Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
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 17, Problem 20P

(a)

To determine

Design a Gates Rubber V-belt drive to transmit power from the motor to the compressor flywheel.

(a)

Expert Solution
Check Mark

Explanation of Solution

Consider a C270 belt.

Write the expression for center to center distance between pulleys.

C=0.25{[Lpπ2(D+d)]+[LPπ2(D+d)]22(Dd)2} (I)

Here, length of the belt is LP, diameter of bigger pulley is D, diameter of smaller pulley is d.

Write the expression for contact angle for smaller pulley.

θd=180°2sin1(Dd2C) (II)

Here, the center distance between pulley is C, diameter of bigger pulley is D, diameter of smaller pulley is d.

Write the expression for contact angle for bigger pulley.

θd=180°+2sin1(Dd2C) (III)

Here, the center distance between pulley is C, diameter of bigger pulley is D, diameter of smaller pulley is d.

Write the expression for contact correction factor

K1=0.143543+0.007468θd0.000015052(θd)2 (III)

Here, the contact angle for smaller pulley is θd.

Write the expression for velocity of the smaller pulley.

V=πdn12 (IV)

Here, the rotational speed of the smaller pulley is n.

Write the expression for allowable power per belt

Ha=K1K2Htab (V)

Here, the angle of wrap correction factor is K1, belt correction factor is K2, and table horse power ratings of standards v-belts is Htab.

Write the expression for design power.

Hd=KsHnomnd (VI)

Here, nominal power is Hnom, the service factor is Ks and the design factor is nd.

Write the expression for number of belts.

Nb=HdHa (VII)

Write the expression for torque

T=63025Hdn (VIII)

Here, the design power is Hd, and the speed of the rotation is n.

Write the expression for difference in tension in tight side an slack side of the belt.

ΔF=2Td (IX)

Here, the torque is T and diameter of the smaller pulley is d.

Write the expression for centrifugal tension

Fc=Kc(V1000)2 (X)

Here, the speed of the belt  is V and constant term is Kc.

Write the expression for initial tension in belt.

Fi=(Td)[efθd+1efθd1] (XI)

Here, the coefficient of the friction is f and angle of the contact for the smaller pulley is θd, the torque developed by the belt is T and the diameter of the smaller of the pulley is d.

Write the expression for tension in tight side of the belt.

F1=Fc+Fi(2efθdefθd+1) (XII)

Here, the centrifugal tension in the belt is Fc and initial tension in the belt is Fi.

Write the expression for tension in slack side of the belt.

F2=F1ΔF (XIII)

Here, the difference in tension in tight side an slack side of the belt is ΔF and tension in tight side of the belt is F1.

Write the expression for factor of the safety.

nfs=HaNbHd (XIV)

Here, allowable power per belt is Ha and design power is Hd and the number of the belt is Nb.

Write the expression for number of the belt passes.

NP=[(KF1+Kbd)b+(KF2+KbD)b]1 (XV)

Here, belt parameter is Kb and durability parameter for the V-belt is K and b.

Write the expression for the belt life.

t=NpLp720V (XVI)

Conclusion:

Substitute.270in+2.9in=272.9in for LP 60in for D and 11in for d in Equation (I).

C=0.25{[272.9inπ2(60in+11in)]+[272.9inπ2(60in+11in)]22(6011)2}=0.25×307.12in=76.78in

Substitute 60in for D and 11in for d and 76.78in C in Equation.(II).

θd=180°2sin1(60in11in2×76.78in)=180°37.2°=142.8°×π180°rad=2.492rad

Substitute 60in for D and 11in for d and 76.78in C in Equation.(III).

θd=180°+2sin1(60in11in2×76.78in)=180°+37.2°=217.2°×π180°rad=3.79rad

Substitute 142.8° for θd in Equation.(IV).

K1=0.143543+0.007468(142.8°)0.000015052(142.8°)2=0.143543+0.759457=0.903

Substitute 11in for d and 875rev/min for n in Equation (V).

V=π×11in×875rev/min12=9625π12ft/min=2520ft/min

Substitute 0.903 for K1, 1.15 for K2, and 7.83hp for Htab in Equation.(VI).

Ha=0.903×1.15×7.83hp=0.903×9.0045=8.13hp

Substitute 1.2 for Ks, 1.1 for nd and 50hp for Hnom in Equation (VII).

Hd=1.2×50hp×1.1=1.32×50hp=66hp

Substitute 8.13hp for Ha and 66hp for Hd in Equation (VIII)

Nb=66hp8.13hp=8.1189

Substitute 8.13hp per belt for Hd and 875rev/min for n in Equation (IX).

T=63025×8.13hp875rev/min=512393.25875lbfin=585.6lbfin

Substitute 585.6lbfin for T and 11in for d in Equation (X).

ΔF=2×585.6lbfin11in=1171.211lbf=106.5lbf

Substitute 1.716 for Kc and 2520ft/min for V in Equation (XI).

Fc=1.716×(2520ft/min1000)2=1.716×6.3504lbf=10.9lbf

Substitute 586.9lbfin for T, 11in for d, 0.5123 for f and 2.492rad for θd in Equation (XII).

Fi=(586.9lbfin11in)[e0.5123×2.492rad+1e0.5123×2.492rad1]=53.35lbf×[3.5846+13.58461]=94.6lbf

Substitute 10.9lbf for Fc, 94.6lbf for Fi, 0.5123 for f and 2.492rad for θd in Equation (XIII).

F1=10.9lbf+(94.6lbf)×(2e0.5123×2.492rade0.5123×2.492rad+1)=10.9lbf+(94.6lbf)×2×3.58464.5846=158.8lbf

Substitute 158.8lbf for F1, 106.5lbf for ΔF, in Equation (XIII).

F2=158.8lbf106.7lbf=52.1lbf

Substitute 84hp for Hd, 8.13hp for Ha and 9  for Nb in Equation (XIV).

nfs=8.13hp×966hp=1.11

Thus, the factor of the safety is 1.11.

Substitute 158.8lbf for F1, 52.1lbf for F2, 1600 for Kb, 60in for D and 11in for d, 2038 for K and 11.173 for b in Equation (XV).

NP=[(2038158.8lbf+160011in)111.173+(203852.1lbf+160060in)11.173]1=[(2038304.25)111.173+(203878.76)11.173]1=1.68×109

Thus, the belt life in passes is 1.68×109.

Substitute 109 for Np , 272.9in for LP and 2520ft/min in Equation (XVI).

t=109×272.9in720×2520ft/min=109×272.91814400hr=150407.84hr

Thus, the tool life in hour is 150407.84hr.

Thus, the selected belt is C-270.

(b)

To determine

Whether the cutting of the V-belt grooves in the flywheel is avoided by using a V-flat drive.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

The cutting of the V-belt grooves in the flywheel cannot be avoided by using a V-flat drive.

Explanation of Solution

Write the expression for the fully developed friction torque on the flywheel using the flats of the V-belts.

Tflat=FiD[efθD1efθD+1] (XVII)

Here the friction coefficient of flat on flywheel is f.

Write the expression for the flywheel torque.

Tfly=mGT (XVIII)

Here, the torque constant is mG.

Conclusion:

Substitute 94.6lbf for Fi, 60in for D, 0.13 for f and 3.791rad for θD in Equation (XVII).

Tflat=94.6lbf×60in×[e0.13×3.7911e0.3×3.791+1]=5676×0.2415lbfin=1371lbfin

Substitute 5.147 for mG, 586.9lbfin for T , in Equation (XVIII).

Tfly=5.147×586.9lbfin=3021lbfin

Since, the torque produced by V-belt is lower than the torque given by flywheel. Thus, he cutting of the V-belt grooves in the flywheel cannot be avoided by using a V-flat drive.

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