Mechanics.Of.Materials.Saouma Episode 5 ppsx

Draft 4.3 Strain Decomposition 25 Given x 1 = X 1 , x 2 = −3X 3 , x 3 =2X 2 , find the deformation gradient F, the right stretch tensor U, the rotation tensor R, and the left stretch tensor V. Solution: From Eq. 4.25 F =    ∂x 1 ∂X 1 ∂x 1 ∂X 2 ∂x 1 ∂X 3 ∂x 2 ∂X 1 ∂x 2 ∂X 2 ∂x 2 ∂X 3 ∂x 3 ∂X 1 ∂x 3 ∂X 2 ∂x 3 ∂X 3    =   10 0 00−3 02 0   (4.130) From Eq. 4.126 U 2 = F T F =   100 002 0 −30     10 0 00−3 02 0   =   100 040 009   (4.131) thus U =   100 020 003   (4.132) From Eq. 4.127 R = FU −1 =   10 0 00−3 02 0     100 0 1 2 0 00 1 3   =   10 0 00−1 01 0   (4.133) Finally, from Eq. 4.128 V = FR T =   10 0 00−3 02 0     100 001 0 −10   =   100 030 002   (4.134) Example 4-11: Polar Decomposition II For the following deformation: x 1 = λ 1 X 1 , x 2 = −λ 3 X 3 ,andx 3 = λ 2 X 2 , find the rotation tensor. Solution: [F]=   λ 1 00 00−λ 3 0 λ 2 0   (4.135) [U] 2 =[F] T [F] (4.136) =   λ 1 00 00λ 2 0 −λ 3 0     λ 1 00 00−λ 3 0 λ 2 0   =   λ 2 1 00 0 λ 2 2 0 00λ 2 3   (4.137) [U]=   λ 1 00 0 λ 2 0 00λ 3   (4.138) [R]=[F][U] −1 =   λ 1 00 00−λ 3 0 λ 2 0     1 λ 1 00 0 1 λ 2 0 00 1 λ 3   =   10 0 00−1 01 0   (4.139) Thus we note that R corresponds to a 90 o rotation about the e 1 axis. Victor Saouma Mechanics of Materials II Draft 26 KINEMATIC Example 4-12: Polar Decomposition III Victor Saouma Mechanics of Materials II Draft 4.3 Strain Decomposition 27 Polar Decomposition Using Mathematica Given x 1 =X 1 +2X 2 , x 2 =X 2 , x 3 =X 3 , a) Obtain C, b) the principal values of C and the corresponding directions, c) the matrix U and U -1 with respect to the principal directions, d) Obtain the matrix U and U -1 with respect to the e i bas obtain the matrix R with respect to the e i basis. Determine the F matrix In[1]:= F = 881, 2, 0<, 80, 1, 0<, 80, 0, 1<< Out[1]= i k j j j j j j j 120 010 001 y { z z z z z z z Solve for C In[2]:= CST = Transpose@FD .F Out[2]= i k j j j j j j j 120 250 001 y { z z z z z z z Determine Eigenvalues and Eigenvectors In[3]:= N@Eigenvalues@CSTDD Out[3]= 81., 0.171573, 5.82843< In[4]:= 8v1, v2, v3< = N@Eigenvectors@CSTD,4D Out[4]= i k j j j j j j j 001. -2.414 1. 0 0.4142 1. 0 y { z z z z z z z In[5]:= << LinearAlgebra‘Orthogonalization‘ In[6]:= vnormalized = GramSchmidt@8v3, −v2, v1<D Out[6]= i k j j j j j j j 0.382683 0.92388 0 0.92388 -0.382683 0 001. y { z z z z z z z In[7]:= CSTeigen = Chop@N@vnormalized . CST . vnormalized, 4DD Out[7]= i k j j j j j j j 5.828 0 0 0 0.1716 0 001. y { z z z z z z z Determine U with respect to the principal directions In[8]:= Ueigen = N@Sqrt@CSTeigenD,4D Out[8]= i k j j j j j j j 2.414 0 0 0 0.4142 0 001. y { z z z z z z z In[9]:= Ueigenminus1 = Inverse@UeigenD Out[9]= i k j j j j j j j 0.414214 0. 0. 0. 2.41421 0. 0. 0. 1. y { z z z z z z z 2 m− Determine U and U -1 with respect to the e i basis In[10]:= U_e = N@vnormalized . Ueigen . vnormalized, 3D Out[10]= i k j j j j j j j 0.707 0.707 0. 0.707 2.12 0. 0. 0. 1. y { z z z z z z z In[11]:= U_einverse = N@Inverse@%D,3D Out[11]= i k j j j j j j j 2.12 -0.707 0. -0.707 0.707 0. 0. 0. 1. y { z z z z z z z Determine R with respect to the e i basis In[12]:= R = N@F.%, 3D Out[12]= i k j j j j j j j 0.707 0.707 0. -0.707 0.707 0. 0. 0. 1. y { z z z z z z z m−polar.nb Victor Saouma Mechanics of Materials II Draft 28 KINEMATIC 4.4 Summary and Discussion 92 From the above, we deduce the following observations: 1. If both the displacement gradients and the displacements themselves are small, then ∂u i ∂X j ≈ ∂u i ∂x j and thus the Eulerian and the Lagrangian infinitesimal strain tensors may be taken as equal E ij = E ∗ ij . 2. If the displacement gradients are small, but the displacements are large, we should use the Eulerian infinitesimal representation. 3. If the displacements gradients are large, but the displacements are small, use the Lagrangian finite strain representation. 4. If both the displacement gradients and the displacements are large, use the Eulerian finite strain representation. 4.5 Compatibility Equation 93 If ε ij = 1 2 (u i,j + u j,i ) then we have six differential equations (in 3D the strain tensor has a total of 9 terms, but due to symmetry, there are 6 independent ones) for determining (upon integration) three unknowns displacements u i . Hence the system is overdetermined, and there must be some linear relations between the strains. 94 It can be shown (through appropriate successive differentiation of the strain expression) that the compatibility relation for strain reduces to: ∂ 2 ε ik ∂x j ∂x j + ∂ 2 ε jj ∂x i ∂x k − ∂ 2 ε jk ∂x i ∂x j − ∂ 2 ε ij ∂x j ∂x k =0. or ∇ x ×L×∇ x =0 (4.140) There are 81 equations in all, but only six are distinct ∂ 2 ε 11 ∂x 2 2 + ∂ 2 ε 22 ∂x 2 1 =2 ∂ 2 ε 12 ∂x 1 ∂x 2 (4.141-a) ∂ 2 ε 22 ∂x 2 3 + ∂ 2 ε 33 ∂x 2 2 =2 ∂ 2 ε 23 ∂x 2 ∂x 3 (4.141-b) ∂ 2 ε 33 ∂x 2 1 + ∂ 2 ε 11 ∂x 2 3 =2 ∂ 2 ε 31 ∂x 3 ∂x 1 (4.141-c) ∂ ∂x 1  − ∂ε 23 ∂x 1 + ∂ε 31 ∂x 2 + ∂ε 12 ∂x 3  = ∂ 2 ε 11 ∂x 2 ∂x 3 (4.141-d) ∂ ∂x 2  ∂ε 23 ∂x 1 − ∂ε 31 ∂x 2 + ∂ε 12 ∂x 3  = ∂ 2 ε 22 ∂x 3 ∂x 1 (4.141-e) ∂ ∂x 3  ∂ε 23 ∂x 1 + ∂ε 31 ∂x 2 − ∂ε 12 ∂x 3  = ∂ 2 ε 33 ∂x 1 ∂x 2 (4.141-f) In 2D, this results in (by setting i =2,j =1andl = 2): ∂ 2 ε 11 ∂x 2 2 + ∂ 2 ε 22 ∂x 2 1 = ∂ 2 γ 12 ∂x 1 ∂x 2 (4.142) Victor Saouma Mechanics of Materials II Draft 4.5 Compatibility Equation 29 I , i 33 2 I , i 2 I , i 11 22 X , x u P P 0 t=0 t=t X O U o x X , x X , x 33 11 Material/Spatial b=0 x 2 , X 3 x 3 , 2 X X 1 O x 1 , u x t=0 +d X X X +d d d x t=t X u u Q 0 Q P P 0 LAGRANGIAN EULERIAN Material Spatial Position Vector x = x(X,t) X = X(x,t) GRADIENTS Deformation F = x∇ X ≡ ∂x i ∂X j H = X∇ x ≡ ∂X i ∂x j H = F −1 Displacement ∂u i ∂X j = ∂x i ∂X j − δ ij or ∂u i ∂x j = δ ij − ∂X i ∂x j or J = u∇ X = F − I K ≡ u∇ x = I − H TENSOR dX 2 =dx·B −1 ·dx dx 2 =dX·C·dX Cauchy Green Deformation B −1 ij = ∂X k ∂x i ∂X k ∂x j or C ij = ∂x k ∂X i ∂x k ∂X j or B −1 = ∇ x X·X∇ x = H c ·H C = ∇ X x·x∇ X = F c ·F C −1 = B −1 STRAINS Lagrangian Eulerian/Almansi dx 2 − dX 2 =dX·2E·dX dx 2 − dX 2 =dx·2E ∗ ·dx Finite Strain E ij = 1 2  ∂x k ∂X i ∂x k ∂X j − δ ij  or E ∗ ij = 1 2  δ ij − ∂X k ∂x i ∂X k ∂x j  or E = 1 2 (∇ X x·x∇ X    F c ·F −I) E ∗ = 1 2 (I − ∇ x X·X∇ x    H c ·H ) E ij = 1 2  ∂u i ∂X j + ∂u j ∂X i + ∂u k ∂X i ∂u k ∂X j  or E ∗ ij = 1 2  ∂u i ∂x j + ∂u j ∂x i − ∂u k ∂x i ∂u k ∂x j  or E = 1 2 (u∇ X + ∇ X u + ∇ X u·u∇ X )    J+J c +J c ·J E ∗ = 1 2 (u∇ x + ∇ x u − ∇ x u·u∇ x )    K+K c −K c ·K Small E ij = 1 2  ∂u i ∂X j + ∂u j ∂X i  E ∗ ij = 1 2  ∂u i ∂x j + ∂u j ∂x i  Deformation E = 1 2 (u∇ X + ∇ X u)= 1 2 (J + J c ) E ∗ = 1 2 (u∇ x + ∇ x u)= 1 2 (K + K c ) ROTATION TENSORS Small [ 1 2  ∂u i ∂X j + ∂u j ∂X i  + 1 2  ∂u i ∂X j − ∂u j ∂X i  ]dX j  1 2  ∂u i ∂x j + ∂u j ∂x i  + 1 2  ∂u i ∂x j − ∂u j ∂x i  dx j deformation [ 1 2 (u∇ X + ∇ X u)    E + 1 2 (u∇ X − ∇ X u)    W ]·dX [ 1 2 (u∇ x + ∇ x u)    E ∗ + 1 2 (u∇ x − ∇ x u)    Ω ]·dx Finite Strain F = R·U = V·R STRESS TENSORS Piola-Kirchoff Cauchy First T 0 = (det F)T  F −1  T Second ˜ T = (det F)  F −1  T  F −1  T Table 4.1: Summary of Major Equations Victor Saouma Mechanics of Materials II Draft 30 KINEMATIC (recall that 2ε 12 = γ 12 .) 95 When he compatibility equation is written in term of the stresses, it yields: ∂ 2 σ 11 ∂x 2 2 − ν ∂σ 22 2 ∂x 2 2 + ∂ 2 σ 22 ∂x 2 1 − ν ∂ 2 σ 11 ∂x 2 1 =2(1+ν) ∂ 2 σ 21 ∂x 1 ∂x 2 (4.143) Example 4-13: Strain Compatibility For the following strain field    − X 2 X 2 1 +X 2 2 X 1 2(X 2 1 +X 2 2 ) 0 X 1 2(X 2 1 +X 2 2 ) 00 000    (4.144) does there exist a single-valued continuous displacement field? Solution: ∂E 11 ∂X 2 = − (X 2 1 + X 2 2 ) − X 2 (2X 2 ) (X 2 1 + X 2 2 ) 2 = X 2 2 − X 2 1 (X 2 1 + X 2 2 ) 2 (4.145-a) 2 ∂E 12 ∂X 1 = (X 2 1 + X 2 2 ) − X 1 (2X 1 ) (X 2 1 + X 2 2 ) 2 = X 2 2 − X 2 1 (X 2 1 + X 2 2 ) 2 (4.145-b) ∂E 22 ∂X 2 1 = 0 (4.145-c) ⇒ ∂ 2 E 11 ∂X 2 2 + ∂ 2 E 22 ∂X 2 1 =2 ∂ 2 E 12 ∂X 1 ∂X 2 √ (4.145-d) Actually, it can be easily verified that the unique displacement field is given by u 1 = arctan X 2 X 1 ; u 2 =0; u 3 = 0 (4.146) to which we could add the rigid body displacement field (if any). 4.6 Lagrangian Stresses; Piola Kirchoff Stress Tensors 96 In Sect. 2.2 the discussion of stress applied to the deformed configuration dA (using spatial coordiantes x), that is the one where equilibrium must hold. The deformed configuration being the natural one in which to characterize stress. Hence we had df = tdA (4.147-a) t = Tn (4.147-b) (note the use of T instead of σ). Hence the Cauchy stress tensor was really defined in the Eulerian space. 97 However, there are certain advantages in referring all quantities back to the undeformed configuration (Lagrangian) of the body because often that configuration has geometric features and symmetries that are lost through the deformation. 98 Hence, if we were to define the strain in material coordinates (in terms of X), we need also to express the stress as a function of the material point X in material coordinates. Victor Saouma Mechanics of Materials II Draft 4.6 Lagrangian Stresses; Piola Kirchoff Stress Tensors 31 4.6.1 First 99 The first Piola-Kirchoff stress tensor T 0 is defined in the undeformed geometry in such a way that it results in the same total force as the traction in the deformed configuration (where Cauchy’s stress tensor was defined). Thus, we define df ≡ t 0 dA 0 (4.148) where t 0 is a pseudo-stress vector in that being based on the undeformed area, it does not describe the actual intensity of the force, however it has the same direction as Cauchy’s stress vector t. 100 The first Piola-Kirchoff stress tensor (also known as Lagrangian Stress Tensor) is thus the linear transformation T 0 such that t 0 = T 0 n 0 (4.149) and for which df = t 0 dA 0 = tdA ⇒ t 0 = dA ddA 0 t (4.150) using Eq. 4.147-b and 4.149 the preceding equation becomes T 0 n 0 = dA dA 0 Tn = T dA dA 0 n (4.151) and using Eq. 4.36 dAn =dA 0 (det F)  F −1  T n 0 we obtain T 0 n 0 = T(det F)  F −1  T n 0 (4.152) the above equation is true for all n 0 , therefore T 0 = (det F)T  F −1  T (4.153) T = 1 (det F) T 0 F T or T ij = 1 (det F) (T 0 ) im F jm (4.154) 101 The first Piola-Kirchoff stress tensor is not symmetric in general, and is not energitically correct. That is multiplying this stress tensor with the Green-Lagrange tensor will not be equal to the product of the Cauchy stress tensor multiplied by the deformation strain tensor. 102 To determine the corresponding stress vector, we solve for T 0 first, then for dA 0 and n 0 from dA 0 n 0 = 1 det F F T n (assuming unit area dA), and finally t 0 = T 0 n 0 . 4.6.2 Second 103 The second Piola-Kirchoff stress tensor, ˜ T is formulated differently. Instead of the actual force df on dA, it gives the force d ˜ f related to the force df in the same way that a material vector dX at X is related by the deformation to the corresponding spatial vector dx at x. Thus, if we let d ˜ f = ˜ tdA 0 (4.155-a) and df = Fd ˜ f (4.155-b) where d ˜ f is the pseudo differential force which transforms, under the deformation gradient F,the (actual) differential force df at the deformed position (note similarity with dx = FdX). Thus, the pseudo vector t is in general in a differnt direction than that of the Cauchy stress vector t. 104 The second Piola-Kirchoff stress tensor is a linear transformation ˜ T such that ˜ t = ˜ Tn 0 (4.156) Victor Saouma Mechanics of Materials II Draft 32 KINEMATIC thus the preceding equations can be combined to yield df = F ˜ Tn 0 dA 0 (4.157) we also have from Eq. 4.148 and 4.149 df = t 0 dA 0 = T 0 n 0 ddA 0 (4.158) and comparing the last two equations we note that ˜ T = F −1 T 0 (4.159) which gives the relationship between the first Piola-Kirchoff stress tensor T 0 and the second Piola- Kirchoff stress tensor ˜ T. 105 Finally the relation between the second Piola-Kirchoff stress tensor and the Cauchy stress tensor can be obtained from the preceding equation and Eq. 4.153 ˜ T = (det F)  F −1  T  F −1  T (4.160) and we note that this second Piola-Kirchoff stress tensor is always symmetric (if the Cauchy stress tensor is symmetric). It can also be shown that it is energitically correct. 106 To determine the corresponding stress vector, we solve for ˜ T first, then for dA 0 and n 0 from dA 0 n 0 = 1 det F F T n (assuming unit area dA), and finally ˜ t = ˜ Tn 0 . Example 4-14: Piola-Kirchoff Stress Tensors 4.7 Hydrostatic and Deviatoric Strain 85 The lagrangian and Eulerian linear strain tensors can each be split into spherical and deviator tensor as was the case for the stresses. Hence, if we define 1 3 e = 1 3 tr E (4.161) then the components of the strain deviator E  are given by E  ij = E ij − 1 3 eδ ij or E  = E − 1 3 e1 (4.162) We note that E  measures the change in shape of an element, while the spherical or hydrostatic strain 1 3 e1 represents the volume change. Victor Saouma Mechanics of Materials II Draft 4.7 Hydrostatic and Deviatoric Strain 33 Piola−Kirchoff Stress Tensors The deformed configuration of a body is described by x 1 =X 1 ê 2, x 2 =−X 2 /2, x 3 =4X 3 ; If the Cauchy stress tensor is given by i k j j j j j j j j 100 0 0 0 00 0 0 0 y { z z z z z z z z MPa; What are the corresponding first and second Piola−Kirchoff stress tensors, and calculate the respective stress tensors on the e 3 plane in the deformed state. ‡ F tensor CST = 880, 0, 0<, 80, 0, 0<, 80, 0, 100<< 880, 0, 0<, 80, 0, 0<, 80, 0, 100<< F = 881 ê 2, 0, 0<, 80, 0, −1 ê 2<, 80, 4, 0<< 99 1 ÄÄÄÄ Ä 2 ,0,0=, 90, 0, - 1 ÄÄÄÄ Ä 2 =, 80, 4, 0<= Finverse = Inverse@FD 982, 0, 0<, 90, 0, 1 ÄÄÄÄ Ä 4 =, 80, - 2, 0<= ‡ First Piola−Kirchoff Stress Tensor Tfirst = Det@FD CST . Transpose@FinverseD 880, 0, 0<, 80, 0, 0<, 80, 25, 0<< MatrixForm@%D i k j j j j j j 000 000 0250 y { z z z z z z ‡ Second Piola−Kirchoff Stress Tensor Tsecond = Inverse@FD . Tfirst 980, 0, 0<, 90, 25 ÄÄÄÄÄÄÄÄ 4 ,0=, 80, 0, 0<= MatrixForm@%D i k j j j j j j j j 000 0 25 ÄÄÄÄÄÄ 4 0 000 y { z z z z z z z z ‡ Cuchy stress vector Can be obtained from t=CST n tcauchy = MatrixForm@CST . 80, 0, 1<D i k j j j j j j 0 0 100 y { z z z z z z ‡ Pseudo−Stress vector associated with the First Piola−Kirchoff stress tensor For a unit area in the deformed state in the e 3 direction, its undeformed area dA 0 n 0 is given by dA 0 n 0 = F T n ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ det F detF = Det@FD 1 n = 80, 0, 1< 80, 0, 1< 2 m−piola.nb MatrixForm@Transpose@FD .nê detFD i k j j j j j j 0 4 0 y { z z z z z z Thus n 0 =e 2 and using t 0 =T 0 n 0 we obtain t01st = MatrixForm@Tfirst . 80, 1, 0<D i k j j j j j j 0 0 25 y { z z z z z z We note that this vector is in the same direction as the Cauchy stress vector, its magnitude is one fourth of that of the Cauchy stress vector, because the undeformed area is 4 times that of the deformed area ‡ Pseudo−Stress vector associated with the Second Piola−Kirchoff stress tensor t0second = MatrixForm@Tsecond . 80, 1, 0<D i k j j j j j j j j 0 25 ÄÄÄÄÄÄ 4 0 y { z z z z z z z z We see that this pseudo stress vector is in a different direction from that of the Cauchy stress vector (and we note that the tensor F transforms e 2 into e 3 ). m−piola.nb 3 Victor Saouma Mechanics of Materials II Draft 34 KINEMATIC ε ε I εε γ 2 III II Figure 4.8: Mohr Circle for Strain 4.8 Principal Strains, Strain Invariants, Mohr Circle 86 Determination of the principal strains (E (3) <E (2) <E (1) , strain invariants and the Mohr circle for strain parallel the one for stresses (Sect. 2.3) and will not be repeated here. λ 3 − I E λ 2 − II E λ − III E =0 (4.163) where the symbols I E , II E and III E denote the following scalar expressions in the strain components: I E = E 11 + E 22 + E 33 = E ii =trE (4.164) II E = −(E 11 E 22 + E 22 E 33 + E 33 E 11 )+E 2 23 + E 2 31 + E 2 12 (4.165) = 1 2 (E ij E ij − E ii E jj )= 1 2 E ij E ij − 1 2 I 2 E (4.166) = 1 2 (E : E − I 2 E ) (4.167) III E = detE = 1 6 e ijk e pqr E ip E jq E kr (4.168) 87 In terms of the principal strains, those invariants can be simplified into I E = E (1) + E (2) + E (3) (4.169) II E = −(E (1) E (2) + E (2) E (3) + E (3) E (1) ) (4.170) III E = E (1) E (2) E (3) (4.171) 88 The Mohr circle uses the Engineering shear strain definition of Eq. 4.91, Fig. 4.8 Example 4-15: Strain Invariants & Principal Strains Victor Saouma Mechanics of Materials II [...]... (5. 3) If A were a force, then this integral would represent the corresponding work 3 If the contour is closed, then we define the contour integral as A·dr = C C A1 dx + A2 dy + A3 dz (5. 4) Draft 2 4 MATHEMATICAL PRELIMINARIES; Part III VECTOR INTEGRALS †It can be shown that if A = ∇φ (i.e a vector), then P2 is independent of the path C connecting P1 to P2 (5. 5-a) along a closed contour line A·dr (5. 5-b)... P1 A·dr = 0 C 5. 3 5 Integration by Parts The integration by part formula is b a 5. 4 6 b b u(x)v (x)dx = u(x)v(x)|a − v(x)u (x)dx (5. 6) a Gauss; Divergence Theorem In the most general case we have δF = Ω F (5. 7) δΩ The divergence theorem (also known as Ostrogradski’s Theorem) comes repeatedly in solid mechanics and can be stated as follows: 7 ∇·vdΩ = Ω v.ndΓ or vi,i dΩ = Γ Ω vi ni dΓ (5. 8) Γ The flux... be needed to properly express the conservation laws in the next chapter 5. 1 1 Integral of a Vector The integral of a vector R(u) = R1 (u)e1 + R2 (u)e2 + R3 (u)e3 is defined as R(u)du = e1 if a vector S(u) exists such that R(u) = R1 (u)du + e2 d du R3 (u)du (5. 1) (S(u)), then d (S(u)) du = S(u) + c du R(u)du = 5. 2 R2 (u)du + e3 (5. 2) Line Integral Given r(u) = x(u)e1 + y(u)e2 + z(u)e3 where r(u) is a... of the divergence of that function over the volume enclosed by the surface 8 For 2D-1D transformations, we have ∇·qdA = A 9 qT nds (5. 9) s This theorem is sometime refered to as Green’s theorem in space 5. 4.1 †Green-Gauss ΦvT ndΓ − Φ∇·vdΩ = Ω Γ (∇Φ)T vdΩ (5. 10) ∂Φ ΨdΩ ∂x (5. 11) Ω If we select vT = [ Ψ 0 0 ], we obtain Φ Ω Victor Saouma ∂Ψ dΩ = ∂x ΦΨnx dΓ − Γ Ω Mechanics of Materials II ... is the calibration factor in units of strain per volt For common values where 4 F = 2.07, G = 1000, E = 5, the calibration factor is simply (2.07)(1000) (5) or 386.47 microstrain per volt Victor Saouma Mechanics of Materials II Draft KINEMATIC Victor Saouma Mechanics of Materials II 42 Draft Chapter 5 MATHEMATICAL PRELIMINARIES; Part III VECTOR INTEGRALS In a preceding chapter, we have reviewed vector... factor is defined as the fractional change in resistance divided by the fractional ∆R R change in length along the axis of the gage GF = ∆L Common gage factors are in the range of 1 .5- 2 L for most resistive strain gages 95 Common strain gages utilize a grid pattern as opposed to a straight length of wire in order to reduce the gage length This grid pattern causes the gage to be sensitive to deformations... 2.3 −0.8  =  0 0 0 0 0 1 (4.179) 0 1 0  0 0  −1.3 (4.181) Example 4-16: Mohr’s Circle Construct the Mohr’s circle for the following  0  0 0 plane strain case:  0 √ 0 5 3  √ 3 3 (4.182) Solution: εs F 3 2 B 1 2 εn 60o 1 2 3 4 5 6 D E We note that since E(1) = 0 is a principal value for plane strain, ttwo of the circles are drawn as shown Victor Saouma Mechanics of Materials II Draft 4.9 Initial... (1) (1)   (1)    1 − 1+2 13 n1 + 3n2    0  n1   √ √  (1) (1) (1) = 3n1 − 1+2 13 n2 0√  n2  (1)   √     (1) n3 1 − 1+2 13  1 − 1+2 13 n3         0  0 =     0   (4.1 75) which gives (1) n1 (1) n3 = √ 1 + 13 (1) √ n2 2 3 = 0 n(1) ·n(1) = ⇒ n(1) (4.176-a) = 0.8 0.6 1√ 1 −  3 0 √ 3 −1 0 (1) n2 2 = 1 ⇒ n1 = 0.8; 2 0 For the second eigenvector λ(2) = 1:  (4.176-b) √ 1...Draft 4.8 Principal Strains, Strain Invariants, Mohr Circle 35 Determine the planes of principal strains for the following strain tensor √   3 0 1 √  3 0 0  0 0 1 (4.172) Solution: The strain invariants are given by IE IIE IIIE = Eii = 2 1 (Eij Eij − Eii Ejj... angle to the strain at a point in a plane are: 92 a b c = = = x cos2 θa + 2 cos θb + cos2 θc + x x y sin2 θa + γxy sin θa cos θa 2 sin θb + γxy sin θb cos θb sin2 θc + γxy sin θc cos θc y y (4.184) (4.1 85) (4.186) When the measured strains a , b , and c , are measured at their corresponding angles from the reference axis and substituted into the above equations the state of strain at a point may be solved, . bridge configuration of Figure 4.12. The formula then is Victor Saouma Mechanics of Materials II Draft 40 KINEMATIC Figure 4.12: Wheatstone Bridge Configurations Victor Saouma Mechanics of Materials II Draft 4.10. 90 o rotation about the e 1 axis. Victor Saouma Mechanics of Materials II Draft 26 KINEMATIC Example 4-12: Polar Decomposition III Victor Saouma Mechanics of Materials II Draft 4.3 Strain Decomposition. definition of Eq. 4.91, Fig. 4.8 Example 4- 15: Strain Invariants & Principal Strains Victor Saouma Mechanics of Materials II Draft 4.8 Principal Strains, Strain Invariants, Mohr Circle 35 Determine