Columns, Strain Energy and Theories of Failure
Euler buckling load and effective lengths, slenderness ratio and Rankine's formula, strain energy, impact loading, Castigliano's theorem, and the theories of failure.
A short column fails by crushing; a long slender one fails by buckling at a much lower load. This lesson covers buckling, strain energy methods, and how to predict yielding under combined stresses.
Euler buckling
For an ideal straight column with pinned ends:
P_cr = π²EI/L_e²
Use the smallest I of the section (it buckles about its weakest axis). The effective length L_e depends on the end conditions:
| End conditions | L_e |
|---|---|
| Both ends pinned | L |
| One fixed, one free | 2L |
| One fixed, one pinned | L/√2 ≈ 0.707L |
| Both ends fixed | L/2 |
Slenderness ratio λ = L_e/k, with I = Ak². Euler's formula holds only for long columns, where P_cr/A is below the proportional limit. Rankine's formula covers all lengths: 1/P = 1/P_crushing + 1/P_Euler.
Strain energy
- Axial load: U = P²L/(2AE) = σ²(AL)/(2E). Resilience per unit volume = σ²/(2E).
- Bending: U = ∫M² dx/(2EI). Torsion: U = T²L/(2GJ).
Impact loading: a load W dropped through height h onto a bar:
σ = (W/A)[1 + √(1 + 2hAE/(WL))]
A suddenly applied load (h = 0) causes twice the stress of the same load applied gradually.
Castigliano's theorem: the deflection under a load P is δ = ∂U/∂P. It gives deflections of curved bars, frames and springs.
Theories of failure
For a ductile material with yield strength S_y, under principal stresses σ₁, σ₂ (σ₃ = 0):
| Theory | Failure when |
|---|---|
| Maximum principal stress (Rankine) | σ₁ = S_y (used for brittle materials) |
| Maximum shear stress (Tresca) | σ₁ − σ₃ = S_y (largest difference of principal stresses) |
| Maximum strain (St. Venant) | σ₁ − νσ₂ = S_y |
| Distortion energy (von Mises) | σ₁² − σ₁σ₂ + σ₂² = S_y² |
In pure shear, Tresca predicts yield at τ = 0.5 S_y and von Mises at τ = 0.577 S_y.
Worked example
A steel column (E = 200 GPa), 3 m long, pinned at both ends, is a solid rod of diameter 50 mm. Find its Euler load.
- I = π × 50⁴/64 = 306,796 mm⁴.
- P_cr = π² × 200,000 × 306,796/3000² = 67.3 kN.
Practice
A column fixed at both ends has an Euler load how many times that of the same column pinned at both ends?
Show answer and explanation
Answer: 4L_e falls from L to L/2, and P_cr ∝ 1/L_e², so the load becomes 4 times.
A material has yield strength 250 MPa. By the maximum shear stress theory, at what shear stress (MPa) does it yield in pure shear?
Show answer and explanation
Answer: 125Pure shear gives σ₁ = τ, σ₃ = −τ, so 2τ = S_y and τ = 125 MPa.
Steel with E = 200 GPa is stressed to 200 MPa. What is its strain energy per unit volume in kJ/m³?
Show answer and explanation
Answer: 100σ²/(2E) = (2 × 10⁸)²/(4 × 10¹¹) = 10⁵ J/m³ = 100 kJ/m³.
A load is applied suddenly to a bar instead of gradually. The maximum stress is:
Show answer and explanation
Answer: BWith h = 0 the impact formula gives σ = 2W/A.
Which failure theories are suited to ductile materials? (Select all that apply.)
Show answer and explanation
Answer: A, BTresca and von Mises predict yielding of ductile metals well; Rankine's theory suits brittle materials.