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.

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.

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:

$$\sigma_1 = \frac{79.5}{2} + \sqrt{\left(\frac{79.5}{2}\right)^2 + 30.8^2} = 90.2$$

against 85.9 at the extreme fibre. 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 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.

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.

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.

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.