Question

8) Given: bar of brass alloy with ơ- diagram given in Fig. 6.12. a-25 mm, b-80 mm, L-200 a) F-200 kN is applied. What is the final length of the bar AFTER the force is released? b) F-800 kN is applied. What is the final length of the bar AFTER the force is released?

Answer 1

Due to the applied force the bar will be in tension and due to which tensile stress will act in the longitudinal direction causing elongation of bar so the final length will be the actual length + the elongated length.

Calculation is as shown below.

Given, dimension, a = 25 mm, b = 80 mm, L = 200 mm. change in length of bar due to applied force is given by: Delta L = FL/AE Where, F = applied force L = Length of bar A = cross sectional area = a times b = 25 times 80 mm^2 A = 2000 mm^2 E = modulus of elasticity of brass E = 105 Gpa = 105 times 10^3 M pa E = 105 times 10^3 N/mm^2. Now substitute the given value in above equation: F = 200 kN = 200 times 10^3 N. Delta L = 200 times 10^3 N times 200 mm/200 mm^2 times 105 times 10^3 N/mm^2 Delta L = 0.19 mm so final length will be L + Delta L = (200 + 0.19) mm Final length = 200.19 mm G = 800 kN = 800 times 10^3 N Delta L = 800 times 10^3 N times 200 mm/200 mm^2 times 105