Question:
A steel column has a length of 4 meters and a square cross-section with moment of inertia $1.6 \times 10^{-4} \text{ m}^4$. If the column is pinned at both ends and the modulus of elasticity of steel is 200 GPa, what is the critical buckling load? (take $\pi^2$ value as 10)
A steel column has a length of 4 meters and a square cross-section with moment of inertia $1.6 \times 10^{-4} \text{ m}^4$. If the column is pinned at both ends and the modulus of elasticity of steel is 200 GPa, what is the critical buckling load? (take $\pi^2$ value as 10)
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Euler's Critical Buckling Load ($P_{cr$) is used for long, slender columns where failure occurs due to buckling rather than yielding. The formula is $P_{cr = \frac{\pi^2 E I{(KL)^2$.
- $E$: Modulus of Elasticity
- $I$: Least Moment of Inertia of the cross-section
- $L$: Actual length of the column
- $K$: Effective length factor, which depends on the end conditions:
- Pinned-Pinned: $K=1$
- Fixed-Fixed: $K=0.5$
- Fixed-Pinned: $K=0.707$
- Fixed-Free: $K=2$
Updated On: May 27, 2025
- 200 kN
- 10000 kN
- 40000 kN
- 20000 kN
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The Correct Option is D
Solution and Explanation
Step 1: Identify the given parameters.
Length of the column, $L = 4 \text{ meters}$
Moment of inertia, $I = 1.6 \times 10^{-4} \text{ m}^4$
End conditions: Pinned at both ends. For pinned-pinned ends, the effective length factor $K = 1$.
Modulus of elasticity of steel, $E = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa}$
Take $\pi^2 = 10$
Step 2: State Euler's Critical Buckling Load formula.
Euler's formula for the critical buckling load ($P_{cr}$) for a column is:
$P_{cr} = \frac{\pi^2 E I}{(KL)^2}$
Step 3: Substitute the values into the formula.
Since the column is pinned at both ends, $K=1$.
$P_{cr} = \frac{10 \times (200 \times 10^9 \text{ Pa}) \times (1.6 \times 10^{-4} \text{ m}^4)}{(1 \times 4 \text{ m})^2}$
$P_{cr} = \frac{10 \times 200 \times 10^9 \times 1.6 \times 10^{-4}}{16}$
Step 4: Perform the calculation.
$P_{cr} = \frac{10 \times 200 \times 1.6 \times 10^{(9-4)}}{16}$
$P_{cr} = \frac{3200 \times 10^5}{16}$
$P_{cr} = 200 \times 10^5 \text{ N}$
$P_{cr} = 200 \times 10^2 \times 10^3 \text{ N}$
$P_{cr} = 20000 \times 10^3 \text{ N}$
$P_{cr} = 20000 \text{ kN}$
Step 5: Evaluate the options.
The calculated critical buckling load is 20000 kN, which matches option 4.
The final answer is $\boxed{\text{4}}$.
Length of the column, $L = 4 \text{ meters}$
Moment of inertia, $I = 1.6 \times 10^{-4} \text{ m}^4$
End conditions: Pinned at both ends. For pinned-pinned ends, the effective length factor $K = 1$.
Modulus of elasticity of steel, $E = 200 \text{ GPa} = 200 \times 10^9 \text{ Pa}$
Take $\pi^2 = 10$
Step 2: State Euler's Critical Buckling Load formula.
Euler's formula for the critical buckling load ($P_{cr}$) for a column is:
$P_{cr} = \frac{\pi^2 E I}{(KL)^2}$
Step 3: Substitute the values into the formula.
Since the column is pinned at both ends, $K=1$.
$P_{cr} = \frac{10 \times (200 \times 10^9 \text{ Pa}) \times (1.6 \times 10^{-4} \text{ m}^4)}{(1 \times 4 \text{ m})^2}$
$P_{cr} = \frac{10 \times 200 \times 10^9 \times 1.6 \times 10^{-4}}{16}$
Step 4: Perform the calculation.
$P_{cr} = \frac{10 \times 200 \times 1.6 \times 10^{(9-4)}}{16}$
$P_{cr} = \frac{3200 \times 10^5}{16}$
$P_{cr} = 200 \times 10^5 \text{ N}$
$P_{cr} = 200 \times 10^2 \times 10^3 \text{ N}$
$P_{cr} = 20000 \times 10^3 \text{ N}$
$P_{cr} = 20000 \text{ kN}$
Step 5: Evaluate the options.
The calculated critical buckling load is 20000 kN, which matches option 4.
The final answer is $\boxed{\text{4}}$.
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