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)
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.
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.
The Torsion Formula If the shaft has a solid circular cross section, If a shaft has a tubular cross section,
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
Solution: Using the torsion formula to determine the maximum stress in the shaft, we have Substituting this into Eq. 1 yields
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
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
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 τ allow =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.
Angle of Twist Integrating over the entire length L of the shaft, we have Φ = angle of twist T(x) = internal torque J(x) = shaft s polar moment of inertia G = shear modulus of elasticity for the material Assume material is homogeneous, G is constant, thus Sign convention is determined by right hand rule,
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,
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
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,
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 T A = -345 Nm and T B = 645 Nm.
Solid Noncircular Shafts The maximum shear stress and the angle of twist for solid noncircular shafts are tabulated as below:
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 τ allow = 56 MPa and the angle of twist at its end is restricted to Φ allow = 0.02 rad. How much torque can be applied to a shaft of circular cross section made from the same amount of material? G al = 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.
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.
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 A m = mean area enclosed boundary
Example 5.14 Calculate the average shear stress in a thin-walled tube having a circular cross section of mean radius r m 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,
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 G al = 26 GPa. Solution: By inspection, the internal resultant torque is T = 85 Nm. The shaded area is For average shear stress,
Solution: For angle of twist, Integral represents the length around the centreline boundary of the tube, thus
Stress Concentration Torsional stress concentration factor, K, is used to simplify complex stress analysis. The maximum shear stress is then determined from the equation
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
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.
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