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8 Morphogenesis
8.3.3 A Few Examples
The following examples illustrate the utility of the linear theory for growing beams
and plates. As always, the emphasis is on understanding fundamental behavior.
Example 8.1 Consider an unloaded rectangular beam (not a plate) of length L,
thickness h, and width b. The growth strain has the parabolic distribution
g = a
1 −
4y 2
h 2
across the thickness, with a being a constant. Determine the stress distribution in
the beam for the following cases: (a) the left end is fixed and the right end is free;
and (b) both ends are fixed.
Solution
For the given growth distribution, the growth force is
N g = bE
h/2
−h/2
g (x, y) dy = 2EAa/3,
where A = bh is the cross-sectional area. In part (a), the beam grows without
external loads or constraints. Thus, equilibrium demands that N = 0 and (8.10) 1
gives
0 =
N g
EA
=
2a
3
(part a).
In part (b), the axial displacement must be zero throughout the beam to satisfy
symmetry and the end conditions. In this case, 0 = 0 and (8.10) 1 gives
N = −N g = −2EAa/3
(part b).
Note that positive growth (a > 0) yields net compression in the beam, as expected.
This completes the stretching part of the solution.
Since there are no transverse loads and growth is symmetric relative to the middle
surface, we expect that the beam will stretch but not bend. Another way to see this
is to note that moment equilibrium gives M = 0, and integrating Eq. (8.11) 2 gives
M g = 0. The lack of both loading and growth moments implies no bending. In part
(a), it also is easy to show that integrating Eq. (8.14) with N = q = M g = 0 and
the given boundary conditions (v = v = 0 at x = 0; M = V = 0 or v = v = 0
at x = L) yields v = 0, consistent with this result. 8
8 Some readers may want to verify this result for themselves.
8 Morphogenesis
8.3.3 A Few Examples
The following examples illustrate the utility of the linear theory for growing beams
and plates. As always, the emphasis is on understanding fundamental behavior.
Example 8.1 Consider an unloaded rectangular beam (not a plate) of length L,
thickness h, and width b. The growth strain has the parabolic distribution
g = a
1 −
4y 2
h 2
across the thickness, with a being a constant. Determine the stress distribution in
the beam for the following cases: (a) the left end is fixed and the right end is free;
and (b) both ends are fixed.
Solution
For the given growth distribution, the growth force is
N g = bE
h/2
−h/2
g (x, y) dy = 2EAa/3,
where A = bh is the cross-sectional area. In part (a), the beam grows without
external loads or constraints. Thus, equilibrium demands that N = 0 and (8.10) 1
gives
0 =
N g
EA
=
2a
3
(part a).
In part (b), the axial displacement must be zero throughout the beam to satisfy
symmetry and the end conditions. In this case, 0 = 0 and (8.10) 1 gives
N = −N g = −2EAa/3
(part b).
Note that positive growth (a > 0) yields net compression in the beam, as expected.
This completes the stretching part of the solution.
Since there are no transverse loads and growth is symmetric relative to the middle
surface, we expect that the beam will stretch but not bend. Another way to see this
is to note that moment equilibrium gives M = 0, and integrating Eq. (8.11) 2 gives
M g = 0. The lack of both loading and growth moments implies no bending. In part
(a), it also is easy to show that integrating Eq. (8.14) with N = q = M g = 0 and
the given boundary conditions (v = v = 0 at x = 0; M = V = 0 or v = v = 0
at x = L) yields v = 0, consistent with this result. 8
8 Some readers may want to verify this result for themselves.