Chapter 2.1, Problem 20P

The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that P = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point B.

Fig. P2.19 and P2.20

To determine

The deflection of the point A $\left({\delta }_A\right)$.

Answer to Problem 20P

The deflection of the point A $\left({\delta }_A\right)$ is $\underset{\_}{0.0189\text{mm}}\uparrow$.

Explanation of Solution

Given information:

The Young’s modulus of the aluminium (E) is $70\text{GPa}$.

The force at the point A (P) is $6\text{kN}$.

The force at the point B (Q) is $42\text{kN}$.

The diameter of the rod AB $\left(d_{AB}\right)$ is $20\text{mm}$.

The diameter of the rod BC $\left(d_{BC}\right)$ is $60\text{mm}$.

The length of the rod AB $\left(L_{AB}\right)$ is $0.4\text{m}$.

The length of the rod BC $\left(L_{BC}\right)$ is $0.5\text{m}$.

Calculation:

Calculate the cross-sectional area of the rod AB $\left(A_{AB}\right)$ using the formula:

$A_{AB}=\frac{\pi }{4}{d}_{AB}^2$

Substitute $20\text{mm}$ for $d_{AB}$.

$\begin{array}{c}{A}_{AB}=\frac{\pi }{4}\times {20}^2\\ =100\pi {\text{mm}}^2\end{array}$

Calculate the cross-sectional area of the rod BC $\left(A_{BC}\right)$ using the formula:

$A_{BC}=\frac{\pi }{4}{d}_{BC}^2$

Substitute $60\text{mm}$ for $d_{BC}$.

$\begin{array}{c}{A}_{AB}=\frac{\pi }{4}\times {60}^2\\ =900\pi {\text{mm}}^2\end{array}$

Calculate the defection of the rod AB $\left({\delta }_{AB}\right)$ using the formula:

${\delta }_{AB}=\frac{P{L}_{AB}}{A_{AB}E}$

Substitute $6\text{kN}$ for P, $0.4\text{m}$ for $L_{AB}$, $100\pi {\text{mm}}^2$ for $A_{AB}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{AB}=\frac{6\times 0.4\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{100\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =109.135\times {10}^{-3}\text{mm}\end{array}$

Calculate the defection of the rod BC $\left({\delta }_{BC}\right)$ using the formula:

${\delta }_{BC}=\frac{\left(P-Q\right)L_{BC}}{A_{BC}E}$

Substitute $42\text{kN}$ for Q, $6\text{kN}$ for P, $0.5\text{m}$ for $L_{BC}$, $900\pi {\text{mm}}^2$ for $A_{BC}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{BC}=\frac{\left(6-42\right)\times 0.5\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{900\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =-36\times 2.5263\times {10}^{-3}\\ =-90.947\times {10}^{-3}\text{mm}\end{array}$

Calculate the deflection of the point A $\left({\delta }_A\right)$:

${\delta }_A={\delta }_{AB}+{\delta }_{BC}$

Substitute $109.135\times {10}^{-3}\text{mm}$ for ${\delta }_{AB}$ and $-90.947\times {10}^{-3}\text{mm}$ for ${\delta }_{BC}$.

$\begin{array}{c}{\delta }_A=109.135\times {10}^{-3}-90.947\times {10}^{-3}\\ =0.01819\text{mm}\\ =0.01819\text{mm}\uparrow \end{array}$

Hence, the deflection of the point A $\left({\delta }_A\right)$ is $\underset{\_}{0.01819\text{mm}}\uparrow$.

To determine

The deflection of point the B $\left({\delta }_B\right)$

Answer to Problem 20P

The deflection of the B $\left({\delta }_B\right)$ is $\underset{\_}{0.0909\text{mm}}\downarrow$.

Explanation of Solution

Given information:

The Young’s modulus of the aluminium (E) is $70\text{GPa}$.

The force at the point A (P) is $6\text{kN}$.

The force at the point B (Q) is $42\text{kN}$.

The diameter of the rod AB $\left(d_{AB}\right)$ is $20\text{mm}$.

The diameter of the rod BC $\left(d_{BC}\right)$ is $60\text{mm}$.

The length of the rod AB $\left(L_{AB}\right)$ is $0.4\text{m}$.

The length of the rod BC $\left(L_{BC}\right)$ is $0.5\text{m}$.

Calculation:

Calculate the cross-sectional area of the rod AB $\left(A_{AB}\right)$ using the formula:

$A_{AB}=\frac{\pi }{4}{d}_{AB}^2$

Substitute $20\text{mm}$ for $d_{AB}$.

$\begin{array}{c}{A}_{AB}=\frac{\pi }{4}\times {20}^2\\ =100\pi {\text{mm}}^2\end{array}$

Calculate the cross-sectional area of the rod BC $\left(A_{BC}\right)$ using the formula:

$A_{BC}=\frac{\pi }{4}{d}_{BC}^2$

Substitute $60\text{mm}$ for $d_{BC}$.

$\begin{array}{c}{A}_{AB}=\frac{\pi }{4}\times {60}^2\\ =900\pi {\text{mm}}^2\end{array}$

Calculate the defection of the rod AB $\left({\delta }_{AB}\right)$ using the formula:

${\delta }_{AB}=\frac{P{L}_{AB}}{A_{AB}E}$

Substitute $6\text{kN}$ for P, $0.4\text{m}$ for $L_{AB}$, $100\pi {\text{mm}}^2$ for $A_{AB}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{AB}=\frac{6\times 0.4\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{100\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =109.135\times {10}^{-3}\text{mm}\end{array}$

Calculate the defection of the rod BC $\left({\delta }_{BC}\right)$ using the formula:

${\delta }_{BC}=\frac{\left(P-Q\right)L_{BC}}{A_{BC}E}$

Substitute $42\text{kN}$ for Q, $6\text{kN}$ for P, $0.5\text{m}$ for $L_{BC}$, $900\pi {\text{mm}}^2$ for $A_{BC}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{BC}=\frac{\left(6-42\right)\times 0.5\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{900\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =-36\times 2.5263\times {10}^{-3}\\ =-90.947\times {10}^{-3}\text{mm}\\ =0.0909\text{mm}\downarrow \end{array}$

The deflection of the point B $\left({\delta }_B\right)$ is same as the defection of the rod BC $\left({\delta }_{BC}\right)$.

Therefore, the deflection of the point B $\left({\delta }_B\right)$ is $\underset{\_}{0.0909\text{mm}}\downarrow$.