6.4 Fundamental Growth Mechanics
261
Solution
For the specified growth, each point in the bar moves from position X to x(X, t). If
the left end of the bar is fixed at X = x = 0, the length of the bar at any instant of
time is x(L 0 , t). Equations (6.3) and (6.5) give
λ x =
∂x
∂X
= G x λ
∗
x .
(6.16)
Since the specified growth is independent of Y and Z, the stress is uniform over
any cross section, and summing forces on an arbitrary section reveals that the bar
remains stress-free. Thus, no elastic deformation occurs, i.e., λ ∗
x = 1 and λ x = G x .
Plugging Eq. (6.15) into (6.16) and integrating over X yields
x(X, t) = X +
aX +
b
3
X 3
L 2
0
(1 − e
−βt ) + C(t).
The boundary condition x(0, t) = 0 gives C = 0, and, therefore, the current length
of the bar is
L = x(L 0 , t) = L 0
1 +
a +
b
3
(1 − e
−βt )
.
(6.17)
Note that L = L 0 at t = 0 and L → L 0
1 +
a +
b
3
as t → ∞.
This example shows that unconstrained axial growth that varies only along the
length of the bar induces no stress. This is also the case if growth is of the form
G y (Y, t) or G z (Z, t). Later, we will see that gradients normal to the direction of
growth are more interesting, e.g., G x (Y, t).
6.4.2 Growth of a Constrained Bar
In Chap. 5, we showed that a contracting element generates tension when its ends are
fixed. Similarly, negative growth causes tension to develop in a bar with constrained
ends, while positive growth generates compression. Among other things, the next
two examples illustrate the solution procedure for problems involving growthinduced stress.
Example 6.2 Consider a bar composed of incompressible neo-Hookean material
with the strain-energy density function
W = c (λ
2
x + λ
2
y + λ
2
z − 3),
261
Solution
For the specified growth, each point in the bar moves from position X to x(X, t). If
the left end of the bar is fixed at X = x = 0, the length of the bar at any instant of
time is x(L 0 , t). Equations (6.3) and (6.5) give
λ x =
∂x
∂X
= G x λ
∗
x .
(6.16)
Since the specified growth is independent of Y and Z, the stress is uniform over
any cross section, and summing forces on an arbitrary section reveals that the bar
remains stress-free. Thus, no elastic deformation occurs, i.e., λ ∗
x = 1 and λ x = G x .
Plugging Eq. (6.15) into (6.16) and integrating over X yields
x(X, t) = X +
aX +
b
3
X 3
L 2
0
(1 − e
−βt ) + C(t).
The boundary condition x(0, t) = 0 gives C = 0, and, therefore, the current length
of the bar is
L = x(L 0 , t) = L 0
1 +
a +
b
3
(1 − e
−βt )
.
(6.17)
Note that L = L 0 at t = 0 and L → L 0
1 +
a +
b
3
as t → ∞.
This example shows that unconstrained axial growth that varies only along the
length of the bar induces no stress. This is also the case if growth is of the form
G y (Y, t) or G z (Z, t). Later, we will see that gradients normal to the direction of
growth are more interesting, e.g., G x (Y, t).
6.4.2 Growth of a Constrained Bar
In Chap. 5, we showed that a contracting element generates tension when its ends are
fixed. Similarly, negative growth causes tension to develop in a bar with constrained
ends, while positive growth generates compression. Among other things, the next
two examples illustrate the solution procedure for problems involving growthinduced stress.
Example 6.2 Consider a bar composed of incompressible neo-Hookean material
with the strain-energy density function
W = c (λ
2
x + λ
2
y + λ
2
z − 3),