Question

A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected

to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an

elongation of 0.50 mm?

Answer 1

"7.654 mm" is the correct solution.

Step-by-step explanation:

According to the question,

• 
  `E=110* 10^3 \ N/mm^2`
• 
  `\sigma_y = 240 \ mPa`
• 
  `P = 6660 \ N`
• 
  `L = 380 \ mm`
• 
  `\delta = 0.5 \ mm`

Now,

As we know,

The Elongation,

⇒
`E=(\sigma)/(e)`

`=((P)/(A) )/((\delta)/(L) )`

or,

⇒
`\delta=(PL)/(AE)`

By substituting the values, we get

`0.5=(6660* 380)/(((\pi)/(4)D^2)(110* 10^3))`

then,

⇒
`D^2=58.587`

`D=√(58.587)`

`=7.654 \ mm`