## Dx

 d ( d£23 + d£31 dx1 dx1 dx2 d ( d£23 d£31 dx2 dx1 dx2 d ( d£23 + d£31 dx3 \ dx1 dx2

In 2D, this results in (by setting i

dx1dx2

dx1dx2

(4.141-a) (4.141-b) (4.141-c) (4.141-d) (4.141-e) (4.141-f)

dx2 dx1dx2

 LAGRANGIAN Material EULERIAN Spatial Position Vector x = x(X,t) X = X(x,t) GRADIENTS Deformation F = xVx = dxj H = XVx = It H = F-1 OXj dXj U1 J = uVX = F - I oxj ij oxj K = uVx = I - H TENSOR Deformation dX2 = dx-B-1 -dx dx2 = dX-C-dX B-1 = VxX-XVx = HCH C = Vxx-xVx = Fc-F C-1 = B-1 STRAINS Lagrangian Eulerian/Almansi Finite Strain dx2 - dX2 = dX-2E-dX dx2 -dX2 = dx-2E*-dX Hc-H E = 1 (uVx + Vxu + Vxu-uVx) E* = 1 ( dui 1 duj duk duk \ or Eij 2 \dxj "T" dxi dxi dx, J ui E* = 2 (uVx + Vxu - Vxu-uVx) J+Jc+Jc -J E = 1 (uVx + Vxu) = 1 (J + Jc) E* = 1 (uVx + Vx u) = 1 (K + Kc) ROTATION TFNSORS VSjCtorj Saouma deformation Í2 \dXj dXi J ^ 2 ^ dXj OXiJiUXj [g (uVx + Vxu) + 1 (uVx - Vxu)]-dX >ducticun joeontinuum Mie^oamcs [2 ^ dxj "T" dxi J 2 ^ dxj dxi J] j ^ (uVx + Vxu) + 1 (uVx - Vxu)]-dx E W E* O

95 When he compatibility equation is written in term of the stresses, it yields:

d2a i i da 22 dx2

d2a22 d2a i i d2 a2 i

dxidx2

Example 4-13: Strain Compatibility

For the following strain field

xJ+xJ Xi 2

does there exist a single-valued continuous displacement field? Solution:

dEi 2

dXi dE22

dXf d2Ei i + d2E22

(X2 + X22) - X2(2X2) (X 2 + X22)2 (X 2 + X22) - X i (2Xi ) = (X2 + X22)2

d2 E

Actually, it can be easily verified that the unique displacement field is given by

u = arctan

to which we could add the rigid body displacement field (if any).

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