The worst stress is not where the worst bending is

A beam is checked twice. Once for bending, at the extreme fibre where $My/I$ is largest; once for shear, at the neutral axis where $VQ/It$ is largest. The two checks are made at different points, using different formulae, and between them they are supposed to cover the section. On a slender beam they do. On a short one they miss the point that governs, and the point they miss is at a height where neither of the two quantities is anywhere near its own maximum.

The stress on a plane is not a property of the point. It is a property of the point and the plane, and there are infinitely many planes.

The circle, and why it turns twice as fast #

Take a small wedge at the point, with one face on the $x$ plane and the hypotenuse at angle $\phi$. Summing forces on the wedge — a free body two millimetres across — gives the stresses on the inclined face:

$$\sigma_\phi = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos2\phi + \tau\sin2\phi$$

and the companion expression for $\tau_\phi$. Both contain $2\phi$ and neither contains $\phi$, which is the whole reason the picture works: a plane turned by $\theta$ moves the state of stress round the circle by $2\theta$. Turn the plane by 90° and the point returns to itself on the other side of the circle, which is correct — the $x$ face and the $y$ face are two descriptions of one state.

Two facts fall straight out of the geometry and both are worth having without any algebra.

The sum of the normal stresses on any perpendicular pair is invariant, because it is twice the centre and the centre does not move: $\sigma_1 + \sigma_2 = \sigma_x + \sigma_y = 140$ here.

The largest shear on any plane is the radius, and it sits 90° round the circle from the principal points, so 45° from the principal planes in real space. That is why a ductile metal in tension slips at 45° to the pull, and it is the reason a mild steel test coupon necks the way it does.

The centre is the pressure and the radius is the distortion #

The circle’s two defining numbers have physical readings, and having them makes half the subject legible without any algebra.

The centre sits at $(\sigma_x + \sigma_y)/2$, the mean of the normal stresses. It is the part of the state that is the same on every plane — a hydrostatic pressure, pushing equally in all directions — and it produces no shear on any plane whatever, because a circle of zero radius centred anywhere has $\tau = 0$ everywhere.

The radius is everything else: the part that varies with the plane, that produces all of the shear, and that distorts the element rather than merely compressing it. It is the deviatoric part, and its magnitude is the largest shear the state contains.

So the decomposition that every yield criterion performs — separate the hydrostatic part, keep the deviatoric part, decide on that — is literally the circle’s centre and radius. Both criteria discard the centre, which is why a metal can be squeezed to enormous pressures without yielding, and both are statements about the radius.

Three states are then recognisable on sight.

Uniaxial stress puts the circle through the origin, so its radius is half its centre and the maximum shear is half the normal stress — which is the check the extreme fibre of a beam satisfies exactly.

Pure shear centres the circle on the origin, so the two principal stresses are $\pm\tau$: a state that is equally a tension and a compression at 45° to each other, which is why a shear failure in concrete is a tension crack at 45° and why the neutral axis of a beam is where its diagonal cracking starts.

Hydrostatic stress collapses the circle to a point on the axis: no shear on any plane, no distortion, and no yielding at any magnitude.

The middle of those three is not a curiosity. It is the state at the neutral axis of the beam this page is about, and it is worth drawing before the beam arrives, because everything the later figures do with the neutral axis is already contained in it.

The reading also settles a question that comes up whenever two load cases are combined. Two stress states cannot be added by adding their circles — a circle is not a vector — but their components can, because $\sigma_x$, $\sigma_y$ and $\tau$ are what superpose. So combining two cases means adding three numbers and redrawing, and the resulting circle’s centre is the sum of the two centres while its radius is emphatically not the sum of the two radii. A designer who adds two principal stresses from two load cases has added two radii, and the error is largest exactly where the two states are oriented differently — which is the case worth combining.

Reading it off the drawing: the pole #

The doubled angle is the one awkward feature of the construction, and there is a device that removes it entirely.

Mark a point on the circle called the pole, or origin of planes, found this way: from the point representing the stresses on a known physical plane, draw a line parallel to that plane; where it cuts the circle again is the pole.

Once it is marked, the rule is direct. A line drawn through the pole parallel to any physical plane cuts the circle at the stresses on that plane — no angles, no doubling, no trigonometry. Join the pole to the two principal points and the two lines are the principal planes, in their true directions on the drawing. Join it to the top and bottom of the circle and the lines are the planes of maximum shear.

That is the construction as it was actually used, and it is why a stress transformation could be executed on a drawing board faster than it could be computed. It also makes the doubled angle stop being a fact to remember: the drawing does the doubling, because a chord subtends twice the angle at the centre that it does at the circumference, and the pole is a point on the circumference. The circle’s most confusing property is the one that makes the trick work.

The habit is worth having even with a calculator to hand, because it is a check. A principal direction computed from an arctangent is a number with a quadrant ambiguity in it and a sign convention behind it; a principal direction read off a pole is a line on a picture, and it is either plausible or it is not.

Which free body produced the number #

The beam figures below are read at a station a quarter of the way along a 3,500 mm span carrying 400 kN at mid-span. Cut there and take the free body to the left: the reaction of 200 kN, no load yet, so $V = 200$ kN and $M = 200 \times 0.875 = 175$ kNm.

Every stress on the page comes from those two numbers and the section’s own properties, computed from its rectangles by the parallel-axis theorem: $I = 542.9\times10^6$ mm⁴ for a 533 × 209 section with 15.6 mm flanges and a 10.1 mm web.

Those two numbers and that one section property produce the two component distributions, and neither is drawn alone here, because the profile below draws both at once and the comparison is the whole point. The bending stress is $My/I$: linear across the depth, zero at the neutral axis, largest at the extreme fibre. That distribution is the whole of a bending check and the whole of what a bending check knows. The shear stress is $VQ/It$: zero at the extreme fibre, largest at the neutral axis, and — the part that matters here — discontinuous at the web-flange junction, because the same first moment $Q$ is divided by two different widths on the two sides of it.

The point neither check looks at #

The junction is where the two components are both large without either being largest. At $y = 251$ mm the bending stress is 79.5 — 93% of its peak — and the shear stress in the web there is 30.8, which is 73% of its peak at the neutral axis. Combine them:

$$\sigma_1 = \frac{79.5}{2} + \sqrt{\left(\frac{79.5}{2}\right)^2 + 30.8^2} = 90.0$$

against 85.9 at the extreme fibre — 90.2 once the two components are taken unrounded, which is the number the profile above reports. The worst principal tension in the section is 5% above the number a bending check produces, at a point 15 mm inside the flange.

The junction has a circle of its own, and setting it beside the first one on this page is the shortest statement of what has happened.

Two things in that figure appear in neither standard check. The tension of 90.0 is on a plane tilted 18.9° out of the cross-section, so no cut a designer draws would reveal it. And the compression of −10.5 exists at all, at a height where the bending calculation says the section is entirely in tension: a circle whose radius exceeds its centre has a point on both sides of the origin, and here 50.3 exceeds 39.8.

The shear jump at the junction is worth its own sentence, because it is exact and it is the reason the effect exists at all. $VQ/It$ has the same $Q$ on both sides of the junction — the area beyond that height has not changed — and a different $t$: 209 mm of flange above, 10.1 mm of web below. So the shear stress jumps by exactly $b_f/t_w = 20.69$, from 1.49 to 30.8, at a plane where nothing about the material changes. An I-section concentrates its shear into the web by a factor equal to the width ratio, which is also the reason the web carries essentially all of it.

Where the crossover is #

At a span-to-depth ratio of 2 the junction is 1.74 times the extreme fibre. At 5, it is 1.14. At 10 it crosses one. At 20 it has fallen to 0.956 and is still falling.

The same section rotated would give a different answer again, because the two components scale differently with the axis. Ten is a threshold a reader can see across a room, and it is the same neighbourhood as the one shear deflection becomes significant at — which is not a coincidence. Both are ratios of a shear effect to a bending effect, both scale as $(d/L)$ to some power, and both say the same thing: a beam is a slender object, and shortening it makes it something else.

The other end of the sweep is worth reading too. The ratio never reaches zero; it settles just under one and stays there. The junction never stops being nearly as stressed as the extreme fibre — it simply stops being more so, and a designer who never checks it is relying on a margin of 4% at $L/d = 20$.

The lines the stress runs along #

This is the picture the whole subject is for, and it is the one that makes a concrete beam’s crack pattern predictable without any calculation. Concrete cracks across the tension trajectory. So:

At mid-span, where the shear is zero and the state is pure bending, the tension trajectory is horizontal and the cracks are vertical.

Near a support, where the shear is large and the moment small, the trajectory tilts and the cracks lean toward the load at something approaching 45°.

At the neutral axis anywhere, the bending stress vanishes, the state is pure shear, and the trajectories cross the beam at exactly 45° in both directions.

Which is what every diagonal shear crack in every reinforced concrete beam has always looked like, and it is also the geometry that decides where the links go. The strut-and-tie model of the same region is these trajectories straightened into lines, which is the sense in which “follow the elastic field” is a concrete instruction rather than a vague one.

Two components, one criterion #

The comparison above used the principal tension, which is the criterion for a brittle material. A ductile one fails on distortion rather than on tension, and the standard measure is von Mises:

$$\sigma_{vM} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2} = \sqrt{\sigma^2 + 3\tau^2}$$

for the plane state a beam has. At the junction that gives 96.5 N/mm² against the extreme fibre’s 85.9 — a 12% excess rather than 5%, so the ductile criterion makes the junction govern more decisively than the brittle one does, and by a wider margin over a wider range of spans.

That is the general shape of the answer whenever two actions arrive together. One strength, two demands on it, and the boundary between safe and unsafe is a curve rather than a pair of independent limits. A section checked separately for bending and separately for shear has been checked against a rectangle drawn round that curve, and the corner of the rectangle lies outside the material. Codes write the shear-and-bending check as a curve for exactly this reason, and it is the same reason the bending-and-axial check is written as one.

Where the two checks came from #

That is the honest limit on the argument. The junction check gets interesting below $L/d = 10$; the section calculation that supplies its inputs starts failing below about $L/d = 4$; and between those two the effect is real and computable. Below 4 the whole apparatus has to be replaced by a model with no section in it.

The same combination, three more places #

Once the habit of combining is established, it turns up everywhere a section carries two things at once.

A hole turns a uniform field into a trajectory field. The concentration factor of three at the edge of a circular hole in a wide plate is a statement about the principal stress there, and the direction of the peak is along the boundary rather than across it — because a free surface can carry no stress across itself, so one principal direction at every free edge runs along the edge and the other is zero.

At every free surface, one principal stress is zero. That is not an approximation; it follows from the surface having nothing on the other side of it. So the state of stress at the extreme fibre of a beam is uniaxial, its circle passes through the origin, and its maximum shear is exactly half its normal stress — which is the check the generator makes and which comes out at 1.000000.

That circle is why the extreme fibre is checked the way it is, and also why a mild steel coupon pulled in tension slips at 45° rather than tearing straight across. The bending calculation evaluates the point at the far right of the circle; the material gives way at the top of it.

A brittle material is judged on the principal tension, and on a length. A flaw does not care about von Mises; it opens under the tension normal to its own plane, so the trajectory field decides which flaws in a member are dangerous and which are harmlessly aligned. The same flaw, at the same depth, in the same steel, is a crack or a scratch depending only on how it sits relative to the lines that figure draws.

A weld is a plane, and its orientation relative to the trajectories decides its life. A fillet weld is stronger across than along for a reason that is the same reason: the two loadings put different points of the same circle onto the throat.

And an anisotropic material has its own preferred planes, which may have nothing to do with the trajectories at all. Timber splits along the grain, and a shear stress that would be harmless in steel is a splitting stress if the grain runs the wrong way — which makes the horizontal shear at the ends of a timber beam the governing check rather than an afterthought.

Where the model stops #

The stress state is taken as plane. A beam’s web is close to plane stress and its flanges are not — near a load or a support the transverse stress $\sigma_y$ is substantial, and it moves the circle’s centre without appearing in either standard check.

$VQ/It$ is itself an approximation. It assumes the shear stress is uniform across the width at any height, which is exactly false in a wide flange and only nearly true in a narrow web. At the junction, where the two are being compared, the flange’s shear is a lateral flow rather than a vertical one, and the 1.49 quoted for it is a number the formula produces rather than a stress anybody would find there.

Nothing here is local. Near the point load itself, or over a support, the stress field bears no relation to any of this, and the peak stresses are set by the bearing rather than by the section.

And no residual stress is included. A rolled section carries locked-in stresses of the order of 0.3 of yield before anything is applied, at the flange tips and at the junction — which is to say at both of the two points this page has been comparing.

What the pictures cannot show #

Mohr’s circle is a drawing in stress space, not in the beam. The two axes are stresses, the angle on the circle is twice the angle in the material, and nothing on the figure occupies a position — so a reader who tries to locate the circle inside the section is being misled by a diagram that has no location.

The trajectory figure is worse in a specific way. It draws smooth curves for a stress field computed from beam theory, which is an approximation that has no business near the load, near the supports, or near the boundary of the drawing. The lines are exact for the model and the model is a caricature at both ends of the beam — which is exactly where the trajectories look most interesting.

And the depth profile draws the shear stress as a curve with a jump in it. That jump is real in the model and not real in the section: the actual transition happens over the fillet radius, where the geometry changes continuously and the formula does not apply at all. The most striking feature of the drawing is the one it is least entitled to.

The ladder from here #

Later rungs on this anchor: the three-dimensional Mohr construction, with its three circles and the fact that the largest shear in a triaxial state involves the largest and smallest principal stresses and never the intermediate one. Yield criteria compared — Tresca against von Mises, the 15% they differ by in pure shear, and which of them a code is quietly using. Principal stress in a plate under combined in-plane actions, which is 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 this field visible before it could be computed and is still the fastest way to see one. The interaction of shear and bending as a code check, and the reason it is written as a curve rather than as two independent limits. And the same combination in a web under tension field action, where the shear has already buckled the panel and the state of stress is not what any of this assumes.

Otto Mohr published the circle in 1882 for exactly the reason it is still used: he wanted a construction a person could carry out with a compass, on a drawing board, and read off. That the construction happens to make the invariants obvious — the centre, the radius, the doubled angle — is the sort of accident that only looks like an accident.