Euler's column theory: critical (crippling) load for a column of length l with one end fixed and the other end hinged Choose the correct expression in terms of E, I, and l.

Introduction / Context:
Euler's buckling theory provides the elastic critical load at which a slender column becomes unstable. The result depends on the flexural rigidity E * I and the effective length, which varies with end conditions. Here the column has one end fixed and the other end hinged (pinned).

Given Data / Assumptions:

• Column length = l, prismatic, linearly elastic.
• End conditions: fixed–hinged.
• Slenderness high enough for Euler theory to apply (no yielding).

Concept / Approach:
Euler formula: Pcr = (pi^2 * E * I) / (Le^2). The effective length Le = K * l depends on the rotational restraint at ends. For fixed–hinged, K ≈ 0.7 (often taken as 0.7 or 1/sqrt(2)). Using K = 1/sqrt(2) gives Le = l / sqrt(2).

Step-by-Step Solution:
Le = l / sqrt(2).Pcr = (pi^2 * E * I) / (Le^2).Pcr = (pi^2 * E * I) / ( (l / sqrt(2))^2 ).Pcr = (pi^2 * E * I) / ( l^2 / 2 ).Pcr = 2 * pi^2 * E * I / l^2.

Verification / Alternative check:
Tabulated K-factors: pinned–pinned K = 1.0, fixed–fixed K = 0.5, fixed–free K = 2.0, fixed–hinged K ≈ 0.7. Substituting K = 0.707 reproduces approximately the same result as 2 * pi^2 * E * I / l^2 when expressed exactly with 1/sqrt(2).

Why Other Options Are Wrong:
(pi^2 * E * I) / l^2 corresponds to pinned–pinned. 4 * pi^2 * E * I / l^2 corresponds to fixed–fixed. The fractional variants with 1/2 reduce capacity improperly; fixed–hinged has a higher capacity than pinned–pinned, not lower.

Common Pitfalls:
Treating all end conditions as the same; forgetting that increasing fixity decreases Le and increases Pcr; using inconsistent K values between design references.

Final Answer:
2 * pi^2 * E * I / l^2