STRUCTURAL MECHANICS: CE203

1. STRUCTURAL MECHANICS: CE203 Chapter 5 Torsion Notes are based on Mechanics of Materials: by R. C. Hibbeler, 7th Edition, Pearson Dr B. Achour & Dr Eng. K. El-kashif Civil Engineering Department, University of Hail, KSA (Spring 2011) Chapter 5: Torsion

2. Torsional Deformation of a Circular Shaft • Torque is a moment that twists a member about its longitudinal axis. • If the angle of rotation is small, the length of the shaft and its radius will remain unchanged. Chapter 5: Torsion

3. 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. Chapter 5: Torsion

4. The Torsion Formula • If the shaft has a solid circular cross section, • If a shaft has a tubular cross section, Chapter 5: Torsion

5. Example 5.2 • The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is resisted by the material contained within the outer region of the shaft, which has an inner radius of c/2 and outer radius c. Solution: Stress in the shaft varies linearly, thus The torque on the ring (area) located within the lighter-shaded region is For the entire lighter-shaded area the torque is Chapter 5: Torsion

6. Solution: • Using the torsion formula to determine the maximum stress in the shaft, we have Substituting this into Eq. 1 yields Chapter 5: Torsion

7. Example 5.3 • 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. 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 Chapter 5: Torsion

8. Power Transmission • Power is defined as the work performed per unit of time. • For a rotating shaft with a torque, the power is • Since , the power equation is • For shaft design, the design or geometric parameter is Chapter 5: Torsion

9. Example 5.5 • A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it is attached. If the shaft rotates at w =175 rpm and the steel has an allowable shear stress of allow τallow =100 MPa, determine the required diameter of the shaft to the nearest mm. Solution: The torque on the shaft is Since As 2c = 21.84 mm, select a shaft having a diameter of 22 mm. Chapter 5: Torsion

10. Angle of Twist • Integrating over the entire length L of the shaft, we have • Assume material is homogeneous, G is constant, thus • Sign convention is determined by right hand rule, Φ = angle of twist T(x) = internal torque J(x) = shaft’s polar moment of inertia G = shear modulus of elasticity for the material Chapter 5: Torsion

11. Example 5.8 • The two solid steel shafts are coupled together using the meshed gears. Determine the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D. Each shaft has a diameter of 20 mm. Solution: From free body diagram, Angle of twist at C is Since the gears at the end of the shaft are in mesh, Chapter 5: Torsion

12. Solution: • Since the angle of twist of end A with respect to end B of shaft AB caused by the torque 45 Nm, The rotation of end A is therefore Chapter 5: Torsion

13. Example 5.10 • The tapered shaft is made of a material having a shear modulus G. Determine the angle of twist of its end B when subjected to the torque. Solution: From free body diagram, the internal torque is T. Thus, at x, For angle of twist, Chapter 5: Torsion

14. Example 5.11 • The solid steel shaft has a diameter of 20 mm. If it is subjected to the two torques, determine the reactions at the fixed supports A and B. Solution: By inspection of the free-body diagram, Since the ends of the shaft are fixed, Using the sign convention, Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm. Chapter 5: Torsion

15. Solid Noncircular Shafts • The maximum shear stress and the angle of twist for solid noncircular shafts are tabulated as below: Chapter 5: Torsion

16. Example 5.13 The 6061-T6 aluminum shaft has a cross-sectional area in the shape of an equilateral triangle. Determine the largest torque T that can be applied to the end of the shaft if the allowable shear stress is τallow = 56 MPa and the angle of twist at its end is restricted to Φallow = 0.02 rad. How much torque can be applied to a shaft of circular cross section made from the same amount of material? Gal = 26 GPa. Solution: By inspection, the resultant internal torque at any cross section along the shaft’s axis is also T. By comparison, the torque is limited due to the angle of twist. Chapter 5: Torsion

17. Solution: • For circular cross section, we have The limitations of stress and angle of twist then require Again, the angle of twist limits the applied torque. Chapter 5: Torsion

18. Thin-Walled Tubes Having Closed Cross Sections • Shear flow q is the product of the tube’s thickness and the average shear stress. • Average shear stress for thin-walled tubes is • For angle of twist, = average shear stress T = resultant internal torque at the cross section t = thickness of the tube Am = mean area enclosed boundary Chapter 5: Torsion

19. Example 5.14 • Calculate the average shear stress in a thin-walled tube having a circular cross section of mean radius rm and thickness t, which is subjected to a torque T. Also, what is the relative angle of twist if the tube has a length L? Solution: The mean area for the tube is For angle of twist, Chapter 5: Torsion

20. Example 5.16 • A square aluminum tube has the dimensions. Determine the average shear stress in the tube at point A if it is subjected to a torque of 85 Nm. Also compute the angle of twist due to this loading. Take Gal = 26 GPa. Solution: By inspection, the internal resultant torque is T = 85 Nm. The shaded area is For average shear stress, Chapter 5: Torsion

21. Solution: • For angle of twist, Integral represents the length around the centreline boundary of the tube, thus Chapter 5: Torsion

22. Stress Concentration • Torsional stress concentration factor, K, is used to simplify complex stress analysis. • The maximum shear stress is then determined from the equation Chapter 5: Torsion

23. Example 5.18 The stepped shaft is supported by bearings at A and B. Determine the maximum stress in the shaft due to the applied torques. The fillet at the junction of each shaft has a radius of r = 6 mm. Solution: By inspection, moment equilibrium about the axis of the shaft is satisfied The stress-concentration factor can be determined by the graph using the geometry, Thus, K = 1.3 and maximum shear stress is Chapter 5: Torsion

24. Inelastic Torsion • Considering the shear stress acting on an element of area dA located a distance p from the center of the shaft, • Shear–strain distribution over a radial line on a shaft is always linear. • Perfectly plastic assumes the shaft will continue to twist with no increase in torque. • It is called plastic torque. Chapter 5: Torsion

25. Example 5.20 • A solid circular shaft has a radius of 20 mm and length of 1.5 m. The material has an elastic–plastic diagram as shown. Determine the torque needed to twist the shaft Φ = 0.6 rad. Solution: The maximum shear strain occurs at the surface of the shaft, The radius of the elastic core can be obtained by Based on the shear–strain distribution, we have Chapter 5: Torsion