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Civil
Bending strain energy under a triangular moment diagram
Elastic energy stored in a length L of prismatic beam in which the bending moment varies linearly from M0 at one end to zero at the other.
- Ref
- CIV-36
- Rev
- v1
- Field
- Civil
- Result
- U [J]
The strain energy stored in bending is U = ∫M²/(2EI) dx. For a moment that varies linearly from M0 to zero over length L the integral of M² is M0²L/3, so U = M0²L/(6EI). The square removes the sign, so positive and negative moments both store energy. This is the basis of Castigliano’s method for deflections.
Elastic energy stored in a length L of prismatic beam in which the bending moment varies linearly from M0 at one end to zero at the other. The integral of M² over the triangle is M0²L/3, and U = ∫M²dx/(2EI). Constant EI, elastic bending only; the square removes the sign of the moment.
Calculator
- Energy under a constant moment M₀, M₀²L/(2EI)60 J
- Mean-square moment, M₀²/31.333×108 N²·m²
- Rotation of the ends from a constant M₀, M₀L/(EI)0.006 rad
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The beam stores 20 J, one third of the 60 J a constant moment M0 would store over the same length.
- Doubling the moment would quadruple the energy to 80 J.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| M0 Peak moment | 100 to 1,000,000 N·m | 5e-4 | 50,000 | 50,000 | +10% | +21 % |
| EI Flexural rigidity | 1,000,000 to 1e9 N·m² | 0.2 | 200 | 199.8 | +10% | −9.09 % |
| L Length | 0.5 to 20 m | 3.333 | 133.3 | 130 | +10% | +10 % |
Sweep of Peak moment
| Peak moment (N·m) | Strain energy (J) |
|---|---|
| 100 | 5e-4 |
| 316.2 | 0.005 |
| 1,000 | 0.05 |
| 3,162 | 0.5 |
| 10,000 | 5 |
| 31,620 | 50 |
| 100,000 | 500 |
| 316,200 | 5,000 |
| 1,000,000 | 50,000 |
Sweep one variable
Variables and how each one contributes
Strain energy
What the equation gives you. Everything else feeds it.
Peak moment
The peak bending moment at one end, in N·m. Raises the stored energy with the square of the moment. Raising it by 10 % (to 22,000) moves the result by +21 %, from 20 to 24.2 J.
Length
The length over which the moment falls to zero, in m. Raises the stored energy in direct proportion. Raising it by 10 % (to 3.3) moves the result by +10 %, from 20 to 22 J.
Flexural rigidity
The flexural rigidity, in N·m². A stiffer beam stores less energy in inverse proportion. Raising it by 10 % (to 1.1×107) moves the result by −9.09 %, from 20 to 18.18 J.
Worked example with the values above
- Integral of M² over the length: M0²L/3 = 4×108 × 3 / 3 = 4×108 N²·m³.
- Divide by 2EI = 2×107.
- U = 4×108 / 2×107 = 20 J.
Assumptions and limits
- Pin joints, loads at the joints, small displacements and linear elastic members.
- Tension positive in both the real and virtual force systems.
Common mistakes
- Using compression positive in one system and tension positive in the other.
- Mixing the real force and the virtual force between members.
- Forgetting that unused members must have zero real or virtual force.
Statistics and uncertainty
Truss displacements are exact for pin-jointed bars with axial deformation only. Real joints are slightly flexible and members have length tolerances, so measured displacements are usually 5 % to 20 % higher than the pin-jointed prediction.
Where it is used
- Deflection and stiffness checks for trusses, braces and frames.
- Estimating thermal movement and strain energy of members.
Maths you need
- Virtual work: external virtual work equals internal virtual work
- Axial deformation: ΔL = NL/EA
- Strain energy: the work stored by elastic deformation