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 = 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