Chapter 16.2, Problem 16.138P

Solve Prob. 16.137 when θ = 90°.

16.137    In the engine system shown, l = 250 mm and b = 100 mm. The connecting rod BD is assumed to be a 1.2-kg uniform slender rod and is attached to the 1.8-kg piston P. During a test of the system, crank AB is made to rotate with a constant angular velocity of 600 rpm clockwise with no force applied to the face of the piston. Determine the forces exerted on the connecting rod at B and D when θ = 180°. (Neglect the effect of the weight of the rod.)

Fig. P16.137

To determine

The forces exerted on the connecting rod at B and D when $\theta =90°$.

Answer to Problem 16.138P

The forces exerted on the connecting rod at B is $\underset{\_}{525\text{N}}$.

The forces exerted on the connecting rod at D is $\underset{\_}{322\text{N}}$.

Explanation of Solution

Given information:

The length of the rod BD is $l=250\text{mm}$.

The length of the rod AB is $b=100\text{mm}$.

The mass of the rod BD is $m_{BD}=1.2\text{kg}$.

The mass of the piston P is $m_P=1.8\text{kg}$.

The angular velocity of AB is ${\omega }_{AB}=600\text{rpm}$.

Calculation:

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

Calculate the angular velocity in $\text{rad}/\text{s}$ as shown below.

$\begin{array}{c}{\omega }_{AB}=600\text{rpm}\times \frac{\frac{2\pi }{60}\text{rad}/\text{s}}{1\text{rpm}}\\ =62.832\text{rad}/\text{s}\end{array}$

Sketch the Free body Diagram of the system as shown in Figure 1.

Refer to Figure 1.

Calculate the distance $\left(x\right)$ as shown below.

$\begin{array}{l}{x}^2=l^2-b^2\\ x=\sqrt{l^2-b^2}\end{array}$

Substitute $250\text{mm}$ for l and $100\text{mm}$ for b.

$\begin{array}{c}x=\sqrt{{250}^2-{100}^2}\\ =\sqrt{52,500}\\ =229.1\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\\ =0.2291\text{m}\end{array}$

Calculate the position vectors as shown below.

Position of B with respect to A.

$\begin{array}{c}{\text{r}}_{B/A}=\left(100\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\text{j}\\ =\left(0.1\text{m}\right)\text{j}\end{array}$

Position of D with respect to B.

$\begin{array}{c}{\text{r}}_{D/B}=-\left(0.2291\text{m}\right)\text{i}-\left(100\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\text{j}\\ =-\left(0.2291\text{m}\right)\text{i}-\left(0.1\text{m}\right)\text{j}\end{array}$

Position of mass center G with respect to D.

$\begin{array}{c}{\text{r}}_{G/D}=\frac{1}{2}\left(\left(0.2291\text{m}\right)\text{i}+\left(100\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)\text{j}\right)\\ =\frac{1}{2}\left(\left(0.2291\text{m}\right)\text{i}+\left(0.1\text{m}\right)\text{j}\right)\\ =\left(0.11455\text{m}\right)\text{i}+\left(0.05\text{m}\right)\text{j}\end{array}$

Calculate the velocity at B $\left({\text{v}}_B\right)$ as shown below.

$v_B={\omega }_{AB}\times r_{B/A}$

Substitute $-\left(62.832\text{rad}/\text{s}\right)\text{k}$ for ${\omega }_{AB}$ and $\left(0.1\text{m}\right)\text{j}$ for $r_{B/A}$.

$\begin{array}{c}{\text{v}}_B=-62.832\text{k}\times 0.1\text{j}\\ =\left(6.2832\text{m}/\text{s}\right)\text{i}\end{array}$

Calculate the velocity at D $\left({\text{v}}_D\right)$ as shown below.

${\text{v}}_D={\text{v}}_B+{\omega }_{BD}\times r_{B/D}$

Substitute $\left(6.2832\text{m}/\text{s}\right)\text{i}$ for ${\text{v}}_B$ and $-\left(0.2291\text{m}\right)\text{i}-\left(0.1\text{m}\right)\text{j}$ for $r_{B/D}$.

$\begin{array}{c}{\text{v}}_D\text{i}=6.2832\text{i}+{\omega }_{BD}\text{k}\times \left(-0.2291\text{i}-0.1\text{j}\right)\\ =6.2832\text{i}-0.2291{\omega }_{BD}\text{j}+0.1{\omega }_{BD}\text{i}\\ =\left(6.2832+0.1{\omega }_{BD}\right)\text{i}-0.2291{\omega }_{BD}\text{j}\end{array}$

Resolving the i and j components as shown below.

For j component.

$\begin{array}{l}0.2291{\omega }_{BD}=0\\ {\omega }_{BD}=0\end{array}$

For i component.

$6.2832+0.1{\omega }_{BD}={\text{v}}_D$

Substitute $0$ for ${\omega }_{BD}$.

${\text{v}}_D=6.2832\text{m}/\text{s}$

Consider that the angular acceleration as ${\alpha }_{AB}=0$.

Calculate the acceleration at B $\left({\text{a}}_B\right)$ as shown below.

${\text{a}}_B={\alpha }_{AB}\times {\text{r}}_{B/A}-{\omega }_{AB}^2{\text{r}}_{B/A}$

Substitute $0$ for ${\alpha }_{AB}$, $62.832\text{rad}/\text{s}$ for ${\omega }_{AB}$, and $\left(0.1\text{m}\right)\text{j}$ for ${\text{r}}_{B/A}$.

$\begin{array}{c}{\text{a}}_B=0-{\left(62.832\text{rad}/\text{s}\right)}^2\times \left(0.1\text{m}\right)\text{j}\\ =-\left(394.786\text{m}/{\text{s}}^2\right)\text{j}\end{array}$

Calculate the acceleration $\left(a_D\right)$ and the angular acceleration $\left({\alpha }_{BD}\right)$ using the relation as shown below.

${\text{a}}_D={\text{a}}_B+{\alpha }_{BD}\times {\text{r}}_{B/D}-{\omega }_{BD}^2{\text{r}}_{D/B}$

Substitute $-\left(394.786\text{m}/{\text{s}}^2\right)\text{j}$ for ${\text{a}}_B$, $0$ for ${\omega }_{BD}$, and $-\left(0.2291\text{m}\right)\text{i}-\left(0.1\text{m}\right)\text{j}$ for $r_{B/D}$.

$\begin{array}{c}{a}_D\text{i}=-394.786\text{j}+{\alpha }_{BD}\text{k}\times \left(-0.2291\text{i}-0.1\text{j}\right)-0\\ =-394.786\text{j}-0.2291{\alpha }_{BD}\text{j}+0.1{\alpha }_{BD}\text{i}\\ =\left(-394.786-0.2291{\alpha }_{BD}\right)\text{j}+0.1{\alpha }_{BD}\text{i}\end{array}$

Resolving i and j components as shown below.

For j component.

$\begin{array}{l}-394.786-0.2291{\alpha }_{BD}=0\\ 0.2291{\alpha }_{BD}=-394.786\\ {\alpha }_{BD}=-1,723\text{rad}/{\text{s}}^{\text{2}}\end{array}$

For i component.

$0.1{\alpha }_{BD}=a_D$

Substitute $-1,723\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BD}$.

$\begin{array}{c}{a}_D=0.1\times \left(-1,723\right)\\ =-172.3\text{m}/{\text{s}}^2\end{array}$

Calculate the acceleration of mass center G of bar BD $\left(a_G\right)$ as shown below.

${\text{a}}_G={\text{a}}_D+{\alpha }_{BD}\times {\text{r}}_{G/D}-{\omega }_{BD}^2{\text{r}}_{G/D}$

Substitute $-\left(172.3\text{m}/{\text{s}}^2\right)\text{i}$ for ${\text{a}}_D$, $-\left(1,723\text{rad}/{\text{s}}^{\text{2}}\right)\text{k}$ for ${\alpha }_{BD}$, $0$ for ${\omega }_{BD}$, and $\left(0.11455\text{m}\right)\text{i}+\left(0.05\text{m}\right)\text{j}$ for $r_{B/D}$.

$\begin{array}{c}{\text{a}}_G\text{i}=-172.3\text{i}+\left(-1,723\right)\text{k}\times \left(\text{0}\text{.11455i}+0.05\text{j}\right)-0\\ =-172.3\text{i}-197.3695\text{j}+86.15\text{i}\\ =-86.15\text{i}-197.3695\text{j}\end{array}$

Resolving the components as shown below.

${\text{a}}_G=-86.15\text{m}/{\text{s}}^{\text{2}}$

Calculate the mass moment of inertia for BD $\left({\overline{I}}_{BD}\right)$ as shown below.

${\overline{I}}_{BD}=\frac{1}{12}{m}_{BD}{l}^2$

Substitute $1.2\text{kg}$ for $m_{BD}$ and $250\text{mm}$ for l.

$\begin{array}{c}{\overline{I}}_{BD}=\frac{1}{12}\times \left(1.2\text{kg}\right)\times {\left(250\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =0.00625\text{kg}\cdot {\text{m}}^2\end{array}$

Sketch the Free Body Diagram of the piston with the bar BD as shown in Figure 2.

Refer to Figure 2.

Apply the Equilibrium of forces along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=m{a}_x\\ B_x=m_P{a}_D+{\left(m_{BD}{a}_G\right)}_x\end{array}$

Substitute $1.8\text{kg}$ for $m_P$, $-172.3\text{m}/{\text{s}}^2$ for ${\text{a}}_D$, $1.2\text{kg}$ for $m_{BD}$, and $-86.15\text{m}/{\text{s}}^{\text{2}}$ for ${\text{a}}_G$.

$\begin{array}{c}{B}_x=1.8\times \left(-172.3\right)+1.2\times \left(-86.15\right)\\ =-413.52\text{N}\end{array}$

Apply the Equilibrium of moment about B as shown below.

$\begin{array}{l}{\displaystyle \sum M_B}=I_G\alpha +mad\\ -\left(xN\right)\text{k}={\overline{I}}_{BD}{\alpha }_{BD}\text{k}+{\text{r}}_{D/B}\left(m_P{\text{a}}_D\right)+{\text{r}}_{G/B}\left(m_{BD}{\text{a}}_G\right)\end{array}$

Substitute $0.2291\text{m}$ for x, $0.00625\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_{BD}$, $-1,723\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_{BD}$, $\left(-0.11455\text{m}\right)\text{i}-\left(0.05\text{m}\right)\text{j}$ for $r_{G/B}$, $0.1\text{m}$ for ${\text{r}}_{D/B}$, $1.8\text{kg}$ for $m_P$, $-172.3\text{m}/{\text{s}}^2$ for ${\text{a}}_D$, $1.2\text{kg}$ for $m_{BD}$, and $\left(-86.15\text{i}-197.3695\text{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\text{a}}_G$.

$\begin{array}{l}-\left(0.2291N\right)\text{k}=\left[\begin{array}{l}\left(0.00625\times \left(-1,723\right)\right)\text{k}+0.1\times \left(1.8\times \left(-172.3\right)\right)\text{k}\\ +\left(-\text{0}\text{.11455i}-0.05\text{j}\right)\left(-86.15\text{i}-197.3695\text{j}\right)\times 1.2\end{array}\right]\\ -0.2291N\text{k}=-10.76875\text{k}-31.014\text{k}+27.1304\text{k}-5.169\text{k}\\ -0.2291N=-19.82135\\ N=86.518\text{N}\end{array}$

Apply the Equilibrium of forces along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=m{a}_x\\ N+B_y={\left(m_{BD}{a}_G\right)}_y\end{array}$

Substitute $86.518\text{N}$ for $N$, $1.2\text{kg}$ for $m_{BD}$, and $-197.3695\text{m}/{\text{s}}^{\text{2}}$ for ${\text{a}}_G$.

$\begin{array}{l}86.518+B_y=1.2\times \left(-197.3695\right)\\ B_y=-323.36\text{N}\end{array}$

Calculate the force acting at B as shown below.

$B=\sqrt{B_x^2+B_y^2}$

Substitute $-413.52\text{N}$ for $B_x$ and $-323.6\text{N}$ for $B_y$.

$\begin{array}{c}B=\sqrt{{\left(-413.52\right)}^2+{\left(-323.6\right)}^2}\\ =\sqrt{275,715.7504}\\ =525\text{N}\end{array}$

Hence, the forces exerted on the connecting rod at B is $\underset{\_}{525\text{N}}$.

Sketch the Free Body Diagram of the piston as shown in Figure 3.

Refer to Figure 3.

Calculate the force acting on the rod at D as shown below.

$\begin{array}{l}{\displaystyle \sum F}D=ma\\ D+N\text{j}=m_P{a}_D\text{i}\end{array}$

Substitute $86.518\text{N}$ for $N$, $1.8\text{kg}$ for $m_P$, and $-172.3\text{m}/{\text{s}}^2$ for ${\text{a}}_D$.

$\begin{array}{l}D+86.518\text{j}=1.8\times \left(-172.3\right)\text{i}\\ D=-310.14\text{i}-86.518\text{j}\end{array}$

Calculate the magnitude of force at D as shown below.

$\begin{array}{c}D=\sqrt{{\left(-310.14\right)}^2+{\left(-86.518\right)}^2}\\ =\sqrt{103,672.1839}\\ =322\text{N}\end{array}$

Therefore, the forces exerted on the connecting rod at D is $\underset{\_}{322\text{N}}$.