Columns and Struts Questions and Answers

Answer: B

The buckling temperature rise is proportional to \(d^{2}\).

With both ends fixed, heating builds up a compressive thermal force \(P = EA\alpha\Delta T\). The beam buckles when this reaches the Euler load for a fixed-fixed column:

\[ \begin{aligned} EA\alpha\Delta T &= \frac{4\pi^{2}EI}{L^{2}} \\ \Delta T &= \frac{4\pi^{2}}{\alpha L^{2}}\cdot\frac{I}{A} \end{aligned} \]

For a circle \(I/A = (\pi d^{4}/64)/(\pi d^{2}/4) = d^{2}/16\), the square of the radius of gyration, so \(\Delta T \propto d^{2}\). E cancels out.

\(d^{4}\) is the dependence of the buckling load itself (through I), forgetting that the thermal force also grows with the area \(d^{2}\); d and \(d^{3}\) do not follow.

Answer: B

The critical load is \(P = \dfrac{\pi^{2}EI}{L^{2}}\).

Euler's buckling load is \(P = \dfrac{\pi^{2}EI}{L_{e}^{2}}\), where the effective length \(L_{e}\) depends on the end conditions. For a column pinned at both ends \(L_{e} = L\), which gives the result above.

The other options belong to other end conditions. \(\pi^{2}EI/4L^{2}\) is for a column fixed at one end and free at the other (\(L_{e} = 2L\)). \(4\pi^{2}EI/L^{2}\) is for both ends fixed (\(L_{e} = L/2\)). \(3\pi^{2}EI/4L^{2}\) matches no standard case.

Answer: C

The ratio is 4.

Euler's buckling load is \(P_{cr} = \pi^{2}EI/L_{e}^{2}\), where L_e is the effective length. A column hinged at both ends has \(L_{e}=L\); one clamped at both ends has \(L_{e}=L/2\). So:

\[ \frac{P_{clamped}}{P_{hinged}} = \frac{L^{2}}{(L/2)^{2}} = 4 \]

A ratio of 2 compares the effective lengths rather than their squares, and 8 has no basis in the Euler formula. 1 would mean end fixity does not matter, but clamping the ends shortens the buckled half-wave and so raises the load.

Answer: C

The minimum buckling load is in case (iii).

Euler's load is \(P = \pi^{2}EI/(KL)^{2}\) with effective length factor K = 2 for a fixed-free column and 1 for pinned-pinned. In units of \(\pi^{2}EI/L^{2}\):

\[ \begin{aligned} \text{(ii)}: &\ \frac{2}{(2 \times 0.7)^{2}} = \frac{2}{1.96} = 1.02 \\ \text{(iii)}: &\ \frac{1}{(2 \times 0.51)^{2}} = \frac{1}{1.0404} = 0.96 \\ \text{(iv)}: &\ \frac{1}{1^{2}} = 1.00 \end{aligned} \]

Case (i) is case (iv) with an extra spring support part way along, which can only raise the buckling load above 1.00. So (iii), at 0.96, is the lowest.

Answer: D

The effective lengths are \(L\), \(2L\), \(L/2\).

The effective length is the length of a pin-ended column that buckles at the same load, measured between points of zero moment (inflection points) in the buckled shape:

• Hinged-hinged: the whole column is one half sine wave, so \(L_{e} = L\).
• Fixed-free: the column is half of a pin-ended half wave, so \(L_{e} = 2L\).
• Fixed-fixed: the inflection points lie at \(L/4\) from each end, leaving \(L/2\) between them, so \(L_{e} = L/2\).

The options with \(4L\) or \(L/4\) confuse the effective length with the factor in the buckling load (\(P_{cr}\) changes by 1/4 and by 4 for these cases), and \(L, L, L\) ignores the end conditions.

Answer: A

The smallest force is \(\frac{2}{\sqrt{3}}\frac{\pi^{2}EI}{L^{2}}\).

The column is pinned at both ends: the top pin runs in a vertical guide that carries the horizontal component of F. Only the axial component \(F\cos\theta\) compresses the column, and buckling starts when it reaches the Euler load for a pinned-pinned column:

\[ \begin{aligned} F\cos 30^{\circ} &= \frac{\pi^{2}EI}{L^{2}} \\ F &= \frac{2}{\sqrt{3}}\frac{\pi^{2}EI}{L^{2}} \end{aligned} \]

\(\frac{\sqrt{3}}{2}\frac{\pi^{2}EI}{L^{2}}\) multiplies by \(\cos\theta\) instead of dividing. \(\pi^{2}EI/(2L^{2})\) and \(2\pi^{2}EI/L^{2}\) come from other end conditions (fixed-pinned gives about \(2.05\pi^{2}EI/L^{2}\)) and also ignore the inclination.

Answer: C

The strongest column is the cantilever column with a length L and cross-section diameter 2d.

Euler's buckling load is \(P_{cr} = \pi^{2}EI/L_{e}^{2}\), where \(I = \pi d^{4}/64\) for a circular section and \(L_{e}\) is the effective length: 2L for a cantilever (fixed-free) and L for a column pinned at both ends (hinge and roller).

In units of \(\pi^{2}EI_{d}/L^{2}\), where \(I_{d}\) is the second moment of area for diameter d:

• Cantilever, length L, diameter d: \(1/(2)^{2} = 0.25\).
• Pinned-pinned, length 2L, diameter d: \(1/(2)^{2} = 0.25\).
• Cantilever, length L, diameter 2d: I is 16 times larger, so \(16/4 = 4\).
• Pinned-pinned, length L, diameter d: 1.

Doubling the diameter multiplies I by \(2^{4} = 16\), which outweighs the factor of 4 lost to the cantilever's effective length.

Answer: B

The ratio is \(\frac{q}{p}\).

Both columns have the same material, length and end conditions, so their Euler loads are in the ratio of their least second moments of area. Buckling happens about the weaker axis:

\[ \begin{aligned} I_{rect,min} &= \frac{pq^{3}}{12},\quad I_{sq} = \frac{r^{4}}{12} \\ r^{2} &= pq \text{ (equal areas)} \end{aligned} \]

\[ \begin{aligned} \frac{F_{rect}}{F_{sq}} &= \frac{pq^{3}}{r^{4}} \\ &= \frac{pq^{3}}{p^{2}q^{2}} = \frac{q}{p} \end{aligned} \]

Since q is less than p, the ratio is below 1: for the same area the square column is the stronger. p/q uses the stiffer axis of the rectangle, which is not where buckling occurs.