The deflection that is not bending

A beam loaded across the middle sags, and the shape it sags into is the second integral of its bending moment diagram. That sentence has carried every deflection on this site so far. It is the reason a span to the fourth power decides so much, the reason one deflection can be had without solving the rest of the structure, and it is short by a term.

The missing term is not small print. A beam under load does two things at once: it curves, and it goes out of square. The second of those is a movement in its own right, it obeys a different diagram, and nothing in the double integral of a moment contains it.

The lower panel is the finding. It has a kink in it and no curvature anywhere, which is not a shape any amount of moment-diagram integration will ever produce.

The term that was dropped, and where it went #

Engineer’s beam theory rests on one assumption, stated in two halves. Plane sections stay plane, and they stay perpendicular to the axis. The first half is the whole of bending theory — it is what makes strain linear in depth and everything after it arithmetic. The second half is quieter and is a claim about shear: a section that stays perpendicular to a deflected axis has not been racked at all, which is to say the beam has no shear strain in it.

Every beam has shear strain in it. A simply supported beam under a central load carries $P/2$ of shear over each half, that shear is a real stress on a real area, and $\gamma = \tau/G$ is a real strain. Assuming it away makes it invisible rather than zero.

So the honest deflection is a sum of two terms:

$$\delta = \underbrace{\int \frac{M m}{EI}\,dx}_{\text{bending}} + \underbrace{\int \frac{V v}{G A_s}\,dx}_{\text{shear}}$$

and the second is computed by exactly the same virtual work as the first, over the shear diagram instead of the moment diagram, with $G A_s$ where $EI$ was. Nothing new is needed. The term was never hard; it was simply never written down.

A slice that goes out of square #

Naming the free body settles what kind of movement this is.

Cut a slice of beam of length $dx$ between two adjacent sections. The two cut faces carry equal and opposite shears $V$. Take that shear as a uniform stress $\tau = V/A_s$ over an effective shear area $A_s$, and the material responds with $\gamma = \tau/G$. That strain is an angle: the two faces of the slice rotate relative to one another, and the slice becomes a parallelogram. It has not curved. Its top and bottom are still the same length as each other, which is exactly what bending is not.

Accumulate that angle along the beam and the result is a displacement:

$$\frac{dv_s}{dx} = \gamma = \frac{V}{GA_s}$$

The shear deflection is therefore an accumulated slope, not an accumulated curvature. That single fact produces every shape on this page.

$A_s$ is the one quantity in the sum that is a rule rather than a measurement, and it comes from a distribution this collection has already drawn.

For a rectangle, then, $A_s = \tfrac{5}{6}bd$, and the ratio that matters below is

$$\frac{I}{A_s} = \frac{bd^3/12}{\tfrac{5}{6}bd} = \frac{d^2}{10}$$

for every rectangle there has ever been, whatever its proportions.

Which free body produced the number #

Both halves of the sum are virtual work, and both are taken over diagrams the site already draws.

For the central point load the shear integral is small enough to do by hand and worth doing, because it shows how little there is to it. The real shear is $P/2$ over each half of the span, and the unit load’s shear is $1/2$ over each half, with both changing sign together at mid-span so their product never does. Then

$$\delta_s = \frac{1}{GA_s}\int_0^L V v\,dx = \frac{1}{GA_s}\left(\frac{P}{4}\cdot\frac{L}{2}\right)\times 2 = \frac{PL}{4GA_s}$$

against the familiar $PL^3/48EI$ for the bending half. The generator behind these figures uses neither closed form: it integrates both products over the sampled diagrams and then checks the answers against the algebra, holding the shear integral to machine precision. A sign slip in the new term has somewhere to show.

Two things fall out of that pair before any arithmetic is done. The bending term goes as the cube of the span and the shear term as the first power, so they do not scale together at all. And the shear term contains no $I$ — the property that rewards putting material far from the middle does nothing here.

Q over lambda squared #

Dividing one by the other kills the load and leaves a pure statement about shape:

$$\frac{\delta_s}{\delta_b} = \frac{k\,(E/G)\,(I/A_s)}{L^2} = \frac{Q}{\lambda^2}, \qquad \lambda = \frac{L}{d}$$

where $k$ belongs to the load case — 12 for a central point load on a simple span, 9.6 for a uniformly loaded one, 3 for a cantilever under a tip load, 4 for a uniformly loaded cantilever — and $Q$ belongs to the section and the material together. For a steel rectangle, $E/G = 70/27$ and $I/A_s = d^2/10$, so

$$Q = 12 \times \frac{70}{27} \times \frac{1}{10} = \frac{28}{9} = 3.111$$

and the shear share of the total deflection is $Q/(Q + \lambda^2)$.

Those three crossings are the practical content of the whole essay. An ordinary floor beam is right to ignore this term and a transfer beam is not, and the boundary between them is a span-to-depth ratio around five — a proportion a reader can see across a room.

Two of the crossings are exact rather than fitted. The ten per cent crossing is at $\lambda = \sqrt{28} = 5.2915$, because $9Q = 28$ exactly. And at $\lambda = 2$ the shear share is $\tfrac{7}{16}$ — not approximately, exactly: the ratio there is $Q/4 = 7/9$, so the split is 28 parts shear to 36 parts bending, and $28/64$ is seven sixteenths.

The second curve on that figure decides what kind of quantity this is. Steel and concrete differ by a tenth in $E/G$ and the two sweeps are indistinguishable, while doubling the depth multiplies the ratio by four. The material is very nearly irrelevant and the geometry decides, which is the verdict this collection also reaches about section shape and about the one number a stronger steel does not change.

The shape is the argument #

Being a percentage is the least interesting thing about the shear term. It is a different curve, and the difference is visible in a single glance once the two are drawn apart.

Bending deflection is the moment diagram integrated twice. Shear deflection is the shear diagram integrated once. So the shear shape carries one degree of curvature less than the bending shape does — and since the integral of the shear diagram is the moment diagram, something rather better than an analogy holds:

$$v_s(x) = \frac{1}{GA_s}\int_0^x V\,dx = \frac{M(x)}{GA_s}$$

for a simply supported span, where $M(0) = M(L) = 0$ and the boundary conditions need no help. The shear deflected shape of a simple beam is the bending moment diagram, to a scale of $1/GA_s$. One curve on the page is the moment diagram divided by a stiffness; the other is that same diagram integrated twice more and divided by a different one. The generator’s two routes to that shape — a running integral along the beam, and a virtual-work product at every station — agree to one part in $10^{19}$.

Under a central point load the moment diagram is a triangle, so the shear deflection is two straight lines meeting under the load, which is the kink in the hero figure. Under a uniformly distributed load the moment diagram is a parabola, and so is the shear deflection.

A cantilever under a tip load takes the third case: constant shear over the whole length, so a single straight line from root to tip, with the whole of the tip movement arriving as an accumulated racking with no curvature in it at all. Its $Q$ is $3 \times \tfrac{70}{27} \times \tfrac{1}{10} = \tfrac{7}{9}$, four times smaller than the simple span’s, so a cantilever is more forgiving of this term at the same slenderness than a beam is — 4.64% at $\lambda = 4$ against the hero figure’s 16.3%.

Two beams at the same slenderness are not the same beam #

$Q$ contains $I/A_s$, and slenderness does not know anything about it.

The mechanism reverses the ranking every other page here gives. An I-section is the section that wins at bending: the flanges are far from the neutral axis and carry almost the whole moment, which is why $I$ is enormous for the area. Those same flanges carry almost no shear, because the shear flow that reaches them runs horizontally — the shear centre essay’s picture of where the flow goes is the same picture read for a different purpose. So the section with the largest $I$ has the smallest $A_s$, and $I/A_s$ — the whole of $Q$ — is worst for exactly the shape that is best at bending.

Optimising a section for bending makes its shear deflection worse, not as a side effect but as the direct arithmetic consequence of what the optimisation does.

That last number is the one to carry. A rectangle has to be squatter than about 5.3 spans-to-depths before it loses a tenth of its movement to shear; a 600 mm I-section on a 10 mm web is still there at 9.3, which is an entirely conventional proportion for a steel beam over a wide opening.

The member for which bending theory is the wrong model #

At the deep end the term stops being a correction at all.

Nothing about that panel’s section is unusual. Its $Q$ is 3.111 — the identical number the 300 × 600 beam had, because it is a rectangle and every rectangle has $I/A_s = d^2/10$. What is unusual is what it has been asked to cross. At $\lambda = 1$ the shear term is more than three times the bending one, and a calculation that stopped at $5wL^4/384EI$ has under-reported the movement by a factor of four.

For that member bending theory is not inaccurate, it is the wrong model, and the reason is visible in the assumption rather than in the answer.

The two figures are the same finding twice. Admitting shear strain means admitting that a section warps out of plane — the shear stress is a parabola, so the shear strain is a parabola, so the section takes up an S-shape rather than a tilt. Every number on this page was computed with $A_s$ and a uniform racking, which is the tidy approximation to that warping rather than the warping itself, and past $\lambda \approx 2$ neither the tidy version nor the beam theory beneath it has any claim on the answer.

Deep members are not a curiosity. A column that stops and hands its load to something beneath usually hands it to a wall or a storey-deep girder, and that essay’s transfer wall deflects 4.34 mm of which 41% is shear — the same argument arriving as a settlement everything above inherits. The core of a tall building is the same member stood on end, and its sway is substantially shear rather than the flexural curve a hand calculation draws.

The same question asked of a truss #

The generalisation runs sideways rather than deeper, into a structure that has no continuum in it at all.

A truss deflects because its members change length, and each member’s share is F·f·L/EA. Split that sum into chords and web and the two halves are the same two terms as this page: the chords are carrying the moment and the diagonals are carrying the shear. A shallow truss is nearly all chord — 86% of the movement at a depth of 0.4 — and a deep one is nearly all web, at 64%. The crossover is a span-to-depth argument in a structure with no sections in it, and it lands in the same place, for the same reason: deepening a member cheapens its moment path and leaves its shear path where it was.

That is the strongest available statement of what the dropped term is. It is not a property of continuous beams at all. It is what a structure does when the load has to travel across the member rather than along it, and every way of building a member pays for that journey.

Where the model stops #

The shear area is a strain-energy equivalence, not a stress. $A_s = \tfrac{5}{6}A$ does not mean five sixths of a rectangle carries shear; it means a uniform stress on that area stores the same energy as the real parabola does on the whole of it. For the I-sections above the web is credited at the overall depth rather than the clear depth, which errs generously — every shear deflection quoted here is, if anything, an underestimate.

Plane sections do not stay plane once shear is admitted. The correction used above tilts the section and keeps it flat, which is Timoshenko’s model rather than the truth. Past a span-to-depth ratio of about two the error in that is comparable to the thing being computed.

The load case sets $k$, and no other load case was solved. The four constants quoted belong to four standard arrangements. A beam under a real pattern of loads has its own $k$, obtained by doing the two integrals and dividing.

Concentrated loads and supports are local. Near a point load the stress field has nothing to do with any beam theory, and a shear deflection computed from a diagram that jumps is a smooth approximation to a very unsmooth thing.

None of this is inelastic. Everything above is $\gamma = \tau/G$ with a constant $G$. A cracked concrete member’s shear stiffness is a small and uncertain fraction of $GA_s$, which makes its shear share larger than these figures and much less knowable — and a member that has yielded has left the calculation entirely.

What the pictures cannot show #

Every deflected shape on this page is drawn at an exaggeration printed in its own caption, and the factors are enormous: 11,932 times in the hero, 274,225 in the wall panel. Nothing here is a picture of a beam. The honest reading is the ratio in the caption — 1 in 79,116 of the span for the first, 1 in 1.5 million for the last — and the drawing is a diagram of a proportion.

The kink under the point load is the second distortion and the more misleading one. It is drawn sharp because the model says the shear diagram is discontinuous there, and a real beam has no such point: the jump is smeared over a length of the order of the depth, the section warps rather than tilting, and the corner is rounded. The two straight limbs are real to within two parts in $10^{16}$ of the model, which is a statement about the arithmetic and not about the beam.

The assumption every one of these figures rests on has been named twice above because it decides everything: the section is racked uniformly through $\gamma = V/GA_s$. A figure that showed the true warped section would show no single deflection curve at all, because different fibres would have moved by different amounts — and the whole idea of the deflection of a beam would go with it.

The ladder from here #

Later rungs on this anchor: the shear coefficient $\kappa$ derived by strain energy for the standard families of section. Timoshenko beam theory set out properly, with the rotation of the section as an independent variable. The effective flexural stiffness $EI/(1 + Q/\lambda^2)$, which depends on the span and is therefore not a section property at all. Shear deflection in redundant structures, where softening one member in shear redistributes force to another. Deep beam design by strut-and-tie, which abandons the section entirely and is the right answer at $\lambda < 2$. Shear deformation in coupled shear walls, where it governs the coupling beam. Shear lag, the same failure of the plane-section assumption seen along the flange instead of through the depth. And sandwich construction, where a weak core makes the shear term dominant at slendernesses no solid member would ever reach.

The term itself is old: Rankine had it in the 1850s, and Timoshenko put the rotation of the section into the differential equation in 1921, which is why the two-term beam carries his name. What is not old is the habit of leaving it out. That arrived with the teaching of the simple theory, and it survives because the members it is learned on are slender, the crossings sit where they do, and a term worth 1% of the answer for the first ten years of a career is indistinguishable from a term that does not exist.