Chapter 11.3, Problem 10P

Using E = 200 GPa, determine (a) the strain energy of the steel rod ABC when P = 25 kN, (b) the corresponding strain-energy density in portions AB and BC of the rod.

Fig. P11.10

To determine

The strain energy of the steel rod ABC.

Answer to Problem 10P

The strain energy of the steel rod ABC is $\underset{\_}{12.185\text{J}}$.

Explanation of Solution

Given information:

The diameter of the steel rod AB is $d_{AB}=20\text{mm}$.

The diameter of the steel rod BC is $d_{BC}=16\text{mm}$.

The length of the rod AB is $L_{AB}=1.2\text{m}$.

The length of the rod BC is $L_{BC}=0.8\text{m}$.

The modulus of elasticity of the steel is $E=200\text{GPa}$

The applied load $P=25\text{kN}$.

Calculation:

Calculate the area of the rod (A) as shown below.

$A=\frac{\pi d^2}{4}$ (1)

For the steel rod AB.

Substitute $20\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{AB}=\frac{\pi \times {20}^2}{4}\\ =314.16{\text{mm}}^2\end{array}$

For the steel rod BC.

Substitute $16\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{BC}=\frac{\pi \times {16}^2}{4}\\ =201.06{\text{mm}}^2\end{array}$

Calculate the strain energy (U) as shown below.

$U={\displaystyle \sum \frac{P^{2}L}{2EA}}$

Calculate the strain energy for rod ABC as shown below.

$\begin{array}{c}U=U_{AB}+U_{BC}\\ =\frac{P^2{L}_{AB}}{2E{A}_{AB}}+\frac{P^2{L}_{BC}}{2E{A}_{BC}}\\ =\frac{P^2}{2E}\left(\frac{L_{AB}}{A_{AB}}+\frac{L_{BC}}{A_{BC}}\right)\end{array}$

Substitute $25\text{kN}$ for P, $200\text{GPa}$ for E, $1.2\text{m}$ for $L_{AB}$, $314.16{\text{mm}}^2$ for $A_{AB}$, $0.8\text{m}$ for $L_{BC}$, and $201.06{\text{mm}}^2$ for $A_{BC}$.

$\begin{array}{c}U=\frac{{\left(25\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\right)}^2}{2\times 200\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}}\left(\begin{array}{l}\frac{1.2\text{m}}{314.16{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ +\frac{0.8\text{m}}{201.06{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\end{array}\right)\\ =\frac{625\times {10}^6}{400\times {10}^9}\left(3,819.7+3,978.9\right)\\ =1.5625\times {10}^{-3}\times 7,798.6\\ =12.185\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\end{array}$

$=12.185\text{J}$

Therefore, the strain energy for the steel rod ABC is $\underset{\_}{12.185\text{J}}$.

To determine

The strain energy density in rod AB and rod BC

Answer to Problem 10P

The strain energy density in rod AB is $\underset{\_}{u_{AB}=15.83\text{kJ}/{\text{m}}^3}$.

The strain energy density in rod BC is $\underset{\_}{u_{BC}=38.65\text{kJ}/{\text{m}}^3}$.

Explanation of Solution

Given information:

The diameter of the steel rod AB is $d_{AB}=20\text{mm}$

The diameter of the steel rod BC is $d_{BC}=16\text{mm}$

The length of the rod AB is $L_{AB}=1.2\text{m}$

The length of the rod BC is $L_{BC}=0.8\text{m}$

The modulus of elasticity of the steel is $E=200\text{GPa}$

The applied load $P=25\text{kN}$.

Calculation:

Refer to part (a).

The area of rod AB is $A_{AB}=314.16{\text{mm}}^2$.

The area of the rod BC is $A_{BC}=201.06{\text{mm}}^2$.

Calculate the stress $\left(\sigma \right)$ as shown below.

$\sigma =\frac{P}{A}$ (2)

For the rod AB.

Substitute $25\text{kN}$ for P and $314.16{\text{mm}}^2$ for A in Equation (2).

$\begin{array}{c}{\sigma }_{AB}=\frac{25\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{314.16{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ =79.58\times {10}^6\text{N}/{\text{mm}}^2\times \frac{1\text{Pa}}{{10}^6\text{N}/{\text{mm}}^2}\\ =79.58\text{Pa}\end{array}$

For the rod BC.

Substitute $25\text{kN}$ for P and $201.06{\text{mm}}^2$ for A in Equation (2).

$\begin{array}{c}{\sigma }_{BC}=\frac{25\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{201.06{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ =124.34\times {10}^6\text{N}/{\text{mm}}^2\times \frac{1\text{Pa}}{{10}^6\text{N}/{\text{mm}}^2}\\ =124.34\text{Pa}\end{array}$

Calculate the strain energy density (u) as shown below.

$u=\frac{{\sigma }^2}{2E}$ (3)

For the rod AB.

Substitute $79.58\text{Pa}$ for $\sigma$ and $200\text{GPa}$ for E in Equation (3).

$\begin{array}{c}{u}_{AB}=\frac{{\left(79.58\text{Pa}\right)}^2}{2\times 200\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}}\\ =\frac{6,332.9764}{400\times {10}^9}\\ =15.83\times {10}^{-9}\text{Pa}\times \frac{{10}^6\text{N}/{\text{mm}}^2}{1\text{Pa}}\\ =15.83\times {10}^{-3}\text{N}/{\text{mm}}^2\times {\left(\frac{1,000\text{mm}}{1\text{m}}\right)}^2\end{array}$

$\begin{array}{c}=15.83\times {10}^3\text{N}\cdot \text{m}/{\text{m}}^3\times \frac{1\text{kJ}}{{10}^3\text{N}\cdot \text{m}}\\ =15.83\text{kJ}\end{array}$

Hence, the strain energy density in rod AB is $\underset{\_}{u_{AB}=15.83\text{kJ}}$.

For the rod BC.

Substitute $124.34\text{Pa}$ for $\sigma$ and $200\text{GPa}$ for A in Equation (3).

$\begin{array}{c}{u}_{BC}=\frac{{\left(124.34\text{Pa}\right)}^2}{2\times 200\text{GPa}\times \frac{{10}^9\text{Pa}}{1\text{GPa}}}\\ =\frac{15,460.4356}{400\times {10}^9}\\ =38.65\times {10}^{-9}\text{Pa}\times \frac{{10}^6\text{N}/{\text{mm}}^2}{1\text{Pa}}\\ =38.65\times {10}^{-3}\text{N}/{\text{mm}}^2\times {\left(\frac{1,000\text{mm}}{1\text{m}}\right)}^2\end{array}$

$\begin{array}{c}=38.65\times {10}^3\text{N}\cdot \text{m}/{\text{m}}^3\times \frac{1\text{kJ}}{{10}^3\text{N}\cdot \text{m}}\\ =38.65\text{kJ}\end{array}$

Therefore, the strain energy density in rod BC is $\underset{\_}{u_{BC}=38.65\text{kJ}}$.