Shafts Introduction. Shafts 509

Shafts 509 C H A P T E R 14 Shafts 1. Introduction.. Material Used for Shafts.. Manufacturing of Shafts. 4. Types of Shafts. 5. Standard Sizes of Transmission Shafts. 6. Stresses in Shafts. 7. Maximum Permissible Working Stresses for Transmission Shafts. 8. Design of Shafts. 9. Shafts Subjected to Twisting Moment Only. 10. Shafts Subjected to Bending Moment Only. 11. Shafts Subjected to Combined Twisting Moment and Bending Moment. 1. Shafts Subjected to Fluctuating Loads. 1. Shafts Subjected to Axial Load in addition to Combined Torsion and Bending Loads. 14. Design of Shafts on the Basis of Rigidity. 14.1 Introduction A shaft is a rotating machine element which is used to transmit power from one place to another. The power is delivered to the shaft by some tangential force and the resultant torque (or twisting moment) set up within the shaft permits the power to be transferred to various machines linked up to the shaft. In order to transfer the power from one shaft to another, the various members such as pulleys, gears etc., are mounted on it. These members along with the forces exerted upon them causes the shaft to bending. In other words, we may say that a shaft is used for the transmission of torque and bending moment. The various members are mounted on the shaft by means of keys or splines. Notes: 1. The shafts are usually cylindrical, but may be square or cross-shaped in section. They are solid in cross-section but sometimes hollow shafts are also used. 509

510 A Textbook of Machine Design. An axle, though similar in shape to the shaft, is a stationary machine element and is used for the transmission of bending moment only. It simply acts as a support for some rotating body such as hoisting drum, a car wheel or a rope sheave.. A spindle is a short shaft that imparts motion either to a cutting tool (e.g. drill press spindles) or to a work piece (e.g. lathe spindles). 14. Material Used for Shafts The material used for shafts should have the following properties : 1. It should have high strength.. It should have good machinability.. It should have low notch sensitivity factor. 4. It should have good heat treatment properties. 5. It should have high wear resistant properties. The material used for ordinary shafts is carbon steel of grades 40 C 8, 45 C 8, 50 C 4 and 50 C 1. The mechanical properties of these grades of carbon steel are given in the following table. Table 14.1. Mechanical properties of steels used for shafts. Indian standard designation Ultimate tensile strength, MPa Yield strength, MPa 40 C 8 560-670 0 45 C 8 610-700 50 50 C 4 640-760 70 50 C 1 700 Min. 90 When a shaft of high strength is required, then an alloy steel such as nickel, nickel-chromium or chrome-vanadium steel is used. 14. Manufacturing of Shafts Shafts are generally manufactured by hot rolling and finished to size by cold drawing or turning and grinding. The cold rolled shafts are stronger than hot rolled shafts but with higher residual stresses. The residual stresses may cause distortion of the shaft when it is machined, especially when slots or keyways are cut. Shafts of larger diameter are usually forged and turned to size in a lathe. 14.4 Types of Shafts The following two types of shafts are important from the subject point of view : 1. Transmission shafts. These shafts transmit power between the source and the machines absorbing power. The counter shafts, line shafts, over head shafts and all factory shafts are transmission shafts. Since these shafts carry machine parts such as pulleys, gears etc., therefore they are subjected to bending in addition to twisting.. Machine shafts. These shafts form an integral part of the machine itself. The crank shaft is an example of machine shaft. 14.5 Standard Sizes of Transmission Shafts The standard sizes of transmission shafts are : 5 mm to 60 mm with 5 mm steps; 60 mm to 110 mm with 10 mm steps ; 110 mm to 140 mm with 15 mm steps ; and 140 mm to 500 mm with 0 mm steps. The standard length of the shafts are 5 m, 6 m and 7 m.

Shafts 511 14.6 Stresses in Shafts The following stresses are induced in the shafts : 1. Shear stresses due to the transmission of torque (i.e. due to torsional load).. Bending stresses (tensile or compressive) due to the forces acting upon machine elements like gears, pulleys etc. as well as due to the weight of the shaft itself.. Stresses due to combined torsional and bending loads. 14.7 Maximum Permissible Working Stresses for Transmission Shafts According to American Society of Mechanical Engineers (ASME) code for the design of transmission shafts, the maximum permissible working stresses in tension or compression may be taken as (a) 11 MPa for shafts without allowance for keyways. (b) 84 MPa for shafts with allowance for keyways. For shafts purchased under definite physical specifications, the permissible tensile stress (σ t ) may be taken as 60 per cent of the elastic limit in tension (σ el but not more than 6 per cent of the ultimate tensile strength (σ u ). In other words, the permissible tensile stress, σ t 0.6 σ el or 0.6 σ u, whichever is less. The maximum permissible shear stress may be taken as (a) 56 MPa for shafts without allowance for key ways. (b) 4 MPa for shafts with allowance for keyways. For shafts purchased under definite physical specifications, the permissible shear stress (τ) may be taken as 0 per cent of the elastic limit in tension (σ el ) but not more than 18 per cent of the ultimate tensile strength (σ u ). In other words, the permissible shear stress, τ 0. σ el or 0.18 σ u, whichever is less. 14.8 Design of Shafts The shafts may be designed on the basis of 1. Strength, and. Rigidity and stiffness. In designing shafts on the basis of strength, the following cases may be considered : (a) Shafts subjected to twisting moment or torque only, (b) Shafts subjected to bending moment only, (c) Shafts subjected to combined twisting and bending moments, and (d) Shafts subjected to axial loads in addition to combined torsional and bending loads. We shall now discuss the above cases, in detail, in the following pages. 14.9 Shafts Subjected to Twisting Moment Only When the shaft is subjected to a twisting moment (or torque) only, then the diameter of the shaft may be obtained by using the torsion equation. We know that T τ...(i) J r where T Twisting moment (or torque) acting upon the shaft, J Polar moment of inertia of the shaft about the axis of rotation, τ Torsional shear stress, and

51 A Textbook of Machine Design r Distance from neutral axis to the outer most fibre d / ; where d is the diameter of the shaft. We know that for round solid shaft, polar moment of inertia, π 4 J d The equation (i) may now be written as T τ π π 4 d or T τ d d 16 From this equation, we may determine the diameter of round solid shaft ( d ). We also know that for hollow shaft, polar moment of inertia,...(ii) π 4 4 J ( do) ( di) where d o and d i Outside and inside diameter of the shaft, and r d o /. Substituting these values in equation (i we have T π 4 4 ( do) ( di) τ d o or T Now the equation (iii) may be written as 4 4 π ( do) ( di) τ 16 do k Ratio of inside diameter and outside diameter of the shaft d i / d o T 4 4 π ( d ) o di π 4 τ 1 ( do ) (1 k ) 16 d τ o d o 16...(iii)...(iv) Shafts inside generators and motors are made to bear high torsional stresses.

Shafts 51 From the equations (iii) or (iv the outside and inside diameter of a hollow shaft may be determined. It may be noted that 1. The hollow shafts are usually used in marine work. These shafts are stronger per kg of material and they may be forged on a mandrel, thus making the material more homogeneous than would be possible for a solid shaft. When a hollow shaft is to be made equal in strength to a solid shaft, the twisting moment of both the shafts must be same. In other words, for the same material of both the shafts, 4 4 di T 4 4 π ( do) ( di) π τ τ d 16 do 16 ( do) ( ) d or (d d o ) (1 k 4 ) d o. The twisting moment (T) may be obtained by using the following relation : We know that the power transmitted (in watts) by the shaft, P π N T P 60 or T 60 π N where T Twisting moment in N-m, and N Speed of the shaft in r.p.m.. In case of belt drives, the twisting moment ( T ) is given by T (T 1 T ) R where T 1 and T Tensions in the tight side and slack side of the belt respectively, and R Radius of the pulley. Example 14.1. A line shaft rotating at 00 r.p.m. is to transmit 0 kw. The shaft may be assumed to be made of mild steel with an allowable shear stress of 4 MPa. Determine the diameter of the shaft, neglecting the bending moment on the shaft. Solution. Given : N 00 r.p.m. ; P 0 kw 0 10 W; τ 4 MPa 4 N/mm d Diameter of the shaft. We know that torque transmitted by the shaft, P 60 0 10 60 T 955 N-m 955 10 N-mm πn π 00 We also know that torque transmitted by the shaft ( T π 955 10 π τ d 4 d 8.5 d 16 16 d 955 10 / 8.5 115 7 or d 48.7 say 50 mm Ans. Example 14.. A solid shaft is transmitting 1 MW at 40 r.p.m. Determine the diameter of the shaft if the maximum torque transmitted exceeds the mean torque by 0%. Take the maximum allowable shear stress as 60 MPa. Solution. Given : P 1 MW 1 10 6 W; N 40 r.p.m. ; T max 1. T mean ; τ 60 MPa 60 N/mm d Diameter of the shaft. We know that mean torque transmitted by the shaft, 6 P 60 1 10 60 T mean 9 784 N-m 9 784 10 N-mm πn π 40

514 A Textbook of Machine Design Maximum torque transmitted, T max 1. T mean 1. 9 784 10 47 741 10 N-mm We know that maximum torque transmitted (T max 47 741 10 π π τ d 60 d 11.78 d 16 16 d 47 741 10 / 11.78 405 10 or d 159.4 say 160 mm Ans. Example 14.. Find the diameter of a solid steel shaft to transmit 0 kw at 00 r.p.m. The ultimate shear stress for the steel may be taken as 60 MPa and a factor of safety as 8. If a hollow shaft is to be used in place of the solid shaft, find the inside and outside diameter when the ratio of inside to outside diameters is 0.5. Solution. Given : P 0 kw 0 10 W; N 00 r.p.m. ; τ u 60 MPa 60 N/mm ; F.S. 8 ; k d i / d o 0.5 We know that the allowable shear stress, Diameter of the solid shaft τ τu 60 45 N/mm FS.. 8 d Diameter of the solid shaft. We know that torque transmitted by the shaft, P 60 0 10 60 T 955 N-m 955 10 N-mm πn π 00 We also know that torque transmitted by the solid shaft (T 955 10 π π τ d 45 d 8.84 d 16 16 d 955 10 / 8.84 108 0 or d 47.6 say 50 mm Ans. Diameter of hollow shaft and d i Inside diameter, and d o Outside diameter. We know that the torque transmitted by the hollow shaft ( T π 955 10 4 τ ( do ) (1 k ) 16 4 π 45 ( d ) [1 o (0.5) ] 8. 16 (do ) (d o ) 955 10 / 8. 115 060 or d o 48.6 say 50 mm Ans. d i 0.5 d o 0.5 50 5 mm Ans. 14.10 Shafts Subjected to Bending Moment Only When the shaft is subjected to a bending moment only, then the maximum stress (tensile or compressive) is given by the bending equation. We know that M σb...(i) I y where M Bending moment, I Moment of inertia of cross-sectional area of the shaft about the axis of rotation,

σ b Bending stress, and y Distance from neutral axis to the outer-most fibre. We know that for a round solid shaft, moment of inertia, π 4 d I d and y 64 Substituting these values in equation (i we have M σb π π 4 d or M σ b d d 64 From this equation, diameter of the solid shaft (d) may be obtained. We also know that for a hollow shaft, moment of inertia, Shafts 515 π 4 4 4 4 I π ( do) ( di) ( do) (1 k )...(where k d 64 64 i / d o ) and y d o / Again substituting these values in equation (i we have M σb π π 4 4 d or M σ 4 b ( do) (1 k ) o ( do ) (1 k ) 64 From this equation, the outside diameter of the shaft (d o ) may be obtained. Note: We have already discussed in Art. 14.1 that the axles are used to transmit bending moment only. Thus, axles are designed on the basis of bending moment only, in the similar way as discussed above. Steam emerges from tower Water heats up and turns to steam Nuclear reactor Condenser Concrete shell Generator Cooling tower Transformer Steam spins the turbine which powers the generator In a neuclear power plant, stearm is generated using the heat of nuclear reactions. Remaining function of steam turbines and generators is same as in theraml power plants.

516 A Textbook of Machine Design Example 14.4. A pair of wheels of a railway wagon carries a load of 50 kn on each axle box, acting at a distance of 100 mm outside the wheel base. The gauge of the rails is 1.4 m. Find the diameter of the axle between the wheels, if the stress is not to exceed 100 MPa. Solution. Given : W 50 kn 50 10 N; L 100 mm ; x 1.4 m ; σ b 100 MPa 100 N/mm Fig. 14.1 The axle with wheels is shown in Fig. 14.1. A little consideration will show that the maximum bending moment acts on the wheels at C and D. Therefore maximum bending moment, *M W.L 50 10 100 5 10 6 N-mm d Diameter of the axle. We know that the maximum bending moment (M π 5 10 6 π σ b d 100 d 9.8 d d 5 10 6 / 9.8 0.51 10 6 or d 79.8 say 80 mm Ans. 14.11 Shafts Subjected to Combined Twisting Moment and Bending Moment When the shaft is subjected to combined twisting moment and bending moment, then the shaft must be designed on the basis of the two moments simultaneously. Various theories have been suggested to account for the elastic failure of the materials when they are subjected to various types of combined stresses. The following two theories are important from the subject point of view : 1. Maximum shear stress theory or Guest's theory. It is used for ductile materials such as mild steel.. Maximum normal stress theory or Rankine s theory. It is used for brittle materials such as cast iron. τ Shear stress induced due to twisting moment, and σ b Bending stress (tensile or compressive) induced due to bending moment. According to maximum shear stress theory, the maximum shear stress in the shaft, 1 τ max ( σ b ) + 4τ * The maximum B.M. may be obtained as follows : R C R D 50 kn 50 10 N B.M. at A, M A 0 B.M. at C, M C 50 10 100 5 10 6 N-mm B.M. at D, M D 50 10 1500 50 10 1400 5 10 6 N-mm B.M. at B, M B 0

Substituting the values of τ and σ b from Art. 14.9 and Art. 14.10, we have Shafts 517 τ max M T 4 M T + + 1 16 16 πd πd πd π τ max d M T or +...(i) 16 The expression M + T is known as equivalent twisting moment and is denoted by T e. The equivalent twisting moment may be defined as that twisting moment, which when acting alone, produces the same shear stress (τ) as the actual twisting moment. By limiting the maximum shear stress (τ max ) equal to the allowable shear stress (τ) for the material, the equation (i) may be written as π T e M + T τ d...(ii) 16 From this expression, diameter of the shaft ( d ) may be evaluated. Now according to maximum normal stress theory, the maximum normal stress in the shaft, σ b(max) 1 1 σ b + ( σ b) + 4τ...(iii) 1 1 M 16 4 + d + π πd πd or π σ b ( max) d 1 ( + + ) πd M M T 1 M + M + T...(iv) The expression 1 ( M + M + T ) is known as equivalent bending moment and is denoted by M e. The equivalent bending moment may be defined as that moment which when acting alone produces the same tensile or compressive stress (σ b ) as the actual bending moment. By limiting the maximum normal stress [σ b(max) ] equal to the allowable bending stress (σ b then the equation (iv) may be written as 1 M e M + M + T π σ b d...(v) From this expression, diameter of the shaft ( d ) may be evaluated. Notes: 1. In case of a hollow shaft, the equations (ii) and (v) may be written as π 4 T e M + T τ ( do) (1 k ) 16 1 and M e ( ) π 4 M + M + T σ ( ) (1 ) b do k. It is suggested that diameter of the shaft may be obtained by using both the theories and the larger of the two values is adopted. Example 14.5. A solid circular shaft is subjected to a bending moment of 000 N-m and a torque of 10 000 N-m. The shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa and a ultimate shear stress of 500 MPa. Assuming a factor of safety as 6, determine the diameter of the shaft.

518 A Textbook of Machine Design Solution. Given : M 000 N-m 10 6 N-mm ; T 10 000 N-m 10 10 6 N-mm ; σ tu 700 MPa 700 N/mm ; τ u 500 MPa 500 N/mm We know that the allowable tensile stress, and allowable shear stress, σ t or σ b τ σtu 700 116.7 N/mm FS.. 6 τu 500 8. N/mm FS.. 6 d Diameter of the shaft in mm. According to maximum shear stress theory, equivalent twisting moment, T e 6 6 M + T ( 10 ) + (10 10 ) 10.44 10 6 N-mm We also know that equivalent twisting moment (T e π 10.44 10 6 π τ d 8. d 16.6 d 16 16 d 10.44 10 6 / 16.6 0.66 10 6 or d 86 mm Reactor vessel Water and steam separator Control rod Steam outlet Core (nuclear fuel assembly) Pump Water inlet Control rod drive Concrete shield Nuclear Reactor Note : This picture is given as additional information and is not a direct example of the current chapter.

Shafts 519 According to maximum normal stress theory, equivalent bending moment, M e ( ) 1 1 M + M + T ( M + T ) e 6 6 1 ( 10 10.44 10 ) + 6.7 106 N-mm We also know that the equivalent bending moment (M e π 6.7 10 6 π σ b d 116.7 d 11.46 d d 6.7 10 6 / 11.46 0.586 10 6 or d 8.7 mm Taking the larger of the two values, we have d 86 say 90 mm Ans. Example 14.6. A shaft supported at the ends in ball bearings carries a straight tooth spur gear at its mid span and is to transmit 7.5 kw at 00 r.p.m. The pitch circle diameter of the gear is 150 mm. The distances between the centre line of bearings and gear are 100 mm each. If the shaft is made of steel and the allowable shear stress is 45 MPa, determine the diameter of the shaft. Show in a sketch how the gear will be mounted on the shaft; also indicate the ends where the bearings will be mounted? The pressure angle of the gear may be taken as 0. Solution. Given : P 7.5 kw 7500 W ; N 00 r.p.m. ; D 150 mm 0.15 m ; L 00 mm 0. m ; τ 45 MPa 45 N/mm ; α 0 Fig. 14. shows a shaft with a gear mounted on the bearings. Fig. 14. We know that torque transmitted by the shaft, P 60 7500 60 T 8.7 N-m πn π 00 Tangential force on the gear, F t T 8.7 18.7 N D 0.15

50 A Textbook of Machine Design and the normal load acting on the tooth of the gear, F 18.7 18.7 W t 87 N cos α cos 0 0.997 Since the gear is mounted at the middle of the shaft, therefore maximum bending moment at the centre of the gear, WL. 87 0. M 169.4 N-m 4 4 d Diameter of the shaft. We know that equivalent twisting moment, T e M + T (169.4) + (8.7) 9.7 N-m 9.7 10 N-mm We also know that equivalent twisting moment (T e π 9.7 10 π τ d 45 d 8.84 d 16 16 d 9.7 10 / 8.84 10 or d say 5 mm Ans. Example 14.7. A shaft made of mild steel is required to transmit 100 kw at 00 r.p.m. The supported length of the shaft is metres. It carries two pulleys each weighing 1500 N supported at a distance of 1 metre from the ends respectively. Assuming the safe value of stress, determine the diameter of the shaft. Solution. Given : P 100 kw 100 10 W; N 00 r.p.m. ; L m ; W 1500 N We know that the torque transmitted by the shaft, P 60 100 10 60 T 18 N-m πn π 00 The shaft carrying the two pulleys is like a simply supported beam as shown in Fig. 14.. The reaction at each support will be 1500 N, i.e. R A R B 1500 N A little consideration will show that the maximum bending moment lies at each pulley i.e. at C and D. Fig. 14. Maximum bending moment, M 1500 1 1500 N-m d Diameter of the shaft in mm. We know that equivalent twisting moment, T e M + T (1500) + (18) 519 N-m 519 10 N-mm We also know that equivalent twisting moment (T e π 519 10 π τ d 60 d 11.8 d 16 16...(Assuming τ 60 N/mm ) d 519 10 / 11.8 98 10 or d 66.8 say 70 mm Ans. Example 14.8. A line shaft is driven by means of a motor placed vertically below it. The pulley on the line shaft is 1.5 metre in diameter and has belt tensions 5.4 kn and 1.8 kn on the tight side and slack side of the belt respectively. Both these tensions may be assumed to be vertical. If the pulley be

Shafts 51 overhang from the shaft, the distance of the centre line of the pulley from the centre line of the bearing being 400 mm, find the diameter of the shaft. Assuming maximum allowable shear stress of 4 MPa. Solution. Given : D 1.5 m or R 0.75 m; T 1 5.4 kn 5400 N ; T 1.8 kn 1800 N ; L 400 mm ; τ 4 MPa 4 N/mm A line shaft with a pulley is shown in Fig 14.4. We know that torque transmitted by the shaft, T (T 1 T ) R (5400 1800) 0.75 700 N-m 700 10 N-mm Fig. 14.4 Neglecting the weight of shaft, total vertical load acting on the pulley, W T 1 + T 5400 + 1800 700 N Bending moment, M W L 700 400 880 10 N-mm d Diameter of the shaft in mm. We know that the equivalent twisting moment, T e M + T (880 10 ) + (700 10 ) 950 10 N-mm Steel shaft

5 A Textbook of Machine Design We also know that equivalent twisting moment (T e π 950 10 π τ d 4 d 8.5 d 16 16 d 950 10 /8.5 479 10 or d 78 say 80 mm Ans. Example 14.9. A shaft is supported by two bearings placed 1 m apart. A 600 mm diameter pulley is mounted at a distance of 00 mm to the right of left hand bearing and this drives a pulley directly below it with the help of belt having maximum tension of.5 kn. Another pulley 400 mm diameter is placed 00 mm to the left of right hand bearing and is driven with the help of electric motor and belt, which is placed horizontally to the right. The angle of contact for both the pulleys is 180 and μ 0.4. Determine the suitable diameter for a solid shaft, allowing working stress of 6 MPa in tension and 4 MPa in shear for the material of shaft. Assume that the torque on one pulley is equal to that on the other pulley. Solution. Given : AB 1 m ; D C 600 mm or R C 00 mm 0. m ; AC 00 mm 0. m ; T 1.5 kn 50 N ; D D 400 mm or R D 00 mm 0. m ; BD 00 mm 0. m ; θ 180 π rad ; μ 0.4 ; σ b 6 MPa 6 N/mm ; τ 4 MPa 4 N/mm The space diagram of the shaft is shown in Fig. 14.5 (a). T 1 Tension in the tight side of the belt on pulley C 50 N...(Given) T Tension in the slack side of the belt on pulley C. We know that 1. log T T μ.θ 0.4 π 0.754 1 0.754. log T T 0.78 or T T 1.17...(Taking antilog of 0.78) T1 50 and T 1058 N.17.17 Vertical load acting on the shaft at C, W C T 1 + T 50 + 1058 08 N and vertical load on the shaft at D 0 The vertical load diagram is shown in Fig. 14.5 (c). We know that torque acting on the pulley C, T (T 1 T ) R C (50 1058) 0. 57.6 N-m The torque diagram is shown in Fig. 14.5 (b). T Tension in the tight side of the belt on pulley D, and T 4 Tension in the slack side of the belt on pulley D. Since the torque on both the pulleys (i.e. C and D) is same, therefore (T T 4 ) R D T 57.6 N-m or T T 4 57.6 57.6 R 0. D 1788 N...(i) T We know that T 1.17 or T T T.17 T 4...(ii) 4

Shafts 5 Fig. 14.5 From equations (i) and (ii we find that T 76 N, and T 4 1588 N Horizontal load acting on the shaft at D, W D T + T 4 76 + 1588 4964 N and horizontal load on the shaft at C 0 The horizontal load diagram is shown in Fig. 14.5 (d). Now let us find the maximum bending moment for vertical and horizontal loading.

54 A Textbook of Machine Design First of all, considering the vertical loading at C. R AV and R BV be the reactions at the bearings A and B respectively. We know that R AV + R BV 08 N Taking moments about A, R BV 1 08 0. or R BV 99.4 N and R AV 08 99.4 15.6 N We know that B.M. at A and B, M AV M BV 0 B.M. at C, M CV R AV 0. 15.6 0. 694.7 N-m B.M. at D, M DV R BV 0. 99.4 0. 198.5 N-m The bending moment diagram for vertical loading in shown in Fig. 14.5 (e). Now considering horizontal loading at D. R AH and R BH be the reactions at the bearings A and B respectively. We know that R AH + R BH 4964 N Taking moments about A, R BH 1 4964 0.8 or R BH 971 N and R AH 4964 971 99 N We know that B.M. at A and B, M AH M BH 0 B.M. at C, M CH R AH 0. 99 0. 97.9 N-m B.M. at D, M DH R BH 0. 971 0. 794. N-m The bending moment diagram for horizontal loading is shown in Fig. 14.5 ( f ). Resultant B.M. at C, M C ( MCV ) + ( MCH ) (694.7) + (97.9) 756 N-m and resultant B.M. at D, M D ( MDV ) + ( MDH ) (198.5) + (794.) 819. N-m The resultant bending moment diagram is shown in Fig. 14.5 (g). We see that bending moment is maximum at D. Maximum bending moment, M M D 819. N-m d Diameter of the shaft. We know that equivalent twisting moment, T e M + T (819.) + (57.6) 894 N-m 894 10 N-mm We also know that equivalent twisting moment (T e 894 10 π π τ d 4 d 8.5 d 16 16 d 894 10 / 8.5 108 10 or d 47.6 mm Again we know that equivalent bending moment, M e ( ) 1 1 1 M + M + T ( M + T ) e (819. + 894) 856.6 N-m 856.6 10 N-mm

Shafts 55 We also know that equivalent bending moment (M e 856.6 10 π π σ b d 6 d 6. d d 856.6 10 /6. 18. 10 or d 51.7 mm Taking larger of the two values, we have d 51.7 say 55 mm Ans. Example 14.10. A shaft is supported on bearings A and B, 800 mm between centres. A 0 straight tooth spur gear having 600 mm pitch diameter, is located 00 mm to the right of the left hand bearing A, and a 700 mm diameter pulley is mounted 50 mm towards the left of bearing B. The gear is driven by a pinion with a downward tangential force while the pulley drives a horizontal belt having 180 angle of wrap. The pulley also serves as a flywheel and weighs 000 N. The maximum belt tension is 000 N and the tension ratio is : 1. Determine the maximum bending moment and the necessary shaft diameter if the allowable shear stress of the material is 40 MPa. Solution. Given : AB 800 mm ; α C 0 ; D C 600 mm or R C 00 mm ; AC 00 mm ; D D 700 mm or R D 50 mm ; DB 50 mm ; θ 180 π rad ; W 000 N ; T 1 000 N ; T 1 /T ; τ 40 MPa 40 N/mm The space diagram of the shaft is shown in Fig. 14.6 (a). We know that the torque acting on the shaft at D, T (T 1 T ) R D T T 1 1 RD T1 1 000 1 50 700 10 N-mm...(Q T 1 /T ) The torque diagram is shown in Fig. 14.6 (b). Assuming that the torque at D is equal to the torque at C, therefore the tangential force acting on the gear C, T F tc R 700 10 N C 00 and the normal load acting on the tooth of gear C, F W C tc 48 N cos α C cos 0 0.997 The normal load acts at 0 to the vertical as shown in Fig. 14.7. Resolving the normal load vertically and horizontally, we get Vertical component of W C i.e. the vertical load acting on the shaft at C, W CV W C cos 0 48 0.997 N and horizontal component of W C i.e. the horizontal load acting on the shaft at C, W CH W C sin 0 48 0.4 849 N Since T 1 / T and T 1 000 N, therefore T T 1 / 000 / 1000 N Camshaft

56 A Textbook of Machine Design Fig. 14.6 Horizontal load acting on the shaft at D, W DH T 1 + T 000 + 1000 4000 N and vertical load acting on the shaft at D, W DV W 000 N

Shafts 57 The vertical and horizontal load diagram at C and D is shown in Fig. 14.6 (c) and (d) respectively. Now let us find the maximum bending moment for vertical and horizontal loading. First of all considering the vertical loading at C and D. R AV and R BV be the reactions at the bearings A and B respectively. We know that R AV + R BV + 000 4 N Taking moments about A, we get R BV 800 000 (800 50) + 00 1 566 600 R BV 1 566 600 / 800 1958 N and R AV 4 1958 75 N We know that B.M. at A and B, M AV M BV 0 Fig. 14.7 B.M. at C, M CV R AV 00 75 00 475 10 N-mm B.M. at D, M DV R BV 50 1958 50 489.5 10 N-mm The bending moment diagram for vertical loading is shown in Fig. 14.6 (e). Now consider the horizontal loading at C and D. R AH and R BH be the reactions at the bearings A and B respectively. We know that R AH + R BH 849 + 4000 4849 N Taking moments about A, we get R BH 800 4000 (800 50) + 849 00 69 800 R BH 69 800 / 800 96 N and R AH 4849 96 1886 N We know that B.M. at A and B, M AH M BH 0 B.M. at C, M CH R AH 00 1886 00 77 00 N-mm B.M. at D, M DH R BH 50 96 50 740 750 N-mm The bending moment diagram for horizontal loading is shown in Fig. 14.6 ( f ). We know that resultant B.M. at C, and resultant B.M. at D, M C ( M ) + ( M ) (475 10 ) + (77 00) CV 606 55 N-mm CH M D ( M ) + ( M ) (489.5 10 ) + (740 750) DV 887 874 N-mm Maximum bending moment The resultant B.M. diagram is shown in Fig. 14.6 (g). We see that the bending moment is maximum at D, therefore Maximum B.M., M M D 887 874 N-mm Ans. DH

58 A Textbook of Machine Design Diameter of the shaft d Diameter of the shaft. We know that the equivalent twisting moment, T e M + T (887 874) + (700 10 ) 111 10 N-mm We also know that equivalent twisting moment (T e 111 10 π π τ d 40 d 7.86 d 16 16 d 111 10 / 7.86 144 10 or d 5.4 say 55 mm Ans. Example 14.11. A steel solid shaft transmitting 15 kw at 00 r.p.m. is supported on two bearings 750 mm apart and has two gears keyed to it. The pinion having 0 teeth of 5 mm module is located 100 mm to the left of the right hand bearing and delivers power horizontally to the right. The gear having 100 teeth of 5 mm module is located 150 mm to the right of the left hand bearing and receives power in a vertical direction from below. Using an allowable stress of 54 MPa in shear, determine the diameter of the shaft. Solution. Given : P 15 kw 15 10 W; N 00 r.p.m. ; AB 750 mm ; T D 0 ; m D 5 mm ; BD 100 mm ; T C 100 ; m C 5 mm ; AC 150 mm ; τ 54 MPa 54 N/mm The space diagram of the shaft is shown in Fig. 14.8 (a). We know that the torque transmitted by the shaft, P 60 15 10 60 T 716 N-m 716 10 N-mm πn π 00 The torque diagram is shown in Fig. 14.8 (b). We know that diameter of gear No. of teeth on the gear module Radius of gear C, T R C C m C 100 5 50 mm and radius of pinion D, T R D D m D 0 5 75 mm Assuming that the torque at C and D is same (i.e. 716 10 N-mm therefore tangential force on the gear C, acting downward, T 716 10 F tc 870 N RC 50 and tangential force on the pinion D, acting horizontally, T 716 10 F td 9550 N RD 75 The vertical and horizontal load diagram is shown in Fig. 14.8 (c) and (d) respectively. Now let us find the maximum bending moment for vertical and horizontal loading. First of all, considering the vertical loading at C. R AV and R BV be the reactions at the bearings A and B respectively. We know that R AV + R BV 870 N Taking moments about A, we get R BV 750 870 150

Shafts 59 Fig. 14.8 R BV 870 150 / 750 574 N and R AV 870 574 96 N We know that B.M. at A and B, M AV M BV 0

50 A Textbook of Machine Design B.M. at C, M CV R AV 150 96 150 44 400 N-mm B.M. at D, M DV R BV 100 574 100 57 400 N-mm The B.M. diagram for vertical loading is shown in Fig. 14.8 (e). Now considering horizontal loading at D. R AH and R BH be the reactions at the bearings A and B respectively. We know that R AH + R BH 9550 N Taking moments about A, we get R BH 750 9550 (750 100) 9550 650 R BH 9550 650 / 750 877 N and R AH 9550 877 17 N We know that B.M. at A and B, M AH M BH 0 B.M. at C, M CH R AH 150 17 150 190 950 N-mm B.M. at D, M DH R BH 100 877 100 87 700 N-mm The B.M. diagram for horizontal loading is shown in Fig. 14.8 ( f ). We know that resultant B.M. at C, and resultant B.M. at D, M C ( M ) + ( M ) (44 400) + (190 950) CV 9 790 N-mm CH M D ( M ) + ( M ) (57 400) + (87 700) DV 89 690 N-mm The resultant B.M. diagram is shown in Fig. 14.8 (g). We see that the bending moment is maximum at D. Maximum bending moment, M M D 89 690 N-mm d Diameter of the shaft. We know that the equivalent twisting moment, or DH T e M + T (89 690) + (716 10 ) 1096 10 N-mm We also know that equivalent twisting moment (T e 1096 10 π π τ d 54 d 10.6 d 16 16 d 1096 10 /10.6 10.4 10 d 47 say 50 mm Ans. 14.1 Shafts Subjected to Fluctuating Loads In the previous articles we have assumed that the shaft is subjected to constant torque and bending moment. But in actual practice, the shafts are subjected to fluctuating torque and bending moments. In order to design such shafts like line shafts and counter shafts, the combined shock and fatigue factors must be taken into account for the computed twisting moment (T ) and bending moment (M ). Thus for a shaft Crankshaft

Shafts 51 subjected to combined bending and torsion, the equivalent twisting moment, and equivalent bending moment, T e ( K M) + ( K + T) M e 1 m t K M K M K T m + ( m ) + ( t ) where K m Combined shock and fatigue factor for bending, and K t Combined shock and fatigue factor for torsion. The following table shows the recommended values for K m and K t. Table 14.. Recommended values for K m and K t. Nature of load K m K t 1. Stationary shafts (a) Gradually applied load 1.0 1.0 (b) Suddenly applied load 1.5 to.0 1.5 to.0. Rotating shafts (a) Gradually applied or 1.5 1.0 steady load (b) Suddenly applied load 1.5 to.0 1.5 to.0 with minor shocks only (c) Suddenly applied load.0 to.0 1.5 to.0 with heavy shocks Example 14.1. A mild steel shaft transmits 0 kw at 00 r.p.m. It carries a central load of 900 N and is simply supported between the bearings.5 metres apart. Determine the size of the shaft, if the allowable shear stress is 4 MPa and the maximum tensile or compressive stress is not to exceed 56 MPa. What size of the shaft will be required, if it is subjected to gradually applied loads? Solution. Given : P 0 kw 0 10 W; N 00 r.p.m. ; W 900 N ; L.5 m ; τ 4 MPa 4 N/mm ; σ b 56 MPa 56 N/mm Size of the shaft d Diameter of the shaft, in mm. We know that torque transmitted by the shaft, P 60 0 10 60 T 955 N-m 955 10 N-mm πn π 00 and maximum bending moment of a simply supported shaft carrying a central load, W L 900.5 M 56.5 N-m 56.5 10 N-mm 4 4 We know that the equivalent twisting moment, T e M + T (56.5 10 ) + (955 10 ) 1108 10 N-mm We also know that equivalent twisting moment (T e 1108 10 π π τ d 4 d 8.5 d 16 16 d 1108 10 / 8.5 14. 10 or d 51. mm

5 A Textbook of Machine Design We know that the equivalent bending moment, 1 1 M e M M T + + ( M + T ) e 1 (56.5 10 1108 10 ) + 85.5 10 N-mm We also know that equivalent bending moment (M e 85.5 10 π π σ b d 56 d 5.5 d d 85.5 10 / 5.5 15 10 or d 5.4 mm Taking the larger of the two values, we have d 5.4 say 55 mm Ans. Size of the shaft when subjected to gradually applied load d Diameter of the shaft. From Table 14., for rotating shafts with gradually applied loads, K m 1.5 and K t 1 We know that equivalent twisting moment, T e ( K M) + ( K T) m (1.5 56.5 10 ) + (1 955 10 ) 174 10 N-mm We also know that equivalent twisting moment (T e 174 10 π π τ d 4 d 8.5 d 16 16 d 174 10 / 8.5 154.6 10 or d 5.6 mm We know that the equivalent bending moment, M e t 1 K ( ) ( ) m M + Km M + Kt T 1 [ K ] m M + Te 1 1.5 56.5 10 + 174 10 1059 10 N-mm We also know that equivalent bending moment (M e 1059 10 π π σ b d 56 d 5.5 d d 1059 10 / 5.5 19.5 10 57.7 mm Taking the larger of the two values, we have d 57.7 say 60 mm Ans. Example 14.1. Design a shaft to transmit power from an electric motor to a lathe head stock through a pulley by means of a belt drive. The pulley weighs 00 N and is located at 00 mm from the centre of the bearing. The diameter of the pulley is 00 mm and the maximum power transmitted is 1 kw at 10 r.p.m. The angle of lap of the belt is 180 and coefficient of friction between the belt and the pulley is 0.. The shock and fatigue factors for bending and twisting are 1.5 and.0 respectively. The allowable shear stress in the shaft may be taken as 5 MPa. Solution. Given : W 00 N ; L 00 mm ; D 00 mm or R 100 mm ; P 1 kw 1000 W ; N 10 r.p.m. ; θ 180 π rad ; μ 0. ; K m 1.5 ; K t ; τ 5 MPa 5 N/mm The shaft with pulley is shown in Fig. 14.9.

Shafts 5 We know that torque transmitted by the shaft, P 60 1000 60 T πn π 10 79.6 N-m 79.6 10 N-mm Fig. 14.9 T 1 and T Tensions in the tight side and slack side of the belt respectively in newtons. Torque transmitted (T 79.6 10 (T 1 T ) R (T 1 T ) 100 T 1 T 79.6 10 / 100 796 N...(i) We know that 1. log T T μ.θ 0. π 0.946 1 log T T 0.946 T1 0.4098 or. T.57...(ii)...(Taking antilog of 0.4098) From equations (i) and (ii we get, T 1 10 N, and T 507 N We know that the total vertical load acting on the pulley, W T T 1 + T + W 10 + 507 + 00 010 N Bending moment acting on the shaft, M W T L 010 00 60 10 N-mm d Diameter of the shaft. We know that equivalent twisting moment, T e ( K M) + ( K + T) m (1.5 60 10 ) + ( 79.6 10 ) 918 10 N-mm We also know that equivalent twisting moment (T e 918 10 π π τ d 5 d 6.87 d 16 16 d 918 10 / 6.87 1.6 10 or d 51.1 say 55 mm Ans. Example 14.14. Fig. 14.10 shows a shaft carrying a pulley A and a gear B and supported in two bearings C and D. The shaft transmits 0 kw at 150 r.p.m. The tangential force F t on the gear B acts vertically upwards as shown. t

54 A Textbook of Machine Design The pulley delivers the power through a belt to another pulley of equal diameter vertically below the pulley A. The ratio of tensions T 1 /T is equal to.5. The gear and the pulley weigh 900 N and 700 N respectively. The permissible shear stress for the material of the shaft may be taken as 6 MPa. Assuming the weight of the shaft to be negligible in comparison with the other loads, determine its diameter. Take shock and fatigue factors for bending and torsion as and 1.5 respectively. Solution. Given : P 0 kw 0 10 W; N 150 r.p.m. ; T 1 /T.5 ; W B 900 N ; W A 700 N ; τ 6 MPa 6 N/mm ; K m ; K t 1.5 ; D B 750 mm or R B 75 mm ; D A 150 mm or R A 65 mm. We know that torque transmitted by the shaft, P 60 0 10 60 T 17 N-m 17 10 N-mm πn π 150 Fig. 14.10 T 1 and T Tensions in the tight side and slack side of the belt on pulley A. Since the torque on the pulley is same as that of shaft (i.e. 17 10 N-mm therefore (T 1 T ) R A 17 10 or T 1 T 17 10 / 65 07 N...(i) Since T 1 / T.5 or T 1.5 T, therefore.5 T T 07 or T 07/1.5 158 N...[From equation (i)] and T 1.5 158 95 N Total vertical load acting downward on the shaft at A T 1 + T + W A 95 + 158 + 700 745 N Assuming that the torque on the gear B is same as that of the shaft, therefore the tangential force acting vertically upward on the gear B, T 17 10 F t 95 N RB 75 Since the weight of gear B (W B 900 N) acts vertically downward, therefore the total vertical load acting upward on the shaft at B F t W B 95 900 495 N Now let us find the reactions at the bearings C and D. R C and R D be the reactions at C and D respectively. A little consideration will show that the reaction R C will act upward while the reaction R D act downward as shown in Fig. 14.11.

Shafts 55 Taking moments about D, we get R C 1000 745 150 + 495 50 10. 10 6 R C 10. 10 6 / 1000 10 00 N Fig. 14.11 For the equilibrium of the shaft, R D + 745 R C + 495 10 00 + 495 1 695 R D 1 695 745 54 N We Know that B.M. at A and B 0 B.M. at C 745 50 186 10 N-mm B.M. at D 495 50 87 10 N-mm We see that the bending moment is maximum at C. Maximum B.M. M M C 186 10 N-mm We know that the equivalent twisting moment, T e ( K M) + ( K T) m t ( 186 10 ) + (1.5 17 10 ) 4187 10 N-mm We also know that equivalent twisting moment (T e π 4187 10 π τ d 6 d 1.7 d 16 16. d 4187 10 / 1.7 8 10 or d 69.6 say 70 mm Ans. Example 14.15. A horizontal nickel steel shaft rests on two bearings, A at the left and B at the right end and carries two gears C and D located at distances of 50 mm and 400 mm respectively from the centre line of the left and right bearings. The pitch diameter of the gear C is 600 mm and that of gear D is 00 mm. The distance between the centre line of the bearings is 400 mm. The shaft

56 A Textbook of Machine Design transmits 0 kw at 10 r.p.m. The power is delivered to the shaft at gear C and is taken out at gear D in such a manner that the tooth pressure F tc of the gear C and F td of the gear D act vertically downwards. Find the diameter of the shaft, if the working stress is 100 MPa in tension and 56 MPa in shear. The gears C and D weighs 950 N and 50 N respectively. The combined shock and fatigue factors for bending and torsion may be taken as 1.5 and 1. respectively. Solution. Given : AC 50 mm ; BD 400 mm ; D C 600 mm or R C 00 mm ; D D 00 mm or R D 100 mm ; AB 400 mm ; P 0 kw 0 10 W; N 10 r.p.m ; σ t 100 MPa 100 N/mm ; τ 56 MPa 56 N/mm ; W C 950 N ; W D 50 N ; K m 1.5 ; K t 1. The shaft supported in bearings and carrying gears is shown in Fig. 14.1. Fig. 14.1 We know that the torque transmitted by the shaft, P 60 0 10 60 T 1590 N-m 1590 10 N-mm πn π 10 Since the torque acting at gears C and D is same as that of the shaft, therefore the tangential force acting at gear C, T 1590 10 F tc 500 N RC 00 Car rear axle.

Shafts 57 and total load acting downwards on the shaft at C F tc + W C 500 + 950 650 N Similarly tangential force acting at gear D, T 1590 10 F td 15 900 N RD 100 and total load acting downwards on the shaft at D F td + W D 15 900 + 50 16 50 N Now assuming the shaft as a simply supported beam as shown in Fig. 14.1, the maximum bending moment may be obtained as discussed below : and Fig. 14.1 R A and R B Reactions at A and B respectively. R A + R B Total load acting downwards at C and D 650 + 16 50 500 N Now taking moments about A, R B 400 16 50 000 + 650 50 4 06.5 10 R B 4 06.5 10 / 400 14 190 N R A 500 14 190 810 N A little consideration will show that the maximum bending moment will be either at C or D. We know that bending moment at C, M C R A 50 810 50 077.5 10 N-mm Bending moment at D, *M D R B 400 14 190 400 5676 10 N-mm Maximum bending moment transmitted by the shaft, M M D 5676 10 N-mm d Diameter of the shaft. We know that the equivalent twisting moment, T e ( K M) + ( K T) m t (1.5 5676 10 ) + (1. 1590 10 ) 875 10 N-mm * The bending moment at D may also be calculated as follows : M D R A 000 ( Total load at C ) 1750

58 A Textbook of Machine Design We also know that the equivalent twisting moment (T e 875 10 π π τ d 56 d 11 d 16 16 d 875 10 / 11 79 10 or d 9.5 mm Again we know that the equivalent bending moment, 1 1 M e K ( ) ( ) ( ) m M Km M Kt T + + K m M + Te 1 1.5 5676 10 + 875 10 860 10 N-mm We also know that the equivalent bending moment (M e 860 10 π π σ b d 100 d 9.8 d...(taking σ b σ t ) d 860 10 / 9.8 878 10 or d 95.7 mm Taking the larger of the two values, we have d 95.7 say 100 mm Ans. Example 14.16. A hoisting drum 0.5 m in diameter is keyed to a shaft which is supported in two bearings and driven through a 1 : 1 reduction ratio by an electric motor. Determine the power of the driving motor, if the maximum load of 8 kn is hoisted at a speed of 50 m/min and the efficiency of the drive is 80%. Also determine the torque on the drum shaft and the speed of the motor in r.p.m. Determine also the diameter of the shaft made of machinery steel, the working stresses of which are 115 MPa in tension and 50 MPa in shear. The drive gear whose diameter is 450 mm is mounted at the end of the shaft such that it overhangs the nearest bearing by 150 mm. The combined shock and fatigue factors for bending and torsion may be taken as and 1.5 respectively. Solution. Given : D 0.5 m or R 0.5 m ; Reduction ratio 1 : 1 ; W 8 kn 8000 N ; v 50 m/min ; η 80% 0.8 ; σ t 115 MPa 115 N/mm ; τ 50 MPa 50 N/mm ; D 1 450 mm or R 1 5 mm 0.5 m ; Overhang 150 mm 0.15 m ; K m ; K t 1.5 Power of the driving motor We know that the energy supplied to the hoisting drum per minute W v 8000 50 400 10 N-m/min Power supplied to the hoisting drum 400 10 6670 W 6.67 kw...(q 1 N-m/s 1 W) 60 Since the efficiency of the drive is 0.8, therefore power of the driving motor 6.67 8. kw Ans. 0.8 Torque on the drum shaft We know that the torque on the drum shaft, T W.R 8000 0.5 000 N-m Ans. Speed of the motor N Speed of the motor in r.p.m. We know that angular speed of the hoisting drum Linear speed v 50 00 rad / min Radius of the drum R 0.5

Shafts 59 Since the reduction ratio is 1 : 1, therefore the angular speed of the electric motor, ω 00 1 400 rad/min and speed of the motor in r.p.m., ω 400 N 8 r.p.m. Ans. π π Diameter of the shaft d Diameter of the shaft. Since the torque on the drum shaft is 000 N-m, therefore the tangential tooth load on the drive gear, T 000 F t R 8900 N 1 0.5 Assuming that the pressure angle of the drive gear in 0, therefore the maximum bending load on the shaft due to tooth load F 8900 t 9470 N cos 0 0.997 Since the overhang of the shaft is 150 mm 0.15 m, therefore bending moment at the bearing, M 9470 0.15 140 N-m We know that the equivalent twisting moment, T e ( K M) + ( K T) m ( 140) + (1.5 000) 410 N-m 410 10 N-mm We also know that equivalent twisting moment (T e 410 10 π π τ d 50 d 9.8 d 16 16 d 410 10 / 9.8 40.6 10 or d 75 mm Again we know that the equivalent bending moment, 1 1 M e K ( ) ( ) ( ) m M Km M Kt T + + K m M + Te 1 ( 140 410) + 485 N-m 485 10 N-mm We also know that equivalent bending moment (M e π 485 10 π σ b d 115 d 11. d d 485 10 / 11. 08.4 10 or d 67.5 mm Taking the larger of the two values, we have d 75 mm Ans. Example 14.17. A solid steel shaft is supported on two bearings 1.8 m apart and rotates at 50 r.p.m. A 0 involute gear D, 00 mm diameter is keyed to the shaft at a distance of 150 mm to the left on the right hand bearing. Two pulleys B and C are located on the shaft at distances of 600 mm and 150 mm respectively to the right of the left hand bearing. The diameters of the pulleys B and C are 750 mm and 600 mm respectively. 0 kw is supplied to the gear, out of which 18.75 kw is taken off at the pulley C and 11.5 kw from pulley B. The drive from B is vertically downward while from C the drive is downward at an angle of 60 to the horizontal. In both cases the belt tension ratio is and the angle of lap is 180. The combined fatigue and shock factors for torsion and bending may be taken as 1.5 and respectively. Design a suitable shaft taking working stress to be 4 MPa in shear and 84 MPa in tension. t

540 A Textbook of Machine Design Solution. Given : PQ 1.8 m ; N 50 r.p.m ; α D 0 ; D D 00 mm or R D 150 mm 0.15 m ; QD 150 mm 0.15 m ; PB 600 mm 0.6 m ; PC 150 mm 1.5 m ; D B 750 mm or R B 75 mm 0.75 m ; D C 600 mm or R C 00 mm 0. m ; P D 0 kw 0 10 W; P C 18.75 kw 18.75 10 W; P B 11.5 kw 11.5 10 W; T B1 /T B T C1 /T C ; θ 180 π rad ; K t 1.5 ; K m ; τ 4 MPa 4 N/mm ; σ t 84 MPa 84 N/mm First of all, let us find the total loads acting on the gear D and pulleys C and B respectively. For gear D We know that torque transmitted by the gear D, PD 60 0 10 60 T D 1146 N-m πn π 50 Tangential force acting on the gear D, TD 1146 F td 7640 N RD 0.15 and the normal load acting on the gear tooth, F D 7640 W D t 810 N cos 0 0.997 The normal load acts at 0 to the vertical as shown in Fig. 14.14. Resolving the normal load vertically and horizontally, we have Vertical component of W D W D cos 0 810 0.997 7640 N Fig. 14.14 Horizontal component of W D W D sin 0 810 0.4 780 N For pulley C We know that torque transmitted by pulley C, T C PC 60 18.75 10 60 716 N-m πn π 50 T C1 and T C Tensions in the tight side and slack side of the belt for pulleyc. We know that torque transmitted by pulley C (T C 716 (T C1 T C ) R C (T C1 T C ) 0. T C1 T C 716 / 0. 87 N...(i) Since T C1 / T C or T C1 T C, therefore from equation (i we have T C 87 N ; and T C1 4774 N Total load acting on pulley C, W C T C1 + T C 4774 + 87 7161 N...(Neglecting weight of pulley C) This load acts at 60 to the horizontal as shown in Fig. 14.15. Resolving the load W C into vertical and horizontal components, we have Vertical component of W C W C sin 60 7161 0.866 600 N Fig. 14.15

Shafts 541 Trainwheels and Axles and horizontal component of W C W C cos 60 7161 0.5 580 N For pulley B We know that torque transmitted by pulley B, PB 60 11.5 10 60 T B 40 N-m πn π 50 T B1 and T B Tensions in the tight side and slack side of the belt for pulley B. We know that torque transmitted by pulley B (T B 40 (T B1 T B ) R B (T B1 T B ) 0.75 T B1 T B 40 / 0.75 1147 N...(ii) Since T B1 / T B or T B1 T B, therefore from equation (ii we have T B 1147 N, and T B1 94 N Total load acting on pulley B, W B T B1 + T B 94 + 1147 441 N This load acts vertically downwards. From above, we may say that the shaft is subjected to the vertical and horizontal loads as follows : Type of loading Load in N At D At C At B Vertical 7640 600 441 Horizontal 780 580 0

54 A Textbook of Machine Design The vertical and horizontal load diagrams are shown in Fig. 14.16 (c) and (d). First of all considering vertical loading on the shaft. R PV and R QV be the reactions at bearings P and Q respectively for vertical loading. We know that R PV + R QV 7640 + 600 + 441 17 81 N Fig. 14.16

Shafts 54 Taking moments about P, we get R QV 1.8 7640 1.65 + 600 1.5 + 441 0.6 041 R QV 041 / 1.8 1 800 N and R PV 17 81 1 800 4481 N We know that B.M. at P and Q, M PV M QV 0 B.M. at B, M BV 4481 0.6 690 N-m B.M. at C, M CV 4481 1.5 441 0.75 470 N-m and B.M. at D, M DV 1 800 0.15 190 N-m The bending moment diagram for vertical loading is shown in Fig. 14.16 (e). Now considering horizontal loading. R PH and R QH be the reactions at the bearings P and Q respectively for horizontal loading. We know that R PH + R QH 780 + 580 660 N Taking moments about P, we get R QH 1.8 780 1.65 + 580 1.5 940 N R QH 940 / 1.8 5 N and R PH 660 5 117 N We know that B.M. at P and Q, M PH M QH 0 B.M. at B, M BH 117 0.6 676 N-m B.M. at C, M CH 117 1.5 151 N-m and B.M. at D, M DH 5 0.15 785 N-m The bending moment diagram for horizontal loading is shown in Fig. 14.16 ( f ). The resultant bending moments for the points B, C and D are as follows : Resultant B.M. at B ( M ) + ( M ) (690) + (676) 774 N-m BV BH Resultant B.M. at C ( M ) + ( M ) (470) + (151) 790 N-m CV Resultant B.M. at D ( MDV ) + ( MDH ) (190) + (785) 074 N-m From above we see that the resultant bending moment is maximum at C. M M C 790 N-m and maximum torque at C, T Torque corresponding to 0 kw T D 1146 N-m d Diameter of the shaft in mm. We know that equivalent twisting moment, T e ( K M) + ( K T) ( 790) + (1.5 1146) m 777 N-m 777 10 N-mm We also know that the equivalent twisting moment (T e 777 10 π π τ d 4 d 8.5 d 16 16 d 777 10 /8.5 94 10 or d 98 mm CH t

544 A Textbook of Machine Design Again, we know that equivalent bending moment, M e 1 1 K ( ) ( ) ( ) m M Km M Kt T + + K m M + Te 1 ( 790 777) + 7676 N-m 7676 10 N-mm We also know that the equivalent bending moment (M e 7676 10 π π σ b d 84 d 8.5 d d 7676 10 / 8.5 90 10 or d 97.6 mm Taking the larger of the two values, we have d 98 say 100 mm Ans. 14.1 Shafts Subjected to Axial Load in addition to Combined Torsion and Bending Loads When the shaft is subjected to an axial load (F) in addition to torsion and bending loads as in propeller shafts of ships and shafts for driving worm gears, then the stress due to axial load must be added to the bending stress (σ b ). We know that bending equation is M σ M. y M d / M b or σ I b y I π 4 d π d and stress due to axial load 64 F 4 F...(For round solid shaft) π d π d 4 F 4 F...( For hollow shaft) π ( d ) ( ) ( do) ( di) o d π i 4 F... (Q k d π ( do ) (1 k ) i /d o ) Resultant stress (tensile or compressive) for solid shaft, M 4 F F d σ 1 + M +...(i) πd πd πd 8 M1 F d... Substituting M1 M + π d 8 In case of a hollow shaft, the resultant stress, σ 1 M 4 F + 4 π ( d ) (1 k ) π ( d ) (1 k ) o o Fd (1 ) o + k M1 + 4 M 4 8 o o Fd (1 ) o + k... Substituting for hollow shaft, M1 M + 8 π ( d ) (1 k ) π( d ) (1 k ) In case of long shafts (slender shafts) subjected to compressive loads, a factor known as column factor (α) must be introduced to take the column effect into account. Stress due to the compressive load, α 4 F σ c...( For round solid shaft) πd

Shafts 545 α 4 F...(For hollow shaft) π( do ) (1 k ) The value of column factor (α) for compressive loads* may be obtained from the following relation : 1 Column factor, α 1 0.0044 ( L/ K) This expression is used when the slenderness ratio (L / K) is less than 115. When the slenderness ratio (L / K) is more than 115, then the value of column factor may be obtained from the following relation : σ y ( L/ K) **Column factor, α Cπ E where L Length of shaft between the bearings, K Least radius of gyration, σ y Compressive yield point stress of shaft material, and C Coefficient in Euler's formula depending upon the end conditions. The following are the different values of C depending upon the end conditions. C 1, for hinged ends,.5, for fixed ends, 1.6, for ends that are partly restrained as in bearings. Note: In general, for a hollow shaft subjected to fluctuating torsional and bending load, along with an axial load, the equations for equivalent twisting moment (T e ) and equivalent bending moment (M e ) may be written as T e α Fdo(1 + k ) Km M + + ( Kt T) 8 π 4 τ ( do) (1 k ) 16 1 (1 ) (1 ) and M e α Fdo + k α Fdo + k Km M + + Km M + + ( Kt T) 8 8 π 4 σ b ( do) (1 k ) It may be noted that for a solid shaft, k 0 and d 0 d. When the shaft carries no axial load, then F 0 and when the shaft carries axial tensile load, then α 1. Example 14.18. A hollow shaft is subjected to a maximum torque of 1.5 kn-m and a maximum bending moment of kn-m. It is subjected, at the same time, to an axial load of 10 kn. Assume that the load is applied gradually and the ratio of the inner diameter to the outer diameter is 0.5. If the outer diameter of the shaft is 80 mm, find the shear stress induced in the shaft. Solution. Given : T 1.5 kn-m 1.5 10 N-m ; M kn-m 10 N-m ; F 10 kn 10 10 N; k d i / d o 0.5 ; d o 80 mm 0.08 m τ Shear stress induced in the shaft. Since the load is applied gradually, therefore from Table 14., we find that K m 1.5 ; and K t 1.0 * The value of column factor (α) for tensile load is unity. ** It is an Euler s formula for long columns.

546 A Textbook of Machine Design We know that the equivalent twisting moment for a hollow shaft, T e α Fdo (1 + k ) Km M + + ( Kt T) 8 1 10 10 0.08 (1 + 0.5 ) 1.5 10 + + (1 1.5 10 ) 8... (Q α 1, for axial tensile loading) (4500 + 15) + (1500) 486 N-m 486 10 N-mm We also know that the equivalent twisting moment for a hollow shaft (T e 486 10 π 4 π 4 τ( do ) (1 k ) τ(80) (1 0.5 ) 94 60 τ 16 16 τ 486 10 / 94 60 51.6 N/mm 51.6 MPa Ans. Crankshaft inside the crank-case Example 14.19. A hollow shaft of 0.5 m outside diameter and 0. m inside diameter is used to drive a propeller of a marine vessel. The shaft is mounted on bearings 6 metre apart and it transmits 5600 kw at 150 r.p.m. The maximum axial propeller thrust is 500 kn and the shaft weighs 70 kn. Determine : 1. The maximum shear stress developed in the shaft, and. The angular twist between the bearings. Solution. Given : d o 0.5 m ; d i 0. m ; P 5600 kw 5600 10 W; L 6 m ; N 150 r.p.m. ; F 500 kn 500 10 N; W 70 kn 70 10 N 1. Maximum shear stress developed in the shaft τ Maximum shear stress developed in the shaft. We know that the torque transmitted by the shaft, P 60 5600 10 60 T 56 460 N-m πn π 150 and the maximum bending moment, W L 70 10 6 M 5 500 N-m 8 8

Now let us find out the column factor α. We know that least radius of gyration, K π 4 4 ( do) ( d ) I i 64 A π ( do) ( di) 4 [( do) + ( di) ] [( do) ( di) ] 16 [( d ) ( d ) ] o i Shafts 547 1 1 ( do) + ( di) (0.5) + (0.) 0.1458 m 4 4 Slenderness ratio, L / K 6 / 0.1458 41.15 1 and column factor, α... Q L < 115 L K 1 0.0044 K 1 1 1 0.0044 41.15 1 0.18 1. Assuming that the load is applied gradually, therefore from Table 14., we find that K m 1.5 and K t 1.0 Also k d i / d o 0. / 0.5 0.6 We know that the equivalent twisting moment for a hollow shaft, T e α Fdo (1 + k ) Km M + + ( Kt T) 8 1. 500 10 0.5 (1 + 0.6 ) 1.5 5 500 + + (1 56 460) 8 (78 750 + 51 850) + (56 460) 80 10 N-m We also know that the equivalent twisting moment for a hollow shaft (T e π 80 10 4 π 4 τ( do ) (1 k ) τ(0.5) [1 (0.6) ] 0.0 τ 16 16 τ 80 10 / 0.0 19 10 6 N/m 19 MPa Ans.. Angular twist between the bearings θ Angular twist between the bearings in radians. We know that the polar moment of inertia for a hollow shaft, π 4 4 π 4 4 J [( do) ( di) ] [(0.5) (0.) ] 0.005 4 m 4 From the torsion equation, T G θ, we have J L T L 56 460 6 θ 9 0.0048 rad G J 84 10 0.00 54... (Taking G 84 GPa 84 10 9 N/m ) 180 0.0048 π 0.75 Ans.

548 A Textbook of Machine Design Example 14.0. A hollow steel shaft is to transmit 0 kw at 00 r.p.m. The loading is such that the maximum bending moment is 1000 N-m, the maximum torsional moment is 500 N-m and axial compressive load is 15 kn. The shaft is supported on rigid bearings 1.5 m apart. The maximum permissible shear stress on the shaft is 40 MPa. The inside diameter is 0.8 times the outside diameter. The load is cyclic in nature and applied with shocks. The values for the shock factors are K t 1.5 and K m 1.6. Solution. Given : *P 0 kw ; *N 00 r.p.m. ; M 1000 N-m 1000 10 N-mm ; T 500 N-m 500 10 N-mm ; F 15 kn 15 000 N ; L 1.5 m 1500 mm ; τ 40 MPa 40 N/mm ; d i 0.8 d o or k d i /d o 0.8 ; K t 1.5 ; K m 1.6 d o Outside diameter of the shaft, and d i Inside diameter of the shaft 0.8 d o...(given) We know that moment of inertia of a hollow shaft, π 4 4 I ( do) ( di) 64 and cross-sectional area of the hollow shaft, π A ( do) ( di) 4 Radius of gyration of the hollow shaft, K I A π 4 4 [( do) ( di) ] 64 π [( do) ( di) ] 4 o + i o i o + i 16 [( do) ( di) ] 16 [( d ) ( d ) ] [( d ) ( d ) ] ( d ) ( d ) o i o d d d 1+ 1 (0.8) 4 d + o 4 and column factor for compressive loads, 0. d o 1 1 α 1 0.0044 ( L/ K) 1 0.0044 (1500 / 0. d o ) 1 do 1 0.6 / do do 0.6 We know that equivalent twisting moment for a hollow shaft, T e α Fdo (1 + k ) Km M + + ( Kt T) 8 do 15000 do (1 0.8 ) do 0.6 + 1.6 1000 10 + + (1.5 500 10 ) 8 * Superfluous data. 075 ( d ) o 1600 10 + + (750 10 ) do 0.6...(i)

Shafts 549 We also know that equivalent twisting moment for a hollow shaft, T e π τ 4 ( do ) (1 k ) 16 and 16 Equating equations (i) and (ii we have 4.65 (d o ) π 4 40 ( d o ) (1 0.8 ) 4.65 (do )...(ii) 075 ( d ) o do 0.6 1600 10 + + (750 10 ) Solving this expression by hit and trial method, we find that d o 76. say 80 mm Ans. d i 0.8 d o 0.8 80 64 mm Ans....(iii) Note : In order to find the minimum value of d o to be used for the hit and trial method, determine the equivalent twisting moment without considering the axial compressive load. We know that equivalent twisting moment, T e ( K M) + ( K T) (1.6 1000 10 ) + (1.5 500 10 )... (iv) m 1767 10 N-mm Equating equations (ii) and (iv 4.65(d o ) 1767 10 or (d o ) 1767 10 /4.65 80 10 d o 7.4 mm Thus the value of d o to be substituted in equation (iii) must be greater than 7.4 mm. 14.14 Design of Shafts on the basis of Rigidity Sometimes the shafts are to be designed on the basis of rigidity. We shall consider the following two types of rigidity. 1. Torsional rigidity. The torsional rigidity is important in the case of camshaft of an I.C. engine where the timing of the valves would be effected. The permissible amount of twist should not exceed 0.5 per metre length of such shafts. For line shafts or transmission shafts, deflections.5 to degree per metre length may be used as limiting value. The widely used deflection for the shafts is limited to 1 degree in a length equal to twenty times the diameter of the shaft. The torsional deflection may be obtained by using the torsion equation, T G. θ T. L or θ J L J. G where θ Torsional deflection or angle of twist in radians, T Twisting moment or torque on the shaft, J Polar moment of inertia of the cross-sectional area about the axis of rotation, π 4 d...(for solid shaft) π 4 4 ( do) ( di)...(for hollow shaft) G Modulus of rigidity for the shaft material, and L Length of the shaft.. Lateral rigidity. It is important in case of transmission shafting and shafts running at high speed, where small lateral deflection would cause huge out-of-balance forces. The lateral rigidity is also important for maintaining proper bearing clearances and for correct gear teeth alignment. If the shaft is of uniform cross-section, then the lateral deflection of a shaft may be obtained by using the deflection formulae as in Strength of Materials. But when the shaft is of variable cross-section, then t

550 A Textbook of Machine Design Rotor blades Rotor blades Engine Gearbox Air acclerating downwards, pushed by the rotating blades, produced an upwards reaction that lifts the helicopter. Note : This picture is given as additional information and is not a direct example of the current chapter. the lateral deflection may be determined from the fundamental equation for the elastic curve of a beam, i.e. d y M dx EI Example 14.1. A steel spindle transmits 4 kw at 800 r.p.m. The angular deflection should not exceed 0.5 per metre of the spindle. If the modulus of rigidity for the material of the spindle is 84 GPa, find the diameter of the spindle and the shear stress induced in the spindle. π Solution. Given : P 4 kw 4000 W ; N 800 r.p.m. ; θ 0.5 0.5 0.0044 rad ; 180 L 1 m 1000 mm ; G 84 GPa 84 10 9 N/m 84 10 N/mm Diameter of the spindle d Diameter of the spindle in mm. We know that the torque transmitted by the spindle, P 60 4000 60 T 47.74 N-m 47 740 N-mm πn π 800 T We also know that G θ T l or J J L G θ π 4 47 740 1000 or d 19 167 84 10 0.0044 d 4 19 167 / π 1. 10 6 or d.87 say 5 mm Ans. Shear stress induced in the spindle τ Shear stress induced in the spindle. We know that the torque transmitted by the spindle (T π π 47 740 τ d τ (5) 840 τ 16 16 τ 47 740 / 840 5.67 N/mm 5.67 MPa Ans. Example 14.. Compare the weight, strength and stiffness of a hollow shaft of the same external diameter as that of solid shaft. The inside diameter of the hollow shaft being half the external diameter. Both the shafts have the same material and length. Solution. Given : d o d ; d i d o / or k d i / d o 1 / 0.5

Shafts 551 Comparison of weight We know that weight of a hollow shaft, W H Cross-sectional area Length Density π ( do) ( di) Length Density...(i) 4 and weight of the solid shaft, W S π d Length Density...(ii) 4 Since both the shafts have the same material and length, therefore by dividing equation (i) by equation (ii we get WH W ( do) ( di) ( do) ( di)...(q d d S d ( do ) o ) ( di ) 1 1 k 1 (0.5) 0.75 Ans. ( do ) Comparison of strength We know that strength of the hollow shaft, and strength of the solid shaft, T H π τ 4 ( do ) (1 k )...(iii) 16 π T S τ d...(iv) 16 Dividing equation (iii) by equation (iv we get 4 4 TH ( do) (1 k ) ( do) (1 k ) T 1 k d ( d ) 4...(Q d d o ) Comparison of stiffness We know that stiffness S 1 (0.5) 4 0.975 Ans. T θ G J L o The propeller shaft of this heavy duty helicopter is subjected to very high torsion.