Chapter 7, Problem 7.6.8P

(a)

To determine

The bulk modulus for the nylon.

Answer to Problem 7.6.8P

The bulk modulus of the nylon is $4.95\text{GPa}$ .

Explanation of Solution

Given information:

The normal stress of the nylon element along the $x$ axis is $-3.9\text{MPa}$ , normal stress along the $y$ axis is $-3.2\text{MPa}$ , stress along the $z$ axis is $-1.8\text{MPa}$ , the strain in the $x$ direction is $-640\times {10}^{-6}$ and the strain in the $y$ direction is $-310\times {10}^{-6}$ .

Explanation:

Write the expression for the bulk modulus.

  $K=\frac{{\sigma }_h}{\frac{\Delta V}{V}}$ ...... (I)

Here, the bulk modulus is $K$ , the hydrostatic stress is ${\sigma }_h$ , the change in volume is $\Delta V$ and the original volume of the element is $V$ .

Write the expression for the change in volume in terms of strains.

  $\frac{\Delta V}{V}={\epsilon }_x+{\epsilon }_y+{\epsilon }_z$ ...... (II)

Here, the strain in the $x$ direction is ${\epsilon }_x$ , strain in the $y$ direction is ${\epsilon }_y$ and the strain in the $z$ direction is ${\epsilon }_z$ .

Write the expression for the hydrostatic stress in terms of stresses.

  ${\sigma }_h=\frac{{\sigma }_x+{\sigma }_y+{\sigma }_z}{3}$ ...... (III)

Here,the stress along $x$ axis is ${\sigma }_x$ , stress along $y$ axis is ${\sigma }_y$ and stress along $z$ axis is ${\sigma }_z$ .

Write the expression for strain in the $x$ direction using Hooke’s law.

  ${\epsilon }_x=\frac{1}{E}\left({\sigma }_x-\nu \left({\sigma }_y+{\sigma }_z\right)\right)$ ...... (IV)

Here, the modulus of elasticity is $E$ and the Poisson’s ratio is $\nu$ .

Write the expression for strain in the $y$ direction using Hooke’s law.

  ${\epsilon }_y=\frac{1}{E}\left({\sigma }_y-\nu \left({\sigma }_x+{\sigma }_z\right)\right)$ ...... (V)

Write the expression for strain in the $z$ direction using Hooke’s law.

  ${\epsilon }_z=\frac{1}{E}\left({\sigma }_z-\nu \left({\sigma }_x+{\sigma }_y\right)\right)$ ...... (VI)

Calculation:

Substitute $-640\times {10}^{-6}$ for ${\epsilon }_x$ , $-3.9\text{MPa}$ for ${\sigma }_x$ , $-1.8\text{MPa}$ for ${\sigma }_z$ and $-3.2\text{MPa}$ for ${\sigma }_y$ in Equation (IV).

  $\begin{array}{l}-640\times {10}^{-6}=\frac{1}{E}\left[-3.9\text{MPa}-\nu \left(-3.2\text{MPa}-1.8\text{MPa}\right)\right]\\ -640\times {10}^{-6}\times E=\left(\left(-3.9\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}\right)+\left(5\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\times \nu \right)\\ -640\times {10}^{-6}E-5\text{Pa}\times \nu \times {10}^6=-3.9\text{Pa}\times {10}^6\end{array}$ ...... (VII)

Substitute $-310\times {10}^{-6}$ for ${\epsilon }_y$ , $-3.9\text{MPa}$ for ${\sigma }_x$ , $-1.8\text{MPa}$ for ${\sigma }_z$ and $-3.2\text{MPa}$ for ${\sigma }_y$ in Equation (V).

  $\begin{array}{l}-310\times {10}^{-6}=\frac{1}{E}\left[-3.2\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}-\nu \left(\begin{array}{l}-3.9\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ -1.8\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\end{array}\right)\right]\\ \left(-310\times {10}^{-6}\right)E=\left(-3.2\text{Pa}+5.7\text{Pa}\times \nu \right)\times {10}^6\\ -310\times {10}^{-6}\times E-5.7\text{Pa}\times \nu \times {10}^6=-3.2\text{Pa}\times {10}^6\end{array}$ ...... (VIII)

Solve the Equations (VII) and (VIII) for the value of $E$ and $\nu$ .

  $\begin{array}{l}2.098\times {10}^{-3}E=6.23\text{Pa}\times {10}^6\\ E=\frac{6.23\text{Pa}\times {10}^6}{2.098\times {10}^{-3}}\\ E=2.9695\times {10}^9\text{Pa}\end{array}$

Substitute $2.9695\times {10}^9\text{Pa}$ for $E$ in Equation (VII).

  $\begin{array}{l}-640\times \left(2.9695\times {10}^9\right)\text{Pa}\times {10}^{-6}-5\text{Pa}\times \nu \times {10}^6=\left(-3.9\times {10}^6\right)\text{Pa}\\ 5\text{Pa}\times \nu \times {10}^6=1999520\text{Pa}\\ \nu =0.3999\end{array}$

Substitute $2.9695\times {10}^9\text{Pa}$ for $E$ , $0.3999$ for $\nu$ , $-3.9\text{MPa}$ for ${\sigma }_x$ , $-1.8\text{MPa}$ for ${\sigma }_z$ and $-3.2\text{MPa}$ for ${\sigma }_y$ in Equation (VI).

  $\begin{array}{c}{\epsilon }_z=\frac{1}{2.9695\times {10}^9\text{Pa}}\left[\left(-1.8\text{Pa}-0.3999\left(-3.9\text{Pa}-3.2\text{Pa}\right)\right)\times {10}^6\right]\\ =\frac{1.03929\times {10}^6\text{Pa}}{2.9695\times {10}^9\text{Pa}}\\ =350\times {10}^{-6}\end{array}$

Substitute $-3.9\text{MPa}$ for ${\sigma }_x$ , $-1.8\text{MPa}$ for ${\sigma }_z$ and $-3.2\text{MPa}$ for ${\sigma }_y$ in Equation (III).

  $\begin{array}{c}{\sigma }_h=\frac{\left(-3.9\text{MPa}\right)+\left(-3.2\text{MPa}\right)+\left(-1.8\text{MPa}\right)}{3}\\ =\frac{-8.9\text{MPa}}{3}\\ =-2.9667\text{MPa}\end{array}$

Substitute $-640\times {10}^{-6}$ for ${\epsilon }_x$ , $-310\times {10}^{-6}$ for ${\epsilon }_y$ and $350\times {10}^{-6}$ for ${\epsilon }_z$ in Equation (II).

  $\begin{array}{c}\frac{\Delta V}{V}=\left(-640\times {10}^{-6}\right)+\left(-310\times {10}^{-6}\right)+\left(350\times {10}^{-6}\right)\\ =\left(-640-310+350\right)\times {10}^{-6}\\ =-600\times {10}^{-6}\end{array}$

Substitute $-2.9667\text{MPa}$ for ${\sigma }_h$ and $-600\times {10}^{-6}$ for $\frac{\Delta V}{V}$ in Equation (I).

  $\begin{array}{c}K=\frac{-2.967\times {10}^6\text{Pa}}{-600\times {10}^{-6}}\\ =\left(4.95\times {10}^9\text{Pa}\right)\times \left(\frac{1\text{GPa}}{{10}^9\text{Pa}}\right)\\ =\text{4}\text{.95}\text{GPa}\end{array}$

Conclusion:

The bulk modulus of the nylon is $4.95\text{GPa}$ .

(b)

To determine

The modulus of elasticity.

The Poisson’s ratio.

Answer to Problem 7.6.8P

The modulus of elasticity is $1.297\text{GPa}$ .

The Poisson’s ratio is $0.399$ .

Explanation of Solution

Given information:

The bulk modulus of the element is $2162\text{MPa}$ , the normal stress along the $x$ axis is $-3.6\text{MPa}$ , normal stress along the $y$ axis is $-2.1\text{MPa}$ , normal stress along the $z$ axis is $-2.1\text{MPa}$ and normal strain in the $x$ direction is $-1480\times {10}^{-6}$ .

Explanation:

Write the expression for the bulk modulus.

  $\frac{{\sigma }_h}{K}=\frac{1-2\nu }{E}\left({\sigma }_x+{\sigma }_y+{\sigma }_z\right)$ ...... (IX)

Calculation:

Substitute $-3.6\text{MPa}$ for ${\sigma }_x$ , $-2.1\text{MPa}$ for ${\sigma }_y$ and $-2.1\text{MPa}$ for ${\sigma }_z$ in Equation (III).

  $\begin{array}{c}{\sigma }_h=\frac{\left(-3.6\text{MPa}\right)+\left(-2.1\text{MPa}\right)+\left(-2.1\text{MPa}\right)}{3}\\ =\frac{-7.8\text{MPa}}{3}\\ =-2.6\text{MPa}\end{array}$

Substitute $-3.6\text{MPa}$ for ${\sigma }_x$ , $-2.1\text{MPa}$ for ${\sigma }_y$ , $-2.1\text{MPa}$ for ${\sigma }_z$ and $2162\text{MPa}$ for $K$ and $-2.6\text{MPa}$ for ${\sigma }_h$ in Equation (IX).

  $\begin{array}{l}\frac{-2.6\text{MPa}}{2162\text{MPa}}=\frac{1-2\nu }{E}\left(-3.6\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-2.1\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-2.1\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right)\\ \frac{-2.6}{2162}=\frac{1-2\nu }{E}\left(-3.6\text{Pa}-2.1\text{Pa}-2.1\text{Pa}\right)\times {10}^6\\ -2.6E=2162\times \left(1-2\nu \right)\left(-7.8\text{Pa}\times {10}^6\right)\end{array}$ ...... (X)

Substitute $-3.6\text{MPa}$ for ${\sigma }_x$ , $-2.1\text{MPa}$ for ${\sigma }_y$ , $-2.1\text{MPa}$ for ${\sigma }_z$ and $-1480\times {10}^{-6}$ for ${\epsilon }_x$ in Equation (IV).

  $\begin{array}{l}-1480\times {10}^{-6}=\frac{1}{E}\left(-3.6\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-\nu \left(\begin{array}{l}-2.1\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ -2.1\text{MPa}\times \left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\end{array}\right)\right)\\ -1480\times {10}^{-6}=\frac{1}{E}\left(-3.6\text{Pa}-\nu \left(-4.2\text{Pa}\right)\right)\times {10}^{-6}\\ -1480\times {10}^{-6}\times E=\left(-3.6\text{Pa}+4.2\nu \text{Pa}\right)\times {10}^6\end{array}$ ...... (XI)

Solve the Equations (X) and (XI) for the value of $E$ and $\nu$ .

  $\begin{array}{l}\frac{2.6\times {10}^6}{3199760\times {10}^{-6}}=\frac{-7.8\times {10}^6+15.6\times {10}^6\nu }{-3.6+4.2\nu }\\ -38996256\nu =-15598128\\ \nu =0.399\end{array}$

Substitute $0.399$ for $\nu$ in Equation (X).

  $\begin{array}{c}E=\frac{-3.6\text{Pa}\times {10}^6+0.4\times 4.2\text{Pa}\times {10}^6}{-1480\times {10}^{-6}}\\ =\left(1.297\times {10}^9\text{Pa}\right)\times \left(\frac{1\text{GPa}}{{10}^9\text{Pa}}\right)\\ =1.297\text{GPa}\end{array}$

Conclusion:

The modulus of elasticity is $1.297\text{GPa}$ .

The Poisson’s ratio is $0.399$ .