Euler buckling — permissible crippling load for a circular column (hinged ends) A prismatic column has a uniform circular cross-section of diameter d = 7.5 cm and effective length L = 5 m (both ends hinged). The material has modulus of elasticity E = 2100 tonnes/cm^2 (consistent units). Using Euler’s formula, compute the permissible maximum crippling load.

Introduction / Context:
Long, slender columns fail by elastic instability (buckling). For ideal columns with hinged ends and within the Euler regime, the critical load depends on the modulus of elasticity, the least second moment of area, and the effective length. Correct unit handling and section properties are essential to avoid large numerical errors.

Given Data / Assumptions:

• Diameter d = 7.5 cm; circular cross-section.
• Effective length L = 5 m = 500 cm; hinged–hinged ends.
• Elastic modulus E = 2100 tonnes/cm^2 (use consistent tonne–cm units).
• Euler formula applies (column is slender; no inelastic effects).

Concept / Approach:
Euler critical load for pinned–pinned column:
P_cr = (pi^2 * E * I) / (L_eff^2).
For a circle, I = (pi * d^4) / 64. Use centimetre units throughout to remain consistent with E and express the final load in tonnes.

Step-by-Step Solution:

Verification / Alternative check:
Slenderness ratio λ = L/r where r = sqrt(I/A). For d = 7.5 cm, r ≈ 7.5/4 = 1.875 cm; λ ≈ 500 / 1.875 ≈ 267 > 100, confirming Euler applicability for a long column.

Why Other Options Are Wrong:

• 128.8 t and 288.0 t: an order-of-magnitude too high, typical of unit mistakes (metres vs. centimetres) or squaring length incorrectly.
• 1.288 t: too small, from dividing by an extra power of 10.
• None of these: unnecessary since 12.88 t is correct.

Common Pitfalls:
Mixing SI and gravitational units; forgetting to square the effective length; using wrong end conditions (which change L_eff). Also, using radius r incorrectly for a circle (r = d/4 for the radius of gyration is only approximate; the exact value is d/4 for thin rings; for solid circles r = d/4 * sqrt(2) is incorrect—always compute r = sqrt(I/A)).

Final Answer:
12.88 tonnes