A strut 4 m long is 80 mm in diameter. One end of the strut is fixed while the other end is hinged. What will be the crippling load? [assume, E = 2 × 10⁵ N/mm², π³ = 31]

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BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

The crippling load for a strut with one end fixed and the other hinged is calculated using Euler's formula: \( P = \frac{\pi^2 E I}{L_e^2} \)

Where, \( L_e = \frac{L}{\sqrt{2}} \) for this end condition.

Given:

L = 4 m = 4000 mm, d = 80 mm, E = 2 × 10⁵ N/mm², π³ = 31

Calculation:

Effective length, \( L_e = \frac{4000}{\sqrt{2}} = 2828.4 \, \text{mm} \)

Moment of Inertia, \( I = \frac{\pi d^4}{64} = \frac{3.14 \times 80^4}{64} = 2010613 \, \text{mm}^4 \)

Crippling Load, \( P = \frac{\pi^2 \times 2 \times 10^5 \times 2010613}{(2828.4)^2} = 495142.4 \, \text{N} \approx 496 \, \text{kN} \)