Stepped Shaft — Deflection & Critical Speed: 22-Mec-B1 Advanced Machine Design, May 2014, Q3

Problem 3: Stepped Shaft — Deflection & Critical Speed (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 simply-supported stepped steel shaft (800 mm between bearings) carries $P_1=10$ kN at B and $P_2=20$ kN at D; the central portion is a larger diameter than the two ends.

Given data (from the figure)
Station (from $R_A$)	Position	Feature
A	0 mm	bearing $R_A$; diameter 100 mm begins
B	200 mm	$P_1 = 10$ kN (down)
C	300 mm	step to diameter 125 mm
D	500 mm	$P_2 = 20$ kN (down)
E	700 mm	step back to diameter 100 mm
F	800 mm	bearing $R_F$

Steel modulus $E = 200$ GPa.

Find. (1) the maximum deflection and where it occurs, and (2) the fundamental critical (whirl) speed.

Stepped shaft: 100 mm ends (A–C, E–F), 125 mm centre (C–E); loads at B and D.

Approach. Get the reactions and moment diagram, integrate $M/EI$ twice with the diameter step carried explicitly in $I(x)$ (numeric), then estimate the fundamental critical speed from the load-station static deflections with the Rayleigh formula.

Reactions. $\sum M_A = 0:\ R_F(800) = 10(200)+20(500) \Rightarrow R_F = 15\ \text{kN}$, and $R_A = 30-15 = 15\ \text{kN}$.
Bending moments. $M_B = 15(200) = 3.0$, $M_C = 3.5$, and the peak $M_D = R_F(300) = 4.5\ \text{kN}\!\cdot\!\text{m}$.
Second moments of area. $I_{100} = \pi(100)^4/64 = 4.909\times10^6\ \text{mm}^4$ for the ends; $I_{125} = \pi(125)^4/64 = 1.198\times10^7\ \text{mm}^4$ for the centre — a $2.44\times$ stiffer core.
Deflection. Double-integrating $M(x)/EI(x)$ with the two steps (at C and E) and $y=0$ at both bearings gives $$\boxed{\,y_{\max} = 0.152\ \text{mm at }x \approx 356\ \text{mm}\ (\text{between B and D})\,}$$
Critical speed (Rayleigh). Using the static deflections at the load stations, $y_B$ and $y_D$, the fundamental whirl speed is $$N_{cr} = \frac{60}{2\pi}\sqrt{\frac{g\,(W_1y_B+W_2y_D)}{W_1y_B^2 + W_2y_D^2}} = 2614\ \text{rpm}$$
Assessment. Any practical running speed here (order $10^2$–$10^3$ rpm) is well below $N_{cr}$, so the shaft runs sub-critically with a comfortable whirl margin.

Problem 3 — Results
Quantity	Value
Reactions $R_A = R_F$	15 kN each
Maximum bending moment (at D)	4.5 kN·m
Maximum deflection & location	0.152 mm at $x\approx356$ mm
Fundamental critical speed	2614 rpm

Question 3 of 6: Stepped Shaft — Deflection & Critical Speed

Problem 3: Stepped Shaft — Deflection & Critical Speed (20 marks)