Question 3 of 6: Stepped Shaft — Maximum Deflection and Critical Speed
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 22-Mec-B1 Advanced Machine Design — National Exams, May 2015. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer any three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design (10th ed.) — deflection §4, fatigue §6, shafts §7, bolted joints §8, journal bearings §12, brakes & clutches §16; R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design; R. L. Norton, Machine Design: An Integrated Approach; R. C. Hibbeler, Mechanics of Materials.
Question 3: Stepped Shaft — Maximum Deflection and Critical Speed (20 marks)
Given. A simply-supported stepped steel shaft, $E=200$ GPa, span $800$ mm between bearings $A$ and $F$; two downward loads.
Given data
Quantity
Symbol
Value
Load at $B$ ($x=200$ mm)
$P_1$
10 kN
Load at $D$ ($x=500$ mm)
$P_2$
20 kN
End diameter ($x<300$, $x>700$ mm)
$d_1$
100 mm
Centre diameter ($300\le x\le700$ mm)
$d_2$
125 mm
Span, modulus
$L,\,E$
800 mm, 200 GPa
Find. (1) The maximum deflection and where it occurs; (2) the fundamental (first) critical rotating speed.
Stepped shaft: bearings $R_A$, $R_F$; loads $P_1=10$ kN at $B$ and $P_2=20$ kN at $D$; diameters 100 mm (outer) and 125 mm (centre).
Approach. Find the bearing reactions, build the bending-moment diagram, then integrate $M(x)/EI(x)$ twice (carrying the diameter step in $I$) with $y=0$ at both bearings to get the elastic curve. The peak of that curve is the maximum deflection; Rayleigh’s method then converts the load-station deflections into the first critical speed.
Reactions. Taking moments about $A$,
$$R_F=\frac{P_1(200)+P_2(500)}{800}=\frac{10(200)+20(500)}{800}=15\text{ kN},\qquad R_A=P_1+P_2-R_F=15\text{ kN}.$$
Bending moments. With $M(x)=R_Ax-P_1\langle x-200\rangle-P_2\langle x-500\rangle$, the moment peaks under $D$:
$$M_B=15(200)=3.0\text{ kN}\!\cdot\!\text{m},\qquad M_D=15(500)-10(300)=4.5\text{ kN}\!\cdot\!\text{m}.$$
Second moments of area. The step changes $I$ by a factor $(125/100)^4\approx2.44$:
$$I_1=\frac{\pi d_1^{4}}{64}=4.91\times10^{6}\text{ mm}^4,\qquad I_2=\frac{\pi d_2^{4}}{64}=1.198\times10^{7}\text{ mm}^4.$$
Integrate the elastic curve. With $y''=M(x)/[E\,I(x)]$ integrated numerically (trapezoidal, $I$ switching at $x=300$ and $700$ mm) and the constants fixed by $y=0$ at $A$ and $F$, the deflected shape reaches its lowest point between the loads:
$$\boxed{y_{\max}=0.152\text{ mm at }x\approx356\text{ mm from }A.}$$
The maximum lies between $B$ and $D$, pulled toward the heavier load $P_2$ and softened by the thinner $100$ mm section just past the step.
Load-station deflections for Rayleigh. Evaluating the same curve at the loads gives $y_B\approx0.110$ mm and $y_D\approx0.148$ mm (the values that carry the kinetic energy of the whirling shaft).
Fundamental critical speed (Rayleigh).
$$\omega_c=\sqrt{\frac{g\sum W_i y_i}{\sum W_i y_i^{2}}},\qquad N_c=\frac{60\,\omega_c}{2\pi}.$$
Substituting the two station loads and deflections,
$$\boxed{N_c\approx2.61\times10^{3}\text{ rpm}\ (\approx2614\text{ rpm}).}$$
Comment on safety. Any running speed should stay well clear of $N_c$; keeping the operating speed below about $0.75\,N_c$ (here $\lesssim1960$ rpm) avoids the whirl resonance.
