Torsion of non-circular bars by Saint-Venant

Semi-inverse method

In the semi-inverse method some features of the solution are assumed (from the displacements), and then it is checked that all equations of elasticity and boundary conditions are satisfied. The solution is then unique.

Assumptions (Saint-Venant):

1. Each cross-section rotates as a rigid body in its plane through angle $\theta z$, with $\theta$ = twist per unit length.
2. The section also warps (out-of-plane displacement), the same for all sections: $w = \theta\,\psi(x,y)$, where $\psi$ is the warping function.

For a circular shaft $\psi = 0$ (no warping), which is why plane sections stay plane there.

Displacements and strains

$$u = -\theta z\,y,\qquad v = \theta z\,x,\qquad w = \theta\,\psi(x,y)$$ $$\gamma_{xz} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} = \theta\left(\frac{\partial \psi}{\partial x} - y\right),\qquad \gamma_{yz} = \theta\left(\frac{\partial \psi}{\partial y} + x\right)$$

All other strains are zero, so $\tau_{xz} = G\gamma_{xz}$, $\tau_{yz} = G\gamma_{yz}$ are the only stresses.

Prandtl stress function

With no body force, the only non-trivial equilibrium equation is $\dfrac{\partial\tau_{xz}}{\partial x} + \dfrac{\partial\tau_{yz}}{\partial y} = 0$. This is identically satisfied by

$$\tau_{xz} = \frac{\partial \phi}{\partial y},\qquad \tau_{yz} = -\frac{\partial \phi}{\partial x}$$

Compatibility: eliminate $\psi$ from the strains:

$$\frac{\partial\gamma_{xz}}{\partial y} - \frac{\partial\gamma_{yz}}{\partial x} = -2\theta \ \Rightarrow\ \frac{\partial\tau_{xz}}{\partial y} - \frac{\partial\tau_{yz}}{\partial x} = -2G\theta$$ $$\boxed{\nabla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} = -2G\theta}$$

Boundary condition: the lateral surface is stress free, so the resultant shear stress must be tangent to the boundary. This requires $d\phi/ds = 0$ along the boundary, i.e. $\phi =$ constant, taken as zero for a solid (simply connected) section.

Torque

$$\begin{aligned} T &= \iint (x\,\tau_{yz} - y\,\tau_{xz})\,dA = -\iint\left(x\frac{\partial\phi}{\partial x} + y\frac{\partial\phi}{\partial y}\right)dA \end{aligned}$$

Integrating by parts and using $\phi = 0$ on the boundary gives each term $= -\iint \phi\,dA$, so

$$\boxed{T = 2\iint \phi\,dA}$$

The torsional constant follows from $T = GJ\theta$.

Check: solid circular shaft of radius $R$

Try $\phi = \dfrac{G\theta}{2}\left(R^2 - x^2 - y^2\right)$. Then $\nabla^2\phi = \dfrac{G\theta}{2}(-2-2) = -2G\theta$ and $\phi = 0$ at $r = R$. $\checkmark$

$$\tau_{xz} = -G\theta y,\quad \tau_{yz} = G\theta x \Rightarrow \tau = G\theta r$$ $$T = 2\int_0^R \frac{G\theta}{2}(R^2 - r^2)\,2\pi r\,dr = 2\pi G\theta\left(\frac{R^4}{2} - \frac{R^4}{4}\right) = G\theta\,\frac{\pi R^4}{2} = GJ\theta$$

which is the familiar result $J = \pi R^4/2$.