Question 9.6: Critical Speed of a Stepped Shaft Figure 9.9(a) shows a step......
Critical Speed of a Stepped Shaft
Figure 9.9(a) shows a stepped round shaft supported by two bearings and carrying the flywheel weight W . Calculate the critical speed in rpm.
Given: The moment of inertia ( 2I ) of the shaft in its central region is twice that of the moment of inertia ( I ) in the end parts and:
W=400 N , \quad L=1 m , \quad I=0.3 \times 10^{-6} m ^4 .
Assumptions: The shaft is made of steel with E =200 GPa. The shaft weight is ignored. Bearings act as simple supports.
The application of the moment-area method (Section 4.6) to obtain the static deflection at the midspan C is illustrated in Figure 9.9. The bending moment diagram is given in Figure 9.9(b) and the M/EI diagram in Figure 9.9(c). Note that, in the latter figure, C_1 and C_2 denote the centroids of the triangular and trapezoidal areas, respectively.
The first moment of the various parts of the M/EI diagram are used to find the deflection. From the symmetry of the beam, the tangent to the deflection curve at C is horizontal. Hence, according to the second moment-area theorem defined by Equation (4.24), the deflection \delta _C is obtained by taking the moment of the M/EI area diagram between A and C about point A . That is,
t_{A B}=\int_A^B x_1 \frac{M d x}{E I}=\left[\text { area of } \frac{M}{E I} \text { diagram between } A \text { and } B\right] \bar{x}_1 (4.24)
\begin{aligned} \delta_C & =(\text { first moment of triangle })+(\text { first moment of trapezoid }) \\ & =\left(\frac{L}{6}\right)\left(\frac{W L^2}{64 E I}\right)+\left(\frac{L}{4}+\frac{5 L}{36}\right)\left(\frac{3 W L^2}{128 E I}\right)=\frac{3 W L^3}{256 E I} \end{aligned} (d)
Substituting this, m=1, \text { and } \delta_1=\delta_C into Equation (9.18), we have
n_{c r}=\frac{1}{2 \pi}\left[\frac{g\left(W_1 \delta_1+W_2 \delta_2+\cdots+W_m \delta_m\right)}{W_1 \delta_1^2+W_2 \delta_2^2+\cdots+W_m \delta_m^2}\right]^{1 / 2}=\frac{1}{2 \pi} \sqrt{\frac{g \Sigma W \delta}{\Sigma W \delta^2}} (9.18)
\begin{aligned} n_{c r} & =\frac{1}{2 \pi} \sqrt{\frac{g}{\delta_C}} \\ & =\frac{1}{2 \pi}\left[\frac{9.81(256)\left(200 \times 10^9\right)\left(0.3 \times 10^{-6}\right)}{3(400)(1)^3}\right]^{1 / 2} \\ & =56.40 cps =3384 rpm \end{aligned}