Explanation Write the expression for the number of worm threads. N G = m G ( N W ) (I) Here, the number of worm threads is N W , the number of teeth on gear is N G and the gear ratio is m G . Write the expression for the pitch of the worm gear ( P t ) G = π p x (II) Here, the pitch of the worm gear is ( P t ) G and axial pitch of worm is p x . Write the expression for worm gear pitch diameter. D = N G P t (III) Write the expression for the addendum. a = p x π (IV) Here, the addendum of gear is a . Write the expression for the dedendum of the gear. b = 1.157 p x π (V) Here, the dedendum is b . Write the expression for the center to center distance. C = d + D 2 (VI) Here, the center to center distance is C . Write the expression for the lead of the worm. L = p x N W (VII) Here, the lead of worm is L . Write the expression for the lead angle. λ = tan − 1 ( L π d w ) (VIII) Here, the lead angle is λ and mean diameter of worm gear is d w . Write the expression for the sliding velocity of worm. V s = π d w n w 12 cos λ (IX) Here, the speed of worm is n w , and the sliding velocity is V s . Write the expression for velocity of worm. V w = π d w n w 12 (X) Here, the velocity of worm is V w . Write the expression for the velocity of gear. V G = π D ( n w m G ) 12 (XI) Here, the velocity of gear is V G Write the expression for the coefficient of friction. f = 0.103 exp [ − 0.110 ( V s ) 0.450 ] + 0.012 (XII) Write the expression for the mechanical efficiency. e w = cos ϕ n − f tan λ cos ϕ n + f cot λ (XIII) Here, the normal pressure angle is ϕ n and mechanical efficiency is e w . Write the expression for the gear transmitted force. W G t = 33000 n d H o K a V G ( e w ) (XIV) Here, the gear transmitted force is W G t , the horsepower is H o , the application factor is K a and design factor is n d . Write the expression for the tangential component of worm force. W w t = W G t ( cos ϕ n sin λ + f cos λ cos ϕ n cos λ − f sin λ ) (XV) Here, the tangential component of worm force is W w t . Write the expression for the power of worm gear. H w = W w t ( V w ) 33000 (XVI) Here, the power of worm is H w . Write the expression for the power of gear. H G = W G t ( V G ) 33000 (XVII) Here, the power of gear is H G . Write the expression for the frictional force. W f = f W G t f sin λ − cos ϕ n cos λ (XVIII) Here, the frictional force is W f . Write the expression for the power of frictional force. H f = W f ( V s ) 33000 hp (XIX) Here, the power for the frictional force is H f . Write the expression for the effective face width. F e = W G t K w d G (XX) Here, the effective face width is F e , the diameter of gear is d G and wear factor is K w . Write the expression for the allowable tangential force on the worm gear tooth. ( W t ) a l l = K w d G F e (XXI) Here, the allowable tangential force on the worm gear tooth is ( W t ) a l l . Write the expression for the normal diametral pitch. P n = P t cos λ (XXII) Here, the normal diametral pitch is P n . Write the expression for the normal circular pitch. P = π P n (XXIII) Here, the normal circular pitch is P . Write the expression for the gear bending stress. σ G = W t G p n F e y (XXIV) Here, the gear bending stress is σ G . Conclusion: Substitute 21 for N G and 10 for m G in Equation (I). 21 = 10 ( N W ) N W = 21 10 N W = 2.1 N W ≈ 3 Thus, the number of teeth on worm is 3 . Substitute 3 for N W and 10 for m G in Equation (I). N G = 10 ( 3 ) N G = 30 Thus, the number of teeth on gear is 30 . Substitute 1.75 in for p x in Equation (II). ( P t ) G = π 1.75 in = 1.795 in Thus, the axial pitch of worm is 1.795 in . Substitute 30 for N G and 1.795 in for P t in Equation (III). D = 30 1.795 in = 16.71 in Thus, the worm gear pitch diameter is 16.71 in . Substitute 1.75 in for p x in Equation (IV). a = 1.75 in π = 0.557 in Substitute 1.75 in for p x in Equation (V). b = 1.157 ( 1.75 in ) π = 2.02475 in π = 0.644 in Substitute 3.6 in for d and 16.71 in for D in Equation (VI). C = 3.6 in + 16.71 in 2 = 20.31 in 2 = 10.15 in Thus, the center to center distance is 10.15 in . Substitute 1.75 in for p x and 3 for N W in Equation (VII). L = ( 1.75 in ) ( 3 ) = 5.25 in Thus, the lead is 5.25 in . Substitute 5.25 in for L and 3.6 in for d w in Equation (VIII). λ = tan − 1 ( 5.25 in π ( 3.6 in ) ) = tan − 1 ( 0.4642 ) = 24.9 ° Thus, the lead angle is 24.9 ° . Substitute 3.6 in for d w , 1125 rev / min for n w and 24.9 ° for λ in Equation (IX). V s = π ( 3.6 in ) ( 1125 rev / min ) 12 ( cos 24.9 ° ) = 12723.45 in ( rev / min ) ( cos 24

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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Chapter 15, Problem 22P

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The design parameter of worm gear.

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Chapter 15, Problem 21P

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Chapter 15 Solutions

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

Ch. 15 - An uncrowned straight-bevel pinion has 20 teeth, a...Ch. 15 - For the gearset and conditions of Prob. 15-1, find...Ch. 15 - An uncrowned straight-bevel pinion has 30 teeth, a...Ch. 15 - For the gearset and conditions of Prob. 15-3, find...Ch. 15 - An uncrowned straight-bevel pinion has 22 teeth, a...Ch. 15 - Prob. 6P Ch. 15 - In straight-bevel gearing, there are some analogs...Ch. 15 - Refer to your solution to Probs. 15-1 and 15-2. If...Ch. 15 - Prob. 9P Ch. 15 - Prob. 10P

Ch. 15 - Apply the relations of Prob. 15-7 to Ex. 15-1 and...Ch. 15 - Prob. 14P Ch. 15 - 15-15 to 15-22 As in Ex. 15-4, design a...Ch. 15 - 15-15 to 1 5-22 As in Ex. 154, design a...Ch. 15 - 15-15 to 15-22 As in Ex. 15-4, design a...Ch. 15 - 15-15 to 15-22 As in Ex. 15-4, design a...Ch. 15 - 15-15 to 15-22 As in Ex. 15-4, design a...Ch. 15 - 15-15 to 15-22 As in Ex. 15-4, design a...Ch. 15 - 15-15 to 15-22 As in Ex. 15-4, design a...Ch. 15 - Prob. 22P

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The design parameter of worm gear.

Solution Summary: The author analyzes the design parameters of worm gear, such as the number of teeth, the axial pitch, and the center to center distance.

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Chapter 15 Solutions