Chapter 2.1, Problem 19P

Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.

Fig. P2.19 and P2.20

To determine

The value of (Q) when the deflection at A is zero.

Answer to Problem 19P

The value of (Q) when the deflection at A is zero is $\underset{\_}{32.8\text{kN}}$.

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 $4\text{kN}$.

The force at the point B is Q.

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 $4\text{kN}$ for P, $0.4\text{m}$ for $L_{AB}$, $100\pi {\text{mm}}^2$ for $A_{AB}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{AB}=\frac{4\times 0.4\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{100\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =72.756\times {10}^{-3}\text{mm}\end{array}$

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

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

Substitute $4\text{kN}$ for P, $0.5\text{m}$ for $L_{BC}$, $900\pi {\text{mm}}^2$ for $A_{BC}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{BC}=\frac{\left(Q-4\right)\times 0.5\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{900\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =(Q-4)\times 2.5263\times {10}^{-3}\\ =2.5263\times {10}^{-3}Q-10.1052\times {10}^{-3}\end{array}$

Calculate the force at the point B (Q):

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

Substitute $72.756\times {10}^{-3}\text{mm}$ for ${\delta }_{AB}$ and $\left(2.5263\times {10}^{-3}Q-10.1052\times {10}^{-3}\right)$ for ${\delta }_{BC}$.

$\begin{array}{l}72.756\times {10}^{-3}=2.5263\times {10}^{-3}Q-10.1052\times {10}^{-3}\\ 2.5263\times {10}^{-3}Q=72.756\times {10}^{-3}+10.1052\times {10}^{-3}\\ Q=\frac{82.8612\times {10}^{-3}}{2.5263\times {10}^{-3}}\\ Q=32.8\text{kN}\end{array}$

Hence, the value of (Q) when the deflection at A is zero is $\underset{\_}{32.8\text{kN}}$.

To determine

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

Answer to Problem 19P

The deflection of B $\left({\delta }_B\right)$ is $\underset{\_}{0.0728\text{mm}}\downarrow$.

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 $4\text{kN}$.

The force at the point B is Q.

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 $4\text{kN}$ for P, $0.4\text{m}$ for $L_{AB}$, $100\pi {\text{mm}}^2$ for $A_{AB}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{AB}=\frac{4\times 0.4\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{100\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =72.756\times {10}^{-3}\text{mm}\end{array}$

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

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

Substitute $4\text{kN}$ for P, $0.5\text{m}$ for $L_{BC}$, $900\pi {\text{mm}}^2$ for $A_{BC}$, and $70\text{GPa}$ for E.

$\begin{array}{c}{\delta }_{BC}=\frac{\left(Q-4\right)\times 0.5\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}}{900\pi \times 70\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}}\\ =(Q-4)\times 2.5263\times {10}^{-3}\\ =2.5263\times {10}^{-3}Q-10.1052\times {10}^{-3}\end{array}$ (1)

Calculate the force at the point B (Q):

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

Substitute $72.756\times {10}^{-3}\text{mm}$ for ${\delta }_{AB}$ and $\left(2.5263\times {10}^{-3}Q-10.1052\times {10}^{-3}\right)$ for ${\delta }_{BC}$.

$\begin{array}{l}72.756\times {10}^{-3}=2.5263\times {10}^{-3}Q-10.1052\times {10}^{-3}\\ 2.5263\times {10}^{-3}Q=72.756\times {10}^{-3}+10.1052\times {10}^{-3}\\ Q=\frac{82.8612\times {10}^{-3}}{2.5263\times {10}^{-3}}\\ Q=32.8\text{kN}\end{array}$

Calculate the deflection of B $\left({\delta }_B\right)$:

Substitute $32.8\text{kN}$ for Q in Equation (1).

$\begin{array}{c}{\delta }_B={\delta }_{BC}=2.5263\times {10}^{-3}\left(32.8\text{kN}\right)-10.1052\times {10}^{-3}\\ =-0.0728\text{mm}\\ =0.0728\text{mm}\downarrow \end{array}$

Hence, the deflection of B $\left({\delta }_B\right)$ is $\underset{\_}{0.0728\text{mm}}\downarrow$.