Design of Power Transmission System S K Mondal’s Chapter 3 2T (viii) P = t d P = Ptanα r t P P = t N cosα (ix) Minimum number of teeth to avoid interference 2A T = w For 20° full depth T =18 to 20 min sin2φ (x) 2 × A T = w min,pinion ⎡ 1 ⎛1 ⎞ ⎤ G⎢ 1+ ⎜ +2⎟sin2φ −1⎥ ⎢⎣ G⎝G ⎠ ⎥⎦ z ω G = Gear ratio = g = p z ω p g h A = fraction of addendum to module = a for pinion p m h A = fraction of addendum to module = f for gear (generally 1) w m (xi) Face width 8m <b <12m; usually b = 10m (xii) Beam strength σ = mbσ Y → Lewis equation b b σ Where σ = ult b 3 ⎛ 0.912⎞ Lewis form factor, Y = 0.154- for 20° full depth gear. ⎜ ⎟ ⎝ z ⎠ (xiii) Wear strength (σ ) = bQdpK ω 2T Where Q = g for external gear T +T g p 2T = g for internal gear T -T g p σ2 sinφcosφ⎛ 1 1 ⎞ load-stress factor(k) = c ⎜ + ⎟ 1.4 ⎝E1 E2⎠ 2 ⎛BHN⎞ = 0.16 ⎜ ⎟ ⎝ 100 ⎠ (xiv) Page 141 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 C 3 P = s P whereC = , for ordinary cut gear v <10m/s eff C t v 3+V v 6 = , for hobbed generated v > 20m/s 6+v 5.6 = , for precision gear v > 20m/s 5.6+ v (xv) Spott's equation,(P ) = C P +P eff s t d en T brr where P = p p 1 2 for steel pinion and steel gear d 2530 r2 +r2 1 2 e =16.00+1.25φ for grade 8 φ= m+0.29 d for all. Page 142 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 (b) Helical Gear (1) p = pcosα where p = transverse circular pitch. n p = normal circular pitch. n P (2) P = where P = normal diametral pitch. n cosα n P = transverse diametral pitch. α=helixangle. (3) pP = π ⎡ 1 ⎤ (4) m = mcosα P = m = transverse module. n ⎢⎣ m⎥⎦ m = normal module. n p πm (5) p = = ; p = axial pitch. a tanα tanα a tanφ (6) cosα= n φ = normal pressure angle (usually 20°). tanφ n φ =transverse pressure angle. TP Tm (7) d = = zm = n ; d = pitch circle diameter. π cosα m (8) a = n {T +T } → centre –to-centre distance. 2 cosα 1 2 T d (9) T′ = ; d′ = cos3α cos2α (10) An imaginary spur gear is considered with a pitch circle diameter of d′ and module m is n called ‘formative’ or ‘virtual’ spur gear. (11) Helix angle α varies from 15 to 25°. (12) Preference value of normal modus (m ) = 1, 1.25, 1.5, 2, 3, 4, 5, 6, 8 n (13) Addendum (h )=m ;dedendum(h )=1.25m ,clearence=0.25m a n f n n (14) 2M 60x106x(KW) P = t =Pcosφ cosα; M = N-mm t d n t 2πN ⎛tanφ ⎞ Pr = Pt⎜⎝ cosαn ⎟⎠ = Psinαφn P = Ptanα= Pcosφ sinα a t n (15) Beam strength S = m bσ Y′ b n b bQd K (16) Wear strength S = p w cos2α (17) Page 143 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 C P 5.6 en T brr P = s t ;C = ; P = p p 1 2 eff Cv v 5.6+ v d 2530 r12 +r22 P = C P +P cosα cosψ eff s t d n (c) Worm Gear (i) Specified and designated by T /T /q/m 1 2 d Where: q is diametric quotient = 1 m (ii) The threads of the worm have an involute helicoids profile. (iii) Axial pitch (p ) = distance between two consecutive teeth-measured along the axis of the x worm. (iv) The lead (l) = when the worm is rotated one revolution, a distance that a point on the helical profile will move. (v) l = p × T; d = mT x 1 z 2 (vi) Axial pitch of the worm = circular pitch of the worm wheel πd P = 2 = πm [ICS-04] x T 2 l = P T = πmT x 1 1 ⎛T ⎞ ⎛ l ⎞ (vii) Lead angle (δ)= tan-1⎜ 1⎟ =tan-1⎜ ⎟ ⎝ q ⎠ ⎝πd1⎠ 1 1 (viii) centre-to-centre distance (a) = (d +d ) = m(T +T ) 2 1 2 2 1 2 (ix) Preferred values of q: 8, 10, 12.5, 16, 20, 25 (x) Number of starts T usually taken 1, 2, or 4 1 (xi) ( ) addendum h =m h =m(2cosδ-1) a1 a2 dedendum(h )=(2.2cosδ-1)m h =m(1+0.2cosδ) f1 f2 clearance(c)=0.2m cosδ c =0.2mcosδ (xii) F = 2m (q+1) effective face width of the root of the worm wheel. (xiii) δ = sin-1 ⎛⎜ F ⎞⎟ ; l = (d +2C)sin-1⎛⎜ F ⎞⎟ = length of the root of the worm wheel teeth ⎜d +2C⎟ r a1 ⎜d +2C⎟ ⎝ a ⎠ ⎝ a ⎠ 1 1 Page 144 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 2m cosαcosδ-μsinδ sinα (xiv) (P ) = t ; (P ) = (P ) ; (P ) = (P ) 1 t d 1 a 1 t cosαsinδ+μcosδ 1 r 1 t cosαsinδ+μcosδ 1 Power output cosα-μtanδ (xv) Efficiency (η) = = Power input cosα+μtanδ πd n (xvi) Rubbing velocity (V ) = 1 1 m/s (remaining 4 cheak) s 60000 cosδ (xvii) Thermal consideration H =1000(1-η)×(KW) g H = K(t-t )A d 0 K(t-t )A KW = 0 1000(1-η) Page 145 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 Objective Questions (IES, IAS, GATE) Previous 20-Years GATE Questions Spur gear GATE-1. Match the type of gears with their most appropriate description. [GATE-2008] Type of gear Description P Helical 1. Axes non parallel and intersecting Q Spiral 2. Axes parallel and teeth are inclined to the axis R Hypoid 3. Axes parallel and teeth are parallel to the axis S Rack and pinion 4. Axes are perpendicular and intersecting, and teeth are inclined to the axis 5. Axes are perpendicular and used for large speed reduction 6. Axes parallel and one of the gears has infinite radius (a) P-2, Q- 4, R- 1, S- 6 (c) P-2, Q- 6, R- 4, S- 2 (b) P-1, Q- 4, R- 5, S- 6 (d) P-6, Q- 3, R- 1, S- 5 GATE-1Ans. (a) GATE-2. One tooth of a gear having 4 module and 32 teeth is shown in the figure. Assume that the gear tooth and the corresponding tooth space make equal intercepts on the pitch circumference. The dimensions 'a' and 'b', respectively, are closest to [GATE-2008] (a) 6.08 mm, 4 mm (b) 6.48 mm, 4.2 mm (c) 6.28 mm, 4.3 mm (d) 6.28 mm, 4.1 GATE-2Ans. (a) Classification of Gears GATE-3. Match the following [GATE-2004] Type of gears Arrangement of shafts P. Bevel gears 1. Non-parallel off-set shafts Q. Worm gears 2. Non-parallel intersecting shafts R. Herringbone gears 3. Non-parallel non-intersecting shafts S. Hypoid gears 4. Parallel shafts (a) P-4 Q-2 R-1 S-3 (b) P-2 Q-3 R-4 S-1 Page 146 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 (c) P-3 Q-2 R-1 S-4 (d) P-1 Q-3 R-4 S-2 GATE-3Ans. (b) Pitch point GATE-4. In spur gears, the circle on which the involute is generated is called the (a) Pitch circle (b) clearance circle [GATE-1996] (c) Base circle (d) addendum circle GATE-4Ans. (a) Minimum Number of Teeth GATE-5. The minimum number of teeth on the pinion to operate without interference in standard full height involute teeth gear mechanism with 20° pressure angle is [GATE-2002] (a) 14 (b) 12 (c) 18 (d) 32 GATE-5Ans. (c) Interference GATE-6 Tooth interference in an external in volute spur gear pair can be reduced by [GATE-2010] (a) Decreasing center distance between gear pair (b) Decreasing module (c) Decreasing pressure angle (d) Increasing number of gear teeth GATE-6Ans. (d) There are several ways to avoid interfering: i. Increase number of gear teeth ii. Modified involutes iii. Modified addendum iv. Increased centre distance GATE-7. Interference in a pair of gears is avoided, if the addendum circles of both the gears intersect common tangent to the base circles within the points of tangency. [GATE-1995] (a) True (b) False (c) Insufficient data (d) None of the above GATE-7Ans. (a) Page 147 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 GATE-8. Twenty degree full depth involute profiled 19-tooth pinion and 37-tooth gear are in mesh. If the module is 5 mm, the centre distance between the gear pair will be [GATE-2006] (a) 140 mm (b) 150 mm (c) 280 mm (d) 300 mm GATE-8Ans. (a) D +D mT +mT 5(19+37) Centre distance= 1 2 = 1 2 = =140mm 2 2 2 Beam Strength of Gear Tooth GATE-9. A spur gear has a module of 3 mm, number of teeth 16, a face width of 36 mm and a pressure angle of 20°. It is transmitting a power of 3 kW at 20 rev/s. Taking a velocity factor of 1.5, and a form factor of 0.3, the stress in the gear tooth is about [GATE-2008] (a) 32 MPa (b) 46 MPa (c) 58 MPa (d) 70MPa GATE-9Ans. (c) Statement for Linked Answer GATE-10 and GATE-11: A 20o full depth involute spur pinion of 4 mm module and 21 teeth is to transmit 15 kW at 960 rpm. Its face width is 25 mm. GATE-10. The tangential force transmitted (in N) is [GATE -2009] (a) 3552 (b) 261 1 (c) 1776 (d) 1305 GATE-10Ans. (a) GATE-11. Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is [GATE -2009] (a) 242.0 (b) 166.5 (c) 121.0 (d) 74.0 GATE-11Ans. (b) Simple Gear train Note: - Common Data for GATE-12 & GATE-13. A gear set has a opinion with 20 teeth and a gear with 40 teeth. The pinion runs at 0 rev/s and transmits a power of 20 kW. The teeth are on the 20° full –depth system and have module of 5 mm. The length of the line of action is 19 mm. GATE-12. The center distance for the above gear set in mm is [GATE-2007] (a) 140 (b) 150 (c) 160 (d) 170. GATE-12Ans. (b) GATE-13 The contact ratio of the contacting tooth [GATE-2007] (a) 1.21 (b) 1.25 (c) 1.29 (d) 1.33 GATE-13Ans. (c) GATE-14. The resultant force on the contacting gear tooth in N is: [GATE-2007] (a) 77.23 (b) 212.20 (c) 225.80 (d) 289.43 Page 148 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 GATE-14Ans. (c) Compound gear train Data for GATE-15 & GATE-16 are given below. Solve the problems and choose correct answers. A compacting machine shown in the figure below is used to create a desired thrust force by using a rack and pinion arrangement. The input gear is mounted on tile motor shaft. The gears have involute teeth of 2 mm module. GATE-15. If the drive efficiency is 80%, then torque required on the input shaft to create 1000 N output thrust is [GATE-2004] (a) 20 Nm (b) 25 Nm (c) 32 Nm (d) 50 Nm GATE-15Ans. (b) D Given : Module m = 2, =2 T ∴ D = 80 × 2 = 160 mm 2F = 1000, or F = 500 N Let T be the torque applied by motor. 1 T be the torque applied by gear. 2 ∴ Power transmission = 80% 2T ×ω Now, Tω = 2 1 1 1 0.8 2×F×(D/2) ω or T = × 1 1 0.8 ω 2 0.16 1 1 =2×500× × × 2 0.8 4 =25N−m. GATE-16. If the pressure angle of the rack is 20°, then force acting along the line of action between the rack and the gear teeth is [GATE-2004] (a) 250 N (b) 342 N (c) 532 N (d) 600 N GATE-16Ans. (c) Pcosφ=F ∴ Force acting along the line of action, Page 149 of 263 Design of Power Transmission System S K Mondal’s Chapter 3 F P= cosφ 500 = cos20(cid:68) =532N Reverted gear train Data for GATE-17 & GATE-18 are given below. Solve the problems and choose correct answers. The overall gear ratio in a 2 stage speed reduction gear box (with all spur gears) is 12. The input and output shafts of the gear box are collinear. The countershaft which is parallel to the input and output shafts has a gear (Z teeth) and pinion (Z = 2 3 15 teeth) to mesh with pinion (Z = 16 teeth) on the input shaft and gear (Z teeth) on 1 4 the output shaft respectively. It was decided to use a gear ratio of 4 with 3 module in the first stage and 4 module in the second stage. GATE-17. Z2 and Z4 are [GATE-2003] (a) 64 and 45 (b) 45 and 64 (c) 48 and 60 (d) 60 and 48 GATE-17Ans. (a) N N D Given, 1 =12, 1 =4= 2 N N D 2 2 1 m =3, m =4 1 2 D D Now, 1 = 2 Z Z 1 2 Z D N 1 ⇒ 1 = 1 = 2 = Z D N 4 2 2 1 ⇒ Z =Z ×4=64 2 1 D ⇒ 12= 4 D 3 D ⇒ 4 =3 D 3 Z D Also, 3 = 3 Z D 4 4 D ⇒ Z =Z 4 =Z ×3=15×3 4 3 D 3 3 =45 GATE-18. The centre distance in the second stage is [GATE-2003] (a) 90 mm (b) 120 mm (c) 160 mm (d) 240mm GATE-18Ans. (b) Page 150 of 263