Torsion of a noncircular cylinder
|
- Solution first found by St. Venant.
- Tractions at the ends are statically equivalent to equal and opposite torques
.
- Lateral surfaces are traction-free.
- An axis passes through the center of twist (
axis).
- Each c.s. projection on to the
plane rotates,but remains undistorted.
- The rotation of each c.s. (
) is proportional to
.
![{\displaystyle \phi =\alpha x_{3}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ce9c99884f5a40371d144b839341b73649630f4)
where
is the twist per unit length.
- The out-of-plane distortion (warping) is the same for each c.s. and is proportional to
.
- Torsional rigidity (
).
- Maximum shear stress.
![{\displaystyle {\begin{aligned}u_{1}&=r\cos(\phi +\theta )-r\cos \theta =x_{1}(\cos \phi -1)-x_{2}\sin \phi \\u_{2}&=r\sin(\phi +\theta )-r\sin \theta =x_{1}\sin \phi +x_{2}(\cos \phi -1)\\u_{3}&=\alpha \psi (x_{1},x_{2})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88e50107b49483616a846382d0d6f3aad2fac0b8)
where
is the warping function.
If
(small strain),
![{\displaystyle {\text{(10)}}\qquad {u_{1}\approx -\alpha x_{2}x_{3}~;~~u_{2}\approx \alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48a2dd84bf65d80c63ed0e759f1b415b0773f313)
![{\displaystyle \varepsilon _{ij}={\frac {1}{2}}\left(u_{i,j}+u_{j,i}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8690ec58acfd48f4422527025faad17d1160bc38)
Therefore,
![{\displaystyle {\begin{aligned}\varepsilon _{11}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{22}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{33}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{kk}&=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0\\\varepsilon _{12}&={\frac {1}{2}}\left(-\alpha x_{3}+\alpha x_{3}\right)=0\\\varepsilon _{23}&={\frac {1}{2}}\left(\alpha \psi _{,2}+\alpha x_{1}\right){\text{(11)}}\qquad \\\varepsilon _{31}&={\frac {1}{2}}\left(\alpha \psi _{,1}-\alpha x_{2}\right){\text{(12)}}\qquad \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cb46e5e09e326ff3db4ec206c3c7e1dde5c47da)
![{\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b0e8f291dbb233890c824dcc1ddb5d0a921847f)
Therefore,
![{\displaystyle {\begin{aligned}\sigma _{11}&=0\\\sigma _{22}&=0\\\sigma _{33}&=0\\\sigma _{kk}&=0\\\sigma _{12}&=0\\\sigma _{23}&=\mu \alpha (\psi _{,2}+x_{1}){\text{(13)}}\qquad \\\sigma _{31}&=\mu \alpha (\psi _{,1}-x_{1}){\text{(14)}}\qquad \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05b34d7091e67b35e6358d40cc751298f904f316)
![{\displaystyle \sigma _{ji,j}=0~~~~{\text{no body forces.}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e72156a107cd7db8ae8b84b9f8fc99d40101b50)
Therefore,
![{\displaystyle {\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+\sigma _{31,3}=0&\Rightarrow ~~0=0\\\sigma _{12,1}+\sigma _{22,2}+\sigma _{32,3}=0&\Rightarrow ~~0=0\\\sigma _{13,1}+\sigma _{23,2}+\sigma _{33,3}=0&\Rightarrow ~~\mu \alpha (\psi _{,11}+\psi _{,22})=\mu \alpha \nabla ^{2}{\psi }=0{\text{(15)}}\qquad \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae4a116057a8ccc141ad3f42bb724aa923cfbd46)
- Normal to cross sections is
.
- Normal traction
.
- Projected shear traction is
.
- Traction vector at a point in the cross section is tangent to the cross section.
- Lateral surface traction-free.
- Unit normal to lateral surface appears as an in-plane unit normal to the boundary
.
We parameterize the boundary curve
using
![{\displaystyle \mathbf {x} ={\tilde {\mathbf {x} }}(s)~,~~0\leq s\leq l~~;~~~{\tilde {\mathbf {x} }}(0)={\tilde {\mathbf {x} }}(l)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/919f0ce2ecd82fe5ac0c18c1a21366b2cf8d4ca2)
The tangent vector to
is
![{\displaystyle {\widehat {\boldsymbol {\nu }}}={\frac {d\mathbf {x} }{ds}}~{\text{and}}~~{\widehat {\mathbf {n} }}{}={\widehat {\boldsymbol {\nu }}}\times {\widehat {\mathbf {e} }}_{3}~~\Rightarrow ~~~{\widehat {\mathbf {n} }}{}={\frac {dx_{2}}{ds}}{\widehat {\mathbf {e} }}{1}-{\frac {dx_{1}}{ds}}{\widehat {\mathbf {e} }}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/666a63423d718abfaaf2295eb6b6a2615d1bfde3)
The tractions
and
on the lateral surface are identically zero.
However, to satisfy the BC
, we need
![{\displaystyle t_{3}=n_{1}\sigma _{13}+n_{2}\sigma _{23}=0~~\Rightarrow ~~~\left(\psi _{,1}-x_{2}\right)n_{1}+\left(\psi _{,2}+x_{1}\right)n_{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b91f12c15d03216d7f04277208c1aadda14a06a)
or,
![{\displaystyle {\text{(16)}}\qquad \left(\psi _{,1}-x_{2}\right){\frac {dx_{2}}{ds}}+\left(\psi _{,2}+x_{1}\right){\frac {dx_{1}}{ds}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6fbc487f6da95ec45cf8f2a907219b00d28cb62)
The traction distribution is statically equivalent to the torque
.
At
,
![{\displaystyle t_{1}=\sigma _{13}~;~~t_{2}=\sigma _{23}~;~~t_{3}=\sigma _{33}=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82cf6c757048851ce3782809d95ef88e65ab9cad)
Therefore,
![{\displaystyle F_{1}=\int _{S}\sigma _{13}~dS=\mu \alpha \int _{S}(\psi _{,1}-x_{2})~dS}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bf7d44552b4eb7bb3a87726ed067225b73663fa)
From equilibrium,
![{\displaystyle {\begin{aligned}\nabla ^{2}{\psi }=0~~\Rightarrow ~~~\psi _{,1}-x_{2}&=(\psi _{,1}-x_{2})+x_{1}(\psi _{,11}+\psi _{,22})\\&=\psi _{,1}+x_{1}\psi _{,11}-x_{2}+x_{1}\psi _{,22}\\&=(x_{1}\psi _{,1}-x_{1}x_{2})_{,1}+(x_{1}\psi _{,2}+x_{1}x_{1})_{,2}\\&=\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8bed34efacacf430e325550e03911dbd435c5ad)
Hence,
![{\displaystyle {\text{(17)}}\qquad F_{1}=\mu \alpha \int _{S}\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}dS}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c1c14a586337ac010b1fbfc331de835c896668c)
If
and
then
![{\displaystyle \int _{S}(Q_{,1}-P_{,2})dS=\oint _{\partial S}(Pdx_{1}+Qdx_{2})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d28110ab07043b5dfe90da09020c3f04e2dfe251)
with the integration direction such that
is to the left.
Applying the Green-Riemann theorem to equation (17), and using
equation (16)
![{\displaystyle {\text{(18)}}\qquad F_{1}=\mu \alpha \oint _{\partial S}-x_{1}(\psi _{,2}+x_{1})dx_{1}+x_{1}(\psi _{,1}-x_{2})dx_{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eeb97f1a05daa6061145a43915f6d5c5c2dd57e6)
Similarly, we can show that
.
since
.
The moments about the
and
axes are also zero.
The moment about the
axis is
![{\displaystyle M_{3}=\int _{S}(x_{1}\sigma _{23}-x_{2}\sigma _{13})dS=\mu \alpha \int _{S}(x_{1}\psi _{,2}+x_{1}^{2}-x_{2}\psi _{1}+x_{2}^{2})dS=\mu \alpha {\tilde {J}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e47565dd5255dd2c4aba26e9a7a5ca8538e64edd)
where
is the torsion constant. Since
, we have
![{\displaystyle \alpha ={\frac {T}{\mu {\tilde {J}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43958144770a659e26d7aab9d6fc6e1f5567bfda)
If
, then
, the polar moment of inertia.
- Find a warping function
that is harmonic. and satisfies the traction BCs.
- Compatibility is not an issue since we start with displacements.
- The problem is independent of applied torque and the material properties of the cylinder.
- So it is just a geometrical problem. Once
is known, we can calculate
- The displacement field.
- The stress field.
- The twist per unit length.