GATE 2021 AE – Question 59
A rigid horizontal bar ABC, with roller support at A, is pinned to columns BD and CE at B and C. BD is fixed at D, CE pinned at E. BC=100 cm, BD=75 cm, CE=125 cm. Both steel columns have E=200 GPa and cross section 1.5 cm ×1.5 cm. Load P acts a distance a from B. For simultaneous Euler buckling, a is ___ cm (one decimal).
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Correct answer: 14.93 to 15.07
Explanation
**Euler buckling load** of a column with effective-length factor $K$ (pinned-pinned $K=1$, fixed-pinned $K\approx0.699$):
$$P_{cr}=\frac{\pi^2EI}{(KL)^2}.$$
Both columns are steel with the same section ($E=200$ GPa, $1.5\times1.5$ cm), so $EI$ is the same.
- **BD** is fixed at D and pinned at B: $K=0.699$, $L=75$ cm, so $KL=0.699\times75=52.4$ cm.
- **CE** is pinned at both ends: $K=1$, $L=125$ cm, so $KL=125$ cm.
**Ratio of the buckling loads:**
$$\frac{P_{CE}}{P_{BD}}=\left(\frac{52.4}{125}\right)^2=\left(\frac{0.699\times75}{125}\right)^2=0.1759 .$$
**Moment equilibrium of the rigid bar** (taking moments about the roller at A, with the load $P$ at a distance $a$ from B and the two column forces at B and C, BC = 100 cm): for both columns to reach their critical loads at once, the load divides between B and C as the lever arms, so
$$\frac{a}{100-a}=\frac{P_{CE}}{P_{BD}}=0.1759\;\Rightarrow\;a=\frac{0.1759\times100}{1.1759}=\mathbf{14.96\ cm}.$$