Question

A copper billet of diameter 100 mm is to be extruded by an extrusion load of 2.72 MN. If the extrusion constant is 250 MPa. Then, the final diameter of the billet after final extrusion is

• A. 40.03 mm
• B. 50.03 mm
• C. 38.05 mm
• D. 42.53 mm

Answer 1

The correct option is B 50.03 mm

$F=k{A}_{0}ln\left(\frac{A_0}{A_f}\right)$

$\Rightarrow 2.72\times {10}^6=250\times \frac{\pi }{4}\times {\left(100\right)}^2\times ln\left(\frac{{100}^2}{d_f^2}\right)$

$\Rightarrow ln\left(\frac{100}{d_f}\right)=0.69264$

$\therefore d_f=50.03mm$