Question

A beam weighing 50 N is held in horizontal position by three wires. The outer wires are of brass of 1.8 mm dia and attached to each end of the beam. The central wire is of steel of 0.9 mm diameter and attached to the middle of the beam. The beam is rigid and the wires are of the same length and unstressed before the beam is attached. Determine the stress induced in each of the wire. Take Young's modulus for brass as 80 GN/m² and for steel as 200 GN/m².

Answer 1

If Pb denotes load taken by each brass wire and Ps denotes load taken by steel wire, then 

Total load P = 2P_b + P_s = 2\(\sigma\)_bA_b + \(\sigma\)_sA_s (i) 

As the beam is horizontal, all the wire extend by the same amount. Further since each wire is of same length, the wires would experience the same amount of strain, thus

e_s = e_b 

\(\sigma\)_s/E_s = \(\sigma\)_b/E_b 

\(\sigma\)_s = (E_s × \(\sigma\)_b)/Eb = (200 × \(\sigma\)_b)/80 = 2.5\(\sigma\)_b.....(iii)

Putting the value of equation (ii) in equation (i)

50 = 2\(\sigma\)_b(\(\pi\)/4)(1.8)² + 2.5\(\sigma\)b (\(\pi\)/4)(0.9)²

50 = 6.678 \(\sigma\)_b 

\(\sigma\)_b = 7.49 N/mm² 

\(\sigma\)_s = 2.5\(\sigma\)_b = 18.71 N/mm²