Chapter 8, Problem 8.54P

Interpretation Introduction

(a)

Interpretation:

The original diameter of 3105 aluminum wire needs to be determined.

Concept Introduction:

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

Answer to Problem 8.54P

The original diameter of the 3105 aluminum wire is 0.04792 inch

Explanation of Solution

Given Information:

  

Final diameter of 3105 aluminum wire = 1mm

Yield strength = 20,000psi

% cold work is given by

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

Here,

$\begin{array}{l}\text{\%cw = cold work percentage}\hfill \\ {\text{A}}_{\text{0}}\text{=Initial area}\hfill \\ {\text{A}}_{\text{F}}\text{=Final area}\hfill \\ {\text{d}}_{\text{0}}\text{=Initial diameter}\hfill \\ {\text{d}}_{\text{F}}\text{=Final diameter=1mm}\hfill \end{array}$

From the graph at yield strength = 20,000 Psi

% cw = 32.5 %

Equation (1) becomes,

$\begin{array}{l}32.5=[\frac{d_0{}^2-1^2}{d_0{}^2}]\\ 32.5d_0{}^2=100d_0{}^2-100\\ 67.5d_0{}^2=100\\ d_0=\sqrt{\frac{100}{67.5}}\\ d_0=1.217mm\\ d_0=0.04792in\end{array}$

Interpretation Introduction

(b)

Interpretation:

The drama force required for 3105 aluminum wire needs to be determined.

Concept Introduction:

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

Answer to Problem 8.54P

Drag force required when stress acting of 3105 aluminum wire is 14.431b

Explanation of Solution

Given Information:

  

$\begin{array}{l}\text{Initial diameter = 0}\text{.04792 inch}\\ \text{From graph cold work = 32}\text{.5 \%,}\text{σ=8000psi}\end{array}$

Since,

$\begin{array}{l}\sigma =\frac{F}{A}\\ \sigma =\frac{F}{\frac{\pi }{4}\times d^2}\end{array}$.........(1)

Here,

$\begin{array}{l}\text{F=Draw force}\hfill \\ \text{σ=Tensile stress=8000Psi}\hfill \\ \text{d=Initial diameter=0}\text{.04792inch}\hfill \end{array}$

Equation (1) becomes,

$\begin{array}{l}F=\sigma \times \frac{\pi }{4}\times d^2\\ =8000\times \frac{\pi }{4}\times {0.04792}^2\\ F=14.43lb\end{array}$

Drag force = 14.43lb

Interpretation Introduction

(c)

Interpretation:

The actual stress needs to be determined. Whether the 3105 aluminum wire will break or not needs to be determined.

Concept Introduction:

Stress is defined as the force acting on the material per unit area of the material stress used to determine the load carrying capacity of the material.

Answer to Problem 8.54P

Actual stress = 10853 Psi which is less than yield strength 20,000 Psi of drawn wire. Hence wire will not break.

Explanation of Solution

Given Information:

  

Drag force = 14.43lb

Final diameter = 1mm

Stress acting on the wire has a final diameter of 1 mm subjected to drag force is given by,

$\begin{array}{l}\sigma =\frac{F}{A}\\ \sigma =\frac{F}{\frac{\pi }{4}\times d^2}\\ =\frac{14.43}{\frac{\pi }{4}\times \frac{1}{2.54\times 10}}\end{array}$

$(\because 1mm=25.4in)$

$\sigma =10853Psi<20,000\text{Psi}\text{of}\text{yield}\text{strength}$

Thus, the wire will not break.