3.3 Analysis of Deformation
75
To write F in terms of the displacement vector u(R), we substitute r = R + u to
get
F = [∇(R + u)]
T
= (∇R)
T
+ (∇u)
T ,
where ∇u is the displacement gradient tensor. The quantity ∇R can be simplified
by working in Cartesian coordinates and then reverting to tensor form. Because
Cartesian base vectors are constant, derivations frequently are easier in Cartesian
coordinates, and we will use this approach often in this book. With ∇ = e i ∂/∂X i ,
we have
∇R = e i
∂
∂X i
(X j e j ) =
∂X j
∂X i
e i e j = δ ij e i e j = e i e i = I.
Since I T = I, the expression for F becomes
F = I + (∇u) T ,
(3.47)
which is valid for any coordinate system.
In terms of scalar components, Eq. (3.46) yields
F = F ij e i e j = (∇r)
T
=
e j
∂
∂X j
x i e i
T
=
∂x i
∂X j
e i e j ,
where the order of the base vectors has been switched to give the transpose. Thus,
the Cartesian components of F are
F ij =
∂x i
∂X j
.
(3.48)
To write these components in terms of displacements, we substitute x i = X i + u i to
get F ij = (∂/∂X j )(X i + u i ) or
F ij = δ ij +
∂u i
∂X j
.
(3.49)
Example 3.10 Consider a disk that undergoes a rigid-body rotation by the angle θ
about the axis X 3 = x 3 , which is normal to the disk and passes through its center.
In Cartesian coordinates, a point in the disk moves from R = X 1 e 1 + X 2 e 2 to
r = x 1 e 1 + x 2 e 2 . (a) Write x 1 and x 2 in terms of X 1 , X 2 , and θ . (b) Determine the
deformation gradient tensor F.