Question

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a load of 11,100 N (2500 lb f ). If the length of the rod is 500 mm (20.0 in.), what must be the diameter to allow an elongation of 0.38 mm (0.015 in.)?

Answer 1

Diameter of the cylinder will be
`d=2.998* 10^4m`

Step-by-step explanation:

We have given young's modulus of steel
`E=207GPa=207* 10^9Pa`

Change in length
`\Delta l=0.38mm`

Length of rod
`l=500mm`

Load F = 11100 KN

Strain is given by
`strain=(\Delta l)/(l)=(0.38)/(500)=7.6* 10^(-4)`

We know that young's modulus
`E=(stress)/(strain)`

So
`207* 10^9=(stress)/(7.6* 10^(-4))`

`stress=1573.2* 10^(-5)N/m^2`

We know that stress
`=(force)/(artea )`

So
`1573.2* 10^(-5)=(11100* 1000)/(area)`

`area=7.055* 10^(8)m^2`

So
`(\pi )/(4)d^2=7.055* 10^(8)`

`d=2.998* 10^4m`