Mechanical vibration best presentation.

In these slide we study about whirling of rotating shaft and numerical problems related to it

1. Mechanical Vibrations

2. Whirling of Rotating Shafts Group Members: Hasnain Abdullah 21-ME-24 Abdul Wahab 21-ME-52 Javad Alam 21-ME-100 Daniyal 21-ME-104 Abdul Hanan 21-ME-144 Abdul Rafay 21-ME-164

3. Whirling of rotating shaft is an undesirable phenomenon where the shaft undergoes lateral or transverse vibrations. (b) When the shaft is rotating. When the shaft is stationary. 3 Definition:

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6. Problem No. 1 Calculate the whirling speed of the shaft 20mm diameter and 0.6m long carrying a mass of 1 kg at its mid point. The density of shaft material is 40 Mg/m 3 and Young’s Modulus of shaft is 200 GPa . Assume the shaft to be freely supported. 6

7. Solution: 7 Weight of the shaft W s = = W s = 73.97 N/m Moment of Inertia I = = = 7.85 * 10 -9 m 4 Deflection for load at Midpoint is given by Deflection for uniform load is given by  

8. Solution Continued … 8   By Dunkarley’s Method       Whirling Speed of the shaft (N) N = N = 52.37 * 60 N = 3142 rpm  

9. 9 What is a critical frequency of a shaft? 8 The critical frequency of a shaft, also known as critical speed, is the rotational speed at which it begins to experience significant vibrations or resonance.

10. 10 Problem No. 2 A shaft is 30 mm diameter and 4 m long and may be regarded as simply supported. The density is 7 830 kg/m³. E = 205 GPa . Calculate the first three critical frequencies. Given d = 30 mm = 0.03m E = 205 GPa = 205 10 9 N/m 2 ρ = 7830 kg/m 3 l = 4 m   To Find First three critical frequencies = ?

11. Solution: First calculate the distributed weight w by calculating the weight of 1 m length. A A = 706.9 10 -6 m² Volume = A 1=706.9 10 6 m³ Weight = A g =7830 706.9 10 6 9.81 = 54.3 N/m   11

12. 12 Solution Continued … 2. Now calculate the Moment of Inertia. I I =39.76 x 10 -9 3. Now calculate the fundamental frequency. f f = 3.77 rev/s If the shaft took up the second mode the frequ e ncy would be 3.77 x 2² = 15.1 rev/s If the shaft took up the third mode the frequ e ncy would be 3.77 x 3² = 33.9 rev/s  

13. Problem No. 3 A shaft 50 mm diameter and 3 meters long is simply supported at the ends and carries three loads of 1000 N, 1500 N and 750 N at 1 m, 2 m and 2.5 m from the left support. The Young's modulus for shaft material is 200 GN/m2. Find the frequency of transverse vibration. 13

14. Solution: D = 50 mm = 0.05 m L = 3 m W1 = 1000 N W2 = 1500 N W3 = 750 N E = 200 * N/   14

15. Solution Continued … 15

16. Problem No. 4 A shaft 1.5 m long, supported in flexible bearings at the ends caries two wheels each of 50 kg mass. One wheel is situated at the center of the shaft and the other at a distance of 375 mm from the center towards left. The shaft is hollow of external diameter 75 mm and internal diameter 40 mm. The density of the shaft material is 7700 kg/m 3 and its modulus of elasticity is 200 GN/m 2 . Find the lowest whirling speed of the shaft, taking into account the mass o f the shaft. 16 Given Data: L = 1.5m m 1 = m 2 = 50kg d 1 = 75mm = 0.075m d 2 = 40mm = 0.04m =7700kg/m 3 E = 200GN/m 2 = 200×10 9 N/m 2  

17. Solution: 17 Moment of inertia of shaft, I = I = 1.4 ×10 -6 m 4   Since the density of shaft is 7700kg/m 3 , therefore mass of shaft per meter length,   m s = Area × length × density = =23.34kg/m S tatic deflection due to load W,  

18. Static deflection due to mass of 50kg at C,  ….(Here a = 0.375m, and b = 1.125m)   18 Similarly, static deflection due to mass of 50kg at D, ….(Here a = b = 0.75m)   Static deflection due to uniformly distributed load or mass of shaft, …(Substituting, w= m s g)   Solution Continued …

19. Solution Continued … 19 We know the frequency of transverse vibration,   Since the whirling speed of shaft (N c ) in r.p.s is equal to the frequency of transverse vibration in Hz. Therefore, N c = 32.4 r.p.s = 32.4 60 = 1944 r.p.m Ans.  

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