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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Textbook Question
Chapter 16, Problem 4P

The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the hinge pins A and B supports a pair of shoes. The actuating mechanism is to be arranged to produce the same force F on each shoe. The face width of the shoes is 75 mm. The material used permits a coefficient of friction of 0.24 and a maximum pressure of 1000 kPa.

Chapter 16, Problem 4P, The figure shows a 400-mm-diameter brake drum with four internally expanding shoes. Each of the

Problem 16–4

The dimensions in millimeters are a = 150, c = 165, R = 200, and d = 50.

(a)    Determine the maximum actuating force.

(b)    Estimate the brake capacity.

(c)    Noting that rotation may be in either direction, estimate the hinge-pin reactions.

(a)

Expert Solution
Check Mark
To determine

The maximum actuating force.

Answer to Problem 4P

The maximum actuating force is 5.7kN.

Explanation of Solution

Write the expression for moment of frictional forces.

    Mf=fpabrsinθaθ1θ2sinθ(racosθ)dθ                                       (I)

Here, coefficient of friction is f, face width of shoe is b, maximum pressure on the shoe is pa, outer radius is a, angle at which frictional material is located from hinge is θ and inner radius is r.and moment of frictional forces is Mf.

Write the expression for moment of normal forces.

    MN=pabrasinθaθ1θ2sin2θdθ                                                      (II)

Here, moment of normal forces is MN.

Write the expression for actuating force.

    F=MNMfc                                                                  (III)

Here, actuating force is F.

Conclusion:

Calculate θ2.

    θ2=90°(15°+25°)=90°40°=50°

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 1000kPa for pa and 0.24 for f in Equation (I).

    Mf=(0.24)(1000kPa)(75mm)(200mm)sin(75°)1075sinθ(200mm150mmcosθ)dθ=[(0.24)(1000kPa)((15000mm2)(1m1000mm)2)sin(75°)][1075sinθ((200mm)(1m1000mm)(150mm)(1m1000mm)cosθ)dθ]=3727[(0.2cosθ)1075+0.15(14cos2θ)1075]=288.9Nm

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 1000kPa for pa and 0.24 for f in Equation (II).

    MN=(1000kPa)(75mm)(200mm)(150mm)sin(70°)10°75°sin2θdθ=(1000kPa)(1000Pa1kPa)(2250000mm3)(1m1000mm)3[θ214sin2θ]1075=2329.4[13π7214(0.158)]=1229.3Nm

Substitute 1229.3Nm for MN, 165mm for c and 288.9Nm for Mf in Equation (III).

    F=1229.3Nm288.9Nm165mm=940.4Nm(165mm)(1m1000mm)=(5699.39Nm)(1kN1000N)5.7kN

Thus, the maximum actuating force is 5.7kN.

(b)

Expert Solution
Check Mark
To determine

The total braking capacity.

Answer to Problem 4P

The total braking capacity is 1752.4Nm.

Explanation of Solution

Write the expression for torque applied by primary shoe.

    TP=fpabr2(cosθ1cosθ2)sinθa                     (IV)

Here, braking torque applied by primary shoe is TP.

Write the expression for moment of frictional forces.

    Mfs=Mf106pa                          (V)

Here, moment of frictional forces for secondary shoe is Mfs.

Write the expression for normal forces for secondary shoes.

    MNs=MN106pa                        (VI)

Here, normal forces for secondary shoes is MNs.

Write the expression for actuating forces.

    F=MNs+MFsc                       (VII)

Here actuating force is F.

Write the expression\n for braking torque on left hand side shoe.

    TL=fpabr2(cosθ1cosθ2)sinθa                           (VIII)

Write the expression for total braking torque.

    TTotal=Tp+Ts                   (IX)

Here, total braking torque is TTotal.

Conclusion:

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 1000kPa for pa and 0.24 for f in Equation (IV).

    TP=0.24(1000kPa)(75mm)(200mm)2(cos10°cos75°)sin75=0.24(1000kPa)(1000Pa1kPa)(3000000mm3)(1m1000mm)3(0.9848(0.2588))0.9659=522.720.9659=541.17Nm

Substitute 288.9Nm for Mf in Equation (V).

    Mfs=288.9Nm106pa=288.9×106pa

Substitute 1229.3Nm for Mn in Equation (VI).

    MNs=1229.3Nm106pa=1229.3×106pa

Substitute 5.7kN for F, 288.9×106pa for Mfs and 1229.3×106pa for MNs in Equation (VII).

    5.7kN=1229.3×106pa+288.9×106pa165mm(5.7kN)(1000N1kN)=1518.2×106pa(165mm)(1m1000mm)pa=940.51518.2×106pa=619.5kPa

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 619.5kPa for pa and 0.24 for f in Equation (VIII).

    Ts=0.24(619.5kPa)(75mm)(200mm)2(cos10°cos75°)sin75=0.24(619.5kPa)(1000Pa1kPa)(3000000mm3)(1m1000mm)3(0.9848(0.2588))0.9659=323.820.9659=335.32Nm

Substitute 335.32Nm for Ts and 541.17Nm for Tp in Equation (IX).

    TTotal=2(541.17Nm)+2(335.32Nm)=1752.4Nm

Thus, the total braking capacity is 1752.4Nm.

(c)

Expert Solution
Check Mark
To determine

The horizontal reaction on primary shoes.

The vertical reaction on primary shoe.

The horizontal reaction on secondary shoes.

The vertical reaction on secondary shoe.

The resultant reaction.

Answer to Problem 4P

The horizontal reaction on primary shoes is 0.66kN.

The, the vertical reaction on primary shoe is 9.88kN.

The horizontal reaction on secondary shoes is 0.137kN.

The, the vertical reaction on secondary shoe is 3.9kN.

The resultant reaction is 6.03kN.

Explanation of Solution

Write the expression for horizontal reaction on hinge pin for primary shoe.

    Rx1=pabrSinθa(θ1θ2sinθcosθdθfθ1θ2sin2θdθ)Fx                  (X)

Here, horizontal reaction is Rx1.

Write the expression for vertical reaction on hinge pin for primary shoe.

    Ry1=pabrSinθa(θ1θ2sin2θdθ+fθ1θ2sinθcosθdθ)Fy              (XI)

Here, vertical reaction is Ry1.

Write the expression for horizontal reaction on hinge pin for secondary shoe.

    Rx2=pabrSinθa(θ1θ2sinθcosθdθ+fθ1θ2sin2θdθ)Fx        (XII)

Here, horizontal reaction on hinge pin for secondary shoe is Rx2.

Write the expression for vertical reaction on hinge pin for secondary shoe.

    Ry2=pabrSinθa(θ1θ2sin2θdθ+fθ1θ2sinθcosθdθ)Fy          (XIII)

Here, vertical reaction on hinge pin for secondary shoe is Ry2.

Write the expression for net horizontal reaction.

    RH=Rx1+Rx2                                                             (XIV)

Here, net horizontal reaction is RH.

Write the expression for net vertical reaction.

    RV=Ry1Ry2                                                           (XV)

Here, net vertical reaction is RV.

    R=RH2+RV2                                                          (XVI)

Here, resultant reaction is R.

Conclusion

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 5.7kN for Fx, 1000kPa for pa and 0.24 for f in Equation (X).

    Rx1=(1000kPa)(75mm)(200mm)Sin75°(1075sinθcosθdθ0.241075sin2θdθ)(5.7kN)=(1000kPa)(15000mm2)Sin75°[(14cos2θ)10750.24(θ214sin2θ)075](5.7kN)=(1000kPa)(15000mm2)(1m1000mm)2Sin75°[0.45150.24(0.528)]5.7kN=0.66kN

Thus horizontal reaction on primary shoes is 0.66kN.

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 0 for Fy, 1000kPa for pa and 0.24 for f in Equation (XI).

    Ry1=(1000kPa)(75mm)(200mm)Sin75°(1075sin2θdθ+0.241075sinθcosθdθ)0=(1000kPa)(15000mm2)Sin75°[(θ214sin2θ)075+0.24(14cos2θ)1075]=(1000kPa)(15000mm2)(1m1000mm)2Sin75°[0.528+0.24(0.4515)]=9.88kN

Thus, the vertical reaction on primary shoe is 9.88kN.

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 5.7kN for Fx, 619.5kPa for pa and 0.24 for f in Equation (XII).

    Rx2=(619.5kPa)(75mm)(200mm)Sin75°(1075sinθcosθdθ+0.241075sin2θdθ)5.7kN=(619.5kPa)(15000mm2)Sin75°[(14cos2θ)1075+0.24(θ214sin2θ)075]5.7kN=(619.5kPa)(15000mm2)(1m1000mm)2Sin75°[0.4515+0.24(0.528)]5.7kN=0.137kN

Thus horizontal reaction on secondary shoes is 0.137kN.

Substitute 10° for θ1, 50° for θ2, 75° for θa, 150mm for a, 200mm for r, 0 for Fy, 619.5kPa  for pa and 0.24 for f in Equation (XIII).

    Ry2=(619.5kPa)(75mm)(200mm)Sin75°(1075sin2θdθ+0.241075sinθcosθdθ)0=(619.5kPa)(15000mm2)Sin75°[(θ214sin2θ)075+0.24(14cos2θ)1075]=(619.5kPa)(15000mm2)(1m1000mm)2Sin75°[0.5280.24(0.4515)]=3.9kN

Thus, the vertical reaction on secondary shoe is 3.9kN.

Substitute 0.66kN for Rx1  and 0.137kN fro Rx2 in Equation (XIV).

    RH=0.66kN+(0.137kN)=0.797kN0.8kN

Substitute 9.88kN for Ry1 and 3.9kN for Ry2 in Equation (XV).

    RV=9.88kN3.9kN=5.98kN

Substitute 5.98kN for Rv and 0.8kN for RH in Equation (XVI).

    R=(0.8kN)2+(5.98kN)2=0.64kN2+35.7604kN2=36.4004kN2=6.03kN

Thus, the resultant reaction is 6.03kN.

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