---
title: "Torsion of non-circular bars by Saint-Venant"
subject: "Mechanics of Solids"
chapter: 9
key: "semester_5/mechanics_of_solids:saint-venant-prandtl-stress-function"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/saint-venant-prandtl-stress-function
site: BCT Engineering
---

# Torsion of non-circular bars by Saint-Venant

Mechanics of Solids · Chapter 9: Torsion

- Asked in: year not given, 8 marks
- Video: [03-6: Noncircular sections in torsion](https://www.youtube.com/watch?v=ACII0rpHahU) — SEM LAB TV
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/saint-venant-prandtl-stress-function
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-9

## Question

Explain Saint-Venant's semi-inverse method for the torsion of a prismatic bar of non-circular section. Introduce Prandtl's stress function and show that it satisfies $\nabla^2\phi = -2G\theta$ with $\phi = 0$ on the boundary and that the torque is $T = 2\iint \phi\,dA$. Verify the result for a solid circular shaft.

## Short answer

Saint-Venant’s method assumes each section rotates rigidly and warps as w=θ ψ(x,y). Prandtl introduced φ with τ_xz=∂φ/∂y, τ_yz=−∂φ/∂x. Equilibrium gives ∇²φ=−2Gθ, boundary φ=0, torque T=2∬φ dA. For a solid circle, φ=Gθ(R²−r²)/2 yields τ=Gθr and T=GJθ.

## Answer

#### 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 $J = \pi R^4/2$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In Saint-Venant's semi-inverse method
   A. All stresses are assumed and displacements derived
   B. Features of the displacements are assumed, then equations of elasticity and boundary conditions are checked
   C. Only boundary conditions are used
   D. A numerical solution is guessed
   Answer: B. Rigid rotation theta z plus a warping w = theta psi(x,y) is assumed and then verified to satisfy all equations. (Key point: Semi-inverse method)

2. Prandtl's function with tau_xz = dphi/dy and tau_yz = -dphi/dx
   A. Satisfies equilibrium identically
   B. Satisfies compatibility identically
   C. Gives zero torque
   D. Is valid only for circular shafts
   Answer: A. d(tau_xz)/dx + d(tau_yz)/dy = 0 is automatically satisfied; compatibility then gives the Poisson equation. (Key point: Stress function and equilibrium)

3. Eliminating warping gives
   A. Laplacian phi = 0 with phi = 0 on boundary
   B. Laplacian phi = -2G theta with phi = constant (zero) on boundary
   C. Laplacian phi = 2G theta with free boundary
   D. Laplacian phi = -T
   Answer: B. Compatibility gives -2G theta; a stress-free lateral surface requires shear tangent to the boundary, so phi is constant (zero for solid sections). (Key point: Governing equation and boundary condition)

4. The torque is T = 2 ∬ phi dA. For a circular shaft with phi = (G theta/2)(R^2 - x^2 - y^2) this gives
   A. J = pi R^4/4
   B. J = pi R^4/2
   C. J = pi R^2
   D. J = 2 pi R^4
   Answer: B. Integration gives T = G theta pi R^4/2, the familiar J = pi R^4/2. (Key point: Torque and circular check)

## Related in Mechanics of Solids

- [Elastic-plastic torsion of a circular shaft](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aplastic-torsion-circular-shaft) — Mechanics of Solids, chapter 12
- [Prandtl membrane analogy and narrow sections](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aprandtl-membrane-analogy-notes) — Mechanics of Solids, chapter 9
- [Torsion of a thin open channel section](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aopen-channel-torsion-numerical) — Mechanics of Solids, chapter 9
- [Stress in a crane hook section](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Acrane-hook-curved-beam-stress) — Mechanics of Solids, chapter 6
- [Stress in a section with product of inertia](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aunsymmetrical-section-product-inertia) — Mechanics of Solids, chapter 7
- [Torsion of a rectangular hollow thin-wall box](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Athin-walled-box-torsion-numerical) — Mechanics of Solids, chapter 9

## Related in other subjects

- [Stress concentration and Saint-Venant's principle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Astress-concentration-saint-venant) — Strength of Materials, chapter 3
- [Torsion vs bending; hollow shaft replaced by solid](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-to-solid-shaft-replacement) — Strength of Materials, chapter 6
- [Plastic torque of solid shaft is 4/3](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplastic-torque-solid-shaft-4-3) — Strength of Materials, chapter 6
- [Saint-Venant, second-order explicit scheme](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asaint-venant-second-order-explicit) — Computational Techniques in Civil Engineering, chapter 5, asked 2×