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

Answer:

"7.654 mm" is the correct solution.

Explanation:

According to the question,

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

Now,

As we know,

The Elongation,

⇒ $E=\frac{\sigma}{e}$

       $=\frac{\frac{P}{A} }{\frac{\delta}{L} }$

or,

⇒ $\delta=\frac{PL}{AE}$

By substituting the values, we get

 $0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}$

then,

⇒ $D^2=58.587$

     $D=\sqrt{58.587}$

         $=7.654 \ mm$