Question 4 (paper Question IV): 04-BS-3, December 2014, Q4

Question 4 (paper Question IV) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A and E are fixed pivots; bar AB is horizontal (19 in). From the dimension chain, taking A as the origin: $A=(0,0)$, $B=(19,0)$, $D=(26,-24)$, $E=(11,-16)$ in. $\omega_{AB}=4\ \text{rad/s}$ CCW, constant ($\alpha_{AB}=0$).

Link lengths (from the coordinates above)
Link	Length (in)
AB	19.00
BD	25.00 (7–24–25 triangle)
DE	17.00 (8–15–17 triangle)
AE (fixed frame)	19.42

Find. $\omega_{BD}$, $\omega_{DE}$, $\alpha_{BD}$, $\alpha_{DE}$.

Figure 4 — four-bar linkage in the position shown.

Approach. Use the relative-velocity and relative-acceleration equations for rigid links, $\mathbf v_D=\mathbf v_B+\boldsymbol\omega_{BD}\times\mathbf r_{B/D}\cdots$ written twice (once through B, once about the fixed pivot E) and match components; repeat with the acceleration equation, using $\alpha_{AB}=0$.

Velocity of B. $$\mathbf v_B = \boldsymbol\omega_{AB}\times\mathbf r_{B/A} = 4\hat k\times(19,0,0) = (0,\,76,\,0)\ \text{in/s}$$
Velocity loop B→D and E→D. With $\mathbf r_{D/B}=(7,-24,0)$ and $\mathbf r_{D/E}=(15,-8,0)$: $$\mathbf v_D = \mathbf v_B+\omega_{BD}\hat k\times(7,-24,0) = (24\omega_{BD},\ 76+7\omega_{BD},\,0)$$ $$\mathbf v_D = \omega_{DE}\hat k\times(15,-8,0) = (8\omega_{DE},\ 15\omega_{DE},\,0)$$ Matching components: $24\omega_{BD}=8\omega_{DE}\Rightarrow\omega_{DE}=3\omega_{BD}$, and $76+7\omega_{BD}=15\omega_{DE}=45\omega_{BD}\Rightarrow 76=38\omega_{BD}$. $$\boxed{\omega_{BD} = 2.00\ \text{rad/s CCW}, \qquad \omega_{DE} = 6.00\ \text{rad/s CCW}}$$
Acceleration of B ($\alpha_{AB}=0$, so only centripetal remains): $$\mathbf a_B = -\omega_{AB}^2\,\mathbf r_{B/A} = -16(19,0,0) = (-304,\,0,\,0)\ \text{in/s}^2$$
Acceleration loop B→D and E→D. $$\mathbf a_D = \mathbf a_B+\alpha_{BD}\hat k\times(7,-24,0)-\omega_{BD}^2(7,-24,0) = (-332+24\alpha_{BD},\ 96+7\alpha_{BD},\,0)$$ $$\mathbf a_D = \alpha_{DE}\hat k\times(15,-8,0)-\omega_{DE}^2(15,-8,0) = (8\alpha_{DE}-540,\ 15\alpha_{DE}+288,\,0)$$ Matching components gives $3\alpha_{BD}-\alpha_{DE}=-26$ and $7\alpha_{BD}-15\alpha_{DE}=192$; solving simultaneously: $$\boxed{\alpha_{BD} = -15.32\ \text{rad/s}^2\ (15.3\ \text{rad/s}^2\ \text{CW}), \qquad \alpha_{DE} = -19.95\ \text{rad/s}^2\ (19.9\ \text{rad/s}^2\ \text{CW})}$$

Four-bar linkage kinematics
Quantity	Value	Sense
$\omega_{BD}$	2.00 rad/s	CCW
$\omega_{DE}$	6.00 rad/s	CCW
$\alpha_{BD}$	15.32 rad/s²	CW
$\alpha_{DE}$	19.95 rad/s²	CW

Topic: Relative motion of rigid links (velocity and acceleration analysis of a four-bar linkage).

Key relations: $\mathbf v_P=\mathbf v_{P_0}+\boldsymbol\omega\times\mathbf r_{P/P_0}$; $\mathbf a_P=\mathbf a_{P_0}+\boldsymbol\alpha\times\mathbf r_{P/P_0}-\omega^2\mathbf r_{P/P_0}$; for a link pivoted to a fixed point, $\mathbf v_{P_0}=\mathbf a_{P_0}=0$.

Why this works: Point D belongs to both link BD and link DE, so its velocity/acceleration computed via either chain must agree — this yields exactly two scalar equations per stage (velocity, then acceleration) for the two unknown angular rates/accelerations. The clean 7-24-25 and 8-15-17 right-triangle link lengths confirm the coordinates were read correctly from the figure.

Common pitfall: Forgetting the $-\omega^2\mathbf r$ centripetal term in the acceleration equation, or reusing $\omega_{AB}$ (rather than the just-solved $\omega_{BD}$, $\omega_{DE}$) in the corresponding centripetal terms for links BD and DE.

Source: Hibbeler, Engineering Mechanics: Dynamics, Ch. 16 (Planar Kinematics of a Rigid Body — Relative-Motion Analysis: Velocity and Acceleration).

Question 4 of 6: Question 4 (paper Question IV)

Question 4 (paper Question IV) (20 marks)