Question 6 of 10

Question 4 (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. Single-throw radial arrangement, three cylinders sharing one crank pin, cylinder axes 120° apart. $r=0.06\,\text{m}$ (crank), $l=0.6\,\text{m}$ (rod, so $r/l=0.1$), $\omega=100\,\text{rad/s}$ (ccw, constant), reciprocating mass per cylinder $m=2.5\,\text{kg}$; rotating masses are already fully balanced.

Find. (i) The net shaking-force magnitude when the crank angle coincides with each of the three cylinder axes; (ii) a counterweight scheme and the recalculated magnitudes; (iii) comment on effectiveness.

Approach. Each cylinder's piston accelerates along its own axis by the standard slider-crank approximation; because all three pistons share one crank angle $\theta$ (offset only by their 120° mounting angle $\beta_i$), sum the three inertia-force vectors (each along its own cylinder axis) and evaluate at $\theta=\beta_i$ for $i=1,2,3$.

1. Piston acceleration along cylinder $i$'s own axis (primary + secondary terms), with $\theta_i=\theta-\beta_i$, $\beta_1{=}0^\circ,\beta_2{=}120^\circ,\beta_3{=}240^\circ$: $$a_i(\theta) = r\omega^2\Big(\cos\theta_i + \tfrac{r}{l}\cos2\theta_i\Big),\qquad F_i(\theta)=m\,a_i(\theta)\ \hat e(\beta_i)$$
2. Sum the three force vectors and evaluate at $\theta=0^\circ,120^\circ,240^\circ$ (crank aligned with each cylinder axis in turn): $$\vec F(0^\circ) = (2475.0,\,0)\ \text{N},\quad \vec F(120^\circ)=(-1237.5,\,2143.4)\ \text{N},\quad \vec F(240^\circ)=(-1237.5,\,-2143.4)\ \text{N}$$ $$\boxed{|\vec F| = 2475.0\ \text{N at all three special positions}}$$ (over the whole revolution $2025\,\text{N}\le|\vec F|\le2475\,\text{N}$ — the three named positions sit at the ripple's maxima.)
3. Decompose into primary and secondary resultants. Because the three cylinders are evenly spaced ($\sum\cos^2\beta_i=\sum\sin^2\beta_i=\tfrac32$, $\sum\sin\beta_i\cos\beta_i=0$), the primary ($\cos\theta_i$) terms sum to a constant-magnitude vector that rotates in phase with the crank, not to zero: $$F_{\text{primary}} = \tfrac32\,m\,r\,\omega^2 = 1.5(2.5)(0.06)(100)^2 = 2250\ \text{N}$$ The doubled mounting angles $2\beta_i=\{0^\circ,240^\circ,120^\circ\}$ are the same evenly-spaced set, so the secondary term is also a rotating (at $2\omega$) resultant of the same closed form with $r/l$ in place of 1: $F_{\text{secondary}}=\tfrac32\,m\,r\,(r/l)\,\omega^2=225\,\text{N}$. Because these two rotating vectors happen to be in phase at $\theta=0^\circ,120^\circ,240^\circ$, they add directly there: $2250+225=2475\,\text{N}$.✓
4. Balancing scheme. Because the (dominant) primary resultant is already a pure rotating vector locked to the crank angle, it can be cancelled exactly — unlike an in-line engine's purely-reciprocating imbalance — by a single counterweight mass $M_c$ at radius $R_c$ on the crank web, diametrically opposite, sized so its centrifugal force matches $F_{\text{primary}}$: $$M_cR_c\,\omega^2 = F_{\text{primary}} = 1.5\,m\,r \;\omega^2 \;\Rightarrow\; \boxed{M_cR_c = 1.5\,m\,r = 0.225\ \text{kg}\cdot\text{m}}$$ (e.g. $M_c=1.0\,\text{kg}$ at $R_c=225\,\text{mm}$, or any product giving the same $0.225\,\text{kg}\cdot\text{m}$.)
5. Recompute the shaking force with the counterweight added at the same three crank positions: $$\vec F_{\text{bal}}(0^\circ)=(225.0,0)\ \text{N},\ \ \vec F_{\text{bal}}(120^\circ)=(-112.5,194.9)\ \text{N},\ \ \vec F_{\text{bal}}(240^\circ)=(-112.5,-194.9)\ \text{N}$$ $$\boxed{|\vec F_{\text{bal}}| = 225.0\ \text{N at all three positions} = F_{\text{secondary}}\text{ exactly}}$$

(iii) Effectiveness. Adding a $0.225\,\text{kg}\cdot\text{m}$ counterweight on the crank cuts the net shaking force from 2475 N to 225 N — a 90.9% reduction — because it exactly cancels the dominant, purely-rotating primary resultant that a single-throw radial (three cylinders sharing one crank) always produces. The residual 225 N secondary force rotates at $2\omega$ and cannot be removed by a simple crank counterweight (it would need a second counter-rotating mass train at $2\omega$, e.g. a Lanchester-type balancer); in practice the small residual is usually accepted given the tenfold reduction achieved.