Force Methods and Consistent Deformation
Method of consistent deformation, propped cantilever and fixed beam results, three-moment theorem, unit load method, Castigliano's theorems and Maxwell's reciprocal theorem.
In a force method, you remove the redundant restraints, find the deflections they would have prevented, and choose the redundant forces that make those deflections zero (or equal to the support movement).
Consistent deformation
- Choose redundants and remove them to get a determinate primary structure.
- Find the deflection at each removed restraint due to the loads.
- Find the deflection due to a unit value of each redundant (flexibility coefficients).
- Write compatibility: total deflection = 0, and solve.
Example: propped cantilever with UDL w. Remove the prop B. The cantilever tip deflects wL⁴/(8EI) down; a force R_B lifts it R_B L³/(3EI). Setting them equal: R_B = 3wL/8. The fixed-end moment is wL²/8 (hogging).
Standard results
| Beam | Result |
|---|---|
| Propped cantilever, UDL | Prop reaction 3wL/8; fixed-end moment wL²/8 |
| Propped cantilever, central load P | Prop reaction 5P/16; fixed-end moment 3PL/16 |
| Fixed beam, UDL | End moments wL²/12; mid-span moment wL²/24 |
| Fixed beam, central load P | End moments PL/8; mid-span moment PL/8 |
| Fixed beam, load P at a from A | M_A = Pab²/L², M_B = Pa²b/L² |
| Fixed beam, support B sinks by Δ | End moments 6EIΔ/L² |
Three-moment theorem (Clapeyron)
For continuous beams with spans L₁ and L₂ meeting at support 2 (constant EI, no settlement):
M₁L₁ + 2M₂(L₁ + L₂) + M₃L₂ = −6A₁x̄₁/L₁ − 6A₂x̄₂/L₂
where A and x̄ are the area of each span's free (simply supported) BMD and its centroid distance from the outer support. For a UDL on a span: 6Ax̄/L = wL³/4.
Energy methods
- Unit load method: deflection Δ = ∫ M m dx/(EI) (m = moment due to a unit load at the point). For trusses: Δ = Σ P k L/(AE).
- Castigliano's first theorem: ∂U/∂Δ = P. Second theorem: ∂U/∂P = Δ; for redundants, ∂U/∂R = 0 (least work).
- Maxwell's reciprocal theorem: the deflection at A due to a unit load at B equals the deflection at B due to a unit load at A (δ_AB = δ_BA). Betti's theorem is its general form.
Worked example
A two-span continuous beam, spans 4 m each, UDL 12 kN/m, simply supported at the ends. Find the support moment at the middle.
- M₁ = M₃ = 0: 2M₂(8) = −2 × (12 × 4³/4) = −384, so M₂ = −24 kN·m (hogging). That is wL²/8, the known result for two equal spans.
Practice
A propped cantilever of span 6 m carries a UDL of 8 kN/m. What is the prop reaction in kN?
Show answer and explanation
Answer: 18R_B = 3wL/8 = 3 × 8 × 6/8 = 18 kN.
A fixed beam of span 6 m carries a UDL of 10 kN/m. What is the fixed-end moment in kN·m?
Show answer and explanation
Answer: 30M = wL²/12 = 10 × 36/12 = 30 kN·m.
A fixed beam of span 5 m (EI = 20,000 kN·m²) has one support sinking 10 mm. What is the induced end moment in kN·m?
Show answer and explanation
Answer: 48M = 6EIΔ/L² = 6 × 20,000 × 0.01/25 = 48 kN·m.
Which theorem states that δ_AB = δ_BA?
Show answer and explanation
Answer: BMaxwell's reciprocal theorem: the deflection at A due to a unit load at B equals that at B due to a unit load at A.