mod 3 Clutch h.pdf

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1. MODULE 3.1
CLUTCH DESIGN

2. FRICTION CLUTCHES

3. MATERIAL FOR FRICTION CLUTCHES

8. STEP 1

Consider an elementary ring of radius r and thickness dr as shown in Fig.
We know that area of the contact surface or friction surface
=Mr.dr
. Normal or axial force on the ring,
SI = Pressure x Area =p x 2 の

the frictional force on the ring acting tangentially at radius r,
F =x OW =p * inr.dr
Frictional torque acting on the ring,
T,=F,xr=upx Mr.drxr=2nup.r.dr

17. STEP 1
Maximum pressure
Let Paro; = Maximum pressure.
Since the intensity of pressure is maximum at the inner radius (r,), therefore
Poy 1) =€ OF C=30p u
We also know that total force on the contact surface (17),
4000 = 2nC (r, -1,)= 2m x50p, (100-50)=15 710p
Pym = 4000 / 15 710=02546Nmm Ans.

18. STEP 2
Minimum pressure
Let Pmin = Minimum pressure.
Since the intensity of pressure is minimum at the outer radius (7, ), therefore,
Pin "= Or C=100p .

We know that the total force on the contact surface (1),
4000 = 2nC (r, -r,)=21 x 100 p,,,, (100 50)=31 420 p,,,,
Prin = 4000/31 420=0.1273 Nimm? Ans.

19. STEP 3

Average pressure
We know that average pressure,

Total normal force on contact surface _ W

Pov ~ Cross-sectional area of contact surface — 지어 Y (5 *]

4000

= = 0.17 Nimm? Ans.
© x{(100)? - (50)]

21. STEP 1

(a) Inner diameter of the friction surfaces
Let d, = Inner diameter of the contact or friction surfaces, and
7, = Inner radius of the contact or friction surfaces.
We know that the torque transmitted by the clutch,
Px60 110x10° x 60 _
T= Inn 21x1250
= 840 x 10° N-mm

840 N-m

22. STEP 2
Axial thrust with which the contact surfaces are held together,
W = Pressure x Area =p XT [(r,P = (1
= 0.17 « [(150) - (r,)"] = 0.534 [(150) - (r.)]
and mean radius of the contact surface for uniform pressure conditions,
2 A = 2 m - ar
alar] 3 Laso (Y

23. STEP 3 7 _
Torque transmitted by the clutch (7),
840 x 107 = nu. WR

silos n= 2)

3 3
= 2X 0.4x 0.534 [(150)° - (5) ]x E e

3| (150) - (m)?

SÍ

= 0.285 [(150) — (r,)]
(150) — (r,)? = 840 x 10%/ 0.285 = 2.95 x 10°
(1)? = (150) —2.95 x 10°= 0.425 x 10° or 7,=75mm
d, =2r,=2%75=150mm Ans.

24. STEP 4
(b) Maximum torque transmitted
We know that the axial thrust,
W = 0.534 [(150) - (r,)] |
= 0.534 [(150)? - (75)] = 9011 N
and mean radius ofthe contact surfaces for uniform wear conditions,

150 +75

+
pa oth ーー =1125 mm

2
:. Maximum torque transmitted,
T=n4WR=2x0.4x9011 x 112.5 = 811 x 10 N-mm

= 811 N-m Ans.

25. STEP 5
Maximum intensity of pressure

Foruniform wear conditions, p.r=C (a constant). Since the intensity of pressure is maximum at
the inner radius (7), therefore
5=C ot C=p, 5 Nimm
We know that the axial thrust (17),
DI =25.C (1, -r))=28 xp, «15 (180-75) =35 347
Pira: = 011/35 347 = 0.255 Nmm An.

27. STEP 1

Let r, = Outside radius of the contact surface.
We know that the torque transmitted,
y PX60 _ 25X 10% x 60
ınN 27X1575

= 151.6 Nm = 151 600 N-mm

28. STEP 2
For uniform wear, we know thatp.r = C. Since the intensity of pressure is maximum at the inner
radius (r,), therefore,
Pug =F or C=01*60=6Nmm
We know that the axial force on each friction surface,
W = DNC (1, 1) = 20 * 6(r, - 60) = 37.7 (r, - 60) 0
For uniform wear, mean radius of the contact surface,
R= qth yt 60

= 057,430
2

29. STEP 3 _ _
We know that number of pairs of contact surfaces,
n=n+m-1=3+2-1=4
+. Torque transmitted (7),
151 600 = n.u.W.R=4 x 0.3 x 37.7 (r, - 60) (0.5 r, + 30)
… [Substituting the value of W from equation 0)]
= 22.62 (1, P 81 432

„_ 151.600 + 81432
サニー ne

r = 101.5 mm Ans.

= 10 302

31. Given: 1, =3:1,=2:n=4: d, = 240 mm or r, = 120 mm: d, = 120 mm or
7, = 60 mm; t= 0.3 ; P= 25 kW =25 x 10° W: N= 1575 rpm.

STEP 1

Total spring load
Let W =Total spring load.
We know that the torque transmitted,

fe PX 60 _25x10° x 60

DEN 2nx1575
= 151.5 x 10 N-mm

= 151.5 N-m

32. STEP 2

Mean radius of the contact surface, for uniform pressure,

2 E ‘tt 2 ue osm

3-7 31020) - 60)
and torque transmitted (T ),
1515 x 10 =n.MR=4% 03x Wx 933= 1127
W=1515 x 10/ 112= 1353 N Ans.

33. STEP 3
Maximum power transmitted
Given: No. of springs = 6
+. Contact surfaces of the spring = 8
Wear on each contact surface = 1.25 mm
Total wear = 8 x 125=10mm=0.01 m
Stiffness of each spring = 13 KN/m = 13 x 10% Nm
시 Reduction in spring force
= Total wear x Stiffness per spring x No. of springs
= 0.01 x 13 x 10 x 6=780N

34. STEP 4
W = 1353-780 = 573 N

new axial load,
We know that mean radius of > contact surfaces for uniform wear,
+r, 120+60
=- =90 mm = 0.09 m

torque transmitted, T=n. WW. R=4 x03 x 573 x 0.09 = 62 N-m

시 Power transmitted, = P= y = Je — =10 227 W =10.227 KW Ans.

38. p, = Intensity of pressure with which the conical friction surfaces are held STEP 1
together (i.e. normal pressure between the contact surfaces),

1, = Outer radius of friction surface,
r, = Inner radius of friction surface,

. . . n+r
R = Mean radius of friction surface = + 5 제

Friction surface

0. = Semi-angle of the cone (also called face angle of the cone) or angle of ：
the friction surface with the axis of the clutch,

it = Coefficient of friction between the contact surfaces, and
b = Width of the friction surfaces (also known as face width or cone face).

39. STEP 2

Friction surface
Consider a small ring of radius r and thickness dr as shown in Fig. 24.7. Let dl is the length of
ring of the friction surface, such that,
dl = dr cosec 0.
Area of ring = 2m r. dl = 2m r.dr cosec 0. dl
We shall now consider the following two cases : dr
1. When there is a uniform pressure, and
2. When there is a uniform wear.

40. STEP 3
1. Considering uniform pressure
We know that the normal force acting on the ring,
dW, = Normal pressure * Area of ring = p, * 2 r.dr cosec 0
and the axial force acting on the ring,
5 = Horizontal component of 67, (i.e. in the direction of W)
=6 x sin =p, x 2T r.dr cosec x sin = 21 x p,.rdr
+. Total axial load transmitted to the clutch or the axial spring force required.

27 2 2
1 r 9 の)
W = [2rxp,rdr=2r p, 回 =2np, E 7 2 |

= mp, (02-09 |

41. STEP 4 u W
Palin? -0r]
We know that frictional force on the ring acting tangentially at radius r,
F, = 1.91, = up, x 2Rr.dr cosec 0.
+. Frictional torque acting on the ring,
T, = Fx r= |p, * 2Ttr.dr cosec メア
= 21 Lp, cosec 0.1? dr

Integrating this expression within the limits from r, to r, for the total frictional torque on the
clutch.

이 Total frictional torque.
ñ 2
T= | 2TL.p,. cosec our? dr=2n Hp, cosec o | —
5
2 n

ay - |

= 20 H.p, cosec 0 | =

42. STEP 4 Continued

Substituting the value of p, from equation (7), we get

T= 2mux 5

rx) -05) ]

00000 |
(my

(i) al |
3

x cosec 0 |

2
= 더 x LW cosec 0 [eae

43. 2. Considering uniform wear

In Fig. .letp, be the normal intensity of pressure at a distance r from the axis of the clutch.
We know that, in case of uniform wear, the intensity of pressure varies inversely with the distance.

a p,r = C(aconstant) or p,=C/r
We know that the normal force acting on the ring.
91, = Normal pressure x Area of ring =p, * 2Tr.dr cosec &
and the axial force acting on the ring,
SW = ST, x sin =p, * 2T r.dr cosec OL ~ 50 0
= 20x p, 1 dr

= 2 X 区 Xr. dr =2n Car mm E)
E

+. Total axial load transmitted to the clutch,

W= [2m car = 2 Cri = 2n € (x =n)

44. STEP 6
. Total axial load transmitted to the clutch,
w= ["2nCadr=2nC [r]i=2n€ (1 - 19)

W
~ 2001 1)
We know that frictional force on the ring acting tangentially at radius r,
F, = LOW, =U.p, x 2T r.dr cosec 0

ど

+. Frictional torque acting on the ring.
T, =F, x r= (Up, x 2T r.dr cosec 0 x r

E
ニル メー x 2 rdr cosec ax r = 27 UC cosec 0 x r dr
の

45. STEP 7

Integrating this expression within the limits from r, to r, for the total frictional torque on the

clutch.
+. Total frictional torque,
2 i
T= j “2m ILC cosec LX 1 dr = 2nu.C cosec 0 回
4 rn
(Y - |
2
Substituting the value of C from equation (iii), we have

2 2
T = 2 재 0 따 쁘 - 쁘 |
2001 -75) 2

= 2m u.Ccosec 0 |

= WW cosec 0 proa] = WR cosec 0

th 。 a
where R= re = Mean radius of friction surface.

(iv)

46. STEP 7 Continued

Total frictional torque. T = u WR cosec 0

Since the normal force acting on the friction surface, JP, = )) cosec 0, therefore the equation
may be written as

T=uUWR

47. Special case: The forces on a friction surface, for steady operation of the clutch and after the clutch is
engaged, is shown in Fig. 24.8 (a) and (b) respectively.

Friction surface

TN HW,
Friction v

surface

Le,

トーーー ——
W= W, sina
(b) During engagement of the clutch.

Fig. 24.8. Forces on a friction surface.

48. Special case:

From Fig. 24.8 (a), we find that
ith
MEN
+. From equation (7). normal pressure acting on the friction surface.

カー ち ニカ sm and R or r,+r,=2R

W _ W _ W
Pa” ri -M)] na+n)m-n) 2nRbsina
W =p, * 27 Rb sin & = W, sm
TV, = Normal load acting on the friction surface =p, ~ 2TR.b
Now the equation (iv) may be written as
T=.U(p, * 27 R.b sin 0) R cosec 0 = 20 Wp, R?.b

50. STEP 1

Given : D= 80. mm or R= 40 mm: a = 15°: t= 0.3: = 200N: N= 900rp.m.
or = 2m x 900/60 = 94.26 rad/s : m = 14 kg : k=160 mm = 0.16 m

Torque required to produce slipping of the clutch
We know that the torque required to produce slipping of the clutch,
T = WR cosec u = 0.3 x 200 x 40 cosec 15° = 9273 N-mm
= 9.273 N-m Ans.

51. STEP 2

Time required for the flywheel to attain full-speed
Let t = Time required for the flywheel to attain full speed from the stationary
position, and
o = Angular acceleration of the flywheel.
We know that mass moment of inertia of the flywheel,
T= m.R = 14 (0.16) = 0.3584 kg-m?
We also know that the torque (7),
9.273 = 1 «= 0.3584 0.
dl = 9.273 / 0.3584 = 25.87 rad / s?
and angular speed (0),
94.26 = Wy + O.t= 0 + 25.87 x 1=25.87 1 (7 0=0)
t = 94.26 / 25.87 = 3.64 5 Ans.

52. STEP 3

Energy lost in slipping of the clutch
We know that angular displacement,

06 +0
2

0 = Average angular speed x time = xt

ーー

- Energy lost in slipping of the clutch,
= T.0=9.273 x 171.6 = 1591 N-m Ans.

54. STEP 1

. Given : P = 30 KW = 30 x 10% W: N= 750 rp.m.: a= 103% 7-02;
D=6to 10d: p, = 0.075 to 0.1 N/mm’: K, = 1.75; D/b=6
First of all, let us find the diameter of shaft (d). We know that the torque transmitted by the shaft,
Ne > ーー XKp= ne x1.75 = 668.4 Nan
= 668.4 x 10% N-mm
We also know that the torque transmitted by the shaft (7 )、

T T
6684 x 10 = 을지대 の = rex @=825d Taking t= 42 N/mm?)
の = 668.4 x 103/8.25=81 x 10? or d=43.3 say 50 mm Ans.

55. STEP 2
Now let us find the principal dimensions of a cone clutch.
Let D = Mean diameter of the clutch,
R = Mean radius of the clutch, and
b = Face width of the clutch.

Since the allowable normal pressure (p,,) for leather and cast iron is 0.075 to 0.1 N/mm’,
therefore let us take p, = 0.1 N/mm?.

We know that the torque developed by the clutch ( 7),
R
668.4 x 10? = 2 0p. p,. R2.b=27 x 0.2 x 0.1 x Rx 37 0.042 À

…(・ D/b=6 or 2R/b=6 or R/b=3)
; R = 668.4 x 107 / 0.042 = 15.9 x 10° or R=250 mm
and D =2R=2 « 250 = 500 mm Ans.

56. STEP 3

57. STEP 4
Since this calculated value of the mean diameter of the clutch (D) is equal to 10 d and the given
value of D is 6 to 10d, therefore the calculated value of D is safe.
We know that face width of the clutch,
b = D/6=500/6=83.3 mm Ans.
From Fig. 24.9, we find that outer radius of the clutch,

bo 833,1
1, = Rosin a= 250 + sin 12-9 = 259 mm Ans.

inner radius of the clutch,

r = fe te os ain 1220 = 24120
a 2 2 2

67. STEP 1
1. Mass of the shoes
Let m = Mass of the shoes.
We know that the angular running speed,
y= ZEN _ 20900
60 60

Since the speed at which the engagement begins is 3/4 th of the running speed, therefore angular
speed at which engagement begins is

= 94.26 rad/s

0, = 3 @ = 2 x 9426 = 707 rad/s
1 4 4

Assuming that the centre of gravity of the shoe lies at a distance of 120 mm (30 mm less than R)
from the centre of the spider, i.e.

r=120mm=0.12m

68. STEP 2
r=120mm=0.12m
We know that the centrifugal force acting on each shoe,
P, = m.o.r=m (94.26) 0.12 = 1066 mN
and the inward force on each shoe exerted by the spring i.e, the centrifugal force at the engagement
speed, (0).
P, = m(@,)°r=m (70.7) 0.12 = 600 mN
We know that the torque transmitted at the running speed,

_ Px 60 _ 15x10 x60

一 一 = =159 Nm
2nN 2XX900

69. STEP 3

_Px0 15x10 X60 _
2140 27x90
We also know that the torque transmitted (7),
159 = 1 (P._—P)R n= 0.25 (1066 m - 600 m) 0.15 x 4= 70 m
m = 159/10 = 2.27 kg Ans.

70. STEP 4
2. Size of the shoes
Let 7 = Contact length of shoes in mm, and
b = Width of the shoes in mm.

Assuming that the arc of contact of the shoes subtend an angle of 8 = 60° or x / 3 radians, at the
centre of the spider, therefore

1= @R= "x 180=157 mm

Area of contact of the shoes
A =1D=157 bom

71. STEPS
Assuming that the intensity of pressure ( p) exerted on the shoes is 0.1 N/mm’, therefore force
with which the shoe presses against the rim
= Ap=157bx0.1=15.7 DN 20
We also know that the force with which the shoe presses against the rim
= PP = 1066 m- 600 m = 466 m
= 466 x 2.27= 1058N .. (11)
From equations (i) and (ii), we find that
b = 1058/15.7=674mm Ans.

72. a Problems for practice

A single disc clutch with both sides of the disc effective is used to transmit 10 KW power at 900 rp.m.
The axial pressure is limited to 0.085 N/mm. If the external diameter of the friction lining is 1.25 times
the internal diameter, find the required dimensions of the friction lining and the axial force exerted by the
springs. Assume uniform wear conditions. The coefficient of friction may be taken as 0.3.

[Ans. 132.5 mm ; 106 mm ; 1500 N]
Q2

A multiple disc clutch has three discs on the driving shaft and two on the driven shaft, providing four
pairs of contact surfaces. The outer diameter of the contact surfaces 1s 250 mm and the inner diameter
is 150 mm. Determine the maximum axial intensity of pressure between the discs for transmitting
18.75 KW at 500 rp.m. Assume uniform wear and coefficient of friction as 0.3.

73. Q3 Problems for practice

An engine developing 22 kW at 1000 r.p.m. is fitted with a cone clutch having mean diameter of 300
mm. The cone has a face angle of 12°. If the normal pressure on the clutch face is not to exceed 0.07
N/mm? and the coefficient of friction is 0.2, determine :

(a) the face width of the clutch, and [Ans. 106 mm ; 1796 N]

(b) the axial spring force necessary to engage the clutch.

04

A centrifugal friction clutch has a driving member consisting of a spider carrying four shoes which are
kept from contact with the clutch case by means of flat springs until increase of centrifugal force
overcomes the resistance of the springs and the power is transmitted by the friction between the shoes
and the case.

Determine the necessary mass and size of each shoe if 22.5 kW is to be transmitted at 750 r.p.m.
with engagement beginning at 75% of the running speed. The inside diameter of the drum is 300
mm and the radial distance of the centre of gravity of each shoe from the shaft axis is 125 mm. Assume
u=0.25. [Ans. 5.66 kg ; / = 157.1 mm ; b = 120 mm]