Question 1 of 6: Six-bar mechanism — instant centres, mechanical advantage & output speed

Question 1: Six-bar mechanism — instant centres, mechanical advantage & output speed (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 linkage with three frame pivots. Reading the pin topology directly off the printed drawing, the chain is a Stephenson-type six-bar built on two ternary links:

• crank 2 (the “Driver”): frame pivot O₂ to pin A, driven at $\omega_2 = 100\ \text{rad/s}$ CW;
• ternary link 3 (the lower shaded triangle): pins A, C and D;
• binary rocker 4: frame pivot O₄ (the upper ground bracket) to pin D — the bar that crosses link 6 in the drawing;
• ternary link 5 (the upper shaded triangle): pin C, pin E and the free Tracer point;
• output rocker 6: frame pivot O₆ (the lower ground bracket) to pin E.

The mechanism is therefore two four-bars in cascade sharing the ternary link 3: loop 1 is O₂–A–D–O₄, and link 3 rigidly carries pin C into loop 2, C–E–O₆. Joint coordinates measured from the printed figure of the 1:5 drawing (drawing units, $x$ right, $y$ up; 1 unit = 0.423 mm at full size):

Find. (i) the $n(n-1)/2 = 15$ instant centres, (ii) the mechanical advantage $\mathrm{MA}=\omega_{in}/\omega_{out}$, and (iii) $\omega_6$ (magnitude and sense).

[Figure not reproduced: The six-bar redrawn from the measured joint coordinates. Link 3 (A–C–D) and link 5 (C–E–Tracer) are ternary; rocker 4 runs from the upper bracket O 4 to pin D and crosses rocker 6, which runs from the lower bracket O 6 to pin E. The input–o. See the official exam paper.]

Approach. Establish the pin topology from the drawing first, locate the primary instant centres by inspection (pin joints and frame pivots), complete the set with Kennedy’s three-centre theorem, then use the common velocity of I₂₆ for $\omega_6$ and the mechanical advantage; cross-check $\omega_6$ with a two-loop relative-velocity solution.

1. Fix the pin topology before anything else. The two ground brackets sit almost one above the other and the two bars leaving them cross, so it is easy to attach rocker 4 to the wrong pin. There is therefore no triple pin here: links 3 and 5 meet at C, links 3 and 4 meet at D, and $I_{45}$ must be found by Kennedy like any other secondary centre.
2. Count and classify the instant centres. With $n=6$ links, $$N = \frac{n(n-1)}{2} = \frac{6\cdot 5}{2} = 15.$$ Seven are read straight off the mechanism: the three frame pivots $I_{12}=O_2$, $I_{14}=O_4$, $I_{16}=O_6$, and the four pins $I_{23}=A$, $I_{34}=D$, $I_{35}=C$, $I_{56}=E$. The remaining eight follow from Kennedy’s three-centre theorem — for any three links the three shared centres are collinear, so each unknown centre is the intersection of two such lines.
3. The complete set of 15 centres (drawing units):

4. Output angular velocity from I₂₆. Intersecting the Kennedy lines that share links 2 and 6 puts the input–output centre at $$I_{26} = (-518,\ 317)\ \text{drawing units},$$ which does lie on the line joining $I_{12}$ and $I_{16}$, as Kennedy requires. $I_{26}$ is the one point whose velocity is the same whether it is taken on link 2 or on link 6, so $$\omega_2\,\overline{I_{12}I_{26}} = \omega_6\,\overline{I_{16}I_{26}}.$$ With $\overline{I_{12}I_{26}} = 645.9$ and $\overline{I_{16}I_{26}} = 1242.1$ units, $$\omega_6 = \omega_2\,\frac{\overline{I_{12}I_{26}}}{\overline{I_{16}I_{26}}} = 100\times\frac{645.9}{1242.1} = \boxed{52.0\ \text{rad/s}}$$ The 1:5 scale cancels in the length ratio. $I_{26}$ lies outside the segment $I_{12}I_{16}$, so links 2 and 6 turn the same way: link 6 rotates clockwise, as the input does.
5. Mechanical advantage. Neglecting friction, input power equals output power, $T_2\omega_2 = T_6\omega_6$, so $$\mathrm{MA} = \frac{T_{out}}{T_{in}} = \frac{\omega_{in}}{\omega_{out}} = \frac{\omega_2}{\omega_6} = \frac{100}{52.0} = \boxed{1.92}$$ In this position the linkage multiplies torque about 1.9:1 (the output moves 1.9× slower than the input).
6. Independent relative-velocity check. Solving loop 1 (four-bar O₂–A–D–O₄) gives $\omega_3 = 44.4\ \text{rad/s}$ CCW and $\omega_4 = 14.8\ \text{rad/s}$ CW. Carrying $\omega_3$ across the rigid link 3 gives the velocity of pin C, and loop 2 (C–E–O₆) then returns $\omega_5 = 93.6\ \text{rad/s}$ CW and $$\omega_6 = 52.0\ \text{rad/s}\ \text{CW},$$ identical to the instant-centre value — confirming both the topology and the measurements.