Chapter 6, Problem 6.2P

(a)

To determine

The change in radius of specimen.

Answer to Problem 6.2P

The change in radius of specimen is not significant and the final radius of specimen is $0.005\text{ m}$ .

Explanation of Solution

Given:

Poisson’s ratio is $0.29$ .

Radius of specimen is $0.005\text{ m}$ .

Elongation in steel rod is $0.616\times {10}^{-2}\text{ m}$ .

Write the expression for tensile stress in rod.

  $\sigma =\frac{F}{A_0}$

Here, $\sigma$ is the tensile stress in rod, $F$ is the applied force and $A_0$ is the original cross-sectional area of specimen.

Substitute $\pi r_0^2$ for $A_0$ in above expression.

  $\sigma =\frac{F}{\pi r_0^2}$ .......... (1)

Here, $r_0$ is the original radius of specimen.

Write the expression for axial strain in steel rod.

  ${\epsilon }_x=\frac{\Delta l}{l_0}$ .......... (2)

Here, ${\epsilon }_x$ is the axial strain, $\Delta l$ is the elongation in rod and $l_0$ is the original length of rod.

Write the expression for axial strain in $y$ direction.

  ${\epsilon }_y=-v{\epsilon }_x$ .......... (3)

Here, ${\epsilon }_y$ is the axial strain in $y$ direction and $v$ is Poisson’s ratio.

Write the expression for final radius of specimen.

  $r=r_0+{\epsilon }_y{r}_0$ .......... (4)

Here, $r$ is the final radius of steel rod.

Substitute $0.005\text{ m}$ for $r_0$ and $0.250\times {10}^5\text{}\text{N}$ for $F$ in equation (1).

  $\begin{array}{c}\sigma =\frac{0.250\times {10}^5}{\pi {\left(0.005\right)}^2}\\ =3.183\times {10}^8{\text{N/m}}^2\end{array}$

Substitute $0.616\times {10}^{-2}\text{}\text{m}$ for $\Delta l$ and $4.000\text{m}$ for $l_0$ in equation (2).

  $\begin{array}{c}{\epsilon }_x=\frac{0.616\times {10}^{-2}\text{}}{4.000}\\ =1.54\times {10}^{-3}\end{array}$

Substitute $0.29$ for $v$ and $1.54\times {10}^{-3}$ for ${\epsilon }_x$ in equation (3).

  $\begin{array}{c}{\epsilon }_y=-\left(0.29\right)\left(1.54\times {10}^{-3}\right)\\ =-0.045\times {10}^{-2}\end{array}$

Substitute $-0.045\times {10}^{-2}$ for ${\epsilon }_y$ and $0.005\text{m}$ for $r_0$ in equation (4).

  $\begin{array}{c}r=0.005+\left(-0.045\times {10}^{-2}\right)\left(0.005\right)\\ =0.005-2.25\times {10}^{-6}\\ =0.005-0.00000225\\ \approx 0.005\end{array}$

Thus, the change in radius of specimen is not significant and the final radius of specimen is $0.005\text{ m}$ .

(b)

To determine

The difference between true stress and engineering stress.

Answer to Problem 6.2P

There is no difference between true stress and engineering stress for this specimen.

Explanation of Solution

As the change in radius of specimen is not significant, the actual and original radius of specimen can be considered as same. The true stress is evaluated on the basis of actual cross-sectional area of specimen and engineering stress is calculated on original cross-sectional area of specimen. As the value of area is constant for same final radius there will be no difference in Engineering stress and true stress.

The calculate engineering stress or true stress of specimen is $3.183\times {10}^8{\text{N/m}}^2$ .

Thus, there is no difference between true stress and engineering stress for this specimen.

(c)

To determine

The value of true strain and compare it with engineering strain.

Answer to Problem 6.2P

The true strain for the specimen is $0.154\times {10}^{-2}$ .

Explanation of Solution

Write the expression for true strain of specimen.

  ${\epsilon }_t=\ln \left(1+\epsilon \right)$ .......... (5)

Here, ${\epsilon }_t$ is true strain.

Substitute $1.54\times {10}^{-3}$ for $\epsilon$ in equation (5).

  $\begin{array}{c}{\epsilon }_t=\ln \left[1+\left(1.54\times {10}^{-3}\right)\right]\\ =0.154\times {10}^{-2}\end{array}$

The engineering strain and true strain will remain same for the specimen.

Thus, the true strain for the specimen is $0.154\times {10}^{-2}$ .