Question

A 70 cm length of aluminium alloy bar is suspended from the ceiling so as to provide a clearance of 0.03 cm between it and a 25 cm length of steel as shown in following figure. 

A_al = 12.5 cm² , E_al = 70 GN/m² ; A_s = 25 cm² E_s = 210 GN/m²

Determine the stress in the aluminium and in the steel due to a 300 kN load applied 50 cm from the ceiling.

Answer 1

Elongation of AB under 300 kN load will be

\( \frac{300\times10^3\times50\times10^{-2}}{12.5\times10^{-4}\times70\times10^9}\) = 1.71428 x 10^-3 m = 0.171428 cm.

This elongation is more than the clearance 0.03 cm between the bars. Hence bars DE and BC will be subjected to compression whereas AB will remain in tension. Let P be compressive force is BC and DE then the tensile force in AB will be (300-P) kN. Then,

\( \frac{(300-p)\times10^3\times50\times10^{-2}}{12.5\times10^{-4}\times70\times10^9}\) - \( \frac{p\times10^3\times20\times10^{-2}}{12.5\times10^{-4}\times70\times10^9}\) - \( \frac{p\times10^3\times25\times10^{-2}}{12.5\times10^{-4}\times210\times10^9}\)

= 0.03 x 10^-2 

\(\therefore\) P = 166.854 kN