Structural Engineering & Statics
where the load goes
Sections and stress

The circle nobody draws

A plane stress state has three principal stresses and the third is zero. When the two on the drawing share a sign, the largest shear in the state involves the one that is not there — and the circle a designer has drawn is not the circle that governs.

Assumes The worst stress is not where the worst bending is, The shear strength nobody measured and The shear nobody draws.

16 min read 7 figures Which failure arrives firstDrawing as calculationGeometry beats material

Mohr’s circle turns every plane through a point into one drawing, and the drawing is complete for the planes it contains. It contains the planes perpendicular to the sheet, and there are others.

One point, every plane through it, one circle. A point carrying 180 N/mm² across one face, 0 across the other and 60 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 108.2 centred at 90.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 198.2 and -18.2, on planes 16.8° from the face the 180 acts on; the largest shear on any plane is 108.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 207.8.
Fig. 1 A point carrying 180 N/mm² across one face, nothing across the other and 60 of shear. The principal stresses are 198.2 and −18.2, the largest shear on any plane in the sheet is 108.2 — exactly the radius — and the von Mises stress is 207.8. This is the case in which the drawn circle is the whole answer, and the reason is that its two principal stresses have opposite signs.

Three stresses, and one of them is zero

At any point in a solid there are three mutually perpendicular planes on which the shear vanishes, and three principal stresses acting on them. That is true in three dimensions and it is true of a state that happens to be flat.

Plane stress means σ3=0\sigma_3 = 0, not that σ3\sigma_3 is absent. A thin plate has no stress across its thickness, so the third principal stress is zero — and zero is a number that takes its place in the ordering with the other two.

The three principal stresses give three Mohr’s circles, one for each pair, and the state occupies the region between the largest and the two smaller ones. The largest shear stress on any plane at all is the radius of the largest of the three circles, which is

τmax⁡=σmax⁡−σmin⁡2\tau_{\max} = \frac{\sigma_{\max} - \sigma_{\min}}{2}

with the maximum and minimum taken over all three.

In the figure above that makes no difference: the two in-plane principals are 198.2 and −18.2, they straddle zero, so the third stress is between them and the in-plane circle is the largest of the three. The drawn radius of 108.2 is the answer.

When both of them share a sign

One point, every plane through it, one circle. A point carrying 180 N/mm² across one face, 90 across the other and 40 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 60.2 centred at 135.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 195.2 and 74.8, on planes 20.8° from the face the 180 acts on; the largest shear on any plane is 60.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 170.6.
Fig. 2 A point at 180 and 90 N/mm² with 40 of shear. The principal stresses are 195.2 and 74.8, the drawn circle has a radius of 60.2, and both principals are positive. The third is zero, so the true extremes of the state are 195.2 and 0 — and the largest shear on any plane through the point is 97.6, which is 1.62 times the radius on the drawing.

Nothing about the drawing is wrong. It reports the shear on every plane perpendicular to the sheet, and 60.2 is the largest of those. The planes it omits are the ones inclined out of the sheet, and one of them — at 45° between the σ1\sigma_1 direction and the thickness direction — carries 97.6.

A 62 per cent error, in the unsafe direction, from a construction that is exactly right about what it draws.

One point, every plane through it, one circle. A point carrying -140 N/mm² across one face, -40 across the other and 30 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 58.3 centred at -90.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are -31.7 and -148.3, on planes 74.5° from the face the -140 acts on; the largest shear on any plane is 58.3, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 135.3.
Fig. 3 The same trap under compression: −140 and −40 with 30 of shear, giving principals of −31.7 and −148.3 and a drawn radius of 58.3. Both are negative, so the largest of the three principal stresses is the zero across the thickness, and the true maximum shear is (0 − (−148.3))/2 = 74.2 — 27 per cent above the drawn value.

The rule is short enough to carry: draw the third circle whenever the two in-plane principal stresses have the same sign. If they straddle zero, the in-plane circle is the largest and nothing is missing.

Which free body produced the number

The free body is a tetrahedron cut out of the material around the point: three faces on the coordinate planes and one on the plane being asked about.

Equilibrium of that tetrahedron gives Cauchy’s relation — the traction on the inclined face is the stress tensor acting on its normal — and it is what makes the whole subject possible: knowing the stresses on three perpendicular planes determines them on every plane through the point, with no further information.

That statement is three-dimensional and it always was. Mohr’s circle is what happens when the inclined face is restricted to be perpendicular to the sheet, which reduces the tetrahedron to a triangle and the family of planes to a one-parameter family. The construction is a section through a three-dimensional result, and taking a section through something is exactly the operation that loses whatever is not in the plane.

The three circles are the three such sections, one for each pair of principal directions, and the shear on a fully general plane lies somewhere in the shaded region between them rather than on any of the three.

How the three circles fit together

It is worth describing the full construction, because once seen it makes the rule obvious rather than remembered.

Order the three principal stresses σ1≥σ2≥σ3\sigma_1 \ge \sigma_2 \ge \sigma_3 and plot them on the normal-stress axis. Draw a circle on each pair as a diameter: the big one on (σ1,σ3)(\sigma_1, \sigma_3), and two smaller ones on (σ1,σ2)(\sigma_1, \sigma_2) and (σ2,σ3)(\sigma_2, \sigma_3) that sit inside it and touch it at its ends.

Every plane through the point maps to a stress pair somewhere in the region between the big circle and the two small ones — a crescent, not a curve. The three circles themselves correspond to planes containing one of the principal directions, and the interior of the crescent to planes containing none.

Two facts follow immediately. The largest shear is the radius of the big circle, since nothing in the crescent is above it. And the plane carrying it contains the σ2\sigma_2 direction and bisects the other two — so the intermediate principal stress never appears in the maximum shear at all, which is exactly the property that makes Tresca’s criterion insensitive to it and von Mises’s not.

For plane stress one of the three is zero, and which of the three circles is the big one depends only on where zero sits in the ordering. That is the whole of the rule in the previous section, stated as a picture.

Which criterion notices

The two criteria in use disagree about whether it matters, and the disagreement is instructive.

One criterion inside the other, touching at six points. The two yield criteria in principal stress space with the third principal stress zero, both normalised by the yield stress. Von Mises is the ellipse — σ₁² − σ₁σ₂ + σ₂² = f_y², which is a circle seen at an angle — and Tresca is the hexagon inscribed in it, touching at the six points where one principal stress is zero or the two are equal. Everywhere else Tresca is the smaller, by up to 15.5 per cent, and the widest gap is at pure shear, where σ₁ = −σ₂ and the two answers are 159 and 138 N/mm². The ratio there is exactly 2/√3, computed rather than quoted, and it is the whole reason a web is checked against f_y over root three.
Fig. 4 The two yield criteria in principal stress space with the third principal stress zero. Von Mises is the ellipse — σ₁² − σ₁σ₂ + σ₂² = f_y² — and Tresca is the hexagon inscribed in it, touching at the six points where one principal stress is zero or the two are equal. Everywhere else Tresca is the smaller, by up to 15.5 per cent, and the widest gap is at pure shear where the two answers are 159 and 138 N/mm².

Tresca is a statement about the largest shear, so it depends directly on which circle is the largest and is wrong by the whole of the error above if the third stress is forgotten. Its hexagon has corners exactly where the ordering of the three principal stresses changes — which is the geometric signature of a criterion that cares which is biggest.

Von Mises is a statement about the second deviatoric invariant, and it takes all three principal stresses whether anybody drew them or not. Writing σ12−σ1σ2+σ22\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 for plane stress is already the three-stress formula with σ3=0\sigma_3 = 0 substituted in, so the third stress is present in the algebra even when it is absent from the picture.

That is a real advantage of the smooth criterion and it is not the one usually given. Von Mises is normally preferred because it fits data slightly better and because it has no corners to differentiate at; the more useful property in practice is that it is hard to use it wrongly, because the formula has nowhere to omit a stress from.

Where this actually bites in a member

A designer meets same-sign principal stresses more often than the algebra suggests.

A biaxially compressed point — in a slab over a column, in a wall, under a bearing — has both in-plane principals negative and the third zero. The governing circle is the one nobody drew.

A web at the flange junction under bending and shear has a large longitudinal stress and a shear, so its principals straddle zero and the in-plane circle governs. That is the ordinary beam case and it is safe.

A flange in biaxial tension — in a deck slab spanning two ways, in a plate under combined in-plane actions — has both principals positive, and again the third circle governs. Orthogonally reinforced plates carry exactly such a field, and the reinforcement’s own directions have nothing to do with the principal ones.

Neither component is worst where the combination is. A 610 deep I-section on a 4200 mm span under 600 kN at mid-span, read at the quarter point where the moment is 252.0 kNm and the shear 300 kN. The bending stress runs from 87.2 N/mm² at the extreme fibre to zero at the neutral axis; the shear stress does the opposite, and jumps by a factor of 19.2 at the web-flange junction because VQ/It has the same Q on both sides and a different t — 1.7 in the flange against 33.4 in the web. The principal tension is 93.5 at that junction against 87.2 at the extreme fibre, so on this beam the worst point in the section is one that neither of the two standard checks evaluates.
Fig. 5 A 610 deep I-section on a 4,200 mm span under 600 kN, read at the quarter point. The bending stress runs from 87.2 N/mm² at the extreme fibre to zero at the neutral axis; the shear does the opposite and jumps by 19.2 at the web–flange junction. The principal tension is 93.5 at that junction against 87.2 at the extreme fibre, so the worst point in the section is one that neither standard check evaluates.
The junction governs a short beam and nothing else does. The principal tension at the web-flange junction divided by the bending stress at the extreme fibre, against span-to-depth ratio, for the same section under a central point load. Above one, the worst stress in the section is not where the bending calculation puts it. The curve crosses one at a span-to-depth ratio of 9.0, so a beam shorter than about 9 times its depth is governed at a point no bending check looks at, and a beam longer than that is not. At a span-to-depth ratio of 2 the ratio is 1.63, and at 20 it is 0.96 and still falling — the junction never goes away, it simply stops mattering.
Fig. 6 The principal tension at the junction divided by the bending stress at the extreme fibre, against span-to-depth ratio. It crosses one at 9.0, so a beam shorter than about nine times its depth is governed at a point no bending check looks at. At a ratio of 2 it is 1.63 and at 20 it is 0.96 and still falling.

Those two figures are the previous rung’s argument and they are worth having here for the contrast: at the web–flange junction the two in-plane principals have opposite signs, so that case — the one this collection has spent most effort on — is the one where the omission does not matter.

Why the omission survives

It is worth asking how a mistake this size persists in a subject this old, and the answer is about what a construction is for rather than about anybody’s carelessness.

Mohr’s circle is a graphical device. It was invented so that a person with a compass could read off principal stresses and their directions from a drawing, and the two-dimensional case is the one that fits on a page and can be constructed with two arcs. Everything about it — the doubled angle, the pole, the radius as the maximum shear — is optimised for that use.

The three-circle version is not constructible in the same way. It requires the three principal stresses to be known already, so it cannot find them; it can only display them once found. It is a diagram of an answer rather than a method of getting one, which is why it appears in textbooks as an illustration and in nobody’s hand calculation.

So the memorable object is the one that is incomplete, and the complete one is the one nobody has a use for. The practical resolution is not to draw more circles but to stop using the picture as the check — evaluate the criterion from the three principal stresses arithmetically, and keep the circle for what it is genuinely good at, which is finding the directions.

The third stress is not always zero

Everything above is plane stress. The other idealisation swaps which quantity is zero and changes the answer in the opposite direction.

Plane strain — a long body restrained along its length, a tunnel lining, the interior of a thick plate — has ε3=0\varepsilon_3 = 0 rather than σ3=0\sigma_3 = 0. Poisson’s ratio then makes σ3=ν(σ1+σ2)\sigma_3 = \nu(\sigma_1 + \sigma_2), which for two tensile in-plane stresses is a tensile third stress lying between them.

That makes the largest circle smaller, so a plane-strain point yields at a higher in-plane stress than a plane-stress one. Restraint against the third direction raises the yield load, which is the mechanism behind two quite different phenomena: the higher apparent strength of the middle of a thick plate, and the brittleness that comes with it, since a state closer to hydrostatic tension has less shear available to relieve it by yielding.

For a metal the third stress matters only through the differences it makes: von Mises contains stress differences alone, so adding the same amount to all three principal stresses changes nothing whatever. That is why a metal has no strength dependence on hydrostatic pressure and why its yield locus is a cylinder along the hydrostatic axis rather than a closed surface.

For concrete, soil and rock it is not a cylinder but a cone. Confinement raises the strength because those materials have friction in them, and the third principal stress is then a variable in its own right rather than a term in a difference — which is a much stronger dependence than anything on this page and is what a yield criterion assumes about its material rather than about its stresses.

The check where it matters most, in numbers

The clearest place to see the size of the effect is a web carrying shear and direct stress together, which is the one combination every steel code writes a rule for.

Bending and shear together, which is where the criterion is actually used. The direct stress a section may carry against the shear stress alongside it, both as fractions of the yield stress. Von Mises gives an ellipse — σ² + 3τ² = f_y² — whose intercept on the shear axis is 0.577f_y, or 159 N/mm² here. Tresca gives σ² + 4τ² = f_y², intercepting at half f_y, 138. The point marked is 200 N/mm² of direct stress with 100 of shear: an equivalent stress of 265 N/mm² and a utilisation of 0.96. The whole of the shear-moment interaction on this site is this curve, and the 0.577 in it came from a decision about distortion energy rather than from any test of a web.
Fig. 7 The direct stress a section may carry against the shear alongside it, both as fractions of yield. Von Mises gives σ² + 3τ² = f_y², intercepting the shear axis at 0.577f_y — 159 N/mm² here — and Tresca gives σ² + 4τ² = f_y², intercepting at half of f_y, 138. The marked point is 200 N/mm² of direct stress with 100 of shear: an equivalent stress of 265 and a utilisation of 0.96.

The point marked is in the safe case: 200 and 100 give principals of 241 and −41, opposite signs, so the in-plane circle governs and the utilisation of 0.96 is the honest one.

Change the state to 200 direct in both directions with the same 100 of shear and the principals become 300 and 100 — both positive. The in-plane radius is still 100, so a Tresca check done on the drawn circle returns 2×100=2002 \times 100 = 200 against 275 and passes at 0.73. The true Tresca stress is 300−0=300300 - 0 = 300 against 275 and fails at 1.09. The same state on von Mises gives 3002−300×100+1002=265\sqrt{300^2 - 300\times100 + 100^2} = 265, and passes at 0.96, because the formula never let σ3\sigma_3 out of its sight.

Three answers — 0.73, 1.09, 0.96 — for one state of stress, and only the first is wrong. The gap between the two criteria is 15 per cent and the gap opened by the omission is 49, which is the argument for checking with the formula rather than with the picture.

What to carry away

Plane stress has three principal stresses and one of them is zero. Zero is a value in the ordering, not an absence.

Draw the third circle when the two in-plane principals share a sign. Then the true maximum shear involves the zero and is larger than the drawn radius — by 62 per cent in one example here and 27 in another.

And the criterion protects against the mistake only if it is von Mises. Tresca is a statement about the largest shear and inherits the error whole; the smooth criterion has all three stresses in its formula whether or not anybody thought about the third.

A rule of thumb, and when to distrust it

Everything on this page condenses into one question asked before any check: do the two in-plane principal stresses have the same sign?

If they straddle zero — which is the ordinary case for a beam web, a bending member’s extreme fibre with any shear on it, or anything in shear at all — the in-plane circle is the largest of the three, the picture is complete, and nothing is owed.

If they share a sign, the third circle is the largest. Its radius is half the larger of the two in-plane principals in magnitude, and the check has to use that.

The states that share a sign are worth listing because they are all common and none of them looks dangerous: biaxial compression under a bearing or in a wall; biaxial tension in a two-way slab or a pressure vessel; a compression flange with little shear; and a point in a plate near a stiffener. In all four the state is mild-looking, the drawn radius is small, and the omission is largest exactly where a designer is least likely to be looking hard.

Where the model stops

The stress state is assumed to come from a member calculation. Where there is no section to design the field is obtained from a model rather than from a formula, and the three principal stresses at a node are whatever the model chose to put there.

The material is isotropic. Every criterion here treats the three principal directions as interchangeable, which rolled steel is not, timber is emphatically not, and a reinforced concrete slab is not at all.

The criteria are for yielding, not for fracture. A state with three tensile principal stresses has very little shear in it and yields late, and what it does instead is cleave — which is not on any of these surfaces.

Plane stress and plane strain are idealisations of a thickness. A real plate is in plane stress at its surfaces and closer to plane strain at its middle, and the transition across the thickness is what makes a thicker plate behave differently from a thinner one of the same steel.

Nothing here is a stability statement. A point in biaxial compression is also a point in a plate that may buckle, and the two limits are unrelated.

And the stress state is assumed known. In a real member it is computed from beam theory, which is exact for bending, approximate for shear and silent about the third direction — so the input to this arithmetic is usually less certain than the arithmetic.

The circle is a construction for one point in one section, and the worst stress is not where the worst bending is is what happens when the same construction is repeated at every point and the results are compared.

The ladder from here

Later rungs on this anchor: the three-dimensional Mohr construction drawn properly, with the shaded region between the three circles and the planes that map into it. Yield criteria compared against test data rather than against each other. Principal stress in a plate under combined in-plane actions, where the trajectories become a design tool for reinforcement direction. Stress trajectories and topology optimisation, where a computed field becomes a shape. Photoelasticity, which made the field visible before it could be computed. Plane strain and the constraint that comes with thickness, which is where this subject meets fracture. And the interaction surface a member is checked against, which is this argument moved from a point to a whole cross-section.

The same combination of bending and shear appears as a member check rather than as a point check, and the translation between the two is not obvious: a section’s interaction curve is an integral of point criteria over a depth, and the integral is what codes tabulate.

Mohr published the circle in 1882 and the three-circle construction in the same body of work, so the omission this page is about has never been a limitation of the method. It is a limitation of the drawing: one circle fits on a page and is memorable, three circles require the reader to keep track of an ordering, and what survived into general use is the one that could be drawn with a compass in the margin.

Where this goes next

One point, every plane through it, one circle. A point carrying 140 N/mm² across one face, 0 across the other and 45 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 83.2 centred at 70.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 153.2 and -13.2, on planes 16.4° from the face the 140 acts on; the largest shear on any plane is 83.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 160.2. The worst stress is not where the worst bending is Every stress this collection has quoted is a stress on a particular plane, and neither the bending stress nor the shear stress is a property of the point. Turn the plane and both change; one pair of numbers does not, and on a short beam it peaks where neither of them does. One criterion inside the other, touching at six points. The two yield criteria in principal stress space with the third principal stress zero, both normalised by the yield stress. Von Mises is the ellipse — σ₁² − σ₁σ₂ + σ₂² = f_y², which is a circle seen at an angle — and Tresca is the hexagon inscribed in it, touching at the six points where one principal stress is zero or the two are equal. Everywhere else Tresca is the smaller, by up to 15.5 per cent, and the widest gap is at pure shear, where σ₁ = −σ₂ and the two answers are 205 and 178 N/mm². The ratio there is exactly 2/√3, computed rather than quoted, and it is the whole reason a web is checked against f_y over root three. The shear strength nobody measured Every web on this site is checked against the yield stress divided by the square root of three, and no test produced that number. It is a consequence of a decision about what makes a metal yield, and the alternative decision gives a different answer by fifteen per cent. The same concrete, held sideways. Two stress-strain curves for one concrete. The lower is a cylinder test: it peaks at 30 N/mm² near a strain of 0.002 and has nothing left by 0.0035, because it fails by splitting apart sideways. The upper is the same material inside a 12 mm hoop at 100 mm centres, which cannot stop it expanding but can make the expansion stretch steel: the lateral pressure of 2.48 N/mm² — 8% of the strength it is multiplying — takes the peak to 44.4 and the ultimate strain to 0.028. The strength gain is 1.48 times and the strain gain 8.0; the area under the curve, which is the toughness, goes up by 11. It is the third number the confinement is provided for. Squeezed sideways into a different material Concrete in a cylinder test fails by splitting apart sideways under a load pushing it down. Put a hoop round it and the splitting has to stretch steel — and a lateral pressure of a twelfth of the strength raises the strength by half and the ultimate strain by eight. Two ways to fail, and the curve between them. The exact plastic interaction between axial force and moment for two sections of identical area, both normalised by their own squash load and their own plastic moment. The rectangle stands 25.0% of its plastic moment outside the straight line at an axial ratio of 0.50; The I-section stands 6.0% of its plastic moment outside the straight line at an axial ratio of 0.12. Every section here is symmetric about its centroid, so the equal-area axis and the centroid coincide and it makes no difference which the moments are taken about. The straight line is the rule that says the two capacities share out in proportion, and everything between it and a curve is capacity that rule gives away. Two ways to fail, and the curve between them A column carrying both compression and bending has two capacities and a rule for sharing them out. The rule is a straight line, the truth is a curve, and for a rectangle the straight line gives away a quarter of the plastic moment at half the squash load.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

ConfinementFree bodyHydrostatic stressMohr circlePlane stressPrincipal stressShear moment interactionShear stressStress transformationTrescaVon misesYield criterion