Chapter 2, Problem 2.81P

(a)

To determine

The work done in expanding a $2\text{ mm}$ thick spherical shell.

Answer to Problem 2.81P

The work done in expanding a $2\text{ mm}$ thick spherical shell is $106.3\text{}\text{kN}\cdot \text{m}$ .

Explanation of Solution

Given:

Thickness of shell is $2\text{ mm}$ .

Initial diameter is $100\text{ mm}$ .

Final diameter is $150\text{mm}$ .

The stress for material is $200+50{\epsilon }^{0.5}\text{ MPa}$ .

Concept used:

Write the expression for true strain for shell.

  $\epsilon =\ln \left(\frac{d_f}{d_0}\right)$   ............. (1)

Here, $\epsilon$ is the true strain for shell, $d_f$ is the final diameter of shell and $d_0$ is the original diameter of shell.

Write the expression for specific energy for balanced biaxial plane stress condition.

  $u=2{\sigma }_1\epsilon$

Substitute $\ln \left(\frac{d_f}{d_0}\right)$ for $\epsilon$ in above expression.

  $u=2{\sigma }_1\ln \left(\frac{d_f}{d_0}\right)$

Here, $u$ is the specific energy and ${\sigma }_1$ is the stress in axial direction.

Write the expression for work done.

  $W=Vu$

Here, $W$ is the work done and $V$ is the volume of shell.

Substitute $2{\sigma }_1\ln \left(\frac{d_f}{d_0}\right)$ for $u$ and $4\pi r_0^2{t}_0$ for $V$ in above expression.

  $W=\left(4\pi r_0^2{t}_0\right)\left(2{\sigma }_1\ln \left(\frac{d_f}{d_0}\right)\right)$

Simplify above expression.

  $W=8\pi r_0^2{t}_0{\sigma }_1\ln \left(\frac{d_f}{d_0}\right)$   ............. (2)

Write the expression for stress for material.

  $\sigma =200+50{\epsilon }^{0.5}$

The true strain is obtained by integrating the above expression.

  ${\displaystyle \int \sigma }={\displaystyle \int 200+50{\epsilon }^{0.5}}d\epsilon$

After integration simplify above expression.

  ${\sigma }_1=200+\frac{50{\epsilon }^{1.5}}{1.5}$  ............. (3)

Calculation:

Substitute $100\text{mm}$ for $d_0$ and $150\text{}\text{mm}$ for $d_f$ in equation (1).

  $\begin{array}{c}\epsilon =\ln \left(\frac{150}{100}\right)\\ =0.4055\end{array}$

Substitute $0.4055$ for $\epsilon$ in equation (3).

  $\begin{array}{c}{\sigma }_1=200+\frac{50{\left(0.4055\right)}^{1.5}}{1.5}\\ =208.6\text{MPa}\end{array}$

Substitute $0.4055$ for $\epsilon$ , $50\text{ mm}$ for $r_0$ , $2\text{ mm}$ for $t_0$ and $208.6\text{MPa}$ for ${\sigma }_1$ in equation (2).

  $\begin{array}{c}W=8\pi {\left(50\text{ mm}\left(\frac{1\text{}\text{m}}{1000\text{}\text{mm}}\right)\right)}^2\left(2\text{mm}\left(\frac{1\text{}\text{m}}{1000\text{}\text{mm}}\right)\right)\left(208.6\text{}\text{MPa}\left(\frac{{10}^6\text{ Pa}}{1\text{ MPa}}\right)\right)\left(0.4055\right)\\ =10629.55\text{N}\cdot \text{m}\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =106.3\text{}\text{kN}\cdot \text{m}\end{array}$

Conclusion:

Thus, the work done in expanding a $2\text{ mm}$ thick spherical shell is $106.3\text{}\text{kN}\cdot \text{m}$ .

(b)

To determine

Explain the dependency of answer on yield criterion.

Explanation of Solution

In this case, the amount of tension applied on biaxial stress plane is equal in magnitude and in biaxial plane with equal stresses the yield criterion for both maximum shear stress theory and distortion energy gives same result.

Thus, the answer is not dependent on any yield criterion.