Question 1 of 6: Six-bar function generator — velocities & accelerations

Question 1: Six-bar function generator — velocities & accelerations (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. Three collinear frame pivots A₀, B₀, C₀. Input crank A₀A drives coupler AB to pin B; B₀B is a rocker that is ternary — the pin C lies on it between B₀ and B (the drawing gives B₀C + CB = B₀B, i.e. C is on the line), so links B₀B and BC are one rigid body (link 4). Coupler CD then drives rocker C₀D. Joint coordinates scaled from the 1:5 drawing (mm, x right, y up, origin at A₀):

Derived link lengths: A₀A = 65.9, AB = 202.3, B₀B = 134.0 (B₀C = 82.1), CD = 99.5, C₀D = 100.5 mm.

Find. ω and α of links AB, B₀B, CD, C₀D, and the linear accelerations of B and D.

[Figure not reproduced: Six-bar mechanism reconstructed to scale from the exam drawing (input crank in red). Link 4 (B₀B) is ternary and carries pin C. See the official exam paper.]

Approach. Treat the mechanism as two four-bars in series that share the ternary rocker (link 4): solve loop A₀ABB₀ for ω₃,ω₄ with the relative-velocity equation, carry pin C’s motion onto link 4, then solve loop B₀CDC₀ for ω₅,ω₆; repeat the identical loop closures for accelerations.

1. Velocity of crank pin A. With $\boldsymbol\omega_2=20\,\hat{\mathbf k}$ rad/s and $\mathbf r_{A/A_0}=(15.2,\,64.1)$ mm, $$\mathbf v_A=\boldsymbol\omega_2\times\mathbf r_{A/A_0}=20(-64.1,\,15.2)=(-1282,\;304)\ \text{mm/s}.$$
2. Loop 1 — solve ω_AB and ω_B0B. The pin B is shared by coupler AB (link 3, about A) and rocker B₀B (link 4, about B₀): $$\mathbf v_A+\boldsymbol\omega_3\times\mathbf r_{B/A}=\boldsymbol\omega_4\times\mathbf r_{B/B_0}.$$ With $\mathbf r_{B/A}=(97.1,-177.5)$ and $\mathbf r_{B/B_0}=(-71.4,-113.4)$ mm, the two scalar components give ω_AB = 2.41 rad/s (CCW) and ω_B₀B = 7.54 rad/s (CW). Then $\mathbf v_B=(-855,\,538)$ mm/s ($|\mathbf v_B|=1.01$ m/s).
3. Velocity of pin C (on link 4). Because C is rigidly on rocker 4, $\mathbf v_C=\boldsymbol\omega_4\times\mathbf r_{C/B_0}$ with $\mathbf r_{C/B_0}=(-43.1,-69.9)$ mm, giving $\mathbf v_C=(-527,\,325)$ mm/s.
4. Loop 2 — solve ω_CD and ω_C0D. Pin D closes coupler CD (about C) with rocker C₀D (about C₀): $$\mathbf v_C+\boldsymbol\omega_5\times\mathbf r_{D/C}=\boldsymbol\omega_6\times\mathbf r_{D/C_0},$$ with $\mathbf r_{D/C}=(97.4,-20.0)$ and $\mathbf r_{D/C_0}=(-44.9,-89.9)$ mm ⇒ ω_CD = 0.57 rad/s (CW) and ω_C₀D = 6.00 rad/s (CW); $\mathbf v_D=(-539,\,269)$ mm/s.
5. Acceleration of A. $\mathbf a_A=\boldsymbol\alpha_2\times\mathbf r_{A/A_0}-\omega_2^{2}\mathbf r_{A/A_0}=100(-64.1,15.2)-400(15.2,64.1)=(-12.5,\,-24.1)\times10^{3}$ mm/s².
6. Loop 1 accelerations ⇒ a_B. Applying $\mathbf a_A+\boldsymbol\alpha_3\times\mathbf r_{B/A}-\omega_3^{2}\mathbf r_{B/A}=\boldsymbol\alpha_4\times\mathbf r_{B/B_0}-\omega_4^{2}\mathbf r_{B/B_0}$ and separating components gives α_AB = 193 rad/s² (CCW), α_B₀B = 151 rad/s² (CCW), and $\mathbf a_B=(21.2,\,-4.35)\times10^{3}$ mm/s² ⇒ $a_B=21.7$ m/s².
7. Acceleration of C, then loop 2 ⇒ a_D. $\mathbf a_C=\boldsymbol\alpha_4\times\mathbf r_{C/B_0}-\omega_4^{2}\mathbf r_{C/B_0}$; the loop 2 acceleration closure then yields α_CD ≈ 0.7 rad/s² (CCW, negligible), α_C₀D = 127 rad/s² (CCW) and $\mathbf a_D=(13.0,\,-2.47)\times10^{3}$ mm/s² ⇒ $a_D=13.2$ m/s².