Chapter 1, Problem 1.5.1P

To determine

(a)

The ultimate tensile stress of metal specimen.

Answer to Problem 1.5.1P

$120\text{ksi}$.

Explanation of Solution

Given:

The diameter of metal specimen is $0.550\text{inch}$.

The load at facture is $28,500\text{pounds}$.

Concept Used:

Write the equation to calculate the ultimate tensile stress.

$f=\frac{P}{A}$ ...... (I)

Here, ultimate tensile stress is $f$, fractured load is $P,$ and the cross-sectional area is $A$.

Calculation:

Calculate the cross-sectional area of specimen.

$A=\frac{\text{π}}{4}\times d^2$ ...... (II)

Here, diameter of the specimen is $d$.

Substitute $0.550\text{inch}$ for $d$ in the Equation (II).

$\begin{array}{c}A=\frac{\text{π}}{4}\times {\left(0.550\text{inch}\right)}^2\\ =\frac{0.95033{\text{inch}}^2}{4}\\ =0.2375{\text{inch}}^2\end{array}$

Calculate the ultimate tensile stress.

Substitute $28,500\text{pounds}$ for $P$ and $0.2375{\text{inch}}^2$ for $A$ in the Equation (I).

$\begin{array}{c}f=\frac{28,500\text{pounds}}{0.2375{\text{inch}}^2}\times \left(\frac{1\text{klb}}{1000\text{pounds}}\right)\\ =\frac{28.5\text{klb}}{0.2375{\text{inch}}^2}\\ =120\text{klb}/{\text{inch}}^2\times \left(\frac{1\text{ksi}}{\text{1}\text{klb}/{\text{inch}}^2}\right)\\ =120\text{ksi}\end{array}$

Conclusion:

Thus, the ultimate tensile stress on the metal specimen is $120\text{ksi}$.

To determine

(b)

The elongation of the metal specimen.

Answer to Problem 1.5.1P

$13.3\%$.

Explanation of Solution

Given:

The original gage length is $2.03\text{inches}$.

The change in gage length is $2.3\text{inches}$.

Concept Used:

Write the equation to calculate the elongation.

$e=\frac{L_{\text{f}}-L_{\text{0}}}{L_{\text{0}}}\times 100$ ...... (III)

Here, the elongation is $e$, the length of the specimen at fracture is $L_{\text{f}},$ and the original length is $L_{\text{0}}$.

Calculation:

Calculate the elongation of the metal specimen.

Substitute $2.03\text{inches}$ for $L_{\text{0}}$ and $2.3\text{inches}$ for $L_{\text{f}}$ in Equation (III).

$\begin{array}{c}e=\left[\frac{\left(2.3\text{inches}\right)-\left(2.03\text{inches}\right)}{\left(2.03\text{inches}\right)}\right]\times 100\\ =0.1330\times 100\\ =13.3\%\end{array}$

Conclusion:

Thus, the elongation of the metal specimen is $13.3\%$.

To determine

(c)

The reduction in the cross-sectional area of the metal specimen.

Answer to Problem 1.5.1P

$38.86\%$.

Explanation of Solution

Given:

The original diameter of metal specimen is $0.550\text{inch}$.

The diameter after fracture load is $0.430\text{inch}$.

Concept Used:

Write the equation to calculate reduction in cross-sectional area.

$R=\frac{A_0-A_{\text{f}}}{A_0}\times 100$ ...... (IV)

Here, the reduction in cross-sectional area is $R$, original cross-sectional area is $A_0,$ and the cross-sectional area after fracture load is $A_{\text{f}}$.

Calculation:

Calculate the cross-sectional area after fracture load.

Substitute $0.430\text{inch}$ for $d$ in the Equation (II).

$\begin{array}{c}A=\frac{\text{π}}{4}\times {\left(0.430\text{inch}\right)}^2\\ =\frac{0.5808{\text{inch}}^2}{4}\\ =0.1452{\text{inch}}^2\end{array}$

Calculate the reduction in the cross-sectional area.

Substitute $0.1452{\text{inch}}^2$ for $A_{\text{f}}$ and $0.2375{\text{inch}}^2$ for $A_0$ in Equation (IV).

$\begin{array}{c}R=\frac{0.2375{\text{inch}}^2-0.1452{\text{inch}}^2}{0.2375{\text{inch}}^2}\times 100\\ =\frac{0.0923{\text{inch}}^2}{0.2375{\text{inch}}^2}\times 100\\ =0.3886\times 100\\ =38.86\%\end{array}$

Conclusion:

Thus, the reduction in the cross-sectional area is $38.86\%$.