Chapter 11, Problem 123RP

Rod AB is made of a steel for which the yield strength is σ_Y = 450 MPa and E = 200 GPa; rod BC is made of an aluminum alloy for which σ_Y = 280 MPa and E = 73 GPa. Determine the maximum strain energy that can be acquired by the composite rod ABC without causing any permanent deformations.

Fig. P11.123

To determine

Find the maximum strain energy that can be acquired by the composite rod ABC.

Answer to Problem 123RP

The maximum strain energy of the composite rod ABC is $\underset{\_}{136.6\text{J}}$.

Explanation of Solution

Given information:

The diameter of the composite rod AB is $d_{AB}=10\text{mm}$.

The diameter of the composite rod BC is $d_{BC}=14\text{mm}$.

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

The length of the rod BC is $L_{BC}=1.6\text{m}$.

The yield strength of the steel rod AB is ${\sigma }_Y=450\text{MPa}$.

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

The yield strength of the aluminum alloy BC is ${\sigma }_Y=280\text{MPa}$.

The modulus of elasticity of the aluminum alloy is $E=73\text{GPa}$

Calculation:

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

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

For the steel rod AB.

Substitute $10\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{AB}=\frac{\pi \times {10}^2}{4}\\ =78.54{\text{mm}}^2\end{array}$

For the aluminum alloy BC.

Substitute $14\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{BC}=\frac{\pi \times {14}^2}{4}\\ =153.94{\text{mm}}^2\end{array}$

Calculate the applied load $\left(P\right)$ as shown below.

$P={\sigma }_{Y}A$ (2)

For the steel rod AB.

Substitute $450\text{MPa}$ for ${\sigma }_Y$ and $78.54{\text{mm}}^2$ for A in Equation (2).

$\begin{array}{c}P=450\text{MPa}\times \frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\times 78.54{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =35.343\times {10}^3\text{N}\end{array}$

For the aluminum alloy BC.

Substitute $280\text{MPa}$ for ${\sigma }_Y$ and $153.94{\text{mm}}^2$ for A in Equation (2).

$\begin{array}{c}P=280\text{MPa}\times \frac{{10}^6\text{N}/{\text{m}}^2}{1\text{MPa}}\times 153.94{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =43.103\times {10}^3\text{N}\end{array}$

Take the smaller value as the applied load, $P=35.343\times {10}^3\text{N}$.

Calculate the strain energy (U) as shown below.

$U=\frac{P^2{L}_{AB}}{2E_{AB}{A}_{AB}}+\frac{P^2{L}_{BC}}{2E_{BC}{A}_{BC}}$

Substitute $35.343\times {10}^3\text{ N}$ for P, $200\text{GPa}$ for $E_{AB}$, $1.2\text{m}$ for $L_{AB}$, $78.54{\text{mm}}^2$ for $A_{AB}$, $1.6\text{m}$ for $L_{BC}$, $73\text{GPa}$ for $E_{BC}$, and $153.94{\text{mm}}^2$ for $A_{BC}$.

$\begin{array}{c}U=\left(\begin{array}{l}\frac{{\left(35.343\times {10}^3\text{N}\right)}^2\times 1.2\text{m}}{2\times 200\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}\times 78.54{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ +\frac{{\left(35.343\times {10}^3\text{N}\right)}^2\times 1.6\text{m}}{2\times 73\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}\times 153.94{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\end{array}\right)\\ =47.7+88.9\\ =136.6\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =136.6\text{J}\end{array}$

Therefore, the maximum strain energy of the composite rod ABC is $\underset{\_}{136.6\text{J}}$.