Question

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.

Answer 1

Answer:

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