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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=
SECTION 16 – BRAKES AND CLUTCHES
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 =
SECTION 16 – BRAKES AND CLUTCHES
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===
∆−=
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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 =
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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
==
SECTION 16 – BRAKES AND CLUTCHES
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)
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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 =
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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 =
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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=
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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=
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
1sinsin θθθθ crfPbrM HF
∫−=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 ==
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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=
φφdfPbrfdNdF sin==
11 90 θφ −=
22 90 θφ +=
∫= rdFT f
SECTION 16 – BRAKES AND CLUTCHES
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
1sinsin θθθθ crfPbrM HF
( ) ( ) ( )[ ]
−−−−+= 1
2
2
2
21 sin1sin12
1sinsin θθθθ crfPbrM HF
( ) ( )
−++= 1
2
2
2
21 sinsin2
1sinsin θθθθ crfPbrM HF
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
SECTION 16 – BRAKES AND CLUTCHES
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
1sinsin θθθθ crfPbrM HF
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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 ===
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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)
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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 =
−=
+===
SECTION 16 – BRAKES AND CLUTCHES
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 =
SECTION 16 – BRAKES AND CLUTCHES
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
φφφφ −−−=brRP
M HN
2sin2 2
1
θfPbrT f =
2sin2
1
θfr
TPbr
f=
inr 10=
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
=−+−′HNHFVH MMdRaR
( )( ) ( )( )[ ]( ) 0960.2705.125.65.2133.1722
=−+− ff TT
lbinT f −= 405,102
CL:
022
=−−−′HNHFVH MMdRaR
( )( ) ( )( )[ ]( ) 0960.2705.125.65.2133.1722
=−−− ff TT
lbinT f −= 51502
(a) Braking Torque 21 ff TT +=
SECTION 16 – BRAKES AND CLUTCHES
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 =
SECTION 16 – BRAKES AND CLUTCHES
Page 54 of 97
WS 33.17=
ina 75.23=
( )( ) 096.275.2333.1711
=−− ff TTW
WT f 9.1031
=
022
=−+−′HNHFVH MMdRaR
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
=−−−′HNHFVH MMdRaR
( )( ) ( )( )[ ]( ) 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=
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
Page 56 of 97
( ) ( )
−−−= 1
2
2
2
21 sinsin2
coscos φφφφR
rfbrPM HF
( ) ( )[ ]1212 2sin2sin24
φφφφ −−−=brRP
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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===
SECTION 16 – BRAKES AND CLUTCHES
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
=
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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=
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
Page 66 of 97
( )
−−=
2
2sin2sin
2
12 φφθ
PabrM BN
( ) edFMd BF =
φφdfPbrfdNdF sin==
( ) 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
SECTION 16 – BRAKES AND CLUTCHES
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===
SECTION 16 – BRAKES AND CLUTCHES
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
2 coscos φφ −= fPbrT OF
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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
2 coscos φφ −= fPbrT OF
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:
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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:
SECTION 16 – BRAKES AND CLUTCHES
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 =−
=
SECTION 16 – BRAKES AND CLUTCHES
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.
SECTION 16 – BRAKES AND CLUTCHES
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 ==
SECTION 16 – BRAKES AND CLUTCHES
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 ===
SECTION 16 – BRAKES AND CLUTCHES
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)
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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
8=n pairs in contact
(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:
SECTION 16 – BRAKES AND CLUTCHES
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
4=n pairs in contact
( )( )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.
SECTION 16 – BRAKES AND CLUTCHES
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 +=
SECTION 16 – BRAKES AND CLUTCHES
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 +=
SECTION 16 – BRAKES AND CLUTCHES
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
SECTION 16 – BRAKES AND CLUTCHES
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 -