Chapter 11.3, Problem 15P

The assembly ABC is made of a steel for which E = 200 GPa and σ_Y = 320 MPa. Knowing that a strain energy of 5 J must be acquired by the assembly as the axial load P is applied, determine the factor of safety with respect to permanent deformation when (a) x = 300 mm, (b) x = 600 mm.

Fig. P11.15

To determine

Find the factor of safety with respect to permanent deformation when $x=300\text{mm}$.

Answer to Problem 15P

The factor of safety with respect to permanent deformation is $\underset{\_}{3.28}$.

Explanation of Solution

Given information:

The diameter of the steel rod AB is $d_{AB}=12\text{mm}$.

The diameter of the steel rod BC is $d_{BC}=18\text{mm}$.

The length of the rod AB is $L_{AB}=300\text{mm}$.

The length of the rod BC is $L_{BC}=600\text{mm}$.

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

The yield strength of steel is ${\sigma }_Y=320\text{MPa}$.

The strain energy acquired by the assembly is $U=5\text{J}$.

Calculation:

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

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

For the steel rod AB.

Substitute $12\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{AB}=\frac{\pi \times {12}^2}{4}\\ =113.097{\text{mm}}^2\end{array}$

For the steel rod BC.

Substitute $18\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{BC}=\frac{\pi \times {18}^2}{4}\\ =254.469{\text{mm}}^2\end{array}$

Hence, the minimum area of the rod $A_{\min }=113.097{\text{mm}}^2$.

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

$P={\sigma }_Y{A}_{\min }$

Substitute $320\text{MPa}$ for ${\sigma }_Y$ and $113.097{\text{mm}}^2$ for $A_{\min }$.

$\begin{array}{c}P=320\text{MPa}\times \frac{1\text{N}/{\text{mm}}^{\text{2}}}{1\text{MPa}}\times 113.097{\text{mm}}^2\\ =36.191\times {10}^3\text{N}\end{array}$

Calculate the strain energy $\left(U_Y\right)$ as shown below.

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

Calculate the strain energy for rod ABC as shown below.

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

Substitute $36.191\times {10}^3\text{N}$ for P, $200\text{GPa}$ for E, $300\text{mm}$ for $L_{AB}$, $113.097{\text{mm}}^2$ for $A_{AB}$, $600\text{mm}$ for $L_{BC}$, and $254.469{\text{mm}}^2$ for $A_{BC}$ in Equation (2).

$\begin{array}{c}{U}_Y=\frac{{\left(36.191\times {10}^3\text{N}\right)}^2}{2\times 200\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}}\left(\begin{array}{l}\frac{300\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}{113.097{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ +\frac{600\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}{254.469{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\end{array}\right)\\ =3.27\times {10}^{-3}\left(2,652.59+2,357.85\right)\\ =16.38\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =16.38\text{J}\end{array}$

Calculate the factor of safety $\left(\text{F}\text{.S}\text{.}\right)$ as shown below.

$\text{F}\text{.S}\text{.}=\frac{U_Y}{U}$ (3)

Substitute $16.38\text{J}$ for $U_Y$ and $5\text{J}$ for U in Equation (3).

$\begin{array}{c}\text{F}\text{.S}\text{.}=\frac{16.38}{5}\\ =3.28\end{array}$

Therefore, the factor of safety with respect to permanent deformation is $\underset{\_}{3.28}$.

To determine

Find the factor of safety with respect to permanent deformation when $x=600\text{mm}$.

Answer to Problem 15P

The factor of safety with respect to permanent deformation is $\underset{\_}{4.24}$.

Explanation of Solution

Given information:

The diameter of the steel rod AB is $d_{AB}=12\text{mm}$.

The diameter of the steel rod BC is $d_{BC}=18\text{mm}$.

The length of the rod AB is $L_{AB}=600\text{mm}$.

The length of the rod BC is $L_{BC}=300\text{mm}$.

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

The yield strength of steel is ${\sigma }_Y=320\text{MPa}$.

The strain energy acquired by the assembly is $U=5\text{J}$.

Calculation:

Refer to part (a).

The area of the steel rod AB is $A_{AB}=113.097{\text{mm}}^2$

The area of the steel rod BC is $A_{BC}=254.469{\text{mm}}^2$

The load acting on the assembly is $P=36.191\times {10}^3\text{N}$

Calculate the strain energy $\left(U_Y\right)$ as shown below.

Substitute $36.191\times {10}^3\text{N}$ for P, $200\text{GPa}$ for E, $600\text{mm}$ for $L_{AB}$, $113.097{\text{mm}}^2$ for $A_{AB}$, $300\text{mm}$ for $L_{BC}$, and $254.469{\text{mm}}^2$ for $A_{BC}$ in Equation (2).

$\begin{array}{c}{U}_Y=\frac{{\left(36.191\times {10}^3\text{N}\right)}^2}{2\times 200\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}}\left(\begin{array}{l}\frac{600\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}{113.097{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\\ +\frac{300\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}}{254.469{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2}\end{array}\right)\\ =3.27\times {10}^{-3}\left(5,305.18+1,178.925\right)\\ =21.203\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =21.203\text{J}\end{array}$

Calculate the factor of safety $\left(\text{F}\text{.S}\text{.}\right)$ as shown below.

Substitute $21.203\text{J}$ for $U_Y$ and $5\text{J}$ for U in Equation (3).

$\begin{array}{c}\text{F}\text{.S}\text{.}=\frac{21.203}{5}\\ =4.24\end{array}$

Therefore, the factor of safety with respect to permanent deformation is $\underset{\_}{4.24}$.