Question 2 of 6: Stepped shaft — maximum deflection and fundamental critical speed (30 marks)

Question 2: Stepped shaft — maximum deflection and fundamental critical speed (30 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 simply-supported stepped shaft in bearings at its two ends, carrying two transverse loads, with the station positions, segment lengths and the two diameters read off the figure:

Find. (1) The largest transverse deflection $y_{\max}$ and the station $x$ at which it occurs; (2) the fundamental critical (whirl) speed $N_{cr}$ of the rotating shaft, and whether a normal operating speed clears it.

[Figure not reproduced: Problem 2 — stepped shaft: station positions, the two diameters and the applied loads, as printed on page 2 of the paper. Bearings at A and F. See the official exam paper.]

Approach. Take the reactions from statics, write the bending moment as a piecewise-linear function of $x$, divide it by the locally varying flexural rigidity $EI\left(x\right)$ to get the curvature, integrate that curvature twice with the two zero-deflection boundary conditions at the bearings to obtain the deflection curve, then feed the deflections at the two load stations into the Rayleigh energy estimate for the first whirl speed.

1. Reactions from statics. Moments about A, then vertical equilibrium: $$R_F=\frac{P_1x_B+P_2x_D}{L}=\frac{10\times200+20\times500}{800}=15.0\ \text{kN},\qquad R_A=P_1+P_2-R_F=15.0\ \text{kN}$$ The two reactions happen to be equal because the load resultant of 30 kN acts at $\left(10\times200+20\times500\right)/30 = 400$ mm, exactly mid-span.
2. Bending moment along the shaft. With sagging positive and $x$ measured from bearing A, $$M\left(x\right)=R_Ax-P_1\left\langle x-200\right\rangle-P_2\left\langle x-500\right\rangle \quad\left[\text{N}\cdot\text{mm}\right]$$ where the Macaulay bracket is zero until its argument turns positive. This is piecewise linear with breaks under each load, so the ordinates at the stations are all that is needed: $M_B=3.00$, $M_C=3.50$, $M_D=4.50$, $M_E=1.50\ \text{kN}\cdot\text{m}$, giving $\boxed{M_{\max}=4.50\ \text{kN}\cdot\text{m}\ \text{at station D}}$. The peak moment therefore sits inside the large-diameter portion, which is exactly where a designer would want it.
3. Second moments of area for the two diameters. For a solid round section $I=\pi d^{4}/64$: $$I_1=\frac{\pi\left(100\right)^{4}}{64}=4.909\times10^{6}\ \text{mm}^{4},\qquad I_2=\frac{\pi\left(125\right)^{4}}{64}=1.198\times10^{7}\ \text{mm}^{4}$$ The 25 % diameter increase raises $I$ by a factor $\left(125/100\right)^{4}=2.44$, so the middle portion is nearly two-and-a-half times as stiff as the ends. Note that both the moment and the rigidity change along the shaft, so no single closed-form simply-supported formula applies — the step must be carried through the integration.
4. Curvature and successive integration. The governing relation is $$\frac{d^{2}y}{dx^{2}}=\frac{M\left(x\right)}{E\,I\left(x\right)},\qquad I\left(x\right)=\begin{cases} I_1, & 0\le x\lt 300\\ I_2, & 300\le x\lt 700\\ I_1, & 700\le x\le 800\end{cases}$$ Integrating once gives the slope and twice the deflection, each integration adding a constant. Because $M/EI$ is piecewise linear with jump discontinuities in value at $x=300$ and $x=700$ mm (the moment is continuous, the rigidity is not), the integration is done numerically by the trapezoidal rule on a fine grid — here 200 000 intervals of 0.004 mm, which is more than enough for five-figure convergence. Slope and deflection must remain continuous across each step: it is the curvature that jumps, not the shape of the shaft.
5. Applying the boundary conditions. The bearings enforce $y\left(0\right)=0$ and $y\left(800\right)=0$. Integrating from A with a trial slope of zero and then correcting linearly — legitimate because the second constant enters the solution as a straight line $C_1x$ — gives the true slope at A: $$\theta_A=-\frac{1}{L}\int_0^{L}\!\!\int_0^{x}\frac{M}{EI}\,dx\,dx = -6.966\times10^{-4}\ \text{rad}$$ Adding $\theta_Ax$ to the trial curve yields the deflection at every station.
6. Maximum deflection and its location. Scanning the corrected curve for the extremum: $$\boxed{\ y_{\max}=0.1465\ \text{mm downward, at }x=356\ \text{mm from bearing A}\ }$$ The peak lies between the two loads and about 44 mm on the A-side of station D, pulled that way because the flexible 100 mm portion sits on the A-side. Sanity check on magnitude: an equivalent uniform 100 mm shaft carrying the same loads would deflect roughly $0.24$ mm, and a uniform 125 mm shaft about $0.10$ mm, so a stepped shaft landing at $0.147$ mm is the right order. As a fraction of span this is $L/5460$, comfortably inside the customary $L/3000$ bending limit for a general machinery shaft and far inside the $0.025$ mm-per-metre slope limit at a plain bearing.
7. Station deflections needed for the whirl estimate. Reading the same curve at the load stations:

   Static deflection at each station (downward, mm)

   Station	B (200 mm)	C (300 mm)	D (500 mm)	E (700 mm)
   $y$	0.1196	0.1442	0.1296	0.0532

   Only B and D matter for the Rayleigh estimate, since those are the two stations carrying concentrated load.
8. Fundamental critical speed by the Rayleigh method. Equating the peak strain energy of the deflected shaft to the peak kinetic energy of the lumped loads and assuming the whirl mode has the same shape as the static deflection curve gives Shigley’s Eq. (7-23), $$\omega_1=\sqrt{\frac{g\sum w_iy_i}{\sum w_iy_i^{2}}}$$ in which the $w_i$ are the station loads treated as weights and the $y_i$ are their static deflections. Substituting (forces in N, deflections in mm, $g = 9810\ \text{mm/s}^2$): $$\sum w_iy_i=10\,000\left(0.1196\right)+20\,000\left(0.1296\right)=3789\ \text{N}\cdot\text{mm}$$ $$\sum w_iy_i^{2}=10\,000\left(0.1196\right)^{2}+20\,000\left(0.1296\right)^{2}=479.1\ \text{N}\cdot\text{mm}^{2}$$ $$\omega_1=\sqrt{\frac{9810\times3789}{479.1}}=278.5\ \text{rad/s}$$ Converting to rev/min, $$\boxed{\ N_{cr}=\frac{60\,\omega_1}{2\pi}=\frac{60\times278.5}{2\pi}=2660\ \text{rev/min}\ }$$ Because the two station deflections are so nearly equal, this is close to what a single-mass Dunkerley estimate on the total load would give; the Rayleigh figure is the upper bound, and the true first whirl speed will be slightly lower once distributed shaft mass is admitted.
9. Interpreting the margin. Standard practice keeps the running speed clear of the first critical speed by at least 25 %, that is below about $0.75\,N_{cr}=1995$ rev/min for subcritical operation, or above roughly $1.4\,N_{cr}=3720$ rev/min if the machine is deliberately run supercritical through the resonance. A typical four-pole motor drive at 1000–1800 rev/min gives a speed ratio of 0.38 to 0.68 and is safely subcritical; a two-pole drive at 3600 rev/min would sit only 35 % above $N_{cr}$ and would have to be accelerated briskly through the whirl.