Any structure will carry the unit load

The unit-load method put an imaginary force of one unit on a beam, drew the moment it caused, multiplied that diagram by the real one and integrated. The deflection arrived without the deflected shape ever being computed, and the justification was a sentence about work: a set of forces in equilibrium, acting through a set of displacements that fit together, does as much work outside the body as inside it.

That sentence has two halves, and the method used only one of them loosely. The forces must be in equilibrium. The displacements must be compatible. Nothing requires the forces and the displacements to belong to the same problem, and nothing requires the forces to belong to the same structure — only that they balance on something the real displacements can be measured against.

Read that way the principle has two uses, and each is free in exactly the place the other is constrained. Choose the forces and take the displacements from the real structure, and the answer is a movement. Choose the displacements and take the forces from the real structure, and the answer is a force. The first reading turns out to be able to put its unit load on a structure that does not exist. The second turns out to be unable to find a force that statics cannot, however cleverly the displacement is chosen — and seeing why is the clearest statement there is of what compatibility is.

A deflection statics cannot give, from a beam statics can #

Take a beam of 8 m, built into a wall at its left end and resting on a prop at its right, carrying 4 kN/m over its whole length. It has one support too many, so statics alone cannot give its moments. The standard result — from the force method, or from the table — is that the prop carries $3wL/8$, which is 12 kN, and the moment at the wall is $wL^2/8$ of hogging, 32 kNm.

Ask how far the point 4 m from the wall moves.

The method as first stated would put the unit load at 4 m on the propped beam, which means solving a second indeterminate problem to draw its moment diagram. That is legal and it is extra work. The alternative is to take the prop away in the virtual system only: put the unit load at 4 m on a plain cantilever, whose moment diagram is a triangle of hogging from the wall to the load and zero beyond it.

Then multiply that triangle by the true moment of the propped beam, and integrate.

Measure $u$ back from the load toward the wall, so $u = 4 - x$. The true moment is $M = 12(8 - x) - 2(8 - x)^2$, which in $u$ is $16 - 4u - 2u^2$. The cantilever’s unit moment is $m = -u$. The product integral runs only over the first 4 m, because the unit diagram is zero beyond the load:

$$EI\,\delta = \int_0^4 -u\,(16 - 4u - 2u^2)\,du = -\left(128 - \tfrac{256}{3} - 128\right) = 85.33.$$

The closed form for a propped cantilever under a uniform load, $wa^2(L-a)(3L-2a)/48EI$, gives $4 \times 16 \times 4 \times 16 / 48 = 85.33$.

That figure is the calculation drawn. Its product diagram is nothing like the one the first essay on this method showed for a simple span: most of its area is piled against the wall, where both diagrams are large and hogging, and a small negative lobe sits between the point of contraflexure and the load. It is the shape of a question asked of a cantilever and answered about a propped beam.

Why the virtual system may stand on anything #

The derivation that justifies this is three lines, and every line is doing something.

The virtual system is in equilibrium. A unit load at 4 m, and the wall’s reaction to it on the cantilever — an upward force of one and a hogging couple of four — balance. They would balance wherever they were drawn.

The real displacements are compatible, and they are the real beam’s. The propped beam does not move vertically or rotate at the wall, and does not move vertically at the prop. Those are exactly the places the virtual reactions act, plus one place (the prop) where the virtual system has no reaction at all.

So the external virtual work is one number. The wall’s virtual force and couple act through a displacement and a rotation that are both zero, so they do no work. The unit load does $1 \times \delta$. The internal work is the virtual moment acting through the real curvature, $\int m\,M/EI\,dx$. Setting the two equal gives the deflection.

Nowhere did the argument ask the virtual system to respect the prop. It needed only that every reaction it does have acts at a point the real structure holds still. A cantilever release satisfies that, because its only support is a subset of the real beam’s supports. So does a simple span, whose pin at the wall exerts no couple and so asks nothing of the real beam’s rotation there, and whose roller is at the real prop. So does any other determinate structure made by taking restraints away, and so does the real beam itself.

There is a second way to say the same thing, and it is the more illuminating one. Subtract the cantilever’s unit diagram from the simple span’s. Each balances a unit load, so their difference balances nothing: it is a moment diagram in equilibrium with no load at all, which is precisely a self-stress state of the propped beam — the moment a prop force and a wall couple can put into the beam without anything being applied. The two releases give the same answer if and only if the integral of that self-stress against the real curvature is zero.

And that integral being zero is not a coincidence the releases happen to satisfy. It is the compatibility condition. It is the equation the force method writes to find the prop force in the first place — the gap at the released restraint must close — in the form of work rather than of geometry. The theorem that lets the unit load stand on any release, often called the reduction theorem, is the force method’s own equation read backwards: once the true moments have been found by making every self-stress do no work, every release is as good as every other for asking where the beam went.

Two products that share nothing but their area #

The simple-span release above and the cantilever release in the hero do not look like the same calculation, and they are worth putting side by side before a third.

On the cantilever, the unit diagram is zero beyond the load, so the right half of the beam contributes nothing at all to the answer. The whole deflection is assembled from the 4 m nearest the wall, and most of it from the metre or so closest to the wall where the true hogging moment is largest. On the simple span, the unit diagram is non-zero everywhere and zero at the wall, so the region the cantilever relied on is switched off and the answer is assembled from the sagging moment in the middle of the span instead, less a small negative contribution from the hogging near the wall.

Neither product diagram is where the deflection comes from in the sense the density reading of the method meant. That reading holds for the unit load on the real structure, because then the integrand at a station is that station’s genuine share of the answer — its curvature multiplied by the lever its rotation has onto the point asked about. On a release the integrand is a bookkeeping device. It gets the total right because the self-stress term integrates to nothing, and it distributes that total along the beam in a way that tells nothing about the real beam’s anatomy. Stiffening the wall end of the propped beam does not reduce its deflection by the share the cantilever product diagram assigns to that region.

That is the price of the freedom, and it is worth stating: a release gives the right number and the wrong map.

A hinge where the beam had no moment #

The third release is the one that gives the most away about what a release is.

Put a hinge in the beam and keep both supports. A hinge releases a moment, so the beam becomes determinate: a cantilever from the wall to the hinge, with a simply supported span hanging from the hinge’s end and the prop. Where the hinge goes is a free choice. Put it at 2 m.

That product diagram never crosses zero, and the reason is the position chosen. The true moment of a propped cantilever under a uniform load is zero at a quarter of the span from the wall — 2 m here — so a hinge at 2 m has been put exactly where the real beam carries no moment.

Which means that for this load, the hinged beam is not a release at all. Solve the hinged girder under the real 4 kN/m by statics and it returns the propped beam’s own moment diagram, because the one condition the hinge imposes is one the real beam already satisfies. The redundancy has been taken away at no cost.

That is the whole idea of the Gerber girder: real hinges inserted at the points of contraflexure make a continuous beam determinate while keeping, for its governing load, the moments it had when continuous. Heinrich Gerber patented it in 1866 and railway bridges used it for a century, because a determinate girder does not care whether its piers settle. What the virtual-work reading adds is the reason the arrangement is only exactly right for one load: the point of contraflexure moves when the load does, and a hinge cannot follow it.

What the pairing that is not allowed actually computes #

The rule has a second half, and it is the half that is easy to break. The virtual moment may come from any release; the real moment may not. It has to be the one the real beam, with all its restraints, actually carries.

The two wrong numbers are not noise, and working out what they are shows the theorem from the outside.

Take the first. The real moment is the cantilever’s, $-w(L-x)^2/2$, which is the moment of the beam with its prop gone — a beam whose tip, at 8 m, comes down $wL^4/8EI = 2{,}048$. The unit load stands on the simple span, whose roller at 8 m pushes up by half a unit to balance it. Run the same three-line derivation with those two systems. The unit load does $1 \times \delta$ through the cantilever’s own deflection at 4 m, which is $725.33/EI$. The roller’s half-unit reaction acts at the tip, which on the cantilever is not held still: it does $-\tfrac12 \times 2{,}048 = -1{,}024$. The sum is $725.33 - 1{,}024 = -298.67$.

The second is the same accident at the other end. The real moment is the simple span’s, a beam free to rotate at the wall, by $wL^3/24EI$, which is $85.33/EI$. The unit load stands on the cantilever, whose wall exerts a couple of 4 kNm to hold it. The simple span moves $213.33/EI$ at 4 m, and the wall’s virtual couple does $-4 \times 85.33 = -341.33$ through a rotation the real beam is not allowed to have. The sum is $-128$.

Each wrong answer is the right answer for a real movement, contaminated by a virtual reaction doing work through a displacement that a real support forbids. The derivation assumed the virtual reactions act at points the real structure holds still; pairing a released moment diagram with a different release’s unit diagram is exactly the case where that assumption fails.

It also gives a check that can be run on any moment diagram. Evaluate the product integral with two different releases. If the moment diagram is the true one, the two agree. If it is merely in equilibrium — any statically admissible diagram, such as a guess, a stale analysis or a model with a support the drawings do not have — they disagree, and the disagreement is the work of the self-stress difference through the incompatibility. The simple span’s moment gives 213.33 with its own unit diagram and −128 with the cantilever’s. A gap of 341 on a check that costs two table lookups is how a wrong continuity assumption announces itself.

The other reading, which pushes the structure #

Now read the principle the other way round: choose the displacements, take the forces from the real structure, and the answer is a force.

The displacement has to be compatible, which on a structure made of rigid pieces joined at hinges means it has to be a mechanism. The virtual work inside the structure is then zero, because nothing bends, and the equation contains only the loads and the one reaction the mechanism was made to move.

The arithmetic is short enough to follow in full. Taking the roller at 6 m away leaves a structure with one freedom: the piece from 0 to 8 m turns about the pin, the piece from 8 to 12 m turns about the far roller, and they share the hinge. Lift the removed support by one unit. The left piece rises in proportion to distance from the pin, so the 20 kN load at 3 m rises 0.5 and the hinge rises $8/6 = 1.333$. The right piece falls back linearly from the hinge to the far roller, so the 30 kN load at 10 m rises $1.333 \times 2/4 = 0.667$. The uniform load’s work is 5 kN/m times the area of the displaced shape, which is two triangles of height 1.333 on bases of 8 and 4, so 40.

The reaction does $R \times 1$ of work against the loads’ $10 + 20 + 40$. So $R = 70$ kN.

Statics, the long way: the suspended piece from 8 to 12 m carries 30 kN at 2 m from the far roller and 20 kN of distributed load centred at 2 m, so the hinge passes 25 kN down onto the left piece. Moments about the pin for the left piece — 20 kN at 3 m, 40 kN at 4 m, the hinge’s 25 kN at 8 m — give $420/6 = 70$ kN.

What the virtual displacement bought is that no internal force was ever computed. The hinge force appeared in statics as an intermediate and in the work equation not at all, because the two pieces move together at the hinge and the force it passes does equal and opposite work on each. That is compatibility doing the job equilibrium did in the other reading: the hinge’s internal force vanishes from the equation because the displacement is continuous across it, exactly as the virtual reactions vanished before because the real displacement was zero under them.

Pushing a redundant beam finds nothing #

Try the second reading on the propped beam, to find the prop force.

Take away the prop. The cantilever that remains is not a mechanism, so a displacement that lifts the prop point must bend the beam, and the internal work is no longer zero: it is the real moment acting through the virtual curvature. The real moment is unknown, but it is not entirely unknown — for a trial prop force $R$, statics gives $M(x;R) = R(8-x) - 2(8-x)^2$, and that diagram is in equilibrium with the load for any $R$ at all. So choose a virtual displacement $v(x)$ that is zero with zero slope at the wall and one at the prop, and write the work balance. The external work is the load’s through $v$, less the prop force’s through one unit. The internal work is $M(x;R)$ against the virtual curvature. Set them equal and solve for $R$.

There is nothing to solve. The residual is zero for every trial force, with every one of the three displacement fields tried, to a few parts in $10^{13}$. The equation reduces to $0 = 0$.

It has to. Integrate the internal work by parts twice. The virtual curvature against the real moment becomes the virtual displacement against the second derivative of the moment — which equilibrium says is the load — plus boundary terms, of which the only survivor is the moment’s slope at the prop, which equilibrium says is the prop force. The internal work is therefore the load’s work through $v$ minus $R$, identically, and it cancels the external work term for term. A virtual displacement statement is equilibrium written as work, and equilibrium is exactly the thing that could not fix $R$.

The virtual-force reading has no such trouble, and the figure’s other line shows why. A unit load at the prop, on the cantilever, gives the gap the prop would have to close as $wL^4/8EI - RL^3/3EI$ — a compatibility statement — and it is zero at one force only.

So the two readings are not interchangeable, and the asymmetry is exact:

The last column is the useful one. A virtual displacement can find a redundant only if the real internal forces are not taken from statics — if they are written instead as a stiffness times the real curvature, $M = EI\kappa$, which brings compatibility in through the back door. That substitution turns the second reading into the displacement method, and when the displacement field is chosen from a finite set of shapes it is the weak form under every finite-element program — which is why the answer from such a model depends on how the field was divided, and errs on the stiff side, while the virtual-force answer depends on nothing but the true moment.

The one place pushing does find a redundant #

There is a case where the second reading finds forces in a redundant structure with nothing but rigid mechanisms and statics, and it is the case this essay has so far excluded: collapse.

A redundant beam loaded until plastic hinges form becomes a mechanism, and at each hinge the moment is no longer unknown. It is the plastic moment, fixed by the section and not by compatibility. Push the mechanism through a virtual rotation and the internal work is the plastic moment times the hinge rotations — a number, not a function of the redundant — so the equation contains the collapse load and nothing else.

That is why the mechanism method can collapse an indeterminate frame with the same rigid-body arithmetic as the hinged girder above, and why the instantaneous centres of its pieces are all it needs. The redundancy that defeated pushing in the elastic beam has not been solved; it has been removed, by a material law that pins the moment at the places the mechanism rotates. Plasticity supplies the missing equation that elasticity takes from compatibility, and the theorem that the answer is an upper bound is the price of guessing where those places are.

A check a computer’s deflection can be given by hand #

The reduction theorem has a use that outlasts hand analysis, because it separates two things a model’s output contains: its moments and its movements.

A frame analysed by a stiffness program returns a moment diagram and a deflected shape, and the deflection is the number most likely to govern a floor. To check it by hand, take the program’s moment diagram as given, choose the simplest release of the member in question — nearly always a cantilever or a simple span — and evaluate one product integral from the table of standard shapes. The deflection comes out without the redundancies being solved again.

And evaluating it with two releases checks the moments themselves. If the program’s moment diagram is compatible with the stiffness it was given, the two agree; if a support was modelled as fixed that is really pinned, or an element has the wrong second moment of area, the two differ, and the gap is the self-stress that should not be there doing work through a rotation the model allowed. The illegal pairing above, 213.33 against −128, was that check failing on a simple-span moment diagram passed off as a propped beam’s.

What the check cannot see is a model that is compatible with the wrong stiffness everywhere. A beam given half its real $EI$ throughout has a true-looking moment diagram — moments in a single member depend on stiffness ratios, not magnitudes — and every release agrees on a deflection twice the real one.

What the figures leave out #

Each band is drawn at its own scale. The true moment, the unit moment and their product are fitted to their strips independently, so no height on one strip can be compared with a height on another. What the product strip communicates is its shape and its sign; the number beside it is the only magnitude.

The residual that is identically zero is drawn as a line on zero, which is the one figure where a correct result looks like nothing has been plotted. Three fields are there, dashed differently, lying on top of one another.

The hinged girder’s displaced shape is drawn at a scale chosen so the hinge rises 110 px. The virtual displacement has no size; only the ratios of its ordinates enter the work equation, which is why any scale is correct and none is physical.

Where the model stops #

Linear elasticity for the first reading. The reduction theorem needs the true moment and the curvature it causes to be related by a fixed $EI$, and it needs superposition, so it holds for a cracked concrete member only to the extent a single effective stiffness does.

Small displacements for both. The virtual work of the loads is computed on the undeformed geometry. A mechanism’s ordinates are first-order in its rotation, and a real girder that has moved enough for its loads to shift sideways relative to the pieces is outside the arithmetic.

Bending only. Every product integral here omits shear and axial terms. For a slender beam that is the usual approximation; a release that puts large axial force in a member the real structure does not — a tie taken away from an arch — would need the axial term, and would get it wrong without it.

Rigid pieces for the mechanism. The work equation of the hinged girder has no internal term because the pieces are assumed not to bend. That is exact for the reaction, since the real bending does no work through a rigid virtual displacement; it would not be exact if the virtual displacement were deformable, which is the case the identity figure shows cannot help.

A redundant structure’s true moment has to come from somewhere. The reduction theorem shortens the second half of a calculation; the first half, finding the moments, is still the force method or a stiffness solution, and the theorem’s only claim is that it does not have to be done twice.

Still open: a frame, where the release decides which terms appear #

Everything here has been one straight member in bending. In a portal frame the unit load produces bending, axial force and shear in every member, and the product integral grows a term for each. The reduction theorem then offers something a beam cannot: a choice of release that makes the virtual system’s axial forces vanish in the members whose axial stiffness is least certain, or its shear vanish where the shear area is hard to define. Which release does that for which frame — and whether it can always be done — is a question about the structure’s topology rather than its numbers, and it is the next thing the principle’s freedom is worth.