Question

The simply supported shaft shown in the figure with a 600 mm span between bearings, is carrying three gears. If the first critical speed of the shaft equals 883 rad/sec, apply Rayleigh's equation to choose a suitable material and diameter for the shaft. 12 N 10 N 12 N +100 mm -200 mm -200 mm + - 100 mm

Answer 1

Solution:

Given that:

12N 10N 12N 100 mm 200 mm 200 mm 100 mm 8 82 83 R1 R2

To find the reaction forces, taking moments about left end.

R2 = 17N

Balancing vertical forces,

R1 = 171

The  Rayleigh equation for the 1st critical speed in rpm is

$N=\sqrt{\frac{g\Sigma w_{i}\delta _{i}}{\Sigma w_{i}\delta _{i}^{2}}}\left ( \frac{60}{2\pi} \right )$

where $\delta _{i}$ & are deflections at the loads $w_{i}$

Superposition rule is applied to get individual deflections at each loading point. Each loading point will have three deflections due to each of the three loads .The total deflection is to be considered:

$\delta _{1}=-\frac{(12) (0.5) (0.1) (0.6^2 -0.5^2 -0.1^2)}{ (6) (0.6) ) EI}-\frac{ (10) (0.3) (0.1) (0.6^2 -0.3^2 -0.1^2)}{ (6) (0.6) EI}-\frac{(12) (0.5) (0.1) (0.6^2 -0.5^2 -0.1^2)}{ (6) (0.6) ) EI}=-\frac{ 0.04967}{EI}$

Similarly,

$\delta _{2}=-\frac{(12) (0.1) (0.3) (0.6^2 -0.1^2 -0.3^2)}{ (6) (0.6) ) EI}-\frac{ (10) (0.3) (0.3) (0.6^2 -0.3^2 -0.3^2)}{ (6) (0.6) EI}-\frac{(12) (0.1) (0.3) (0.6^2 -0.^2 -0.3^2)}{ (6) (0.6) ) EI}=-\frac{ 0.097}{EI}$

By symmetry,

$\delta _{1}=\delta _{3}=-\frac{0.04967}{EI}$

Applying Rayleigh's equation,

$N=\left ( \frac{60}{2\pi} \right )\sqrt{(9.81)EI\frac{[(12)(0.4967)+(10)(0.097)+(12)(0.04967)]}{[(12)(0.4967)^{2}+(10)(0.097)^{2}+(12)(0.04967)^{2}]}}=112.32\sqrt{EI}=883\,rad/s$

$EI=61.803\,Nm^{2}$

For a shaft, $I=\frac{\pi d^{4}}{32}$ A suitable material for the shaft would be AISI 4140. Then E = 200 GPa.

$(200\times 10 ^{9}\left ( \frac{\pi d^{4}}{32} \right )= 61.803 Nm^2$

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