Torsion

The strength of material play and important role in any structure. The effect of Torsion is always considerable.

1. Government Engineering College, Bhavnagar. Civil Engineering Department

2. Strength Of Materials Topic:- Torsion By, Bhavik Shah – 130210106049

3. Contents Introduction Torsion Formula Assumptions Power Transmitted by shaft Torsional Rigidity

4. Introduction Torsion is a moment that twists/deforms a member about its longitudinal axis. By observation, if angle of rotation is small , length of shaft and its radius remain unchanged.

5. The Torsion Formula When material is linear-elastic, Hooke’s law applies. A linear variation in shear strain leads to a corresponding linear variation in shear stress along any radial line on the cross section.

6. The Torsion Formula If the shaft has a solid circular cross section, If a shaft has a tubular cross section,

7. Example The shaft is supported by two bearings and is subjected to three torques. Determine the shear stress developed at points A and B , located at section a – a of the shaft.

8. Solution From the free-body diagram of the left segment, The polar moment of inertia for the shaft is Since point A is at ρ = c = 75 mm, Likewise for point B , at ρ = 15 mm, we have

9. Assumptions The material is homogeneous, i.e. of uniform elastic properties throughout. The material is elastic, following Hooke's law with shear stress proportional to shear strain. The stress does not exceed the elastic limit or limit of proportionality. Circular sections remain circular.

10. Assumptions Cross-sections remain plane. (This is certainly not the case with the torsion of non- circular sections.) Cross-sections rotate as if rigid, i.e. every diameter rotates through the same angle. Practical tests carried out on circular shafts have shown that the theory developed below on the basis of these assumptions shows excellent correlation with experimental results.

11. Power Transmission Power is defined as work performed per unit of time Instantaneous power is Since shaft’s angular velocity  = d  / dt , we can also express power as Frequency f of a shaft’s rotation is often reported. It measures the number of cycles per second and since 1 cycle = 2 radians, and  = 2  f T , then power P = T ( d  / dt ) P = T  P = 2  f T

12. Power Transmission Shaft Design If power transmitted by shaft and its frequency of rotation is known, torque is determined from Eqn 5-11 Knowing T and allowable shear stress for material,  allow and applying torsion formula , For solid shaft, substitute J = ( /2) c 4 to determine c For tubular shaft, substitute J = ( /2)( c o 2  c i 2 ) to determine c o and c i

13. Example Solid steel shaft shown used to transmit 3750 W from attached motor M . Shaft rotates at  = 175 rpm and the steel  allow = 100 MPa . Determine required diameter of shaft to nearest mm.

14. Solution Torque on shaft determined from P = T , Thus, P = 3750 N · m /s Thus, P = T , T = 204.6 N · m Since 2 c = 21.84 mm , select shaft with diameter of d = 22 mm

15. Torsional Rigidity

16. Thank You For Bearing