Section 16

SECTION 16 – BRAKES AND CLUTCHES

Page 1 of 97

ENERGY TO BRAKES

881. A motor operates a hoist through a pair of spur gears, with a velocity ratio of 4.

The drum on which the cable wraps is on the same shaft as the gear, and the

torque cause by the weight of the load and hoist is 12,000 ft-lb. The pinion is on

the motor shaft. Consider first on which shaft to mount the brake drum; in the

process make trial calculations, and try to think of pros and cons. Make a

decision and determine the size of a drum that will not have a temperature rise

greater than Fto150=∆ when a 4000-lb. load moves down 200 ft. at a constant

speed. Include a calculation for the frp/sq. in. of the drum’s surface.

Solution:

Consider that brake drum is mounted on motor shaft that has lesser torque.

lbinlbftlbft

T f −=−=−

= 000,3630004

000,12

From Table AT 29,

Assume 35.0=f , psip 75= , max. fpmvm 5000=

2

FDT f =

D

TfNF

f2==

Df

TN

f2=

A

Np =

DbA π=

( )( )

7535.0

000,362222

====bDbfD

T

Db

Np

f

πππ

8732 =bD

use 8732 =bD

2

873

Db =

Then,

cW

lbftUFt

m

f −=∆ o

Assume a cast-iron, 3253.0 inlb=ρ

101=c

VWm ρ=


Page 2 of 97

+=+=

44

22 D

DbttDDbtV ππ

π

( )( ) lbftU f −== 000,8002004000

Fto150=∆

tc

UVW

f

m ∆== ρ

( )( )101150

000,800253.0 =V

37.208 inV =

But

+=

4

2DDbtV π

2

873

Db =

+=

4

873 2D

DtV π

For minimum V :

02

8732

=

+

−=

D

Dt

dD

dVπ

( )87323 =D

inD 12=

For t :

( )

+==

4

12

12

8737.208

2

tV π

int 611.0=

say int8

5=

( )ininb

16

160625.6

12

8732

===

Therefore use inD 12= , int8

5= , inb

16

16=

For A

fhpinsqfhp =..

000,33

mFvfhp =


Page 3 of 97

( )lb

D

TF

f6000

12

000,3622===

fpmvm 5000= (max.)

( )( )hpfhp 909

000,33

50006000==

( ) 2

16

1612 inDbA

== ππ

98.355.228

909.. ===

A

fhpinsqfhp (peak value)

882. A 3500-lb. automobile moving on level ground at 60 mph, is to be stopped in a

distance of 260 ft. Tire diameter is 30 in.; all frictional energy except for the

brake is to be neglected. (a) What total averaging braking torque must be

applied? (b) What must be the minimum coefficient of friction between the tires

and the road in order for the wheels not to skid if it is assumed that weight is

equally distributed among the four wheels (not true)? (c) If the frictional energy

is momentarily stored in 50 lb. of cast iron brake drums, what is the average

temperature rise of the drums?

Solution:

(a) Solving for the total braking torque.

( )22

212ssf vv

g

WKEU −=∆−=

lbW 3500=

fpsmphvs 88601

==

fpsmphvs 002

==

22.32 fpsg =

( )( ) lbftU f −=−= 000,421088

2.322

3500 22

( ) ( )000,63000,33

nlbinTlbftTfhp

fmf −=

−=

ω

( )( )

2

222

892.142602

880

2

12 fpss

vva

ss −=−

=−

=

sec91.5892.14

88012 =

−

−=

−=

a

vvt

ss

( )( ) ( )hp

t

U

t

KEfhp

f130

91.5550

000,421

550550===

∆−=


Page 4 of 97

( )( )rpm

ft

fps

D

vn m 336

12

30

minsec60882

1

=

==

ππ

000,63

nTfhp

f=

( )lbinT f −== 375,24

336

130000,63

(b) N

Ff =

for each wheel, lbN 8754

3500==

lbinT f −== 60944

375,24

( )lbin

D

TF

f −=== 40630

609422

464.0875

406===

N

Ff

(c) cW

Ut

m

f=∆

lbftU f −= 000,421

lbWm 50=

Flblbftc −−=101 for cast-iron

( )( )Ft o4.83

10150

000,421==∆

884. An overhead traveling crane weighs 160,000 lb. with its load and runs 253 fpm.

It is driven by a 25-hp motor operating at 1750 rpm.The speed reduction from the

motor to the 18-in. wheels is 32 to 1. Frictional energy other than at the brake is

negligible. (a) How much energy must be absorbed by the brake to stop this crane

in a distance of 18 ft.? (b) Determine the constant average braking torque that

must be exerted on the motor shaft. (c) If all the energy is absorbed by the rim of

the cast-iron brake drum, which is 8 in. in diameter, 1 ½ in. thick, with a 3 ¼-in.

face, what will be its temperature rise? (d) Compute the average rate at which the

energy is absorbed during the first second (fhp). Is it reasonable?

Solution:


Page 5 of 97

( )22

212ssf vv

g

WKEU −=∆−=

lbW 000,160= 22.32 fpsg =

fpsfpmvs 22.42531

==

fpsvs 02

=

( )( )[ ] lbftU f −=−= 245,44022.4

2.322

000,160 22

(b) ( )

n

fhpT f

000,63=

( )( )

2

222

495.0182

22.40

2

12 fpss

vva

ss −=−

=−

=

sec53.8495.0

22.4012 =

−

−=

−=

a

vvt

ss

( )hp

t

Ufhp

f43.9

53.8550

245,44

550===

( ) ( )( )

( )lbin

n

fhpT f −=== 68

17502

1

000,6343.9000,63 on the motor shaft.

(c) cW

Ut

m

f=∆

DbtV π= (rim only) on the motor shaft

inD 8=

inb 25.3=

int 5.0=

( )( )( ) 384.405.025.38 inV == π

VWm ρ=

3253.0 inlb=ρ for cast iron

Flblbftc −−=101 for cast-iron

( )( ) lbWm 33.1084.40253.0 ==

( )( )Ft o4.42

10133.10

245,44==∆

(d) First second:


Page 6 of 97

fpsvs 22.41

=

2495.0 fpsa −=

( ) fpsatvv ss 73.31495.022.412

=−=+=

( )( ) ( )[ ] lbftKEU f −=−=∆−= 968073.322.4

2.322

000,160 22

( )hphp

t

Ufhp

f256.17

1550

9680

550<=== , therefore reasonable.

885. The diagrammatic hoist shown with its load weighs 6000 lb. The drum weighs

8000 lb., has a radius of gyration ftk 8.1= ; ftD 4= . A brake on the drum shaft

brings the hoist to rest in 10 ft. from fpsvs 8= (down). Only the brake frictional

energy is significant, and it can be reasonably assumed that the acceleration is

constant. (a) From the frictional energy, compute the average braking torque. (b)

If the average fhp/sq. in. is limited to 0.15 during the first second, what brake

contact area is needed?

Problems 885, 886

Solution:

n

fhpT f

000,63=

( ) ( )2222

2

2

11

21 1122ssf vv

g

WIKEKEU −+−=∆−∆−= ωω

fpsvs 81

= , fpsvs 02

=

( )srad

D

vs4

4

8221

1 ===ω , srad02 =ω

g

kWI

2

11 =

lbW 80001 =

ftk 8.1=

lbW 60002 =


Page 7 of 97

22.32 fpsg =

( ) ( ) ( )( )

( )( )

( ) lbftvvg

WIU ssf −=+=−+−= 400,128

2.322

600004

2.322

8.180000

22

222

2222

2

2

11

11ωω

s

vva

ss

2

22

12−

=

fts 10=

( )2

22

2.3102

80fpsa −=

−=

sec5.22.3

8012 =

−

−=

−=

a

vvt

ss

( )hp

t

Ufhp

f9

5.2550

400,12

550===

rpmnπω

2

60=

( ) 02042

1−=−= sradsradω

( )rpmn 1.19

2

260==

π

( )lbin

n

fhpT f −=== 700,29

1.19

9000,63000,63

(b) 15.0.. =insqfhp (first second)

( ) fpsatvv ss 8.412.3812

=−=+=

( )sec4.2

4

8.4222

2 radD

vs ===ω

( )( )

( ) ( )[ ]( )

( ) ( )[ ] lbftU f −=−+−= 61068.482.322

600004.24

2.322

8.180000 22222

( )hp

t

Ufhp

f10.11

1550

6106

550===

27415.0

10.11

..in

insqfhp

fhpA ===

887. The same as 885, except that a traction drive, arranged as shown, is used; the

counterweight weighs 4000 lb. The ropes pass twice about the driving sheave; the

brake drum is on this same shaft.


Page 8 of 97

Problem 887.

Solution:

(a) ( )22

212ss

Tf vv

g

WKEU −=∆−=

lblblbWT 000,1060004000 =+=

KE∆− of pulley is negligible

fpsvs 81

= , fpsvs 02

=

( )( ) lbftU f −== 940,98

2.322

000,10 2

( )2

2222

2.3102

80

2

12 fpss

vva

ss −=−

=−

=

sec5.22.3

8012 =

−

−=

−=

a

vvt

ss

( )hp

t

Ufhp

f23.7

5.2550

9940

550===

ftD 4=

( )sec4

4

8221

1 radD

vs ===ω

( )sec0

4

0222

2 radD

vs ===ω

( ) ( ) sec2042

1

2

121 rad=+=+= ωωω

( )rpmn 1.19

2

260

2

60

ππω

==


Page 9 of 97

Braking torque, ( )

lbinn

fhpT f −=== 850,23

1.19

23.7000,63000,63

(b) 15.0.. =insqfhp (first second)

fpsvs 81

=

atvv ss =−12

( )12.382

−=−sv

fpsvs 8.42

=

( )( ) ( )[ ] lbftU f −=−= 63608.48

2.322

000,10 22

( )hp

t

Ufhp

f56.11

1550

6360

550===

Contact area = 21.7715.0

56.11

..in

insqfhp

fhpA ===

SINGLE-SHOE BRAKES

888. For the single-shoe, short-block brake shown (solid lines) derive the expressions

for brake torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) In

which direction of rotation does the brake have self-actuating properties? If

25.0=f , for what proportions of e and c would the brake be self-actuating?

Problems 888 – 891, 893.

Solution:

(a) Clockwise rotation (as shown)


Page 10 of 97

2

FDT f =

fNF =

[ ]∑ = 0HM

cNWaefN =+

WaefNcN =−

fec

WaN

−=

fec

fWaF

−=

( )fec

fWaDT f −

=2

(b) Counter Clockwise Rotation

2

FDT f =

fNF =

[ ]∑ = 0HM


Page 11 of 97

cNefNWa +=

fec

WaN

+=

fec

fWaF

+=

( )fec

fWaDT f +

=2

(c) Clockwise rotation is self-actuating

fec >

with 25.0=f

ec 25.0>

889. The same as 888, except that the wheel and brake shoe are grooved, θ2 degrees

between the sides of the grooves (as in a sheave, Fig. 17.38, Text).

Solution:

[ ]∑ = 0VF

NN =θsin2 1

12 NfF =


Page 12 of 97

θθ sinsin22

NfNfF =

=

(a) Clockwise rotation

fec

WaN

−=

( ) θsinfec

fWaF

−=

( ) θsin2 fec

fWaDT f −

=

(b) Counter clockwise rotation

fec

WaN

+=

( ) θsinfec

fWaF

+=

( ) θsin2 fec

fWaDT f +

=

(c) Clockwise rotation is self-actuating

fec >

with 25.0=f

ec 25.0>

890. Consider the single-shoe, short-block brake shown (solid lines) with the drum

rotating clockwise; let e be positive measured downward and cD 6.1= . (a) Plot

the mechanical advantage MA (ordinate) against f values of 0.1, 0.2, 0.3, 0.4,

0.5 (abscissa) when ce has values 2, 0.5, 0, -0.5, -1. (b) If f may vary from 0.3

to 0.4, which proportions give the more nearly constant brake response? Are

proportions good? (c) What proportions are best if braking is needed for both

directions of rotation?

Solution:


Page 13 of 97

(a) Wa

TMA

f= , Clockwise rotation

( )fec

DfMA

−=

2

cD 6.1=

( )fec

fcMA

−=

2

6.1

−=

c

fe

fMA

1

8.0

Tabulation:

Values of MA

ce

f 2 0.5 0 -0.5 -1

0.1 0.100 0.084 0.08 0.076 0.073

0.2 0.267 0.178 0.16 0.145 0.133

0.3 0.600 0.284 0.24 0.209 0.185

0.4 1.600 0.400 0.32 0.267 0.229

0.5 ∞ 0.533 0.40 0.320 0.267


Page 14 of 97

Plot:

(b) 4.03.0 tof = , 1−=ce , with ttanconsMA ≈ .

They are good because c

fe>1 except 2=ce .

(c) 0=ce is the best if braking is needed for both directions of rotation with MA the

same.

891. A single-block brake has the dimensions: cast-iron wheel of inD 15= .,

ina2

132= ., inc

8

39= ., ine

16

114= ., width of contact surface = 2 in. The brake

block lined with molded asbestos, subtends 80o, symmetrical about the center

line; it is permitted to absorb energy at the rate of 0.4 hp/in.2; rpmn 200= .

Assume that p is constant, that F and N act at K , and compute (a) mpv and

the approximate braking torque, (b) the force W to produce this torque, (c) the

mechanical advantage, (d) the temperature rise of the 3/8-in.-thick rim, if it

absorbs all the energy with operation as specified, in 1 min. (e) How long could

this brake be so applied for Fto400=∆ ? See 893.

Solution:


Page 15 of 97

inD 15=

ina 5.32=

inc 375.9=

ine 6875.4=

inb 2=

(a) Solving for mpv

minlbftfpAvFv mm −=

24.0 inhpA

Fvm =

( )( )22

min200,13min000,334.0

in

lbft

in

hplbfthp

A

Fvm −=

−−=

mm fpv

A

Fv=

35.0=f from Table AT 29, molded asbestos on cast iron

mm pv

A

Fv35.0200,13 ==

min700,37 −−= insqlbftpvm

Solving for braking torque

min..200,13 −−= insqlbftA

Fvm

( ) fpmDnvm 78520012

15=

== ππ

2

DbA

θ=

( ) rad3963.1180

80 =

=

πθ


Page 16 of 97

( )( )( )..21

2

2153963.1

2insq

DbA ===

θ

( )200,13

21

785=

F

lbF 353=

( )( )lbin

FDT f −=== 2650

2

15353

2

(b) Solving for W

fec

WafF

−=

lbF 353=

35.0=f

ina 5.32=

ine 6875.4=

inc 375.9=

( ) ( ) ( )( )[ ]( )( )

lbfa

fecFW 240

5.3235.0

6875.435.0375.9353=

−=

−=

(c) Solving for MA

( )( )( )

( )( )[ ]34.0

6875.435.0375.92

1535.0

2=

−=

−=

fec

DfMA

(d) Solving for t∆

cW

lbftUFt

m

f −=∆

,o

DbtWm ρπ=

inD 15=

inb 2=

inint 375.08

3==

3253.0 inlb=ρ for cast iron

( )( )( )( )( ) lbWm 942.8375.0215253.0 == π

Flblbftc −−= 101 for cast iron


Page 17 of 97

( )fhptU f′= 550

sec60min1 ==′t

( )( ) fhpfhpU f 000,3360550 ==

( )( )hp

nTfhp

f4127.8

000,63

2002650

000,63===

( ) lbftU f −== 619,2774127.8000,33

( )( )F

cW

Ut

m

f o310101942.8

619,277===∆

(e) Solving for t′ , (time) with Fto400=∆

tcWU mf ∆=

( )( )( ) lbftU f −== 260,361400101942.8

( )( )fUtfhp =′550

( )( ) 260,3614127.8550 =′t

min3.1sec78 ==′t

892. For a single-block brake, as shown, ina 26= ., inc2

17= ., ine 75.3= .,

inD 15= ., drum contact width inb2

13= . The molded asbestos lining subtends

o60=θ , symmetrical about the vertical axis; force lbW 300= .; rpmn 600= .

Assume that p is constant, that F and N act at K , and compute (a) mpv and

the braking torque, (b) the energy rate in fhp/in.2 of contact surface. (c) the

mechanical advantage, (d) the temperature of the 3/8-in.-thick rim, if it absorbs

all the energy with the operation as specified in 1 min. (e) How long could this

brake be so applied for Ftrim

o400=∆ ? See 894.


Page 18 of 97

Problems 892, 894.

Solution:

For greater braking torque, fT , use counterclockwise rotation

[ ]∑ = 0AM

cNefNaW =+

efc

WaN

−=

efc

WafF

−=

From Table AT 29, 35.0=f for molded asbestos

lbW 300=

ina 26=

inc 5.7=

75.3=e

( )( )( )( )( )

lbF 44235.075.35.7

2630035.0=

−=

(a) Solving for mpv

mm fpAvFv =

( )( )fpm

Dnvm 2536

12

60015

12===

ππ

2

DbA

θ=

rad047.1180

60 =

=

πθ

( )( )( ) 25.272

5.315047.1inA ==

( )( ) ( )( ) mm pvFv 5.2735.02536442 ==

min..500,116 −−= insqlbftpvm


Page 19 of 97

Solving for the braking torque,

( )( )lbin

FDT f −=== 3315

2

15442

2

(b) Energy rate, fhp.in2.

( )( )hp

nTfhp

f6.31

000,63

6003315

000,63===

25.27 inA =

2

2

2 15.15.27

6.31inhp

in

hpinfhp ==

(c) ( )( )

425.026300

3315===

Wa

TMA

f

(d) cW

lbftUFt

m

f −=∆

,o

DbtWm ρπ=

inint 375.08

3==

inD 15=

inb 5.3= 3253.0 inlb=ρ for cast iron

Flblbftc −−= 101 for cast iron

( )( )( )( )( ) lbWm 648.15375.05.315253.0 == π

For 1 min

( )( ) ( )( ) lbftfhpU f −=== 800,042,16.311000,331000,33

( )( )Ft o660

101648.15

800,042,1==∆

(e) Ftrim

o400=∆

( )( )( ) lbftU f −== 179,632101648.15400

( )min61.0

6.31000,33

179,632

000,33min ===′

fhp

Ut

f


Page 20 of 97

LONG-SHOE BRAKES

FIXED SHOES

893. The brake is as described in 891 and is to absorb energy at the same rate but the

pressure varies as θsinPp = . Derive the equations needed and compute (a) the

maximum pressure, (b) the moment HFM of F about H , (c) the moment HNM

of N about H , (d) the force W , (e) the braking torque, (f) the x and y

components of the force at H .

Solution:

φθ sinsin PPp ==

2

Dr =

φpbrddN =

φfpbrddF =


Page 21 of 97

∫= rdFT f

∫= φdfpbrT f

2

∫= φφdPfbrT f sin2

( )21

2 coscos φφ −= PfbrT f

(a) Solving for P

( )21

2 coscos φφ −=

fbr

TP

f

2

Dr =

er

c

−=αtan

inc 375.9=

inr 5.72

15==

ine 6875.4=

6875.45.7

375.9tan

−=α

o3.73=α o80=θ

o3.332

803.73

21 =−=−=

θαφ

o3.1132

803.73

22 =+=+=

θαφ

35.0=f

inb 2=

inr 5.7=

( )21

2 coscos φφ −=

fbr

TP

f

( )( )( ) ( )psi

TTP

ff

5.483.113cos3.33cos5.7235.02

=−

=

n

fhpT f

000,63=

( )( )Ainfhpfhp 2=

2

DbA

θ=

rad396.1180

80 =

=

πθ


Page 22 of 97

( )( )( ) 2212

215396.1inA ==

4.02 =infhp

( )( ) hpfhp 4.8214.0 ==

rpmn 200=

( )lbinT f −== 2646

200

4.8000,63

( )o90.max555.48

2646

5.482 >==== φPpsi

TP

f

(b) ( )∫ −= dFRrM HF φcos

( )∫ −=2

1

sincosφ

φφφφ dfbrPRrM HF

( )∫ −=2

1

cossinsinφ

φφφφφ dRrfbrPM HF

2

1

2sin2

cos

φ

φ

φφ

−−=

RrfbrPM HF

( ) ( )

−−−= 1

2

2

2

21 sinsin2

coscos φφφφR

rfbrPM HF

( ) ( ) ( ) inercR 788.96875.45.7375.92222 =−+=−+=

( )( )( )( ) ( ) ( )

−−−= 3.33sin3.113sin

2

788.93.113cos3.33cos5.7555.7235.0 22

HFM

lbinM HF −=1900

(c) ∫= dNRM HN φsin

∫=2

1

2sinφ

φφφbrdRPM HN

∫=2

1

2sinφ

φφφdbrRPM HN

( )∫ −=2

1

2cos12

φ

φφφ d

brRPM HN

2

1

2sin2

1

2

φ

φ

φφ

−=

brRPM HN

( ) ( )[ ]1212 2sin2sin24

φφφφ −−−=brRP

M HN

rad396.112 ==− θφφ

( ) o6.2263.11322 2 ==φ

( ) o6.663.3322 1 ==φ


Page 23 of 97

( )( )( )( ) ( ) ( )[ ]6.66sin6.226sin396.124

55788.95.72−−=HNM

lbinM HN −= 8956

(d) ∑ = 0HM

0=−+ HNHF MMWa

ina 5.32=

( ) 0895619005.32 =−+W

lbW 217=

(e) lbinT f −= 2646

(f) ∑ = 0xF

∫ ∫ =++−− 0cossincos φφα dFdNWH x

∫∫ −−=−2

1

2

1

cossinsincos 2φ

φ

φ

φφφφφφα dfPbrdPbrWH x

( ) ( )[ ] ( )1

2

2

2

1212 sinsin2

2sin2sin24

cos φφφφφφα −−−−−−=−fbrPbrP

WH x

( )( )( ) ( ) ( )[ ]

( )( )( )( ) ( )3.113sin3.113sin2

555.7235.0

6.66sin6.226sin396.124

555.723.73cos217

22 −−

−−−=− xH

lbH x 931−=−

lbH x 931=

∑ = 0yF

∫ ∫ =+−+− 0sincossin φφα dFdNWH y

αφφφφφφ

φ

φ

φsinsincossin

2

1

2

1

2 WdfbrPdbrPH y −−=− ∫∫

( ) ( ) ( )[ ] αφφφφφφ sin2sin2sin24

sinsin2

12121

2

2

2W

fbrPbrPH y −−−−−−=−

( )( )( ) ( )( )( )( )( ) ( ) ( )[ ] 3.73sin2176.66sin6.226sin396.12

4

555.7235.0

3.33sin3.113sin2

555.72 22

−−−−

−=− yH

lbH y 305−=−

lbH y 305=


Page 24 of 97

894. The brake is as described in 892, but the pressure varies as φsinPp = . Assume

the direction of rotation for which a given W produces the greater fT , derive the

equations needed, and compute (a) the maximum pressure, (b) the moment of F

about A , (c) The moment of N about A , (d) the braking torque, (e) the x and y

components of the force at A .

Solution:

φsinPp =

φpbrddN =

φφdPbrdN sin=

φφdfPbrfdNdF sin==

Solving for 1φ and 2φ


Page 25 of 97

er

c

+=αtan

inD

r 5.72

==

75.35.7

5.7tan

+=α

o69.33=α

o69.32

6069.33

21 =−=−=

θαφ

o69.632

6069.33

21 =+=+=

θαφ

( )∫ −= dFrRM AF φcos

( ) φφφφ

φdfPbrrRM AF sincos

2

1∫ −=

( ) φφφφφ

φdrRfPbrM AF ∫ −=

2

1

sincossin

( ) ( )

−+−= 121

2

2

2 coscossinsin2

φφφφ rR

fPbrM AF

( ) ( ) ( ) inrecR 52.135.775.35.72222 =++=++=

( ) ( )( ) ( ) ( )

−+−

= 69.3cos69.63cos5.769.3sin69.63sin

2

52.135.75.335.0 22

PM AF

PM AF 43.11=

∫= dNRM AN φsin

∫=2

1

2sinφ

φφφdRPbrM AN

( )∫ −=2

1

2cos12

φ

φφφ d

brPRM AN

( ) ( )[ ]1212 2sin2sin24

φφφφ −−−=brPR

M AN

rad047.112 ==− θφφ

( ) o38.12769.6322 2 ==φ

( ) o38.769.322 1 ==φ

( )( ) ( ) ( ) ( )[ ]38.7sin38.127sin047.124

52.135.75.3−−=

PM AN

PM AN 68.126=

(a) ∑ = 0AM

0=−+ ANAF MMWa

lbW 300=


Page 26 of 97

ina 26=

( )( ) 068.12643.1126300 =−+ PP

psiP 68.67=

max. psiPp 67.6069.63sin68.67sin 2 === φ

(b) ( ) lbinM AF −== 77468.6743.11

(c) ( ) lbinM AN −== 857568.6768.126

(d) ∫= rdFT f

∫=2

1

sin2φ

φφφdfPbrT f

( )21

2 coscos φφ −= fPbrT f

( )( )( )( ) ( )69.63cos69.3cos5.75.368.6035.02 −=fT

lbinT f −= 2587

(e) [ ]∑ = 0xF

∫ ∫ =−+−− 0cossincos φφα dFdNWH x

∫∫ +−=−2

1

2

1

cossinsincos 2φ

φ

φ

φφφφφφα dfPbrdPbrWH x

( ) ( )[ ] ( )1

2

2

2

1212 sinsin2

2sin2sin24

cos φφφφφφα −+−−−−=−fPbrPbr

WH x

( )( )( ) ( ) ( )[ ]

( )( )( )( ) ( )69.3sin69.63sin2

5.75.368.6735.0

38.7sin38.127sin047.124

5.75.368.6769.33cos300

22 −+

−−−=− xH

lbH x 136−=−

lbH x 136=

[ ]∑ = 0yF

∫ ∫ =−−+ 0sincossin φφα dFdNWH y

αφφφφφφ

φ

φ

φsinsincossin

2

1

2

1

2 WdfPbrdPbrH y −+= ∫∫

( ) ( ) ( )[ ] αφφφφφφ sin2sin2sin24

sinsin2

12121

2

2

2W

fPbrPbrH y −−−−+−=

( )( )( ) ( )( )( )( )( ) ( ) ( )[ ] 69.33sin30038.7sin38.127sin047.12

4

5.75.368.6735.0

69.3sin69.63sin2

5.75.368.67 22

−−−+

−=yH


Page 27 of 97

lbH y 766=

895. (a) For the brake shown, assume αcosPp = and the direction of rotation for

which a given force W results in the greater braking torque, and derive equations

for fT in terms of W , f , and the dimensions of the brake. (b) Under what

circumstances will the brake be self-acting? (c) Determine the magnitude and

location of the resultant forces N and F .

Solution:

(a) Clockwise rotation has greatest braking torque.

αcosPp =

ααα dPbrpbrddN cos==

ααα dfPbrfpbrdfdNdF cos===

( )∫− +=2

1

sinθ

θα dFcrM HF

( )∫− +=2

1

cossinθ

θααα dfPbrcrM HF


Page 28 of 97

( )∫− +=2

1

cossincosθ

θαααα dcrfPbrM HF

2

1

2sin2

1sin

θ

θ

αα−

+= crfPbrM HF

( ) ( )[ ] ( ) ( )[ ]

−−+−−= 1

2

2

2

12 sinsin2

1sinsin θθθθ crfPbrM HF

( ) ( )

−++= 1

2

2

2

12 sinsin2


∫−=2

1

cosθ

θαdNM HN

∫−=2

1

2cosθ

θααdcPbrM HN

( )∫− +=2

1

2cos12

θ

θαα d

cPbrM HN

[ ] 2

12sin2

4

θθα −+=

cPbrM HN

( ) ( )[ ]1212 2sin2sin24

θθθθ +++=cPbr

M HN

[ ]∑ = 0HM

0=−+ HNHF MMWa

( ) ( ) ( ) ( )[ ]12121

2

2

2

12 2sin2sin24

sinsin2

1sinsin θθθθθθθθ +++=

−+++

cPbrcrfPbrWa

( ) ( )[ ] ( ) ( )[ ]1

2

2

2

121212 sinsinsinsin22

2sin2sin24

θθθθθθθθ −++−+++=

crfbrcbr

WaP

( ) ( )[ ] ( ) ( )[ ]{ }1

2

2

2

121212 sinsinsinsin222sin2sin2

4

θθθθθθθθ −++−+++=

crfcbr

WaP

∫= rdFT f

∫−=2

1

cos2θ

θααdfPbrT f

[ ] 2

1sin2 θ

θα −= fPbrT f

( )12

2 sinsin θθ += fPbrT f

( )( ) ( )[ ] ( ) ( )[ ]{ }1

2

2

2

121212

12

2

sinsinsinsin222sin2sin2

sinsin4

θθθθθθθθθθ

−++−+++

+=

crfcbr

fWabrT f

( )( ) ( )[ ] ( ) ( )[ ]1

2

2

2

121212

12

sinsinsinsin222sin2sin2

sinsin4

θθθθθθθθθθ

−++−+++

+=

crfc

fWarT f

where 2

Der ==


Page 29 of 97

(b) ( ) ( )[ ] ( ) ( )[ ]1

2

2

2

121212 sinsinsinsin222sin2sin2 θθθθθθθθ −++>+++ crfc

( )( ) ( ) ( )1

2

2

2

1212

12

sinsin22sin2sin2

sinsin4

θθθθθθθθ

−−+++

+>

f

frc

(c) ∫= dNN

∫−=2

1

cosθ

θααdPbrN

[ ] 2

1sin

θθα −= PbrN

( )12 sinsin θθ += PbrN

fNF =

( )12 sinsin θθ += fPbrF

Solving for the location of F and N .

Let A = vertical distance from O .

( )∫∑ −−=

2

1

cos.

θ

θα dFrAM LocF

( )∫∑ −−=

2

1

2

. coscosθ

θααα fbrdrAPM LocF

( )∫∑ −−=

2

1

2

. coscosθ

θααα drAPfbrM LocF

( )∫∑ −

+−=

2

1

2cos12

1cos.

θ

θααα drAPfbrM LocF

2

1

2sin2

1

2

1sin.

θ

θ

ααα−

+−=∑ rAPfbrM LocF

( )[ ] ( ) ( )

+++−+=∑ 121212. 2sin2sin

2

1

2

1sinsin θθθθθθ rAPfbrM LocF

Then 0. =∑ LocFM

( )[ ] ( ) ( ) 02sin2sin2

1

2

1sinsin 121212 =

+++−+ θθθθθθ rA

( ) ( ) ( )

+++=+ 121212 2sin2sin

2

1

2

1sinsin θθθθθθ rA

( ) ( )

( )12

1212

sinsin

2sin2sin2

1

2

1

θθ

θθθθ

+

+++=

r

A

( ) ( )[ ]( )12

1212

sinsin4

2sin2sin2

θθθθθθ

+

+++=

rA


Page 30 of 97

896. For the brake shown with 21 θθ ≠ , assume that the direction of rotation is such

that a given W results in the greater braking torque and that φsinPp = . (a)

Derive equations in terms of 1θ and 2θ for the braking torque, for the moment

HFM and for HNM . (b) Reduce the foregoing equations for the condition

21 θθ = . (c) Now suppose that θ , taken as 21 θθθ += , is small enough that

θθ ≈sin , 1cos ≈θ , 2

21

θθθ == . What are the resulting equations?

Solution:

(a) Use clockwise rotation

φsinPp =

φφdPbrdN sin=


11 90 θφ −=

22 90 θφ +=

∫= rdFT f


Page 31 of 97

∫=2

1

sin2φ

φφφdfPbrT f

( )21

2 coscos φφ −= fPbrT f

( ) ( )[ ]21

2 90cos90cos θθ +−−= fPbrT f

( )21

2 sinsin θθ += fPbrT f

( )dFcrM HF ∫ −= φcos

( ) φφφφ

φdcrPrfbM HF sincos

2

1∫ −=

( ) φφφφφ

φdcrfPbrM HF ∫ −=

2

1

cossinsin

2

1

2sin2

1cos

φ

φ

φφ

−−= crfPbrM HF

( ) ( )

−−−= 1

2

2

2

21 sinsin2

1coscos φφφφ crfPbrM HF

( ) ( )[ ] ( ) ( )[ ]

−−+−+−−= 1

2

2

2

21 90sin90sin2

190cos90cos θθθθ crfPbrM HF

( ) ( )

−−+= 1

2

2

2

21 coscos2


( ) ( ) ( )[ ]

−−−−+= 1

2

2

2

21 sin1sin12


( ) ( )

−++= 1

2

2

2

21 sinsin2


dNrM HN φ∫= sin

φφφ

φdPrbM HN ∫=

2

1

22 sin

( ) φφφ

φd

PbrM HN ∫ −=

2

1

2cos12

2

[ ] 2

12sin2

4

2φφφ−=

PbrM HN

( ) ( )[ ]1212

2

2sin2sin24

φφφφ −−−=Pbr

M HN

( ) ( )[ ] ( ) ( )[ ]{ }1212

2

902sin902sin909024

θθθθ −−+−−−+=Pbr

M HN

( ) ( )[ ]1212

2

2sin2sin24

θθθθ −−−+=Pbr

M HN


Page 32 of 97

( ) ( )[ ]1212

2

2sin2sin24

θθθθ +++=Pbr

M HN

(b) 21 θθ =

( )21

2 sinsin θθ += PrfbT f

1

2 sin2 θPrfbT f =

( ) ( )

−++= 1

2

2

2

21 sinsin2


1

2 sin2 θfPbrM HF =

( ) ( )[ ]1212

2

2sin2sin24

θθθθ −−−+=Pbr

M HN

( )11

2

2sin244

θθ +=bPr

M HN

( )111

2

cossin444

θθθ +=bPr

M HN

( )111

2 cossin θθθ += bPrM HN

(c) 21 θθθ +=

θθ ≈sin

1cos ≈θ

221

θθθ ==

1

2 sin2 θPrfbT f =

θθθ 222

22

2sin2 PrfbPrfbPrfbT f =

=

=

1

2 sin2 θfPbrM HF =

θθθ 222

22

2sin2 PrfbPrfbPrfbM HF =

=

=

( )111

2 cossin θθθ += bPrM HN

( ) θθθ 2

HN bPrbPrM =

+= 1

22

2


Page 33 of 97

897. The brake shown is lined with woven asbestos; the cast-iron wheel is turning at

60 rpm CC; width of contact surface is 4 in. A force lbW 1300= . is applied via

linkage systemnot shown; o90=θ . Let φsinPp = . (a) With the brake lever as a

free body, take moments about the pivot J and determine the maximum pressure

and compare with permissible values. Compute (b) the braking torque, (c) the

frictional energy in fhp. (d) Compute the normal force N , the average pressure

on the projected area, and decide if the brake application can safely be

continuous.

Solution:

(a)

fdNdF =

φsinPp =

φφφ dPbrpbrddN sin==

φφdfPbrdF sin=

( )dFrRM JF ∫ −= φcos

( )∫ −=2

1

sincosφ

φφφφ drRfPbrM JF


Page 34 of 97

( )∫ −=2

1

sincossinφ

φφφφφ drRfPbrM JF

2

1

cossin2

1 2

φ

φ

φφ

+= rRfPbrM JF

( ) ( )

−+−= 121

2

2

2 coscossinsin2

1φφφφ rRfPbrM JF

10

5.12tan =β

o34.51=β

21

θβφ −=

o90=θ

o34.62

9034.511 =−=φ

o34.962

9034.51

21 =+=+=

θβφ

inb 4=

inr 10=

for woven asbestos 4.0=f (Table At 29)

( ) ( ) inR 16105.1222 =+=

( ) ( )

−+−= 121

2

2

2 coscossinsin2

1φφφφ rRfPbrM JF

( ) ( )( ) ( ) ( )

−+−= 34.6cos34.96cos1034.6sin34.96sin

2

161044.0 22

PM JF

PM JF 81.51−=

∫= dNRM JN φsin

∫=2

1

2sinφ

φφφdPbrRM JN

[ ] 2

12cos1

2

φφφ−=

PbrRM JN

( ) ( )[ ]1212 2sin2sin24

φφφφ −−−=PbrR

M JN

( )( )( ) ( ) ( ) ( )( )

−−

−= 34.62sin34.962sin

18034.634.962

4

16104 πPM JN

PM JN 9.572=

0=−+=∑ JNJFJ MMWaM


Page 35 of 97

( )( ) ( ) 09.57281.51251300 =−−+ PP

psiP 52=

max. psiPp 52== , 902 >φ

From Table AT 29, permissible psip 50=

Therefore epermissiblpp ≈max

(b) ∫= rdFT f

∫=2

1

sinφ

φφφdfPbrT f

( )21 coscos φφ −= fPbrT f

( )( )( )( )( ) lbinT f −=−= 918834.96cos34.6cos104524.0

(c) 000,63

nTfhp

f= , rpmn 60=

( )( )hpfhp 75.8

000,63

609188==

(d) ∫= dNN

∫=2

1

sinφ

φφφdPbrN

( )21 coscos φφ −= PbrN

( )( )( )( ) lbN 229734.96cos34.6cos10452 =−=

2sin2

.θ

br

Npave =

o90=θ

( )( )psipave 6.40

2

90sin1042

2297. ==

( ) ( )( ) min..755,12602012

6.4012

−−=

=

= insqlbft

Dnppvm

ππ

since min..000,28 −−< insqlbftpvm (§18.4)

Application is continuous.


Page 36 of 97

PIVOTED-SHOE BRAKES

898. In the brake shown, the shoe is lined with flexible woven asbestos, and pivoted at

point K in the lever; face width is 4 in.; o90=θ . The cast-iron wheel turns 60

rpm CL; let the maximum pressure be the value recommended in Table At 29.

On the assumption that K will be closely at the center of pressure, as planned,

compute (a) the brake torque, (b) the magnitude of force W , (c) the rate at which

frictional energy grows, (d) the time of an application if it is assumed that all this

energy is stored in the 1-in. thick rim with Ftrim 350=∆ , (e) the average pressure

on projected area. May this brake be applied for a “long time” without damage?

(f) What would change for CC rotation?

Problem 898.

Solution:

ina 27= , inb 4= , rpmn 60= CL

θθ

θ

sin

2sin2

+=

D

c

inD 20= , inr 10=

rad571.190 == oθ

( )inc 0.11

90sin571.1

2

90sin202

=+

=

(a) 2

sin2 2 θfPbrT f =

For woven asbestos, Table AT 29, 4.0=f

psiP 50=

( )( )( )( ) lbinT f −== 314,112

90sin104504.02

2


Page 37 of 97

(b)

( )( )( ) lbPbrN 25712

90sin571.110450

2

sin=

+=

+=

θθ

[ ]∑ = 0JM

NWa 12=

( ) ( )25711215 =W

lbW 2057=

(c) ( )( )

hpnT

fhpf

78.10000,63

60314,11

000,63===

rate of frictional energy ( ) min740,35578.10000,33000,33 lbftfhp −===

(d) Time (min) fhp

U f

000,33=

cW

lbftUFt

m

f −=∆ o

DbtWm ρπ=

For cast iron 3253.0 inlb=ρ

Flblbftc −−= 101

int 1=

( ) ( )( )( ) lbWm 6.631420253.0 == π

( )( )Flblbftlb

lbftUFt

f

−−

−==∆

1016.63350o

lbftU f −= 260,248,2

Time (min) ( )

min32.678.10000,33

260,248,2==

(e) Ave.

( )( )psi

br

Np 45.45

2

90sin1042

2571

2sin2

===θ


Page 38 of 97

( )( )( )( )Finsqlbft

Dnppvm −−=== ..280,14

12

602045.45

12

ππ

since 000,28<mpv , this brake may be applied for a long time.

(f) Since the moment arn of F is zero, no change or CC rotation.

899. The pivoted-shoe brake shown is rated at 450 ft-lb. of torque; o90=θ ; contact

width is 6.25 in.; cast-iron wheel turns at 600 rpm; assume a symmetric

sinusoidal distribution of pressure. (a) Locate the center of pressure and compute

with the location of K. Compute (b) the maximum pressure and compare with

allowable value, (c) the value of force W , (d) the reaction at the pin H , (e) the

average pressure and mpv , and decide whether or not the application could be

continuous at the rated torque. (f) Compute the frictional work from ωT and

estimate the time it will take for the rim temperature to reach 450 F (ambient, 100

F).

Problem 899.

Solution:

(a) θθ

θ

sin

2sin2

+=

D

c

inD 18=

rad571.190 == oθ

( )inc 9011.9

90sin571.1

2

90sin182

=+

=

but location of K = 9.8125 in

then, Klocationc ≈

(b) 2

sin2 2 θfPbrT f =

lbinlbftT f −=−= 5400450


Page 39 of 97

inb 25.6=

inr 9=

use 4.0=f (on cast-iron)

2sin2 2 θ

fPbrT f =

( ) ( )( )2

90sin925.64.025400

2P=

psiP 86.18= < allowable (Table AT 9)

(c) ( ) ( )375.10375.20 NW =

( )( )( ) lbPbrN 13642

90sin571.1925.686.18

2

sin=

+=

+=

θθ

( )( )lbW 695

375.20

375.101364==

(d) ↓=−=−= lbWNH 6696951364

(e) Ave.

( )( )psi

br

Np 15.17

2

90sin925.62

1364

2sin2

===θ

rpmn 600=

( )( )( )( )Finsqlbft

Dnppvm −−=== ..490,48

12

6001815.17

12

ππ

since 000,28>mpv , not continuous

(f) Frictional work ( ) ( )sec275,28

minsec60

6002450 perlbft

rpmlbftT −=

−==

πω

cW

lbftUFt

m

f −=∆ o

DbtWm ρπ=

For cast iron 3253.0 inlb=ρ

Flblbftc −−= 101

( ) ( )( ) ( )tttWm 154

4

1825.618253.0

2

=

+=

ππ

Ft 350100450 =−=∆

( )( )( ) lbftttctWU mf −==∆= 900,443,5101154350


Page 40 of 97

lbftU f −= 260,248,2

sec5.192275,28

900,443,5t

tTime ==

Assume int2

1=

sec96=Time

TWO-SHOE BRAKES

PIVOTED SHOES

900. The double-block brake shown is to be used on a crane; the force W is applied

by a spring, and the brake is released by a magnet (not shown); o90=θ ; contact

width = 2.5 in. Assume that the shoes are pivoted at the center of pressure. The

maximum pressure is the permissible value of Table AT 29. Compute (a) the

braking torque, (b) the force W , (c) the rate of growth of frictional energy at 870

rpm, (d) the time it would take to raise the temperature of the 0.5-in.-thick rim by

Ft 300=∆ (usual assumption of energy storage), (e) mpv . (f) Where should the

pivot center be for the calculations to apply strictly?

Problem 900.

Solution:

( )in

D

c 5.5

90sin2

2

90sin102

sin

2sin2

=+

=+

=πθθ

θ


Page 41 of 97

[ ]∑ = 0

1RM

( ) 11 75.675.12875.05.5 NWF =+−

( ) 11 75.675.12625.4 NWfN =+

f

WN

625.425.6

75.121 −

=

[ ]∑ = 0

2RM

( ) 22 75.6875.05.575.12 NFW +−=

22 75.6625.475.12 NfNW +=

f

WN

625.425.6

75.122 +

=

Assume flexible woven asbestos,

40.0=f , psip 50=

( )W

WN 898.2

40.0625.425.6

75.121 =

−=

( )( ) WWfNF 16.1898.24.011 ===


Page 42 of 97

( )W

WN 574.1

40.0625.425.6

75.122 =

+=

( )( ) WWfNF 63.0574.14.022 ===

1.max ff TT =

cFfPbr 1

2

2sin2 =

θ

( )( )( ) ( )( )5.516.12

90sin

2

105.25040.02

2

W=

lbW 277=

(a) Braking torque = ( ) ( )( )( ) lbincFFTT ff −=+=+=+ 27275.527763.016.12121

(b) lbW 277=

(c) ( )( )

hpnT

fhpf

66.37000,63

8702727

000,63===

(d) Solving for tine:

cW

lbftUFt

m

f −=∆ o

FFtoo 300=∆

101=c , 253.0=ρ for cast iron

VWm ρ=

( )( )( ) ( ) ( ) 3

22

54.784

5.0105.05.210

4in

tDDbtV =+=+=

ππ

ππ

( )( ) lbWm 87.1954.78253.0 ==

( )( )( ) lbftU f −== 061,60210187.19300

( )sec29min4844.0

66.37000,33

061,602

000,33====

fhp

UTime

f

(e) mpv :

( )( )fpm

Dnvm 2278

12

87010

12===

ππ

( )( ) 900,113227850 ==mpv

(f) inc 5.5=

901. A pivoted-shoe brake, rated at 900 ft-lb. torque, is shown. There are 180 sq. in. of

braking surface; woven asbestos lining; 600 rpm of the wheel; 90o arc of brake

contact on each shoe. The effect of spring A is negligible. (a) Is the pin for the


Page 43 of 97

shoe located at the center of pressure? (b) How does the maximum pressure

compare with that in Table AT 29? (c) What load W produces the rated torque?

(d) At what rate is energy absorbed? Express in horsepower. Is it likely that this

brake can operate continuously without overheating? (e) Does the direction of

rotation affect the effectiveness of this brake?

Problem 901.

Solution:

(a)

( )in

D

c 9.9

90sin2

2

90sin182

sin

2sin2

=+

=+

=πθθ

θ

and in9.92

16

1319

≈ , therefore the pin located at the center of pressure

(b)

16

1319

4tan =α

o4.11=α

[ ]∑ = 0QM

WFA 5.8cos4 =α


Page 44 of 97

WFA 5.84.11cos4 =

WFA 168.2=

[ ]∑ = 0VF and [ ]∑ = 0HF

( ) WWWWFQ Av 429.14.11sin168.2sin =+=+= α

( ) WWFQ Ah 125.24.11cos168.2cos === α

[ ]∑ = 0

1RM

( ) hQN 375.20375.101 =

( ) ( )WN 125.2375.20375.101 =

WN 173.41 =

11 NfF =

For woven asbestos lining, 40.0=f , psip 50=

( )( ) WWF 67.1173.440.01 == (either direction)


Page 45 of 97

[ ]∑ = 02RM

αcos375.20375.10 2 AFN =

( ) WWN 174.44.11cos168.2375.10

375.202 ==

( )( ) WWF 67.1174.440.02 == (either direction)

( )cFFT f 21 +=

( )( ) ( )( )( )9.967.167.112900 W+=

lbW 6.326=

2sin2 2

21

θfPbrFcTT ff ===

but brA θ=

θAr

br =2

( )( )( )( )( )( )( )

2

2

90sin91804.02

9.96.32667.1π

P

=

psipsiP 5026.9 <=

(c) lbW 6.326=

(d) ( )( )( )

hpnT

fhpf

103000,63

60012900

000,63===

( )( )fpm

Dnvm 2827

12

60018

12===

ππ

( )( ) Finsqlbftpvm −−== ..178,26282726.9

since 000,28<mpv , it is likely to operate continuously.

(e) Since the value of F is independent of rotation, the direction doesn’t affect the

effectiveness of this brake.

902. Refer to the diagrammatic representation of the brake of Fig. 18.2, Text, and let

the dimensions be: 16

94==== tmba , 14=c , 15=D , inh

16

99= ., and the

contact width is 4 in.; arc of contact = 90o; lining is asbestos in resin binder,

wheel rotation of 100 rpm CC; applied load lbW 2000= . (a) Locate the center of

pressure for a symmetrical sinusoidal pressure distribution and compare with the

actual pin centers. Assume that this relationship is close enough for approximate


Page 46 of 97

results and compute (b) the dimensions k and e if the braking force on each

shoe is to be the same, (c) the normal force and the maximum pressure, (d) the

braking torque, (e) mpv . Would more-or-less continuous application be

reasonable?

Figure 18.2

Solution:

(a)

( )in

D

c 25.8

90sin2

2

90sin152

sin

2sin2

=+

=+

=πθθ

θ

On Centers:

cinmtK >=+=+ 125.916

94

16

94:

cinbaB >=+=+ 125.916

94

16

94:

[ ]∑ = 0CRM


Page 47 of 97

( )WceeRF +=

We

ceRF

+=

WRR FC −=

e

cWWW

e

ceRC =−

+=

[ ]∑ = 0

HRM

aRbFhN F=− 11

aRbfNhN F=− 11

fbh

aRN F

−=1

fbh

afRF F

−=1

( )( )fbhe

WcefaF

−

+=1

[ ]∑ = 0

ERM

kRtFhN C=+ 22


Page 48 of 97

kRtfNhN C=+ 22

fth

kRN C

+=2

fth

kfRF C

−=2

( )fthe

fkcWF

+=2

(b) 21 ff TT =

cFcF 21 =

21 FF =

( )( ) ( )fthe

fkcW

fbhe

Wcefa

+=

−

+

( )fth

kc

fbh

cea

+=

−

+

For asbestos in resin binder,

35.0=f , Table AT 29

inina 5625.416

94 ==

ininb 5625.416

94 ==

ininm 5625.416

94 ==

inint 5625.416

94 ==

inc 14=

ininh 5625.916

99 ==

( )( )

( )( )5625.435.05625.9

14

5625.435.05625.9

145625.4

+=

−

+ ke

ke 1903.214 =+

but emk =+

or 5625.4+= ke

then kk 1903.2145625.4 =++

ink 6.15=

ine 1625.205625.46.15 =+=

(c) ( )

( )( )( )( ) ( )[ ]

lbfthe

kcWNNN 2720

5625.435.05625.91625.20

2000146.1521 =

−=

+===


Page 49 of 97

(d) ( ) ( )( )( ) lbincNNfTTT fff −==+=+= 708,1525.82720235.02121

(e) ( )( )

fpmDn

vm 39312

10015

12===

ππ

( )( ) Finsqlbftpvm −−== ..195,2539311.64

since 000,28<mpv , continuous application is reasonable.

FIXED SHOES

903. A double-block brake has certain dimensions as shown. Shoes are lined with

woven asbestos; cast-iron wheel turns 60 rpm; applied force lbW 70= . For each

direction of rotation, compute (a) the braking torque, (b) the rate of generating

frictional energy (fhp). (c) If the maximum pressure is to be psiP 50= (Table

AT 29), what contact width should be used? (d) With this width, compute mpv

and decide whether or not the applications must be intermittent.

Problems 903, 904.

Solution:

[ ]0=∑ BM

WQ 264 =


Page 50 of 97

WQ 5.6=

[ ]0=∑ RM

( )WQS 5.66625.2 ==

WS 33.17=

WSRH 33.17==

WQRV 5.6==

ine 10=

inR 5.12=

ina 75.235.12925.2 =++=

011

=−−=∑ HNHFH MMSaM (CC)

011

=−+=∑ HNHFH MMSaM (CL)

( ) ( )

−−−= 1

2

2

2

21 sinsin2

coscos1

φφφφR

rfbrPM HF

( ) ( )[ ]1212 2sin2sin241


M HN

2sin2 2

1

θfPbrT f =

2sin2

1

θfr

TPbr

f=

inr 10=


Page 51 of 97

inr 112

sin2 =θ

( ) in112

sin102 =θ

rad165.143.66 == oθ

4.0=f for woven asbestos

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

211

1 θ

φφφφ

fr

RrfT

Mf

HF

−−−=

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

211

1 θ

φφφφ

r

RrT

Mf

HF

−−−=

rad9886.064.562

73.6690

2901 ==−=−= oθ

φ

o28.1132 1 =φ

rad1530.236.1232

73.6690

2902 ==+=+= oθ

φ

o72.2462 2 =φ

θφφ =− 12

( ) ( )

( )1

1

1

2

73.66sin102

64.56sin36.123sin2

5.1236.123cos64.56cos10 22

f

f

HF T

T

M =

−−−=

( ) ( )[ ]

−−−=

2sin24

2sin2sin2 12121

1 θ

φφφφ

fr

RTM

f

HN

( ) ( )[ ]

( )( )1

1

196.2

2

73.66sin104.08

28.113sin72.246sin165.125.12f

f

HN TT

M =−−

=

CC:

011

=−−=∑ HNHFH MMSaM

( )( )( ) 0960.275.237033.1711

=−− ff TT

lbinT f −= 72761

CL:

011

=−+=∑ HNHFH MMSaM


Page 52 of 97

( )( )( ) 0960.275.237033.1711

=−+ ff TT

lbinT f −= 700,141

ine 10=

ind 5.12=

CC: [ ]∑ = 0HM

022

=−+−′HNHFVH MMdRaR

CL: [ ]∑ = 0HM

022

=−−−′HNHFVH MMdRaR

2

2

12

1

f

f

f

HFHF TT

TMM =

=

2

2

12960.2

1

f

f

f

HNHN TT

TMM =

=

CC:

022


( )( ) ( )( )[ ]( ) 0960.2705.125.65.2133.1722

=−+− ff TT

lbinT f −= 405,102

CL:

022


( )( ) ( )( )[ ]( ) 0960.2705.125.65.2133.1722

=−−− ff TT

lbinT f −= 51502

(a) Braking Torque 21 ff TT +=


Page 53 of 97

CC:

lbinTTT fff −=+=+= 681,17405,10727621

CL:

lbinTTT fff −=+=+= 850,195150700,1421

(b) Rate of generating frictional energy

000,63

nTfhp

f=

CC: ( )( )

hpfhp 84.16000,63

60681,17==

CL: ( )( )

hpfhp 90.18000,63

60850,19==

(c) psip 50=

2sin2 2

21

θfPbrTorT ff =

CC:

( )( )( )in

Prf

Tb

f73.4

2

73.66sin10504.02

405,10

2sin2

22

1 ===θ

CL:

( )( )( )in

Prf

Tb

f68.6

2

73.66sin10504.02

700,14

2sin2

22

2 ===θ

(d) mpv

( )( )fpm

Dnvm 314

12

6020

12===

ππ

( )( ) 000,55700,1531450 <==mpv

( )( ) 000,28700,1531450 <==mpv

application can be continuous or intermittent.

904. If the brake shown has a torque rating of 7000 lb-in. for counter-clockwise

rotation, what braking torque would it exert for clockwise rotation, force W the

same?

Solution:

CC:

011

=−− HNHF MMSa

11 fHF TM =

11960.2 fHN TM =


Page 54 of 97

WS 33.17=

ina 75.23=

( )( ) 096.275.2333.1711

=−− ff TTW

WT f 9.1031

=

022


WRH 33.17=

WRV 53.6=

ina 5.21=′

22 fHF TM =

22960.2 fHN TM =

( )( ) ( )( ) 0960.25.125.65.2133.1722

=−+− ff TTWW

WT f 65.1482

=

21 fff TTT +=

WW 65.1489.1037000 +=

lbW 7.27=

CL:

011

=−+ HNHF MMSa

( )( )( ) 096.275.237.2733.1711

=−− ff TT

lbinT f −= 58171

022


( )( ) ( )( )[ ]( ) 0960.27.275.125.65.2133.1722

=−+− ff TT

lbinT f −= 20382

lbinTTT fff −=+=+= 78552038581721

(CL)

905. A double-block brake is shown for which o90=θ , inb 5= ., rpmn 300= , rim

thickness = ¾ in., and lbW 400= . The shoes are lined with asbestos in resin

binder. Determine the frictional torque for (a) clockwise rotation, (b)

counterclockwise rotation. (c) How much energy is absorbed by the brake?

Express in horsepower. (d) Will the brake operate continuously without danger of

overheating? How long for a Ftrim 300=∆ ? How does mpv compare with Text

values?

ind 5.12=


Page 55 of 97

Problem 905

Solution:

44

4tan

+=α

o565.26=α

[ ]0=∑ RM

( )( ) WQ 164cos =α

( )( ) ( )400164565.26cos =Q

lbQ 1789=

lbQRH 1600565.26cos1789cos === α

lbWQRV 1200400565.26sin1789sin =+=+= α


Page 56 of 97

( ) ( )

−−−= 1

2

2

2

21 sinsin2

coscos φφφφR

rfbrPM HF

( ) ( )[ ]1212 2sin2sin24


M HN

2sin2 2 θ

fPbrT f =

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

21

θ

φφφφ

r

RrT

Mf

HF

−−−=

( ) ( )[ ]

2sin8

2sin2sin2 1212

θφφφφ

fr

RTM

f

HN

−−−=

inr 102

20==

12

4tan =β

o435.18=β

rad571.190 == oθ

rad464.0565.26435.182

9090

2901 ==−−=−−= oβ

θφ

( ) o13.53565.2622 1 ==φ

rad034.2565.116435.182

9090

2902 ==−+=−+= oβ

θφ

( ) o13.233565.11622 2 ==φ

inR 65.12124 22 =+=

Asbestos in resin binder 35.0=f


Page 57 of 97

( ) ( )

( )f

f

HF T

T

M 6803.0

2

90sin102

565.26sin565.116sin2

65.125656.116cos565.26cos10 22

=

−−−=

( ) ( ) ( )[ ]

( )( )f

f

HN TT

M 03.3

2

90sin1035.08

13.53sin13.233sin464.0034.2265.12=

−−−=

(a) Clockwise

[ ]∑ = 0

1HM

( )( ) ( )( ) 024cos5.2sin1111

=−++ HNHF MMQQ αα

( )( ) ( )( ) 003.36803.024565.26cos17895.2565.26sin178911

=−++ ff TT

lbinT f −= 195,171

[ ]∑ = 0

2HM

0245.22222

=++− HFHNHV MMRR

( ) ( ) 06803.003.316002412005.222

=++− ff TT

lbinT f −= 95412


Page 58 of 97

lbinTTT fff −=+=+= 736,269541195,1721

(b) Counterclockwise

[ ]∑ = 0

1HM

0sin5.2cos241111

=−−+ HNHF MMQQ αα

( )( ) ( )( ) 003.36803.0565.26sin17895.2565.26cos17892411

=−−+ ff TT

lbinT f −= 890,101

[ ]∑ = 0

2HM

0245.22222

=+−− HNHFHV MMRR

( ) ( ) 003.36803.016002412005.222

=+−− ff TT

lbinT f −= 066,152

lbinTTT fff −=+=+= 956,25066,15890,1021


Page 59 of 97

(c) CL:

( )( )hp

nTfhp

f3.127

000,63

300736,26

000,63===

CC:

( )( )hp

nTfhp

f6.123

000,63

300956,25

000,63===

(d) ( )( )

fpmDn

vm 157112

30020

12===

ππ

For p :

12sin2 2

ff TfPbrT ==θ

(CL)

( )( )( )( )2

90sin10535.02195,17

2P=

psiP 48.69=

( )( ) 000,28153,109157148.69 >==mpv

the brake operate continuously with danger of overheating.

For time:

cW

lbftUFt

m

f −=∆ o

101=c , 253.0=ρ

VWm ρ=

4

2tD

DbtVπ

π +=

( )( ) ( ) 3

2

24.4714

3

4

20

4

3520 inV =

+

=π

π

( )( ) lbVWm 22.11924.471253.0 === ρ

( )( )( ) lbfttcWU mf −==∆= 366,612,330010122.119

Time = fhp

U f

000,33

CL: Time = ( )

sec53min886.06.123000,33

366,612,3

000,33===

fhp

U f

CC: Time = ( )

sec52min860.03.127000,33

366,612,3

000,33===

fhp

U f

000,28>mpv , not good for continuous application.


Page 60 of 97

906. The double-block brake for a crane has the dimensions: 3.14=a , 37.2=b ,

10=D , 05.11=e , 1.7=g , 12=h , 6.6=j , 55.10=k , inm 5.3= ., the width of

shoes is 4 in., and the subtended angle is o90=θ ; wocen asbestos lining. Its

rated braking torque is 200 ft-lb. The shoes contact the arms in such a manner

that they are virtually fixed to the arms. What force W must be exerted by a

hydraulic cylinder to develop the rated torque for (a) counterclockwise rotation,

(b) clockwise rotation? Is the torque materially affected by the direction of

rotation? (c) Compute the maximum pressure and compare with that in Table AT

29. (Data courtesy of Wagner Electric Corporation.)

Problem 906.

Solution:

83.005.11

37.2tan

−=

−=

ce

bα

o056.13=α


Page 61 of 97

[ ]∑ = 0RM

eWcQbQ =+ αα sincos

( )( ) ( )( ) WQQ 3.14056.13sin83.0056.13cos37.2 =+

WQ 7286.5=

WWQRH 58.5056.13cos7286.5cos === α

WWWWQRV 294.0056.13sin7286.5sin =−=−= α

6.6

275.5

2tan ==

j

kβ

o63.38=β

rad1112.037.663.382

9090

2901 ==−−=−−= oβ

θφ

o74.122 1 =φ

rad6820.137.9663.382

9090

2902 ==−+=−+= oβ

θφ

o74.1922 2 =φ

( ) injk

R 449.86.62

55.10

2

2

2

2

2

=+

=+

=

inD

r 52

10

2===


Page 62 of 97

( ) ( )

2sin2

sinsin2

coscos 1

2

2

2

21

θ

φφφφ

r

RrT

Mf

HF

−−−=

( ) ( )[ ]

2sin8

2sin2sin2 1212

θφφφφ

fr

RTM

f

HN

−−−=

For woven asbestos lining, 40.0=f

( ) ( )

( )f

f

HF T

T

M 1985.0

2

90sin52

37.6sin37.96sin2

449.837.96cos37.6cos5 22

=

−−−=

( ) ( )[ ]

( )( )f

f

HN TT

M 6755.2

2

90sin54.08

74.12sin74.192sin1112.0682.12449.8=

−−−=

(a) CC:

[ ]∑ = 01HM

( ) ( ) 025.0121111

=−−− HNHFVH MMRR

( )( ) ( )( ) 06755.21985.025.0294.01258.511

=−−− ff TTWW

WT f 3.231

=


Page 63 of 97

[ ]∑ = 0

2HM

05.3sin25.0cos122222

=−−++ WMMQQ HNHFαα

( ) ( )22

1985.06755.25.3056.13sin3.1425.0056.13cos3.1412 ff TTWWW −=−+

WT f 4.662

=

21 fff TTT +=

lbinlbftT f −=−= 2400200

WW 4.663.232400 +=

lbW 8.26=

(b) CL:


Page 64 of 97

[ ]∑ = 01HM

( ) ( ) 025.0121111

=−+− HNHFVH MMRR

( )( ) ( )( ) 06755.21985.025.0294.01258.511

=−+− ff TTWW

WT f 0.271

=

[ ]∑ = 0

2HM

05.3sin25.0cos122222

=−−−+ WMMQQ HNHFαα

( ) ( )22

1985.06755.25.3056.13sin3.1425.0056.13cos3.1412 ff TTWWW +=−+

WT f 2.572

=

21 fff TTT +=

WW 2.570.272400 +=

lbW 5.28=

Since W has different values, torque is materially affected by the direction of rotation.

(c) 2

sin2 2 θfPbrT f =

For woven asbestos lining, 40.0=f

Use ( ) lbinWT f −=== 17808.264.664.66

inb 4=

inr 5= o90=θ

( ) ( )( )2

90sin544.021780

2PT f ==

psiP 47.31=


Page 65 of 97

From Table AT 29, psip 50max =

psipsi 5047.31 <

INTERNAL-SHOE BRAKES

908. Assuming that the distribution of pressure on the internal shoe shown is given by

φsinPp = , show that the moments BNM , BFM , and OFT of N with respect to

B and of F with respect to B and to O are (b = face width)

( ) ( )[ ]22sin2sin2 12 φφθ −−= PbarM BN ,

( ) ( )[ ]2sinsincoscos 1

2

2

2

21 φφφφ −−−= arfPbrM BF ,

( )21

2 coscos φφ −= fPbrT OF .

Problems 908 – 910.

Solution:

φsinPp =

( ) kdNMd BN =

( ) φφφφ dPbrbrdPdN sinsin ==

( ) φφ sin90cos aak =−=

( ) ( )( ) φφφφφ dPabrdPbraMd BN

2sinsinsin ==

( )∫∫ −==2

1

2

1

2cos12

sin2φ

φ

φ

φφφφφ d

PabrdPabrM BN

( ) ( )

−−−=

−=

2

2sin2sin

22sin

2

1

2

1212

2

1

φφφφφφ

φ

φ

PabrPabrM BN

but θφφ =− 12


Page 66 of 97

( )

−−=

2

2sin2sin

2

12 φφθ

PabrM BN

( ) edFMd BF =


( ) φφ cos90sin arare −=−+=

( ) ( )( ) ( ) φφφφφφφ darfPbrdfPbrarMd BF cossinsinsincos −=−=

[ ] 2

1

2sincosφ

φφφ arfPbrM BF −−=

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF

( ) φφdfPbrrdFTd OF sin2==

[ ] 2

1cos2 φ

φφ−= fPbrT OF

( )21

2 coscos φφ −= fPbrT OF

909. The same as 908, except that a pressure distribution of αcosPp = is assumed.

( ) ( )[ ]2sinsin42sin2sin2 1

2

2

2

12 ααααθ −+++= chPbrM BN ,

( ) ( ) ( )[ ]42sin2sin22sinsinsinsin 121

2

2

2

12 ααθαααα ++−−++= chrfPbrM BF

( )12

2 sinsin αα += fPbrM OF .

Solution:

αα sincos chk +=

αα cossin chre −+=

ααα dPbrpbrddN cos==

ααdfPbrfdNdF cos==

( )( )αααα dPbrchkdNdM BN cossincos +==

( ) αααα dchPbrdM BN cossincos2 +=

( )∫− +=2

1

cossincos2α

ααααα dchPbrM BN

( ) 2

1

2

sin

4

2sin2 2α

α

ααα

−

+

+=

chPbrM BN

but 21 ααθ +=

( ) ( )[ ] ( )[ ]

−−

+−−++

=2

sinsin

4

2sin2sin2 1

2

2

2

1212 αααααα chPbrM BN


Page 67 of 97

( ) ( )

−+

++=

2

sinsin

4

2sin2sin2 1

2

2

2

12 ααααθ chPbrM BN

( )( )αααα dfPbrchredFdM BF coscossin −+==

( ) ααααα dchrfPbrdM BF

2coscossincos −+=

( ) 2

1

4

2sin2

2

sinsin

2α

α

αααα

−

+−+=

chrfPbrM BF

( )[ ] ( )[ ] ( ) ( )[ ]

−−++

−−−

+−−=4

2sin2sin2

2

sinsinsinsin 12121

2

2

2

12

αααααααα

chrfPbrM BF

( ) ( ) ( )

++−

−++=

4

2sin2sin2

2

sinsinsinsin 121

2

2

2

12

ααθαααα

chrfPbrM BF

( ) αααα dfPbrdfPbrrrdFdM OF coscos 2===

[ ] ( )[ ]12

22 sinsinsin 2

1ααα α

α −−=−= − fPbrfPbrM OF

( )12

2 sinsin αα += fPbrM OF

910. The same as 909, except that the α is to be measured from OG , a perpendicular

to OB ; limits from 1α− to 2α+ .

Solution:

αcosak =

αsinare +=

( ) ( ) ααααααα dPbar

dPbardPbrakdNdM BN 2cos12

coscoscos 2 +====

( ) ( )[ ]1212 2sin2sin242

2sin2

2

2

1

αααααα

α

α

−−++=

+=

−

PbarPbarM BN

( )12 2sin2sin24

ααθ −+=Pbar

M BN

( )( ) ( ) ααααααα darfPbrdfPbraredFdM BF cossincoscossin +=+==

( )[ ] ( )[ ]

−−

+−−=

+=

−2

sinsinsinsin

2

sinsin 1

2

2

2

12

2 2

1

αααα

αα

α

α

arfPbr

arfPbrM BF

( ) ( )

−++=

2

sinsinsinsin 1

2

2

2

12

αααα

arfPbrM BF

( ) αααα dfPbrdfPbrrrdFdM OF coscos 2===


Page 68 of 97

[ ] ( )[ ]12

22 sinsinsin 2

1ααα α

α −−=−= − fPbrfPbrM OF

( )12

2 sinsin αα += fPbrM OF

911. The following dimensions apply to a two-shoe truck brake somewhat as shown:

face 5=b , 8=r , 1.5=h , 6.2=c , inuw 4.6== ., o110=θ , o151 =φ . Lining is

asbestos in rubber compound. For a maximum pressure on each shoe of 100 psi,

determine the force Q , and the braking torque for (a) clockwise rotation, (b)

counterclockwise rotation. See 908. (Data courtesy of Wagner Electric

Corporation.)

Problems 911, 912.

Solution: See 908.

( )

−−=

2

2sin2sin

2

12 φφθ

PbarM BN

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF ,

( )21


o151 =φ , o302 1 =φ , o1251101512 =+=+= θφφ , o2502 1 =φ

psip 100=

inb 5=

inr 8=

( ) ( ) incha 7245.56.21.52222 =+=+=

rad92.1110 == oθ

For asbestos in rubber compound

35.0=f


Page 69 of 97

(a) Both sides (clockwise rotation)

( ) 0=−++ BNBF MMwhQ

( )( )( )( ) ( )lbinM BN −=

−−= 224,30

2

30sin250sin92.1

2

87245.55100

( )( )( )( ) ( ) ( )lbinM BF −=

−−−= 436,14

2

30sin125sin7245.5125cos30cos88510035.0

22

inh 1.5= , inw 4.6=

( ) 0224,30436,144.61.5 =−++Q

lbQ 1373=

( )( )( )( ) ( ) lbinT OF −=−= 242,17125cos15cos8510035.02

( ) lbinTT OFf −=== 484,34242,1722

(b) Counterclockwise rotation

( ) 0=−−+ BFBN MMwhQ


Page 70 of 97

( ) 0436,14224,304.61.5 =−−+Q

lbQ 3883=

( ) lbinTT OFf −=== 484,34242,1722

913. The data are the same as 911, but the shoe arrangement is as shown for this

problem. For a maximum pressure on the shoes of 100 psim determine the force

Q and OFT for (a) Cl rotation, (b) CC rotation, See 908.

Problem 913.

Solution:

( )21

2 coscos φφ −== fPbrTT fOF

( )21 coscos φφ −=

fr

TPbr

f

( )( )

( )

−−

−=

−−=

2

2sin2sin

coscos22

2sin2sin

2

12

21

12 φφθ

φφφφ

θfr

aTPbarM

f

BN

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF

( )( ) ( )

−−−

−=

2

sinsincoscos

coscos2

1

2

2

2

21

21

φφφφ

φφa

rr

TM

f

BF

From 911: o151 =φ , o302 1 =φ , o1251101512 =+=+= θφφ , o2502 1 =φ

rad92.1110 == oθ

( ) ( ) incha 7245.56.21.52222 =+=+=

35.0=f


Page 71 of 97

( )( )( )( )

f

f

BN TT

M 753.12

30sin250sin92.1

125cos15cos835.02

7245.5=

−−

−=

( )( ) ( )

f

f

BF TT

M 43.02

15sin125sin7245.5125cos15cos8

125cos15cos28

22

=

−−−

−=

(a) CL rotation:

Left Side

[ ]∑ = 0BM

( ) 011

=−−+ BFBN MMwhQ

( ) 043.0753.14.61.511

=−−+ ff TTQ

QT f 268.51

=

Right Side:

[ ]∑ = 0BM


Page 72 of 97

( ) 022

=−++ BNBF MMwhQ

( ) 0753.143.04.61.522

=−++ ff TTQ

QT f 6924.82

=

QTT ff 6924.82max

==

( )21

2 coscosmax

φφ −= fPbrT f

( )( )( )( ) ( )125cos15cos8510035.06924.82 −=Q

lbQ 1984=

( ) lbinQT f −=== 452,101984268.5268.51

( ) lbinQT f −=== 246,1719846924.86924.82

Total lbinTTT ffOF −=+=+= 698,27246,17452,1021

(b) CC rotation

Left Side

[ ]∑ = 0BM

( ) 011

=−++ BNBF MMwhQ

( ) 0753.143.04.61.511

=−++ ff TTQ

QT f 6924.81

=

Right Side:


Page 73 of 97

[ ]∑ = 0BM

( ) 022

=−−+ BNBF MMwhQ

( ) 0753.143.04.61.522

=−−+ ff TTQ

QT f 268.52

=

Since values are just interchanged

lbQ 1984=

Total lbinT OF −= 698,27 as in (a)

914. A double-shoe internal brake is actuated by an involute cam as shown, where RQ

is the force on the right shoe at a radius Rw and LQ is the force on the left shoe at

a radius Lw . The pressure of each shoe is proportional to the rotation of the shoe

about B which is inversely proportional to w ; therefore, the ratio of the

maximum pressures is LRRL wwPP = . The dimensions are: face width 4=b ,

6=r , 16

94=h ,

8

11=c ,

16

59=Lw , inwR

16

58= .: for each shoe, o120=θ ,

o301 =φ . The lining is asbestos in rubber compound, Determine the braking

torque and forces RQ and LQ for the maximum permissible pressure for (a)

clockwise rotation, (b) counterclockwise rotation.


Page 74 of 97

Problem 914.

Solution:

( )

−−=

2

2sin2sin

2

12 φφθ

PbarM BN

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arfPbrM BF ,

( )21


incha 70.48

11

16

94

22

22 =

+

=+=

8926.0

16

59

16

58

===L

R

R

L

w

w

p

p

For asbestos in rubber compound, 35.0=f , psip 75=

psipR 75=

( ) psipL 67758926.0 ==

(a) Clockwise rotation

Left Side:


Page 75 of 97

[ ]∑ = 0

LBM

0=−−LLLL BNBFLL MMwQ

( ) ( )

−−−=

2

sinsincoscos 1

2

2

2

21

φφφφ

arbrfPM LBF LL

o301 =φ o602 1 =φ

o1503012012 =+=+= φθφ o3002 2 =φ

rad094.2120 == oθ

( )( )( )( ) ( ) ( )lbinM

LL BF −=

−−−= 5849

2

30sin150sin70.4150cos30cos6646735.0

22

( )

−−=

2

2sin2sin

2

12 φφθ

barPM L

BNLL

( )( )( )( ) ( )lbinM

LL BN −=

−−= 185,11

2

60sin300sin094.2

2

67.4467

0185,11584916

59 =−−

LQ

lbQL 1829=

( ) ( )21

2 coscos φφ −= brfPT LOFL

( ) ( )( )( )( ) ( ) lbinTL

OF −=−= 5849150cos30cos646735.02

Right side:


Page 76 of 97

[ ]∑ = 0

RBM

0=−+RRRR BNBFRR MMwQ

( )( )lbin

P

PMM

L

RBF

BFLL

RR−=== 6547

67

755849

( )( )lbin

P

PMM

L

RBN

BNLL

RR−=== 520,12

67

75185,11

0520,12654716

58 =−+

RQ

lbQR 719=

( )( ) ( )( )

lbinP

PTT

L

ROF

OFL

R−=== 6547

67

755849

( ) ( ) ( ) lbinTTTLR

OFOFOF −=+=+= 396,1258496547

(b) Counterclockwise rotation

Left side:


Page 77 of 97

[ ]∑ = 0LBM

0=−+LLLL BNBFLL MMwQ

0185,11584916

59 =−+

LQ

lbQL 573=

( ) lbinTL

OF −= 5849

Right Side:

[ ]∑ = 0

RBM

0=−−RRRR BNBFRR MMwQ

0520,12654716

58 =−−

RQ

lbQR 2294=

( ) lbinTR

OF −= 6547


Page 78 of 97

( ) ( ) ( ) lbinTTTLR

OFOFOF −=+=+= 396,1258496547

BAND BRAKES

915. The steel band for the brake shown is lined with flexible asbestos and it is

expected tha the permissible pressure of Table AT 29 is satisfactory; o245=θ ,

ina 20= ., inm2

13= ., inD 18= ., and face width inb 4= .; rotation CL. The

cast-iron wheel turns 200 rpm. Set up suitable equations, use the average f

given and compute (a) the force in each end of the band, (b) the brake torque and

fhp. (c) Determine the mechanical advantage for the limit values of f in Table

AT 29 and its percentage variation fron that for the average f . (d) Investigate

the overheating problem using relevant information given in the Text.

Problem 915.

Solution:

(1) θfe

F

F=

2

1

[ ]∑ = 0intpoFixedM

mFWa 2=

(2) m

WaF =2


Page 79 of 97

(3) fpAFFF =−= 21

(4) 2

FDT f =

From Table AT 29, flexible asbestos

Ave. 40.0=f , psip 50=

(a) For 1F and 2F :

2

DbA

θ=

rad276.4245 == oθ

( )( )( ) 21542

418276.4inA ==

( )( )( ) lbfpAFFF 30801545040.021 ===−=

( )( ) 5312.5276.440.0

2

1 === eeF

F fθ

21 5312.5 FF =

30805312.5 22 =−= FFF

lbF 6802 =

( ) lbF 37606805312.51 ==

(b) fT and fhp

( )( )lbin

FDT f −=== 720,27

2

183080

2

( )( )hp

nTfhp

f88

000,63

200720,27

000,63===

(c) For MA

mF

T

Wa

TMA

ff

2

==

2

FDT f =

12 −

= θfe

FF

( )m

eD

e

Fm

FD

MAf

f

2

1

1

2 −=

−

=θ

θ

inD 18=

inm 5.3=

rad276.4=θ


Page 80 of 97

Limit values (Table AT 29) 35.0=f to 45.0 .

35.0=f ( )( )[ ]

( )914.8

5.32

118 276.435.0

=−

=e

MA

45.0=f ( )( )[ ]

( )042.15

5.32

118 276.445.0

=−

=e

MA

with 40.0=f (average) ( )( )[ ]

( )652.11

5.32

118 276.440.0

=−

=e

MA

Percentage variation from 40.0=f .

35.0=f

( ) %5.23%100652.11

914.8652.11var% =

−=

45.0=f

( ) %1.29%100652.11

652.11042.15var% =

−=

(d) Overheating problem

257.0154

88infhp

A

fhp==

Therefore, a problem of overheating is expected as Rasmussen recommends 0.2 to 0.3

fhp per square inch of brake contact area.

916. (a) For the band brake shown, derive the expressions for the braking torque in

terms of W , etc., for CL rotation and for CC rotation, and specify the ratio bc

for equal effectiveness in both directions of rotation. Are there any proportions of

b and c as shown that would result in the brae being self locking? (b) When o270=θ , ina 16= ., incb 3== ., and inD 12= ., it was found that a

force lbW 50= . Produced a frictional torque of 1000 in-lb. Compute the

coefficient of friction.


Page 81 of 97

Problem 916.

Solution:

(a)

CL:

[ ]∑ = 0OM

cFbFaW 21 += θfeFF 21 =

cFbeFaW f

22 += θ

cbe

aWF

f += θ2

cbe

aWeeFF

f

ff

+== θ

θθ

21

( )cbe

eaW

cbe

aWaWeFFF

f

f

f

f

+

−=

+

−=−= θ

θ

θ

θ 121


Page 82 of 97

+

−==

cbe

eWaDFDT

f

f

f θ

θ 1

22

CC:

[ ]∑ = 0OM

cFbFaW 12 +=

+

−=

bce

eWaDT

f

f

f θ

θ 1

2

No proportions of b and c as shown that would result in the brake being self-locking.

(b) lbW 50=

lbinT f −=1000

inD 12=

ina 16=

incb 3==

rad7124.4270 == oθ

( )( )( )

+

−==

33

1

2

1216501000 θ

θ

f

f

fe

eT

625.01

1=

+

−θ

θ

f

f

e

e

333.47124.4 == ffee

θ

311.0=f

917. (a) For the brake shown, assume the proper direction of rotation of the cast-iron

wheel for differential acion and derive expressions for the braking torque. (b) Let

inD 14= ., inn4

31= ., inm 4= ., o235=θ , and assume the band to be lined with

woven asbestos. Is there a chance that this brake will be self-acting? If true, will


Page 83 of 97

it always be for the range of values of f given in Table AT 29? (c) The ratio

mn should exceed what value in order for the brake to be self-locking? (d) If the

direction of rotation of the wheel is opposite to that taken in (a), what is the

braking torque with a force lbW 10= . at ina 8= .? (e) Suppose the brake is

used as a stop to prevent reverse motion on a hoist. What is the frictional

horsepower for the forward motion if the wheel turns 63 rpm?

Problems 917, 918.

Solution:

(a) Assume CL

[ ]∑ = 0OM

mFnFWa 21 =+ θfeFF 21 =

( )θθ ff nemFneFmFWa −=−= 222

( )θfnem

WaF

−=2


Page 84 of 97

( )θ

θ

f

f

nem

WaeF

−=1

( )θ

θ

f

f

nem

eWaFFF

−

−=−=

121 , Braking force.

−

−== θ

θ

f

f

fnem

eWaDFDT

1

22, Braking torque.

(b) inD 14=

inn4

31=

inm 4=

rad10.4235 == oθ

Table AT 29, woven asbestos

35.0=f to 45.0

There is a chance of self-acting if

mnef >θ

inm 4=

use 40.0=f ( )( )

menef >== 0.975.1 10.440.0θ

use 35.0=f ( )( )

menef >== 35.775.1 10.435.0θ

use 45.0=f ( )( )

menef >== 07.1175.1 10.445.0θ

Therefore true for the range of values of f .

(c) mnef >θ , 40.0=f (average)

θfem

n 1>

( )( )10.44.0

1

em

n>

2.0>m

n

(d) For CC:


Page 85 of 97

[ ]∑ = 0OM

mFnFWa 12 =+

nFmFWa 21 −= θfeFF 21 =

( )nmeFnFmeFWa ff −=−= θθ222

nme

WaF

f −= θ2

nme

WaeF

f

f

−= θ

θ

1

( )nme

eWaFFF

f

f

−

−=−= θ

θ 121

−

−==

nme

eWaDFDT

f

f

f θ

θ 1

22

( )( )( ) ( )( )

( )( ) lbine

eT f −=

−

−= 3.123

75.14

1

2

1481010.440.0

10.440.0

(e) ( )( )

hpnT

fhpf

1233.0000,63

633.123

000,63===

918. A differential band brake similar to that shown and lined with woven asbestos,

has the dimensions: inD 18= ., inn 2= ., inm 12= ., o195=θ . (a) Is there a

chance that this brake will be self-acting? (b) If lbW 30= . and ina 26= . ,

compute the maximum braking torque and the corresponding mechanical

advantage. (c) What is the ratio of the braking torque for CL rotation to the

braking torque for CC rotation? (d) A 1/16-in.-thick steel band, SAE 1020 as

rolled, carries the asbestos lining. What should be its width for a factor of safety

of 8, based on the ultimate stress? What should be the face width if the average

pressure is 50 psi?

Solution:


Page 86 of 97

(a) For CL:

mnef >θ

rad4.3195 == oθ

inm 12=

inn 2=

4.0=f

( )

me <= 8.72 4.34.0 , not self-acting

For CC: θf

men >

nmef <θ

( )ne >= 8.4612 4.34.0 , not self-acting

Therefore, there is no change that this brake will be self-acting.

(b) ff TT =max

(CL)

( )( )( ) ( )

( ) lbine

e

nem

eWadT

f

f

f −=

−

−

=

−

−

= 4832

212

1

2

1826301

2 4.34.0

4.34.0

θ

θ

( )( )2.6

2630

4832===

Wa

TMA

f

(c) ( ) lbinCLT f −= 4832

( ) ( )( )( ) ( )

( ) lbine

e

nme

eWadCCT

f

f

f −=

−

−

=

−

−

= 454

212

1

2

1826301

2 4.34.0

4.34.0

θ

θ

( )( )

64.10454

4832===

CCT

CLTRatio

f

f

(d) For SAE 1020, as rolled.

ksisu 65=

psiksiN

ss u 8125125.8

8

65====

bt

Fs 1=

inint 0625.016

1==

max. θ

θ

f

f

nem

WaeF

−=1 (CL)

( )( ) ( )

( ) lbe

eF 3.722

212

26304.34.0

4.34.0

1 =−

=


Page 87 of 97

( )0625.0

3.7228125

bs ==

inb 422.1=

With psip 50=

fpAF =

2

bDA

θ=

max. ( ) ( )( ) ( )( )

( ) lbe

e

nem

eWaF

f

f

9.536212

1263014.34.0

4.34.0

=−

−=

−

−= θ

θ

2

bDfpF

θ=

( )( )( )( )( )2

184.3504.09.536

b=

inb 88.0=

919. A differential band brake is to be design to absorb 10 fhp at 250 rpm. (a)

Compute the maximum and minimum diameters from both equations (z) and (a),

p. 495, Text. Decide on a size. (b) The band is to be lined with woven asbestos.

The Rasmussen recommendation (§18.4) will help in deciding on the face width.

Also check the permissible pressure in Table AT 29. Choose dimensions of the

lever, its location and shape and the corresponding θ . Be sure the brake is not

self locking. What is the percentage variation of the mechanical advantage from

the minimum value ( minf ) for the f limits in Table AT 29?

Solution:

( )lbin

n

hpT f −=== 2520

250

10000,63000,63

(a) Eq. (z)

inT

Df

96.75

2520

5

3

1

3

1

min =

=

=

inT

Df

57.84

2520

4

3

1

3

1

max =

=

=

Eq. (a)

( ) ( )[ ] infhpD 44.8106060 3

1

3

1

min ===

( ) ( )[ ] infhpD 28.9108080 3

1

3

1

min ===

use inD 5.8=

(b) By Rasmussen

Energy absorption capacity = 0.2 to 0.3 fhp per sq. in. of brake contact area.


Page 88 of 97

Say 0.25 fhp per sq. in.

2

bDA

θ=

hpfhp 10=

inD 5.8=

assume radπθ == o180

A

fhpinfhp =2

( )

=

2

5.8

1025.0

bπ

inb 3=

From Table AT 29, 40.0=f , psipper 50. =

fA

Fp =

( )( ) 2402

5.83inA ==

π

( )lb

D

TF

f593

5.8

252022===

( )psipsip 501.37

404.0

593<== (OK)

For MA :

( )( )θ

θ

f

ff

bec

eD

WA

TMA

−

−==

2

1

Not self-locking θf

bec >

θfe

b

c>

π4.0e

b

c>

5.3>b

c

say 4=b

c or bc 4=

For 40.0=f

( )( )

( )( ) bbeb

e

bec

eD

WA

TMA

f

ff 96.21

42

15.8

2

14.0

4.0

=−

−=

−

−== π

π

θ

θ

For min35.0 ff ==


Page 89 of 97

( )( )

( )( ) bbeb

e

bec

eD

WA

TMA

f

ff 54.8

42

15.8

2

135.0

35.0

=−

−=

−

−== π

π

θ

θ

( ) %157%10054.8

54.896.21% =

−=iationvar

DISK CLUTCHES

920. An automobile engine develops its maximum brake torque at 2800 rpm when the

bhp = 200. A design value of 25.0=f is expected to be reasonable for the

asbestos facing and it is desired that the mean diameter not exceed 8.5 in.;

permissible pressure is 35 psi. Designing for a single plate clutch, Fig. 18.10,

Text, determine the outer and inner diameters of the disk.

Solution:

( ) inDDD iom 5.82

1=+=

inrr io 5.8=+

( )lbin

n

hpT f −=== 4500

2800

200000,63000,63

psip 35=

( )2

iof

rrNfT

+=

( )( )( )2

5.825.04500

N=

lbN 4235=

ave. ( )22

io rr

Np

−=

π

( )22

423535

io rr −=

π

5.3822 =− io rr

io rr −= 5.8

( ) 5.385.8 22 =−− ii rr

5.381725.72 22 =−+− iii rrr

inri 985.1=

say inri 0.2=

inro 5.60.25.8 =−=

( ) inrD oo 135.622 ===

( ) inrD ii 40.222 ===


Page 90 of 97

921. An automobile engine can develop a maximum brake torque of 2448 in-lb.

Which of the following plate clutches, which make up a manufacturer’s standard

“line,” should be chosen for this car? Facing sizes: (a) 8

78=oD , inDi

8

16= ., (b)

10=oD , inDi8

16= ., (c)

16

111=oD , inDi

8

16= . In each case, assume 3.0=f .

The unit pressures are (a) 34 psi, (b) 30 psi, and (c) 26.2 psi.

Solution:

( )2

iof

rrNfT

+=

( )4

iof

DDNfT

+=

( )22

4ioave DDpN −

=

π

( )( )16

22

ioioavef

DDDDpT

+−=

π

(a) inDo 875.8= ; inDi 125.6= , psip 34= , 3.0=f

( )( )16

22

ioioavef

DDDDpT

+−=

π

( )( ) ( ) ( )[ ]( )lbinT f −=

+−= 1239

16

125.6875.8125.6875.8343.022π

(b) inDo 10= ; inDi 125.6= , psip 30= , 3.0=f

( )( )16

22

ioioavef

DDDDpT

+−=

π

( )( ) ( ) ( )[ ]( )lbinT f −=

+−= 1780

16

125.610125.610303.022π

(c) inDo 0625.11= ; inDi 125.6= , psip 2.26= , 3.0=f

( )( )16

22

ioioavef

DDDDpT

+−=

π

( )( ) ( ) ( )[ ]( )lbinT f −=

+−= 2251

16

125.60625.11125.60625.112.263.022π

use (c)


Page 91 of 97

922. A single-disk clutch for an industrial application, similar to that in Fig. 18.11,

Text, except that there are two disks attached to one shaft and one attached to the

other. The clutch is rated at 50 hp at 500 rpm. The asbestos-in-resin-binder facing

has a inDo2

18= . and inDi

4

34= . What must be the axial force and average

pressure? How does this pressure compare with that recommended by Table AT

29?

Solution:

2=n pairs in contact

35.0=f (Table AT 29)

psip 75=

( )lbin

n

hpT

m

f −=== 6300500

50000,63000,63

inDo 5.8=

inDi 75.4=

( )2

iof

rrNnfT

+=

( )4

iof

DDNnfT

+=

( )( )( )( )4

75.45.835.026300

+=

N

lbN 2717=

ave. ( )( )

( ) ( )[ ] psipsiDD

Np

io

756.6975.45.8

2717442222

<=−

=−

=ππ

923. A multiple-disk clutch similar to Fig. 18.11, Text, is rated at 22 hp at 100 rpm.

The outside and inside diameters of the disks are 14 and 7 ½ in., respectively. If

25.0=f , find (a) the axial force required to transmit the rated load, and (b) the

unit pressure between the disks.

Solution:

(a) Fig. 18-11, 4=n pairs in contact

( )lbin

n

hpT

m

f −=== 860,13100

22000,63000,63

( )4

iof

DDNnfT

+=

inDo 14=

inDi 5.7=

( )( )( )4

5.71425.04860,13

+=

N


Page 92 of 97

lbN 2579=

(b) ( )( )

( ) ( )[ ] psiDD

Np

io

5.235.714

2579442222

=−

=−

=ππ

924. A multiple-disk clutch for a machine tool operation has 4 phosphor-bronze

driving disks and 5 hardened-steel driven disks. This clutch is rated at 5.8 hp at

100 rpm when operated dry. The outside and inside diameters of the disks are 5

½ and 4 3/16 in., respectively. (a) If the pressure between the disks is that

recommended for metal on metal in Table AT 29, what coefficient of friction is

required to transmit the rated power? (b) What power may be transmitted for f

and p as recommended in Table AT 29?

Solution:

inDo 5.5=

inDi 1875.4=

( )lbin

n

hpT

m

f −=== 3654100

8.5000,63000,63


(a) Table AT 29, psip 150= , metal to metal

( )4

iof

DDNnfT

+=

( ) ( ) ( )[ ] lbDDpN io 14981875.45.54

1504

2222 =−

=−

=

ππ

( )( )( )( )4

1875.45.5149883654

+==

fT f

126.0=f

(b) from Table AT 29, psip 150= , 2.0=f

( )( )( )( )lbinT f −=

+= 5805

4

1875.45.514982.08

( )( )hp

nThp

mf2.9

000,63

1005805

000,63===

925. A multiple-disk clutch with three disks on one shaft and two on the other, similar

to that in Fig. 18.11, Text, is rated at 53 hp at 500 rpm. (a) What is the largest

value of iD if f and p are given by Table AT 29 for asbestos in resin binder

and inDo 5.10= . (b) For the diameter used of inDi 7= .,what is the required

axial force and the average pressure?

Solution:


Page 93 of 97

Table AT 29, asbestos in resin binder, 3.0=f , psip 75=

( )2

iof

rrNnfT

+=

( )4

iof

DDNnfT

+=

( )io

f

DDnf

TN

+=

4

but

( )22

4io DDpN −

=

π

( )( )ioiof DDDDnfpT +−

= 22

44

π

( )lbin

n

hpT

m

f −=== 6678500

53000,63000,63


( )( )ioiof DDDDnfpT +−

= 22

44

π

( ) ( )( )( ) ( )[ ]( )ii DD +−

= 5.105.10

4753.0466784 22π

inDi 5607.9=

(b) inDi 7=

( )4

iof

DDNnfT

+=

( )( )( )( )4

75.103.046678

+=

N

lbN 1272=

ave. ( )( )

( ) ( )[ ] psiDD

Np

io

44.2675.10

1272442222

=−

=−

=ππ

MISCELLANEOUS CLUTCHES AND BRAKES

926. For the cone brake shown, find an expression for the braking torque for a given

applied force W on the bell crank. Consider the force F ′ , Fig. 18.12, Text, in

obtaining the expression.


Page 94 of 97

Problems 926-928.

Solution:

927. For the cone brake similar to that shown, certain dimensions are: inDm 15= .,

inc2

12= ., o12=α , inb 9= ., and ina 20= . The contact surfaces are metal and

asbestos. (a) For an applied force lbW 80= ., what braking torque may be

expected of this brake? Consider the resistance F ′ , Fig. 18.12, Text. (b) If the

rotating shaft comes to rest from 300 rpm during 100 revolutions, what frictional

work has been done? (c) What must be the diameter of the steel pin P , SAE

1020 as rolled, for a factor fo safety of 6 against being sheard off? The diameter

of the hub ind2

14= . (d) What is the unit pressure on the face of the brake?

Solution:

(a) ( ) ( )αααα cossin2cossin2 fb

aWDf

f

RDfT mm

f +=

+=

Table AT 29, asbestos on metal, 40.0=f

( )( )( )( )( )( )

lbinT f −=+

= 89012cos4.012sin92

80201540.0

(b) ( )

sec42.3160

30021 rad==

πω

sec02 rad=ω

( ) rad3.6282100 == πθ

( )t212

1ωωθ +=

( )t042.312

13.628 +=


Page 95 of 97

sec40=t

( ) rpmnm 15003002

1=+=

( )( )hp

nTfhp

mf119.2

000,63

150890

000,63===

( )( ) ( )( ) lbfttfhpU f −=== 618,4640119.2550550

(c) For SAE 1020, as rolled,

ksissu 49=

ksiN

ss su

s 17.86

49===

2

4

d

Rss π

=

( )( )lb

b

aWR 8.177

9

8020===

( )2

8.17748170

dss π

==

ind 1665.0=

say ind16

3=

(d) cD

Np

mπ=

lbf

RN 297

12cos4.012sin

8.177

cossin=

+=

+=

αα

( )( )psip 52.2

5.215

297==

π

928. A cone clutch for industrial use is to transmit 15 hp at 400 rpm. The mean

diameter of the clutch is 10 in. and the face angle o10=α ; let 3.0=f for the

cast-iron cup and the asbestos lined cone; permissible psip 35= . Compute (a)

the needed axial force, (b) the face width, (c) the minimum axial force to achieve

engagement under load.

Solution:

( )lbin

n

hpT f −=== 5.2362

400

15000,63000,63

(a) ( )αα cossin2 f

RDfT m

f +=


Page 96 of 97

( )( )( )10cos3.010sin2

103.05.2362

+=

R

lbR 739=

(b) cD

Np

mπ=

lbf

RN 1575

10cos3.010sin

739

cossin=

+=

+=

αα

( )c15

157535

π=

inc 44.1=

(c) max. 4.0=f (Table AT 29)

( )αα cossin2 f

RDfT m

f +=

( )( )( )10cos4.010sin2

104.05.2362

+=

R

lbR 670= , minimum.

929. An “Airflex” clutch, Fig. 18.15, Text, has a 16-in drum with a 5-in. face. This

clutch is rated at 110 hp at 100 rpm with an air pressure of 75 psi. What must be

the coefficient of friction if the effect of centrifugal force is neglected? (Data

courtesy of Federal Fawick Corporation.)

Solution:

inD 16=

inb 5=

hphp 110=

rpmrpm 100=

psip 75=

( )lbin

n

hpT f −=== 300,69

100

110000,63000,63

2

FDT f =

( )2

16300,69

F=

lbF 5.8662=

( ) ( )( )( )( ) lbDbpN 850,1851675 === ππ

46.0850,18

5.8662===

N

Ff


Page 97 of 97

930. The same as 929 except that the diameter is 6 in., the face width is 2 in., and the

rated horsepower is 3.

Solution:

( )lbin

n

hpT f −=== 1890

100

3000,63000,63

2

FDT f =

( )2

6300,69

F=

lbF 630=

( ) ( )( )( )( ) lbDbpN 28272675 === ππ

22.02827

630===

N

Ff

- end -

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Section 16