Shear, moment and the Müller-Breslau principle

Take a simply supported span of length L and pick a section at distance a from the left support. Now walk a unit load across and watch the shear at that section.

Load to the left of the section (x < a). Consider the part of the beam to the left of the cut. It carries R_A upwards and the unit load downwards, so the shear at the cut is R_A − 1 = (L − x)/L − 1 = −x/L.

Load to the right of the section (x > a). Now the load is not on the left-hand part at all, so the shear is simply R_A = (L − x)/L.

Both are straight lines. At x = a they do not meet: the value jumps from −a/L to (L − a)/L, a jump of exactly 1.0 — the unit load itself passing over the cut.

That discontinuity is the signature of a shear influence line, and it makes physical sense: as the load crosses the section, it switches from being carried by the left-hand part to being carried by the right-hand part, and the shear at the cut changes by the whole of it.

The bending moment influence line at the same section is a triangle:

Load to the left (x < a): take the right-hand part as a free body — M = R_B(L − a) = (x/L)(L − a).

Load to the right (x > a): take the left-hand part — M = R_A × a = a(L − x)/L.

Both are straight lines, and this time they do meet, at x = a, where both give a(L − a)/L. That peak occurs with the load standing directly on the section, which is exactly what intuition says.

The influence line is therefore a triangle: zero at both supports, rising linearly to a(L − a)/L under the section. It has no discontinuity, because the bending moment is continuous as the load crosses.

Now the shortcut. The Müller-Breslau principle states:

The influence line for any force effect is, to a scale factor, the deflected shape the structure takes when the corresponding restraint is released and a unit displacement is imposed in the direction of that effect.

In practice:

• For a reaction: remove that support and push the structure up by one unit there. The shape it takes is the influence line for that reaction.
• For shear at a section: cut the beam there and slide the two faces past one another by one unit, keeping them parallel. The resulting shape is the shear influence line — and the parallel-sliding is what produces the jump.
• For bending moment at a section: insert a hinge there and rotate the two faces apart through unit relative rotation. The shape is the moment influence line.

For a statically determinate structure the released structure is a mechanism, so the shape is made of straight lines and you can draw it exactly with no calculation at all.

For an indeterminate structure the released structure still stands up, so the shape is a smooth curve. You cannot read exact values off it by eye — but you can see immediately where to put the load, which is usually the question that matters.

One more result makes influence lines genuinely powerful. Suppose instead of a point load you have a distributed load of intensity w covering part of the span.

Each small element of that load, w dx, contributes w dx × η. Adding them all up:

effect = ∫ w η dx = w × (area under the influence line over the loaded length)

So the effect of a UDL is the load intensity multiplied by the area under the influence line beneath it. This immediately answers a question that catches people out: to maximise an effect, load the parts of the span where the influence line is positive, and leave the negative parts unloaded. On a continuous beam, the pattern of loading that produces the worst moment at one section often leaves whole spans empty — a fact that looks wrong until you look at the influence line.

Müller-Breslau's real power appears on indeterminate structures, where deriving an influence line algebraically would mean re-solving the structure for every position of the load.

Release the effect you want and impose a unit displacement, exactly as before. The difference is that the released structure still stands up, so the shape it takes is a smooth curve rather than a set of straight lines.

For the prop reaction of a propped cantilever built in at x = 0 and propped at x = L, releasing the prop leaves a cantilever. Pushing its tip up by one unit gives the standard cantilever deflected shape, normalised:

η(x) = (3Lx² − x³)/(2L³)

It is zero at the built-in end, rises smoothly, and reaches exactly 1 at the prop — which it must, since a unit load standing directly on the prop is carried entirely by it.

Two consequences follow. The influence line is entirely positive, so every part of the span adds to the prop reaction and the worst case is a fully loaded span — unlike a continuous beam, where the influence line changes sign and pattern loading matters. And the area under it must reproduce the known answer for a UDL, which is the check worth doing.