---
title: "Prandtl membrane analogy and narrow sections"
subject: "Mechanics of Solids"
chapter: 9
key: "semester_5/mechanics_of_solids:prandtl-membrane-analogy-notes"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/prandtl-membrane-analogy-notes
site: BCT Engineering
---

# Prandtl membrane analogy and narrow sections

Mechanics of Solids · Chapter 9: Torsion

- Asked in: year not given, 5 marks
- Video: [Module 6: Membrane analogy & equilibrium equation](https://www.youtube.com/watch?v=kdxD5zndqJQ) — jeevan ealias
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/prandtl-membrane-analogy-notes
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-9

## Question

Write short notes on Prandtl's elastic membrane (soap film) analogy for torsion. State the results for a narrow rectangular section and explain why open thin-walled sections are far weaker in torsion than closed sections.

## Short answer

Membrane analogy: soap-film deflection equals φ, slope equals shear stress, contour lines show stress direction, twice the volume gives torque. For a narrow rectangle, τ_max=3T/(bt²) and J=bt³/3. Open sections (J≈Σbt³/3) are weak because stiffness depends on t³, while closed sections (J=4A²/∮ds/t) depend on enclosed area squared, making them much stiffer.

## Answer

#### Membrane analogy

Prandtl observed that the equation for the torsion stress function, $\nabla^2\phi = -2G\theta$, has the same form as the equation for the small deflection $z$ of a uniformly stretched membrane under pressure $p$ and tension $S$:

$$
\nabla^2 z = -\frac{p}{S}
$$

So if a thin membrane (soap film) is stretched over a hole of the same shape as the cross-section and inflated slightly, and $p/S$ is made equal to $2G\theta$, the deflection $z$ is equal to $\phi$.

```text
   hole shaped like the section
   +--------------------+
   |   .--------.       |   membrane bulges by z
   |  /  contours \     |   (contour lines = lines of
   |  \          /      |    shear stress direction)
   |   '--------'       |
   +--------------------+
```

#### Analogy rules

| Membrane | Torsion bar |
|---|---|
| Deflection $z$ | Stress function $\phi$ |
| Slope of membrane at a point | Resultant shear stress at that point |
| Contour line (constant $z$) | Direction of shear stress at that point (stress is tangent to contour) |
| Twice the volume under membrane | Torque $T = 2\iint\phi\,dA$ |
| Steepest slope | $\tau_{max}$; occurs at the boundary closest to the centre, not at corners (for re-entrant corners there is a stress concentration; at convex corners the stress is zero) |

The analogy gives the stress picture without solving the equation, and explains why the stress is zero at the sharp convex corner of a square and highest at the mid-point of the sides.

#### Narrow rectangle ($b \gg t$)

The membrane is nearly cylindrical, with a parabolic profile across the thickness $t$. Then

$$
\tau_{max} = \frac{3T}{b\,t^2},\qquad \theta = \frac{3T}{G\,b\,t^3},\qquad J = \frac{b\,t^3}{3}
$$

The shear stress varies linearly across the thickness, from zero at the mid-plane to $\tau_{max}$ at the long edges.

#### Open and closed sections

An open thin-walled section (angle, channel, I) behaves as a set of narrow rectangles: $J = \tfrac13\sum b_it_i^3$. The torsional stiffness depends on $t^3$, and the shear flow circulates only within each thin strip.

A closed tube carries a shear flow around the whole wall, enclosing area $A$, with $J = 4A^2/\oint ds/t$, which depends on $t$ only to the first power and on the enclosed area squared. A closed tube is therefore typically tens to hundreds of times stiffer and gives much lower stress than an open section of the same material, which is why box or tubular sections are used where torsion is important.

## Check yourself

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

1. The membrane analogy works because
   A. Both involve circular geometry
   B. Torsion's Laplacian phi = -2G theta has the same form as the membrane's Laplacian z = -p/S
   C. Membrane tension equals torque
   D. Both are non-linear
   Answer: B. Equal equations with phi = z = 0 on the boundary, with p/S = 2G theta, make z equal to phi. (Key point: Analogy basis)

2. In the analogy, the torque corresponds to
   A. The slope of the membrane
   B. Twice the volume under the membrane
   C. The area of the hole
   D. The maximum deflection
   Answer: B. T = 2 ∬ phi dA; the slope gives shear stress and contours give the stress direction. (Key point: Interpreting the analogy)

3. For a narrow rectangle b >> t, J and tau_max are
   A. J = b t^3/3, tau_max = 3T/(b t^2)
   B. J = b t^2/3, tau_max = 3T/(b t^3)
   C. J = b t/3, tau_max = T/(b t)
   D. J = b^3 t/3, tau_max = 3T/(b^2 t)
   Answer: A. The profile is parabolic across t giving J = b t^3/3 and tau_max = T t/J = 3T/(b t^2). (Key point: Narrow rectangle results)

4. Closed tubes are far stiffer than open sections because
   A. Shear flow circulates around the whole enclosed area, J = 4A^2/∮ds/t, rather than depending on t^3
   B. They use more material
   C. Open sections have no shear stress
   D. Open sections have larger A
   Answer: A. Open J varies with t^3 and shear flow stays within each strip; closed J depends on enclosed area squared and t to the first power. (Key point: Open vs closed)

## Related in Mechanics of Solids

- [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
- [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
- [Thin closed ring under diametral compression](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aclosed-ring-diametral-load) — Mechanics of Solids, chapter 6
- [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
- [Shear flow and shear centre in thin-wall beams](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Ashear-flow-shear-centre-thin-wall) — Mechanics of Solids, chapter 10

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