A rectangular bar 40 mm wide and 30 mm thick is 0.5 m long. It is acted upon by axial tensile force of 120 KN. E = 200 Gpa and p-0.25. What is the change in dimensions of the bar?

A rectangular bar 40 mm wide and 30 mm thick is 0.5 m long. It is acted upon by axial tensile force of 120 KN. E = 200 Gpa and p-0.25. What is the change in dimensions of the bar?

Solution

 Though procedure is simple you have to be cautious about units of different quantities. Convert all of them into m , sq m and kN.

Length l=0.5 m let change in length be dl ; width w = 40 mm =0.04 m, let change in width =dw ; thickness t = 30 mm =0.03 m ,let change in thickness =dt.; cross sectional area A = 0.04 *0.03 = 0.0012 sq m

F, applied longitudinal force = 120 kN ;

 Modulus of elasticity  E=200 GPa =200,000000kN/sq m. ; poisson’s ratio ,p , lateral strain /longitudinal strain =-0.25

Longitudinal stress =F/A=120/0.0012 

=100000 kN/sq m; longitudinal strain 

=dl/0.5 ; E= stress /strain = 200,000000 = 100000 *0.5/dl ; 

dl =2.5/10000. m 

=0.25 mm

Lateral strain in width =dw/w =p*longitudinal strain

 =-0.25 *2.5/10000 ; change in width =dw =-0.25 *2.5*0.04/10000 m 

=-1/400000 m =-1/400 mm

= - 0.0025mm, negative sign indicates compression in width.

Lateral strain in thickness =dt/t =p *longitudinal strain 

=-0.25*2.5/10000 ; change in thickness =dt =-0.25 *2.5 *0.03 /10000 

=-3/1600000 m 

=- 3/1600 mm

 =-0.001875mm, compression