Question 4 of 6: Stepped Round Shaft — Maximum Deflection and Critical Speed
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 07-Mec-B1 Advanced Machine Design, December 2016. Open book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; three of the four Part II problems (3–6) are required. All six problems are solved as a complete study resource.
Given. Bearings (simple supports) at A ($x=0$) and F ($x=800$ mm); downward loads $P_1=10$ kN at $x=200$ mm (B) and $P_2=20$ kN at $x=500$ mm (D); the central portion ($300\!\le\! x\!\le\!700$ mm) is $d_2=125$ mm diameter, the end portions $d_1=100$ mm; steel $E=200$ GPa.
Given data
Station
$x$ (mm)
Feature
A
0
bearing $R_A$
B
200
$P_1=10\ \text{kN}$ down
C
300
step $d_1\!\to\!d_2$
D
500
$P_2=20\ \text{kN}$ down
E
700
step $d_2\!\to\!d_1$
F
800
bearing $R_F$
Find. (1) the maximum transverse deflection and where it occurs; (2) the fundamental (first) critical rotating speed.
Stepped simply-supported shaft: 100 mm ends, 125 mm central band, with 10 kN at B and 20 kN at D.
Approach. Get the bearing reactions from statics, build the bending-moment diagram, then integrate $M/EI$ twice with the diameter step carried explicitly in $I(x)$ (numerically) subject to zero deflection at both bearings to get $y(x)$ and its extremum. For the critical speed apply the Rayleigh energy method using the static deflections at the two load stations.
Reactions. $\sum M_A=0$: $R_F(800)=P_1(200)+P_2(500)\Rightarrow R_F=\dfrac{10(200)+20(500)}{800}=15\ \text{kN}$, and $R_A=P_1+P_2-R_F=15\ \text{kN}.$
Bending moments. $M(x)=R_A x-P_1\langle x-200\rangle-P_2\langle x-500\rangle$ (N·mm). The peak is at D: $M(500)=15\,000(500)-10\,000(300)=4.5\times10^{6}\ \text{N}\cdot\text{mm}=4.5\ \text{kN}\cdot\text{m}.$
Second moments of area. $I_1=\dfrac{\pi d_1^4}{64}=4.91\times10^{6}\ \text{mm}^4$ (ends), $I_2=\dfrac{\pi d_2^4}{64}=1.20\times10^{7}\ \text{mm}^4$ (centre).
Double integration with the step. Integrating $y''=M/(E\,I(x))$ numerically with $y(0)=y(800)=0$ gives the elastic curve. Its lowest point is
$$\boxed{\ y_{\max}=0.152\ \text{mm}\ \text{at}\ x\approx356\ \text{mm}\ }$$
i.e. between the two loads, biased toward the heavier $P_2$.
Rayleigh critical speed. Using the static deflections at the load stations ($y_B$ under $P_1$, $y_D$ under $P_2$), the first critical angular speed is
$$\omega_{cr}=\sqrt{\frac{g\,\sum W_i y_i}{\sum W_i y_i^{2}}}\quad\Rightarrow\quad N_{cr}=\frac{60\,\omega_{cr}}{2\pi}=2.61\times10^{3}\ \text{rpm}.$$
Whirl margin. A typical operating speed (order 1000 rpm) is below $0.7\,N_{cr}\approx1830$ rpm, so the shaft runs safely below its first critical speed.
