TORSION OBJECTIVES: This chapter starts with torsion theory in the circular cross section followed by the behaviour of torsion member. The calculation of the stress stress and the angle of twist will be also showed here. Lastly, the topics that include the members having noncircular cross sections will be discussed. The examples and exercises will be included to better understanding. 5.1 Introduction Torsion refers to the twisting of a structural member that is loaded by couples or torque. It will produce rotation about the member s longitudinal axis. Couples that produce twisting of a bar are called torques, twisting couples or twisting moments and may be represented in several ways. The example of shafts; a) Rotating Machinery; Propeller shaft, Drive shaft b) Structural Systems; Landing gear strut, Flap drive mechanism Figure.1: Example of driving shaft Mechanics of Materials 177
5. Torsion theory of circular cross section Figure. : The circular cross section Shaft connecting gears and pulleys are a common application involving torsion members. To examine the stress and deformation of a torsion member, a prismatic bar with a circular cross section that is twisted by torques, T acting at the ends will be considered as shown in Figure.. (a) Before deformation (b) After deformation Figure.3: Deformation of bar To better understand the behaviour of a torsion member, superimpose a grid on the surface of shaft and observe the deformation of the shaft as a result of the applied torque, T. Hold the left end of the bar fixed in position. When a bar or shaft of circular section is twisted by moment, its Mechanics of Materials 178
called pure tension and the deformed element shown in Figure.3 are said to be in a state of pure shear. Figure. Referring to Figure.3 and., several characteristics of torsional loading on the circular prismatic bar is observed; 1. The longitudinal axis of the shaft remains straight.. The shaft does not increase or decrease in length. 3. Radial lines remain straight and radial as the cross section rotates.. Cross sections rotate about the axis of the member. The warping does not occur in circular section. The theory of torsion with the circular section is relatively simple. 5..1 Shear stress in circular section, τ (tau) Tr J Where; τ : Shear stress in the shaft if τ max =maximum shear stress occurs at the outer surface @ tegasan ricih. (N/m ) J : Polar moment of inertia of the cross sectional area @ moment sifatekun kutub (m ) T : Applied torque acting at the cross section (Nm) r : Radius of the shaft (m) Mechanics of Materials 179
5.. Types of circular section 5..3 Angles of twist, ( phi) @ sudut piuhan Where; TL JG Ө = angle of twist (radians) T = applied torque L = length of member G = shear modulus of material/ modulus of rigidity@modulus ketegaran (N/m ) J = polar moment of inertia 5.. Power transmission Circular bars or shafts are commonly used for transmission of power. From design stand point, it is necessary for the shaft to be strong enough to transmit power safely without exceeding the shaft material's elastic limit. Power is defined as the work performed per unit of time. The work transmitted by a rotating shaft equals to the torque applied times the angle of rotation. Work = Torque x Angular Displacement Mechanics of Materials 180
Power = d/dt (Work) If torque is not a function of time, then the equation for power simply becomes: P = T ω Where; ω is the angular velocity of the shaft (rad/s) T : Applied torque acting at the cross section (Nm) P : Power (W)(1W = 1 Nm/s) For machinery, the frequency of a shaft s rotation, f is often reported. This is a measure of the revolutions number@cycles of the shaft. P = π f T Where; f : frequency (Hz @ hertz) (1 Hz = 1 cycle/s) ω: π f It is important that we use consistent units for P, T, and ω. Power is commonly specified in horsepower, HP. Angular velocity is usually given in revolutions per minute or RPM. It should then be converted to rad/sec. To do this multiply the value in RPM by π and divide by 0. EXAMPLE.1 Determine the maximum torque of a hollow circular shaft with inside diameter of 0mm and an outside diameter of 100mm without exceeding the maximum shearing stress of 70MPa. Solution; Given; d i = 0mm, τ max = 70MPa d o = 100mm J (100 3 0 ) 8.55x10 mm Mechanics of Materials 181
Tmax r max J max J Tmax r (70x10 )(8.55x10 0.05 ) 11.97kNm Remember: τ max =maximum shear stress occurs at the outer surface/radius. EXAMPLE. A hollow steel shaft has an outside diameter of 150mm and an inside diameter 100mm. The shaft is subjected to a torque of 35kNm. The modulus of rigidity for the steel is 80GPa. Determine; a) the shearing stress at the outside surface of the shaft. b) the shearing stress at the inside surface of the shaft. c) the magnitude of the angle of twist in a.5m length. Solution; J (150 3 100 ) 39.89x10 mm a) the shearing stress at the outside surface of the shaft. Tr out J 3 35x10 (0.075) 5.81MPa 39.89x10 b) the shearing stress at the inside surface of the shaft Tr in J 3 35x10 (0.05) 3.9MPa 39.89x10 c) the magnitude of the angle of twist in a.5m length. TL JG 3 35x10 (.5) 9 80x10 (39.89x10 0.07rad ) Mechanics of Materials 18
EXAMPLE.3 A 3m long hollow steel shaft has an outside diameter of 100mm and an inside diameter of 0mm. The shear modulus of steel is 80GPa. The maximum shearing stress in the shaft is 80MPa and the angular velocity is 00rpm. Determine; a) the power being transmitted by the shaft. b) the magnitude of the angle of twist in the shaft. Solution; J (100 3 0 ) 8.55x10 mm Tmax r max J max J Tmax r (80x10 )(8.55x10 0.05 ) 13.8kNm a) the power being transmitted by the shaft. From the previous notes stated that angular velocity is usually given in revolutions per minute or RPM. It should then be converted to rad/sec. To do this multiply the value in RPM by π and divide by 0. 3 T (00)(13.8x10 ) P Hp( horse power ) 8. 55kW 0 0 b) the magnitude of the angle of twist in the shaft. TL JG 3 13.8x10 (3) 0.0rad 9 80x10 (8.55x10 ) Mechanics of Materials 183
EXAMPLE. A solid circular steel shaft 1.5m long transmits 00kW at a speed of 00rpm. If the allowable shearing stress is 70MPa and the allowable angle of twist is 0.05 rad. The shear modulus of steel is 80GPa. a) the minimum permissible diameter for the shaft. b) the speed at the same power that can be delivered if the stress is not exceed 50MPa in a diameter of 75mm. Solution; Given L =1.5m P = 00kW ω = 00 rpm τ allow = 70MPa allow =.º or 0.05 rad G = 80GPa a) P T T 0 (00) 00x10 0 T. 77kNm 3 T i) Tr allow J.77x10 allow d 3 385d 70x10 0.098d 3 d ii) allow TL JG.77 d 3 3 x10 (1.5) 0.05 9 (80x10 ) 7 9.109x10 = d d 0.07mm@7. 1mm 70x 10 d 3 87. Pick d = 70.7mm b) Given d = 75mm d 0.07m@70. 7mm J d (75) 3 3 Tr J 3.11x10 mm Mechanics of Materials 18
T(0.0375) 3.11x10 50x 10 T =.15kNm P T T 0 3 3 (.15x10 ) 00x10 0 = 1rpm EXERCISE.1 A hollow circular shaft has an outside diameter D of 100mm and an inside diameter D 1 of 0mm. Given the maximum allowable shearing stress is 55 N/mm. a) Show that its polar moment of inertia, J is 8.55x10 mm. b) Calculate the maximum torque, T that can applied to the shaft c) Determine the shear stress at the inner surface of the hollow shaft EXERCISE. A tubular shaft having an inner diameter of 30mm and outer diameter of mm is to be used to transmit 90kW of power. Determine the frequency of rotation of the shaft so that the shear stress will not exceed 50MPa. [Ans: f =.Hz] Mechanics of Materials 185
5.3 Composite Bars A B C Shaft 1 Shaft To solve the problem of combined/composite shaft; a) T = T shaft 1 + T shaft +..T shaft n n = number of shaft b) 1 c) TL JG EXAMPLE.5 The composite bars with the different material is subjected to the torque is shown in figure. Determine the maximum shear stress and the position. Determine the angle of twist at C. 10kNm A B C knm m 0.3m BAR r (mm) G(Pa) AB 50 3x10 10 BC 5 8x10 10 Solution; J AB d 3 (100 ) 3 9.8x10 mm J BC d 3 (50 ) 3 0.x10 mm 10kNm A B C knm T = outward ( ) T = inward ( ) m 0.3m Or applied Right Hand Rule Mechanics of Materials 18
knm 10kNm A B C knm T A m 0.3m knm T AB = knm (T) knm 10kNm T BC = knm(c) Maximum shear stress; TABrAB x10 (50) AB 30.55N / mm J 9.8x10 AB BC TBC r J BC BC x10 (5) 0.x10 11.3N / mm The maximum shear stress occurs in the bar of BC. Angle of twist at the end of C; TL TL TL JG JG JG AB BC (x10 )(000) ( x10 )(300) 9.8x10 (3x10 ) 0.x10 (8x10 ) 0.0158rad Mechanics of Materials 187
EXERCISE.3 The horizontal shaft AD is attached to a fixed support at D and is subjected to point torques as shown in figure. A 0mm diameter hole has been drilled into the portion CD of the shaft. By assuming that the entire shaft is made of steel and G =70GN/m, determine the angle of twist at end A. 5. Torsion of non-cylindrical member Generally, we deal with axisymmetric bodies and the shear strain is linear through the entire body. However, non-circular cross-sections are not axisymmetric causing complex behaviors, which may cause bulging or warping when the shaft is twisted. Figure.5: Condition of bulging of non-circular shaft Mechanics of Materials 188
The mathematics is beyond the scope of the course, however there are empirical formulas for various shapes. Table.1: Formula for various shapes EXAMPLE. The aluminum shaft shown in figure 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, allow is = 5MPa and the angle of twist at its end is restricted to allow= 0.0 rad. Given G al = GPa. Mechanics of Materials 189
Solution; By inspection, the type of shaft is equilateral triangle. Refer to Table.1, i) 0T allow 3 a 0T 5N / mm 3 0 T 179. knmm TL ii) allow a Gal T (100mm) T (1.m) 0.0rad or 0.0rad 3 N (0 mm)(x10 ) (0.0 m)(x10 mm T. 1kNmm T. 1Nm 9 N m ) By comparison, the torque is limited due to the angle of twist. T. 1kNmm 5.5 Thin-walled Having Closed Cross Sections Thin walled of noncircular shape are often used to construct lightweight frameworks which is used in aircraft. In this section, we will analyze the effects of applying a torque to a thin walled having a closed cross section. The section does not have any breaks or slits along its length. 5.5.1 Shear flow, q. Due the applied torque, T, shear stress is developed on the front face of the element. Shear flow in a solid body is the gradient of a shear stress through the body. Shear flow is the product of the tube s thickness and the average shear stress. This value is constant at all points along the tube s cross section. As a result, the largest average shear stress on the cross section occurs where the tube s thickness is small. q t avg Mechanics of Materials 190
5.5. Average shear stress, ave In non-circular thin walled shafts for closed segments. We assume that the stress is uniformly distributed across the thickness and that we can assume an average shear stress. The average shear stress in the body is; ave T ta m where, ave - average shear stress t - the thickness of the shaft at the point of interest A m - mean area enclosed within the boundary of the centerline of the shaft thickness. T - the applied torque A m is shown in shaded area. The relationship between shear stress, τ and torque, T; df ds τ ave t h τ ave T dt h(df) Mechanics of Materials 191
h( ave ave t ave ave t t ds) t h ds d A m A m ave T ta m Since q avg section using the equation; t, we can determine the shear flow throughout the cross q T A m 5.5.3 Angle of twist, This angle can be determined by using the energy method. The angle given in radians, can be expressed as; TL A m G ds t Here the integration must be performed around the entire boundary of the tube s cross sectional area. where, t - thickness of the interior segment L - length of the section G - modulus of rigidity of the section/shear modulus Mechanics of Materials 19
EXAMPLE.7 The tube is made of bronze and has a rectangular section as shown in figure. If it is subjected to the two torques; a) determine the average shear stress in the tube at points A and B. b) determine the angle of twist at end C. Given G = 38GPa. Solution; a) 0Nm 5Nm C D E 0Nm 0Nm T CD = 0Nm (T) 0Nm 5Nm 35Nm T DE = 35Nm (T) OR The tube is subjected to the two torques at C and D. The free body diagram is shown in figure. The internal torque is 35 Nm. T = T 0 = 5 ----> to balance (0-5)=35Nm Mechanics of Materials 193
The area, A m ; A m ( 0.035)(0.057) 0.00m 57mm The shaded area = A m 35mm The average shear stress at point A with t A = 5mm; T ave tam 35 A 1. 75MPa (0.005)(0.00) The average shear stress at point B with t B = 3mm; 35 B. 9MPa (0.003)(0.00) b) The internal torques in regions DE and CD are 35Nm and 0Nm, respectively. TL ds TL ds A t t m G A CD m G DE 0(0.5) 57mm 35mm 9 (0.00 )(38x10 ) 5mm 3mm Mechanics of Materials 19
35(1.5) 57mm 35mm 9 (0.00 )(38x10 ) 5mm 3mm 30 08000.8x10.x10.8 3.33.8 3.33 3 3 3.98x10 rad 3 5.5 08000 EXAMPLE.8 A square aluminum tube has the dimensions as shown in figure. a) determine the average shear stress in the tube at point A if it is subjected to a torque of 85Nm. b) compute the angle of twist due to this loading. Given G al = GPa. 1.5m Solution; a) The area, A m ; A m ( 50)(50) 500mm 50mm The shaded area = A m avg T ta m 50mm 3 85x10 (10)(500) 1.7N / mm Mechanics of Materials 195
Since t is a constant because of the square tube, the average shear stress is the same at all points on the cross section. b) Angle of twist; TL A m G ds t 3 (85x10 )(1500) 50mm 3 (500) (x10 ) 10mm 1.9x10 (0) 3.9x10 3 rad Here, the integral represents the length around the centerline boundary of the tube. EXERCISE. If a =5mm and b = 15mm, determine the maximum shear stress in the circular and elliptical shafts when applied torque is T=80Nm. By what percentage is the shaft of circular section more efficient at withstanding the torque than the shaft of elliptical cross section? [Ans: b) τ (max)c = 3.MPa, τ (max)e = 9.05MPa, 178%] Mechanics of Materials 19
EXERCISE.5 The square shaft is used at the end of a drive cable in order to registrar the rotation of the cable on a gauge. If it has the dimensions shown and is subjected to a torque of 8Nm, determine the shear stress in the shaft at point A. Mechanics of Materials 197
TUTORIAL 5 1. a) If the shearing stress is not exceed 70 N/mm, determine the maximum torque that may be transmitted by a solid circular shaft of diameter 98.5mm and length of 1.m. Take shear modulus of steel is 80kN/mm. b) The square thin walled tube is subjected to a torque of 150Nm. Determine the average shear stress in the tube if the mean dimension a = 00mm. Each side has a thickness of t = 3mm. [Ans: a) τ ave = 1.5MPa] a. a) A hollow circular shaft has an outside diameter D of 100mm and an inside diameter D 1 of 75mm. Show that its polar moment of inertia, J is.71x10 mm. Given the maximum allowable shearing stress is 5 N/mm. Calculate; i) the maximum torque, T that can applied to the shaft. ii) the shear stress at the inner surface of the hollow shaft. b) A continuous circular shaft is made of two segments AB and BC with diameters 05mm and 105mm respectively. The shaft is fixed at the end A and subjected to anticlockwise torques of 0kNm at B and 30kNm at C as shown in figure. Take G = 80kN/mm. Find the total angle of twist at the end C. Mechanics of Materials 198
3. a) The figure shows the cross sectional of a hollow shaft. Prove the torsion equation for concentrically hollow circular shafts is given by expression; T J R max where; T = applied torque R = outer radius of the shaft R 1 = inner radius of the shaft τ max = shear stress at outer radius R J = polar moment of inertia dr = an elementary ring of a thickness at r, radius. b) A hollow steel shaft is subjected to receive a torque of 30kNm. The shaft has an outside diameter of 150mm and inside diameter of 100mm. The Modulus of Rigidity for the steel is 80GPa. Determine; i) the shear stress on the outside surface of the shaft. ii) the shear stress on the inside surface of the shaft. iii) the maximum shear stress of the shaft. iv) the twist angle for a 3.5m length of the shaft.. a) Discuss THREE (3) assumptions to be made in the derivation of torsion equation for a circular shaft. b) A steel circular bar in torsion consists of two parts, as shown in Figure Q(a). Part AB has diameter 0 mm and length 1 m, and part BC has diameter 30 mm and length 1. m. What is the allowable torque T if the angle of twist between the ends of the bar is not to exceed 0.035 radians and the shear stress is not to exceed 30 MPa? Assume Modulus of Rigidity, G steel = 80 GPa. c) A torque of 5 knm is applied to the rectangular section shown in Figure Q(b). Determine the wall thickness t so as not to exceed a shear stress of 80 MPa. Mechanics of Materials 199
0 mm 30 mm T A B C T 1 m 1. m Figure Q(a) t 50 mm 100 mm Figure Q(c) (FINAL EXAM SEM II 007/008 UTHM) 5. A hollow section as shown in Figure Q5 is subjected to a torque of 100Nm at point A. a) Sketch a thin wall tube to show elements subjected to torsion. b) Derive the equation of average shear stress, τ ave in terms of internal torque, T, the thickness of hollow section, t and the mean area enclosed within the boundary of the centre line of the thickness of the hollow section. c) Determine the average shear stress in the hollow section at point A. Mechanics of Materials 00
d) Calculate the angle of twist assuming G = 00GPa. 0 A 00 L = 3m 0 150 Figure Q5 All units in mm (FINAL EXAM SEM I 007/008 UTHM). a) Determine the torque T that can be applied to the rectangular tube if the average shear stress is no exceed 85MPa. The mean dimensions of the tube are shown and the tube has thickness of 3mm. 50mm 100mm [Ans: a) T =.55kNm] b) The solid compound rod made of three different materials carried the two torques as shown in Figure Q3. Given G aluminum = 8GPa, G bronze = 35GPa dan G steel = 83GPa. (Answer in unit N/mm ). i) Calculate the maximum shear stress in each material. ii) Find the angle of rotation of the free end of the rod. Mechanics of Materials 01
5 kn.m Aluminum Steel Bronze kn.m 100 mm 75 mm 3 m m 1.5 m Mechanics of Materials 0