Torsion

• Circular Cylinder.
• Centroidal axis thru the center of each c.s.
• Length ${\displaystyle L}$, Outer radius ${\displaystyle c}$.
• Applied torque ${\displaystyle T}$.
• Angle of twist ${\displaystyle \phi }$.

• Each c.s. remains plane and undistorted.
• Each c.s. rotates through the same angle.
• No warping or change in shape.
• Amount of displacement of each c.s. is proportional to distance from end.

• Shear strains in the cylinder (${\displaystyle \gamma }$).
• Shear stress in the cylinder (${\displaystyle \tau }$).
• Relation between torque (${\displaystyle T}$) and angle of twist (${\displaystyle \phi }$).
• Relation between torque (${\displaystyle T}$) and shear stress (${\displaystyle \tau }$).

If ${\displaystyle \gamma }$ is small, then

${\displaystyle {\text{(1)}}\qquad L\gamma =r\phi ~~\Rightarrow ~~{\gamma ={\frac {r\phi }{L}}}}$

Therefore,

${\displaystyle {\text{(2)}}\qquad \gamma _{\text{max}}={\frac {c\phi }{L}}~~\Rightarrow ~~\gamma ={\frac {r}{\gamma }}_{\text{max}}}$

If the deformation is elastic,

${\displaystyle {\text{(3)}}\qquad \tau =G\gamma ~~\Rightarrow ~~{\tau ={\frac {r\phi G}{L}}}}$

Therefore,

${\displaystyle {\text{(4)}}\qquad \tau _{\text{max}}={\frac {r\phi G}{L}}~~\Rightarrow ~~\tau ={\frac {r}{\tau }}_{\text{max}}}$

The torque on each c.s. is given by

${\displaystyle {\text{(5)}}\qquad T=\int _{A}\tau rdA={\frac {\phi G}{L}}\int _{A}r^{2}dA={\frac {G\phi J}{L}}}$

where ${\displaystyle J}$ is the polar moment of inertia of the c.s.

${\displaystyle {\text{(6)}}\qquad J={\begin{cases}{\frac {1}{2}}\pi c^{4}&{\text{solid circular c.s.}}\\{\frac {1}{2}}\pi (c_{2}^{~4}-c_{1}^{~4})&{\text{ annular circular c.s.}}\end{cases}}}$

Therefore,

${\displaystyle {\text{(7)}}\qquad {\tau ={\frac {Tr}{J}}}~~\Rightarrow \tau _{\text{max}}={\frac {Tc}{J}}}$

and

${\displaystyle {\text{(8)}}\qquad {\phi ={\frac {TL}{JG}}}}$

• Solution first found by St. Venant.
• Tractions at the ends are statically equivalent to equal and opposite torques ${\displaystyle \pm \mathbf {T} =\pm T{\widehat {\mathbf {e} }}{3}}$.
• Lateral surfaces are traction-free.

• An axis passes through the center of twist (${\displaystyle x_{3}}$ axis).
• Each c.s. projection on to the ${\displaystyle x_{1}-x_{2}}$ plane rotates,but remains undistorted.
• The rotation of each c.s. (${\displaystyle \phi }$) is proportional to ${\displaystyle x_{3}}$.  :${\displaystyle {\text{(9)}}\qquad \phi =\alpha x_{3}}$ where ${\displaystyle \alpha }$ is the twist per unit length.
• The out-of-plane distortion (warping) is the same for each c.s. and is proportional to ${\displaystyle \alpha }$.

• Torsional rigidity (${\displaystyle T/\alpha }$).
• Maximum shear stress.

${\displaystyle {\begin{aligned}u_{1}&=r\cos(\phi +\theta )-r\cos \theta =x_{1}(\cos \phi -1)-x_{2}\sin \phi \\u_{2}&=r\sin(\phi +\theta )-r\sin \theta =x_{1}\sin \phi +x_{2}(\cos \phi -1)\\u_{3}&=\alpha \psi (x_{1},x_{2})\end{aligned}}}$

where ${\displaystyle \psi (x_{1},x_{2})}$ is the { warping function}.\\ If ${\displaystyle \phi =\alpha x_{3}<<1}$ (small strain),

${\displaystyle {\text{(10)}}\qquad {u_{1}\approx -\alpha x_{2}x_{3}~;~~u_{2}\approx \alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}}$

${\displaystyle \varepsilon _{ij}={\frac {1}{2}}\left(u_{i,j}+u_{j,i}\right)}$

Therefore,

${\displaystyle {\begin{aligned}\varepsilon _{11}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{22}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{33}&={\frac {1}{2}}\left(0+0\right)=0\\\varepsilon _{kk}&=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0\\\varepsilon _{12}&={\frac {1}{2}}\left(-\alpha x_{3}+\alpha x_{3}\right)=0\\\varepsilon _{23}&={\frac {1}{2}}\left(\alpha \psi _{,2}+\alpha x_{1}\right){\text{(11)}}\qquad \\\varepsilon _{31}&={\frac {1}{2}}\left(\alpha \psi _{,1}-\alpha x_{2}\right){\text{(12)}}\qquad \end{aligned}}}$

${\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}}$

Therefore,

${\displaystyle {\begin{aligned}\sigma _{11}&=0\\\sigma _{22}&=0\\\sigma _{33}&=0\\\sigma _{kk}&=0\\\sigma _{12}&=0\\\sigma _{23}&=\mu \alpha (\psi _{,2}+x_{1}){\text{(13)}}\qquad \\\sigma _{31}&=\mu \alpha (\psi _{,1}-x_{1}){\text{(14)}}\qquad \end{aligned}}}$

${\displaystyle \sigma _{ji,j}=0~~~~{\text{no body forces.}}}$

Therefore,

${\displaystyle {\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+\sigma _{31,3}=0&\Rightarrow ~~0=0\\\sigma _{12,1}+\sigma _{22,2}+\sigma _{32,3}=0&\Rightarrow ~~0=0\\\sigma _{13,1}+\sigma _{23,2}+\sigma _{33,3}=0&\Rightarrow ~~\mu \alpha (\psi _{,11}+\psi _{,22})=\mu \alpha \nabla ^{2}{\psi }=0{\text{(15)}}\qquad \end{aligned}}}$

• Normal to cross sections is ${\displaystyle {\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{3}}$.
• Normal traction ${\displaystyle t^{n}=\mathbf {t} \bullet {\widehat {\mathbf {n} }}{}=0}$.
• Projected shear traction is ${\displaystyle t^{s}={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}}$.
• 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 ${\displaystyle \partial S}$.

We parameterize the boundary curve ${\displaystyle \partial S}$ using

${\displaystyle \mathbf {x} ={\tilde {\mathbf {x} }}(s)~,~~0\leq s\leq l~~;~~~{\tilde {\mathbf {x} }}(0)={\tilde {\mathbf {x} }}(l)}$

The tangent vector to ${\displaystyle s}$ is

${\displaystyle {\widehat {\boldsymbol {\nu }}}={\frac {d\mathbf {x} }{ds}}~{\text{and}}~~{\widehat {\mathbf {n} }}{}={\widehat {\boldsymbol {\nu }}}\times {\widehat {\mathbf {e} }}_{3}~~\Rightarrow ~~~{\widehat {\mathbf {n} }}{}={\frac {dx_{2}}{ds}}{\widehat {\mathbf {e} }}{1}-{\frac {dx_{1}}{ds}}{\widehat {\mathbf {e} }}_{2}}$

The tractions ${\displaystyle t_{1}}$ and ${\displaystyle t_{2}}$ on the lateral surface are identically zero. However, to satisfy the BC ${\displaystyle t_{3}=0}$, we need

${\displaystyle t_{3}=n_{1}\sigma _{13}+n_{2}\sigma _{23}=0~~\Rightarrow ~~~\left(\psi _{,1}-x_{2}\right)n_{1}+\left(\psi _{,2}+x_{1}\right)n_{2}=0}$

or,

${\displaystyle {\text{(16)}}\qquad \left(\psi _{,1}-x_{2}\right){\frac {dx_{2}}{ds}}+\left(\psi _{,2}+x_{1}\right){\frac {dx_{1}}{ds}}=0}$

The traction distribution is statically equivalent to the torque ${\displaystyle \mathbf {T} }$. At ${\displaystyle x_{3}=L}$,

${\displaystyle t_{1}=\sigma _{13}~;~~t_{2}=\sigma _{23}~;~~t_{3}=\sigma _{33}=0}$

Therefore,

${\displaystyle F_{1}=\int _{S}\sigma _{13}~dS=\mu \alpha \int _{S}(\psi _{,1}-x_{2})~dS}$

From equilibrium,

${\displaystyle {\begin{aligned}\nabla ^{2}{\psi }=0~~\Rightarrow ~~~\psi _{,1}-x_{2}&=(\psi _{,1}-x_{2})+x_{1}(\psi _{,11}+\psi _{,22})\\&=\psi _{,1}+x_{1}\psi _{,11}-x_{2}+x_{1}\psi _{,22}\\&=(x_{1}\psi _{,1}-x_{1}x_{2})_{,1}+(x_{1}\psi _{,2}+x_{1}x_{1})_{,2}\\&=\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}\end{aligned}}}$

Hence,

${\displaystyle {\text{(17)}}\qquad F_{1}=\mu \alpha \int _{S}\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}dS}$

If ${\displaystyle P=f(x_{1},x_{2})}$ and ${\displaystyle Q=q(x_{1},x_{2})}$ then

${\displaystyle \int _{S}(Q_{,1}-P_{,2})dS=\oint _{\partial S}(Pdx_{1}+Qdx_{2})}$

with the integration direction such that ${\displaystyle S}$ is to the left.

Applying the Green-Riemann theorem to equation (17), and using equation (16)

${\displaystyle {\text{(18)}}\qquad F_{1}=\mu \alpha \oint _{\partial S}-x_{1}(\psi _{,2}+x_{1})dx_{1}+x_{1}(\psi _{,1}-x_{2})dx_{2}=0}$

Similarly, we can show that ${\displaystyle F_{2}=0}$. ${\displaystyle F_{3}=0}$ since ${\displaystyle t_{3}=0}$.

The moments about the ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ axes are also zero.

The moment about the ${\displaystyle x_{3}}$ axis is

${\displaystyle M_{3}=\int _{S}(x_{1}\sigma _{23}-x_{2}\sigma _{13})dS=\mu \alpha \int _{S}(x_{1}\psi _{,2}+x_{1}^{2}-x_{2}\psi _{1}+x_{2}^{2})dS=\mu \alpha {\tilde {J}}}$

where ${\displaystyle J}$ is the torsion constant. Since ${\displaystyle M_{3}=T}$, we have

${\displaystyle \alpha ={\frac {T}{\mu {\tilde {J}}}}}$

If ${\displaystyle \psi =0}$, then ${\displaystyle {\tilde {J}}=J}$, the polar moment of inertia.

• Find a warping function ${\displaystyle \psi (x_{1},x_{2})\,}$ 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 ${\displaystyle \psi }$ is known, we can calculate
  • The displacement field.
  • The stress field.
  • The twist per unit length.

Introduction to Elasticity