Rectangular Shaft in Torsion (Saint-Venant)

stressmass-cost

Twist a solid circular shaft and nothing surprises you: the shear stress grows straight out from the axis and peaks all the way around the rim (that is the torsion shaft page). Twist a solid rectangular bar and two things happen that no first-year intuition predicts — the peak shear stress sits at the midpoint of the long side, and the corners carry exactly zero stress. This page is about those two facts, the coefficients that quantify them, and why a rectangular shaft is an inefficient way to carry torque.

Convention (fixed for the whole page): $a$ is the long side, $b$ the short side, so the aspect ratio $a/b \ge 1$. (Sources label these differently — some swap $a$ and $b$, some use $2a\times 2b$ half-sides — so pin it once and read everything below against it.)

$\tau_{max} = \frac{T}{c_1\,a\,b^2}, \qquad \theta' = \frac{T}{c_2\,a\,b^3\,G}$

Here $\tau_{max}$ is the peak shear stress (at the long-side midpoint), $\theta'$ is the twist per unit length, and $c_1(a/b)$, $c_2(a/b)$ are two dimensionless coefficients that depend only on the shape ratio $a/b$, not on the absolute size. They come from the classical Saint-Venant solution and are read from a cited table (below).

Why the corners are dead and the long-side middle is hottest

The clean way to see it is Prandtl’s membrane analogy (Timoshenko & Goodier §107): imagine a soap film stretched over a hole the shape of the cross-section and pushed up by a uniform pressure. The height of that film is the torsion stress function, and the shear stress at any point is the slope of the film, running along its contour lines. Two consequences fall straight out:

• At a convex corner the film is pinned to zero along both edges meeting there, so it flattens and its slope — the stress — goes to zero. The corners of a twisted rectangular bar are unloaded.
• The film is steepest where the boundary comes closest to the bulging centre, which is the midpoint of the long side. That is where $\tau_{max}$ lives, not at the corners and not on the short side (whose midpoint sees a smaller peak).

Drive the aspect ratio in the widget and watch the shear humps in the cross-section: the long-side hump is always the tallest, the corners always dead.

One material property, and the efficiency trap

Look at what each material property touches. $G$ (the shear modulus) sets the twist $\theta'$ and nothing else; $\sigma_y$ sets the margin; and $\tau_{max}$ — geometry and torque alone — is blind to the material. Swap steel for aluminium ($G \approx 79 \to 26$ GPa) and the same bar twists about three times as much per metre at an identical peak stress. That is the same stiffness-versus-strength split the round torsion shaft makes.

Now the comparison that makes the page’s point. Replace the rectangle with a solid round shaft of the same cross-sectional area (same material, same weight per metre): $\pi r_{eq}^2 = a\,b$. That round shaft carries the same torque at a lower peak stress and a smaller twist — the rectangle loses on both counts, and loses by more as it gets flatter. A square section carries about 36 % more peak stress than the equal-area circle; a 2 : 1 rectangle, about 62 % more. The corners are dead weight and the flats are underworked: that is why round shafts win, and why a square shaft is a bad deal. The widget shows the equal-area circle as a dashed overlay with its own stress readout.

The coefficients, and where they stop

$c_1$ and $c_2$ are cited data: the classical Saint-Venant coefficients as tabulated by Timoshenko & Goodier (§109). The widget looks them up by $a/b$ and interpolates linearly between the published rows. As the section flattens toward an infinitely thin strip both coefficients climb to the same limit,

$c_1 = c_2 = \tfrac{1}{3} \qquad (a/b \to \infty),$

which is the exact membrane solution for an infinite strip — the consistency check the build’s physics test pins the whole table against. Two boundaries are enforced, each a global refusal rather than a fabricated number:

• $a/b < 1$ — you have labelled the shorter dimension as $a$. Swap them; $a$ is the long side.
• $a/b > 10$ — the published table stops at $10$, and interpolating past the last row would invent a coefficient nobody measured. Beyond that, use the thin-strip limit $1/3$ above.

Every authored coefficient is cross-checked two independent ways — against the exact Fourier-series solution and against Roark’s separately published closed form — and the results are on the /verification/ page, which states exactly what is machine-proven here (the lookup, the interpolation, the round-shaft algebra, the two refusals) and what rests on citation (the tabulated coefficients themselves). For the thin-walled cousin of this problem — where the section is a closed tube and the shear runs uniformly around the wall — see thin-tube torsion.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget answers one narrow, elastic question — given this rectangular section and this torque, what is the peak shear stress and the twist? — and reports a static shear-yield margin on that peak. Most of the ways a bar like this actually gets into trouble live in what the page deliberately does not model.

• Free warping is assumed — restrain it and the whole picture changes. The Saint-Venant solution behind $c_1$ and $c_2$ lets every cross-section warp freely out of its plane (rectangles do not stay plane under torsion the way circles do). Clamp an end, or load the bar so warping is prevented, and you get warping normal stresses and a stiffer response that this page has no term for. The effect is mild for a chunky rectangle but dominant for thin open sections (a channel, an angle, an I-beam in torsion), which is exactly why closed thin-walled tubes — see thin-tube torsion — are so much better at carrying torque than open ones of the same weight.

• Thin sections are torsionally awful, and the $b^3$ tells you so. Both the stiffness and the strength ride on the short side cubed ($\theta' \propto 1/b^3$) or squared ($\tau_{max}\propto 1/b^2$). Halve the thickness (holding $a$) and the twist rate jumps by roughly eightfold — exactly eightfold in the thin-strip limit, where $c_2$ stops changing. As $a/b\to\infty$ the coefficients saturate at $1/3$, so a long thin strip is essentially a membrane with almost no torsional rigidity — the reason sheet metal is stiff in bending but hopeless in twist.

• Rectangular- and square-wire springs are this problem, coiled. A helical spring wound from rectangular (or square) wire carries its load as torsion of a rectangular bar wrapped into a helix, so its stress and its rate are governed by exactly these $c_1$, $c_2$ coefficients rather than the round-wire formulas on the helical spring page. Designers reach for rectangular wire precisely because it packs more material into a given coil envelope — but the sharp corners are stress raisers under the superposed curvature, and the flats warp, so the round-wire intuition does not transfer cleanly.

• The safety factor is a static shear-yield illustration, not a fatigue life. $\mathrm{SF} = (\sigma_y/2)/\tau_{max}$ uses the maximum-shear-stress (Tresca) criterion, the same conservative shear yield the round torsion shaft uses. Real torsion members usually fail by fatigue, and they crack at the long-side midpoints — the very spots this page marks hottest — cycle after cycle. A comfortable static $\mathrm{SF}$ here says nothing about cycles survived; that is a later phase built on top of this peak stress.

• A “rectangle” with a sharp re-entrant feature is a different problem. The corners of a plain rectangle are dead in elastic torsion — but cut a keyway, a spline, or an internal notch and that concave corner becomes a stress concentration, the opposite of the unloaded convex corner. That is the stepped-shaft-fillet story ($K_t$ from a cited chart) transplanted onto a torsion member, and it is not included here.

• One torque, one prismatic bar, linear-elastic throughout. No bending combined with the torsion (that superposition is the combined shaft page), no taper along the length, no local plastic redistribution once the long-side midpoint yields, and manufactured corner radii are idealised as perfectly sharp. Read the page as where the stress goes and why the round shaft wins, not as a finished design margin.

Sources

• Timoshenko, S. P., & Goodier, J. N., Theory of Elasticity, 3rd ed., McGraw-Hill, 1970 — §107 (membrane analogy) and §108-109 (torsion of a bar of rectangular cross section: the coefficient table for maximum stress and torsional rigidity, and the thin-strip limit 1/3).
• Young, W. C., & Budynas, R. G., Roark's Formulas for Stress and Strain, 7th ed., McGraw-Hill, 2002, Ch. 10 (Table 10.7 in the 8th ed.) — solid rectangular section in torsion: the closed forms K = a·b^3·[1/3 − 0.21(b/a)(1 − b^4/12a^4)] and tau_max at the midpoint of the longer side.
• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3-3.5 (circular shafts in torsion: tau = 16T/(πd^3), theta = TL/(GJ), J = πd^4/32), used here for the equal-area round-shaft comparison.
• Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §5-4 (maximum-shear-stress theory, S_sy = 0.5 S_y).