Graphical velocity analysis of a six-bar mechanism: 22-Mec-A2 Kinematics and Dynamics of Machines, Undated paper, Q2

Question 2: Graphical velocity analysis of a six-bar mechanism (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 six-bar chain (from the figure): input slider at $A$ translating horizontally with $V_A=10$ m/s (to the right); a binary link 3 to $C$; link 4 pinned to the fixed pivot $D$; link 5 from $C$ to $E$; link 6 pinned to the fixed pivot $F$.

Member	Length	Orientation (figure)
Link 2 arm ($A\!\to\!B$)	1.0 m	30° from horizontal
Link 3 ($B\!\to\!C$)	2.0 m	— (closes the loop)
Link 4 ($D\!\to\!C$)	1.5 m	60° from horizontal
Link 5 ($C\!\to\!E$)	2.5 m	— (closes the loop)
Link 6 ($F\!\to\!E$)	1.0 m	45° (figure; see note)
Fixed pivots $D,F$	3.0 m apart	horizontal

Find. The angular velocities $\omega_4,\ \omega_5,\ \omega_6$.

Check — input-link and geometry reconstruction. Treating the “slider (input link)” as link 2 translating makes it a proper single-DOF six-bar ($n=6$, one prismatic + six revolutes, $M=1$); the pin $B$ therefore moves with the slider, $V_B=V_A=10\,\hat\imath$ m/s and $\omega_2=0$. The printed dimensions do not close a unique polygon (with $|DF|=3.0$ and $L_4$ at $60^\circ$, honouring $L_5=2.5$ forces link 6 to $\approx77^\circ$, not the drawn $45^\circ$). A self-consistent configuration — $D=(0,0)$, $F=(3,0)$, link 4 at $60^\circ$, closed with $L_5=2.5$, and the slider on the ground line — is used below; the method (relative-velocity loops / velocity polygon) is exact, and the reported $\omega$’s are representative of that reconstructed geometry.

Approach. Two relative-velocity loops: loop 1 ($B\!\to\!C$ via link 3, and $C$ about $D$ via link 4) gives $\omega_3,\omega_4$; carry $V_C$ into loop 2 ($C\!\to\!E$ via link 5, and $E$ about $F$ via link 6) for $\omega_5,\omega_6$.

Fig. 2. Reconstructed self-consistent configuration (schematic). Input slider $A$ drives pin $B$; loops $B\!-\!C\!-\!D$ and $C\!-\!E\!-\!F$ close on the fixed pivots $D,F$.

Input velocity. Link 2 translates, so $V_B=V_A=(10,\,0)$ m/s and $\omega_2=0$.
Loop 1 — solve $\omega_3,\omega_4$. With $\mathbf r_{BC}=C-B$ and $\mathbf r_{DC}=C-D$, the velocity of $C$ from either path must agree: $$V_C = V_B + \omega_3\,\hat k\times\mathbf r_{BC} = \omega_4\,\hat k\times\mathbf r_{DC}.$$ Solving the two scalar (x,y) equations gives $\omega_3=+2.52\ \text{rad/s}$ and $\boxed{\omega_4=-6.15\ \text{rad/s}}$ (negative $=$ clockwise), with $V_C=(7.99,\,-4.61)$ m/s, $|V_C|=9.23$ m/s.
Loop 2 — solve $\omega_5,\omega_6$. With $\mathbf r_{CE}=E-C$ and $\mathbf r_{FE}=E-F$: $$V_E = V_C + \omega_5\,\hat k\times\mathbf r_{CE} = \omega_6\,\hat k\times\mathbf r_{FE}.$$ The two scalar equations give $\boxed{\omega_5=+1.07\ \text{rad/s}}$ (CCW) and $\boxed{\omega_6=-8.57\ \text{rad/s}}$ (CW).

Graphically, the same result is read off a velocity polygon: lay off $V_B=10$ m/s; the tip of $V_C$ lies at the intersection of a line through $b$ perpendicular to $BC$ (the $\omega_3$ direction) and a line through the pole perpendicular to $DC$ (since $C$ is on link 4, $V_C\perp DC$). Then $\omega_4=V_C/|DC|$, and $\omega_5,\omega_6$ follow from a second polygon on $E$.

Quantity	Value (reconstructed geometry)
$\omega_4$ (link 4, about $D$)	6.15 rad/s CW
$\omega_5$ (link 5)	1.07 rad/s CCW
$\omega_6$ (link 6, about $F$)	8.57 rad/s CW

Topic. Planar velocity analysis of a multi-loop mechanism by the relative-velocity (velocity-polygon) method.

Key relations. $V_{P}=V_{Q}+\boldsymbol\omega\times\mathbf r_{QP}$; for a point on a link grounded at $O$, $V_P=\boldsymbol\omega\times\mathbf r_{OP}$ (i.e. $V_P\perp OP$, magnitude $\omega\,|OP|$).

Why this works. A rigid link’s velocity field is fully set by one point’s velocity and the link’s single angular velocity. Writing the closure of each independent loop as a relative-velocity equation gives exactly two scalar equations per loop — enough to solve the two unknown $\omega$’s introduced by that loop. Solving loop 1 first supplies $V_C$, which becomes the known “input” for loop 2.

Common pitfall. Forgetting that a translating slider has $\omega=0$ (its pin moves at the slide velocity), and mixing up the perpendicular directions: $\omega\times\mathbf r$ is $90^\circ$ CCW from $\mathbf r$, so a velocity component must be taken normal to the link, not along it.

Source. Norton, Design of Machinery, §6.3–6.7 (velocity of any point on a link; graphical & analytic velocity analysis).

Question 2 of 7: Graphical velocity analysis of a six-bar mechanism

Question 2: Graphical velocity analysis of a six-bar mechanism (20 marks)