Module 11 · Lesson 11.4

Non-circular sections, warping and the cost of a slit

Why circular-shaft theory fails everywhere else, and how much a single cut costs.

Why this matters

The torsion formula is exact for a circular shaft and wrong for everything else. Structural sections are almost never circular, so you need to know what replaces it, why the replacement is empirical rather than derived, and — the practical headline — just how catastrophic an open section is in torsion compared with a closed one.

By the end of this lesson you should be able to

Explain why plane sections warp in a non-circular bar, and why that breaks the derivation
Use the Saint-Venant coefficients for a solid rectangular bar
Calculate the torsion constant of a thin open section, J = (1/3)Σbt³
Compare open and closed sections quantitatively
Distinguish uniform (Saint-Venant) torsion from warping torsion

What you should already know

The torsion formula T/J = τ/r = Gθ/L (this module)
Bredt shear flow for closed thin-walled sections (this module)
Shear stress and shear strain (Module 7)

Go back to the derivation of the circular torsion formula and find the assumption that fails.

It was this: plane cross-sections remain plane. That held only because a circle looks identical from every direction about its centre. Every point on the boundary is the same distance from the axis, so there is no reason for any part of the section to move along the shaft more than any other.

A rectangle has no such symmetry. Its corners are further from the centre than the mid-sides. Twist it and the section does not stay flat: it warps out of plane, dishing in some places and bulging in others.

Two consequences follow immediately, and the second one catches people out:

1.The polar second moment J is no longer the right constant. The correct torsion constant is smaller — usually much smaller — and for most shapes cannot be written down in closed form at all.
2.The maximum shear stress is not at the corner, which is the point furthest from the centre. It is at the middle of the longest side. At the corner the stress is exactly zero, because two free surfaces meet there and neither can supply a complementary shear stress.

That second point is worth pausing on. The intuition carried over from circular shafts — furthest from the axis means highest stress — is not merely inaccurate here, it points at the one location where the stress vanishes.

Predict first

Where is the shear stress greatest in a solid rectangular bar under torsion?

For a solid rectangle the exact solution is an infinite series, so design practice uses tabulated coefficients that condense it into two numbers:

τ_max = T/(k₁ a b²) and θ = TL/(k₂ a b³ G)

with a the longer side and b the shorter. Both coefficients depend only on the aspect ratio a/b:

a/b	k₁	k₂
1.0	0.208	0.141
1.5	0.231	0.196
2.0	0.246	0.229
3.0	0.267	0.263
6.0	0.299	0.299
10.0	0.312	0.312
∞	1/3	1/3

Notice that both tend to 1/3 as the rectangle becomes a long thin strip, and that they converge on each other. That limit is the useful one, because it generalises.

Torsion constant of a thin open section

J = \frac{1}{3} \sum b t^{3}

What it calculates: the torsional stiffness of a section built from narrow rectangles

b: length of each wall element along its mid-line (mm)
t: thickness of that element (mm)

This assumes

Each element is long and thin, so b/t is large
Uniform (Saint-Venant) torsion — the ends are free to warp
The elements are joined so they twist together

In plain terms: The thickness is cubed. Halving every wall thickness divides the torsional stiffness by eight, while the area only halves. That single exponent is why open sections are so poor in torsion.

Take care

What a longitudinal slit costs

Take a thin tube of mean radius 100 mm and wall 8 mm. As a closed section its torsion constant is 4A_m²t/s = 5.03 × 10⁷ mm⁴.

Now cut it along its whole length. Nothing has been removed — the same material is still there, and its bending stiffness is barely changed. But it is now one long narrow rectangle, so J = (1/3)st³ = 1.07 × 10⁵ mm⁴.

The torsional stiffness has fallen by a factor of about 470.

The reason is physical, not algebraic. A closed section carries torque as a shear flow running continuously round the loop, with a large lever arm about the centre. Cut the loop and the flow has nowhere to go; the section is reduced to each wall twisting about its own mid-line, with a lever arm of half its own thickness.

This is why a channel or an angle is a poor choice wherever torsion is present, why box sections are used for curved bridge decks, and why a slit in a torsion member is never a small detail.

One refinement completes the picture. Everything above is uniform torsion, also called Saint-Venant torsion: the section is free to warp, warping is the same at every cross-section, and the torque is carried entirely by shear stresses circulating in the section.

If warping is restrained — at a fixed end, at a stiff connection, or wherever the torque changes along the member — the situation differs. The section wants to warp but is prevented, so direct stresses appear along the member, and their variation provides an additional torsional resistance. This is warping torsion, and the total is:

T = T_Saint-Venant + T_warping

For closed sections warping torsion is negligible, because Saint-Venant torsion is so stiff that it takes almost the whole torque. For open sections it is the reverse: Saint-Venant torsion is so weak that warping torsion often carries most of the load, and ignoring it can be badly unconservative for a member with restrained ends.

A full warping analysis is beyond this course. What matters here is knowing that it exists, and that an I-beam restrained against warping is considerably stiffer in torsion than J = (1/3)Σbt³ alone suggests.

Worked example

A rectangular bar, and the cost of slitting a tube

Given

Case A: solid rectangular steel bar, 60 mm × 20 mm, length 1.00 m, carrying T = 2.00 kNm
Case B: thin tube of mean radius 100 mm and wall thickness 8 mm, compared closed and slit
G = 79 000 N/mm²

Find

The peak stress and twist of the bar, and the ratio of closed to open torsional stiffness.

Practice

A solid rectangular steel bar 60.0 mm × 20.0 mm carries a torque of 2.00 kNm. With k₁ = 0.267 for its aspect ratio, what is the maximum shear stress, in N/mm²?

Practice

A thin open section is built from three narrow rectangles: two flanges 150 mm × 12 mm and a web 276 mm × 8 mm. What is its torsion constant J, in mm⁴?

Practice

If every wall thickness in that section were halved, by what factor would the torsion constant change?

Summary

Non-circular sections warp out of plane, which breaks the circular derivation
Peak shear stress is at the middle of the longest side; at the corners it is zero
Solid rectangle: τ = T/k₁ab² and θ = TL/k₂ab³G, with tabulated coefficients
Thin open section: J = (1/3)Σbt³ — the thickness is cubed
Closed section: J = 4A_m²t/s, hundreds of times stiffer than the same material slit open
Uniform torsion assumes free warping; restrained warping adds resistance, and matters most for open sections

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint