Question 5 of 7: Shaft Alignment and Engine Balancing

Question 5: Shaft Alignment and Engine Balancing (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.

Part (a) – Propeller shaft alignment. The standard shop/dockside method is sag-and-gap (rim-and-face) alignment, applied progressively bearing-by-bearing while the shaft line is built up from the engine or gearbox coupling aft to the stern tube. With the shaft coupling halves brought close but not bolted, a dial gauge measures (i) the gap between the two coupling faces at four positions (top, bottom, port, starboard) as the shaft is rotated together, and (ii) the sag (rim offset) between the two coupling rims at the same four positions. Equal gap readings all round mean the two shaft axes intersect with no angular misalignment; equal rim readings mean no parallel offset. The bearing supporting the misaligned end is then packed up or down (shimmed) and the readings repeated until both gap and sag fall within the class-society tolerance (typically a few hundredths of a millimetre per bearing), after which the coupling is match-bolted. On larger, longer shaft lines this is increasingly supplemented or replaced by optical or laser alignment (a laser transmitter and target carried progressively along the shaft centreline) and by strain-gauging the coupling bolts after final bolt-up to confirm the as-built bending stress at each coupling is within limits – useful because sag-and-gap alone does not directly verify the bearing loads once the hull has deflected under the ship's own weight and the engine's operating temperature.

Part (b) – Given. An in-line 8-cylinder engine, cylinder spacing 180 mm, crank angles $0^{\circ},270^{\circ},90^{\circ},180^{\circ},135^{\circ},225^{\circ},45^{\circ},315^{\circ}$ for cylinders 1–8, crank radius $r=120$ mm, connecting-rod length $l=500$ mm, speed 1500 rpm, reciprocating weight 150 N/cylinder. Evaluated at $\theta=0$ (cylinder 1's piston at TDC).

Find. The resultant primary and secondary unbalanced reciprocating force and moment for the whole engine, at the instant cylinder 1 is at TDC.

Approach. Compute the per-cylinder primary and secondary force amplitudes from $m,r,\omega$ and the rod ratio, then vector-sum the eight cylinders' contributions (each phased by its crank angle) for the resultant force, and sum the same terms weighted by axial position $z_i$ for the resultant moment about cylinder 1's plane.

1. Kinematics and per-cylinder amplitudes. $\omega=2\pi(1500)/60=157.08$ rad/s; $m=W_{rec}/g=150/9.81=15.29$ kg. Primary force amplitude $F_p=mr\omega^2=15.29(0.120)(157.08)^2=45{,}273$ N; secondary force amplitude $F_s=mr\omega^2(r/l)=45{,}273(0.120/0.500)=10{,}866$ N.
2. Resultant primary force. Cylinder $i$ contributes $F_p\cos(\theta+\phi_i)$ along the cylinder axis; at $\theta=0$, summing $\cos\phi_i$ over the eight given angles gives $\sum\cos\phi_i\approx0$ (exactly zero to machine precision) — the crank star is symmetric enough that the primary forces cancel completely: $$\boxed{\Sigma F_{p} \approx 0\ \text{(fully balanced)}}$$
3. Resultant secondary force. Cylinder $i$ contributes $F_s\cos(2(\theta+\phi_i))$; summing $\cos(2\phi_i)$ over the eight angles also gives zero: $$\boxed{\Sigma F_{s} \approx 0\ \text{(fully balanced)}}$$
4. Resultant primary moment. With $z_i=(i-1)(0.180)$ m measured from cylinder 1, the moment has components $\sum z_i\cos\phi_i=-1398.2$ and $\sum z_i\sin\phi_i=-3375.5$ (units N·m once multiplied by $F_p$); the resultant magnitude is $$M_p=F_p\sqrt{\left(\textstyle\sum z_i\cos\phi_i\right)^2+\left(\textstyle\sum z_i\sin\phi_i\right)^2}=\boxed{3.65\ \text{kN}\!\cdot\!\text{m}}$$ acting in a plane at $247.5^{\circ}$ from the cylinder-1 reference direction (i.e. NOT cancelled, even though the force itself is balanced).
5. Resultant secondary moment. The same weighted sum using $\cos(2\phi_i),\sin(2\phi_i)$ comes out at essentially zero: $$\boxed{M_s \approx 0\ \text{(fully balanced)}}$$

This crank arrangement is a textbook example of a "perfectly" force-balanced 8-cylinder in-line layout – both the primary and secondary reciprocating forces cancel exactly, and even the secondary moment cancels, but a genuine primary moment of 3.65 kN·m remains unbalanced. In practice this residual couple is absorbed by the crankshaft's own torsional/bending stiffness and the engine mounts; it cannot be removed by reciprocating balance alone and would require a balance shaft if it proved troublesome.