kelompok 1 ( 24.1 - 24.10 )

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PR

ELEMEN MESIN 2

= KELOMPOK 1 =

SUHENDRA SYARIF( 1010912063 )

M. ABDUL HAFIZH ( 1210912032 )

ARDI DESFIANDRA ( 1310911030 )

TRISNA MISWAR ( 13109110! )

IRFAN FAHREZA ( 13109120!3 )

"URUSAN TEKNIK MESIN

FAKULTAS TEKNIK

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UNI#ERSITAS ANDALAS

PADAN$

2016

E%&'* 2!.1

Determine the maximum, minimum and average pressure in a place clutch when the

axial force is 4 kN. The inside radius of the contact surface is 50 mm and the outside

radius is 100 mm. ssume uniform wear.

S+,-+/.

!iven " # $ 4 kN $ 4000N

r % $ 50 mm

r 1 $ 100 mm

M&%',' *,*

&et 'max $ maximum pressure

(ince the intensit) of pressure is maximum at the inner radius *r %+, therefore

'max x r % $ or $ 50 'max

#e also know that total force on the contact surface *#+

4000 $ %- *r 1r %+ $ %- 50 'max *10050+ $ 15/10 'max

'max $ 400015/10 $ 0.%54 Nmm% A/.

M/',' *,*

&et 'min $ minimum pressure.

(ince the intensit) of pressure is minimum at the outer radius *r1+, therefore,

'min x r 1 $ or $ 100 'min

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#e know that the total force on the contact surface *#+

4000 $ %-*r 1r %+ $ %- x 100 'min *10050+ $ 214%0 'min.

'min $ 4000214%0 $ 0.1%/2 Nmm%

A/.

A*&* *,*

#e know that average pressure,

'ave $Total normal forecon the contact surface

Cross−sectionarea of contact surface $

r 1

(¿)2−(r 2 )2¿

π ¿W

¿

$4000

π [ (100)2−(50)2] $ 0.1/ Nmm% A/.

E%&'* 2!.2

plate clutch having a single driving plate with contact surface on each side is

re3uired to transmit 110 k# at 1%50 rpm. The outer diameter of the contact surface is

to e 200 mm. the coefficient of friction is 0.4.

*a+ ssume a uniform pressure of 0.1/ Nmm% determine the inner diameter of

the friction surfaces.$n*+ ssuming the same dimensions and the same total axial thrust, determine the

maximum tor3ue that can e transmitted and the maximum intensit) of

pressure when uniform wear conditions have een reached.

S+,-+/.

!iven " ' $ 110 k# $ 110 x 102#

N $ 1%50 rpm

d1 $ 200 mm

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r 1 $ 150 mm

6 $ 0.4

p $ 0.1/ Nmm%

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(&) I//* 4&'*-* +5 -* 57-+/ ,5&7*

&et d% $ inner diameter of the contact or friction surfaces, and

r2 $ inner radius of the contact or friction surfaces.

#e know that the tor3ue transmitted ) the clutch,

T $ P x60

2πN $110 x 10

3 x60

2π x1250 $ 740 Nm

$ 740 x 102 Nmm

xial thust with whitch the contact surface are held together,

# $ 'ressure x rea $ p x - 8*r 1+% 9 *r %+

%:

$ 0.1/ x - 8*150+% 9 *r %+%: $ 0.524 8*150+%*r %+

%:

nd mean radius on the contact surface for uniform pressure conditions.

; $2

3 8(r 1 )3−(r 2)3

(r1 )2−(r 2)2 : $2

3 8(150 )3−(r 2 )3

(150 )2−(r 2 )2 :

Tor3ue transmitted ) the clutch *T+

740 x 102 $ n. 6.#.;

$ % x 0.4 x 0.524 8*150+% 9*r %+%: x

2

3 8(150 )3−(r 2)3

(150 )2−(r 2)2 :

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nd mean radius of the contact surface for uniform wear conditions,

; $r1+¿r 22

¿ $

150+752 $ 11%.5 mm

>aximum tor3ue transmitted,

T $ n.6.#.; $ % x 0.4 x =011 x 11%.5 $ 711 x 102 Nmm $ 711 Nm A/.

M&%',' /-*/-: +5 *,*

?or uniform wear conditioning p.r $ *a constant+. (ince the intensit) of pressure is

maximum at the inner radius * r2 +, therefore

'max $ x r % $ or $ 'max x /5 Nmm

#e know that the axial thust *#+

=000 $ %- *r 1r %+ $ %- x 'max x /5 *150/5+ $ 2524/ 'max'max$ =011 25 24/ $ 0.%55 Nmm

% A/.

E%&'* 2!.3

single plate clutch, effective on oth sides, is re3uired to transmit %5 k# at 2000

r.p.m. Determine the outer and inner diameters of frictional surface if the coefficient

of friction is 0.%55, ratio of diameters is 1.%5 and the maximum pressure is not to

exceed 0.1 Nmm%. lso, determine the axial thrust to e provided ) springs.

ssume the theor) of uniform wear.

S+,-+/

!iven " n $ %

P $ %5 k# $ %5 @ 102 #

N $ 2000 r.p.m.

A $ 0.%55

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d 1 / d 2 $ 1.%5

r 1 / r 2 $ 1.%5

pmax $ 0.1 Nmm%

O,-* &/4 //* 4&'*-* +5 57-+/& ,5&7*

&et d 1 and d 2 $ Buter and inner diameters *in mm+ of frictional surface, and

r 1 and r 2 $ orresponding radii *in mm+ of frictional surface.


T = P x60

2π N =

25 x103 x 60

2 π x 3000=79.6 N −m=79600 N −mm

?or uniform wear conditions, p.r = C *a constant+. (ince the intensit) of pressure is

maximum at the inner radius *r 2+, therefore. pmax × r 2 = C

or C $ 0.1 r 2 Nmm

and normal or axial load acting on the friction surface,

W = %π C (r 1 – r 2 ) = %π × 0.1 r 2 ( 1.%5 r 2 – r 2 )

= 0.15/ (r 2 )2 ... * r 1 / r 2 $ 1.%5+

#e know that mean radius of the frictional surface *for uniform wear+,

R=

r1+r

2

2 =

1.25r1+r

2

2 =1.125 r2

and the tor3ue transmitted *T +,

/= 00 $ n.μ.W.R $ % @ 0.%55 @ 0.15/ *r 2+% 1.1%5 r 2 $ 0.0= *r 2+

2

∴ *r 2+2 $ /=. @ 102 0.0= $ 774 @ 102 or r 2 $ = mm

and r 1 $ 1.%5 r 2 $ 1.%5 @ = $ 1%0 mm

Buter diameter of frictional surface,∴

d 1 $ %r 1 $ % @ 1%0 $ %40 mm A/.

and inner diameter of frictional surface,

d 2 $ %r 2 $ % @ = $ 1=% mm A/.

A%& -,- -+ 8* +4*4 8: /

#e know that axial thrust to e provided ) springs,

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W = %- C (r 1 – r 2 ) = %- × 0.1 r 2 ( 1.%5 r 2 – r 2 )

$ 0.15/ *r 2+% $ 0.15/ *=+% $ 144/ N A/.

E%&'* 2!.!

dr) single plate clutch is to e designed for an automotive vehicle whose engine is

rated to give 100 k# at %400 r.p.m. and maximum tor3ue 500 Nm. The outer radius

of the friction plate is %5C more than the inner radius. The intensit) of pressure

etween the plate is not to exceed 0.0/ Nmm%. The coefficient of friction ma) e

assumed e3ual to 0.2. The helical springs re3uired ) this clutch to provide axialforce necessar) to engage the clutch are eight. f each spring has stiffness e3ual to 40

Nmm, determine the dimensions of the friction plate and initial compression in the

springs.

S+,-+/.

!iven " P $ 100 k# $ 100 @ 102 #

E N $ %400 r.p.m.

T $ 500 Nm $ 500 @ 102 Nmm

p $ 0.0/ Nmm%

A $ 0.2

No. of springs $ 7

(tiffnessspring $ 40 Nmm

D'*/+/ +5 -* 57-+/ &-*

&et r 1 $ Buter radius of the friction plate, and

r 2 $ nner radius of the friction plate.

(ince the outer radius of the friction plate is %5C more than the inner radius,

therefore

r 1 $ 1.%5 r 2

?or uniform wear conditions, p.r = C *a constant+. (ince the intensit) of pressure is

maximum at

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the inner radius *r 2+, therefore

p.r 2 = C or C = 0.0/ r 2 Nmm

and axial load acting on the friction plate,

W = %- C (r 1 – r 2 ) = %- × 0.0/ r 2 ( 1.%5 r 2 – r 2 ) = 0.11 (r 2 )2 N ...()

#e know that mean radius of the friction plate, for uniform wear,

R=r1+r

2

2=1.25r

1+r

2

2=1.125 r

2

Tor3ue transmitted *∴ T +,

500 @ 102 $ n.A.W.R $ % @ 0.2 @ 0.11 *r 2+% 1.1%5 r 2 $ 0.0/4 *r 2+

2 ...* n $ %+

*r 2+2 $ 500 @ 102 0.0/4 $ /5/ @ 102 or r 2 $ 1=0 mm A/.

and r 1 $ 1.%5 r 2 $ 1.%5 @ 1=0 $ %2/.5 mm A/.

I/-& 7+'*+/ / -* /

#e know that total stiffness of the springs,

s $ (tiffness per spring @ No. of springs $ 40 @ 7 $ 2%0 Nmm

xial force re3uired to engage the clutch,

W $ 0.11 *r 2+% $ 0.11 *1=0+% $ 2=/0 N ... 8?rom e3uation ():

nitial compression in the springs∴

$ W / s $ 2=/0 2%0 $ 1%.4 mm A/.

E%&'* 2!.;

single dr) plate clutch is to e designed to transmit /.5 k# at =00 r.p.m. ?ind "

1. Diameter of the shaft,

%. >ean radius and face width of the friction lining assuming the ratio of the mean

radius to the face width as 4,

2. Buter and inner radii of the clutch plate, and

4. Dimensions of the spring, assuming that the numer of springs are and spring

index $ . The allowale shear stress for the spring wire ma) e taken as 4%0 >'a.

S+,-+/.

!iven " P $ /.5 k# $ /500 #

N $ =00 r.p.m.

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r b $ 4

No. of springs $

C $ Dd $

1τ $ 4%0 >'a $ 4%0 Nmm%

1. Diameter of the shaft

&et d S $ Diameter of the shaft, and

1τ $ (hear stress for the shaft material. t ma) e assumed as 40 Nmm

#e know that the tor3ue transmitted,

T $ N

P

π %

0×

$=00%

0/500

×

×

π $

Nm./= $

Nmm/=00 ..(i )

#e also know that the tor3ue transmitted *T +,

79!! $1

π

2

1 +* sd τ ×$

1

π

2+*40 sd ×

$

2+*755./ sd

10124755.//=00+* 2 == sd

or

.%1= sd

sa) mm A/.

2. Mean radius and face width of the friction lining

&et R $ >ean radius of the friction lining, and

b $ ?ace width of the friction lining $ R4 ... *!iven+

#e know that the area of the friction faces,

" $ Rbπ %

Normal or the axial force acting on the friction faces,

W $ " @ p $ Rbπ %

and tor3ue transmitted,

n R p Rbn RW T .+..%*.. π µ µ ==

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n p Rn R p R

R ...%

.+..4

%* 2 µ

π π µ ==

ssuming the intensit) of pressure * p+ as 0.0/ Nmm% and coefficient of friction * µ+

as 0.%5, wehave from e3uations (i ) and (ii ),

22 .055.0%0/.0%5.0%

/=00 R R =××××= π

... * n $ %, for oth sides of plate effective+

2 1045.1055.0/=00 ×== R #r sa$ R %.112= mm114A/.

nd

mm Rb 5.%741144 ===

A/.3. Outer and inner radii of the clutch plate

&et r 1 and r 2 $ Buter and inner radii of the clutch plate respectivel).

(ince the face width *or radial width+ of the plate is e3ual to the difference of the

outer and inner radii, therefore,

%1 r r b −= #r mmr r 5.%7%1 =−...(iii )

#e know that for uniform wear, mean radius of the clutch plate,

%

%1 r r R −=

#r mm Rr r %77114%%%1 =×==+. ..(iv)

?rom e3uations (iii ), and (iv), we find that

mmr %5.1%7= and mmr /5.==% =A/.

!. Dimensions of the spring

&et D $ >ean diameter of the spring, and

d $ Diameter of the spring wire.

#e know that the axial force on the friction faces,

N pb RW %.14%=0/.05.%7114%..% =×××== π π

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n order to allow for adFustment and for maximum engine tor3ue, the spring is

designed for anoverload of %5C.

Total load on the springs

N W 5.1/7%.14%=%5.1%5.1 =×==

(ince there are springs, therefore maximum load on each spring,

N W s /5.%=/5.1/7 ==

#e know that #ahlGs stress factor,

%5%5.1

15.0

44

1415.0

44

14=+

−×

−×=+

−

−=

C C

C %

#e also know that maximum shear stress induced in the wire *τ

+,

%%%

5=//5.%=/7%5%5.1

74%0

d d d

C W % s =

×××=×=

π π

5.124%05=/%

==d #r mmd 7.2=

#e shall take a standard wire of siHe SW& 7 having diameter *d + $ 4.04 mm A/.

and mean diameter of the spring,

D $ C.d $ @ 4.04 $ %4.274 sa) %4.4 mm A/.

&et us assume that the spring has 4 active turns *'.. n $ 4+. Therefore compression of

the spring,

mm&d

nC W s 02.04.41074

4/5.%=/7.72

2

=××

×××==δ

... *Taking & $ 74 @ 102 Nmm%+

ssuming s3uared and ground ends, total numer of turns,

n $ n I % $ 4 I % $

#e know that free length of the spring,

δ δ 15.0G ++= d n * +

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mm2%.2102.15.002.04.4 =×++×=A/.

and pitch of the coils

mmn

* + %4.1

2%.21

1G

=

−

=

−

=

A/.

E%&'* 2!.6

Design a single plate automoile clutch to transmit a maximum tor3ue of %50Nm at

%000 r.p.m. The outside diameter of the clutch is %50 mm and the clutch is engaged at

55 kmh.?ind " 1. the numer of revolutions of the clutch slip during engagement and

%. heat to e dissipated) the clutch for each engagement.

The following additional data is availale"

Jngine tor3ue during engagement $ 100 Nm >ass of the automoile $ 1500 kg

Diameter ofthe automoile wheel $ 0./ m >oment of inertia of comined engine

rotating parts, fl)wheel an input side of the clutch $ 1 kgm% !ear reduction ratio at

differential $ 5 Tor3ue at rear wheelsavailale for accelerating automoile $ 1/5 N

m oefficient of friction for the clutch material$ 0.2 'ermissile pressure $ 0.12

Nmm%.

S+,-+/. !iven " T $ %50 Nm $ %50 @ 10 Nmm

N $ %000 r.p.m.

d $ %50 mm or r 1$ 1%5 mm

, $ 55 km- $ 15.2 ms

T $ 100 Nm m $ 1500 kg

D $ 0./ m

R $ 0.25 m

$ 1 kgm%

Ta $ 1/5 Nm

!ear ratio $ 5

µ $ 0.2

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p $ 0.12 Nmm%

1. Number of revolutions of the clutch slip during engagement

?irst of all, let us find the inside radius of the clutch *r %+. #e know that, for uniform

wear, mean radius of the clutch,

%

%%1 5.05.%%

1%5

%r

r r r R +=

+=

+=

and axial force on the clutch,

+:*+1%58*12.0:+*+8*. %%%

%

%

1 r r r pW −×=−= π π

#e know that the tor3ue transmitted *T +,

:5.05.%:8+*+1%58*12.02.0%...10%50%

%

%

%2 r r RW n +−××==× π µ

:+*5.0+*5.%5./71%105.=/8%45.0 2%%

%%

2 r r r −−+×=

(olving ) hit and trial, we find that

r %$ /0 mm

#e know that angular velocit) of the engine,

%100%000%0%2 =×== π π ω N srad

and angular velocit) of the wheel,

=.

ω ((0 rad'1s#2.-

.-((0 3(0#4'5$#2 /.42

25.0

2.15=

. R

,

srad

(ince the gear ratio is 5, therefore angular velocit) of the clutch follower shaft,

5.%175/.4250 =×=×= .ω ω srad

#e know that angular acceleration of the engine during the clutch slip period of the

clutch,

1501

%50100−=

−=

−=

/

T T ((α % srad

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Heat to be dissipated by the clutch for each engagement

#e know that heat to e dissipated ) the clutch for each engagement

$ T .θ

$ %50 @ 0.7 $ 1/0 K A/.

E%&'* 2!.

multiple disc clutch has five plates having four pairs of active friction surfaces. f

the intensit) of pressure is not to exceed 0.1%/ Nmm%, find the power transmitted at

500 r.p.m. The outer and inner radii of friction surfaces are 1%5 mm and /5 mm

respectivel). ssume uniform wear and take coefficient of friction $ 0.2.

S+,-+/.

!iven " n1 I n% $ 5

n $ 4

p $ 0.1%/ Nmm%

N $ 500 r.p.m.

r 1 $ 1%5 mm

r % $ /5 mm

A $ 0.2

#e know that for uniform wear, p.r $ C *a constant+. (ince the intensit) of pressure is

maximum at the inner radius *r %+, therefore,

p.r % $ C or C $ 0.1%/ @ /5 $ =.5%5 Nmm

and axial force re3uired to engage the clutch,

W $ %-C *r 1 9 r %+ $ %- @ =.5%5 *1%5 9 /5+ $ %==2 N

>ean radius of the friction surfaces,

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R=r1+r

2

2=125+75

2=100mm=0.1m


T $ n.A.W.R $ 4 @ 0.2 @ %==2 @ 0.1 $ 25= Nm

'ower transmitted,

P=T x2 π N

60=359 x2 π x 500

60=18800W =18.8 kW Ans.

E%&'* 2!.<

>ultidisc clutch has three discs on the driving shaft and two on the drivenshaft. The inside diameter of the contact surface is 1%0 mm. The maximum pressure

etween the surface is limited to 0.1 Nmm%. Design the clutch for transmitting %5

k# at 15/5 r.p.m. ssume uniform wear condition and coefficient of friction as 0.2.

S+,-+/.

!iven " n1 $ 2

n% $ %

d % $ 1%0 mm

r % $ 0 mm

P max $ 0.1 Nmm%

P $ %5 k# $ %5 @ 102 #

N $ 15/5 r.p.m.

A $ 0.2

&et r 1 $ Butside radius of the contact surface.


T = P x60

2π N =

25 x103 x 60

2 π x 1575=151.6 N −m=151600 N −mm

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?or uniform wear, we know that p.r $ C . (ince the intensit) of pressure is maximum

at the inner radius *r %+, therefore,

P max @ r % $ C or C $ 0.1 @ 0 $ Nmm

#e know that the axial force on each friction surface,

W $ %-C *r 1 9 r %+ $ %- @ *r 1 9 0+ $ 2/./ *r 1 9 0+ ... (i )

?or uniform wear, mean radius of the contact surface,

R=r1+r

2

2=

r1+60

2=0.5r

1+30

#e know that numer of pairs of contact surfaces,

n $ n1 I n% 9 1 $ 2 I % 9 1 $ 4

Tor3ue transmitted *T +,

151 00 $ n.A.W.R $ 4 @ 0.2 @ 2/./ *r 1 9 0+ *0.5 r 1 I 20+

... 8(ustituting the value of W from e3uation (i ):

$ %%.% *r 1+% 9 71 42%

(r1)2=

151600+8143222.62

=10302∨r1=101.5mm Ans.

E%&'* 2!.9

multiple disc clutch, steel on ronHe, is to transmit 4.5 k# at /50 r.p.m. The inner

radius of the contact is 40 mm and outer radius of the contact is /0 mm. The clutch

operates in oil with an expected coefficient of 0.1. The average allowale pressure is

0.25Nmm%. ?ind " 1.the total numer of steel and ronHe discs %.the actual axial

force re3uired 2. the actual average pressure and 4. the actual maximum pressure.

S+,-+/.

!iven " '$ 4.5k# $ 4500 #

N$ /50r.p.m.

r %$ 40 mm

r 1$ /0 mm

6$ 0.1

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P av$ 0.25 Nmm%

1. T+-& /,'8* +5 -** &/4 8+/* 47

&et n $ Numer of pairs of contact surfaces.


T $ P ×60

2πN =

4500×60

2π ×750=57.3 N − M =57300 N −mm

?or uniform wear, mean radius of the contact surfaces,

; $r1+r

2

2=70+40

2=55mm

and average axial force re3uired,

# $

r

r2¿40

¿(70 )

2

−(¿¿2 ]=3630 N

(¿¿ 1)2−(¿¿ 2 ]=0.35× π ¿¿

Pav × π ¿

#e also know that the tor3ue transmitted *T+,

5/ 200 $ n . π . W . R=n ×0.1×3630×55=19965n

n $ 5/ 200 1= =5 $ %.7/

(ince the numer of pairs of contact surfaces must e even, therefore we shall

use 4 pairs of contact surface with 2 steel discs and % ronHe discs * ecause the

numer of pairs of contact surfaces is one less than the total numer of discs +.

A/.

2. A7-,& &%& 5+7* *,*4&et #G $ ctual axial force re3uire.

(ince the actual numer of pairs of contact surface is 4, therefore actual tor3ue

developed ) the clutch for one pair of contact surface,

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TG $T

n=57300

4=14325 N −mm

#e know that tor3ue developed for one pair of conact surface *TG+

14 2%5 $ μ .W ' . R=0.1×W

' ×55=5.5W

'

#G $ 14 2%5 5.5 $ %04.5 N ns.

3. A7-,& &*&* *,*

#e know that the actual average pressure,

'Gav $

r1

¿¿r2

¿¿2−¿¿70

¿¿40

¿¿2−¿¿¿

π ¿

W

'

¿

ns.

!. A7-,& '&%',' *,*

&et 'max$ ctual maximum pressure

?or uniform wear, p.r $ . (ince the intensit) of pressure is maximum at the

inner radius, therefore,

Pmax ×r2=C ∨C =40 Pmax N

mm

#e know that the actual axial force *#G+,

%04.5 $ 2πC ( r1−r2 )=2π ×40 Pmax (70−40 )=7541 Pmax

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'max $ 2604.5/7541=0.345 N /mm2

ns.

E%&'* 2!.10

plate clutch has three discs on the driving shaft and two discs on the driven shaft, providing four pairs of contact surface. The outside diameter of the contact surfaces is

%40 mm and inside diameter 1%0 mm. ssuming uniform pressure and 6 $ 0.2, find

the total spring load pressing the plates together to transmit %5 k# at 15/5 r.p.m.

if there are springs each of stiffnes 12 kNm and each of the contact surfaces has

worn awa) ) 1.%5 mm, find the maximum power that can e transmitted, assuming

uniform wear.

S+,-+/.

!iven " n1$ 2

n%$ %

n$4

d$ %40 mm

r 1$1%0 mm

d%$1%0 mm

r %$ 0 mm

6 $ 0.2

'$%5k#$%5 L102 #

N$ 15/5 r.p.m.

T+-& / +&4

&et #$ Total spring load.


T = P ×60

2πN =

25×103

×60

2 π ×1575=151.5 N − M

>ean radius of the contact surface ,for uniform pressure,

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r

r2¿¿¿2

¿120

¿¿60

¿¿¿3¿

120

¿¿60

¿¿¿2

( r1)2

−¿

(¿¿1)3−(r2 )3

¿

R=2

3⌊¿

and tor3ue transmitted *T+,

151.5×103

=n . μ . W . R=4×0.3×W ×93.3=112W

W =151.5×103/112=1353 N ns.

M&%',' +>* -&/'--*4

!iven " No. of springs $

contact surfaces of the spring $ 7

wear on each contact surface $ 1.%5 mm

Total wear $ 8×1.25=10mm=0.01m

(tiffness of each spring $ 13kN /m=13×103 N / M

;eduction in spring force

8/17/2019 KELOMPOK 1 ( 24.1 - 24.10 )

23/23

¿ total ear × stiffness !er s!rin" × No . of s!rin"s

¿0.01×13×103

×6=780 N

and new axial # $ 1353−780=573 N

#e know that mean radius of the contact surfaces for uniform wear,

R=r1+r

2

2=120+60

2=90mm=0.09m

and tor3ue transmitted T =n . μ . W . R=4×0.3×573×0.09=62 N −m

'ower Transmitted, P=T ×2 πN

60=62×2π ×1575

60=10227W =10.227 kW

A/.

kelompok 1 ( 24.1 - 24.10 )

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