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).

0 0

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