980 A Textbook of Machine Design
ZN 3 K = p ∴ Bearing modulus at the minimum point of friction, 1.⎛⎞ZN =× 1 = K = ⎜⎟ 28 9.33 33⎝⎠p ⎛⎞ZN. = Since the calculated value of bearing characteristic number ⎜⎟12.24 is more than 9.33, ⎝⎠p therefore the bearing will operate under hydrodynamic conditions. 4. From Table 26.3, we find that for centrifugal pumps, the clearance ratio (c/d) = 0.0013 5. We know that coefficient of friction,
33⎛⎞⎛⎞ZN d += 33 × × 1 + μ = ⎜⎟⎜⎟ k 12.24 0.002 1088⎝⎠pc⎝⎠ 10 0.0013 = 0.0031 + 0.002 = 0.0051 ... [From Art. 26.13, k = 0.002] 6. Heat generated, ⎛⎞πdN. ⎛⎞πdN. Q = μ=μWV W⎜⎟W ...⎜⎟QV = g ⎝⎠60 ⎝⎠60 ⎛⎞π×0.1 × 900 = 0.0051×= 20000⎜⎟ 480.7 W ⎝⎠60 ... (d is taken in metres) 7. Heat dissipated,
Qd = C.A (tb – ta) = C.l.d (tb – ta) W ... ( QA = l × d) We know that
1 1 (tb – ta)= 2 (t0 – ta) = 2 (55°– 15.5°) = 19.75°C ∴ Qd = 1232 × 0.16 × 0.1 × 19.75 = 389.3 W ... (l and d are taken in metres) We see that the heat generated is greater than the heat dissipated which indicates that the bear- ing is warming up. Therefore, either the bearing should be redesigned by taking t0 = 63°C or the bearing should be cooled artificially. We know that the amount of artificial cooling required
= Heat generated – Heat dissipated = Qg – Qd = 480.7 – 389.3 = 91.4 W Mass of lubricating oil required for artificial cooling Let m = Mass of the lubricating oil required for artificial cooling in kg / s. We know that the heat taken away by the oil,
Qt = m.S.t = m × 1900 × 10 = 19 000 m W ... [Q Specific heat of oil (S) = 1840 to 2100 J/kg/°C] Equating this to the amount of artificial cooling required, we have 19 000 m = 91.4 ∴ m = 91.4 / 19 000 = 0.0048 kg / s = 0.288 kg / min Ans. Example 26.2. The load on the journal bearing is 150 kN due to turbine shaft of 300 mm diameter running at 1800 r.p.m. Determine the following : Sliding Contact Bearings 981
1. Length of the bearing if the allowable bearing pressure is 1.6 N/mm2, and 2. Amount of heat to be removed by the lubricant per minute if the bearing temperature is 60°C and viscosity of the oil at 60°C is 0.02 kg/m-s and the bearing clearance is 0.25 mm. Solution. Given : W = 150 kN = 150 × 103 N; d = 300 mm = 0.3 m ; N = 1800 r.p.m. ; p = 1.6 N/mm2 ; Z = 0.02 kg / m-s ; c = 0.25 mm 1. Length of the bearing Let l = Length of the bearing in mm. We know that projected bearing area, A = l × d = l × 300 = 300 l mm2 and allowable bearing pressure ( p), W 150× 103 500 1.6 = == All300 ∴ l = 500 / 1.6 = 312.5 mm Ans. Axle bearing 2. Amount of heat to be removed by the lubricant We know that coefficient of friction for the bearing, × 33⎛⎞⎛⎞ZN . d += 33 ⎛ 0.02 1800 ⎞⎛⎞ 300 + μ = ⎜⎟⎜⎟k ⎜ ⎟⎜ ⎟ 0.002 1088⎝⎠pc⎝⎠ 10 ⎝1.6 ⎠⎝ 0.25 ⎠ = 0.009 + 0.002 = 0.011 Rubbing velocity, ππ××dN. 0.3 1800 V = ==28.3 m/s 60 60 ∴ Amount of heat to be removed by the lubricant, μ 3 Qg = .W.V = 0.011 × 150 × 10 × 28.3 = 46 695 J/s or W = 46.695 kW Ans. ... ( 1 J/s = 1 W) Example 26.3. A full journal bearing of 50 mm diameter and 100 mm long has a bearing pressure of 1.4 N/mm2. The speed of the journal is 900 r.p.m. and the ratio of journal diameter to the diametral clearance is 1000. The bearing is lubricated with oil whose absolute viscosity at the operating temperature of 75°C may be taken as 0.011 kg/m-s. The room temperature is 35°C. Find : 1. The amount of artificial cooling required, and 2. The mass of the lubricating oil required, if the difference between the outlet and inlet temperature of the oil is 10°C. Take specific heat of the oil as 1850 J / kg / °C. Solution. Given : d = 50 mm = 0.05 m ; l = 100 mm = 0.1 m; p = 1.4 N/mm2 ; N = 900 r.p.m. ; d / c = 1000 ; Z = 0.011 kg / m-s ; t0 = 75°C ; ta = 35°C ; t = 10°C ; S = 1850 J/kg / °C 1. Amount of artificial cooling required We know that the coefficient of friction, × 33⎛⎞⎛⎞ZN d += 33 ⎛ 0.011 900 ⎞ + μ = ⎜⎟⎜⎟k ⎜ ⎟(1000) 0.002 1088⎝⎠pc⎝⎠ 10 ⎝1.4 ⎠ = 0.002 33 + 0.002 = 0.004 33 Load on the bearing, W = p × d.l = 1.4 × 50 × 100 = 7000 N 982 A Textbook of Machine Design and rubbing velocity, ππ××dN.0.05900 V = ==2.36 m/s 60 60 ∴ Heat generated, μ Qg = .W.V = 0.004 33 × 7000 × 2.36 = 71.5 J/s Let tb = Temperature of the bearing surface. We know that 1 1 (t – t )= (t – t ) = (75 – 35) = 20°C b a 2 0 a 2 Since the value of heat dissipation coefficient (C ) for unventilated bearing varies from 140 to 420 W/m2/°C, therefore let us take C = 280 W/m2 /° C We know that heat dissipated,
Qd = C.A (tb – ta ) = C.l.d (tb – ta) = 280 × 0.05 × 0.1 × 20 = 28 W = 28 J/s ∴ Amount of artificial cooling required
= Heat generated – Heat dissipated = Qg – Qd = 71.5 – 28 = 43.5 J/s or W Ans. 2. Mass of the lubricating oil required Let m = Mass of the lubricating oil required in kg / s. We know that heat taken away by the oil,
Qt = m.S.t = m × 1850 × 10 = 18 500 m J/s Since the heat generated at the bearing is taken away by the lubricating oil, therefore equating
Qg = Qt or 71.5 = 18 500 m ∴ m = 71.5 / 18 500 = 0.003 86 kg / s = 0.23 kg / min Ans. Example 26.4. A 150 mm diameter shaft supporting a load of 10 kN has a speed of 1500 r.p.m. The shaft runs in a bearing whose length is 1.5 times the shaft diameter. If the diametral clearance of the bearing is 0.15 mm and the absolute viscosity of the oil at the operating temperature is 0.011 kg/m-s, find the power wasted in friction. Solution. Given : d = 150 mm = 0.15 m ; W = 10 kN = 10 000 N ; N = 1500 r.p.m. ; l = 1.5 d ; c = 0.15 mm ; Z = 0.011 kg/m-s We know that length of bearing, l = 1.5 d = 1.5 × 150 = 225 mm ∴ Bearing pressure, WW 10000 p = == =0.296 N/mm2 Ald. 225× 150 We know that coefficient of friction, × 33⎛⎞⎛⎞ZN d += 33 ⎛ 0.011 1500 ⎞⎛⎞ 150 + μ = ⎜⎟⎜⎟k ⎜ ⎟⎜ ⎟ 0.002 1088⎝⎠pc⎝⎠ 10 ⎝0.296 ⎠⎝ 0.15 ⎠ = 0.018 + 0.002 = 0.02 ππ××dN. 0.15 1500 and rubbing velocity, V = ==11.78 m / s 60 60 Sliding Contact Bearings 983
We know that heat generated due to friction, μ Qg = .W.V = 0.02 × 10 000 × 11.78 = 2356 W ∴ Power wasted in friction
= Qg = 2356 W = 2.356 kW Ans. Example 26.5. A 80 mm long journal bearing supports a load of 2800 N on a 50 mm diameter shaft. The bearing has a radial clearance of 0.05 mm and the viscosity of the oil is 0.021 kg / m-s at the operating temperature. If the bearing is capable of dissipating 80 J/s, determine the maximum safe speed. Solution. Given : l = 80 mm ; W = 2800 N ; d = 50 mm ; = 0.05 m ; c / 2 = 0.05 mm or c = 0.1 mm ; Z = 0.021 kg/m-s ; Qd = 80 J/s Let N = Maximum safe speed in r.p.m. We know that bearing pressure, W 2800 p = ==0.7 N/mm2 ld.8050× and coefficient of friction, 33⎛⎞⎛⎞ZN d+= 33 ⎛ 0.021 N ⎞⎛⎞ 50 + μ = ⎜⎟⎜⎟0.002 ⎜ ⎟⎜⎟ 0.002 1088⎝⎠pc⎝⎠ 10 ⎝0.7 ⎠⎝⎠ 0.1 495 N = + 0.002 108
Front hub-assembly bearing ⎛⎞π dN ∴ Heat generated, Q = μ=μ..WV . W ⎜⎟ J/s g ⎝⎠60 ⎛⎞⎛⎞495NNπ× 0.05 = ⎜⎟+ 0.002 2800⎜⎟ ⎝⎠108 ⎝⎠60 3628 N 2 = + 0.014 66 N 108 Equating the heat generated to the heat dissipated, we have 3628 N 2 +=0.014 66N 80 108 984 A Textbook of Machine Design or N2 + 404 N – 2.2 × 106 = 0
−±404 (404)26 +×× 4 2.2 10 ∴ N = 2 −±404 2994 = = 1295 r.p.m Ans. ... (Taking +ve sign) 2 Example 26.6. A journal bearing 60 mm is diameter and 90 mm long runs at 450 r.p.m. The oil used for hydrodynamic lubrication has absolute viscosity of 0.06 kg / m-s. If the diametral clearance is 0.1 mm, find the safe load on the bearing. Solution. Given : d = 60 mm = 0.06 m ; l = 90 mm = 0.09 m ; N = 450 r.p.m. ; Z = 0.06 kg / m-s ; c = 0.1 mm First of all, let us find the bearing pressure ( p ) by using Sommerfeld number. We know that 2 Z Nd⎛⎞ ⎜⎟= 14.3× 106 pc⎝⎠ 2 × 6 0.06× 450⎛⎞ 60 669.72 10 ⎜⎟= 14.3×=× 10 or 14.3 10 p ⎝⎠0.1 p ∴ p = 9.72 × 106 / 14.3 × 106 = 0.68 N/mm2 We know that safe load on the bearing, W=p.A = p.l.d = 0.68 × 90 × 60 = 3672 N Ans. 26.20 Solid Journal Bearing A solid bearing, as shown in Fig. 26.9, is the simplest form of journal bearing. It is simply a block of cast iron with a hole for a shaft providing running fit. The lower portion of the block is extended to form a base plate or sole with two holes to receive bolts for fastening it to the frame. An oil hole is drilled at the top for lubrication. The main disadvantages of this bearing are
Fig. 26.9. Solid journal bearing. Fig. 26.10. Bushed bearing. 1. There is no provision for adjustment in case of wear, and 2. The shaft must be passed into the bearing axially, i.e. endwise. Since there is no provision for wear adjustment, therefore this type of bearing is used when the shaft speed is not very high and the shaft carries light loads only. 26.21 Bushed Bearing A bushed bearing, as shown in Fig. 26.10, is an improved solid bearing in which a bush of brass or gun metal is provided. The outside of the bush is a driving fit in the hole of the casting whereas the inside is a running fit for the shaft. When the bush gets worn out, it can be easily replaced. In small bearings, the frictional force itself holds the bush in position, but for shafts transmitting high power, grub screws are used for the prevention of rotation and sliding of the bush. Sliding Contact Bearings 985
Bronze bushed bearing assemblies 26.22 Split Bearing or Plummer Block A split-bearing is used for shafts running at high speeds and carrying heavy loads. A split- bearing, as shown in Fig. 26.11, consists of a cast iron base (also called block or pedestal), gunmetal or phosphor bronze brasses, bushes or steps made in two-halves and a cast iron cap. The two halves of the brasses are held together by a cap or cover by means of mild steel bolts and nuts. Sometimes thin shims are introduced between the cap and the base to provide an adjustment for wear. When the bottom wears out, one or two shims are removed and then the cap is tightened by means of bolts.
Fig. 26.11. Split bearing or plummer block. 986 A Textbook of Machine Design
The brasses are provided with collars or flanges on either side in order to prevent its axial movement. To prevent its rotation along with the shaft, the following four methods are usually used in practice. 1. The sungs are provided at the sides as shown in Fig. 26.12 (a). 2. A sung is provided at the top, which fits inside the cap as shown in Fig. 26.12 (b). The oil hole is drilled through the sung. 3. The steps are made rectangular on the outside and they are made to fit inside a corresponding hole, as shown in Fig. 26.12 (c). 4. The steps are made octagonal on the outside and they are made to fit inside a corresponding hole, as shown in Fig. 26.12 (d). The split bearing must be lubricated properly.
Fig. 26.12. Methods of preventing rotation of brasses. 26.23 Design of Bearing Caps and Bolts When a split bearing is used, the bearing cap is tightened on the top. The load is usually carried by the bearing and not the cap, but in some cases e.g. split connecting rod ends in double acting steam engines, a considerable load comes on the cap of the bearing. Therefore, the cap and the holding down bolts must be designed for full load. The cap is generally regarded as a simply supported beam, supported by holding down bolts and loaded at the centre t as shown in Fig. 26.13. W Let W = Load supported at the centre, α = Distance between centres a of holding down bolts, Fig. 26.13. Bearing cap. l = Length of the bearing, and t = Thickness of the cap. We know that maximum bending moment at the centre, M = W.a /4 and the section modulus of the cap, Z = l.t2 /6 Sliding Contact Bearings 987
∴ Bending stress, MWa.63. Wa σ = =×= b Z 4 lt.2.22 lt 3.Wa and t = σ 2.b l Note : When an oil hole is provided in the cap, then the diameter of the hole should be subtracted from the length of the bearing. The cap of the bearing should also be investigated for the stiffness. We know that for a simply supported beam loaded at the centre, the deflection,
Wa...33 Wa Wa 3 ⎛⎞3 δ == = lt. = 33 ...⎜⎟Q I 48EI . lt.4. Elt ⎝⎠12 48 E × 12 1/3 ⎡⎤W ∴ t = 0.63 a ⎣⎦⎢⎥El..δ The deflection of the cap should be limited to about 0.025 mm. In order to design the holding down bolts, the load on each bolt is taken 33% higher than the 4 W normal load on each bolt. In other words, load on each bolt is taken , where n is the number of 3n bolts used for holding down the cap.
Let dc = Core diameter of the bolt, and σ t = Tensile stress for the material of the bolt. π 4 W ∴ ()d 2 σ = × 4 ct3 n
From this expression, the core diameter (dc) may be calculated. After finding the core diameter, the size of the bolt is fixed. 26.24 Oil Groves The oil grooves are cut into the plain bearing surfaces to assist in the distribution of the oil between the rubbing surfaces. It prevents squeezing of the oil film from heavily loaded low speed journals and bearings. The tendency to squeeze out oil is greater in low speed than in high speed bearings, because the oil has greater wedging A self-locking nut used in bearing assemblies. action at high speeds. At low speeds, the journal rests upon a given area of oil film for a longer period of time, tending to squeeze out the oil over the area of greatest pressure. The grooves function as oil reservoirs which holds and distributes the oil especially during starting or at very low speeds. The oil grooves are cut at right angles to the line of the load. The circumferential and diagonal grooves should be avoided, if possible. The effectiveness of the oil grooves is greatly enhanced if the edges of grooves are chamfered. The shallow and narrow grooves with chamfered edges distributes the oil more evenly. A chamfered edge should always be provided at the parting line of the bearing. Example 26.7. A wall bracket supports a plummer block for 80 mm diameter shaft. The length of bearing is 120 mm. The cap of bearing is fastened by means of four bolts, two on each side of the shaft. The cap is to withstand a load of 16.5 kN. The distance between the centre lines of the bolts is 988 A Textbook of Machine Design
150 mm. Determine the thickness of the bearing cap and the diameter of the bolts. Assume safe stresses in tension for the material of the cap, which is cast iron, as 15 MPa and for bolts as 35 MPa. Also check the deflection of the bearing cap taking E = 110 kN / mm2. Solution : Given : d = 80 mm ; l = 120 mm ; n = 4 ; W = 16.5 kN = 16.5 × 103 N; a = 150 mm ; σ 2 σ 2 2 3 2 b = 15 MPa = 15 N/mm ; t = 35 MPa = 35 N/mm ; E = 110 kN/mm = 110 × 10 N/mm Thickness of the bearing cap We know that thickness of the bearing cap, 3Wa . 3××× 16.5 103 150 ==2062.5 t = σ×× 2.b l 215120 = 45.4 say 46 mm Ans. Diameter of the bolts
Let dc = Core diameter of the bolts. We know that π 4 W ()d 2 σ= × 43ct n π×416.5103 or ()35d 23=× = 5.510 × 434c 5.5×× 103 4 ∴ (d )2 = ==200 ord 14.2 mm Ans. c π×35 c Deflection of the cap We know that deflection of the cap, Wa.333 16.5× 10 (150) δ = ==0.0108 mm Ans. 4Elt . .333 4××× 110 10 120(46) Since the limited value of the deflection is 0.025 mm, therefore the above value of deflection is within limits. 26.25 Thrust Bearings A thrust bearing is used to guide or support the shaft which is subjected to a load along the axis of the shaft. Such type of bearings are mainly used in turbines and propeller shafts. The thrust bearings are of the following two types : 1. Foot step or pivot bearings, and 2. Collar bearings. In a foot step or pivot bearing, the loaded shaft is vertical and the end of the shaft rests within the bearing. In case of collar bearing, the shaft continues through the bearing. The shaft may be vertical or horizontal with single collar or many collars. We shall now discuss the design aspects of these bearings in the following articles. 26.26 Footstep or Pivot Bearings A simple type of footstep bearing, suitable for a slow running and lightly loaded shaft, is shown in Fig. 26.14. If the shaft is not of steel, its end Footstep bearing Sliding Contact Bearings 989 must be fitted with a steel face. The shaft is guided in a gunmetal bush, pressed into the pedestal and prevented from turning by means of a pin. Since the wear is proportional to the velocity of the rubbing surface, which (i.e. rubbing velocity) increases with the distance from the axis (i.e. radius) of the bearing, therefore the wear will be different at different radii. Due to this wear, the distribution of pressure over the bearing surface is not
Fig. 26.14. Footstep or pivot bearings. uniform. It may be noted that the wear is maximum at the outer radius and zero at the centre. In order to compensate for end wear, the following two methods are employed. 1. The shaft is counter-bored at the end, as shown in Fig. 26.14 (a). 2. The shaft is supported on a pile of discs. It is usual practice to provide alternate discs of different materials such as steel and bronze, as shown in Fig. 26.14 (b), so that the next disc comes into play, if one disc seizes due to improper lubrication. It may be noted that a footstep bearing is difficult to lubricate as the oil is being thrown outwards from the centre by centrifugal force. In designing, it is assumed that the pressure is uniformly distributed throughout the bearing surface. Let W = Load transmitted over the bearing surface, R = Radius of the bearing surface (or shaft), A = Cross-sectional area of the bearing surface, p = Bearing pressure per unit area of the bearing surface between rubbing surfaces, μ = Coefficient of friction, and N = Speed of the shaft in r.p.m. When the pressure in uniformly distributed over the bearing area, then WW p = = A π R2 and the total frictional torque, 2 T = μ..WR 3 ∴ Power lost in friction, 2.π NT P = watts ... (T being in N-m) 60 990 A Textbook of Machine Design
Notes : 1. When the counter-boring of the shaft is considered, then the bearing pressure, W p = , where r = Radius of counter-bore, π(R22 − r ) and the total frictional torque, ⎛⎞33− 2 μ R r T = .W ⎜⎟22 3 ⎝⎠R − r 2. The allowable bearing pressure (p) for the footstep bearings may be taken as follows : (a) For rubbing speeds (V) from 15 to 60 m/min, the bearing pressure should be such that p.V ≤ 42, when p is in N/mm2 and V in m/min. (b) For rubbing speeds over 60 m/min., the pressure should not exceed 0.7 N/mm2. (c) For intermittent service, the bearing pressure may be taken as 10.5 N/mm2. (d) For very slow speeds, the bearing pressure may be taken as high as 14 N/mm2. 3. The coefficient of friction for the footstep bearing may be taken as 0.015. 26.27 Collar Bearings We have already discussed that in a collar bearing, the shaft continues through the bearing. The shaft may be vertical or horizontal, with single collar or many collars. A simple multicollar bearing for horizontal shaft is shown in Fig. 26.15. The collars are either integral parts of the shaft or rigidly fastened to it. The outer diameter of the collar is usually taken as 1.4 to 1.8 times the inner diameter of the collar (i.e. diameter of the shaft). The thickness of the collar is kept as one-sixth diameter of the shaft and clearance between collars as one-third diameter of the shaft. In designing collar bearings, it is assumed that the pressure is uniformly distributed over the bearing surface. Collar bearings Let W = Load transmitted over the bearing surface, n = Number of collars, R = Outer radius of the collar, r = Inner radius of the collar, A = Cross-sectional area of the bearing surface = n π (R2 – r2), p = Bearing pressure per unit area of the bearing surface, between rubbing surfaces, μ = Coefficient of friction, and N = Speed of the shaft in r.p.m. When the pressure is uniformly distributed over the bearing surface, then bearing pressure, WW p = = A nRr.(π−22 ) and the total frictional torque, 2 ⎛⎞R33− r μ. W T = ⎜⎟22 3 ⎝⎠R − r Sliding Contact Bearings 991
Fig. 26.15. Collar bearing. ∴ Power lost in friction, 2.π NT P = watts ... (when T is in N-m) 60 Notes : 1. The coefficient of friction for the collar bearings may be taken as 0.03 to 0.05. 2. The bearing pressure for a single collar and water cooled multi-collared bearings may be taken same as for footstep bearings. Example 26.8. A footstep bearing supports a shaft of 150 mm diameter which is counter- bored at the end with a hole diameter of 50 mm. If the bearing pressure is limited to 0.8 N/mm2 and the speed is 100 r.p.m.; find : 1. The load to be supported; 2. The power lost in friction; and 3. The heat generated at the bearing. Assume coefficient of friction = 0.015. Solution. Given : D = 150 mm or R = 75 mm ; d = 50 mm or r = 25 mm ; p = 0.8 N/mm2 ; N = 100 r.p.m. ; μ = 0.015 1. Load to be supported Let W = Load to be supported. Assuming that the pressure is uniformly distributed over the bearing surface, therefore bearing pressure ( p), WWW 0.8 = == π−(Rr22 ) π [(75) 2 − (25) 2 ] 15 710 ∴ W = 0.8 × 15 710 = 12 568 N Ans. 2. Power lost in friction We know that total frictional torque, 2 ⎛⎞R33− r μ.W T = ⎜⎟22 3 ⎝⎠R − r 2⎡⎤ (75)33− (25) ××0.015 12568 N-mm = ⎢⎥22 3 ⎣⎦(75)− (25)
= 125.68 × 81.25 = 10 212 N-mm = 10.212 N-m ∴ Power lost in friction, 2ππ××NT 2 100 10.212 P = ===107 W 0.107 kW Ans. 60 60 992 A Textbook of Machine Design
3. Heat generated at the bearing We know that heat generated at the bearing = Power lost in friction = 0.107 kW or kJ / s = 0.107 × 60 = 6.42 kJ/min Ans. Example 26.9. The thrust of propeller shaft is absorbed by 6 collars. The rubbing surfaces of these collars have outer diameter 300 mm and inner diameter 200 mm. If the shaft runs at 120 r.p.m., the bearing pressure amounts to 0.4 N/mm2. The coefficient of friction may be taken as 0.05. Assuming that the pressure is uniformly distributed, determine the power absorbed by the collars. Solution. Given : n = 6 ; D = 300 mm or R = 150 mm ; d = 200 mm or r = 100 mm ; N = 120 r.p.m. ; p = 0.4 N/mm2 ; μ = 0.05 First of all, let us find the thrust on the shaft (W). Since the pressure is uniformly distributed over the bearing surface, therefore bearing pressure ( p), WWW 0.4 = == nRπ−π−(22 r ) 6 [(150) 2 (100) 2 ] 235650 ∴ W = 0.4 × 235 650 = 94 260 N We know that total frictional torque,
2⎛⎞Rr33−− 2 ⎡ (150) 3 (100) 3 ⎤ T = μ=××.W 0.05 94260 N-mm ⎜⎟22⎢⎥ 2 2 33⎝⎠Rr−− ⎣(150) (100) ⎦ = 597 000 N-mm = 597 N-m ∴ Power absorbed by the collars, 2ππ×× .NT . 2 120 597 P = ===7503 W 7.503 kW Ans. 60 60 Example 26.10. The thrust of propeller shaft in a marine engine is taken up by a number of collars integral with the shaft which is 300 mm is diameter. The thrust on the shaft is 200 kN and the speed is 75 r.p.m. Taking μ constant and equal to 0.05 and assuming the bearing pressure as uniform and equal to 0.3 N/mm2, find : 1. Number of collars required, 2. Power lost in friction, and 3. Heat generated at the bearing in kJ/min. Solution. Given : d = 300 mm or r = 150 mm ; W = 200 kN = 200 × 103 N; N = 75 r.p.m. ; μ = 0.05 ; p = 0.3 N/mm2 1. Number of collars required Let n = Number of collars required. Since the outer diameter of the collar (D) is taken Industrial bearings. as 1.4 to 1.8 times the diameter of shaft (d ), therefore let us take D = 1.4 d = 1.4 × 300 = 420 mm or R = 210 mm We know that the bearing pressure ( p), × 3 W ==200 10 2.947 0.3 = nRrπ−(22 ) n π [(210) 2 − (150) 2 ] n ∴ n = 2.947 / 0.3 = 9.8 say 10 Ans. Sliding Contact Bearings 993
2. Power lost in friction We know that total frictional torque, 2⎛⎞Rr33−− 2 ⎡ (210) 3 (150) 3 ⎤ μ=×××W 0.05 200 103 N-mm T = ⎜⎟22⎢⎥ 2 2 33⎝⎠Rr−− ⎣(210) (150) ⎦ =1817 × 103 N-mm = 1817 N-m ∴ Power lost in friction, 2ππ××NT . 2 75 1817 P = ===14 270 W 14.27 kW Ans. 60 60 3. Heat generated at the bearing We know that heat generated at the bearing = Power lost in friction = 14.27 kW or kJ/s = 14.27 × 60 = 856.2 kJ/min Ans.
EXERCISES 1. The main bearing of a steam engine is 100 mm in diameter and 175 mm long. The bearing supports a load of 28 kN at 250 r.p.m. If the ratio of the diametral clearance to the diameter is 0.001 and the absolute viscosity of the lubricating oil is 0.015 kg/m-s, find : 1. The coefficient of friction ; and 2. The heat generated at the bearing due to friction. [Ans. 0.002 77 ; 101.5 J/s] 2. A journal bearing is proposed for a steam engine. The load on the journal is 3 kN, diameter 50 mm, length 75 mm, speed 1600 r.p.m., diametral clearance 0.001 mm, ambient temperature 15.5°C. Oil SAE 10 is used and the film temperature is 60°C. Determine the heat generated and heat dissipated. Take absolute viscosity of SAE10 at 60°C = 0.014 kg/m-s. [Ans. 141.3 J/s ; 25 J/s] 3. A 100 mm long and 60 mm diameter journal bearing supports a load of 2500 N at 600 r.p.m. If the room temperature is 20°C, what should be the viscosity of oil to limit the bearing surface temperature to 60°C? The diametral clearance is 0.06 mm and the energy dissipation coefficient based on projected area of bearing is 210 W/m2/°C. [Ans. 0.0183 kg/m-s] 4. A tentative design of a journal bearing results in a diameter of 75 mm and a length of 125 mm for supporting a load of 20 kN. The shaft runs at 1000 r.p.m. The bearing surface temperature is not to exceed 75°C in a room temperature of 35°C. The oil used has an absolute viscosity of 0.01 kg/m-s at the operating temperature. Determine the amount of artificial cooling required in watts. Assume d/c = 1000. [Ans. 146 W] 5. A journal bearing is to be designed for a centrifugal pump for the following data : Load on the journal = 12 kN ; Diameter of the journal = 75 mm ; Speed = 1440 r.p.m ; Atmospheric temperature of the oil = 16°C ; Operating temperature of the oil = 60°C; Absolute viscosity of oil at 60°C = 0.023 kg/m-s. Give a systematic design of the bearing. 6. Design a journal bearing for a centrifugal pump running at 1440 r.p.m. The diameter of the journal is 100 mm and load on each bearing is 20 kN. The factor ZN/p may be taken as 28 for centrifugal pump bearings. The bearing is running at 75°C temperature and the atmosphere temperaturic is 30°C. The energy dissipation coefficient is 875 W/m2/°C. Take diametral clearance as 0.1 mm. 7. Design a suitable journal bearing for a centrifugal pump from the following available data : Load on the bearing = 13.5 kN; Diameter of the journal = 80 mm; Speed = 1440 r.p.m.; Bearing characterisitic number at the working temperature (75°C) = 30 ; Permissible bearing pressure intensity 994 A Textbook of Machine Design
= 0.7 N/mm2 to 1.4 N/mm2; Average atmospheric temperature = 30°C. Calculate the cooling requirements, if any. 8. A journal bearing with a diameter of 200 mm and length 150 mm carries a load of 20 kN, when the journal speed is 150 r.p.m. The diametral clearance ratio is 0.0015. If possible, the bearing is to operate at 35°C ambient temperature without external cooling with a maximum oil temperature of 90°C. If external cooling is required, it is to be as little as possible to minimise the required oil flow rate and heat exchanger size. 1. What type of oil do you recommend ? 2. Will the bearing operate without external cooling? 3. If the bearing operates without external cooling, determine the operating oil temperature? 4. If the bearing operates with external cooling, determine the amount of oil in kg/min required to carry away the excess heat generated over heat dissipated, when the oil temperature rises from 85°C to 90°C, when passing through the bearing.
QUESTIONS 1. What are journal bearings? Give a classification of these bearings. 2. What is meant by hydrodynamic lubrication? 3. List the basic assumptions used in the theory of hydrodynamic lubrication. 4. Explain wedge film and squeeze film journal bearings. 5. Enumerate the factors that influence most the formation and maintenance of the thick oil film in hydrodynamic bearings. 6. Make sketches to show the pressure distribution in a journal bearing with thick film lubrication in axial and along the circumference. 7. List the important physical characteristics of a good bearing material. 8. What are the commonly used materials for sliding contact bearings? 9. Write short note on the lubricants used in sliding contact bearings. 10. Explain the following terms as applied to journal bearings : (a) Bearing characteristic number ; and (b) Bearing modulus. 11. What are the various terms used in journal bearings analysis and design? Give their definitions in brief. 12. Explain with reference to a neat plot the importance of the bearing characteristic curve. 13. What is the procedure followed in designing a journal bearing? 14. Explain with sketches the working of different types of thrust bearing. OBJECTIVE TYPE QUESTIONS 1. In a full journal bearing, the angle of contact of the bearing with the journal is (a) 120° (b) 180° (c) 270° (d) 360° 2. A sliding bearing which can support steady loads without any relative motion between the journal and the bearing is called (a) zero film bearing (b) boundary lubricated bearing (c) hydrodynamic lubricated bearing (d) hydrostatic lubricated bearing Sliding Contact Bearings 995
3. In a boundary lubricated bearing, there is a ...... of lubricant between the journal and the bearing. (a) thick film (b) thin film 4. When a shaft rotates in anticlockwise direction at slow speed in a bearing, then it will (a) have contact at the lowest point of bearing (b) move towards right of the bearing making metal to metal contact (c) move towards left of the bearing making metal to metal contact (d) move towards right of the bearing making no metal to metal contact 5. The property of a bearing material which has the ability to accommodate small particles of dust, grit etc., without scoring the material of the journal, is called (a) bondability (b) embeddability (c) comformability (d) fatigue strength 6. Teflon is used for bearings because of (a) low coefficient of friction (b) better heat dissipation (c) smaller space consideration (d) all of these 7. When the bearing is subjected to large fluctuations of load and heavy impacts, the bearing characteristic number should be ...... the bearing modulus. (a) 5 times (b) 10 times (c) 15 times (d) 20 times 8. When the length of the journal is equal to the diameter of the journal, then the bearing is said to be a (a) short bearing (b) long bearing (c) medium bearing (d) square bearing 9. If Z = Absolute viscosity of the lubricant in kg/m-s, N = Speed of the journal in r.p.m., and p = Bearing pressure in N/mm2, then the bearing characteristic number is Z N Z p (a) (b) p N Z pN (c) (d) pN Z 10. In thrust bearings, the load acts (a) along the axis of rotation (b) parallel to the axis of rotation (c) perpendicular to the axis of rotation (d) in any direction
ANSWERS 1. (d) 2. (d) 3. (b) 4. (c) 5. (b) 6. (a) 7. (c) 8. (d) 9. (a) 10. (a) 996 A Textbook of Machine Design
C H A P T E R 27 Rolling Contact Bearings
1. Introduction. 2. Advantages and Disadvantages of Rolling Contact Bearings Over Sliding Contact Bearings. 3. Types of Rolling Contact Bearings. 4. Types of Radial Ball Bearings. 5. Standard Dimensions and Designation of Ball Bearings. 6. Thrust Ball Bearings. 7. Types of Roller Bearings. 8. Basic Static Load Rating of Rolling Contact Bearings. 9. Static Equivalent Load for 27.1 Introduction Rolling Contact Bearings. In rolling contact bearings, the contact between the 10. Life of a Bearing. bearing surfaces is rolling instead of sliding as in sliding 11. Basic Dynamic Load Rating contact bearings. We have already discussed that the of Rolling Contact Bearings. ordinary sliding bearing starts from rest with practically 12. Dynamic Equivalent Load metal-to-metal contact and has a high coefficient of friction. for Rolling Contact It is an outstanding advantage of a rolling contact bearing Bearings. over a sliding bearing that it has a low starting friction. 13. Dynamic Load Rating for Rolling Contact Bearings Due to this low friction offered by rolling contact bearings, under Variable Loads. these are called antifriction bearings. 14. Reliability of a Bearing. 27.2 Advantages and Disadvantages of 15. Selection of Radial Ball Rolling Contact Bearings Over Sliding Bearings. Contact Bearings 16. Materials and Manufacture of Ball and Roller Bearings. The following are some advantages and disadvantages 17. Lubrication of Ball and of rolling contact bearings over sliding contact bearings. Roller Bearings.
996 Rolling Contact Bearings 997
Advantages 1. Low starting and running friction except at very high speeds. 2. Ability to withstand momentary shock loads. 3. Accuracy of shaft alignment. 4. Low cost of maintenance, as no lubrication is required while in service. 5. Small overall dimensions. 6. Reliability of service. 7. Easy to mount and erect. 8. Cleanliness. Disadvantages 1. More noisy at very high speeds. 2. Low resistance to shock loading. 3. More initial cost. 4. Design of bearing housing complicated. 27.3 Types of Rolling Contact Bearings Following are the two types of rolling contact bearings: 1. Ball bearings; and 2. Roller bearings.
Fig. 27.1. Ball and roller bearings. Fig. 27.2. Radial and thrust ball bearings. The ball and roller bearings consist of an inner race which is mounted on the shaft or journal and an outer race which is carried by the housing or casing. In between the inner and outer race, there are balls or rollers as shown in Fig. 27.1. A number of balls or rollers are used and these are held at proper distances by retainers so that they do not touch each other. The retainers are thin strips and is usually in two parts which are assembled after the balls have been properly spaced. The ball bearings are used for light loads and the roller bearings are used for heavier loads. The rolling contact bearings, depending upon the load to be carried, are classified as : (a) Radial bearings, and (b) Thrust bearings. The radial and thrust ball bearings are shown in Fig. 27.2 (a) and (b) respectively. When a ball bearing supports only a radial load (WR), the plane of rotation of the ball is normal to the centre line of the bearing, as shown in Fig. 27.2 (a). The action of thrust load (WA) is to shift the plane of rotation of the balls, as shown in Fig. 27.2 (b). The radial and thrust loads both may be carried simultaneously. 27.4 Types of Radial Ball Bearings Following are the various types of radial ball bearings: 1. Single row deep groove bearing. A single row deep groove bearing is shown in Fig. 27.3 (a). 998 A Textbook of Machine Design
Fig. 27.3. Types of radial ball bearings. During assembly of this bearing, the races are offset and the maximum number of balls are placed between the races. The races are then centred and the balls are symmetrically located by the use of a retainer or cage. The deep groove ball bearings are used due to their high load carrying capacity and suitability for high running speeds. The load carrying capacity of a ball bearing is related to the size and number of the balls. 2. Filling notch bearing. A filling notch bearing is shown in Fig. 27.3 (b). These bearings have notches in the inner and outer races which permit more balls to be inserted than in a deep groove ball bearings. The notches do not extend to the bottom of the race way and therefore the balls inserted through the notches must be forced in position. Since this type of bearing contains larger number of balls than a corresponding unnotched one, Radial ball bearing therefore it has a larger bearing load capacity. 3. Angular contact bearing. An angular contact bearing is shown in Fig. 27.3 (c). These bearings have one side of the outer race cut away to permit the insertion of more balls than in a deep groove bearing but without having a notch cut into both races. This permits the bearing to carry a relatively large axial load in one direction while also carrying a relatively large radial load. The angular contact bearings are usually used in pairs so that thrust loads may be carried in either direction. 4. Double row bearing. A double row bearing is shown in Fig. 27.3 (d). These bearings may be made with radial or angular contact between the balls and races. The double row bearing is appreciably narrower than two single row bearings. The load capacity of such bearings is slightly less than twice that of a single row bearing. 5. Self-aligning bearing. A self-aligning bearing is shown in Fig. 27.3 (e). These bearings permit shaft deflections within 2-3 degrees. It may be noted that normal clearance in a ball bearing are too small to accommodate any appreciable misalignment of the shaft relative to the housing. If the unit is assembled with shaft misalignment present, then the bearing will be subjected to a load that may be in excess of the design value and premature failure may occur. Following are the two types of self-aligning bearings : (a) Externally self-aligning bearing, and (b) Internally self-aligning bearing. In an externally self-aligning bearing, the outside diameter of the outer race is ground to a spherical surface which fits in a mating spherical surface in a housing, as shown in Fig. 27.3 (e). In case of internally self-aligning bearing, the inner surface of the outer race is ground to a spherical Rolling Contact Bearings 999 surface. Consequently, the outer race may be displaced through a small angle without interfering with the normal operation of the bearing. The internally self-aligning ball bearing is interchangeable with other ball bearings. 27.5 Standard Dimensions and Designations of Ball Bearings The dimensions that have been standardised on an international basis are shown in Fig. 27.4. These dimensions are a function of the bearing bore and the series of bearing. The standard dimensions are given in millimetres. There is no standard for the size and number of steel balls. The bearings are designated by a number. In general, the number consists of atleast three digits. Additional digits or letters are used to indicate special features e.g. deep groove, filling notch etc. The last three digits give the series and the bore of the bearing. The last two digits from 04 onwards, when multiplied by 5, give the bore diameter in millimetres. The third from the last digit designates the series of the bearing. The most common ball bearings are available in four series as follows : 1. Extra light (100), 2. Light (200), 3. Medium (300), 4. Heavy (400) Notes : 1. If a bearing is designated by the number 305, it means that the bearing is of medium series whose bore is 05 × 5, i.e., 25 mm. Fig. 27.4. Standard designations 2. The extra light and light series are used where the loads are of ball bearings. moderate and shaft sizes are comparatively large and also where available space is limited. 3. The medium series has a capacity 30 to 40 per cent over the light series. 4. The heavy series has 20 to 30 per cent capacity over the medium series. This series is not used extensively in industrial applications.
Oilless bearings made using powder metallergy. 1000 A Textbook of Machine Design
The following table shows the principal dimensions for radial ball bearings. Table 27.1. Principal dimensions for radial ball bearings.
Bearing No. Bore (mm) Outside diameter Width (mm)
200 10 30 9 300 35 11 201 12 32 10 301 37 12 202 15 35 11 302 42 13 203 17 40 12 303 47 14 403 62 17 204 20 47 14 304 52 14 404 72 19 205 25 52 15 305 62 17 405 80 21 206 30 62 16 306 72 19 406 90 23 207 35 72 17 307 80 21 407 100 25 208 40 80 18 308 90 23 408 110 27 209 45 85 19 309 100 25 409 120 29 210 50 90 20 310 110 27 410 130 31 211 55 100 21 311 120 29 411 140 33 212 60 110 22 312 130 31 412 150 35 Rolling Contact Bearings 1001
Bearing No. Bore (mm) Outside diameter Width (mm) 213 65 120 23 313 140 33 413 160 37 214 70 125 24 314 150 35 414 180 42 215 75 130 25 315 160 37 415 190 45 216 80 140 26 316 170 39 416 200 48 217 85 150 28 317 180 41 417 210 52 218 90 160 30 318 190 43 418 225 54 27.6 Thrust Ball Bearings The thrust ball bearings are used for carrying thrust loads exclusively and at speeds below 2000 r.p.m. At high speeds, centrifugal force causes the balls to be forced out of the races. Therefore at high speeds, it is recommended that angular contact ball bearings should be used in place of thrust ball bearings.
Fig. 27.5. Thrust ball bearing. A thrust ball bearing may be a single direction, flat face as shown in Fig. 27.5 (a) or a double direction with flat face as shown in Fig. 27.5 (b). 27.7 Types of Roller Bearings Following are the principal types of roller bearings : 1. Cylindrical roller bearings. A cylindrical roller bearing is shown in Fig. 27.6 (a). These bearings have short rollers guided in a cage. These bearings are relatively rigid against radial motion 1002 A Textbook of Machine Design and have the lowest coefficient of friction of any form of heavy duty rolling-contact bearings. Such type of bearings are used in high speed service.
Radial ball bearing
2. Spherical roller bearings. A spherical roller bearing is shown in Fig. 27.6 (b). These bearings are self-aligning bearings. The self-aligning feature is achieved by grinding one of the races in the 1 ° form of sphere. These bearings can normally tolerate angular misalignment in the order of ± 1 and 2 when used with a double row of rollers, these can carry thrust loads in either direction.
Fig. 27.6. Types of roller bearings. 3. Needle roller bearings. A needle roller bearing is shown in Fig. 27.6 (c). These bearings are relatively slender and completely fill the space so that neither a cage nor a retainer is needed. These bearings are used when heavy loads are to be carried with an oscillatory motion, e.g. piston pin bearings in heavy duty diesel engines, where the reversal of motion tends to keep the rollers in correct alignment. 4. Tapered roller bearings. A tapered roller bearing is shown in Fig. 27.6 (d). The rollers and race ways of these bearings are truncated cones whose elements intersect at a common point. Such type of bearings can carry both radial and thrust loads. These bearings are available in various combinations as double row bearings and with different cone angles for use with different relative magnitudes of radial and thrust loads. Cylindrical roller bearings Rolling Contact Bearings 1003
Spherical roller Needle roller bearings Tapered roller bearings bearings 27.8 Basic Static Load Rating of Rolling Contact Bearings The load carried by a non-rotating bearing is called a static load. The basic static load rating is defined as the static radial load (in case of radial ball or roller bearings) or axial load (in case of thrust ball or roller bearings) which corresponds to a total permanent deformation of the ball (or roller) and race, at the most heavily stressed contact, equal to 0.0001 times the ball (or roller) diameter. In single row angular contact ball bearings, the basic static load relates to the radial component of the load, which causes a purely radial displacement of the bearing rings in relation to each other. Note : The permanent deformation which appear in balls (or rollers) and race ways under static loads of moderate magnitude, increase gradually with increasing load. The permissible static load is, therefore, dependent upon the permissible magnitude of permanent deformation. Experience shows that a total permanent deformation of 0.0001 times the ball (or roller) diameter, occurring at the most heavily loaded ball (or roller) and race contact can be tolerated in most bearing applications without impairment of bearing operation. In certain applications where subsequent rotation of the bearing is slow and where smoothness and friction requirements are not too exacting, a much greater total permanent deformation can be permitted. On the other hand, where extreme smoothness is required or friction requirements are critical, less total permanent deformation may be permitted.
According to IS : 3823–1984, the basic static load rating (C0) in newtons for ball and roller bearings may be obtained as discussed below :
1. For radial ball bearings, the basic static radial load rating (C0) is given by 2 α C0 = f0.i.Z.D cos where i = Number of rows of balls in any one bearing, Z = Number of ball per row, D = Diameter of balls, in mm, α = Nominal angle of contact i.e. the nominal angle between the line of action of the ball load and a plane perpendicular to the axis of bearing, and
f0 = A factor depending upon the type of bearing. The value of factor ( f0 ) for bearings made of hardened steel are taken as follows : f0 = 3.33, for self-aligning ball bearings = 12.3, for radial contact and angular contact groove ball bearings. 2. For radial roller bearings, the basic static radial load rating is given by α C0 = f0.i.Z.le.D cos where i = Number of rows of rollers in the bearing, Z = Number of rollers per row,
le = Effective length of contact between one roller and that ring (or washer) where the contact is the shortest (in mm). It is equal to the overall length of roller minus roller chamfers or grinding undercuts, 1004 A Textbook of Machine Design
D = Diameter of roller in mm. It is the mean diameter in case of tapered rollers, α = Nominal angle of contact. It is the angle between the line of action of the roller resultant load and a plane perpendicular to the axis of the bearing, and
f0 = 21.6, for bearings made of hardened steel. 3. For thrust ball bearings, the basic static axial load rating is given by 2 α C0 = f0.Z.D sin where Z = Number of balls carrying thrust in one direction, and
f0 = 49, for bearings made of hardened steel. 4. For thrust roller bearings, the basic static axial load rating is given by α C0 = f0.Z.le.D.sin where Z = Number of rollers carrying thrust in one direction, and
f0 = 98.1, for bearings made of hardened steel. 27.9 Static Equivalent Load for Rolling Contact Bearings The static equivalent load may be defined as the static radial load (in case of radial ball or roller bearings) or axial load (in case of thrust ball or roller bearings) which, if applied, would cause the same total permanent deformation at the most heavily stressed ball (or roller) and race contact as that which occurs under the actual conditions of loading.
More cylindrical roller bearings
The static equivalent radial load (W0R) for radial or roller bearings under combined radial and axial or thrust loads is given by the greater magnitude of those obtained by the following two equations, i.e.
1. W0R = X0.WR + Y0.WA ; and 2. W0R = WR where WR = Radial load, WA = Axial or thrust load, X0 = Radial load factor, and Y0 = Axial or thrust load factor.
According to IS : 3824 – 1984, the values of X0 and Y0 for different bearings are given in the following table : Rolling Contact Bearings 1005
Table 27.2. Values of X0 and Y0 for radial bearings. S.No. Type of bearing Single row bearing Double row bearing
X0 Y0 X0 Y0 1. Radial contact groove ball bearings 0.60 0.50 0.60 0.50 2. Self aligning ball or roller bearings 0.50 0.22 cot θ 1 0.44 cot θ and tapered roller bearing 3. Angular contact groove bearings : α = 15° 0.50 0.46 1 0.92 α = 20° 0.50 0.42 1 0.84 α = 25° 0.50 0.38 1 0.76 α = 30° 0.50 0.33 1 0.66 α = 35° 0.50 0.29 1 0.58 α = 40° 0.50 0.26 1 0.52 α = 45° 0.50 0.22 1 0.44
Notes : 1. The static equivalent radial load (W0R) is always greater than or equal to the radial load (WR). 2. For two similar single row angular contact ball bearings, mounted ‘face-to-face’ or ‘back-to-back’, use the values of X0 and Y0 which apply to a double row angular contact ball bearings. For two or more similar single row angular contact ball bearings mounted ‘in tandem’, use the values of X0 and Y0 which apply to a single row angular contact ball bearings.
3. The static equivalent radial load (W0R) for all cylindrical roller bearings is equal to the radial load (WR).
4. The static equivalent axial or thrust load (W0A) for thrust ball or roller bearings with angle of contact α ≠ 90º, under combined radial and axial loads is given by α W0A = 2.3 WR.tan + WA This formula is valid for all ratios of radial to axial load in the case of direction bearings. For single ≤ α direction bearings, it is valid where WR / WA 0.44 cot . 5. The thrust ball or roller bearings with α = 90º can support axial loads only. The static equivalent axial load for this type of bearing is given by
W0A = WA 27.10 Life of a Bearing The life of an individual ball (or roller) bearing may be defined as the number of revolutions (or hours at some given constant speed) which the bearing runs before the first evidence of fatigue develops in the material of one of the rings or any of the rolling elements. The rating life of a group of apparently identical ball or roller bearings is defined as the number of revolutions (or hours at some given constant speed) that 90 per cent of a group of bearings will complete or exceed before the first evidence of fatigue develops (i.e. only 10 per cent of a group of bearings fail due to fatigue). The term minimum life is also used to denote the rating life. It has been found that the life which 50 per cent of a group of bearings will complete or exceed is approximately 5 times the life which 90 per cent of the bearings will complete or exceed. In other words, we may say that the average life of a bearing is 5 times the rating life (or minimum life). It may be noted that the longest life of a single bearing is seldom longer than the 4 times the average life and the maximum life of a single bearing is about 30 to 50 times the minimum life. 1006 A Textbook of Machine Design
The life of bearings for various types of machines is given in the following table. Table 27.3. Life of bearings for various types of machines.
S. No. Application of bearing Life of bearing, in hours
1. Instruments and apparatus that are rarely used (a) Demonstration apparatus, mechanism for operating 500 sliding doors (b) Aircraft engines 1000 – 2000 2. Machines used for short periods or intermittently and whose 4000 – 8000 breakdown would not have serious consequences e.g. hand tools, lifting tackle in workshops, and operated machines, agricultural machines, cranes in erecting shops, domestic machines. 3. Machines working intermittently whose breakdown would have 8000 – 12 000 serious consequences e.g. auxillary machinery in power stations, conveyor plant for flow production, lifts, cranes for piece goods, machine tools used frequently. 4. Machines working 8 hours per day and not always fully utilised 12 000 – 20 000 e.g. stationary electric motors, general purpose gear units. 5. Machines working 8 hours per day and fully utilised e.g. 20 000 – 30 000 machines for the engineering industry, cranes for bulk goods, ventilating fans, counter shafts. 6. Machines working 24 hours per day e.g. separators, compressors, 40 000 – 60 000 pumps, mine hoists, naval vessels. 7. Machines required to work with high degree of reliability 100 000 – 200 000 24 hours per day e.g. pulp and paper making machinery, public power plants, mine-pumps, water works.
27.11 Basic Dynamic Load Rating of Rolling Contact Bearings The basic dynamic load rating is defined as the constant stationary radial load (in case of radial ball or roller bearings) or constant axial load (in case of thrust ball or roller bearings) which a group of apparently identical bearings with stationary outer ring can endure for a rating life of one million revolutions (which is equivalent to 500 hours of operation at 33.3 r.p.m.) with only 10 per cent failure. The basic dynamic load rating (C) in newtons for ball and roller bearings may be obtained as discussed below : 1. According to IS: 3824 (Part 1)– 1983, the basic dynamic radial load rating for radial and angular contact ball bearings, except the filling slot type, with balls not larger than 25.4 mm in diameter, is given by α 0.7 2/3 1.8 C = fc (i cos ) Z . D and for balls larger than 25.4 mm in diameter, α 0.7 2/3 1.4 C = 3.647 fc (i cos ) Z . D Rolling Contact Bearings 1007 where fc = A factor, depending upon the geometry of the bearing components, the accuracy of manufacture and the material used. and i, Z, D and α have usual meanings as discussed in Art. 27.8.
Ball bearings 2. According to IS: 3824 (Part 2)–1983, the basic dynamic radial load rating for radial roller bearings is given by α 7/9 3/4 29/27 C = fc (i.le cos ) Z . D 3. According to IS: 3824 (Part 3)–1983, the basic dynamic axial load rating for single row, single or double direction thrust ball bearings is given as follows : (a) For balls not larger than 25.4 mm in diameter and α = 90º, 2/3 1.8 C = fc . Z . D (b) For balls not larger than 25.4 mm in diameter and α ≠ 90º, α 0.7 α 2/3 1.8 C = fc (cos ) tan . Z . D (c) For balls larger than 25.4 mm in diameter and α = 90º 2/3 1.4 C = 3.647 fc . Z . D (d) For balls larger than 25.4 mm in diameter and α ≠ 90º, α 0.7 α 2/3 1.4 C = 3.647 fc (cos ) tan . Z . D 4. According to IS: 3824 (Part 4)–1983, the basic dynamic axial load rating for single row, single or double direction thrust roller bearings is given by 7/9 3/4 29/27 α C = fc . le . Z . D ... (when = 90º) α 7/9 α 3/4 29/27 α ≠ = fc (le cos ) tan .Z . D ... (when 90º) 27.12 Dynamic Equivalent Load for Rolling Contact Bearings The dynamic equivalent load may be defined as the constant stationary radial load (in case of radial ball or roller bearings) or axial load (in case of thrust ball or roller bearings) which, if applied to a bearing with rotating inner ring and stationary outer ring, would give the same life as that which the bearing will attain under the actual conditions of load and rotation. 1008 A Textbook of Machine Design
The dynamic equivalent radial load (W ) for radial and angular contact bearings, except the filling slot types, under combined constant radial load (WR) and constant axial or thrust load (WA) is given by
W = X . V. WR + Y . WA where V = A rotation factor, = 1, for all types of bearings when the inner race is rotating, = 1, for self-aligning bearings when inner race is stationary, = 1.2, for all types of bearings except self-aligning, when inner race is stationary. The values of radial load factor (X ) and axial or thrust load factor (Y ) for the dynamically loaded bearings may be taken from the following table: Table 27.4. Values of X and Y for dynamically loaded bearings.
WA WA Type of bearing Specifications ≤ e > e e WR WR XYXY
WA Deep groove = 0.025 2.0 0.22 C0 ball bearing = 0.04 1.8 0.24 = 0.07 1.6 0.27 = 0.13 1 0 0.56 1.4 0.31 = 0.25 1.2 0.37 = 0.50 1.0 0.44 Angular contact Single row 0 0.35 0.57 1.14 ball bearings Two rows in tandem 0 0.35 0.57 1.14 Two rows back to back 1 0.55 0.57 0.93 1.14 Double row 0.73 0.62 1.17 0.86 Self-aligning Light series : for bores bearings 10 – 20 mm 1.3 2.0 0.50 25 – 35 1 1.7 6.5 2.6 0.37 40 – 45 2.0 3.1 0.31 50 – 65 2.3 3.5 0.28 70 – 100 2.4 3.8 0.26 105 – 110 2.3 3.5 0.28 Medium series : for bores 12 mm 1.0 0.65 1.6 0.63 15 – 20 1.2 1.9 0.52 25 – 50 1.5 2.3 0.43 55 – 90 1.6 2.5 0.39 Spherical roller For bores : bearings 25 – 35 mm 2.1 3.1 0.32 40 – 45 1 2.5 0.67 3.7 0.27 50 – 100 2.9 4.4 0.23 100 – 200 2.6 3.9 0.26 Taper roller For bores : bearings 30 – 40 mm 1.60 0.37 45 – 110 1 0 0.4 1.45 0.44 120 – 150 1.35 0.41 Rolling Contact Bearings 1009 27.13 Dynamic Load Rating for Rolling Contact Bearings under Variable Loads The approximate rating (or service) life of ball or roller bearings is based on the fundamental equation, k ⎛⎞C 6 L = ⎜⎟× 10 revolutions ⎝⎠W 1/k ⎛⎞L or C = W ⎜⎟ ⎝⎠106 where L = Rating life, C = Basic dynamic load rating, W = Equivalent dynamic load, and k = 3, for ball bearings, = 10/3, for roller bearings. The relationship between the life in revolutions (L) and Roller bearing the life in working hours (LH) is given by
L = 60 N . LH revolutions where N is the speed in r.p.m.
Now consider a rolling contact bearing subjected to variable loads. Let W1, W2, W3 etc., be the loads on the bearing for successive n1, n2, n3 etc., number of revolutions respectively.
If the bearing is operated exclusively at the constant load W1, then its life is given by k C ⎛⎞ 6 L1 = ⎜⎟ × 10 revolutions ⎝⎠W1 ∴ Fraction of life consumed with load W1 acting for n1 number of revolutions is n k 1 ⎛⎞W1 × 1 = n ⎜⎟ 6 L1 1 ⎝⎠C 10
Similarly, fraction of life consumed with load W2 acting for n2 number of revolutions is k n2 = n ⎛⎞W2 × 1 L 2 ⎜⎟ 2 ⎝⎠C 106 and fraction of life consumed with load W3 acting for n3 number of revolutions is n k 3 ⎛⎞W3 × 1 = n ⎜⎟ 6 L3 3 ⎝⎠C 10
nn12n3 But +++ .... = 1 LL123 L kkk ⎛⎞WW12×+111 ⎛ ⎞ ×+⎛⎞W3 × or nn12⎜⎟ ⎜ ⎟ n 3 ⎜⎟ + ...... = 1 ⎝⎠CCC10666 ⎝ ⎠ 10⎝⎠ 10 ∴ k k k k 6 n1(W1) + n2 (W2) + n3 (W3) + ...... = C × 10 ...(i) If an equivalent constant load (W) is acting for n number of revolutions, then k ⎛⎞C n = ⎜⎟ × 106 ⎝⎠W 1010 A Textbook of Machine Design
∴ n (W)k = C k × 106 ...(ii) where n = n1 + n2 + n3 + ..... From equations (i) and (ii), we have k k k k n1 (W1) + n2 (W2) + n3 (W3) + ..... = n (W) kkk+++1/k ⎡⎤nW11( ) n 2 ( W 2 ) nW 3 ( 3 ) .... ∴ W = ⎢⎥ ⎣⎦⎢⎥n Substituting n = n1 + n2 + n3 + ...... , and k = 3 for ball bearings, we have 333+++1/3 ⎡⎤nW11( ) n 2 ( W 2 ) nW 3 ( 3 ) .... W = ⎢⎥+++ ⎣⎦⎢⎥nn123 n.... Note : The above expression may also be written as
333+++1/3 ⎡⎤LW11( ) LW 2 ( 2 ) LW 33 ( ) ... W = ⎢⎥+++ ⎣⎦LLL123.... See Example 27.6. 27.14 Reliability of a Bearing We have already discussed in the previous article that the rating life is the life that 90 per cent of a group of identical bearings will complete or exceed before the first evidence of fatigue develops. The reliability (R) is defined as the ratio of the number of bearings which have successfully completed L million revolutions to the total number of bearings under test. Sometimes, it becomes necessary to select a bearing having a reliability of more than 90%. According to Wiebull, the relation between the bearing life and the reliability is given as b 1/b ⎛⎞1 ⎛⎞L L = ⎡⎤⎛⎞1 log ⎜⎟= ⎜⎟ or ⎢⎥loge ⎜⎟ ...(i) e ⎝⎠R ⎝⎠a aR⎣⎦⎝⎠ where L is the life of the bearing corresponding to the desired reliability R and a and b are constants whose values are a = 6.84, and b = 1.17
If L90 is the life of a bearing corresponding to a reliability of 90% (i.e. R90), then 1/b L90 ⎡⎤⎛⎞1 = ⎢⎥loge ⎜⎟ ...(ii) a ⎣⎦⎝⎠R90 Dividing equation (i) by equation (ii), we have
1/b L ⎡⎤log (1/R ) e 1/1.17 = ⎢⎥ = *6.85 [loge (1/R)] ... ( b = 1.17) L90 ⎣⎦loge (1/R90 ) Q This expression is used for selecting the bearing when the reliability is other than 90%. Note : If there are n number of bearings in the system each having the same reliability R, then the reliability of the complete system will be
RS = Rp where RS indicates the probability of one out of p number of bearings failing during its life time.
1/b 1/1.17 0.8547 * [loge (1 / R90)] = [loge (1/0.90)] = (0.10536) = 0.146
1/b LR==[loge (1/ )] 1/1.17 ∴ 6.85 [loge (1/R )] L90 0.146 Rolling Contact Bearings 1011
Example 27.1. A shaft rotating at constant speed is subjected to variable load. The bearings supporting the shaft are subjected to stationary equivalent radial load of 3 kN for 10 per cent of time, 2 kN for 20 per cent of time, 1 kN for 30 per cent of time and no load for remaining time of cycle. If the total life expected for the bearing is 20 × 106 revolutions at 95 per cent reliability, calculate dynamic load rating of the ball bearing.
Solution. Given : W1 = 3 kN ; n1 = 0.1 n ; W2 = 2 kN ; n2 = 0.2 n ; W3 = 1 kN ; n3 = 0.3 n ; 6 W4 = 0 ; n4 = (1 – 0.1 – 0.2 – 0.3) n = 0.4 n ; L95 = 20 × 10 rev
Let L90 = Life of the bearing corresponding to reliability of 90 per cent, L95 = Life of the bearing corresponding to reliability of 95 per cent = 20 × 106 revolutions ... (Given) We know that L 1/b 1/1.17 95 ⎡⎤⎡⎤logee (1/R95 )= log (1/ 0.95) = ⎢⎥⎢⎥ ... ( b = 1.17) L90 ⎣⎦⎣⎦logee (1/R90 ) log (1/ 0.90) Q 0.8547 ⎛⎞0.0513 = ⎜⎟= 0.54 ⎝⎠0.1054 ∴ 6 6 L90 = L95 / 0.54 = 20 × 10 / 0.54 = 37 × 10 rev We know that equivalent radial load, 3333+++1/3 ⎡⎤nW11() n 2 () W 2 nW 33 () nW 44 () W = ⎢⎥+++ ⎣⎦⎢⎥nn1234 nn 1/3 ⎡⎤0.1nnnn×+ 33333 0.2 ×+ 2 0.3 ×+ 1 0.4 × 0 = ⎢⎥+++ ⎣⎦0.1nnnn 0.2 0.3 0.4 = (2.7 + 1.6 + 0.3 + 0)1/3 = 1.663 kN We also know that dynamic load rating, 1/k × 6 1/3 ⎛⎞L90 ⎛⎞37 10 C = W ⎜⎟6 = 1.663 ⎜⎟6 = 5.54 kN Ans. ⎝⎠10 ⎝⎠10 ... ( Q k = 3, for ball bearing) Example 27.2. The rolling contact ball bearing are to be selected to support the overhung countershaft. The shaft speed is 720 r.p.m. The bearings are to have 99% Ball bearings in reliability corresponding to a life of 24 000 a race Oil hours. The bearing is subjected to an equivalent radial load of 1 kN. Consider life adjustment factors for operating condition and material as 0.9 and 0.85 respectively. Find the basic dynamic load rating of the bearing from manufacturer's catalogue, Ball bearing specified at 90% reliability. Solution. Given : N = 720 r.p.m. ;
LH = 24 000 hours ; W = 1 kN
Another view of ball-bearings 1012 A Textbook of Machine Design
We know that life of the bearing corresponding to 99% reliability, 6 L99 = 60 N. LH = 60 × 720 × 24 000 = 1036.8 × 10 rev
Let L90 = Life of the bearing corresponding to 90% reliability. Considering life adjustment factors for operating condition and material as 0.9 and 0.85 respectively, we have 1/b 1/1.17 LR99= ⎡⎤loge (1/ 99 ) ⎡⎤loge (1/ 0.99) ⎢⎥× 0.9 × 0.85 = ⎢⎥ × 0.9 × 0.85 LR90⎣⎦loge (1/ 90 ) ⎣⎦loge (1/ 0.9) 0.8547 ⎡⎤0.01005 = ⎢⎥ × 0.9 × 0.85 = 0.1026 ⎣⎦0.1054 ∴ 6 6 L90 = L99 / 0.1026 = 1036.8 × 10 / 0.1026 = 10 105 × 10 rev We know that dynamic load rating, 1/k ⎛⎞L90 C = W ⎜⎟ ⎝⎠106 1/3 ⎛⎞10 105× 106 = 1 ⎜⎟6 kN ⎝⎠10 ... ( Q k = 3, for ball bearing) = 21.62 kN Ans. 27.15 Selection of Radial Ball Bearings In order to select a most suitable ball bearing, first of all, the basic dynamic radial load is calculated. It is then multiplied by the service factor (KS) to get the design basic dynamic radial load capacity. The service Radial ball bearings factor for the ball bearings is shown in the following table.
Table 27.5. Values of service factor (KS).
S.No. Type of service Service factor (KS) for radial ball bearings
1. Uniform and steady load 1.0 2. Light shock load 1.5 3. Moderate shock load 2.0 4. Heavy shock load 2.5 5. Extreme shock load 3.0
After finding the design basic dynamic radial load capacity, the selection of bearing is made from the catalogue of a manufacturer. The following table shows the basic static and dynamic capacities for various types of ball bearings. Rolling Contact Bearings 1013
Table 27.6. Basic static and dynamic capacities of various types of radial ball bearings.
Bearing Basic capacities in kN No. Single row deep Single row angular Double row angular Self-aligning groove ball bearing contact ball bearing contact ball bearing ball bearing Static Dynamic Static Dynamic Static Dynamic Static Dynamic
(C0 ) (C) (C0 ) (C) (C0 ) (C) (C0 ) (C) (1) (2) (3) (4) (5) (6) (7) (8) (9)
200 2.24 4 — — 4.55 7.35 1.80 5.70 300 3.60 6.3 ——————
201 3 5.4 — — 5.6 8.3 2.0 5.85 301 4.3 7.65 ————3.09.15
202 3.55 6.10 3.75 6.30 5.6 8.3 2.16 6 302 5.20 8.80 — — 9.3 14 3.35 9.3
203 4.4 7.5 4.75 7.8 8.15 11.6 2.8 7.65 303 6.3 10.6 7.2 11.6 12.9 19.3 4.15 11.2 4031118——————
204 6.55 10 6.55 10.4 11 16 3.9 9.8 304 7.65 12.5 8.3 13.7 14 19.3 5.5 14 404 15.6 24 ——————
205 7.1 11 7.8 11.6 13.7 17.3 4.25 9.8 305 10.4 16.6 12.5 19.3 20 26.5 7.65 19 4051928——————
206 10 15.3 11.2 16 20.4 25 5.6 12 306 14.6 22 17 24.5 27.5 35.5 10.2 24.5 406 23.2 33.5 ——————
207 13.7 20 15.3 21.2 28 34 8 17 307 17.6 26 20.4 28.5 36 45 13.2 30.5 407 30.5 43 ——————
208 16 22.8 19 25 32.5 39 9.15 17.6 308 22 32 25.5 35.5 45.5 55 16 35.5 408 37.5 50 ——————
209 18.3 25.5 21.6 28 37.5 41.5 10.2 18 309 30 41.5 34 45.5 56 67 19.6 42.5 4094460——————
210 21.2 27.5 23.6 29 43 47.5 10.8 18 310 35.5 48 40.5 53 73.5 81.5 24 50 4105068—————— 1014 A Textbook of Machine Design
(1) (2) (3) (4) (5) (6) (7) (8) (9) 211 26 34 30 36.5 49 53 12.7 20.8 311 42.5 56 47.5 62 80 88 28.5 58.5 4116078—————— 212 32 40.5 36.5 44 63 65.5 16 26.5 312 48 64 55 71 96.5 102 33.5 68 412 67 85 — — — — — — 213 35.5 44 43 50 69.5 69.5 20.4 34 313 55 72 63 80 112 118 39 75 413 76.5 93 — — — — — — 214 39 48 47.5 54 71 69.5 21.6 34.5 314 63 81.5 73.5 90 129 137 45 85 414 102 112 — — — — — — 215 42.5 52 50 56 80 76.5 22.4 34.5 315 72 90 81.5 98 140 143 52 95 415 110 120 — — — — — — 216 45.5 57 57 63 96.5 93 25 38 316 80 96.5 91.5 106 160 163 58.5 106 416 120 127 — — — — — — 217 55 65.5 65.5 71 100 106 30 45.5 317 88 104 102 114 180 180 62 110 417 132 134 — — — — — — 218 63 75 76.5 83 127 118 36 55 318 98 112 114 122 — — 69.5 118 418 146 146 — — — — — — 219 72 85 88 95 150 137 43 65.5 319 112 120 125 132 — — — — 220 81.5 96.5 93 102 160 146 51 76.5 320 132 137 153 150 — — — — 221 93 104 104 110 — — 56 85 321 143 143 166 160 — — — — 222 104 112 116 120 — — 64 98 322 166 160 193 176 — — — —
Note: The reader is advised to consult the manufacturer's catalogue for further and complete details of the bearings. Example 27.3. Select a single row deep groove ball bearing for a radial load of 4000 N and an axial load of 5000 N, operating at a speed of 1600 r.p.m. for an average life of 5 years at 10 hours per day. Assume uniform and steady load.
Solution. Given : WR = 4000 N ; WA = 5000 N ; N = 1600 r.p.m. Rolling Contact Bearings 1015
Since the average life of the bearing is 5 years at 10 hours per day, therefore life of the bearing in hours,
LH = 5 × 300 × 10 = 15 000 hours ... (Assuming 300 working days per year) and life of the bearing in revolutions, 6 L = 60 N × LH = 60 × 1600 × 15 000 = 1440 × 10 rev We know that the basic dynamic equivalent radial load,
W = X.V.WR + Y.WA ...(i)
In order to determine the radial load factor (X) and axial load factor (Y), we require WA/ WR and WA / C0. Since the value of basic static load capacity (C0) is not known, therefore let us take WA / C0 = 0.5. Now from Table 27.4, we find that the values of X and Y corresponding to WA / C0 = 0.5 and WA/WR = 5000 / 4000 = 1.25 (which is greater than e = 0.44) are X = 0.56 and Y = 1 Since the rotational factor (V) for most of the bearings is 1, therefore basic dynamic equivalent radial load, W = 0.56 × 1 × 4000 + 1 × 5000 = 7240 N
From Table 27.5, we find that for uniform and steady load, the service factor (KS) for ball bearings is 1. Therefore the bearing should be selected for W = 7240 N. We know that basic dynamic load rating,
1/k 1/3 ⎛⎞L ⎛⎞1440× 106 = 7240 C = W ⎜⎟66⎜⎟= 81 760 N ⎝⎠10⎝⎠ 10 = 81.76 kN ... (Q k = 3, for ball bearings) From Table 27.6, let us select the bearing No. 315 which has the following basic capacities,
C0 = 72 kN = 72 000 N and C = 90 kN = 90 000 N
Now WA / C0 = 5000 / 72 000 = 0.07 ∴ From Table 27.4, the values of X and Y are X = 0.56 and Y = 1.6 Substituting these values in equation (i), we have dynamic equivalent load, W = 0.56 × 1 × 4000 + 1.6 × 5000 = 10 240 N ∴ Basic dynamic load rating, 1/3 ⎛⎞1440× 106 C = 10 240 ⎜⎟6 = 115 635 N = 115.635 kN ⎝⎠10 From Table 27.6, the bearing number 319 having C = 120 kN, may be selected. Ans. Example 27.4. A single row angular contact ball bearing number 310 is used for an axial flow compressor. The bearing is to carry a radial load of 2500 N and an axial or thrust load of 1500 N. Assuming light shock load, determine the rating life of the bearing.
Solution. Given : WR = 2500 N ; WA = 1500 N From Table 27.4, we find that for single row angular contact ball bearing, the values of radial factor (X) and thrust factor (Y ) for WA / WR = 1500 / 2500 = 0.6 are X = 1 and Y = 0 Since the rotational factor (V ) for most of the bearings is 1, therefore dynamic equivalent load,
W = X.V.WR + Y.WA = 1 × 1 × 2500 + 0 × 1500 = 2500 N 1016 A Textbook of Machine Design
From Table 27.5, we find that for light shock load, the service factor (KS) is 1.5. Therefore the design dynamic equivalent load should be taken as W = 2500 × 1.5 = 3750 N From Table 27.6, we find that for a single row angular contact ball bearing number 310, the basic dynamic capacity, C = 53 kN = 53 000 N We know that rating life of the bearing in revolutions, k 3 ⎛⎞C 66 ⎛53 000 ⎞ L = ⎜⎟×=10 ⎜ ⎟ × 10 = 2823 × 106 rev Ans. ⎝⎠W ⎝3750 ⎠ ... (Q k = 3, for ball bearings) Example 27.5. Design a self-aligning ball bearing for a radial load of 7000 N and a thrust load of 2100 N. The desired life of the bearing is 160 millions of revolutions at 300 r.p.m. Assume uniform and steady load, 6 Solution. Given : WR = 7000 N ; WA = 2100 N ; L = 160 × 10 rev ; N = 300 r.p.m. From Table 27.4, we find that for a self-aligning ball bearing, the values of radial factor (X ) and thrust factor (Y) for WA / WR = 2100 / 7000 = 0.3, are as follows : X = 0.65 and Y = 3.5 Since the rotational factor (V ) for most of the bearings is 1, therefore dynamic equivalent load,
W = X.V.WR + Y.WA = 0.65 × 1 × 7000 + 3.5 × 2100 = 11 900 N
From Table 27.5, we find that for uniform and steady load, the service factor KS for ball bearings is 1. Therefore the bearing should be selected for W = 11 900 N. We know that the basic dynamic load rating,
1/k 1/3 ⎛⎞L ⎛⎞160× 106 = 11 900 C = W ⎜⎟66⎜⎟ = 64 600 N = 64.6 kN ⎝⎠10⎝⎠ 10 ... (Q k = 3, for ball bearings) From Table 27.6, let us select bearing number 219 having C = 65.5 kN Ans. Example 27.6. Select a single row deep groove ball bearing with the operating cycle listed below, which will have a life of 15 000 hours. Fraction of Type of load Radial Thrust Speed Service factor cycle (N) (N) (R.P.M.) 1/10 Heavy shocks 2000 1200 400 3.0 1/10 Light shocks 1500 1000 500 1.5 1/5 Moderate shocks 1000 1500 600 2.0 3/5 No shock 1200 2000 800 1.0 Assume radial and axial load factors to be 1.0 and 1.5 respectively and inner race rotates.
Solution. Given : LH = 15 000 hours ; WR1 = 2000 N ; WA1 = 1200 N ; N1 = 400 r.p.m. ; KS1 = 3 ; WR2 = 1500 N ; WA2 = 1000 N ; N2 = 500 r.p.m. ; KS2 = 1.5 ; WR3 = 1000 N ; WA3 = 1500 N ; N3 = 600 r.p.m. ; KS3 = 2 ; WR4 = 1200 N ; WA4 = 2000 N ; N4 = 800 r.p.m. ; KS4 = 1 ; X = 1 ; Y = 1.5 Rolling Contact Bearings 1017
We know that basic dynamic equivalent radial load considering service factor is
W =[X.V.WR + Y.W A] KS ...(i) It is given that radial load factor (X ) = 1 and axial load factor (Y ) = 1.5. Since the rotational factor (V) for most of the bearings is 1, therefore equation (i) may be written as
W =(WR + 1.5 WA) KS
Now, substituting the values of WR, WA and KS for different operating cycle, we have
W1 =(WR1 + 1.5 WA1) KS1 = (2000 + 1.5 × 1200) 3 = 11 400 N
W2 =(WR2 + 1.5 WA2) KS2 = (1500 + 1.5 × 1000) 1.5 = 4500 N
W3 =(WR3 + 1.5 WA3) KS3 = (1000 + 1.5 × 1500) 2 = 6500 N and W4 =(WR4 + 1.5 WA4) KS4 = (1200 + 1.5 × 2000) 1 = 4200 N We know that life of the bearing in revolutions 6 L = 60 N.LH = 60 N × 15 000 = 0.9 × 10 N rev ∴ Life of the bearing for 1/10 of a cycle, 1 1 L = × 0.9 × 106 N = × 0.9 × 106 × 400 = 36 × 106 rev 1 10 1 10 Similarly, life of the bearing for the next 1/10 of a cycle, 1 1 L = × 0.9 × 106 N = × 0.9 × 106 × 500 = 45 × 106 rev 2 10 2 10 Life of the bearing for the next 1/5 of a cycle, 1 1 L = × 0.9 × 106 N = × 0.9 × 106 × 600 = 108 × 106 rev 3 5 3 5 and life of the bearing for the next 3/5 of a cycle, 3 3 L = × 0.9 × 106 N = × 0.9 × 106 × 800 = 432 × 106 rev 4 5 4 5 We know that equivalent dynamic load,
3333+++ 1/3 ⎡⎤LW11() LW 22 () LW 33 () L 44 () W W = ⎢⎥+++ ⎣⎦⎢⎥LLLL1234
1/3 ⎡36×+×+×+× 1063 (11 400) 45 10 63 (4500) 108 10 63 (6500) 432 10 63 (4200) ⎤ = ⎢ 66 6 6 ⎥ ⎣ 36×+×+×+× 10 45 10 108 10 423 10 ⎦ 1/3 ⎡⎤1.191×× 10812 10 12 1/3 = ⎢⎥6 = (0.1918 × 10 ) = 5767 N ⎣⎦621× 10 and L = L1 + L2 + L3 + L4 = 36 × 106 + 45 × 106 +108 × 106 + 432 × 106 = 621 × 106 rev We know that dynamic load rating,
1/k 1/3 ⎛⎞L ⎛⎞621× 106 C = W ⎜⎟6 = 5767 ⎜⎟6 ⎝⎠10 ⎝⎠10 = 5767 × 8.53 = 49 193 N = 49.193 kN From Table 27.6, the single row deep groove ball bearing number 215 having C = 52 kN may be selected. Ans. 1018 A Textbook of Machine Design 27.16 Materials and Manufacture of Ball and Roller Bearings Since the rolling elements and the races are subjected to high local stresses of varying magnitude with each revolution of the bearing, therefore the material of the rolling element (i.e. steel) should be of high quality. The balls are generally made of high carbon chromium steel. The material of both the balls and races are heat treated to give extra hardness and toughness.
Ball and Roller Bearings
The balls are manufactured by hot forging on hammers from steel rods. They are then heat- treated, ground and polished. The races are also formed by forging and then heat-treated, ground and polished. 27.17 Lubrication of Ball and Roller Bearings The ball and roller bearings are lubricated for the following purposes : 1. To reduce friction and wear between the sliding parts of the bearing, 2. To prevent rusting or corrosion of the bearing surfaces, 3. To protect the bearing surfaces from water, dirt etc., and 4. To dissipate the heat. In general, oil or light grease is used for lubricating ball and roller bearings. Only pure mineral oil or a calcium-base grease should be used. If there is a possibility of moisture contact, then potassium or sodium-base greases may be used. Another additional advantage of the grease is that it forms a seal to keep out dirt or any other foreign substance. It may be noted that too much oil or grease cause the temperature of the bearing to rise due to churning. The temperature should be kept below 90ºC and in no case a bearing should operate above 150ºC.
EXERCISES
1. The ball bearings are to be selected for an application in which the radial load is 2000 N during 90 per cent of the time and 8000 N during the remaining 10 per cent. The shaft is to rotate at 150 r.p.m. Determine the minimum value of the basic dynamic load rating for 5000 hours of operation with not more than 10 per cent failures. [Ans. 13.8 kN] Rolling Contact Bearings 1019
2. A ball bearing subjected to a radial load of 5 kN is expected to have a life of 8000 hours at 1450 r.p.m. with a reliability of 99%. Calculate the dynamic load capacity of the bearing so that it can be selected from the manufacturer's catalogue based on a reliability of 90%. [Ans. 86.5 kN] 3. A ball bearing subjected to a radial load of 4000 N is expected to have a satisfactory life of 12 000 hours at 720 r.p.m. with a reliability of 95%. Calculate the dynamic load carrying capacity of the bearing, so that it can be selected from manufacturer's catalogue based on 90% reliability. If there are four such bearings each with a reliability of 95% in a system, what is the reliability of the complete system? [Ans. 39.5 kN ; 81.45%] 4. A rolling contact bearing is subjected to the following work cycle : (a) Radial load of 6000 N at 150 r.p.m. for 25% of the time; (b) Radial load of 7500 N at 600 r.p.m. for 20% of the time; and (c) Radial load of 2000 N at 300 r.p.m. for 55% of the time. The inner ring rotates and loads are steady. Select a bearing for an expected average life of 2500 hours. 5. A single row deep groove ball bearing operating at 2000 r.p.m. is acted by a 10 kN radial load and 8 kN thrust load. The bearing is subjected to a light shock load and the outer ring is rotating. Deter- mine the rating life of the bearing. [Ans. 15.52 × 106 rev] 6. A ball bearing operates on the following work cycle : Element No. Radial load Speed Element time (N) (R.P.M.) (%) 1 3000 720 30 2. 7000 1440 40 3. 5000 900 30 The dynamic load capacity of the bearing is 16 600 N. Calculate 1. the average speed of rotation ; 2. the equivalent radial load ; and 3. the bearing life. [Ans. 1062 r.p.m. ; 6.067 kN ; 20.5 × 106 rev]
QUESTIONS
1. What are rolling contact bearings? Discuss their advantages over sliding contact bearings. 2. Write short note on classifications and different types of antifriction bearings. 3. Where are the angular contact and self-aligning ball bearings used? Draw neat sketches of these bearings. 4. How do you express the life of a bearing? What is an average or median life? 5. Explain how the following factors influence the life of a bearing: (a) Load (b) Speed (c) Temperature (d) Reliability 6. Define the following terms as applied to rolling contact bearings: (a) Basic static load rating (b) Static equivalent load (c) Basic dynamic load rating (d) Dynamic equivalent load. 7. Derive the following expression as applied to rolling contact bearings subjected to variable load cycle 333+++ 3 NW11( ) N 2 ( W 2 ) NW 33 ( ) .... We = +++ NN123 N....
where We = Equivalent cubic load,
W1, W2 and W3 = Loads acting respectively for N1, N2, N3 .... 8. Select appropriate type of rolling contact bearing under the following condition of loading giving reasons for your choice. 1. Light radial load with high rotational speed. 2. Heavy axial and radial load with shock. 3. Light load where radial space is very limited. 4. Axial thrust only with medium speed. 1020 A Textbook of Machine Design
OBJECTIVE TYPE QUESTIONS 1. The rolling contact bearings are known as (a) thick lubricated bearings (b) plastic bearings (c) thin lubricated bearings (d) antifriction bearings 2. The bearings of medium series have capacity ...... over the light series. (a) 10 to 20% (b) 20 to 30% (c) 30 to 40% (d) 40 to 50% 3. The bearings of heavy series have capacity ...... over the medium series. (a) 10 to 20% (b) 20 to 30% (c) 30 to 40% (d) 40 to 50% 4. The ball bearings are usually made from (a) low carbon steel (b) medium carbon steel (c) high speed steel (d) chrome nickel steel 5. The tapered roller bearings can take (a) radial load only (b) axial load only (c) both radial and axial loads (d) none of the above 6. The piston pin bearings in heavy duty diesel engines are (a) needle roller bearings (b) tapered roller bearings (c) spherical roller bearings (d) cylindrical roller bearings 7. Which of the following is antifriction bearing? (a) journal bearing Ball bearing (b) pedestal bearing (c) collar bearing (d) needle bearing 8. Ball and roller bearings in comparison to sliding bearings have (a) more accuracy in alignment (b) small overall dimensions (c) low starting and running friction (d) all of these 9. A bearing is designated by the number 405. It means that a bearing is of (a) light series with bore of 5 mm (b) medium series with bore of 15 mm (c) heavy series with bore of 25 mm (d) light series with width of 20 mm 10. The listed life of a rolling bearing, in a catalogue, is the (a) minimum expected life (b) maximum expected life (c) average life (d) none of these
ANSWERS 1. (d) 2. (c) 3. (b) 4. (d) 5. (c) 6. (a) 7. (d) 8. (d) 9. (c) 10. (a) C H A P T E R 28 Spur Gears
1. Introduction. 2. Friction Wheels. 3. Advantages and Disadvantages of Gear Drives. 4. Classification of Gears. 5. Terms used in Gears. 6. Condition for Constant Velocity Ratio of Gears–Law of Gearing. 7. Forms of Teeth. 8. Cycloidal Teeth. 9. Involute Teeth. 10. Comparison Between Involute and Cycloidal Gears. 11. Systems of Gear Teeth. 12. Standard Proportions of Gear Systems. 13. Interference in Involute Gears. 28.1 Introduction 14. Minimum Number of Teeth on the Pinion in order to Avoid We have discussed earlier that the slipping of a belt or Interference. rope is a common phenomenon, in the transmission of 15. Gear Materials. motion or power between two shafts. The effect of slipping 16. Design Considerations for a is to reduce the velocity ratio of the system. In precision Gear Drive. machines, in which a definite velocity ratio is of importance 17. Beam Strength of Gear Teeth- (as in watch mechanism), the only positive drive is by gears Lewis Equation. or toothed wheels. A gear drive is also provided, when 18. Permissible Working Stress for the distance between the driver and the follower is very Gear Teeth in Lewis Equation. small. 19. Dynamic Tooth Load. 20. Static Tooth Load. 28.2 Friction Wheels 21. Wear Tooth Load. The motion and power transmitted by gears is 22. Causes of Gear Tooth Failure. kinematically equivalent to that transmitted by frictional 23. Design Procedure for Spur wheels or discs. In order to understand how the motion can Gears. 24. Spur Gear Construction. be transmitted by two toothed wheels, consider two plain 25. Design of Shaft for Spur Gears. circular wheels A and B mounted on shafts. The wheels have 26. Design of Arms for Spur Gears. sufficient rough surfaces and press against each other as shown in Fig. 28.1. 1021 1022 A Textbook of Machine Design
Fig. 28.1. Friction wheels. Fig. 28.2. Gear or toothed wheel. Let the wheel A is keyed to the rotating shaft and the wheel B to the shaft to be rotated. A little consideration will show that when the wheel A is rotated by a rotating shaft, it will rotate the wheel B in the opposite direction as shown in Fig. 28.1. The wheel B will be rotated by the wheel A so long as the tangential force exerted by the wheel A does not exceed the maximum frictional resistance between the two wheels. But when the tangential force (P) exceeds the *frictional resistance (F), slipping will take place between the two wheels. In order to avoid the slipping, a number of projections (called teeth) as shown in Fig. 28.2 are provided on the periphery of the wheel A which will fit into the corresponding recesses on the periphery of the wheel B. A friction wheel with the teeth cut on it is known as gear or toothed wheel. The usual connection to show the toothed wheels is by their pitch circles. Note : Kinematically, the friction wheels running without slip and toothed gearing are identical. But due to the possibility of slipping of wheels, the friction wheels can only be used for transmission of small powers. 28.3 Advantages and Disadvantages of Gear Drives The following are the advantages and disadvantages of the gear drive as compared to other drives, i.e. belt, rope and chain drives : Advantages 1. It transmits exact velocity ratio. 2. It may be used to transmit large power. 3. It may be used for small centre distances of shafts. 4. It has high efficiency. 5. It has reliable service. 6. It has compact layout. Disadvantages In bicycle gears are used to transmit 1. Since the manufacture of gears require special motion. Mechanical advantage can tools and equipment, therefore it is costlier than be changed by changing gears. other drives. μ * We know that frictional resistance, F = . RN where μ = Coefficient of friction between the rubbing surfaces of the two wheels, and
RN = Normal reaction between the two rubbing surfaces. Spur Gears 1023
2. The error in cutting teeth may cause vibrations and noise during operation. 3. It requires suitable lubricant and reliable method of applying it, for the proper operation of gear drives. 28.4 Classification of Gears The gears or toothed wheels may be classified as follows : 1. According to the position of axes of the shafts. The axes of the two shafts between which the motion is to be transmitted, may be (a) Parallel, (b) Intersecting, and (c) Non-intersecting and non-parallel. The two parallel and co-planar shafts connected by the gears is shown in Fig. 28.2. These gears are called spur gears and the arrangement is known as spur gearing. These gears have teeth parallel to the axis of the wheel as shown in Fig. 28.2. Another name given to the spur gearing is helical gearing, in which the teeth are inclined to the axis. The single and double helical gears connecting parallel shafts are shown in Fig. 28.3 (a) and (b) respectively. The object of the double helical gear is to balance out the end thrusts that are induced in single helical gears when transmitting load. The double helical gears are known as herringbone gears. A pair of spur gears are kinematically equivalent to a pair of cylindrical discs, keyed to a parallel shaft having line contact. The two non-parallel or intersecting, but coplaner shafts connected by gears is shown in Fig. 28.3 (c). These gears are called bevel gears and the arrangement is known as bevel gearing. The bevel gears, like spur gears may also have their teeth inclined to the face of the bevel, in which case they are known as helical bevel gears.
Fig. 28.3 The two non-intersecting and non-parallel i.e. non-coplanar shafts connected by gears is shown in Fig. 28.3 (d). These gears are called skew bevel gears or spiral gears and the arrangement is known as skew bevel gearing or spiral gearing. This type of gearing also have a line contact, the rotation of which about the axes generates the two pitch surfaces known as hyperboloids. Notes : (i) When equal bevel gears (having equal teeth) connect two shafts whose axes are mutually perpendicu- lar, then the bevel gears are known as mitres. (ii) A hyperboloid is the solid formed by revolving a straight line about an axis (not in the same plane), such that every point on the line remains at a constant distance from the axis. (iii) The worm gearing is essentially a form of spiral gearing in which the shafts are usually at right angles. 2. According to the peripheral velocity of the gears. The gears, according to the peripheral velocity of the gears, may be classified as : (a) Low velocity, (b) Medium velocity, and (c) High velocity. 1024 A Textbook of Machine Design
The gears having velocity less than 3 m/s are termed as low velocity gears and gears having velocity between 3 and 15 m / s are known as medium velocity gears. If the velocity of gears is more than 15 m / s, then these are called high speed gears. 3. According to the type of gearing. The gears, according to the type of gearing, may be classified as : (a) External gearing, (b) Internal gearing, and (c) Rack and pinion.
Fig. 28.4 In external gearing, the gears of the two shafts mesh externally with each other as shown in Fig. 28.4 (a). The larger of these two wheels is called spur wheel or gear and the smaller wheel is called pinion. In an external gearing, the motion of the two wheels is always unlike, as shown in Fig. 28.4 (a). In internal gearing, the gears of the two shafts mesh internally with each other as shown in Fig. 28.4 (b). The larger of these two wheels is called annular wheel and the smaller wheel is called pinion. In an internal gearing, the motion of the wheels is always like as shown in Fig. 28.4 (b). Sometimes, the gear of a shaft meshes externally and internally with the gears in a *straight line, as shown in Fig. 28.5. Such a type of gear is called rack and pinion. The straight line gear is called rack and the circular wheel is called pinion. A little consideration will show that with the help of a rack and pinion, we can convert linear motion into rotary motion and vice-versa as shown in Fig. 28.5. 4. According to the position of teeth on the gear surface. The teeth on the gear surface may be (a) Straight, (b) Inclined, and (c) Curved. We have discussed earlier that the spur gears have straight teeth whereas helical gears have their teeth inclined to the wheel rim. In case of spiral gears, the teeth are curved Fig. 28.5. Rack and pinion. over the rim surface. 28.5 Terms used in Gears The following terms, which will be mostly used in this chapter, should be clearly understood at this stage. These terms are illustrated in Fig. 28.6. 1. Pitch circle. It is an imaginary circle which by pure rolling action, would give the same motion as the actual gear.
* A straight line may also be defined as a wheel of infinite radius. Spur Gears 1025
2. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually specified by the pitch circle diameter. It is also called as pitch diameter. 3. Pitch point. It is a common point of contact between two pitch circles. 4. Pitch surface. It is the surface of the rolling discs which the meshing gears have replaced at the pitch circle. 5. Pressure angle or angle of obliquity. It is the angle between the common normal to two gear teeth at the point of contact and the common tangent at the pitch point. It is usually denoted by φ. The 1 standard pressure angles are 14 /2° and 20°. 6. Addendum. It is the radial distance of a tooth from the pitch circle to the top of the tooth. 7. Dedendum. It is the radial distance of a tooth from the pitch circle to the bottom of the tooth. 8. Addendum circle. It is the circle drawn through the top of the teeth and is concentric with the pitch circle. 9. Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also called root circle. Note : Root circle diameter = Pitch circle diameter × cos φ, where φ is the pressure angle. 10. Circular pitch. It is the distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc. Mathematically, π Circular pitch, pc = D/T where D = Diameter of the pitch circle, and T = Number of teeth on the wheel. A little consideration will show that the two gears will mesh together correctly, if the two wheels have the same circular pitch.
Note : If D1 and D2 are the diameters of the two meshing gears having the teeth T1 and T2 respectively; then for them to mesh correctly, ππ DD12= DT11= pc = or TT12 DT22
Fig. 28.6. Terms used in gears. 1026 A Textbook of Machine Design
Spur gears
11. Diametral pitch. It is the ratio of number of teeth to the pitch circle diameter in millimetres.
It denoted by pd. Mathematically, T π ⎛⎞πD = ... p = Diametral pitch, pd = ⎜⎟Q c Dpc ⎝⎠T where T = Number of teeth, and D = Pitch circle diameter. 12. Module. It is the ratio of the pitch circle diameter in millimetres to the number of teeth. It is usually denoted by m. Mathematically, Module, m = D / T Note : The recommended series of modules in Indian Standard are 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40 and 50. The modules 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5,5.5, 7, 9, 11, 14, 18, 22, 28, 36 and 45 are of second choice. 13. Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, in a meshing gear. A circle passing through the top of the meshing gear is known as clearance circle. 14. Total depth. It is the radial distance between the addendum and the dedendum circle of a gear. It is equal to the sum of the addendum and dedendum. 15. Working depth. It is radial distance from the addendum circle to the clearance circle. It is equal to the sum of the addendum of the two meshing gears. 16. Tooth thickness. It is the width of the tooth measured along the pitch circle. 17. Tooth space. It is the width of space between the two adjacent teeth measured along the pitch circle. 18. Backlash. It is the difference between the tooth space and the tooth thickness, as measured on the pitch circle. Spur Gears 1027
19. Face of the tooth. It is surface of the tooth above the pitch surface. 20. Top land. It is the surface of the top of the tooth. 21. Flank of the tooth. It is the surface of the tooth below the pitch surface. 22. Face width. It is the width of the gear tooth measured parallel to its axis. 23. Profile. It is the curve formed by the face and flank of the tooth. 24. Fillet radius. It is the radius that connects the root circle to the profile of the tooth. 25. Path of contact. It is the path traced by the point of contact of two teeth from the beginning to the end of engagement. 26. Length of the path of contact. It is the length of the common normal cut-off by the addendum circles of the wheel and pinion. 27. Arc of contact. It is the path traced by a point on the pitch circle from the beginning to the end of engagement of a given pair of teeth. The arc of contact consists of two parts, i.e. (a) Arc of approach. It is the portion of the path of contact from the beginning of the engagement to the pitch point. (b) Arc of recess. It is the portion of the path of contact from the pitch point to the end of the engagement of a pair of teeth. Note : The ratio of the length of arc of contact to the circular pitch is known as contact ratio i.e. number of pairs of teeth in contact. 28.6 Condition for Constant Velocity Ratio of Gears–Law of Gearing Consider the portions of the two teeth, one on the wheel 1 (or pinion) and the other on the wheel 2, as shown by thick line curves in Fig. 28.7. Let the two teeth come in contact at point Q, and the wheels rotate in the directions as shown in the figure. Let TT be the common tangent and MN be the common normal to the curves at point of contact
Q. From the centres O1 and O2, draw O1M and O2N perpendicular to MN. A little consideration will show that the point Q moves in the direction QC, when considered as a point on wheel 1, and in the direction QD when considered as a point on wheel 2.
Let v1 and v2 be the velocities of the point Q on the wheels 1 and 2 respectively. If the teeth are to remain in contact, then the components of these velocities along the common normal MN must be equal. ∴ α β v1 cos = v2 cos ω α ω β or ( 1 × O1Q) cos =( 2 × O2Q) cos ON ω× OM1 ω× 2 ()11OQ = ()22OQ OQ1 OQ2 ∴ωω 1.O1M = 2 . O2N ω 1 ON2 or ω = ...(i) 2 OM1 Also from similar triangles O1MP and O2NP, ON OP 2 = 2 ...(ii) OM1 OP1 Combining equations (i) and (ii), we have ω 1 ON2 OP2 ω = = ...(iii) 2 OM1 OP1 We see that the angular velocity ratio is inversely Fig. 28.7. Law of gearing. proportional to the ratio of the distance of P from the centres 1028 A Textbook of Machine Design
O1 and O2, or the common normal to the two surfaces at the point of contact Q intersects the line of centres at point P which divides the centre distance inversely as the ratio of angular velocities.
Aircraft landing gear is especially designed to absorb shock and energy when an aircraft lands, and then release gradually.
Therefore, in order to have a constant angular velocity ratio for all positions of the wheels, P must be the fixed point (called pitch point) for the two wheels. In other words, the common normal at the point of contact between a pair of teeth must always pass through the pitch point. This is fundamental condition which must be satisfied while designing the profiles for the teeth of gear wheels. It is also known as law of gearing. Notes : 1. The above condition is fulfilled by teeth of involute form, provided that the root circles from which the profiles are generated are tangential to the common normal. 2. If the shape of one tooth profile is arbitrary chosen and another tooth is designed to satisfy the above condition, then the second tooth is said to be Gear trains inside a mechanical watch conjugate to the first. The conjugate teeth are not in common use because of difficulty in manufacture and cost of production.
3. If D1 and D2 are pitch circle diameters of wheel 1 and 2 having teeth T1 and T2 respectively, then velocity ratio, ω 1 OP222== D T ω = 2 OP111 D T Spur Gears 1029 28.7 Forms of Teeth We have discussed in Art. 28.6 (Note 2) that conjugate teeth are not in common use. Therefore, in actual practice, following are the two types of teeth commonly used. 1. Cycloidal teeth ; and 2. Involute teeth. We shall discuss both the above mentioned types of teeth in the following articles. Both these forms of teeth satisfy the condition as explained in Art. 28.6.
28.8 Cycloidal Teeth A cycloid is the curve traced by a point on the circumference of a circle which rolls without slipping on a fixed straight line. When a circle rolls without slipping on the outside of a fixed circle, the curve traced by a point on the circumference of a circle is known as epicycloid. On the other hand, if a circle rolls without slipping on the inside of a fixed circle, then the curve traced by a point on the circumference of a circle is called hypocycloid.
Fig. 28.8. Construction of cycloidal teeth of a gear. In Fig. 28.8 (a), the fixed line or pitch line of a rack is shown. When the circle C rolls without slipping above the pitch line in the direction as indicated in Fig. 28.8 (a), then the point P on the circle traces the epicycloid PA . This represents the face of the cycloidal tooth profile. When the circle D rolls without slipping below the pitch line, then the point P on the circle D traces hypocycloid PB which represents the flank of the cycloidal tooth. The profile BPA is one side of the cycloidal rack tooth. Similarly, the two curves P' A' and P' B' forming the opposite side of the tooth profile are traced by the point P' when the circles C and D roll in the opposite directions. In the similar way, the cycloidal teeth of a gear may be constructed as shown in Fig. 28.8 (b). The circle C is rolled without slipping on the outside of the pitch circle and the point P on the circle C traces epicycloid PA , which represents the face of the cycloidal tooth. The circle D is rolled on the inside of pitch circle and the point P on the circle D traces hypocycloid PB, which represents the flank of the tooth profile. The profile BPA is one side of the cycloidal tooth. The opposite side of the tooth is traced as explained above. The construction of the two mating cycloidal teeth is shown in Fig. 28.9. A point on the circle D will trace the flank of the tooth T1 when circle D rolls without slipping on the inside of pitch circle of wheel 1 and face of tooth T2 when the circle D rolls without slipping on the outside of pitch circle of wheel 2. Similarly, a point on the circle C will trace the face of tooth T1 and flank of tooth T2. The rolling circles C and D may have unequal diameters, but if several wheels are to be interchangeable, they must have rolling circles of equal diameters. 1030 A Textbook of Machine Design
Fig. 28.9. Construction of two mating cycloidal teeth. A little consideration will show that the common normal XX at the point of contact between two cycloidal teeth always passes through the pitch point, which is the fundamental con- dition for a constant velocity ratio. 28.9 Involute Teeth An involute of a circle is a plane curve generated by a point on a tangent, which rolls on the circle without slipping or by a point on a taut string which is unwrapped from a reel as shown in Fig. 28.10 (a). In connection with toothed wheels, the circle is known as base circle. The involute is traced as follows : Let A be the starting point of the involute. The base circle is divided into equal number of parts e.g. AP1, P1 P2, P2 P3 etc.The tangents at P1, P2, P3 etc., are drawn and the lenghts P1A1, P2A2, P3A3 equal to the arcs AP1, AP2 and AP3 are set off. Joining the points A, A1, A2, A3 etc., we obtain the involute curve AR. A little consideration will show that at any instant
A3, the tangent A3T to the involute is perpendicular to P3A3 and P3A3 is the normal to the involute. In other words, normal at any point of an involute is a tangent to the circle.
Now, let O1 and O2 be the fixed centres of the two base circles as shown in Fig. 28.10(b). Let the corresponding involutes AB and A'B' be in contact at point Q. MQ and NQ are normals to the involute at Q and are tangents to base circles. Since the normal for an involute at a given point is the tangent drawn from that point to the base circle, therefore the common normal MN at Q is also the common tangent to the two base circles. We see that the common normal MN intersects the line of centres O1O2 at the fixed point P (called pitch point). Therefore the involute teeth satisfy the fundamental condition The clock built by Galelio of constant velocity ratio. used gears. Spur Gears 1031
From similar triangles O2 NP and O1 MP, ω OM1 OP12= = ω ...(i) ON2 OP21 which determines the ratio of the radii of the two base circles. The radii of the base circles is given by φ φ O1M = O1 P cos , and O2N = O2 P cos where φ is the pressure angle or the angle of obliquity. Also the centre distance between the base circles OM O N OM+ O N = OP+= O P 1212 + = 12cosφφ cos cos φ
Fig. 28.10. Construction of involute teeth. A little consideration will show, that if the centre distance is changed, then the radii of pitch circles also changes. But their ratio remains unchanged, because it is equal to the ratio of the two radii of the base circles [See equation (i)]. The common normal, at the point of contact, still passes through the pitch point. As a result of this, the wheel continues to work correctly*. However, the pressure angle increases with the increase in centre distance. 28.10 Comparison Between Involute and Cycloidal Gears In actual practice, the involute gears are more commonly used as compared to cycloidal gears, due to the following advantages : Advantages of involute gears Following are the advantages of involute gears : 1. The most important advantage of the involute gears is that the centre distance for a pair of involute gears can be varied within limits without changing the velocity ratio. This is not true for cycloidal gears which requires exact centre distance to be maintained. 2. In involute gears, the pressure angle, from the start of the engagement of teeth to the end of the engagement, remains constant. It is necessary for smooth running and less wear of gears. But in cycloidal gears, the pressure angle is maximum at the beginning of engagement, reduces to zero at pitch point, starts increasing and again becomes maximum at the end of engagement. This results in less smooth running of gears. 3. The face and flank of involute teeth are generated by a single curve whereas in cycloidal gears, double curves (i.e. epicycloid and hypocycloid) are required for the face and flank respectively. * It is not the case with cycloidal teeth. 1032 A Textbook of Machine Design Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, the basic rack has straight teeth and the same can be cut with simple tools. Note : The only disadvantage of the involute teeth is that the interference occurs (Refer Art. 28.13) with pinions having smaller number of teeth. This may be avoided by altering the heights of addendum and dedendum of the mating teeth or the angle of obliquity of the teeth. Advantages of cycloidal gears Following are the advantages of cycloidal gears : 1. Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger than the involute gears for the same pitch. Due to this reason, the cycloidal teeth are preferred specially for cast teeth. 2. In cycloidal gears, the contact takes place between a convex flank and concave surface, whereas in involute gears, the convex surfaces are in contact. This condition results in less wear in cycloidal gears as compared to involute gears. However the difference in wear is negligible. 3. In cycloidal gears, the interference does not occur at all. Though there are advantages of cycloidal gears but they are outweighed by the greater simplicity and flexibility of the involute gears. 28.11 Systems of Gear Teeth The following four systems of gear teeth are commonly used in practice. 1 1 1. 14 /2° Composite system, 2. 14 /2° Full depth involute system, 3. 20° Full depth involute system, and 4. 20° Stub involute system. 1 The 14 /2° composite system is used for general purpose gears. It is stronger but has no inter- changeability. The tooth profile of this system has cycloidal curves at the top and bottom and involute curve at the middle portion. The teeth are produced by formed milling cutters or hobs. The tooth 1 profile of the 14 /2° full depth involute system was developed for use with gear hobs for spur and helical gears. The tooth profile of the 20° full depth involute system may be cut by hobs. The increase of the 1 pressure angle from 14 /2° to 20° results in a stronger tooth, because the tooth acting as a beam is wider at the base. The 20° stub involute system has a strong tooth to take heavy loads. 28.12 Standard Proportions of Gear Systems The following table shows the standard proportions in module (m) for the four gear systems as discussed in the previous article. Table 28.1. Standard proportions of gear systems.
1 ° S. No. Particulars 14 /2 composite or full 20° full depth 20° stub involute depth involute system involute system system
1. Addendum 1m 1m 0.8 m 2. Dedendum 1.25 m 1.25 m 1 m 3. Working depth 2 m 2 m 1.60 m 4. Minimum total depth 2.25 m 2.25 m 1.80 m 5. Tooth thickness 1.5708 m 1.5708 m 1.5708 m 6. Minimum clearance 0.25 m 0.25 m 0.2 m 7. Fillet radius at root 0.4 m 0.4 m 0.4 m Spur Gears 1033 28.13 Interference in Involute Gears A pinion gearing with a wheel is shown in Fig. 28.11. MN is the common tangent to the base circles and KL is the path of contact between the two mating teeth. A little consideration will show, that if the radius of the addendum circle of pinion is increased to O1N, the point of contact L will move from L to N. When this radius is further increased, the point of contact L will be on the inside of base circle of wheel and not on the involute profile of tooth on wheel. The tip of tooth on the pinion will then undercut the tooth on the wheel at the root and remove part of the involute profile of tooth on the wheel. This effect is known as interference and occurs when the teeth are being cut. In brief, the phenomenon when the tip of a tooth undercuts the root on its mating gear is known as interference. A drilling machine drilling holes for lamp retaining screws
Fig. 28.11. Interference in involute gears.
Similarly, if the radius of the addendum circle of the wheel increases beyond O2M, then the tip of tooth on wheel will cause interference with the tooth on pinion. The points M and N are called interference points. Obviously interference may be avoided if the path of contact does not extend beyond interference points. The limiting value of the radius of the addendum circle of the pinion is
O1N and of the wheel is O2M. From the above discussion, we conclude that the interference may only be avoided, if the point of contact between the two teeth is always on the involute profiles of both the teeth. In other words, interference may only be prevented, if the addendum circles of the two mating gears cut the common tangent to the base circles between the points of tangency. 1034 A Textbook of Machine Design
Note : In order to avoid interference, the limiting value of the radius of the addendum circle of the pinion (O1 N) and of the wheel (O2 M), may be obtained as follows : From Fig. 28.11, we see that
222+=++φ 2 O1N = ()()OM1 MN ()[()sin] rb r R φ φ where rb = Radius of base circle of the pinion = O1P cos = r cos 22+=++ 2φ 2 Similarly O2M = ()()ON2 MN ()[()sin] Rb r R φ φ where Rb = Radius of base circle of the wheel = O2P cos = R cos 28.14 Minimum Number of Teeth on the Pinion in Order to Avoid Interference We have seen in the previous article that the interference may only be avoided, if the point of contact between the two teeth is always on the involute profiles of both the teeth. The minimum number of teeth on the pinion which will mesh with any gear (also rack) without interference are given in the following table. Table 28.2. Minimum number of teeth on the pinion in order to avoid interference.
S. No. Systems of gear teeth Minimum number of teeth on the pinion
1 ° 1. 14 /2 Composite 12 1 ° 2. 14 /2 Full depth involute 32 3. 20° Full depth involute 18 4. 20° Stub involute 14
The number of teeth on the pinion (TP) in order to avoid interference may be obtained from the following relation :
2 AW TP = ⎡⎤11⎛⎞2 G ⎢⎥12sin–1++⎜⎟ φ ⎣⎦GG⎝⎠ where AW = Fraction by which the standard addendum for the wheel should be multiplied,
G = Gear ratio or velocity ratio = TG / TP = DG / DP, φ = Pressure angle or angle of obliquity. 28.15 Gear Materials The material used for the manufacture of gears depends upon the strength and service conditions like wear, noise etc. The gears may be manufactured from metallic or non-metallic materials. The metallic gears with cut teeth are commercially obtainable in cast iron, steel and bronze. The non- metallic materials like wood, rawhide, compressed paper and synthetic resins like nylon are used for gears, especially for reducing noise. The cast iron is widely used for the manufacture of gears due to its good wearing properties, excellent machinability and ease of producing complicated shapes by casting method. The cast iron gears with cut teeth may be employed, where smooth action is not important. The steel is used for high strength gears and steel may be plain carbon steel or alloy steel. The steel gears are usually heat treated in order to combine properly the toughness and tooth hardness. Spur Gears 1035
The phosphor bronze is widely used for worm gears in order to reduce wear of the worms which will be excessive with cast iron or steel. The following table shows the properties of commonly used gear materials. Table 28.3. Properties of commonly used gear materials. Material Condition Brinell hardness Minimum tensile number strength (N/mm2) (1) (2) (3) (4) Malleable cast iron (a) White heart castings, Grade B — 217 max. 280 (b) Black heart castings, Grade B — 149 max. 320 Cast iron (a) Grade 20 As cast 179 min. 200 (b) Grade 25 As cast 197 min. 250 (c) Grade 35 As cast 207 min. 250 (d) Grade 35 Heat treated 300 min. 350 Cast steel — 145 550 Carbon steel (a) 0.3% carbon Normalised 143 500 (b) 0.3% carbon Hardened and 152 600 tempered (c) 0.4% carbon Normalised 152 580 (d) 0.4% carbon Hardened and 179 600 tempered (e) 0.35% carbon Normalised 201 720 ( f ) 0.55% carbon Hardened and 223 700 tempered Carbon chromium steel (a) 0.4% carbon Hardened and 229 800 tempered (b) 0.55% carbon ” 225 900 Carbon manganese steel (a) 0.27% carbon Hardened and 170 600 tempered (b) 0.37% carbon ” 201 700 Manganese molybdenum steel (a) 35 Mn 2 Mo 28 Hardened and 201 700 tempered (b) 35 Mn 2 Mo 45 ” 229 800 Chromium molybdenum steel (a) 40 Cr 1 Mo 28 Hardened and 201 700 tempered (b) 40 Cr 1 Mo 60 ” 248 900 1036 A Textbook of Machine Design
(1) (2) (3) (4) Nickel steel 40 Ni 3 ” 229 800 Nickel chromium steel 30 Ni 4 Cr 1 ” 444 1540 Nickel chromium molybdenum steel Hardness and 40 Ni 2 Cr 1 Mo 28 tempered 255 900 Surface hardened steel (a) 0.4% carbon steel — 145 (core) 551 460 (case) (b) 0.55% carbon steel — 200 (core) 708 520 (case) (c) 0.55% carbon chromium steel — 250 (core) 866 500 (case) (d) 1% chromium steel — 500 (case) 708 (e) 3% nickel steel — 200 (core) 708 300 (case) Case hardened steel (a) 0.12 to 0.22% carbon — 650 (case) 504 (b) 3% nickel — 200 (core) 708 600 (case) (c) 5% nickel steel — 250 (core) 866 600 (case) Phosphor bronze castings Sand cast 60 min. 160 Chill cast 70 min. 240 Centrifugal cast 90 260
28.16 Design Considerations for a Gear Drive In the design of a gear drive, the following data is usually given : 1. The power to be transmitted. 2. The speed of the driving gear, 3. The speed of the driven gear or the velocity ratio, and 4. The centre distance. The following requirements must be met in the design of a gear drive : (a) The gear teeth should have sufficient strength so that they will not fail under static loading or dynamic loading during normal running conditions. (b) The gear teeth should have wear characteristics so that their life is satisfactory. (c) The use of space and material should be economical. (d) The alignment of the gears and deflections of the shafts must be considered because they effect on the performance of the gears. (e) The lubrication of the gears must be satisfactory. Spur Gears 1037 28.17 Beam Strength of Gear Teeth – Lewis Equation The beam strength of gear teeth is determined from an equation (known as *Lewis equation) and the load carrying ability of the toothed gears as determined by this equation gives satisfactory results. In the investigation, Lewis assumed that as the load is being transmitted from one gear to another, it is all given and taken by one tooth, because it is not always safe to assume that the load is distributed among several teeth. When contact begins, the load is assumed to be at the end of the driven teeth and as contact ceases, it is at the end of the driving teeth. This may not be true when the number of teeth in a pair of mating gears is large, because the load may be distributed among several teeth. But it is almost certain that at some time during the contact of teeth, the proper distribution of load does not exist and that one tooth must transmit the full load. In any pair of gears having unlike number of teeth, the gear which have the fewer teeth (i.e. pinion) will be the weaker, because the tendency toward undercutting of the teeth becomes more pronounced in gears as the number of teeth becomes smaller. Consider each tooth as a cantilever beam loaded by a normal load (WN) as shown in Fig. 28.12. It is resolved into two components i.e. tangential component (W ) and radial component T Fig. 28.12. Tooth of a gear. (WR) acting perpendicular and parallel to the centre line of the tooth respectively. The tangential component (WT) induces a bending stress which tends to break the tooth. The radial component (WR) induces a compressive stress of relatively small magnitude, therefore its effect on the tooth may be neglected. Hence, the bending stress is used as the basis for design calculations. The critical section or the section of maximum bending stress may be obtained by drawing a parabola through A and tangential to the tooth curves at B and C. This parabola, as shown dotted in Fig. 28.12, outlines a beam of uniform strength, i.e. if the teeth are shaped like a parabola, it will have the same stress at all the sections. But the tooth is larger than the parabola at every section except BC. We therefore, conclude that the section BC is the section of maximum stress or the critical section. The maximum value of the bending stress (or the permissible working stress), at the section BC is given by σ w = M.y / I ...(i) where M = Maximum bending moment at the critical section BC = WT × h,
WT = Tangential load acting at the tooth, h = Length of the tooth, y = Half the thickness of the tooth (t) at critical section BC = t/2, I = Moment of inertia about the centre line of the tooth = b.t3/12, b = Width of gear face. Substituting the values for M, y and I in equation (i), we get ()/2()6Wht××× Wh σ = TT= w bt./1232 bt . σ 2 or WT = w × b × t /6 h In this expression, t and h are variables depending upon the size of the tooth (i.e. the circular pitch) and its profile.
* In 1892, Wilfred Lewis investigated for the strength of gear teeth. He derived an equation which is now extensively used by industry in determining the size and proportions of the gear. 1038 A Textbook of Machine Design
Let t = x × pc , and h = k × pc ; where x and k are constants. 22 2 ∴ σ× ×xp. c =σ×× ×x WT = wwcbbp 6.kpc 6 k Substituting x2 /6k = y, another constant, we have σ σ π π WT = w . b . pc . y = w . b . m . y ...(Q pc = m)
The quantity y is known as Lewis form factor or tooth form factor and WT (which is the tangential load acting at the tooth) is called the beam strength of the tooth. xt22p t 2 y == ×=c , Since 2 therefore in order to find the value of y, the 666.khhp()pc c quantities t, h and pc may be determined analytically or measured from the drawing similar to Fig. 28.12. It may be noted that if the gear is enlarged, the distances t, h and pc will each increase proportionately. Therefore the value of y will remain unchanged. A little consideration will show that the value of y is independent of the size of the tooth and depends only on the number of teeth on a gear and the system of teeth. The value of y in terms of the number of teeth may be expressed as follows : 0.684 y = 0.124 – , for 141/ ° composite and full depth involute system. T 2 0.912 = 0.154 – , for 20° full depth involute system. T 0.841 = 0.175 – , for 20° stub system. T 28.18 Permissible Working Stress for Gear Teeth in the Lewis Equation σ The permissible working stress ( w) in the Lewis equation depends upon the material for which σ an allowable static stress ( o) may be determined. The allowable static stress is the stress at the
Gear cable Chain pulls on mechanism Hub Sprocket set
Tensioner Derailleur mechanism
Idler Going up hill On the level sprocket
Bicycle gear mechanism switches the chain between different sized sprockets at the pedals and on the back wheel. Going up hill, a small front and a large rear sprocket are selected to reduce the push required for the rider. On the level, a large front and small rear. sprocket are used to prevent the rider having to pedal too fast. Spur Gears 1039 elastic limit of the material. It is also called the basic stress. In order to account for the dynamic σ effects which become more severe as the pitch line velocity increases, the value of w is reduced. According to the Barth formula, the permissible working stress, σ σ w = o × Cv σ where o = Allowable static stress, and
Cv = Velocity factor.
The values of the velocity factor (Cv) are given as follows : 3 C = , for ordinary cut gears operating at velocities upto 12.5 m / s. v 3 + v 4.5 = , for carefully cut gears operating at velocities upto 12.5 m/s. 4.5 + v 6 = , for very accurately cut and ground metallic gears 6 + v operating at velocities upto 20 m / s. 0.75 = , for precision gears cut with high accuracy and 0.75 + v operating at velocities upto 20 m / s.
⎛⎞0.75 + = ⎜⎟0.25, for non-metallic gears. ⎝⎠1 + v In the above expressions, v is the pitch line velocity in metres per second. The following table shows the values of allowable static stresses for the different gear materials. Table 28.4. Values of allowable static stress.
σ 2 Material Allowable static stress ( o) MPa or N/mm Cast iron, ordinary 56 Cast iron, medium grade 70 Cast iron, highest grade 105 Cast steel, untreated 140 Cast steel, heat treated 196 Forged carbon steel-case hardened 126 Forged carbon steel-untreated 140 to 210 Forged carbon steel-heat treated 210 to 245 Alloy steel-case hardened 350 Alloy steel-heat treated 455 to 472 Phosphor bronze 84 Non-metallic materials Rawhide, fabroil 42 Bakellite, Micarta, Celoron 56 σ Note : The allowable static stress ( o) for steel gears is approximately one-third of the ultimate tensile stregth σ σ σ ( u) i.e. o = u /3. 1040 A Textbook of Machine Design 28.19 Dynamic Tooth Load In the previous article, the velocity factor was used to make approximate allowance for the effect of dynamic loading. The dynamic loads are due to the following reasons : 1. Inaccuracies of tooth spacing, 2. Irregularities in tooth profiles, and 3. Deflections of teeth under load. A closer approximation to the actual conditions may be made by the use of equations based on extensive series of tests, as follows :
WD = WT + WI where WD = Total dynamic load, WT = Steady load due to transmitted torque, and WI = Increment load due to dynamic action. The increment load (WI) depends upon the pitch line velocity, the face width, material of the gears, the accuracy of cut and the tangential load. For average conditions, the dynamic load is determined by using the following Buckingham equation, i.e. 21vbC ( .+ W ) W = WWW+= + T ...(i) D TIT ++ 21vbCW . T where WD = Total dynamic load in newtons,
WT = Steady transmitted load in newtons, v = Pitch line velocity in m/s, b = Face width of gears in mm, and C = A deformation or dynamic factor in N/mm. A deformation factor (C) depends upon the error in action between teeth, the class of cut of the gears, the tooth form and the material of the gears. The following table shows the values of deformation factor (C) for checking the dynamic load on gears. Table 28.5. Values of deformation factor (C ).
Material Involute Values of deformation factor (C) in N-mm tooth Tooth error in action (e) in mm form Pinion Gear 0.01 0.02 0.04 0.06 0.08 Cast iron Cast iron 55 110 220 330 440 1 ° Steel Cast iron 14 /2 76 152 304 456 608 Steel Steel 110 220 440 660 880 Cast iron Cast iron 57 114 228 342 456 Steel Cast iron 20° full 79 158 316 474 632 Steel Steel depth 114 228 456 684 912 Cast iron Cast iron 59 118 236 354 472 Steel Cast iron 20° stub 81 162 324 486 648 Steel Steel 119 238 476 714 952
The value of C in N/mm may be determined by using the following relation : Ke. C = ... (ii) 11+ EEPG Spur Gears 1041 where K = A factor depending upon the form of the teeth.
1 ° = 0.107, for 14 2 full depth involute system. = 0.111, for 20° full depth involute system. = 0.115 for 20° stub system. 2 EP = Young's modulus for the material of the pinion in N/mm . 2 EG = Young's modulus for the material of gear in N/mm . e = Tooth error action in mm. The maximum allowable tooth error in action (e) depends upon the pitch line velocity (v) and the class of cut of the gears. The following tables show the values of tooth errors in action (e) for the different values of pitch line velocities and modules. Table 28.6. Values of maximum allowable tooth error in action (e) verses pitch line velocity, for well cut commercial gears.
Pitch line Tooth error in Pitch line Tooth error in Pitch line Tooth error in velocity (v) m/s action (e) mm velocity (v) m/s action (e) mm velocity (v) m/s action (e) mm
1.25 0.0925 8.75 0.0425 16.25 0.0200 2.5 0.0800 10 0.0375 17.5 0.0175 3.75 0.0700 11.25 0.0325 20 0.0150 5 0.0600 12.5 0.0300 22.5 0.0150 6.25 0.0525 13.75 0.0250 25 and over 0.0125 7.5 0.0475 15 0.0225
Table 28.7. Values of tooth error in action (e) verses module. Tooth error in action (e) in mm Module (m) in mm First class Carefully cut gears Precision gears commercial gears Upto 4 0.051 0.025 0.0125 5 0.055 0.028 0.015 6 0.065 0.032 0.017 7 0.071 0.035 0.0186 8 0.078 0.0386 0.0198 9 0.085 0.042 0.021 10 0.089 0.0445 0.023 12 0.097 0.0487 0.0243 14 0.104 0.052 0.028 16 0.110 0.055 0.030 18 0.114 0.058 0.032 20 0.117 0.059 0.033 1042 A Textbook of Machine Design 28.20 Static Tooth Load The static tooth load (also called beam strength or endurance strength of the tooth) is obtained σ by Lewis formula by substituting flexural endurance limit or elastic limit stress ( e) in place of σ permissible working stress ( w). ∴ Static tooth load or beam strength of the tooth, σ σ π WS = e.b.pc.y = e.b. m.y σ The following table shows the values of flexural endurance limit ( e) for different materials. Table 28.8. Values of flexural endurance limit. Material of pinion and Brinell hardness number Flexural endurance σ gear (B.H.N.) limit ( e) in MPa Grey cast iron 160 84 Semi-steel 200 126 Phosphor bronze 100 168 Steel 150 252 200 350 240 420 280 490 300 525 320 560 350 595 360 630 400 and above 700
For safety, against tooth breakage, the static tooth load (WS) should be greater than the dynamic load (WD). Buckingham suggests the following relationship between WS and WD. ≥ For steady loads, WS 1.25 WD ≥ For pulsating loads, WS 1.35 WD ≥ For shock loads, WS 1.5 WD σ Note : For steel, the flexural endurance limit ( e) may be obtained by using the following relation : σ e = 1.75 × B.H.N. (in MPa) 28.21 Wear Tooth Load The maximum load that gear teeth can carry, without premature wear, depends upon the radii of curvature of the tooth profiles and on the elasticity and surface fatigue limits of the materials. The maximum or the limiting load for satisfactory wear of gear teeth, is obtained by using the following Buckingham equation, i.e.
Ww = DP.b.Q.K where Ww = Maximum or limiting load for wear in newtons, DP = Pitch circle diameter of the pinion in mm, b = Face width of the pinion in mm, Q = Ratio factor 2..× VR 2T = G , = ++ for external gears VR.. 1 TGP T 2..× VR 2T = = G , for internal gears. VR..–1 TGP – T V. R . = Velocity ratio = TG / TP, K = Load-stress factor (also known as material combination factor) in N/mm2. Spur Gears 1043
The load stress factor depends upon the maximum fatigue limit of compressive stress, the pressure angle and the modulus of elasticity of the materials of the gears. According to Buckingham, the load stress factor is given by the following relation : σφ2 ()sines ⎛⎞11+ K = ⎜⎟ 1.4 ⎝⎠EEPG σ 2 where es = Surface endurance limit in MPa or N/mm , φ = Pressure angle, 2 EP = Young's modulus for the material of the pinion in N/mm , and 2 EG = Young's modulus for the material of the gear in N/mm . σ The values of surface endurance limit ( es) are given in the following table. Table 28.9. Values of surface endurance limit. Material of pinion Brinell hardness number Surface endurance limit σ 2 and gear (B.H.N.) ( es) in N/mm Grey cast iron 160 630 Semi-steel 200 630 Phosphor bronze 100 630 Steel 150 350 200 490 240 616 280 721 300 770 320 826 350 910 400 1050
Clutch lever
Roller
Intermediate gear wheels
Blades Gearwheel to turn the blades
Driving gear wheel An old model of a lawn-mower 1044 A Textbook of Machine Design
Notes : 1. The surface endurance limit for steel may be obtained from the following equation : σ 2 es = (2.8 × B.H.N. – 70) N/mm 2. The maximum limiting wear load (Ww) must be greater than the dynamic load (WD). 28.22 Causes of Gear Tooth Failure The different modes of failure of gear teeth and their possible remedies to avoid the failure, are as follows : 1. Bending failure. Every gear tooth acts as a cantilever. If the total repetitive dynamic load acting on the gear tooth is greater than the beam strength of the gear tooth, then the gear tooth will fail in bending, i.e. the gear tooth will break. In order to avoid such failure, the module and face width of the gear is adjusted so that the beam strength is greater than the dynamic load. 2. Pitting. It is the surface fatigue failure which occurs due to many repetition of Hertz contact stresses. The failure occurs when the surface contact stresses are higher than the endurance limit of the material. The failure starts with the formation of pits which continue to grow resulting in the rupture of the tooth surface. In order to avoid the pitting, the dynamic load between the gear tooth should be less than the wear strength of the gear tooth. 3. Scoring. The excessive heat is generated when there is an excessive surface pressure, high speed or supply of lubricant fails. It is a stick-slip phenomenon in which alternate shearing and welding takes place rapidly at high spots. This type of failure can be avoided by properly designing the parameters such as speed, pressure and proper flow of the lubricant, so that the temperature at the rubbing faces is within the permissible limits. 4. Abrasive wear. The foreign particles in the lubricants such as dirt, dust or burr enter between the tooth and damage the form of tooth. This type of failure can be avoided by providing filters for the lubricating oil or by using high viscosity lubricant oil which enables the formation of thicker oil film and hence permits easy passage of such particles without damaging the gear surface. 5. Corrosive wear. The corrosion of the tooth surfaces is mainly caused due to the presence of corrosive elements such as additives present in the lubricating oils. In order to avoid this type of wear, proper anti-corrosive additives should be used. 28.23 Design Procedure for Spur Gears In order to design spur gears, the following procedure may be followed : 1. First of all, the design tangential tooth load is obtained from the power transmitted and the pitch line velocity by using the following relation : P W = × C ...(i) T v S where WT = Permissible tangential tooth load in newtons, P = Power transmitted in watts, π DN *v = Pitch line velocity in m / s = , 60 D = Pitch circle diameter in metres,
* We know that circular pitch, π π pc = D / T = m ...(Q m = D / T) ∴ D = m.T Thus, the pitch line velocity may also be obtained by using the following relation, i.e. ππDN... mTN pTN.. v = ==c 60 60 60 where m = Module in metres, and T = Number of teeth. Spur Gears 1045
N = Speed in r.p.m., and
CS = Service factor. The following table shows the values of service factor for different types of loads : Table 28.10. Values of service factor. Type of service Type of load Intermittent or 3 hours 8-10 hours per day Continuous 24 hours per day per day Steady 0.8 1.00 1.25 Light shock 1.00 1.25 1.54 Medium shock 1.25 1.54 1.80 Heavy shock 1.54 1.80 2.00
Note : The above values for service factor are for enclosed well lubricated gears. In case of non-enclosed and grease lubricated gears, the values given in the above table should be divided by 0.65. 2. Apply the Lewis equation as follows : σ σ π WT = w.b.pc.y = w.b. m.y σ π σ σ =( o.Cv) b. m.y ...(Q w = o.Cv) Notes : (i) The Lewis equation is applied only to the weaker of the two wheels (i.e. pinion or gear). (ii) When both the pinion and the gear are made of the same material, then pinion is the weaker. σ σ (iii) When the pinion and the gear are made of different materials, then the product of ( w × y) or ( o × y) σ σ is the *deciding factor. The Lewis equation is used to that wheel for which ( w × y) or ( o × y) is less.
A bicycle with changeable gears.
* We see from the Lewis equation that for a pair of mating gears, the quantities like WT, b, m and Cv are σ σ constant. Therefore ( w × y) or ( o × y) is the only deciding factor. 1046 A Textbook of Machine Design σ (iv) The product ( w × y) is called strength factor of the gear.
(v) The face width (b) may be taken as 3 pc to 4 pc (or 9.5 m to 12.5 m) for cut teeth and 2 pc to 3 pc (or 6.5 m to 9.5 m) for cast teeth.
3. Calculate the dynamic load (WD) on the tooth by using Buckingham equation, i.e.
WD = WT + WI 21vbCW ( .+ ) = W + T T ++ 21vbCW . T In calculating the dynamic load (WD), the value of tangential load (WT) may be calculated by neglecting the service factor (CS) i.e.
WT = P / v, where P is in watts and v in m / s. 4. Find the static tooth load (i.e. beam strength or the endurance strength of the tooth) by using the relation, σ σ π WS = e.b.pc.y = e.b. m.y
For safety against breakage, WS should be greater than WD. 5. Finally, find the wear tooth load by using the relation,
Ww = DP.b.Q.K
The wear load (Ww) should not be less than the dynamic load (WD). Example 28.1. The following particulars of a single reduction spur gear are given : Gear ratio = 10 : 1; Distance between centres = 660 mm approximately; Pinion transmits 500 kW at 1800 r.p.m.; Involute teeth of standard proportions (addendum = m) with pressure angle of 22.5°; Permissible normal pressure between teeth = 175 N per mm of width. Find : 1. The nearest standard module if no interference is to occur; 2. The number of teeth on each wheel; 3. The necessary width of the pinion; and 4. The load on the bearings of the wheels due to power transmitted. 3 Solution : Given : G = TG / TP = DG / DP = 10 ; L = 660 mm ; P = 500 kW = 500 × 10 W; φ NP = 1800 r.p.m. ; = 22.5° ; WN = 175 N/mm width 1. Nearest standard module if no interference is to occur Let m = Required module,
TP = Number of teeth on the pinion,
TG = Number of teeth on the gear,
DP = Pitch circle diameter of the pinion, and
DG = Pitch circle diameter of the gear. We know that minimum number of teeth on the pinion in order to avoid interference,
2 AW TP = ⎡⎤11⎛⎞2 G ⎢⎥12sin–1++⎜⎟ φ ⎣⎦GG⎝⎠ 21× 2 = ==13.3 say 14 0.15 ⎡⎤11⎛⎞2 10⎢⎥ 1++⎜⎟ 2 sin 22.5 ° – 1 ⎣⎦10⎝⎠ 10 ... (Q AW = 1 module) ∴ TG = G × TP = 10 × 14 = 140 ...(Q TG / TP = 10) Spur Gears 1047
DDDD10 We know that L = GG+=+PP =5.5 D ...( D / D = 10) 222 2 P Q G P ∴ 660 = 5.5 DP or DP = 660 / 5.5 = 120 mm
We also know that DP = m . TP ∴ m = DP / TP = 120 / 14 = 8.6 mm Since the nearest standard value of the module is 8 mm, therefore we shall take m = 8 mm Ans. 2. Number of teeth on each wheel We know that number of teeth on the pinion,
TP = DP / m = 120 / 8 = 15 Ans. and number of teeth on the gear,
TG = G × TP = 10 × 15 = 150 Ans. 3. Necessary width of the pinion We know that the torque acting on the pinion, P ×××60 500 103 60 ==2652 N-m T = ππ× 221800NP T 2652 ∴ ==44 200 N Tangential load, WT = ...(Q DP is taken in metres) DP /2 0.12/2 and normal load on the tooth, W 44 200 W = T ==47 840 N N cosφ° cos 22.5 Since the normal pressure between teeth is 175 N per mm of width, therefore necessary width of the pinion, 47 840 b = = 273.4 mm AAns. 175 4. Load on the bearings of the wheels We know that the radial load on the bearings due to the power transmitted, φ WR = WN . sin = 47 840 × sin 22.5° = 18 308 N = 18.308 kN Ans. Example 28.2. A bronze spur pinion rotating at 600 r.p.m. drives a cast iron spur gear at a transmission ratio of 4 : 1. The allowable static stresses for the bronze pinion and cast iron gear are 84 MPa and 105 MPa respectively. The pinion has 16 standard 20° full depth involute teeth of module 8 mm. The face width of both the gears is 90 mm. Find the power that can be transmitted from the standpoint of strength. σ 2 Solution. Given : NP = 600 r.p.m. ; V.R. = TG / TP = 4 ; OP = 84 MPa = 84 N / mm ; σ 2 OG = 105 MPa = 105 N/mm ; TP = 16 ; m = 8 mm ; b = 90 mm We know that pitch circle diameter of the pinion,
DP = m.TP = 8 × 16 = 128 mm = 0.128 m ∴ Pitch line velocity, π DN. π×0.128 × 600 v = PP==4.02 m/s 60 60 Since the pitch line velocity (v) is less than 12.5 m/s, therefore velocity factor, 33 C = ==0.427 v 334.02++v 1048 A Textbook of Machine Design
We know that for 20° full depth involute teeth, tooth form factor for the pinion, 0.912 0.912 y = 0.154 –== 0.154 – 0.097 P T 16 and tooth form factor for the gear, P 0.912 0.912 y = 0.154 –== 0.154 – 0.14 ... ( T / T = 4) G × Q G P TG 416 ∴σ OP × yP = 84 × 0.097 = 8.148 σ and OG × yG = 105 × 0.14 = 14.7 σ σ Since ( OP × yP) is less than ( OG × yG), therefore the pinion is weaker. Now using the Lewis equation for the pinion, we have tangential load on the tooth (or beam strength of the tooth), σ π σ π σ σ WT = wP.b. m.yP = ( OP × Cv) b. m.yP (Q WP = OP.Cv) = 84 × 0.427 × 90 × π × 8 × 0.097 = 7870 N ∴ Power that can be transmitted
= WT × v = 7870 × 4.02 = 31 640 W = 31.64 kW Ans. Example 28.3. A pair of straight teeth spur gears is to transmit 20 kW when the pinion rotates at 300 r.p.m. The velocity ratio is 1 : 3. The allowable static stresses for the pinion and gear materials are 120 MPa and 100 MPa respectively. The pinion has 15 teeth and its face width is 14 times the module. Determine : 1. module; 2. face width; and 3. pitch circle diameters of both the pinion and the gear from the standpoint of strength only, taking into consideration the effect of the dynamic loading. The tooth form factor y can be taken as 0912. y = 0.154 – No. of teeth and the velocity factor Cv as 3 C = , where v is expressed in m / s. v 3+v 3 Solution. Given : P = 20 kW = 20 × 10 W; NP = 300 r.p.m. ; V. R . = TG / TP =3 ; σ 2 σ 2 OP = 120 MPa = 120 N/mm ; OG = 100 MPa = 100 N/mm ; TP = 15 ; b = 14 module = 14 m 1. Module Let m = Module in mm, and
DP = Pitch circle diameter of the pinion in mm. We know that pitch line velocity, ππ DNPP mTN.. P P ν = = ... ( D = m.T ) 60 60 Q P P π×m 15 × 300 = = 236mm mm/s = 0.236 m/s 60 Assuming steady load conditions and 8-10 hours of service per day, the service factor (CS) from Table 28.10 is given by
CS =1 We know that design tangential tooth load, P 20× 103 84 746 W = ×=C ×=1N T vmmS 0.236 33 and velocity factor, C = = v 3++vm 3 0.236 Spur Gears 1049
We know that tooth form factor for the pinion, 0.912 0.912 0.154 –= 0.154 – yP = TP 15 = 0.154 – 0.0608 = 0.0932 and tooth form factor for the gear, 0.912 0.912 y = 0.154 –= 0.154 – G × TG 315
= 0.154 – 0.203 = 0.1337 ... (Q TG = 3TP) ∴σ OP × yP = 120 × 0.0932 = 11.184 σ and OG × yG = 100 × 0.1337 = 13.37 σ σ Since ( OP × yP) is less than ( OG × yG), therefore the pinion is weaker. Now using the Lewis equation to the pinion, we have σ π σ π WT = wP.b. m.yP = ( OP × Cv) b. m . yP 2 84 746 ⎛⎞3×π × = 1476 m ∴ = 120⎜⎟ 14mm 0.0932 m ⎝⎠3++ 0.236mm 3 0.236 or 3 + 0.236 m = 0.0174 m3 Solving this equation by hit and trial method, we find that m = 6.4 mm The standard module is 8 mm. Therefore let us take m = 8 mm Ans. 2. Face width We know that the face width, b = 14 m = 14 × 8 = 112 mm Ans.
Spur gear Handle Cutting blade
Semi- circular rack
Pinion
Kitchen Gear : This 1863 fruit and vegetable peeling machine uses a rack and pinion to drive spur gears that turn an apple against a cutting blade. As the handle is pushed round the semi-circular base, the peel is removed from the apple in a single sweep. 1050 A Textbook of Machine Design
3. Pitch circle diameter of the pinion and gear We know that pitch circle diameter of the pinion,
DP = m.TP = 8 × 15 = 120 mm Ans. and pitch circle diameter of the gear,
DG = m.TG = 8 × 45 = 360 mm Ans. ... (Q TG = 3 TP) Example 28.4. A gear drive is required to transmit a maximum power of 22.5 kW. The velocity ratio is 1:2 and r.p.m. of the pinion is 200. The approximate centre distance between the shafts may be taken as 600 mm. The teeth has 20° stub involute profiles. The static stress for the gear material (which is cast iron) may be taken as 60 MPa and face width as 10 times the module. Find the module, face width and number of teeth on each gear. Check the design for dynamic and wear loads. The deformation or dynamic factor in the Buckingham equation may be taken as 80 and the material combination factor for the wear as 1.4.
Solution. Given : P = 22.5 kW = 22 500 W ; V. R .= DG/DP = 2 ; NP = 200 r.p.m. ; L = 600 mm ; σ σ 2 OP = OG = 60 MPa = 60 N/mm ; b = 10 m ; C = 80 ; K = 1.4 Module Let m = Module in mm,
DP = Pitch circle diameter of the pinion, and
DG = Pitch circle diameter of the gear. We know that centre distance between the shafts (L), DDDD 2 600 = PPP+=+G =1.5 D ... ( D = V.R. × D ) 222 2 P Q G P
Arm of a material handler In addition to gears, hydraulic rams as shown above, play important role in transmitting force and energy. Spur Gears 1051 ∴ DP = 600 / 1.5 = 400 mm = 0.4 m and DG =2 DP = 2 × 400 = 800 mm = 0.8 m Since both the gears are made of the same material, therefore pinion is the weaker. Thus the design will be based upon the pinion. We know that pitch line velocity of the pinion, π DN. π×0.4 × 200 v = PP==4.2 m / s 60 60 Since v is less than 12 m / s, therefore velocity factor, 33 C = ==0.417 v 334.2++v We know that number of teeth on the pinion,
TP = DP / m = 400 / m ∴ Tooth form factor for the pinion, × 0.841= 0.841 m yP = 0.175 – 0.175 – ... (For 20° stub system) TP 400 = 0.175 – 0.0021 m ...(i)
Assuming steady load conditions and 8–10 hours of service per day, the service factor (CS) from Table 28.10 is given by
CS =1 We know that design tangential tooth load, P 22 500 W = ×=C ×=1 5357 N T v S 4.2
We also know that tangential tooth load (WT), σ π σ π 5357 = wP . b. m.yP =( OP × Cv) b. m.yP = (60 × 0.417) 10 m × π m (0.175 – 0.0021 m) = 137.6 m2 – 1.65 m3 Solving this equation by hit and trial method, we find that m = 0.65 say 8 mm Ans. Face width We know that face width, b = 10 m = 10 × 8 = 80 mm Ans. Number of teeth on the gears We know that number of teeth on the pinion,
TP = DP / m = 400 / 8 = 50 Ans. and number of teeth on the gear,
TG = DG / m = 800 / 8 = 100 Ans. Checking the gears for dynamic and wear load We know that the dynamic load, 21vbCW ( .+ ) W = W + T D T ++ 21vbCW . T 21××+ 4.2 (80 80 5357) = 5357 + 21×+ 4.2 80 ×+ 80 5357 1052 A Textbook of Machine Design
1.037× 106 = 5357+=+= 5357 5273 10 630 N 196.63 From equation (i), we find that tooth form factor for the pinion,
yP = 0.175 – 0.0021 m = 0.175 – 0.0021 × 8 = 0.1582 σ 2 From Table 28.8, we find that flexural endurance limit ( e) for cast iron is 84 MPa or 84 N/mm . ∴ Static tooth load or endurance strength of the tooth, σ π π WS = e . b. m. yP = 84 × 80 × × 8 × 0.1582 = 26 722 N We know that ratio factor, 2..22××VR Q = ==1.33 VR..++ 1 2 1 ∴ Maximum or limiting load for wear,
Ww = DP . b. Q . K = 400 × 80 × 1.33 × 1.4 = 59 584 N
Since both WS and Ww are greater than WD, therefore the design is safe. Example 28.5. A pair of straight teeth spur gears, having 20° involute full depth teeth is to transmit 12 kW at 300 r.p.m. of the pinion. The speed ratio is 3 : 1. The allowable static stresses for gear of cast iron and pinion of steel are 60 MPa and 105 MPa respectively. Assume the following: 4.5 Number of teeth of pinion = 16; Face width = 14 times module; Velocity factor (C ) = , v 4.5 + v 0.912 v being the pitch line velocity in m / s; and tooth form factor (y) = 0.154 – No. of teeth Determine the module, face width and pitch diameter of gears. Check the gears for wear; given σ 2 2 es = 600 MPa; EP = 200 kN/mm and EG = 100 kN/mm . Sketch the gears. φ 3 Solution : Given : = 20°; P = 12 kW = 12 × 10 W; NP = 300 r.p.m ; V.R. = TG / TP = 3 ; σ 2 σ 2 σ OG = 60 MPa = 60 N/mm ; OP = 105 MPa = 105 N/mm ; TP = 16; b = 14 module = 14 m ; es = 600 2 2 3 2 2 3 2 MPa = 600 N/mm ; EP = 200 kN/mm = 200 × 10 N/mm ; EG = 100 kN/mm = 100 × 10 N/mm Module Let m = Module in mm, and
DP = Pitch circle diameter of the pinion in mm. We know that pitch line velocity, ππDN... mTN v = PP= PP ... ( D = m.T ) 60 60 Q P P π×m 16 × 300 = = 251mm mm / s = 0.251 m / s 60 Assuming steady load conditions and 8–10 hours of service per day, the service factor (CS) from Table 28.10 is given by CS = 1. We know that the design tangential tooth load, P 12×× 1033 47.8 10 W = ×=C ×=1N T vmmS 0.251 4.5 4.5 and velocity factor, C = = v 4.5++vm 4.5 0.251 We know that tooth form factor for pinion, 0.912== 0.912 yP = 0.154 – 0.154 – 0.097 TP 16 Spur Gears 1053 and tooth form factor for gear, 0.912 0.912 y = 0.154 –== 0.154 – 0.135 ... ( T = 3 T ) G × Q G P TG 316 ∴σ OP × yP = 105 × 0.097 = 10.185 σ and OG × yG = 60 × 0.135 = 8.1 σ σ Since ( OG × yG) is less than ( OP × yP), therefore the gear is weaker. Now using the Lewis equation to the gear, we have σ π σ π σ σ WT = wG . b. m.yG = ( OG × Cv) b. m.yG ... (Q wG = OG.Cv) × 3 2 47.8 10 ⎛⎞4.5×π × = 1603.4 m = 60⎜⎟ 14mm 0.135 m ⎝⎠4.5++ 0.251mm 4.5 0.251 or 4.5 + 0.251 m = 0.0335 m3 Solving this equation by hit and trial method, we find that m = 5.6 say 6 mm Ans. Face width We know that face width, b = 14 m = 14 × 6 = 84 mm Ans. Pitch diameter of gears We know that pitch diameter of the pinion,
DP = m.TP = 6 × 16 = 96 mm Ans. and pitch diameter of the gear,
DG = m.TG = 6 × 48 = 288 mm Ans. ...(Q TG = 3 TP)
This is a close-up photo (magnified 200 times) of a micromotor’s gear cogs. Micromotors have been developed for use in space missions and microsurgery. 1054 A Textbook of Machine Design
Checking the gears for wear We know that the ratio factor, 2..23××VR Q = ==1.5 VR..++ 1 3 1 σφ2 ()sines ⎛⎞11+ and load stress factor, K = ⎜⎟ 1.4 ⎝⎠EEPG 2 ° (600) sin 20⎡⎤ 1+ 1 = ⎢⎥33 1.4 ⎣⎦200×× 10 100 10 = 0.44 + 0.88 = 1.32 N/mm2 We know that the maximum or limiting load for wear,
Ww = DP.b.Q.K = 96 × 84 × 1.5 × 1.32 = 15 967 N and tangential load on the tooth (or beam strength of the tooth), 47.8×× 1033 47.8 10 W = ==7967 N T m 6 Since the maximum wear load is much more than the tangential load on the tooth, therefore the design is satisfactory from the standpoint of wear. Ans. Example 28.6. A reciprocating compressor is to be connected to an electric motor with the help of spur gears. The distance between the shafts is to be 500 mm. The speed of the electric motor is 900 r.p.m. and the speed of the compressor shaft is desired to be 200 r.p.m. The torque, to be transmitted is 5000 N-m. Taking starting torque as 25% more than the normal torque, determine : 1. Module and face width of the gears using 20 degrees stub teeth, and 2. Number of teeth and pitch circle diameter of each gear. Assume suitable values of velocity factor and Lewis factor.
Solution. Given : L = 500 mm ; NM = 900 r.p.m. ; NC = 200 r.p.m. ; T = 5000 N-m ; Tmax = 1.25 T 1. Module and face width of the gears Let m = Module in mm, and b = Face width in mm. Since the starting torque is 25% more than the normal torque, therefore the maximum torque, 3 Tmax = 1.25 T = 1.25 × 5000 = 6250 N-m = 6250 × 10 N-mm We know that velocity ratio, N 900 V.R. = M ==4.5 NC 200 Let DP = Pitch circle diameter of the pinion on the motor shaft, and DG = Pitch circle diameter of the gear on the compressor shaft. We know that distance between the shafts (L), D D 500 = P + G or D + D = 500 × 2 = 1000 ...(i) 22 P G D G = 4.5 and velocity ratio, V. R .= or DG = 4.5 DP ...(ii) DP Substituting the value of DG in equation (i), we have DP + 4.5 DP = 1000 or DP = 1000 / 5.5 = 182 mm and DG = 4.5 DP = 4.5 × 182 = 820 mm = 0.82 m We know that pitch line velocity of the drive, π DN. π×0.82 × 200 v = GC==8.6 m / s 60 60 Spur Gears 1055
∴ Velocity factor, 33 C = ==0.26 ...( v is less than 12.5 m/s) v 338.6++v Q Let us assume than motor pinion is made of forged steel and the compressor gear of cast steel. Since the allowable static stress for the cast steel is less than the forged steel, therefore the design should be based upon the gear. Let us take the allowable static stress for the gear material as σ 2 OG = 140 MPa = 140 N/mm We know that for 20° stub teeth, Lewis factor for the gear, × 0.841 0.841 m ⎛⎞D 0.175 –= 0.175 – ... T = G yG = ⎜⎟Q G TDGG ⎝⎠m 0.841 m = 0.175 –= 0.175 – 0.001 m 820 and maximum tangential force on the gear, 2 T 2625010××3 max ==15 244 N WT = DG 820 We also know that maximum tangential force on the gear, σ π σ π σ σ WT = wG.b. m.yG = ( OG × Cν) b × m × yG ...(Q wG = OG. Cν) 15 244 = (140 × 0.26) × 10 m × π m (0.175 – 0.001 m) = 200 m2 – 1.144 m3 ...(Assuming b = 10 m) Solving this equation by hit and trial method, we find that m = 8.95 say 10 mm Ans. and b = 10 m = 10 × 10 = 100 mm Ans. 2. Number of teeth and pitch circle diameter of each gear We know that number of teeth on the pinion, D 182 T = P ==18.2 P m 10 D 820 T = G ==82 G m 10 1056 A Textbook of Machine Design
In order to have the exact velocity ratio of 4.5, we shall take
TP = 18 and TG = 81 Ans. ∴ Pitch circle diameter of the pinion,
DP = m × TP = 10 × 18 = 180 mm Ans. and pitch circle diameter of the gear,
DG = m × TG = 10 × 81 = 810 mm Ans. 28.24 Spur Gear Construction The gear construction may have different designs depending upon the size and its application. When the dedendum circle diameter is slightly greater than the shaft diameter, then the pinion teeth are cut integral with the shaft as shown in Fig. 28.13 (a). If the pitch circle diameter of the pinion is less than or equal to 14.75 m + 60 mm (where m is the module in mm), then the pinion is made solid with uniform thickness equal to the face width, as shown in Fig. 28.13 (b). Small gears upto 250 mm pitch circle diameter are built with a web, which joins the hub and the rim. The web thickness is generally equal to half the circular pitch or it may be taken as 1.6 m to 1.9 m, where m is the module. The web may be made solid as shown in Fig. 28.13 (c) or may have recesses in order to reduce its weight.
Fig. 28.13. Construction of spur gears.
Fig. 28.14. Gear with arms. Spur Gears 1057
Large gears are provided with arms to join the hub and the rim, as shown in Fig. 28.14. The number of arms depends upon the pitch circle diameter of the gear. The number of arms may be selected from the following table. Table 28.11. Number of arms for the gears.
S. No. Pitch circle diameter Number of arms
1. Up to 0.5 m 4 or 5 2. 0.5 – 1.5 m 6 3. 1.5 – 2.0 m 8 4. Above 2.0 m 10 The cross-section of the arms is most often elliptical, but other sections as shown in Fig. 28.15 may also be used.
Fig. 28.15. Cross-section of the arms. The hub diameter is kept as 1.8 times the shaft diameter for steel gears, twice the shaft diameter for cast iron gears and 1.65 times the shaft diameter for forged steel gears used for light service. The length of the hub is kept as 1.25 times the shaft diameter for light service and should not be less than the face width of the gear. The thickness of the gear rim should be as small as possible, but to facilitate casting and to avoid sharp changes of section, the minimum thickness of the rim is generally kept as half of the circular pitch (or it may be taken as 1.6 m to 1.9 m, where m is the module). The thickness of rim (tR) may also be calculated by using the following relation, i.e. T t = m R n where T = Number of teeth, and n = Number of arms. The rim should be provided with a circumferential rib of thickness equal to the rim thickness. 1058 A Textbook of Machine Design 28.25 Design of Shaft for Spur Gears In order to find the diameter of shaft for spur gears, the following procedure may be followed.
1. First of all, find the normal load (WN), acting between the tooth surfaces. It is given by φ WN = WT / cos where WT = Tangential load, and φ = Pressure angle.
A thrust parallel and equal to WN will act at the gear centre as shown in Fig. 28.16. Fig. 28.16. Load acting on the gear. 2. The weight of the gear is given by 2 WG = 0.001 18 TG.b.m (in N) where TG = No. of teeth on the gear, b = Face width in mm, and m = Module in mm. 3. Now the resultant load acting on the gear, 22++× φ WR = ()WWNG () 2 WW NG cos 4. If the gear is overhung on the shaft, then bending moment on the shaft due to the resultant load,
M = WR × x where x = Overhang i.e. the distance between the centre of gear and the centre of bearing. 5. Since the shaft is under the combined effect of torsion and bending, therefore we shall determine the equivalent torque. We know that equivalent torque, 22+ Te = MT where T = Twisting moment = WT × DG /2 6. Now the diameter of the gear shaft (d ) is determined by using the following relation, i.e. π 3 T = ×τ×d e 16 where τ = Shear stress for the material of the gear shaft. Note : Proceeding in the similar way as discussed above, we may calculate the diameter of the pinion shaft. 28.26 Design of Arms for Spur Gears The cross-section of the arms is calculated by assuming them as a cantilever beam fixed at the hub and loaded at the pitch circle. It is also assumed that the load is equally distributed to all the arms. It may be noted that the arms are designed for the stalling load. The stalling load is a load that will develop the maximum stress in the arms and in the teeth. This happens at zero velocity, when the drive just starts operating. The stalling load may be taken as the design tangential load divided by the velocity factor.
Design tangential load = WT Let WS = Stalling load = , Velocity factor Cv DG = Pitch circle diameter of the gear, n = Number of arms, and σ b = Allowable bending stress for the material of the arms. Spur Gears 1059
Now, maximum bending moment on each arm, WD××/2 WD M = SG= SG nn2 and the section modulus of arms for elliptical cross-section, π ()ab2 Z = 11 32 where a1 = Major axis, and b1 = Minor axis. σ The major axis is usually taken as twice the minor axis. Now, using the relation, b = M / Z, we can calculate the dimensions a1 and b1 for the gear arm at the hub end. Note : The arms are usually tapered towards the rim about 1/16 per unit length of the arm (or radius of the gear). ∴ Major axis of the section at the rim end
11DDGG = aa– Taper = –××= Length of the arm = a – a – 1116 1 16 2 1 32 Example 28.7. A motor shaft rotating at 1500 r.p.m. has to transmit 15 kW to a low speed shaft 1 ° with a speed reduction of 3:1. The teeth are 14 /2 involute with 25 teeth on the pinion. Both the pinion and gear are made of steel with a maximum safe stress of 200 MPa. A safe stress of 40 MPa may be taken for the shaft on which the gear is mounted and for the key. Design a spur gear drive to suit the above conditions. Also sketch the spur gear drive. Assume starting torque to be 25% higher than the running torque. 3 φ 1 ° Solution : Given : NP = 1500 r.p.m. ; P = 15 kW = 15 × 10 W; V.R. = TG/TP = 3 ; = 14 /2 ; σ σ 2 τ 2 TP = 25 ; OP = OG = 200 MPa = 200 N/mm ; = 40 MPa = 40 N/mm Design for spur gears Since the starting torque is 25% higher than the running torque, therefore the spur gears should be designed for power, 3 P1 = 1.25 P = 1.25 × 15 × 10 = 18 750 W
We know that the gear reduction ratio (TG / TP) is 3. Therefore the number of teeth on the gear,
TG =3 TP = 3 × 25 = 75 Let us assume that the module (m) for the pinion and gear is 6 mm. ∴ Pitch circle diameter of the pinion,
DP = m.TP = 6 × 25 = 150 mm = 0.15 m and pitch circle diameter of the gear,
DG = m. TG = 6 × 75 = 450 mm We know that pitch line velocity, π DN. π×0.15 × 1500 v = PP==11.8 m / s 60 60 Assuming steady load conditions and 8–10 hours of service per day, the service factor (CS) from Table 28.10 is given by
CS =1 ∴ Design tangential tooth load, P 18 750 W = 1 ×=C ×=11590N T v S 11.8 We know that for ordinary cut gears and operating at velocities upto 12.5 m/s, the velocity factor, 33 C = ==0.203 v 3++v 3 11.8 1060 A Textbook of Machine Design
Since both the pinion and the gear are made of the same material, therefore the pinion is the weaker. 1 ° We know that for 14 /2 involute teeth, tooth form factor for the pinion,
0.684== 0.684 yP = 0.124 – 0.124 – 0.0966 TP 25 Let b =Face width for both the pinion and gear. We know that the design tangential tooth load
(WT), σ π σ π 1590 = wP.b. m.yP = ( OP.Cv) b. m.yP = (200 × 0.203) b × π × 6 × 0.0966 = 74 b ∴ b = 1590 / 74 = 21.5 mm In actual practice, the face width (b) is taken as 9.5 m to 12.5 m, but in certain cases, due to space limitations, it may be taken as 6 m. Therefore let us take the face width, b =6 m = 6 × 6 = 36 mm Ans. This mathematical machine called From Table 28.1, the other proportions, for the difference engine, assembled in 1832, 1 ° pinion and the gear having 14 /2 involute teeth, are as used 2,000 levers, cams and gears. follows : Addendum = 1 m = 6 mm Ans. Dedendum = 1.25 m = 1.25 × 6 = 7.5 mm Ans. Working depth = 2 m = 2 × 6 = 12 mm Ans. Minimum total depth = 2.25 m = 2.25 × 6 = 13.5 mm Ans. Tooth thickness = 1.5708 m = 1.5708 × 6 = 9.4248 mm Ans. Minimum clearance = 0.25 m = 0.25 × 6 = 1.5 mm Ans. Design for the pinion shaft We know that the normal load acting between the tooth surfaces, W 1590 1590 T ===1643 N WN = φ 1 coscos14 /2° 0.9681 and weight of the pinion, 2 2 WP = 0.00118 TP.b.m = 0.001 18 × 25 × 36 × 6 = 38 N ∴ Resultant load acting on the pinion, 22++ φ *WR = ()WWWWNP () 2..cos NP
22 1 ° = (1643)++××× (38) 2 1643 38 cos 14 /2 = 1680 N Assuming that the pinion is overhung on the shaft and taking overhang as 100 mm, therefore Bending moment on the shaft due to the resultant load,
M = WR × 100 = 1680 × 100 = 168 000 N-mm
* Since the weight of the pinion (WP) is very small as compared to the normal load (WN), therefore it may be neglected. Thus the resultant load acting on the pinion (WR) may be taken equal to WN. Spur Gears 1061 and twisting moment on the shaft, D 150 T = W ×=P 1590 ×= 119 250 N-mm T 22 ∴ Equivalent twisting moment, 22+= 2 + 2 = × 3 Te = MT (168 000) (119 250) 206 10 N-mm
Let dP = Diameter of the pinion shaft.
We know that equivalent twisting moment (Te), ππ233 206 × 103 = ×τ(ddd ) = × 40 ( ) = 7.855 ( ) 16PPP 16 ∴ 3 3 3 (dP) = 206 × 10 / 7.855 = 26.2 × 10 or dP = 29.7 say 30 mm Ans. We know that the diameter of the pinion hub
= 1.8 dP = 1.8 × 30 = 54 mm Ans. and length of the hub = 1.25 dP = 1.25 × 30 = 37.5 mm Since the length of the hub should not be less than that of the face width i.e. 36 mm, therefore let us take length of the hub as 36 mm. Ans. Note : Since the pitch circle diameter of the pinion is 150 mm, therefore the pinion should be provided with a web and not arms. Let us take thickness of the web as 1.8 m, where m is the module. ∴ Thickness of the web = 1.8 m = 1.8 × 6 = 10.8 mm Ans. Design for the gear shaft We have calculated above that the normal load acting between the tooth surfaces,
WN = 1643 N We know that weight of the gear, 2 2 WG = 0.001 18 TG.b.m = 0.001 18 × 75 × 36 × 6 = 115 N ∴ Resulting load acting on the gear, 22++× φ WR = ()WWNG () 2 WW NG cos
22++××1 °= = (1643) (115) 2 1643 115 cos14 /2 1755 N Assuming that the gear is overhung on the shaft and taking the overhang as 100 mm, therefore bending moment on the shaft due to the resultant load,
M = WR × 100 = 1755 × 100 = 175 500 N-mm and twisting moment on the shaft, D 450 T = W ×=G 1590 ×= 357 750 N-mm T 22 ∴ Equivalent twisting moment, 22+= 2 + 2 = × 3 Te = MT (175 500) (357 750) 398 10 N-mm
Let dG = Diameter of the gear shaft.
We know that equivalent twisting moment (Te), ππ333 398 × 103 = ×τ(ddd ) = × 40 ( ) = 7.855 ( ) 16GGG 16 ∴ 3 3 3 (dG) = 398 × 10 / 7.855 = 50.7 × 10 or dG = 37 say 40 mm Ans. 1062 A Textbook of Machine Design
We know that diameter of the gear hub
= 1.8 dG = 1.8 × 40 = 72 mm Ans. and length of the hub = 1.25 dG = 1.25 × 40 = 50 mm Ans. Design for the gear arms Since the pitch circle diameter of the gear is 450 mm, therefore the gear should be provided with four arms. Let us assume the cross-section of the arms as elliptical with major axis (a1) equal to twice the minor axis (b1). ∴ Section modulus of arms, ππ()ab22 () a a Z = 11=×= 1 10.05 (a )3 ... ( b = a /2) 32 32 2 1 Q 1 1 Since the arms are designed for the stalling load and stalling load is taken as the design tangential load divided by the velocity factor, therefore stalling load, W 1590 T ==7830 N WS = ... (Q Cv = 0.203) Cv 0.203 ∴ Maximum bending moment on each arm, WD 7830 450 M = SG×= × =440 440 N-mm n 242 σ We know that bending stress ( b), M 440 440 9× 106 == σ 2 42 = 33 ... (Taking b = 42 N/mm ) Z 0.05()aa11 () ∴ 3 6 6 (a1) = 9 × 10 / 42 = 0.214 × 10 or a1 = 60 mm Ans.
and b1 = a1 / 2 = 60 / 2 = 30 mm Ans. These dimensions refer to the hub end. Since the arms are tapered towards the rim and the taper is 1 / 16 per unit length of the arm (or radius of the gear), therefore Major axis of the arm at the rim end, 1 D a = aa–Taper= – × G 2 1116 2 1450 = 60 –×= 46 mm Ans. 16 2 and minor axis of the arm at the rim end, Major axis 46 b2= ==23 mm Ans. 22 Design for the rim
The thickness of the rim for the pinion (tRP) may be taken as 1.6 m to 1.9 m, where m is the module. Let us take thickness of the rim for the pinion,
tRP = 1.6 m = 1.6 × 6 = 9.6 say 10 mm Ans. The thickness of the rim for the gear (tRG) may be obtained by using the relation, T 45 t = m G ==620mmAns. RG n 4 EXERCISES
1. Calculate the power that can be transmitted safely by a pair of spur gears with the data given below. Calculate also the bending stresses induced in the two wheels when the pair transmits this power. Number of teeth in the pinion = 20 Number of teeth in the gear = 80 Spur Gears 1063
Module = 4 mm Width of teeth = 60 mm Tooth profile = 20° involute Allowable bending strength of the material = 200 MPa, for pinion = 160 MPa, for gear Speed of the pinion = 400 r.p.m. Service factor = 0.8 0.912 Lewis form factor = 0.154 – T 3 Velocity factor = [Ans. 13.978 kW ; 102.4 MPa ; 77.34 MPa] 3 + v 1 2. A spur gear made of bronze drives a mid steel pinion with angular velocity ratio of 3/2 : 1. The 1 ° pressure angle is 14 /2 . It transmits 5 kW at 1800 r.p.m. of pinion. Considering only strength, design the smallest diameter gears and find also necessary face width. The number of teeth should not be less than 15 teeth on either gear. The elastic strength of bronze may be taken as 84 MPa and of steel as 105 1 ° MPa. Lewis factor for 14 /2 pressure angle may be taken as 0.684 0.124 – y = No. of teeth
[Ans. m = 3 mm ; b = 35 mm ; DP = 48 mm ; DG = 168 mm] 3. A pair of 20° full-depth involute tooth spur gears is to transmit 30 kW at a speed of 250 r.p.m. of the σ pinion. The velocity ratio is 1 : 4. The pinion is made of cast steel having an allowable static stress, o σ = 100 MPa, while the gear is made of cast iron having allowable static stress, o = 55 MPa. The pinion has 20 teeth and its face width is 12.5 times the module. Determine the module, face width and pitch diameters of both the pinion and gear from the standpoint of strength only taking velocity factor into consideration. The tooth form factor is given by the expression 0.912 0.154 – y = No. of teeth and velocity factor is given by 3 C = , where v is the peripheral speed of the gear in m/s. v 3 + v
[Ans. m = 20 mm ; b = 250 mm ; DP = 400 mm ; DG = 1600 mm] 4. A micarta pinion rotating at 1200 r.p.m. is to transmit 1 kW to a cast iron gear at a speed of 192 r.p.m. Assuming a starting overload of 20% and using 20° full depth involute teeth, determine the module, number of teeth on the pinion and gear and face width. Take allowable static strength for micarta as 40 MPa and for cast iron as 53 MPa. Check the pair in wear. 5. A 15 kW and 1200 r.p.m. motor drives a compressor at 300 r.p.m. through a pair of spur gears having 20° stub teeth. The centre to centre distance between the shafts is 400 mm. The motor pinion is made of forged steel having an allowable static stress as 210 MPa, while the gear is made of cast steel having allowable static stress as 140 MPa. Assuming that the drive operates 8 to 10 hours per day under light shock conditions, find from the standpoint of strength, 1. Module; 2. Face width and 3. Number of teeth and pitch circle diameter of each gear. Check the gears thus designed from the consideration of wear. The surface endurance limit may be
taken as 700 MPa.[Ans. m = 6 mm ; b = 60 mm ; TP = 24 ; TG = 96 ; DP = 144mm ; DG = 576 mm] 6. A two stage reduction drive is to be designed to transmit 2 kW; the input speed being 960 r.p.m. and overall reduction ratio being 9. The drive consists of straight tooth spur gears only, the shafts being spaced 200 mm apart, the input and output shafts being co-axial. 1064 A Textbook of Machine Design
(a) Draw a layout of a suitable system to meet the above specifications, indicating the speeds of all rotating components. (b) Calculate the module, pitch diameter, number of teeth, blank diameter and face width of the gears for medium heavy duty conditions, the gears being of medium grades of accuracy. (c) Draw to scale one of the gears and specify on the drawing the calculated dimensions and other data complete in every respect for manufacturing purposes. 7. A motor shaft rotating at 1440 r.p.m. has to transmit 15 kW to a low speed shaft rotating at 500 r.p.m. The teeth are 20° involute with 25 teeth on the pinion. Both the pinion and gear are made of cast iron with a maximum safe stress of 56 MPa. A safe stress of 35 MPa may be taken for the shaft on which the gear is mounted. Design and sketch the spur gear drive to suit the above conditions. The starting torque may be assumed as 1.25 times the running torque. 8. Design and draw a spur gear drive transmitting 30 kW at 400 r.p.m. to another shaft running approxi- mately at 100 r.p.m. The load is steady and continuous. The materials for the pinion and gear are cast steel and cast iron respectively. Take module as 10 mm. Also check the design for dynamic load and wear. 0.912 [Hint : Assume : σ = 140 MPa ; σ = 56 MPa ; T = 24 ; y = 0.154 – ; OP OG P No. of teeth
3 σ σ 2 2 Cv = ; e = 84 MPa ; e = 0.023 mm ; es = 630 MPa ; EP = 210 kN/mm ; EG = 100 kN/mm ] 3 + v 9. Design a spur gear drive required to transmit 45 kW at a pinion speed of 800 r.p.m. The velocity ratio is 3.5 : 1. The teeth are 20° full-depth involute with 18 teeth on the pinion. Both the pinion and gear are made of steel with a maximum safe static stress of 180 MPa. Assume a safe stress of 40 MPa for the material of the shaft and key. 10. Design a pair of spur gears with stub teeth to transmit 55 kW from a 175 mm pinion running at 2500 r.p.m. to a gear running at 1500 r.p.m. Both the gears are made of steel having B.H.N. 260. Approximate the pitch by means of Lewis equation and then adjust the dimensions to keep within the limits set by the dynamic load and wear equation. QUESTIONS 1. Write a short note on gear drives giving their merits and demerits. 2. How are the gears classified and what are the various terms used in spur gear terminology ? 3. Mention four important types of gears and discuss their applications, the materials used for them and their construction. 4. What condition must be satisfied in order that a pair of spur gears may have a constant velcoity ratio? 5. State the two most important reasons for adopting involute curves for a gear tooth profile. 6. Explain the phenomenon of interference in involute gears. What are the conditions to be satisfied in order to avoid interference ? 7. Explain the different causes of gear tooth failures and suggest possible remedies to avoid such fail- ures. 8. Write the expressions for static, limiting wear load and dynamic load for spur gears and explain the various terms used there in. 9. Discuss the design procedure of spur gears. 10. How the shaft and arms for spur gears are designed ? OBJECTIVE TYPE QUESTIONS
1. The gears are termed as medium velocity gears, if their peripheral velocity is (a) 1–3 m / s (b) 3–15 m / s (c) 15–30 m / s (d) 30–50 m / s Spur Gears 1065
2. The size of gear is usually specified by (a) pressure angle (b) pitch circle diameter (c) circular pitch (d) diametral pitch 3. A spur gear with pitch circle diameter D has number of teeth T. The module m is defined as (a) m = d / T (b) m = T / D (c) m = π D / T (d) m = D.T 4. In a rack and pinion arrangement, the rack has teeth of ...... shape. (a) square (b) trepazoidal 5. The radial distance from the ...... to the clearance circle is called working depth. (a) addendum circle (b) dedendum circle 6. The product of the diametral pitch and circular pitch is equal to (a)1 (b)1/π (c) π (d) π × No. of teeth 7. The backlash for spur gears depends upon (a) module (b) pitch line velocity (c) tooth profile (d) both (a) and (b) 8. The contact ratio for gears is (a) zero (b) less than one (c) greater than one (d) none of these 9. If the centre distance of the mating gears having involute teeth is increased, then the pressure angle (a) increases (b) decreases (c) remains unchanged (d) none of these 10. The form factor of a spur gear tooth depends upon (a) circular pitch only (b) pressure angle only (c) number of teeth and circular pitch (d) number of teeth and the system of teeth 11. Lewis equation in spur gears is used to find the (a) tensile stress in bending (b) shear stress (c) compressive stress in bending (d) fatigue stress 12. The minimum number of teeth on the pinion in order to avoid interference for 20° stub system is (a)12 (b)14 (c)18 (d)32 13. The allowable static stress for steel gears is approximately ...... of the ultimate tensile stress. (a) one-fourth (b) one-third (c) one-half (d) double 14. Lewis equation in spur gears is applied (a) only to the pinion (b) only to the gear (c) to stronger of the pinion or gear (d) to weaker of the pinion or gear 15. The static tooth load should be ...... the dynamic load. (a) less than (b) greater than (c) equal to ANSWERS 1. (b) 2. (b) 3. (a) 4. (b) 5. (a) 6. (c) 7. (d) 8. (c) 9. (a) 10. (d) 11. (c) 12. (b) 13. (b) 14. (d) 15. (b) 1066 A Textbook of Machine Design
C H A P T E R 29 Helical Gears
1. Introduction. 2. Terms used in Helical Gears. 3. Face Width of Helical Gears. 4. Formative or Equivalent Number of Teeth for Helical Gears. 5. Proportions for Helical Gears. 6. Strength of Helical Gears.
29.1 Introduction A helical gear has teeth in form of helix around the gear. Two such gears may be used to connect two parallel shafts in place of spur gears. The helixes may be right handed on one gear and left handed on the other. The pitch surfaces are cylindrical as in spur gearing, but the teeth instead of being parallel to the axis, wind around the cylinders helically like screw threads. The teeth of helical gears with parallel axis have line contact, as in spur gearing. This provides gradual engagement and continuous contact of the engaging teeth. Hence helical gears give smooth drive with a high efficiency of transmission. We have already discussed in Art. 28.4 that the helical gears may be of single helical type or double helical type. In case of single helical gears there is some axial thrust between the teeth, which is a disadvantage. In order to eliminate this axial thrust, double helical gears (i.e.
1066 Helical Gears 1067 herringbone gears) are used. It is equivalent to two single helical gears, in which equal and opposite thrusts are provided on each gear and the resulting axial thrust is zero. 29.2 Terms used in Helical Gears The following terms in connection with helical gears, as shown in Fig. 29.1, are important from the subject point of view. 1. Helix angle. It is a constant angle made by the helices with the axis of rotation. 2. Axial pitch. It is the distance, parallel to the axis, between similar faces of adjacent teeth. It is the same as circular pitch and is therefore denoted by pc. The axial pitch may also be defined as the circular pitch in the plane of rotation or the diametral plane. Fig. 29.1. Helical gear 3. Normal pitch. It is the distance between similar faces of adjacent (nomenclature). teeth along a helix on the pitch cylinders normal to the teeth. It is denoted by pN. The normal pitch may also be defined as the circular pitch in the normal plane which is a plane perpendicular to the teeth. Mathematically, normal pitch, α pN = pc cos Note : If the gears are cut by standard hobs, then the pitch (or module) and the pressure angle of the hob will apply in the normal plane. On the other hand, if the gears are cut by the Fellows gear-shaper method, the pitch and pressure angle of the cutter will apply to the plane of rotation. The relation between the normal pressure φ φ angle ( N) in the normal plane and the pressure angle ( ) in the diametral plane (or plane of rotation) is given by φ φ α tan N = tan × cos 29.3 Face Width of Helical Gears In order to have more than one pair of teeth in contact, the tooth displacement (i.e. the ad- vancement of one end of tooth over the other end) or overlap should be atleast equal to the axial pitch, such that α Overlap = pc = b tan ...(i) The normal tooth load (WN) has two components ; one is tangential component (WT) and the other axial component (WA), as shown in Fig. 29.2. The axial or end thrust is given by α α WA = WN sin = WT tan ...(ii) From equation (i), we see that as the helix angle increases, then the tooth overlap increases. But at the same time, the end thrust as given by equation (ii), also increases, which is undesirable. It is usually recom- mended that the overlap should be 15 percent of the circular pitch. ∴ α Overlap = b tan = 1.15 pc 1.15 p 1.15 ×πm or b = c = ... ( p = π m) tanαα tan Q c where b = Minimum face width, and m = Module. Notes : 1. The maximum face width may be taken as 12.5 m to 20 m, where m is the module. In terms of pinion diameter (D ), the face width should be 1.5 D to P P Fig. 29.2. Face width of 2 D , although 2.5 D may be used. P P helical gear. 2. In case of double helical or herringbone gears, the minimum face width is given by 2.3pm 2.3 ×π b = c = tanαα tan The maximum face width ranges from 20 m to 30 m. 1068 A Textbook of Machine Design
3. In single helical gears, the helix angle ranges from 20° to 35°, while for double helical gears, it may be made upto 45°. 29.4 Formative or Equivalent Number of Teeth for Helical Gears The formative or equivalent number of teeth for a helical gear may be defined as the number of teeth that can be generated on the surface of a cylinder having a radius equal to the radius of curvature at a point at the tip of the minor axis of an ellipse obtained by taking a section of the gear in the normal plane. Mathematically, formative or equivalent number of teeth on a helical gear, 3 α TE = T / cos where T = Actual number of teeth on a helical gear, and α = Helix angle. 29.5 Proportions for Helical Gears Though the proportions for helical gears are not standardised, yet the following are recommended by American Gear Manufacturer's Association (AGMA). Pressure angle in the plane of rotation, φ = 15° to 25° Helix angle, α = 20° to 45° Addendum = 0.8 m (Maximum) Dedendum = 1 m (Minimum) Minimum total depth = 1.8 m Minimum clearance = 0.2 m Thickness of tooth = 1.5708 m
In helical gears, the teeth are inclined to the axis of the gear. Helical Gears 1069 29.6 Strength of Helical Gears In helical gears, the contact between mating teeth is gradual, starting at one end and moving along the teeth so that at any instant the line of contact runs diagonally across the teeth. Therefore in order to find the strength of helical gears, a modified Lewis equation is used. It is given by σ π WT =( o × Cv) b. m.y' where WT = Tangential tooth load, σ o = Allowable static stress,
Cv = Velocity factor, b = Face width, m = Module, and y' = Tooth form factor or Lewis factor corresponding to the formative or virtual or equivalent number of teeth.
Notes : 1. The value of velocity factor (Cv) may be taken as follows : 6 C = , for peripheral velocities from 5 m / s to 10 m / s. v 6 + v 15 = , for peripheral velocities from 10 m / s to 20 m / s. 15 + v 0.75 = , for peripheral velocities greater than 20 m / s. 0.75 + v 0.75 = + 0.25, for non-metallic gears. 1 + v 2. The dynamic tooth load on the helical gears is given by 21vbC ( . cos2 α+ W ) cos α W = W + T D T +α+2 21vbC . cos WT where v, b and C have usual meanings as discussed in spur gears. 3. The static tooth load or endurance strength of the tooth is given by σ π WS = e.b. m.y' 4. The maximum or limiting wear tooth load for helical gears is given by
DbQKP .. . W = w cos2 α where DP, b, Q and K have usual meanings as discussed in spur gears. σφ2 ()sin1es N ⎡⎤+ 1 In this case, K = ⎢⎥ 1.4 ⎣⎦EEPG φ where N = Normal pressure angle. Example 29.1. A pair of helical gears are to transmit 15 kW. The teeth are 20° stub in diametral plane and have a helix angle of 45°. The pinion runs at 10 000 r.p.m. and has 80 mm pitch diameter. The gear has 320 mm pitch diameter. If the gears are made of cast steel having allowable static strength of 100 MPa; determine a suitable module and face width from static strength considerations σ and check the gears for wear, given es = 618 MPa. 3 φ α Solution. Given : P = 15 kW = 15 × 10 W; = 20° ; = 45° ; NP = 10 000 r.p.m. ; DP = 80 mm σ σ 2 σ 2 = 0.08 m ; DG = 320 mm = 0.32 m ; OP = OG = 100 MPa = 100 N/mm ; es = 618 MPa = 618 N/mm Module and face width Let m = Module in mm, and b = Face width in mm. 1070 A Textbook of Machine Design
Since both the pinion and gear are made of the same material (i.e. cast steel), therefore the pinion is weaker. Thus the design will be based upon the pinion. We know that the torque transmitted by the pinion, P ×××60 15 103 60 ==14.32 N-m T = ππ× 2NP 2 10 000 ∴ *Tangential tooth load on the pinion, T 14.32 ==358 N WT = DP / 2 0.08/ 2 We know that number of teeth on the pinion,
TP = DP / m = 80 / m and formative or equivalent number of teeth for the pinion,
TP 80 /mm 80 / 226.4 T = === E cos33α° cos 45 (0.707) 3m ∴ Tooth form factor for the pinion for 20° stub teeth, −=−0.841 0.841 =− y'P = 0.175 0.175 0.175 0.0037 m TmE 226.4 / We know that peripheral velocity, π DN. π×0.08 × 10 000 v = PP==42 m / s 60 60 ∴ Velocity factor, 0.75 0.75 C = ==0.104 ...( v is greater than 20 m/s) v 0.75++v 0.75 42 Q Since the maximum face width (b) for helical gears may be taken as 12.5 m to 20 m, where m is the module, therefore let us take b = 12.5 m
We know that the tangential tooth load (WT), σ π 358 = ( OP . Cv) b. m.y'P = (100 × 0.104) 12.5 m × π m (0.175 – 0.0037 m) = 409 m2 (0.175 – 0.0037 m) = 72 m2 – 1.5 m3 Solving this expression by hit and trial method, we find that m = 2.3 say 2.5 mm Ans. and face width, b = 12.5 m = 12.5 × 2.5 = 31.25 say 32 mm Ans. Checking the gears for wear We know that velocity ratio, D 320 V. R .= G ==4 DP 80 ∴ Ratio factor, 2..24××VR Q = ==1.6 VR..++ 1 4 1 φ φ α We know that tan N = tan cos = tan 20° × cos 45° = 0.2573 ∴φ N = 14.4°
* The tangential tooth load on the pinion may also be obtained by using the relation, PDNπ . W = ,wherev = PP (inm/s) T v 60 Helical Gears 1071
The picture shows double helical gears which are also called herringbone gears. Since both the gears are made of the same material (i.e. cast steel), therefore let us take 2 3 2 EP = EG = 200 kN/mm = 200 × 10 N/mm ∴ Load stress factor, σφ2 ()sines N ⎛⎞11+ K = ⎜⎟ 1.4 ⎝⎠EEPG (618)2 sin 14.4° ⎛⎞ 1 1 +=0.678 N/mm2 = ⎜⎟33 1.4 ⎝⎠200×× 10 200 10 We know that the maximum or limiting load for wear, DbQK.. . 80×× 32 1.6 × 0.678 W = P ==5554 N w cos22α° cos 45 Since the maximum load for wear is much more than the tangential load on the tooth, therefore the design is satisfactory from consideration of wear. Example 29.2. A helical cast steel gear with 30° helix angle has to transmit 35 kW at 1500 r.p.m. If the gear has 24 teeth, determine the necessary module, pitch diameter and face width for 20° full depth teeth. The static stress for cast steel may be taken as 56 MPa. The width of face may be taken as 3 times the normal pitch. What would be the end thrust on the gear? The tooth factor for 20° − 0. 912 full depth involute gear may be taken as 0., 154 where TE represents the equivalent number TE of teeth. α 3 φ Solution. Given : = 30° ; P = 35 kW = 35 × 10 W; N = 1500 r.p.m. ; TG = 24 ; = 20° ; σ 2 o = 56 MPa = 56 N/mm ; b = 3 × Normal pitch = 3 pN Module Let m = Module in mm, and
DG = Pitch circle diameter of the gear in mm. We know that torque transmitted by the gear, P ×××60 35 103 60 T = ===×223 N-m 223 103 N-mm 2ππ×N 2 1500 1072 A Textbook of Machine Design
Formative or equivalent number of teeth,
TG 24 24 T' = ===37 E cos33α° cos 30 (0.866) 3 0.912 0.912 ∴ Tooth factor, y' = 0.154−=−= 0.154 0.129 TE 37 We know that the tangential tooth load, TTT22 W = == ... ( D = m.T ) T × Q G G DDmTGG/2 G 2×× 223 103 18 600 = = N mm× 24 and peripheral velocity, ππDN.... mTN v = GG= mm / s ...(D and m are in mm) 60 60 G π×m ×24 × 1500 = ==1885mm mm / s 1.885 m / s 60 Let us take velocity factor, 15 15 C = = v 15++vm 15 1.885 We know that tangential tooth load, σ π σ π ...( b = 3 p ) WT =( o × Cv) b. m.y' = ( o × Cv) 3pN × m × y' Q N σ α π α =( o × Cv) 3 × pc cos × m × y' ...(Q pN = pc cos ) σ π α π π =( o × Cv) 3 m cos × m × y' ...(Q pc = m) 18 600⎛⎞ 15 ∴ =π×°×π×56 3mm cos 30 0.129 mm⎝⎠⎜⎟15+ 1.885 2780 m2 = 15+ 1.885 m or 279 000 + 35 061 m = 2780 m3 Solving this equation by hit and trial method, we find that m = 5.5 say 6 mm Ans. Pitch diameter of the gear We know that the pitch diameter of the gear,
DG = m × TG = 6 × 24 = 144 mm Ans. Face width It is given that the face width, α π α b =3 pN = 3 pc cos = 3 × m cos = 3 × π × 6 cos 30° = 48.98 say 50 mm Ans. End thrust on the gear We know that end thrust or axial load on the gear, 18 600 18 600 W = W tanα= × tan 30 ° = × 0.577 =1790 N Ans. A T m 6 Helical Gears 1073
Example 29.3. Design a pair of helical gears for transmitting 22 kW. The speed of the driver gear is 1800 r.p.m. and that of driven gear is 600 r.p.m. The helix angle is 30° and profile is corresponding to 20° full depth system. The driver gear has 24 teeth. Both the gears are made of cast steel with allowable static stress as 50 MPa. Assume the face width parallel to axis as 4 times the circular pitch and the overhang for each gear as 150 mm. The allowable shear stress for the shaft material may be taken as 50 MPa. The form factor may be taken as 0.154 – 0.912 / TE, where TE is 350 , the equivalent number of teeth. The velocity factor may be taken as 350+ v where v is pitch line velocity in m / min. The gears are required to be designed only against bending failure of the teeth under dynamic condition. 3 α Solution. Given : P = 22 kW = 22 × 10 W; NP = 1800 r.p.m.; NG = 600 r.p.m. ; = 30° ; φ σ 2 τ 2 = 20° ; TP = 24 ; o = 50 MPa = 50 N/mm ; b = 4 pc ; Overhang = 150 mm ; = 50 MPa = 50 N/mm
Gears inside a car Design for the pinion and gear We know that the torque transmitted by the pinion, P ×××60 22 103 60 ===116.7 N-m 116 700 N-mm T = ππ× 2NP 2 1800 Since both the pinion and gear are made of the same material (i.e. cast steel), therefore the pinion is weaker. Thus the design will be based upon the pinion. We know that formative or equivalent number of teeth, T 24 24 T = P ===37 E cos33α° cos 30 (0.866) 3 0.912 0.912 ∴ Form factor, y' = 0.154−=−= 0.154 0.129 TE 37 1074 A Textbook of Machine Design
First of all let us find the module of teeth. Let m = Module in mm, and
DP = Pitch circle diameter of the pinion in mm. We know that the tangential tooth load on the pinion, TTT22 W = == ...( D = m.T ) T × Q P P DDmTPP/2 P 2× 116 700 9725 = N = mm× 24 π π and peripheral velocity, v = DP.NP = m.TP.NP = π m × 24 × 1800 = 135 735 m mm / min = 135.735 m m / min 350 350 ∴ Velocity factor, C = = v 350++vm 350 135.735 We also know that the tangential tooth load on the pinion, σ π σ π ... ( b = 4 p ) WT =( o.Cv) b. m.y' = ( o.Cv) 4 pc × m × y' Q c σ π π ' ... ( p = π m) =( o.Cv) 4 × m × m × y Q c 2 ∴ 9725=×π×=⎛⎞ 35022 89 126 m 50⎜⎟ 4m 0.129 mm⎝⎠350++ 135.735 350 135.735 m 3.4 × 106 + 1.32 × 106 m = 89 126 m3 Solving this expression by hit and trial method, we find that m = 4.75 mm say 6 mm Ans.
Helical gears. We know that face width, π π b =4 pc = 4 m = 4 × 6 = 75.4 say 76 mm Ans. and pitch circle diameter of the pinion,
DP = m × TP = 6 × 24 = 144 mm Ans. Since the velocity ratio is 1800 / 600 = 3, therefore number of teeth on the gear,
TG =3 TP = 3 × 24 = 72 and pitch circle diameter of the gear,
DG = m × TG = 6 × 72 = 432 mm Ans. Helical Gears 1075
Design for the pinion shaft
Let dP = Diameter of the pinion shaft. We know that the tangential load on the pinion, 9725 9725 W= ==1621 N T m 6 and the axial load of the pinion, α WA = WT tan = 1621 tan 30° = 1621 × 0.577 = 935 N Since the overhang for each gear is 150 mm, therefore bending moment on the pinion shaft due to the tangential load,
M1 = WT × Overhang = 1621 × 150 = 243 150 N-mm and bending moment on the pinion shaft due to the axial load, D 144 M = W ×=×P 935 = 67320 N-mm 2 A 22 Since the bending moment due to the tangential load (i.e. M1) and bending moment due to the axial load (i.e. M2) are at right angles, therefore resultant bending moment on the pinion shaft, 22+= 2 + 2 = M = (MM12 ) ( ) (243 150) (67 320) 252293 N-mm The pinion shaft is also subjected to a torque T = 116 700 N-mm, therefore equivalent twisting moment, 22+= 2 + 2 = Te = MT (252293) (116700) 277 975 N-mm
We know that equivalent twisting moment (Te), ππ 277 975 = ×τ()ddd333 = × 50()9.82() = 16PPP 16 ∴ 3 (dP) = 277 975 / 9.82 = 28 307 or dP = 30.5 say 35 mm Ans. Let us now check for the principal shear stress. We know that the shear stress induced, 16 T 16× 277975 τ = e ===33 N/mm2 33 MPa ππ33 ()dP (35) and direct stress due to axial load,
WA2==935 = σ = ππ0.97 N/mm 0.97 MPa ()d 22 (35) 44P
Helical gears 1076 A Textbook of Machine Design
∴ Principal shear stress, 11 = ⎡⎤⎡σ+224 τ = (0.97) 2 + 4(33) 2 ⎤ = 33 MPa 22⎣⎦⎣ ⎦ Since the principal shear stress is less than the permissible shear stress of 50 MPa, therefore the design is satisfactory. We know that the diameter of the pinion hub
= 1.8 dP = 1.8 × 35 = 63 mm Ans. and length of the hub = 1.25 dP = 1.25 × 35 = 43.75 say 44 mm Since the length of the hub should not be less than the face width, therefore let us take length of the hub as 76 mm. Ans. Note : Since the pitch circle diameter of the pinion is 144 mm, therefore the pinion should be provided with a web. Let us take the thickness of the web as 1.8 m, where m is the module. ∴ Thickness of the web = 1.8 m = 1.8 × 6 = 10.8 say 12 mm Ans. Design for the gear shaft
Let dG = Diameter of the gear shaft. We have already calculated that the tangential load,
WT = 1621 N and the axial load, WA = 935 N ∴ Bending moment due to the tangential load,
M1 = WT × Overhang = 1621 × 150 = 243 150 N-mm and bending moment due to the axial load, D 432 M = W ×=×=G 935 201 960 N-mm 2 A 22 ∴ Resultant bending moment on the gear shaft, 22+= 2 + 2 = M = (MM12 ) ( ) (243 150) (201 960) 316 000 N-mm Since the velocity ratio is 3, therefore the gear shaft is subjected to a torque equal to 3 times the torque on the pinion shaft. ∴ Torque on the gear shaft, T = Torque on the pinion shaft × V. R. = 116 700 × 3 = 350 100 N-mm We know that equivalent twisting moment, 22+= 2 + 2 = Te = MT (316 000) (350 100) 472 000 N-mm
We also know that equivalent twisting moment (Te),
ππ333 472 000 = ×τ×(ddd ) = × 50 ( ) = 9.82 ( ) 16GGG 16 ∴ 3 (dG) = 472 000 / 9.82 = 48 065 or dG = 36.3 say 40 mm Ans. Let us now check for the principal shear stress. We know that the shear stress induced, 16 T 16× 472000 τ = e ===37.6 N/mm2 37.6 MPa ππ33 ()dG (40) Helical Gears 1077 and direct stress due to axial load, W 935 σ A2== = = ππ0.744 N/mm 0.744 MPa 22 (dG ) (40) ∴ Principal shear stress 44 11⎡⎤⎡22 2 2 ⎤ = σ+4 τ = (0.744) + 4 (37.6) = 37.6 MPa 22⎣⎦⎣ ⎦ Since the principal shear stress is less than the permissible shear stress of 50 MPa, therefore the design is satisfactory. We know that the diameter of the gear hub
= 1.8 dG = 1.8 × 40 = 72 mm Ans. and length of the hub = 1.25 dG = 1.25 × 40 = 50 mm We shall take the length of the hub equal to the face width, i.e. 76 mm. Ans. Since the pitch circle diameter of the gear is 432 mm, therefore the gear should be provided with four arms. The arms are designed in the similar way as discussed for spur gears. Design for the gear arms
Let us assume that the cross-section of the arms is elliptical with major axis (a1) equal to twice the minor axis (b1). These dimensions refer to hub end. ∴ Section modulus of arms, ππ23 ba11()== () a 1 3 ⎛⎞= a1 Z = 0.05 (a1 ) ⎜⎟Qb1 32 64 ⎝⎠2 Since the arms are designed for the stalling load and it is taken as the design tangential load divided by the velocity factor, therefore + WT = ⎛⎞350 135.735m Stalling load, WS = 1621 ⎜⎟ Cv ⎝⎠350 ⎛⎞350+× 135.735 6 = 1621 ⎜⎟= 5393 N ⎝⎠350 ∴ Maximum bending moment on each arm, WD 5393 432 M = SG×= × =291 222 N-mm n 242 σ We know that bending stress ( b), M 291 222 5824× 103 == σ 2 42 = 33 ... (Taking b = 42 N/mm ) Z 0.05 (aa11 ) ( ) ∴ 3 3 3 (a1) = 5824 × 10 / 42 = 138.7 × 10 or a1 = 51.7 say 54 mm Ans. and b1 = a1 / 2 = 54 / 2 = 27 mm Ans. Since the arms are tapered towards the rim and the taper is 1/16 mm per mm length of the arm (or radius of the gear), therefore Major axis of the arm at the rim end, 1 D a = a – Taper = a −×G 2 1 1 16 2 1 432 = 54−× = 40 mm Ans. 16 2 and minor axis of the arm at the rim end,
b2 = a2 / 2 = 40 / 2 = 20 mm Ans. 1078 A Textbook of Machine Design
Design for the rim The thickness of the rim for the pinion may be taken as 1.6 m to 1.9 m, where m is the module. Let us take thickness of the rim for pinion,
tRP = 1.6 m = 1.6 × 6 = 9.6 say 10 mm Ans.
The thickness of the rim for the gear (tRG) is given by T 72 t = m G ==625.4say26mm Ans. RG n 4 EXERCISES
1. A helical cast steel gear with 30° helix angle has to transmit 35 kW at 2000 r.p.m. If the gear has 25 teeth, find the necessary module, pitch diameters and face width for 20° full depth involute teeth. The static stress for cast steel may be taken as 100 MPa. The face width may be taken as 3 times the normal pitch. The tooth form factor is given by the expression
y' = 0.154 – 0.912/TE , where TE represents the equivalent num- 6 ber of teeth. The velocity factor is given by C = , where v 6 + v v is the peripheral speed of the gear in m/s. [Ans. 6 mm ; 150 mm ; 50 mm] 2. A pair of helical gears with 30° helix angle is used to transmit 15 kW at 10 000 r.p.m. of the pinion. The velocity ratio is 4 : 1. Both the gears are to be made of hardened steel of static strength 100 N/mm2. The gears are 20° stub and the pinion is to have 24 teeth. The face width may be taken as 14 times the module. Find the module and face width from the standpoint of strength and check the gears for wear. [Ans. 2 mm ; 28 mm]
Gears inside a car engine. Helical Gears 1079
3. A pair of helical gears consist of a 20 teeth pinion meshing with a 100 teeth gear. The pinion rotates at 720 r.p.m. The normal pressure angle is 20° while the helix angle is 25°. The face width is 40 mm and the normal module is 4 mm. The pinion as well as gear are made of steel having ultimate strength of 600 MPa and heat treated to a surface hardness of 300 B.H.N. The service factor and factor of safety are 1.5 and 2 respectively. Assume that the velocity factor accounts for the dynamic load and calculate the power transmitting capacity of the gears. [Ans. 8.6 kW] 4. A single stage helical gear reducer is to receive power from a 1440 r.p.m., 25 kW induction motor. The gear tooth profile is involute full depth with 20° normal pressure angle. The helix angle is 23°, number of teeth on pinion is 20 and the gear ratio is 3. Both the gears are made of steel with allowable beam stress of 90 MPa and hardness 250 B.H.N. (a) Design the gears for 20% overload carrying capacity from standpoint of bending strength and wear. (b) If the incremental dynamic load of 8 kN is estimated in tangential plane, what will be the safe power transmitted by the pair at the same speed? QUESTIONS 1. What is a herringbone gear? Where they are used? 2. Explain the following terms used in helical gears : (a) Helix angle; (b) normal pitch; and (c) axial pitch. 3. Define formative or virtual number of teeth on a helical gear. Derive the expression used to obtain its value. 4. Write the expressions for static strength, limiting wear load and dynamic load for helical gears and explain the various terms used therein. OBJECTIVE TYPE QUESTIONS
1. If T is the actual number of teeth on a helical gear and φ is the helix angle for the teeth, the formative number of teeth is written as (a) T sec3 φ (b) T sec2 φ (c) T/sec3φ (d) T cosec φ 2. In helical gears, the distance between similar faces of adjacent teeth along a helix on the pitch cylin- ders normal to the teeth, is called (a) normal pitch (b) axial pitch (c) diametral pitch (d) module 3. In helical gears, the right hand helices on one gear will mesh ...... helices on the other gear. (a) right hand (b) left hand 4. The helix angle for single helical gears ranges from (a) 10° to 15° (b) 15° to 20° (c) 20° to 35° (d) 35° to 50° 5. The helix angle for double helical gears may be made up to (a) 45° (b)60° (c) 75° (d)90°
ANSWERS 1. (a) 2. (a) 3. (b) 4. (c) 5. (a) 1080 A Textbook of Machine Design
C H A P T E R 30 Bevel Gears
1. Introduction. 2. Classification of Bevel Gears. 3. Terms used in Bevel Gears. 4. Determination of Pitch Angle for Bevel Gears. 5. Proportions for Bevel Gears. 6. Formative or Equivalent Number of Teeth for Bevel Gears—Tredgold's Approximation. 7. Strength of Bevel Gears. 8. Forces Acting on a Bevel Gear. 9. Design of a Shaft for Bevel Gears. 30.1 Introduction The bevel gears are used for transmitting power at a constant velocity ratio between two shafts whose axes intersect at a certain angle. The pitch surfaces for the bevel gear are frustums of cones. The two pairs of cones in contact is shown in Fig. 30.1. The elements of the cones, as shown in Fig. 30.1 (a), intersect at the point of intersection of the axis of rotation. Since the radii of both the gears are proportional to their distances from the apex, therefore the cones may roll together without sliding. In Fig. 30.1 (b), the elements of both cones do not intersect at the point of shaft intersection. Consequently, there may be pure rolling at only one point of contact and there must be tangential sliding at all other points of contact. Therefore, these cones, cannot be used as pitch surfaces because it is impossible to have positive driving and sliding in the same direction at the same time. We, thus, conclude that the elements of bevel
1080 Bevel Gears 1081 gear pitch cones and shaft axes must intersect at the same point.
Fig. 30.1. Pitch surface for bevel gears.
The bevel gear is used to change the axis of rotational motion. By using gears of differing numbers of teeth, the speed of rotation can also be changed.
30.2 Classification of Bevel Gears The bevel gears may be classified into the following types, depending upon the angles between the shafts and the pitch surfaces. 1. Mitre gears. When equal bevel gears (having equal teeth and equal pitch angles) connect two shafts whose axes intersect at right angle, as shown in Fig. 30.2 (a), then they are known as mitre gears. 2. Angular bevel gears. When the bevel gears connect two shafts whose axes intersect at an angle other than a right angle, then they are known as angular bevel gears. 1082 A Textbook of Machine Design
3. Crown bevel gears. When the bevel gears connect two shafts whose axes intersect at an angle greater than a right angle and one of the bevel gears has a pitch angle of 90º, then it is known as a crown gear. The crown gear corresponds to a rack in spur gearing, as shown in Fig. 30.2 (b).
Fig. 30.2. Classification of bevel gears. 4. Internal bevel gears. When the teeth on the bevel gear are cut on the inside of the pitch cone, then they are known as internal bevel gears. Note : The bevel gears may have straight or spiral teeth. It may be assumed, unless otherwise stated, that the bevel gear has straight teeth and the axes of the shafts intersect at right angle. 30.3 Terms used in Bevel Gears
Fig. 30.3. Terms used in bevel gears.
A sectional view of two bevel gears in mesh is shown in Fig. 30.3. The following terms in connection with bevel gears are important from the subject point of view : Bevel Gears 1083
1. Pitch cone. It is a cone containing the pitch elements of the teeth. 2. Cone centre. It is the apex of the pitch cone. It may be defined as that point where the axes of two mating gears intersect each other. θ 3. Pitch angle. It is the angle made by the pitch line with the axis of the shaft. It is denoted by ‘ P’. 4. Cone distance. It is the length of the pitch cone element. It is also called as a pitch cone radius. It is denoted by ‘OP’. Mathematically, cone distance or pitch cone radius, Pitch radius ==DP /2 DG / 2 OP = θθ sin θPP1P2sin sin 5. Addendum angle. It is the angle subtended by the addendum of the tooth at the cone centre. It is denoted by ‘α’ Mathematically, addendum angle, ⎛⎞a α = tan–1 ⎜⎟ ⎝⎠OP where a = Addendum, and OP = Cone distance. 6. Dedendum angle. It is the angle subtended by the dedendum of the tooth at the cone centre. It is denoted by ‘β’. Mathematically, dedendum angle, ⎛⎞d β = tan–1 ⎜⎟ ⎝⎠OP where d = Dedendum, and OP = Cone distance. 7. Face angle. It is the angle subtended by the face of the tooth at the cone centre. It is denoted by ‘φ’. The face angle is equal to the pitch angle plus addendum angle. 8. Root angle. It is the angle subtended by the root of the tooth at the cone centre. It is denoted θ by ‘ R’. It is equal to the pitch angle minus dedendum angle. 9. Back (or normal) cone. It is an imaginary cone, perpendicular to the pitch cone at the end of the tooth.
10. Back cone distance. It is the length of the back cone. It is denoted by ‘RB’. It is also called back cone radius. 11. Backing. It is the distance of the pitch point (P) from the back of the boss, parallel to the pitch point of the gear. It is denoted by ‘B’. 12. Crown height. It is the distance of the crown point (C) from the cone centre (O), parallel to the axis of the gear. It is denoted by ‘HC’. 13. Mounting height. It is the distance of the back of the boss from the cone centre. It is denoted by ‘HM’. 14. Pitch diameter. It is the diameter of the largest pitch circle. 15. Outside or addendum cone diameter. It is the maximum diameter of the teeth of the gear. It is equal to the diameter of the blank from which the gear can be cut. Mathematically, outside diameter, θ DO = DP + 2 a cos P where DP = Pitch circle diameter, a = Addendum, and θ P = Pitch angle. 16. Inside or dedendum cone diameter. The inside or the dedendum cone diameter is given by θ Dd = DP – 2d cos P where Dd = Inside diameter, and d = Dedendum. 1084 A Textbook of Machine Design 30.4 Determination of Pitch Angle for Bevel Gears Consider a pair of bevel gears in mesh, as shown in Fig. 30.3. θ Let P1 = Pitch angle for the pinion, θ P2 = Pitch angle for the gear, θ S = Angle between the two shaft axes,
DP = Pitch diameter of the pinion,
DG = Pitch diameter of the gear, and DTGG N P V. R . = Velocity ratio = == Mitre gears DTNPP G From Fig. 30.3, we find that θ θ θ θ θ θ S = P1 + P2 or P2 = S – P1 ∴ θ θ θ θ θ θ θ sin P2 =sin ( S – P1) = sin S . cos P1 – cos S . sin P1 ...(i) We know that cone distance, sin θ D DP /2 = DG / 2 P2 = G OP = θθ or θ = V. R. sinP1 sin P2 sin P1D P ∴ θ θ sin P2 = V. R . × sin P1 ...(ii) From equations (i) and (ii), we have θ θ θ θ θ V. R. × sin P1 = sin S . cos P1 – cos S . sin P1 θ Dividing throughout by cos P1 we get θ θ θ θ V.R. tan P1 = sin S – cos S. tan P1 sin θ θ S or tan P1 = +θ V.R cos S sin θ ∴θ–1 ⎛⎞S P1 = tan ⎜⎟+θ ...(iii) ⎝⎠V.R cos S Similarly, we can find that sin θ tan θ = S P2 1 +θcos V.R S ⎛⎞ ⎜⎟θ sin S ∴θ= tan–1 ⎜⎟ ...(iv) P2 ⎜⎟1 +θ ⎜⎟cos S ⎝⎠V.R θ Note : When the angle between the shaft axes is 90º i.e. S = 90º, then equations (iii) and (iv) may be written as 1 D T θ –1 ⎛⎞–1 ⎛⎞P –1 ⎛⎞P –1 ⎛⎞NG P1 = tan ⎜⎟ = tan ⎜⎟ = tan ⎜⎟ = tan ⎜⎟ ⎝⎠V.R ⎝⎠DG ⎝⎠TG ⎝⎠NP
⎛⎞DG ⎛⎞TG ⎛⎞NP θ –1 –1 –1 –1 and P2 = tan (V. R .) = tan ⎜⎟ = tan ⎜⎟ = tan ⎜⎟ ⎝⎠DP ⎝⎠TP ⎝⎠NG
30.5 Proportions for Bevel Gear The proportions for the bevel gears may be taken as follows : 1. Addendum, a = 1 m Bevel Gears 1085
2. Dedendum, d = 1.2 m 3. Clearance = 0.2 m 4. Working depth = 2 m 5. Thickness of tooth = 1.5708 m where m is the module. Note : Since the bevel gears are not interchangeable, therefore these are designed in pairs. 30.6 Formative or Equivalent Number of Teeth for Bevel Gears – Tredgold’s Approximation We have already discussed that the involute teeth for a spur gear may be generated by the edge of a plane as it rolls on a base cylinder. A similar analysis for a bevel gear will show that a true section of the resulting involute lies on the surface of a sphere. But it is not possible to represent on a plane surface the exact profile of a bevel gear tooth lying on the surface of a sphere. Therefore, it is important to approximate the bevel gear tooth profiles as accurately as possible. The approximation (known as Tredgold’s approximation) is based upon the fact that a cone tangent to the sphere at the pitch point will closely approximate the surface of the sphere for a short distance either side of the pitch point, as shown in Fig. 30.4 (a). The cone (known as back cone) may be developed as a plane surface and spur gear teeth corresponding to the pitch and pressure angle of the bevel gear and the radius of the developed cone can be drawn. This procedure is shown in Fig. 30.4 (b).
Fig. 30.4 θ Let P = Pitch angle or half of the cone angle, R = Pitch circle radius of the bevel pinion or gear, and
RB = Back cone distance or equivalent pitch circle radius of spur pinion or gear. Now from Fig. 30.4 (b), we find that θ RB = R sec P We know that the equivalent (or formative) number of teeth, 2 R ⎛⎞Pitch circle diameter T = B ... ⎜⎟Q Number of teeth = E m ⎝⎠Module 2secR θ = P =θT sec m P where T = Actual number of teeth on the gear. 1086 A Textbook of Machine Design
Notes : 1. The action of bevel gears will be same as that of equivalent spur gears. 2. Since the equivalent number of teeth is always greater than the actual number of teeth, therefore a given pair of bevel gears will have a larger contact ratio. Thus, they will run more smoothly than a pair of spur gears with the same number of teeth. 30.7 Strength of Bevel Gears The strength of a bevel gear tooth is obtained in a similar way as discussed in the previous articles. The modified form of the Lewis equation for the tangential tooth load is given as follows: ⎛⎞Lb− W =(σ × C ) b.π m.y' ⎜⎟ T o v ⎝⎠L σ where o = Allowable static stress,
Cv = Velocity factor, 3 = , for teeth cut by form cutters, 3 + v 6 = , for teeth generated with precision machines, 6 + v v = Peripheral speed in m / s, b = Face width, m = Module, y' = Tooth form factor (or Lewis factor) for the equivalent number of teeth, L = Slant height of pitch cone (or cone distance),
2 2 ⎛⎞⎛⎞DG + DP = ⎜⎟⎜⎟ ⎝⎠22⎝⎠
Ring gear Input Pinion
Drive shaft
Hypoid bevel gears in a car differential Bevel Gears 1087
DG = Pitch diameter of the gear, and
DP = Pitch diameter of the pinion. ⎛⎞Lb− Notes : 1. The factor ⎝⎠⎜⎟L may be called as bevel factor. 2. For satisfactory operation of the bevel gears, the face width should be from 6.3 m to 9.5 m, where m is the module. Also the ratio L / b should not exceed 3. For this, the number of teeth in the pinion must not less than 48 , where V.R. is the required velocity ratio. 1(..)+ VR2
3. The dynamic load for bevel gears may be obtained in the similar manner as discussed for spur gears. 4. The static tooth load or endurance strength of the tooth for bevel gears is given by ⎛⎞Lb– W = σ .b.π m.y' ⎜⎟ S e ⎝⎠L σ The value of flexural endurance limit ( e) may be taken from Table 28.8, in spur gears. 5. The maximum or limiting load for wear for bevel gears is given by
DbQKP .. . Ww = θ cos P1 where DP, b, Q and K have usual meanings as discussed in spur gears except that Q is based on formative or equivalent number of teeth, such that
2 TEG Q = + TTEG EP
30.8 Forces Acting on a Bevel Gear
Consider a bevel gear and pinion in mesh as shown in Fig. 30.5. The normal force (WN) on the tooth is perpendicular to the tooth profile and thus makes an angle equal to the pressure angle (φ) to the pitch circle. Thus normal force can be resolved into two components, one is the tangential component
(WT) and the other is the radial component (WR). The tangential component (i.e. the tangential tooth load) produces the bearing reactions while the radial component produces end thrust in the shafts. The magnitude of the tangential and radial components is as follows : φ φ φ WT = WN cos , and WR = WN sin = WT tan ...(i) 1088 A Textbook of Machine Design
These forces are considered to act at the mean radius (Rm). From the geometry of the Fig. 30.5, we find that ⎛⎞D / 2 ⎛⎞−θ=−bb ⎛⎞DP sinθ=P Rm = ⎜⎟LLsin P1 ⎜⎟ ... ⎜⎟Q P1 L ⎝⎠222 ⎝⎠L ⎝⎠ Now the radial force (WR) acting at the mean radius may be further resolved into two components, WRH and WRV, in the axial and radial directions as shown in Fig. 30.5. Therefore the axial force acting on the pinion shaft, θ φ θ WRH = WR sin P1 = WT tan . sin P1 ...[From equation (i)] and the radial force acting on the pinion shaft, θ φ θ WRV = WR cos P1 = WT tan . cos P1
Fig. 30.5. Forces acting on a bevel gear. A little consideration will show that the axial force on the pinion shaft is equal to the radial force on the gear shaft but their directions are opposite. Similarly, the radial force on the pinion shaft is equal to the axial force on the gear shaft, but act in opposite directions. 30.9 Design of a Shaft for Bevel Gears In designing a pinion shaft, the following procedure may be adopted : 1. First of all, find the torque acting on the pinion. It is given by P × 60 T = π N-m 2 NP where P = Power transmitted in watts, and
NP = Speed of the pinion in r.p.m.
2. Find the tangential force (WT) acting at the mean radius (Rm) of the pinion. We know that
WT = T / Rm
3. Now find the axial and radial forces (i.e. WRH and WRV) acting on the pinion shaft as discussed above. 4. Find resultant bending moment on the pinion shaft as follows :
The bending moment due to WRH and WRV is given by
M1 = WRV × Overhang – WRH × Rm and bending moment due to WT,
M2 = WT × Overhang Bevel Gears 1089
∴ Resultant bending moment, 22+ M = ()MM12 ( ) 5. Since the shaft is subjected to twisting moment (T ) and resultant bending moment (M), therefore equivalent twisting moment, 22+ Te = MT 6. Now the diameter of the pinion shaft may be obtained by using the torsion equation. We know that π T = × τ (d )3 e 16 P where dP = Diameter of the pinion shaft, and τ = Shear stress for the material of the pinion shaft. 7. The same procedure may be adopted to find the diameter of the gear shaft. Example 30.1. A 35 kW motor running at 1200 r.p.m. drives a compressor at 780 r.p.m. through a 90° bevel gearing arrangement. The pinion has 30 teeth. The pressure angle of teeth is 1 14 /2 ° . The wheels are capable of withstanding a dynamic stress, ⎛⎞280 σ = 140 ⎜⎟ MPa, where v is the pitch line speed in m / min. w ⎝⎠280 +v The form factor for teeth may be taken as 0.686 0.124 – , where TE is the number of teeth TE equivalent of a spur gear. 1 The face width may be taken as of the 4 slant height of pitch cone. Determine for the pinion, the module pitch, face width, addendum, dedendum, outside diameter and slant height. Solution : Given : P = 35 kW = 35 × 103 θ W; NP = 1200 r.p.m. ; NG = 780 r.p.m. ; S = 90º ; φ 1 TP = 30 ; = 14 /2º; b = L / 4 Module and face width for the pinion Let m = Module in mm, High performance 2- and 3 -way bevel gear b = Face width in mm boxes = L / 4, and ...(Given)
DP = Pitch circle diameter of the pinion. We know that velocity ratio, NP 1200 V.R. = ==1.538 NG 780 ∴ Number of teeth on the gear,
TG = V.R. × TP = 1.538 × 30 = 46 Since the shafts are at right angles, therefore pitch angle for the pinion,
⎛⎞11−−11 ⎛ ⎞ θ = tan–1 ⎜⎟==tan ⎜ ⎟ tan (0.65) = 33º P1 ⎝⎠VR. . ⎝ 1.538 ⎠ and pitch angle for the gear, θ P2 = 90º – 33º = 57º 1090 A Textbook of Machine Design
We know that formative number of teeth for pinion, θ TEP = TP.sec P1 = 30 × sec 33º = 35.8 and formative number of teeth for the gear, θ TEG = TG.sec P2 = 46 × sec 57º = 84.4 Tooth form factor for the pinion
0.686=− 0.686 = y'P = 0.124 – 0.124 0.105 TEP 35.8 and tooth form factor for the gear,
0.686=− 0.686 = y'G = 0.124 – 0.124 0.116 TEG 84.4 σ Since the allowable static stress ( o) for both the pinion and gear is same (i.e. 140 MPa or 2 N/mm ) and y'P is less than y'G, therefore the pinion is weaker. Thus the design should be based upon the pinion. We know that the torque on the pinion, ×××3 P 60= 35 10 60 T = ππ× = 278.5 N-m = 278 500 N-mm 2NP 2 1200 ∴ Tangential load on the pinion, × 2TT== 2 2 278 500 = 18 567 WT = × N DmTmPP.30 m We know that pitch line velocity, ππDN... mTN π×m 30 × 1200 v = PP== PP m / min 1000 1000 1000 = 113.1 m m / min ∴ Allowable working stress, 280 280 ⎛⎞⎛= ⎞ 2 σ = 140 ⎜⎟⎜140 ⎟ MPa or N / mm w ⎝⎠⎝280++vm 280 113.1 ⎠ We know that length of the pitch cone element or slant height of the pitch cone, × × DmTPP==m 30 L = θθ = 27.54 m mm 2sinP1 2sin P1 2sin33º Since the face width (b) is 1/4th of the slant height of the pitch cone, therefore Lm27.54 b = = = 6.885 m mm 44 We know that tangential load on the pinion, ⎛⎞Lb− W =(σ × C ) b.π m.y' ⎜⎟ T OP v P ⎝⎠L ⎛⎞Lb− σ π ′ ... ( σ = σ × C ) = w.b. m.y P ⎜⎟ Q w OP v ⎝⎠L 18 567 ⎛⎞280 ⎛⎞27.54mm− 6.885 or = 140 ⎜⎟ 6.885 m × π m × 0.105 ⎜⎟ m ⎝⎠280+ 113.1m ⎝⎠27.54m 66 780 m2 = 280+ 113.1m m 2 3 or 280 + 113.1 m = 66 780 m × 18 567 = 3.6 m Solving this expression by hit and trial method, we find that Bevel Gears 1091
m = 6.6 say 8 mm Ans. and face width, b = 6.885 m = 6.885 × 8 = 55 mm Ans. Addendum and dedendum for the pinion We know that addendum, a =1 m = 1 × 8 = 8 mm Ans. and dedendum, d = 1.2 m = 1.2 × 8 = 9.6 mm Ans. Outside diameter for the pinion We know that outside diameter for the pinion, θ θ ( D = m . T ) DO = DP + 2 a cos P1 = m.TP + 2 a cos P1 ... Q P P = 8 × 30 + 2 × 8 cos 33º = 253.4 mm Ans. Slant height We know that slant height of the pitch cone, L = 27.54 m = 27.54 × 8 = 220.3 mm Ans. Example 30.2. A pair of cast iron bevel gears connect two shafts at right angles. The pitch diameters of the pinion and gear are 80 mm and 100 mm respectively. The tooth profiles of the gears 1 are of 14 /2º composite form. The allowable static stress for both the gears is 55 MPa. If the pinion transmits 2.75 kW at 1100 r.p.m., find the module and number of teeth on each gear from the stand- point of strength and check the design from the standpoint of wear. Take surface endurance limit as 630 MPa and modulus of elasticity for cast iron as 84 kN/mm2. θ φ 1 ° Solution. Given : S = 90º ; DP = 80 mm = 0.08 m ; DG = 100 mm = 0.1 m ; = 14 2 ; σ σ 2 σ OP = OG = 55 MPa = 55 N/mm ; P = 2.75 kW = 2750 W ; NP = 1100 r.p.m. ; es = 630 MPa = 630 2 2 3 2 N/mm ; EP = EG = 84 kN/mm = 84 × 10 N/mm Module Let m = Module in mm. Since the shafts are at right angles, therefore pitch angle for the pinion, ⎛⎞1 ⎛⎞D ⎛⎞80 θ –1 –1 P –1 P1 = tan ⎜⎟ = tan ⎜⎟ = tan ⎜⎟ = 38.66º ⎝⎠VR.. ⎝⎠DG ⎝⎠100 and pitch angle for the gear, θ P2 = 90º – 38.66º = 51.34º We know that formative number of teeth for pinion, 80 102.4 T = T . sec θ = × sec 38.66º = ... ( T = D / m ) EP P P1 m m Q P P and formative number of teeth on the gear, 100 160 T = T . sec θ = × sec 51.34º = ... ( T = D / m) EG G P2 m m Q G G Since both the gears are made of the same material, therefore pinion is the weaker. Thus the design should be based upon the pinion. 1 We know that tooth form factor for the pinion having 14 /2º composite teeth, 0.684 0.684 × m y'P = 0.124 – = 0.124 – TEP 102.4 = 0.124 – 0.006 68 m and pitch line velocity, π DN. π×0.08 × 1100 v = PP== 4.6 m/s 60 60 1092 A Textbook of Machine Design
Taking velocity factor, 66 C = = = 0.566 v 664.6++v We know that length of the pitch cone element or slant height of the pitch cone,
22 22 ⎛⎞DG + ⎛⎞DP ⎛⎞⎛⎞100+ 80 *L = ⎜⎟⎜⎟ = ⎜⎟⎜⎟ = 64 mm ⎝⎠22⎝⎠ ⎝⎠⎝⎠22 Assuming the face width (b) as 1/3rd of the slant height of the pitch cone (L), therefore b = L / 3 = 64 / 3 = 21.3 say 22 mm We know that torque on the pinion, ×× P 60= 2750 60 T = π× π× = 23.87 N-m = 23 870 N-mm 2NP 2 1100 ∴ Tangential load on the pinion, T = 23 870 WT = = 597 N DP / 280/2 We also know that tangential load on the pinion, ⎛⎞Lb− W =(σ × C ) b × π m × y' ⎜⎟ T OP v P ⎝⎠L ⎛⎞64− 22 or 597 = (55 × 0.566) 22 × π m (0.124 – 0.00 668 m) ⎜⎟ ⎝⎠64 = 1412 m (0.124 – 0.006 68 m) = 175 m – 9.43 m2 Solving this expression by hit and trial method, we find that m = 4.5 say 5 mm Ans. Number of teeth on each gear We know that number of teeth on the pinion,
TP = D P / m = 80 / 5 = 16 Ans. and number of teeth on the gear,
TG = DG / m = 100 / 5 = 20 Ans. Checking the gears for wear We know that the load-stress factor, σφ2 ()sines ⎡⎤11+ K = ⎢⎥ The bevel gear turbine 1.4 ⎣⎦EEPG 21° ⎡⎤ (630) sin 14 /2 11+ = ⎢⎥33 = 1.687 1.4 ⎣⎦84×× 10 84 10 × 2 TEG = 2160/m and ratio factor,Q = ++ = 1.22 TTEG EP 160 / m 102.4 / m
* The length of the pitch cone element (L) may also obtained by using the relation θ L = DP / 2 sin P1 Bevel Gears 1093
∴ Maximum or limiting load for wear, ×× × DbQKP .. . = 80 22 1.22 1.687 Ww = θ = 4640 N cosP1 cos 38.66º Since the maximum load for wear is much more than the tangential load (WT), therefore the design is satisfactory from the consideration of wear. Ans. Example 30.3. A pair of bevel gears connect two shafts at right angles and transmits 9 kW. Determine the required module and gear diameters for the following specifications : Particulars Pinion Gear Number of teeth 21 60 Material Semi-steel Grey cast iron Brinell hardness number 200 160 Allowable static stress 85 MPa 55 MPa Speed 1200 r.p.m. 420 r.p.m. 1 ° 1 ° Tooth profile 14 2 composite 14 2 composite Check the gears for dynamic and wear loads. θ σ 2 Solution. Given : S = 90º ; P = 9 kW = 9000 W ; TP = 21 ; TG = 60 ; OP = 85 MPa = 85 N/mm ; σ 2 φ 1 OG = 55 MPa = 55 N/mm ; NP = 1200 r.p.m. ; NG = 420 r.p.m. ; = 14 /2º Required module Let m = Required module in mm. Since the shafts are at right angles, therefore pitch angle for the pinion, ⎛⎞1 ⎛⎞T ⎛⎞21 θ –1 –1 P –1 P1 = tan ⎜⎟ = tan ⎜⎟ = tan ⎜⎟ = 19.3º ⎝⎠VR.. ⎝⎠TG ⎝⎠60 and pitch angle for the gear, θ θ θ P2 = S – P1 = 90º – 19.3º = 70.7º We know that formative number of teeth for the pinion, θ TEP = TP . sec P1 = 21 sec 19.3º = 22.26 and formative number of teeth for the gear, θ TEG = TG . sec P2 = 60 sec 70.7º = 181.5 We know that tooth form factor for the pinion, 0.684 0.684 y'P = 0.124 – = 0.124 – = 0.093 TEP 22.26 1 ... (For 14 /2º composite system) and tooth form factor for the gear, 0.684 0.684 y'G = 0.124 – = 0.124 – = 0.12 TEG 181.5 ∴σ OP × y'P = 85 × 0.093 = 7.905 σ and OG × y’G = 55 × 0.12 = 6.6 σ σ Since the product OG × y'G is less than OP × y'P, therefore the gear is weaker. Thus, the design should be based upon the gear. We know that torque on the gear, ×× P 60= 9000 60 T = ππ× = 204.6 N-m = 204 600 N-mm 22420NG 1094 A Textbook of Machine Design
∴ Tangential load on the gear, × TT==2 2 204 600 = 6820 W = × N ... ( DG = m .TG) T DmTmGG/2 . 60 m Q We know that pitch line velocity, ππDN... mTN π×m 60 × 420 v = GG== GG mm / s 60 60 60 = 1320 m mm / s = 1.32 m m / s Taking velocity factor, 66 C = = v 661.32++vm We know that length of pitch cone element, × DmTGG==. m 60 *L = θ× = 32 m mm 2 sinP2 2 sin 70.7º 2 0.9438 Assuming the face width (b) as 1/3rd of the length of the pitch cone element (L), therefore Lm32 b = ==10.67 m mm 33 We know that tangential load on the gear, ⎛⎞Lb– W =(σ × C ) b.π m.y' ⎜⎟ T OG v G ⎝⎠L 6820 ⎛⎞6 ⎛⎞32mm – 10.67 ∴ = 55 ⎜⎟ 10.67 m × π m × 0.12 ⎜⎟ m ⎝⎠61.32+ m ⎝⎠32 m 885m2 = 61.32+ m or 40 920 + 9002 m = 885 m3 Solving this expression by hit and trial method, we find that m = 4.52 say 5 mm Ans. and b = 10.67 m = 10.67 × 5 = 53.35 say 54 mm Ans. Gear diameters We know that pitch diameter for the pinion,
DP = m.TP = 5 × 21 = 105 mm Ans. and pitch circle diameter for the gear,
DG = m.TG = 5 × 60 = 300 mm Ans. Check for dynamic load We know that pitch line velocity, v = 1.32 m = 1.32 × 5 = 6.6 m / s and tangential tooth load on the gear, 6820 6820 W = = = 1364 N T m 5 From Table 28.7, we find that tooth error action for first class commercial gears having module 5 mm is e = 0.055 mm
* The length of pitch cone element (L) may be obtained by using the following relation, i.e.
DmT22Dm2 . 2 ⎛⎞GG+=⎛⎞P ⎛ ⎞ +mT. P =22 + L = ⎜⎟⎜⎟ ⎜ ⎟ () ()TTGP () ⎝⎠22⎝⎠ ⎝ 2 ⎠ 2 2 Bevel Gears 1095
1 3 2 3 2 Taking K = 0.107 for 14 /2º composite teeth, EP = 210 × 10 N/mm ; and EG = 84 × 10 N/mm , we have Deformation or dynamic factor, Ke. 0.107× 0.055 C = = = 353 N / mm 11++ 1 1 33 EEPG210108410×× We know that dynamic load on the gear, 21vbCW ( .+ ) W = W + T D T ++ 21vbCW . T 21××+ 6.6 (54 353 1364) = 1364 + 21×+ 6.6 54 × 353 + 1364 = 1364 + 10 054 = 11 418 N σ From Table 28.8, we find that flexural endurance limit ( e) for the gear material which is grey cast iron having B.H.N. = 160, is σ 2 e = 84 MPa = 84 N/mm We know that the static tooth load or endurance strength of the tooth, σ π π WS = e.b. m.y'G = 84 × 54 × × 5 × 0.12 = 8552 N
Since WS is less that WD, therefore the design is not satisfactory from the standpoint of dynamic load. We have already discussed in spur gears (Art. 28.20) that WS ≥ 1.25 WD for steady loads. For a satisfactory design against dynamic load, let us take the precision gears having tooth error in action (e = 0.015 mm) for a module of 5 mm. ∴ Deformation or dynamic factor, 0.107× 0.015 C = = 96 N/mm 11+ 210×× 1033 84 10 and dynamic load on the gear, 21××+ 6.6 (54 96 1364) W = 1364 + = 5498 N D 21×+ 6.6 54 ×+ 96 1364
From above we see that by taking precision gears, WS is greater than WD, therefore the design is satisfactory, from the standpoint of dynamic load. Check for wear load From Table 28.9, we find that for a gear of grey cast iron having B.H.N. = 160, the surface endurance limit is, σ 2 es = 630 MPa = 630 N/mm ∴ Load-stress factor, σφ2 ()sines ⎡⎤11+ K = ⎢⎥ 1.4 ⎣⎦EEPG
(630)2 sin14 1 ° ⎡⎤11 2 + 2 = ⎢⎥33 = 1.18 N/mm 1.4 ⎣⎦210×× 10 84 10
× 2 TEG = 2 181.5 and ratio factor, Q = ++ = 1.78 TTEG EP 181.5 22.26 1096 A Textbook of Machine Design
We know that maximum or limiting load for wear,
Ww = DP.b.Q.K = 105 × 54 × 1.78 × 1.18 = 11 910 N
Since Ww is greater then WD, therefore the design is satisfactory from the standpoint of wear. Example 30.4. A pair of 20º full depth involute teeth bevel gears connect two shafts at right angles having velocity ratio 3 : 1. The gear is made of cast steel having allowable static stress as 70 MPa and the pinion is of steel with allowable static stress as 100 MPa. The pinion transmits 37.5 kW at 750 r.p.m. Determine : 1. Module and face width; 2. Pitch diameters; and 3. Pinion shaft diameter. Assume tooth form factor, 0912. y = 0.154 – , where TE is the TE 1 formative number of teeth, width = /3 rd the length of pitch cone, and pinion shaft overhangs by 150 mm. φ θ Involute teeth bevel gear Solution. Given : = 20º ; S = 90º ; σ 2 V.R . = 3 ; OG = 70 MPa = 70 N/mm ; σ 2 OP = 100 MPa = 100 N/mm ; P = 37.5 kW = 37 500 W ; NP = 750 r.p.m. ; b = L / 3 ; Overhang = 150 mm Module and face width Let m = Module in mm, b = Face width in mm = L / 3, ...(Given)
DG = Pitch circle diameter of the gear in mm. Since the shafts are at right angles, therefore pitch angle for the pinion, ⎛⎞1 ⎛⎞1 θ = tan–1 ⎜⎟ = tan–1 ⎜⎟ = 18.43º P1 ⎝⎠VR.. ⎝⎠3 and pitch angle for the gear, θ θ θ P2 = S – P1 = 90º – 18.43º = 71.57º Assuming number of teeth on the pinion (TP) as 20, therefore number of teeth on the gear, = TG = V.R. × TP = 3 × 20 = 60 ... (../)QVR TGP T We know that formative number of teeth for the pinion, θ TEP = TP . sec P1 = 20 × sec 18.43º = 21.08 and formative number of teeth for the gear, θ TEG = TG . sec P2 = 60 sec 71.57º = 189.8 We know that tooth form factor for the pinion, 0.912 0.912 =−0.154 y'P = 0.154 – = 0.111 TEP 21.08 and tooth form factor for the gear, 0.912 0.912 =−0.154 y'G = 0.154 – = 0.149 TEG 189.8 ∴σ OP × y'P = 100 × 0.111 = 11.1 σ and OG × y'G = 70 × 0.149 = 10.43 Bevel Gears 1097 σ σ Since the product OG × y'G is less than OP × y'P, therefore the gear is weaker. Thus, the design should be based upon the gear and not the pinion. We know that the torque on the gear, PP××60 60 = ... ( V.R. = N / N = 3) T = ππ× Q P G 22/3NNGP 37 500× 60 3 = 2750/3π× = 1432 N-m = 1432 × 10 N-mm ∴ Tangential load on the gear, 22TT= W = ... ( D = m.T ) T DmTGG. Q G G 2×× 1432 1033 47.7 × 10 = = N mm× 60 We know that pitch line velocity, ππDN.../3 mTN v = GG= GP 60 60 π×m 60 × 750 / 3 = = 785.5 m mm / s = 0.7855 m m/s 60 Taking velocity factor, 33 C = = v 3++vm 3 0.7855 We know that length of the pitch cone element, × DmTGG==. m 60 L = θ× = 31.62 m mm 2 sinP2 2 sin 71.57º 2 0.9487 Since the face width (b) is 1/3rd of the length of the pitch cone element, therefore Lm31.62 b = = = 10.54 m mm 33 We know that tangential load on the gear, − σ π ⎛⎞Lb WT =( OG × Cv) b. m.y'G ⎜⎟ ⎝⎠L
Racks 1098 A Textbook of Machine Design
× 3 ⎛⎞31.62mm− 10.54 ∴ 47.7 10⎛⎞ 3 π = 70⎜⎟ 10.54 m × m × 0.149 ⎜⎟ mm⎝⎠3+ 0.7855 ⎝⎠31.62m 691m2 = 30.7855+ m 143 100 + 37 468 m = 691 m3 Solving this expression by hit and trial method, we find that m = 8.8 say 10 mm Ans. and b = 10.54 m = 10.54 × 10 = 105.4 mm Ans. Pitch diameters We know that pitch circle diameter of the larger wheel (i.e. gear),
DG = m.TG = 10 × 60 = 600 mm Ans. and pitch circle diameter of the smaller wheel (i.e. pinion),
DP = m.TP = 10 × 20 = 200 mm Ans. Pinion shaft diameter
Let dP = Pinion shaft diameter. We know that the torque on the pinion, ×× P 60= 37 500 60 T = π× π× = 477.4 N-m = 477 400 N-mm 22750NP and length of the pitch cone element, L = 31.62 m = 31.62 × 10 = 316.2 mm ∴ Mean radius of the pinion,
⎛⎞− b DP =−⎛⎞105.4 200 Rm = ⎜⎟L ⎜⎟316.2 = 83.3 mm ⎝⎠2 222316.2L ⎝⎠× We know that tangential force acting at the mean radius, T = 477 400 WT = = 5731 N Rm 83.3 Axial force acting on the pinion shaft, φ θ WRH = WT tan . sin P1 = 5731 × tan 20º × sin 18.43º = 5731 × 0.364 × 0.3161 = 659.4 N and radial force acting on the pinion shaft, φ θ WRV = WT tan . cos P1 = 5731 × tan 20º × cos 18.43º = 5731 × 0.364 × 0.9487 = 1979 N ∴ Bending moment due to WRH and WRV,
M1 = WRV × Overhang – WRH × Rm = 1979 × 150 – 659.4 × 83.3 = 241 920 N-mm and bending moment due to WT,
M2 = WT × Overhang = 5731 × 150 = 859 650 N-mm ∴ Resultant bending moment, 22+= 2 + 2 M = (MM12 ) ( ) (241 920) (859 650) = 893 000 N-mm Since the shaft is subjected to twisting moment (T ) and bending moment (M ), therefore equivalent twisting moment, Bevel Gears 1099
22+ Te = MT
=+(893 000)22 (477 400) = 1013 × 103 N-mm
We also know that equivalent twisting moment (Te), π 1013 × 103 = × τ (d )3 16 P π = × 45 (d )3 = 8.84 (d )3 16 P P ... (Taking τ = 45 N/mm2) ∴ 3 3 3 (dP) = 1013 × 10 / 8.84 = 114.6 × 10 Bevel gears or dP = 48.6 say 50 mm Ans.
EXERCISES
1. A pair of straight bevel gears is required to transmit 10 kW at 500 r.p.m. from the motor shaft to another shaft at 250 r.p.m. The pinion has 24 teeth. The pressure angle is 20°. If the shaft axes are at right angles to each other, find the module, face width, addendum, outside diameter and slant height. The gears are capable of withstanding a static stress of 60 MPa. The tooth form factor may be taken as 4.5 0.154 – 0.912/T , where T is the equivalent number of teeth. Assume velocity factor as , E E 4.5 + v 1 where v the pitch line speed in m/s. The face width may be taken as of the slant height of the pitch 4 cone. [Ans. m = 8 mm ; b = 54 mm ; a = 8 mm ; DO = 206.3 mm ; L = 214.4 mm] 2. A 90º bevel gearing arrangement is to be employed to transmit 4 kW at 600 r.p.m. from the driving shaft to another shaft at 200 r.p.m. The pinion has 30 teeth. The pinion is made of cast steel having a static stress of 80 MPa and the gear is made of cast iron with a static stress of 55 MPa. The tooth 1 profiles of the gears are of 14 /2º composite form. The tooth form factor may be taken as 3 y' = 0.124 – 0.684 / TE, where TE is the formative number of teeth and velocity factor, Cv = 3 + v , where v is the pitch line speed in m/s. 1 The face width may be taken as /3 rd of the slant height of the pitch cone. Determine the module, face width and pitch diameters for the pinion and gears, from the standpoint of strength and check the design from the standpoint of wear. Take surface endurance limit as 630 MPa and modulus of 2 2 elasticity for the material of gears is EP = 200 kN/mm and EG = 80 kN/mm .
[Ans. m = 4 mm ; b = 64 mm ; DP = 120 mm ; DG = 360 mm] 3. A pair of bevel gears is required to transmit 11 kW at 500 r.p.m. from the motor shaft to another shaft, the speed reduction being 3 : 1. The shafts are inclined at 60º. The pinion is to have 24 teeth with a pressure angle of 20º and is to be made of cast steel having a static stress of 80 MPa. The gear is to be made of cast iron with a static stress of 55 MPa. The tooth form factor may be taken as
y = 0.154 – 0.912/TE, where TE is formative number of teeth. The velocity factor may be taken as 3 , where v is the pitch line velocity in m/s. The face width may be taken as 1/ th of the slant 3 + v 4 height of the pitch cone. The mid-plane of the gear is 100 mm from the left hand bearing and 125 mm from the right hand bearing. The gear shaft is to be made of colled-rolled steel for which the allowable tensile stress may be taken as 80 MPa. Design the gears and the gear shaft. 1100 A Textbook of Machine Design
QUESTIONS 1. How the bevel gears are classified ? Explain with neat sketches. 2. Sketch neatly the working drawing of bevel gears in mesh. 3. For bevel gears, define the following : (i) Cone distance; (ii) Pitch angle; (iii) Face angle; (iv) Root angle; (v) Back cone distance; and (vi) Crown height. 4. What is Tredgold's approximation about the formative number of teeth on bevel gear? 5. What are the various forces acting on a bevel gear ? 6. Write the procedure for the design of a shaft for bevel gears. OBJECTIVE TYPE QUESTIONS
1. When bevel gears having equal teeth and equal pitch angles connect two shafts whose axes intersect at right angle, then they are known as (a) angular bevel gears (b) crown bevel gears (c) internal bevel gears (d) mitre gears 2. The face angle of a bevel gear is equal to (a) pitch angle – addendum angle (b) pitch angle + addendum angle (c) pitch angle – dedendum angle (d) pitch angle + dedendum angle 3. The root angle of a bevel gear is equal to (a) pitch angle – addendum angle (b) pitch angle + addendum angle (c) pitch angle – dedendum angle (d) pitch angle + dedendum angle 4. If b denotes the face width and L denotes the cone distance, then the bevel factor is written as (a) b / L (b) b / 2L (c) 1 – 2 b.L (d)1 – b / L θ 5. For a bevel gear having the pitch angle , the ratio of formative number of teeth (TE) to actual number of teeth (T) is 1 1 (a) sin θ (b) cos θ
1 θ θ (c) tan θ (d) sin cos
ANSWERS
1. (d) 2. (b) 3. (c) 4. (d) 5. (b) Worm Gears 1101
C H A P T E R 31 Worm Gears
1. Introduction 2. Types of Worms 3. Types of Worm Gears. 4. Terms used in Worm Gearing. 5. Proportions for Worms . 6. Proportions for Worm Gears. 7. Efficiency of Worm Gearing. 8. Strength of Worm Gear Teeth . 9. Wear Tooth Load for Worm Gear. 10. Thermal Rating of Worm Gearing. 11. Forces Acting on Worm Gears. 12. Design of Worm Gearing. 31.1 Introduction The worm gears are widely used for transmitting power at high velocity ratios between non-intersecting shafts that are generally, but not necessarily, at right angles. It can give velocity ratios as high as 300 : 1 or more in a single step in a minimum of space, but it has a lower efficiency. The worm gearing is mostly used as a speed reducer, which consists of worm and a worm wheel or gear. The worm (which is the driving member) is usually of a cylindrical form having threads of the same shape as that of an involute rack. The threads of the worm may be left handed or right handed and single or multiple threads. The worm wheel or gear (which is the driven member) is similar to a helical gear with a face curved to conform to the shape of the worm. The worm is generally made of steel while the worm gear is made of bronze or cast iron for light service.
1101 1102 A Textbook of Machine Design
The worm gearing is classified as non-interchangeable, because a worm wheel cut with a hob of one diameter will not operate satisfactorily with a worm of different diameter, even if the thread pitch is same. 31.2 Types of Worms The following are the two types of worms : 1. Cylindrical or straight worm, and 2. Cone or double enveloping worm. The cylindrical or straight worm, as shown in Fig. 31.1 (a), is most commonly used. The shape of the thread is involute helicoid of pressure angle 14 ½° for single and double threaded worms and 20° for triple and quadruple threaded worms. The worm threads are cut by a straight sided milling cutter having its diameter not less than the outside diameter of worm or greater than 1.25 times the outside diameter of worm. The cone or double enveloping worm, as shown in Fig. 31.1 (b), is used to some extent, but it requires extremely accurate alignment.
Fig. 31.1. Types of worms. 31.3 Types of Worm Gears The following three types of worm gears are important from the subject point of view : 1. Straight face worm gear, as shown in Fig. 31.2 (a), 2. Hobbed straight face worm gear, as shown in Fig. 31.2 (b), and 3. Concave face worm gear, as shown in Fig. 31.2 (c).
Fig. 31.2. Types of worms gears. The straight face worm gear is like a helical gear in which the straight teeth are cut with a form cutter. Since it has only point contact with the worm thread, therefore it is used for light service. The hobbed straight face worm gear is also used for light service but its teeth are cut with a hob, after which the outer surface is turned. Worm Gears 1103
The concave face worm gear is the accepted standard form and is used for all heavy service and general industrial uses. The teeth of this gear are cut with a hob of the same pitch diameter as the mating worm to increase the contact area.
Worm gear is used mostly where the power source operates at a high speed and output is at a slow speed with high torque. It is also used in some cars and trucks. 31.4 Terms used in Worm Gearing The worm and worm gear in mesh is shown in Fig. 31.3. The following terms, in connection with the worm gearing, are important from the subject point of view : 1. Axial pitch. It is also known as linear pitch of a worm. It is the distance measured axially (i.e. parallel to the axis of worm) from a point on one thread to the corresponding point on the adjacent thread on the worm, as shown in Fig. 31.3. It may be noted that the axial pitch (pa) of a worm is equal to the circular pitch ( pc ) of the mating worm gear, when the shafts are at right angles.
Fig. 31.3 . Worm and Worm gear. 1104 A Textbook of Machine Design
2. Lead. It is the linear distance through which a point on a thread moves ahead in one revolution of the worm. For single start threads, lead is equal to the axial pitch, but for multiple start threads, lead is equal to the product of axial pitch and number of starts. Mathematically,
Lead, l = pa . n where pa = Axial pitch ; and n = Number of starts. 3. Lead angle. It is the angle between the tangent to the thread helix on the pitch cylinder and the plane normal to the axis of the worm. It is denoted by λ. A little consideration will show that if one complete turn of a worm thread be imagined to be unwound from the body of the worm, it will form an inclined plane whose base is equal to the pitch circumference of the worm and altitude equal to lead of the worm, as shown in Fig. 31.4. Fig. 31.4. Development of a helix thread. From the geometry of the figure, we find that Lead of the worm tan λ = Pitch circumference of the worm l = pna . = ππ ...(Q l = pa . n) DDWW pn. π mn.. mn c == π = ππ ...(Q pa = pc ; and pc = m) DDDWWW where m = Module, and
DW = Pitch circle diameter of worm. The lead angle (λ) may vary from 9° to 45°. It has been shown by F.A. Halsey that a lead angle less than 9° results in rapid wear and the safe value of λ is 12½°.
Hydraulic or Pneumatic Speed Ratio Detection Switches Change Actuator Round Housing With O-ring Seated Cooling Jacket
Motor Flange
Hollow Through Bore for Drawbar Integration OUTPUT- External Spline to Spindle
Housing OD Designed to meet INPUT RAM Bore Dia, and Share Motor Spline to Accept Coolant Supply Motor Shaft Model of sun and planet gears. Worm Gears 1105
For a compact design, the lead angle may be determined by the following relation, i.e. 1/3 ⎛⎞NG tan λ = ⎜⎟, ⎝⎠NW where NG is the speed of the worm gear and NW is the speed of the worm. 4. Tooth pressure angle. It is measured in a plane containing the axis of the worm and is equal to one-half the thread profile angle as shown in Fig. 31.3. The following table shows the recommended values of lead angle (λ) and tooth pressure angle (φ). Table 31.1. Recommended values of lead angle and pressure angle.
Lead angle (λ) 0 – 16 16 – 25 25 – 35 35 – 45 in degrees
Pressure angle(φ) 14½ 20 25 30 in degrees For automotive applications, the pressure angle of 30° is recommended to obtain a high efficiency and to per- mit overhauling. 5. Normal pitch. It is the distance measured along the normal to the threads between two corresponding points on two adjacent threads of the worm. Mathematically, λ Normal pitch, pN = pa.cos Note. The term normal pitch is used for a worm having single start threads. In case of a worm having multiple start threads, the term normal lead (lN) is used, such that λ lN = l . cos Worm gear teeth generation on gear hobbing machine. 6. Helix angle. It is the angle between the tangent to the thread helix on the pitch cylinder and the axis of the worm. It is denoted by α W, in Fig. 31.3. The worm helix angle is the complement of worm lead angle, i.e. α λ W + = 90° It may be noted that the helix angle on the worm is generally quite large and that on the worm λ α gear is very small. Thus, it is usual to specify the lead angle ( ) on the worm and helix angle ( G) on the worm gear. These two angles are equal for a 90° shaft angle.
7. Velocity ratio. It is the ratio of the speed of worm (NW) in r.p.m. to the speed of the worm gear (NG) in r.p.m. Mathematically, velocity ratio, N V.R. = W NG Let l = Lead of the worm, and
DG = Pitch circle diameter of the worm gear. We know that linear velocity of the worm, lN. v = W W 60 1106 A Textbook of Machine Design and linear velocity of the worm gear, π DN v = GG G 60 Since the linear velocity of the worm and worm gear are equal, therefore lN. ππDN. N D W = GGor W= G 60 60 NlG We know that pitch circle diameter of the worm gear,
DG = m . TG where m is the module and TG is the number of teeth on the worm gear. NDmTππ. ∴ V.R. = WG== G NlG l pT.. pT T caGGG== ... ( p = π m = p ; and l = p . n) = Q c a a lpnna . where n = Number of starts of the worm. From above, we see that velocity ratio may also be defined as the ratio of number of teeth on the worm gear to the number of starts of the worm. The following table shows the number of starts to be used on the worm for the different velocity ratios : Table 31.2. Number of starts to be used on the worm for different velocity ratios.
Velocity ratio (V. R.) 36 and above 12 to 36 8 to 12 6 to 12 4 to 10
Number of starts or threads on the worm Single Double Triple Quadruple Sextuple
(n = Tw) 31.5 Proportions for Worms The following table shows the various porportions for worms in terms of the axial or circular pitch ( pc ) in mm. Table 31.3. Proportions for worm.
S. No. Particulars Single and double Triple and quadruple threaded worms threaded worms 1. Normal pressure angle (φ) 14½° 20°
2. Pitch circle diameter for 2.35 pc + 10 mm 2.35 pc + 10 mm worms integral with the shaft
3. Pitch circle diameter for 2.4 pc + 28 mm 2.4 pc + 28 mm worms bored to fit over the shaft
4. Maximum bore for shaft pc + 13.5 mm pc + 13.5 mm
5. Hub diameter 1.66 pc + 25 mm 1.726 pc + 25 mm
6. Face length (LW) pc (4.5 + 0.02 TW) pc (4.5 + 0.02 TW)
7. Depth of tooth (h) 0.686 pc 0.623 pc
8. Addendum (a) 0.318 pc 0.286 pc
Notes: 1. The pitch circle diameter of the worm (DW) in terms of the centre distance between the shafts (x) may be taken as follows : ()x 0.875 D = ... (when x is in mm) W 1.416 Worm Gears 1107
2. The pitch circle diameter of the worm (DW ) may also be taken as
DW =3 pc, where pc is the axial or circular pitch. 3. The face length (or length of the threaded portion) of the worm should be increased by 25 to 30 mm for the feed marks produced by the vibrating grinding wheel as it leaves the thread root. 31.6 Proportions for Worm Gear The following table shows the various proportions for worm gears in terms of circular pitch
( pc ) in mm. Table 31.4. Proportions for worm gear. S. No. Particulars Single and double threads Triple and quadruple threads
1. Normal pressure angle (φ) 14½° 20°
2. Outside diameter (DOG) DG + 1.0135 pc DG + 0.8903 pc
3. Throat diameter (DT) DG + 0.636 pc DG + 0.572 pc
4. Face width (b) 2.38 pc + 6.5 mm 2.15 pc + 5 mm
5. Radius of gear face (Rf) 0.882 pc + 14 mm 0.914 pc + 14 mm
6. Radius of gear rim (Rr) 2.2 pc + 14 mm 2.1 pc + 14 mm 31.7 Efficiency of Worm Gearing The efficiency of worm gearing may be defined as the ratio of work done by the worm gear to the work done by the worm. Mathematically, the efficiency of worm gearing is given by tanλ (cosφ− μ tanλ ) η = ...(i) cosφλ+μ tan where φ = Normal pressure angle, μ = Coefficient of friction, and λ = Lead angle. The efficiency is maximum, when tan λ = 1 + μ2 − μ In order to find the approximate value of the efficiency, assuming square threads, the following relation may be used : tanλ (1 –μ tanλ ) Efficiency, η = tan λ+μ − μ λ = 1tan 1/tan+ μ λ λ = tan λ+φ tan (1 ) ...(Substituting in equation (i), φ = 0, for square threads) φ where 1 = Angle of friction, such that tan φ = μ. 1 A gear-cutting machine is used to cut gears. 1108 A Textbook of Machine Design
The coefficient of friction varies with the speed, reaching a minimum value of 0.015 at a π ⎛⎞= DNWW. rubbing speed ⎜⎟vr between 100 and 165 m/min. For a speed below 10 m/min, take ⎝⎠cos λ μ = 0.015. The following empirical relations may be used to find the value of μ, i.e. 0.275 μ , = 0.25 for rubbing speeds between 12 and 180 m/min ()vr v = 0.025 + r for rubbing speed more than 180 m/min 18000 Note : If the efficiency of worm gearing is less than 50%, then the worm gearing is said to be self locking, i.e. it cannot be driven by applying a torque to the wheel. This property of self locking is desirable in some applications such as hoisting machinery. Example 31.1. A triple threaded worm has teeth of 6 mm module and pitch circle diameter of 50 mm. If the worm gear has 30 teeth of 14½° and the coefficient of friction of the worm gearing is 0.05, find 1. the lead angle of the worm, 2. velocity ratio, 3. centre distance, and 4. efficiency of the worm gearing.
Solution. Given : n = 3 ; m = 6 ; Hardened and ground worm shaft and worm wheel φ DW = 50 mm ; TG = 30 ; = 14.5° ; pair μ = 0.05. 1. Lead angle of the worm Let λ = Lead angle of the worm. mn.63× We know that tan λ = ==0.36 DW 50 ∴λ= tan–1 (0.36) = 19.8° Ans. 2. Velocity ratio We know that velocity ratio,
V. R .=TG / n = 30 / 3 = 10 Ans. 3. Centre distance We know that pitch circle diameter of the worm gear
DG = m.TG = 6 × 30 = 180 mm ∴ Centre distance, DD+ 50+ 180 x = WG==115 mm Ans. 22 4. Efficiency of the worm gearing We know that efficiency of the worm gearing. tanλ (cosφ − μ tanλ ) η = cosφλ+μ . tan tan 19.8°°−×° (cos 14.5 0.05 tan 19.8 ) = cos 14.5°× tan 19.8 °+ 0.05 0.36 (0.9681−× 0.05 0.36) 0.342 = ==0.858 or 85.8% Ans. 0.9681×+ 0.36 0.05 0.3985 Worm Gears 1109
Note : The approximate value of the efficiency assuming square threads is 1 –μλ tan 1 − 0.05 × 0.36 0.982 η = ===0.86 or 86% Ans. 1+μ / tan λ 1 + 0.05/ 0.36 1.139 31.8 Strength of Worm Gear Teeth In finding the tooth size and strength, it is safe to assume that the teeth of worm gear are always weaker than the threads of the worm. In worm gearing, two or more teeth are usually in contact, but due to uncertainty of load distribution among themselves it is assumed that the load is transmitted by one tooth only. We know that according to Lewis equation, σ π WT =( o . Cv) b. m . y where WT = Permissible tangential tooth load or beam strength of gear tooth, σ o = Allowable static stress, Cv = Velocity factor, b = Face width, m = Module, and y = Tooth form factor or Lewis factor. Notes : 1. The velocity factor is given by 6 C = , where v is the peripheral velocity of the worm gear in m/s. v 6 + v 2. The tooth form factor or Lewis factor (y) may be obtained in the similar manner as discussed in spur gears (Art. 28.17), i.e. 0.684 y = 0.124− , for 14½° involute teeth. TG 0.912 = 0.154− , for 20° involute teeth. TG 3. The dynamic tooth load on the worm gear is given by Wv⎛⎞6 + T = W WD = T ⎜⎟ Cv ⎝⎠6 where WT = Actual tangential load on the tooth. The dynamic load need not to be calculated because it is not so severe due to the sliding action between the worm and worm gear. 4. The static tooth load or endurance strength of the tooth
(WS) may also be obtained in the similar manner as discussed in spur gears (Art. 28.20), i.e. σ π WS = e.b m.y σ where e = Flexural endurance limit. Its value may be taken as 84 MPa for cast iron and 168 MPa for phosphor bronze gears. 31.9 Wear Tooth Load for Worm Gear
The limiting or maximum load for wear (WW) is given by
WW = DG . b . K
where DG = Pitch circle diameter of the worm gear, Worm gear assembly. 1110 A Textbook of Machine Design
b = Face width of the worm gear, and K = Load stress factor (also known as material combination factor). The load stress factor depends upon the combination of materials used for the worm and worm gear. The following table shows the values of load stress factor for different combination of worm and worm gear materials. Table 31.5. Values of load stress factor (K ). Material S.No. Load stress factor (K) Worm Worm gear N/mm2 1. Steel (B.H.N. 250) Phosphor bronze 0.415 2. Hardened steel Cast iron 0.345 3. Hardened steel Phosphor bronze 0.550 4. Hardened steel Chilled phosphor bronze 0.830 5. Hardened steel Antimony bronze 0.830 6. Cast iron Phosphor bronze 1.035
Note : The value of K given in the above table are suitable for lead angles upto 10°. For lead angles between 10° and 25°, the values of K should be increased by 25 per cent and for lead angles greater than 25°, increase the value of K by 50 per cent. 31.10 Thermal Rating of Worm Gearing In the worm gearing, the heat generated due to the work lost in friction must be dissipated in order to avoid over heating of the drive and lubricating oil. The quantity of heat generated (Qg) is given by η Qg = Power lost in friction in watts = P (1 – ) ...(i) where P = Power transmitted in watts, and η = Efficiency of the worm gearing. The heat generated must be dissipated through the lubricating oil to the gear box housing and then to the atmosphere. The heat dissipating capacity depends upon the following factors : 1. Area of the housing (A),
2. Temperature difference between the housing surface and surrounding air (t2 – t1), and 3. Conductivity of the material (K). Mathematically, the heat dissipating capacity,
Qd = A (t2 – t1) K ...(ii)
From equations (i) and (ii), we can find the temperature difference (t2 – t1). The average value of K may be taken as 378 W/m2/°C.
Notes : 1. The maximum temperature (t2 – t1) should not exceed 27 to 38°C. 2. The maximum temperature of the lubricant should not exceed 60°C. 3. According to AGMA recommendations, the limiting input power of a plain worm gear unit from the standpoint of heat dissipation, for worm gear speeds upto 2000 r.p.m., may be checked from the following relation, i.e. 3650 x1.7 P = VR..+ 5 where P = Permissible input power in kW, x = Centre distance in metres, and V.R. = Velocity ratio or transmission ratio. Worm Gears 1111 31.11 Forces Acting on Worm Gears When the worm gearing is transmitting power, the forces acting on the worm are similar to those on a power screw. Fig. 31.5 shows the forces acting on the worm. It may be noted that the forces on a worm gear are equal in magnitude to that of worm, but opposite in direction to those shown in Fig. 31.5.
Fig. 31.5. Forces acting on worm teeth. The various forces acting on the worm may be determined as follows : 1. Tangential force on the worm, 2× Torque on worm WT = Pitch circle diameter of worm (DW ) = Axial force or thrust on the worm gear
The tangential force (WT) on the worm produces a twisting moment of magnitude (WT × DW / 2) and bends the worm in the horizontal plane. 2. Axial force or thrust on the worm, λ WA = WT / tan = Tangential force on the worm gear 2× Torque on the worm gear = Pitch circle diameter of worm gear(DG ) The axial force on the worm tends to move the worm axially, induces an axial load on the bearings and bends the worm in a vertical plane with a bending moment of magnitude (WA × DW / 2). 3. Radial or separating force on the worm, φ WR = WA . tan = Radial or separating force on the worm gear The radial or separating force tends to force the worm and worm gear out of mesh. This force also bends the worm in the vertical plane. Example 31.2. A worm drive transmits 15 kW at 2000 r.p.m. to a machine carriage at 75 r.p.m. The worm is triple threaded and has 65 mm pitch diameter. The worm gear has 90 teeth of 6 mm module. The tooth form is to be 20° full depth involute. The coefficient of friction between the mating teeth may be taken as 0.10. Calculate : 1. tangential force acting on the worm ; 2. axial thrust and separating force on worm; and 3. efficiency of the worm drive. 3 Solution. Given : P = 15 kW = 15 × 10 W ; NW = 2000 r.p.m. ; NG = 75 r.p.m. ; n = 3 ; φ μ DW = 65 mm ; TG = 90 ; m = 6 mm ; = 20° ; = 0.10 1. Tangential force acting on the worm We know that the torque transmitted by the worm P ×××60 15 103 60 ===71.6 N-m 71 600 N-mm = ππ× 2NW 2 2000 1112 A Textbook of Machine Design
∴ Tangential force acting on the worm, Torque on worm 71 600 W = ==2203 N Ans. T Radius of worm 65 / 2 2. Axial thrust and separating force on worm Let λ = Lead angle. mn.63× We know that tan λ = ==0.277 DW 65 or λ = tan–1 (0.277) = 15.5° ∴ Axial thrust on the worm, λ WA = WT / tan = 2203 / 0.277 = 7953 N Ans. and separating force on the worm φ WR = WA . tan = 7953 × tan 20° = 7953 × 0.364 = 2895 N Ans. 3. Efficiency of the worm drive We know that efficiency of the worm drive, tanλ (cosφ − μ . tanλ ) η = cosφλ+μ .tan
tan 15.5°°−×° (cos 20 0.10 tan 15.5 ) = cos 20°× tan 15.5 °+ 0.10
0.277 (0.9397−× 0.10 0.277) 0.2526 = ==0.701 or 70.1% Ans. 0.9397×+ 0.277 0.10 0.3603
31.12 Design of Worm Gearing In designing a worm and worm gear, the quantities like the power transmitted, speed, velocity ratio and the centre distance between the shafts are usually given and the quantities such as lead angle, lead and number of threads on the worm are to be determined. In order to determine the satisfactory combination of lead angle, lead and centre distance, the following method may be used: From Fig. 31.6 we find that the centre distance, DD+ x = WG 2
Worm gear boxes are noted for reliable power transmission.
Fig. 31.6. Worm and worm gear. Worm Gears 1113
The centre distance may be expressed in terms of the axial lead (l), lead angle (λ) and velocity ratio (V.R.), as follows : l (cotλ+VR . .) x = 2 π λ In terms of normal lead (lN = l cos ), the above expression may be written as :
lN ⎛⎞1..+ VR x = 2sincosπλ⎝⎠⎜⎟ λ
xVR=+11⎛⎞ .. or πλ⎜⎟ λ ...(i) lN 2sincos⎝⎠ Since the velocity ratio (V.R.) is usually given, therefore the equation (i) contains three variables λ i.e. x, lN and . The right hand side of the above expression may be calculated for various values of velocity ratios and the curves are plotted as shown in Fig. 31.7. The lowest point on each of the curves gives the lead angle which corresponds to the minimum value of x / lN. This minimum value repre- sents the minimum centre distance that can be used with a given lead or inversely the maximum lead that can be used with a given centre distance. Now by using Table 31.2 and standard modules, we can determine the combination of lead angle, lead, centre distance and diameters for the given design specifications.
Fig. 31.7. Worm gear design curves.
Note : The lowest point on the curve may be determined mathematically by differentiating the equation (i) with respect to λ and equating to zero, i.e. (..)sinVR 33λ− cos λ = 0 or V.R. = cot3 λ sin22λλ . cos Example 31.3. Design 20° involute worm and gear to transmit 10 kW with worm rotating at 1400 r.p.m. and to obtain a speed reduction of 12 : 1. The distance between the shafts is 225 mm. φ Solution. Given : = 20° ; P = 10 kW = 10 000 W ; NW = 1400 r.p.m. ; V.R.= 12 ; x = 225 mm The worm and gear is designed as discussed below : 1. Design of worm
Let lN = Normal lead, and λ = Lead angle. 1114 A Textbook of Machine Design
Worm gear of a steering mechanism in an automobile.
We have discussed in Art. 31.12 that the value of x / lN will be minimum corresponding to cot3 λ = V. R . = 12 or cot λ = 2.29 ∴λ= 23.6° x 11⎛⎞+ VR .. We know that = πλ⎜⎟ λ lN 2sincos⎝⎠ 225 11⎛⎞ 121 +=+=(2.5 13.1) 2.5 = π°⎜⎟ °π lN 2⎝⎠ sin 23.6 cos 23.6 2 ∴ lN = 225 / 2.5 = 90 mm λ and axial lead, l = lN / cos = 90 / cos 23.6° = 98.2 mm From Table 31.2, we find that for a velocity ratio of 12, the number of starts or threads on the worm,
n = TW = 4 ∴ Axial pitch of the threads on the worm,
pa = l / 4 = 98.2 / 4 = 24.55 mm ∴ π π m = pa / = 24.55 / = 7.8 mm Let us take the standard value of module, m = 8 mm ∴ Axial pitch of the threads on the worm, π pa = m = p × 8 = 25.136 mm Ans. Axial lead of the threads on the worm,
l = pa . n = 25.136 × 4 = 100.544 mm Ans. and normal lead of the threads on the worm, λ lN = l cos = 100.544 cos 23.6° = 92 mm Ans. We know that the centre distance,
lN ⎛⎞⎛1..921+=VR + 12 ⎞ x = ⎜⎟⎜ ⎟ 2πλ⎝⎠⎝ sin cos λ 2 π sin 23.6 ° cos 23.6 ° ⎠ = 14.64 (2.5 + 13.1) = 230 mm Ans.
Let DW = Pitch circle diameter of the worm. λ l We know that tan = π DW l 100.544 ∴ D = ==73.24 mm Ans. W πλπtan tan 23.6 ° Worm Gears 1115
Since the velocity ratio is 12 and the worm has quadruple threads (i.e. n = TW = 4), therefore number of teeth on the worm gear,
TG = 12 × 4 = 48 From Table 31.3, we find that the face length of the worm or the length of threaded portion is
LW = pc (4.5 + 0.02 TW) ...( p = p ) = 25.136 (4.5 + 0.02 × 4) = 115 mm Q c a This length should be increased by 25 to 30 mm for the feed marks produced by the vibrating grinding wheel as it leaves the thread root. Therefore let us take
LW = 140 mm Ans. We know that depth of tooth,
h = 0.623 pc = 0.623 × 25.136 = 15.66 mm Ans. ...(From Table 31.3) and addendum, a = 0.286 pc = 0.286 × 25.136 = 7.2 mm Ans. ∴ Outside diameter of worm,
DOW = DW + 2a = 73.24 + 2 × 7.2 = 87.64 mm Ans. 2. Design of worm gear We know that pitch circle diameter of the worm gear,
DG = m . TG = 8 × 48 = 384 mm = 0.384 m Ans. From Table 31.4, we find that outside diameter of worm gear,
DOG = DG + 0.8903 pc = 384 + 0.8903 × 25.136 = 406.4 mm Ans. Throat diameter,
DT = DG + 0.572 pc = 384 + 0.572 × 25.136 = 398.4 mm Ans. and face width, b = 2.15 pc + 5 mm = 2.15 × 25.136 + 5 = 59 mm Ans. Let us now check the designed worm gearing from the standpoint of tangential load, dynamic load, static load or endurance strength, wear load and heat dissipation. (a) Check for the tangential load
Let NG = Speed of the worm gear in r.p.m. We know that velocity ratio of the drive, NNWW==1400 = V. R .= orNG 116.7 r.p.m NVRG .. 12 ∴ Torque transmitted, P ××60 10 000 60 ==818.2 N-m T = ππ× 2NG 2 116.7 and tangential load acting on the gear, 2×× Torque 2 818.2 ==4260 N WT = DG 0.384 We know that pitch line or peripheral velocity of the worm gear, π ..DN π×0.384 × 116.7 v = GG==2.35 m/s 60 60 ∴ Velocity factor, 66 C = ==0.72 v 6++v 6 2.35 1116 A Textbook of Machine Design
Gears are usually enclosed in boxes to protect them from environmental pollution and provide them proper lubrication. and tooth form factor for 20° involute teeth, 0.912 0.912 y = 0.154−=−= 0.154 0.135 TG 48 Since the worm gear is generally made of phosphor bronze, therefore taking the allowable static σ 2 stress for phosphor bronze, o = 84 MPa or N/mm . We know that the designed tangential load, σ π π WT =( o . Cv) b. m . y = (84 × 0.72) 59 × × 8 × 0.135 N = 12 110 N Since this is more than the tangential load acting on the gear (i.e. 4260 N), therefore the design is safe from the standpoint of tangential load. (b) Check for dynamic load We know that the dynamic load,
WD = WT / Cv = 12 110 / 0.72 = 16 820 N
Since this is more than WT = 4260 N, therefore the design is safe from the standpoint of dynamic load. (c) Check for static load or endurance strength We know that the flexural endurance limit for phosphor bronze is σ 2 e = 168 MPa or N/mm ∴ Static load or endurance strength, σ π π WS = e . b. m . y = 168 × 59 × × 8 × 0.135 = 33 635 N
Since this is much more than WT = 4260 N, therefore the design is safe from the standpoint of static load or endurance strength. (d) Check for wear Assuming the material for worm as hardened steel, therefore from Table 31.5, we find that for hardened steel worm and phosphor bronze worm gear, the value of load stress factor, K = 0.55 N/mm2 Worm Gears 1117
∴ Limiting or maximum load for wear,
WW = DG . b . K = 384 × 59 × 0.55 = 12 461 N Since this is more than WT = 4260 N, therefore the design is safe from the standpoint of wear. (e) Check for heat dissipation First of all, let us find the efficiency of the worm gearing (η). We know that rubbing velocity, π π× × DNWW. ==0.07324 1400 v = 351.6 m / min r cosλ° cos 23.6
...(DW is taken in metres) ∴ Coefficient of friction, v 351.6 μ 0.025+=+=r 0.025 0.0445 = 18 000 18 000 ∴ ...( vr is greater than 180 m/min) φ –1 μ –1 and angle of friction, 1 = tan = tan (0.0445) = 2.548° We know that efficiency, tanλ° tan 23.6 0.4369 η ===0.89 or 89% = λ+φ + tan (1 ) tan (23.6 2.548) 0.4909 Assuming 25 per cent overload, heat generated, η Qg = 1.25 P (1 – ) = 1.25 × 10 000 (1 – 0.89) = 1375 W We know that projected area of the worm, ππ A = (D )222== (73.24) 4214 mm W 44W and projected area of the worm gear, ππ A = (D )22== (384) 115 827 mm 2 G 44G ∴ Total projected area of worm and worm gear, 2 A = AW + AG = 4214 + 115 827 = 120 041 mm = 120 041 × 10–6 m2 We know that heat dissipating capacity, –6 Qd = A (t2 – t1) K = 120 041 × 10 (t2 – t1) 378 = 45.4 (t2 – t1) The heat generated must be dissipated in order to avoid over heating of the drive, therefore equating Qg = Qd, we have t2 – t1 = 1375 / 45.4 = 30.3°C Since this temperature difference (t2 – t1) is within safe limits of 27 to 38°C, therefore the design is safe from the standpoint of heat. 3. Design of worm shaft
Let dW = Diameter of worm shaft. We know that torque acting on the worm gear shaft, 1.25P ××× 60 1.25 10000 60 ==1023 N-m Tgear = ππ× 2NG 2 116.7 = 1023 × 103 N-mm ...(Taking 25% overload) ∴ Torque acting on the worm shaft,
Tgear 1023 3 T = ===×96 N-m 96 10 N-mm worm VR. .×η 12 × 0.89 1118 A Textbook of Machine Design
Differential inside an automobile. We know that tangential force on the worm,
WT = Axial force on the worm gear 2 × T 29610××3 = worm ==2622 N DW 73.24 Axial force on the worm,
WA = Tangential force on the worm gear × 3 2 Tgear 2×× 1023 10 = ==5328 N DG 384 and radial or separating force on the worm
WR = Radial or separating force on the worm gear φ = WA . tan = 5328 × tan 20° = 1940 N
Let us take the distance between the bearings of the worm shaft (x1) equal to the diameter of the worm gear (DG), i.e.
x1 = DG = 384 mm ∴ Bending moment due to the radial force (WR ) in the vertical plane Wx× 1940× 384 = R1==186240 N-mm 44 and bending moment due to axial force (WA) in the vertical plane WD× 5328× 73.24 = AW==97556 N-mm 44 ∴ Total bending moment in the vertical plane,
M1 = 186 240 + 97 556 = 283 796 N-mm
We know that bending moment due to tangential force (WT) in the horizontal plane, WD× 2622× 384 M = TG==251 712 N-mm 2 44 ∴ Resultant bending moment on the worm shaft, 22+= 2 + 2 = Mworm = (MM12 ) ( ) (283 796) (251 712) 379340 N-mm Worm Gears 1119
We know that equivalent twisting moment on the worm shaft, 22+=×+ 322 Tew = (TMworm ) ( worm ) (96 10 ) (379 340) N-mm = 391 300 N-mm
We also know that equivalent twisting moment (Tew),
ππ333 391 300 = ×τ()ddd = × 50()9.82() = 16WWW 16 ...(Taking τ = 50 MPa or N/mm2) ∴ 3 (dW) = 391 300 / 9.82 = 39 850 or dW = 34.2 say 35 mm Ans. Let us now check the maximum shear stress induced. We know that the actual shear stress,
16 T 16× 391 300 2 τ = ew ==46.5 N/mm ππ33 ()dW (35) and direct compressive stress on the shaft due to the axial force,
W 5328 σ A ==5.54 N/mm2 c = ππ ()d 22 (35) 44W ∴ Maximum shear stress, 11 τ = (σ+τ= )22 4 (5.54) 2 + 4 (46.5) 2 = 46.6 MPa max 22c Since the maximum shear stress induced is less than 50 MPa (assumed), therefore the design of worm shaft is satisfactory. 4. Design of worm gear shaft
Let dG = Diameter of worm gear shaft. We have calculated above that the axial force on the worm gear = 2622 N Tangential force on the worm gear = 5328 N and radial or separating force on the worm gear = 1940 N We know that bending moment due to the axial force on the worm gear Axial force × D 2622× 384 = G ==251 712 N-mm 44 The bending moment due to the axial force will be in the vertical plane.
Let us take the distance between the bearings of the worm gear shaft (x2) as 250 mm. ∴ Bending moment due to the radial force on the worm gear Radial force × x 1940× 250 = 2 ==121 250 N-mm 44 The bending moment due to the radial force will also be in the vertical plane. ∴ Total bending moment in the vertical plane
M3 = 251 712 + 121 250 = 372 962 N-mm 1120 A Textbook of Machine Design
We know that the bending moment due to the tangential force in the horizontal plane Tangential force × x 5328× 250 M = 2 ==333 000 N-mm 4 44 ∴ Resultant bending moment on the worm gear shaft, 22+= 2 + 2 Mgear = (MM34 ) ( ) (372 962) (333 000) N-mm = 500 × 103 N-mm We have already calculated that the torque acting on the worm gear shaft, 3 Tgear = 1023 × 10 N-mm ∴ Equivalent twisting moment on the worm gear shaft, 2+=×+× 2 32 32 Teg = (TMgear ) ( gear ) (1023 10 ) (500 10 ) N-mm = 1.14 × 106 N-mm
We know that equivalent twisting moment (Teg), ππ333 1.14 × 106 = ×τ()ddd = × 50()9.82() = 16GGG 16 ∴ 3 6 3 (dG) = 1.14 × 10 / 9.82 = 109 × 10 or dG = 48.8 say 50 mm Ans. Let us now check the maximum shear stress induced. We know that actual shear stress, 6 16 Teg 16×× 1.14 10 2 τ = ===46.4 N/mm 46.4 MPa ππ33 ()dG (50) and direct compressive stress on the shaft due to the axial force, Axial force 2622 σ ==1.33 N/mm3 = 1.33 MPa c = ππ ()d 22 (50) 44G ∴ Maximum shear stress, 11 τ = (σ+τ= )22 4 (1.33) 2 + 4 (46.4) 2 = 46.4MPa max 22c Since the maximum shear stress induced is less than 50 MPa Speed change Oil collector (assumed), therefore the design for shift axis worm gear shaft is satisfactory. Bearing housing Planet Example 31.4. A speed output belt pulley gears reducer unit is to be designed for an input of 1.1 kW with a transmission ratio 27. The speed of Input sun the hardened steel worm is 1440 gear r.p.m. The worm wheel is to be made of phosphor bronze. The tooth form is to be 20° involute.
Solution. Given : P = 1.1 kW Motor = 1100 W ; V.R. = 27 ; NW = 1440 flange r.p.m. ; φ = 20° Slide dog clutch A speed reducer unit (i.e., Output sun worm and worm gear) may be gear designed as discussed below. Sun and Planet gears. Worm Gears 1121
Since the centre distance between the shafts is not known, therefore let us assume that for this size unit, the centre distance (x) = 100 mm. We know that pitch circle diameter of the worm, ()x 0.875 (100) 0.875 D = ==39.7 say 40 mm W 1.416 1.416 ∴ Pitch circle diameter of the worm gear,
DG =2x – DW = 2 × 100 – 40 = 160 mm From Table 31.2, we find that for the transmission ratio of 27, we shall use double start worms. ∴ Number of teeth on the worm gear,
TG = 2 × 27 = 54
We know that the axial pitch of the threads on the worm (pa) is equal to circular pitch of teeth on the worm gear ( pc). π D π× ∴ ==G 160 = pa = pc 9.3 mm TG 54 p 9.3 c ==2.963 say 3 mm and module, m = ππ ∴ Actual circular pitch, π π pc = m = × 3 = 9.426 mm Actual pitch circle diameter of the worm gear, pT. 9.426× 54 D = c G ==162 mm Ans. G ππ and actual pitch circle diameter of the worm,
DW =2x – DG = 2 × 100 – 162 = 38 mm Ans. The face width of the worm gear (b) may be taken as 0.73 times the pitch circle diameter of worm (DW). ∴ b = 0.73 DW = 0.73 × 38 = 27.7 say 28 mm Let us now check the design from the standpoint of tangential load, dynamic load, static load or endurance strength, wear load and heat dissipation. 1. Check for the tangential load
Let NG = Speed of the worm gear in r.p.m. We know that velocity ratio of the drive, NNWW==1440 = V. R .= orNG 53.3 r.p.m NVRG .. 27 ∴ Peripheral velocity of the worm gear, π DN. π×0.162 × 53.3 v = GG==0.452 m/s 60 60
... (DG is taken in metres) 66 and velocity factor, C = ==0.93 v 660.452++v We know that for 20° involute teeth, the tooth form factor, 0.912 0.912 y = 0.154−=−= 0.154 0.137 TG 54 1122 A Textbook of Machine Design
From Table 31.4, we find that allowable static stress for phosphor bronze is σ 2 o = 84 MPa or N/mm ∴ Tangential load transmitted, σ π π WT =( o . Cv) b . m . y = (84 × 0.93) 28 × × 3 × 0.137 N = 2825 N and power transmitted due to the tangential load,
P = WT × v = 2825 × 0.452 = 1277 W = 1.277 kW Since this power is more than the given power to be transmitted (1.1 kW), therefore the design is safe from the standpoint of tangential load. 2. Check for the dynamic load We know that the dynamic load,
WD = WT / Cv = 2825 / 0.93 = 3038 N and power transmitted due to the dynamic laod,
P = WD × v = 3038 × 0.452 = 1373 W = 1.373 kW Since this power is more than the given power to be transmitted, therefore the design is safe from the standpoint of dynamic load. 3. Check for the static load or endurance strength From Table 31.8, we find that the flexural endurance limit for phosphor bronze is σ 2 e = 168 MPa or N/mm ∴ Static load or endurance strength, σ π π WS = e . b. m . y = 168 × 28 × × 3 × 0.137 = 6075 N and power transmitted due to the static load,
P = WS × v = 6075 × 0.452 = 2746 W = 2.746 kW Since this power is more than the power to be transmitted (1.1 kW), therefore the design is safe from the standpoint of static load. 4. Check for the wear load From Table 31.5, we find that the load stress factor for hardened steel worm and phosphor bronze worm gear is K = 0.55 N/mm2 ∴ Limiting or maximum load for wear,
WW = DG . b . K = 162 × 28 × 0.55 = 2495 N and power transmitted due to the wear load,
P = WW × v = 2495 × 0.452 = 1128 W = 1.128 kW Since this power is more than the given power to be transmitted (1.1 kW), therefore the design is safe from the standpoint of wear. 5. Check for the heat dissipation We know that permissible input power, 3650 (x )1.7 3650 (0.1) 1.7 P = ==2.27 kW ... (x is taken in metres) VR.5++ 275 Since this power is more than the given power to be transmitted (1.1 kW), therefore the design is safe from the standpoint of heat dissipation. Worm Gears 1123
EXERCISES
1. A double threaded worm drive is required for power transmission between two shafts having their axes at right angles to each other. The worm has 14½° involute teeth. The centre distance is approxi- mately 200 mm. If the axial pitch of the worm is 30 mm and lead angle is 23°, find 1. lead; 2. pitch circle diameters of worm and worm gear; 3. helix angle of the worm; and 4. efficiency of the drive if the coefficient of friction is 0.05. [Ans. 60 mm ; 45 mm ; 355 mm ; 67° ; 87.4%]
The worm in its place. One can also see the two cubic worm bearing blocks and the big gear. 2. A double threaded worm drive has an axial pitch of 25 mm and a pitch circle diameter of 70 mm. The torque on the worm gear shaft is 1400 N-m. The pitch circle diameter of the worm gear is 250 mm and the tooth pressure angle is 25°. Find : 1. tangential force on the worm gear, 2. torque on the worm shaft, 3. separating force on the worm, 4. velocity ratio, and 5. efficiency of the drive, if the coefficient of friction between the worm thread and gear teeth is 0.04. [Ans. 11.2 kN ; 88.97 N-m ; 5220 N ; 82.9%] 3. Design a speed reducer unit of worm and worm wheel for an input of 1 kW with a transmission ratio of 25. The speed of the worm is 1600 r.p.m. The worm is made of hardened steel and wheel of phosphor bronze for which the material combination factor is 0.7 N/mm2. The static stress for the wheel material is 56 MPa. The worm is made of double start and the centre distance between the axes of the worm and wheel is 120 mm. The tooth form is to be 14½° involute. Check the design for strength, wear and heat dissipation. 4. Design worm and gear speed reducer to transmit 22 kW at a speed of 1440 r.p.m. The desired velocity ratio is 24 : 1. An efficiency of atleast 85% is desired. Assume that the worm is made of hardened steel and the gear of phosphor bronze. QUESTIONS 1. Discuss, with neat sketches, the various types of worms and worm gears. 2. Define the following terms used in worm gearing : (a) Lead; (b) Lead angle; (c) Normal pitch; and (d) Helix angle. 3. What are the various forces acting on worm and worm gears ? 4. Write the expression for centre distance in terms of axial lead, lead angle and velocity ratio. 1124 A Textbook of Machine Design
OBJECTIVE TYPE QUESTIONS
1. The worm gears are widely used for transmitting power at ...... velocity ratios between non-inter- secting shafts. (a) high (b) low 2. In worm gears, the angle between the tangent to the thread helix on the pitch cylinder and the plane normal to the axis of worm is called (a) pressure angle (b) lead angle (c) helix angle (d) friction angle 3. The normal lead, in a worm having multiple start threads, is given by λ λ (a) lN = l / cos (b) lN = l . cos
(c) lN = l (d) lN = l tan
where lN = Normal lead, l = Lead, and λ = Lead angle. 4. The number of starts on the worm for a velocity ratio of 40 should be (a) single (b) double (c) triple (d) quadruple
5. The axial thrust on the worm (WA) is given by φ φ (a) WA = WT . tan (b) WA = WT / tan λ λ (c) WA = WT . tan (d) WA = WT / tan
where WT = Tangential force acting on the worm, φ = Pressure angle, and λ = Lead angle.
ANSWERS
1. (a) 2. (b) 3. (b) 4. (a) 5. (d) Internal Combustion Engine Parts 1125
C H A P T E R 32 Internal Combustion Engine Parts
1. Introduction. 2. Principal Parts of an I. C. Engine. 3. Cylinder and Cylinder Liner. 4. Design of a Cylinder. 5. Piston. 6. Design Considerations for a Piston. 7. Material for Pistons. 8. Piston Head or Crown . 9. Piston Rings. 10. Piston Barrel. 11. Piston skirt. 12. Piston Pin. 13. Connecting Rod. 14. Forces Acting on the Connecting Rod. 32.1 Introduction 15. Design of Connecting Rod. 16. Crankshaft. As the name implies, the internal combustion engines 17. Material and Manufacture (briefly written as I. C. engines) are those engines in which of Crankshafts. the combustion of fuel takes place inside the engine cylinder. 18. Bearing Pressures and The I.C. engines use either petrol or diesel as their fuel. In Stresses in Crankshafts. petrol engines (also called spark ignition engines or S.I 19. Design Procedure for engines), the correct proportion of air and petrol is mixed Crankshaft. in the carburettor and fed to engine cylinder where it is 20. Design for Centre ignited by means of a spark produced at the spark plug. In Crankshaft. diesel engines (also called compression ignition engines 21. Side or Overhung or C.I engines), only air is supplied to the engine cylinder Crankshaft. during suction stroke and it is compressed to a very high 22. Valve Gear Mechanism. 23. Valves. pressure, thereby raising its temperature from 600°C to 24. Rocker Arm. 1000°C. The desired quantity of fuel (diesel) is now injected into the engine cylinder in the form of a very fine spray and gets ignited when comes in contact with the hot air. The operating cycle of an I.C. engine may be completed either by the two strokes or four strokes of the 1125 1126 A Textbook of Machine Design piston. Thus, an engine which requires two strokes of the piston or one complete revolution of the crankshaft to complete the cycle, is known as two stroke engine. An engine which requires four strokes of the piston or two complete revolutions of the crankshaft to complete the cycle, is known as four stroke engine. The two stroke petrol engines are generally employed in very light vehicles such as scooters, motor cycles and three wheelers. The two stroke diesel engines are generally employed in marine propulsion. The four stroke petrol engines are generally employed in light vehicles such as cars, jeeps and also in aeroplanes. The four stroke diesel engines are generally employed in heavy duty vehicles such as buses, trucks, tractors, diesel locomotive and in the earth moving machinery. 32.2 Principal Parts of an Engine The principal parts of an I.C engine, as shown in Fig. 32.1 are as follows : 1. Cylinder and cylinder liner, 2. Piston, piston rings and piston pin or gudgeon pin, 3. Connecting rod with small and big end bearing, 4. Crank, crankshaft and crank pin, and 5. Valve gear mechanism. The design of the above mentioned principal parts are discussed, in detail, in the following pages.
Fig. 32.1. Internal combustion engine parts. 32.3 Cylinder and Cylinder Liner The function of a cylinder is to retain the working fluid and to guide the piston. The cylinders are usually made of cast iron or cast steel. Since the cylinder has to withstand high temperature due to the combustion of fuel, therefore, some arrangement must be provided to cool the cylinder. The single cylinder engines (such as scooters and motorcycles) are generally air cooled. They are provided with fins around the cylinder. The multi-cylinder engines (such as of cars) are provided with water jackets around the cylinders to cool it. In smaller engines. the cylinder, water jacket and the frame are Internal Combustion Engine Parts 1127 made as one piece, but for all the larger engines, these parts are manufactured separately. The cylinders are provided with cylinder liners so that in case of wear, they can be easily replaced. The cylinder liners are of the following two types : 1. Dry liner, and 2. Wet liner.
Fig. 32.2. Dry and wet liner. A cylinder liner which does not have any direct contact with the engine cooling water, is known as dry liner, as shown in Fig. 32.2 (a). A cylinder liner which have its outer surface in direct contact with the engine cooling water, is known as wet liner, as shown in Fig. 32.2 (b). The cylinder liners are made from good quality close grained cast iron (i.e. pearlitic cast iron), nickel cast iron, nickel chromium cast iron. In some cases, nickel chromium cast steel with molybdenum may be used. The inner surface of the liner should be properly heat-treated in order to obtain a hard surface to reduce wear. 32.4 Design of a Cylinder In designing a cylinder for an I. C. engine, it is required to determine the following values : 1. Thickness of the cylinder wall. The cylinder wall is subjected to gas pressure and the piston side thrust. The gas pressure produces the following two types of stresses : (a) Longitudinal stress, and (b) Circumferential stress.
Cam Piston ring seals the piston to Camshaft controls the prevent gases escaping valves
Air intake
Valve lets fuel and air in and exhaust gases out
Alternator Belt drives alternator to supply electricity to spark plugs.
Crankshaft turns the Dip stick to check oil level piston action into rotation Sump is filled with oil to reduce friction Oil is pumped up into Piston cylinders to lubricate pistons The above picture shows crankshaft, pistons and cylinder of a 4-stroke petrol engine. 1128 A Textbook of Machine Design
Since these two stressess act at right angles to each other, therefore, the net stress in each direction is reduced. The piston side thrust tends to bend the cylinder wall, but the stress in the wall due to side thrust is very small and hence it may be neglected.
Let D0 = Outside diameter of the cylinder in mm, D = Inside diameter of the cylinder in mm, p = Maximum pressure inside the engine cylinder in N/mm2, t = Thickness of the cylinder wall in mm, and 1/m = Poisson’s ratio. It is usually taken as 0.25. The apparent longitudinal stress is given by π ××2 ForceDp Dp2 . σ== 4 = l Area π 22− [(DD )22− ] ()DD0 4 0 and the apparent circumferential stresss is given by Force Dl×× p D × p σ= = = c Area 2tl× 2 t ... (where l is the length of the cylinder and area is the projected area) σ ∴ Net longitudinal stress = σ− c l m σ and net circumferential stress = σ− l c m The thickness of a cylinder wall (t) is usually obtained by using a thin cylindrical formula,i.e., × pD+ t = σ C 2 c where p = Maximum pressure inside the cylinder in N/mm2, D = Inside diameter of the cylinder or cylinder bore in mm, σ c = Permissible circumferential or hoop stress for the cylinder material in MPa or N/mm2. Its value may be taken from 35 MPa to 100 MPa depending upon the size and material of the cylinder. C = Allowance for reboring. The allowance for reboring (C ) depending upon the cylinder bore (D) for I. C. engines is given in the following table : Table 32.1. Allowance for reboring for I. C. engine cylinders.
D (mm) 75 100 150 200 250 300 350 400 450 500 C (mm) 1.5 2.4 4.0 6.3 8.0 9.5 11.0 12.5 12.5 12.5 The thickness of the cylinder wall usually varies from 4.5 mm to 25 mm or more depending upon the size of the cylinder. The thickness of the cylinder wall (t) may also be obtained from the following empirical relation, i.e. t = 0.045 D + 1.6 mm The other empirical relations are as follows : Thickness of the dry liner = 0.03 D to 0.035 D Internal Combustion Engine Parts 1129
Thickness of the water jacket wall = 0.032 D + 1.6 mm or t /3 m for bigger cylinders and 3t /4 for smaller cylinders Water space between the outer cylinder wall and inner jacket wall = 10 mm for a 75 mm cylinder to 75 mm for a 750 mm cylinder or 0.08 D + 6.5 mm 2. Bore and length of the cylinder. The bore (i.e. inner diameter) and length of the cylinder may be determined as discussed below : 2 Let pm = Indicated mean effective pressure in N/mm , D = Cylinder bore in mm, A = Cross-sectional area of the cylinder in mm2, = π D2/4 l = Length of stroke in metres, N = Speed of the engine in r.p.m., and n = Number of working strokes per min = N, for two stroke engine = N/2, for four stroke engine. We know that the power produced inside the engine cylinder, i.e. indicated power, plAn×× × IP..= m watts 60 From this expression, the bore (D) and length of stroke (l) is determined. The length of stroke is generally taken as 1.25 D to 2D. Since there is a clearance on both sides of the cylinder, therefore length of the cylinder is taken as 15 percent greater than the length of stroke. In other words, Length of the cylinder, L = 1.15 × Length of stroke = 1.15 l η Notes : (a) If the power developed at the crankshaft, i.e. brake power (B. P.) and the mechanical efficiency ( m) of the engine is known, then BP.. I.P. = η m
(b) The maximum gas pressure ( p ) may be taken as 9 to 10 times the mean effective pressure ( pm). 3. Cylinder flange and studs. The cylinders are cast integral with the upper half of the crank- case or they are attached to the crankcase by means of a flange with studs or bolts and nuts. The cylinder flange is integral with the cylinder and should be made thicker than the cylinder wall. The flange thickness should be taken as 1.2 t to 1.4 t, where t is the thickness of cylinder wall. The diameter of the studs or bolts may be obtained by equating the gas load due to the maximum pressure in the cylinder to the resisting force offered by all the studs or bolts. Mathematically, π π × Dp2 . = nd×σ()2 4 sct4 where D = Cylinder bore in mm, p = Maximum pressure in N/mm2,
ns = Number of studs. It may be taken as 0.01 D + 4 to 0.02 D + 4
dc = Core or minor diameter, i.e. diameter at the root of the thread in mm, 1130 A Textbook of Machine Design σ t = Allowable tensile stress for the material of studs or bolts in MPa or N/mm2. It may be taken as 35 to 70 MPa.
The nominal or major diameter of the stud or bolt (d ) usually lies between 0.75 tf to tf, where tf is the thickness of flange. In no case, a stud or bolt less than 16 mm diameter should be used. The distance of the flange from the centre of the hole for the stud or bolt should not be less than d + 6 mm and not more than 1.5 d, where d is the nominal diameter of the stud or bolt. In order to make a leak proof joint, the pitch of the studs or bolts should lie between 19 d to 28.5d , where d is in mm. 4. Cylinder head. Usually, a separate cylinder head or cover is provided with most of the engines. It is, usually, made of box type section of considerable depth to accommodate ports for air and gas passages, inlet valve, exhaust valve and spark plug (in case of petrol engines) or atomiser at the centre of the cover (in case of diesel engines).
The cylinder head may be approximately taken as a flat circular plate whose thickness (th) may be determined from the following relation : Cp. D th = σ c where D = Cylinder bore in mm, p = Maximum pressure inside the cylinder in N/mm2, σ 2 c = Allowable circumferential stress in MPa or N/mm . It may be taken as 30 to 50 MPa, and C = Constant whose value is taken as 0.1. The studs or bolts are screwed up tightly alongwith a metal gasket or asbestos packing to provide a leak proof joint between the cylinder and cylinder head. The tightness of the joint also depends upon the pitch of the bolts or studs, which should lie between 19 d to 28.5d . The pitch circle diameter (Dp) is usually taken as D + 3d. The studs or bolts are designed in the same way as discussed above. Example 32.1. A four stroke diesel engine has the following specifications : Brake power = 5 kW ; Speed = 1200 r.p.m. ; Indicated mean effective pressure = 0.35 N / mm 2 ; Mechanical efficiency = 80 %. Determine : 1. bore and length of the cylinder ; 2. thickness of the cylinder head ; and 3. size of studs for the cylinder head.
Ignition 4-Stroke Petrol Engine system Hot gases expand and force Terminal Inlet causes a Spark plug Piston the piston down valve spark Exhaust Ceramic Crankshaft valve insulator Spark plug casing Central electrode Screw fitting Earth electrode 1. Intake 2. Compression 3. Power 4. Exhaust, 5. Spark plug Internal Combustion Engine Parts 1131
Solution. Given: B.P. = 5kW = 5000 W ; N = 1200 r.p.m. or n = N / 2 = 600 ; 2 η pm = 0.35 N/mm ; m = 80% = 0.8 1. Bore and length of cylinder Let D = Bore of the cylinder in mm, π A = Cross-sectional area of the cylinder = × D22mm 4 l=Length of the stroke in m. = 1.5 D mm = 1.5 D / 1000 m ....(Assume) We know that the indicated power, η I.P = B.P. / m = 5000 / 0.8 = 6250 W We also know that the indicated power (I.P.), plAn.. . 0.35××π× 1.5DD2 600 6250 = m = = 4.12 × 10–3 D3 60 60×× 1000 4 ( ... Q For four stroke engine, n = N/2) ∴ D3 = 6250 / 4.12 × 10–3 = 1517 × 103 or D = 115 mm Ans. and l = 1.5 D = 1.5 × 115 = 172.5 mm Taking a clearance on both sides of the cylinder equal to 15% of the stroke, therefore length of the cylinder, L = 1.15 l = 1.15 × 172.5 = 198 say 200 mm Ans. 2. Thickness of the cylinder head Since the maximum pressure ( p) in the engine cylinder is taken as 9 to 10 times the mean effective pressure ( pm), therefore let us take 2 p =9pm = 9 × 0.35 = 3.15 N/mm We know that thickness of the cyclinder head, Cp. 0.1× 3.15 D = 115 th = σ = 9.96 say 10 mm Ans. t 42 σ 2 ...(Taking C = 0.1 and t = 42 MPa = 42 N/mm ) 3. Size of studs for the cylinder head Let d = Nominal diameter of the stud in mm,
dc = Core diameter of the stud in mm. It is usually taken as 0.84 d. σ t = Tensile stress for the material of the stud which is usually nickel steel.
ns = Number of studs. We know that the force acting on the cylinder head (or on the studs) ππ = ××=Dp22(115) 3.15 = 32 702 N ...(i) 44 The number of studs (ns ) are usually taken between 0.01 D + 4 (i.e. 0.01 × 115 + 4 = 5.15) and 0.02 D + 4 (i.e. 0.02 × 115 + 4 = 6.3). Let us take ns = 6. We know that resisting force offered by all the studs ππ = nd×σ=×( )222 6 (0.84 d ) 65 = 216 d N ...(ii) sct44 σ 2 ...(Taking t = 65 MPa = 65 N/mm ) From equations (i) and (ii), d 2 = 32 702 / 216 = 151 or d = 12.3 say 14 mm 1132 A Textbook of Machine Design
The pitch circle diameter of the studs (Dp) is taken D + 3d. ∴ Dp = 115 + 3 × 14 = 157 mm We know that pitch of the studs π× D p π×157 = ==82.2 mm ns 6 We know that for a leak-proof joint, the pitch of the studs should lie between 19 d to 28.5d , where d is the nominal diameter of the stud. ∴ Minimum pitch of the studs = 19 d = 19 14 = 71.1 mm and maximum pitch of the studs = 28.5d == 28.5 14 106.6mm Since the pitch of the studs obtained above (i.e. 82.2 mm) lies within 71.1 mm and 106.6 mm, therefore, size of the stud (d ) calculated above is satisfactory. ∴ d=14 mm Ans. 32.5 Piston The piston is a disc which reciprocates within a cylinder. It is either moved by the fluid or it moves the fluid which enters the cylinder. The main function of the piston of an internal combustion engine is to receive the impulse from the expanding gas and to transmit the energy to the crankshaft through the connecting rod. The piston must also disperse a large amount of heat from the combustion chamber to the cylinder walls.
Fig. 32.3. Piston for I.C. engines (Trunk type). Internal Combustion Engine Parts 1133
The piston of internal combustion engines are usually of trunk type as shown in Fig. 32.3. Such pistons are open at one end and consists of the following parts : 1. Head or crown. The piston head or crown may be flat, convex or concave depending upon the design of combustion chamber. It withstands the pressure of gas in the cylinder. 2. Piston rings. The piston rings are used to seal the cyliner in order to prevent leakage of the gas past the piston. 3. Skirt. The skirt acts as a bearing for the side thrust of the connecting rod on the walls of cylinder. 4. Piston pin. It is also called gudgeon pin or wrist pin. It is used to connect the piston to the connecting rod. 32.6 Design Considerations for a Piston In designing a piston for I.C. engine, the following points should be taken into consideration : 1. It should have enormous strength to withstand the high gas pressure and inertia forces. 2. It should have minimum mass to minimise the inertia forces. 3. It should form an effective gas and oil sealing of the cylinder. 4. It should provide sufficient bearing area to prevent undue wear. 5. It should disprese the heat of combustion quickly to the cylinder walls. 6. It should have high speed reciprocation without noise. 7. It should be of sufficient rigid construction to withstand thermal and mechanical distortion. 8. It should have sufficient support for the piston pin. 32.7 Material for Pistons The most commonly used materials for pistons of I.C. engines are cast iron, cast aluminium, forged aluminium, cast steel and forged steel. The cast iron pistons are used for moderately rated
Spark plug
Carburettor Twin-cylinder aeroplane engine
Cylinder head
Propeller 2. Side view
1. Front view
Twin cylinder airplane engine of 1930s. 1134 A Textbook of Machine Design engines with piston speeds below 6 m / s and aluminium alloy pistons are used for highly rated en- gines running at higher piston sppeds. It may be noted that 1. Since the *coefficient of thermal expansion for aluminium is about 2.5 times that of cast iron, therefore, a greater clearance must be provided between the piston and the cylinder wall (than with cast iron piston) in order to prevent siezing of the piston when engine runs continuously under heavy loads. But if excessive clearance is allowed, then the piston will develop ‘piston slap’ while it is cold and this tendency increases with wear. The less clearance between the piston and the cylinder wall will lead to siezing of piston. 2. Since the aluminium alloys used for pistons have high **heat conductivity (nearly four times that of cast iron), therefore, these pistons ensure high rate of heat transfer and thus keeps down the maximum temperature difference between the centre and edges of the piston head or crown.
Notes: (a) For a cast iron piston, the temperature at the centre of the piston head (TC) is about 425°C to 450°C under full load conditions and the temperature at the edges of the piston head (TE) is about 200°C to 225°C.
(b) For aluminium alloy pistons, TC is about 260°C to 290°C and TE is about 185°C to 215°C. 3. Since the aluminium alloys are about ***three times lighter than cast iron, therfore, its mechanical strength is good at low temperatures, but they lose their strength (about 50%) at temperatures above 325°C. Sometimes, the pistons of aluminium alloys are coated with aluminium oxide by an electrical method. 32.8 Piston Head or Crown The piston head or crown is designed keeping in view the following two main considerations, i.e. 1. It should have adequate strength to withstand the straining action due to pressure of explosion inside the engine cylinder, and 2. It should dissipate the heat of combustion to the cylinder walls as quickly as possible. On the basis of first consideration of straining action, the thickness of the piston head is determined by treating it as a flat circular plate of uniform thickness, fixed at the outer edges and subjected to a uniformly distributed load due to the gas pressure over the entire cross-section.
The thickness of the piston head (tH ), according to Grashoff’s formula is given by 3.pD2 tH = σ (in mm) ...(i) 16 t where p = Maximum gas pressure or explosion pressure in N/mm2, D = Cylinder bore or outside diameter of the piston in mm, and σ t = Permissible bending (tensile) stress for the material of the piston in MPa or N/mm2. It may be taken as 35 to 40 MPa for grey cast iron, 50 to 90 MPa for nickel cast iron and aluminium alloy and 60 to 100 MPa for forged steel. On the basis of second consideration of heat transfer, the thickness of the piston head should be such that the heat absorbed by the piston due combustion of fuel is quickly transferred to the cylinder walls. Treating the piston head as a flat ciucular plate, its thickness is given by H t = − (in mm) ...(ii) H 12.56kT (CE T )
* The coefficient of thermal expansion for aluminium is 0.24 × 10–6 m / °C and for cast iron it is 0.1 × 10–6 m / °C. ** The heat conductivity for aluminium is 174.75 W/m/°C and for cast iron it is 46.6 W/m /°C. *** The density of aluminium is 2700 kg / m3 and for cast iron it is 7200 kg / m3. Internal Combustion Engine Parts 1135
where H = Heat flowing through the piston head in kJ/s or watts, k=Heat conductivity factor in W/m/°C. Its value is 46.6 W/m/°C for grey cast iron, 51.25 W/m/°C for steel and 174.75 W/m/°C for aluminium alloys.
TC = Temperture at the centre of the piston head in °C, and
TE = Temperature at the edges of the piston head in °C.
The temperature difference (TC – TE) may be taken as 220°C for cast iron and 75°C for aluminium. The heat flowing through the positon head (H) may be deternined by the following expression, i.e., H = C × HCV × m × B.P. (in kW) where C = Constant representing that portion of the heat supplied to the engine which is absorbed by the piston. Its value is usually taken as 0.05. HCV = Higher calorific value of the fuel in kJ/kg. It may be taken as 45 × 103 kJ/kg for diesel and 47 × 103 kJ/ kg for petrol, m = Mass of the fuel used in kg per brake power per second, and B.P. = Brake power of the engine per cylinder
Notes : 1. The thickness of the piston head (tH) is calculated by using equations (i) and (ii) and larger of the two values obtained should be adopted.
2. When tH is 6 mm or less, then no ribs are required to strengthen the piston head against gas loads. But when tH is greater then 6 mm, then a suitable number of ribs at the centre line of the boss extending around the skirt should be provided to distribute the side thrust from the connecting rod and thus to prevent distortion of the skirt. The thickness of the ribs may be takes as tH / 3 to tH /2. 3. For engines having length of stroke to cylinder bore (L / D) ratio upto 1.5, a cup is provided in the top of the piston head with a radius equal to 0.7 D. This is done to provide a space for combustion chamber. 32.9 Piston Rings The piston rings are used to impart the necessary radial pressure to maintain the seal between the piston and the cylinder bore. These are usually made of grey cast iron or alloy cast iron because of their good wearing properties and also they retain spring characteristics even at high temperatures. The piston rings are of the following two types : 1. Compression rings or pressure rings, and 2. Oil control rings or oil scraper. The compression rings or pressure rings are inserted in the grooves at the top portion of the piston and may be three to seven in number. These rings also transfer heat from the piston to the cylinder liner and absorb some part of the piston fluctuation due to the side thrust. The oil control rings or oil scrapers are provided below the compression rings. These rings provide proper lubrication to the liner by allowing sufficient oil to move up during upward stroke and at the same time scraps the lubricating oil from the surface of the liner in order to minimise the flow of the oil to the combustion chamber. The compression rings are usually made of rectangular cross-section and the diameter of the ring is slightly larger than the cylinder bore. A part of the ring is cut- off in order to permit it to go into the cylinder against the liner wall. The diagonal cut or step cut ends, as shown in Fig. 32.4 (a) and (b) respectively, may be used. The gap between the ends should be sufficiently large when the ring is put cold so that even at the highest temperature, the ends do not touch each other when the ring expands, otherwise there might be buckling of the ring. 1136 A Textbook of Machine Design
Fig. 32.4. Piston rings.
The radial thickness (t1) of the ring may be obtained by considering the radial pressure between the cylinder wall and the ring. From bending stress consideration in the ring, the radial thickness is given by 3p D w t1 = σ t where D = Cylinder bore in mm, 2 pw = Pressure of gas on the cylinder wall in N/mm . Its value is limited from 0.025 N/mm2 to 0.042 N/mm2, and σ t = Allowable bending (tensile) stress in MPa. Its value may be taken from 85 MPa to 110 MPa for cast iron rings.
The axial thickness (t2 ) of the rings may be taken as 0.7 t1 to t1.
The minimum axial thickness (t2 ) may also be obtained from the following empirical relation: D t = 2 10nR where nR = Number of rings. The width of the top land (i.e. the distance from the top of the piston to the first ring groove) is made larger than other ring lands to protect the top ring from high temperature conditions existing at the top of the piston, ∴ Width of top land,
b1 = tH to 1.2 tH The width of other ring lands (i.e. the distance between the ring grooves) in the piston may be made equal to or slightly less than the axial thickness of the ring (t2). ∴ Width of other ring lands,
b2 = 0.75 t2 to t2 The depth of the ring grooves should be more than the depth of the ring so that the ring does not take any piston side thrust.
The gap between the free ends of the ring is given by 3.5 t1 to 4 t1. The gap, when the ring is in the cylinder, should be 0.002 D to 0.004 D. 32.10 Piston Barrel
It is a cylindrical portion of the piston. The maximum thickness (t3) of the piston barrel may be obtained from the following empirical relation :
t3 = 0.03 D + b + 4.5 mm Internal Combustion Engine Parts 1137 where b = Radial depth of piston ring groove which is taken as 0.4 mm larger
than the radial thickness of the piston ring (t1)
= t1 + 0.4 mm Thus, the above relation may be written as
t3 = 0.03 D + t1 + 4.9 mm
The piston wall thickness (t4) towards the open end is decreased and should be taken as 0.25 t3 to 0.35 t3. 32.11 Piston Skirt The portion of the piston below the ring section is known as piston skirt. In acts as a bearing for the side thrust of the connecting rod. The length of the piston skirt should be such that the bearing pressure on the piston barrel due to the side thrust does not exceed 0.25 N/mm2 of the projected area for low speed engines and 0.5 N/mm2 for high speed engines. It may be noted that the maximum thrust will be during the expansion stroke. The side thrust (R) on the cylinder liner is usually taken as 1/10 of the maximum gas load on the piston. Cylinder head Valve Cooling fins above piston Push rod
Magneto Ignition leads
Exhaust pipe
Ignition timing
Gearbox Spur gear on end of crankshaft 1000 cc twin -cylinder motorcycle engine.
We know that maximum gas load on the piston, πD2 P = p × 4 ∴ Maximum side thrust on the cylinder, πD2 R=P/10 = 0.1 p × ...(i) 4 where p = Maximum gas pressure in N/mm2, and D=Cylinder bore in mm. The side thrust (R) is also given by R=Bearing pressure × Projected bearing area of the piston skirt
= pb × D × l where l = Length of the piston skirt in mm. ...(ii) 1138 A Textbook of Machine Design
From equations (i) and (ii), the length of the piston skirt (l) is determined. In actual practice, the length of the piston skirt is taken as 0.65 to 0.8 times the cylinder bore. Now the total length of the piston (L) is given by L = Length of skirt + Length of ring section + Top land The length of the piston usually varies between D and 1.5 D. It may be noted that a longer piston provides better bearing surface for quiet running of the engine, but it should not be made unnecessarily long as it will increase its own mass and thus the inertia forces. 32.12 Piston Pin The piston pin (also called gudgeon pin or wrist pin) is used to connect the piston and the connecting rod. It is usually made hollow and tapered on the inside, the smallest Fig.32.5. Piston pin. inside diameter being at the centre of the pin, as shown in Fig. 32.5. The piston pin passes through the bosses provided on the inside of the piston skirt and the bush of the small end of the connecting rod. The centre of piston pin should be 0.02 D to 0.04 D above the centre of the skirt, in order to off-set the turning effect of the friction and to obtain uniform distribution of pressure between the piston and the cylinder liner. The material used for the piston pin is usually case hardened steel alloy containing nickel, chromium, molybdenum or vanadium having tensile strength from 710 MPa to 910 MPa.
Fig. 32.6. Full floating type piston pin. The connection between the piston pin and the small end of the connecting rod may be made either full floating type or semi-floating type. In the full floating type, the piston pin is free to turn both in the *piston bosses and the bush of the small end of the connecting rod. The end movements of the piston pin should be secured by means of spring circlips, as shown in Fig. 32.6, in order to prevent the pin from touching and scoring the cylinder liner. In the semi-floating type, the piston pin is either free to turn in the piston bosses and rigidly secured to the small end of the connecting rod, or it is free to turn in the bush of the small end of the connecting rod and is rigidly secured in the piston bosses by means of a screw, as shown in Fig. 32.7 The piston pin should be designed for the maximum gas load or the inertia force of the piston, whichever is larger. The bearing area of the piston pin should be about equally divided between the piston pin bosses and the connecting rod bushing. Thus, the length of the pin in the connecting rod bushing will be about 0.45 of the cylinder bore or piston diameter (D), allowing for the end clearance
* The mean diameter of the piston bosses is made 1.4 d0 for cast iron pistons and 1.5 d0 for aluminium pistons, where d0 is the outside diameter of the piston pin. The piston bosses are usually tapered, increasing the diameter towards the piston wall. Internal Combustion Engine Parts 1139 of the pin etc. The outside diameter of the piston pin (d0 ) is determined by equating the load on the piston due to gas pressure (p) and the load on the piston pin due to bearing pressure ( pb1 ) at the small end of the connecting rod bushing.
(a) Piston pin secured to the small (b) Piston pin secured to the boss end of the connecting rod. of the piston. Fig. 32.7. Semi-floating type piston pin.
Let d0 = Outside diameter of the piston pin in mm
l1 = Length of the piston pin in the bush of the small end of the connecting rod in mm. Its value is usually taken as 0.45 D.
pb1 = Bearing pressure at the small end of the connecting rod bushing in N/mm2. Its value for the bronze bushing may be taken as 25 N/mm2. We know that load on the piston due to gas pressure or gas load π D 2 = × p ..(i) 4 and load on the piston pin due to bearing pressure or bearing load
= Bearing pressure × Bearing area = pb1 × d0 × l1 ...(ii)
From equations (i) and (ii), the outside diameter of the piston pin (d0) may be obtained. The piston pin may be checked in bending by assuming the gas load to be uniformly distributed over the length l1 with supports at the centre of the bosses at the two ends. From Fig. 32.8, we find that the length between the supports, Dl−+ l D l = l +=11 2 1 22 Now maximum bending moment at the centre of the pin,
PPlll211 M = ×−×× 22l1 24 PPll = ×−×21 2224 + Fig. 32.8 PP⎛⎞lD11−× l = ⎜⎟ 222⎝⎠× 24 Pl..PD.. Pl PD = 11+−= ...(iii) 8888 1140 A Textbook of Machine Design
We have already discussed that the piston pin is made hollow. Let d0 and di be the outside and inside diameters of the piston pin. We know that the section modulus, π 44− ⎡⎤()dd0 ()i Z= ⎢⎥ 32 ⎣⎦⎢⎥d0 We know that maximum bending moment, π ⎡⎤44− ×σ =()dd0 ()i σ M = Z bb⎢⎥ 32 ⎣⎦⎢⎥d0 σ where b = Allowable bending stress for the material of the piston pin. It is usually taken as 84 MPa for case hardened carbon steel and 140 MPa for heat treated alloy steel.
Assuming di = 0.6 d0, the induced bending stress in the piston pin may be checked.
Spring Cam chain is driven by crankshaft Camshaft is pushed round by chain Waste gases out
Valve
Carburettor Spark plug Air in
Piston
Fuel flows in when Crankshaft needle rises Fuel in Another view of a single cylinder 4-stroke petrol engine. Example 32.2. Design a cast iron piston for a single acting four stroke engine for the following data: Cylinder bore = 100 mm ; Stroke = 125 mm ; Maximum gas pressure = 5 N/mm2 ; Indicated mean effective pressure = 0.75 N/mm2 ; Mechanical efficiency = 80% ; Fuel consumption = 0.15 kg per brake power per hour ; Higher calorific value of fuel = 42 × 103 kJ/kg ; Speed = 2000 r.p.m. Any other data required for the design may be assumed. 2 2 Solution. Given : D = 100 mm ; L = 125 mm = 0.125 m ; p = 5 N/mm ; pm = 0.75 N/mm ; η –6 3 m = 80% = 0.8 ; m = 0.15 kg / BP / h = 41.7 × 10 kg / BP / s; HCV = 42 × 10 kJ / kg ; N = 2000 r.p.m. The dimensions for various components of the piston are determined as follows : 1. Piston head or crown The thickness of the piston head or crown is determined on the basis of strength as well as on the basis of heat dissipation and the larger of the two values is adopted. Internal Combustion Engine Parts 1141
We know that the thickness of piston head on the basis of strength, 22× 3.pD = 3 5(100) tH = σ× = 15.7 say 16 mm 16t 16 38 σ 2 ...(Taking t for cast iron = 38 MPa = 38 N/mm ) Since the engine is a four stroke engine, therefore, the number of working strokes per minute, n = N / 2 = 2000 / 2 = 1000 and cross-sectional area of the cylinder, ππD 22(1 00 ) A = ==7855 mm2 44 We know that indicated power, pLAn... 0.75××× 0.125 7855 1000 IP = m ==12 270 W 60 60 = 12.27 kW ∴ η η Brake power, BP = IP × m = 12.27 × 0.8 = 9.8 kW ...(Q m = BP / IP) We know that the heat flowing through the piston head, H = C × HCV × m × BP = 0.05 × 42 × 103 × 41.7 × 10–6 × 9.8 = 0.86 kW = 860 W ....(Taking C = 0.05) ∴Thickness of the piston head on the basis of heat dissipation, H 860 ===0.0067 m 6.7 mm tH = −×× 12.56kT (CE T ) 12.56 46.6 220 ...( For cast iron , k = 46.6 W/m/°C, and T – T = 220°C) Q C E Taking the larger of the two values, we shall adopt
tH = 16 mm Ans. Since the ratio of L/D is 1.25, therefore a cup in the top of the piston head with a radius equal to 0.7 D (i.e. 70 mm) is provided. 2. Radial ribs
The radial ribs may be four in number. The thickness of the ribs varies from tH / 3 to tH /2. ∴ Thickness of the ribs, tR = 16 / 3 to 16 / 2 = 5.33 to 8 mm
Let us adopt tR = 7 mm Ans. 3. Piston rings
Let us assume that there are total four rings (i.e. nr = 4) out of which three are compression rings and one is an oil ring. We know that the radial thickness of the piston rings, 3 p 30.035× t = w ==mm 1 D σ 100 3.4 t 90 2 σ ...(Taking pw = 0.035 N/mm , and t = 90 MPa) and axial thickness of the piston rings
t2 = 0.7 t1 to t1 = 0.7 × 3.4 to 3.4 mm = 2.38 to 3.4 mm
Let us adopt t2 =3 mm 1142 A Textbook of Machine Design
We also know that the minimum axial thickness of the pistion ring, D 100 t = ==2.5 mm 2 × 10nr 10 4 Thus the axial thickness of the piston ring as already calculated (i.e. t2 = 3 mm)is satisfactory. Ans. The distance from the top of the piston to the first ring groove, i.e. the width of the top land,
b1 = tH to 1.2 tH = 16 to 1.2 × 16 mm = 16 to 19.2 mm and width of other ring lands,
b2 = 0.75 t2 to t2 = 0.75 × 3 to 3 mm = 2.25 to 3 mm
Let us adopt b1 = 18 mm ; and b2 = 2.5 mm Ans. We know that the gap between the free ends of the ring,
G1 = 3.5 t1 to 4 t1 = 3.5 × 3.4 to 4 × 3.4 mm = 11.9 to 13.6 mm and the gap when the ring is in the cylinder,
G2 = 0.002 D to 0.004 D = 0.002 × 100 to 0.004 × 100 mm = 0.2 to 0.4 mm
Let us adopt G1 = 12.8 mm ; and G2 = 0.3 mm Ans. 4. Piston barrel Since the radial depth of the piston ring grooves (b) is about 0.4 mm more than the radial thickness of the piston rings (t1), therefore,
b = t1 + 0.4 = 3.4 + 0.4 = 3.8 mm We know that the maximum thickness of barrel,
t3 = 0.03 D + b + 4.5 mm = 0.03 × 100 + 3.8 + 4.5 = 11.3 mm and piston wall thickness towards the open end,
t4 = 0.25 t3 to 0.35 t3 = 0.25 × 11.3 to 0.35 × 11.3 = 2.8 to 3.9 mm
Let us adopt t4 = 3.4 mm 5. Piston skirt Let l = Length of the skirt in mm. We know that the maximum side thrust on the cylinder due to gas pressure ( p ), ππD22(100) R=μ× ×p =0.1 × × 5 = 3928 N 44 ...(Taking μ = 0.1)
We also know that the side thrust due to bearing pressure on the piston barrel ( pb ), R = pb × D × l = 0.45 × 100 × l = 45 l N 2 ...(Taking pb = 0.45 N/mm ) From above, we find that 45 l = 3928 or l = 3928 / 45 = 87.3 say 90 mm Ans. ∴ Total length of the piston , L = Length of the skirt + Length of the ring section + Top land
= l + (4 t2 + 3b2) + b1 = 90 + (4 × 3 + 3 × 3) + 18 = 129 say 130 mm Ans. 6. Piston pin
Let d0 = Outside diameter of the pin in mm, l1 = Length of pin in the bush of the small end of the connecting rod in mm, and Internal Combustion Engine Parts 1143
pb1 = Bearing pressure at the small end of the connecting rod bushing in N/mm2. It value for bronze bushing is taken as 25 N/mm2. We know that load on the pin due to bearing pressure
= Bearing pressure × Bearing area = pb1 × d0 × l1 = 25 × d0 × 0.45 × 100 = 1125 d0 N ...(Taking l1 = 0.45 D) We also know that maximum load on the piston due to gas pressure or maximum gas load ππD22(100) =×=×=p 5 39 275 N 44 From above, we find that
1125 d0 = 39 275 or d0 = 39 275 / 1125 = 34.9 say 35 mm Ans.
The inside diameter of the pin (di) is usually taken as 0.6 d0. ∴ di = 0.6 × 35 = 21 mm Ans. σ Let the piston pin be made of heat treated alloy steel for which the bending stress ( b )may be taken as 140 MPa. Now let us check the induced bending stress in the pin. We know that maximum bending moment at the centre of the pin, PD. 39 275× 100 M = ==×491 103 N-mm 88 We also know that maximum bending moment (M), ππ−⎡⎤()dd44− () ⎡⎤(35)44 (21) 3 0 i σ= σ= σ 491 × 10 = ⎢⎥bbb⎢⎥3664 32⎣⎦⎢⎥d0 32⎣⎦ 35 ∴ σ 3 2 b = 491 × 10 / 3664 = 134 N/mm or MPa Since the induced bending stress in the pin is less than the permissible value of 140 MPa (i.e. 2 140 N/mm ), therefore, the dimensions for the pin as calculated above (i.e. d0 = 35 mm and di = 21 mm) are satisfactory.
Fan blows air over engine to cool it
Air filter stops dust and dirt from being sucked into engine
Driveshaft
Disk brake Spark plug Twin rotors
German engineer Fleix Wankel (1902-88) built a rotary engine in 1957. A triangular piston turns inside a chamber through the combustion cycle. 1144 A Textbook of Machine Design 32.13 Connecting Rod The connecting rod is the intermediate member between the piston and the crankshaft. Its primary function is to transmit the push and pull from the piston pin to the crankpin and thus convert the reciprocating motion of the piston into the rotary motion of the crank. The usual form of the connecting rod in internal combustion engines is shown in Fig. 32.9. It consists of a long shank, a small end and a big end. The cross-section of the shank may be rectangular, circular, tubular, I-section or H-section. Generally circular section is used for low speed engines while I-section is preferred for high speed engines.
Fig. 32.9. Connecting rod. The *length of the connecting rod ( l ) depends upon the ratio of l / r, where r is the radius of crank. It may be noted that the smaller length will decrease the ratio l / r. This increases the angularity of the connecting rod which increases the side thrust of the piston against the cylinder liner which in turn increases the wear of the liner. The larger length of the connecting rod will increase the ratio l / r. This decreases the angularity of the connecting rod and thus decreases the side thrust and the resulting wear of the cylinder. But the larger length of the connecting rod increases the overall height of the engine. Hence, a compromise is made and the ratio l / r is generally kept as 4 to 5. The small end of the connecting rod is usually made in the form of an eye and is provided with a bush of phosphor bronze. It is connected to the piston by means of a piston pin. The big end of the connecting rod is usually made split (in two **halves) so that it can be mounted easily on the crankpin bearing shells. The split cap is fastened to the big end with two cap bolts. The bearing shells of the big end are made of steel, brass or bronze with a thin lining (about 0.75 mm) of white metal or babbit metal. The wear of the big end bearing is allowed for by inserting thin metallic strips (known as shims) about 0.04 mm thick between the cap and the fixed half of the connecting rod. As the wear takes place, one or more strips are removed and the bearing is trued up. * It is the distance between the centres of small end and big end of the connecting rod. ** One half is fixed with the connecting rod and the other half (known as cap) is fastened with two cap bolts. Internal Combustion Engine Parts 1145
The connecting rods are usually manufactured by drop forging process and it should have adequate strength, stiffness and minimum weight. The material mostly used for connecting rods varies from mild carbon steels (having 0.35 to 0.45 percent carbon) to alloy steels (chrome-nickel or chrome- molybdenum steels). The carbon steel having 0.35 percent carbon has an ultimate tensile strength of about 650 MPa when properly heat treated and a carbon steel with 0.45 percent carbon has a ultimate tensile strength of 750 MPa. These steels are used for connecting rods of industrial engines. The alloy steels have an ultimate tensile strength of about 1050 MPa and are used for connecting rods of aeroengines and automobile engines. The bearings at the two ends of the connecting rod are either splash lubricated or pressure lubricated. The big end bearing is usually splash lubricated while the small end bearing is pressure lubricated. In the splash lubrication system, the cap at the big end is provided with a dipper or spout and set at an angle in such a way that when the connecting rod moves downward, the spout will dip into the lubricating oil contained in the sump. The oil is forced up the spout and then to the big end bearing. Now when the connecting rod moves upward, a splash of oil is produced by the spout. This splashed up lubricant find its way into the small end bearing through the widely chamfered holes provided on the upper surface of the small end. In the pressure lubricating system, the lubricating oil is fed under pressure to the big end bearing through the holes drilled in crankshaft, crankwebs and crank pin. From the big end bearing, the oil is fed to small end bearing through a fine hole drilled in the shank of the connecting rod. In some cases, the small end bearing is lubricated by the oil scrapped from the walls of the cyinder liner by the oil scraper rings. 32.14 Forces Acting on the Connecting Rod The various forces acting on the connecting rod are as follows : 1. Force on the piston due to gas pressure and inertia of the reciprocating parts, 2. Force due to inertia of the connecting rod or inertia bending forces, 3. Force due to friction of the piston rings and of the piston, and 4. Force due to friction of the piston pin bearing and the crankpin bearing. We shall now derive the expressions for the forces acting on a vertical engine, as discussed below. 1. Force on the piston due to gas pressure and inertia of reciprocating parts Consider a connecting rod PC as shown in Fig. 32.10.
Fig. 32.10. Forces on the connecting rod. 1146 A Textbook of Machine Design
Spark Compressed plug Rotor fuel-air mix
Fuel-air mix in
Exhaust out 1. Induction : turning 2. Compression: Fuel-air 3. Ignition: Compressed 4. Exhuast : the rotor rotor sucks in mixture of mixture is compressed as fuel-air mixture is ignited continues to turn and petrol and air. rotor carriers it round. by the spark plug. pushed out waste gases.
Let p = Maximum pressure of gas, D = Diameter of piston, π D2 A = Cross-section area of piston = , 4 mR = Mass of reciprocating parts, 1 = Mass of piston, gudgeon pin etc. + rd mass of connecting rod, 3 ω = Angular speed of crank, φ = Angle of inclination of the connecting rod with the line of stroke, θ = Angle of inclination of the crank from top dead centre, r = Radius of crank, l = Length of connecting rod, and n = Ratio of length of connecting rod to radius of crank = l / r. We know that the force on the piston due to pressure of gas, π 2 FL = Pressure × Area = p . A = p × D /4 and inertia force of reciprocating parts, θ ω2 ⎛⎞θ+cos 2 FI = Mass × *Acceleration = mR . . r ⎜⎟cos ⎝⎠n It may be noted that the inertia force of reciprocating parts opposes the force on the piston when it moves during its downward stroke (i. e. when the piston moves from the top dead centre to bottom dead centre). On the other hand, the inertia force of the reciprocating parts helps the force on the piston when it moves from the bottom dead centre to top dead centre. ∴ Net force acting on the piston or piston pin (or gudgeon pin or wrist pin),
FP = Force due to gas pressure m Inertia force
= FL m FI The –ve sign is used when piston moves from TDC to BDC and +ve sign is used when piston moves from BDC to TDC.
When weight of the reciprocating parts (WR = mR . g) is to be taken into consideration, then
FP = FL m F1 ± WR
⎛⎞cos 2θ * Acceleration of reciprocating parts = ωθ+2 ·cosr ⎜⎟ ⎝⎠n Internal Combustion Engine Parts 1147
The force FP gives rise to a force FC in the connecting rod and a thrust FN on the sides of the cylinder walls. From Fig. 32.10, we see that force in the connecting rod at any instant, * FFPP= F = φ C cos sin 2 θ 1 − n 2 The force in the connecting rod will be maximum when the crank and the connecting rod are perpendicular to each other (i.e. when θ = 90°). But at this position, the gas pressure would be de- creased considerably. Thus, for all practical purposes, the force in the connecting rod (FC) is taken equal to the maximum force on the piston due to pressure of gas (FL), neglecting piston inertia effects. 2. Force due to inertia of the connecting rod or inertia bending forces Consider a connecting rod PC and a crank OC rotating with uniform angular velocity ω rad / s. In order to find the accleration of various points on the connecting rod, draw the Klien’s acceleration diagram CQNO as shown in Fig. 32.11 (a). CO represents the acceleration of C towards O and NO represents the accleration of P towards O. The acceleration of other points such as D, E, F and G etc., on the connecting rod PC may be found by drawing horizontal lines from these points to intresect CN at d, e, f, and g respectively. Now dO, eO, fO and gO respresents the acceleration of D, E, F and G all towards O. The inertia force acting on each point will be as follows: Inertia force at C = m × ω2 × CO Inertia force at D = m × ω2 × dO Inertia force at E = m × ω2 × eO, and so on.
Fig. 32.11. Inertia bending forces. The inertia forces will be opposite to the direction of acceleration or centrifugal forces. The inertia forces can be resolved into two components, one parallel to the connecting rod and the other perpendicular to rod. The parallel (or longitudinal) components adds up algebraically to the force * For derivation, please refer ot Authors’ popular book on ‘Theory of Machines’. 1148 A Textbook of Machine Design acting on the connecting rod (FC) and produces thrust on the pins. The perpendicular (or transverse) components produces bending action (also called whipping action) and the stress induced in the connecting rod is called whipping stress. It may be noted that the perpendicular components will be maximum, when the crank and connecting rod are at right angles to each other. The variation of the inertia force on the connecting rod is linear and is like a simply supported beam of variable loading as shown in Fig. 32.11 (b) and (c). Assuming that the connecting rod is of uniform cross-section and has mass m1 kg per unit length, therefore, Inertia force per unit length at the crankpin ω2 = m1 × r and inertia force per unit length at the piston pin = 0 Inertia force due to small element of length dx at a distance x from the piston pin P, x dF = m × ω2r × × dx 1 1 l ∴ Resultant inertia force,
l l xxmr×ω2 ⎡⎤2 F = mr×ω2 × × dx = 1 I ∫ 1 ll⎢⎥2 0 ⎣⎦0 ml. m = 1 ×ω22rr = ×ω ...(Substituting m · l = m) 22 1 This resultant inertia force acts at a distance of 2l / 3 from the piston pin P. 1 Since it has been assumed that rd mass of the connecting rod is concentrated at piston pin P 3 2 (i.e. small end of connecting rod) and rd at the crankpin (i.e. big end of connecting rod), therefore, 3 the reaction at these two ends will be in the same proportion. i.e. 12 R = FR,and = F P 33ICI
Honeycomb To exhaust coated with metal catalysts
Co Carbon monoxide
CO2 Carbon dioxide
NOx Nitrous oxides HC Hydrocarbons
N2 Nitrogen H O Water Metal casing 2
Emissions of an automobile. Internal Combustion Engine Parts 1149
Now the bending moment acting on the rod at section X – X at a distance x from P, xx1 M = Rxm×−* ×ω××2 r x × X P1 l 23 3 1 ×−ml1. ×ω×2 x = FxI r 323l 2 ...(Multiplying and dividing the latter expression by l) Fx× xx33 F⎛⎞ II−×Fx = − = I 22⎜⎟ ...(i) 333ll⎝⎠
For maximum bending moment, differentiate MX with respect to x and equate to zero, i.e. d F ⎡⎤3x2 MX = 0 or I 10−= ⎢⎥2 dx 3 ⎣⎦l 3 x 2 l ∴ 1 − = 0 or 3orxl22== x l 2 3 Maximum bending moment, 3 ⎡⎤⎛⎞l ⎢⎥⎜⎟ M = FI ⎢⎥l − ⎝⎠3 ...[From equation (i)] max ⎢⎥2 3 ⎣⎦3 l ×× FFlFlIII⎡⎤ll−= ×=2 2 = ⎢⎥ 33⎣⎦333 33 93 ml l = 2 × ×ω×22rmr = ×ω× 2 93 93 and the maximum bending stress, due to inertia of the connecting rod, M σ = max max Z where Z = Section modulus. From above we see that the maximum bending moment varies as the square of speed, therefore, the bending stress due to high speed will be dangerous. It may be noted that the maximum axial force and the maximum bending stress do not occur simultaneously. In an I.C. engine, the maximum gas load occurs close to top dead centre whereas the maximum bending stress occurs when the crank angle θ = 65° to 70° from top dead centre. The pressure of gas falls suddenly as the piston moves from dead centre. Thus the general practice is to design a connecting rod by assuming the force in the connecting rod (FC) equal to the maximum force due to pressure (FL), neglecting piston inertia effects and then checked for bending stress due to inertia force (i.e. whipping stress). 3. Force due to friction of piston rings and of the piston The frictional force ( F ) of the piston rings may be determined by using the following expression : π μ F = D · tR · nR · pR · where D = Cylinder bore,
tR = Axial width of rings,
⎛⎞x * B.M. due to variable force from ⎜⎟0tomrω×2 is equal to the area of triangle multiplied by the distance ⎝⎠1 l ⎛⎞x of C.G. from XXie− ⎜⎟.. . ⎝⎠3 1150 A Textbook of Machine Design
nR = Number of rings, 2 pR = Pressure of rings (0.025 to 0.04 N/mm ), and μ = Coefficient of friction (about 0.1). Since the frictional force of the piston rings is usually very small, therefore, it may be neglected. The friction of the piston is produced by the normal component of the piston pressure which varies from 3 to 10 percent of the piston pressure. If the coefficient of friction is about 0.05 to 0.06, then the frictional force due to piston will be about 0.5 to 0.6 of the piston pressure, which is very low. Thus, the frictional force due to piston is also neglected. 4. Force due to friction of the piston pin bearing and crankpin bearing The force due to friction of the piston pin bearing and crankpin bearing, is to bend the connecting rod and to increase the compressive stress on the connecting rod due to the direct load. Thus, the maximum compressive stress in the connecting rod will be σ c (max) = Direct compressive stress + Maximum bending or whipping stress due to inertia bending stress 32.15 Design of Connecting Rod In designing a connecting rod, the following dimensions are required to be determined : 1. Dimensions of cross-section of the connecting rod, 2. Dimensions of the crankpin at the big end and the piston pin at the small end, 3. Size of bolts for securing the big end cap, and 4. Thickness of the big end cap. The procedure adopted in determining the above mentioned dimensions is discussed as below :
Evaporators change liquid hydrogen to gas Fuel tank Engine
This experimental car burns hydrogen fuel in an ordinary piston engine. Its exhaust gases cause no pollution, because they contain only water vapour. Internal Combustion Engine Parts 1151
1. Dimensions of cross-section of the connecting rod A connecting rod is a machine member which is subjected to alternating direct compressive and tensile forces. Since the compressive forces are much higher than the tensile forces, therefore, the cross-section of the connecting rod is designed as a strut and the Rankine’s formula is used. A connecting rod, as shown in Fig. 32.12, subjected to an axial load W may buckle with X-axis as neutral axis (i.e. in the plane of motion of the connecting rod) or Y-axis as neutral axis (i.e. in the plane perpendicular to the plane of motion). The connecting rod is considered like both ends hinged for buckling about X-axis and both ends fixed for buckling about Y-axis. A connecting rod should be equally strong in buckling about both the axes. Let A = Cross-sectional area of the connecting rod, l = Length of the connecting rod, σ c = Compressive yield stress, WB = Buckling load, Ixx and Iyy = Moment of inertia of the section about X-axis and Y-axis respectively, and
kxx and kyy = Radius of gyration of the section about X-axis and Y-axis respectively. According to Rankine’s formula, σσ..AA WXabout−= axis = cc B 22...( For both ends hinged, L = l) ⎛⎞Ll ⎛⎞ Q 11++aa⎜⎟ ⎜⎟ ⎝⎠kkxx ⎝⎠ xx σσ..AA WYabout−= axis cc = l and B 22...[Q For both ends fixed, L = ] ⎛⎞Ll ⎛ ⎞ 2 11++aa⎜⎟ ⎜ ⎟ ⎝⎠kkyy ⎝2 yy ⎠ where L = Equivalent length of the connecting rod, and a = Constant = 1 / 7500, for mild steel = 1 / 9000, for wrought iron = 1 / 1600, for cast iron
Fig. 32.12. Buckling of connecting rod. In order to thave a connecting rod equally strong in buckling about both the axes, the buckling loads must be equal, i.e. 2 2 σσ..AA⎛⎞ll⎛⎞ cc==or 22⎜⎟kk⎜⎟2 ⎛⎞l ⎛l ⎞ ⎝⎠xx⎝⎠ yy 1 + a 1 + a ⎜⎟k ⎜⎟ ⎝⎠xx ⎝⎠2kyy ∴ 22== ...( I = A · k2) kkxx4or4 yy I xx I yy Q 1152 A Textbook of Machine Design
This shows that the connecting rod is four times strong in buckling about Y-axis than about X-axis. If
Ixx > 4 Iyy, then buckling will occur about Y- axis and if Ixx < 4 Iyy, buckling will occur about X-axis. In actual practice, Ixx is kept slightly less than 4 Iyy. It is usually taken between 3 and 3.5 and the connecting rod is designed for bucking about X-axis. The design will always be satisfactory for buckling about Y-axis. The most suitable section for the connecting rod is I-section with the proportions as shown in Fig. 32.13 (a). Fig. 32.13. I-section of connecting rod. Let thickness of the flange and web of the section = t Width of the section, B = 4 t and depth or height of the section, H = 5t From Fig. 32.13 (a), we find that area of the section, A = 2 (4 t × t) + 3 t × t = 11 t2 Moment of inertia of the section about X-axis, 1 419 ⎡⎤4(5)tt334−= 3(3) tt t Ixx = 12⎣⎦ 12 and moment of inertia of the section about Y-axis, ⎡⎤1 1 131 2(4)(3)××tt334 + tt = t Iyy = ⎢⎥ ⎣⎦12 12 12 I 419 12 ∴ xx =×=3.2 I yy 12 131 I Since the value of xx lies between 3 and 3.5, therefore, I-section chosen is quite satisfactory. I yy After deciding the proportions for I-section of the connecting rod, its dimensions are determined by considering the buckling of the rod about X-axis (assuming both ends hinged) and applying the Rankine’s formula. We know that buckling load, σ c .A W = 2 B ⎛⎞L 1 + a⎜⎟ ⎝⎠kxx
The buckling load (WB) may be calculated by using the following relation, i.e.
WB = Max. gas force × Factor of safety The factor of safety may be taken as 5 to 6. Notes : (a) The I-section of the connecting rod is used due to its lightness and to keep the inertia forces as low as possible specially in case of high speed engines. It can also withstand high gas pressure. (b) Sometimes a connecting rod may have rectangular section. For slow speed engines, circular section may be used. (c) Since connecting rod is manufactured by forging, therefore the sharp corner of I-section are rounded off as shown in Fig. 32.13 (b) for easy removal of the section from dies. Internal Combustion Engine Parts 1153
The dimensions B = 4 t and H = 5 t, as obtained above by applying the Rankine’s formula, are at the middle of the connecting rod. The width of the section (B) is kept constant throughout the length of the connecting rod, but the depth or height varies. The depth near the small end (or piston end) is taken as H1 = 0.75 H to 0.9H and the depth near the big end (or crank end) is taken H2 = 1.1H to 1.25H. 2. Dimensions of the crankpin at the big end and the piston pin at the small end Since the dimensions of the crankpin at the big end and the piston pin (also known as gudgeon pin or wrist pin) at the small end are limited, therefore, fairly high bearing pressures have to be allowed at the bearings of these two pins. The crankpin at the big end has removable precision bearing shells of brass or bronze or steel with a thin lining (1 mm or less) of bearing metal (such as tin, lead, babbit, copper, lead) on the inner surface of the shell. The allowable bearing pressure on the crankpin depends upon many factors such as material of the bearing, viscosity of the lubricating oil, method of lubrication and the space limitations.The value of bearing pressure may be taken as 7 N/mm2 to 12.5 N/mm2 depending upon the material and method of lubrication used. Engine of a motorcyle. The piston pin bearing is usually a phosphor bronze bush of about 3 mm thickness and the allowable bearing pressure may be taken as 10.5 N/mm2 to 15 N/mm2. Since the maximum load to be carried by the crankpin and piston pin bearings is the maximum force in the connecting rod (FC), therefore the dimensions for these two pins are determined for the maximum force in the connecting rod (FC) which is taken equal to the maximum force on the piston due to gas pressure (FL) neglecting the inertia forces. We know that maximum gas force, π D2 F = × p ...(i) L 4 where D = Cylinder bore or piston diameter in mm, and p = Maximum gas pressure in N/mm2 Now the dimensions of the crankpin and piston pin are determined as discussed below :
Let dc = Diameter of the crank pin in mm, lc = Length of the crank pin in mm, 2 pbc = Allowable bearing pressure in N/mm , and dp,lp and pbp = Corresponding values for the piston pin, We know that load on the crank pin = Projected area × Bearing pressure
= dc . lc . pbc ...(ii) Similarly, load on the piston pin
= dp . lp . pbp ...(iii) 1154 A Textbook of Machine Design
Equating equations (i) and (ii), we have
FL = dc · lc · pbc Taking lc = 1.25 dc to 1.5 dc, the value of dc and lc are determined from the above expression. Again, equating equations (i) and (iii), we have
FL = dp · lp · pbp Taking lp = 1.5 dp to 2 dp, the value of dp and lp are determined from the above expression. 3. Size of bolts for securing the big end cap The bolts and the big end cap are subjected to tensile force which corresponds to the inertia force of the reciprocating parts at the top dead centre on the exhaust stroke. We know that inertia force of the reciprocating parts, θ ωθ+2 ⎛⎞cos 2 ∴ F = mrR ..cos⎜⎟ I ⎝⎠lr/ We also know that at the top dead centre, the angle of inclination of the crank with the line of stroke, θ = 0
=ω2 ⎛⎞ +r ∴ FmIR..1 r⎜⎟ ⎝⎠l where mR = Mass of the reciprocating parts in kg, ω = Angular speed of the engine in rad / s, r = Radius of the crank in metres, and l = Length of the connecting rod in metres. The bolts may be made of high carbon steel or nickel alloy steel. Since the bolts are under repeated stresses but not alternating stresses, therefore, a factor of safety may be taken as 6.
Let dcb = Core diameter of the bolt in mm, σ t = Allowable tensile stress for the material of the bolts in MPa, and
nb = Number of bolts. Generally two bolts are used. ∴ Force on the bolts π = ()dn2 σ× 4 cb t b Internal Combustion Engine Parts 1155
Equating the inertia force to the force on the bolts, we have π F = ()dn2 σ× I 4 cb t b From this expression, dcb is obtained. The nominal or major diameter (db) of the bolt is given by d d = cb b 0.84 4. Thickness of the big end cap
The thickness of the big end cap (tc) may be determined by treating the cap as a beam freely supported at the cap bolt centres and loaded by the inertia force at the top dead centre on the exhaust θ = 0 stroke (i.e. FI when ). This load is assumed to act in between the uniformly distributed load and the centrally concentrated load. Therefore, the maximum bending moment acting on the cap will be taken as Fx× M = * I C 6 where x = Distance between the bolt centres.
= Dia. of crankpin or big end bearing (dc) + 2 × Thickness of bearing liner (3 mm) + Clearance (3 mm)
Let bc = Width of the cap in mm. It is equal to the length of the crankpin or big end bearing (lc), and σ b = Allowable bending stress for the material of the cap in MPa. We know that section modulus for the cap, bt()2 Z = cc C 6 MFx××6 Fx ∴ σ C =×II = Bending stress, b = Z 6 22 C btcc() bt cc ()
From this expression, the value of tc is obtained. Note: The design of connecting rod should be checked for whipping stress (i.e. bending stress due to inertia force on the connecting rod). Example 32.3. Design a connecting rod for an I.C. engine running at 1800 r.p.m. and developing a maximum pressure of 3.15 N/mm2. The diameter of the piston is 100 mm ; mass of the reciprocating parts per cylinder 2.25 kg; length of connecting rod 380 mm; stroke of piston 190 mm and compression ratio 6 : 1. Take a factor of safety of 6 for the design. Take length to diameter ratio for big end bearing as 1.3 and small end bearing as 2 and the corresponding bearing pressures as 10 N/mm2 and 15 N/mm2. The density of material of the rod may be taken as 8000 kg/m3 and the allowable stress in the bolts as 60 N/mm2 and in cap as 80 N/mm2. The rod is to be of I-section for which you can choose your own proportions. Draw a neat dimensioned sketch showing provision for lubrication. Use Rankine formula for which the numerator constant may be taken as 320 N/mm2 and the denominator constant 1 / 7500.
* We know that the maximum bending moment for a simply or freely supported beam with a uniformly
distributed load of FI over a length x between the supports (In this case, x is the distance between the cap Fx× bolt centres) is I . When the load F is assumed to act at the centre of the freely supported beam, then 8 I Fx× the maximum bending moment is I . Thus the maximum bending moment in between these two bending 4 ⎛⎞Fx××× Fx Fx moments ⎜⎟ie..III and is . ⎝⎠846 1156 A Textbook of Machine Design
2 Solution. Given : N = 1800 r.p.m. ; p = 3.15 N/mm ; D = 100 mm ; mR = 2.25 kg ; l = 380 mm = 0.38 m ; Stroke = 190 mm ; *Compression ratio = 6 : 1 ; F. S . = 6. The connecting rod is designed as discussed below : 1. Dimension of I- section of the connecting rod Let us consider an I-section of the connecting rod, as shown in Fig. 32.14 (a), with the following proportions : Flange and web thickness of the section = t Width of the section, B = 4t and depth or height of the section, H= 5t First of all, let us find whether the section Fig. 32.14 chosen is satisfactory or not. We have already discussed that the connecting rod is considered like both ends hinged for buckling about X-axis and both ends fixed for buckling about Y-axis. The connecting rod should be equally strong in buckling about both the axes. We know that in order to have a connecting rod equally strong about both the axes,
Ixx = 4 Iyy where Ixx = Moment of inertia of the section about X-axis, and
Iyy = Moment of inertia of the section about Y-axis.
In actual practice, Ixx is kept slightly less than 4 Iyy. It is usually taken between 3 and 3.5 and the connecting rod is designed for buckling about X-axis. Now, for the section as shown in Fig. 32.14 (a), area of the section, A = 2 (4 t × t ) + 3t × t = 11 t2 1 419 I = ⎡⎤4(5)tt334−× 3 t (3) t = t xx 12⎣⎦ 12 1 1 131 and I = 2(4)3××tt334 +××= tt t yy 12 12 12 I 419 12 ∴ xx = ×=3.2 I yy 12 131 I Since xx = 3.2, therefore the section chosen in quite satisfactory. I yy Now let us find the dimensions of this I-section. Since the connecting rod is designed by taking the force on the connecting rod (FC) equal to the maximum force on the piston (FL) due to gas pressure, therefore, ππD 22(100) F = Fp=×=×=3.15 24 740 N C L 44 We know that the connecting rod is designed for buckling about X-axis (i.e. in the plane of motion of the connecting rod) assuming both ends hinged. Since a factor of safety is given as 6, therefore the buckling load,
WB = FC × F. S. = 24 740 × 6 = 148 440 N * Superfluous data Internal Combustion Engine Parts 1157
We know that radius of gyration of the section about X-axis, I 419t 4 1 k = xx =×=1.78 t xx A 12 11t 2 Length of crank, Stroke of piston 190 r = ==95 mm 22 Length of the connecting rod, l = 380 mm ...(Given) ∴ Equivalent length of the connecting rod for both ends hinged, L = l = 380 mm
Now according to Rankine’s formula, we know that buckling load (WB), σ × 2 c .A = 320 11 t 148 440 = 22 + ⎛⎞L + 1 ⎛ 380 ⎞ 1 a ⎜⎟1 ⎜⎟ ⎝⎠kxx 7500⎝⎠ 1.78 t σ 2 ... (It is given that c = 320 MPa or N/mm and a = 1 / 7500) 24 148 440== 11tt 11 320 6.1 2 + 1 + t 6.1 2 t 464 (t2 + 6.1) = 11 t4 or t 4 – 42.2 t 2 – 257.3 = 0 42.2±+× (42.2)2 4 257.3 ± ∴ t2 = = 42.2 53 = 47.6 22 ... (Taking +ve sign) or t = 6.9 say 7 mm Thus, the dimensions of I-section of the connecting rod are : Thickness of flange and web of the section = t = 7 mm Ans. Width of the section, B = 4 t = 4 × 7 = 28 mm Ans. and depth or height of the section, H = 5 t = 5 × 7 = 35 mm Ans.
Piston and connecting rod. 1158 A Textbook of Machine Design
These dimensions are at the middle of the connecting rod. The width (B) is kept constant through- out the length of the rod, but the depth (H) varies. The depth near the big end or crank end is kept as 1.1H to 1.25H and the depth near the small end or piston end is kept as 0.75H to 0.9H. Let us take Depth near the big end,
H1 = 1.2H = 1.2 × 35 = 42 mm and depth near the small end,
H2 = 0.85H = 0.85 × 35 = 29.75 say 30 mm ∴ Dimensions of the section near the big end = 42 mm × 28 mm Ans. and dimensions of the section near the small end = 30 mm × 28 mm Ans. Since the connecting rod is manufactured by forging, therefore the sharp corners of I-section are rounded off, as shown in Fig. 32.14 (b), for easy removal of the section from the dies. 2. Dimensions of the crankpin or the big end bearing and piston pin or small end bearing
Let dc = Diameter of the crankpin or big end bearing,
lc = length of the crankpin or big end bearing = 1.3 dc ...(Given) 2 pbc = Bearing pressure = 10 N/mm ...(Given) We know that load on the crankpin or big end bearing = Projected area × Bearing pressure 2 = dc .lc . pbc = dc × 1.3 dc × 10 = 13 (dc)
Since the crankpin or the big end bearing is designed for the maximum gas force (FL), therefore, equating the load on the crankpin or big end bearing to the maximum gas force, i.e. 2 13 (dc) = FL = 24 740 N ∴ 2 (dc ) = 24 740 / 13 = 1903 or dc = 43.6 say 44 mm Ans. and lc = 1.3 dc = 1.3 × 44 = 57.2 say 58 mm Ans. The big end has removable precision bearing shells of brass or bronze or steel with a thin lining (1mm or less) of bearing metal such as babbit.
Again, let dp = Diameter of the piston pin or small end bearing,
lp = Length of the piston pin or small end bearing = 2dp ...(Given) 2 pbp = Bearing pressure = 15 N/mm ..(Given) We know that the load on the piston pin or small end bearing = Project area × Bearing pressure 2 = dp . lp . pbp = dp × 2 dp × 15 = 30 (dp)
Since the piston pin or the small end bearing is designed for the maximum gas force (FL), therefore, equating the load on the piston pin or the small end bearing to the maximum gas force, i.e. 2 30 (dp) = 24 740 N ∴ 2 (dp) = 24 740 / 30 = 825 or dp = 28.7 say 29 mm Ans. and lp =2dp = 2 × 29 = 58 mm Ans. The small end bearing is usually a phosphor bronze bush of about 3 mm thickness. Internal Combustion Engine Parts 1159
3. Size of bolts for securing the big end cap
Let dcb = Core diameter of the bolts, σ t = Allowable tensile stress for the material of the bolts = 60 N/mm2 ...(Given) and nb = Number of bolts. Generally two bolts are used. We know that force on the bolts ππ = ()dnd22σ× = ()60294.26() × = d 2 44cb t b cb cb The bolts and the big end cap are subjected to tensile force which corresponds to the inertia force of the reciprocating parts at the top dead centre on the exhaust stroke. We know that inertia force of the reciprocating parts, θ 2 ⎛⎞cos 2 F = m . ω . r ⎜⎟cos θ+ I R ⎝⎠lr/ We also know that at top dead centre on the exhaust stroke, θ = 0. π× 2 ∴ ω2 ⎛⎞+=r ⎛21800 ⎞ ⎛ + 0.095 ⎞ FI = mR . . r ⎜⎟1 2.25 ⎜ ⎟ 0.095 ⎜ 1 ⎟ N ⎝⎠l ⎝60 ⎠ ⎝ 0.38 ⎠ = 9490 N Equating the inertia force to the force on the bolts, we have 2 2 9490 = 94.26 (dcb) or (dcb) = 9490 / 94.26 = 100.7 ∴ dcb = 10.03 mm and nominal diameter of the bolt, d 10.03 d ==cb =11.94 b 0.84 0.84 say 12 mm Ans. 4. Thickness of the big end cap
Let tc = Thickness of the big end cap,
bc = Width of the big end cap. It is taken equal to the length of the
crankpin or big end bearing (lc) = 58 mm (calculated above) σ b = Allowable bending stress for the material of the cap = 80 N/mm2 ...(Given) The big end cap is designed as a beam freely supported at the cap bolt centres and loaded by the inertia force at the top dead centre on the exhaust stroke (i.e. FI when θ = 0). Since the load is assumed to act in between the uniformly distributed load and the centrally concentrated load, therefore, maximum bending moment is taken as Fx× M = I C 6 where x = Distance between the bolt centres 1160 A Textbook of Machine Design
= Dia. of crank pin or big end bearing + 2 × Thickness of bearing liner + Nominal dia. of bolt + Clearance
= (dc + 2 × 3 + db + 3) mm = 44 + 6 + 12 + 3 = 65 mm ∴ Maximum bending moment acting on the cap,
Fx× 9490× 65 M = I ==102 810 N-mm C 66 Section modulus for the cap
bt()22 58() t Z = cc== c 9.7 (t )2 C 66 c σ We know that bending stress ( b ),
M 102 810 10 600 80 ==C = Z 22 C 9.7 (ttcc ) ( ) ∴ 2 (tc) = 10 600 / 80 = 132.5 or tc = 11.5 mm Ans. Let us now check the design for the induced bending stress due to inertia bending forces on the connecting rod (i.e. whipping stress). We know that mass of the connecting rod per metre length,
m1 = Volume × density = Area × length × density ρ 2 ρ 2 = A × l × = 11t × l × ...(Q A = 11t ) = 11(0.007)2 (0.38) 8000 = 1.64 kg ρ 3 ...[Q = 8 000 kg /m (given)] ∴ Maximum bending moment, l l2 M = m . ω2 . r × = m . ω2 . r × ...( m = m · l) max 93 1 93 Q 1
2 ⎛⎞2π× 1800 (0.38)2 = 1.64⎜⎟ (0.095)= 51.3 N-m ⎝⎠60 93 = 51 300 N-mm
I 419t4 2 and section modulus, Z ==xx ×=13.97 t3 =13.97 × 73 = 4792 mm3 xx 5/2tt 12 5 ∴ Maximum bending stress (induced ) due to inertia bending forces or whipping stress,
σ=Mmax =51 300 = 2 bmax() 10.7 N / mm Zxx 4792 Since the maximum bending stress induced is less than the allowable bending stress of 80 N/mm2, therefore the design is safe. Internal Combustion Engine Parts 1161 32.16 Crankshaft A crankshaft (i.e. a shaft with a crank) is used to convert reciprocating motion of the piston into rotatory motion or vice versa. The crankshaft consists of the shaft parts which revolve in the main bearings, the crankpins to which the big ends of the connecting rod are connected, the crank arms or webs (also called cheeks) which connect the crankpins and the shaft parts. The crankshaft, depending upon the position of crank, may be divided into the following two types : 1. Side crankshaft or overhung crankshaft, as shown in Fig. 32.15 (a), and 2. Centre crankshaft, as shown in Fig. 32. 15 (b).
Fig. 32.15. Types of crankshafts. The crankshaft, depending upon the number of cranks in the shaft, may also be classfied as single throw or multi-throw crankshafts. A crankhaft with only one side crank or centre crank is called a single throw crankshaft whereas the crankshaft with two side cranks, one on each end or with two or more centre cranks is known as multi-throw crankshaft. The side crankshafts are used for medium and large size horizontal engines. 32.17 Material and manufacture of Crankshafts The crankshafts are subjected to shock and fatigue loads. Thus material of the crankshaft should be tough and fatigue resistant. The crankshafts are generally made of carbon steel, special steel or special cast iron. In industrial engines, the crankshafts are commonly made from carbon steel such as 40 C 8, 55 C 8 and 60 C 4. In transport engines, manganese steel such as 20 Mn 2, 27 Mn 2 and 37 Mn 2 are generally used for the making of crankshaft. In aero engines, nickel chromium steel such as 35 Ni 1 Cr 60 and 40 Ni 2 Cr 1 Mo 28 are extensively used for the crankshaft. The crankshafts are made by drop forging or casting process but the former method is more common. The surface of the crankpin is hardened by case carburizing, nitriding or induction hardening. 32.18 Bearing Pressures and Stresses in Crankshaft The bearing pressures are very important in the design of crankshafts. The *maximum permissible bearing pressure depends upon the maximum gas pressure, journal velocity, amount and method of lubrication and change of direction of bearing pressure. The following two types of stresses are induced in the crankshaft. 1. Bending stress ; and 2. Shear stress due to torsional moment on the shaft. * The values of maximum permissible bearing pressures for different types of engines are given in Chapter 26, Table 26.3. 1162 A Textbook of Machine Design
Most crankshaft failures are caused by a progressive fracture due to repeated bending or reversed torsional stresses. Thus the crankshaft is under fatigue loading and, therefore, its design should be based upon endurance limit. Since the failure of a crankshaft is likely to cause a serious engine destruction and neither all the forces nor all the stresses acting on the crankshaft can be determined accurately, therefore a high factor of safety from 3 to 4, based on the endurance limit, is used. The following table shows the allowable bending and shear stresses for some commonly used materials for crankshafts : Table 32.2. Allowable bending and shear stresses.
Material Endurance limit in MPa Allowable stress in MPa Bending Shear Bending Shear Chrome nickel 525 290 130 to 175 72.5 to 97 Carbon steel and cast steel 225 124 56 to 75 31 to 42 Alloy cast iron 140 140 35 to 47 35 to 47
32.19 Design Procedure for Crankshaft The following procedure may be adopted for designing a crankshaft. 1. First of all, find the magnitude of the various loads on the crankshaft. 2. Determine the distances between the supports and their position with respect to the loads. 3. For the sake of simplicity and also for safety, the shaft is considered to be supported at the centres of the bearings and all the forces and reactions to be acting at these points. The distances between the supports depend on the length of the bearings, which in turn depend on the diameter of the shaft because of the allowable bearing pressures.
4. The thickness of the cheeks or webs is assumed to be from 0.4 ds to 0.6 ds, where ds is the diameter of the shaft. It may also be taken as 0.22D to 0.32 D, where D is the bore of cylinder in mm. 5. Now calculate the distances between the supports. 6. Assuming the allowable bending and shear stresses, determine the main dimensions of the crankshaft. Notes: 1. The crankshaft must be designed or checked for at least two crank positions. Firstly, when the crank- shaft is subjected to maximum bending moment and secondly when the crankshaft is subjected to maximum twisting moment or torque. 2. The additional moment due to weight of flywheel, belt tension and other forces must be considered. 3. It is assumed that the effect of bending moment does not exceed two bearings between which a force is considered. 32.20 Design of Centre Crankshaft We shall design the centre crankshaft by considering the two crank possitions, i.e. when the crank is at dead centre (or when the crankshaft is subjected to maximum bending moment) and when the crank is at angle at which the twisting moment is maximum. These two cases are discussed in detail as below : 1. When the crank is at dead centre. At this position of the crank, the maximum gas pressure on the piston will transmit maximum force on the crankpin in the plane of the crank causing only bending of the shaft. The crankpin as well as ends of the crankshaft will be only subjected to bending moment. Thus, when the crank is at the dead centre, the bending moment on the shaft is maximum and the twisting moment is zero. Internal Combustion Engine Parts 1163
Fig. 32.16. Centre crankshaft at dead centre. Consider a single throw three bearing crankshaft as shown in Fig. 32.16. Let D = Piston diameter or cylinder bore in mm, p = Maximum intensity of pressure on the piston in N/mm2, W = Weight of the flywheel acting downwards in N, and
*T1 + T2 = Resultant belt tension or pull acting horizontally in N. The thrust in the connecting rod will be equal to the gas load on the piston (FP ). We know that gas load on the piston,
π 2 F = ××Dp P 4
Due to this piston gas load (FP) acting horizontally, there will be two horizontal reactions H1 and H2 at bearings 1 and 2 respectively, such that Fb× Fb× H = P1; and H = P2 1 b 2 b Due to the weight of the flywheel (W) acting downwards, there will be two vertical reactions V2 and V3 at bearings 2 and 3 respectively, such that Wc× Wc× V = 1 ; and V = 2 2 c 3 c Now due to the resultant belt tension (T1 + T2), acting horizontally, there will be two horizontal ′ ′ reactions H2 and H3 at bearings 2 and 3 respectively, such that ()TTc+ ()TTc+ H ′= 121; and H ′= 122 2 c 3 c The resultant force at bearing 2 is given by =+′+22 RHHV2222()()
* T1 is the belt tension in the tight side and T2 is the belt tension in the slack side. 1164 A Textbook of Machine Design and the resultant force at bearing 3 is given by =+22 RH333() () V Now the various parts of the centre crankshaft are designed for bending only, as discussed below: (a) Design of crankpin
Let dc = Diameter of the crankpin in mm, lc = Length of the crankpin in mm, σ 2 b = Allowable bending stress for the crankpin in N/mm . We know that bending moment at the centre of the crankpin,
MC = H1 · b2 ...(i) We also know that π M = ()d 3 σ ...(ii) C 32 cb From equations (i) and (ii), diameter of the crankpin is determined. The length of the crankpin is given by
FP I = c dpcb. 2 where pb = Permissible bearing pressure in N/mm . (b) Design of left hand crank web The crank web is designed for eccentric loading. There will be two stresses acting on the crank web, one is direct compressive stress and the other is bending stress due to piston gas load (FP).
Water cooled 4-cycle diesel engine Internal Combustion Engine Parts 1165
The thickness (t) of the crank web is given empirically as
t = 0.4 ds to 0.6 ds = 0.22D to 0.32D
= 0.65 dc + 6.35 mm where ds = Shaft diameter in mm, D = Bore diameter in mm, and
dc = Crankpin diameter in mm, The width of crank web (w) is taken as
w = 1.125 dc + 12.7 mm We know that maximum bending moment on the crank web, =−−⎛⎞lc t MHb12⎜⎟ ⎝⎠22 1 and section modulus, Z =×wt. 2 6 ⎛⎞l t 6Hb−−c 12⎜⎟22 σ=M = ⎝⎠ ∴ Bending stress, b Z wt. 2 and direct compressive stress on the crank web, σ=H1 c wt. ∴ Total stress on the crank web σ+σ = Bending stress + Direct stress = bc ⎛⎞l t 6Hb−−c 12⎝⎠⎜⎟22 H = + 1 wt. 2 wt. This total stress should be less than the permissible bending stress. (c) Design of right hand crank web The dimensions of the right hand crank web (i.e. thickness and width) are made equal to left hand crank web from the balancing point of view. (d) Design of shaft under the flywheel
Let ds = Diameter of shaft in mm. We know that bending moment due to the weight of flywheel,
MW = V3 . c1 and bending moment due to belt tension, ′ MT = H3 . c1 These two bending moments act at right angles to each other. Therefore, the resultant bending moment at the flywheel location, =+=+22 2 2 MMSW()() M T (·)(·) VcHc 3131 ... (i) We also know that the bending moment at the shaft, π Md=σ()3 ... (ii) S 32 sb σ 2 where b = Allowable bending stress in N/mm .
From equations (i) and (ii), we may determine the shaft diameter (ds). 1166 A Textbook of Machine Design
2. When the crank is at an angle of maximum twisting moment The twisting moment on the crankshaft will be maximum when the tangential force on the crank
(FT) is maximum. The maximum value of tangential force lies when the crank is at angle of 25° to 30º from the dead centre for a constant volume combustion engines (i.e., petrol engines) and 30º to 40º for constant pressure combustion engines (i.e., diesel engines). Consider a position of the crank at an angle of maximum twisting moment as shown in Fig. 32.17 (a). If p′ is the intensity of pressure on the piston at this instant, then the piston gas load at this position of crank, π F = ××Dp2 ' P 4 and thrust on the connecting rod, F F = P Q cosφ where φ = Angle of inclination of the connecting rod with the line of stroke PO.
The *thrust in the connecting rod (FQ) may be divided into two components, one perpendicular to the crank and the other along the crank. The component of FQ perpendicular to the crank is the tangential force (FT) and the component of FQ along the crank is the radial force (FR) which produces thrust on the crankshaft bearings. From Fig. 32.17 (b), we find that
Fig. 32.17. (a) Crank at an angle of maximum twisting moment. (b) Forces acting on the crank.
( θ φ FT = FQ sin + ) θ φ and FR = FQ cos ( + ) It may be noted that the tangential force will cause twisting of the crankpin and shaft while the radial force will cause bending of the shaft.
* For further details, see Author’s popular book on ‘Theory of Machines’. Internal Combustion Engine Parts 1167
Due to the tangential force (FT), there will be two reactions at bearings 1 and 2, such that Fb× Fb× H = T1; and H = T2 T1 b T2 b Due to the radial force (FR), there will be two reactions at the bearings 1 and 2, such that Fb× Fb× H = R1; and H = R2 R1 b R2 b
Pull-start motor in an automobile
The reactions at the bearings 2 and 3, due to the flywheel weight (W) and resultant belt pull
(T1 + T2) will be same as discussed earlier. Now the various parts of the crankshaft are designed as discussed below : (a) Design of crankpin
Let dc = Diameter of the crankpin in mm. We know that bending moment at the centre of the crankpin,
MC = HR1 × b2 and twisting moment on the crankpin,
TC = HT1 × r ∴ Equivalent twisting moment on the crankpin, =+=×+×22 2 2 TMe ()CC () T ( HbHr R12T1 ) ( ) ...(i) We also know that twisting moment on the crankpin, π T = ()d 3 τ ...(ii) e 16 c where τ = Allowable shear stress in the crankpin. From equations (i) and (ii), the diameter of the crankpin is determined. 1168 A Textbook of Machine Design
(b) Design of shaft under the flywheel
Let ds = Diameter of the shaft in mm. We know that bending moment on the shaft,
MS = R3 × c1 and twisting moment on the shaft,
TS = FT × r ∴ Equivalent twisting moment on the shaft, 22+=×+× 2 2 Te = ()MTSS () ( RcFr 31T ) ( ) ... (i) We also know that equivalent twisting moment on the shaft, π T = ()d 3 τ ... (ii) e 16 s where τ = Allowable shear stress in the shaft. From equations (i) and (ii), the diameter of the shaft is determined. (c) Design of shaft at the juncture of right hand crank arm
Let ds1 = Diameter of the shaft at the juncture of right hand crank arm. We know that bending moment at the juncture of the right hand crank arm,
⎛⎞⎛⎞++llcctt − + M = Rb12⎜⎟⎜⎟ F Q S1 ⎝⎠⎝⎠22 22 and the twisting moment at the juncture of the right hand crank arm,
TS1 = FT × r ∴ Equivalent twisting moment at the juncture of the right hand crank arm, 22+ Te = ()()MTS1 S1 ... (i) We also know that equivalent twisting moment, π T = ()d 3 τ ... (ii) e 16 s1 where τ = Allowable shear stress in the shaft. From equations (i) and (ii), the diameter of the shaft at the juncture of the right hand crank arm is determined. (d) Design of right hand crank web The right hand crank web is subjected to the following stresses: (i) Bending stresses in two planes normal to each other, due to the radial and tangential
components of FQ,
(ii) Direct compressive stress due to FR, and (iii) Torsional stress.
The bending moment due to the radial component of FQ is given by, ⎛⎞l t M = Hb−−c ... (i) R R2⎝⎠⎜⎟ 1 22 1 We also know that M = σ×=σ××Z wt. 2 ... (ii) R bbRR6 Internal Combustion Engine Parts 1169 σ where bR = Bending stress in the radial direction, and 1 Z = Section modulus = × wt· 2 6 σ From equations (i) and (ii), the value of bending stress bR is determined.
The bending moment due to the tangential component of FQ is maximum at the juncture of crank and shaft. It is given by ⎡⎤− ds1 M = FrT ⎢⎥ ... (iii) T ⎣⎦2 where ds1 = Shaft diameter at juncture of right hand crank arm, i.e. at bearing 2. 1 We also know that M = σ×=σ××Z tw. 2 ... (iv) T bbTT6 σ where bT = Bending stress in tangential direction. σ From equations (iii) and (iv), the value of bending stress bT is determined. The direct compressive stress is given by, F σ = R d 2.wt σ The maximum compressive stress ( c) will occur at the upper left corner of the cross-section of the crank. ∴σσ σ σ c = bR + bT + d Now, the twisting moment on the arm, ⎛⎞ll ⎛⎞ l Hb+−×=cc F H b − c T = T1⎝⎠⎜⎟ 222 T T2 ⎝⎠⎜⎟ 1 2 We know that shear stress on the arm, TT4.5 τ = = 2 ZP wt. wt. 2 where Z = Polar section modulus = P 4.5 ∴ Maximum or total combined stress, σ 1 (σ ) = c +σ+τ()22 4 c max 22 c
Snow blower on a railway track 1170 A Textbook of Machine Design σ The value of ( c)max should be within safe limits. If it exceeds the safe value, then the dimension w may be increased because it does not affect other dimensions. (e) Design of left hand crank web Since the left hand crank web is not stressed to the extent as the right hand crank web, therefore, the dimensions for the left hand crank web may be made same as for right hand crank web. ( f ) Design of crankshaft bearings The bearing 2 is the most heavily loaded and should be checked for the safe bearing pressure. We know that the total reaction at the bearing 2, FTTW + R = P12++ 2 22 2
R2 ∴ Total bearing pressure = ld21. s where l2 = Length of bearing 2. 32.21 Side or Overhung Crankshaft The side or overhung crankshafts are used for medium size and large horizontal engines. Its main advantage is that it requires only two bearings in either the single or two crank construction. The design procedure for the side or overhung crankshaft is same as that for centre crankshaft. Let us now design the side crankshaft by considering the two crank positions, i.e. when the crank is at dead centre (or when the crankshaft is subjected to maximum bending moment) and when the crank is at an angle at which the twisting moment is maximum. These two cases are discussed in detail as below: 1. When the crank is at dead centre. Consider a side crankshaft at dead centre with its loads and distances of their application, as shown in Fig. 32.18.
Fig. 32.18. Side crankshaft at dead centre. Internal Combustion Engine Parts 1171
Let D = Piston diameter or cylinder bore in mm, p = Maximum intensity of pressure on the piston in N/mm2, W = Weight of the flywheel acting downwards in N, and
T1 + T2 = Resultant belt tension or pull acting horizontally in N. We know that gas load on the piston, π F = ××Dp2 P 4
Due to this piston gas load (FP) acting horizontally, there will be two horizontal reactions H1and H2 at bearings 1 and 2 respectively, such that Fa()+ b Fa× H = P ; and H = P 1 b 2 b
Due to the weight of the flywheel (W) acting downwards, there will be two vertical reactions V1 and V2 at bearings l and 2 respectively, such that Wb. Wb. V = 1 ; and V = 2 1 b 2 b
Now due to the resultant belt tension (T1 + T2) acting horizontally, there will be two horizontal ′ ′ reactions H1 and H2 at bearings 1 and 2 respectively, such that ()TTb+ ()TTb+ H ′ = 121; and H ′ = 122 1 b 2 b The various parts of the side crankshaft, when the crank is at dead centre, are now designed as discussed below: (a) Design of crankpin. The dimensions of the crankpin are obtained by considering the crank- pin in bearing and then checked for bending stress.
Let dc = Diameter of the crankpin in mm,
lc = Length of the crankpin in mm, and 2 pb = Safe bearing pressure on the pin in N/mm . It may be between 9.8 to 12.6 N/mm2.
We know that FP = dc . lc . pb
From this expression, the values of dc and lc may be obtained. The length of crankpin is usually from 0.6 to 1.5 times the diameter of pin. The crankpin is now checked for bending stress. If it is assumed that the crankpin acts as a cantilever and the load on the crankpin is uniformly distributed, then maximum bending moment will Fl× be P c . But in actual practice, the bearing 2 pressure on the crankpin is not uniformly distributed and may, therefore, give a greater value Fl× of bending moment ranging between P c and 2 × FlP c . So, a mean value of bending moment, i.e. 3 Fl× may be assumed. 4 P c Close-up view of an automobile piston 1172 A Textbook of Machine Design
∴ Maximum bending moment at the crankpin, 3 M = Fl× ... (Neglecting pin collar thickness) 4 P c Section modulus for the crankpin, π Z = ()d 3 32 c ∴ Bending stress induced, σ b = M / Z This induced bending stress should be within the permissible limits. (b) Design of bearings. The bending moment at the centre of the bearing 1 is given by
M = FP (0.75 lc + t + 0.5 l1) ...(i) where lc = Length of the crankpin,
t = Thickness of the crank web = 0.45 dc to 0.75 dc, and
l1 = Length of the bearing = 1.5 dc to 2 dc. We also know that π ()d 3 σ M= 32 1 b ...(ii) From equations (i) and (ii), the diameter of the bearing 1 may be determined. Note : The bearing 2 is also made of the same diameter. The length of the bearings are found on the basis of allowable bearing pressures and the maximum reactions at the bearings. (c) Design of crank web. When the crank is at dead centre, the crank web is subjected to a bending moment and to a direct compressive stress. We know that bending moment on the crank web,
M = FP (0.75 lc + 0.5 t) 1 and section modulus, Z = × wt. 2 6 M ∴ Bending stress, σ = b Z We also know that direct compressive stress, F σ = P d wt. ∴ Total stress on the crank web, σ σ σ T = b + d This total stress should be less than the permissible limits. (d) Design of shaft under the flywheel. The total bending moment at the flywheel location will be the resultant of horizontal bending moment due to the gas load and belt pull and the vertical bending moment due to the flywheel weight.
Let ds = Diameter of shaft under the flywheel. We know that horizontal bending moment at the flywheel location due to piston gas load,
M1 = FP (a + b2) – H1 .b2 = H2 .b1 Internal Combustion Engine Parts 1173 and horizontal bending moment at the flywheel location due to belt pull, ().TTbb+ M = H ′.b = H ′.b = 1212 2 1 2 2 1 b ∴ Total horizontal bending moment,
MH = M1 + M2 We know that vertical bending moment due to flywheel weight, Wb b M = V .b = V .b = 12 V 1 2 2 1 b ∴ Resultant bending moment, 22+ MR = ()()MMHV ...(i) We also know that π ()d 3 σ MR = 32 sb ...(ii)
From equations (i) and (ii), the diameter of shaft (ds) may determined. 2. When the crank is at an angle of maximum twisting moment. Consider a position of the crank at an angle of maximum twisting moment as shown in Fig. 32.19. We have already discussed in the design of a centre crankshaft that the thrust in the connecting rod (FQ) gives rise to the tangential force (FT) and the radial force (FR).
Fig. 32.19. Crank at an angle of maximum twisting moment.
Due to the tangential force (FT), there will be two reactions at the bearings 1 and 2, such that Fa()+ b Fa× H = T ; and H = T T1 b T2 b Due to the radial force (FR), there will be two reactions at the bearings 1 and 2, such that Fa()+ b Fa× H = R ; and H = R R1 b R2 b The reactions at the bearings 1 and 2 due to the flywheel weight (W) and resultant belt pull
(T1 + T2) will be same as discussed earlier. Now the various parts of the crankshaft are designed as discussed below: (a) Design of crank web. The most critical section is where the web joins the shaft. This section is subjected to the following stresses :
(i) Bending stress due to the tangential force FT ; 1174 A Textbook of Machine Design
(ii) Bending stress due to the radial force FR ; (iii) Direct compressive stress due to the radial force FR ; and (iv) Shear stress due to the twisting moment of FT. We know that bending moment due to the tangential force, ⎛⎞− d1 MbT = FrT ⎜⎟ ⎝⎠2 where d1 = Diameter of the bearing 1.
Q
Diesel, petrol and steam engines have crank shaft
∴ Bending stress due to the tangential force, MMbbTT= 6 1 σ = ...( Z = × tw· 2 ) ...(i) bT Z tw. 2 Q 6 We know that bending moment due to the radial force,
MbR = FR (0.75 lc + 0.5 t) ∴ Bending stress due to the radial force, MM6 1 σ = bbRR= ...(Here Z = × wt· 2 ) ...(ii) bR Z wt. 2 6 We know that direct compressive stress, F σ = R ...(iii) d wt. ∴ Total compressive stress, σ σ σ σ c = bT + bR + d ...(iv) We know that twisting moment due to the tangential force,
T = FT (0.75 lc + 0.5 t) TT= 4.5 ∴ Shear stress, τ = 2 ZP wt. wt. 2 where Z = Polar section modulus = P 4.5 Internal Combustion Engine Parts 1175
Now the total or maximum stress is given by σ 1 σ = c +σ+τ()22 4 ...(v) max 22 c This total maximum stress should be less than the maximum allowable stress. (b) Design of shaft at the junction of crank
Let ds1 = Diameter of the shaft at the junction of the crank. We know that bending moment at the junction of the crank,
M = FQ (0.75lc + t) and twisting moment on the shaft
T = FT × r ∴ Equivalent twisting moment, 22+ Te = MT ...(i) We also know that equivalent twisting moment, π T = ()d 3 τ ...(ii) e 16 s1
From equations (i) and (ii), the diameter of the shaft at the junction of the crank (ds1) may be determined. (c) Design of shaft under the flywheel
Let ds = Diameter of shaft under the flywheel. The resultant bending moment (MR) acting on the shaft is obtained in the similar way as Q discussed for dead centre position. We know that horizontal bending moment acting on the shaft due to piston gas load, ⎡⎤22 M = Fa()()()+− b H + H b 1 P2⎣⎦⎢⎥ R1T12 and horizontal bending moment at the flywheel location due to belt pull, ().TTbb+ M = Hb′=′=.. Hb 1212 2 12 21 b ∴ Total horizontal bending moment,
MH = M1 + M2 Vertical bending moment due to the flywheel weight, Wb b M = V · b = V · b = 12 V 1 2 2 1 b ∴ Resultant bending moment, 22+ MR = ()()MMHV We know that twisting moment on the shaft,
T = FT × r ∴ Equivalent twisting moment, 22+ Te = ()MTR ...(i) We also know that equivalent twisting moment, π T = ()d 3 τ ...(ii) e 16 s
From equations (i) and (ii), the diameter of shaft under the flywheel (ds) may be determined. Example 32.4. Design a plain carbon steel centre crankshaft for a single acting four stroke single cylinder engine for the following data: 1176 A Textbook of Machine Design
Bore = 400 mm ; Stroke = 600 mm ; Engine speed = 200 r.p.m. ; Mean effective pressure = 0.5 N/mm2; Maximum combustion pressure = 2.5 N/mm2; Weight of flywheel used as a pulley = 50 kN; Total belt pull = 6.5 kN. When the crank has turned through 35° from the top dead centre, the pressure on the piston is 1N/mm2 and the torque on the crank is maximum. The ratio of the connecting rod length to the crank radius is 5. Assume any other data required for the design. 2 2 Solution. Given : D = 400 mm ; L = 600 mm or r = 300 mm ; pm = 0.5 N/mm ; p = 2.5 N/mm ; θ ′ 2 W = 50 kN ; T1 + T2 = 6.5 kN ; = 35° ; p = 1N/mm ; l / r = 5 We shall design the crankshaft for the two positions of the crank, i.e. firstly when the crank is at the dead centre ; and secondly when the crank is at an angle of maximum twisting moment.
Part of a car engine
1. Design of the crankshaft when the crank is at the dead centre (See Fig. 32.18) We know that the piston gas load, ππ F = ××=Dp22(400) 2.5 = 314200 N = 314.2 kN P 44 Assume that the distance (b) between the bearings 1 and 2 is equal to twice the piston diameter (D). ∴ b = 2D = 2 × 400 = 800 mm Internal Combustion Engine Parts 1177
b 800 and b = b = ==400 mm 1 2 22
We know that due to the piston gas load, there will be two horizontal reactions H1 and H2 at bearings 1 and 2 respectively, such that Fb× 314.2× 400 H = P1==157.1 kN 1 b 800 Fb× 314.2× 400 and H = P2==157.1 kN 2 b 800
Assume that the length of the main bearings to be equal, i.e., c1 = c2 = c / 2. We know that due to the weight of the flywheel acting downwards, there will be two vertical reactions V2 and V3 at bearings 2 and 3 respectively, such that Wc× Wc× / 250 W V = 1 === = 25 kN 2 cc22 Wc× Wc× / 250 W and V = 2 === = 25 kN 3 cc22
Due to the resultant belt tension (T1 + T2) acting horizontally, there will be two horizontal ′ ′ reactions H2 and H3 respectively, such that ()()/2TTc++ TTc TT +6.5 H ′ = 12112==== 12 3.25kN 2 cc22 ()()/2TTc++ TTc TT +6.5 and H ′ = 12212==== 12 3.25kN 3 cc22 Now the various parts of the crankshaft are designed as discussed below: (a) Design of crankpin
Let dc = Diameter of the crankpin in mm ; lc = Length of the crankpin in mm ; and σ b = Allowable bending stress for the crankpin. It may be assumed as 75 MPa or N/mm2. We know that the bending moment at the centre of the crankpin,
MC = H1 · b2 = 157.1 × 400 = 62 840 kN-mm ...(i) We also know that ππ M = (dd )33σ= ( ) 75 = 7.364( d ) 3N-mm C 32cb 32 c c –3 3 = 7.364 × 10 (dc) kN-mm ...(ii) Equating equations (i) and (ii), we have 3 –3 6 (dc) = 62 840 / 7.364 × 10 = 8.53 × 10 or dc = 204.35 say 205 mm Ans. We know that length of the crankpin, F 314.2× 103 P ==153.3 lc = × say 155 mm Ans. dpcb·20510 2 ...(Taking pb = 10 N/mm ) (b) Design of left hand crank web We know that thickness of the crank web,
t = 0.65 dc + 6.35 mm = 0.65 × 205 + 6.35 = 139.6 say 140 mm Ans. 1178 A Textbook of Machine Design and width of the crank web, w = 1.125 dc + 12.7 mm = 1.125 × 205 + 12.7 = 243.3 say 245 mm Ans. We know that maximum bending moment on the crank web,
⎛⎞−−lc t M = Hb12⎜⎟ ⎝⎠22 ⎛⎞−−155 140 = = 157.1⎜⎟ 400 39 668 kN-mm ⎝⎠22 11 Section modulus, Z = ×=×wt.2233 245 (140) =× 800 10 mm 66 M 39 668 − ∴ Bending stress, σ = ==×49.6 1032 kN/mm = 49.6 N/mm 2 b Z 800× 103 We know that direct compressive stress on the crank web,
H1 157.1 −32 2 σ = ==×4.58 10 kN/mm = 4.58 N/mm c wt. 245× 140 ∴ Total stress on the crank web σ σ 2 = b + c = 49.6 + 4.58 = 54.18 N/mm or MPa Since the total stress on the crank web is less than the allowable bending stress of 75 MPa, therefore, the design of the left hand crank web is safe. (c) Design of right hand crank web From the balancing point of view, the dimensions of the right hand crank web (i.e. thickness and width) are made equal to the dimensions of the left hand crank web. (d) Design of shaft under the flywheel
Let ds = Diameter of the shaft in mm. Since the lengths of the main bearings are equal, therefore
⎛⎞b −−=lc ⎛⎞ −155 − l1 = l2 = l3 = 2 ⎜⎟t 2⎜⎟ 400 140 = 365 mm ⎝⎠22⎝⎠ 2 Assuming width of the flywheel as 300 mm, we have c = 365 + 300 = 665 mm
Hydrostatic transmission inside a tractor engine Internal Combustion Engine Parts 1179
Allowing space for gearing and clearance, let us take c = 800 mm. c 800 ∴ c = c = ==400 mm 1 2 22 We know that bending moment due to the weight of flywheel, 6 MW = V3 · c1 = 25 × 400 = 10 000 kN-mm = 10 × 10 N-mm and bending moment due to the belt pull, ′ 6 MT = H3 · c1 = 3.25 × 400 = 1300 kN-mm = 1.3 × 10 N-mm ∴ Resultant bending moment on the shaft, 2+=×+× 2 62 62 MS = (MMWT ) ( ) (10 10 ) (1.3 10 ) = 10.08 × 106 N-mm
We also know that bending moment on the shaft (MS), ππ 10.08 × 106 = ()dd33σ= ()424.12() = d 3 32sb 32 s s ∴ 3 6 6 (ds) = 10.08 × 10 / 4.12 = 2.45 × 10 or ds = 134.7 say 135 mm Ans. 2. Design of the crankshaft when the crank is at an angle of maximum twisting moment We know that piston gas load, ππ F = ××′=Dp22(400) 1 = 125 680 N = 125.68 kN P 44
In order to find the thrust in the connecting rod (FQ), we should first find out the angle of inclination of the connecting rod with the line of stroke (i.e. angle φ). We know that sinθ° sin 35 sin φ = ==0.1147 lr/5 ∴φ= sin–1 (0.1147) = 6.58° We know that thrust in the connecting rod, F 125.68 125.68 F = P == = 126.5 kN Q cosφ cos6.58º 0.9934 Tangential force acting on the crankshaft, θ φ FT = FQ sin ( + ) = 126.5 sin (35° + 6.58°) = 84 kN θ φ and radial force, FR = FQ cos ( + ) = 126.5 cos (35° + 6.58°) = 94.6 kN
Due to the tangential force (FT), there will be two reactions at bearings 1 and 2, such that Fb× 84× 400 H = T1= = 42 kN T1 b 800 Fb× 84× 400 and H = T2= = 42 kN T2 b 800 Due to the radial force (FR), there will be two reactions at bearings 1 and 2, such that Fb× 94.6× 400 H = R1= = 47.3 kN R1 b 800 Fb× 94.6× 400 H = R2= = 47.3 kN R2 b 800 Now the various parts of the crankshaft are designed as discussed below: (a) Design of crankpin
Let dc = Diameter of crankpin in mm.