Torsional Vibration

1. Torsional vibrations

2. Linear Vibrations Vs Torsional vibrations • Frequency of torsional vibrations • Where, q = Torsional Stiffness (N-m) (Torque to rotate the spring by 1 rad) I = Mass moment of inertia (kg-m2) (Resistance to rotary motion) • Frequency of linear vibrations • Where, s = spring Stiffness (N/m) (Force to deflect the spring by 1 m) m = mass (kg) (Resistance to linear motion)

3. Torsional stiffness • Torsional Stiffness Torque per unit angle(rad) q = Where, T = Torque (N-m) q = angular displacement (rad) C or G = Rigidity modulus (N/m2) J = Polar moment of inertia (m4) • Stiffness Force per unit length s = W/d Where, W = Load (N) d = Static deflection (m)

4. Given • d = 50 mm = 0.05 m ; • m = 500 kg ; • k = 0.5m; • C = 80 GN/m2 • = 80 × 109 N/m2 • Polar moment of inertia A flywheel is mounted on a vertical shaft as shown in figure. The both ends of a shaft are fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg and itsradius of gyration is 0.5 m. Find the natural frequency of torsional vibrations, if the modulus of rigidity for the shaft material is 80 GN/m2.

5. Torsional stiffness of length l1 • Torsional stiffness of length l2 Total torsional stiffness: q = q1 + q2 = 56 x 103 + 84 x 103 = 140 x 103 N-m

6. Mass Moment of inertia of shaft • Natural frequency of torsional vibrations f n = 5.35 Hz

7. Two Rotor system • In a two Rotor system, a shaft held in bearings carries a rotor at each end, it can vibrate torsionally such that the two rotors move in the opposite directions. • Thus some length of the shaft is twisted in one direction while the rest is twisted in the other. • The section which does not undergo any twist is called the node.

8. Two Rotor system • In a two rotor system,the frequency of both the parts of the shaft are same. And so,

9. Three rotor system • Consider , the two rotors A and B are fixed to the ends of the shaft. and the rotor C is in between. Let the rotors A !nd B rotate in the same direction and C in the opposite direction and the node points lie at D and E as shown in

10. Three rotor system • When A and C are rotating in same direction and B is rotating in opposite direction, node is formed in between C and B

11. Three rotor system • When B and C are rotating in same direction and A is rotating in opposite direction, node is formed in between C and A

12. Torsionally equivalent shaft • To find natural frequency of rotors attached to shaft with varying diameter, it is converted into an equivalent shaft of a nominal diameter. • A torsionally equivalent shaft is one which has the same torsional stiffness as that of the stepped shaft We know that T/q = GJ/l => q = Tl/GJ

13. Torsionally equivalent shaft • Total twist in shaft =>

14. Torsionally equivalent shaft • If the diameter of the first section of the stepped shaft ‘d1’ is taken as the diameter of the ‘torsionally equivalent shaft’ , d, then Put d = d1

15. 1. A steel shaft 1.5 m long is 95 mm in diameter for the first 0.6 m of its length, 60 mm in diameter for the next 0.5 m of the length and 50 mm in diameter for the remaining 0.4m of its length. The shaft carries two flywheels at. two ends, the first having a mass of 900 kg and 0.85 m radius of gyration located at the 95 mm diameter end and the second having a mass of 700 kg and 0.55 m radius of gyration Located at the other end. Determine the location of the node and the natural frequency of free torsional vibration of the system. The modulus af rigidity of shaft material may be taken as 80 GN/m2. Given: L = 1.5 m; d1 = 95 mm= 0.095 m; l1 = 0.6 m; d2 = 60 mm= 0.06 m. l2 = 0.5 m ; d3 = 50 mm= 0.05 m ; 13 = 0.4 m ; mA = 900 kg; kA = 0.85 m; mb = 700 kg; kB= 0.5 m C = 80 GN/m2 = 80 x 109 N/m2

16. Length of torsionally equivalent shaft of diameter d1 • Location of node Mass moment of inertia of A Mass Moment of inertia of B

17. Location of node • Position of node on original shaft

18. Natural frequency of torsional vibrations • Polar Moment of inertia • Natural frequency fn = 3.35 Hz

19. Torsional shaft to Stepped shaft 0.85m Stepped shaft to Torsional shaft 2.2m S.S T.S T.S S.S 2.2/(0.095)4 = 0.6/(0.095)4 +l2/(0.06)4 27,010.2286 = 7,366.43 +l2/(0.06)4 19,643.8 x (0.06)4 = l2 l2= 0.255 m Total distance from A in stepped shaft 0.6+0.255 = 0.855 m la/(0.095)4 = 0.6/(0.095)4 +0.25/(0.06)4 la= (7,366.43 + 19,290.1235) x (0.095)4 la = 2.17 m

20. 2. A steel shaft ABCD 1.5 m long has flywheel at its ends A and D. The mass ofthe flywheel A is 600 kg and has a radius of gyration of 0.6 m. The mass of the flywheel D is 800 kg and has a radius of gyration of 0.9 m. The connecting shaft has a diameter of 50 mm for the Portion AB which is 0.4 m long ; and has a diameter of 60 mm for the portion BC which is 0.5 m long ; and has a diameter of d mm for the portion CD which is 0.6 m long. Detennine : • the diameter 'd' of the portion CD so that the node of the torsional vibration of the system will be at the centre of the length BC; and • The natural frequency of the torsional vibrations. The modulus of rigidity for the shaft material is 80 GN/m2 Given : L = 1.5 m ; mA = 600 kg ; kA = 0.6 m ; mD= 800 kg ; kD = 0.9 m ; d1 = 50 mm = 0.05 m ; L1 = 0.4 m ; d2 = 60 mm = 0.06 m ; l2= 0.5 m ; d3 = d ; 13 = 0.6 m ; C = 80 GN/m2 = 80 x109 N/m2

21. Length of equivalent shaft • Mass moment of inertia of Flywheel A • Mass moment of inertia of Flywheel A

22. Location of node But, we know that length of torsionally equivalent shaft is, l = lA + lD = 0.52 + 0.173 = 0.693m

23. The diameter of shaft CD 0.693 d = 0.0917 m = 91.7 mm • Natural frequency of shaft • Polar moment of inertia • Natural frequency fnA

24. 40 mm 30 mm 50 mm 0.25 m 0.6 m 0.4 m