An A-36 steel column has a length of 12 feet and is pinned at both ends. If the cross-sectional area has the dimension shown, determine the critical load.
The correct answer and explanation is:
To determine the critical load for an A-36 steel column pinned at both ends, use Euler’s Buckling Formula: Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(KL)^2}
Given:
- Length L=12L = 12 feet = 144144 inches
- Pinned at both ends, so effective length factor K=1K = 1
- E (modulus of elasticity for A-36 steel) = 29,000,000 psi
- I = moment of inertia (value depends on the cross-sectional shape)
Since the moment of inertia depends on the cross-sectional dimensions and you mentioned “has the dimension shown” but did not attach an image, we will assume a typical W8 × 40 wide flange steel column often used with A-36 steel.
For a W8 × 40:
- Area, A = 11.7 in²
- I (least moment of inertia), I = 91.8 in⁴
Now plug into Euler’s formula: Pcr=π2⋅29,000,000⋅91.8(1⋅144)2P_{cr} = \frac{\pi^2 \cdot 29,000,000 \cdot 91.8}{(1 \cdot 144)^2} Pcr=3.14162⋅29,000,000⋅91.820736P_{cr} = \frac{3.1416^2 \cdot 29,000,000 \cdot 91.8}{20736} Pcr≈26.7×10920736P_{cr} \approx \frac{26.7 \times 10^9}{20736} Pcr≈1,287,031 pounds=1,287 kipsP_{cr} \approx 1,287,031 \text{ pounds} = 1,287 \text{ kips}
Final Answer:
The critical load is approximately 1,287 kips (1 kip = 1,000 pounds).
Explanation:
Euler’s formula determines the axial load at which a long slender column will buckle. The formula considers the length, end conditions, and stiffness (moment of inertia and modulus of elasticity) of the column. Since the ends are pinned, the column is more prone to buckling than one with fixed ends. The smaller the moment of inertia or longer the length, the lower the critical load. This analysis assumes linear elastic behavior and no initial imperfections. Always select the least moment of inertia (about the weaker axis) to ensure safety.