Chapter 9, Problem 9.20P

(a)

To determine

The strain in the tube at time $t=0\text{ s}$ .

Answer to Problem 9.20P

The strain in the tube at time $t=0\text{ s}$ is $0.25\times {10}^{-3}$ .

Explanation of Solution

Given:

Temperature of soda-lime glass is $395°\text{C}$ .

Tensile stress is $10\text{}\text{MPa}$ .

Elastic modulus of glass is $40\text{GPa}$ .

Concept used:

Write the expression for elastic strain rate in material at $t=0\text{ s}$ .

  ${\epsilon }_e=\frac{\sigma }{E}$ …… (1)

Here, ${\epsilon }_e$ is the elastic strain rate in material at $t=0\text{ s}$ , $\sigma$ is the stress in material and $E$ is the Elastic modulus of glass.

Calculation:

Substitute $10\text{MPa}$ for $\sigma$ and $40\text{}\text{GPa}$ for $E$ in equation (1).

  $\begin{array}{c}{\epsilon }_e=\frac{10\text{MPa}}{40\text{}\text{GPa}\left(\frac{{10}^3\text{MPa}}{1\text{ GPa}}\right)}\\ =0.25\times {10}^{-3}\end{array}$

Conclusion:

Thus, the strain in the tube at time $t=0\text{ s}$ is $0.25\times {10}^{-3}$ .

(b)

To determine

Total strain in the tube after one year.

Answer to Problem 9.20P

Total strain in the tube after one year is $1.05025$ .

Explanation of Solution

Given:

Value of constant $n$ is $1$ .

Concept used:

Refer to Figure 9.3 “The temperature dependence of the viscosity of silica glass $\left({\text{SiO}}_{\text{2}}\right)$ soda-lime glass, and boron oxide $\left({\text{B}}_{\text{2}}{\text{O}}_{\text{3}}\right)$ ” to obtain the viscosity of soda lime glass at $668\text{ K}$ as ${10}^{14}\text{ Pa}\cdot \text{s}$ .

Write the expression for plastic strain in soda-lime glass.

  $\frac{d{\epsilon }_p}{dt}=\frac{\sigma }{3\eta }$ …… (2)

Here, $\frac{d{\epsilon }_p}{dt}$ is the plastic strain in soda-lime glass and $\eta$ is the viscosity of material.

Write the expression for total plastic strain in material.

  ${\epsilon }_p=\frac{d{\epsilon }_p}{dt}\left(t\right)$ …… (3)

Here, ${\epsilon }_p$ is the total plastic strain in material and $t$ is the time.

Write the expression for total strain.

  ${\epsilon }_t={\epsilon }_e+{\epsilon }_p$ …… (4)

Here, ${\epsilon }_t$ is the total strain.

Calculation:

Substitute ${10}^{14}\text{ Pa}\cdot \text{s}$ for $\eta$ and $10\text{ MPa}$ for $\sigma$ in equation (2).

  $\begin{array}{c}\frac{d{\epsilon }_p}{dt}=\frac{10\text{ MPa}\left(\frac{{10}^6\text{}\text{Pa}}{1\text{}\text{MPa}}\right)}{3\left({10}^{14}\text{ Pa}\cdot \text{s}\right)}\\ =3.33\times {10}^{-8}{\text{s}}^{-1}\end{array}$

Substitute $3.33\times {10}^{-8}{\text{s}}^{-1}$ for $\frac{d{\epsilon }_p}{dt}$ and $1\text{ year}$ for $t$ in equation (3).

  $\begin{array}{c}{\epsilon }_p=\left(3.33\times {10}^{-8}{\text{s}}^{-1}\right)\left(1\text{ year}\right)\left(\frac{365\text{ days}}{1\text{ year}}\right)\left(\frac{24\text{h}}{1\text{ day}}\right)\left(\frac{3600\text{ s}}{1\text{ h}}\right)\\ =1.05\end{array}$

Substitute $1.05$ for ${\epsilon }_p$ and $0.25\times {10}^{-3}$ for ${\epsilon }_e$ in equation (4).

  $\begin{array}{c}{\epsilon }_t=0.25\times {10}^{-3}+1.05\\ =1.05025\end{array}$

Conclusion:

Thus, the total strain in the tube after one year is $1.05025$ .