Question 2 of 6: Stepped shaft — maximum deflection & critical speed
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-Mec-B1 Advanced Machine Design, May 2017. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; only three of the four Part II problems (Problems 3–6) are required. All six problems are solved as a study resource. “State all assumptions clearly… assume any missing data and properly state it” is an explicit exam instruction, invoked in Problem 2.
[Figure not reproduced: Figure 2.1 — Stepped simply-supported shaft: bearings at A and F (span 800 mm); $P_1=10$ kN at 100 mm from A; $P_2=20$ kN at 125 mm from C. Diameters are not printed on the exam figure and are assumed. See the official exam paper.]
Check (assumed data): The exam figure gives no diameters, only that segments AB, CD and EF are the thick sections and BC, DE the thin sections. Following the exam’s “assume any missing data” instruction and the diameters used on the companion B1 stepped-shaft problems, we take the thick sections as $d=125$ mm and the thin sections as $d=100$ mm, steel $E=200$ GPa. The method and the safety conclusion are insensitive to the exact diameters; only the numerical deflection and speed scale with them.
Given. Simply-supported stepped steel shaft, span $L=800$ mm (bearings A, F). Loads $P_1=10$ kN at $x=100$ mm, $P_2=20$ kN at $x=425$ mm. Assumed diameters: thick $d_1=125$ mm (AB, CD, EF), thin $d_2=100$ mm (BC, DE); $E=200$ GPa. Operating speed 1000 rpm.
Given data
Quantity
Symbol
Value
Span between bearings
$L$
800 mm
Load 1 (at 100 mm from A)
$P_1$
10 kN
Load 2 (at 425 mm from A)
$P_2$
20 kN
Thick-section diameter (assumed)
$d_1$
125 mm
Thin-section diameter (assumed)
$d_2$
100 mm
Young’s modulus (steel)
$E$
200 GPa
Find. (1) the maximum lateral deflection and where it occurs; (2) the fundamental (whirling) critical speed; and a judgement on whether running at 1000 rpm is safe.
Approach. Get the bearing reactions from statics, build the bending-moment field $M(x)$, then integrate $M/EI$ twice with the diameter step carried inside $I(x)$ to obtain the elastic curve (BCs $y=0$ at both bearings); the extreme value gives the maximum deflection. Feed the static deflections at the load stations into the Rayleigh–Ritz formula for the first critical speed, and compare 1000 rpm to it.
Bearing reactions. Moments about A: $R_F L = P_1(100)+P_2(425)$, so $$R_F=\frac{10(100)+20(425)}{800}=11.875\text{ kN},\qquad R_A=P_1+P_2-R_F=18.125\text{ kN}.$$
Bending-moment field. With singularity functions, $$M(x)=R_A\,x-P_1\langle x-100\rangle-P_2\langle x-425\rangle\quad(\text{N, mm}).$$ The moment peaks under $P_2$: $M_{max}=18.125(425)-10(325)=\boxed{4.45\text{ kN}\cdot\text{m}}$.
Section stiffness. $I_1=\dfrac{\pi d_1^{4}}{64}=1.198\times10^{7}\ \text{mm}^4$ (thick), $I_2=\dfrac{\pi d_2^{4}}{64}=4.909\times10^{6}\ \text{mm}^4$ (thin). $I(x)$ switches at each station: AB thick, BC thin, CD thick, DE thin, EF thick.
Elastic curve (double integration). Integrating $y''=M(x)/[E\,I(x)]$ numerically with $y(0)=y(800)=0$ and the piecewise $I(x)$ (continuity of slope and deflection enforced at every step) gives the downward elastic curve. The extreme deflection is $$\boxed{y_{max}=0.159\text{ mm at }x\approx411\text{ mm}},$$ i.e. just to the A-side of the heavier load $P_2$, as expected for a shaft that is stiffer over CD but carries most of the load there.
Deflection at the load stations. From the same curve, $y_1=y(100)=0.0606$ mm under $P_1$ and $y_2=y(425)=0.1587$ mm under $P_2$. These station deflections drive the critical-speed estimate.
Fundamental critical speed (Rayleigh–Ritz). Treating the loads as lumped weights $W_i=P_i$ deflecting their own static amount, $$\omega_{cr}=\sqrt{\frac{g\sum W_i y_i}{\sum W_i y_i^{2}}},\qquad N_{cr}=\frac{60\,\omega_{cr}}{2\pi}.$$ With $g=9810\ \text{mm/s}^2$, $\sum W_i y_i=10(0.0606)+20(0.1587)=3.78$ and $\sum W_i y_i^{2}=10(0.0606)^2+20(0.1587)^2=0.540$, $$N_{cr}=\boxed{2501\text{ rpm}}.$$
Safety at 1000 rpm. The operating-to-critical ratio is $$\frac{N}{N_{cr}}=\frac{1000}{2501}=0.40.$$ This is well below the usual whirl guard band (keep $N\le 0.75\,N_{cr}$, or $N\ge 1.4\,N_{cr}$ for super-critical running). Running at 1000 rpm is therefore safe: the shaft operates comfortably in the sub-critical regime with a large margin to resonance.
