Discussion
For the beam-column configurations shown in figure, the minimum Euler buckling load is obtained for the case (Young's modulus and second moment of cross-sectional area are as indicated)
- A.
- B.
- C.
- D.
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.
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