere about the centre of the specimen until the specimen ruptures, as indicated by the asterisk. Note that, during loading into the plastic region, the yield stress increases . For example, if one unloads and re-loads (as in Fig. 8.1.1) , the material stays elastic up until a stress higher than the original yield stress Y. In this respect, the stress-strain curve can be regarded as a yield stress versus strain curve. Section 8.1 Solid Mechanics Part II Kelly 245 Figure 8.1.2: typical stress/strain curves; (a) engineering stress and strain, (b) true stress and strain Compression Test A compression test will lead to similar results as the tensile stress. The yield stress in compression will be approximately the same as (the negative of) the yield stress in tension. If one plots the true stress versus true strain curve for both tension and compression (absolute values for the compression), the two curves will more or less coincide. This would indicate that the behaviour of the material under compression is broadly similar to that under tension. If one were to use the nominal stress and strain, then the two curves would not coincide; this is one of a number of good reasons for using the true definitions. The Bauschinger Effect If one takes a virgin sample and loads it in tension into the plastic range, and then unloads it and continues on into compression, one finds that the yield stress in compression is not the same as the yield strength in tension, as it would have been if the specimen had not first been loaded in tension. In fact the yield point in this case will be significantly less than the corresponding yield stress in tension. This reduction in yield stress is known as the Bauschinger effect . The effect is illustrated in Fig. 8.1.3. The solid line depicts the response of a real material. The dotted lines are two extreme cases which are used in plasticity models; the first is the isotropic hardening model, in which the yield stress in tension and compression are main tained equal, the second being kinematic hardening , in which the total elastic range is maintain ed constant throughout the deformation. Y Ynσ ε∗∗ σ tεA A )a() b( Section 8.1 Solid Mechanics Part II Kelly 246 Figure 8.1.3: The Bauschinger effect The presence of the Bauschinger effect complicates any plasticity theory. However, it is not an issue provided there are no reversal s of stress in the problem under study. Hydrostatic Pressure Careful experiments show that, for metals, the yield behaviour is independent of hydrostatic pressure. That is, a stress state pzz yy xx −===σσσ has negligible effect on the yield stress of a material, right up to very high pressures. Note however that this is not true for soils or rocks. 8.1.3 Assumptions of Plasticity Theory Regarding the above test results then, in formulating a basic plasticity theory with which to begin, the following assumptions are usually made: (1) the response is independent of rate effects (2) the material is incompressible in the plastic range (3) there is no Bauschinger effect (4) the yield stress is independent of hydrostatic pressure (5) the material is isotropic The first two of these will usually be ve ry good approximations, the other three may or may not be, depending on the material and circumstances. For example, most metals can be regarded as isotropic. After large plastic deformation however, for example in rolling, the material will have become anisotropic: there will be distinct material directions and asymmetries. Together with these, assumptions can be made on the type of hardening and on whether elastic deformations are significant. For example, consider the hierarchy of models illustrated in Fig. 8.1.4 below, commonly used in theoretical analyses. In (a) both the elastic and plastic curves are assumed linear. In (b) work-hardening is neglected and the σ tε0Y1Y 1Y02Y kinematic hardening isotropic hardening Section 8.1 Solid Mechanics Part II Kelly 247yield stress is constant after initial yield. Such perfectly-plastic models are particularly appropriate for studying processes where the metal is worked at a high temperature – such as hot rolling – where work hardening is small. In many areas of applications the strains involved are large, e.g. in metal working processes such as extrusion, rolling or drawing, where up to 50% reduction ratios are common. In such cases the elastic strains can be neglected altogether as in the two models (c) and (d). The rigid/perfectly-plastic model (d) is the crudest of all – and hence in many ways the most useful. It is widely used in analysing metal forming processes, in the design of steel and concrete structures and in the analysis of soil and rock stability. 00σ σ ε εYY (a) Linear Elastic-Plastic (b) Elastic/Perfectly-Plastic 0 0σ σ ε εY Y (c) Rigid/Linear Hardening (d) Rigid-Perfectly-Plastic Figure 8.1.4: Simple models of elastic and plastic deformation 8.1.4 The Tangent and Plastic Modulus Stress and strain are related through εσE= in the elastic region, E being the Young’s modulus, Fig. 8.1.5. The tangent modulus K is the slope of the stress-strain curve in the plastic region and will in general change during a deformation. At any instant of strain, the increment in stress σd is related to the increment in strain εd through2 εσKd d= (8.1.7) 2 the symbol ε here represents the true strain (the subscript t has been dropped for clarity); as mentioned, when the strains are small, it is not necessary to specify which strain is in use since all strain measures are then equivalent Section 8.1 Solid Mechanics Part II Kelly 248 Figure 8.1.5: The tangent modulus After yield, the strain increment consists of both elastic, eε, and plastic, pdε, strains: p ed d d εεε+= (8.1.8) The stress and plastic strain increments are related by the plastic modulus H: pdH dεσ= (8.1.9) and it follows that { ▲Problem 4} HE K1 1 1+= (8.1.10) 8.1.5 Friction Block Models Some additional insight into the way plastic materials respond can be obtained from friction block models. The rigid perfectly pl astic model can be simulated by a Coulomb friction block, Fig. 8.1.6. No strain occurs until σ reaches the yield stress Y. Then there is movement – although the amount of movement or plastic strain cannot be determined without more information being available. The stress cannot exceed the yield stress in this model: Y≤σ (8.1.11) If unloaded, the block stops moving and the stress returns to zero, leaving a permanent strain, Fig. 8.1.6b. σ E•• εdpdεK edε ε Section 8.1 Solid Mechanics Part II Kelly 249 Figure 8.1.6: (a) Friction block model for the rigid perfectly plastic material, (b) response of the rigid-perfectly plastic model The linear elastic perfectly plastic model incorporates a free spring with modulus E in series with a friction block, Fig. 8.1.7. The spring stretches when loaded and the block also begins to move when the stress reaches Y, at which time the spring stops stretching, the maximum possible stress again being Y. Upon unloading, the block stops moving and the spring contracts. Figure 8.1.7: Friction block model for the elastic perfectly plastic material The linear elastic plastic model with linear strain hardening incorporates a second, hardening, spring with stiffness H, in parallel with the friction block, Fig. 8.1.8. Once the yield stress is reached, an ever increasing stre ss needs to be applied in order to keep the block moving – and elastic strain continues to occur due to further elongation of the free spring. The stress is then split into the yield stress, which is carried by the moving block, and an overstress Y−σ carried by the hardening spring. Upon unloading, the block “locks” – the stress in the hardening spring remains constant whilst the free spring contracts. At zero stress, there is a negative stress taken up by the friction block, equal and opposite to the stress in the hardening spring. The slope of the elastic loading line is E. For the plastic hardening line, HY Ep e −+=+=σσεεε → HEEH ddK+==εσ (8.1.12) It can be seen that H is the plastic modulus. σE YYσσ εpermanent deformation unload (a) (b) Section 8.1 Solid Mechanics Part II Kelly 250 Figure 8.1.8: Friction block model for a linear elastic-plastic material with linear strain hardening; (a) stress-free, (b) elasti c strain, (c) elastic and plastic strain, (d) unloading 8.1.6 Problems 1. Give two differences between plas tic and viscoelastic materials. 2. A test specimen of initial length 01.0m is extended to length 0101.0 m. What is the percentage difference between the engineering and true strains? What is this difference when the specimen is extended to length 015.0 m? 3. Derive the relation 8.1.6, 0/ / lln=σσ . 4. Derive Eqn. 8.1.10. 5. Which is larger, H or K? In the case of a perfectly-plastic material? 6. The Ramberg-Osgood model of plasticity is given by n p e b E⎟ ⎠⎞⎜ ⎝⎛+=+=σσεεε where E is the Young’s modulus and b and n are model constants (material parameters) obtained from a curve-fitting of the uniaxial stress-strain curve. (i) Find the tangent and plastic moduli in terms of plastic strain pε (and the material constants). Yσ E eε pεeε pε0=eεε σ ε σ ε σ ε(a) (b) (c) (d)H Section 8.1 Solid Mechanics Part II Kelly 251(ii) Which of the simple models of Fig. 8.1.4 does the model reduce to in the case of 1=n ? (iii) A material with model parameters 4=n , GPa70=E and MPa800=b is strained in tension to 02.0=pε and is subsequently unloaded and put into compression. Find the stress at the initiation of compressive yield assuming isotropic hardening [Note that the yield stress is actually zero in this model] 7. Consider the plasticity model shown below. (i) What is the elastic modulus? (ii) What is the yield stress? (iii) What are the tangent and plastic moduli? Draw a typical loading and unloading curve. 8. Draw the stress-strain diagram for a cycle of loading and unloading to the rigid - plastic model shown here. Take the maximum load reached to be 1 max 4Y=σ and 1 22Y Y= . What is the permanent deformation after complete removal of the load? [Hint: split the cycle into the following regions: (a) 1 0 Y≤≤σ , (b) 1 1 3Y Y≤≤σ , (c) 1 1 4 3 Y Y≤≤σ , then unload, (d) 1 1 3 4 Y Y≤≤σ , (e) 1 1 2 3 Y Y≤≤σ , (f) 0 21≤≤σY .] 1Y2E1E 2Yσ 1E Y2E Section 8.2 Solid Mechanics Part II Kelly 2528.2 Stress Analysis for Plasticity This section follows on from the analysis of th ree dimensional stress carried out in §7.2. The plastic behaviour of materi als is often independent of a hydrostatic stress and this feature necessitates the study of the deviatoric stress . 8.2.1 Deviatoric Stress Any state of stress can be decomposed into a hydrostatic (or mean ) stress Imσ and a deviatoric stress s, according to ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 33 32 3123 22 2113 12 11 33 32 3123 22 2113 12 11 0 00 00 0 s s ss s ss s s mmm σσσ σσσσσσσσσ (8.2.1) where 333 22 11σσσσ++=m ( 8.2.2) and () () ()⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−−−−− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ 22 11 33 31 23 1323 33 11 22 31 1213 12 33 22 11 31 33 32 3123 22 2113 12 11 222 σσσ σ σσ σσσ σσ σ σσσ s s ss s ss s s (8.2.3) In index notation, ij ij m ij s+=δσσ (8.2.4) In a completely analogous manner to the derivation of the principal stresses and the principal scalar invariants of the stress matrix, §7.2.4, one can determine the principal stresses and principal scalar invariants of the deviatoric stress matrix. The former are denoted 3 2 1,,sss and the latter are denoted by 3 2 1,, JJJ . The characteristic equation analogous to Eqn. 7.2.23 is 03 22 13=−−− JsJ sJ s (8.2.5) and the deviatoric invariants are (compare with 7.2.24, 7.2.26)1 1 unfortunately, there is a convention (adhered to by most authors) to write the characteristic equation for stress with a σ2I+ term and that for deviatoric stress with a sJ2− term; this means that the formulae for J2 in Eqn. 8.2.5 are the negative of those for 2I in Eqn. 7.2.24 Section 8.2 Solid Mechanics Part II Kelly 253 () () 3213123122 12332 31222 2311 332211 313 32 212 312 232 12 1133 3322 2211 23 2 133 22 11 1 2 ssssss ss ss ss sss Jss ss sss s s ss ss ss Js s ss s s J =+−−− =++−=−−−++−=++=++= ( 8.2.6) Since the hydrostatic stress remains unchanged with a change of coordinate system, the principal directions of stress coincide with th e principal directions of the deviatoric stress, and the decomposition can be expressed with respect to the principal directions as ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ + ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡321 321 000 000 0 00 00 0 0 00 00 0 sss mmm σσσ σσσ (8.2.7) Note that, from the definition Eqn. 8.2.3, the first invariant of the deviatoric stress, the sum of the normal stresses, is zero: 01=J ( 8.2.8) The second invariant can also be expressed in the useful forms { ▲Problem 3} ()2 32 22 1 21 2 s s s J ++= , (8.2.9) and, in terms of the principal stresses, { ▲Problem 4} () () ()[]2 1 32 3 22 2 1 261σσσσσσ −+−+−=J . (8.2.10) Further, the deviatoric invariants are relate d to the stress tensor invariants through {▲Problem 5} ()( )3 213 1 271 3 22 1 31 2 27 9 2 ,3 I II I J I I J +−= −= (8.2.11) A State of Pure Shear The stress state at a point is one of pure shear if for any one coordinate axes through the point one has only shear stress acting, i.e. the stress matrix is of the form [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = 000 23 1323 1213 12 σσσ σσσ σij (8.2.12) Section 8.2 Solid Mechanics Part II Kelly 254Applying the stress transformation rule 7.2.16 to this stress matrix and using the fact that the transformation matrix Q is orthogonal, i.e. IQQ QQ ==T T, one finds that the first invariant is zero, 033 22 11 =′+′+′σσσ . Hence the deviatoric st ress is one of pure shear. 8.2.2 The Octahedral Stresses Examine now a material element subjected to principal stresses 3 2 1 ,,σσσ as shown in Fig. 8.2.1. By definition, no shear stresses act on the planes shown. Figure 8.2.1: stresses acting on a material element Consider next the octahedral plane ; this is the plane shown shaded in Fig. 8.2.2, whose normal an makes equal angles with the principal directions. It is so-called because it cuts a cubic material element (with faces perpendicular to the principal directions) into a triangular plane and eight of these triangle s around the origin form an octahedron. Figure 8.2.2: the octahedral plane Next, a new Cartesian coordinate system is constructed with axes parallel and perpendicular to the octahedral plane, Fig. 8.2.3. One axis runs along the unit normal an; 1σ2σ3σ 2σ1σ 3σ2 13 an 2 13 Section 8.2 Solid Mechanics Part II Kelly 255this normal has components ( )3/1,3/1,3/1 with respect to the principal axes. The angle 0θ the normal direction makes with the 1 direction can be obtained from 0 1cosθ=⋅ena , where ()0,0,11=e is a unit vector in the 1 direction, Fig. 8.2.3. To complete the new coordinate system, any two perpendicular unit vectors which lie in (parallel to) the octahedral plane can be chosen. Choose one which is along the projection of the 1 axis down onto the octahedral plane. The components of this vector are {▲Problem 6} ( )6/1,6/1,3/2 −−=cn . The final unit vector bn is chosen so that it forms a right hand Cartesian coordinate system with an and cn, i.e. c b a n n n=× . In summary, ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = 112 61, 110 21, 111 31 c b a n n n (8.2.13) Figure 8.2.3: a new Cartesian coordinate system To express the stress state in terms of components in the cba,, directions, construct the stress transformation matrix: ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⋅⋅⋅⋅⋅⋅⋅⋅⋅ = 6/1 2/13/16/1 2/1 3/16/2 0 3/1 3 3 32 2 21 1 1 c b ac b ac b a ne ne nene ne nene ne ne Q (8.2.14) and the new stress components are () () () () () () () () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ++ − −−− + −−−− −−++ =⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ 3 2 1 61 3 2 321 3 2 1 2313 2 321 3 2 21 3 2 613 2 1 231 3 2 61 3 2 1 31321 4 220 00 00 0 σσσ σσ σσσσσ σσ σσσσσ σσ σσσσσσ σσσσσσσσσ Q QT cc cb cabc bb baac ab aa (8.2.15) a an0θb c1e2e3e bn cn0θ 0θ2 13 Section 8.2 Solid Mechanics Part II Kelly 256 Now consider the stress components acting on the octahedral plane, ac ab aaσσσ ,,, Fig. 8.2.4. Recall from Cauchy’s law, Eqn. 7.2.9, that these are the components of the traction vector )(ant acting on the octahedral plane, with respect to the ( a,b,c) axes: c ac b ab a aaan n n tnσσσ ++=)( (8.2.16) Figure 8.2.4: the stress vector σ and its components The magnitudes of the normal and shear stresses acting on the octahedral plane are called the octahedral normal stress octσ and the octahedral shear stress octτ. Referring to Fig. 8.2.4, these can be expressed as { ▲Problem 7} () () () ()32 3131 31 2 2 1 32 3 22 2 12 21 3 2 1 JI ac ab octaa oct =−+−+−=+==++== σσσσσσσστσσσσσ (8.2.17) The octahedral normal and shear stresses on all 8 octahedral planes around the origin are the same. Note that the octahedral normal stress is simply the hydrostatic stress. This implies that the deviatoric stress has no normal component in the direction an and only contributes to shearing on the octahedral plane. Indeed, from Eqn. 8.2.15, ()() () () () () () () ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −− − −−− −−−−−−− −− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ 3 2 1 61 3 2 321 3 2 1 2313 2 321 3 2 1 61 3 2 613 2 1 231 3 2 61 2 222 0 σσσ σσ σσσσσ σσσ σσσσσ σσ cc cb cabc bb baac ab aa s s ss s ss s s (8.2.18) )(ant abσ acσ• octσσ=aa octτ 2 13 Section 8.2 Solid Mechanics Part II Kelly 257The σ’s on the right here can be replaced with s’s since j i j i s s−=−σσ . 8.2.3 Problems 1. What are the hydrostatic and deviatoric stresses for the uniaxial stress 0 11σσ= ? What are the hydrostatic and deviatoric stresses for the state of pure shear τσ=12 ? In both cases, verify that the first invariant of the deviatoric stress is zero: 01=J . 2. For the stress state ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ 314122421 33 32 3123 22 2113 12 11 σσσσσσσσσ , calculate (a) the hydrostatic stress (b) the deviatoric stresses (c) the deviatoric invariants 3. The second invariant of the deviatoric stress is given by Eqn. 8.2.6, ( )13 32 21 2 ss ss ss J ++−= By squaring the relation 03 2 1 1 =++= s s s J , derive Eqn. 8.2.9, ()2 32 22 1 21 2 s s s J ++= 4. Use Eqns. 8.2.9 (and your work from Problem 3) and the fact that 2 1 2 1 s s−=−σσ , etc. to derive 8.2.10, ()()() [ ]2 1 32 3 22 2 1 61 2 σσσσσσ −+−+−=J 5. Use the fact that 03 2 1 1 =++= s s s J to show that 3 13 32 21 321 32 13 32 21 21 ) (3) (3 m mmm ss ss ss sss Iss ss ss II σ σσσ +++ +=+++== Hence derive Eqns. 8.2.11, ()( )3 213 1 271 3 22 1 31 2 27 9 2 ,3 I II I J I I J +−= −= 6. Show that a unit normal cn in the octahedral plane in the direction of the projection of the 1 axis down onto the octahedral plane has coordinates ()61 61 32, ,−− , Fig. 8.2.3. To do this, note the geometry shown below and the fact that when the 1 axis is projected down, it remains at equal angles to the 2 and 3 axes. cnan 1x 0θ 31 0 cos=θ project down Octahedral plane Section 8.2 Solid Mechanics Part II Kelly 258 7. Use Eqns. 8.2.15 to derive Eqns. 8.2.17. 8. For the stress state of problem 2, calculat e the octahedral normal stress and the octahedral shear stress Section 8.3 Solid Mechanics Part II Kelly 2598.3 Yield Criteria in Three Dimensional Plasticity The question now arises: a material yields at a stress level Y in a uniaxial tension test, but when does it yield when subjected to a complex three-dimensional stress state? It can be assumed that yield will occur at a particle when some combination of the stress components reaches some critical value, when k F =),,,,,(33 23 22 13 12 11 σσσσσσ (8.3.1) say. Here, F is some function of the 6 independent components of the stress tensor and k is some material property which can be determined experimentally. Alternatively, since the stress state at a point is characterised by the principal stresses, one could say that k Fi=),,,(3 2 1 nσσσ (8.3.2) where in represent the principal directions. If the material is isotropic, the response is independent of any material direction – independent of any “direction” the stress acts in, and so the yield criterion can be expressed in the simple form k F =),,(3 2 1 σσσ (8.3.3) Further, since it should not matter which direction is labelled ‘1’, which ‘2’ and which ‘3’, F must be a symmetric function of the three principal stresses. Alternatively, since the three principal invariants of stress are independent of material orientation, one can write k IIIF =),,(3 2 1 (8.3.4) or, more usually, k JJIF =),,(3 2 1 (8.3.5) where 3 2,JJ are the non-zero principal invariants of the deviatoric stress. With the further restriction that the yield stress is independent of the hydrostatic stress, one has k JJF =),(3 2 (8.3.6) 8.3.1 The Tresca and Von Mises Yield Conditions The two most commonly used and successful yield criteria for isotropic metallic materials are the Tresca and Von Mises criteria. Section 8.3 Solid Mechanics Part II Kelly 260 The Tresca Yield Condition The Tresca yield criterion states that a material will yield if the maximum shear stress reaches some critical value, that is, Eqn. 8.3.3 takes the form k= ⎭⎬⎫ ⎩⎨⎧− − −1 3 3 2 2 121,21,21max σ σ σ σ σ σ (8.3.7) The value of k can be obtained from a simple experiment. For example, in a tension test, 0 ,3 2 0 1 = = = σ σσ σ , and failure occurs when 0σ reaches Y, the yield stress in tension. It follows that 2Yk=. (8.3.8) In a shear test, τ σ στ σ −= = =3 2 1 ,0 , , and failure occurs when τ reaches Yτ, the yield stress of a material in pure shear, so that Y kτ=. The Von Mises Yield Condition The Von Mises criterion states that yield occurs when the principal stresses satisfy the relation () () ()k=− + − + − 62 1 32 3 22 2 1 σ σ σ σ σ σ (8.3.9) Again, from a uniaxial tension test, one finds that the k in Eqn. 8.3.9 is 3Yk= . (8.3.10) Writing the Von Mises condition in terms of Y, one has () () () Y= − + − + −2 1 32 3 22 2 121σ σ σ σ σ σ (8.3.11) The quantity on the left is called the Von Mises Stress , sometimes denoted by VMσ. When it reaches the yield stress in pure tension, the material begins to deform plastically. In the shear test, one again finds that Y kτ=, the yield stress in pure shear. Sometimes it is preferable to work with arbitrary stress components; for this purpose, the Von Mises condition can be expressed as { ▲Problem 2} () () ( )( )2 2 312 232 122 11 332 33 222 22 11 6 6 k= + + + − + − + − σ σ σ σ σ σ σ σ σ (8.3.12) Section 8.3 Solid Mechanics Part II Kelly 261The piecewise linear nature of the Tresca yield condition is sometimes a theoretical advantage over the quadratic Mises condition. However, the fact that in many problems one often does not know which principal stress is the maximum and which is the minimum causes difficulties when working with the Tresca criterion. The Tresca and Von Mises Yield Criteria in terms of Invariants From Eqn. 8.2.10 and 8.3.9, the Von Mises criterion can be expressed as 0 )( 2 2 2 = − ≡ k J Jf (8.3.13) Note the relationship between 2J and the octahedral shear stress, Eqn. 8.2.17; the Von Mises criterion can be interpreted as predicti ng yield when the octahedral shear stress reaches a critical value. With 3 2 1 σ σ σ ≥ ≥ , the Tresca condition can be expressed as 0 64 96 36 27 4),(6 24 2 22 2 33 2 3 2 = − + − − ≡ k Jk Jk J J JJf (8.3.14) but this expression is too cumbersome to be of much use. Experiments of Taylor and Quinney In order to test whether the Von Mises or Tresca criteria best modelled the real behaviour of metals, G I Taylor & Quinney (1931), in a series of classic experiments, subjected a number of thin-walled cylinders made of copper and steel to combined tension and torsion, Fig. 8.3.1. σ σ ττ Figure 8.3.1: combined tension and torsion of a thin-walled tube The cylinder wall is in a state of plane stress, with σ σ=11 , τ σ=12 and all other stress components zero. The principal stresses corresponding to such a stress-state are (zero and) {▲Problem 3} 2 2 41 21τ σ σ + ± (8.3.15) and so Tresca's condition reduces to 2 2 24 4 k= +τ σ or 12/2 2 =⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛ Y Yτ σ (8.3.16) Section 8.3 Solid Mechanics Part II Kelly 262The Mises condition reduces to { ▲Problem 4} 2 2 23 3 k= +τ σ or 1 3/2 2 =⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛ Y Yτ σ (8.3.17) Thus both models predict an elliptical yield locus in ()τσ, stress space , but with different ratios of principal axes, Fig. 8.3.2. The origin in Fig. 8.3.2 corresponds to an unstressed state. The horizontal axes refer to uniaxial tension in the absence of shear, whereas the vertical axis refers to pure torsion in the absence of tension. When there is a combination of σ and τ, one is off-axes. If the combination remains “inside” the yield locus, the material remains elastic; if the combination is such that one reaches anywhere along the locus, then plasticity ensues. Figure 8.3.2: the yield locus for a thin-walled tube in combined tension and torsion Taylor and Quinney, by varying the amount of tension and torsion, found that their measurements were closer to the Mises ellipse than the Tresca locus, a result which has been repeatedly confirmed by other workers1. 2D Principal Stress Space Fig. 8.3.2 gives a geometric interpretation of the Tresca and Von Mises yield criteria in ()τσ, space. It is more usual to interpret yield criteria geometrically in a principal stress space . The Taylor and Quinney tests are an example of plane stress, where one principal stress is zero. Following the convention for plane stress, label now the two non-zero principal stresses 1σ and 2σ, so that 03=σ (even if it is not the minimum principal stress). The criteria can then be displayed in ( )2 1,σσ 2D principal stress space. With 03=σ , one has Tresca: {} Y= −1 2 2 1 , , max σ σ σ σ (8.3.18) Von Mises: 2 2 2 212 1 Y= + − σ σσ σ 1 the maximum difference between the predicted stresses from the two criteria is about 15%. The two criteria can therefore be made to agree to within ± 7.5% by choosing k to be half-way between 2/Y and 3/Y στ 3/YY=τ 2/YY=τ YMises Tresca Section 8.3 Solid Mechanics Part II Kelly 263These are plotted in Fig. 8.3.3. The Tresca criterion is a hexagon and the Von Mises criterion is an ellipse with axes inclined at 045 to the principal axes. Some stress states are shown in the stress space: point A corresponds to a uniaxial tension, B to a equi- biaxial tension and C to a pure shear τ. Figure 8.3.3: yield loci in 2D principal stress space Again, points inside these loci represent an elastic stress state. Any combination of principal stresses which push the point out to the yield loci results in plastic deformation. 8.3.2 Three Dimens ional Principal Stress Space The 2D principal stress space has limited use. For example, a stress state that might start out two dimensional can develop into a fully three dimensional stress state as deformation proceeds. In three dimensional principal stress space, one has a yield surface ( )0 ,,3 2 1 =σσσf , Fig. 8.3.4 2. In this case, one can draw a line at equal angles to all three principal stress axes, the space diagonal . Along the space diagonal 3 2 1 σ σ σ = = and so points on it are in a state of hydrostatic stress. Assume now, for the moment, that hydrostatic stress does not affect yield and consider some arbitrary point A, () ( )cba,, ,,3 2 1 =σσσ , on the yield surface, Fig. 8.3.4. A pure hydrostatic stress hσ can be superimposed on this stress state without affecting yield, so any other point ( )( )h h h c b a σ σ σ σσσ + + += , , ,,3 2 1 will also be on the yield surface. Examples of such points are shown at B, C and D, which are obtained from A by moving along a line parallel to the space diagonal. The yield behaviour of the material is therefore specified by a yield locus on a plane perpendicular to the space diagonal, and the yield surface is generated by sliding this locus up and down the space diagonal. 2 as mentioned, one has a six dimensional stress space for an anisotropic material and this cannot be visualised Y2σ •• • 1σ2σ τCAB ••EDY1σ Section 8.3 Solid Mechanics Part II Kelly 264 Figure 8.3.4: Yield locus/surface in three dimensional stress-space The π-plane Any surface in stress space can be described by an equation of the form ( )const ,,3 2 1 =σσσf (8.3.19) and a normal to this surface is the gradient vector 3 32 21 1e e eσ σ σ ∂∂+∂∂+∂∂ f f f (8.3.20) where 3 2 1,,eee are unit vectors along the stress space axes. In particular, any plane perpendicular to the space diagonal is described by the equation const3 2 1 = + + σ σ σ (8.3.21) Without loss of generality, one can choose as a representative plane the π – plane , which is defined by 03 2 1 = + + σ σ σ . For example, the point ( )( )0,1,1 ,,3 2 1 −=σσσ is on the π – plane and, with yielding independent of hydrostatic stress, is equivalent to points in principal stress space which differ by a hydrostatic stress, e.g. the points () ( ) 1,2,0,1,0,2 −− , etc. The stress state at any point A represented by the vector ( )3 2 1,,σσσ=σ can be regarded as the sum of the stress state at the corresponding point on the π – plane, D, represented by the vector ()3 2 1,,sss=s together with a hydrostatic stress represented by the vector ()m m m σσσ ,, =ρ : () ( )( )m m m m m m σσσ σ σσ σσ σ σσσ ,, , , ,,3 2 1 3 2 1 + − − − = (8.3.22) 1σ2σ3σ •hydrostatic stress deviatoric stress the π - plane • •( )cba,,A B C•Dyield locus σρ s Section 8.3 Solid Mechanics Part II Kelly 265 The components of the first term/vector on the right here sum to zero since it lies on the π – plane, and this is the deviatoric stress, whilst the hydrostatic stress is () 3/3 2 1 σ σ σ σ + + =m . Projected view of the π-plane Fig. 8.3.5a shows principal stress space and Fig. 8.3.5b shows the π – plane. The heavy lines 3 2 1 ,, σσσ ′′′ in Fig. 8.3.5b represent the projections of the principal axes down onto the −πplane (so one is “looking down” the space diagonal). Some points, CBA ,, in stress space and their projections onto the −πplane are also shown. Also shown is some point D on the −πplane. It should be kept in mind that the deviatoric stress vector s in the projected view of Fig. 8.3.5b is in reality a three dimensional vector (see the corresponding vector in Fig. 8.3.5a). Figure 8.3.5: Stress space; (a) principal stress space, (b) the π – plane Consider the more detailed Fig. 8.3.6 below. Point A here represents the stress state () 0,1,2− , as indicated by the arrows in the figure. It can also be “reached” in different ways, for example it represents ()1,0,3 and ( )1,2,1 −− . These three stress states of course differ by a hydrostatic stress. The actual −πplane value for A is the one for which 03 2 1 = + + σ σ σ , i.e. () ( )( )31 34 35 3 2 1 3 2 1 ,, ,, ,, −− = = sss σσσ . Points B and C also represent multiple stress states { ▲Problem 7}. 1σ′2σ′3σ′ C• 1σ2σ3σ space diagonal •• BA ••• CBA )a() b(• sD •D s Section 8.3 Solid Mechanics Part II Kelly 266 Figure 8.3.6: the π-plane The bisectors of the principal plane projections , such as the dotted line in Fig. 8.3.6, represent states of pure shear. For example, the −πplane value for point D is () 2,2,0−, corresponding to a pure shear in the 3 2σ σ− plane. The dashed lines in Fig. 8.3.6 are helpful in th at they allow us to plot and visualise stress states easily. The distance between each dashed line along the directions of the projected axes represents one unit of principal stress. Note, however, that these “units” are not consistent with the actual magnitudes of the deviatoric vectors in the −πplane. To create a more complete picture, note first that a unit vector along the space diagonal is [ ]31 31 31,, =ρn , Fig. 8.3.7. The components of this normal are the direction cosines; for example, a unit normal along the ‘1’ principal axis is []0,0,11=e and so the angle 0θ between the ‘1’ axis and the space diagonal is given by 31 0 1cos = =⋅ θρen . From Fig. 8.3.7, the angle θ between the ‘1’ axis and the −πplane is given by 32cos =θ , and so a length of 1σ units gets projected down to a length 132σ ==ss . For example, point E in Fig. 8.3.6 represents a pure shear ( )( ) 0,2,2 ,,3 2 1 −=σσσ , which is on the −πplane. The length of the vector out to E in Fig. 8.3.6 is 32 “units”. To convert to actual magnitudes, multiply by 32 to get 22=s , which agrees with 22 2 22 2 2 32 22 1 = + = + + == s s s ss . 1σ′ 2σ′3σ′ ()1σ′− ()2σ′− ()3σ′−• •• AB C bisector •D•E Section 8.3 Solid Mechanics Part II Kelly 267 Figure 8.3.7: principal stress projected onto the π-plane Typical π-plane Yield Loci Consider next an arbitrary point ),,( cba on the −πplane yield locus . If the material is isotropic, the points ),,( bca , ),,( cab , ),,( acb , ),,( bac and ),,( abc are also on the yield locus. If one assumes the same yield behaviour in tension as in compression, e.g. neglecting the Bauschinger effect, then so also are the points ),,( cba −−− , ),,( bca −−− , etc. Thus 1 point becomes 12 and one need only consider the yield locus in one 30o sector of the −πplane, the rest of the locus being generated through symmetry. One such sector is shown in Fig. 8.3.8, the axes of symmetry being the three projected principal axes and their (pure shear) bisectors. Figure 8.3.8: A typical sector of the yield locus The Tresca and Von Mises Yield Loci in the π-plane The Tresca criterion, Eqn. 8.3. 7, is a regular hexagon in the −πplane as illustrated in Fig. 8.3.9. Which of the six sides of the locus is relevant depends on which of 3 2 1 ,, σσσ is the maximum and which is the minimum, and whether they are tensile or compressive. For example, yield at the pure shear 2 / ,0 ,2/3 2 1 Y Y −= = = σ σ σ is indicated by point A in the figure. 1 π-plane • •s 132σ=sθspace diagonal 0θ1σ ρn 1n yield locus Section 8.3 Solid Mechanics Part II Kelly 268Point B represents yield under uniaxial tension, Y=1σ . The distance oB, the “magnitude” of the hexagon, is therefore Y32; the corresponding point on the π – plane is () () Y Y Y sss31 31 32 32 1 , , , − − = . A criticism of the Tresca crit erion is that there is a sudden change in the planes upon which failure occurs upon a small change in stress at the sharp corners of the hexagon. Figure 8.3.9: The Tresca criterion in the π-plane Consider now the Von Mises criterion. From Eqns. 8.3.10, 8.3.13, the criterion is 3/2Y J= . From Eqn. 8.2.9, this can be re-written as Y s s s322 32 22 1 = + + (8.3.23) Thus, the magnitude of the deviatoric stress ve ctor is constant and one has a circular yield locus with radius k Y 232= , which transcribes the Tresca hexagon, as illustrated in Fig. 8.3.10. 1σ′ 2σ′3σ′ ()1σ′− ()2σ′− ()3σ′−Y= −2 1σσ Y=−1 2σ σ Y= −3 2σ σY= −3 1σσY=−1 3σσ Y= −2 3σσ •A•Bo Y32 Section 8.3 Solid Mechanics Part II Kelly 269 Figure 8.3.10: The Von Mises criterion in the π-plane The yield surface is a circular cylinder with axis along the space diagonal, Fig. 8.3.11. The Tresca surface is a similar hexagonal cylinder. Figure 8.3.11: The Von Mises and Tresca yield surfaces 8.3.3 Haigh-Westergaar d Stress Space Thus far, yield criteria have been desc ribed in terms of principal stresses ) ,,(3 2 1 σσσ . It is often convenient to work with ( )θρ,,s coordinates, Fig. 8.3.12; these cylindrical coordinates are called Haigh-Westergaard coordinates . They are particularly useful for describing and visualising geometrically pressure-dependent yield-criteria. 1σ2σ3σ plane stress yield locus π - plane yield locus Von Mises yield surface )0 (3=σ )0 (3 2 1 = + + σ σσTresca yield surface1σ′ 2σ′3σ′ ()1σ′− ()2σ′− ()3σ′−Tresca Von Mises Y s32= Section 8.3 Solid Mechanics Part II Kelly 270The coordinates ()s,ρ are simply the magnitudes of, respectively, the hydrostatic stress vector ()m m m σσσ ,, =ρ and the deviatoric stress vector ( )3 2 1,,sss=s . These are given by (scan be obtained from Eqn. 8.2.9) 2 1 2 ,3/ 3 J s Im == = == s ρ σ ρ (8.3.24) Figure 8.3.12: A point in stress space θ is measured from the 1σ′ )(1s axis in the −πplane. To express θ in terms of invariants, consider a unit vector e in the −πplane in the direction of the 1σ′ axis; this is the same vector cn considered in Fig. 8.2.3 in connection with the octahedral shear stress, and it has coordinates () ( )61 61 32 2 2 1 , , ,, − − =σσσ , Fig. 8.3.12. The angle θ can now be obtained from θcoss=⋅es {▲Problem 9}: 21 23cos Js=θ (8.3.25) Further manipulation leads to the relation { ▲Problem 10} 2/3 23 2333cosJJ=θ (8.3.26) Since 2J and 3J are invariant, it follows that θ3cos is also. Note that 3J enters through θ3cos , and does not appear in ρ or s; it is 3J which makes the yield locus in the π-plane non-circular. From Eqn. 8.3.25 and Fig. 8.3.12b, the deviat oric stresses can be expressed in terms of the Haigh-Westergaard coordinates through 1σ2σ3σ •ρ s 1σ′2σ′3σ′ θ1σ′ θse )a() b( Section 8.3 Solid Mechanics Part II Kelly 271() () ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ +− = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ θ πθ πθ 3/2cos3/2coscos 32 2 321 J sss (8.3.27) The principal stresses and the Haigh-Westergaard coordinates can then be related through { ▲Problem 12} () () ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ +− + ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ = ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ 3/2 cos3/2 coscos 32 31 321 π θπ θθ ρρρ σσσ s (8.3.28) In terms of the Haigh-Westergaard coordinates, the yield criteria are Von Mises: 0 )(2 21=− = ks sf (8.3.29) Tresca ( ) 0 sin2 ),(3 =−+ = Y s sfπθ θ 8.3.4 Pressure Depende nt Yield Criteria The Tresca and Von Mises criteria are inde pendent of hydrostatic pressure and are suitable for the modelling of plasticity in metals. For materials such as rock, soils and concrete, however, there is a strong dependence on the hydrostatic pressure. The Drucker-Prager Criteria The Drucker-Prager criterion is a simple modification of the Von Mises criterion, whereby the hydrostatic-dependent first invariant 1I is introduced to the Von Mises Eqn. 8.3.13: 0 ),(2 1 2 1 =− + ≡ k J I JIf α (8.3.30) with α is a new material parameter. On the π – plane, 01=I , and so the yield locus there is as for the Von Mises criterion, a circle of radius k2, Fig. 8.3.13a. Off the π – plane, the yield locus remains circular but the radius changes. When there is a state of pure hydrostatic stress, the magnitude of the hydrostatic stress vector is { ▲Problem 13} α ρ 3/k==ρ , with 0==ss . For large pressures, 03 2 1 < = = σ σ σ , the 1I term in Eqn. 8.3.30 allows for large deviatoric stresses. This effect is shown in the meridian plane in Fig. 8.3.13b, that is, the ),(sρ plane which includes the 1σ axis. Section 8.3 Solid Mechanics Part II Kelly 272 Figure 8.3.13: The Drucker-Prager criterion; (a) the π-plane, (b) the Meridian Plane The Drucker-Prager surface is a right-circular cone with apex at α ρ 3/k= , Fig. 8.3.14. Note that the plane stress locus, where the cone intersects the 03=σ plane, is an ellipse, but whose centre is off-axis, at some )0 ,0 (2 1 < <σ σ . Figure 8.3.14: The Drucker-Prager yield surface In terms of the Haigh-Westergaard coordinates, the yield criterion is 0 2 6 ),( = −+ = k s s f αρ ρ (8.3.31) The Mohr Coulomb Criteria The Mohr-Coulomb criterion is based on Coulomb’s 1773 friction equation, which can be expressed in the form φ σ τ tann c−= (8.3.32) ρs α3kk2π - plane )a(1σ′ 2σ′3σ′ )b(s meridian plane 3σ− 1σ−2σ−ρs Section 8.3 Solid Mechanics Part II Kelly 273where φ,c are material constants; c is called the cohesion3 and φ is called the angle of internal friction . τ and nσ are the shear and normal stresses acting on the plane where failure occurs (through a shearing effect), Fig. 8.3.15, with φtan playing the role of a coefficient of friction. The criterion states that the larger the pressure nσ− , the more shear the material can sustain. Note that the Mohr-Coulomb criterion can be considered to be a generalised version of the Tresca criterion, since it reduces to Tresca’s when 0=φ with kc=. Figure 8.3.15: Coulomb friction over a plane This criterion not only includes a hydrostatic pr essure effect, but also allows for different yield behaviours in tension and in compression. Maintaining isotropy, there will now be three lines of symmetry in any deviatoric plan e, and a typical sector of the yield locus is as shown in Fig. 8.3.16 (compare with Fig. 8.3.8) Figure 8.3.16: A typical sector of the yield locus for an isotropic material with different yield behaviour in tension and compression Given values of c and φ, one can draw the failure locus (lines) of the Mohr-Coulomb criterion in ) ,(τσn stress space, with intercepts c±=τ and slopes φtanm , Fig. 8.3.17. Given some stress state 3 2 1 σ σ σ ≥ ≥ , a Mohr stress circle can be drawn also in ) ,(τσn space (see §7.2.6). When the stress state is such that this circle reaches out and touches the failure lines, yield occurs. 3 0=c corresponds to a cohesionless material such as sand or gravel, which has no strength in tension τnσ− yield locus Section 8.3 Solid Mechanics Part II Kelly 274 Figure 8.3.17: Mohr-Coulomb failure criterion From Fig. 8.3.17, and noting that the large Mohr circle has centre ( ) ( )0,3 1 21σ σ+ and radius ()3 1 21σ σ− , one has φσ σ σ σσφσ στ sin2 2cos2 3 1 3 13 1 −++=−= n (8.3.33) Thus the Mohr-Coulomb criterion in terms of principal stresses is ( ) ( )φ σ σφ σ σ sin cos23 1 3 1 + − = − c (8.3.34) The strength of the Mohr-Coulomb material in uniaxial tension, Ytf, and in uniaxial compression, Ycf, are thus φφ φφ sin1cos2,sin1cos2 −=+=cfcfYc Yt (8.3.35) In terms of the Haigh-Westergaard coordinates, the yield criterion is ( ) ( ) 0 cos6 sin cos sin3 sin2 ),,(3 3 = − + ++ + = φ φ θ θ φ ρ θρπ πc s s s f (8.3.36) The Mohr-Coulomb yield surface in the π – plane and meridian plane are displayed in Fig. 8.3.18. In the π – plane one has an irregular hexagon which can be constructed from two lengths: the magnitude of the deviatoric stress in uniaxial tension at yield, 0ts, and the corresponding (larger) value in compression, 0cs; these are given by: ( ) ( ) φφ φφ sin3sin1 6,sin3sin1 6 0 0−−=+−=Yc cYc tfsfs (8.3.37) τ nσc cφ φfailure line • • • 3σ2σ1στφ Section 8.3 Solid Mechanics Part II Kelly 275In the meridian plane, the failure surface cuts the 0=s axis at φ ρ cot3c = {▲Problem 14}. Figure 8.3.18: The Mohr-Coulomb criterion; (a) the π-plane, (b) the Meridian Plane The Mohr-Coulomb surface is thus an irregular hexagonal pyramid, Fig. 8.3.19. Figure 8.3.19: The Mohr-Coulomb yield surface By adjusting the material parameters φ α ,,,ck , the Drucker-Prager cone can be made to match the Mohr-Coulomb hexagon, either inscribing it at the minor vertices, or circumscribing it at the major vertices, Fig. 8.3.20. 1σ′ 2σ′3σ′3σ− 1σ−2σ−ρs φcot3cπ - plane )a(1σ′ 2σ′3σ′ )b(meridian plane 0ts 0cs0ts 0cs Section 8.3 Solid Mechanics Part II Kelly 276Figure 8.3.20: The Mohr-Coulomb and Drucker-Prager criteria matched in the π- plane Capped Yield Surfaces The Mohr-Coulomb and Drucker-Prager surfaces are open in that a pure hydrostatic pressure can be applied without affecting yi eld. For many geomaterials, however, for example soils, a large enough hydrostatic pressure will induce permanent deformation. In these cases, a closed (capped) yield surface is more appropriate, for example the one illustrated in Fig. 8.3.21. Figure 8.3.21: a capped yield surface An example is the modified Cam-Clay criterion: ()131 2 131 2 2 3 I p MI Jc+ −= or ( ) 0 , 231 2 332< + −= ρ ρ ρcp M s (8.3.38) with M and cp material constants. In terms of the standard geomechanics notation, it reads ( )p ppM qc− = 22 2 (8.3.39) where s J q I p233 , 31 31 2 1 = = −= −= ρ (8.3.40) The modified Cam-Clay locus in the meridian plane is shown in Fig. 8.3.22. Since s is constant for any given ρ, the locus in planes parallel to the π - plane are circles. The material parameter cp is called the critical state pressure , and is the pressure which carries the maximum deviatoric stress. M is the slope of the dotted line shown in Fig. 8.3.22, known as the critical state line . 3σ− 1σ−2σ− Section 8.3 Solid Mechanics Part II Kelly 277 Figure 8.3.22: The modified Cam-Clay criterion in the Meridian Plane 8.3.5 Anisotropy Many materials will display anisotropy. For example metals which have been processed by rolling will have characteristic material directions, the tensile yield stress in the direction of rolling being typically 15% greater than that in the transverse direction. The form of anisotropy exhibited by rolled sheets is such that the material properties are symmetric about three mutually orthogonal plan es. The lines of intersection of these planes form an orthogonal set of axes known as the principal axes of anisotropy . The axes are (a) in the rolling direction, (b) normal to the sheet, (c) in the plane of the sheet but normal to rolling direction. This form of anisotropy is called orthotropy (see Part I, §6.2.2). Hill (1948) proposed a yield conditi on for such a material which is a natural generalisation of the Mises condition: () ( ) ( ) 01 2 2 2)(2 2 122 312 232 22 112 11 332 33 22 =− + + +− + − + − = σ σ σσ σ σ σ σ σ σ N M LH G F fij (8.3.41) where F, G, H, L, M, N are material constants. It reduc es to the Mises condition 8.3.12 when 261 3 3 3 kN MLHGF = = == == (8.3.42) The 1, 2, 3 axes of reference in 8.3.41 are the principal axes of anisotropy. The form appropriate for a general choice of axes can be derived by using the usual stress transformation formulae. It is compli cated and involves cross-terms such as 23 11σσ , etc. 8.3.6 Problems 1. A material is to be loaded to a stress state [] ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −− = 0 0 00 90 300 30 50 ijσ MPa pq π - plane Mpc cp2cpCritical state line Section 8.3 Solid Mechanics Part II Kelly 278What should be the minimum uniaxial yield stress of the material so that it does not fail, according to the (a) Tresca criterian (b) Von Mises criterion What do the theories predict when the yield stress of the material is 80MPa? 2. Use Eqn. 8.2.6, ( )1133 3322 22112 312 232 12 2 ss ss ss s s s J + + − + + = to derive Eqn. 8.3.12, () () ( )( )2 2 312 232 122 11 332 33 222 22 11 6 6 k= + + + − + − + − σ σ σ σ σ σ σ σ σ , for the Von Mises criterion. 3. Use the plane stress principal stress formula 2 122 22 11 22 11 2,12 2σσ σ σ σσ +⎟ ⎠⎞⎜ ⎝⎛ −±+= to derive Eqn. 8.3.15 for the Taylor-Quinney tests. 4. Derive Eqn. 8.3.17 for the Taylor-Quinney tests. 5. Describe the states of stress represented by the points D and E in Fig. 8.3.3. (the complete stress states can be visualised with the help of Mohr’s circles of stress, Fig. 7.2.16.) 6. Suppose that, in the Taylor and Quinney tension-torsion tests, one has 2/Y=σ and 4/3Y =τ . Plot this stress state in the 2D pr incipal stress state, Fig. 8.3.3. (Use Eqn. 8.3.15 to evaluate the principal stresses.) Keeping now the normal stress at 2/Y=σ , what value can the shear stress be increased to before the material yields, according to the von Mises criterion? 7. What are the −πplane principal stress values for the points B and C in Fig. 8.3.6? 8. Sketch on the −πplane Fig. 8.3.6 a line corresponding to 2 1σ σ= and also a region corresponding to 3 2 1 0σ σ σ >> > 9. Using the relation θcoss=⋅es and 1 3 2 s ss −=+ , derive Eqn. 8.3.25, 21 23cos Js=θ . 10. Using the trigonometric relation θ θ θ cos3 cos4 3cos3− = and Eqn. 8.3.25, 21 23cos Js=θ , show that ()22 1 2/3 21 2333cos J sJs− =θ . Then using the relations 8.2.6, ()13 32 21 2 ssssss J + + −= , with 01=J , derive Eqn. 8.3.26, 2/3 23 2333cos JJ=θ 11. Consider the following stress states. For each one, evaluate the space coordinates ),,( θρs and plot in the −πplane (see Fig. 8.3.12b): (a) triaxial tension: 02 3 2 1 1 > = = >= T T σ σ σ (b) triaxial compression: 02 2 1 1 3 < −= = <−= p p σ σ σ (this is an important test for geomaterials, which are dependent on the hydrostatic pressure) Section 8.3 Solid Mechanics Part II Kelly 279(c) a pure shear τ σ=xy : τ σ στ σ −= = +=3 2 1 ,0 , (d) a pure shear τ σ=xy in the presence of hydrostatic pressure p: τ σ στ σ −−= −= +−= p p p3 2 1 , ,, i . e . 3 2 2 1 σ σ σ σ − = − 12. Use relations 8.3.24, 2 1 2 ,3/ J s I = =ρ and Eqns. 8.3.27 to derive Eqns. 8.3.28. 13. Show that the magnitude of the hydrostatic stress vector is α ρ 3/k==ρ for the Drucker-Prager yield criterion when the deviatoric stress is zero 14. Show that the magnitude of the hydrostatic stress vector is φ ρ cot3c = for the Mohr-Coulomb yield criterion when the deviatoric stress is zero 15. Show that, for a Mohr-Coulomb material, )1 /()1( sin + −= r r φ , where Yt Ycffr / = is the compressive to tensile strength ratio 16. A sample of concrete is subjected to a stress Ap p −= −= =33 22 11 ,σ σ σ where the constant 1>A . Using the Mohr-Coulomb criterion and the result of Problem 15, show that the material will not fail provided rpfAYc + < / Section 8.4 Solid Mechanics Part II Kelly 2808.4 Elastic Perfectly Plastic Materials Once yield occurs, a material will deform plastically. Predicting and modelling this plastic deformation is the topic of this section. For the most part, in this section, the material will be assumed to be perfectly plastic, that is, there is no work hardening. 8.4.1 Plastic Stra in Incr ements When examining the strains in a plastic material, it should be emphasised that one works with increments in strain rather than a total accumulated strain. One reason for this is that when a material is subjected to a cert ain stress state, the corresponding strain state could be one of many. Similarly, the strain state could correspond to many different stress states. Examples of this state of affairs are shown in Fig. 8.4.1. Figure 8.4.1: stress-strain curve; (a) different strains at a certain stress, (b) different stress at a certain strain One cannot therefore make use of stress-strain relations in plastic regions (except in some special cases), since there is no unique rela tionship between the current stress and the current strain. However, one can rela te the current stress to the current increment in strain , and these are the “stress-strain” laws which are used in plasticity theory. The total strain can be obtained by summing up, or integrating, the strain increments. 8.4.2 The Prandtl- Reuss Equations An increment in strain εd can be decomposed into an elastic part edε and a plastic part pdε. If the material is isotropic, it is re asonable to suppose that the principal plastic strain increments p idε are proportional to the principal deviatoric stresses is: 0 33 22 11≥=== λεεεdsd sd sdp p p (8.4.1) This relation only gives the ratios of the plastic strain increments to the deviatoric stresses. To determine the precise relati onship, one must specify the positive scalar λd (see later). Note that the plastic volume constancy is inherent in this relation: 03 2 1 =++p p pd d d εεε . ∗ ∗ ∗∗ σ σ ε ε Section 8.4 Solid Mechanics Part II Kelly 281 Eqns. 8.4.1 are in terms of the principal devi atoric stresses and principal plastic strain increments. In terms of Cartesian coordinates, one has λεεεεεεdsd sd sd sd sd sd yzp yz xzp xz xyp xy zzp zz yyp yy xxp xx====== (8.4.2) or, succinctly, λε ds dijp ij= (8.4.3) These equations are often expressed in the alternative forms λσσεε σσεε εεεεdd d d d s sd d s sd d zz yyp zzp yy yy xxp yyp xx zz yyp zzp yy yy xxp yyp xx==−−=−−==−−=−−L L (8.4.4) or, dividing by dt to get the rate equations, λσσεε εε&L&& L&& ==−−==−− yy xxp yyp xx yy xxp yyp xx s s (8.4.5) In terms of actual stresses, one has, from 8.2.3, () [ ] ()[] ()[] zxp zxyzp yzxyp xyyy xx zzp zzxx zz yyp yyzz yy xxp xx d dd dd dd dd dd d λσελσελσεσσσλεσσσλεσσσλε ===+− =+− =+− = 21 3221 3221 32 (8.4.6) This plastic stress-strain law is known as a flow rule . Other flow rules will be considered later on. Note that one cannot propose a flow rule which gives the plastic strain increments as explicit functions of the stre ss, otherwise the yield criterion might not be met (in particular, when there is strain hardening); one must include the to-be-determined scalar plastic multiplier λ. The plastic multiplier is determined by ensuring the stress- state lies on the yield surface during plastic flow. The full elastic-plastic stress-strain relations are now, using Hooke’s law, Section 8.4 Solid Mechanics Part II Kelly 282()[] () ()[] () ()[] () zx zx zxyz yz yzxy xy xyyy xx zz yy xx zz zzxx zz yy zz xx yy yyzz yy xx zz yy xx xx d dEdd dEdd dEdd d d dEdd d d dEdd d d dEd λσσνελσσνελσσνεσσσλ σσνσεσσσλ σσνσεσσσλ σσνσε ++=++=++=⎥⎦⎤ ⎢⎣⎡+− ++− =⎥⎦⎤ ⎢⎣⎡+− ++− =⎥⎦⎤ ⎢⎣⎡+− ++− = 11121 32 121 32 121 32 1 (8.4.7) or ij kk ij ij ij sd dEdEd λσδνσνε + −+=1 These expressions are called the Prandtl-Reuss equations . If the first, elastic, terms are neglected, they are known as the Lévy-Mises equations. 8.4.3 Application: Plane St rain Compression of a Block Consider the plane strain compression of a thic k block, Fig. 8.4.2. The block is subjected to an increasing pressure pxx−=σ , is constrained in the z direction, so 0=zzε , and is free to move in the y direction, so 0=yyσ . Figure 8.4.2: Plane strain compression of a thick block The solution to the elastic problem is obtained from 8.4.7 (disregarding the plastic terms). One finds that { ▲Problem 1} () ()ννεν ενσ σ ++= −−= −=−= 1 , 1 , ,2 Ep Epp pyy xx zz xx (8.4.8) Rigid Walls xy zp Section 8.4 Solid Mechanics Part II Kelly 283and all other stress and strain components are zero. In this elastic phase, the principal stresses are clearly xx zz yy σσσσσσ =>=>==3 2 1 0 (8.4.9) The Prandtl-Reuss equations are [] () () []⎥⎦⎤ ⎢⎣⎡+−+− =+ −+ −=⎥⎦⎤ ⎢⎣⎡− +− = zz xx xx zz zzzz xx zz xx yyzz xx zz xx xx d d dEdd d dEdd d dEd σσλσνσεσσλσσνεσσλσνσε 21 32 13121 32 1 (8.4.10) The magnitude of the plastic straining is determined by the multiplier λd. This can be evaluated by noting that plastic deformation proceeds so long as the stress state remains on the yield surface, the so-called consistency condition . By definition, a perfectly plastic material is one whose yield surface remains unchanged during deformation. A Tresca Material Take now the Tresca yield criterion, which states that yield occurs when Yxx−=σ , where Y is the uniaxial yield stress (in compression). Assume further perfect plasticity, so that Yxx−=σ holds during all subsequent plastic flow. Thus, with 0=xxdσ , and since 0=zzdε , 8.4.10 reduce to () ⎥⎦⎤ ⎢⎣⎡+ +=−+−=⎥⎦⎤ ⎢⎣⎡+−−= zz zzzz zz yyzz zz xx Y d dEYd dEdYd dEd σλσσλσνεσλσνε 21 32 103121 32 (8.4.11) Thus Yd Ed zzzz +−=σσλ23 (8.4.12) and, eliminating λd from Eqns. 8.4.11 { ▲Problem 2}, zz zzzz zz zzzz yyzz zzzz zz zzzz xx dYdYYd EddYdYY d Ed σσσσσσνεσσσσσσνε 2/ 21 2/1 22/ 21 2/1 +++−−=++++−= (8.4.13) Using the relation Section 8.4 Solid Mechanics Part II Kelly 284 ()axax dxaxx+−=+∫ln (8.4.14) and the initial (yield point) conditions, i.e. Eqns. 8.4.8 with Yp=, one can integrate 8.4.13 to get { ▲Problem 2} () () () νσνσνενσνσνε 232121 / 2121ln432212121 / 2121ln43 +−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ +−+=−−−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ +−−= Y Y YEY Y YE zz zzyyzz zzxx , νσ−<Yzz (8.4.15) The stress-strain curves are shown in Fig. 8.4.3 below for 3.0=ν . Note that, for a typical metal, 310~/YE , and so the strains are very small right through the plastic compression; the plastic strains are of comparable size to th e elastic strains. There is a rapid change of stress and then little change once zzσ has approached close to its limiting value of 2/Y− . The above plastic analysis was based on xxσ remaining the minimum principal stress. This assumption has proved to be valid, since zzσ remains between 0 and Y− in the plastic region. Figure 8.4.3: Stress-strain results for plan e strain compression of a thick block for 3.0=ν A Von Mises Material Slightly different results are obtained with the Von Mises yield criterion, Eqn. 8.4.11, which for this problem reads -3-2.5-2-1.5-1-0.50-0.5 -0.4 -0.3 -0.2 -0.1Yzzσ εYE xxεyyε−elasticν− ()21ν−−()νν+−1 Section 8.4 Solid Mechanics Part II Kelly 2852 2 2Yzz zz xx xx =+− σσσσ (8.4.16) The Prandtl-Reuss equations can be solved by making the substitution θ σ cos 32Y xx−= (8.4.17) in the plastic region. In what follows, use is made of the trigonometric relations θθ θπθ θ θπ sin21cos23 6cossin23cos21 6sin + =⎟ ⎠⎞⎜ ⎝⎛−−=⎟ ⎠⎞⎜ ⎝⎛− (8.4.18) From Eqn. 8.4.16, ⎟ ⎠⎞⎜ ⎝⎛− −= θπσ6sin 32Y zz (8.4.19) Substituting into 8.4.10 then leads to ⎭⎬⎫ ⎩⎨⎧+⎥⎦⎤ ⎢⎣⎡−⎟ ⎠⎞⎜ ⎝⎛− =⎭⎬⎫ ⎩⎨⎧ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛−−−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛−+−=⎭⎬⎫ ⎩⎨⎧ ⎟ ⎠⎞⎜ ⎝⎛− −⎥⎦⎤ ⎢⎣⎡⎟ ⎠⎞⎜ ⎝⎛− − = θλ θθνθπεθπθλθθπθν εθπλ θθπνθ ε sin 31sin6cos 326sin cos31 6cos sin 326cos 31 6cos sin 32 Ed d dYEEd d dYEEd d dYE zzyyxx (8.4.20) Using 0=zzdε leads to θθθνθπ λ dEdsinsin6cos3−⎟ ⎠⎞⎜ ⎝⎛− −= (8.4.21) and () θθ θπν ε d dYE xx⎭⎬⎫ ⎩⎨⎧+⎟ ⎠⎞⎜ ⎝⎛− −= cosec43 6cos21 32 (8.4.22) An integration gives () CYE xx + +⎟ ⎠⎞⎜ ⎝⎛− −−=2tanln23 6sin21 32 θθπν ε (8.4.23) Section 8.4 Solid Mechanics Part II Kelly 286To determine the constant of integration, c onsider again the conditions at first yield. Suppose the block first yields when xxσ reaches Y xxσ. Then Y xxY zzνσσ= and 21ννσ +−−=YY xx (8.4.24) Note that in this case it is predicted that first yield occurs when Yxx−<σ . From Eqn. 8.4.17, the value of θ at first yield is 2123cos ννθ +−=Y or 321tanνθ−=Y (8.4.25) Thus, with ()EY xx xx / 12νσε −= at yield, 2tanln2312Y Cθνν−+−−= (8.4.26) and so ()212cot2tanln23 6sin21 32ννθθθπν ε +−+ +⎟ ⎠⎞⎜ ⎝⎛− −=−Y xxYE (8.4.27) This leads to a similar stress-strain curve as for the Tresca criterion, only now the limiting value of zzσ is Y Y 58.0 3/−≈− . 8.4.4 Application: Combined Tension/ Torsion of a thin walled tube Consider now the combined tension/torsion of a thin-walled tube as in the Taylor/Quinney tests. The only stresses in the tube are σσ=xx due to the tension along the axial direction and τσ=xy due to the torsion. The Prandtl-Reuss equations reduce to xy xy xyxx xx zz yyxx xx xx d dEdd dEd dd dEd λσσνελσσνεελσσε ++=−−==+= 13132 1 (8.4.28) Consider the case where the tube is twisted up to the yield point. Torsion is then halted and tension is applied, holding the angle of twist constant. In that case, during the tension, 0 =xydε and so {▲Problem 3} Section 8.4 Solid Mechanics Part II Kelly 287σττνσεd EdEdxx+−=1 32 1 (8.4.29) If one takes the Von Mises criterion, then 2 2 23 Y=+τσ (see Eqn. 8.3.17). Assuming perfect plasticity, one has { ▲Problem 4}, 2 221 32 1 σσσνσε−++=Yd EdEdxx (8.4.30) Using the relation ⎟ ⎠⎞⎜ ⎝⎛ −++−=−∫xaxa ax dxx axln22 22 , (8.4.31) an integration leads to { ▲Problem 5} () () ⎭⎬⎫ ⎩⎨⎧⎟ ⎠⎞⎜ ⎝⎛ −+++−=YY Y YExx/ 1/ 1ln 1 2131 σσνσν ε (8.4.32) This result is plotted in Fig. 8.4.4. Note that, with 2 2 23 Y=+τσ , as σ increases (rapidly) to its limiting value Y, τ decreases from its yield value of 3/Y to zero. Figure 8.4.4: Stress-strain results for comb ined tension/torsion of a thin walled tube for 3.0=ν 8.4.5 The Tresca Flow Rule The flow rule used in the preceding applications was the Prandtl-Reuss rule 8.4.7. Many other flow rules have been proposed. For example, the Tresca flow rule is simply (for 3 2 1σσσ>> ) 00.511.522.5 0.2 0.4 0.6 0.8 1YσxxYEε Section 8.4 Solid Mechanics Part II Kelly 288λεελε d ddd d ppp −==+= 321 0 (8.4.33) This flow rule will be used in the next section, which details the classic solution for the plastic deformation and failure of a thick cylinder under internal pressure. A unifying theory of flow rules will be presen ted in a later section, in which the reason for the name “Treca flow rule” will become clear. 8.4.6 Problems 1. Derive the elastic strains for the plane strain compression of a thick block, Eqns. 8.4.8. 2. Derive Eqns. 8.4.13 and 8.4.15 3. Derive Eqn. 8.4.29 4. Use Eqn. 8.3.17 to show that ()2 2 2/ / σσσττσ − −= Y d d and hence derive Eqn. 8.4.30 5. Derive Eqns. 8.4.32 6. Does the axial stress-stress curve of Fig. 8.4.4 differ when the Tresca criterion is used? 7. Consider the uniaxial straining of a perfectly plastic isotropic Von Mises metallic block. There is only one non-zero strain, xxε. One only need consider two stresses, yy xxσσ, since yy zzσσ= by isotropy. (i) Write down the two relevant Prandtl-Reuss equations (ii) Evaluate the stresses and strains at first yield (iii) For plastic flow, show that yy xxd dσσ= and that the plastic modulus is ()ν εσ 213−=E dd xxxx 8. Consider the combined tension-torsion of a thin-walled cylindrical tube. The tube is made of a perfectly plastic Von Mises metal and Y is the uniaxial yield strength in tension. The only stresses are σσ=xx and τσ=xy and the Prandtl-Reuss equations reduce to Section 8.4 Solid Mechanics Part II Kelly 289xy xy xyxx xx zz yyxx xx xx d dEdd dEd dd dEd λσσνελσσνεελσσε ++=−−==+= 13132 1 The axial strain is increased from zero until yielding occurs (with 0=xyε ). From first yield, the axial strain is held constant and the shear strain is increased up to its final value of E Y 3/) 1(ν+ (i) Write down the yield criterion in terms of σ and τ only and sketch the yield locus in τσ− space (ii) Evaluate the stresses and strains at first yield (iii)Evaluate λd in terms of σσd, (iv) Relate σσd, to ττd, and hence derive a differentia l equation for shear strain in terms of τ only (v) Solve the differential equation and evaluate any constant of integration (vi) Evaluate the shear stress when xyε reaches its final value of E Y 3/) 1(ν+ . Taking 2/1=v , put in the form Yατ= with α to 3 d.p. Section 8.5 Solid Mechanics Part II Kelly 2908.5 The Internally Pressurised Cylinder 8.5.1 Elastic Solution Consider the problem of a long thick hollow cylinder, with internal and external radii a and b, subjected to an internal pressure p. This can be regarded as a plane problem, with stress and strain independent of the axial direction z. The solution to the axisymmetric elastic problem is (see §4.3.5) () zzzz rr zzrr EabpEabrbpabrbp ενεσσνσσσ θθθθ +−=++=−++=−−−= 1 /21 /1 /1 /1 / 2 22 22 22 22 2 (8.5.1) There are no shear stresses and these are the principal stresses. Axial Force The axial force in the tube is the resultant of the zzσ stress: ()[]∫∫∫∫++ = =π θθπ θσσνεθσ2 02 0drdr E drdr Pb arr zzb azz (8.5.2) Assuming the strain zzε to be constant over any cross section, () () ()2 2 22 02 2 2pa a b Edrdr a b EP zzb arr zz νπ πεθσσν πεπ θθ +− =+ +− = ∫∫ (8.5.3) Axial Strain and End Conditions There are three possible end-conditions, assuming zzε to be constant (from which it follows that zzσ is constant): (1) open-ended : the resultant axial force is zero and so, since zzσ is constant, 0=zzσ . This is equivalent to plane stress. (2) closed-ended : the resultant axial force is zzσ times the cylinder’s cross-sectional area ()2 2a b−π and this is balanced by the internal pressure p acting over the end area 2aπ, so that ()()θθσσ σ +=− =rr zz abp21 2 21 // . This is a strain state known as generalised plane strain , where zzε is constant but non-zero. Section 8.5 Solid Mechanics Part II Kelly 291(3) plane strain : it is assumed that 0=zzε , so that ()()θθσσν νσ +=− =rr zz abp 1 //22 2 The end-conditions can be summarised as 1 // 2 2−=abEpzzαε (8.5.4) where ⎪⎩⎪⎨⎧ −− = end open 2strain plane 0end closed 21 νν α (8.5.5) 8.5.2 Plastic Solution The pressure is now increased so that the cylinde r begins to deform plastically. It will be assumed that the material is isotropic and elastic perfectly-plastic and that it satisfies the Tresca criterion. First Yield It can be seen from 8.5.1, 8.5.4-5, that rr zzσσσθθ >>> 0 and so the Tresca criterion reads kabrbprr rr 21 //22 22 2 ≡−=−=− σσσσθθ θθ (8.5.6) This expression has its maximum value at the inner surface, ar=, and hence it is here that plastic flow first begins. From the above, plastic deformation begins when ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−=22 1 bak pflow (8.5.7) irrespective of the end conditions. Confined Plastic Flow and Collapse As the pressure increases above flowp, the plastic region spreads out from the inner face; suppose that it reaches out to cr=. With the material perfectly plastic, the material in the annulus cra<< satisfies the yield condition 8.5.6 at all times. Consider now the equilibrium of this plastic material. Since this is an axi-symmetric problem, there is only one equilibrium equation: Section 8.5 Solid Mechanics Part II Kelly 292() 01=−+θθσσσ rrrr r drd. (8.5.8) It follows that 1 ln2 02Crkrk drd rrrr+=→=− σσ. (8.5.9) The constant of integration can be obtained from the pressure boundary condition at ar=, leading to ()()cra ark prr ≤≤ +−= /ln2 σ (8.5.10) The stresses in the elastic region are again give n by the elastic stress solution 8.5.1, only with a replaced by c and the pressure p is now replaced by the pressure exerted by the plastic region at cr=, i.e. ()ack p /ln2− . The precise location of the boundary c can be obtained by noting that the elastic stresses must satisfy the yield criterion at cr=. Since in the elastic region, () () brccbrback prr ≤≤−−=−1 //)/ln(2 22 22 2 σσθθ (8.5.11) one has from () kcrrr 2=−=σσθθ that )/ 1()/ln(22 2bc k ack p −+ = (8.5.12) Fig. 8.5.1 shows a plot of Eqn. 8.5.12. Figure 8.5.1: Extent of the plastic region cr= during confined plastic flow The complete cylinder will become plastic when c reaches b, or when the pressure reaches the collapse pressure (or ultimate pressure ) )/ln(2 abk pU= . (8.5.13) c ac= bc=kp 2 2/ 1/ ba kp−=()ab kp /ln2 /= Section 8.5 Solid Mechanics Part II Kelly 293This problem illustrates a number of features of elastic-plastic problems in general. First, confined plastic flow occurs. This is where the plastic region is surrounded by an elastic region, and so the plastic strains are of the same order as the elastic strains. It is only when the pressure reaches the collapse pressure does catastrophic failure occur. Stress Field Using 8.5.12, the stresses in the elastic region can be shown to be zz zzrr Ebckrb bckrb bck ενσσσ θθ ++=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ +=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −= 2222 2222 22 211 , brc≤≤ (8.5.14) For the plastic region, the radial and hoop stresses can be obtained from 8.5.10 and 8.5.6. The Tresca flow rule, 8.4.33, implies that 0 =p zzε and zzε is purely elastic. Thus the elastic relation ()zz rr zz Eεσσνσθθ++= holds also in the plastic region, and zz zzrr Erc bckrc bckrc bck ε νσσσ θθ +⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−++=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−−= ln2 2ln2 1ln2 1 222222 , cra≤≤ (8.5.15) To determine the axial strain, consider again the axial force. First, using the equation of equilibrium 8.5.8, ()() ()rrrrrrrr rr rr rdrdrdrdrr rr σσσσσσσσθθ θθ 2222 =+ =+−=+ (8.5.16) Then, from 8.5.2, ()[] ()2 2 22 2 2 22 pa a b Er a b EP zzb arr zz νπ πεσπν πε +− =+− = (8.5.17) This axial force is the same as Eqn. 8.5.3. In other words, although zzσ in general varies in the plastic zone, the axial force is independent of the plastic zone size c. Eqns. 8.5.3-4 are therefore again valid here and Section 8.5 Solid Mechanics Part II Kelly 294 ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−−=ac bc abk Ezz ln2 11 /22 2 2αε (8.5.18) From the third of 8.5.15, 0≥zzσ for the closed-end and plane strain conditions. For the open-end condition, the axial stress is tensile in some parts and negative in other parts (so that the resultant force is 0=P ) – note that this is not now a condition of plane stress. As an example, consider the case of 1=a , 2=b , with 5.1=c . The stresses for this case are plotted in Fig. 8.5.2. . Figure 8.5.2: Stress field in the cylinder for the case of 5.1 ,2 ,1 === c b a Displacement In the elastic region, the strains are given by Hooke’s law ()[] ()[]zz rrzz rr rr EE νενσσννενενσσννε θθ θθθθ −−−+=−−−+= 1111 (8.5.19) From the definition of strain rudrdu rr rr == θθεε (8.5.20) and 8.5.14, ---00.20.40.60.8 1 1.2 1.4 1.6 1.8 2ar/ krr2/σk2/θθσ kzz2/σ plastic elastic closed plane strain open Section 8.5 Solid Mechanics Part II Kelly 295()zz r rrbr bc Ek u εν νν−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−+=2 22 211, brc≤≤ (8.5.21) Now, from 8.4.33, () ( )[] ( ) () ( ) ()[] () ( )()zz rrrr zz rrzz rre e rr rr EEEE νεσσννσσνενσσννσσσνεεεε θθθθ θθθθθθ θθ 221 12 112 11 −+−+=++ −+−=−+−=+=+ (8.5.22) Using 8.5.16 and 8.5.19, one has ()()()()zz rr r r rdrd Erudrdενσνν221 12−−+= (8.5.23) which integrates to ()() rCr rEuzz rr r +−−+= ενσνν 21 1 (8.5.24) Equations 8.5.22-24 are valid in both the elastic and plastic regions. The constant of integration can be obtained from the condition 0=rrσ at br=, when ru equals the elastic displacement 8.5.21, and so () Ec k C / 122 2ν−= and () ( ) () zz rr r rErck rEu εννσνν−−+−+=2 21221 1, cra≤≤ (8.5.25) 8.5.3 Unloading Residual Stress Suppose that the cylinder is loaded beyond flowp but not up to the collapse pressure, to a pressure 0p say. It is then unloaded completely . After unloading the cylinder is still subjected to a stress field – these stresses which are locked into the cylinder are called residual stresses . If the unloading process is fully elastic, the new stresses are obtained by subtracting 8.5.1 from 8.5.14-15. Using Eqn. 8.5.7 { ▲Problem 1}, Section 8.5 Solid Mechanics Part II Kelly 29622 0 2222 22 0 2222 22 0 22 2 ba pp ackba ra pp ackba ra pp ack flowzzflowflowrr ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −−= νσσσ θθ , brc≤≤ (8.5.26) ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −− −=⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −=⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− −= ar ppkar ra ppkar ra ppk flowzzflowflowrr ln21 2ln22 1ln2 1 022 022 0 νσσσ θθ , cra≤≤ (8.5.27) Again consider as an example the case 1=a , 2=b , with 5.1=c , for which 624.02,375.020= =kp kpflow (8.5.28) The residual stresses are as shown in Fig. 8.5.3. Figure 8.5.3: Residual stresses in th e unloaded cylinder for the case of 5.1 ,2 ,1 === c b a Note that the axial strain, being purely elastic, is completely removed, and the axial stress is independent of the end condition. -0.6-0.4-0.200.2 1 1.2 1.4 1.6 1.8 2ar/ krr2/σ k2/θθσkzz2/σ plastic elastic Section 8.5 Solid Mechanics Part II Kelly 297There is the possibility that if the original pressure 0p is very large, the unloading will lead to compressive yield. The maximum value of rrσσθθ− occurs at ar=, where it equals ( )1 / 20−flowppk and so, neglecting any Bauschinger effect, yield will occur if flowp p 20≥ . Yielding will not occur right up to the collapse pressure Up if the wall ratio ab/ is such that flow U p p p 20<= . From 8.5.7 and 8.5.13, this reads as ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−<22 12 lnba ab (8.5.29) The largest wall ratio for which the unloading is completely elastic is 22.2 /≈ab . For larger wall ratios, a new plastic zone will develop at the inner wall in which krr 2−=−σσθθ . Shakedown When the cylinder is initially loaded, plasticity begins at a pressure flowpp= . If it is loaded to some pressure 0p,with flow flow p p p 20<< , then unloading will be completely elastic. When the cylinder is reloaded again it will remain elastic up to pressure 0p. In this way, it is possible to strengthen the cylinder by an initial loading; theoretically it is possible to increase the flow pressure by a factor of 2. This maximum possible new flow pressure is called the shakedown pressure ()U flow s p p p , 2min= . Shakedown is said to have occurred when any subsequent loading/unloading cycles are purely elastic. The strengthening of the cylinder is due to the compressive residual hoop stresses at the inner wall – similar to the way a barrel can be strengthened with hoops. This method of strengthening is termed autofrettage , a French term meaning “self-hooping”. 8.5.4 Validity of the Solution One needs to check whether the assumption of the ordering of the principal stresses, rr zzσσσθθ>> , holds through the deformation. It can be confirmed that the inequality rr zzσσ≥ always holds. For the inequality zzσσθθ≥ , consider the inequality 0≥−zzσσθθ . The quantity on the left is a minimum when ar=, where it equals () ⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+− −−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−+−ac bc ab ac bck ln2 1 1 /2 ln2 12122 2 2 22αν ν (8.5.30) This quantity must be positive for all values of c up to the maximum value b, where it takes its minimum value, and so one must have () 0 ln21 /2 ln12122 2≥⎟ ⎠⎞⎜ ⎝⎛ −−+⎟ ⎠⎞⎜ ⎝⎛−−ab ab ab αν ν (8.5.31) Section 8.5 Solid Mechanics Part II Kelly 298The solution is thus valid only for limited values of ab/. For 3.0=ν , one must have 43.5 /<ab (closed ends), 75.5 /<ab (plane strain), 19.6 /<ab (open ends). For higher wall ratios, the axial stress becomes equal to the hoop stress. In this case, a solution based on large changes in geometry is necessary for higher pressures. 8.5.5 Problems 1. Derive Eqns. 8.5.26-27. Section 8.6 Solid Mechanics Part II 299 Kelly Hardening In the applications discussed in the preceding two sections, the material was assumed to be perfectly plastic. The issue of hardening (softening) materials is addressed in this section. 8.6.1 Hardening In the one-dimensional (uniaxial test) case, a specimen will deform up to yield and then generally harden, Fig. 8.6.1. Also shown in the figure is the perfectly-plastic idealisation. In the perfectly plastic case, once the stress reaches the yield point (A), plastic deformation ensues, so long as the stress is maintained at Y. If the stress is reduced, elastic unloading occurs. In the hardening case, once yield occurs, the stress needs to be continually increased in order to drive the plastic deformation. If the stress is held constant, for example at B, no further plastic deformation will occur; at the same time, no elastic unloading will occur. Note that this condition cannot occur in the perfectly-plastic case, where there is one of plastic deformation or elastic unloading. Figure 8.6.1: uniaxial stress-strain curve (for a typical metal) These ideas can be extended to the multiaxial case, where the initial yield surface will be of the form 0)(0=ijfσ (8.6.1) In the perfectly plastic case, the yield surface remains unchanged.. In the more general case, the yield surface may change size, shape and position, and can be described by 0),(=Κi ijfσ (8.6.2) Here,iK represents one or more hardening parameters , which change during plastic deformation and determine the evolution of the yield surface. They may be scalars or hardening 0ABstress strain Yield point Y •perfectly-plastic • elastic unload Section 8.6 Solid Mechanics Part II 300 Kelly higher-order tensors. At first yield, the hardening parameters are zero, and ) ( )0,(0 ij ij f f σ σ= . The description of how the yield surface changes with plastic deformation, Eqn. 8.6.2, is called the hardening rule . Strain Softening Materials can also strain soften , for example soils. In this case, the stress-strain curve “turns down”, as in Fig. 8.6.2. The yield surface for such a material will in general decrease in size wi th further straining. Figure 8.6.2: uniaxial stress-strain curve for a strain-softening material 8.6.2 Hardening Rules A number of different hardening rules are discussed in this section. Isotropic Hardening Isotropic hardening is where the yield surface remains the same shape but expands with increasing stress, Fig. 8.6.3. In particular, the yield function takes the form 0 )( ),( 0 =Κ−=Κij i ij f f σ σ (8.6.3) The shape of the yield function is specified by the initial yield function and its size changes as the hardening parameter Κ changes. 0stress strain Section 8.6 Solid Mechanics Part II 301 Kelly Figure 8.6.3: isotropic hardening For example, consider the Von Mises yield surface. At initial yield, one has () () () () Y ssY JY f ijijij − =−=−−+−+− = 2322 1 32 3 22 2 1 0 321σσσσσσ σ (8.6.4) where Y is the yield stress in uniaxial tension. Subsequently, one has () 0 3 ,2=Κ−−=Κ Y J fi ijσ (8.6.5) The initial cylindrical yield surface in stress-space with radius Y32 (see Fig. 8.3.11) develops with radius ()Κ+Y32. The details of how the hardening parameter Κ actually changes with plastic deformation have not yet been specified. As another example, consider the Drucker-Prager criterion, Eqn. 8.3.30, () 02 1 0 =−+= k J I fijασ . In uniaxial tension, YI=1 , 3/2Y J= , so ()Y k 3/1+=α . Isotropic hardening can then be expressed as () () 0 3/11,2 1 =Κ−−+ +=Κ Y J I fi ij α ασ (8.6.6) Kinematic Hardening The isotropic model implies that, if the yield strength in tension and compression are initially the same, i.e. the yield surface is symmetric about the stress axes, they remain equal as the yield surface develops with plastic strain. In order to model the Bauschinger effect, and similar responses, wh ere a hardening in tension will lead to a initial yield surface subsequent yield surface 1σ2σ •• stress at initial yield elastic loading elastic unloading plastic deformation (hardenin g) Section 8.6 Solid Mechanics Part II 302 Kelly softening in a subsequent compression, one can use the kinematic hardening rule. This is where the yield surface remains the same shape and size but merely translates in stress space, Fig. 8.6.4. Figure 8.6.4: kinematic hardening The yield function now takes the general form 0) ( ),(0 =−=Κij ij i ij f f ασ σ (8.6.7) The hardening parameter here is the stress ijα, known as the back-stress or shift- stress ; the yield surface is shifted relative to the stress-space axes by ijα, Fig. 8.6.5. Figure 8.6.5: kinematic hardening; a shift by the back-stress For example, again considering the Von Mises material, one has, from 8.6.4, and using the deviatoric part of ασ− rather than the deviatoric part of σ, () 0 ) )( ( ,23=−−−=Κ Y s s fd ij ijd ij ij i ij αα σ (8.6.8) where dα is the deviatoric part of α. Again, the details of how the hardening parameter ijα might change with deformation will be discussed later. 1σ2σ ••initial yield surface subsequent loading surface ijαinitial yield surface subsequent yield surface 1σ2σ •• stress at initial yield elastic loading plastic deformation (hardenin g) elastic unloading Section 8.6 Solid Mechanics Part II 303 Kelly Other Hardening Rules More complex hardening rules can be used. For example, the mixed hardening rule combines features of both the isotropic and kinematic hardening models, and the loading function takes the general form () 0 ) ( ,0 =Κ−−=Κij ij i ij f f ασ σ (8.6.9) The hardening parameters are now the scalar Κ and the tensor ijα. 8.6.3 The Flow Curve In order to model plastic deformation and hardening in a complex three-dimensional geometry, one will generally have to us e but the data from a simple test. For example, in the uniaxial tension test, one will have the data shown in Fig. 8.6.6a, with stress plotted against plastic strain. The idea now is to define a scalar effective stress σˆ and a scalar effective plastic strain pεˆ, functions respectively of the stresses and plastic strains in the loaded body. The following hypothesis is then introduced: a plot of effective stress against effective plastic strain follows the same universal plastic stress-strain curve as in the uniaxial case. This assumed universal curve is known as the flow curve . The question now is: how should one define the effective stress and the effective plastic strain? Figure 8.6.6: the flow curve; (a) uniaxial stress – plastic strain curve, (b) effective stress – effective plastic strain curve 8.6.4 A Von Mises Material with Isotropic Hardening Consider a Von Mises material. Here, it is appropriate to define the effective stress to be ()23 ˆ Jij=σσ (8.6.10) 0Y pεσ pddHεσ≡()phεσ= 0Y pεˆσˆ pddHεσ ˆˆ≡()phεσ ˆ ˆ= )a() b( Section 8.6 Solid Mechanics Part II 304 Kelly This has the essential property that, in the uniaxial case, ()Yij=σσˆ . (In the same way, for example, the effective stress fo r the Drucker-Prager material, Eqn. 8.6.6, would be ()() )3/1 /( ˆ2 1 + += α ασσ J Iij .) For the effective plastic strain, one possibility is to define it in the following rather intuitive, non-rigorous, way. The deviatoric stress s and plastic strain (increment) tensor pdε are of a similar character. In particular, their traces are zero, albeit for different physical reasons; 01=J because of independence of hydrostatic pressure, 0=p iidε because of material incompressibility in the plastic range. For this reason, one chooses the effective pl astic strain (increment) pdεˆ to be a similar function of p ijdε as σˆ is of the ijs. Thus, in lieu of ijijss23ˆ=σ , one chooses p ijp ijpddC d εε ε=ˆ . One can determine the constant C by ensuring that the expression reduces to p pd d1ˆεε= in the uniaxial case. Considering this uniaxial case, p p p p pd d d d d1 21 33 22 1 11 , εεεεε −== = , one finds that () () ()2 1 32 3 22 2 132 32ˆ p p p p p pp ijp ijp d d d d d ddd d εεεεεεεεε −+−+− == (8.6.11) Let the hardening in the uniaxial tension case be described using a relationship of the form (see Fig. 8.6.6) ()phεσ= (8.6.12) The slope of this flow curve is the plastic modulus, Eqn. 8.1.9, pddHεσ= (8.6.13) The effective stress and effective plastic strain for any conditions are now assumed to be related through ()phεσ ˆ ˆ= (8.6.14) and the effective plastic modulus is given by pddHεσ ˆˆ= (8.6.15) Isotropic Hardening Assuming isotropic hardening, the yield surface is given by Eqn. 8.6.5, and with the definition of the effective stress, Eqn. 8.6.10, Section 8.6 Solid Mechanics Part II 305 Kelly () 0 ˆ , =Κ−−=Κ Y fi ijσσ (8.6.16) Differentiating with respect to the effective plastic strain, p pH εεσ ˆ ˆˆ ∂Κ∂= ∂∂= (8.6.17) One can now see how the hardening parameter evolves with deformation: Κ here is a function of the effective plastic strain, and its functional dependence on the effective plastic strain is given by the plastic modulus H of the universal flow curve. Loading Histories Each material particle undergoes a plastic strain history. One such path is shown in Fig. 8.6.7. At point q, the plastic strain is ) (q p iε . The effective plastic strain at q must be evaluated through an integration over the complete history of deformation: ∫∫==q p ip iq p pdd d q 032 0ˆ )(ˆ εε ε ε (8.6.18) Note that the effective plastic strain at q is not simply )()(32q qp ip iεε , hence the definition of an effective plastic strain increment in Eqn. 8.6.11. Figure 8.6.7: plastic strain space Prandtl-Reuss Relations in terms of Effective Parameters Using the Prandtl-Reuss (Levy-Mises) flow rule 8.4.1, and the definitions 8.6.10-11 for effective stress and effective plastic strain, one can now express the plastic multiplier as{ ▲Problem 1} p 3ε • )(qpε p 2ε p 1εqstrain path Section 8.6 Solid Mechanics Part II 306 Kelly σελˆˆ 23pdd= (8.6.19) and the plastic strain increments, Eqn. 8.4.6, now read ()() [ ] () ()[] () ()[] () () ()zxp p zxyzp p yzxyp p xyyy xx zzp p zzxx zz yyp p yyzz yy xxp p xx d dd dd dd dd dd d σσεεσσεεσσεεσσσσεεσσσσεεσσσσεε ˆ/ˆˆ/ˆˆ/ˆˆ/ˆˆ/ˆˆ/ˆ 232323212121 ===+− =+− =+− = (8.6.20) or ijp p ij sddσεεˆˆ 23= . (8.6.21) Knowledge of the plastic modulus, E qn. 8.6.15, now makes equations 8.6.21 complete. Note here that the plastic modulus in the Prandtl-Reuss equations is conveniently expressible in a simple way in terms of the effective stress and plastic strain increment, Eqn. 8.6.19. It will be shown in the next section that this is no coincidence, and that the Prandtl-Reuss flow-rule is indeed naturally associated with the Von-Mises criterion. 8.6.5 Application: Combined Tension/Torsion of a thin walled tube with Isotropic Hardening Consider again the thin-walled tube under combined tension and torsion. The Von Mises yield function in terms of the axial stress σ and the shear stress τ is, as in §8.3.1, () 0 32 2 0 =−+= Y fij τσσ . This defines the ellipse of Fig. 8.3.2. Subsequent yield surfaces are defined by () () () 0ˆ3 , 02 2 =+−=−=−−+=Κ KYK fKY f iji ij σστσ σ (8.6.22) Whereas the initial yield surface is the ellipse with major and minor axes Y and 3/Y , subsequent yield ellipses have axes KY+ and 3/) ( KY+ , Fig. 8.6.8. Section 8.6 Solid Mechanics Part II 307 Kelly Figure 8.6.8: expansion of the yield locu s (ellipse) for a thin-walled tube under isotropic hardening The Prandtl-Reuss equations in terms of effective stress and effective plastic strain, 8.6.20-21, reduce to τσετνεσσεσνεεσσεσε ˆˆ 23 1ˆˆ 21ˆˆ 1 p xyp zz yyp xx ddEdddEd dddEd ++=−−==+= (8.6.23) Consider the case where the material is brought to first yield through tension only, in which case the Von Mises condition reduces to Y=σ . Let the material then be subjected to a twist whilst maintaining the axial stress constant. The expansion of the yield surface is then as shown in Fig. 8.6.9. Figure 8.6.9: expansion of the yield locus for a thin-walled tube under constant axial loading Introducing the plastic modulus, then, one has στ 3/Y Y KY+3/) (Κ+Y στ Yplastic loading Section 8.6 Solid Mechanics Part II 308 Kelly τσστνεσσεεσσε ˆˆ1 23 1ˆˆ1 21ˆˆ1 d HdEdYd Hd dYd Hd xyzz yyxx ++=−=== (8.6.24) Using 2 23 ˆ τ σ+=Y , 3/1 23 13/ 213/ 2 222 22 2 Yd HdEdYd HYd dYd HYd xyzz yyxx +++=+−==+= ττττνετττεετττε (8.6.25) These equations can now be integrated. If the material is linear hardening , so H is constant, then they can be integrated exactly using () ⎟ ⎠⎞⎜ ⎝⎛−=++=+ ∫ ∫axax dxa xxa x dxa xxarctan , ln21 2 22 2 2 2 2 (8.6.26) leading to { ▲Problem 2} ⎥⎦⎤ ⎢⎣⎡⎟ ⎠⎞⎜ ⎝⎛− +⎟ ⎠⎞⎜ ⎝⎛+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ −==⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ += Y YHE Y YEY HE YE YEY HE YE xyzz yyxx τ τ τνετεετε 3 arctan 31 23) 1(31ln431ln211 22 022 (8.6.27) Results are presented in Fig. 8.6.10 for the case of 10 /,3.0 = = HEν . The axial strain grows logarithmically and is eventu ally dominated by the faster-growing shear strain. Section 8.6 Solid Mechanics Part II 309 Kelly Figure 8.6.10: Stress-strain curves for thin-walled tube with isotropic linear strain hardening 8.6.6 Kinematic Hardening Rules A typical uniaxial kinematic hardening curve is shown in Fig. 8.6.11a (see Fig. 8.1.3). During cyclic loading, the elastic zone always remains at Y2. Depending on the stress history, one can even have the situation shown in Fig. 8.6.11b, where yielding occurs upon unloading, even though the stress is still tensile. Figure 8.6.11: Kinematic Hardening; (a) load-unload, (b) cyclic loading The multiaxial yield function for a kinematic hardening Von Mises is given by Eqn. 8.6.8, () 0 ) )( ( ,23=−−−=Κ Y s s fd ij ijd ij ij i ij αα σ The deviatoric shift stress d ijα describes the shift in the centre of the Von Mises cylinder, as viewed in the π-plane, Fig. 8.6.12. This is a generalisation of the σ Y2first yield Y ε•• •begin unload yield in compression )a() b(σ ε• yield 02468 0.2 0.4 0.6 0.8 1 YτεYE xxε xyε Section 8.6 Solid Mechanics Part II 310 Kelly uniaxial case, in that the radius of the Von Mises cylinder remains constant, just as the elastic zone in the uniaxial case remains constant (at Y2). Figure 8.6.12: The Von Mises cylinder shifted in the π-plane One needs to specify, by specifying the evolution of the hardening paremter α, how the yield surface shifts with deformation. In the multiaxial case, one has the added complication that the direction in which the yield surface shifts in stress space needs to be specified. The simplest model is the linear kinematic (or Prager’s ) hardening rule. Here, the back stress is assumed to depend on the plastic strain according to p ij ijp ij ij cd d c εα εα = = or (8.6.28) where c is a material parameter, which might change with deformation. Thus the yield surface is translated in the same directi on as the plastic strain increment. This is illustrated in Fig. 8.6.13, where the principa l directions of stress and plastic strain are superimposed. Figure 8.6.13: Linear kinematic hardening rule One can use the uniaxial (possibly cyclic) curve to again define a universal plastic modulus H. Using the effective plastic strain, one can relate the constant c to H. This will be discussed in §8.8, where a more general formulation will be used. Ziegler’s hardening rule is ()()ij ijp ij ijda d ασεα − = (8.6.29) 1σ′ 2σ′3σ′ s dαdαs− pd1 1,εσpd2 2,εσ ••pdε pcd dεα= Section 8.6 Solid Mechanics Part II 311 Kelly where a is some scalar function of the plastic strain. Here, then, the loading function translates in the direction of ij ijασ− , Fig. 8.6.14. Figure 8.6.14: Ziegler’s kinematic hardening rule 8.6.7 Strain Hardening and Work Hardening In the models considered above, the hardening parameters have been functions of the plastic strains. For example, in the Von Mises isotropic hardening model, the hardening parameter Κ is a function of the effective plastic strain, pεˆ. Hardening expressed in this way is called strain hardening . Another means of generalising the uniaxial results to multiaxial conditions is to use the plastic work (per unit volume), also known as the plastic dissipation , p ij ijpd dW εσ= (8.6.30) The total plastic work is the area under the stress – plastic strain curve of Fig. 8.6.6a, ∫=p ij ijpd W εσ (8.6.31) A plot of stress against the plastic work can therefore easily be generated, as in Fig. 8.6.15. Figure 8.6.15: uniaxial stress – plastic work curve (for a typical metal) 0Y p pd W εσ∫=σ pdWdσ()pWw=σ1σ2σ ••αd• ασ−σ Section 8.6 Solid Mechanics Part II 312 Kelly The stress is now expressed in the form (compare with Eqn. 8.6.12) ()()p pd w Ww εσ σ ∫== (8.6.32) Again defining an effective stress σˆ, the universal flow curve to be used for arbitrary loading conditions is then (compare with Eqn. 8.6.14) ()pWw=σˆ (8.6.33) where now pW is the plastic work during the multiaxial deformation. This is known as a work hardening formulation. Equivalence of Strain and Work Hardening for the Isotropic Hardening Von Mises Material Consider the Prandtl-Reuss flow rule, Eqn. 8.4.1, λε ds dip i= (other flow rules will be examined more generally in §8.7). In this case, working with principal stresses, the plastic work increment is (see Eqns. 8.2.7-10) () () ()[] λσσσσσσλσεσ ddsd dW iip i ip 2 1 32 3 22 2 131−+−+−=== (8.6.34) Using the Von Mises effective stress 8.6.10, and Eqn. 8.6.19, pp dd dW εσλσ ˆˆˆ2 32 == (8.6.35) where pεˆ is the very same effective plastic strain as used in the strain hardening isotropic model, Eqn. 8.6.11. Although true for the Von Mises yield condition, this will not be so in general. 8.6.8 Problems 1. Staring with the definition of the effective plastic strain, Eqn. 8.6.11, and using Eqn. 8.4.1, ip i sd dλε= , derive Eqns. 8.6.19, σελˆˆ 23pdd= 2. Integrate Eqns. 8.6.25 and use the initia l (first yield) conditions to get Eqns. 8.6.27. Section 8.6 Solid Mechanics Part II 313 Kelly 3. Consider the combined tension-torsion of a thin-walled cylindrical tube. The tube is made of an isotropic hardening Von Mises metal with uniaxial yield stress Y. The strain-hardening is linear with plastic modulus H. The tube is loaded, keeping the ratio 3 /=τσ at all times throughout the elasto-plastic deformation, until Y=σ . (i) Show that the stresses and strains at first yield are given by EYv EYY YY xyY xxY Y 61, 21, 61, 21 += = = = ε ε τ σ (ii) The Prandtl-Reuss equations in terms of the effective stress and effective plastic strain are given by Eqns. 8.6.23. Eliminate τ from these equations (using 3 /=τσ ). (iii) Eliminate the effective plastioc strain using the plastic modulus. (iv) The effective stress is defined as 2 23 ˆ τσσ+= (see Eqn. 8.6.22). Eliminate the effective stress to obtain σ σνεσσε dHdEddHdEd xyxx 1 23 1 311 1 ++=+= (v) Integrate the differential equations and evaluate any constants of integration (vi) Hence, show that the strains at the final stress values Y=σ , 3/Y=τ are given by ⎟ ⎠⎞⎜ ⎝⎛− ++=⎟ ⎠⎞⎜ ⎝⎛−+= 21123 31211 1 HE YEHE YE xyxx νεε (vii) Sketch the initial yield (elliptical) locus and the final yield locus in ()τσ, space and the loading path. (viii) Plot σ against xxε. Section 8.7 Solid Mechanics Part II Kelly 3148.7 Associated and Non-associated Flow Rules Recall the Levy-Mises flow rule, Eqn. 8.4.3, ijp ij sd dλε= (8.7.1) The plastic multiplier can be determined from the hardening rule. Given the hardening rule one can more generally, instead of the particular flow rule 8.7.1, write ijp ij Gd dλε= , (8.7.2) where ijG is some function of the stresses and perhaps other quantities, for example the hardening parameters. It is symmetric because the strains are symmetric. A wide class of material behaviour (perha ps all that one would realistically be interested in) can be modelled using the general form ijp ijgd dσλε∂∂= . (8.7.3) Here, g is a scalar function which, when differentiated with respect to the stresses, gives the plastic strains. It is called the plastic potential . The flow rule 8.7.3 is called a non-associated flow rule . Consider now the sub-class of materials whose plastic potential is the yield function, fg=: ijp ijfd dσλε∂∂= . (8.7.4) This flow rule is called an associated flow-rule , because the flow rule is associated with a particular yield criterion. 8.7.1 Associated Flow Rules The yield surface ()0=ijfσ is displayed in Fig 8.7.1. The axes of principal stress and principal plastic strain are also shown; the material being isotropic, these are taken to be coincident. The normal to the yield surface is in the direction ijfσ/∂ and so the associated flow rule 8.7.4 can be interpreted as saying that the plastic strain increment vector is normal to the yield surface , as indicated in the figure. This is called the normality rule . Section 8.7 Solid Mechanics Part II Kelly 315 Figure 8.7.1: Yield surface The normality rule has been confirmed by many experiments on metals. However, it is found to be seriously in error for soils and rocks, where, for example, it overestimates plastic volume expansion. For these materials, one must use a non- associative flow-rule. Next, the Tresca and Von Mises yield criteria will be discussed. First note that, to make the differentiation easier, the associated flow-rule 8.7.4 can be expressed in terms of principal stresses as ip ifd dσλε∂∂= . (8.7.5) Tresca Taking 3 2 1σσσ>> , the Tresca yield criterion is k f −−=23 1σσ (8.7.6) One has 21,0 ,21 3 2 1−=∂∂=∂∂+=∂∂ σ σ σf f f (8.7.7) so, from 8.7.5, the flow-rule associated with the Tresca criterion is ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −+ = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 2121 321 0λ εεε d ddd ppp . (8.7.8) This is the flow-rule of Eqns. 8.4.33. The plastic strain increment is illustrated in Fig. 8.7.2 (see Fig. 8.3.9). All plastic deformation occurs in the 31− plane. Note that 8.7.8 is independent of stress. ••pdε pd1 1,εσpd2 2,εσ pd3 3,εσσd Section 8.7 Solid Mechanics Part II Kelly 316 Figure 8.7.2: The plastic strain increment vector and the Tresca criterion in the π-plane (for the associated flow-rule) Von Mises The Von Mises yield criterion is 02 2=−= k Jf . With () () ()[] ()⎥⎦⎤ ⎢⎣⎡+−=−+−+−∂∂=∂∂ 3 2 12 1 32 3 22 2 1 1 12 21 32 61σσσ σσσσσσσσJ (8.7.9) one has () ( ) ()() ()() ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ +−+−+− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ 2 1 21 3 323 1 21 2 323 2 21 1 32 321 σσσσσσσσσ λ εεε d ddd ppp. (8.7.10) This are none other than the Levy-Mises flow rule 8.4.61. The associative flow-rule is very appealing, connecting as it does the yield surface to the flow-rule. Many attempts have been made over the years to justify this rule, both mathematically and physically. However, it should be noted that the associative flow- rule is not a law of nature by any means. It is simply very convenient. That said, it 1 note that if one were to use the alternative but equivalent expression 02=−= k J f , one would have a 22/1 J term common to all three principal strain increments, which could be “absorbed” into the λd giving the same flow-rule 8.7.10 1σ′ 2σ′3σ′ 023 1=−−= k fσσ• pdε Section 8.7 Solid Mechanics Part II Kelly 317does agree with experimental observations of many plastically deforming materials, particularly metals. In order to put the notion of associative flow-rules on a sounder footing, one can define more clearly the type of material for which the associative flow-rule applies; this is tied closely to the notion of stable and unstable materials. 8.7.2 Drucker’s Postulate Stress Cycles First, consider the one-dimensional loading of a hardening material. The material may have undergone any type of deformat ion (e.g. elastic or plastic) and is now subjected to the stress *σ, point A in Fig. 8.7.3. An additional load is now applied to the material, bringing it to the current yield stress σ at point B (if *σ is below the yield stress) and then plastically (great ly exaggerated in the figure) through the infinitesimal increment σd to point C. It is conventional to call these additional loads the external agency . The external agency is then removed, bringing the stress back to *σ and point D. The material is said to have undergone a stress cycle . Figure 8.7.3: A stress cycle for a hardening material Consider now a softening material, Fig. 8.7.4. The external agency first brings the material to the current yield stress σ at point B. To reach point C, the loads must be reduced. This cannot be achieved with a stress (force) control experiment, since a reduction in stress at B will induce el astic unloading towards A. A strain (displacement) control must be used, in wh ich case the stress required to induce the (plastic) strain will be seen to drop to σσd+ ( 0<σd ) at C. The stress cycle is completed by unloading from C to D. εσd *σA DB σCpdε Section 8.7 Solid Mechanics Part II Kelly 318 Figure 8.7.4: A stress cycle for a softening material Suppose now that *σσ= , so the material is at point B, on the yield surface, before action by the external agency. It is now not possible for the material to undergo a stress cycle, since the stress cannot be increased. This provides a means of distinguishing between strain hardening and softening materials: Strain-hardening … Material ca n always undergo a stress-cycle Strain-softening … Material can not always undergo a stress-cycle Drucker’s Postulate The following statements define a stable material : (these statements are also known as Drucker’s postulate ): (1) Positive work is done by the external agency during the application of the loads (2) The net work performed by the external agency over a stress cycle is nonnegative By this definition, it is clear that a strain hardening material is stable (and satisfies Drucker’s postulates). For example, considering plastic deformation ( *σσ= in the above), the work done during an increment in stress is εσdd . The work done by the external agency is the area shaded in Fig. 8.7.5a and is clearly positive (note that the work referred to here is not the total work, ∫+εε εεσdd, but only that part which is done by the external agency2). Similarly, the net work over a stress cycle will be positive. On the other hand, note that plastic loading of a softening (or perfectly plastic) material results in a non-positive work, Fig. 8.7.5b. 2 the laws of thermodynamics insist that the total work is positive (or zero) in a complete cycle. εσd *σA DB σ C pdε Section 8.7 Solid Mechanics Part II Kelly 319 Figure 8.7.5: Stable (a) and un stable (b) stress-strain curves The work done (per unit volume) by the additional loads during a stress cycle A-B-C-D is given by: ()()∫ −−−− = DCBAd W εσεσ* (8.7.11) This is the shaded work in Fig. 8.7.6. Writing p ed d d εεε+= and noting that the elastic work is recovered, i.e. the net work due to the elastic strains is zero, this work is due to the plastic strains, ()()∫ −−= CBpd W εσεσ* (8.7.12) With σd infinitesimal, this equals ()p pdd d W εσεσσ21 *+−= (8.7.13) Figure 8.7.6: Work W done during a stress cycle of a strain-hardening material The requirement (2) of a stable material is that this work be non-negative, () 021 *≥ +−=p pdd d W εσεσσ (8.7.14) εσd *σA DBσCpdε Wεσd εd 0>εσddσd εd 0<εσdd ε )a() b( Section 8.7 Solid Mechanics Part II Kelly 320 Making σσσ d>>−*, this reads () 0*≥−pdεσσ (8.7.15) On the other hand, making *σσ= , it reads 0≥pddεσ (8.7.16) The three dimensional case is illustrated in Fig. 8.7.7, for which one has () 0 ,0*≥ ≥−p ij ijp ij ij ij dd d εσ εσσ (8.7.17) Figure 8.7.7: Stresses during a loading/unloading cycle 8.7.3 Consequences of the Drucker’s Postulate The criteria that a material be stable have very interesting consequences. Normality In terms of vectors in principal stress (plastic strain increment) space, Fig. 8.7.8, Eqn. 8.7.17 reads () 0*≥⋅−pdεσσ (8.7.18) These vectors are shown with the solid lines in Fig. 8.7.8. Since the dot product is non-negative, the angle between the vectors *σσ− and pdε (with their starting points coincident) must be less than 90o. This implies that the plastic strain increment vector must be normal to the yield surface since, if it were not, an initial stress state *σ could be found for which the angle was greater than 90o (as with the dotted vectors in Fig. 8.7.8). Thus a consequence of a material satisfying the stability requirements is that the normality rule holds, i.e. the flow rule is associative, Eqn. 8.7.4. initial yield surfaceijσ ••• * ijσnew yield surface Section 8.7 Solid Mechanics Part II Kelly 321 Figure 8.7.8: Normality of the pl astic strain increment vector When the yield surface has sharp corners, as with the Tresca criterion, it can be shown that the plastic strain increment vector must lie within the cone bounded by the normals on either side of the corner, as illustrated in Fig. 8.7.9. Figure 8.7.9: The plastic strain increment vector for sharp corners Convexity Using the same arguments, one cannot have a yield surface like the one shown in Fig. 8.7.10. In other words, the yield surface is convex : the entire elastic region lies to one side of the tangent plane to the yield surface3. Figure 8.7.10: A non-convex surface 3 note that when the plastic deformation affects the elastic response of the material, it can be shown that the stability postulate again ensures normality, but that the convexity does not necessarily hold •pdε *σσ− •*σ tangent plane convex surface •pdε*σσ ••pdε *σσ−•*σ Section 8.7 Solid Mechanics Part II Kelly 322In summary then, Drucker’s Postulate, which is satisfied by a stable, strain-hardening material, implies normality (associative flow rule) and convexity4. 8.7.4 The Principle of Ma ximum Plastic Dissipation The rate form of Eqn. 8.7.18 is () 0*≥−p ij ij ijεσσ & (8.7.19) The quantity p ijijεσ& is called the plastic dissipation , and is a measure of the rate at which energy is being dissipated as deformation proceeds. Eqn. 8.7.19 can be written as p ijijp ijijεσεσ &&*≥ or p pεσεσ &&⋅≥⋅* (8.7.20) and in this form is known as the principle of maximum plastic dissipation : of all possible stress states * ijσ (within or on the yield surface), the one which arises is that which requires the maximum plastic work. Although the principle of maximum plastic dissipation was “derived” from Drucker’s postulate in the above, it is more general, holding also for the case of perfectly plastic and softening materials. To see this, disregard stress cycles and consider a stress state *σ which is at or below th e current (yield) stress σ, and apply a strain 0>εd . For a perfectly plastic material, 0*≥−σσ and 0 >=pd dεε . For a softening material, again 0*≥−σσ and 0<edε , 0>>εε d dp. It follows that the normality rule and convexity hold also for the perfectly plastic and softening materials which satisfy the principle of maximum plastic dissipation. In summary: Drucker’s postulate leads to the Principle of maximum plastic dissipation For hardening materials Principle of maximum plastic dissipation leads to Drucker’s postulate For softening materials Principle of maximum plastic dissipation does not lead to Drucker’s postulate Finally, note that, for many materials, hardening and softening, a non-associative flow rule is required, as in Eqn. 8.7.3. Here, the plastic strain increment is no longer normal to the yield surface and the principle of maximum plastic dissipation does not hold in general. In this case, when there is hardening, i.e. the stress increment is directed out from the yield surface, it is easy to see that one can have 0<p ij ijddεσ , Fig. 8.7.11, contradicting the stability postu late (1),. With hardening, there is no 4 it also ensures the uniqueness of solution to the boundary value elastoplastic problem Section 8.7 Solid Mechanics Part II Kelly 323obvious instability, and so it could be argued that the use of the term “stability” in Drucker’s postulate is inappropriate. Figure 8.7.11: plastic strain increment vector not normal to the yield surface; non-associated flow-rule 8.7.5 Problems 1. Derive the flow-rule associated with the Drucker-Prager yield criterion k J I f −+=2 1α 2. Derive the flow-rule associated with the Mohr-Coulomb yield criterion, i.e. with 3 2 1σσσ>> , k=− 23 1σασ Here, 20 ,1sin1sin1 πφφφα <<>−+= Evaluate the volumetric plastic strain increment, that is p p pp d d dVV 3 2 1 εεε++=ΔΔ, and hence show that the model predicts dilatancy (expansion). 3. Consider the plastic potential k g −−=23 1σβσ Derive the non-associative flow-rule corre sponding to this poten tial. Hence show that compaction of material can be modell ed by choosing an appropriate value of β. σ••pdε σd A1Answers to Selected Problems: Part II, Chapter 1 1.1 2. Yes 3. One of the equations of e quilibrium is not satisfied. 4. ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− = 1323 xxx b 5. ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−− = 81.9) 2() 2() 2( 2 3 3 3402 2 2 3402 1 1 340 x xx xx x a 1.2 1. ()()1 , 1 , 2 ,32 21 2 21+= +−= = = yA y A Axy Az xy yy xx ω ε ε ε 2. C CxBx u CyA x uz y x += ++= −+= αω α α , 2 ,2 21 3. 42 814 23 61 2 3 31 4 32 21, , C Ax AxxC C Ax Ax Ay uyC CyAx u zy x +−=++−+= −+= ω 1.3 1. [ ]2 2 2 21 213 , , ,2 , y x y Ax Ax Ay AyA z xy yy xx +± = = = = ω ε ε ε A2Answers to Selected Problems: Part II, Chapter 2 2.1 1. () 2/xLxEg−ρ 2. ()()() [ ] () () () () ( ) () xLxLxL xu −×+×=−×+×=−×+×= − −− − 4 87 37 3 1085.3 100.71083.1 103.31083.1 103.3 σε&& 2.2 2. []∑∞ =++Ω⎟ ⎠⎞⎜ ⎝⎛ ΩΩ+Ω= 1) sin() sin( ) cos(sin sin tan cos ),( nn n n n n x ct B ct At xcLcxctxu λλ λα K,2,1 ,2)12(=−== nLc ncn nπλω 3. []∑∞ =++Ω⎟ ⎠⎞⎜ ⎝⎛ Ω+ΩΩ−Ω= 1) cos() sin( ) cos(cos sin cos tan ),( nn n n n n x ct B ct At xcxcLc Ectxu λ λ λα K,2,1 ,2)12(=−== nLc ncn nπλω A3Answers to Selected Problems: Part II, Chapter 3 3.1 1. ()() () ()() () 0 1 1 2 120 1 1 2 12 22 2 22 222 2 22 2 =+ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ∂∂−+∂∂∂++ ∂∂ −=+ ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ∂∂−+∂∂∂++ ∂∂ − yy x yxx y x b xu yxu yu Eb yu yxu xu E ν ν νν ν ν 2. () ()()()yAExAExAE xy xx 2 2 212, 2 12, 12 νσν ν νσ +−= + += += 3. (a) Yes, (b) Yes, (c) y xuu, non-zero along the base (xxε is non-zero which in itself is inconsistent with 0 =xu along base) 3.2 3. B A5= , ()()[]3 210 3021 x xyEB VV− −=Δν 4. [ ][][ ]2 3 45 3 2 45 3 2 4524 4 , 11 3 , 17 15 xy x y yx y yxE xy E yy E xx −−= + = − = ε ε ε [ ] [ ] A Cx y yx x uB Cyxy yx uE y E x ++++−=++− =4 411 22 23 4 413 45 3 3 45, 17 5 6. () () ⎥⎦⎤ ⎢⎣⎡−+−−=⎥⎦⎤ ⎢⎣⎡−+++− = 23 22 23 2222 23 22 2 3 343 16 2 34 bLxbL bx bxy EbFuybLyby byx EbFu yx νν ν 8. xgyxggyxy yy xxβρσββρσρσ2 2tan,tan2 tan, −=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+− = −= A4Answers to Selected Problems: Part II, Chapter 4 4.3 2. () prEup p pzz rr νννσ σ σθθ 211,2 , , −+−=−=−=−= 3. ⎥⎦⎤ ⎢⎣⎡ −+−−=−++−−−=−−−−−= i zzii rr pabpababpabrbpabab rbpabrbpabab rb 1 /1 1 //21 /1 / 1 // /1 /1 / 1 // / 2 2 0 2 22 22 22 2 0 2 22 2 2 22 22 2 0 2 22 2 2 2 νσσσ θθ 4. () rra Ep u ⎥⎦⎤ ⎢⎣⎡+−+−=22 211νν, ()ν−= 12p P 5. () () () () () ()()r bara Ep ubapbarapbarap zzrr ⎥⎦⎤ ⎢⎣⎡ −+−−+−=−+−=−+−−−=−+−+−= 2 22 22 22 22 22 22 2 / 211/ 1211/ 2112/ 211/ 211/ 211/ 211 νννννσννσννσ θθ 6. za bbpa bapEuo i z⎭⎬⎫ ⎩⎨⎧ −−−−=2 22 2 222ν A7Answers to Selected Problems: Part II, Chapter 7 7.1 1. iivv, 1v, kv 3. ikδ,3,3 6. No 7. jiicba 9. kj ijBA , jijivAv , kl jk ji BAB 10. ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ − −− 2/1 2/12/12/12/1 2/12/1 0 2/1 11. ⎥ ⎦⎤ ⎢ ⎣⎡ +−+ 2/3312/33 12. ⎥ ⎦⎤ ⎢ ⎣⎡ −+− + 327 10363 325 41 7.2 1. iivv, 1v, kv 3. Section 7.2 1. 3 /4=Nσ , 62 .2≈Sσ 2. ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−−− 2 2 22 2 02 0 4 4. The 2D stress transformation equations 5. 3 ,2,1=iσ 2 1 121 21e e n −= , 2 1 221 21e e n += , 3 3e n= 1max=τ