Question 16.SP.13: In the engine system from Sample Prob. 15.15, the crank AB h...
In the engine system from Sample Prob. 15.15, the crank AB has a constant clockwise angular velocity of 2000 rpm. Knowing that the connecting rod BD weighs 4 lb and the piston P weighs 5 lb, determine the forces on the connecting rod at B and D. Assume the center of mass of BD is at its geometric center and it can be treated as a uniform, slender rod.
STRATEGY: Since you are asked to find forces at the instant shown, use Newton’s second law.
MODELING: Since you want to determine the forces at B and D, start by choosing the connecting rod BD as your system. The pin forces at B and D are represented by horizontal and vertical components, and since the rod is undergoing general plane motion, you can represent the acceleration of the center of mass in the kinetic diagram as having a vertical and a horizontal component. The free-body and kinetic diagrams for this system are shown in Fig. 1, where \ell = 8 in. = 0.6667 ft and β = 13.95°.
ANALYSIS: Using Fig. 1, applying Newton’s second law in the x-direction and y-direction, and summing moments about point G gives
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad B_x + D_x=m_{B D}\bar{a}_x (1)
+\uparrow \Sigma F_y=m \bar{a}_y: \quad B_y + D_y-W_{B D}=m_{B D} \bar{a}_y (2)
\begin{aligned}+\circlearrowleft M_G=\bar{I} \alpha:&-B_y(\ell / 2) \cos \beta-B_x(\ell / 2) \sin \beta+D_y(\ell / 2) \cos \beta \\&+D_x(\ell / 2) \sin \beta=\bar{I}_{B D} \alpha_{B D}\end{aligned} (3)
where
\begin{aligned}m_{B D} &=\frac{W_{B D}}{g}=\frac{4 \mathrm{lb}}{32.2 \mathrm{ft} / \mathrm{s}^2}=0.1242 \mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft} \\\bar{I}_{B D} &=\frac{1}{12} m_{B D} \ell^2=\frac{1}{12}\left(0.1242 \mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}\right)(0.6667 \mathrm{ft})^2=0.004601 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2\end{aligned}
In Eqs. (1) through (3), you have seven unknowns: B_x, B_y, D_x, D_y, \bar{a}_x, \bar{a}_y, and \alpha_{B D}. Therefore, you need more equations. You can get them from kinematics or by choosing another system. Choose the piston to be your system, model it as a particle, and draw its free-body and kinetic diagrams (Fig. 2).
Note that you must draw D_x and D_y in the opposite directions to what you drew for the connecting rod. Using Fig. 2 and applying Newton’s second law in the x-direction and y-direction gives
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad-D_x=m_P a_D (4)
+\uparrow \Sigma F_y=m \bar{a}_y: \quad-D_y + N – W_P=0 (5)
where
m_P=\frac{W_P}{g}=\frac{5 \mathrm{lb}}{32.2 \mathrm{ft} / \mathrm{s}^2}=0.1553 \mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}
You now have five equations and nine unknowns: N, B_x, B_y, D_x, D_y, \bar{a}_x \bar{a}_y, \alpha_{B D}, and a_D. You could choose crank AB as another system, but since this will introduce three additional unknowns (the reactions at A and the driving torque) and you are not provided its mass, you should turn to kinematics for additional equations. From Sample Prob. 15.15, you obtained \boldsymbol{ \omega}_{B D} =62.0 \mathrm{rad} / \mathrm{s} \circlearrowleft , \mathbf{a}_D=9290 \mathrm{ft} / \mathrm{s}^2 \leftarrow, and \boldsymbol{\alpha}_{B D}=9940 \mathrm{rad} / \mathrm{s}^2 \circlearrowleft . These reduce the number of unknowns by two, so you have five equations and seven unknowns: N, B_x, B_y, D_x, D_y, \bar{a}_x, and \bar{a}_y . You can find two more equations by relating the acceleration of the center of mass of the connecting rod to the acceleration of D,
\mathbf{a}_G=\mathbf{a}_D + \mathbf{a}_{G / D}=\mathbf{a}_D + \boldsymbol{\alpha} \times \mathbf{r}_{G / D}-\omega_{B D}^2 \mathbf{r}_{G / D}
Substituting in known and assumed values (Fig. 1) \mathbf{a}_D=a_D \mathbf{i}, where a_D=-9290 \mathrm{ft} / \mathrm{s}^2 , and \boldsymbol{\alpha}_{B D}=\alpha_{B D} \mathbf{k}, where \alpha_{B D}=9940 \mathrm{rad} / \mathrm{s}^2, gives
\begin{aligned}\bar{a}_x \mathbf{i}+\bar{a}_y \mathbf{j} &=a_D \mathbf{i} + \alpha_{B D} \mathbf{k} \times\left[-\frac{\ell}{2} \cos \beta \mathbf{i} + \frac{\ell}{2} \sin \beta \mathbf{j}\right] – \omega_{B D}^2\left[-\frac{\ell}{2} \cos \beta \mathbf{i} + \frac{\ell}{2} \sin \beta \mathbf{j}\right] \\&=a_D \mathbf{i} – \alpha_{B D} \frac{\ell}{2} \cos \beta \mathbf{j} – \alpha_{B D} \frac{\ell}{2} \sin \beta \mathbf{i} + \omega_{B D}^2 \frac{\ell}{2} \cos \beta \mathbf{i} – \omega_{B D}^2 \frac{\ell}{2} \sin \beta \mathbf{j}\end{aligned}
Equating components, you have
\text { i: } \bar{a}_x=a_D – \alpha_{B D} \frac{\ell}{2} \sin \beta + \omega_{B D}^2 \frac{\ell}{2} \cos \beta (6)
\text { j: } \bar{a}_y=-\alpha_{B D} \frac{\ell}{2} \cos \beta – \omega_{B D}^2 \frac{\ell}{2} \sin \beta (7)
You now have seven equations and seven unknowns. Substituting in numerical values and solving these equations using your calculator or software such as MathCad, Maple, Matlab, or Mathematica gives you B_x=-2541 \mathrm{lb}, B_y=207.2 \mathrm{lb}, D_x=1442 \mathrm{lb}, D_y=-641 \mathrm{lb}, N=-636 \mathrm{lb}, \bar{a}_x= -8845 \mathrm{ft} / \mathrm{s}^2 and \bar{a}_y=-3524 \mathrm{ft} / \mathrm{s}^2.
\begin{aligned}B_x=2541 \mathrm{lb} \leftarrow & B_y=207 \mathrm{lb} \uparrow \\D_x=1442 \mathrm{lb} \rightarrow & D_y=641 \mathrm{lb} \downarrow\end{aligned}
REFLECT and THINK: The calculated forces are much larger than the weight of the piston and the connecting rod. This problem required multiple systems and rigid body kinematics to solve, most of which was done in Sample Prob. 15.15. In problems like this, it is a good practice to focus on the problem formulation and to keep track of equations and unknowns. Once you have enough equations to solve for all the unknowns, using a computer or calculator to solve the resulting equations is often the easiest approach.
