Which one of the following vertical columns, of circular cross-section, sustains the highest load without buckling?
Show Hint
- Cantilever column with a length \( L \) and cross-section diameter \( d \)
- Column with hinge at one end and roller at the other end with a length \( 2L \) and cross-section diameter \( d \)
- Cantilever column with a length \( L \) and cross-section diameter \( 2d \)
- Column with hinge at one end and roller at the other end with a length \( L \) and cross-section diameter \( d \)
The Correct Option is C
Solution and Explanation
The buckling load for a column is given by the formula: \[ P_{cr} = \frac{\pi^2 E I}{(K L)^2} \] where:
- \( P_{cr} \) is the critical buckling load,
- \( E \) is the Young’s modulus of the material,
- \( I \) is the moment of inertia of the column’s cross-section,
- \( L \) is the length of the column,
- \( K \) is the effective length factor, which depends on the type of column and how it is supported.
Step 2: For a cantilever column (fixed at one end), the effective length factor \( K \) is 2. For a column with a hinge at one end and roller at the other end, the effective length factor \( K \) is \( \sqrt{2} \).
Step 3: The moment of inertia \( I \) for a circular cross-section is given by: \[ I = \frac{\pi d^4}{64} \] where \( d \) is the diameter of the circular cross-section.
Thus, for a column to carry the highest load without buckling, we need to consider the combination of effective length factor and the moment of inertia. The larger the moment of inertia and the shorter the effective length, the greater the load the column can withstand.
Step 4: In Option (C), the column has the largest diameter \( 2d \) and a fixed length \( L \), which maximizes both the moment of inertia (because \( I \propto d^4 \)) and minimizes the effective length factor (which is 2 for a cantilever column). Therefore, it will sustain the highest load compared to the other options.
Step 5: Therefore, the correct option is (C).
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