Question

A rod of 150 cm long and of diameter 2 cm

is subjected to an axial pull of 20 KN.

Determine : the Stress.​

Answer 1

Answer:

Each of the segment of this composite bar is subjected to axial pull P =150 kN. Axial Stress in the middle portion σ2 = Axial pull/Area = 150 × 103/[(Π/4).(d22)] Since stress is limited to 125 N/mm2, in the middle portion 125 = 150 × 103/[(Π/4).(d22)] d22 = 1528.66 mm Diameter of middle portion d2 = 39.1mm (ii) Stress in the end portions, σ1 = σ3 = 150 × 103/[(Π/4).(502)] = 76.43 N/m2  Total change in length of the bar, = change in length of end portions + change in length of mid portion δL = δL1 + δL2 + δL3 = σ1L1/E + σ2L2/E + σ3L3/E Since E is same for all portions = σ1(L1 + L3)/E + σ2L2/E L1 + L3 = 300 – L2 Now putting all the values; 0.15 = [76.43(300 – L2)]/2 × 105 + 125L2/2 × 105 L2 = 145.58 mm.Read more on Sarthaks.com - https://www.sarthaks.com/513423/the-shown-subjected-axial-pull-150kn-determine-diameter-the-middle-portion-stress-there