Torsion

Torsion of Circular Cylinders

Torsion of a cylinder with a circular cross section

About the problem:

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

Assumptions:

• 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.

Find:

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

Solution:

If $\gamma$ is small, then

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

Therefore,

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

If the deformation is elastic,

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

Therefore,

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

The torque on each c.s. is given by

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

where $J$ is the polar moment of inertia of the c.s.

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

Therefore,

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

and

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

Torsion of Non-Circular Cylinders

Torsion of a noncircular cylinder

About the problem

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

Assumptions:

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

Find:

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

Solution:

Displacements

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

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

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

Strains

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

Therefore,

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

Stresses

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

Therefore,

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

Equilibrium

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

Therefore,

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

Internal Tractions

• Normal to cross sections is ${\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{3}$.
• Normal traction $t^{n}=\mathbf {t} \bullet {\widehat {\mathbf {n} }}{}=0$.
• Projected shear traction is $t^{s}={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}$.
• Traction vector at a point in the cross section is { tangent} to the cross section.

Boundary Conditions on Lateral Surfaces

• Lateral surface traction-free.
• Unit normal to lateral surface appears as an in-plane unit normal to the boundary $\partial S$.

We parameterize the boundary curve $\partial S$ using

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

The tangent vector to $s$ is

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

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

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

or,

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

Boundary Conditions on End Surfaces

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

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

Therefore,

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

From equilibrium,

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

Hence,

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

The Green-Riemann Theorem

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

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

with the integration direction such that $S$ is to the left.

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

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

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

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

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

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

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

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

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

The Torsion Problem Summarized

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

Related Content

Introduction to Elasticity