TORSION GROUP 2.pptx

torsion

1. Fuels and Energy Engineering department Fuels and Energy Engineering Module : Engineering materials Group presentation : Topic Torsion Lecturer : Engineer L Maregedze

2. Topic : course outline objectives Identify material cross section. determine both the stress distribution within the material Determine the angle of twist. Determine stress concentrations and residual stress caused by torsional loadings

3. Definition   Torsion is the twisting of an object due to an applied torque Torsion is the moment applied in a plane containing the longitudinal axis of the beam or shaft. e.g. Shaft transmitting torque or power, I beams, Portico beams, curved beams, closed coil springs circular cross section is an efficient shape for resisting torsional loads hence commonly used to transmit power in rotating machinery.

4. Assumptions The material of the shaft is uniform throughout Circular sections remain circular even after twisting Plane sections remain plane and do not twist or warp Stresses do not exceed the proportionality limit Shaft is loaded by twisting couples in planes that are perpendicular to the axis of the shaft Twist along the shaft is uniform

5. Torsion of circular shafts In factories and workshops circular shafts are commonly used to transmit energy from one end to another Moment of couple acting on shaft is torque/turning moment/twisting moment Torque,T =turning force X diameter of shaft T= F X 2R , R-radius Units for T are N.mm or kN.mm

6. Angle of twist( θ ) When a shaft is subjected to Torque(T),point A on shaft surface comes to A’ position. The angle at the Centre ,AOA’ is the angle of twist( θ ) It is measured in radians

7. Shear stress in shafts(   Shear stress -- type of stress that acts coplanar with a given cross-section of the material When a shaft is subjected to equal and opposite end couples whose axes coincides with shaft axis, the shaft is said to be in pure torsion at any point in shaft shear stress is induced

8. Strength of shafts Strength of shaft is the maximum torque /power that a shaft can transmit from one pulley to another For solid circular shafts : Maximum Torque ,T = ,D –diameter of shaft -shear stress in the shaft For a hollow circular shafts : Maximum Torque ,T = , D-outer diameter of shaft , d-inner diameter of shaft -shear stress in the shaft  

9. Polar moment inertia :J Is the moment of inertia of plane area with respect to axis perpendicular to plane of figure For a solid circular shaft : J= For hollow circular shaft : J =  

10. Power transmitted by shaft Work done by torque = (N-m/min) T - mean torque ; N - rotational speed of shaft Power transmitted by shaft= N-m/sec N-m/min Equating Work done by torque to Power transmitted by shaft For determining diameter of shafts, maximum torque (not average torque) is taken into account because maximum shear stress developed is ensured to be in safe limits  

11. Example 1 Calculate diameter of shaft to transmit 10 KW at a speed of 15 Hz. The maximum shear stress should not exceed 60 mpa . s/n P=10 kw ; N=15 Hz =15 cycles/sec = 15 x 60 rpm = 900 rpm = 60 m ра =60 x pa T = = , = 9006.0 D = 20.80mm  

12. Example 1 External and internal diameter of a propeller shaft are 400 mm and 200 mm respectively. Find maximum shear stress developed in the cross section when a twisting moment of 50kn.M is applied. Take modulus of rigidity C=0.8×103 N/mm^2.if span of shaft is 4m,also find twisting angle of shaft. Solution D=400 mm D=200 mm T=50KN⋅m L=4 m

13. Working ​T = 50 = = 4.24M/ Now, = R = = = 200mm  

14. Example 2 Calculate the diameter of the shaft required to transmit 45kw at 120 rpm. The maximum torque is likely to exceed the mean by 30% for a maximum permissible shear stress of 55 N/mm^2. Calculate also the angle of twist for a length of 2 m. P =45Kw N=120rpm =55 N/mm2​  

15. Tmax = 1.30 Tmin =1.30 3580.98 =4655.28 N.m 465528 N.mm Tmax = D = 75.54mm Now using the relation  

16. solution = =  

17. Example 3 A shaft has to transmit 105kw power at 160rpm. If the shear stress is not to exceed 65 N/mm^2 & the twist in a length of 3.5 m must not to exceed 1 degree. Find suitable diameter. Take T=6266.72×103 N.mm P=105kw N=160rpm =65 N/mm^2 L=3500 mm G=8×104 N/mm^2​ Now,  

18. Solution T = 6266.72 1 1.) For strength T = D = 78.89mm  

19. 2.) For Stiffness = = Now, J = 15.756 = D = 112.55mm  

20. Composite shafts Shafts in series If shafts are connected in series : i) Each shaft transmit the same torque ii) The angle of twist is the sum of angel of twist of each shaft = + +… If shafts are made of same material moduli of rigidity is the same; = = = + +…)  

21. Example A composite shaft ABC is as shown in fig. A clockwise torque of 2000 Nm is applied at section B common to both the shafts. Plot stress distribution for shaft AB and for shaft BC. Take Gs =80 GPa and Gc =40 GPa . Calculate the twist of section B

22. b) Shafts in parallel If shafts are connected in parallel: i) Angle of twist is the same for each shaft ii) The applied toque is divided between shafts = = … If shafts are made of same material moduli of rigidity is the same; = = = = … ; = If torque is shared equally among two shafts ; = =  

23. A compound bar is made of steel rod 19 mm in diameter surrounded by the closely fitting brass tube of 32mm outside diameter and two are securely fixed together at ends. Calculate the value of 'G' for brass if the angle of the twist over a length of 1 m is 7.2 degree when compound bar is subjected to twisting couple of 520 Nm. Also calculate the maximum shear stress in the two material. Gs =80GPa (steel).