Question

A steel wire is drawn from an initial diameter $\left(d_i\right)$ of 10 mm to a final diameter $\left(d_f\right)$ of 7.5 mm. The half cone angle $\left(\alpha \right)$ of the die is $5^{\circ }$ and the coefficient of friction $\left(\mu \right)$ between the die and the wire is 0.1. The average of the initial and final yield stress $\left[\right({\sigma }_y{)}_{avg}]$ is 350 MPa. The equation for drawing stress ${\sigma }_f$. (in MPa) is given as:


${\sigma }_f=({\sigma }_y{)}_{avg}\{1+\frac{1}{\mu cot\alpha }\}[1-{\left(\frac{d_f}{d_i}\right)}^{2\mu cot\alpha }]$

The drawing stress (in MPa) required to carry out this operation is (correct to two decimal places).

Answer 1

Given : $d_i=10mm;d_f=7.5mm,\alpha =5^{\circ }$

$\mu =0.1,({\sigma }_y{)}_{avg}=350MPa$

${\sigma }_f=({\sigma }_y{)}_{avg}\times \{1+\frac{1}{\mu cot\alpha }\}[1-{\left(\frac{d_f}{d_i}\right)}^{2\mu cot\alpha }]$

$=350\times [1+\frac{1}{0.1cot{5}^{\circ }}][1-{\left(\frac{7.5}{10}\right)}^{2\times 0.1\times cot{5}^{\circ }}]$

= 316.2472 MPa $\approx$ 316.25 MPa