Chapter 8, Problem 8.53P

Interpretation Introduction

(a)

Interpretation:

The draw force needs to be calculated considering no friction of copper wire.

Concept Introduction:

Cold working is the mechanical process in which material is plastically deformed under recrystallization temperature.

Answer to Problem 8.53P

The draw force required for original copper wire is 6096.65 N.

Explanation of Solution

Given Information:

1 inch = 2.54 cm

$\begin{array}{l}\begin{array}{l}\text{Initial diameter of bar=0}\text{.75cm}\hfill \\ \text{Final diameter of bar=0}\text{.625cm}\hfill \end{array}\\ \text{Yield strength=20,000Psi}\end{array}$

% cold work is given by the following relation:

$\begin{array}{l}\%cw=[\frac{A_0-A_F}{A_0}]\times 100\\ \%cw=[\frac{\frac{\pi }{4}{d}_0{}^2-\frac{\pi }{4}{d}_F{}^2}{\frac{\pi }{4}{d}_0{}^2}]\\ \%cw=[\frac{d_0{}^2-d_F{}^2}{d_0{}^2}]\end{array}$ ---------------------- (1)

Here,

$\begin{array}{l}{A}_0=Initialofbar\hfill \\ A_F=Finalareaofbar\hfill \\ d_0=Initial areaofbar=0.75cm\hfill \\ d_F=Finalareaofbar=0.625cm\hfill \end{array}$

Equation (1) becomes,

$\begin{array}{l}\%cw=[\frac{{(0.75)}^2-{(0.625)}^2}{{(0.75)}^2}]\\ \%cw=(\frac{0.1719}{0.5625})\times 100\\ \%cw=30.56\%\end{array}$

Thus at 30.56 % cold work from the graph yield strength of copper = 138 MPa.

Now, drag force (F) is calculated from the following formula,

$F=ys(\frac{\pi }{4}{d}_0{}^2)$ -------------------- (2)

Here,

$\begin{array}{l}{Y}_s=\text{Yield strength of copper=138MPa}\\ d_0=\text{Initial diameter=0}\text{.75cm}\\ \text{=7}\text{.5mm}\end{array}$

Equation (2) becomes,

$F=138\times \frac{\pi }{4}\times {(7.5)}^2$

$F=6096.6\text{5 N}$

Interpretation Introduction

(b)

Interpretation:

Whether the drawn wire break in the drawing wire or not needs to be determined.

Concept Introduction:

Cold working is the mechanical process in which matrial is plastically deformed under recrystallization temperature.

Answer to Problem 8.53P

The stress after the wire is drawn under cold work is 198.71MPa which is less than actual yield strength 370MP. Hence wire will not break.

Explanation of Solution

Given Information:

1 inch = 2.54 cm

$\begin{array}{l}\begin{array}{l}\text{Initial diameter of bar=0}\text{.75cm}\hfill \\ \text{Final diameter of bar=0}\text{.625cm}\hfill \end{array}\\ \text{Yield strength=20,000Psi}\end{array}$

Thus drag force F of wire of copper is 6096.65 N

Final diameter = 0.625 cm

The stress (s) on the wire after cold work is given by the following relation:

$\begin{array}{l}s=\frac{F}{\frac{\pi }{4}{d}_F{}^2}\\ s=\frac{6096.6}{\frac{\pi }{4}\times {(0.625\times 10)}^2}\end{array}$

$(\because 1\text{0mm=1cm)}$

$s=198.71MPa<370MPa$

Yield strength material will not break.