A column that can support same load in compression as it can in tension is called

Question

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

Answer 1

A column's compressive capacity equals its tensile capacity only when buckling plays no role at all, i.e. when the column is too short and stocky to buckle before it crushes.

Intermediate column: buckling already reduces capacity somewhat below the pure material strength.
Long column: buckling governs strongly, cutting compressive capacity far below tensile capacity.
Short column: too stocky to buckle, so it crushes at full material strength, the same basis as its tensile capacity.
Cannot be determined: the slenderness-based classification makes this determinable, not unknown.

So the column that carries the same load in compression as in tension is a short column.

Answer 2

A third way to see this is to compare a rough numeric estimate of the Euler buckling stress against the material's crushing stress for representative values of slenderness ratio, rather than only discussing the formula qualitatively.

Using $P_{cr} = \dfrac{\pi^2 EI}{L_e^2}$ and expressing this in terms of the slenderness ratio $\lambda = L_e/r$ (where r is the radius of gyration), the buckling stress becomes $\sigma_{cr} = \dfrac{\pi^2 E}{\lambda^2}$. For a very small slenderness ratio, say $\lambda = 20$, and typical structural steel with $E \approx 2\times10^5\ \text{N/mm}^2$, this gives $\sigma_{cr} \approx \dfrac{\pi^2(2\times10^5)}{400} \approx 4900\ \text{N/mm}^2$, a value far higher than the yield stress of ordinary structural steel (around 250 N/mm2). This confirms the column would crush at its yield stress long before this enormous buckling stress could ever be reached.

Intermediate column: at moderate slenderness ratios (say $\lambda \approx 80$ to $120$), the same formula gives a buckling stress much closer to, or below, the material's yield stress, so buckling starts to control the design instead of pure crushing.
Long column: at large slenderness ratios, the computed buckling stress falls well below the yield stress, so buckling clearly governs and compressive capacity is much lower than tensile capacity.
Short column: the numeric estimate above, at a small slenderness ratio, shows the buckling stress vastly exceeds the yield stress, confirming crushing (not buckling) governs, so compressive capacity equals the tensile capacity based on the same yield strength.
Cannot be determined: the numeric comparison above shows the outcome is entirely predictable from the slenderness ratio, so this option does not apply.

Plugging in representative numbers confirms that only at low slenderness ratios (short columns) does the compressive capacity match the tensile capacity.

Therefore, the correct answer is short column.

Answer 3

A third way to see this is to compare a rough numeric estimate of the Euler buckling stress against the material's crushing stress for representative values of slenderness ratio, rather than only discussing the formula qualitatively.

Using $P_{cr} = \dfrac{\pi^2 EI}{L_e^2}$ and expressing this in terms of the slenderness ratio $\lambda = L_e/r$ (where r is the radius of gyration), the buckling stress becomes $\sigma_{cr} = \dfrac{\pi^2 E}{\lambda^2}$. For a very small slenderness ratio, say $\lambda = 20$, and typical structural steel with $E \approx 2\times10^5\ \text{N/mm}^2$, this gives $\sigma_{cr} \approx \dfrac{\pi^2(2\times10^5)}{400} \approx 4900\ \text{N/mm}^2$, a value far higher than the yield stress of ordinary structural steel (around 250 N/mm2). This confirms the column would crush at its yield stress long before this enormous buckling stress could ever be reached.

Intermediate column: at moderate slenderness ratios (say $\lambda \approx 80$ to $120$), the same formula gives a buckling stress much closer to, or below, the material's yield stress, so buckling starts to control the design instead of pure crushing.
Long column: at large slenderness ratios, the computed buckling stress falls well below the yield stress, so buckling clearly governs and compressive capacity is much lower than tensile capacity.
Short column: the numeric estimate above, at a small slenderness ratio, shows the buckling stress vastly exceeds the yield stress, confirming crushing (not buckling) governs, so compressive capacity equals the tensile capacity based on the same yield strength.
Cannot be determined: the numeric comparison above shows the outcome is entirely predictable from the slenderness ratio, so this option does not apply.

Plugging in representative numbers confirms that only at low slenderness ratios (short columns) does the compressive capacity match the tensile capacity.

Therefore, the correct answer is short column.

A column that can support same load in compression as it can in tension is called

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The Correct Option is C

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Approach Solution -2

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