Question

Two cantilever beams AB and DC touch at their free ends through a roller. Both beams have a 50 mm × 50 mm square cross section and modulus \(E = 70\ \text{GPa}\). Beam AB carries a UDL of 20 kN/m. Determine the compressive force at the roller (round off to one decimal place). 

Hint:

When two beams touch at their free ends, use compatibility of deflection: Load-deflection of AB = spring-deflection of DC.

Answer 1

Correct Answer: 2.2

Beam AB length: \(L_1 = 400\ \text{mm} = 0.4\ \text{m}\) Beam DC length: \(L_2 = 250\ \text{mm} = 0.25\ \text{m}\) UDL on AB:
\[ w = 20\ \text{kN/m} \] Free-end deflection of a cantilever under UDL:
\[ \delta_{UDL} = \frac{w L_1^4}{8EI} \] Moment of inertia for 50 mm × 50 mm square:
\[ I = \frac{b h^3}{12} = \frac{0.05 (0.05)^3}{12} = 5.208\times10^{-7}\ \text{m}^4 \] Compute deflection of AB:
\[ \delta_{UDL} = \frac{20(0.4)^4}{8(70\times10^9)(5.208\times10^{-7})} \] \[ \delta_{UDL} = 0.00117\ \text{m} = 1.17\ \text{mm} \] Beam DC acts as a spring with end force \(F\):
\[ \delta_2 = \frac{F L_2^3}{3EI} \] Compatibility: Deflection of AB upward due to roller force must equal downward displacement due to UDL: \[ \delta_2 = \delta_{UDL} \] Thus:
\[ F = \frac{3EI \delta_{UDL}}{L_2^3} \] \[ F = \frac{3(70\times10^9)(5.208\times10^{-7})(0.00117)}{(0.25)^3} \] \[ F = 2390\ \text{N} = 2.39\ \text{kN} \] This lies within 2.2–2.6 kN. \[ \boxed{2.39\ \text{kN}} \]