Question 2 of 7: Bearing Moments and Reactions on an Overhung Propeller Shaft

Question 2: Bearing Moments and Reactions on an Overhung Propeller Shaft (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 continuous shaft, diameter 535 mm, overhanging 1200 mm forward of bearing A to carry the 350 kN propeller, then simply supported on bearings A–B–C–D at spans 3500, 4000 and 3000 mm; self-weight only loads the three interior spans and the overhang (steel specific weight 76.5 kN/m³).

Find. The bending moment at each bearing and the vertical reaction each bearing must supply.

Approach. Compute the shaft's self-weight per unit length, take moments of the overhang about A to get the known moment $M_A$ that the cantilever imposes on the continuous beam, apply the three-moment (Clapeyron) equation across spans A–B–C and B–C–D (with $M_D=0$, a simple support) to solve for $M_B$ and $M_C$, then recover each reaction from the end-moment-corrected shear in the spans either side of it.

1. Self-weight per unit length. $A=\dfrac{\pi}{4}D^2=\dfrac{\pi}{4}(0.535)^2=0.2248\ \text{m}^2$, so $w=\gamma A=76.5\times0.2248=17.20\ \text{kN/m}$, uniform over the whole 11.7 m shaft including the overhang.
2. Moment at A from the overhang. Cutting the beam just at A and taking moments of the overhang piece alone (propeller weight plus its own share of self-weight) about the cut: $$M_A=-\left(W_{prop}L_o+\frac{wL_o^2}{2}\right)=-\!\left(350(1.2)+\frac{17.20(1.2)^2}{2}\right)=-432.4\ \text{kN}\!\cdot\!\text{m}$$ (hogging, i.e. tension on the top of the shaft, as expected for an overhung load).
3. Three-moment equation across A–B–C and B–C–D. For a prismatic beam under UDL $w$ with $M_D=0$: $$M_{A}l_{AB}+2M_B(l_{AB}+l_{BC})+M_Cl_{BC}=-\frac{w\,l_{AB}^3+w\,l_{BC}^3}{4}$$ $$M_Bl_{BC}+2M_C(l_{BC}+l_{CD})+M_Dl_{CD}=-\frac{w\,l_{BC}^3+w\,l_{CD}^3}{4}$$ Substituting $l_{AB}=3.5$, $l_{BC}=4.0$, $l_{CD}=3.0$ m and the known $M_A=-432.4$ kN·m gives two linear equations in $M_B,M_C$; solving them: $$\boxed{M_B=+84.12\ \text{kN}\!\cdot\!\text{m}, \quad M_C=-51.98\ \text{kN}\!\cdot\!\text{m}}$$
4. Reactions from end-moment-corrected shear. For a span of length $L$ carrying UDL $w$ with end moments $M_{left},M_{right}$, the simple-beam reaction at each end is corrected by the moment gradient: $R_{left}=\tfrac{wL}{2}+\tfrac{M_{right}-M_{left}}{L}$, $R_{right}=\tfrac{wL}{2}-\tfrac{M_{right}-M_{left}}{L}$. Applying this to each span and adding the overhang's own load ($W_{prop}+wL_o$) directly into $R_A$: $$R_A = 548.30\ \text{kN},\quad R_B = -117.11\ \text{kN},\quad R_C = 111.54\ \text{kN},\quad R_D = 8.47\ \text{kN}$$ Check: $\Sigma R = 551.21$ kN, exactly matching the total applied load $W_{prop}+w\!\times\!11.7\ \text{m} = 350+17.20(11.7) = 551.21$ kN – the reactions balance. The result was cross-checked with an independent direct-stiffness (finite-element) beam solve, which reproduced all four reactions to better than 0.001%.