8.3 Linear Theory for Growing Beams and Plates
421
Solution
For all cases, the growth force is given by
N g =
E
1 − ν 2
h/2
−h/2
g dy =
E
1 − ν 2
0
−h/2
0 dy +
h/2
0
a dy
=
Eah
2(1 − ν 2 )
.
(8.25)
Similarly, the growth moment is
M g =
E
1 − ν 2
h/2
−h/2
yy g dy =
Eah 2
8(1 − ν 2 )
.
(8.26)
Stretching A roller support is the same as a pinned support, with one exception.
Both prevent vertical displacement while allowing free rotation. But whereas a pin
also prevents horizontal displacement, a roller does not. This difference directly
affects the stretching analysis. For cases (a) and (b), axial equilibrium gives N = 0
and Eq. (8.16) 1 yields
0 = (1 − ν
2 )
N g
Eh
=
a
2
(cases a & b).
For case (c), the axial displacement is zero at both ends, giving 0 = 0 and
N = −N g = −
Eah
2(1 − ν 2 )
(case c).
(8.27)
In agreement with intuition, N > 0 for a < 0 (atrophy) and N < 0 for a > 0
(positive growth).
Bending For p = M
g = 0, the governing equation (8.20) can be written in the
form
v
− β
2 v
= 0,
(8.28)
where
β =
N
D
.
(8.29)
For cases (a) and (b), N = β = 0 and Eq. (8.28) reduces to v = 0, which has
the general solution (8.22). The boundary conditions for case (a) are the same ones
considered in Example 8.2. Thus, the solution (8.24) applies here as well, i.e.,
v(x) = −
M g
2D
x
2
(case a).
(8.30)