8.3 Linear Theory for Growing Beams and Plates
429
Next, we substitute Eqs. (8.43) into (8.41) to get
¯
v n =
¯
M gn
nπ
¯
L
2
+ ¯
N
,
n= 1, 2, . . . .
(8.44)
With ¯
N provided by Eq. (8.40), substitution into (8.43) 1 yields ¯
v p ( ¯
x).
Consider now the boundary conditions (8.31). Since ¯
M g = 0 at x = L, these
boundary conditions become
¯
v(0) = ¯
v
(0) = 0
¯
v( ¯
L) = ¯
v
( ¯
L) = 0.
(8.45)
Other than ¯
v (0) = 0, the series (8.43) 1 for ¯
v p satisfies these conditions term by
term. If all four conditions were satisfied, e.g., if the left end were also pinned, then
we would be done, i.e., the particular solution would be the total solution. Here,
however, we need to add the complementary solution ¯
v c ( ¯
x) by solving Eq. (8.41)
with ¯
M
g = 0 and the boundary conditions
¯
v(0) = ¯
v p (0) + ¯
v c (0) = 0
→
¯
v c (0) = 0
¯
v
(0) = ¯
v
p (0) + ¯
v
c (0) = 0
→
¯
v
c (0) = −¯ v
p (0) = −
n
¯
v n
nπ
¯
L
¯
v( ¯
L) = ¯
v p ( ¯
L) + ¯
v c ( ¯
L) = 0
→
¯
v c ( ¯
L) = 0
¯
v
( ¯
L) = ¯
v
p ( ¯
L) + ¯
v
c ( ¯
L) = 0
→
¯
v
c ( ¯
L) = 0.
Following the procedure in Example 8.3 yields
¯
v c ( ¯
x) = c 3 (sinh ¯
β ¯
x − ¯
β ¯
x) + c 4 (cosh ¯
β ¯
x − 1) −
n
¯
v n
nπ ¯
x
¯
L
c 3 = −c 4 coth ¯
β ¯
L
c 4 =
−π
n n ¯
v n
1 − ¯
β ¯
L coth ¯
β ¯
L
.
(8.46)
Substituting these relations, along with the particular solution (8.43) 1 , into (8.42)
provides the total deflection, which depends on a, ¯
d, and ¯
L.
Results
For L/ h = 10, deflection curves are shown for four fractional lengths (d/L) of the
growth region (Fig. 8.15c). For illustration, we take a = 0.01, ν = 0.4, and 100
terms in the series solution. The curve for d = L agrees closely with the exact
solution for this problem, given by the lower solid blue curve in Fig. 8.12c. As
discussed in Example 8.3, positive growth over the entire lower layer causes most
429
Next, we substitute Eqs. (8.43) into (8.41) to get
¯
v n =
¯
M gn
nπ
¯
L
2
+ ¯
N
,
n= 1, 2, . . . .
(8.44)
With ¯
N provided by Eq. (8.40), substitution into (8.43) 1 yields ¯
v p ( ¯
x).
Consider now the boundary conditions (8.31). Since ¯
M g = 0 at x = L, these
boundary conditions become
¯
v(0) = ¯
v
(0) = 0
¯
v( ¯
L) = ¯
v
( ¯
L) = 0.
(8.45)
Other than ¯
v (0) = 0, the series (8.43) 1 for ¯
v p satisfies these conditions term by
term. If all four conditions were satisfied, e.g., if the left end were also pinned, then
we would be done, i.e., the particular solution would be the total solution. Here,
however, we need to add the complementary solution ¯
v c ( ¯
x) by solving Eq. (8.41)
with ¯
M
g = 0 and the boundary conditions
¯
v(0) = ¯
v p (0) + ¯
v c (0) = 0
→
¯
v c (0) = 0
¯
v
(0) = ¯
v
p (0) + ¯
v
c (0) = 0
→
¯
v
c (0) = −¯ v
p (0) = −
n
¯
v n
nπ
¯
L
¯
v( ¯
L) = ¯
v p ( ¯
L) + ¯
v c ( ¯
L) = 0
→
¯
v c ( ¯
L) = 0
¯
v
( ¯
L) = ¯
v
p ( ¯
L) + ¯
v
c ( ¯
L) = 0
→
¯
v
c ( ¯
L) = 0.
Following the procedure in Example 8.3 yields
¯
v c ( ¯
x) = c 3 (sinh ¯
β ¯
x − ¯
β ¯
x) + c 4 (cosh ¯
β ¯
x − 1) −
n
¯
v n
nπ ¯
x
¯
L
c 3 = −c 4 coth ¯
β ¯
L
c 4 =
−π
n n ¯
v n
1 − ¯
β ¯
L coth ¯
β ¯
L
.
(8.46)
Substituting these relations, along with the particular solution (8.43) 1 , into (8.42)
provides the total deflection, which depends on a, ¯
d, and ¯
L.
Results
For L/ h = 10, deflection curves are shown for four fractional lengths (d/L) of the
growth region (Fig. 8.15c). For illustration, we take a = 0.01, ν = 0.4, and 100
terms in the series solution. The curve for d = L agrees closely with the exact
solution for this problem, given by the lower solid blue curve in Fig. 8.12c. As
discussed in Example 8.3, positive growth over the entire lower layer causes most