Chapter 9, Problem 9.22P

(a)

To determine

The maximum possible elastic modulus in the polymer that will result in the maximum allowable clamping force.

Answer to Problem 9.22P

The maximum possible elastic modulus in the polymer that will result in the maximum allowable clamping force is $2\text{GPa}$ .

Explanation of Solution

Given:

The length of the polymer is $0.05\text{m}$ .

The change in length of the polymer is $0.0001\text{m}$ .

The maximum clamping force is $1\times {10}^4\text{N}$ .

Formula Used:

Write the expression for the strain in the polymer as:

  $\epsilon =\frac{\Delta l}{l}$ …… (I)

Here, $\epsilon$ is the elastic strain in the material, $\Delta l$ is the change in length of the polymer and $l$ is the length of the polymer.

Write the expression for the stress acting on the polymer as:

  $\sigma =\frac{F}{A}$ …… (II)

Here, $F$ is the clamping force and $A$ is the cross sectional area.

Write the expression for the elastic modulus in the polymer as:

  $E=\frac{\sigma }{\epsilon }$ …… (III)

Here, $\sigma$ is the stress acting on the polymer.

Calculation:

Substitute $0.05\text{m}$ for $l$ and $0.0001\text{m}$ for $\Delta l$ in equation (I)

  $\begin{array}{c}\epsilon =\frac{0.0001\text{m}}{0.05\text{m}}\\ =0.02\end{array}$

Substitute $1\times {10}^4\text{N}$ for $F$ and ${\left(0.05\text{m}\right)}^2$ for $A$ in equation (II)

  $\begin{array}{c}\sigma =\frac{1\times {10}^4\text{N}}{{\left(0.05\text{m}\right)}^2}\\ =4\times {10}^6\text{Pa}\\ \text{=4}\text{MPa}\end{array}$

Substitute $4\times {10}^6\text{Pa}$ for $\sigma$ and $0.02$ for $\epsilon$ in equation (III)

  $\begin{array}{c}E=\frac{4\times {10}^6\text{Pa}}{0.002}\\ =2\times {10}^9\text{Pa}\times \frac{{10}^{-9}\text{GPa}}{1\text{Pa}}\\ =\text{2}\text{GPa}\end{array}$

Conclusion:

Thus, the maximum possible elastic modulus in the polymer that will result in the maximum allowable clamping force is $2\text{GPa}$ .

(b)

To determine

The minimum allowable viscosity of the polymer.

Answer to Problem 9.22P

The minimum allowable viscosity of the polymer is $30\times {10}^{15}\text{Pa}\cdot \text{s}$ .

Explanation of Solution

Given:

The length of the polymer is $0.05\text{m}$ .

The change in length of the polymer is $0.0001\text{m}$ .

The maximum clamping force is $1\times {10}^4\text{N}$ .

Formula Used:

Write the expression for the relation of viscosity and decay of stress.

  $\frac{\sigma }{{\sigma }_0}=\exp \left(-\frac{E}{3\eta }t\right)$ …… (IV)

Here, $\sigma$ is the stress in the polymer after one year, ${\sigma }_0$ is the initial stress in the polymer, $\eta$ is the viscosity of the polymer and $t$ is the expected life of the polymer.

Calculation:

Substitute $4\times {10}^6\text{Pa}$ for ${\sigma }_0$ , $2\times {10}^6\text{Pa}$ for $\sigma$ , $2\times {10}^9\text{Pa}$ for $E$ and $365\text{days}$ for $t$ in equation (IV).

  $\begin{array}{c}\frac{2\times {10}^6\text{Pa}}{4\times {10}^6\text{Pa}}=\exp \left(-\frac{2\times {10}^9\text{Pa}}{3\eta }\times \left(365\text{days}\right)\times \left(\frac{24\text{h}}{1\text{day}}\right)\times \left(\frac{60\text{min}}{1\text{h}}\right)\times \left(\frac{60\text{s}}{1\min }\right)\right)\\ 0.5=\exp \left(\frac{-2.1\times {10}^{16}}{\eta }\right)\\ \ln \left(0.5\right)=\ln \left[\exp \left(\frac{-2.1\times {10}^{16}}{\eta }\right)\right]\\ \eta =30\times {10}^{15}\text{Pa}\cdot \text{s}\end{array}$

Conclusion:

Thus, the minimum allowable viscosity of the polymer is $30\times {10}^{15}\text{Pa}\cdot \text{s}$ .