How To Solve A Problem Of Torque And Shear Stress

Transcription

1 ENGINEERING COUNCI IPOMA EVE MECHANICS OF SOIS 20 TUTORIA 10 - TORSION You should judge your progress by completing the self assessment exercises. On completion of this tutorial you should be able to do the following. erive the torsion equation erive polar second moment of area. Solve problems involving torque, shear stress and angle of twist. erive the formula for the power transmitted by a shaft Relate power transmission to torsion. Outline the method of solution for rectangular cross sections. Solve problems with shafts of rectangular cross section. It is assumed that students doing this tutorial are already familiar with the concepts of second moments of area and shear stress..j.unn 1

2 1. TORSION OF SHAFTS Torsion occurs when any shaft is subjected to a torque. This is true whether the shaft is rotating (such as drive shafts on engines, motors and turbines) or stationary (such as with a bolt or screw). The torque makes the shaft twist and one end rotates relative to the other inducing shear stress on any cross section. Failure might occur due to shear alone or because the shear is accompanied by stretching or bending TORSION EQUATION The diagram shows a shaft fixed at one end and twisted at the other end due to the action of a torque T. The radius of the shaft is R and the length is. Figure 1 Imagine a horizontal radial line drawn on the end face. When the end is twisted, the line rotates through an angle. The length of the arc produced is R. Now consider a line drawn along the length of the shaft. When twisted, the line moves through an angle. The length of the arc produced is. If we assume that the two arcs are the same it follows that R = Rθ Hence by equating = R we get γ...(1a) If you refer to basic stress and strain theory, you will appreciate that is the shear strain on the outer surface of the shaft. The relationship between shear strain and shear stress is G...(1B) γ is the shear stress and G the modulus of rigidity. G is one of the elastic constants of a material. The equation is only true so long as the material remains elastic. Gθ Substituting (1A) into (1B) we get...(1c) R Since the derivation could be applied to any radius, it follows that shear stress is directly proportional to radius 'r' and is a maximum on the surface. Equation (1C) could be written as Gθ...(1) r Now let's consider how the applied torque 'T' is balanced by the internal stresses of the material..j.unn freestudy.co.uk 2

3 Consider an elementary ring of material with a shear stress acting on it at radius r. The area of the ring is da = 2 r dr The shear force acting on it tangential is df = da = 2r dr This force acts at radius r so the torque produced is dt = 2r2 dr Gθ r Gθ Since from equation (1) then dt 2π r dr The torque on the whole cross section resulting from the shear stress is The expression reduces to Gθ T 2π R 0 r dr Figure 2 R 2π r dr is called the polar second moment of area and denoted as 'J'. The Torque equation 0 Gθ T J and this is usually written as Combining (1) and (1E) we get the torsion equation 1.2 POAR SECON MOMENTS OF AREA T Gθ...(1E) J T Gθ...(1F) J r This tutorial only covers circular sections. The formula for J is found by carrying out the integration or may be found in standard tables. π For a shaft of diameter the formula is J 2 This is not to be confused with the second moment of area about a diameter, used in bending of beams (I) but it should be noted that J = 2 I. WORKE EXAMPE No.1 A shaft 50 mm diameter and 0.7 m long is subjected to a torque of 1200 Nm. Calculate the shear stress and the angle of twist. Take G = 0 GPa. Important values to use are = 0.05 m, = 0.7 m, T = 1200 Nm, G = 0 x10 Pa π π x J 1.5 x 10 m 2 2 TR 1200 x max 8.8 x 10 Pa or 8.8 MPa - J 1.5 x 10 T 1200 x 0.7 θ radian - J 0 x 10 x 1.5 x x 10 x 0.7 Alternately θ radian GR 0 x 10 x o Converting to degrees θ x J.unn freestudy.co.uk

4 1. HOOW SHAFTS Since the shear stress is small near the middle, then if there are no other stress considerations other than torsion, a hollow shaft may be used to reduce the weight. π d The formula for the polar second moment of area is J. 2 is the outside diameter and d the inside diameter. WORKE EXAMPE No.2 Repeat the previous problem but this time the shaft is hollow with an internal diameter of 0 mm. π d π x J 5.07 x 10 m 2 2 TR 1200 x max 5.17 x 10 Pa or 5.17 MPa - J 5.07 x 10 T 1200 x 0.7 θ radian - J 0 x 10 x 5.07 x x 10 x 0.7 Alternately θ radian GR 0 x 10 x o Converting to degrees θ x 1 Note that the answers are nearly the same even though there is much less material in the shaft. WORKE EXAMPE No. A shaft 0 mm diameter is made from steel and the maximum allowable shear stress for the material is 50 MPa. Calculate the maximum torque that can be safely transmitted. Take G = 0 GPa. Important values to use are: = 0.0 m, R = 0.02 m, = 50 x10 Pa and G = 0 x10 Pa T Gθ J r π π x J x 10 m 2 2 T Gθ The complete torsion equation is Rearrange and ignore the middle term. J R - max J 50 x 10 x x 10 T 28. Nm R 0.02.J.unn freestudy.co.uk

5 SEF ASSESSMENT EXERCISE No.1 1. A shaft is made from tube 25 mm outer diameter and 20 mm inner diameter. The shear stress must not exceed 150 MPa. Calculate the maximum torque that should be placed on it. (Ans Nm). 2. A shaft is made of solid round bar 0 mm diameter and 0.5 m long. The shear stress must not exceed 200 MPa. Calculate the following. i. The maximum torque that should be transmitted. ii. The angle of twist which will occur. Take G = 0 GPa. (Ans. 100 Nm and.2o) 1. MECHANICA POWER TRANSMISSION BY A SHAFT In this section you will derive the formula for the power transmitted by a shaft and combine it with torsion theory. Mechanical power is defined as work done per second. Work done is defined as force times distance moved. Hence P = Fx/t where P is the Power F is the force x is distance moved. t is the time taken. Since distance moved/time taken is the velocity of the force we may write P = F v...(2a) where v is the velocity. When a force rotates at radius R it travels one circumference in the time of one revolution. Hence the distance moved in one revolution is x = 2R If the speed is N rev/second then the time of one revolution is 1/N seconds. The mechanical power is hence P = F 2R/(1/N) = 2NFR Since FR is the torque produced by the force this reduces to P = 2NT...(2B) Since 2N is the angular velocity radians/s it further reduces to P = T...(2C) Note that equations (2C) is the angular equivalent of equation (2A) and all three equations should be remembered..j.unn freestudy.co.uk 5

6 WORKE EXAMPE No. A shaft is made from tube. The ratio of the inside diameter to the outside diameter is 0.. The material must not experience a shear stress greater than 500 kpa. The shaft must transmit 1.5 MW of mechanical power at 1500 rev/min. Calculate the shaft diameters. The important quantities are P = 1.5 x 10 Watts, = 500 x 10 Pa, N = 1500 rev/min and d = 0.. N 1500 rev/min 1500/0 25 rev/s P 2 π N T hence T π J T J d π 0. π 0. R 2 P 2π N x Nm 2π x 25 hence x 500 x m 81. mm x 2 x 500 x 10 d mm SEF ASSESSMENT EXERCISE No.2 1. A shaft is made from tube 25 mm outer diameter and 20 mm inner diameter. The shear stress must not exceed 150 MPa. Calculate the maximum power that should be transmitted at 500 rev/min. (Ans kw) 2. A shaft must transmit 20 kw of power at 00 rev/min. The shear stress must not exceed 150 MPa. Calculate a suitable diameter. (Ans.27.8 mm) A steel shaft 5 m long, having a diameter of 50 mm, is to transmit power at a rotational speed of 00 rev/min. If the maximum shear stress is limited to 0MN/m2. etermine the following. (i) The maximum power that can be transmitted. (2.5 kw) (ii) The corresponding angle of twist. (8.5o) Assume the modulus of rigidity for steel is 80 GN/m2.. A hollow steel shaft with a diameter ratio of 0.75 and a length of m is required to transmit 1 MW at 120 rev/min. The maximum shear stress is not to exceed 70 MN/m2 nor is the overall angle of twist to exceed 1.75o. etermine the following. (i) The necessary outside diameter of the shaft so that both of the above limitations are satisfied. (222 mm) (ii) The actual maximum shear stress and the actual angle of twist. (1.75o) Assume the modulus of rigidity for steel is 80 GPa.J.unn freestudy.co.uk

7 2 TORSION OF NON CIRCUAR SECTIONS Shear stress cannot act in a direction normal to a free surface. It follows that at a corner the shear stress is zero. A way of visualising the shear stress over a section is found by the use of the elastic membrane analogy. Imagine a hollow section with the same dimensions covered in a thin elastic membrane. The inside is then pressurised so that the membrane expands. The shape obtained for a rectangle is like this. Figure The gradient of the membrane represents the shear stress and this is zero in the corners, zero in the centre and a maximum at the centre of the longest edge. This theory also works with other shapes such as a tee section. RECTANGUAR SECTIONS Figure Without proof the following formulae are offered based on various sources. The maximum shear stress in a rectangular section is T max and it occurs at the mid point of the longest 2 αb edge. T The angle of twist is θ is the length of the βb G shaft. Figure 5 α and β are figures that depend on the ratio of the dimensions and are given by the following formulae. 1 K α β 1.8/n 1 n = B/ Here is a table based on these formulae. n n 12n n 1.0 (square) α β Note that as the ratio n increases the values of α and β tend towards a value of 1/ and for ratios larger than 10 this figure is used..j.unn freestudy.co.uk 7

8 Some sources use a constant called the apparent polar second moment of area K which for a rectangle is 1 1 K B If B = n, K n B 12 B 1 n 12n n 1 1 When n is large K When square n = 1 and K n = B/ 1.0 (square) K Another source gives the maximum shear stress as: TB 1.8.8T m For a square B = 2 2 m which is consistent with the previous work. B T The same source gives the angle of twist as θ KG 7.1T For a square θ and this is consistent with the previous work. G WORKE EXAMPE No.5 Calculate the maximum shear stress and angle of twist for a square shaft 50 mm side and 1 m long when it transmits a torque of 000 Nm. The modulus of rigidity is 8 GPa.8T.8 x MPa 0.05 m 7.1T θ G 7.1x 000 x x 8 x 10 o 0.01radian or 2.5 WORKE EXAMPE No. Calculate the maximum shear stress and angle of twist for a rectangular shaft 0 mm x 20 mm and long when it transmits a torque of 000 Nm. The modulus of rigidity is 8 GPa 1 m K B x B 12 B 0 TB x x 0.02 m 50 MPa B T 000 x 1 o θ or 1.8 KG 12. x 10 x 8 x x 10.J.unn freestudy.co.uk 8

9 SEF ASSESSMENT EXERCISE No. 1. Calculate the maximum shear stress and angle of twist for a rectangular shaft 1 m long and section 0 mm x 0 mm when it transmits a torque of 200 Nm. The modulus of rigidity is 80 GPa (Answers 2 MPa and 0.7 o ) 2. Calculate the maximum shear stress and angle of twist for a rectangular shaft 1 m long and section 0 mm x 20 mm when it transmits a torque of 00 Nm. The modulus of rigidity is 80 GPa (Answers 0 MPa and 2. o ).J.unn freestudy.co.uk