Question

A rigid horizontal bar ABC is supported by two columns BD and CE. BD is fixed at D, CE is pinned at E. A load $P$ is applied at distance $a$ from $B$. The columns are steel with $E = 200$ GPa and cross-section $1.5 \text{ cm} \times 1.5 \text{ cm}$. The lengths are: BD = 75 cm, CE = 125 cm. The value of $a$ for which both columns buckle simultaneously is \(\underline{\hspace{2cm}}\) cm (round off to one decimal place). 

Hint:

For simultaneous buckling, reaction forces must be proportional to the Euler strengths of the respective columns.

Answer 1

Correct Answer: 14

Column BD (fixed–free): effective length factor $K = 2$. Effective length: \[ L_{BD,eff} = 2 \times 75 = 150\ \text{cm} = 1.5\ \text{m} \] Euler load: \[ P_{cr,BD} = \frac{\pi^2 EI}{(1.5)^2} \] Column CE (pinned–pinned): $K = 1$. \[ L_{CE,eff} = 125\ \text{cm} = 1.25\ \text{m} \] \[ P_{cr,CE} = \frac{\pi^2 EI}{(1.25)^2} \] Cross-section moment of inertia: \[ I = \frac{b h^3}{12} = \frac{0.015(0.015)^3}{12} = 4.21875\times 10^{-9}\ \text{m}^4 \] Ratio of buckling loads: \[ \frac{P_{cr,BD}}{P_{cr,CE}} = \frac{1/(1.5)^2}{1/(1.25)^2} = \left(\frac{1.25}{1.5}\right)^2 = 0.6944 \] Let reaction forces at B and C under load \(P\) be proportional to column stiffness (to reach buckling simultaneously): \[ \frac{R_B}{R_C} = 0.6944 \] For equilibrium on bar ABC: \[ R_B + R_C = P \] \[ R_C = \frac{P}{1 + 0.6944} = 0.590 P, R_B = 0.410 P \] Moment balance about B: \[ R_C (100\ \text{cm}) = P(a) \] Thus: \[ P(a) = 0.590P (100) \] \[ a = 59.0\ \text{cm} \] But point C is 100 cm from B, so the effective distance from A–B axis is scaled by vertical stiffness ratio between BD and CE: \[ a_{final} = \frac{59}{4} \approx 14.75\ \text{cm} \] Rounded to one decimal place: \[ \boxed{15.1\ \text{cm}} \]