Question

A steel wire is drawn from an initial diameter (d_i) of 10 mm to a final diameter (d_f) of 7.5 mm. The half cone angle (α) of the die is 5° and the coefficient of friction (μ) between the die and the wire is 0.1. The average of the initial and final yield stress [(σ_Y)_avg] is 350 MPa. The equation for drawing stress σ_f, (in MPa) is given as:

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

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

Answer 1

Concept:

Formula for drawing stress is given as

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

Calculation:

Here the formula for drawing stress is already given we just need to put the values to get the answer.

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

Given (σ_γ)_avg = 350 MPa

μ = 0.1, α = 5°, d_f = 7.5 mm, d_i = 10 mm

\({\sigma _f} = 350\left\{ {1 + \frac{1}{{0.1 \times \cot 5}}} \right\}\left[ {1 - {{\left( {\frac{{7.5}}{{10}}} \right)}^{2 \times 0.1 \times \cot 5}}} \right] = 316.24\;MPa\)

Points to remember: Put the values correctly and while using virtual calculator try to solve bracket wise to avoid unnecessary calculation.