CONTINUUM MECHANICS for ENGINEERS

form consistent with deformation analysis. In component notation, the material
description is
and the spatial description is
u i(X A) = x i(X A) – X i
From the first of these, Eq 4.6-13 becomes
which reduces to
(4.6-20a)
u A(x i) = x A – X A (x i) (4.6-20b)
2E AB = x i,Ax i,B – δ AB = (u i,A + δ iA)(u i,B – δ iB) – δ AB
2E AB = u A,B + u B,A + u i,A u i,B
and from the second, Eq 4.6-17 becomes
which reduces to
2e ij = δ ij – X A,iX A,j = δ ij – (δ Ai – u A,i)(δ Aj – u A,j)
2e ij = u i,j + u j,i – u A,iu A,j
(4.6-21)
(4.6-22)
Example 4.6-1
Let the simple shear deformation x 1 = X 1; x 2 = X 2 + kX 3; x 3 = X 3 + kX 2, where
k is a constant, be applied to the small cube of edge dimensions dL shown
in the sketch. Draw the deformed shape of face ABGH of the cube and
determine the difference (dx) 2 – (dX) 2 for the diagonals AG, BH and OG of
the cube.
Solution
From the mapping equations directly, the origin O is seen to remain in place,
and the particles originally at points A, B, G and H are displaced to the points
a(dL,O,O), b(dL, dL,kdL), g(dL, (1+k)dL, (1+ k)dL) and h(dL, kdL, dL), respectively,
so that particles in planes parallel to the X 2X 3 remain in those planes,
and the square face ABGH becomes the diamond-shaped parallelogram abgh
shown below. Also from the mapping equations and Eq 4.6-8, we see that
the deformation gradient F has the matrix form
[ FiA]= ⎡1
0 0⎤
⎢ ⎥
⎢
0 1 k
⎥
⎣
⎢0
k
1⎦
⎥