According to Euler's column theory, the crippling (buckling) load is given by P = (π^2 * E * I) / (C * l^2). For a column with one end fixed and the other end hinged (pinned), state whether the value of C is 1/2.

Introduction:
Euler's column theory relates the critical buckling load to end conditions through an effective length factor. This question checks consistency between the constant C and end condition “fixed–hinged”.

Given Data / Assumptions:

• Long, slender column (Euler regime).
• Modulus E and second moment I are constants.
• End conditions: one end fixed, the other hinged (pinned).

Concept / Approach:
General Euler form: P_cr = π^2 * E * I / (L_e^2), where L_e = K * l. If the problem is written as P_cr = π^2 * E * I / (C * l^2), then C = (L_e / l)^2 = K^2. For fixed–hinged, K ≈ 0.7, so C ≈ 0.7^2 ≈ 0.49 ≈ 1/2.

Step-by-Step Solution:
1) Take K (fixed–hinged) ≈ 0.7.2) Compute C = K^2 ≈ 0.49.3) Approximating 0.49 as 1/2 is standard in classroom design tables for Euler's ideal case.4) Therefore, the statement that C = 1/2 is correct within Euler theory's conventional approximation.

Verification / Alternative check:
Using numerator form P_cr = C_n * π^2 * E * I / l^2 with C_n values: for fixed–hinged, C_n ≈ 2. Hence in denominator form C = 1 / C_n ≈ 1/2.

Why Other Options Are Wrong:
False: conflicts with the effective length relation.

True only for very short/very long columns: K depends on end restraint, not directly on length regime (though Euler validity requires slenderness).

Common Pitfalls:
Mixing the two conventions (C in numerator vs denominator) and forgetting that C here equals K^2. Also confusing fixed–fixed (C ≈ 1/4) with fixed–hinged (C ≈ 1/2).

Final Answer:
True