Both at once, and neither matters until it does

Every capacity computed so far on this site has been computed alone. A plastic moment with no shear near it; a shear capacity with the flanges ignored. That separation is a convenience, and at the support of a continuous beam it is exactly where the assumption is least defensible: the largest moment and the largest shear are in the same cross-section, a few millimetres apart, in the same steel.

The interaction between them is real, and its shape is unusual enough to be worth the whole page. Nothing happens for the first half of the shear range, and then everything happens at once.

Which free body produced the number #

The mechanism is von Mises and there is nothing else in it.

Take a fibre in the web. It is carrying a shear stress $\tau = V/A_v$ and it is being asked for a normal stress as well. The yield condition in two dimensions is

$$\sigma^2 + 3\tau^2 = f_y^2$$

so what is left for the normal stress is $f_y\sqrt{1 - 3\tau^2/f_y^2}$. Write $v = V/V_{pl}$ and note that $V_{pl} = A_v f_y/\sqrt{3}$ makes $\tau/\tau_y$ exactly $v$. Then

$$\sigma_{\text{left}} = f_y\sqrt{1 - v^2}$$

Now assemble the plastic moment out of the parts. The flanges are outside the web, carry no shear, and reach $f_y$ whatever is happening below them. The web reaches $\sigma_{\text{left}}$. So

$$M_{red} = W_f f_y + W_w f_y \sqrt{1 - v^2}$$

and dividing by $M_{pl} = (W_f + W_w) f_y$,

$$\frac{M_{red}}{M_{pl}} = 1 - \rho\left(1 - \sqrt{1 - v^2}\right), \qquad \rho = \frac{W_w}{W_f + W_w}$$

The whole interaction is one number: the web’s share of the plastic modulus.

The number, for a real section #

For the beam here — 200 mm flanges 16 mm thick, a 468 mm web 10 mm thick, at 355 N/mm² — the two plastic moduli are

$$W_f = 2 b_f t_f \frac{h_w + t_f}{2} = 1.549 \times 10^6\;\text{mm}^3 \qquad W_w = \frac{t_w h_w^2}{4} = 0.548 \times 10^6\;\text{mm}^3$$

so $\rho = 0.261$. A quarter of the section’s bending strength is in the web and three quarters is in the flanges, which is the whole point of an I-section and is also why the interaction is so weak.

The consequences fall out of the algebra:

Half the web’s shear capacity costs three and a half per cent of the moment. That is not a small effect that ought to be checked anyway; it is smaller than the difference between one rolled section and the next in the same serial size.

That share is a fact about the shape rather than about the steel. Material far from the middle is what makes an I-section efficient, and the same arrangement that puts three quarters of the strength into the flanges puts three quarters of it out of reach of the shear. A section has already decided how much this page matters to it before any load is applied, and it decided it in the rolling mill.

Why the far end is so steep #

$\sqrt{1 - v^2}$ has an infinite derivative at $v = 1$. That single fact is the reason the interaction has the shape it does, and it is worth reading physically rather than as calculus.

Near $v = 1$ the web is very nearly fully committed to shear. The normal stress it has left is falling as the square root of the remaining capacity, so a small further increase in shear takes a large fraction of what is left. Between $v = 0.9$ and $v = 0.99$ the moment loss goes from 14.7% to 22.4% — nearly as much as it lost over the whole range from zero to 0.9.

Which gives the practical rule the curve is really about. There is no need to be careful about the interaction anywhere except very close to the shear limit, and there it is not a correction, it is a cliff. That is the standing question of which failure arrives first answered for one particular pair of capacities, and the answer is that the crossing point is a property of the section rather than of the load on it.

Drawing the same section at the far end of the range says the same thing in stresses rather than in a curve.

The one section where it bites #

On a simply supported beam the interaction is worth nothing: the moment peaks where the shear is zero and the shear peaks where the moment is zero. Shear is the derivative of moment, so a maximum of one is a zero of the other, and it takes continuity, or a cantilever, to break that coincidence and put both maxima in the same place. A support with beam on both sides of it does exactly that, by putting a large hogging moment where the shear has not yet turned round.

For a fixed-ended beam under a uniform load, the support carries $wL^2/12$ of moment and $wL/2$ of shear at the same section. On the 9 m beam at 60 kN/m drawn in the span view, the support shear is 270 kN against a $V_{pl}$ of 959, so $v = 0.28$, and the interaction adds 1% to a utilisation of 54%. Not worth the arithmetic.

Shorten and load it, though: 4.5 m at 340 kN/m gives a support shear of 765 kN, $v = 0.80$, and a utilisation that goes from 77% ignoring the interaction to 86% including it. Nine points of utilisation, on the section that governs, from a term the same beam did not need at twice the span.

The same drawing for the short beam is the argument for why the check exists at all. Nothing about the section changes between the two; only the span and the load, and therefore where the support station sits on the curve.

The same shape, one action along #

The curve above is not the only interaction on this site, and it is worth putting beside the one it most resembles.

Axial force and moment interact for the same reason: an axial force commits part of the section, and what is left carries the moment. But the arithmetic differs in a way that changes the shape completely. Axial force commits the material nearest the neutral axis, which is the material contributing least to the moment, so the axial–moment curve is also flat near the origin — but it is flat because of where the material is, not because of a square root, and it reaches zero linearly rather than vertically.

So the two curves are flat for opposite halves of their range. The moment–shear one falls off a cliff at the end; the moment–axial one arrives at its end smoothly and leaves the section with nothing. Two ways to fail at once draws that other curve, and the difference between a geometric argument and a material one is visible in the shape: there the section is divided by position, here it is divided by a yield criterion, and only one of those divisions has a square root in it.

A hinge at high shear is a worse hinge #

The plastic moment above is a strength. Using it in an analysis needs something more: the section has to be able to hold that moment while it rotates, because a collapse mechanism needs several hinges and the first one formed has to survive until the last one does.

A web at high shear is much worse at that than a web at low shear. It is already close to its shear yield, its normal stress capacity is on the steep part of the square root, and small further shear strain takes what is left. Hinges at supports of continuous beams, which is exactly where the shear is largest, are therefore the least reliable hinges in any mechanism — and the redistribution that plastic design depends on is asking most of the section least able to give it.

That is a ductility argument rather than a strength one, and this page’s curve says nothing about it. The curve gives a capacity; whether the section can be relied on to keep delivering it through a rotation is a separate question with a separate answer. The mechanism the capacity is for puts hinges at both supports and one in the span, and the support hinges have to hold while the span one forms — which asks the two sections at the largest shear to be the most patient members of the set.

The elastic picture disagrees about where the worst point is #

Everything above is plastic: the section is at yield through its depth and the question is how much of it is available for bending. The elastic question is different and its answer is different.

Take the same short beam at working load. At the support, three stations are worth measuring:

The worst point is the junction, and it is worst by a margin. It has 94% of the bending stress — the flange is thin, so the junction is close to the extreme fibre — and a shear stress that is nearly the largest in the section, because the web is where the shear flow concentrates. The shear flow jumps where the section narrows at that junction, so the largest shear stress anywhere in the flange is small and the smallest anywhere in the web is not.

This is the same finding as the one on principal stress, and the two pages are the same observation about different questions. That one asks where the material first yields; this one asks what the section as a whole will carry once it has. The elastic worst point and the plastic reduction are not two answers to one question, and neither replaces the other.

A plate girder has a stronger interaction and a weaker web #

Everything scales with $\rho$, and $\rho$ is a design decision.

A rolled beam puts a quarter of its plastic modulus in the web because rolling wants thick material. A welded plate girder, proportioned to carry shear over a long span, puts far more there: for a 600 by 6 web with 150 by 10 flanges, $\rho$ rises to 0.37, and half the shear now costs 5% of the moment instead of 3.5%.

That is the smaller half of the change. The larger half is that a 600 by 6 web has a slenderness of 100 and does not reach $f_y/\sqrt{3}$ at all — it buckles in shear first, and the capacity it does have comes from the diagonal tension field that forms afterwards. At that point the interaction on this page has to be rewritten against a different shear capacity and a web whose behaviour is post-buckling rather than plastic.

The section most likely to meet a real moment–shear interaction is the section this model describes least well, which is a shape that recurs in this subject often enough to be worth naming.

That is the half of the change the curve can draw. The other half is that the web it is drawn for does not behave the way the curve assumes.

In concrete the two help each other #

The interaction on this page is a competition: shear and moment draw on the same web material through one yield criterion, and what one takes the other cannot have. In reinforced concrete without links the relationship is the opposite, and the reason is worth having because it says what kind of quantity a shear capacity is in each material.

A concrete member’s shear capacity without links is an empirical expression whose ingredients are the concrete strength, the effective depth and the longitudinal reinforcement ratio — the flexural steel — entering as a cube root. So

$$V_{Rd,c} \propto (100\rho_l f_{ck})^{1/3}$$

and doubling the tension steel raises the shear capacity by $2^{1/3} = 1.26$: 26 per cent more shear, from steel put in for moment.

The mechanism is the one the four mechanisms essay sets out. More tension steel means a deeper compression zone, which carries more of the shear directly; more dowel area across the crack; and narrower cracks, which means more aggregate interlock across them. All three push the same way.

So the two actions are positively correlated in concrete and negatively correlated in steel, on the same pair of axes, and the difference is that steel’s interaction is a material yield criterion and concrete’s is a set of mechanisms sharing a geometry.

Put a member with links into the same comparison and it changes again. There the shear is carried by a truss, and the coupling is not through the web at all — it is through the tension chord, which carries $M/z$ from the moment and $V\cot\theta/2$ from the truss, and the two add. So the interaction reappears, on the chord rather than in the web, and it is the reason a bar has to be curtailed further along than the moment diagram says.

Three materials-and-arrangements, three completely different couplings between the same two actions, and only one of them is a yield surface.

And torsion joins in linearly #

The second action a web carries alongside shear is a torque, and it interacts differently again — linearly rather than quadratically, because it is the same kind of stress rather than a different one.

Take a 200 × 200 × 8 hollow section. The vertical shear is carried by the two webs at $\tau_V = V/(2 h t)$; the torque is carried by a shear flow circulating round the loop at $\tau_T = T/(2A_m t)$, with $A_m = 36{,}864$ mm². On one of the two webs the two flows run the same way and add; on the other they oppose and subtract.

So the governing wall carries $\tau_V + \tau_T$, and setting that sum to the shear yield stress gives an interaction that is a straight line in $V$ and $T$. That is exactly how the codes write it: a shear capacity reduced by the factor $(1 - \tau_T/(f_y/\sqrt3))$, linear, with none of the curvature the moment interaction has.

The reason for the difference is worth stating because it is the whole logic of interactions. Shear and torsion are the same stress component on the same plane, so they add arithmetically and the interaction is linear. Shear and moment are different components — one shear, one direct — so they combine through a yield criterion that squares them, and the interaction is an ellipse. Two actions of the same kind add; two of different kinds combine as a criterion, and which case applies is decided before any arithmetic.

An open section inverts the picture. Its torsional capacity is feeble to begin with, and the St Venant shear runs up one face of each plate and back down the other rather than round a loop — so it barely competes with the vertical shear at all, and the interaction is negligible for a reason that is a compliment to neither.

The asymmetry between the two webs of the hollow section is worth one more sentence, because it is a fact about the member rather than about the check. One web is at $\tau_V + \tau_T$ and the other at $\tau_V - \tau_T$, so a box carrying shear and torsion together is loaded unevenly across its own section, and the wall that governs depends on the sign of the torque. Reverse the torque — a vehicle passing on the other side of a curved deck, a crane travelling the other way — and the critical wall swaps. A section checked for one direction has checked half of its own geometry, and the two halves are only equal if the torque is.

Where the model stops #

The shear stress is uniform over the web. It is not — the parabolic distribution has its peak at the neutral axis and is about 10% below it at the junction — and the plastic model does not care, because at collapse the web is fully plastic in shear and the distribution really is uniform. The elastic table above uses the true distribution and the plastic curve does not, and that is deliberate rather than inconsistent.

The web is assumed to reach yield in shear. A slender web buckles first, and then the whole of this page is about a section that does not exist: the shear capacity is set by plate buckling rather than by $A_v f_y/\sqrt{3}$, and the moment interaction has to be written against that instead.

And the flanges carry no shear at all. They carry a little, and it is genuinely negligible: the shear flow in a flange runs horizontally rather than vertically, so it contributes almost nothing to the vertical force even though it is what makes the flange work.

What the pictures cannot show #

The interaction curve is drawn for a section. A beam has a length, and what the curve says about the section at the support says nothing about whether the beam can redistribute — a fixed-ended beam that reaches its reduced capacity at the support forms a hinge there and carries on to a span mechanism, so the reduced capacity is a stage rather than an end.

Nor can the drawing show that $\rho$ is a design variable. A deeper, thinner web puts more of the plastic modulus into the web — for a 600 by 6 web with 150 by 10 flanges $\rho$ rises to 0.37 and half the shear costs 5% instead of 3.5% — so a plate girder proportioned for shear has a stronger interaction than a rolled beam, on top of having a slenderer web.

The assumption the figure rests on #

The shear area is the web, full stop: $A_v = h_w t_w$, with the root fillets and the part of the flange over the web ignored. That is a convention rather than a measurement, and it is a conservative one — real shear areas are quoted a few per cent larger. Every $v$ on this page is measured against it, so the curve moves slightly left if a more generous definition is used, and the shape of the curve does not move at all.

The ladder from here #

Later rungs on this anchor: the interaction with axial force as well, which is the same fibre argument with three actions and a surface rather than a curve. Slender webs, where the shear capacity is a buckling load and the interaction is written against tension-field action instead. The interaction in a plastic hinge that has to rotate, where a web at high shear has much less rotation capacity than one at low shear — which is the failure mode this page’s ductility assumption quietly needs. Bolted and welded connections at a support, where the same combination has to cross a joint. And the elastic side developed properly: the von Mises contour over the whole section, and why its peak moves from the junction to the neutral axis as the span shortens.