Explanation Write the expression for diameter of pinion. d p = N p p (I) Here, the number of teeth on pinion is N P and the diametral pitch is p . Write the expression for diameter of the gear. d G = N G p (II) Here, the number of teeth on gear is N G and the diametral pitch is p . Write the expression for velocity of the pinion. V = π d p n p (III) Here, the number of rotation made by pinion is n p . Write the expression for constant of transmission accuracy level number. B = 0.25 ( 12 − Q v ) 2 3 . (IV) Here, the transmission accuracy level number is Q v . Write the expression for constant A . A = 50 + 56 ( 1 − B ) (V) Write the expression for dynamic factor. K v = ( A + V A ) B (VI) Write the expression for allowable bending stress number through hardened steels. S t = 77.3 H B + 12800 (VII) Here, the Brinel hardness number is H B . Write the expression for stress cycle factor for bending. Y N = 1.6831 ( N ) − 0.0323 (VIII) Here, the number of cycles is N . Write the expression for allowable stress. ( σ all ) = S t Y N S F K T K R (IX) Here, the reliability factor is K R , the temperature factor is is K T and the safety factor against failure is S F . Write the expression for load correction factor for uncrowned teeth. C m c = 1 . (X) Write the expression for pinion proportion factor. C p f = F 10 d − 0.0375 + 0.0125 F (XI) Here, the face width is F and the diameter is d . Write the expression for pinion proportion modifier for straddle mounted pinion. C p m = 1 (XII) Write the expression for mesh alignment factor. C m a = A + B F + C F 2 (XIII) Here, the empirical constant is A , B and C . Write the expression for mesh alignment correction factor. C e = 1 (XIV) Write the expression for load distribution factor K m . K m = 1 + C m c ( C p f C p m + C m a C e ) . (XV) Write the expression for overload factor for pinion. ( K o ) p = 1.192 ( F Y p p ) 0.0535 (XVI) Here, the Lewis form factor for pinion is Y P . Write the expression for overload factor for gear. ( K o ) G = 1.192 ( F Y G p ) 0.0535 (XVII) Here, the Lewis form factor for gear is Y G . Write the expression for transmitted load in pinion. W p t = F J p σ all ( K o ) p K v K s p K m K B (XIX) Here, the spur gear geometry factor for pinion is J P and the rim thickness factor is K B . Write the expression for power for pinion. H p = W p t V 33000 (XX) Write the expression for transmitted load for gear. W G t = F J G σ all ( K o ) G K v K s p K m K B (XXI) Here the spur gear geometry factor for gear is J G . Write the expression for power for gear. H G = W G t V 33000 (XXII) Write the expression for gear ratio. m G = N G N P (XXIII) Here, the number of teeth on gear is N G and the number of teeth on pinion is N P Write the expression for pitting resistance stress cycle factor for pinion. ( Z N ) p = 2.466 ( N ) − 0.056 (XXIV) Write the expression for pitting resistance stress cycle factor for gear. ( Z N ) G = 2.466 ( N m G ) − 0.056 (XXV) Write the expression for geometry factor. I = cos ϕ t sin ϕ t 2 m N ( m G m G + 1 ) . (XXVI) Here, the pressure angle is ϕ t . Write the expression for hardness ratio factor C H . C H = H B P H B G = 1 . (XXVII) Write the expression for contact fatigue strength for pinion. 0.99 ( S c P ) 10 7 = 322 ( H B ) + 29100 . (XXVIII) Write the expression for contact fatigue strength for gear. 0.99 ( S c G ) 10 7 = 322 ( H B ) + 29100 . (XXIX) Write the expression for pinion contact endurance strength. ( σ all ) P = 0.99 ( S c ) 10 7 ( Z N ) ( C H ) S H ( K T ) ( K R ) . (XXX) Write the expression for transmitted load. W 3 t = ( ( σ all ) p C P ) 2 F d p I K o K v ( K s ) p K m C f . (XXXI) Here, the elastic coefficient is C P . Write the expression for power of pinion. H 3 = W 3 t V 33000 (XXXII) Write the expression for gear contact strength. ( σ all ) G = S c ( Z N ) ( C H ) S H ( K T ) ( K R ) . (XXXIII) Write the expression for transmitted load. W 4 t = ( ( σ all ) G C P ) 2 F d p I K o K v ( K s ) G K m C f (XXXIV) Write the expression for power of gear. H 4 = W 4 t V 33000 (XXXV) Write the expression for rated power. H r a t e d = min ( H 1 , H 2 , H 3 , H 4 ) . (XXXVI) Conclusion: Substitute 22 teeth for N P and 4 teeth / in for p in Equation (I). d p = 22 teeth 4 teeth / in = 5.5 in Substitute 60 teeth for N P and 4 teeth / in for p in Equation (II) d G = 60 teeth 4 teeth / in = 15 in Substitute 5.5 in for d p and 1145 rpm / min for n p in Equation (III). V = π ( 5.5 in ) ( 1145 rev / min ) = π ( 5.5 in ) ( 1145 rev / min ) 12 in / ft = 19784.17 12 ft / min = 1648.68 ft / min Substitute 6 for Q v in Equation (IV). B = 0.25 ( 12 − 6 ) 2 3 = 0.25 ( 6 ) 2 3 = 0.25 × 3.3019 = 0.8255 Substitute 0.8255 for B in Equation (V). A = 50 + 56 ( 1 − 0.8255 ) = 50 + 56 ( 0.1745 ) = 50 + 9.772 = 59.772 Substitute 59.772 for A , 1648.68 ft / min for V and 0.8255 for B in Equation (VI). K v = ( 59.772 + 1648.68 59.772 ) 0.8255 = ( 59.772 + 40.603 59.772 ) 0.8255 = ( 100.375 59.772 ) 0.8255 = ( 1.679 ) 0.8255 K v = 1.53 Substitute 250 for H B in Equation (VII). S t = ( 77.3 ( 250 ) + 12800 ) psi = ( 19325 + 12800 ) psi = 32125 psi Substitute 3 ( 10 9 ) for N in Equation (VIII). Y N = 1.6831 ( 3 ( 10 9 ) ) − 0.0323 = 1.6831 ( 0.4941 ) = 0.8316 Substitute 32125 psi for S t , 0.8316 for Y N , 1 for K R , 1 for K T , 1 for S F in Equation (IX). ( σ a l l ) P = ( 32125 psi ) × 0.8316 1 × 1 × 1 = 26715.15 psi Substitute 3.25 in for F and 5.5 in for d in Equation (XI). C p f = 3.25 10 ( 5.5 ) − 0.0375 + 0.0125 ( 3.25 ) = 3.25 55 − 0.0375 + 0.04062 = 0 .059 − 0.0375 + 0.04062 = 0.06212 Refer to table 14-9, “Empirical constant A , B and C for Eq.(14-34),face width F in inches”, obtain the empirical constant A , B and C as 0.127 , 0.0158 and 0.930 × 10 − 4 . Substitute 0.127 for A , 0.0158 for B , 3.25 in for F and 0.930 × 10 − 4 for C in Equation (XIII). C m a = 0.127 + 0.0158 ( 3.25 ) − ( 0.930 × 10 − 4 ) ( 3.25 ) 2 = 0.127 + 0.05135 − ( 0

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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A pair of helical gears are machined with a hob. The hob has a 20 degree pressure angle and machines circular pitch of 0.55 inch/tooth on a spur gear. Two gears are machined, the pinion has 18 teeth and the gear has 36 teeth. The measured transverse diametral pitch on the pinion is 5 inch/tooth. The pinion transmits 10 hp at 1,000 rpm. Calculate the helix angle of the gearset, in degrees.

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Chapter 14 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 14 - A steel spur pinion has a pitch of 6 teeth/in, 22...Ch. 14 - A steel spur pinion has a diametral pitch of 10...Ch. 14 - Prob. 3P Ch. 14 - A steel spur pinion has 16 teeth cut on the 20...Ch. 14 - Prob. 5P Ch. 14 - A 20 full-depth steel spur pinion has 20 teeth and...Ch. 14 - A 20 full-depth steel spur pinion has a diametral...Ch. 14 - A steel spur pinion is to transmit 20 hp at a...Ch. 14 - A 20 full-depth steel spur pinion with 18 teeth is...Ch. 14 - Prob. 10P

Ch. 14 - A speed reducer has 20 full-depth teeth and...Ch. 14 - Prob. 12P Ch. 14 - Prob. 13P Ch. 14 - A 20 20-tooth cast-iron spur pinion having a...Ch. 14 - A steel spur pinion and gear have a diametral...Ch. 14 - A milled-teeth steel pinion and gear pair have Sut...Ch. 14 - A 20 full-depth steel spur pinion rotates at 1145...Ch. 14 - A steel spur pinion has a pitch of 6 teeth/in, 17...Ch. 14 - A commercial enclosed gear drive consists of a 20...Ch. 14 - A 20 spur pinion with 20 teeth and a module of 2.5...Ch. 14 - Repeat Prob. 14-19 using helical gears each with a...Ch. 14 - A spur gearset has 17 teeth on the pinion and 51...Ch. 14 - Prob. 23P Ch. 14 - Prob. 24P Ch. 14 - The speed reducer of Prob. 14-24 is to be used for...Ch. 14 - Prob. 26P Ch. 14 - Prob. 27P Ch. 14 - The gearset of Prob. 14-27 is going to be upgraded...Ch. 14 - Prob. 29P Ch. 14 - AGMA procedures with cast-iron gear pairs differ...Ch. 14 - Spur-gear teeth have rolling and slipping contact...Ch. 14 - In Ex. 14-5 use nitrided grade 1 steel (4140)...Ch. 14 - In Ex. 14-5 use carburized and case-hardened gears...Ch. 14 - Prob. 35P Ch. 14 - Prob. 36P Ch. 14 - The countershaft in Prob. 373, p. 152, is part of...Ch. 14 - Build on the results of Prob. 1340, p. 720, to...Ch. 14 - Build on the results of Prob. 1341, p. 721, to...

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Chapter 14 Solutions