In this case, the form of
is not obvious and has to be
derived from the traction-free BCs
![{\displaystyle (\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac8317ce975d296f9fd2e3f0ec4507ac32035fee)
Suppose that
and
are the two sides of the rectangle, and
.
Also
is the side parallel to
and
is the side parallel to
.
Then, the traction-free BCs are
![{\displaystyle \psi _{,1}=x_{2}~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~\psi _{,2}=-x_{1}~~{\text{on}}~~x_{2}=\pm b\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad5cd642699590bca2981130dfd73a6e07e46ac1)
A suitable
must satisfy these BCs and
.
We can simplify the problem by a change of variable
![{\displaystyle {\bar {\psi }}=x_{1}~x_{2}-\psi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/dab96118f1cae75123fb31e3c3ec2568e3c6c69d)
Then the equilibrium condition becomes
![{\displaystyle \nabla ^{2}{\bar {\psi }}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56c1b2d90630d14c214ed4e418cad607bcccdadb)
The traction-free BCs become
![{\displaystyle {\bar {\psi }}_{,1}=0~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~{\bar {\psi }}_{,2}=2x_{1}~~{\text{on}}~~x_{2}=\pm b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b829aa0e98f399d86792c43b85b5ae9598025b4f)
Let us assume that
![{\displaystyle {\bar {\psi }}(x_{1},x_{2})=f(x_{1})g(x_{2})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98e85db5170a537b1b7ecc2a5b020a912f80149d)
Then,
![{\displaystyle \nabla ^{2}{\bar {\psi }}={\bar {\psi }}_{,11}+{\bar {\psi }}_{,22}=f^{''}(x_{1})g(x_{2})+g^{''}(x_{2})f(x_{1})=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc330d1aadeca79dded878a9a50e74b40677f013)
or,
![{\displaystyle {\frac {f^{''}(x_{1})}{f(x_{1})}}=-{\frac {g^{''}(x_{2})}{g(x_{2})}}=\eta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/62de5a00d5e0a4e6928f1a84800cc210ee1d8937)
In both these cases, we get trivial values of
.
Let
![{\displaystyle \eta =-k^{2}~~~;~~k>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a8624a1c6ad27a33dc3a1fdcbbed49084449134)
Then,
![{\displaystyle {\begin{aligned}f^{''}(x_{1})+k^{2}f(x_{1})=0~~\Rightarrow &~~f(x_{1})=C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\\g^{''}(x_{2})-k^{2}g(x_{2})=0~~\Rightarrow &~~g(x_{2})=C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2136b78d63db51115fc793d408619e610479587d)
Therefore,
![{\displaystyle {\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bcb2287f2c30081bbf32884dfdc0a1b041a4235)
Apply the BCs at
~~ (
), to get
![{\displaystyle {\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8f3546d1357cc29b7f3b8e541db9f4809ba2b02)
or,
![{\displaystyle F(x_{1})G^{'}(b)=2x_{1}~~;~~~F(x_{1})G^{'}(-b)=2x_{1}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76dd0533504c1e2e6f9da485a90b0c0b406f4b1f)
The RHS of both equations are odd. Therefore,
is odd. Since,
is an even function, we must have
.
Also,
![{\displaystyle F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7d9023ae4253379b8f94da24302aff5143a00b4)
Hence,
is even. Since
is an odd function, we must
have
.
Therefore,
![{\displaystyle {\bar {\psi }}(x_{1},x_{2})=C_{2}C_{4}\sin(kx_{1})\sinh(kx_{2})=A\sin(kx_{1})\sinh(kx_{2})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e28a2ac43b673f730f2c6d52be03a6b4c04e72f6)
Apply BCs at
(
), to get
![{\displaystyle Ak\cos(ka)\sinh(kx_{2})=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6137bdb86a8f99cd680bbd9bcef2e7d08fa127c)
The only nontrivial solution is obtained when
, which means that
![{\displaystyle k_{n}={\frac {(2n+1)\pi }{2a}}~~,~~~n=0,1,2,...}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c521b09a3ddf5d8fe492f6747df7a3a55c43da1)
The BCs at
are satisfied by every terms of the series
![{\displaystyle {\bar {\psi }}(x_{1},x_{2})=\sum _{n=0}^{\infty }A_{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dc337e93e119dbf65d407b7c0e8bd40f1733db7)
Applying the BCs at
again, we get
![{\displaystyle \sum _{n=0}^{\infty }A_{n}k_{n}\sin(k_{n}x_{1})\cosh(k_{n}b)=2x_{1}~~\Rightarrow ~~~\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})=2x_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7220325b6d91feaac366b3a815d0c0f3dd739bb2)
Using the orthogonality of terms of the sine series,
![{\displaystyle \int _{-a}^{a}\sin(k_{n}x_{1})\sin(k_{m}x_{1})dx_{1}={\begin{cases}0&{\rm {if}}~m\neq n\\a&{\rm {if}}~m=n\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4e6b494afdd92626e6e097520533b7303f5226a)
we have
![{\displaystyle \int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a7507d4b2ef54e1539e8202728c76e9fef29c6e)
or,
![{\displaystyle B_{m}a={\frac {4}{ak_{m}^{2}}}\sin(k_{m}a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/323c501069f8e1d10edcfda93378d0f344302d0c)
Now,
![{\displaystyle \sin(k_{m}a)=\sin \left({\frac {(2m+1)\pi }{2}}\right)=(-1)^{m}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c96da189985153ba211959b0a9be64d42f1cd513)
Therefore,
![{\displaystyle A_{m}={\frac {B_{m}}{k_{m}\cosh(k_{m}b)}}={\frac {(-1)^{m}32a^{2}}{(2m+1)^{3}\pi ^{3}\cosh(k_{m}b)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df839c5d77feec13eea41e48b545cf00331f75a1)
The warping function is
![{\displaystyle \psi =x_{1}x_{2}-{\frac {32a^{2}}{\pi ^{3}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}{(2n+1)^{3}\cosh(k_{n}b)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/685d51c33b4a6c54be9b1670cd91d5d3c8d0b7a5)
The torsion constant and the stresses can be calculated from
.