A cylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Determine the force that will produce an elastic reduction of 2.8 * 10-3 mm (1.102 * 10-4 in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (ν) are, respectively, 207 GPa and 0.30.

given d = 10.2x10-3m

change in diameter d' = 3.4x10-6 m

elastic modulus=207x109pa

1/m=0.30

we know that stress/strain=E

but poissons ratio 1/m = lateral starin/longitudinal starin

0.3 = (d'/d) / (L'/L)

L'/L = 3.4x10-6 / (0.3x10.2x10-3)

= 1.11*10-3

E = (f/A) / (L'/L)

force f = E*A * (L'/L)

f = (1.11*10-3) * 207*109 * (3.1415 * (10.2*10-3) 2) / 4

= 18775.2N = 18.775KN

Explanation:

The force that produce an elastic reduction is 18.775KN