Question

Find the modulus of elasticity for a rod, which tapers uniformly from 30 mm to 15 mm diameter in a length of 350 mm. The rod is subjected to an axial load of 5.5 kN and extension of the rod is 0.025mm.​

Answer 1

SOLUTION :-

Given:

Larger diameter D1 = 30mm

Smaller diameter D2 = 15mm

Length of rod, L = 350mm

Axial load P = 5.5 kN = 5500 N

Extension dL = 0.025mm

Using equation (1.10), we get

dL = 4PL / $\pi E D_{1} D_{2} dL$

⇒E =4PL/ $\pi D_{1} D_{2} dL$  =  4 x 5500 x 350 / π x 30 x 15 x 0.025

= 217865 N/mm² or    $2.17865 \ * 10^{5} N/ mm^{2}$