Columns and Struts Questions and Answers

Answer: C

The effective length is 12m.

The pole is fixed at its base in the foundation, and its top is free to move sideways, so it behaves as a column with one end fixed and the other free. For that case the effective length is twice the actual length.

\[ L_e = 2L = 2 \times 6 = 12\ \text{m} \]

6 m would apply with both ends hinged, 3 m with both ends fixed (0.5L), and 18 m matches no standard end condition.

Answer: D

It is the middle quarter rule for circular section.

When a column carries an eccentric load, the bending stress it causes opposes the direct compressive stress on one side. For a solid circular section of diameter D, the section stays free of tension as long as the eccentricity does not exceed D/8, that is, the load lies within a central circle of diameter D/4. This core is the middle quarter, hence the name.

The rectangular counterpart is the middle third rule. The zeroth law belongs to thermodynamics, and 'third rule' and 'Archimedes rule' are not names for this kern.

Answer: B

Rankine's formula is \(P = \frac{\sigma_c A}{1 + \alpha\left(\frac{L_e}{K}\right)^{2}}\).

Rankine combined the crushing load of a short column, \(P_c = \sigma_c A\), with Euler's buckling load of a long column, \(P_E\), through \(\frac{1}{P} = \frac{1}{P_c} + \frac{1}{P_E}\):

\[ \begin{aligned} P &= \frac{\sigma_c A}{1 + \frac{\sigma_c}{\pi^2 E}\left(\frac{L_e}{K}\right)^2} \\ \alpha &= \frac{\sigma_c}{\pi^2 E} \end{aligned} \]

The slenderness ratio \(L_e/K\) appears squared, which rules out the first option. Squaring \(\sigma_c\) or A, as the other options do, gives a quantity that is not a force.

Answer: D

The assumption that is not true is The bending stress is very small as compared to the direct stress.

Euler's theory deals with long columns that fail by buckling. It assumes the opposite: the direct stress is very small compared with the bending stress, so the direct stress is neglected and failure is by buckling alone.

The other three are genuine Euler assumptions. The column fails by buckling alone, its cross-section is uniform along its length, and it is very long compared with its lateral dimensions. The theory also assumes an initially straight column, an axial load and a homogeneous material.

Answer: C

For buckling about the X axis, both ends are hinged.

A connecting rod swings in the plane of motion of the crank. In that plane, buckling about the X axis, its ends are free to turn on the gudgeon pin and the crank pin, so it behaves as a column with both ends hinged and an equivalent length equal to its actual length.

In the perpendicular plane, buckling about the Y axis, the pins prevent rotation and the ends act as fixed, so the equivalent length is half the actual length. That is why the rod is designed with an I-section whose moment of inertia about X is about four times that about Y. The fixed and mixed end conditions in the other options belong to the Y axis or to neither.

Answer: A

The critical load is \(P_{cr} = \frac{\pi^2 EI}{L^2}\).

Euler's formula for buckling is \(P_{cr} = \frac{\pi^2 EI}{L_e^2}\), where \(L_e\) is the effective length. For a column pinned (hinged) at both ends, the deflected shape is a half sine wave over the full length, so \(L_e = L\).

Other end conditions change only \(L_e\): \(2L\) for a cantilever, \(L/\sqrt{2}\) for one end fixed and one hinged, \(L/2\) for both ends fixed. The options with \(\pi^2\) in the denominator, with \(3L^2\), or with \(\pi\) rather than \(\pi^2\) match none of these cases.

Answer: B

The effective length is L/2.

Effective length is the length of an equivalent pin-ended column with the same buckling load. Fixing both ends forces the column to buckle into a shape whose inflection points lie at quarter span, so the curved part between them is half the length. Euler's theory therefore gives an effective length of L/2 and a buckling load four times that of a pin-ended column.

L is for both ends pinned, 2L for one end fixed and the other free, and L divided by root 2 for one end fixed and one pinned. Neither 4L nor root 2 times L corresponds to any standard end condition.

Answer: B

The radius of gyration is \(\frac{d}{4}\).

Radius of gyration is \(k = \sqrt{I/A}\). For a solid circle \(I = \frac{\pi d^{4}}{64}\) and \(A = \frac{\pi d^{2}}{4}\).

\[ \begin{aligned} k^{2} &= \frac{\pi d^{4}/64}{\pi d^{2}/4} = \frac{d^{2}}{16} \\ k &= \frac{d}{4} \end{aligned} \]

d/2 is the radius, a common slip. The squared options give an area-like quantity, not a length.

Answer: C

For both ends fixed the ratio is \(\frac{1}{2}\).

Effective length is the length of an equivalent pin-ended column that buckles at the same load. With both ends fixed, the column bends between two points of contraflexure a quarter length from each end, so the effective length is \(L/2\).

A ratio of 2 belongs to a column fixed at one end and free at the other, \(1/\sqrt{2}\) to one fixed at one end and hinged at the other, and \(\sqrt{2}\) is not a standard case.

Answer: A

Euler's buckling load for a column with both ends hinged is \(\frac{\pi^{2}EI}{L^{2}}\).

Euler's critical load is \(P_{cr} = \frac{\pi^{2}EI}{L_{e}^{2}}\), where \(L_{e}\) is the effective length. For both ends hinged \(L_{e} = L\), giving \(\frac{\pi^{2}EI}{L^{2}}\).

The other options square E or I or drop a power of \(\pi\). The load is directly proportional to the flexural rigidity EI, not to its square.