Torsion Stresses in Tubes and Rods

Torsion Stresses in Tubes and Rods This initial analysis is valid only for a restricted range of problem for which the assumptions are: Rod is initially straight. Rod twists without bending. Material is homogeneous. Material obeys Hooke s law in shear. Cross-section of the rod or tube is circular. Straight, radial lines on the cross-section remain straight and radial after twisting of the rod or tube. In these notes we go on to develop the Torsion equation which when the above assumption apply may be used for both hollow and solid shafts. The equation is given by: Where Shear stress (N/m 2 ) radius (m) torque (Nm) polar second moment of area (m 4 ) Modulus of rigidity (N/m 2 ) angle of twist (rad) length of shaft (m) Consider in the first instance a thin-walled circular tube: r T L Thin-walled tube, unwrapped showing deformation 1

Deformation: From the above diagram we see that Rearranged this gives strain.. This should be recognised from previous studies as shear Equilibrium: A torque applied to the tube will develop a shear stress in the wall of the tube. Since the tube is thin walled we can assume this stress to be constant across the section. For equilibrium the applied torque and the resisting torque must have the same value, therefore Where A is the cross sectional area of the tube. Hooke's Law: From Hooke's law we obtain the relationship Rearranging and substituting for gives Rearranging still further we obtain Rearranging the equilibrium equation and dividing both sides by r gives Where is the Polar Second Moment of Area and denoted J, since A is a small area concentrated at radius r. Combining the above equations gives the torsion equation for thin walled tubes 2

Now consider a thick-walled tube or solid rod r dr r Solid shaft At radius r a thin-walled elemental tube of thickness dr requires a shear stress through an angle. Therefore from the previous section we have to twist it Deformation: Since we are assuming that radial lines remain radial after twisting, every elemental tube will have the same angle of twist. If follow then that the above equation applies to every radius on the cross-section of the tube or rod. Equilibrium: The shear stress distribution over the cross-section must give a torque equal in magnitude to the applied torque., where Substituting for from the above equation gives The integration is taken over the complete cross-section of the wall of the thick-walled tube, and is by definition the polar second moment of area of the cross-section. We therefore have for all tubes or rods of circular cross-section, Note: For a tube and for solid circular section The above theory is insufficient to cover many practical problems and must be modified or replaced when the basic assumptions are disregarded. Examples of such problems include: 3

Variable Diameter A shaft having a sudden change in diameter, or a tapering shaft, can be dealt with by applying the theory to a short length. The torque will be the same at every section of the shaft (or will vary in a known fashion) and the total angle of twist will be the sum of the effects for each short length of shaft. (1) (2) T Variable Diameter Shaft dx x d 1 d 2 L Tapering Section which gives Applying this to the elemental section Therefore Obtain an expression for J in terms of x and integrate. Compound Cross-Section If the shaft is composed of a tube shrunk onto a solid rod (or smaller tube) of another material, the problem can be divided into two simple sub-problems to which the above theory can be applied. The total torque is the sum of the torques' acting upon the individual tubes and each will have the same angle of twist if they are firmly fixed together. Rigidly fixed 4

Plastic Torsion If the material is deformed beyond the yield point, Hooke s law no longer applies and a different stress-strain relationship must be used to develop a new theory. Worked Examples 1. A shaft 50mm diameter and 0.7m long is subjected to a torque of 1200Nm. Calculate the shear stress and the angle of twist. Take G = 90GPa. Solution Using Using Converting to degrees 2. Repeat the previous problem but this time the shaft is hollow with an internal diameter of 30mm. Solution Therefore And Converting to degrees Note that the answers are almost the same even though there is much less material in the shaft. Power Transmission by a Shaft Since one of the main uses of a shaft is to transmit power it is felt necessary to include a brief reference to this topic here. Mechanical power is defined as work done per unit of time or distance moved, and t is time., where F is force, x is However, distance moved per unit of time is velocity so that the above can be rewritten as. When a force rotates at radius R, in one revolution it will turn through a distance. 5

If the speed of the shaft is N rev/second then the time for one revolution is mechanical transmitted by a rotating shaft is given by. seconds. Hence However,, therefore we can write which can be rewritten as, where angular velocity in rad/second. Worked Example 1. A shaft is made from tube. The ratio of the inside diameter to outside diameter is 0.6. The material must not experience a shear stress greater than 500kPa. The shaft must transmit 1.5MW of mechanical power at 1500rev/min. Calculate the shaft diameter. Solution therefore Using then Therefore The above gives and Problems - Torsion of Circular Shafts 1. A solid circular steel shaft, 75 mm diameter and 2 m long is subjected to a torque of 900 Nm. Calculate (a) the maximum shear stress in the shaft, (b) the angle of twist (in degrees) over the 2 m length. Take the Modulus of Rigidity as 82.5 GN/m 2. (10.85 MN/m 2 ; 0.4 o ) 2. A hollow steel shaft of external diameter 150 mm and internal diameter 60 mm runs at 70 rev/min. If the angle of twist over a length of 4 m is limited to 2 o, calculate (a) the maximum power that the shaft may transmit, (b) the maximum shear stress, (c) the shear stress at the inner radius. Take the Modulus of Rigidity as 80 GN/m 2. (249 kw; 52.25 MN/m 2 ; 20.9 MN/m 2 ) 3. 250 kw is to be transmitted by a solid shaft running at 300 rev/min. If the torsional shear stress is limited to 60 MN/m 2 calculate the necessary shaft diameter. 6

If a hollow shaft with the same maximum shear stress is now substituted, with internal diameter = 1/2 external diameter, calculate the new diameters of the shaft. (87.6 mm; 89.6 mm; 44.8 mm) 4. A hollow steel shaft 3 m long is required to transmit 20 kw at 150 rev/min. Calculate the required outer and inner diameters if the allowable shear stress is 70 MN/m 2 and the angle of twist is limited to 5 o. 5. Take the Modulus of Rigidity as 80 GN/m 2. 20 mm dia. 25 mm dia. 12 mm dia. (60 mm; 52.2 mm) 300 mm 200 mm 150 mm For the stepped shaft shown above calculate (a) the maximum power that may be transmitted when running at 400 rev/min if the allowable shear stress is 80 MN/m 2, (b) the angle of twist of one end relative to the other when transmitting this power, (c) the length of the torsionally equivalent shaft of 25 mm diameter. Take Modulus of Rigidity as 82.5 GN/m 2. (1.14 kw; 1.85 o ; 3.76 m) 6. A hollow shaft of inside diameter 20 mm is subjected to a torque of 1 knm. Find the required outside diameter if the shear stress is limited to 80 MN/m 2. (40.8 mm) 7. A solid aluminium alloy shaft 1 m long tapers uniformly from 40 mm diameter at one end to 80 mm diameter at the other end. Calculate the angle of twist when the shaft is under the action of an axial torque of 200 Nm. Take the Modulus of Rigidity as 38 GN/m 2. (0.35 o ) 8. A hollow steel shaft with an external diameter of 150 mm is required to transmit 1 MW at 300 rev/min. Calculate a suitable internal diameter for the shaft if its shear stress is not to exceed 70 MN/m 2. Compare the torque-carrying capacity of this shaft with a solid steel shaft having the same weight per unit length and limiting shear stress. G = 80 GN/m 2. (112.3 mm; 2.35) (M of E M) 7

9. A steel shaft has to transmit 1 MW at 240 rev/min so that the maximum shear stress does not exceed 55 MN/m 2 and there is not more than 2 o twist on a length of 30 diameters. Determine the required diameter of shaft. G = 80 GN/m 2. (163.3 mm) (M of E M) 10. A hollow steel drive shaft with an outside diameter of 50 mm and an inside diameter of 40 mm is fastened to a coupling using six 8 mm diameter bolts an a pitch circle diameter of 150 mm. If the shear stress in the shaft or bolt material is not to exceed 170 MN/m 2 calculate the maximum power which the system could transmit at a rotational speed of 100 rev/min. What would be the effect of only using three bolts? (25.8 kw; 20.13 kw) (M of E M) 11. A torsional vibration damper consists of a hollow steel shaft fixed at one end and to the other end is attached a solid circular steel shaft which passes concentrically along the inside of the hollow shaft, as shown below. Determine the maximum torque T that can be applied to the free end of the solid shaft so that the angle of twist where the torque is applied does not exceed 5 o. Local effects where the two parts are connected may be ignored. Shear modulus, G = 80 GN/m 2. (5.33 knm) (M of E M) 750 mm 600 mm T 50 mm 70 mm 100 mm 8