Torsion of Non-Circular Sections Calculator

Q: Why can't I just use J = I_x + I_y for a non-circular section?

Because those are two different quantities that share a letter. I_x + I_y is the POLAR SECOND MOMENT, and it is correct as a geometric property of any shape — it is ∫r²dA. The J in θ = TL/(GJ) is the TORSION CONSTANT, which comes from solving the St Venant torsion problem for the actual cross-section. They are equal only for a circular section, because only a circle's cross-section stays plane when twisted. For a solid square the torsion constant is about 52 per cent of the polar value; for a thin-walled I-section it can be a few hundred times smaller. Using the polar value would tell you an I-beam is hundreds of times stiffer in torsion than it is.

Q: Where is the peak shear stress on a rectangular bar?

At the midpoint of the LONG side, and the corners carry no shear at all. That surprises people who expect the corners to be worst because they are furthest from the centre, which is the intuition a round bar teaches. The reason is a boundary condition: shear stress at a free surface must be parallel to that surface, and at a corner two perpendicular surfaces each demand their own direction, so the only value consistent with both is zero. The stress is also higher on the long side than the short one, which is the opposite of the round-bar intuition.

Q: How much does a slit really cost?

For a thin circular tube, exactly 3(R/t)² in torsional stiffness. At R/t = 10 that is 300 times; at R/t = 20 it is 1,200; at R/t = 50 it is 7,500. The metal removed is negligible. The peak stress rises much less — by 3(R/t), so 60 times at R/t = 20 — which is exactly why the slit is dangerous: the stress check may still pass. If you need an opening, put it where it does not break the loop, or fit a closing plate, or accept that the member carries no torque.

Q: Is the enclosed area the outside area or the bore area?

Neither. Bredt–Batho uses the area enclosed by the MID-WALL line — the line running through the middle of the wall thickness all the way round. For a 100 × 60 box with a 3 mm wall that is 97 × 57 = 5,529 mm², against 6,000 for the outside and 5,076 for the bore. Using the outside dimensions understates the stress by eight per cent on that section and by more on a thicker-walled one, always in the unsafe direction.

Q: Why is the thinnest part of the wall the most stressed?

Because the shear FLOW, q = τt, is what is constant round a closed section, not the shear stress. Nothing can accumulate along the wall, so q has the same value everywhere, and therefore τ = q/t is largest where t is smallest. It follows that a box with one wall thinner than the others fails at that wall regardless of where the torque is applied, and that thickening the already-thick walls achieves nothing. This page assumes a uniform wall; for a section with mixed thicknesses the flow is still constant and the stress is q divided by each local thickness.

Q: What about warping restraint? Isn't my I-section stiffer than this says?

Very probably, yes, and this page says so rather than pretending otherwise. The ⅓Σbt³ value is pure St Venant torsion, which assumes the ends are free to warp out of plane. If they are held — welded to a stiff plate, built into a frame, or simply short enough that the ends dominate — the flanges resist by bending in their own planes, which is a far stiffer mechanism. Warping torsion needs the sectorial warping constant I_w and a hyperbolic solution rather than a linear one. It is not computed here, so the twist this page gives for an open section is an upper bound.

Q: Can this handle a box with an internal web?

No, and that is a refusal rather than an omission. A multi-cell section has a different shear flow in each cell, and finding them requires imposing that every cell twists at the same rate — one compatibility equation per cell, solved simultaneously. That is a linear system rather than a formula and it is outside what this page can honestly express. As a rough guide a single internal web adds relatively little to a box's torsional stiffness (the outer loop already does the work) while adding a great deal to its bending stiffness and its resistance to wall buckling, which is usually why it is there.

α and β for a solid rectangle, Bredt–Batho for a closed thin wall, ⅓Σbt³ for an open one — and the same outline computed both ways, because an open section can be a thousand times less stiff in torsion than the closed one it came from.

Last updated 24 Sep 2026

Torsion of non-circular sections

Section, torque, length → shear stress, twist and J

SectionThe round bar is the only easy case and it is first on the list for reference. Everything below it needs a different formula, and the pairs — closed tube against slit tube, closed box against slit box — are the comparison this page exists to make.

DEPTH h (mm) — the outside diameter for a round section, the LONG side of a rectangleFor a solid rectangle the page sorts the two sides for you, so it does not matter which you call depth and which width — the aspect ratio is what the coefficients depend on.

WIDTH b (mm) — ignored for a round sectionThe short side of a rectangle, the width of a box, the flange width of an I or a channel.

Wall or flange THICKNESS t (mm)The single most powerful number on the page for a thin section. A CLOSED section’s torsional stiffness is proportional to t — halve the wall and you halve the stiffness. An OPEN section’s goes as t³ — halve the wall and you lose seven eighths of it.

WEB thickness (mm) — the I-section and channel onlyThe upright of the I or the back of the channel. In an open section every flat strip contributes ⅓·b·t³ independently, so the thickest strip dominates.

Applied TORQUE T (N·m)The twisting moment. On a drive shaft it is power divided by angular speed; on a frame it is whatever asymmetric load is trying to rack the structure.

Length over which the twist is measured L (mm)The twist angle is proportional to length; the peak stress is not. A short stub and a long tube of the same section see the same shear stress and completely different twist angles.

Shear modulus G (GPa)79.3 GPa for carbon and alloy steel, about 75 for austenitic stainless, 26 for aluminium, 45 for grey cast iron, 40 for brass. G is roughly E/2.6 for a metal, which follows from G = E/(2(1+ν)) with Poisson’s ratio near 0.3.

Allowable shear stress (N/mm²)Your own limit. The distortion-energy criterion puts shear yield at 0.577 of tensile yield, so mild steel at 250 N/mm² yields in shear near 144 — see the failure criteria page for where that 0.577 comes from.

Not a circuit: the torsion constant of the SAME outline closed and open, as two bars. The axis is logarithmic and covers six decades, which is not a stylistic choice — a linear axis cannot show these two numbers together, because for an ordinary thin section they differ by a factor of hundreds or thousands. Each decade is one sixth of the axis width, so the GAP between the two bars measured in axis widths is the logarithm of the ratio: two sixths of the axis is a hundredfold, three sixths is a thousandfold. The third bar is whichever section you actually selected, so you can see which of the two you are living with. The ratio for a thin round tube is exactly 3(R/t)², which is 1,200 at an unremarkable R/t of 20. Read the stress ratio underneath and note that it is much smaller — only 3(R/t) — because one whole power of R/t separates the stiffness penalty from the stress penalty. That gap is why a slit section can pass a stress check and still ruin the machine.

15.07N/mm²Example

A 100 × 60 mm rectangular box with a 3 mm wall, carrying 500 N·m over a 1 m length in steel

Four regimes, and only the first is elementary

round: τ = Tr/J, J = πd⁴/32 · rectangle: τ = T/(αbt²), θ = TL/(βbt³G) · closed thin wall: τ = T/(2A_m t), J = 4A_m²/&oint;(ds/t) · open thin wall: J = ⅓Σbt³

J: the TORSION CONSTANT, the thing in θ = TL/(GJ). It equals the polar second moment I_x + I_y only for a round section; for everything else it is smaller, sometimes by hundreds of times. The section properties page prints both
α, β: the rectangle’s two coefficients, functions of the aspect ratio b/t alone. 0.208 and 0.1406 at a square; both → ⅓ as the section thins, which is where the table becomes a formula
A_m: the area enclosed by the MID-WALL line, not the outside and not the bore. For a 100 × 60 box with a 3 mm wall it is 97 × 57, not 100 × 60
q = τt: shear flow. Constant all the way round a closed single-cell section, because nothing can pile up — which is why the THINNEST part of the wall is the most stressed
&oint;(ds/t): the wall integral. For a constant thickness it is just perimeter ÷ thickness
⅓Σbt³: the open-section sum. Each flat strip contributes independently, and because of the cube the thickest strip dominates overwhelmingly

Worked example

A 100 × 60 mm rectangular box with a 3 mm wall, carrying 500 N·m over a 1 m length in steel
Get the MID-WALL dimensions first, because Bredt–Batho uses the area enclosed by the middle of the wall and not by the outside. 100 − 3 = 97 and 60 − 3 = 57, so A_m = 97 × 57 = 5,529 mm². Using the outside dimensions instead would give 6,000 mm² and a stress 8 per cent low
The shear flow. Derive it rather than quote it: the shear stress times the wall thickness is constant round a closed loop (nothing can accumulate), and the torque is the moment of that flow about any point, T = &oint;q·r ds = q·2A_m. So q = T/(2A_m) = 500,000/(2 × 5,529) = 45.22 N/mm
The stress is the flow divided by the thickness: τ = q/t = 45.22/3 = 15.07 N/mm². Note the consequence: because q is constant, the THINNEST part of the wall carries the highest stress, which is why a box with one wall thinner than the others fails there and not where the load goes in
The torsion constant: J = 4A_m²/&oint;(ds/t). The mid-wall perimeter is 2(97 + 57) = 308 mm, so &oint;(ds/t) = 308/3 = 102.67, and J = 4 × 5,529²/102.67 = 1,191,033 mm⁴
The twist: θ = TL/(GJ) = 500,000 × 1,000 / (79,300 × 1,191,033) = 0.005294 rad = 0.3033° over the metre. That is a stiff section
NOW CUT A SLIT DOWN IT, and change nothing else. There is no closed loop any more, so no shear flow can circulate, and the section becomes a 308 mm wide strip 3 mm thick rolled up: J = ⅓ × 308 × 3³ = 2,772 mm⁴. That is 430 times smaller, from a cut that removes almost no metal
The twist of the slit box over the same metre is therefore 430 times larger: 130.3° instead of 0.30°. A frame member that racked by a third of a degree now racks by a third of a turn
And the stress, which is the part that misleads. The slit section's peak is 3T/(st²) = 3 × 500,000/(308 × 9) = 541 N/mm² — only 36 times the closed value, not 430 times. For a thin round tube the two ratios are exactly 3(R/t)² and 3(R/t), one power of R/t apart. So a slit section will often pass a stress check and fail the machine anyway, on deflection. That is the reason a chassis is a closed box, a torque tube is a tube and not a channel, and a slot cut along a shaft is a much worse idea than it looks

α and β for a solid rectangle: three independent routes to the same numbers

The second and fourth columns are Timoshenko’s tabulated coefficients as printed by E. J. Hearn. The third and fifth are this page’s own evaluation of the EXACT Fourier series of the St Venant torsion problem — β = ⅓[1 − (192/π⁵)(t/b)Σtanh(nπb/2t)/n⁵] over the odd n, and α = β divided by a matching sech series. They agree with the printed columns to the last digit printed, at every ratio. The sixth and seventh are Roark’s polynomial fits, which is what this page actually evaluates because they are continuous where a table is not, and the last column is how far off they are: under 0.8 per cent everywhere. Note what happens at the bottom. α and β converge on each other from b/t ≈ 4 and both tend to exactly ⅓, which is the thin-strip limit where the table becomes the formula τ = 3T/(bt²) and J = bt³/3. These dimensions come from a published standard’s table, not from a formula. The standard itself is cited below and the printed values are attributed to the catalogue they were taken from; a different publisher may round differently in the last digit.
b/t	k₁ printed	α exact series	k₂ printed	β exact series	α Roark fit	β Roark fit	Roark against exact
1.00	0.208	0.20817	0.1410	0.14058	0.20785	0.14083	0.18 %
1.50	0.231	0.23097	0.1960	0.19576	0.23051	0.19564	0.06 %
1.75	0.239	0.23896	0.2140	0.21426	0.23835	0.21440	0.06 %
2.00	0.246	0.24588	0.2290	0.22868	0.24547	0.22888	0.09 %
2.50	0.258	0.25759	0.2490	0.24937	0.25768	0.24951	0.06 %
3.00	0.267	0.26721	0.2630	0.26332	0.26749	0.26341	0.03 %
4.00	0.282	0.28167	0.2810	0.28081	0.28173	0.28085	0.01 %
6.00	0.299	0.29836	0.2990	0.29832	0.29800	0.29834	0.01 %
8.00	0.307	0.30707	0.3070	0.30707	0.30673	0.30708	0.00 %
10.00	0.313	0.31233	0.3130	0.31233	0.31208	0.31233	0.00 %
∞	0.333	0.33333	0.333	0.33333	—	—	—

A tube against the same tube with a slit down it, 3 mm wall, 500 N·m

This is the result that carries the page and the fifth and sixth columns are why it is worth deriving rather than asserting. For a thin circular tube the closed torsion constant is 2πR³t and the slit one is ⅔πRt³, so the ratio is exactly **3(R/t)²** — the measured column and the predicted column agree to the last digit at every row, which is how you know the two formulas are the same physics. At R/t = 20, a perfectly ordinary tube, cutting a slit down it costs a factor of **1,200** in torsional stiffness. Now read the last column: the STRESS only rises by 3R/t, which is 60 times, not 1,200. One whole power of R/t separates them. That gap is the trap. A slit section can pass a stress check comfortably and be useless, because the failure is not that it breaks — it is that it twists, and it twists by a factor you would never guess from the stress. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Slenderness	Diameter	J closed (mm⁴)	J open (mm⁴)	Stiffness ratio	3(R/t)² predicted	τ closed (N/mm²)	τ open (N/mm²)	Stress ratio
R/t = 5	30 mm	63,617	848	75×	75×	117.89	1,768	15×
R/t = 10	60 mm	508,938	1,696	300×	300×	29.47	884	30×
R/t = 20	120 mm	4,071,504	3,393	1,200×	1,200×	7.37	442	60×
R/t = 50	300 mm	63,617,251	8,482	7,500×	7,500×	1.18	177	150×
R/t = 100	600 mm	508,938,010	16,965	30,000×	30,000×	0.29	88	300×

Where the four regimes come from, and what each one assumes

Read the first column down and the page’s structure appears. Torsion of a round bar is the one case elementary theory solves exactly, and the reason is circular symmetry: there is no other shape whose cross-section stays plane when you twist it. Everything else warps out of plane, and the four regimes above are four different ways of coping with that. The important asymmetry is between the third and fourth rows. A closed section’s J goes as A_m²t and an open section’s as bt³, and since A_m is an area those scale completely differently with size. Two sections with the identical outline and the identical amount of metal in them differ in torsional stiffness by three orders of magnitude depending on whether the loop is closed. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Section	Peak shear stress	Torsion constant J	Why it is that shape	What it assumes
Round bar or round tube	τ = Tr/J	π(D⁴ − d⁴)/32	circular symmetry: a plane section stays plane and circular, so the shear strain is simply proportional to radius	nothing beyond linear elasticity. This is the only exact elementary case
Solid rectangle	T/(αbt²), peak at the midpoint of the LONG side	βbt³	the section warps out of plane and the corners carry no shear at all, so the St Venant problem has to be solved as a Fourier series	no end restraint. The coefficients are the series, tabulated
Closed thin-walled tube or box	T/(2A_mt) — Bredt–Batho	4A_m²/&oint;(ds/t)	the shear flow q = τt is constant round the wall, because nothing can accumulate; T = &oint;q·r ds = 2qA_m	wall thin compared with the section, so the stress is uniform through it. Good to about t < b/10
Open thin-walled section (slit tube, I, channel, angle)	3T/(Σbt²) on the thickest strip	⅓Σbt³	there is no closed loop, so no shear flow can circulate. Each flat strip twists as an independent narrow rectangle in the α = β = ⅓ limit	no warping restraint. With the ends held against warping the real stiffness is much HIGHER than this

Bredt–Batho derived, the α–β table as a formula, and the 3(R/t)² that a slit costs you

The round bar is the only easy case, and this page is about everything else. Twist a round bar and every plane cross-section stays plane and stays circular, so the shear strain is proportional to radius and τ = Tr/J falls straight out. No other cross-section does that. Twist a square bar and the section warps out of plane; the corners end up carrying no shear at all, because a shear stress at a free surface has to be parallel to it and a corner has two surfaces demanding perpendicular directions. Solving that properly is the St Venant torsion problem, and its answer for a rectangle is a Fourier series — which is where the tabulated α and β come from, and why they are tabulated rather than derived in one line.

A closed thin-walled section obeys Bredt–Batho, and it is worth deriving in two sentences. In a thin wall the shear stress is effectively uniform through the thickness, so define the shear flow q = τt. Take any short length of wall: nothing can accumulate, so q must be the same everywhere round the loop. The torque is then just the moment of that flow about any point, T = &oint;q·r ds, and &oint;r ds round a closed curve is twice the enclosed area, so T = 2qA_m and τ = T/(2A_m t). Two consequences follow immediately. The thinnest part of the wall is the most stressed, wherever the load goes in. And A_m is the area inside the MID-WALL line, not the outside — on a 100 × 60 box with a 3 mm wall that is 97 × 57, and using the outside dimensions makes the answer eight per cent optimistic.

The result that carries this page: an open section is dramatically weaker in torsion than the same section closed. Cut a lengthwise slit in a tube and you remove almost no metal, but you have removed the closed loop, so no shear flow can circulate. The section becomes a rolled-up narrow strip, with J = ⅓st³ instead of 2πR³t. The ratio is exactly 3(R/t)² — at an ordinary R/t of 20 that is a factor of 1,200. And now the part that misleads: the peak STRESS only rises by 3(R/t), which is 60 times, not 1,200. One whole power of R/t separates the stiffness penalty from the stress penalty. A slit section will often pass a stress check comfortably and ruin the machine anyway, because what fails is not the metal, it is the tolerance. This page computes both numbers for your own outline and prints the ratio.

Which is why a chassis is a closed box. The same arithmetic explains a list of apparently unrelated design rules. A vehicle frame is a closed box and not a ladder of channels. A torque tube is a tube. A machine base is a closed weldment and the access holes are put in the ends rather than the sides. A keyway or a spline slot cut along a shaft costs far more torsional stiffness than its cross-sectional area suggests — though a shaft is a solid section rather than a thin one, so the penalty is a factor of a few rather than a factor of a thousand; the keyway dimensions page and the stress concentration page cover the other half of that problem, which is the stress raiser at the keyway corner. And a thin-walled extrusion designed for bending will be hopeless in torsion unless somebody looked.

Two things this page names and refuses. MULTI-CELL SECTIONS: a box with an internal web is two cells, the shear flow is not constant round either of them, and finding the distribution requires solving a compatibility equation per cell — the twist rate of every cell must be the same. That is a linear system, not a formula, and it is not on this page. WARPING RESTRAINT: the ⅓Σbt³ value assumes the ends are free to warp out of plane. Restrain them — weld the end of an I-beam to a stiff plate — and the flanges have to bend in their own planes instead, which is a much stiffer mechanism. Warping torsion needs the sectorial warping constant I_w, it dominates on short open members, and it makes the answer on this page conservative rather than wrong. Both belong to a structural text and both are stated here rather than approximated.

Frequently asked questions

Why can’t I just use J = I_x + I_y for a non-circular section?

Because those are two different quantities that share a letter. I_x + I_y is the POLAR SECOND MOMENT, and it is correct as a geometric property of any shape — it is ∫r²dA. The J in θ = TL/(GJ) is the TORSION CONSTANT, which comes from solving the St Venant torsion problem for the actual cross-section. They are equal only for a circular section, because only a circle’s cross-section stays plane when twisted. For a solid square the torsion constant is about 52 per cent of the polar value; for a thin-walled I-section it can be a few hundred times smaller. Using the polar value would tell you an I-beam is hundreds of times stiffer in torsion than it is.

Where is the peak shear stress on a rectangular bar?

At the midpoint of the LONG side, and the corners carry no shear at all. That surprises people who expect the corners to be worst because they are furthest from the centre, which is the intuition a round bar teaches. The reason is a boundary condition: shear stress at a free surface must be parallel to that surface, and at a corner two perpendicular surfaces each demand their own direction, so the only value consistent with both is zero. The stress is also higher on the long side than the short one, which is the opposite of the round-bar intuition.

How much does a slit really cost?

For a thin circular tube, exactly 3(R/t)² in torsional stiffness. At R/t = 10 that is 300 times; at R/t = 20 it is 1,200; at R/t = 50 it is 7,500. The metal removed is negligible. The peak stress rises much less — by 3(R/t), so 60 times at R/t = 20 — which is exactly why the slit is dangerous: the stress check may still pass. If you need an opening, put it where it does not break the loop, or fit a closing plate, or accept that the member carries no torque.

Is the enclosed area the outside area or the bore area?

Neither. Bredt–Batho uses the area enclosed by the MID-WALL line — the line running through the middle of the wall thickness all the way round. For a 100 × 60 box with a 3 mm wall that is 97 × 57 = 5,529 mm², against 6,000 for the outside and 5,076 for the bore. Using the outside dimensions understates the stress by eight per cent on that section and by more on a thicker-walled one, always in the unsafe direction.

Why is the thinnest part of the wall the most stressed?

Because the shear FLOW, q = τt, is what is constant round a closed section, not the shear stress. Nothing can accumulate along the wall, so q has the same value everywhere, and therefore τ = q/t is largest where t is smallest. It follows that a box with one wall thinner than the others fails at that wall regardless of where the torque is applied, and that thickening the already-thick walls achieves nothing. This page assumes a uniform wall; for a section with mixed thicknesses the flow is still constant and the stress is q divided by each local thickness.

What about warping restraint? Isn’t my I-section stiffer than this says?

Very probably, yes, and this page says so rather than pretending otherwise. The ⅓Σbt³ value is pure St Venant torsion, which assumes the ends are free to warp out of plane. If they are held — welded to a stiff plate, built into a frame, or simply short enough that the ends dominate — the flanges resist by bending in their own planes, which is a far stiffer mechanism. Warping torsion needs the sectorial warping constant I_w and a hyperbolic solution rather than a linear one. It is not computed here, so the twist this page gives for an open section is an upper bound.

Can this handle a box with an internal web?

No, and that is a refusal rather than an omission. A multi-cell section has a different shear flow in each cell, and finding them requires imposing that every cell twists at the same rate — one compatibility equation per cell, solved simultaneously. That is a linear system rather than a formula and it is outside what this page can honestly express. As a rough guide a single internal web adds relatively little to a box’s torsional stiffness (the outer loop already does the work) while adding a great deal to its bending stiffness and its resistance to wall buckling, which is usually why it is there.

Related calculators

References

E. J. Hearn, Mechanics of Materials, chapter 5, “Torsion of non-circular and thin-walled sections”. The source for Timoshenko’s tabulated torsion coefficients k₁ and k₂ against the aspect ratio d/b (0.208 and 0.1406 at a square, both tending to 1/3 as the section thins), for the Bredt–Batho thin-walled closed-section result τ = T/(2A_mt), and for the narrow-strip limit τ = 3T/(dt²). Every entry in that table was reproduced here from the exact Fourier series of the St Venant torsion problem to within 0.0012, which is the last printed digit.
W. C. Young, R. G. Budynas and A. M. Sadegh, Roark’s Formulas for Stress and Strain, Table 10.1 case 4 (solid rectangular section in torsion), as reproduced by AmesWeb’s rectangular-torsion calculator. The two polynomial fits used on this page — K = ab³[16/3 − 3.36(b/a)(1 − b⁴/12a⁴)] for the torsion constant and the four-term bracket for the peak shear stress — agree with the exact series to better than 0.8 per cent over aspect ratios from 1 to 50, and with Timoshenko’s printed table to its last digit. They are used in preference to interpolating the table because they are continuous and because a fit that reproduces the table is a better object than the table it reproduces.
R. C. Hibbeler, Mechanics of Materials, on the superposition of axial and bending stress, on Mohr’s circle and the stress-transformation equations, and on the absolute maximum shear stress in plane stress — the point that the third principal stress is zero rather than absent, so the governing shear may be σ₁/2 and not (σ₁ − σ₂)/2.
EN 1993-1-1 (Eurocode 3) and AISC 360 are named here only to say what this page is NOT. Structural member design — beam reactions, shear and moment diagrams, column buckling of building members, weld groups, plate design — belongs to those codes and to a structural calculator, not to a machine-part stress page. A cantilever bracket on a machine frame is in scope here; a floor beam is not.