428
8 Morphogenesis
For discontinuous M g , there are at least three ways to solve this problem. The
classical method is to solve Eq. (8.41) separately for regions 1 and 2 (with M
g = 0
in each region). This gives v 1 (x) and v 2 (x), with each having four constants of
integration. These eight constants are then determined using the usual two boundary
conditions at each end, as well as four continuity conditions at x = d: v 1 = v 2 ,
θ 1 = θ 2 , M 1 = M 2 , and V 1 = V 2 . Each additional discontinuity leads to four more
equations and unknowns.
A considerably easier approach is to write M g (x) in terms of discontinuity
(singularity) functions (Hibbeler 2010). With this magical method, the problem can
be solved without breaking the plate into regions. Although my personal favorite,
this technique is beyond the scope of this book.
Here, we use Fourier series to solve the problem (see Appendix B for a brief
summary). 9 The general solution to Eq. (8.41) can be written in the form
¯
v( ¯
x) = ¯
v p ( ¯
x) + ¯
v c ( ¯
x),
(8.42)
where ¯
v p ( ¯
x) and ¯
v c ( ¯
x) are the particular (M
g = 0) and complementary (M
g = 0)
solutions, respectively. To obtain the particular solution for a plate of length L, we
first expand the displacement and growth moment as Fourier sine series given by
¯
v p ( ¯
x) =
∞
n=1
¯
v n sin
nπ ¯
x
¯
L
¯
M g ( ¯
x) =
∞
n=1
¯
M gn sin
nπ ¯
x
¯
L
,
(8.43)
in which ¯
v n and ¯
M gn are Fourier coefficients. For the distribution of (8.37) 2 , the
¯
M gn are computed using the standard formula
¯
M gn =
2
¯
L
¯
L
0
¯
M g ( ¯
x) sin
nπ ¯
x
¯
L
d ¯
x
=
2
¯
L
¯
d
0
3a
2
sin
nπ ¯
x
¯
L
d ¯
x
= −
3a
nπ
cos
nπ ¯
d
¯
L
− 1
,
n= 1, 2, . . . .
For d = L/2, Fig. 8.15b shows how the series representation (8.43) 2 improves as
the number of terms increases.
9 To compare results with the “exact” solution, try Problem 8.5.
8 Morphogenesis
For discontinuous M g , there are at least three ways to solve this problem. The
classical method is to solve Eq. (8.41) separately for regions 1 and 2 (with M
g = 0
in each region). This gives v 1 (x) and v 2 (x), with each having four constants of
integration. These eight constants are then determined using the usual two boundary
conditions at each end, as well as four continuity conditions at x = d: v 1 = v 2 ,
θ 1 = θ 2 , M 1 = M 2 , and V 1 = V 2 . Each additional discontinuity leads to four more
equations and unknowns.
A considerably easier approach is to write M g (x) in terms of discontinuity
(singularity) functions (Hibbeler 2010). With this magical method, the problem can
be solved without breaking the plate into regions. Although my personal favorite,
this technique is beyond the scope of this book.
Here, we use Fourier series to solve the problem (see Appendix B for a brief
summary). 9 The general solution to Eq. (8.41) can be written in the form
¯
v( ¯
x) = ¯
v p ( ¯
x) + ¯
v c ( ¯
x),
(8.42)
where ¯
v p ( ¯
x) and ¯
v c ( ¯
x) are the particular (M
g = 0) and complementary (M
g = 0)
solutions, respectively. To obtain the particular solution for a plate of length L, we
first expand the displacement and growth moment as Fourier sine series given by
¯
v p ( ¯
x) =
∞
n=1
¯
v n sin
nπ ¯
x
¯
L
¯
M g ( ¯
x) =
∞
n=1
¯
M gn sin
nπ ¯
x
¯
L
,
(8.43)
in which ¯
v n and ¯
M gn are Fourier coefficients. For the distribution of (8.37) 2 , the
¯
M gn are computed using the standard formula
¯
M gn =
2
¯
L
¯
L
0
¯
M g ( ¯
x) sin
nπ ¯
x
¯
L
d ¯
x
=
2
¯
L
¯
d
0
3a
2
sin
nπ ¯
x
¯
L
d ¯
x
= −
3a
nπ
cos
nπ ¯
d
¯
L
− 1
,
n= 1, 2, . . . .
For d = L/2, Fig. 8.15b shows how the series representation (8.43) 2 improves as
the number of terms increases.
9 To compare results with the “exact” solution, try Problem 8.5.