EBK MATHEMATICS FOR MACHINE TECHNOLOGY
EBK MATHEMATICS FOR MACHINE TECHNOLOGY
8th Edition
ISBN: 9781337798396
Author: SMITH
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 49, Problem 22AR
To determine

(a)

The cutting speed.

Expert Solution
Check Mark

Answer to Problem 22AR

  C=151 ft/min

Explanation of Solution

Given information:

Diameter of shaft: D=2.300 inches

Rotational speed of lathe: N=250 rpm

Calculations:

The cutting speed is calculated using the given formula:

  C=3.1416DN12Substituting values of D and N:C=3.1416( 2.300)( 250)12C=150.535 ft/minRounding the value to the nearer whole number:C=151 ft/min

Conclusion:

The cutting velocity is C=151 ft/min.

To determine

(b)

The RPM of the aluminum cylinder.

Expert Solution
Check Mark

Answer to Problem 22AR

  N=1273 revolutions per minute (RPM).

Explanation of Solution

Given information:

Diameter of the cylinder: D=40.00 mm

Cutting speed: C=160 m/min

Calculations:

The RPM of the cylinder is calculated using the given formula:

  N=1000C3.1416DN=1000( 160)3.1416( 40.0)or,N=1273.23Rounding to the nearest whole number:N=1273 revolutions per minute (RPM)

Conclusion:

The rotational speed of the cylinder is N=1273 RPM.

To determine

(c)

The total cutting time.

Expert Solution
Check Mark

Answer to Problem 22AR

  Ttotal=3.74 min

Explanation of Solution

Given information:

Number of holes n=20

Diameter of each holes D=5.50 mm

Depth of holes d=58.00 mm

Rotational speed of drill N=6500 RPM

Feed F=0.05 mm/rev

Calculations:

The time taken to drill one hole is given by the relation:

  T=LFNwhere,L=d+AP1or,L=d+0.5D=58.00+0.5(5.50)=60.75 mmSubstituting values:T=60.75 mm( 6500 rev/min)( 0.05 mm/rev)T=0.1869 minHence, total time taken to drill 20 holes:Ttotal=nT=20(0.1869)Ttotal=3.738 min=3.74 min

Conclusion:

The total cutting time is Ttotal=3.74 min

To determine

(d)

The pitch of the gear.

Expert Solution
Check Mark

Answer to Problem 22AR

  P=3.50 teeth/inch

Explanation of Solution

Given information:

Number of teeth on the gear: N=44

Pitch diameter: D=12.5714 inches

Calculations:

The pitch of the gear is defined by the relation:

  P=NDsubstituting values:P=4412.5714P=3.50 teeth/inch

Conclusion:

The pitch of the gear is P=3.50 teeth/inch.

To determine

(e)

The whole depth of the gear.

Expert Solution
Check Mark

Answer to Problem 22AR

  WD=0.5393 inch

Explanation of Solution

Given information:

Number of teeth on the gear: N=20

Pitch diameter: D=5.0000 inches

Calculations:

The whole depth of the gear is defined by the relation:

  WD=2.157Pwhere, PPitch of the gearP=ND=205=4 teeth/inchsubstituting:WD=2.1574=0.53925 inchor,WD=0.5393 inch

Conclusion:

The whole depth of the gear is WD=0.5393 inch.

To determine

(f)

The center distance between the gear and the pinion.

Expert Solution
Check Mark

Answer to Problem 22AR

  CD=8.5989 inches

Explanation of Solution

Given information:

Pitch diameter of pinion: Dp=6.4445 inches

Pitch diameter of gear: Dg=10.7533 inches

Calculations:

The center distance between the gear and the pinion is defined by the relation:

  CD=Dp+Dg2substituting:CD=6.4445+10.75332inchesor,CD=8.5989 inches

  P=NDsubstituting values:P=4412.5714P=3.50 teeth/inch

Conclusion:

The center distance between the gear and the pinion is CD=8.5989 inches.

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