The shear that moves the moments

Some of a beam’s deflection is not bending, and the stiffness that results belongs to the span rather than to the section. Both of those are statements about movement. In a simply supported beam, movement is all that shear deformation can change: the reactions came from equilibrium before anything deformed, the moment diagram came from the reactions, and a beam can be made thirteen times more flexible in shear without a single moment in it moving.

A redundant beam is different in kind. It has a support more than equilibrium needs, and how the load divides between its routes to the ground is decided by stiffness. Stiffness enters through movement, so every flexibility that enters the movement enters the forces. Shear deformation is one of them, and for a stubby beam it is not a small one.

Statics settles a determinate beam before it deforms #

In a simply supported beam, taking moments about one support gives the other reaction, and the bending moment at every section follows from the reactions and the load. Nothing in that calculation knows what the beam is made of. It could be steel, timber or foam, rigid in shear or as soft as a sandwich core, and the moment diagram would be the same diagram.

Add a support and the calculation stops one equation short. A beam fixed at one end and propped at the other has three unknown reactions and two useful equations of equilibrium, so equilibrium describes a family of possible moment diagrams, one for every value of the prop’s force. Every member of that family carries the load and balances. The beam picks one of them, and the only thing it can pick by is how it deforms: the prop does not move, so the prop’s force is whatever keeps the beam’s end at the prop’s level.

That condition is compatibility, and it is where the deformation enters. A support that settles by ten millimetres produces a set of moments with no load on the beam at all, for the same reason: the forces in a redundant structure are whatever its movements require. Anything that changes the movement for a given force changes the force for a given movement. Shear deformation changes the movement.

The propped cantilever, counted with both terms #

The calculation is short enough to do in full, and doing it shows exactly where the shear term acts. Choose the prop as the thing to take away, and the beam becomes a cantilever carrying its load, with a free tip that drops. Under a uniform load $w$ the tip drops by a bending part and a shear part:

$$\delta_w = \frac{wL^4}{8EI} + \frac{wL^2}{2GA_s}$$

The prop’s force $R$, pushing up at the tip, lifts it by

$$\delta_R = R\left(\frac{L^3}{3EI} + \frac{L}{GA_s}\right)$$

and the prop’s force is the one that makes the two equal. Writing $\varphi = 12EI/(GA_sL^2)$ for the ratio of the two flexibilities, the wall moment comes out as

$$M_{\text{wall}} = \frac{wL^2}{2(4+\varphi)}$$

which is $wL^2/8$ when $\varphi$ is zero and less for every beam that shears.

The shear term appears twice, in the drop under the load and in the lift under the prop, and it is proportionally larger in the drop. Shear deflection goes as the square of the span and bending as the fourth or the third, so the shear part of the cantilever’s drop under a distributed load is a larger fraction than the shear part of its lift under a concentrated tip force. The prop has more to lift than bending theory says, so it pushes harder; the prop’s extra force comes off the wall.

For a solid rectangle, the shear area is five-sixths of the gross area, and with steel’s $E/G$ of 2.59 the flexibility ratio is

$$\varphi = 1.2\,\frac{E}{G}\left(\frac{d}{L}\right)^2 = \frac{3.11}{(L/d)^2}$$

At a span-to-depth ratio of 4 that is 0.194 and the wall moment is 0.119 $wL^2$, five per cent below the textbook value. At 2 it is 0.778 and the wall carries 0.105. At 1 it is 3.11 and the wall carries 0.070, which is 56 per cent of what bending theory gives it.

Where the wall’s moment goes #

The load has not changed, so the moment the wall gives up has to go somewhere, and equilibrium says where. The prop’s share of the load rises from 0.375 to 0.381, 0.395 and 0.430 across the three beams. A larger prop force means a larger sagging moment in the span. The largest sagging moment is $9wL^2/128$, or 0.0703 $wL^2$, rigid in shear. At a span-to-depth ratio of 2 it is 0.0782, and at 1 it is 0.0923. The span moment has grown by 31 per cent while the wall moment fell by 44, and the section of peak sagging has moved from 0.625 of the span toward the wall, to 0.571.

That is a design consequence, not a refinement. A stubby beam reinforced for the moments bending theory gives it is over-reinforced at the wall and under-reinforced in its span, by amounts that grow as the square of its depth over its span. Both moment diagrams satisfy equilibrium exactly, so no equilibrium check on the drawings can tell them apart. Only a calculation that counts the shear in the compatibility condition can.

Neither half of the shape reaches the prop #

The deflected shape shows the mechanism more directly than the algebra does.

The two parts are drawn from the same moment and shear diagrams, the real ones for this beam. The bending part is the moment diagram integrated twice from the wall. The shear part is the shear diagram integrated once, and it has no curvature where the shear is constant and no kink except where the shear jumps. Neither part meets the prop. The bending part overshoots it upward, because the prop’s force has been raised until it bends the beam back past level. The shear part falls short of it downward, because the shear strain accumulated along the span has nowhere to go. The two misses are equal and opposite, which is the compatibility condition, drawn.

Two further details of the shape are only visible with the shear drawn separately. At the wall the fixed end holds the section square to the wall, and nothing holds the centreline, which leaves the wall at a slope equal to the shear strain there. A fixed end in a beam that shears is a statement about the rotation of the section and not about the slope of the axis, which is the distinction Timoshenko’s beam theory was built on. And the largest movement, 2.71 times the rigid-in-shear value, has moved from 0.579 of the span to 0.531, toward the wall, where the shear is largest.

How stubby a beam must be #

The effect scales with $\varphi$, and $\varphi$ is a property of the proportion and the section together, so the useful question is where on the scale of slenderness it stops being negligible.

The two curves are one formula, $4/(4+\varphi)$, with different coefficients inside $\varphi$. The rectangle’s is 3.11 over the span-to-depth ratio squared. The I-section’s is 9.62, three times larger, and the reason is the ratio of its second moment to its shear area. An I-section puts its material in its flanges to raise $I$ and carries its shear on its web alone: 6,000 mm² of shear area out of 11,700 mm² of section, against 150,000 of 180,000 for the rectangle. Its $I/A_s$ is 111,300 mm² against the rectangle’s 36,000. The shape that makes a section efficient in bending is the shape that makes it flexible in shear, and a redundant I-beam’s forces notice at slendernesses where a rectangle’s do not.

A span-to-depth ratio of 4 is not a curiosity for a steel girder. A transfer girder carrying a column over an opening is often that deep for its span, and it is usually built into the frame at both ends, so it is redundant as well as stubby.

The case where nothing moves #

If shear flexibility moves the forces of every redundant beam, it is fair to ask why the effect is not common knowledge. Part of the answer is that the redundant beam most often used to build intuition is exactly the one where it vanishes.

A beam fixed at both ends has two redundants, and two compatibility conditions to find them: the far end’s section does not turn, and the far end does not move relative to the near one. The first involves only bending, since shear strain does not rotate a section. The second involves both terms, and the shear term in it is the shear strain integrated along the span,

$$\int_0^L \frac{V}{GA_s}\,dx = \frac{M_L - M_0}{GA_s}$$

because shear is the derivative of moment. Under a symmetric load the two end moments are equal, so that integral is zero, and the shear flexibility drops out of both compatibility conditions entirely. The moments are the bending-theory moments at every slenderness. The movement is not: the mid-span deflection is $(1 + 4\varphi)$ times $wL^4/384EI$, which at a span-to-depth ratio of 1 is thirteen times the familiar value.

The beam that would reveal the effect, if anyone tried shear flexibility on it, is the one guaranteed to show nothing.

A point load moves moment from one end to the other #

Break the symmetry and the effect returns, in a form that is easy to state.

Rigid in shear, the end nearer the load takes three times the moment of the far end. As the beam gets stubbier the two converge. The limit is a beam with no shear stiffness worth the name, which cannot afford any integrated shear strain between its ends, so its end moments must be equal. Equal end moments that together with the load satisfy equilibrium are $Pab/2L$ each. At a span-to-depth ratio of 1 the beam is three-quarters of the way there: the near end has lost a quarter of its moment and the far end has gained three-quarters.

So the bending-theory end moments are safe at one end and unsafe at the other. A stubby fixed beam designed with them has a near end with a quarter more capacity than it needs and a far end whose real moment is 75 per cent more than it was designed for.

The number every hand method passes along #

Moment distribution clamps every joint, releases one, and passes half of the released moment along each member to its far end. The half is the carry-over factor, and the stiffness that decides how the released moment is shared out is $4EI/L$. Both numbers come from a member turned through a unit rotation at one end with its far end fixed, and both assume the member is rigid in shear.

Counting shear, the turned end’s stiffness is $(4+\varphi)/(1+\varphi) \cdot EI/L$ and the carry-over factor is $(2-\varphi)/(4+\varphi)$. The first is unsurprising: a member that shears is less stiff, and at large $\varphi$ it tends to $EI/L$, a quarter of its rigid value. The second is the striking one. The carry-over factor passes through zero at φ = 2 and becomes negative, tending to minus one.

The physical reason is the same integral that made the symmetric beam immune. Turning one end with the far end held produces two end moments, and the shear in the member is their sum over the span. That shear strains the member and moves one end relative to the other, and the far end is held against moving. Rigid in shear, the ends stay level by bending the member into double curvature, with the far end’s moment half the near end’s and in the same rotational sense. Soft in shear, keeping the ends level requires keeping the shear small, which requires the far end’s moment to oppose the near end’s. In the limit the member can carry no shear at all, the two end moments are equal and opposite, and the far end is turned back as far as the near end is turned forward.

A concrete coupling beam joining two shear walls is a member of exactly this kind. At a span-to-depth ratio of 1, uncracked, with $E/G = 2.4$, its $\varphi$ is 2.88 and its carry-over factor is −0.128. Diagonal cracking reduces its shear stiffness far more than its flexural stiffness, so the cracked value is further past the sign change than that. The coupled systems a tall building is braced by are analysed with members where the textbook half is not merely inaccurate but has the wrong sign.

A frame program’s beam element carries the same $\varphi$ once shear deformation is switched on. The stiffness matrix that replaced moment distribution then has $(4+\varphi)$, $(2-\varphi)$ and $(1+\varphi)$ in its terms, and an analysis told to ignore shear deformation sets $\varphi$ to zero in every member at once — including the stubby ones, where it is the difference between a half and a negative number.

Where else a redundant structure is soft in shear #

A truss’s web is its shear stiffness. The diagonals and verticals of a truss supply the movement a solid beam’s shear term would, so a redundant truss — continuous over three supports, or fixed at its ends — has its reactions set partly by its web, and a deep truss with a light web redistributes like a stubby beam.

A sandwich panel continuous over supports has the same property at slendernesses no solid beam reaches. Its $E/G$ is the face modulus over the core’s shear modulus, in the thousands, and a two-span panel’s middle support moment is correspondingly smaller than bending theory gives.

A support settlement in a stubby beam produces less moment. A beam fixed at both ends whose ends move apart vertically by Δ carries end moments of $6EI\Delta/L^2$ rigid in shear, and $6EI\Delta/(L^2(1+\varphi))$ counting it. The effect that made a redundant beam sensitive to its supports is itself softened by shear.

Where the model stops #

The shear area is an approximation. Five-sixths for a rectangle and the web alone for an I-section are energy-based coefficients that depend a little on Poisson’s ratio and on how the section is idealised, and they matter more here than in a deflection, because $\varphi$ appears in a denominator with the bending term.

Below a span-to-depth ratio of about 2, beam theory itself is failing. Plane sections do not stay plane near supports and loads, and a member that short has no section to design. The Timoshenko beam is the last model that still treats the member as a line; a strut-and-tie model is the one that replaces it, and the trend the figures show is the right trend even where their numbers stop being reliable.

Cracking changes φ. Reinforced concrete loses shear stiffness after diagonal cracking faster than it loses flexural stiffness, so a cracked stubby beam is further along every curve here than an uncracked calculation places it.

The distribution is elastic. A ductile beam loaded to collapse redistributes moment by forming hinges, and its collapse load does not depend on the elastic distribution at all. The shear-induced shift matters for cracking, deflection, fatigue and the service stresses in members that are not ductile — which is to say for everything a stubby concrete member is actually checked for.

Still open: the coupling beam that is asked for shear it cannot supply #

The carry-over factor’s change of sign is where the open questions start. A coupled shear wall, where the coupling beams are the stubbiest members in the building and their shear flexibility decides how much of the overturning moment the pair of walls carries as a couple, against how much each wall carries alone. The shear coefficient derived by strain energy for the families of section, since every number here inherits it. Timoshenko’s theory set out with the section rotation as an unknown in its own right, which is what the fixed end’s slope has already demanded. And a frame with a transfer girder analysed both ways, to find which column the shear term unloads and which it loads.

The redundant beam that appears first in every table of fixed-end moments is the fixed-ended beam under a uniform load, and it is the one case in which shear flexibility changes no force, so a check of the effect made on that beam finds nothing to see. Every redundant beam with less symmetry than that one shows it, and a stubby one shows it in tens of per cent.