Propeller Shaft Diameter Design: 25-Nav-B2 Marine Engineering and Vibrations, May 2017, Q1

Question 1: Propeller Shaft Diameter Design (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 propulsion shaft transmitting 7350 kW at 95 rpm, carrying a steady axial thrust of 800 kN, with the torque fluctuating ±5% about its mean value; the shaft steel has yield strength 207 MPa, endurance limit 186 MPa and (assumed) ultimate strength 415 MPa – this last value matches ABS Table 3 category 4 (tail/stern-tube shaft, water-lubricated bearings), so the shaft is being sized as a water-lubricated tail shaft.

Given data
Symbol	Value
P	7350 kW
N	95 rpm
F_thrust	800.0 kN (steady)
Torque variation	±5%
σ_Y	207 MPa
σ_e	186 MPa
S_u (assumed)	415 MPa

Find. The minimum solid shaft diameter D that survives the combined torsional-fatigue and thrust loading.

Check – assumption on bending moment: no bearing span or shaft-line arrangement is given for this shaft, so a bending moment from the overhung propeller weight cannot be computed here (Question 2 shows exactly this calculation once bearing spacing is known). This diameter is therefore sized for combined torsion + steady thrust, checked against fatigue using the given endurance limit and yield strength; the resulting D is the theoretical minimum (factor of safety = 1 on the Soderberg line). A real class-society design would apply a service factor of roughly 1.5–2 on top of this value, and would re-check bending once the shaft-line layout (bearing spacing, overhang) is fixed.

Approach. Convert power and speed to mean torque, apply the ±5% fluctuation to get alternating torque, combine the resulting shear stress with the steady thrust's axial (direct) stress through the von Mises equivalent stress, and solve the Soderberg fatigue line for the smallest diameter that just satisfies it.

Mean and alternating torque from power. $\omega=\dfrac{2\pi N}{60}=\dfrac{2\pi(95)}{60}=9.948\ \text{rad/s}$, so $T_m=\dfrac{P}{\omega}=\dfrac{7350\times10^3}{9.948}=738.81\ \text{kN}\!\cdot\!\text{m}$. With a ±5% fluctuation, $T_a=0.05\,T_m=36.94\ \text{kN}\!\cdot\!\text{m}$.
Stresses as a function of diameter D. Torsional shear stress $\tau=\dfrac{16T}{\pi D^3}$ gives $\tau_m=\dfrac{16T_m}{\pi D^3}$, $\tau_a=\dfrac{16T_a}{\pi D^3}$; the steady thrust gives a compressive direct stress $\sigma_{ax,m}=\dfrac{4F_{thrust}}{\pi D^2}$ (mean only – the thrust is stated steady, so it has no alternating component). Bending is taken as zero per the stated assumption.
Von Mises equivalent mean and alternating stress. With only one normal stress and one shear stress present, $\sigma'_m=\sqrt{\sigma_{ax,m}^2+3\tau_m^2}$ and $\sigma'_a=\sqrt{3}\,\tau_a$ (the axial stress is steady, so it contributes to $\sigma'_m$ only).
Soderberg fatigue line, solved for D. $$\frac{\sigma'_a}{\sigma_e}+\frac{\sigma'_m}{\sigma_Y}=1$$ Both terms fall with increasing D ($\sigma'_a\propto D^{-3}$, $\sigma'_m$ dominated by the $D^{-3}$ torsion term), so the left side decreases monotonically with D; solving numerically (bisection) for the diameter where the sum equals 1 gives $$\boxed{D \approx 0.3216\ \text{m} = 321.6\ \text{mm}}$$ Checking at this diameter: $\tau_m=113.08$ MPa, $\tau_a=5.65$ MPa, $\sigma_{ax,m}=9.85$ MPa, so $\sigma'_m=196.10$ MPa and $\sigma'_a=9.79$ MPa, giving $9.79/186+196.10/207=0.0526+0.9474=1.000$ – the boundary is satisfied.

The thrust stress (9.85 MPa) is small next to the torsional shear (113 MPa) – as expected for a propulsion shaft, torque governs the design almost entirely, and the fatigue check (driven by the 5% torque swing) sets the margin against the mean-stress term rather than the other way round.

Final Results
Quantity	Value
Mean torque T_m	738.81 kN·m
Alternating torque T_a (±5%)	36.94 kN·m
Minimum shaft diameter D	321.6 mm
Soderberg utilisation at D	1.000 (boundary, FoS = 1)

Topic: Fatigue design of a rotating shaft under combined torsion and steady axial load.

Key relations: $\tau=16T/(\pi D^3)$; $\sigma_{ax}=4F/(\pi D^2)$; von Mises $\sigma'=\sqrt{\sigma^2+3\tau^2}$; Soderberg line $\sigma'_a/\sigma_e+\sigma'_m/\sigma_Y=1/N$.

Why this works: A propeller shaft never sees a perfectly constant torque – blade-passing and sea-state effects impose a cyclic component on top of the steady mean torque, so the shaft must be checked for fatigue (endurance limit) as well as static yield. The Soderberg line is the standard conservative straight-line combination of the alternating-over-endurance and mean-over-yield ratios, and because both stress terms scale with negative powers of D, the diameter that makes the sum exactly 1 is the smallest diameter that survives infinite life.

Common pitfall: Treating the thrust as contributing to the alternating stress (it is explicitly steady, so it belongs only in $\sigma'_m$), or omitting the factor of 3 on shear inside the von Mises combination.

Source: Shigley, Mechanical Engineering Design, ch. 6 (fatigue failure, Soderberg/Goodman lines) & ch. 7 (shaft design); ABS Rules for Building and Classing Steel Vessels, Pt. 4 Ch. 3 §5.1 gives the class-society empirical formula $D=100K\sqrt[3]{(H/R)(c_1/(U+c_2))}$ for the same duty, used as an industry cross-check once the shaft type factor K is selected from Table 1/2.

Question 1 of 7: Propeller Shaft Diameter Design

Question 1: Propeller Shaft Diameter Design (20 marks)