Question 15.SP.16: The linkage ABDE moves in the vertical plane. Knowing that, ...
The linkage ABDE moves in the vertical plane. Knowing that, in the position shown, crank AB has a constant angular velocity \omega_{1} of 20 rad/s counterclockwise, determine the angular velocities and angular accelerations of the connecting rod BD and of the crank DE.
STRATEGY: The linkage consists of three interconnected rigid bodies. Use multiple velocity and acceleration kinematic equations to relate the motions of each body. You could solve this problem with the method used in Sample Prob. 15.15; however, we illustrate a vector approach, choosing position vectors r_B, r_D, and r_{D/B} as shown in Fig. 1.
\boldsymbol{\omega}_{A B}=\omega_{A B} \mathbf{k}=(20 \mathrm{rad} / \mathrm{s}) \mathbf{k} \quad \boldsymbol{\omega}_{B D}=\omega_{B D} \mathbf{k} \quad \boldsymbol{\omega}_{D E}=\omega_{D E} \mathbf{k}
where k is a unit vector pointing out of the page. We can obtain the velocity of D by relating it to point E, as
\mathbf{v}_D=\mathbf{v}_E + \mathbf{v}_{D / E}=0 + \omega_{A B} \mathbf{k} \times \mathbf{r}_D (1)
We can obtain the velocity of B by relating it to point A, as
\mathbf{v}_B=\mathbf{v}_A + \mathbf{v}_{B / A}=0 + \omega_{A B} \mathbf{k} \times \mathbf{r}_B (2)
The relationship between the velocities of D and B is
\mathbf{v}_D=\mathbf{v}_B + \mathbf{v}_{D / B} (3)
Substituting Eqs. (1) and (2) into Eq. (3) and using v_{D/B} = \omega_{BD}k \times r_{D/B} gives
\begin{aligned}\omega_{D E} \mathbf{k} \times \mathbf{r}_D &=\omega_{A B} \mathbf{k} \times \mathbf{r}_B + \omega_{B D} \mathbf{k} \times \mathbf{r}_{D / B} \\\omega_{D E} \mathbf{k} \times(-17 \mathbf{i} + 17 \mathbf{j}) &=20 \mathbf{k} \times(8 \mathbf{i} + 14 \mathbf{j})+\omega_{B D} \mathbf{k} \times(12 \mathbf{i} + 3 \mathbf{j}) \\-17 \omega_{D E} \mathbf{j} – 17 \omega_{D E} \mathbf{i} &=160 \mathbf{j} – 280 \mathbf{i} + 12 \omega_{B D} \mathbf{j} – 3 \omega_{B D} \mathbf{i}\end{aligned}
Equating the coefficients of the unit vectors i and j, the following two scalar equations are
\begin{aligned}&-17 \omega_{D E}=-280 – 3 \omega_{B D} \\&-17 \omega_{D E}=+160 + 12 \omega_{B D}\end{aligned}
Solving these gives you \boldsymbol{\omega}_{B D}=-(29.33 \mathrm{rad} / \mathrm{s}) \mathbf{k} \quad \omega_{D E}=(11.29 \mathrm{rad} / \mathrm{s}) \mathbf{k}
Accelerations. At the instant considered, crank AB has a constant angular velocity, so you have
\begin{aligned}\boldsymbol{\alpha}_{A B}=0 \quad \boldsymbol{\alpha}_{B D} &=\alpha_{B D} \mathbf{k} \quad \boldsymbol{\alpha}_{D E}=\alpha_{D E} \mathbf{k} \\\mathbf{a}_D &=\mathbf{a}_B+\mathbf{a}_{D / B}\end{aligned} (4)
Evaluate each term of Eq. (4) separately:
\begin{aligned}\operatorname{Bar} D E: \quad \mathbf{a}_D &=\alpha_{D E} \mathbf{k} \times \mathbf{r}_D – \omega_{D E}^2 \mathbf{r}_D \\&=\alpha_{D E} \mathbf{k} \times(-17 \mathbf{i} + 17 \mathbf{j}) – (11.29)^2(-17 \mathbf{i} + 17 \mathbf{j}) \\&=-17 \alpha_{D E} \mathbf{j} – 17 \alpha_{D E} \mathbf{i} + 2170 \mathbf{i} – 2170 \mathbf{j} \\\operatorname{Bar} A B: \quad \mathbf{a}_B &=\alpha_{A B} \mathbf{k} \times \mathbf{r}_B – \omega_{A B}^2 \mathbf{r}_B=0-(20)^2(8 \mathbf{i} + 14 \mathbf{j}) \\&=-3200 \mathbf{i} – 5600 \mathbf{j} \\\operatorname{Bar} B D: \quad \mathbf{a}_{D / B} &=\alpha_{B D} \mathbf{k} \times \mathbf{r}_{D / B} – \omega_{B D}^2 \mathbf{r}_{D / B} \\&=\alpha_{B D} \mathbf{k} \times(12 \mathbf{i} + 3 \mathbf{j}) – (29.33)^2(12 \mathbf{i} + 3 \mathbf{j}) \\&=12 \alpha_{B D} \mathbf{j} – 3 \alpha_{B D} \mathbf{i} – 10,320 \mathbf{i} – 2580 \mathbf{j}\end{aligned}Substituting into Eq. (4) and equating the coefficients of i and j, you obtain
\begin{aligned}-17 \alpha_{D E} + 3 \alpha_{B D} &=-15,690 \\-17 \alpha_{D E} – 12 \alpha_{B D} &=-6010\end{aligned}
Solving these gives you \boldsymbol{\alpha}_{B D}=-\left(645 \mathrm{rad} / \mathrm{s}^2\right) \mathbf{k} \quad \boldsymbol{\alpha}_{D E}=\left(809 \mathrm{rad} / \mathrm{s}^2\right) \mathbf{k}
REFLECT and THINK: The vector approach is preferred when there are more than two linkages. It is a very methodic approach and is easier to program when simulating mechanism movement over time.