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Queensferry

Module 15 · Lesson 15.3

Unit load for beams, and reciprocal theorems

The same idea with an integral instead of a sum — plus a result that saves real work.

Try it

Deflection by the unit-load method — watch the product build it

δ = ∫Mm/EI. The real load makes the moment diagram M; a fictitious UNIT load at the point you care about makes m; their product Mm has an area, and that area over EI IS the deflection. Three diagrams, one answer.

Beam

Real load

50% of span
Real moment, virtual moment, and their productreal moment M100 kN·mvirtual moment m (unit load at the probe)2.00 mproduct M·m — its area ÷ EI is the deflectionproductunit load ↓δ = area(M·m) / EI = 29.09 mm at 4.0 m · closed form 29.09 mm ✓
Beam and load
simply supported, point load
Deflection point
4.00 m
Peak real moment M
100.0 kN·m
Peak virtual moment m
2.000 m
Product area ∫Mm dx
5.33e+2 ×10¹² N·mm²·mm
Deflection δ
29.091 mm
Closed form
29.091 mm
Agreement
100.00 %
The deflection is the AREA of the Mm product diagram, divided by EI: 29.091 mm. The unit load is not a real force — it is a probe that, by virtual work, makes ∫Mm/EI equal exactly the deflection under it. Where M and m have the same sign the product is positive and adds deflection; here it is positive throughout, so the beam deflects the way the unit load points. It matches the memorised formula PL³/48EI (central point load) exactly — which is the point: the method gives the standard results without memorising any of them.

Things worth trying

  • Start with a simply-supported beam, central point load, probe at mid-span. Look at the three panels: M is a triangle (the real moment), m is a smaller triangle (the unit load's moment), and their product is the plum diagram whose area over EI is the deflection.
  • The bottom panel is the whole method made visible. δ isn't a formula to memorise — it's literally the area of that product diagram, divided by EI. Read the 'closed form ✓' at the bottom: the integral reproduces PL³/48EI exactly.
  • The middle panel is the strange one, and the key one. m comes from a UNIT load — a fictitious 1 kN probe placed at the point you want. Its peak is L/4, a LENGTH, not a force. Change the real load and M changes; m does not. It is a mathematical probe, not a real force.
  • Now slide the probe off mid-span. Watch m change shape to follow it — its apex is always UNDER the probe, because that is where the unit load sits. The deflection updates to the new point.
  • Move the probe to the quarter point. The two triangles no longer share an apex, the product diagram becomes lop-sided, and the deflection is smaller than at mid-span — as it must be, since the middle of a beam moves most.
  • Switch to a uniform load. Now M is a parabola and m is still a triangle. Their product is a smooth hump, and its area gives 5wL⁴/384EI — the standard UDL result, again with nothing memorised.
  • Switch to a cantilever. The supports change, both diagrams flip to hogging (below the line), but M and m still share a sign — so the product is positive and the tip deflection comes out as PL³/3EI.
  • The deepest idea: the unit load 'selects' the deflection at its own location by virtual work. External virtual work is 1 × δ; internal is ∫Mm/EI; setting them equal gives δ. That is why a fictitious load can measure a real movement.

Why this matters

A truss stores energy in a finite number of members, so virtual work gives a sum. A beam stores it continuously along its length, so the sum becomes an integral — and everything else stays the same. This one integral replaces the whole table of standard deflection formulae, and it handles loadings that are not in any table.

By the end of this lesson you should be able to

  • Apply δ = ∫Mm/EI dx to beam deflection problems
  • Use a unit moment to find a rotation instead of a deflection
  • State and apply Maxwell's reciprocal theorem

What you should already know

  • Bending moment diagrams (Module 3)
  • Beam deflection and EI (Module 13)
  • The unit-load method for trusses (this module, lesson 2)

The argument is identical to the truss case, with one substitution. In a truss the internal deformation of a member was its extension, NL/AE. In a beam, the internal deformation of a short length dx is its rotation, which from moment–curvature is (M/EI) dx.

So the internal virtual work is the virtual moment m acting through that real rotation, integrated along the beam:

Unit-load method for beams

What it calculates: deflection or rotation at any point on a beam

M
bending moment under the real loading (N·mm)
m
bending moment under a unit load (or unit moment) at the point wanted (dimensionless or mm)
EI
flexural rigidity (N·mm²)

This assumes

  • Bending deformation dominates; shear deflection is neglected
  • Linearly elastic material and small deflections
  • The same sign convention is used for M and m

In plain terms: Apply a unit force and the integral gives a deflection. Apply a unit moment and exactly the same integral gives a rotation. Nothing else changes.

In practice the work is:

  1. 1.Write M(x) for the real loading. Use as many segments as the loading demands.
  2. 2.Remove the real loads. Apply a unit load at the point and in the direction you want, and write m(x) over the same segments.
  3. 3.Integrate the product, segment by segment, and add.

Measure x from whichever end makes the expressions simplest — often the free end of a cantilever, because then there is nothing to the left to worry about.

Worked example

Cantilever tip deflection, derived rather than looked up

Given

  • Cantilever of span L = 3.00 m, fixed at the left, free at the right
  • Point load P = 15.0 kN at the free end
  • E = 205 000 N/mm², I = 1.20 × 10⁸ mm⁴

Find

The vertical deflection of the free end, from first principles by the unit-load method.

    One further result comes almost free from virtual work, and it is worth knowing because it saves genuine effort.

    Maxwell's reciprocal theorem states that for a linearly elastic structure, the deflection at point B caused by a load at point A equals the deflection at point A caused by the same load at point B.

    It is a surprising statement the first time you meet it. The two situations can look completely different — one might load a stiff region and measure at a flexible one, the other the reverse. But the deflections are identical.

    The reason is that both quantities come out of the same virtual-work integral, ∫Mm/EI dx, with the roles of M and m interchanged. Multiplication is commutative, so the integral does not care which is which.

    Betti's theorem is the generalisation to whole sets of loads: the work done by one system moving through the displacements of a second equals the work done by the second moving through those of the first.

    The practical payoff is large: it halves the number of coefficients needed in the flexibility method, and it is the reason the flexibility and stiffness matrices of a structure are symmetric — a property that every structural analysis program relies on.

    Predict first

    A unit load at quarter-span on a beam causes a deflection of 3.0 mm at mid-span. What deflection does a unit load at mid-span cause at quarter-span?

    Practice

    A cantilever spans 3.00 m and carries a 15.0 kN point load at its free end, with E = 205 000 N/mm² and I = 1.20 × 10⁸ mm⁴. What is the tip deflection, in mm?

    Practice

    A simply supported beam spans 6.00 m and carries a 40.0 kN point load at mid-span, with E = 205 000 N/mm² and I = 2.00 × 10⁸ mm⁴. Applying the unit-load method gives δ = PL³/48EI. What is the mid-span deflection, in mm?

    Practice

    On a certain beam, a 10 kN load applied at point A produces a deflection of 2.4 mm at point B. What deflection would the same 10 kN load applied at B produce at A, in mm?

    Summary

    • For beams the virtual-work sum becomes δ = ∫Mm/EI dx
    • A unit force gives a deflection; a unit moment gives a rotation
    • The method reproduces the standard formulae, and keeps working when they run out
    • Maxwell: deflection at B from a load at A equals deflection at A from the same load at B
    • Reciprocity is why stiffness and flexibility matrices are symmetric
    Progress is kept in this browser only.

    This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint