3.3 Analysis of Deformation
69
(b) We compute the deformed length in two ways. The easy way is to realize that
= x(L) − x(0) = (1 + b)L + cL 2 . The second way is to integrate the stretch
ratio, i.e.,
L
0
dx =
L
0
λ(X) dX =
L
0
(1 + b + 2cX) dX
x(L) − x(0) = =
(1 + b)X + cX
2
L
0
= (1 + b)L + cL
2 .
It is useful to examine some special cases. First, for b = c = 0, we get x = X+a,
which represents rigid-body displacement of the bar, and consequently λ = 1. If
c = 0 but b = 0, the bar deforms uniformly with λ = 1 + b = constant = 1. In this
case, the value of b must be > −1; otherwise λ ≤ 0, which is nonsense since the
length of the deformed bar would be zero or negative.
Finally, to set the stage for adding a second dimension, we write the deformation
measures in terms of the displacement u(X) as defined by x = X + u. Substitution
into Eqs. (3.32) gives
λ = 1 +
du
dX
=
du
dX
E =
du
dX
+
1
2
du
dX
2
.
(3.33)
These relations emphasize that strain depends on displacement gradients, which
cause particles to stretch or shorten. If the displacement gradient du/dX is small
compared to unity, then the nonlinear term in E can be neglected, and E reduces
approximately to the linear strain .
3.3.2 Deformation in 2D
Deformation becomes considerably more complicated with the addition of a second
dimension. On the other hand, the problems also become more interesting.
Two main factors contribute to the added complexity. First, as deformations
grow large, displacements in one direction can cause significant stretching in the
orthogonal direction (in addition to Poisson’s ratio effects). For example, if a
horizontal bar is pinned at its left end and its right end moves vertically upward,
69
(b) We compute the deformed length in two ways. The easy way is to realize that
= x(L) − x(0) = (1 + b)L + cL 2 . The second way is to integrate the stretch
ratio, i.e.,
L
0
dx =
L
0
λ(X) dX =
L
0
(1 + b + 2cX) dX
x(L) − x(0) = =
(1 + b)X + cX
2
L
0
= (1 + b)L + cL
2 .
It is useful to examine some special cases. First, for b = c = 0, we get x = X+a,
which represents rigid-body displacement of the bar, and consequently λ = 1. If
c = 0 but b = 0, the bar deforms uniformly with λ = 1 + b = constant = 1. In this
case, the value of b must be > −1; otherwise λ ≤ 0, which is nonsense since the
length of the deformed bar would be zero or negative.
Finally, to set the stage for adding a second dimension, we write the deformation
measures in terms of the displacement u(X) as defined by x = X + u. Substitution
into Eqs. (3.32) gives
λ = 1 +
du
dX
=
du
dX
E =
du
dX
+
1
2
du
dX
2
.
(3.33)
These relations emphasize that strain depends on displacement gradients, which
cause particles to stretch or shorten. If the displacement gradient du/dX is small
compared to unity, then the nonlinear term in E can be neglected, and E reduces
approximately to the linear strain .
3.3.2 Deformation in 2D
Deformation becomes considerably more complicated with the addition of a second
dimension. On the other hand, the problems also become more interesting.
Two main factors contribute to the added complexity. First, as deformations
grow large, displacements in one direction can cause significant stretching in the
orthogonal direction (in addition to Poisson’s ratio effects). For example, if a
horizontal bar is pinned at its left end and its right end moves vertically upward,