Module 15 · Lesson 15.1
Work, strain energy and the missing half
Where the ½ in ½Pδ comes from, and what the structure does with the energy.
Why this matters
Every deflection method you have used so far has been tied to one particular arrangement — a formula for a cantilever, another for a simply supported beam. Energy methods replace that catalogue with a single idea that works for trusses, frames, curved members and structures nobody has tabulated. This lesson builds the foundation: what strain energy is, and why the factor of one half is not a fudge.
By the end of this lesson you should be able to
- Explain why a gradually applied load does work ½Pδ
- Calculate strain energy in an axially loaded member
- Calculate strain energy in bending
- Use conservation of energy to find a deflection directly
What you should already know
- Axial extension δ = PL/AE (Module 7)
- Bending moment diagrams (Module 3)
- Moment–curvature and EI (Module 13)
Push a structure and it moves. You have done work on it, and if the structure is elastic that work has not disappeared — it is stored, and you get all of it back when you release the load. That stored quantity is the strain energy U.
The subtlety is in how much work is done. If you hung the full load P on instantly, the load would move through the full deflection δ at full value, and the work would be Pδ. But that is not what happens when a load is applied gradually. The load builds from zero to P while the deflection builds from zero to δ, in step with it. At every instant the load is only as large as the deflection so far allows.
So the work is the area under the load–deflection line. For a linearly elastic structure that line is straight, and the area under a straight line from the origin is a triangle:
What it calculates: the strain energy stored by a single gradually applied load
- U
- strain energy stored (N·mm (or J))
- P
- final value of the applied load (N)
- δ
- deflection under the load, in its direction (mm)
This assumes
- The structure is linearly elastic, so the load–deflection line is straight
- The load is applied gradually, not suddenly
- Nothing is lost to friction, heat or permanent deformation
In plain terms: The one-half is geometry, not a correction factor. It is the area of a triangle. If the structure were not linear the line would curve and you would have to integrate instead.
Now put that to work. For a bar of length L and area A carrying an axial force P, the extension is δ = PL/AE. Substituting into U = ½Pδ:
What it calculates: energy stored in a bar carrying a constant axial force
- P
- axial force (tension or compression) (N)
- L
- length of the member (mm)
- A
- cross-sectional area (mm²)
- E
- Young's modulus (N/mm²)
This assumes
- The force is constant along the member
- The material stays elastic
In plain terms: P appears squared, so the energy is the same whether the member is in tension or compression, and doubling the force stores four times the energy.
What it calculates: energy stored in a beam from its bending moment distribution
- M
- bending moment, which varies with x (N·mm)
- E
- Young's modulus (N/mm²)
- I
- second moment of area (mm⁴)
- L
- span (mm)
This assumes
- Bending dominates — shear and axial energy are neglected
- EI is constant, or is kept inside the integral if it is not
In plain terms: The direct analogue of P²L/2AE, but the moment varies along the beam so the sum becomes an integral. M is squared, so sagging and hogging both store energy — they never cancel.
Predict first
A gradually applied load P stores strain energy U in an elastic bar. If the load is doubled to 2P, what is the new strain energy?
Worked example
Strain energy and deflection of a hanger
Given
- Steel hanger, length L = 2.00 m, area A = 1200 mm²
- Gradually applied tensile load P = 50.0 kN
- E = 205 000 N/mm²
Find
The extension, the strain energy stored, and a cross-check between the two.
Practice
A steel bar 1.50 m long with a cross-sectional area of 900 mm² carries a gradually applied load of 40.0 kN. With E = 205 000 N/mm², how much strain energy is stored, in joules? (1 J = 1000 N·mm)
Practice
For the same bar and load, what is the extension, in mm?
Practice
The load on the same bar is doubled to 80 kN. By what factor does the stored strain energy increase?
Worked example
Strain energy stored in a bent beam
Given
- Cantilever of span L = 3.00 m carrying a point load P = 15.0 kN at its free end
- E = 205 000 N/mm², I = 1.20 × 10⁸ mm⁴
Find
The strain energy stored in bending, and a check against ½Pδ.
Practice
A cantilever of span 3.00 m carries a 15.0 kN tip load, with E = 205 000 N/mm² and I = 1.20 × 10⁸ mm⁴. Using U = P²L³/6EI, how much strain energy does it store in bending, in N·mm?
Practice
For a cantilever with a tip load, the bending strain energy is U = P²L³/6EI. If the span is doubled with everything else unchanged, by what factor does the stored energy increase?
Summary
- A gradually applied load does work ½Pδ; the half is the area of a triangle
- Elastic structures store that work as recoverable strain energy
- Axial: U = P²L/2AE. Bending: U = ∫M²/2EI dx
- Energy goes as the square of the load, so it does not superpose
- Equating external work to strain energy gives a deflection directly — but only under a single load, in its own direction
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