Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 3.5, Problem 74P

Three shafts and four gears are used to form a gear train that will transmit power from the motor at A to a machine tool at F. (Bearings for the shafts are omitted in the sketch.) The diameter of each shaft is as follows: dAB = 16mm, dCD = 20 mm, dEF = 28 mm. Knowing that the frequency of the motor is 24 Hz and that the allowable shearing stress for each shaft is 75 MPa, determine the maximum power that can be transmitted.

Chapter 3.5, Problem 74P, Three shafts and four gears are used to form a gear train that will transmit power from the motor at

Expert Solution & Answer
Check Mark
To determine

The maximum power that can be transmitted by shaft.

Answer to Problem 74P

The maximum power that can be transmitted by the shaft is 7.11kW_.

Explanation of Solution

Given information:

The frequency of the motor is 24 Hz.

The allowable shearing stress in each shaft is 75 MPa.

The diameter of the shaft AB is dAB=16mm.

The diameter of the shaft CD is dCD=20mm.

The diameter of the shaft EF is dEF=28mm.

Calculation:

The maximum shear stress in the shaft (τmax) is expressed as shown below:

τmax=TcJ (1)

Here, T is the torque transmitted by the shaft, c is the radius of the shaft, and J is the polar moment of inertia of the shaft.

The power transmitted by the shaft (P) is expressed as follows:

P=(2πf)T (2)

For shaft AB:

The polar moment of inertia of shaft AB with radius c is,

JAB=π2(c)4=π2(8mm)4=6.434×103mm4=6.434×109m4

Substitute 75 MPa for τmax, 8 mm for c, and 6.434×109m4 for J in Equation (1).

75MPa=(T)(8mm)6.434×109m475MPa×106Pa1MPa=(T)(8mm×103m1mm)6.434×109m4T=60.319Nm

The frequency of the shaft AB is fAB=24Hz.

Substitute 60.319Nm for T and 24Hz for f in Equation (2).

P=2π(24Hz)(60.319Nm)=9.1×103W=9.1kW

For shaft CD:

The polar moment of inertia of shaft CD with radius c is,

JCD=π2(c)4=π2(10mm)4=1.5708×104mm4=1.5708×108m4

Substitute 75 MPa for τmax, 10 mm for c, and 1.5708×108m4 for J in Equation (1).

75MPa=(T)(10mm)1.5708×108m475MPa×106Pa1MPa=(T)(10mm×103m1mm)1.5708×108m4T=117.81Nm

The radius at gear B is rB=60mm.

The radius at gear C is rC=150mm.

The frequency of the shaft CD is,

fCD=rBrCfAB=60mm150mm(24Hz)=9.6Hz

Substitute 117.81Nm for T and 9.6Hz for f in Equation (2).

P=2π(9.6Hz)(117.81Nm)=7.11×103W=7.11kW

For shaft EF:

The polar moment of inertia of shaft EF with radius c is,

JEF=π2(c)4=π2(14mm)4=6.0344×104mm4=6.0344×108m4

Substitute 75 MPa for τmax, 14 mm for c, and 6.0344×108m4 for J in Equation (1).

75MPa=(T)(14mm)6.0344×108m475MPa×106Pa1MPa=(T)(14mm×103m1mm)6.0344×108m4T=323.27Nm

The radius at gear D is rD=60mm.

The radius at gear E is rE=150mm.

The frequency of the shaft EF is,

fEF=rDrEfCD=60mm150mm(9.6Hz)=3.84Hz

Substitute 323.27Nm for T and 3.84Hz for f in Equation (2).

P=2π(3.84Hz)(323.27Nm)=7.8×103W=7.8kW

The maximum allowable power is the smallest value calculated among the shafts AB, CD, and EF.

The allowable power Pall=7.11kW.

Therefore, the maximum power that can be transmitted by the shaft is 7.11kW_.

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Chapter 3 Solutions

Mechanics of Materials, 7th Edition

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