Question

A cylindrical bar of metal having a diameter of 20.3 mm and a length of 205 mm is deformed elastically in tension with a force of 46400 N. Given that the elastic modulus and Poisson's ratio of the metal are 66.9 GPa and 0.32, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answer 1

Answer:

a) 0.4393 mm

b)  -0.141 mm

Explanation:

Cylindrical bar :  Diameter = 20.3 mm , Length = 205 mm

Force that deforms bar of metal = 46400 N

elastic modulus = 66.9 GPa

Poisson's ratio ( u ) = 0.32

A) Determine the amount by which this specimen will elongate in the direction of applied stress

dl = $\frac{P*L}{A*E}$

where P = 46400 N

          L = 205 mm

          A = $\frac{\pi }{4} * 20.3^2$ = 323.65

          E = 66.9 GPa

dl =  ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 )  = 0.4393 mm

B) determine the change in diameter of the specimen

change in diameter( compressed due to elongation in length )

= - u * dl

 = - 0.32 * 0.4393

=  -0.141 mm