Chapter 11.5, Problem 46P

(a)

To determine

Find the maximum deflection of end C.

Answer to Problem 46P

The maximum deflection of end C is $\underset{\_}{{\Delta }_m=1.656\text{mm}}$.

Explanation of Solution

Given information:

The mass of the collar D is $m=15\text{kg}$.

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

The length of the rod AB is $L_{AB}=2\text{m}$.

The length of the rod BC is $L_{BC}=1.5\text{m}$.

The diameter of rod AB is $d_{AB}=40\text{mm}$

The diameter of rod BC is $d_{BC}=30\text{mm}$

Calculation:

Consider the acceleration due to gravity as $g=9.81\text{m}/{\text{s}}^2$.

Calculate the weight of the collar (m) as shown below.

$W=mg$

Substitute $9.81\text{m}/{\text{s}}^2$ for g and $15\text{kg}$ for m.

$\begin{array}{c}W=15\text{kg}\times 9.81\text{m}/{\text{s}}^2\\ =147.15\text{kg}\cdot \text{m}/{\text{s}}^2\times \frac{\text{1}\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\\ =147.15\text{N}\end{array}$

Calculate the cross sectional area $A$ of the rod as shown below.

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

For rod AB.

Substitute $40\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{AB}=\frac{\pi \times {40}^2}{4}\\ =1,256.64{\text{mm}}^2\times {\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}^2\\ =1.2566\times {10}^{-3}{\text{m}}^2\end{array}$

For rod BC.

Substitute $30\text{mm}$ for d in Equation (1).

$\begin{array}{c}{A}_{BC}=\frac{\pi \times {30}^2}{4}\\ =706.86{\text{mm}}^2\times {\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}^2\\ =0.7069\times {10}^{-3}{\text{m}}^2\end{array}$

The rod BC has the minimum area $A_{\min }=0.7069\times {10}^{-3}{\text{m}}^2$.

Calculate the deflection $\left(\Delta \right)$ as shown below.

$\Delta =\frac{PL}{AE}$ (2)

For rod AB.

Substitute $P_m$ for $P$, $2\text{m}$ for L, $1.2566\times {10}^{-3}{\text{m}}^2$ for A, and $200\text{GPa}$ for E in Equation (2).

$\begin{array}{c}{\Delta }_{AB}=\frac{P_m\left(2\text{m}\right)}{1.2566\times {10}^{-3}{\text{m}}^2\times 200\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}}\\ =7.958\times {10}^{-9}{P}_m\end{array}$

For rod BC.

Substitute $P_m$ for $P$, $1.5\text{m}$ for L, $0.7069\times {10}^{-3}{\text{m}}^2$ for A, and $200\text{GPa}$ for E in Equation (2).

$\begin{array}{c}{\Delta }_{BC}=\frac{P_m\left(1.5\text{m}\right)}{2\times 0.7069\times {10}^{-3}{\text{m}}^2\times 200\text{GPa}\times \frac{{10}^9\text{N}/{\text{m}}^2}{1\text{GPa}}}\\ =10.61\times {10}^{-9}{P}_m\end{array}$

Calculate the maximum deflection $\left({\Delta }_m\right)$ as shown below.

${\Delta }_m={\Delta }_{AB}+{\Delta }_{BC}$

Substitute $7.958\times {10}^{-9}{P}_m^2$ for ${\Delta }_{AB}$ and $10.61\times {10}^{-9}{P}_m$ for ${\Delta }_{BC}$.

$\begin{array}{c}{\Delta }_m=7.958\times {10}^{-9}{P}_m^2+10.61\times {10}^{-9}{P}_m\\ =18.568\times {10}^{-9}{P}_m\end{array}$ (3)

$P_m=53.856\times {10}^6{\Delta }_m$ (4)

Sketch the Free Body Diagram of the rod after deformation as shown in Figure 1.

Refer to Figure 1.

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

$U_m=\frac{1}{2}{P}_m{\Delta }_m$

Substitute $53.856\times {10}^6{\Delta }_m$ for $P_m$.

$\begin{array}{c}{U}_m=\frac{1}{2}\times 53.856\times {10}^6{\Delta }_m\times {\Delta }_m\\ =26.928\times {10}^6{\Delta }_m^2\end{array}$

Consider that the distance $h=0.5\text{m}$.

Calculate the maximum deflection $\left({\Delta }_m\right)$ as shown below.

$W\left(h+{\Delta }_m\right)=U_m$

Substitute $147.15\text{N}$ for W, $0.5\text{m}$ for $h$ , and $26.928\times {10}^6{\Delta }_m^2$ for $U_m$.

$\begin{array}{l}147.15\left(0.5+{\Delta }_m\right)=26.928\times {10}^6{\Delta }_m^2\\ 0.5+{\Delta }_m=182.997\times {10}^3{\Delta }_m^2\\ 182.997\times {10}^3{\Delta }_m^2-{\Delta }_m-0.5=0\\ {\Delta }_m=1.656\times {10}^{-3}\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\end{array}$

${\Delta }_m=1.656\text{mm}$

Hence, the maximum deflection of end C is $\underset{\_}{{\Delta }_m=1.656\text{mm}}$.

(b)

To determine

The equivalent static load.

Answer to Problem 46P

The equivalent static load is $\underset{\_}{P_m=89.186\text{kN}}$.

Explanation of Solution

Given information:

The mass of the collar D is $m=15\text{kg}$.

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

The length of the rod AB is $L_{AB}=2\text{m}$.

The length of the rod BC is $L_{BC}=1.5\text{m}$.

The diameter of rod AB is $d_{AB}=40\text{mm}$

The diameter of rod BC is $d_{BC}=30\text{mm}$

Calculation:

Refer to part (a).

The maximum deflection of end C is ${\Delta }_m=1.656\times {10}^{-3}\text{m}$.

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

Substitute $1.656\times {10}^{-3}\text{m}$ for ${\Delta }_m$ in Equation (4).

$\begin{array}{c}{P}_m=53.856\times {10}^6\times 1.656\times {10}^{-3}\\ =89.186\times {10}^3\text{N}\times \frac{1\text{kN}}{1,000\text{N}}\\ =89.186\text{kN}\end{array}$

Substitute $250\text{MPa}$ for ${\sigma }_{\text{all}}$ and $0.7069\times {10}^{-3}{\text{m}}^2$ for $A_{\min }$.

$\begin{array}{c}{P}_m=250\text{MPa}\times \frac{{10}^6\text{N}/{\text{m}}^{\text{2}}}{1\text{MPa}}\times 0.7069\times {10}^{-3}{\text{m}}^2\\ =176,725\text{N}\times \frac{1\text{kN}}{1,000\text{N}}\\ =176.725\text{kN}\end{array}$

Therefore, the equivalent static load is $\underset{\_}{P_m=89.186\text{kN}}$.

(c)

To determine

The maximum stress occurs in the rod.

Answer to Problem 46P

The maximum stress occurs in the rod is $\underset{\_}{{\sigma }_m=126.2\text{MPa}}$.

Explanation of Solution

Given information:

The mass of the collar D is $m=15\text{kg}$.

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

The length of the rod AB is $L_{AB}=2\text{m}$.

The length of the rod BC is $L_{BC}=1.5\text{m}$.

The diameter of rod AB is $d_{AB}=40\text{mm}$

The diameter of rod BC is $d_{BC}=30\text{mm}$

Calculation:

Refer to part (a).

The minimum area of the rod is $A_{\min }=0.7069\times {10}^{-3}{\text{m}}^2$.

Refer to part (b).

The equivalent static load is $P_m=89.186\text{kN}$.

Calculate the maximum stress $\left({\sigma }_m\right)$ as shown below.

${\sigma }_m=\frac{P_m}{A_{\min }}$

Substitute $89.186\text{kN}$ for $P_m$ and $0.7069\times {10}^{-3}{\text{m}}^2$ for $A_{\min }$.

$\begin{array}{c}{\sigma }_m=\frac{89.186\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{0.7069\times {10}^{-3}{\text{m}}^2}\\ =126.165\times {10}^6\text{N}/{\text{m}}^{\text{2}}\times \frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^{\text{2}}}\\ =126.2\text{MPa}\end{array}$

Therefore, the maximum stress occurs in the rod is $\underset{\_}{{\sigma }_m=126.2\text{MPa}}$.