The stiffest path takes the load

Two steel beams of the same span lie side by side, one twice as deep as the other, and a bolt through both of them at midspan ties them together. A hundred kilonewtons is hung off the bolt. The question is how much of it each beam carries, and there is a strong temptation to answer it by looking at the picture.

The picture has nothing to say. The load is at one point, on one bolt, and the bolt is on both beams. No area can be allocated, no half can be taken, no line can be drawn on the plan that divides anything. What settles it is a condition that has not yet been written down anywhere in this collection as an equation in its own right: the two beams are bolted together, so they move the same distance.

Eleven point one against eighty-eight point nine, from two beams that differ in one dimension by a factor of two. That ratio is the whole subject, and everything below is a consequence of it.

Sharing a force is a decision; sharing a displacement is not #

Most divisions of load in this collection are decisions. The load a beam is given comes from a tributary rule that somebody chose, and two defensible choices differ by sixty per cent on the same floor. Both are legitimate because equilibrium objects to neither: any division that hands over the whole load and puts it in the right place satisfies statics, so statics declines to arbitrate.

Sharing a displacement is a different category of thing. The moment two members are attached at a point, that point has one position, and whatever each member does it does to the same distance. That is not a modelling choice and not a rule of thumb — it is a geometrical fact about a structure that has not come apart, and it supplies exactly the equation statics is short of.

$$P = R_1 + R_2 \qquad\text{(equilibrium: one equation, two unknowns)}$$

$$\delta_1 = \delta_2 \qquad\text{(compatibility: the missing one)}$$

With $\delta = R/k$ for each member, the second line becomes $R_1/k_1 = R_2/k_2$, and the two together give

$$R_i = P\,\frac{k_i}{\sum k}.$$

That is the whole of load sharing. It is the same shortage of equations that makes a propped cantilever indeterminate, met in its smallest possible form: two members, one point, one degree of redundancy. And the resolution is the same one — a statement about movement, imported into a problem that thought it was about force.

Which free body produced the split #

The free body is the bolt. Cut it out of the structure and draw it alone: the applied 100 kN pushes it down, and the two beams push it up with $R_1$ and $R_2$. Vertical equilibrium of that one small body gives $R_1 + R_2 = 100$ and nothing else — no moment equation is available, because everything acts through the same point, and no other cut anywhere in the structure produces a second independent equation.

The second equation comes from a different cut. Take each beam on its own, with $R_i$ applied upward at midspan, and integrate its curvature twice. The standard result is $\delta = RL^3/48EI$, so the point stiffness of a simply supported beam loaded at its centre is

$$k = \frac{48EI}{L^3}.$$

For the pair drawn above, $L$ is the same and $E$ is the same, so the stiffness ratio is the ratio of the second moments — and second moment goes as depth cubed. Twice the depth is eight times the stiffness, and eight times the stiffness is eight times the force: $100 \times 8/9 = 88.9$ kN into the deep beam, $100 \times 1/9 = 11.1$ kN into the shallow one.

Cutting each beam again at midspan gives the moments that follow, $RL/4$: 133.3 kNm in the deep beam and 16.7 kNm in the shallow one. And both midspan points have moved 5.00 mm, which is the check — two independently computed deflections that had to agree, and do.

Nothing about the load entered any of that. The same pair under 40 kN divides it 4.4 and 35.6; under a moving axle it divides every position the same way. The split is a property of the structure, not of the loading.

The case where the drawing has nothing to divide #

A pair of beams bolted together is a slightly artificial arrangement. Two beams crossing each other, with a column landing where they cross, is a floor plan, and it is the case where the tributary habit fails hardest.

The short beam takes three quarters of a load that a plan drawing splits down the middle. The arithmetic is one line: both beams have the same $EI$, so their point flexibilities are $L^3/48EI$ and the stiffnesses go as $1/L^3$, giving

$$\text{share of the 6 m beam} = \frac{1/6^3}{1/6^3 + 1/9^3} = \frac{9^3}{6^3 + 9^3} = \frac{729}{945} = 77.1\%.$$

The cube is the entire content of that result. A three-metre difference in span moves three quarters of the load onto one member, and it does so with both beams made of the same steel in the same section. Geometry beats material outright here: changing $E$ changes nothing at all, because it appears identically in both flexibilities and cancels.

Set beside it what the tributary method would say about the same floor.

That division is honest, closes to the last square metre, and answers a question the crossing-beam case does not pose. The tributary rule works by area, and a point load standing on two members at once has none. Where there is an area the rule is a defensible approximation; where there is not it degenerates into “half each”, the one answer compatibility never gives unless the members are identical.

Pushing the spans further apart makes the plan drawing’s failure absurd.

A designer who sized the long beam for half the load has bought a member that is doing almost nothing, and has under-sized the short one by a factor of nearly two. Both errors come from the same source: a picture that shows where things are and says nothing about how stiff they are.

The whole curve, and the bound it never reaches #

One member’s share is $r/(1+r)$, where $r$ is its stiffness measured in units of everything it shares with. That function is worth drawing in full, because it has a property that is easy to state and easy to forget.

The share always falls short of the stiffness ratio, because a member’s own stiffness is part of the total it is being divided by. Doubling a member’s stiffness never doubles its force; it cannot, since half of what it competes against is itself.

Run backwards, that is the useful half. To halve the force in the shallow beam of the hero figure, its stiffness has to fall not to a half but to 0.47 of what it was — and for two equal members the number is a third. Softening sheds load onto the neighbour, the neighbour deflects more, and some of the shed load comes straight back. Every intervention in a shared-displacement system is partly undone by the system’s response to it.

Load sharing does not need two separate members. A slab spanning both ways is the same argument with the members drawn as strips of a continuum.

The exponent is four rather than three because a strip carries a distributed load rather than a point one, and $5wL^4/384EI$ replaces $PL^3/48EI$. Everything else is identical, including the conclusion: two-way action is worth having at a ratio of one and worth almost nothing by two. No amount of reinforcement in the long direction changes it, because the long strips are not declining the load — they are outvoted by a stiffness they cannot alter without becoming deeper.

The same argument with the load taken away #

Compatibility does not need a load to enforce. Impose a movement on the tie instead of a force and every member develops whatever force that movement demands of it.

The change is small on the page and large in what it means. Under an applied force the total is fixed and the members argue over it. Under an imposed movement the total is whatever the structure happens to add up to, and making a member stiffer increases the total rather than redistributing it. That is the regime of a support that has settled, of a temperature change nobody applied, and of jacking, shrinkage and construction misfit — every case where the input is a distance.

It is also where the bound of the previous section is reached exactly. Doubling the shallow beam’s second moment under an applied force raised its force by 1.80; under an imposed movement it raises it by exactly 2, because $R = k\delta$ and $\delta$ is not free to change.

Stiffening the member that is not working #

Here is the finding, and it inverts the reflex that produced the question in the first place.

The two curves come from the same dimension. Multiplying $I$ by $f$ means multiplying depth by $f^{1/3}$ and section modulus by $f^{2/3}$, so

$$\frac{\text{force after}}{\text{force before}} = \frac{f(k+K)}{fk+K}, \qquad \frac{\text{capacity after}}{\text{capacity before}} = f^{2/3},$$

and the stress multiplier is the first divided by the second. Setting the two equal and solving for the share $s = k/(k+K)$ gives the break-even line:

$$s^{*} = \frac{f^{1/3}-1}{f-1}.$$

For a doubling that is 0.260. For a quadrupling it is 0.196. And in the limit of an infinitesimal stiffening it is exactly one third — so any member carrying less than a third of a shared load is made worse by the first millimetre of extra depth. The number has no material in it, no span in it and no load in it. It is the cube root of two, minus one, and it decides whether a perfectly sensible-looking repair helps or hurts.

The shallow beam of the hero figure sits at 11%, well below the line. Doubling its second moment takes its force up 80%, its capacity up 59%, and its stress up 13%. It has been made deeper, heavier and more expensive, and it is worse off than before.

Two readings follow. First, the members that respond well to stiffening are the ones already carrying most of the load — a member on 80% of a shared load is nearly alone, and its capacity gain outruns the small extra share it can pick up. Second, the way to relieve an overstressed member sharing a displacement is to soften something, or to stiffen its neighbour; adding material to the victim is the intervention with the worst return in the subject.

That is why a strengthening scheme can fail in the direction nobody planned for. A stiff new element added beside a flexible one takes load out of all proportion to its size, which is the mechanism behind the redistribution nobody chose at a connection.

Springs in parallel, springs in series #

The same algebra read from the other end produces the opposite advice, and the pair is worth holding together because confusing them is easy.

Members sharing a displacement are springs in parallel: stiffnesses add, the stiffest one dominates, and the softest is nearly irrelevant. Components carrying the same force through a chain are springs in series: flexibilities add, the softest one dominates, and the stiffest is nearly irrelevant. A joint is the second kind and a floor grillage is the first, and the diagnostic is a single question — do these things share a force, or share a movement?

Answer it wrong and every instinct inverts. In series, stiffening the strongest-looking component buys 5%; in parallel, stiffening the weakest-looking member makes it worse. Both surprises are the same equation with the reciprocal taken.

Where the model stops #

Both beams are elastic and stay that way. The share $k_i/\sum k$ is a statement about a linear system. Once the deep beam yields it stops taking more load and the shallow one starts catching up, so the split at collapse is governed by strength rather than stiffness — which is exactly why plastic analysis is indifferent to the elastic distribution and why a ductile structure is forgiving of the whole of this essay.

The tie is rigid and carries only a vertical force. A real bolt, weld or bearing has its own flexibility in series with the two beams, and if it is soft compared with them it takes over the answer: two very stiff beams joined by a soft link share almost equally, because the link’s flexibility swamps the difference between them. A stiff connection does the opposite — it forces the two rotations to match as well, which adds equations and changes the split. The figures here assume a rigid pin.

Both members have the same modulus. Where they do not — a steel beam sharing with a concrete one, a new member with an old — the ratio is $E_1I_1$ against $E_2I_2$ and the modulus stops cancelling. That is when a stiffer material attracts load, which is the composite case and a different argument.

Nothing here creeps. A concrete member sharing a displacement with a steel one hands load over slowly as it creeps, so the share drifts for years after the load arrives. The elastic split is the day-one answer and, for a mixed-material system, not the long-term one.

What the pictures cannot show #

The figures on this page draw force and stiffness and never draw time, which hides the one thing that makes load sharing hard in practice. Every share plotted here assumes both members were present when the load arrived. A member added afterwards shares only the load applied after it was added, so a stiff member installed late may carry almost nothing while its neighbour carries everything — and the sweep curve says nothing about it, because a structure that was never complete has a different stiffness at every stage of its own building.

The second omission is the tie itself. It is drawn as a line, and its flexibility is set to zero everywhere in this family. That assumption is what makes $\delta_1 = \delta_2$ an equality rather than an approximation, and it is the assumption a real detail is least likely to honour.

The generalisation #

Once the question is “do these share a force or a movement?”, the same answer keeps arriving from unfamiliar directions.

A rigid corner in a frame is a load-sharing problem: the moment that goes round the corner divides between beam and column in proportion to their stiffnesses, and the beam-to-column stiffness ratio is the $r$ of the sweep curve wearing different clothes. A raft is a load-sharing problem along its own length: a beam on the ground shares its load with a bed of springs, and the characteristic length is the distance over which that sharing dies out. A truss is one in which the members are axial: which member moved the roof ranks them by $Ff L/EA$, and the ranking is not the ranking of their forces. A cable brings no bending stiffness to the argument at all, so the stiffness that comes from the shape depends on the tension already in it — a member whose share depends on how hard it has already been pulled.

The closing observation is the one stiffness is not strength makes from the other side. Sizing a member by strength and then discovering what it carries is backwards in a shared-displacement system, because what it carries is decided by the property nobody was sizing for.

History, briefly #

Bridge decks forced the issue first, because a deck is a grillage and a wheel is a point load standing where two members cross. Before computers, distribution across a deck was handled by charts: Guyon’s work in the 1940s and Massonnet’s in the 1950s produced distribution coefficients for an orthotropic plate, which are tables of exactly the share function drawn above, evaluated by hand for a range of stiffness ratios. Computer grillage analysis arrived in the late 1950s and made the charts unnecessary. They are worth remembering anyway — they are the moment the profession accepted that a wheel does not divide by geometry.

The ladder from here #

Later rungs on this anchor. The stiffness matrix, which is this essay’s two equations written for every degree of freedom at once. Shear walls sharing a storey shear, where the split follows $I$ but the twist of the floor plate follows the distance from the centre of rigidity. Piles under a cap, where the group’s stiffness is not the sum of the individual ones because the soil transmits between them. Composite action between a slab and the beam beneath it, and what happens when the shear connection is partial. Load sharing between old and new in a strengthening scheme, where the sequence decides the split. The effect of creep on a share, which drifts for years. Grillage and orthotropic plate models for decks, and what their distribution coefficients actually are. The stiffest path under dynamic load, where the share depends on frequency rather than on stiffness alone. And the design use of deliberate softening — the bearing, the sliding joint, the slotted hole put in to stop a member attracting load it is not there to carry.