stress and stuff good

1 1Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012CHAPTER ONE STRESS Learning objectives 1. Understanding the concept of stress. 2. Understanding the two-step analysis of rela ting stresses to external forces and moments. _______________________________________________ On January 16th, 1943 a World War II tanker S.S. Schenectady, wh ile tied to the pier on Swan Is land in Oregon, fractured just aft of the bridge and broke in two, as shown in Figure 1.1 . The fracture started as a small crac k in a weld and propagated rapidly overcoming the strength of the material. But what exactly is the strength? How do we an alyze it? To answer these questions, we introduce the concept of stress . Defining this variable is the first step toward developing formulas that can be used in strength analysis and the design of structural members. Figure 1.2 shows two links of the logic that will be fully de veloped in Section 3.2. What motivates the construction of these two links is an idea introduced in Statics—analysis is simpler if any distributed forces in the free-body diagram are replaced by equivalent forces and moments before writing equilib rium equations (see Appendix A.6 ). Formulas developed in mechanics of materials relate stresses to internal forces and moments. Free-body diagra ms are used to relate internal forces and moments to external forces and moments. Figure 1.1 Failure of S.S. Schenectady. Figure 1.2 Two-step process of relating stresses to external forces and moments. Static equivalency Equilibrium 1 2Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.1 STRESS ON A SURFACE The stress on a surface is an internally di stributed force system that can be resolv ed into two components: normal (perpendicu - lar) to the imaginary cut surface, called normal stress, and tangent (parallel) to the imaginary cut surface, called shear stress. 1.1.1 Normal Stress In Figure 1.3, the cable of the chandelier and the columns supporting the building must be strong enough to support the weight of the chandelier and the weight of the building, respectively. If we make an imaginary cut an d draw the free-body diagrams, we see that forces normal to the imaginary cut are needed to ba lance the weight. The internal normal force N divided by the area of the cross section A exposed by the imag inary cut gives us the average intensity of an internal normal force distribution, which we call the average normal stress: (1.1) where σ is the Greek letter sigma used to designate normal stress and the subscript av emphasizes that the normal stress is an average value. We may view σav as a uniformly distributed normal force, as shown in Figure 1.3 , which can be replaced by a statically equivalent internal normal for ce. We will develop this viewpoint further in Section 1.1.4 . Notice that N is in boldface italics, as are all internal forces (and moments) in this book. Equation (1.1) is consistent with our intuitive understanding of strength. Consider the following two observations. (i) We know that if we keep increasing the force on a body, then the body will eventually break. Thus we expect the quantifier for strength (stress) to increase in value with the increase of force until it reaches a critical value. In other words, we expect stress to be directly proportional to force, as in Equation (1.1) . (ii) If we compare two bodies that are identical in all respects except that one is thicker than the other, then we expect that the thicker body is stronger. Thus, for a given force, as the body gets thicker (larger cross-sectional area), we mo ve away from the critical breaking value, and the value of the quantifier of streng th should decrease. In other words, stress should vary inversely with the cross-sectional area, as in Equation (1.1) . Equation (1.1) shows that the unit of stress is force per unit ar ea. Table 1.1 lists the various units of stress used in this book. It should be noted that 1 psi is equal to 6.895 kPa, or ap proximately 7 kPa. Alternatively, 1 kPa is equal to 0.145 psi, or σavN A----= Tensile Normal Force Compressive Normal ForceImaginary Cut Chandelier Weight Building Weight Imaginary CutChandelier WeightTensile Normal Stress Building Weight Compressive Normal StressN N NNσavg σavg σavg σavg Figure 1.3 Examples of normal stress distribution. 1 3Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012approximately 0.15 psi. Normal st ress that pulls the imag inary surface away from the material is called tensile stress, as shown on the cable of the chandelier in Figure 1.3 . Normal stress that pushes the imaginar y surface into the material is called compressive stress, as shown on the column. In other words, tensile stress acts in the direction of the outward normal whereas compressive stress is opposite to the direction of the outward normal to the imaginary surface. Norm al stress is usually reported as tensile or compressive a nd not as positive or negative. Thus σ = 100 MPa (T) or σ = 10 ksi (C) are the preferred ways of reporting tensile or compressive normal stresses. The normal stress acting in the direct ion of the axis of a slender member (rod, cable, bar, column) is called axial stress. The compressive normal stress that is produced when one real surface presses against another is called the bearing stress. Thus, the stress that exist between the base of the column an d the floor is a bearing stress but the compressive stress inside the column is not a bearing stress. An important consideration in all analyses is to know whethe r the calculated values of the variables are reasonable. A sim - ple mistake, such as forgetting to convert feet to inches or millimeters to mete rs, can result in values of stress that are inc orrect by orders of magnitude. Less dramatic erro rs can also be caught if one has a sense of the limiting stress values for a material . Table 1.2 shows fracture stress values for a few common materials. Fracture stress is th e experimentally measured value at which a material breaks. The numbers are approximate, and + indicates variations of the stress values in each class of material. The order of magnitude and the relative strength with respect to wood are shown to help you in acquiring a feel for the numbers .TABLE 1.1 Units of stress Abbreviation Units Basic Units psi Pounds per square inch lb/in.2 ksi Kilopounds (kips) per square inch 103 lb/in.2 Pa Pascal N/m2 kPa Kilopascal 103 N/m2 MPa Megapascal 106 N/m2 GPa Gigapascal 109 N/m2 TABLE 1.2 Fracture stress magnitudes Material ksi MPa Relative to Wood Metals 90 + 90% 630 + 90% 7.0 Granite 30 + 60% 210 + 60% 2.5 Wood 12 + 25% 84 + 25% 1.0 Glass 9 + 90% 63 + 90% 0.89 Nylon 8 + 10% 56 + 10% 0.67 Rubber 2.7 + 20% 19 + 20% 0.18 Bones 2 + 25% 14 + 25% 0.16 Concrete 6 + 90% 42 + 90% 0.03 Adhesives 0.3 + 60% 2.1 + 60% 0.02 1 4Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012EXAMPLE 1.1 A girl whose mass is 40 kg is using a swing set. The diamet er of the wire used for constructing the links of the chain is 5 mm. Determine the average normal stress in the links at the bottom of the swing, assuming that the inerti al forces can be neglected . PLAN We make an imaginary cut through the chains, draw a free-body diagram, and find the tension T in each chain. The link is cut at two imaginary surfaces, and henc e the internal normal force N is equal to T/2 from which we obtain th e average normal stress. SOLUTION The cross-sectional area and the we ight of the girl can be found as (E1) Figure 1.5 shows the free body diagram after an imaginary cut is made through the chains. The tens ion in the chain and the norm al force at each surface of the link can be found as shown in Equations (E2) and (E3). (E2) (E3) The average normal stress can be found as shown in Equation (E4). (E4) ANS. COMMENTS 1. The stress calculations had two st eps. First, we found the internal force by e quilibrium; and second we calculated the stress f rom it. 2. An alternative view is to think that the total material area of the link in each chain is The internal normal force in each chain is T = 196.2 N thus the average normal stress is as before. 1.1.2 Shear Stress In Figure 1.6 a the double-sided tape used for sticking a hook on the wall must have sufficient bonding strength to support the weight of the clothes hung from the hook. The free-body diagram shown is created by making an imaginary cut at the wall sur - Figure 1.4 Girl in a swing, Example 1.2. Aπd2 4---------π0.005 m()2 4------------------------------ - 19.6 106–() m2== = W 40 kg() 9.81 m/s2() 392.4 N == T 2N= 2T 392.4 N=o r 4 N 392.4 N=o r N 98.1 N= Figure 1.5 Free-body diagram of swing.N N TT T W σavN A----98.1 N 19.6 10 ×6– m2()----------------------------------------- - 4.996 106× N/m2== = σav 5.0 MPa (T)= 2A 39.2 10 ×6–m2. = σavT 2A------- = 196.2 39.2 106–× ⁄ () 51 06× N/m2, == 1 5Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012face. In Figure 1.6 b the paper in the ring binder will tear out if the pull of the hand overcomes the strength of the paper. The free-body diagram shown is created by making an imaginary cut along the path of the rings as the paper is torn out. In both fre e- body diagrams the internal force necessa ry for equilibrium is parallel (tange nt) to the imaginary cut surface. The internal shear force V divided by the cross sectional area A exposed by the imaginary cut gives us the average intensity of the internal shear force distribution, whic h we call the average shear stress: (1.2) where τ is the Greek letter tau used to design ate shear stress and the subscr ipt av emphasizes that the shear stress is an average value. We may view τav as a uniformly distri buted shear force, which can be replaced by a statically equivale nt internal normal force V. We will develop this viewpoint further in Section 1.1.4 . 1.1.3 Pins Pins are one of the most common exampl e of a structural member in which sh ear stress is assumed uniform on the imag - inary surface perpendicular to the pin ax is. Bolts, screws, nails, and rivets are often approximated as pins if the primary func - tion of these mechanical fasteners is th e transfer of shear forces from one member to another. However, if the primary function of these mechanical fasteners is to press two solid bodies into each othe r (seals) then these fasteners cannot be approximated as pins as the for ces transferred are normal forces. Shear pins are mechanical fuses designed to br eak in shear when the force being tr ansferred exceeds a level that would damage a critical component. In a lawn mower shear pins attach the blades to the transm ission shaft and break if the blades hit a large rock that may bend the transmission shaft.τavV A----= Weight of the Clothes Imaginary cut between the wall and the tape VWeight of theClothes τ τImaginary cut along the possible path of the edge of the ring. Pull of the hand Pullof thehand Pull of thehandτ ττV V VMwall MwallWeight of the Clothes Figure 1.6 Examples of shear stress distribution.(a) (b) 1 6Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012Figure 1.7 shows magnified views of two types of connections at a support. Figure 1.7 a shows pin in single shear as a sin - gle cut between the suppo rt and the member will break the connection. Figure 1.7 b shows a pin in double shear as two cuts are needed to break the connection. For th e same reaction force, th e pin in double shear ha s a smaller shear stress. When more than two members (forces) are acting on a pin, it is im portant to visualize the im aginary surface on which the shear stress is to be calculated. Figure 1.8 a shows a magnified view of a pin conn ection between three members. The shear stress on the imaginary cut surface 1 will be different from that on the imaginary cut surface 2, as shown by the free-body dia - grams in Figure 1.8 b. EXAMPLE 1.2 Two possible configurations for the assembly of a joint in a mach ine are to be evaluated. The ma gnified view of the two configu rations with the forces in the members are shown in Figure 1.9 . The diameter of the pin is 1 in. Determine which joint assembly is preferred by calculating the maximum shear stress in the pin for each case. PLAN We make imaginary cuts between individual members for the two configurations a nd draw free-body diagrams to determine the shear force at each cut. We calculate and compare the shear stresses to determine the maximum shear stress in each configuration. F F V (a) Figure 1.7 Pins in ( a) single and ( b) double shear. (b) F V VF Figure 1.8 Multiple forces on a pin.NBNVBV NC VBVVDV (b) Cut 1 (a)NCNDN NBN Cut 2 NDN VDV Configuration 1Configuration 2NC=2 0k i p s NC=2 0k i p s NB= 15 kips NA=15 kips NA= 15 kips ND=2 0k i p s ND=2 0k i p sNB=1 5k i p s ABCDB CD A Figure 1.9 Forces on a joint and different joining configurations. 1 7Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012SOLUTION The area of the pin is . Making imaginary cuts between member s we can draw the free-body diagrams and calculate the internal shear force at the imaginary cut, as shown in Figure 1.10 . Configuration 1: From the free-body diagrams in Figure 1.10 a (E1) We see that the maximum shear force exists between members C and D. Thus the maximum shear stress is (E2) Configuration 2: From the free-body diagrams in Figure 1.10 b (E3) The maximum shear force exists the between C and B. Thus the maximum shear stress is (E4) Comparing Equations (E2) and (E3) we conclude ANS. COMMENTS 1. Once more note the two steps: we first calculated the internal shear force by equilibrium and then calculated the shear stress from it. 2. The problem emphasizes the importance of visualizing the imaginary cu t surface in the calculation of stresses. 3. A simple change in an assembly sequence can cause a joint to fa il. This observation is true any time more than two members are joined together. Gusset plates are often used at the joints such as in bridge shown in Figure 1.11 to eliminate the problems associated with an assembly sequence.A π0.5 in.()20.7854 in.2== Imaginary cut between members A and B NA= 15 kipsV1 Imaginary cut between members B and C NB= 15 kips NA= 15 kipsV2 Imaginary cut between members C and D ND=2 0k i p sV3 Imaginary cut between members A and C Imaginary cut between members C and B Imaginary cut between members B and DV1 NA= 15 kips NC=2 0k i p s NA= 15 kips(V2)y (V2)x ND=2 0k i p sV3A AA DA A DBC Figure 1.10 Free-body diagrams. (a) Configuration 1. (b) Configuration 2.(a)(b) V115 kips= V20= V320 kips= τmax = V3A =⁄ 25.46 ksi. V115 kips= V2()x15 kips= V2()y20 kips= V2152202+ 25 kips. == V320 kips= τmaxV2A⁄ 31.8 ksi. == The configuration 1 is preferred, as it will result in smaller shear stres Figure 1.11 Use of gusset plates at jo ints in a bridge truss.Gusset plate 1 8Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012EXAMPLE 1.3 All members of the truss show n in Figure 1.12 have a cr oss-sectional area of 500 mm2 and all pins have a diameter of 20 mm. Determine: (a) The axial stresses in members BC and DE, (b) The shear stress in the pin at A, assuming the pin is in double shear. PLAN (a) The free-body diagram of joint D can be us ed to find the internal axial force in member DE. The free body diagram drawn aft er an imaginary cut through BC, CF, and EF can be used to find the in ternal force in member BC. (b) The free-body diagram of the enti re truss can be used to find the support reaction at A, from wh ich the shear stress in th e pin at A can be found. SOLUTION The cross-sectional areas of pins and memb ers can be calculated as in Equation (E1) (E1) (a) Figure 1.13 a shows the free-body diagram of joint D. The internal axial force NDE can be found using equili brium equations as shown in Equation (E3). (E2) (E3) The axial stress in member DE can be found as shown in Equation (E4). (E4) ANS. Figure 1.13 b shows the free-body diagram after an imaginary cut is made through members CB, CF, and EF. By taking the moment about point F we can find the internal axial force in member CB as shown in Equation (E5). (E5) The axial stress in member CB can be found as shown in Equation (E6). (E6) ANS. (b) Figure 1.13 c shows the free-body diagram of the entire truss. By moment equilibrium about point G we obtain (E7) The shear force in the pin will be half the force of NAB as it is in double shear. We obtain the shear stress in the pin as (E8) ANS. P/H11005 21 k NA B F E D GC 2m 2m 2mG2m Figure 1.12 Truss. Apπ0.02 m()2 4--------------------------- - 314.2 106–() m2== Am500 106–() m2= NDC 45o sin 21 kN –0 = o r NDC29.7 kN= 21 kNNDENDC D45° Figure 1.13 Free-body diagrams. 21 kNNEFNNCFNCBAx GxG GyG 21NAB(a) (b)(c)N–DENDC 45o0= cos –o r NDE21– k N= σDENDE Am-----------21 103() N –[] 500 106–() m2[] ---------------------------------------- - 42–1 06() N/m2== = σDE 42 MPa (C)= NCB2 m() 21 kN()–4 m () 0= o r NCB 42 kN= σCBNCB Am---------- 84 106() N/m2 == σCD84 MPa (T)= NAB2 m() 21 kN 6 m () 0= –o r NAB63 kN= τANAB2⁄ Ap-----------------31.5 103() N 314.2 106–() m2------------------------------------ - 100 106() N/m2== = τA100 MPa= 1 9Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012COMMENTS 1. We calculated the internal forces in each member before calculating the axial stresses, emphasizing the two steps Figure 1.2 of relat - ing stresses to external forces. 2. In part (a) we could have solved for the force in BC by noting that EC is a zero force member and by drawing the free-body diagram of joint C. PROBLEM SET 1.1 Tensile stress 1.1 In a tug of war, each person shown in Figure P1.1 exerts a force of 200 lb. If the effective diameter of the rope is determine the axial stress in the rope. 1.2 A weight is being raised using a cable and a pu lley, as shown in Figure P1.2. If the weight W = 200 lb, determine the axial stress assuming: (a) the cable diameter is (b) the cable diameter is 1.3 The cable in Figure P1.2 has a diameter of If the maximum stress in the cable must be limited to 4 ksi (T), what is the maximum weight that can be lifted? 1.4 The weight W = 250 lb in Figure P1.2. If the maximum stre ss in the cable must be limited to 5 ksi (T), determine the minimum diameter of the cable to the nearest 1.5 A 6-kg light shown in Figure P1.5 is hanging from the ceiling by wires of 0.75-mm diameter. Determine the tensile stress in wir es AB and BC. 1.6 An 8-kg light shown in Figure P1.5 is hanging from the ceiling by wires. If the tensile stress in the wires cannot exceed 50 MPa, determine the minimum diameter of the wire, to the nearest tenth of a millimeter.1 2--- in., Figure P1.1 1 8--- in.1 4--- in. W Figure P1.2 1 5--- in. 1 16------in. 2 m LightAA B C2.5 m2.5 m Figure P1.5 1 10Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.7 Wires of 0.5-mm diameter are to be used for hanging lights such as the one shown in Figure P1.5. If the tensile stress in the w ires cannot exceed 80 MPa, determine the maximum mass of the light that can be hung using these wires. 1.8 A 3 kg picture is hung using a wire of 3 mm diameter, as shown in Figure P1.8. What is the average normal stress in the wires? 1.9 A 5 kg picture is hung using a wire, as shown in Figure P1.8. If the tensile stress in the wires cannot exceed 10MPa, determine the minimum required diameter of the wire to the nearest millimeter. 1.10 Wires of 16-mil diameter are used for hanging a picture, as shown in Figure P1.8. If the tensile stress in the wire cannot exce ed 750 psi, determine the maximum weight of the picture that can be hung using these wires. 1 mil = 1.11 A board is raised to lean against the left wall using a cable an d pulley, as shown in Figure P1.11. Determine the axial stress in the cable in terms of the length L of the board, the specific weight γ per unit length of the board, the cable diameter d, and the angles θ and α, shown in Figure P1.11 . Compressive and bearing stresses 1.12 A hollow circular column supporting a building is attached to a metal plate and bolted into the concrete foundation, as shown i n Figure P1.12 . The column outside diameter is 100 mm and an inside diameter is 75 mm. The metal plate dimensions are 200 mm × 200 mm × 10 mm. The load P is estimated at 800 kN. Determine: (a) the compressive stress in the column; (b) the average bearing stress between the metal plate and the concrete. 1.13 A hollow circular column supporting a building is attached to a metal plate and bolted into the concrete foundation, as shown i n Figure P1.12 . The column outside diameter is 4 in. and an inside diameter is 3.5 in. The metal plate dimensions are 10 in. × 10 in. × 0.75 in. If the allow - able average compressive stress in the column is 30 ksi and the allowable average bearing stress in concrete is 2 ksi, determine the maximum load P that can be applied to the column. 54o Figure P1.8 1 1000------------ in. /H9258/H9251 Board Figure P1.11 MetalConcreteP Figure P1.12 1 11Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.14 A hollow square column supporting a building is attached to a metal plate and bolted into the concrete foundation, as shown in Figure P1.14 . The column has outside dimensions of 120 mm × 120 mm and a thickness of 10 mm. The load P is estimated at 600 kN. The metal plate dimensions are 250 mm × 250 mm × 15 mm. Determine: (a) the compressive stress in the column; (b) the average bearing stress between the metal plate and the concrete. 1.15 A column with the cross section shown in Figure P1.15 supports a building. The column is attached to a metal plate and bolted i nto the concrete foundation. The load P is estimated at 750 kN. The metal plate dimensions are 300 mm × 300 mm × 20 mm. Determine: (a) the com - pressive stress in the column; (b) the average bearing stress between the metal plate and the concrete 1.16 A 70-kg person is standing on a bathro om scale that has dimensions of 150 mm × 100 mm × 40 mm (Figures P1.16). Determine the bear - ing stress between the scale and the floor. Assume the weight of the scale is negligible. 1.17 A 30-ft-tall brick chimney has an outside diameter of 3 ft and a wall thickness of 4 in. (Figure P1.17). If the specific weight of the bricks is 80 lb/ft3, determine the average bearing stress at the base of the chimney. MetalConcreteP Figure P1.14 Concrete160 mm MetalP 10 mm10 mm10 mm 10 mm160 mm Figure P1.15 Figure P1.16 Figure P1.17 30 ft 1 12Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.18 Determine the average bearing stress at the bottom of the bl ock shown in Figure P1.18 in terms of the specific weight γ and the length dimen - sions a and h. 1.19 The Washington Monument is an obelisk with a hollow rectangula r cross section that tapers along its length. An approximation of the monument geometry is shown in Figure P1.19 . The thickness at the base is 4.5 m and at top it is 2.5 m. The monument is constructed from marble and granite. Using a specific weight of 28 kN/m3 for these materials, determine the average bearing stress at the base of the monument. 1.20 Show that the average compressive stress due to weight on a cross section at a distance x from the top of the wall in Figure P1.20 b is half that of wall in Figure P1.20 a, thus confirming the wisdom of ancient Egyptians in buil ding inward-sloping walls for the pyramids. (Hint: Using γ the specific weight of wall material, H the height of the wall, t the thickness of the wall, and L the length of the wall, calculate the average com - pressive stress at any cross section at a distance x from the top for the two walls.). 1.21 The Great pyramid of Giza shown in Figure 1.14 d has a base of 757.7 ft x 757.7 ft and a height of 480.9 ft. Assume an average specific weight of γ = 75 lb/ft3. Determine (a) the bearing stress at the base of the pyramid. (b) the average compressive stress at mid height. 1.22 The Bent pyramid shown in Figure 1.14 c has a base of 188 m x 188 m. The initial slopes of the sides is 54o27’44”. After a certain height the slope is 43o22’. The total height of the pyramid is 105 m. Assume an average mass density of 1200 kg/ m3. Determine the bearing stress at the base of the pyramid. 1.23 A steel bolt of 25 mm diameter passes through an aluminum sleeve of thickness 4 mm and outside diameter of 48 mm as shown in Figure ha 100a10h Figure P1.18 10 m 17 m169 m10 m 17 m Figure P1.19 Figure P1.20 (a) Straight wall (b) In ward sloping tapered wall.(a)(b) Hx H tt xL L 1 13Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012P1.23 . Determine the average normal stress in the sleeve if in the assembled position the bolt has an average normal stress of 100 MPa (T). Shear stress 1.24 The device shown in Figure P1.24 is used for determining the shear strength of the wood. The dimensions of the wood block are 6 in. × 8 in. × 2 in. If the force required to break the wood block is 15 kips, determine the average shear strength of the wood. 1.25 The dimensions of the wood block in Figure P1.24 are 6 in. × 8 in. × 1.5 in. Estimate the force P that should be applied to break the block if the average shear strength of the wood is 1.2 ksi. 1.26 The punch and die arrangement shown schematically in Figure P1.26 is used to punch out thin plate objects of different shapes. The cross section of the punch and die shown in Figure P1.26 is a circle of 1-in. diameter. A force P = 6 kips is applied to the punch. If the plate thickness t = determine the average shear stress in the plate along the path of the punch. 1.27 The cross section of the punch and die shown in Figure P1.26 is a square of 10 mm × 10 mm. The plate shown has a thickness t = 3 mm and an average shear strength of 200 MPa. Determine the average force P needed to drive the punch through the plate. 1.28 The schematic of a punch and die for punching washers is shown in Figure P1.28. Determine the force P needed to punch out washers, in terms of the plate thickness t, the average plate shear strength τ, and the inner and outer diameters of the washers di and do Figure P1.23 SleeveRigid washer s 300 mm 25 mm 25 mm P 6 in 2i n6 in Figure P1.24 1 8--- in., P PlatePunch Die Diet Figure P1.26 di dodP Punch Die Die t Figure P1.28 1 14Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.29 The magnified view of a pin joint in a truss are shown in Figure P1.29. The diameter of the pin is 25 mm. Determine the maximum trans - verse shear stress in the pin. Normal and shear stresses 1.30 A weight W = 200 lb. is being raised using a cable and a pulley, as shown in Figure P1.30. The cable effective diameter is and the pin in the pulley has a diameter of Determine the axial stress in the cable and the shea r stress in the pin, assuming the pin is in dou - ble shear. 1.31 The cable in Figure P1.30 has a diameter of and the pin in the pulley has a diameter of If the maximum normal stress in the cable must be limited to 4 ksi (T) and the maximum shear stress in the pin is to be limited to 2 ksi, determine the maximum weight that can be lifted to the nearest lb. The pin is in double shear. 1.32 The manufacturer of the plastic carrier for drywall panels shown in Figure P1.32 prescribes a maximum load P of 200 lb. If the cross-sec - tional areas at sections AA and BB are 1.3 in.2 and 0.3 in.2, respectively, determine the average shear stress at section AA and the average normal stress at section BB at the maximum load P. 1.33 A bolt passing through a piece of wood is sh own in Figure P1.33. Determine: (a) the axial stress in the bolt; (b) the average shear stress in the bolt head; (c) the average bearing stress between the bolt head and the wood; (d) the average shear stress in the wood.40 kN 30 kN50 kN 36.9° Figure P1.29 1 4--- in. 3 8--- in. 55° W Figure P1.30 1 5--- in.3 8--- in. Figure P1.32 PAA B B P/H11005 1.5 ki ps3 8i in3 4 in1 2 Figure P1.33 1 15Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.34 A load of P = 10 kips is transferred by the riveted joint shown in Figure P1.34. Determine (a) the average shear stress in the rivet. (b) the largest average normal stress in the members attached (c) the largest average bearing stress between the pins and members. 1.35 A joint in a wooden structure is shown in Figure P1.35. The dimension and Determine the average normal stress on plane BEF and average shear stress on plane BCD. Assume pl ane BEF and the horizontal plane at AB are a smooth surfaces. 1.36 A metal plate welded to an I-beam is securely fastened to the foundation wall using four bolts of 1/2 in. diameter as shown in Figure P1.36 . If P = 12 kips determine the normal and shear stress in each bolt. Assume the load is equally distributed among the four bolts. 1.37 A metal plate welded to an I-beam is securely fastened to the fo undation wall using four bolts as shown Figure P1.36. The allow able nor - mal stress in the bolts is 100 MPa and the allowable shear stress is 70 MPa. Assume the load is equally distributed among the four bolts. If the beam load P= 50 kN, determine the minimum diameter to the nearest millimeter of the bolts. 1.38 A metal plate welded to an I-beam is securely fasten ed to the foundation wall using four bolts of 1/2 in. diameter as shown Figure P1.36. The allowable normal stress in the bolts is 15 ksi and the allowable shear stress is 12 ksi. Assume the load is equally distributed among the four bolts. Determine the maximum load P to the nearest pound the beam can support. 1.39 An adhesively bonded joint in wood is fabricated as shown in Figure P1.39. The length of the overlap is L= 4 in. and the thickness of the wood is 3/8 in. Determine the average shear stress in the adhesive. . Figure P1.34 1i n . 1i n . P P 2i n . 0.5 in. 0.5 in. 0.5 in.0.5 in. P P h 43 8--- in. = d 11 8--- in. = E4 in F10 kips 30/H11034 h d A B CD Figure P1.35 Figure P1.36 P 60o Figure P1.39 1 16Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.40 A double lap joint adhesively bonds three pieces of wood as sh own in Figure P1.40. The joints transmits a force of P= 20 kips and has the following dimensions: L = 3 in., a = 8 in. and h = 2 in. Determine the maximum average normal stress in the wood and the average shear stress in the adhesive. 1.41 The wood in the double lap joint of Figure P1.40 has a strength of 15 MPa in tension and the strength of the adhesive in shear is 2 MPa. The joint has the following dimensions: L = 75 mm, a =200 mm, and h = 50 mm. Determine the maximum force P the joint can transfer. 1.42 A wooden dowel of diameter d = 20 mm is used for constructing the double lap joint in Figure P1.42. The wooden members have a strength of 10 MPa in tension, the bearing stress between the dowel and the members is to be limited to 18 MPa, the shear strength of the dowel is 25 MPa. The joint has the following dimensions: L = 75 mm, a =200 mm, and h = 50 mm. Determine the maximum force P the joint can transfer . 1.43 A couple is using the chair lift shown in Figure P1.43 to see the Fall colors in Michigan’s Upper Peninsula. The pipes of the c hair frame are 1/16 in. thick. Assuming each person weighs 180 lb, determine the average normal stress at section AA and BB and average shear stress at section CC assuming the chair is moving at a constant speed . 1.44 The axial force P = 12 kips acts on a rectangular member, as shown in Figure P1.44. Determine the average normal and shear stresses on the inclined plane AA. Figure P1.40 h/2 h/2P La hP/2 P/2 Figure P1.42 h/2 h/2P La hP/2 P/2d Figure P1.43 1.5 in.2i n .1/16 in. 1/6 in. Section CC 1.5 in.B B 80oA A 2i n . C C P2i n PA 1.5 in Figure P1.44 1 17Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.45 A wooden axial member has a cross section of 2 in.× 4 in. The member was glued along line AA and transmits a force of P = 80 kips as shown in Figure P1.45 . Determine the average normal and shear stress on plane AA. 1.46 Two rectangular bars of 10-mm thickness are loaded as shown in Figure P1.46. If the normal stress on plane AA is 180 MPa (C), determine the force F1 and the normal and shear stresses on plane BB. 1.47 A butt joint is created by welding two plates to transmits a force of P = 250 kN as shown in Figure P1.47. Determine the average normal and shear stress on the plane AA of the weld. 1.48 A square tube of 1/4 in thickness is welded along the seam and used for transmitting a force of P = 20 kips as shown in Figure P1.48. Determine average normal and shear stress on the plane AA of the weld. 1.49 (a) In terms of P, a, b, and θ determine the average normal and shear stresses on the inclined plane AA shown in Figure P1.49. (b) Plot the normal and shear stresses as a function of θ and determine the maximum values of the normal and shear stresses. (c) At what angles of the inclined plane do the maximum normal and maximum shear stresses occurs. Figure P1.45 A AP 4i n .40o 50 kN50 kN F1 F3A A BB 75° 65°30 mm10 mm 60 mm Figure P1.46 PP900 mm 50 mm200 mm 60o60o Figure P1.47 A A Figure P1.48 P P30o2.5 in. 2.5 in. Pb Pa /H9258A A Figure P1.49 1 18Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.50 An axial load is applied to a 1-in-diameter circular rod (Figure P1.50). The shear stress on section AA was found to be 20 ksi. The section AA is at 45o to the axis of the rod. Determine the applied force P and the average normal stress acting on section AA. 1.51 A simplified model of a child’s arm lifting a weight is shown in Figure P1.51. The cross-sectional area of the biceps muscle is estimated as 2 in2. Determine the average normal stress in the muscle and the average shear force at the elbow joint A. 1.52 Figure P1.52 shows a truss and the sequence of assembly of members at pins H, G, and F. All members of the truss have cross-sec tional areas of 250 mm2 and all pins have diameters of 15 mm. Determine (a) the axial stresses in members HA, HB, HG, and HC. (b) the maximum shear stress in pin H. 1.53 Figure P1.52 shows a truss and the sequence of assembly of members at pins H, G, and F. All members of the truss have cross-sec tional areas of 250 mm2 and all pins have diameters of 15 mm. Determine (a) the axial stresses in members FG, FC, FD, and FE. (b) the maximum shear stress in pin F. 1.54 Figure P1.52 shows a truss and the sequence of assembly of members at pins H, G, and F. All members of the truss have cross-sec tional areas of 200 mm2 and all pins have diameters of 10 mm. Determine (a) the axial stresses in members GH, GC, and GF of the truss shown in Fig- ure P1.52 . (b) the maximum shear stress in pin G 1.55 The pin at C in Figure P1.55 is has a diameter of in. and is in double shear. The cross-sectional areas of members AB and BC are 2 in.2 and 2.5 in.2, respectively. Determine the axial stress in member AB and the shear stress in pin C.P P AA Figure P1.50 5 lb Bone 2i nAABone6 ini 7 in7i nBiceps muscle Figure P1.51 A BCDEFG H 30° 30° 4 kN 2 kN 3 kN3 m 3 m 3 m 3 m Figure P1.52 HAHBHGHC Pin HGHGCGF Pin G Pin FFGFCFDFE4 kN 2 kN 3 kN 1 2--- 60°60 in 66 in80 lb/inA B C Figure P1.55 1 19Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.56 All pins shown in Figure P1.56 are in single shear and have diameters of 40 mm. All members have square cross sections and the surface at E is smooth. Determine the maximum shear stresses in the pins and the axial stress in member BD. 1.57 A student athlete is lifting weight W = 36 lbs as shown in Figure P1.57a. The weight of the athlete is WA = 140 lb. A model of the student pelvis and legs is shown in Figure P1.57 b. The weight of legs and pelvis WL = 32 lb acts at the center of gravity G. Determine the normal stress in the erector spinae muscle that supports the trunk if the aver age muscle area at the time of lifting the weight is 1.75 in2. 1.58 A student is exercising his shoulder muscles using a W = 15 lb dumbbell as shown in Figure P1.58a. The model of the student arm is shown in Figure P1.58 b. The weight of the arm of WA = 9 lb acts at the center of gravity G. Determine the average normal stress in the deltoid muscle if the average area of the muscle is 0.75 in2 at the time the weight is in the horizontal position. Design problems 1.59 The bottom screw in the hook shown in Figure P1.59 supports 60% of the load P while the remaining 40% of P is carried by the top screw. The shear strength of the screws is 50 MPa. Develop a table for the maximum load P that the hook can support for screw diameters that vary from 1 mm to 5 mm in steps of 1 mm. B A 3 m 2.5 m50 mm150 mm50 kN/m 2.5 mDC E200 mm Figure P1.56 Figure P1.57 (a) (b) A B CErector spinaemuscles 1.5i n GA B8.2i n 10i n 20 inHip Joint (W+WA)/2 3i n45o45o WL C Figure P1.58 (a) (b) Shoulder joint15o A O 6i n 6i n 12 in W WAdeltoid muscle A OG B P Figure P1.59 P 1 20Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.60 Determine the maximum force P that can be transferred by the rive ted joint shown in Figure P1.34 if the limits apply: maximum n ormal stress in the attached members can be 30 ksi., maximum bearing stress between the pins and members can be 15 ksi, and the maximum shear stress in the rivet can be 20 ksi. 1.61 A tire swing is suspended using three chains, as shown in Figure P1.61. Each chain makes an angle of 12o with the vertical. The chain is made from links as shown. For design purpos es assume that more than one person may use the swing, and hence the swing is to be designed to carry a weight of 500 lb. If the maximum average normal stress in the links is not to exceed 10 ksi, determine to the nearest the diameter of the wire that should be used for constructing the links. 1.62 Two cast-iron pipes are held together by a bolt, as shown in Fi gure P1.62. The outer diameters of the two pipes are 50 mm and 7 0 mm and the wall thickness of each pipe is 10 mm. The diameter of the bolt is 15 mm. What is the maximum force P this assembly can tr ansmit if the maximum permissible stresses in the bo lt and the cast iron are 200 MPa in shear and 150 MPa in tension, respectively. 1.63 A normal stress of 20 ksi is to be transferred from one plate to another by riveting a plate on top, as shown in Figure P1.63. The shear strength of the in. rivets used is 40 ksi. Assuming all rivets carry equal shear stress, determ ine the minimum even number of rivets that must be used. 1.64 Two possible joining configurations are to be evaluated. The forces on joint in a truss were calculated and a magnified view is shown Figure P1.64 . The pin diameter is 20 mm. Determine wh ich joint assembly is better by calculating the maximum shear stress in the pin for ea ch case. 1 16------ in. Figure P1.61 12o P P Figure P1.62 1 2--- /H9268 /H11005 20 ksi 1 in/H9268 /H11005 20 ksi6 in8 in 1 in 1 in Figure P1.63 ND= 30 kN ND= 30 kN NB= 67.32 kNNB= 67.32 kNNC= 50 kN NC= 50 kN NA= 32.68 kNNA= 32.68 kN 30o 30o30o30o Configuration 1 Configuration 2 Figure P1.64 1 21Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.65 Truss analysis showed the forces at joint A given in Figure P1.65. Determine the sequence in which the three members at joint A should be assembled so that the shear stress in the pin is minimum. 1.66 An 8 in × 8 in reinforced concrete bar needs to be designed to carry a compressive axial force of 235 kips. The reinforcement i s done using round steel bars. Assuming the normal stress in concrete to be a uniform maximum value of 3 ksi and in steel bars to be a unifo rm value of 20 ksi, determine the minimum number of iron bars that are needed. 1.67 A wooden axial member has a cross section of 2 in. × 4 in. The member was glued along line AA, as shown in Figure P1.45. Determine the maximum force P that can be applied to the repaired axial member if th e maximum normal stress in the glue cannot exceed 800 psi and the maxi - mum shear stress in the glue cannot exceed 350 psi. 1.68 An adhesively bonded joint in wood is fabricated as shown in Figure P1.68. The length of the bonded region L = 5 in. Determine the maxi - mum force P the joint can support if the shear strength of the adhesive is 300 psi and the wood strength is 6 ksi in tension. 1.69 The joint in Figure P1.68 is to support a force P = 25 kips. What should be the length L of the bonded region if the adhesive strength in shear is 300 psi? 1.70 The normal stress in the members of the truss shown in Figure P1.70 is to be limited to 160 MPa in tension or compression. All members have circular cross sections. The shear stress in the pins is to be limited to 250 MPa. Determine (a) the minimum diameters to the nearest millimeter of members ED, EG, and EF. (b) the minimum diameter of pin E to the nearest millimeter and the sequence of assembly of members ED, EG, and EF. 1.71 The normal stress in the members of the truss shown in Figure P1.70 is to be limited to 160 MPa in tension or compression. All members have circular cross sections. The shear stress in the pins is to be limited to 250 MPa. Determine (a) the minimum diameters to the nearest millimeter of members CG, CD, and CB. (b) the minimum diameter of pin C to the nearest millimeter and the sequence of assembly of members CG, CD, and CB. Stretch yourself 1.72 Truss analysis showed the forces at joint A given in Figure P1.72. Determine the sequence in which the four members at joint A should be assembled to minimize the shear stress in the pin. NC /H11005 40 kips NB /H11005 32.77 kips35/H11034ND /H11005 22.94 kips A Figure P1.65 1 2----in. P P L8 in 1 in 1 in 1 in Figure P1.68 FED C G HB A40 kN 65 kN4 m 3 m4 m 3 m Figure P1.70 A65/H11034ND /H11005 25 kips NC /H11005 27.58 kips NE /H11005 28.34 kips NB /H11005 40 kips Figure P1.72 1 22Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012MoM in Action: Pyramids The pyramids of Egypt are a remarkab le engineering feat. The size, gr andeur, and age of the pyramids excites the human imagination. Scie nce fiction authors create stories a bout aliens building them. Pyramid design, however, is a story about human engineering in a design process that incorp orates an intuitive under - standing of material strength. Before pyramids were built, Egyptians kings and nobles were buried in tombs called mastaba ( Figure 1.14a). Mastaba have underground chambers that are bloc ked off by dropping heavy stones down vertical shafts. On top of these underground burial chambers ar e rectangular structures with inward-sloping, tapered brick mud walls. The ancient Egyptians had learned by experience that inward-sloping walls that taper towards the top do not crumble as quickly as stra ight walls of uniform thickness (see problem 1.20). Imhotep, the world’s first renowned engineer-architect, took many of the design elements of mastaba to a very large scale in building th e world’s first Step pyramid ( Figure 1.14 b) for his pharaoh Djozer (2667- 2648 BCE). By building it on a bedrock, Imhotep intuitively understood the importance of bearing stresses which were not properly accounted for in building of th e leaning tower of Pisa 4000 years later. The Step pyramid rose in six steps to a height of 197 ft with a base of 397 ft x 358 ft. A 92-ft deep central shaft was dug beneath the base for the granite burial chamber. Th e slopes of the faces of the Step pyramid varied from 72o to 78o. Several pharaohs after Djozer tried to build their own step pyramids but were unsuccessful. The next development in pyramid design took place in the reign of pharaoh Sneferu (2613-2589 BCE). Sneferu architects started by building a step pyrami d but abandoned it because he wanted a pyramid with smooth sides. The pyramid shown in Figure 1.14 c was started with a base of 617 ft x 617 ft and an initial slope of approximately 54o. Signs of structural problem convinced the builders to change the slope to 43o resulting in the unique bent shape seen in Figure 1.14 c (see problem 1.22). Sneferu then ordered a third pyr - amid built. This pyramid had an initial slope of 43o, stood on a base of 722 ft x722 ft, rose to a height of 345 ft, and had smooth sides. This experience wa s used by architects in the reign of Khufu (2589-2566 BCE) to build the largest pyramid in the wo rld called the Great pyramid of Giza ( Figure 1.14 d). The Great Pyramid (see problem 1.21) stands on a base of 756.7 ft x 756.7 ft and has a height of 481 ft. The ancient Egyptians did not have a formal definition of stress, but they had an intuitive understanding of the concept of strength. As happen s often in engineering design they were able to design and construct pyramids through trial and error. Modern engineering can reduce th is costly and time consuming process by developing rigorous methodologies and formulas. In th is book we will be developi ng formulas for strength and stiffness design of stru ctures and machine elements. Figure 1.14 Pyramids of Egypt (a) Mastaba (b) Step pyramid (c) Bent pyramid (d) Great pyramid of Giza.(a)(b) (c) (d) 1 23Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.1.4 Internally Distributed Force Systems In Sections 1.1.1 and 1.1.2 the normal stre ss and the shear stress were introduced as the average intensity of an internal norm al and shear force distribution, resp ectively. But what if there are internal moments at a cross sect ion? Would there be normal an d shear stresses at such sections? How are th e normal and shear stresses related to inte rnal moments? To answer these questions and to get a better understanding of the ch aracter of normal stress and shear stress, we now consider an alternative and more fundamental view. The forces of attraction and repulsion be tween two particles (atoms or molecules) in a body are assumed to act along the line that joins the two particles.1 The forces vary inversely as an exponent of the radial distance separating the two particles. Thus every particle exerts a force on ever y other particle, as shown symbolically in Figure 1.15 a on an imaginary surface of a body. These forces between particles hold the body together and are referred to as internal forces . The shape of the body changes when we apply external forces th us changing distance between particles a nd hence changing the forces between the particles (internal forces). The body breaks when th e change in the internal forces exceeds some char acteristic material value. Thus the strength of the material can be characterized by the measure of change in the intensity of internal forces. This mea - sure of change in the intensity of in ternal forces is what we call stress. In Figure 1.15 b we replace all forces that are exerted on any single particle by the result ants of these forces on that parti - cle. The magnitude and direction of these resultant forces will va ry with the location of the particle (point) implying that th is is an internal distributed force system. The intensity of intern ally distributed forces on an imaginary cu t surface of a body i s called the stress on the surface . The internally distributed forces (stress on a surface) can be resolved into normal (perpendic - ular to the surface) and tangential (paral lel to the surface) distribution. The intens ity of an internally distributed force th at is normal to the surface of an imaginary cut is called the normal stress on the surface. The intensity of an internally distributed force that is parallel to the surface of an imaginary cut surface is called the shear stress on the surface. Normal stress on a surface may be viewed as the internal forces that develo p due to the material resistance to the pulling apart or pushing together of two adjoining planes of an imaginar y cut. Like pressure, normal stress is always perpendicular to the surface of the imaginary cut. But unlike pressure, which can only be comp ressive, normal stress can be tensile. 1Forces that act along the line joining two particles are called central forces. The concept of central forces started with Newt on’s universal gravitation law, which states: “the force between two particles is inversel y proportional to the square of the radial dist ance between two partic les and acts along the l ine joining the two particles.” At atomic levels the cen - tral forces do not vary with the square of the radial di stance but with an exponent, which is a power of 8 or 10.A B C D E (a) F/H9270ANormal to plane(a) FB FC FEFDFA /H9268A/H9270A B EDCA Tangent in plane (b)Normal to plane Figure 1.15 Internal forces between particles on two sides of an imaginary cut. ( a) Forces between particles in a body, shown on par- ticle A. (b) Resultant force on each particle. 1 24Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012Shear stress on a surface may be viewed as the internal forces that develop due to the material resistance to the sliding of two adjoining planes along the imaginary cut. Like friction, sh ear stresses act tangent to the plane in the direction opposite to the impending motion of the surface. But unlike friction, shear stress is not re lated to the normal forces (stresses). Now that we have established that the stress on a surface is an internally distributed force sy stem, we are in a position to answer the questions raised at the beginni ng of the section. If the no rmal and shear stresses are constant in magnitude and direction across the cross section, as shown in Figure 1.16 a and b, then these can be replaced by statically equivalent normal and shear forces. [We obtain the equivalent forms of Equations ( 1.1) and ( 1.2).] But if either the magnitude or the direction of the normal and shear stresses changes across the cross section, then internal bending moments My , Mz and the internal torque T may be necessary for static equivalency, as shown in Figure 1.16 c, d, and e. Figure 1.16 shows some of the stress distribu - tions we will see in this book. But how do we deduce the variation of stress on a surf ace when stress is an internal quantity that cannot be measured directly ? The theories in this book that answer this question were developed over a long period of time using experimental observations, intuition, and logical dedu ction in an iterative manner. Assumptions have to be made regarding loading, geometry, and material pr operties of the structural member in the development of the theory. If the theoret - ical predictions do not conform to experi mental observations, then assumptions have to be modified to include added com - plexities until the theoretical pr edictions are consistent with experimental observations. In Section 3.2 , we will see the logic whose two links are shown in Figure 1.2 . This logic with assumptions regarding lo ading, geometry, and material properties will be used to develop the simplified theories in Chapters 4 through 6. EXAMPLE 1.4 Figure 1.17 shows a fiber pull-out test that is conducted to determine the shear stre ngth of the interface between the fiber an d the resin matrix in a composite material (see Section 3.12.3 ). Assuming a uniform shear stress τ at the interface, derive a formula for the shear stress in terms of the applied force P, the length of fiber L, and the fiber diameter D. PLAN The shear stress is acting on the cylindrical surface area of th e embedded fiber. The shear stress is uniform and hence can be replaced by an equivalent shear force V, which we can equate to P.(a) (b) (c) (d) (e) Figure 1.16 Static equivalency. x y z x y zx y z x y z MzT MyNormal stress linear in yNormal stress linear in zUniformshear stressin tangentialdirection. Uniform Normal Stress σavg N σavgA = V τavgA =Uniform Shear Stress τavg P L DResinFiber Figure 1.17 Fiber pull-out test. 1 25Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012SOLUTION Figure 1.18 a shows the cylindrical surface of the fiber with the uniform shear stress on the surface. The surface area A is equal to the cir - cumference multiplied by the fiber length L as shown by the Equation (E1). (E1) The shear force is the shear stre ss multiplied by the surface area, (E2) By equilibrium of forces in Figure 1.18 b (E3) ANS. COMMENTS 1. First, we replaced an internal distributed force system (shear stress) by an equivalent shear fo rce. Second, we related the int ernal shear force to external force by equilibrium. 2. In the preceding test it is implicitly assumed that the strength of the fiber is greater than the interface strength. Otherwise the fiber would break before it gets pulled out. 3. In a test the force P is increased slowly until the fiber is pulled out. The pull-out force is reco rded, and the shear strength can be calcu - lated. 4. Suppose we have determined th e shear strength from our formula for specific dimensions D and L of the fiber. Now we should be able to predict the force P that a fiber with different dimensions would suppor t. If on conducting the test th e experimental value of P is sig - nificantly different from the value predicted, then our assumption of uniform shear stress in the interface is most likely inco rrect. EXAMPLE 1.5 Figure 1.19 shows a test to determine the shear strength of an adhe sive. A torque (a moment along the axis) is applied to two t hin cylin - ders joined together with the adhe sive. Assuming a uniform shear stress τ in the adhesive, develop a form ula for the adhesive shear stress τ in terms of the applied torque Text, the cylinder radius R, and the cylinder thickness t. PLAN A free body diagram can be constructed after ma king an imaginary cut through the adhesive layer. On a differential area the int ernal shear force can be found and the moment from the internal shear force on the differentia l area obtained. By integrating we can find the total internal moment acting on the adhesive, which we can equate to the a pplied external moment Text. A πDL= P DL Figure 1.18 Free body diagrams of the fiber in Ex ample 1.4 (a) with shear stresses, (b) with equivalent internal shear force.P V(a) (b) V τA πDL() τ == V P=o r πDL() τ P= τ PπDL()⁄ = RT Tt Adhesive Figure 1.19 Adhesive shear strength test.Text Text 1 26Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012SOLUTION We make an imaginary cut through the adhesive and draw the shea r stress in the tangential dire ction, as shown in Figure 1.20 a. The differential area is the differential arc length ds multiplied by the thickness t. The differential tangential shear force dV is the shear stress multiplied by the differential area. The differential internal torque (moment) is the moment arm R multiplied by dV, that is, Noting that , we obtain the total internal torque by integrating over the en tire circumference. (E1) By equilibrium of moment in Figure 1.20 b (E2) ANS. COMMENTS 1. By recording the value of the torque at which the top half of the cylinder separates from the bottom half, we can calculate the shear strength of the adhesive. 2. The assumption of uniform shear stress can on ly be justified for thin cy linders. In Chapter 5 we will see that shear stress for thicker cyl - inders varies linearly in the radial direction. 3. First, we replaced an internal distributed force system (shear stress) by an equivalent inte rnal torque. Second, we related the internal torque to external torque by equilibrium. EXAMPLE 1.6 Figure 1.21 shows a drill being used to make a L = 12-in.-deep hole for placing explosive charge s in a granite rock. The shear strength of the granite is τ = 5 ksi. Determine the minimum torque T that must be applied to a drill of radius R = 1-in., assuming a uniform shear stress along the length of the dri ll. Neglect the taper at the end.dV/H9270() dst T /H9270 /H9270/H9270dsss Figure 1.20 Free-body diagrams in Example 1.5 (a) with shear stress, (b) with equiva lent internal torque.Text(a)(b) TextT dT RdV Rτtds == ds R dθ =[] T RdV∫Rτt()Rθd ∫τtR2θτ tR22π() =d 02π ∫== = T Text=o r 2 πR2tτ Text= τText 2πR2t-------------- - = Figure 1.21 Torque on a drill.12 inA /H9270TText 1 27Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012PLAN The imaginary cut surface is the surface of the hole in the gr anite. The shear stress on the surface of the hole would act like a distributed frictional force on the cylindrical surface of the drill bit. We can find the mo ment from this frictiona l force and relate it t o the applied torque. SOLUTION The shear stress acts tangential to the cylindrical surface of the drill bit, as shown in Figure 1.22 a. Multiplying the shear stress by the differential surface area ds dx we obtain the differenti al tangential shear force dV. Multiplying dV by the moment arm R, we obtain the internal torque , which is due to the shear stress over the differential surface area. Integrating over th e circumference and the length of the drill, we obtain the total internal torque. or (E1) By equilibrium of moment in Figure 1.22 b (E2) ANS. COMMENTS 1. In this example and in Example 1.4 shear stress acted on the outside cylindrical surface. In Example 1.4 we replaced the shear stresses by just an internal shear force, whereas in this example we repl aced the shear stresses by an internal torque. The difference c omes from the direction of the shear stress. 2. In Example 1.5 and in this example the su rfaces on which the shear stresses are acting are different. Yet in both examples we r eplaced the shear stresses by the equi valent internal torque. 3. The two preceding comments emphasize that be fore we can define which internal force or which internal moment is statically equi v- alent to the internal stress distribution, we must specify the direction of stress and the orie ntation of the surface on which the stress is acting. We shall develop this concept further in Section 1.2 .T A dx dsx R/H9270 dV /H11005 /H9270 dx ds Figure 1.22 Free body diagram of drill bit in Ex ample 1.6 (a) with shear stress, (b) with equivalent internal torque.Text (a) (b) xA TText dT RdV Rτdsdx == ds R dθ = T R dV∫Rτ()Rθd dx 02π ∫0L∫τR2θd dx 02π ∫0L∫τR22π xd 0L ∫2π 2πτR2L == = = = = T 2π5 ksi() 1 in.()212 in.() 120π in.·kips == TextT= Text377 in.· kips= Consolidate your knowledge 1. In your own words describe stress on a surface. 1 28Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012PROBLEM SET 1.2 Internally Distributed Force Systems 1.73 The post shown in Figure P1.73 has a rectangular cross section of 2 in. × 4 in. The length L of the post buried in the ground is 12 in. and the average shear strength of the soil is 2 psi. Determine the force P needed to pull the post out of the ground. 1.74 The post shown in Figure P1.73 has a circular cross section of 100-mm diameter. The length L of the post buried in the ground is 400 mm. It took a force of 1250 N to pull the post out of the ground. What wa s the average shear strength of the soil? 1.75 The cross section of the post shown in Figure P1.73 is an equilateral triangle with each side of dimension a. If the average shear strength of the soil is τ, determine the force P needed to pull the post out of the ground in terms of τ, L, and a.QUICK TEST 1.1 Time: 15 minutes Total: 20 points Answer true or false and justify each answer in one senten ce. Grade yourself with the answers given in Appendix E. Give yourself one point for each correct answer (true or false) and one point for every correct explanation. 1. You can measure stress directly w ith an instrument the way you meas ure temperature w ith a thermometer. 2. There can be only one normal stress componen t acting on the surface of an imaginary cut. 3. If a shear stress component on the left surface of an imaginary cu t is upward, then on the right surface it will be downward. 4. If a normal stress component puts the left surface of an imaginary cut in tens ion, then the right surface will be in compression. 5. The correct way of reporting shear stress is τ = 70 kips. 6. The correct way of reporting positive axial stress is σ = +15 MPa. 7. 1 GPa equals 106 Pa. 8. 1 psi is approximately equal to 7 Pa. 9. A common failure stress value for metals is 10,000 Pa. 10. Stress on a surface is the same as pressure on a surface as both quantities have the same units. P Ground PostL Figure P1.73 1 29Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.76 A force P = 10 lb is applied to the handle of a hammer in an effort to pull a nail out of the wood, as shown in Figure P1.76. The nail has a diameter of and is buried in wood to a depth of 2 in. Determine the average shear stress acting on the nail. 1.77 Two cast-iron pipes are adhesively bonded together over a length of 200 mm as shown in Figure P1.77. The outer diameters of the two pipes are 50 mm and 70 mm, and the wall thickness of each pipe is 10 mm. The two pipes separated while transmitting a force of 100 kN. What was the average shear stress in the adhesive just before the two pipes separated? 1.78 Two cast-iron pipes are adhesively bonded to gether over a length of 200 mm (Figure P1.78). The outer diameters of the two pipes are 50 mm and 70 mm, and the wall th ickness of each pipe is 10 mm. The two pipes separated while transmitting a torque of What was the aver - age shear stress in the adhesive just before the two pipes separated? 1.79 Two cast-iron pipes are held together by a bolt, as shown in Fi gure P1.79. The outer diameters of the two pipes are 50 mm and 7 0 mm, and the wall thickness of each pipe is 10 mm. The diameter of the bolt is 15 mm. The bolt broke while transmitting a torque of On what sur - face(s) did the bolt break? What was the average shear stress in the bolt on the surf ace where it broke? 1.80 The can lid in Figure P1.80a gets punched on two sides AB and AC of an equilateral triangle ABC. Figure P1.80b is the top view showing relative location of the points. The thickness of the lid is t = 1/64 in. and the lid material can at most support a shear stress of 1800 psi. Assume a uniform shear stress during punching an d point D acts like a pin joint. Use a= 1/2 in., b = 3 in. and c =1/4 in. Determine the minimum force F that must be applied to the can opener. 1 8--- in. Figure P1.76 2i n .P 12 in. P P Figure P1.77 2 kN m. ⋅ TT Figure P1.78 2 kN ·m. TT Figure P1.79 Figure P1.80 (a) (b) B AC DF E aA A BC BC D E a a cb 1 30Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.81 It is proposed to use diameter bolts in a 10-in.-diameter coupling for transferring a torque of 100 in. · kips from one 4-in.-diameter shaft onto another ( Figure P1.81 ). The maximum average shear stress in the bolts is to be limited to 20 ksi. How many bolts are needed, and at what radius should the bolts be placed on the coupling? (Note there are multiple answers.) 1.82 A human hand can comfortably apply a torsional moment of 15 in.· lb (Figure P1.82). (a) What should be the breaking shear streng th of a seal between the lid and the bottle, assuming the lid has a diameter of in. and a height of in.? (b) If the same sealing strength as in part (a) is used on a lid that is 1 in. in diameter and in. in height, what would be the to rque needed to open the bottle? 1.83 The torsional moment on the lid is applied by hand exerting a force F on the handle of bottle opener as shown in Figure P1.83. Assume the average shear strength of the bond between the lid and the bottle is 10 psi. Determine the minimum force F needed to open the bottle. Use t = 3/ 8 in. d = 2 1/2 in. and a = 4 in. 1.2 STRESS AT A POINT The breaking of a structure starts at th e point where the internal fo rce intensity—that is, where stress exceeds some material characteristic value. This imp lies that we need to refine our defini tion of stress on a surface to that of Stress at a Point . But an infinite number of planes (surfaces) can pass through a point. Which im aginary surface do we shrink to zero? And when we shrink the surface area to zero, wh ich equation should we use, ( 1.1) or ( 1.2)? Both difficulties can be addressed by assigning an orientation to the imag inary surface and to the inte rnal force on this surface. We label these direct ions with subscripts for t he stress components, in the same way that subscripts x, y, and z describe the components of vectors. Figure 1.23 shows a body cut by an imaginary plane that has an outward normal in the i direction. On this surface we have a differential area Δ Ai on which a resultant force acts. Δ Fj is the component of the force in the j direction. A component 1 2----in. T T Figure P1.81 11 2---1 2--- 1 2--- Figure P1.82 Figure P1.83 td Fa 1 31Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012of average stress is Δ Fj /Δ Ai. If we shrink ΔAi to zero we get the definition of a stress component at a point as shown by the Equation (1.3) . (1.3) Now when we look at a stress co mponent, the first subscript tells us the orientation of the imaginary surface and the second th e direction of the internal force. In three dimensions each subscript i and j can refer to an x, y, or z direction. In other words, there are nine possible com - binations of the two subscripts. This is shown in the stress matrix in Equation (1.4) . The diagonal elements in the stress matrix are the normal stresses and all off-diagonal elements represent the shear stresses. (1.4) To specify the stress at a point, we need a magnitude and two directions. In three dimensions we need nine components of stress, and in two dimensions we need four components of stress to completely define stress at a point. Table 1.3 shows the number of components needed to specify a scalar, a vector, and stress. Now force, moment, velocity, and acceleration are all different quantities, but they all ar e called vectors. In a similar manner, stress belongs to a category called tensors . More spe - cifically, stress is a second-order tensor ,2 where ‘second order’ refers to the exponent in the last row. In this terminology, a vector is a tensor of order 1, an d a scalar is a tensor of order 0. 1.2.1 Sign convention To obtain the sign of a stress component in Equation (1.3) we establish the following sign convention. Sign Convention: Differential area Δ Ai will be considered positive if the outward normal to the surface is in the positive i direction. If the outward normal is in the negative i direction, then Δ Ai will be considered negative. We can now deduce the sign for stress. A stress component can be positive in two ways. Both the numerator and the denomina - tor are positive or both the numerator and the denominator are negative in Equation (1.3) . Alternatively, if numerator and the 2To be labeled as tensor, a quantity must also satisfy certain coordina te transformation properties, which will be discussed bri efly in Chapter 8.TABLE 1.3 Comparison of num ber of components Quantity One Dimension Two Dimensions Three Dimensions Scalar 1 = 101 = 20 1 = 30 Vector 1 = 112 = 21 3 = 31 Stress 1 = 124 = 22 9 = 32σijΔFj ΔAi---------⎝⎠⎛⎞ ΔAi0→lim= direction of outward normal to imaginary cut surfacedirection of internal force component i /H9004Ai /H9004FjInternal forceOutward normal Figure 1.23 Stress at a point. σxxτxyτxz τyxσyyτyz τzxτzy σzz 1 32Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012denominator in Equation (1.3) have: the same sign the stress component is positive ; if they have opposite signs the stress com - ponent is negative . We conclude this section with the following points to remembe r. •Stress is an internal quantity that has units of force per unit area. •A stress component at a point is specified by a magnitude and two directions. Stress at a point is a second-order tensor. •Stress on a surface is specified by a magnitude and only one di rection. Stress on a surface thus is a vector. •The first subscript on stress gives the di rection of the outward normal of the imaginary cut surface. The second sub- script gives the direction of the internal force. •The sign of a stress component is determ ined from the direction of the internal force and th e direction of the outward normal to the imaginary cut surface. 1.3 STRESS ELEMENTS The previous section showed that stress at a point is an abstract quantity. Stress on a surface, however, is easier to visualiz e as the intensity of a distributed force on a surface. A stress element is an imaginary object that he lps us visualize stress at a point by constructing surfaces that have outwar d normals in the coordinate directions. In Cartesian coordinates the stress element is a cube; in cylindrical or a spherical coordina tes the stress element is a fr agment of a cylind er or a sphere, respectively. We st art our discussion with the construction of a stress element in Ca rtesian coordinates to emphasize the basic construction principle s. We can use a similar process to draw stre ss elements in cylindrical and spherical coordinate systems as demonstrated in Exam - ple 1.9 . 1.3.1 Construction of a Stress Element for Axial Stress Suppose we wish to visualize a positive stress component σxx at a point that may be generated in an bar under axial forces shown in Figure 1.24 a. Around this point imagine an object that has sides with outward normals in the coordinate direction. The cube has six surfaces with out ward normals that are either in the positive or in the negative coordinate direction, as show n in Figure 1.24 . The first subscript of σxx tells us it must be on the surface that has an outward normal in the x direction. Thus, the two surfaces on which σxx will be shown are at A and B. The direction of the outwa rd normal on surface A is in the positive x direction [the denominator is positive in Equation (1.3) ]. For the stress component to be positive on surface A, the force must be in the positive x direction [the numerator must be positive in Equation (1.3) ], as shown in Figure 1.24 b. The direction of the out ward normal on surface B is in the negative x direction [the denominator is negative in Equation (1.3) ]. For the stress component to be positive on surface B, the force must be in the negative x direction [the numerator must be negative in Equation (1.3) ], as shown in Figure 1.24 b. The positive stress component σxx are pulling the cube in opposite directions; th at is, the cube is in tension due to a posi - tive normal stress component. We can use th is information to dr aw normal stresses in place of subscripts. A tensile normal stress will pull the surface away from the in terior of the element and a compressive no rmal stress will push the surface into t he element. As mentioned earlier, no rmal stresses are usually reported as tension or compression and not as positive or negative. It should be emphasized that the single arrow used to show th e stress component does not im ply that the stress component is a force. Showing the stress components as dist ributed forces on surfaces A and B as in Figure 1.25 is visually more accurate Figure 1.24 (a) Axial bar. (b) Stress element for axial stress.(a) (b) xy zP Py Aσxx B x zσxx 1 33Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012but very tedious to draw every time we ne ed to visualize stress. We will show st ress components using single arrows as in Fig- ure 1.24 , but visualize them as shown in Figure 1.25 . 1.3.2 Construction of a Stress Element for Plane Stress Plane stress is one of the two types of two-dimensional simplifi cations used in mechanics of ma terials. In Chapter 2 we will study the other type, plane strain. In Chapter 3 we will study the difference between the two types. By two dimensional we imply that one of the coordinates does not play a role in the description of the problem. If we choose z to be the coordinate, we set all stresses with subscript z to zero to get (1.5) We assume that the stress components in Equation (1.5) are positive. Let us consid er the first row. The first subscript gives us the direction of the outward normal, which is the x direction. Surfaces A and B in Figure 1.26 a have outward normals in the x direction, and it is on these surfaces that the stress component of the first row will be shown. The direction of the outwa rd normal on surface A is in the positive x direction [the denominator is positive in Equation (1.3) ]. For the stress component to be positive on surface A, the force must be in the positive direction [the numerator must be positive in Equation (1.3) ], as shown in Figure 1.26 a. The direction of the out ward normal on surface B is in the negative x direction [the denominator is negative in Equation (1.3) ]. For the stress componen t to be positive on surface B, the force must be in the negative direction [the numerator must be negative in Equation (1.3) ], as shown in Figure 1.26 a. Now consider row 2 in the stress matrix in Equation (1.5). Fr om the first subscript we know that the normal to the surface is in the y direction. Surface C has an outward normal in the positive y direction, therefor e all forces on surface C are in the positive direction of the sec ond subscript, as shown in Figure 1.26 a. Surface D has an outward normal in the negative y direc - tion, therefore all forces on surface D are in the negative di rection of the second subscript, as shown in Figure 1.26 a. We note that the plane with outward normal in the z direction is stress-free. St ress-free surfaces are also called free sur - faces , and these surfaces play an impor tant role in stress analysis. Figure 1.26b shows the two-dimensional representation of the stress element that will be seen looking down the z axis.y Figure 1.25 Stress components are distri buted forces on a surface.σxxx zσxx σxxτxy0 τyxσyy0 0 0 0 x zy dzdy /H9268xx/H9268 /H9268xx/H9268/H9270xy/H9270/H9268yy/H9268/H9268 /H9268yy/H9268/H9268/H9270yx/H9270/H9270 dx Figure 1.26 Plane stress: (a) 3-dimensional el ement (b) 2-dimensional element.ABC DABC Dxy σxxσxxσyy σyyτxy τxy τyxτyx dy dx 1 34Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.4 SYMMETRIC SHEAR STRESSES If a body is in equilibrium, then all points on the body are in equilibrium. Is the stress element that repr esents a point on t he body in equilibrium? To answer this question we need to conver t the stresses into forces by multiplying by the surface area. We take a simple problem of plane st ress and assume that the cube in Figure 1.26 has lengths of dx, dy, and dz in the coordinate directions. We draw a two-dimensional pi cture of the stress cube af ter multiplying each stress co mponent by the surface area and get the force diagram of Figure 1.27.3 In Figure 1.27 we note that the equations of force equilibr ium are satisfied by the assumed st ate of stress at a point. We consider the moment about point O and obtain (1.6) We cancel the differential volume (dx dy dz) on both sides to obtain (1.7a) In a similar manner we can show that (1.7b) (1.7c) Equations (1.7a) through (1.7c) emphasi ze that shear stress is symmetric. The sy mmetry of shear stress implies that in three dimensions there are only six independent stress components out of the nine componen ts necessary to specify stress at a point. In two dimensions there are only three independent stress components out of the four comp onents necessary to specify stress at a point. In Figure 1.26 notice that the sh ear stress components τxy and τyx point either toward the co rners or away from the cor - ners. This observation can be used in draw ing the symmetric pair of sh ear stresses after drawing the shear stress on one of the surfaces of the stress cube. EXAMPLE 1.7 Show the non-zero stress components on the surfaces of the two cubes shown in different coor dinate systems in Figure 1.28. PLAN We can identify the surface with the outward normal in the dire ction of the first subscript. Using the sign convention and Equa tion (1.3) we draw the force in the dire ction of the second subscript. 3Figure 1.27 is only valid if we assume that the stresses are varying very slowly with the x and y coordinates. If this were not true, we would have to account for the increase in stresses over a differential element. But a more rigorous analysis will also reveal that shear stresses are symmetric, see Prob lem 1.105 . Figure 1.27 Force diagram for plane stress./H9268yy/H9268/H9268(dx dz) /H9268xx/H9268(dy dz)/H9270yx/H9270/H9270(dx dz) /H9270yx/H9270/H9270(dx dz)/H9270xy/H9270(dy dz) /H9268yy/H9268/H9268(dx dz)/H9268xx/H9268(dy dz) /H9270xy/H9270(dy dz) Odxdy τxydy dz() dx τyxdx dz() dy = τxyτyx= τyzτzy= τzxτxz= yz xCube 1Cube 2 Figure 1.28 Cubes in different coordina te systems in Example 1.7. zy xσxx80MPa T () = τxy30MPa=0 τyx30MPa= σyy40MPa C () =0 00 0 1 35Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012SOLUTION Cube 1: The first subscript of σxx and τxy shows that the outward normal is in the x direction; hence these co mponents will be shown on surfaces C and D in Figure 1.29 a. The outward normal on surface C is in the negative x direction; hence the denominator in Equation (1.3) is negative. Therefore on Figure 1.29 a: • The internal force has to be in the negative x direction to produce a positive (tensile) σxx. • The internal force has to be in the negative y direction to produce a positive τxy . The outward normal on surface D is in the positive x direction; hence the denominat or in Equation (1.3) is positive. Therefore on Figure 1.29 a: • The internal force has to be in the positive x direction to produce a positive (tensile) σxx. • The internal force has to be in the positive y direction to produce a positive τxy . The first subscript of τyx and σyy shows that the outward normal is in the y direction; hence this compone nt will be shown on surfaces A and B. The outward normal on surface A is in the positive y direction; hence the denomina tor in Equation (1.3) is posit ive. Therefore on Figure 1.29 a: • The internal force has to be in the positive x direction to produce a positive τyx. • The internal force has to be in the negative y direction to produce a negative (compressive) σyy. The outward normal on surface B is in the negative y direction; hence the denominator in Equation (1.3) is negative. Therefore on Figure 1.29 a: • The internal force has to be in the negative x direction to produce a positive τyx. • The internal force has to be in the positive y direction to produce a negative (compressive) σyy. Cube 2: The first subscript of σxx and τxy shows that the outward normal is in the x direction; hence these co mponents will be shown on surfaces E and F. The outward normal on surface E is in the negative x direction; hence the denomina tor in Equation (1.3) is ne gative. Therefore on Figure 1.29 a: • The internal force has to be in the negative x direction to produce a positive (tensile) σxx. • The internal force has to be in the negative y direction to produce a positive τxy . The outward normal on surface F is in the positive x direction; hence the denominator in Equation (1.3) is positive. Therefore on Figure 1.29 a: • The internal force has to be in the positive x direction to produce a positive (tensile) σxx. • The internal force has to be in the positive y direction to produce a positive τxy . The first subscript of τyx and σyy shows that the outward normal is in the y direction; hence this compone nt will be shown on surfaces A and B. The outward normal on surface A is in the negative y direction; hence the denominator in Equation (1.3) is negative. Therefore on Figure 1.29 a: • The internal force has to be in the negative x direction to produce a positive τyx. • The internal force has to be in the positive y direction to produce a negative (compressive) σyy. The outward normal on surface B is in the positive y direction; hence the denomina tor in Equation (1.3) is posit ive. Therefore on Figure 1.29 a: • The internal force has to be in the positive x direction to produce a positive τyx. • The internal force has to be in the negative y direction to produce a negative (compressive) σyy. COMMENTS 1. Figure 1.30 shows the two-dimensional represen tations of stress cubes sh own in Figure 1.29. These tw o-dimensional representatio ns are easier to draw but it must be kept in mind that the point is in three-dimensio nal space with surfaces with outward normals in the z direction being stress free. Figure 1.29 Solution of Example 1.7.(a ) Cube 1. (b) Cube 2.80 (a) (b) zy xDAC B 8030 30 30404030 EF yzx A DC B 40 30 8080 30 3040 EF Figure 1.30 Two-dimensional depiction of the solution of Example 1.7.(a) Cube 1. (b) Cube 2.(a) (b) zy xAC80 8030 30 30404030 B Dyzx A4030 8080 30 3040 EF 1 36Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20122. We note that σxx pulls the surfaces outwards and σyy pushes the surfaces inwards in Figures 1. 29 and 1.30 as these are tensile and com - pressive stresses, respectively. Hence, we can use this informat ion to draw these stress componen ts without using the subscript s. 3. The shear stress τxx and τyx either point towards the corner or away from the corn er as seen in Figures 1.29 and 1.30. Using this infor - mation we can draw the shear stress on the appropriate surfaces after obtaining the direction on one surface using subscripts. 1.5* CONSTRUCTION OF A STRESS ELEMENT IN 3-DIMENSION We once more visualize a cube with outwar d normals in the coordinate direction around the point we wish to show our stress components. The cube has six surfaces with ou tward normals that are either in the positive or in th e negative coordinate direc - tion, as shown in Figure 1.31 . In other words, we have now accounted for the fi rst subscript in our stre ss definition. We know that force is in the positive or negative direction of the second subscript. We use our sign convention to show the stress in t he direction of the force on each of the six surfaces. To demonstrate the construction of the stre ss element we will assume that all nine stress components in the stress matrix shown in Figure 1.31 a are positive. Let us consider the first row. The fi rst subscript gives us the direction of the outward nor - mal, which is the x direction. Surfaces A and B in Figure 1.31 b have outward normals in the x direction, and it is on these sur - faces that the stress component of the first row will be shown. The direction of the outwa rd normal on surface A is in the positive x direction [the denominator is positive in Equation (1.3) ]. For the stress component to be positive on surface A, the force must be in the positive direction [the numerator must be positive in Equation (1.3) ], as shown in Figure 1.31 b. The direction of the out ward normal on surface B is in the negative x direction [the denominator is negative in Equation (1.3) ]. For the stress componen t to be positive on surface B, the force must be in the negative direction [the numerator must be negative in Equation (1.3) ], as shown in Figure 1.31 b. Now consider row 2 in the stress matrix in Figure 1.31 a. From the first subscript we k now that the normal to the surface is in the y direction. Surface C has an outward normal in the positive y direction, therefor e all forces on surface C are in the positive direction of the sec ond subscript, as shown in Figure 1.31 b. Surface D has an outward normal in the negative y direc - tion, therefore all forces on surface D are in the negative di rection of the second subscript, as shown in Figure 1.31 b. By the same logic, the components of row 3 in the stress matrix are shown on surfaces E and F in Figure 1.31 b. A /H9268xx/H9268xy /H9270xz/H9270/H9270yx/H9270/H9270/H9270yz/H9270/H9270/H9268yy/H9268/H9268 /H9270 /H9270/H9270zx/H9270/H9270zy/H9270 yyyz /H9270/H9270C xy zF Figure 1.31 Stress cube showing all positive stress component s in three dimensions.σxx τyx τzxτxy σyy τzyτxz τyz σzz(a)(b) 1 37Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012EXAMPLE 1.8 Show the nonzero stress components on surfaces A, B, and C of the two cubes shown in differen t coordinate systems in Figure 1.32. PLAN We can identify the surface with the outward normal in the dire ction of the first subscript. Using the sign convention and Equa tion (1.3) we draw the force in the dire ction of the second subscript. SOLUTION Cube 1: The first subscript of σxx, τxy , and τxz shows that the outward normal is in the x direction; hence these components will be shown on surface C. The outward normal on surface C is in the negative x direction; hence the denominator in Equation (1.3) is negative. There - fore on Figure 1.33 a: • The internal force has to be in the negative x direction to produce a positive (tensile) σxx. • The internal force has to be in the negative y direction to produce a positive τxy . • The internal force has to be in the positive z direction to produce a negative τxz. The first subscript of τyx shows that the outward normal is in the y direction; hence this compone nt will be shown on surface B. The out - ward normal on surface B is in the positive y direction; hence the denominator in Equation (1.3) is positive. Therefore on Figure 1.33 a: • The internal force has to be in the positive x direction to produce a positive τyx. The first subscript of τzx, σzz shows that the outward normal is in the z direction; hence th ese components will be shown on surface A. The outward normal on surface A is in the positive z direction; hence the denominator in Equation (1.3) is positive. Therefore on Figure 1.33a: • The internal force has to be in the negative x direction to produce a negative τzx. • The internal force has to be in the negative z direction to produce a negative (compressive) σzz. Cube 2: The first subscript of σxx , τxy , and τxz shows that the outward normal is in the x direction; hence these components will be shown on surface A. The outward normal on surface A is in the negative x direction; hence the denominator in Equation (1.3) is negative. There - fore in Figure 1.33 b: • The internal force has to be in the negative x direction to produce a positive (tensile) σxx. • The internal force has to be in the negative y direction to produce a positive τxy. • The internal force has to be in the positive z direction to produce a negative τxz. The first subscript of τyx shows that the outward normal is in the y direction; hence this compone nt will be shown on surface B. The out - ward normal on surface B is in the negative y direction; hence the denominator in Equation (1.3) is negative. Therefore in Figure 1.33 b: • The internal force has to be in the negative y direction to produce a positive τyx. The first subscript of τzx, σzz shows that the outward normal is in the z direction; hence these compone nts will be shown on surface C. The outward normal on surface C is in the positive z direction; hence the denominator in Equation (1.3) is positive. Therefore in Figure 1.33b: • The internal force has to be in the negative x direction to produce a negative τzx. • The internal force has to be in the negative z direction to produce a negative (compressive) σzz. COMMENTS 1. In drawing the normal stresses we coul d have made use of the fact that σxx is tensile and hence pul ls the surface outward. σzz is com - pressive and hence pushes the surfa ce inward. This is a quicker way of getting th e directions of these stress components than t he argu - ments based on signs and subscripts.Cube 1 z xy ABC Figure 1.32 Cubes in different coordinate systems.xz yCube 2 ABCσxx80 MPa (T)= τxy30 MPa= τxz70 MPa–= τyx30 MPa= σyy0= τyz0= τzx70 MPa–= τzy0= σzz40 MPa (C)= z xyA803033 3070 70 40BC (a) Figure 1.33 Solution of Example 1.8.xz AB y70 3030 807040 C (b) 1 38Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20122. Once we have drawn τxy and τxz using the subscripts, we could draw τyx and τzx using the observation that th e pair of symmetric shear stresses point towards or away from the corn er formed by the two adjoining surfaces, thus saving so me effort in the constructio n of the stress element. EXAMPLE 1.9 Show the following pos itive stress components on a st ress element drawn in the spherical c oordinate system shown in Figure 1.34 . PLAN We construct a stress element with surfaces that have outward normals in the r, θ, and φ directions. The first subscript will identify the surface on which the row of stress components is to be shown. The second subscript then will show the direction of the stress c omponent on the surface. SOLUTION We draw a stress element with lines in the directions of r, θ, and φ, as shown in Figure 1.35. The stresses σrr, τrθ, and τrφ will be on surface A in Figure 1.35 . The outward normal on surface A is in the positive r direction. Thus the forces have to be in the positive r, θ, and φ directions to result in positive σrr, τrθ, and τrφ. The stresses τθr, σθθ, and τθφ will be on surface B in Figure 1.35 . The outward normal on surface B is in the negative θ direction. Thus the forces have to be in the negative r, θ, and φ directions to result in positive τθr, σθθ, and τθφ. The stresses τφr, τφθ, and σφφ will be on surface C in Figure 1.35 . The outward normal on surface C is in the positive φ direction. Thus the forces have to be in the positive r, θ, and φ directions to result in positive τφr, τφθ, and σφφ. COMMENT 1. This example demonstrates that use of subs cripts in determining the direction of st ress components follows the same procedure a s in cartesian coordinates even though the stre ss element is a fra gment of a sphere.z r y x/H9258/H9278 Figure 1.34 Stresses in spherical coordinates.σrrτrθτrφ τθrσθθτθφ τφrτφθσφφ r /H9258/H9278 /H9258 /H9258 /H9270/H9258/H9278/H9270 /H9270/H9258/H9270/H9270r/H9268/H9258/H9258/H9268 Figure 1.35 Stress element in spherical coordinates. Consolidate your knowledge 1. In your own words describe stress at a point and how it differs from stress on a surface 1 39Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012PROBLEM SET 1.3 Plane Stress: Cartesian Coordinates 1.84 Show the stress components of a point in pl ane stress on the square in Figure P1.84. 1.85 Show the stress components of a point in pl ane stress on the square in Figure P1.85. 1.86 Show the stress components of a point in pl ane stress on the square in Figure P1.86. 1.87 Show the stress components of a point in pl ane stress on the square in Figure P1.87.QUICK TEST 1.2 Time: 15 minutes/Total: 20 points Answer true or false and justify each answer in one sentence. Grade yourself with the answers given in Appendix E. Give yourself one point for every correct answer (true or fa lse) and one point for ev ery correct explanation. 1. Stress at a point is a vect or like stress on a surface. 2. In three dimensions stress has nine components. 3. In three dimensions stress has six independent components. 4. At a point in plane stress there are three independent stress components. 5. At a point in plane stress there ar e always six zero stress components. 6. If the shear stress on the left surf ace of an imaginary cut is upward and defined as pos itive, then on the right surface of the imaginary cut it is downward and negative. 7. A stress element can be drawn to any scale. 8. A stress element can be dr awn at any orientation. 9. Stress components are opposite in direction on the two surfaces of an imaginary cut. 10. Stress components have opposite signs on the two surfaces of an imaginary cut. y x Figure P1.84 σxx100 MPa (T) τxy75 MPa–= = τyx75– M P a σyy85 MPa (T)= = y x Figure P1.85 σxx85 MPa (C) τxy75 MPa= = τyx75 MPa σyy100 MPa (T)= = y x Figure P1.86 σxx27 ksi (C) τxy18 ksi= = τyx18 ksi σyy85 ksi (T)= = yx Figure P1.87 σxx27 ksi (C) τxy18 ksi= = τyx18 ksi σyy85 ksi (T)= = 1 40Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.88 Show the nonzero stress components on the A, B, and C faces of the cube in Figure P1.88. 1.89 Show the nonzero stress components on the A, B, and C faces of the cube in Figure P1.89. Plane Stress: Polar Coordinates 1.90 Show the stress components of a point in plane stress on the stress element in polar coordinates in Figure P1.90. 1.91 Show the stress components of a point in plane stress on the stress element in polar coordinates in Figure P1.91. 1.92 Show the stress components of a point in plane stress on the stress element in polar coordinates in Figure P1.92. 1.93 Show the stress components of a point in plane stress on the stress element in polar coordinates in Figure P1.93.z xy ABC Figure P1.88 σxx70 MPa (T) τxy40– M P a τxz0= = = τyx40– M P a σyy85 MPa (C) τyz0= = = τzx0 τzy0 σzz0 = = = x yz ABC Figure P1.89 σxx70 MPa (T) τxy40– M P a τxz0= = = τyx40– M P a σyy85 MPa (C) τyz0= = = τzx0 τzy0 σzz0= = = /H9258y r x Figure P1.90 σrr125 MPa (T) τrθ65 MPa–= = τθr65– M P a σθθ90 MPa (C)= = /H9258y r x Figure P1.91 σrr125 MPa (T) τrθ65 MPa–= = τθr65– M P a σθθ90 MPa (C)= = /H9258y r x Figure P1.92 σrr18 ksi (T) τrθ12– ksi= = τθr12– k s i σθθ25 ksi (C)= = /H9258y r x Figure P1.93 σrr25 ksi (C) τrθ12 ksi= = τθr12 ksi σθθ18 ksi (T)= = 1 41Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012Stress Element in 3-dimensions 1.94 Show the nonzero stress components on the A, B, and C faces of the cube in Figure P1.94. 1.95 Show the nonzero stress components on the A, B, and C faces of the cube in Figure P1.95. 1.96 Show the nonzero stress components on the A, B, and C faces of the cube in Figure P1.96. 1.97 Show the nonzero stress components on the A, B, and C faces of the cube in Figure P1.97. 1.98 Show the nonzero stress components in the r, θ, and x cylindrical coordinate system on the A, B, and C faces of the stress elements shown in Figures P1.98 .zy x ABC Figure P1.94 σxx100 MPa (T) τxy200 MPa= τxz125 MPa–= = τyx200 MPa σyy175 MPa (C)= τyz225 MPa= = τzx125– M P a τzy225 MPa σzz150 MPa (T)= = = yx z ABC Figure P1.95 σxx90 MPa (T) τxy200 MPa τxz0= = = τyx200 MPa σyy175 MPa (T) τyz225– M P a= = = τzx0 τzy225– M P a σzz150 MPa (C)= = = zy x ABC Figure P1.96 σxx0 τxy15ksi τxz0= = = τyx15 ksi σyy10 ksi (T) τyz25– ksi= = = τzx0 τzy25– ksi σzz20 ksi (C)= = = xz y ABC Figure P1.97 σxx0 τxy15– k s i τxz0= = = τyx15– k s i σyy10 ksi (C) τyz25 ksi= = = τzx0 τzy25 ksi σzz20 ksi (T)= = = A B C/H9258r x Figure P1.98 σrr150 MPa (T) τrθ100– M P a τrx125 MPa= = = τθr100– M P a σθθ160 MPa (C) τθx165 MPa= = = τxr125 MPa τxθ165 MPa σxx145 MPa (C) = = = 1 42Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.99 Show the nonzero stress components in the r, θ, and x cylindrical coordinate system on the A, B, and C faces of the stress elements shown in P1.99. 1.100 Show the nonzero stress components in the r, θ, and φ spherical coordinate system on the A, B, and C faces of the stress elements shown in Figure P1.100 . 1.101 Show the nonzero stress components in the r, θ, and φ spherical coordinate system on the A, B, and C faces of the stress elements shown in P1.101 . Stretch yourself 1.102 Show that the normal stress σxx on a surface can be replaced by the equivalent internal normal force N and internal bending moments My and Mz as shown in Figure P1.102 and given by the equations (1.8a) through (1.8c) . r x CCC A Figure P1.99 σrr10 ksi (C) τrθ22 ksi τrx32 ksi= = = τθr22 ksi σθθ0 τθx25 ksi= = = τxr32 ksi τxθ25 ksi σxx20 ksi (T) = = = z r y x AAA Figure P1.100 σrr150 MPa (T) τrθ100 MPa τrφ125 MPa= = = τθr100 MPa σθθ160 MPa (C) τθφ175– M P a= = = τφr125 MPa τφθ175– M P a σφφ135 MPa (C)= = = z r y x A B Figure P1.101 σrr0 τrθ18– k s i τrφ0 = = = τθr18– ksi σθθ10 ksi (C) τθφ25 ksi= = = τφr0 τφθ25 ksi σφφ20 ksi (T) = = = y zx O MzMyN Figure P1.102 (1.8a) (1.8b) (1.8c)N σxx Ad A∫= Myzσxx Ad A∫–= Mzyσxx Ad A∫–= 1 43Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 20121.103 The normal stress on a cross section is given by σxx = a + by, where y is measured from the centroid of the cross section. If A is the cross- sectional area, Izz is the area moment of inertia about the z axis, and N and Mz are the internal axial force and the internal bending moment given by Equations (1.8a) and (1.8c) , respectively, prove the result in Equation (1.8) . (1.8) We will encounter Equation (1.8) in combined ax ial and symmetric bending problems in later Chapter 10. 1.104 The normal stress on a cross section is given by σxx = a + by + cz, where y and z are measured from the centroid of the cross section. Using Equations (1.8a) , (1.8b) , and (1.8c) prove the result of Equation (1.9) . (1.9) where Iyy, Izz, and Iyz are the area moment of inertias. Equation (1.9) is used in the unsymmetrical bending of beams. Note that if either y or z is an axis of symmetry, then Iyz = 0. In such a case Equation (1.9) simplifies considerably. 1.105 An infinitesimal element in plane stress is shown in Figure P1.105. Fx and Fy are the body forces acting at the point and have the dimen - sions of force per unit volume. By converting stresses into fo rces and writing equilibrium equa tions obtain the results in Equa tions (1.10a) through (1.10c) . 1.6* CONCEPT CONNECTOR Formulating the concept of stress took 500 years of struggle, brie fly described in Section 1.6.1. In hindsight, the long evolut ion of quantifier of the strength is not surprisi ng, because stress is not a single idea. It is a package of ideas that may be repa ckaged in many ways, depending on the needs of the analysis. Our chapter dealt wi th only one such package, called Cauchy’ s stress, which is used most in engi neering design and analysis. 1.6.1 History: The Concept of Stress The first formal treatment of strength is seen in the note s of the inventor and artist Leonardo da Vinci (1452–1519). Leonardo conducted several experiments on the strength of struct ural materials. His notes on “testing the strength of iron wires of various lengths” includes a sketch of how to measure the strength of wire experimentally. We now recognize that the dependence of the strength of a material on its length is due to the variations in manufacturing defects along the length. The first indication of a concept of stress is found in Galileo Galilei (1564 –1642). Galileo was born in Pisa and became a professor of mathematics at the age of tw enty-five. For his belief in the Coperni can theory on the motion of planets, which contradicted the interpretation of scriptures at that time, Galileo was put under house arrest for the last eight years of his life. During that period he wrote Two New Sciences, which lays out his contributions to the field of mechanics. Here he discusses the strength of a cantilever beam bending under the action of its own weight. Galileo viewed stre ngth as the resistance to frac - ture, concluding that the strength of a bar depends on its cross- sectional area but is independent of its length. We will discu ss Galileo’s work on beam bending in Section 6.7 . The first person to differentiate between normal stress an d shear stress was Charles-A ugustin Coulomb (1736–1806) born in Angoulême. He was honored by the French Academy of Sciences in 1781 for his memoir Theorie des machines simples, in σxxN A----Mz Izz-------⎝⎠⎛⎞y – = σxxN A----MzIyy My– Iyz IyyIzz Iyz2–----------------------------------- - ⎝⎠⎜⎟⎛⎞y –MyIzz Mz– Iyz IyyIzz Iyz2–----------------------------------- ⎝⎠⎜⎟⎛⎞z – = OFx /H9270yx/H9270xy/H9268xx /H9268yyFy dxdydy/H11128/H9268yy /H11128y/H9268yy /H11001dy/H11128/H9270yx /H11128y/H9270yx /H11001 dx/H11128/H9270xy /H11128x/H9270xy /H11001 dx/H11128/H9268xx /H11128x/H9268xx /H11001 Figure P1.105 (1.10a) (1.10b) (1.10c)∂σxx ∂x-----------∂τyx ∂y---------- Fx++ 0 = ∂τxy ∂x----------∂σyy ∂y----------- Fy++ 0 = τxyτyx= 1 44Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012which he discussed friction between bodies. The theory of dry friction is named after him. Given the similarities between shear stress and friction, it seems only natural that Coulomb would be the first to differen tiate between normal and shear stress. We will see other works of Coulomb when we come to failure theory and on the torsion of circular shafts. Claude Louis Navier (1785–1857) initiated the mathematical de velopment of the concept of stress starting with Newton’s concept of a central force—one th at acts along a line between two particles. His a pproach led to a controversy that took eighty years to resolve, as we shall see in Section 3.12 . Augustin Cauchy (1789–1857) brought the concept of stress to the form we studied it in this chapter (Figure 1.36). Forced to leave Paris, his birt hplace, during the French Revolution, he took refuge in the village of Arcueil along with many other mathematicians and scientists of the period. At the age of twenty-one Cauchy worked en gineering at the port of Cher - bourg, which must have enhanced his understanding of the hydr odynamic concept of pressure. Pr essure acts always normal to a surface, but Cauchy assumed that on an internal surface it acts at an angle hen ce he reasoned it can be resolved into compo - nents, the normal stress and shear stress . Combining this idea with his natural mathematical abilities, Cauchy developed what is now called Cauchy’s stress. We shall see Cauchy’s genius again in chapters on strain, material properties, and stress and strain transformation. We have seen that unlike force, which is indivisible into more elementary ideas, stress is a package of ideas. Other pack - ages will contain related but different elem entary ideas. If instead of the cross-sec tional area of an undeformed body, we use the cross-sectional area of a deformed body, then we get true stress . If we use the cross-sectiona l area of a deformed body and take the component of this area in the undeformed configuration, then we get Kirchhoff’ s stress . Still other stress measures are used in nonlinear analysis. The English physicist James Clerk Maxwell (1831–1879) recogni zed the fact that the symmetry of shear stress given by Equations (1.7a) through (1.7c) is a consequence of there being no body moments. If a body moment is present, as in electro - magnetic fields, then shear stre sses will not be symmetric. In Figure 1.15 we replaced the internal forces on a particle by a resultant fo rce but no moment, because we assumed a central force between two particles. Woldemar V oigt (1850–1919) , a German scientist who worked extensively with crystals is credited with introducing the stress tensor. V oigt recognized that in some cases a couple vect or should be included when representing the interaction between particles by equivalent intern al loads. If stress analysis is conducted at a very small sc ale, as the frontier research in nanostructures, then the moment transmitted by bonds be tween molecules may need to be included. The term couple stress is sometimes used to indicate the presence of a couple vector. As history makes clear, stress has many definition. We choose the definition depending on the problem at hand and the information we are seeking. Most engineering analysis is linear and deals with large bodies, for which Cauchy’s stress gives very good results. Cauchy’s stress is thus sometimes refe rred to as engineering stre ss. Unless stated otherwise, stress always means Cauchy’s stress in mechanic s of materials and in this book. 1.7 CHAPTER CONNECTOR In this chapter we have established the linkage between stresses, internal forces and moments, and external forces and moments. We have seen that to replace st resses by internal forces an d internal moments requires knowledge of how the stress varies at each point on the surf ace. Although we can deduce simple stress behavior on a cross section, we would like to have other alternatives, in particular ones in which the danger of assuming physically impossible deformations is eliminated. This can be achieved if we can establish a re lationship between stresses and deformations. Before we can discu ss this relationship we need to understand the measure of deformation, which is the subject of Chapter 2. We will relate stresses and strains in Figure 1.36 Pioneers of stress concept. Augustin Cauchy. Claude Louis Navier 1 45Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012Chapter 3. In Section 3.2 we will synthesize the links introduced in Chapters 1, 2, and 3 into a logic that is used in mechanics of materials. We will use the logic to obtain simplified theories of one-dimensional structure members in Chapters 4, 5, 6, and 7. All analyses in mechanics are conducted in a coordinate system, which is chosen for simplification whenever possible. Thus the stresses we obtain are in a given coordinate system. Now, our motivation for learning about stress is to define a mea - sure of strength. Thus we can conclude th at a material will fail when the stress at a point reaches so me critical maximum value. There is no reason to expect that the stresses will be maximum in the arbitrarily chosen coordinate system. To deter - mine the maximum stress at a point thus implies that we estab lish a relationship between stresses in different coordinate sys - tems, as we shall do in Chapter 8. We have seen that the concept of stress is a difficult one. If th is concept is to be internalized so that an intuitive under - standing is developed, then it is impera tive that a discipline be developed to visualize the imaginary surface on which the stress is being considered. 1 46Printed from: http://www.me.mtu.edu/~mavable/MoM2ndMechanics of Materials: Stress M. Vable January, 2012POINTS AND FORMULAS TO REMEMBER •Stress is an internal quantity. •The internally distributed for ce on an imaginary cut surface of a body is called stress on a surface. •Stress has units of force per unit area. •1 psi is equal to 6.95 kPa, or approximately 7 kPa. 1 kPa is equal to 0.145 psi, or approximately 0.15 psi. •The internally distributed for ce that is normal (perpendicula r) to the surface of an imagin ary cut is called normal stress on a surface. •Normal stress is usually reported as tensile or compressive and not as positive or negative. •Average stress on a surface: • (1.1) (1.2) •where σav is the average normal stress, τav is the average shear stress, N is the internal normal force, V is the internal shear force, and A is the cross-sectional area of the imaginary cut on which N and V act. •The relationship of external forces (and mo ments) to internal forces and the relatio nship of internal forces to stress distri - butions are two distinct ideas. •Stress at a point: • (1.3) •where i is the direction of the outward nor mal to the imaginary cut surface, and j is the outward normal to the direction of the internal force. •Stress at a point needs a magnitude and two directions to sp ecify it, i.e., stress at a po int is a second-order tensor. •The first subscript on stress denotes th e direction of the outward normal of the imaginary cu t surface. The second sub - script denotes the directio n of the internal force. •The sign of a stress component is determin ed from the direction of the internal force and the direction of the outward nor - mal to the imaginary cut surface. •Stress element is an imaginary object that helps us visualize stress at a point by constructing surfaces that have outward normals in the coordinate directions. • (1.7a) (1.7b) (1.7c) •Shear stress is symmetric. •In three dimensions there are nine stress components, but only six are independent. •In two dimensions there are four stress components, but only three are independent. •The pair of symmetric shear stress compone nts point either toward the corner or away from the corner on a stress ele - ment. •A point on a free surface is said to be in plane stress.σavNA⁄ = τavVA⁄ = σijΔFj ΔAi---------⎝⎠⎛⎞ ΔAi0→lim= τxyτyx= τyzτzy= τzxτxz=