Determine the critical buckling load for each of the columns, using the Euler equation. E = 29,000 ksi. Proportional limit = 36,000 psi. Assume simple ends and maximum permissible L/r = 200. A solid round bar of 2 in diameter: a. L = 4 ft 0 in b. L = 2 ft 3 in c. L = 6 ft 6 in
The Correct Answer and Explanation is:
To determine the critical buckling load using Euler’s formula, we apply the equation:Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 E I}{(KL)^2}Pcr=(KL)2π2EI
Where:
- PcrP_{cr}Pcr = critical buckling load (lb)
- EEE = modulus of elasticity = 29,000 ksi = 29,000,000 psi
- III = moment of inertia of the cross section (in⁴)
- LLL = column length (in)
- KKK = column effective length factor (K = 1.0 for pinned-pinned ends)
- π≈3.1416\pi \approx 3.1416π≈3.1416
Step 1: Compute Moment of Inertia for a solid circular section
For a solid round bar:I=πd464I = \frac{\pi d^4}{64}I=64πd4
Given diameter d=2d = 2d=2 in:I=π(2)464=π⋅1664=π4≈0.7854 in4I = \frac{\pi (2)^4}{64} = \frac{\pi \cdot 16}{64} = \frac{\pi}{4} \approx 0.7854 \text{ in}^4I=64π(2)4=64π⋅16=4π≈0.7854 in4
Step 2: Compute critical loads for each length
a. L = 4 ft 0 in = 48 in
Pcr=π2⋅29,000,000⋅0.7854(48)2=9.8696⋅29,000,000⋅0.78542304≈224,899,102.62304≈97,611.5 lbP_{cr} = \frac{\pi^2 \cdot 29,000,000 \cdot 0.7854}{(48)^2} = \frac{9.8696 \cdot 29,000,000 \cdot 0.7854}{2304} \approx \frac{224,899,102.6}{2304} \approx 97,611.5 \text{ lb}Pcr=(48)2π2⋅29,000,000⋅0.7854=23049.8696⋅29,000,000⋅0.7854≈2304224,899,102.6≈97,611.5 lb
b. L = 2 ft 3 in = 27 in
Pcr=π2⋅29,000,000⋅0.7854(27)2=224,899,102.6729≈308,434.4 lbP_{cr} = \frac{\pi^2 \cdot 29,000,000 \cdot 0.7854}{(27)^2} = \frac{224,899,102.6}{729} \approx 308,434.4 \text{ lb}Pcr=(27)2π2⋅29,000,000⋅0.7854=729224,899,102.6≈308,434.4 lb
c. L = 6 ft 6 in = 78 in
Pcr=224,899,102.6(78)2=224,899,102.66084≈36,957.4 lbP_{cr} = \frac{224,899,102.6}{(78)^2} = \frac{224,899,102.6}{6084} \approx 36,957.4 \text{ lb}Pcr=(78)2224,899,102.6=6084224,899,102.6≈36,957.4 lb
Final Answers:
- (a) L = 4 ft 0 in → 97,612 lb
- (b) L = 2 ft 3 in → 308,434 lb
- (c) L = 6 ft 6 in → 36,957 lb
Explanation:
Euler’s formula estimates the axial load at which long, slender columns buckle. The moment of inertia depends on cross-sectional geometry. The slenderness ratio L/rL/rL/r ensures the column behaves elastically (Euler’s theory applies). In all cases above, the slenderness ratio is below the permissible limit of 200, confirming Euler’s formula is applicable. The shorter the column, the higher the critical buckling load, as buckling is inversely proportional to the square of the column length.
