The shear nobody draws

Take four planks, stack them, and stand on the middle. They sag, and their ends do something a solid beam’s ends do not: they fan out, each plank sliding a little past its neighbour.

Glue the stack and the fanning stops. The four planks become one beam roughly sixteen times stiffer, and whatever the glue is now resisting is a stress that appears nowhere in a bending calculation.

Why anything slides at all #

The reason is that the bending moment changes along the beam, which means the force in the flange changes too.

Take a short length of beam and look only at the material above some horizontal plane through it. On the left face of that block, the bending stresses push with a total force $F_1$; on the right face, with $F_2$. If the moment at the two stations differs — and it does wherever there is shear — then $F_1$ and $F_2$ differ, and the block is out of balance horizontally.

Something has to supply the difference. There is only one surface left: the horizontal plane at the bottom of the block. So a horizontal shear stress acts on that plane, and its total over the length is exactly $F_2 - F_1$.

That is the whole derivation, and it delivers the formula in three lines. With $\sigma = My/I$, the force on a face above height $y_1$ is $\frac{M}{I}\int y\,dA$, and the integral is the first moment of the area above the cut, written $Q$. So the out-of-balance over a length $dx$ is $\frac{dM}{I}Q$, and since $dM/dx = V$, the shear flow per unit length is

$$q = \frac{VQ}{I}, \qquad \tau = \frac{VQ}{It}.$$

Two things about that are worth pausing on. The quantity is derived from equilibrium of a block, not from any new assumption — it is a free body chosen with a horizontal cut instead of a vertical one. And $Q$ is the same integral that locates a centroid, appearing here in a role that has nothing to do with centroids.

The peak is where the bending is zero #

The distribution has a shape that catches people out, and the shape follows directly from $Q$.

At the extreme fibre there is no material above the cut, so $Q$ is zero and the shear stress is zero. Moving inward, $Q$ grows, because more area is being counted. At the neutral axis $Q$ reaches its maximum, and below it $Q$ falls again by symmetry.

So the shear stress is largest at the neutral axis and zero at the faces — exactly the reverse of the bending stress, which is zero at the neutral axis and largest at the faces. The two quantities a cut reveals are complementary in that specific sense: neither is greatest where the other is.

For a rectangle the distribution is a parabola with a peak of

$$\tau_{\text{max}} = \frac{3V}{2A} = 1.5\,\frac{V}{A},$$

fifty per cent above the average. The generator behind these figures returns exactly $1.50$ for a rectangle, which is worth checking rather than asserting, because a “shear stress” quoted as force over area is a number that never occurs anywhere in the section.

The flow is continuous; the stress is not #

The formula produces two quantities, and the difference between them is where the practical consequences live.

The flow $q = VQ/I$ is a force per unit length along the beam. It varies smoothly, because $Q$ varies smoothly.

The stress $\tau = q/t$ is that flow divided by the width available to carry it. Where the width changes abruptly — at the junction of an I-section’s flange and web — the stress jumps by the ratio of the two widths.

That is why an I-section’s shear distribution has a step in it. In the flange, a wide plate carries a small flow at a low stress; in the web, a narrow plate carries almost the same flow at a stress many times higher. For a typical rolled section the web carries something like ninety-five per cent of the total shear, which is the justification for the everyday approximation that shear stress is the shear divided by the web area alone.

The step also explains the shape of the section. The flanges are there to carry bending and the web is there to carry shear, and the reason the division is so clean is that each quantity peaks where the other does not.

Ninety-five per cent, measured #

The claim that the web carries nearly all the shear is quotable and easy to check, and checking it is worth doing because the number turns out to be remarkably insensitive to the section.

Integrating the computed distribution over the section — the stress at each height, times the width there, summed up the depth — must return the applied shear, since that is what the distribution is a distribution of. For the I-section in the hero figure it returns $999.9$ against an applied $1{,}000$, which confirms the sweep rather than the theory.

Splitting that integral at the flange-web junctions gives the share. For this profile the web carries 94.9 per cent of the total, with the two flanges between them carrying five.

The everyday approximation follows: shear stress equals the shear divided by the web area alone, ignoring the flanges entirely. It overestimates by about five per cent, in the safe direction, and it requires none of $Q$, $I$ or the sweep. Practically every steel shear check in existence is done that way.

Whether that is a fact about this profile or a fact about shape is answerable by asking the same sweep of four sections at once.

What makes the approximation robust is that the flange’s contribution is small for a structural reason rather than a numerical accident. The flange is where $Q$ is smallest, because there is little area above it; the web is where $Q$ is largest and the width smallest. Both factors push in the same direction, and they do so for any section shaped to be efficient in bending. A section that is good at bending is automatically one whose shear is concentrated in its web — which is a coincidence worth noticing, and the reason a single section shape serves both purposes.

Two properties of the formula worth having #

$Q$ can be taken from either side of the cut. The first moments of the whole section about its own centroid sum to zero, so the area above a cut and the area below it have first moments equal in magnitude and opposite in sign. $Q$ is used as a magnitude, so it does not matter which side is counted.

That is a convenience rather than a curiosity. For an unsymmetric section — a tee, a channel with a heavy flange, a composite section with a slab on top — one side of a given cut is often a single rectangle and the other is three shapes and a fillet. Taking the easier one is free, and the agreement between the two is the cheapest available check on an arithmetic slip.

It is the same move as choosing which piece of a beam to sum, one scale down: the free body may be cut either way and the answer is the same, so cut it the way that is less work.

And the zeros at the faces are a boundary condition, not an accident of $Q$. A shear stress on the cut face is always accompanied by an equal one on the perpendicular plane — the complementary shear, and the reason the horizontal splitting stress in a beam equals the vertical shear stress at the same point. At the top and bottom faces of the beam that perpendicular plane is the free surface, and a free surface carries no stress across itself. So the complementary shear there is zero, and therefore so is the shear on the cut.

The algebra says $Q = 0$ at the extreme fibre. The physics says the stress had nowhere to pair with. They agree, and the second is the one that generalises: a shear stress cannot run out of a free boundary, which is why it vanishes at the tip of an outstand, at the edge of a flange, and at the corner of a rectangular bar in torsion — three results usually met separately and all of them the same sentence.

Where the diagonal tension goes #

Shear stress on its own is not what cracks concrete or splits timber. What does is the tension that shear implies, and locating it takes one more step.

A small element carrying pure shear — no direct stress — has principal stresses of equal magnitude in tension and compression, at forty-five degrees to the axis. So at the neutral axis of a beam, where the bending stress is zero and the shear is greatest, the material is being pulled apart along a line at forty-five degrees and squeezed along the perpendicular one.

Away from the neutral axis the two stresses combine, and the principal directions rotate. At the extreme tension fibre, where the shear is zero and the bending stress is largest, the principal tension is horizontal. Between them the direction swings smoothly, tracing a family of curves across the beam that were drawn by hand for the first time in the 1860s and are still the clearest single picture of what a beam is doing.

The crack pattern of a reinforced concrete beam is that family of curves made visible. Near mid-span, where bending dominates, the cracks are vertical and start at the bottom face. Toward the supports, where shear dominates, they lean over — the same cracks, rotated, following the principal tension exactly. A photograph of a cracked beam is a plot of principal stress trajectories, drawn by the beam.

That is also the design argument for links. Vertical bars cross the inclined cracks and hold the two sides together, so the beam continues to carry shear after the concrete’s tensile capacity is gone. The resulting model is a truss inside the beam: the concrete between the cracks acts as inclined compression struts and the links as vertical ties, which is strut-and-tie modelling arriving in a region that looked like an ordinary beam. Ritter proposed exactly this in 1899, and every concrete shear provision since is a refinement of it.

What it governs #

The horizontal shear is invisible in ordinary beam design and decisive in several specific places.

Timber. Wood is very weak in shear along the grain — a fraction of its strength across it — and the plane of maximum shear stress in a beam is exactly along the grain. So a timber beam that is short and heavily loaded splits horizontally at mid-depth near its supports, and the failure is a clean separation along the neutral axis rather than anything to do with bending. Deep timber beams and glulam members are routinely governed by this.

Glue lines and shear connectors. A composite member’s components must not slide, and the demand is $q$ per unit length. A glued laminated beam’s glue lines carry it continuously; a steel-and-concrete composite beam carries it through discrete studs, spaced according to $q$ — closely near the supports where the shear is large, widely at mid-span where it is not. Reading a stud layout tells the reader the shear diagram.

Web design in plate girders. A deep thin web can buckle in shear before it yields, in diagonal ripples running corner to corner, and the remedy is transverse stiffeners. A plate’s critical stress goes as the square of its thickness over its width, so a girder deep enough to be efficient in bending is automatically at risk in shear.

Bolted and riveted built-up members. Before welding, a plate girder was assembled from angles and plates joined by rivets, and the rivet pitch along the flange was set by the shear flow at that station. The rivets were doing what the glue does in the stack of planks, and reading their spacing along an old girder gives the shear diagram directly — closely spaced near the supports, opening out toward mid-span. It is one of the few places where a structural calculation is legible from the outside of the finished object.

Concrete. Diagonal tension is what shear does in a material with no tensile capacity. The principal tension at forty-five degrees to the axis cracks the web, and links — vertical bars crossing the crack — are what holds it together. A concrete beam without links fails in shear suddenly and with little warning, which is why the failure is treated with far more caution than bending.

The stack of planks, quantified #

The introduction’s stack is worth returning to with numbers, because it makes the value of the shear connection concrete.

Four planks each $b$ wide and $d$ deep, loose, have four separate second moments of $bd^3/12$, totalling $4 \times bd^3/12 = bd^3/3$. Glued into one member of depth $4d$, the second moment is $b(4d)^3/12 = 16bd^3/3$.

A factor of sixteen in stiffness, from glue. The strength ratio is a factor of four, since the section modulus goes as the square rather than the cube. Nothing was added but a connection.

The reason the gain is so large is the parallel-axis term: each plank in the glued member contributes not only its own small second moment but its area times the square of its distance from the combined neutral axis, and that transfer term is what the sliding was preventing. A stack that slides has no transfer terms at all.

Which reframes the whole subject of shear connection. It is not a detail that stops components moving; it is the mechanism by which several small sections become one large one, and the entire benefit of composite construction rests on carrying $q$.

So the useful question is not what the connection is worth but what it has to carry, and the same figure answers it as soon as it is given a real shear and a connector to size.

That spacing is the price of the factor of sixteen. It is also why the shear connection is the item that decides whether composite construction is worth doing on a given member: the benefit is fixed by the geometry and the cost is fixed by $q$, which rises with the shear rather than with the moment — so the connection is most expensive exactly where the bending gain is least, at the ends of the span.

Where the model stops #

The formula assumes the flange stresses are the beam-theory ones. $q = VQ/I$ was derived by differencing $My/I$ across a short length, so it inherits every assumption behind that expression — plane sections, elasticity, slenderness and distance from loads.

Thin-walled sections need a different reading. For an open thin-walled section the flow runs along the walls rather than across the section, and the formula gives the flow at a point on the wall’s path. For a channel, that flow in the flanges produces a couple, which is why the load must be applied through a point outside the section if the member is not to twist.

Non-rectangular widths. At a point where the section’s boundary is not perpendicular to the neutral axis — the sloping face of a tee’s fillet, or a circular section — the “width” in the denominator is an approximation and the stress varies across it.

Sudden changes. At a re-entrant corner the elastic shear stress is theoretically unbounded, and the real value depends on the fillet radius. Formulae give nominal values there and nothing more.

Shear deformation is not included. The formula gives a stress, not a movement. The deflection due to shear is a separate calculation with $GA$ in it rather than $EI$, and for a deep or short member it is not negligible.

The figures carry a specific limitation. The distribution is drawn as a smooth curve against the section, which implies the stress is uniform across the width at every height. It is not — even in a rectangle the shear stress varies slightly across the breadth, and $VQ/It$ is the average across it. The approximation is excellent for a narrow section and progressively worse for a wide flat one, which is the same slenderness condition every other result on this site depends on, appearing in an unexpected direction.

The ladder from here #

Later rungs on this anchor: shear flow in thin-walled open and closed sections. The shear centre. Shear connection design and stud spacing. Shear buckling of webs and tension-field action. Diagonal tension and link design in concrete. Interface shear between precast and in-situ concrete. Shear in timber and its grain dependence. Rolling shear in cross-laminated panels. And shear deformation, which is the movement this stress produces and which beam theory omits entirely.

Jourawski derived the formula in 1855 while investigating why the timber beams of the St Petersburg to Moscow railway bridges were splitting horizontally. The failures were not bending failures, the theory of the day had nothing to say about them, and the expression he produced is the one used unchanged today.