1.5 STRAIN 7 Tmax =59.6 -Ov Figure 1.8.Mohr's circle for stress state in Exam- =35 Ixy=20 ple 1.2. 03=40 =25.9 01=79.1 0x=70 A ● Figure 1.9.Deformation,translation,and rotation of a line in a material. oB' A SOLUTION:The Mohr's circle is plotted in Figure 1.8.The largest shear stress is tmax=(o1-o3)/2=[79.1-(-40)]/2=59.6MPa. 1.5 STRAIN Strain describes the amount of deformation in a body.When a body is deformed,points in that body are displaced.Strain must be defined in such a way that it excludes effects ofrotation and translation.Figure 1.9 shows a line in a material that has been deformed. The line has been translated,rotated,and deformed.The deformation is characterized by the engineering or nominal strain,e e=(E-o)/eo=△e/eo- (1.19) An alternative definition*is that of true or logarithmic strain,s,defined by dede/e, (1.20) which on integrating gives s In(e/o)=In(1 +e) 8 In(e/o)In(1 +e). (1.21) The true and engineering strains are almost equal when they are small.Expressing s ase=ln(e/eo)=ln(1+e)and expanding,so as e→0，e→e. There are several reasons why true strains are more convenient than engineering strains.The following examples indicate why. True strain was first defined by P.Ludwig,Elemente der Technishe Mechanik,Springer,1909
1.5 STRAIN 7 τ σ σ1 τmax σ3 σy τ xy σ2 σx = 59.6 = -40 = 79.1 = 35 = 20 = 70 = 25.9 Figure 1.8. Mohr’s circle for stress state in Example 1.2. 0 A A′ B B′ Figure 1.9. Deformation, translation, and rotation of a line in a material. SOLUTION: The Mohr’s circle is plotted in Figure 1.8. The largest shear stress is τ max = (σ1 − σ3)/2 = [79.1 − (−40)]/2 = 59.6 MPa. 1.5 STRAIN Strain describes the amount of deformation in a body. When a body is deformed, points in that body are displaced. Strain must be defined in such a way that it excludes effects of rotation and translation. Figure 1.9 shows a line in a material that has been deformed. The line has been translated, rotated, and deformed. The deformation is characterized by the engineering or nominal strain, e e = ( − 0)/0 = /0. (1.19) An alternative definition∗ is that of true or logarithmic strain, ε, defined by dε = d/, (1.20) which on integrating gives ε = ln(/0) = ln(1 + e) ε = ln(/0) = ln(1 + e). (1.21) The true and engineering strains are almost equal when they are small. Expressing ε as ε = ln(/0) = ln(1 + e) and expanding, so as e → 0, ε → e. There are several reasons why true strains are more convenient than engineering strains. The following examples indicate why. ∗ True strain was first defined by P. Ludwig, Elemente der Technishe Mechanik, Springer, 1909.
8 STRESS AND STRAIN EXAMPLE 1.4: (a)A bar of length,o,is uniformly extended until its length,e=2 o.Compute the values of the engineering and true strains. (b)What final length must a bar of length o,be compressed if the strains are the same (except sign)as in part (a)? SOLUTION: (a)e=△e/eo=1.0,e=ln(e/eo)=ln2=0.693 (b)e=-1 =(e-Co)/o,so e=0.This is clearly impossible to achieve. e=-0.693=ln(e/eo),s0e=oexp(0.693)=o/2. EXAMPLE 1.5:A bar 10 cm long is elongated to 20 cm by rolling in three steps: 10 cm to 12 cm,12 cm to 15 cm,and 15 cm to 20 cm. (a)Calculate the engineering strain for each step and compare the sum of these with the overall engineering strain. (b)Repeat for true strains. SOLUTION: (a)e1=2/10=0.20,e2=3/12=0.25,e3=5/15=0.333,eot=0.20+.25+.333 =0.833,eoverall=10/10=1. (b)e1=ln(12/10)=0.182,e2=ln(15/12)=0.223,e3=ln(20/15)=0.288,eot= 0.693,Eoverall=1n(20/10)=0.693. With true strains,the sum of the increments equals the overall strain,but this is not so with engineering strains. EXAMPLE 1.6:A block of initial dimensions,o,wo,to,is deformed to dimensions ofe,w,t. (a)Calculate the volume strain,s=In(v/vo)in terms of the three normal strains, Et,w and Et. (b)Plastic deformation causes no volume change.With no volume change,what is the sum of the three normal strains? SOLUTION: (a)sv=In[(e wt)/(owoto)]In(e/o)+In(w/wo)In(t/to)=E+8w +81. (b)If gy =0,E+Eu +8=0. Examples 1.4,1.5 and 1.6 illustrate why true strains are more convenient than engi- neering strains. 1.True strains for an equivalent amount of tensile and compressive deformation are equal except for sign. 2.True strains are additive. 3.The volume strain is the sum of the three normal strains
8 STRESS AND STRAIN EXAMPLE 1.4: (a) A bar of length, 0, is uniformly extended until its length, = 2 0. Compute the values of the engineering and true strains. (b) What final length must a bar of length 0, be compressed if the strains are the same (except sign) as in part (a)? SOLUTION: (a) e = /0 = 1.0, ε = ln(/0) = ln 2 = 0.693 (b) e = −1 = ( − 0)/0, so = 0. This is clearly impossible to achieve. ε = −0.693 = ln(/0), so = 0 exp(0.693) = 0/2. EXAMPLE 1.5: A bar 10 cm long is elongated to 20 cm by rolling in three steps: 10 cm to 12 cm, 12 cm to 15 cm, and 15 cm to 20 cm. (a) Calculate the engineering strain for each step and compare the sum of these with the overall engineering strain. (b) Repeat for true strains. SOLUTION: (a) e1 = 2/10 = 0.20, e2 = 3/12 = 0.25, e3 = 5/15 = 0.333, etot = 0.20 + .25 + .333 = 0.833, eoverall = 10/10 = 1. (b) ε1 = ln(12/10) = 0.182, ε2 = ln(15/12) = 0.223, ε3 = ln(20/15) = 0.288, εtot = 0.693, εoverall = ln(20/10) = 0.693. With true strains, the sum of the increments equals the overall strain, but this is not so with engineering strains. EXAMPLE 1.6: A block of initial dimensions, 0, w0, t0, is deformed to dimensions of , w, t. (a) Calculate the volume strain, εv = ln(v/v0) in terms of the three normal strains, ε, εw and εt . (b) Plastic deformation causes no volume change. With no volume change, what is the sum of the three normal strains? SOLUTION: (a) εv= ln[( wt)/(0w0t0)] = ln(/0) + ln(w/w0) + ln(t/t0) = ε + εw + εt . (b) If εv = 0, ε + εw + εt = 0. Examples 1.4, 1.5 and 1.6 illustrate why true strains are more convenient than engineering strains. 1. True strains for an equivalent amount of tensile and compressive deformation are equal except for sign. 2. True strains are additive. 3. The volume strain is the sum of the three normal strains.
1.6 SMALL STRAINS 9 If strains are small,true and engineering strains are nearly equal.Expressing true strain as In()=In(1+e/lo)=In(1 +e)and taking the series expansion, e=e-e2/2+e/3!..,it can be seen that as e→0，e→e. EXAMPLE 1.7:Calculate the ratio of s/e for e=0.001,0.01,0.02,0.05,0.1 and 0.2. SOLUTION: Fore=0.001,e=l1n(1.001)=0.0009995;e/e=0.9995. Fore=0.01,e=ln(1.01)=0.00995,e/e=0.995 Fore=0.02,e=n(1.02)=0.0198,e/e=0.99. Fore=0.05,e=ln(1.05)=0.0488,e/e=0.975 Fore=0.1,e=ln(1.1)=0.095,e/e=0.95. Fore=0.2,e=ln(1.2)=0.182,e/e=0.912. As e gets larger the difference between g and e become greater. 1.6 SMALL STRAINS Figure 1.10 shows a small two-dimensional element,ABCD,deformed into 4'B'CD' where the displacements are u and v.The normal strain,ex,is defined as exx =(A'D'-AD)/AD A'D'/AD-1. (1.22) Neglecting the rotation exx A'D'/AD-1= dx-u+u+(au/ax)dxI or dx exx =ou/ox. (1.23) Similarly,eyy=av/ay and e==aw/az for a three-dimensional case. (au/ay)dy P0---- v+(av/ay)dy 业 B dy Q (av/ax)dx D U+(du/ax) dx y Figure 1.10.Distortion of a two-dimensional element
1.6 SMALL STRAINS 9 If strains are small, true and engineering strains are nearly equal. Expressing true strain as ε = ln( 0 + 0 ) = ln(1 + /0) = ln(1 + e) and taking the series expansion, ε = e − e2/2 + e3/3! . . . . , it can be seen that as e → 0, ε → e. EXAMPLE 1.7: Calculate the ratio of ε/e for e = 0.001, 0.01, 0.02, 0.05, 0.1 and 0.2. SOLUTION: For e = 0.001, ε = ln(1.001) = 0.0009995; ε/e = 0.9995. For e = 0.01, ε = ln(1.01) =0.00995, ε/e = 0.995. For e = 0.02, ε = ln(1.02) =0.0198, ε/e = 0.99. For e = 0.05, ε = ln(1.05) = 0.0488, ε/e = 0.975. For e = 0.1, ε = ln(1.1) = 0.095, ε/e = 0.95. For e = 0.2, ε = ln(1.2) = 0.182, ε/e = 0.912. As e gets larger the difference between ε and e become greater. 1.6 SMALL STRAINS Figure 1.10 shows a small two-dimensional element, ABCD, deformed into A B C D where the displacements are u and v. The normal strain, exx, is defined as ex x = (A D − AD)/AD = A D /AD − 1. (1.22) Neglecting the rotation ex x = A D /AD − 1 = dx − u + u + (∂u/∂x) dx dx − 1 or ex x = ∂u/∂x. (1.23) Similarly, eyy = ∂v/∂y and ezz = ∂w/∂z for a three-dimensional case. A B C D A C D B y x x dx y dy v u v + ( v/ y)dy u + (∂u/∂x)dx (∂v/∂x)dx (∂u/∂y)dy Q P ∂ ∂ Figure 1.10. Distortion of a two-dimensional element.
10 STRESS AND STRAIN The shear strain are associated with the angles between AD and 4'D'and between 4B and 4'B'.For small deformations ∠4≈au/ox and∠4≈au/ay (1.24) The total shear strain is the sum of these two angles, au v w==+a脉 (1.25a) Similarly, av ow y Yy:=Yay=82 and (1.25b) ow du Yix=Yx:= ax (1.25c) This definition of shear strain,y,is equivalent to the simple shear measured in a torsion of shear test. 1.7 THE STRAIN TENSOR If tensor shear strains,ij,are defined as e=(1/2)h, (1.26) small shear strains form a tensor, Exx Eyx Ezx Eij= Exy Eyy Ezy (1.27) Exz Eyz Ezz Because small strains form a tensor,they can be transformed from one set ofaxes to another in a way identical to the transformation of stresses.Mohr's circle relations can be used.It must be remembered,however,that s;=yi/2 and that the transformations hold only for small strains.If yy==Y=x=0, Ex=Exlix +eyliy+yxylxxlr'y (1.28) and rty =26xer'xlyt +26yexylyy Ysy(er'xlyy +eyxer',). (1.29) The principal strains can be found from the Mohr's circle equations for strains, 812s x+e±/2[e-eP+y2]P. (1.30) 2 Strains on other planes are given by ex,y=(1/2)(e1+E2)±(1/2)(e1-E2)cos29 (1.31) and Yxy =(81-82)sin 20. (1.32) 1.8 ISOTROPIC ELASTICITY Although the thrust of this book is on plastic deformation,a short treatment ofelasticity is necessary to understand springback and residual stresses in forming processes
10 STRESS AND STRAIN The shear strain are associated with the angles between AD and A D and between AB and A B . For small deformations ∠AD AD ≈ ∂v/∂x and ∠AB AB ≈ ∂u/∂y (1.24) The total shear strain is the sum of these two angles, γx y = γyx = ∂u ∂y + ∂v ∂x . (1.25a) Similarly, γyz = γzy = ∂v ∂z + ∂w ∂y and (1.25b) γzx = γx z = ∂w ∂x + ∂u ∂z . (1.25c) This definition of shear strain, γ , is equivalent to the simple shear measured in a torsion of shear test. 1.7 THE STRAIN TENSOR If tensor shear strains, εi j , are defined as εi j = (1/2)γi j, (1.26) small shear strains form a tensor, εi j = εx x εyx εzx εx y εyy εzy εx z εyz εzz . (1.27) Because small strains form a tensor, they can be transformed from one set of axes to another in a way identical to the transformation of stresses. Mohr’s circle relations can be used. It must be remembered, however, that εi j = γ ij/2 and that the transformations hold only for small strains. If γyz = γzx = 0, εx = εx2 x x + εy2 x y + γx yx xx y (1.28) and γx y = 2εxxxy x + 2εyx yy y + γx y (x xy y + y xx y ). (1.29) The principal strains can be found from the Mohr’s circle equations for strains, ε1,2 = εx + εy 2 ± (1/2) (εx − εy ) 2 + γ 2 x y 1/2 . (1.30) Strains on other planes are given by εx,y = (1/2)(ε1 + ε2) ± (1/2)(ε1 − ε2) cos 2θ (1.31) and γx y = (ε1 − ε2) sin 2θ. (1.32) 1.8 ISOTROPIC ELASTICITY Although the thrust of this book is on plastic deformation, a short treatment of elasticity is necessary to understand springback and residual stresses in forming processes.
1.9 STRAIN ENERGY 11 Hooke's laws can be expressed as ex=(1/E)[ox-v(o,+o)月， ey=(1/E)[oy-v(a:+ox小， (1.33) e:=(1/E)[o2-v(ox+ov)], and :=(1/G)te, y2x=(1/G)r2x, (1.34) Yxy =(1/G)txy, where E is Young's modulus,v is Poisson's ratio and G is the shear modulus.For an isotropic material,E,v and G are inter-related by E=2G(1+v)or (1.35) G=E/[2(1+U)]. (1.36 EXAMPLE 1.8:In Example 1.2 with ox=70 MPa,oy=35 MPa,txy 20,o:= ax ty==0,it was found that o1 79.1 MPa and o2=25.9 MPa.Using E =61 GPa and v=0.3 for aluminum,calculate s and by (a)First calculating ex,&,and yxy using equations 1.33 and then transforming these strains to the 1,2 axes with the Mohr's circle equations. (b)By using equations 1.33 with oi and o2. SOLUTION: (a) ex=(1/61×10)[70×105-0.30(35×106+0)]=0.9754×10-3 ey=(1/61×10)[35×106-0.30(70×106+0)]=0.2295×10-3 yxy=[2(1.3)/61×10](20×10)=0.853×10-3 Now using the Mohr's strain circle equations. e1,2=(e+e)/2±(1/2[er-e2+yJn =0.603×10-3±(1/2[0.1391×10-3p+(0.856×10-321/2 =1.169×10-3,0.0361×10-3 (b) e1=(1/61×109)[79.9×106-0.30(25.9×10)]=1.169, e2=(1/61×109)[25.9×106-0.30(79.9×10)]=0.0361. 1.9 STRAIN ENERGY If a bar of length,x and cross-sectional area,A,is subjected to a tensile force Fx,which caused an increase in length,dr,the incremental work,dw,is dw Fxdx. (1.37) The work per volume,dw,is dwdW/A=Fxdx/(Ax)=oxdex. (1.38)
1.9 STRAIN ENERGY 11 Hooke’s laws can be expressed as ex = (1/E)[σx − υ(σy + σz)], ey = (1/E)[σy − υ(σz + σx )], (1.33) ez = (1/E)[σz − υ(σx + σy )], and γyz = (1/G)τyz, γzx = (1/G)τzx , (1.34) γx y = (1/G)τx y , where E is Young’s modulus, υ is Poisson’s ratio and G is the shear modulus. For an isotropic material, E, υ and G are inter-related by E = 2G(1 + υ) or (1.35) G = E/[2(1 + υ)]. (1.36) EXAMPLE 1.8: In Example 1.2 with σx = 70 MPa, σy = 35 MPa, τx y = 20, σz = τzx = τyz = 0, it was found that σ1 = 79.1 MPa and σ2 = 25.9 MPa. Using E = 61 GPa and υ = 0.3 for aluminum, calculate ε1 and ε1 by (a) First calculating εx , εy and γx y using equations 1.33 and then transforming these strains to the 1, 2 axes with the Mohr’s circle equations. (b) By using equations 1.33 with σ1 and σ2. SOLUTION: (a) ex = (1/61 × 109)[70 × 106 − 0.30(35 × 106 + 0)] = 0.9754 × 10−3 ey = (1/61 × 109)[35 × 106 − 0.30(70 × 106 + 0)] = 0.2295 × 10−3 γ x y = [2(1.3)/61 × 109](20 × 106) = 0.853 × 10−3 Now using the Mohr’s strain circle equations. e1,2 = (ex + ey)/2 ± (1/2)[(ex − ey) 2 + γ 2 x y ] 1/2 = 0.603 × 10−3 ± (1/2)[(0.1391 × 10−3/2 + (0.856 × 10−3) 2] 1/2 = 1.169 × 10−3, 0.0361 × 10−3 (b) e1 = (1/61 × 109)[79.9 × 106– 0.30(25.9 × 106)] = 1.169, e2 = (1/61 × 109)[25.9 × 106– 0.30(79.9 × 106)] = 0.0361. 1.9 STRAIN ENERGY If a bar of length, x and cross-sectional area, A, is subjected to a tensile force Fx, which caused an increase in length, dx, the incremental work, dW, is dW = Fxdx. (1.37) The work per volume, dw, is dw = dW/A = Fxdx/(Ax) = σxdex . (1.38)