CH_17

Page 17-1

Chapter 17 • Brakes and Clutches

17.1 The disk brake shown in sketch a has brake pads in the form of circular sections withinner radius r, outer radius 2r, and section angle π/4. Calculate the brake torque when thepads are applied with normal force P. The brake is worn in so pu is constant, where p isthe contact pressure and u is the sliding velocity. The coefficient of friction is µ.

Notes: If pu is constant, and u=rω, then this gives a relationship for pressure in terms of radius.This relation is then substituted into Equation (17.2) and to eliminate the unknown constant, andinto Equation (17.3) to get an expression for torque.

Solution:Since pu is constant and u=rω, then we can deduce that p=C/rω, where C is a constant.Substituting into Equation (17.2), and noting that the section angle is π/4,

dP = prdθdr = Crω

rdθdr = Cω

dθdr; P = Cω

drdθ

r

2r

∫0

π / 4

∫ = Cπr4ω

;C = 4ωPπr

Note that the torque is generated on both sides of the rotor. Therefore, the torque is obtained fromEquation (17.3) as

T = 2 µpr2drdθ = 2 µ C

rω

r2drdθ = 2µC

ωrdrdθ = 2µC

ωπ4

1

22r( )2 − 1

2r2

= 3πµCr 2

4ω∫∫∫∫∫∫Substituting for C gives T=3µrP.

17.2 An automotive clutch with a single friction surface is to be designed with a maximumtorque of 140Nm. The materials are chosen such that µ=0.35 and pmax=0.35 MPa. Usesafety factor ns=1.3 with respect to slippage at full engine torque and as small an outsidediameter as possible, determine appropriate values of ro, ri and P by using both theuniform pressure and uniform wear models.

Notes: The approach is the same as in Example 17.1 on page 787; an equation is obtained forouter radius in terms of inner radius. Taking the derivative and solving for the optimum gives theanswer. Equations (17.5) and (17.12) are used for the torque for the two wear models.

Solution:I. Uniform Pressure ModelFrom Equation (17.5), and using the safety factor prescribed, the torque is given by

Page 17-2

nsTp = 2πµp0

3ro

3 − ri3( );ro =

3nsTp

2πµp0+ ri

3

1 / 3

= 3 1.3( ) 140Nm( )2π 0.35( ) 0.35MPa( ) + ri

3

1 / 3

= 7.09 ×10−4 m3 + ri3( )1 / 3

Taking deravitive with respect to ri, setting equal to zero and solving for ri gives:∂ro∂ri

= ∂∂r

7.09 ×10−4m3 + ri3( )1 / 3

= 13

7.09 ×10−4m3 + ri3( )−2 / 3

3ri2( ) = 0; ri = 0

Therefore, the outer radius is 0.08917m. From Equation (17.4), the actuating force is

Pp = πp0 ro2 − ri

2( ) = π 0.35MPa( ) 0.08917 m( )2 − 02[ ]= 8743N

II. Uniform Wear ModelFrom Equation (17.12), and using the safety factor gives

nsTw = πµri pmax ro2 − ri

2( )ro = nsTw

πµri pmax+ ri

2

1 / 2

= 1.3( ) 140Nm( )π 0.35( ) 0.35MPa( )

1

ri+ ri

2

1 / 2

= 4.73 ×10−4 m3 1

ri+ ri

2

1 / 2

Taking derivative with respect to ri as before, setting equal to zero and solving for ri gives:

∂ro∂ri

= 12

4.73 ×10−4m3 1ri

+ ri2

−1 / 2

−4.73 ×10−4 m3 1

ri2 + 2ri

= 0

Solving this numerically gives ri=0.0618m, so that ro=0.107m. From Equation (17.13), themaximum actuating force is (remember to use the safety factor)

nsTw =µPw ro + ri( )

2;Pw = 2nsTw

µ ro + ri( ) = 2 1 . 3( ) 140 Nm( )0.35( ) 0.107 m + 0.0618m( ) = 6160N

17.3 The brakes used to stop and turn a tank are built like a multiple disk clutch with threeloose disks connected through splines to the drive shaft and four flat rings connected tothe frame of the tank. The brake has an outer contact diameter of 600mm, an innercontact diameter of 300mm, and six contact surfaces. The coefficient of friction of thebrake is 0.12, and the friction between the caterpillar and the ground is 0.16, which givesa braking torque of 12,800 Nm to block one caterpillar track so that it slides along theground. Calculate the force needed to press the brake disks together to block onecaterpillar track. Also calculate the force when the brake is new.

Notes: When new, the brakes are better characterized by the uniform pressure model of Equation(17.5). After the brakes are in service, however, the brake is better modeled through Equation(17.13) of the uniform wear model.

Solution:Note that there are six contact surfaces. Therefore, the torque per contact surface is 12,800Nm/6=2133Nm. From the uniform wear model, the force needed can be calculated from Equation(17.13) as:

Tw =µPw ro + ri( )

2; Pw = 2Tw

µ ro + ri( ) = 2 2133Nm( )0.12( ) 0.3m + 0.15m( ) = 79kN

For new brakes, the uniform pressure model is better, so Equation (17.5) gives

Tp =2µPp ro

3 − ri3( )

3 ro2 − ri

2( ) ; Pp =3Tp ro

2 − ri2( )

2µ ro3 −ri

3( ) =3 2133Nm( ) 0.3m( )2 − 0.15m( )2( )

2 0.12( ) 0.3m( )3 − 0.15m( )3( ) = 76.2kN

Page 17-3

17.4 A disk brake used in a printing machine is designed as shown in sketch b. The brake padis mounted on an arm that can swivel around point O. Calculate the braking torque whenthe force P=5000N. The friction pad is a circular cross section with the inner radius equalto half the outer radius. Also, a=150mm, b=50mm, D=300mm, and µ=0.25. The wear ofthe brake lining is proportional to the pressure and the sliding distance.

Notes: One equation relates the wear to the product of pressure and sliding distance, anotherrelates wear to the angular rotation. This allows for the pressure to be determined as a function ofφ.

Solution:Since the wear is proportional to the pressure and the sliding distance (which is l=rωt), the wearvolume at some radius r can be written as

dV=C1prωtdAAlso, from the geometry,

dV=C2rcosφTherefore, equating dV/dA gives

dVdA

= C1prωt = C2r cosφ ; p = C2

C1ωtcosφ = C3 cosφ

Note that we have taken ω and t as constants because they do not rely on radius or angularposition. From Equation (17.3), the torque is

T = prdφdrµr

ri

ro

∫ = C3µ r2dr

ri

ro

∫ cos φdφ

−α /2

α / 2

∫ = 23

C3µ ro3 − ri

3( )sinα2

The actuating moment is

Pa = prdφdrr cosφ = C3

6ro

3 − ri3( ) α + sinα( )

−α / 2

α /2

∫ri

ro

∫Substituting into the torque expression gives

T = 4µPasin

α2

α + sinα( ) = 4 0.25( ) 5000N( ) 0.15m( ) sin30°π3

+ s i n 6 0°

=196Nm

Page 17-4

17.5 A disk brake for a flywheel is designed as shown in sketch c. The hydraulic pistonsactuating the brake need to be placed at a radius rp so that the brake pads wear evenlyover the entire contact surface. Calculate the actuating force P and the radius so that theflywheel can be stopped within 4s when it rotates at 1000rpm and has a kinetic energy of5 x 10

5 Nm. The input parameters are µ=0.3, α=30°, ro=120mm, ri=60mm.

Notes: This is solved with the uniform wear model, so that equation (17.7) gives an expressionbetween pressure and radius. Equations (17.2) and (17.3) are then used to obtain a solution.

Solution:First of all, ω=1000rpm=104.7rad/s. From Equation (17.7), p=C/r so C=pr. Therefore, fromEquation (17.2),

P = prdθdr

0

α

∫ri

ro

∫ = Cdθdr

0

α

∫ri

ro

∫ = Cα ro − ri( )

Similarly, from Equation (17.3), noting that there are two friction surfaces,

T = 2 µpr2drdθ

ri

ro

∫0

α

∫ = µCα ro2 − ri

2( )Substituting into the expression for P gives:

P = Tµ ro + ri( )

The kinetic energy of a flywheel is from Equation (11.65), Ke=Jω2/2. The torque causes a

decelleration ofT=-Jdω/dt, or Tdt=-Jdω.

Integrating,

T dt

0

t

∫ = −J dω

ω m

0

∫ ;T = Jωm

t= Jω m

2

22

ω mt= Ke

2ω mt

Therefore

Page 17-5

T dt

0

t

∫ = −J dω

ω m

0

∫ ;T = Jωm

t= Jω m

2

22

ω mt= Ke

2ω mt

= 5×105 Nm( ) 2104.7rad / s( ) 4s( ) = 2390Nm

Therefore, the actuating force is

P = Tµ ro + ri( ) = 2390Nm

0.3( ) 0.12m + 0.06m( ) = 44.3kN

To calculate the position rp, note that

rpP = prdφdrr cosφ

−α / 2

α /2

∫ri

ro

∫ = C ro2 − ri

2( ) sinα2

Since P=Cα(ro-ri),

rp =C ro

2 − ri2( )sin

α2

Cα ro − ri( ) =ro + ri( )sin

α2

α= 0.12m + 0.06m( )sin15°

π / 6= 0.089m

17.6 Three pairs of thrust disk clutches are mounted on a shaft. Each has a pair of frictionalsurfaces. The hardened-steel clutches are identical, with an inside diameter of 100 mmand an outside diameter of 245mm. What is the torque capacity of these clutches basedon a) uniform wear and b) uniform pressure?

Notes: This problem is straightforward and is similar to Problem 17.3. However, one must obtaindata for the contacting materials from Table 17.1 on page 787.

Solution:From Table 17.1 on page 787, for hard steel µ is between 0.15 and 0.25, so we take µ=0.2 as anaverage value. Also from Table 17.1, pmax is between 690 and 1720kPa, so take pmax=1200kPa.There are six friction surfaces. Therefore, for the uniform wear model, the torque is given byEquation (17.12) as

Tw = 6πµripmax ro2 − ri

2( ) = 6π 0.2( ) 0.05 m( ) 1200kPa( ) 0.1225m( )2 − 0.05 m( )2( ) = 2839Nm

For the uniform pressure model, Equation (17.5) gives

Tp = 62πµp0

3ro

3 − ri3( ) = 12π 0.2( ) 1200kPa( )

30.1225m( )3 − 0.005m( )3( ) = 5167Nm

17.7 A pair of disk clutches has an inside diameter of 250mm and an outside diameter of 420mm. A normal force if 18.5kN is applied and the coefficient of friction of the contactingsurfaces is 0.215. Using the uniform wear and uniform pressure assumptions determinethe maximum pressure acting on the clutches. Which of these assumptions wouldproduce results closer to reality?

Notes: Equations (17.4) and (17.10) are needed to solve this problem. Two clutches can bearranged so that each takes the full applied load or that each takes one-half the load. This solutionconsiders the case when each clutch sees the applied load.

Solution:For the uniform pressure model, Equation (17.4) gives

Page 17-6

Pp = πp0 ro2 − ri

2( ); p0 =Pp

π ro2 − ri

2( ) = 18.5kN

π 0.21m( )2 − 0.125m( )2( ) = 206.8kPa

For the uniform wear model, Equation (17.10) gives

Pw = 2πpmaxri ro − ri( ); pmax = Pw

2πri ro − ri( ) = 18,500 N2π 0.125m( ) 0.21m − 0.125m( ) = 277kPa

17.8 A disk clutch is made of cast iron and has a maximum torque of 210 Nm. Because ofspace limitations the outside diameter must be minimized. Using the uniform wearassumption and a safety factor of 1.3, determinea) The inner and outer radii of the clutchb) The maximum actuating force needed.

Notes: The approach is the same as in Example 17.1 on page 787; an equation is obtained forouter radius in terms of inner radius. Taking the derivative and solving for the optimum gives theanswer. Equation (17.12) is used for the torque. Table 17.1 is used to get the friction coefficientand the maximum pressure.

Solution:From Table 17.1 for cast iron, an average value of friction coefficient is 0.20. The low range ofpressure is 690kPa. Equation (17.12) gives the torque as (note the inclusion of the safety factor):

Tw = πµripmax

nsro

2 − ri2( )

ro = nsTwπµri pmax

+ ri2

1 / 2

= 1.3( ) 210 Nm( )π 0.2( ) ri( ) 690kPa( )

+ ri2

1 / 2

= 6.30 ×10−4 m3 1ri

+ ri2

1 / 2

Taking derivative with respect to ri, setting equal to zero and solving gives:

drodri

= 12

6.30 ×10−4m3 1ri

+ ri2

−1 / 2

−6.30 × 10−4 m3 1

ri2 + 2ri

= 0; ri = 0.068m

Therefore, ro=0.118m. Therefore, the maximum actuating force needed is, from Equation (17.10),

Pw = 2π pmax

nsri ro − ri( ) = 2π 690kPa( )

1.30.068m( ) 0.118m − 0.068m( ) =11.3kN

17.9 A disk clutch produces a torque of 125Nm and a maximum pressure of 315kPa. Thecoefficient of friction of the contacting surfaces is 0.28. Assuming a safety factor of 1.8for maximum pressure and design the smallest disk clutch for the above constraints. Whatshould the normal force be?

Notes: The approach is very similar to Problem 17.8.

Solution:Equation (17.12) gives the torque as (note the inclusion of the safety factor):

Page 17-7

Tw = πµripmax

nsro

2 − ri2( )

ro = nsTwπµri pmax

+ ri2

1 / 2

= 1.8( ) 125 Nm( )π 0.28( ) ri( ) 315kPa( )

+ ri2

1 / 2

= 8.12 ×10−4 m3 1ri

+ ri2

1 / 2

Taking derivative with respect to ri, setting equal to zero and solving gives:

drodri

= 12

8.12 ×10−4m3 1ri

+ ri2

−1 / 2

−8.12 ×10−4 m3 1

ri2 + 2ri

= 0;ri = 0.074m

Therefore, ro=0.128m. Therefore, the maximum actuating force needed is, from Equation (17.10),

Pw = 2π pmax

nsri ro − ri( ) = 2π 315kPa( )

1.80.074m( ) 0.128m − 0.074m( ) = 4.39kN

17.10 A leather-face cone clutch is to transmit 1200lbf-in of torque. The half-cone angle α=10°,the mean diameter of the friction surface is 12in, and face width b=2in. For coefficient offriction µ=0.25 find the normal force P and the maximum contact pressure p by usingboth the uniform pressure and uniform wear models.

Notes: Equations (17.23) and (17.21) are used for the uniform pressure model. Equations (17.26)and (17.24) are used for the uniform pressure model. The diameters have to be determined asdefined in Figure 17.4.

Solution:The mean diameter is given as 12 inches. Referring to Figure 17.4 on page 789, it can be seen that

D = dave + 2b2

tan α

= 12in + 2 in( ) tan10° =12.35in

d = dave − 2b2

tan α

= 12in − 2 in( ) tan10° =11.65in

For the uniform pressure model, Equation (17.23) gives

T =µW D3 − d3( )

3sinα D2 − d2( ) ;W =3T sinα D2 − d2( )

µ D3 − d3( ) =3 1 2 0 0inlb( )sin10° 12.35in( )2 − 11.65in( )2( )

0.25( ) 12.35in( )3 − 11.65 in( )3( ) = 139lb

Therefore, from Equation (17.21),

W = πp0

4D2 − d2( ); p0 = 4W

π D2 − d2( ) = 4 1 3 9lb( )π 12.35in( )2 − 11.65in( )2( ) =10.5psi

For the uniform wear model, Equation (17.26) gives

T = µW4 s i nα

D + d( );W = 4T sinαµ D + d( ) = 4 1 2 0 0inlb( )sin10°

0.25( ) 12.35in + 11.65 in( ) =140lb

Note from Equation (17.9) that c=pmaxri=pmaxd/2. Therefore, from Equation (17.24),

W = πc D − d( )= π pmax( ) d2

D− d( ); pmax = 2Wπd D − d( ) = 2 140 lb( )

π 11.65in( ) 12.35 in− 11.65in( ) =10.9 psi

17.11 The synchronization clutch for the second gear of a car has a major cone diameter of50mm and a minor diameter of 40mm. When the stick shift is moved to second gear, thesynchronization clutch is engaged with an axial force of 100N, and the moment of inertia

Page 17-8

of 0.005kg-m2 is accelerated 200rad/s

2 in 1s to make it possible to engage the gear. The

coefficient of friction of the cone clutch is 0.09. Determine the smallest cone clutch widththat still gives large enough torque. Assume the clutches are worn in.

Notes: Since the clutch is worn in, we can use the uniform wear model. Equation (17.26) will beused to determine α, which then allows one to calculate b.

Solution:Note that the torque is the product of the moment of inertia and angular acceleration, so that

T=(0.005kgm2)(200rad/s

2)=1Nm

From Equation (17.26),

T = µW4 s i nα

D + d( );sinα = µW4T

D+ d( ) = 0.09( ) 100N( )4 1Nm( ) 0.05m + 0.04m( ) = 0.2025

or α=11.68°. Therefore, from Figure 17.4,

tan α = D − d2b

;b = D − d2tan α

= 0.05m − 0.04m2tan11.68°

= 0.0242m = 24.2mm

17.12 A safety brake for an elevator is a self-locking cone clutch. The minor diameter is120mm, the width is 60 mm, and the major diameter is 130mm. The force applying thebrake comes from a prestressed spring. Calculate the spring force needed if the 2-tonelevator must stop from a speed of 3m/s in a maximum distance of 3m while the coneclutch rotates five revolutions per meter of elevator motion. The coefficient of friction inthe cone clutch is 0.26.

Notes: Using the sketch of Figure 17.4, the cone angle α must be derived. The torque is obtainedfrom the absorbed energy, and the spring force then follows from Equation (17.23). Note also thatsince the problem statement is in metric units, a ton will be taken as 1000 kg.

Solution:From Figure 17.4, the cone angle is obtained as:

tan α = D − d2b

= 130mm − 120mm( )2 60mm( ) = 0.08333; α = 4.764°

The energy which must be absorbed by the clutch is the product of the torque and angulardisplacement, and this equals the energy change in the elevator over the same interval. Theenergy change in the elevator is the sum of the kinetic and potential energies, so that

Tθ = 12

mav2 + ma gh;T =

12

mav2 + magh

θ=

12

2000kg( ) 3m / s( )2 + 2000kg( ) 9.81m / s2( ) 3m( )15rev 2πrad / rev( )

or T=720Nm. Therefore, from Equation (17.23),

T =µW D3 − d3( )

3sinα D2 − d2( ) ;W =3T sinα D2 − d2( )

µ D3 − d3( ) =3 720Nm( )sin 4.764° 0.13m( )2 − 0.12m( )2( )

0.26( ) 0.13m( )3 − 0.12m( )3( ) = 3678N

17.13 A cone clutch is used in a car automatic gearbox to fix the planet wheel carriers to thegearbox housing when the gear is reversing. The car weighs 1300kg with 53% loading onthe front wheels. The gear ratio from the driven front wheelsto the reversing clutch is

Page 17-9

16.3:1 (i.e., the torque on the clutch is 16.3 times lower than the torque on the wheels ifall friction losses are neglected). The car wheel diameter is 550mm, the cone clutch majordiameter is 85mm and the minor diameter is 80mm, and the coefficients of friction are0.3 in the clutch and 1.0 between the wheel and the ground. Dimension the width of thecone clutch so that it is not self-locking. Calculate the axial force needed when the clutchis worn in.

Notes: Statics on the cone gives a relationship for a cone angle which will disengage whendesired. The torque on the wheel is calculated for spinning tires (the worst case scenario), reducedby the gear ratio, and then Equation (17.26) or (17.23) will allow for determination of W.

Solution:The forces acting on the cone clutch are sketched above for when the inner cone is trying todisengage from the outer cone (thus the direction of the friction force). From horizontalequilibrium, if µNcosα is greater than Nsinα, then the inner cone can’t disengage from the outercone. This means:

N sinα > Nµcosα ; t anα > µ; α > tan−1 0.3 = 0.291rad =16.7°Note from Figure 17.4,

tan α = D − d2b

= 0.085m − 0.080m2b

= 0.0025mb

Combining the two relationships,0.0025m

b> 0.3;b < 0.025m

0.3= 0.0083m

As for the specification of these values, one must consider the likelihood of the frictioncoefficient becoming larger than 0.3, and if so, by how much. If we assume that this is already aworst case scenario, then these values can be used as a design solution, although from amanufacturing perspective it may be better practice to specify a 20° cone angle and b=6mm.However, let us continue the analysis using b=0.0083 and α=16.7°. The torque on the ground asthe front wheels spin (the maximum torque) is:

T = 1300kg( ) 9.81m / s2( ) 0.53( ) 0.550m2

1.0( ) = 1860Nm

The torque at the clutch, because of the gear reduction, is 1860/16.3=114Nm. Therefore, fromEquation (17.23), the axial force required can be found as:

T =µW D3 − d3( )

3sinα D2 − d2( ) ;W =3T sinα D2 − d2( )

µ D3 − d3( ) =3 1 1 4Nm( )s in16.7° 0.085m( )2 − 0.08m( )2( )

0.3( ) 0.085m( )3 − 0.08m( )3( ) = 2646N

while Equation (17.26) would predict

T = µW4 s i nα

D + d( );W = 4T sinαµ D + d( ) = 4 114Nm( )sin16.7°

0.3( ) 0.085m + 0.08m( ) = 2647N

Which is not significantly different from the results of Equation (17.23).

17.14 A cone clutch has a major diameter of 328mm and a minor diameter of 310 mm, is 50mm wide and transfers 250Nm of torque. The coefficient of friction is 0.31. Using theassumptions of uniform pressure and uniform wear, determine the actuating force and thecontact pressure.

Notes: The cone angle is found from Equation (17.4). Equations (17.21) and (17.23) are neededfor the uniform pressure model, Equations (17.24) and (17.26) for the uniform wear model.

Page 17-10

Solution:From Figure 17.4,

tan α = D − d2b

= 328mm − 310mm2 50mm( ) = 0.18; α = 10.2°

For the uniform pressure model, Equation (17.23) can be solved for the force as:

T =µW D3 − d3( )

3sinα D2 − d2( ) ;W =3T sinα D2 − d2( )

µ D3 − d3( ) =3 250Nm( )sin10.2° 0.328m( )2 − 0.310m( )2( )

0.31( ) 0.328m( )3 − 0.310m( )3( )or W=895N. The pressure is obtained from (17.21) as

W = πp0

4D2 − d2( ); p0 = 4W

π D2 − d2( ) = 4 895N( )π 0.328m( )2 − 0.31m( )2( ) = 99.3kPa

For the uniform wear model, Equation (17.26) gives

T = µW4 s i nα

D + d( ); W = 4T sinαµ D + d( ) = 4 250Nm( )sin10.2°

0.31( ) 0.328m + 0.30 m( ) = 896N

Therefore, noting from Equation (17.9) that c=pmaxri=pmaxd/2, the maximum pressure is obtainedfrom Equation (17.24) as

W = πpmaxd D − d( )2

; pmax = 2Wπd D − d( ) = 2 896N( )

π 0.30m( ) 0.318m − 0.30m( ) = 102kPa

17.15 The coefficient of friction of a cone clutch is 0.25. It can support a maximum pressure of410 kPa while transferring a maximum torque of 280Nm. The width of the clutch is65mm. Minimize the major diameter of the clutch. Determine the clutch dimension andthe actuating force.

Notes: The solution is obtained from Equations (17.25) and (17.9).

Solution:From Equation (17.9) note that c=pmaxri=pmaxd/2, so that Equation (17.25) gives:

T = πµpmaxd8sin α

D2 − d2( )Also, since from Figure 17.4,

tan α = D − d2b

; s i nα = D− d

2D − d

2

2+ b2

Substituting into the torque relation gives a complicated function of D and d. Note that at d=0, thetorque transmitted is zero since the maximum pressure is at the center and is infinite for a finitetorque transferred. Therefore, d must be greater than zero. A plot of the required D as a functionof d is shown below. The outer diameter is minimum as it approaches d, so that a approacheszero. Using D=d=0.164 is not practical to manufacture; use D=0.165m and d=0.163m just toillustrate that W in this case is 210N.

Page 17-11

17.16 A block brake is used to stop and hold a rope used to transport skiers from a valley to thetop of a mountain. The distance between cars used to transport the skiers is 100m, thelength of the rope from the valley to the top of the mountain is 4 km, and the altitudedifference is 1.4km. The rope is driven by a V-groove wheel with a diameter of 2m. Therope is stopped and held with a block brake mounted on the shaft of the V-groove wheel,shown in sketch d. Neglect all friction in the different parts of the ropeway except frictionin the driving sheave, and assume that the slope of the mountain is constant. Dimensionthe brake for 20 passengers with each passenger’s equipment weighing 100 kg, andassume that all ropeway cars descending from the top are empty of people. The directionof rotation for the drive motor is shown in the figure. Calculate the braking force Wneeded to hold the ropeway still if all passengers are on the way up. Do the samecalculation if all passengers continue on down to the valley with the ropeway. Thecoefficient of friction in the brake is 0.23.

Notes: The cars and the rope serve as counterweights for each other, so the passengers are theonly load that need to be considered. The sketch shows rotation for the drive motor; whenpassengers going uphill are stopped, the rotation will be in the opposite direction. During theascent of the mountain, the brake is self-energizing; Equation (17.28) allows for calculation of therequired force. During descent, the brake is de-energizing, so one term changes in sign.

Page 17-12

Solution:Note that the forces on the short shoe have been added to the sketch for the angular velocityshown. The slope of the mountain is sinα=1400m/4000m; α=20.49°. If the distance between carsis 100m and the total distance is 4000 m, then there are 40 cars going in each direction. The totalweight of all passengers is

Wt=(20passengers/car)(40cars)(100kg/passenger)(9.81m/s2)=784.6kN.

From statics, the load in the cable is Wtsinα=(784.6kN)sin20.49°=274.6kN. The torque on thedriving wheel is then T=Pr=(274.6kN)(1.0m)=274.6kNm.

If the brake holds the passengers as they are going uphill, the brake rotation will be in theopposite direction as the drive motor, and opposite that shown. It is self-energizing. Momentequilibrium about the pin (see sketch) gives

W(0.45m+0.90m)-P(0.45m)+µP(0.15m)=0The braking force P can be obtained from Equation (17.28),

T = µrP;P =T

µr=

274.6kNm

0.23( ) 0.50m( )= 2.388MN

so that W is 735kN.Going downhill, if there are passengers going both uphill and downhill, the braking force

decreases, since the passengers serve as counterweights for each other. However, if there arepassengers only on the downhill side, the brake needs to hold them in position, and the brakedrum rotation will be opposite that shown. Therefore, the friction force direction changes, but therequired torque is still T=274.6kN, and P=2.388MN. The moment equilibrium equation gives Was:

W 1.35m( )− P 0.45 m( )+ µP 0.15m( );W =P 0.45m + µ 0.15m( )( )

1.35m= 857kN

17.17 The motion of an elevator is controlled by an electric motor and a block brake. On oneside, the rotating shaft of the electric motor is connected to the gearbox driving theelevator, and on the other side it is connected to the block brake. The motor has twomagnetic poles and can be run on either 60- or 50- Hz electricity (3600 or 3000 rpm).When the elevator motor is driven by 50Hz electricity, the braking distance needed tostop is 52cm when going down and 31cm when going up with maximum load in theelevator. To use it with 60-Hz electricity and still be able to stop exactly at the differentfloor levels without changing the electric switch positions, the brake force at the motorshould be changed. How should it be changed for going up and going down? The brakegeometry is like that shown in Fig. 17.5 with d1=0.030m, d3=0.100m, d4=0.400m,r=0.120m, and µ=0.20. Only the inertia of the elevator must be considered, not therotating parts. Make the brake self-energizing when the elevator is going down.

Notes: The energy change in the elevator must be calculated as in Problem 17.12. This problemcan be solved by expressing the ratio of the equations for the elevator going up and going down.In doing so, one must realize that the angular rotation of the drum is linearly proportional to thelinear displacement of the elevator. The best that can be done with the information given is toexpress the required loads at 60Hz in terms of the forces at 50Hz.

Solution:The product of torque and angular displacement of the drum is the energy dissipated, which mustequal the energy of the elevator car. Therefore, while using Equation (17.28) for the descendingcar, energy conservation requires:

Page 17-13

mav2

2+ magh1 = T1φ1 = µrd4W

d3 − µd1φ1

where φ1 is the angular rotation in the brake when the brake is actuated. When raising the car, thesame requirement yields (note: the brake is now de-energizing so Equation (17.30) is used, andthe height change is in the opposite direction):

mav2

2− magh2 = T2φ2 = µrd4W

d3 + µd1φ2

Dividing the first equation by the second:

mav2

2+ magh1

mav2

2− magh2

=

µrd4W

d3 − µd1φ1

µrd4W

d3 + µd1φ2

;

v2

2+ gh1

v2

2− gh2

=

φ1

d3 − µd1φ2

d3 + µd1

= d3 + µd1

d3 − µd1

φ1

φ2

Note that the ratio of angular rotations is the same ratio of the ratio of height changes; thereforeφ1/φ2=h1/h2. Therefore, substituting g=9.81m/s

2, h1=0.52m, h2=0.31m, d3=0.100m µ=0.20, and

d1=0.030m,

v2

2+ 9.81m / s2( ) 0.52m( )

v2

2− 9.81m / s2( ) 0.31m( )

= 0.100m + 0.20( ) 0.030m( )0.10m − 0.20( ) 0.030m( )

0.520.31

;v = 4.93m / s

With a 60Hz power source, the car velocity will be

v2 = 6050

v = 6050

4.93m / s( ) = 5.92m / s

If the stopping distance is the same, the ratio of descending equations for the 50Hz divided by the60Hz cases are:

mav2

2+ magh1

mav22

2+ magh1

=

µrd4W50

d3 − µd1φ1

µrd4W60

d3 − µd1φ1

;

v2

2+ gh1

v22

2+ gh1

=W50

W60

or,

W60 =

v22

2+ gh1

v 2

2+ gh1

W50 =

5.92m / s( )2

2+ 9.81m / s2( ) 0.52m( )

4.93m / s( )2

2+ 9.81m / s2( ) 0.52m( )

W50 = 1.31W50

Similarly, if the car is going up,

mav2

2− magh1

mav22

2− magh1

=

µrd4W50

d3 + µd1φ1

µrd4W60

d3 + µd1φ1

;

v 2

2− gh1

v22

2− gh1

=W50

W60

W60 =

v22

2+ gh1

v 2

2+ gh1

W50 =

5.92m / s( )2

2− 9.81m / s2( ) 0.52 m( )

4.93m / s( )2

2− 9.81m / s2( ) 0.52m( )

W50 = 1.59W50

Page 17-14

17.18 An anchor winding is driven by an oil hydraulic motor with a short-shoe brake to stop theanchor machinery from rotating and letting out too much anchor chain when the windmoves the ship. The maximum force transmitted from the anchor through the chain is1.1MN at a radius of 2m. Figure 17.5 describes the type of block brake used, which isself-energizing. Calculate the brake force W needed when the brake dimensions ared1=0.9m, d3=1.0m, d4=6m, r=3m, and µ=0.31. Also calculate the contact force betweenthe brake shoe and the drum.

Notes: Equations (17.22) and (17.28) are needed to solve this problem.

Solution:The torque which is applied to the shaft containing the brake is

T=(1.1MN)(2m)=2.2MNmTherefore, from Equation (17.28) for a self-energizing shoe,

T = µrd4Wd3 − µd1

;W =T d3 − µd1( )

µrd4=

2.2MNm( ) 1.0m − 0.31( ) 0.9m( )( )0.31( ) 3m( ) 6m( ) = 284kN

Therefore, the normal force can be calculated from Equation (17.27) as:

P = d4Wd3 − µd1

= 6m( ) 284kN( )1.0m − 0.31( ) 0.9m( ) = 2.36MN

17.19 The hand brake shown in sketch e has an average pressure of 600kPa across the shoe andis 50mm wide. The wheel runs at 150rpm and the coefficient of friction is 0.25.Dimensions are in millimeters. Determine the following:a) If x=150mm, what should the actuating force be?b) What value of x causes self-locking?c) What torque is transferred?d) If the direction of motion is reversed, how would the answers to parts a to c change?

Notes: One uses the uniform pressure approach in this problem. After determining the forceacting on the cylinder, the rest of the problem is solved through statics.

Solution:Note that 80°=1.40rad. The normal force acting on the brake is

N = pbrdθ = 600kPa( ) 0.050m( ) 0.17m( ) 1.40 rad( ) = 7121N

−0.70rad

0.70 rad

∫Assume that this force acts at the center of the shoe. Summing moments about the pin:

N 0.5m( ) − µN x( ) − Pn c o s 2 0° 1.250m( ) + Pn sin20° 0.1m + x − 0.150m tan20°( ) = 0Substituting x=0.150m, then Pn=2.97kN.

Page 17-15

For self-locking, Pn is zero, so the moment equation would give

N 0.5m( ) − µNx = 0;x = N 0.5m( )µN

= 0.5m0.25

= 2.0m

The torque transferred isT=µNR=(0.25)(7121N)(0.17m)=303Nm

If the direction is reversed, then self locking cannot occur; N is unchanged, therefore the torquetransferred is unchanged; the moment equation becomes:

N 0.5m( ) + µN x( ) − Pn c o s 2 0° 1.250m( ) + Pn sin20° 0.1m + x − 0.150m tan20°( ) = 0Therefore, for x=0.15m, Pn is found to be 3455N.

17.20 The short-shoe brake shown in sketch fhas an average pressure of 1 MPa and acoefficient of friction of 0.32. The shoe is250 mm long and is 45 mm wide. Thedrum rotates at 310 rpm and has adiameter of 550 mm. Dimensions are inmillimeters.a) Obtain the value of x for the self-locking condition.b) Calculate the actuating force ifx=275mm.c) Calculate the braking torque.d) Calculate the reaction at point A.

Notes: This equation is solved through statics only. The main complication is that this geometryhas both a self-energizing and deenergizing shoe.

Solution:The only shoe that can self-lock is the self-energizing shoe, which in this case is the top one.For self-locking, P=0. Also, N=pbl=(1MPa)(0.25m)(0.045m)=11.25kN. Then, with µ=0.32,moment equilibrium about point A gives:

N 0.58m( ) − µNx − P 1.39m( ) = 0; x = N 0.58m( )µN

= 0.58m0.32

=1.81m

If x=0.275m, solving the moment equilibrium equation for P gives

N 0.58m( ) − µNx − P 1.39m( ) = 0; P =N 0.58m − µx( )

1.39m=

11.25kN( ) 0.58m − 0.32( ) 0.275m( )( )1.39m

= 3982N

With this actuating force, N for the bottom shoe is, from moment equilibrium for the lower shoe,

N 0.58m( ) + µNx − P 1.39m( ) = 0; N =P 1.39m( )

0.58m + µx= 3982N( ) 1.39m( )

0.58m + 0.32( ) 0.275m( ) = 8286N

Therefore the torque isT=µ(Ntop+Nbottom)r=(0.32)(11.25kN+8.286kN)(0.55m)=3.44kNm

The reactions have been added to the sketch in red. The reaction at point A for the top hinge isobtained by applying force equilibrium for the top shoe:

Fx = 0 = Rx + µNtop ;Rx = − 0.32( ) 11.25kN( )= 3.6kN∑Fy = 0 = Ry + Ntop − P;Ry = P − Ntop = 3.982kN −11.25kN = −7.268kN∑

For the bottom shoe:Fx = 0 = Rx − µNbottom ; Rx =∑ µNbottom = 0.32( ) 8.286N( ) = 2.65kN

Fy = 0 = Ry − Nbottom + P; Ry =∑ Nbottom − P = 8.286kN − 3.982kN = 4.30kN

Page 17-16

17.21 The brake on the rear wheel of a car is the long shoe internal type. The brake dimensionsaccording to Figure 17.7 are θ1=10°, θ2=120°, r=95mm, d7=73mm, d6=120mm, andd5=30mm. The brake shoe lining is 38mm wide, and the maximum allowable contactpressure is 5MPa. Calculate the braking torque and the fraction of the torque producedfrom each brake shoe when the brake force is 5000N. Also calculate the safety factoragainst contact pressure that is too high. The coefficient of friction is 0.29.

Notes: Consider the self-energizing and de-energizing shoes separately. MP and MF have to becalculated from (17.34) and (17.35). Equations (17.36) and (17.43) then give the maximumpressure in the shoes. Equation (17.38) gives the torque for each shoe.

Solution:I. Self Energizing ShoeTo apply Equation (17.36), we first need to calculate MF and MP. Note that from the discussion onpage 794, θa=90°. From Equation (17.34),

Mp =brd7 pmax

4sinθa2 θ2 −θ1( ) π

180°− s i n 2θ2 +sin 2θ1

=0.038m( ) 0.095m( ) 0.073m( )

4 s i n 9 0°2 1 2 0° −10°( ) π

180°− sin240 °+ sin20°

pmax = 3.326 ×10−4 m3( )pmax

and from Equaton (17.35),

MF = µpmaxbrsinθa

−r cosθ2 − cosθ1( ) − d72

sin2 θ2 − sin 2θ1( )

= 0.29( ) 0.038( ) 0.095( )s i n 9 0°

− 0.095( ) cos120°− cos10°( )− 0.073m

2sin2 120° −sin210°( )

= 1.202 ×10−4 m3( ) pmax

From Equation (17.36),

−Wd6 − MF + MP = 0; − 5000N( ) 0.12m( ) − 1.202 ×10−4m3( ) pmax + 3.326 ×10−4m3( )pmax

pmax = 5000N( ) 0.12m( )2.124 ×10−4m3 = 2.82MPa

The torque for the self energizing shoe is obtained from Equation (17.38) as

T =µpmaxbr2 cosθ1 − cosθ2( )

sinθa= 0.29( ) 2.82MPa( ) 0.038m( ) 0.095m( )2 cos10° −cos120°( )

sin90°or T=416Nm.II. De-Energizing ShoeThe values of MF and MP do not change. However, the pressure is obtained from Equation (17.43)as:

−Wd6 + MF + MP = 0; − 5000N( ) 0.12m( ) + 1.202 ×10−4m3( ) pmax + 3.326 × 10−4m3( ) pmax

pmax = 5000N( ) 0.12m( )4.528 ×10−4m3 = 1.325MPa

The torque for the deenergizing shoe is, from Equation (17.38),

T =µpmaxbr2 cosθ1 − cosθ2( )

sinθa= 0.29( ) 1.325MPa( ) 0.038m( ) 0.095m( )2 cos10°− cos120°( )

s i n 9 0°or T=196Nm.III. SummaryThe total torque is 416+196=612Nm. The self-energizing shoe contributes 416/612=68.0% of thetorque, the de-energizing shoe only 32%. The safety factor against contact pressure is calculatedfrom the self-energizing shoe as ns=5Mpa/2.82MPa=1.77.

Page 17-17

17.22 The maximum volume of the long-shoe internal brake on a car is given as 10-3m

3. The

brake should have two equal shoes, one self-energizing and one deenergizing, so that thebrake can fit on both the right and left sides of the car. Calculate the brake width andradius for maximum braking power if the space available inside the wheel is 400mm indiameter and 100mm wide. The brake lining material has a maximum allowable contactpressure of 4MPa and a coefficient of friction of 0.18. Also, calculate the maximumbraking torque.

Notes: This problem uses Equation (17.36) to obtain an expression for the actuation force. Thetorque equation (17.38) suggests that θ2-θ1 be maximized to maximize the torque, which leads tothe angular extent of the shoes.

Solution:From Equations (17.34), (17.35) and (17.37),

W = pmaxbrd6 sinθa

d7

42 θ2 −θ1( ) π

180°− sin2θ2 + s i n 2θ1

− µ −r cosθ2 − cosθ1( )− d7

2sin2 θ2 −sin 2θ1( )

Note from Equation (17.38) that the braking torque increases when θ2-θ1 increases. This wouldsuggest setting θ2=180° and θ1=0°, but this leads to obvious mounting problems. Therefore, letθ2=175° and θ1=5°. The brake volume is

V=πr2b=10

-3m

3.

From Equation (17.38),

T =µpmax 10−3( )

π sinθacosθ1 − cosθ2( )

Ttotal = Ts + Td = pmax,s + pmax,d( ) 0.18( ) 0.3183× 10−3( ) 2( ) 0.9962( )[ ]=1.142 ×10−4 pmax,s + pmax,d( )

The actuating force W is the same for the two shoes. Therefore, pmax is largest when r/d7 is thesmallest, so r/d7=1.1. Therefore,

Wd6 sinθbrd7

= 4.702 ×106 = pmax,d 1.570 + 1.992( ) 0.18( ) 1.1( )[ ]; pmax,d = 2.394MPa

Ttotal=1.142x10-4(4+2.394)(10

6)=730Nm.

17.23 A long-shoe brake in a car is designed to give as high a braking torque as possible for agiven force on the brake pedal. The ratio between the actuating force and the pedal forceis given by the hydraulic area ratio between the actuating cylinder and the cylinder underthe pedal. These brake shoe angles are θ1=10°, θ2=170°, and θa=90°. The maximum brakeshoe pressure is 5MPa, the brake shoe width is 40mm and the drum radius is 100mm.Find the distance d7 that gives the maximum braking power for a coefficient of friction of0.2 at any pedal force. What braking torque would result if the coefficient of friction were0.25?

Notes: Equations (17.34) and (17.35) are evaluated to obtain MF and MP. Since the actuating forceis the same for both shoes, Equation (17.37) can be equated to (17.44). Solving for the d7 whichgives W=0 gives a self-locking brake, which is the maximum braking power.

Page 17-18

Solution:Equation (17.34) gives

Mp = brd7 pmax

4sinθa2 θ2 − θ1( ) π

180°− sin2θ2 + sin 2θ1

= 0.04m( ) 0.1m( )d7 5MPa( )4 1( )

2 1 7 0° −10°( ) π180°

− sin340 °+ sin20°

= 31.35kN( )d7

Similarly, from Equation (17.35),

MF =µpmaxbr

sinθa−r cosθ2 − cosθ1( ) −

d7

2sin2 θ2 − sin2 θ1( )

= 0.2( ) 5MPa( ) 0.04( ) 0.1m( )sin90°

− 0.1( ) cos170° −cos10°( ) − d7

2sin2 170° −sin210°( )

= 788Nm

The actuating force is, from Equation (17.37),

W =MP − MF

d6=

31.35kN( )d7 − 788Nm

d6The maximum torque occurs when it self-locks, or when W=0. Solving the numerator for d7 givesd7=0.025m=25mm. For µ=0.25, the brake locks and cannot be rotated.

17.24 A long shoe internal brake similar to that shown in Figure 17.7 should be optimized formaximum wear life. The brake lining wear is proportional to the sliding distance timesthe contact pressure times the coefficient of friction. The actuating force, applied by ahydraulic cylinder, is equally large for the two shoes. The contact pressure is limited to4MPa. The two brake shoes are made geometrically equal to be able to use the brakeequally well for both directions of rotation, but the coefficient of friction might bedifferent. The brake is 120mm in radius and 30 mm wide. The brake torque needed is900Nm. Assume that θ1=20°, θ2=160°, and d7=85mm.

Notes: Equations (17.34) and (17.35) are needed to obtain MF and MP. Since the same actuatingforce is applied to both shoes, Equations (17.37) and (17.44) can be set equal to each other. Theresulting expression, as well as the stated wear model, allows determination of the frictioncoefficients which optimize the design.

Solution:From Equation (17.34),

Mp = brd7 pmax4sinθa

2 θ 2 − θ1( ) π180°

− sin2θ2 + s i n 2θ1

= 0.030m( ) 0.12m( ) 0.085m( )4 s i n 9 0°

2 1 6 0° −20°( ) π180°

− sin320 °+ sin40°

pmax = 4.722 ×10−4 m3( )pmax

From Equation (17.35),

MF =µpmaxbr

sinθa−r cosθ2 − cosθ1( ) −

d7

2sin2 θ2 − sin2 θ1( )

=µ 0.030m( ) 0.120m( )

s i n 9 0°− 0.120m( ) cos160° −c o s 2 0°( ) −

0.085m

2sin2 160° − sin2 20°( )

pmax

= 8.12 × 10−4m3( )µpmax

Page 17-19

Now differentiate the shoes with the subscripts s for self engergizing and d for deenergizing.Since the actuating force is constant, Equations (17.37) and (17.44) are equated as:

MPs − MFs = MPd + MFd

4.722 × 10−4m3( )pmax,s − 8.12 × 10−4 m3( )µspmax,s = 4.722 × 10−4m3( )pmax,d + 8.12 × 10−4 m3( )µd pmax,d

4.722 × 10−4m3( ) − 8.12 × 10−4 m3( )µs[ ] pmax,s = 4.722 × 10−4 m3( ) + 8.12 × 10−4m3( )µd[ ]pmax,d

For the same amount of wear on each shoe, from the given relationship, this meanspmax,dµd=pmax,sµs, so

4.722 ×10−4 m3( ) − 8.12 × 10−4m3( )µs[ ]pmax,s = 4.722 × 10−4m3( ) + 8.12 × 10−4 m3( )µd[ ]pmax,sµs

µdFor equal wear, µpmax is equal for the two shoes, therefore,

µd =µs

1 − 3.439µsIf the maximum pressure is 4MPa, torque equilibrium gives µ s=0.1995 and µd=0.6359.

17.25 An external drum brake assembly (see sketch g) has a normal force P=200lbf acting onthe lever. Dimensions are in inches. Assume that the coefficient of friction µ=0.25 andmaximum contact pressure p=100psi. Determine the following from long-shoecalculations:a) Free-body diagram with the directionality of the forces acting on each component.b) Which shoe is self-energizing and which is deenergizing.c) Total braking torqued) Pad width as obtained from the self energizing shoe (deenergizing shoe width equalsself energizing shoe width).e) Pressure acting on the deenergizing shoe.

Notes: MP and MF are calculated as with internal long shoe brakes. Equation (17.38) gives thetorque. The free body diagrams of components are shown:

Page 17-20

Solution:The free body diagrams are as shown, with forces as determined from static equilibrium. Notefrom the sketch that θ1=30°, θ2=150°, θa=90°, d7=17in, r=15in. d6 for the top shoe is simply 40 in,since the force is vertical. The force is not vertical for the bottom shoe, so we need to calculate d6

for the bottom shoe. Note from the triangle that tanα=1000/200=5, so that α=78.64°. Also,tanα=4/x, so that x=0.8in. Finally, d6=(35.5in)sinα=34.52in. The applied force on the bottomshoe is 1020lb, while that on the top is 1000lb.

Note that on the top shoe, the moment due to friction and the applied torque act in theopposite direction, while on the bottom they act in the same directions. Therefore, the bottomshoe is self-energizing.

From Equation (17.34),

Mp = brd7 pmax4sin θa

2 θ2 −θ1( ) π180°

− sin2θ2 + s i n 2θ1

= 15in( ) 17in( )4 s i n 9 0°

2 150° − 30°( ) π180°

− s in300°+ sin60°

bpmax = 377in2( )bpmax

From Equation (17.35),

MF = µpmaxbrsinθa

−r cosθ2 − cosθ1( ) − d72

sin2 θ2 − sin2 θ1( )

= 0.25( ) 15in( )sin90°

− 15in( ) cos150° −c o s 3 0°( )− 17in

2sin2 150° −sin2 30°( )

bpmax = 97.4in2( )bpmax

Since the bottom shoe is self energizing, and p=100psi on the self energizing shoe, Equation(17.37) gives

Wd6 = MP − MF = 377in2( )bpmax − 97.4 in2( )bpmax = 279.6in2( )bpmax

b = Wd6

279.6 in2( )pmax

= 1020 lb( ) 34.52in( )279.6 in2( ) 100 psi( )

= 1.26in

It is good design practice to prescribe a length which is the next largest one-eight or one-quarterinch increment. However, we will progress with the problem using this value of the brake width.For the deenergizing (top) shoe, Equation (17.4) gives:

Wd6 = MP + MF = 377in2( )bpmax + 97.4 in2( )bpmax = 474.4in2( )bpmax

pmax = Wd6

474.4in2( )b = 1000lb( ) 40 in( )474.4in2( ) 1.26in( )

= 67.0 psi

From Equation (17.38), applied separately to the top and bottom shoes:

Page 17-21

Tb =µpmaxbr2 cosθ1 − cosθ2( )

sinθa=

0.25( ) 100psi( ) 1.26 in( ) 15in( )2 c o s 3 0°− cos150°( )sin90°

= 12.25 kipin

Tt =µpmaxbr2 cosθ1 − cosθ2( )

sinθa=

0.25( ) 67.0psi( ) 1.26in( ) 15in( )2 cos30° −cos150°( )s i n 9 0°

= 8.20kipin

Therefore, the total torque is T=Tb+Tl=12.25kipin+8.20kipin=20.45kipin.

17.26 A long-shoe external brake as shown in Fig. 17.10 has a pivot point such that d7=4r,d6=2r, θ1=5°, and θ2=45°. Find the coefficient of friction needed to make the brake self-lock if the rotation is in the direction shown in Fig. 17.10. If the shaft rotates in theopposite direction, calculate the drum radius needed to get a braking torque of 180 Nmfor the actuating force of 10000N.

Notes: MP and MF are calculated as before, but r and µ are unknowns in the expressions. Equation(17.47) gives a requirement for self locking. Equation (17.38) then gives an expression for thetorque.

Solution:From Equations (17.47), (17.34) and (17.35), self locking occurs when W≤0, which requiresbrpmax

sinθa

4r

42 45° − 5°( ) π

180°− sin90° + sin10°

+ µ −r cos45° −cos5°( ) −

4r

2sin2 45° −sin2 5°( )

≤ 0

or µ≥0.8192. Equation (17.38) then gives

Wd6 =br2 pmax

sinθa0.5699 − µ 0.2891 − 0.9848( )[ ];r = 0.04332m = 43.32 mm

17.27 An external, long brake shoe is mounted on an elastic arm. When a load is applied, thearm and the brake lining bend and redistribute the pressure. Instead of the normal sinepressure distribution, the pressure becomes constant along the length of the lining. For agiven actuating force calculate how the brake torque changes when the pressuredistribution changes from sinusoidal to a constant pressure. Also assume that d7=110mm,r=90mm, b=40mm, θ1=20°, θ2=160°, d6=220mm, µ=0.25 and W=12kN.

Notes: This problem requires reformulation of Equations (17.34) and (17.35) in terms of aconstant pressure. Equation (17.47) will give an equation which can be solved for the brakepressure, and then the torque can be readily calculated.

Solution:Note that θ1=20°=0.349rad, and θ2=160°=2.792rad. From Equation (17.34), using a constantpressure p,

MP = d7 sinθdP =d7pbr sinθdθθ1

θ2

∫∫ = d7pbr cosθ1 − cosθ2( )

= 0.11 m( ) 0.04m( ) 0.09 m( ) c o s 2 0° −cos160°( )p = 7.44 × 10−4 m3( )p

Similarly, from Equation (17.35),

Page 17-22

MF = r − d7 cosθ( )µdP =∫ r − d7 cosθ( )µpbrdθ =∫ µpb r2 θ2 − θ1( ) − rd7 sinθ2 − sinθ1( )[ ]= 0.25( ) 0.040m( ) 0.09m( )2 2.792rad − 0.349rad( ) − 0.09m( ) 0.110m( ) sin160° −s i n 2 0°( )[ ]p

or MF=(1.98x10-4m

3)p. From Equation (17.47),

Wd6 = MF + MP = 7.44 ×10−4 m3( )p + 1.98 × 10−4 m3( )p = 9.42 ×10−4 m3( )pp =

Wd6

9.42 × 10−4 m3( ) =12kN( ) 0.22m( )

9.42 × 10−4m3( ) = 2.80 MPa

The torque is then

T = µpAr = µpbr2 θ2 −θ1( ) = 0.25( ) 2.80MPa( ) 0.04m( ) 0.09m( )2 2.792 − 0.349( ) = 554 NmWith a conventional shoe, Equations (17.34) and (17.35) are applied directly to give:

Mp =brd7 pmax

4sinθa2 θ2 − θ1( ) π

180°− sin2θ2 + sin 2θ1

= 0.04m( ) 0.09m( ) 0.11m( )pmax

4 s i n 9 0°2 1 6 0° −20°( ) π

180°− sin320° +sin40°

= 6.11× 10−4m3( ) pmax

MF =µpmaxbr

sinθa−r cosθ2 − cosθ1( ) −

d7

2sin2 θ2 − sin 2θ1( )

= 1.52 × 10−4 m3( )pmax

Therefore, from (17.47),

Wd6 = MF + MP = 6.11 ×10−4 m3( )pmax + 1.52 × 10−4m3( )pmax = 7.63 ×10−4 m3( )pmax

pmax = Wd6

7.63 × 10−4 m3 = 12000N( ) 0.22m( )7.63 × 10−4 m3 = 3.46MPa

From Equation (17.38),

Tb =µpmaxbr2 cosθ1 − cosθ2( )

sinθa=

0.25( ) 3.46MPa( ) 0.04m( ) 0.09m( )2 cos20° − cos160°( )sin90°

or T=526.6Nm. Note that the torque with a constant pressure is (554/526.6)=1.05 times theconventional shoe, or 5% greater.

17.28 A long-shoe external brake has two identical shoes coupled in series so that the peripheralforce from the first shoe is directly transferred to the second shoe. No radial force istransmitted between the shoes. Each of the two shoes covers 90° of the circumference,and the brake linings cover the central 70° of each shoe, leaving 10° at each end withoutlining as shown in sketch h. The actuating force is applied tangentially to the brake drumat the end of the loose shoe, 180° from the fixed hinge point of the other shoe. Calculatethe braking torques for both rotational directions when d7=150mm, r=125mm, b=50mm,W=14,000N and µ=0.2. Also show a free-body diagram of these forces acting on the twoshoes.

Notes: MP and MF are calculated from Equations (17.34) and (17.35) as in previous problems.Equation (17.47) gives an equation which allows solution for pmax. Force equilibrium gives theconnecting force. The left shoe follows the same approach but with the friction force acting inthe opposite direction.

Page 17-23

Solution:The free body diagram is shown. From the problem statement, θ1=10°, θ2=80°, therefore θa=80°,b=0.050m, r=0.125m, d7=0.15m. Equation (17.34) gives

Mp = brd7 pmax4sin θa

2 θ2 −θ1( ) π180°

− sin2θ2 + s i n 2θ1

= 0.05m( ) 0.125m( ) 0.15m( )pmax

sin80°2 80° −10°( ) π

180°−sin160° +s i n 2 0°

= 5.815 ×10−4 m3( )pmax

From Equation (17.35),

MF =µpmaxbr

sinθ a−r cosθ2 − cosθ1( ) −

d7

2sin2 θ2 − sin2θ1( )

= 0.2( ) 0.05m( ) 0.125m( )sin80°

− 0.125m( ) cos80° − cos10°( ) − 0.15m

2sin2 80° −sin2 10°( )

= 3.924 ×10−5m3( )pmax

Note that d6=d7(2)1/2

. Therefore, from Equation (17.47),

Wd6 = MP + MF = 5.815 ×10−4m3( )pmax + 3.924 × 10−5m3( )pmax = 6.207 × 10−4m3( )pmax

pmax = Wd6

6.207 ×10−4 m3 = 14000N( ) 0.15m( ) 2

2 6.207 × 10−4 m3 = 3.383MPa

The connecting force F is obtained from force equilibrium:

F − pbr cosφdφ10°

80°∫ + µpbr sinφdφ

10°

80°∫ = 0; F = pbr cosφdφ

10°

80°∫ − µpbrsinφdφ

10°

80°∫

Recognizing from Equation (17.31) that p=pmax(sinθ/sinθa), this is integrated as:

F =brpmax

4sin θa2 θ2 − θ1( ) π

180+ sin2θ2 − s i n 2θ1

−

µbrpmax

sinθasin2 θ2 − sin 2θ1[ ]

= 0.05m( ) 0.125m( ) 3.383MPa( )4 s i n 8 0°

2 80° −10°( ) π180

+ sin160° −sin20°

−

0.2( ) 0.05m( ) 0.125m( ) 3.383MPa( )sin80°

sin2 80°− sin2 10°[ ]or F=9.08kN. For the right shoe, Equation (17.47) becomes

Fd6 = MP + MF = 6.207 × 10−4 m3( )pmax

pmax = Fd6

6.207 × 10−4 m3 = 9.08kN( ) 0.15m( )6.207 × 10−4m3 = 2.195MPa

Page 17-24

The torque is given by Equation (17.38):

T =µbr2 cosθ1 − cosθ2( )

sinθapmax,l + pmax,r( )

= 0.2( ) 0.05m( ) 0.125m( )2 cos10° − cos80°( )sin80°

3.38MPa + 2.19Mpa( ) = 717Nm

For the drum rotating in the opposite direction, the friction forces change sign. MP is still(5.815x10

-4m

3)pmax, but MF=-(3.924x10

-5m

3)pmax. Equation (17.47) then gives pmax=3.872MPa. The

connecting force is then 19.63kN. pmax for the right shoe is then 5.429MPa and the torque is1197Nm.

17.29 A special type of brake is used in a car factory to hold the steel panels during drillingoperations so that the forces from the drill bits cannot move the panels. The brake isshown in sketch i. Calculate the braking force PB on the steel panel when it moves to theright with the speed ub between drilling operations. The actuating force is PM. The brakelining is thin relative to the other dimensions.

Notes: This problem is solved through statics only. Recognizing that p=cx, then taking momentequilibrium and force equilibrium gives PB in terms of PM.

Solution:The wear of the brake pad is proportional to x so p=cx. Summing moments about the pin,

PM 2L = pbxdxL

2L

∫ = cbx2dxL

2L

∫ = 73

cbL3 ; c = 6PM

7bL2

Summing horizontal forces,

PB = µpbdxL

2L

∫ = µcbx2dxL

2L

∫ = 32

µcbL2

Substituting for c,

PB = 32

µcbL2 = 32

µbL2 6PM

7bL2

=

97

µPM

Page 17-25

17.30 Redo problem 17.19 for long-shoe assumptions. The average contact pressure occurs at40°. Determine the maximum contact pressure and its location. Assume that the distancex is 150mm. What is the braking torque? Also, repeat this problem while reversing thedirection of rotation. Discuss the changes in the results.

Notes: The values of θ1, θ2, and d7 are obtained through geometry. pmax is found from Equation(17.31). MF and MP are found from (17.34) and (17.35). The torque is obtained from (17.38).

Solution:From the geometry,

tan θ0 =100mm + 170mm( ) − 100mm + 150mm( )

500mm; θ0 = 2.3°

Therefore,θ1=90°-40°-2.3°=47.7°

θ2=90°+40°-2.3°=127.7°d7=500mm/cos2.3°=504mm~0.5m

From Equation (17.31), since the average pressure occurs at 40° and θa=90°,

p = pmaxsinθsinθa

; pmax =

p

sinθ=

600,000Pa

s i n 4 0°= 933kPa

Equations (17.34) and (17.35) give:

Mp =brd7pmax

4sinθa2 θ2 − θ1( ) π

180°− s i n 2θ2 + sin 2θ1

=0.05m( ) 0.17m( ) d7( ) 933kPa( )

4s in90°2 1 2 7 . 7° −47.7°( ) π

180°− sin255.4° + sin95.4 °

= 9433N( )d7

or for d7=0.5m, MP=4717Nm. MF is obtained as

MF =µpmaxbr

sinθa−r cosθ2 − cosθ1( ) −

d7

2sin2 θ2 − sin 2θ1( )

=0.25( ) 933kPa( ) 0.05m( ) 0.17m( )

sin90°− 0.17m( ) cos127.7° − cos47.7 °( ) −

0.5m

2sin2 127.7° −sin 2 47.7°( )

= 433.15Nm − 78.33N( )d7 = 394NmFrom moment equilibrium on the bar,

MP − MF + Pn sin 20° 0.1m + 0.15m − 0.15m t a n 2 0°( ) − 1.25mcos20°[ ]= 0; Pn = 3900NSelf locking occurs if MP-MF=0, or:

MP − MF = 9443N( )d7 − 433.15Nm + 78.33 N( )d7 = 0; d7 = 0.0455mThe torque is obtained from Equation (17.38):

T =µbr2pmax cosθ1 − cosθ2( )

sinθa

= 0.25( ) 0.05m( ) 0.17m( )2 933kPa( ) cos47.7° − cos127.7°( )s i n 9 0°

= 433Nm

If the direction is reversed, there is no self locking. Pn is obtained as before, but with the frictionalmoment of opposite sign:

MP + MF + Pn sin 20° 0.1m + 0.15m − 0.15m t a n 2 0°( ) − 1.25mcos20°[ ]= 0; Pn = 4600NThe torque is the same but in the opposite direction.

Page 17-26

17.31 A symmetrically loaded, pivot-shoe brake has a wrap angle of 180° and the optimumdistance d7, giving a symmetrical pressure distribution. The coefficient of friction of thebrake lining is 0.30. A redesign is considered that will increase the braking torquewithout increasing the actuating force. The wrap angle is decreased to 80° (+40° to -40°),and the d7 distance is decreased to still give a symmetrical pressure distribution. Howmuch does the brake torque change?

Notes: This problem is solved by substituting Equation (17.53) into (17.52) and evaluating thetorque for the two cases.

Solution:Note from Equation (17.53),

Rx = pmaxbr2

2θ2π

180°

+ sin2θ2

; pmaxbr = 2Rx

2θ2π

180°

+ sin2θ2

Substituting this into Equation (17.52) gives:

T = 2µr2bpmax sinθ2 = 4µrRx sinθ2

2θ2π

180°

+ sin2θ2

For the 180° wrap angle, θ2=90° and T=1.27µrRx. For the 80° wrap angle, θ2=40° andT=1.08µrRx. Therefore the brake torque decreases by 15%.

17.32 The band brake shown in sketch j is activated by a compressed-air cylinder with diameterdc. The brake cylinder is driven by air pressure p=0.7MPa. Calculate the maximumpossible brake moment if the coefficient of friction between the band and the drum is0.25. The mass force on the brake arm is neglected, dc=50mm, r=200mm, l1=500mm,l2=200mm, and l3=500mm.

Notes: The force on the cylinder is obtained from the product of pressure and area. One then mustapply equilibrium to the torque arm to get the force on the left side, and Equation (17.58) to getthe other force. This then allows calculation of the braking torque from Equation (17.59).

Page 17-27

Solution:The maximum force exerted by the cylinder is:

Pcyl = pπ4

dcyl2 = 0.7MPa( )π

40.05m( )2 =1374N

Force equilibrium on the moment armPcyll3-P1l2=0; P1=3435N

Note that the wrap angle is φ=π+asin(r/l1)=π+asin(0.4)=3.553rad. Note from the direction ofrotation that P2 is larger than P1. Therefore, from Equation (17.58),

P2

P1= eµφ = e 0.25( ) 3.553( ) = 2.43;P2 = 2.43P1 = 2.43 3435N( ) = 8350N

From Equation (17.59),T = r P2 − P1( ) = 0.2 m( ) 8350N − 3435N( ) = 983Nm

17.33 The band brake shown in sketch k has wrap angle φ=225° and cylinder radius r=80mm.Calculate the brake torque when the lever is loaded by 100N and the coefficient offriction is µ=0.3. How long is the braking time from 1200 rpm if the rotor mass momentof inertia is 2.5 kg-m

2?

Notes: Equilibrium on the bar gives P1; Equation (17.58) gives P2. Equation (17.59) gives thebraking torque. The stopping time is obtained from Tt=Jω.

Solution:From moment equilibrium on the bar,

P1a-P(3a)=0; P1=3P=300NFrom Equation (17.58) with a wrap angle of 225°,

P2

P1= eµφπ / 1 8 0° = e 0.3( ) 225( )π / 1 8 0= 3.25; P2 = 3.25P1 = 975N

The braking torque is given by Equation (17.59) asT=r(P2-P1)=(0.08m)(975N-300N)=54Nm

To slow the rotor from 1200 rpm to 0 rpm, we note that Tt=Jω, so that

Tt = Jω ; t = JωT

=2.5kgm2( ) 1200rpm( ) 2πrad / rev( ) 1min/60sec( )

54Nm= 5.82s

Page 17-28

17.34 The hand brake shown in sketch l has wrap angle φ=215° and cylinder radius r=60mm.Calculate the brake torque when the coefficient of friction µ=0.25. How long is thebraking time from 1500 rpm if the rotor moment of inertia is J=2kg-m2?

Notes: Equilibrium on the bar gives P1; Equation (17.58) gives P2. Equation (17.59) gives thebraking torque. The stopping time is obtained from Tt=Jω.

Solution:From moment equilibrium on the bar,

P1(50mm)-P(120mm)=0; P1=168NFrom Equation (17.58) with a wrap angle of 225°,

P2

P1= eµφπ / 1 8 0° = e 0.25( ) 215( )π /180 = 2.56; P2 = 2.56 P1 = 429 N

The braking torque is given by Equation (17.59) asT=r(P2-P1)=(0.06m)(429N-168N)=15.7Nm

To slow the rotor from 1200 rpm to 0 rpm, we note that Tt=Jω, so that

Tt = Jω ; t = JωT

=2kgm2( ) 1500rpm( ) 2πrad / rev( ) 1 m i n / 6 0 s e c( )

15.7Nm= 20.0s

17.35 A brake (see sketch m) consists of a drum with a brake shoe pressing against it. Drumradius r=80mm. Calculate the brake torque when P=7000N, µ=0.35, and brake pad widthb=40mm. The wear is proportional to the contact pressure times the sliding distance.

Notes: Equation (17.48) gives p=pmaxcosφ. Integrating Equation (17.49) over the brake contactarea gives the force, and Equation (17.52) gives the torque.

Solution:From Equation (17.48), and recognizing that we are using φ as the angular position variable,p=pmaxcosφ. The actuating force is given by Equation (17.49) as

Page 17-29

P = pbrdφ cosφ−α

α

∫ = bpmaxr cos2 φdφ−α

α

∫ = bpmaxr α +sin2α

2

recognizing α=30°=0.523rad, and solving for pmax gives:

pmax =P

br α + s i n 2α2

=7000N

0.04 m( ) 0.08m( ) 0.523 + sin60°2

= 2.29MPa

The torque is given by equation (17.52):

T = µr pdA∫ = µr pmax cosφdA∫ = 205Nm

17.36 For the band brake shown in sketch n the following conditions are given: d=350mm,pmax=1.2MPa, µ=0.25 and b=50mm. All dimensions are in millimeters. Determine thefollowing:a) The braking torque.b) The actuating force.c) The forces acting at hinge O.

Notes: The bottom band will have the larger force. Therefore, Equation (17.62) gives this forcebased on the maximum pressure. The wrap angle is determined from geometry, and the force inthe top band is obtained from Equation (17.58). Equation (17.59) then gives the torque. Momentand force equilibrium gives the actuating force; force equilibrium gives the reactions at O.

Solution:Note the angles and forces labeled in the sketch for the nomenclature used in the problem. Notethat because of the direction of rotation, F1 is the larger force, so the convention is the same as thetextbook derivation. From geometry, note the following:

sinθ1 =d / 2

150mm + 210mm=

350mm / 2

360mm; θ1 = 29.1°

sinθ2 =d / 2

210mm=

350mm / 2

210mm; θ2 = 56.4°

φ1=90°-θ1=90°-29.1°=60.9°φ2=90°-θ2=90°-56.4°=33.6°

Therefore the angle of wrap isφ=360°-φ1-φ2=360°-60.9°-33.6°=265.5°

From Equation (17.62),

pmax =F1

br;F1 = brpmax = 0.05m( ) 0.35m / 2( ) 1.2 MPa( ) = 10.5kN

F2 is found from Equation (17.58):

Page 17-30

F1

F2= eµφπ /180° ; F2 =

F1

eµφπ / 1 8 0° =10.5kN

e 0.25( ) 265.5°( )π / 1 8 0° = 3.30kN

The torque is obtained from Equation (17.59) asT=r(F1-F2)=(0.35m/2)(10.5kN-3.30kN)=1.26kNm

The actuating force is found through moment equilibrium about point O:MO = 0 = P 210mm + 150mm + 130mm( ) − F1sinθ1 210mm + 150mm( ) + F2 sinθ2 210mm( )∑

P = F1 sinθ1 360mm( ) − F2 sinθ2 210mm( )490mm( )

= 10.5kNsin 29.1° 0.36m( ) − 3.3kNs in56.4° 0.21m( )0.49m

or P=2.57kN. The reactions Ox and Oy are found from force equilibrium. For the bar:Fx = 0 = Oxbar − F2 cosθ2 − F1 cosθ1 ; Oxbar = 11kN∑

Fy = 0 = Oybar + F2 sinθ2 − F1 sinθ1 + P; Oybar = 0.21kN∑For the drum:

Fx = 0 = Oxdrum − F2 cosθ2 − F1cosθ1; Oxdrum =11kN∑Fy = 0 = Oydrum − F2 sinθ2 + F1sinθ1; Oydrum = −2.36kN∑

If they are mounted on the same pin, the support reactions need to be subtracted, so that Ox=0kNand Oy=2.57kN.

17.37 The band brake shown in sketch o is 40mm wide and can take a maximum pressure of1.1MPa. All dimensions are in millimeters. The coefficient of friction is 0.3. Determinethe following:a) The maximum allowable actuating forceb) The braking torquec) The reaction supports at O1 and O2

d) Whether it is possible to change the distance O1A in order to have self-locking.Assume point A can be anywhere on line CO1A.

Notes: This is very similar to problem 17.36. As labeled, F1 will be the largest force. Therefore,Equation (17.62) gives this force based on the maximum pressure. The wrap angles aredetermined from geometry, and the force between the drums is obtained from F1 and Equation(17.58). Since the drum mounted on O1 has no applied force, it is an idler, and F3=F2. Equation(17.59) then gives the torque. Moment and force equilibrium gives the actuating force; forceequilibrium gives the reactions at O. Self locking would occur if W=0, and this is investigated bytaking moment equilibrium of bar CB about point O1.

Solution:Note the sketch and the nomenclature used to solve the problem. Note the extra points labeled Gand H. The distance BO2 is obtained from the Pythagorean theorem as

BO2 = 0.18m( )2 + 0.050m( )2 = 0.1868m

Page 17-31

Notice the O2H is the radius of the drum, or 0.1m. Therefore, BH is

BH = 0.1868m( )2 − 0.1m( )2 = 0.1578mθ is obtained from

sinθ =0.1m( )cos θ + 0.05m

0.1578m; θ = 48°

Therefore, φ2=180°+48°=228°. Now note that O1G is the radius of the drum, or 0.1m. Since O1Ais given as 0.135m, and since alternate interior angles are equal, a is

sinα =O1G

O1A=

0.1m

0.135m; α =47.8°

Since F1 is the largest force, it determines the maximum brake lining pressure. From Equation(17.62),

pmax =F1

br;F1 = brpmax = 0.04m( ) 0.1m( ) 1.1MPa( ) = 4.4kN

F2 is found from Equation (17.58):F1

F2= eµφπ /180° ; F2 =

F1

eµφπ / 1 8 0° =4.4kN

e 0.3( ) 228°( )π /180° = 1.33kN

Note that the drum mounted at O1 is an idler; therefore F3=F2=1.33kN, and this drum produces notorque. The torque for the drum mounted at O2 is obtained from Equation (17.59) as

T=r(F1-F2)=(0.1m)(4.4kN-1.33kN)=307NmThe actuating force W is found through moment equilibrium about point O1:

MO1= 0 =∑ W 0.35m( ) − F3sin α 0.135 m( ) − F1sin θ 0.085m + 0.135m( ) − F1 cosθ 0.05m( )

W 0.35 m( )= 1.33kN s in47.8° 0.135m( ) + 4.4kN s i n 4 8° 0.22m( ) + c o s 4 8° 0.05m( )[ ]or W=2.85kN. The reactions O1x and O1y for the bar are found from force equilibrium. For thebar:

Fx = 0 = O1xbar − F3 cosα + F1 cosθ ; O1xbar = −2.05kN∑Fy = 0 = O1ybar − W − F3sinα − F1sinθ ; O1ybar = 7.1kN∑

For the drum:Fx = 0 = O1xdrum − F2 − F3 cosα ; O1xdrum = 2.23kN∑

Fy = 0 = O1ydrum − F3sinα ; O1ydrum = 0.989kN∑If they are mounted on the same supports, then O1x=0.18kN and O1y=8.1kN. For drum 2,

Fx = 0 = O2x − F2 − F1 cosθ ; O2x = 4.27kN∑Fy = 0 = O2y + F1sinθ ; O2y = −3.27kN∑

For the brake to self lock, it would be necessary that F1 be positive for W=0. Taking momentequilibrium for the bar, with the location of A a distance x from point O1:

MO1= 0 = F3 sinα( )x + F1cosθ( )∑ 0.05m( ) + F1sin θ 0.085m + 0.135m( )

This would require x=-0.87m, which is not possible on this bar.