Using the Airy Stress Function : Fourier Series Solutions [ edit | edit source ]
Useful for more general boundary conditions.
Suppose
φ
=
f
(
x
2
)
cos
(
λ
x
1
)
=
o
r
=
φ
=
f
(
x
2
)
sin
(
λ
x
1
)
{\displaystyle \varphi =f(x_{2})\cos(\lambda x_{1})~~=or=~~\varphi =f(x_{2})\sin(\lambda x_{1})}
Substitute into the biharmonic equation. Then,
f
(
x
2
)
=
(
A
+
B
x
2
)
e
λ
x
2
+
(
C
+
D
x
2
)
e
−
λ
x
2
{\displaystyle f(x_{2})=(A+Bx_{2})e^{\lambda x_{2}}+(C+Dx_{2})e^{-\lambda x_{2}}}
or, equivalently,
f
(
x
2
)
=
(
A
+
B
x
2
)
cosh
λ
x
2
+
(
C
+
D
x
2
)
sinh
λ
x
2
{\displaystyle f(x_{2})=(A+Bx_{2})\cosh {\lambda x_{2}}+(C+Dx_{2})\sinh {\lambda x_{2}}\,}
The hyperbolic form allows us to take advantage of symmetry about the
x
2
=
0
{\displaystyle x_{2}=0}
plane.
If
φ
=
f
(
x
2
)
cos
(
λ
x
1
)
{\displaystyle \varphi =f(x_{2})\cos(\lambda x_{1})}
,
σ
11
=
−
λ
2
f
(
x
2
)
cos
(
λ
x
1
)
;
σ
22
=
f
″
(
x
2
)
cos
(
λ
x
1
)
;
σ
12
=
λ
f
′
(
x
2
)
sin
(
λ
x
1
)
{\displaystyle \sigma _{11}=-\lambda ^{2}f(x_{2})\cos(\lambda x_{1})~;~~\sigma _{22}=f^{''}(x_{2})\cos(\lambda x_{1})~;~~\sigma _{12}=\lambda f^{'}(x_{2})\sin(\lambda x_{1})}
Bending of an elastic beam on a foundation
The traction boundary conditions are
σ
12
=
0
;
x
2
=
±
b
σ
22
=
−
p
1
(
x
1
)
;
x
2
=
b
σ
22
=
−
p
2
(
x
1
)
;
x
2
=
−
b
σ
11
=
0
;
x
1
=
±
a
{\displaystyle {\begin{aligned}\sigma _{12}&=0~;~~x_{2}=\pm b\\\sigma _{22}&=-p_{1}(x_{1})~;~~x_{2}=b\\\sigma _{22}&=-p_{2}(x_{1})~;~~x_{2}=-b\\\sigma _{11}&=0~;~~x_{1}=\pm a\end{aligned}}}
The problem is broken up into four subproblems which are superposed.
The subproblems are chosen so that the even/odd properties of
hyperbolic functions can be exploited.
The loads for the four subproblems are chosen to be
f
1
(
x
1
)
=
f
1
(
−
x
1
)
=
1
4
[
p
1
(
x
1
)
+
p
1
(
−
x
1
)
+
p
2
(
x
1
)
+
p
2
(
−
x
1
)
]
f
2
(
x
1
)
=
−
f
2
(
−
x
1
)
=
1
4
[
p
1
(
x
1
)
−
p
1
(
−
x
1
)
+
p
2
(
x
1
)
−
p
2
(
−
x
1
)
]
f
3
(
x
1
)
=
f
3
(
−
x
1
)
=
1
4
[
p
1
(
x
1
)
+
p
1
(
−
x
1
)
−
p
2
(
x
1
)
−
p
2
(
−
x
1
)
]
f
4
(
x
1
)
=
−
f
4
(
−
x
1
)
=
1
4
[
p
1
(
x
1
)
−
p
1
(
−
x
1
)
−
p
2
(
x
1
)
+
p
2
(
−
x
1
)
]
{\displaystyle {\begin{aligned}f_{1}(x_{1})=f_{1}(-x_{1})&={\cfrac {1}{4}}\left[p_{1}(x_{1})+p_{1}(-x_{1})+p_{2}(x_{1})+p_{2}(-x_{1})\right]\\f_{2}(x_{1})=-f_{2}(-x_{1})&={\cfrac {1}{4}}\left[p_{1}(x_{1})-p_{1}(-x_{1})+p_{2}(x_{1})-p_{2}(-x_{1})\right]\\f_{3}(x_{1})=f_{3}(-x_{1})&={\cfrac {1}{4}}\left[p_{1}(x_{1})+p_{1}(-x_{1})-p_{2}(x_{1})-p_{2}(-x_{1})\right]\\f_{4}(x_{1})=-f_{4}(-x_{1})&={\cfrac {1}{4}}\left[p_{1}(x_{1})-p_{1}(-x_{1})-p_{2}(x_{1})+p_{2}(-x_{1})\right]\end{aligned}}}
The new boundary conditions are
σ
12
=
0
;
x
2
=
±
b
σ
22
=
−
f
1
(
x
1
)
−
f
2
(
x
1
)
−
f
3
(
x
1
)
−
f
4
(
x
1
)
;
x
2
=
b
σ
22
=
−
f
1
(
x
1
)
−
f
2
(
x
1
)
+
f
3
(
x
1
)
+
f
4
(
x
1
)
;
x
2
=
−
b
σ
11
=
0
;
x
1
=
±
a
{\displaystyle {\begin{aligned}\sigma _{12}&=0~;~~x_{2}=\pm b\\\sigma _{22}&=-f_{1}(x_{1})-f_{2}(x_{1})-f_{3}(x_{1})-f_{4}(x_{1})~;~~x_{2}=b\\\sigma _{22}&=-f_{1}(x_{1})-f_{2}(x_{1})+f_{3}(x_{1})+f_{4}(x_{1})~;~~x_{2}=-b\\\sigma _{11}&=0~;~~x_{1}=\pm a\end{aligned}}}
Let us look at the subproblem with loads
±
f
3
(
x
1
)
{\displaystyle \pm f_{3}(x_{1})}
applied on the top and bottom of the beam. The problem is even in
x
1
{\displaystyle x_{1}}
and odd in
x
2
{\displaystyle x_{2}}
.
So we use,
φ
=
∑
n
=
1
∞
f
n
(
x
2
)
cos
(
λ
n
x
1
)
=
∑
n
=
1
∞
[
A
n
x
2
cosh
(
λ
n
x
2
)
+
B
n
sinh
(
λ
n
x
2
)
]
cos
(
λ
n
x
1
)
{\displaystyle {\begin{aligned}\varphi &=\sum _{n=1}^{\infty }f_{n}(x_{2})\cos(\lambda _{n}x_{1})\\&=\sum _{n=1}^{\infty }\left[A_{n}x_{2}\cosh(\lambda _{n}x_{2})+B_{n}\sinh(\lambda _{n}x_{2})\right]\cos(\lambda _{n}x_{1})\end{aligned}}}
At
x
1
=
a
{\displaystyle x_{1}=a}
,
σ
11
=
−
∑
n
=
1
∞
λ
n
2
f
n
(
x
2
)
cos
(
λ
n
a
)
{\displaystyle \sigma _{11}=-\sum _{n=1}^{\infty }\lambda _{n}^{2}f_{n}(x_{2})\cos(\lambda _{n}a)}
Hence
σ
11
=
0
{\displaystyle \sigma _{11}=0}
if
λ
n
=
(
2
n
−
1
)
π
/
2
a
{\displaystyle \lambda _{n}=(2n-1)\pi /2a}
.
We can substitute
φ
{\displaystyle \varphi }
and express the stresses in terms of
Fourier series.
Applying the boundary conditions of
x
2
=
±
b
{\displaystyle x_{2}=\pm b}
we get
∑
n
=
1
∞
[
A
n
λ
n
cosh
(
λ
n
b
)
+
A
n
λ
n
2
b
sinh
(
λ
n
b
)
+
B
n
λ
n
2
cosh
(
λ
n
b
)
]
sin
(
λ
n
x
1
)
=
0
∑
n
=
1
∞
[
A
n
λ
n
2
b
cosh
(
λ
n
b
)
+
B
n
λ
n
2
sinh
(
λ
n
b
)
]
cos
(
λ
n
x
1
)
=
f
3
(
x
1
)
{\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }\left[A_{n}\lambda _{n}\cosh(\lambda _{n}b)+A_{n}\lambda _{n}^{2}b\sinh(\lambda _{n}b)+B_{n}\lambda _{n}^{2}\cosh(\lambda _{n}b)\right]\sin(\lambda _{n}x_{1})&=0\\\sum _{n=1}^{\infty }\left[A_{n}\lambda _{n}^{2}b\cosh(\lambda _{n}b)+B_{n}\lambda _{n}^{2}\sinh(\lambda _{n}b)\right]\cos(\lambda _{n}x_{1})&=f_{3}(x_{1})\end{aligned}}}
The first equation is satisfied if
A
m
λ
m
cosh
(
λ
m
b
)
+
A
m
λ
m
2
b
sinh
(
λ
m
b
)
+
B
m
λ
m
2
cosh
(
λ
m
b
)
=
0
(
1
)
{\displaystyle A_{m}\lambda _{m}\cosh(\lambda _{m}b)+A_{m}\lambda _{m}^{2}b\sinh(\lambda _{m}b)+B_{m}\lambda _{m}^{2}\cosh(\lambda _{m}b)=0\qquad (1)}
Integrate the second equation from
−
a
{\displaystyle -a}
to
a
{\displaystyle a}
after multiplying by
cos
(
λ
m
x
1
)
{\displaystyle \cos(\lambda _{m}x_{1})}
.
All the odd functions are zero, except the
case where
n
=
m
{\displaystyle n=m}
.
Therefore, all that remains is
[
A
m
λ
m
2
b
cosh
(
λ
m
b
)
+
B
m
λ
m
2
sinh
(
λ
m
b
)
]
a
=
∫
−
a
a
f
3
(
x
1
)
cos
(
λ
m
x
1
)
d
x
1
(
2
)
{\displaystyle \left[A_{m}\lambda _{m}^{2}b\cosh(\lambda _{m}b)+B_{m}\lambda _{m}^{2}\sinh(\lambda _{m}b)\right]a=\int _{-a}^{a}f_{3}(x_{1})\cos(\lambda _{m}x_{1})dx_{1}\qquad (2)}
We can calculate
A
m
{\displaystyle A_{m}}
and
B
m
{\displaystyle B_{m}}
from equations (1) and (2), substitute them into the expressions for stress to get the solution.
We do the same thing for the other subproblems.
The Fourier series approach is particularly useful if we have discontinuous or point loads.