DOM unit-I

8/16/2019 DOM unit-I

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Dynamics of Machinery Unit I: Force Analysis

Unit-I

Force Analysis

Inertia force

• The inertia force is an imaginary force, which when acts upon a

rigid body, brings it an equilibrium position

• It is equal to the accelerating force in magnitude, but opposite in

direction

Mathematically,

Inertia force ! " Accelerating force

D-Alembert’s Principle

D#Alembert$s principle states that the resultant force acting on a

body together with the reversed effective force (or inertia force! are in

e"uilibrium#

According to %ewton$s second law of motion,

F $ m#a

&here

m ! Mass of the body, and

a ! 'inear acceleration of the centre of gra(ity of the body

F % m#a $ &

If the quantity % m#a be treated as a force, equal, opposite

FI $ % m#a

)ub FI in (alue in equation *+

F ' FI $ &

Application

This principle is used to reduce a dynamic problem into an

equi(alent static problem

Dynamic Analysis of )lider cran* mechanism

*i -raphical method

*ii Analytical method

+raphical methods

+ .lien$s construction,

/ 0itterhaus$s construction, and

1 2ennett$s construction

,ote

• -raphical methods gi(en the accurate answer

• Analytical methods gi(en appro3imate solution

Dynamic analysis of slider-cran* mechanism by analytical

method

After rotation of cran* by angle

Dept.of Mechanical Engg. AAMEC Page 1

(1)


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l # 'ength of connecting rod between the centers,

r # 0adius of cran4 or cran4 pin circle,

5 # Inclination of connecting rod to the line of stro4e

n # l 6r #0atio of length of connecting rod to the radius of cran4

.o find

• 7elocity of piston

• Acceleration of piston

• Angular 7elocity of connecting rod

• Angular Acceleration of connecting rod

/elocity of the piston

From the geometry

3 ! 898 ! 89 ; 8 ! *89

*8> !l cos ?@ >! r cos

3 ! *l = r ; *l cos ? = r cos

,cos#*+,cos+* φ θ l r +−=

+−= ,cos#*+

r ,cos+* φ θ l

r

[ ],cos#n*+,cos+* φ θ +−= r

From triangles

sin? ! 6 8< ! 8< sin?

! l sin?

From triangles

sin ! 6 <

! < sin

! r sin

From equ */ and *1

! l sin? ! r sin

l 6r !sin 6 sin?

&here n ! l 6r

n ! sin 6sin?

sin 5 ! sin 6n

&e 4now that

( ) /+

/sin+cos φ φ −=

/+

/

/sin

+cos

−=

n

θ φ

B3panding the abo(e e3pression by binomial theorem, we get

sin

/

++cos

/

/

+×−=n

θ φ

*%eglecting higher terms

/

/

/

sincos+

n

θ φ =−

)ubstituting the (alue of *+ " cos 5 in equation (0! we have


(1)

(2)

(3)

(4)


3/48


( )

×+−=

/

/

/

sin cos+

nnr x

θ θ

( )

+−=

nr

/

sin cos+

/θ

θ

Differentiating equation *C with respect to

+=n

r d

dx

/

cossin/sin

θ θ θ

θ

+=

nr

d

dx

/

cossin/sin

θ θ θ

θ

+=n

r /

/sinsin

θ θ

( )θ θ θ /sincossin/ =

∴ 7elocity of 8 with respect to *or (elocity of the piston 8

dt

d

d

dx

dt

dxvv p PO

θ

θ ×===

ω θ

×=d

dx

*0atio of change of angular (elocity ! !d 6 dt

)ubstituting the (alue of d36d from equation *E, we ha(e

+== nr vv p PO //sin

sin

θ

θ ω

Acceleration of the piston

)ince the acceleration is the rate of change of (elocity, therefore

acceleration of the piston 8,

ω θ

θ

θ ×=×==

d

dv

dt

d

d

dv

dt

dva

p p p p

Differentiating equation * with respect to ,

×+=n

r d

dv p

/

//coscos

θ θ ω

θ

+=n

r d

dv p θ θ ω

θ

/coscos

)ubstitute d( p6d (alue in equation *G we ha(e

ω θ

θ ω ×

+=

nr a p

/coscos

+=

nr a p

θ θ ω

/coscos

/

Angular /elocity and Angular Acceleration of the 1onnecting 2od

Angular /elocity

From the geometry of the figure,

CQ = l sin Ø = r sin θ

n ! sin 6sin ?

n

θ φ

sinsin =

Differentiating both sides with respect to time t,


(5)

(6)

(7)

(8)

()


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dt

d

ndt

d θ θ φ φ ×=×

coscos

ω θ φ

φ ×=×ndt

d coscos

=ω

θ

dt

d

φ

ω θ φ

cos

cos ×=ndt

d

Angular (elocity of connecting rod

dt

d PC

φ ω =

φ

ω θ

cos

cos×=

n

&e 4now that

( ) /+

/sin+cos φ φ −=

=

n

θ φ

sinsin

/+

/

/sin+cos

−=n

θ φ

/

+

/

/sin+

cos

−

×=

n

n PC

θ

θ ω ω

( ) /+

// sin+

cos

θ

θ ω

−

×=

nn

n

( ) /+

// sin

cos

θ

θ ω ω

−

=

n

PC

Angular acceleration of the connecting rod PC :

H8


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*Di(iding and multiplying by *n/#sin/ +6/

( )

( )

−

−−−=

/1

//

///

sin

cossinsin

θ

θ θ θ ω

n

n

( )( )[ ]θ θ

θ

θ ω ///

/1

//

cossin

sin

sin+−

−

−= n

n

( )[ ]+

sin

sin /

/1

//

−

−

−= n

n θ

θ ω

*sin/=cos/!+

( ) [ ]( ) /

1//

//

sin

+sin

θ

θ ω ω

θ

ω α

−

−−=×=

n

n

d

d PC PC

The negati(e sign shows that the sense of the acceleration of the

connecting rod is such that it tends to reduce the angle 5

,otes

Angular (elocity of connecting rod ( ) /

+//

sin

cos

θ

θ ω ω

−

=

n PC

Angular Acceleration of connecting

( )

( ) /1

//

//

sin

+sin

θ

θ ω ω

θ

ω α

−

−−=×=

n

n

d

d PC PC

)ince sin/ is small as compared to n/, therefore it may be neglected in

angular (elocity equation and angular acceleration equations of the

connecting rod

Therefore, Angular (elocityn

PC

θ ω ω

cos=

Angular Acceleration ( )

1

// +sin

n

n pc

−−=

θ ω α

Also in abo(e equation unity is small as compared to n34 hence the term

unity may be neglected

Angular Accelerationn

pc

θ ω α

sin/−

=

0# In a slider cran* mechanism! the length of the cran* and connecting

rod are 05& mm and 6&& mm respectively# .he cran* position is 6&7

from inner dead centre# .he cran* shaft speed is 85& r#p#m#

(cloc*wise# Using analytical method! determine

i# /elocity and acceleration of the slider! and

ii Angular velocity and angular acceleration of the connecting

rod

+iven


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'ength of the connecting rod l ! E mm ! E m@


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+cos/cos/ =−+ θ θ n

+cos111cos// =−+ θ θ

( ) ( )

//

+/K111111cos

/

×××+±−

=θ

cos !/E !C

=angine

(,eglecting the ?eight of the 1onnecting 2od

F8 # 8iston effort

F % # Thrust on the sides of the cylinder walls *or normal reaction on the

guide bars

F> # Force acting along the connecting rod

FT #


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,resistancefrictionalg*%eglectinFF I' =

,resistancefrictionalng*2 += ( )5Acoscos

F p +×=φ

%# Crank effort (or Turning moment (or) Tor&ue on the crank shaft: (T)

1ran* effort

T ! FT r( )

r ×+

= 5cos5AsinF8

−+×=

Asinn/

sin/AsinAr FT

//8

( )r ×

+=

φ

φ θ

cos

sinFT 8

)in *A=2 ! )inA


9/48


,/sinsincos/* θ θ θ =

r n

F T P ×

−+=

θ

θ θ

//sin/

/sinsin

;# .he cran*-pin circle radius of a hori@ontal engine is ;&& mm# .hemass of the reciprocating part is 35& *g# ?hen the cran* has travelled

6&7 from I#D#1#! the difference between the driving and the bac*

pressures is 5 ,:mm3# .he connecting rod length between centers is

0#3 m and the cylinder bore is m# If the engine runs at 35& rpm

and if the effect of piston rod diameter is neglected! 1alculate

i# Pressure on slide bars!

ii# .hrust in the connecting rod!

iii# .angential force on the cran*-pin! and

iv# .urning moment on the cran* shaft#

+iven

r ! 1 mm ! 1 m@

m0 ! /C 4g@

! EJ@ p+ " p/ ! 1C %6mm

/@

l ! +/ m@

D ! C m ! C mm@

% ! /C rpm)olution

! / L /C6E ! /E/ rad6s8iston effort F p !F' # FI *#(e sign for acceleration of the piston

( ) //+' DK

L p pF ×−= /C

K

L1C ××= % EG1=

0atio of length of connecting rod and cran4

r ln = 1I/+= K=

+=

n

cos/AcosAr mF

/0 I ( )

+××=

K

+/cosEcos1///E/C

FI ! +N1E %

F p ! EG1 " +N1E ! KNK/K % ! KNK/K 4%

(i Pressure on slide bars

tan5FF p % =

n

sinAsin5 =

K

Esin= /+EC=

( )/+ECsin5 +−= +/C=

tan+/CKNK/KF % = %+NE= 4%NE+=

(ii .hrust in the connecting rod

cos5FF p

> =cos+/CKNK/K= %CE/= 4%E/C=

(iii .angential force on the cran*-pin

( )5AsinFF >T += ( )+/CEsinCE/ += ! KG/G4%

(iv.urning moment on the cran* shaft

T ! FT r ! KG/G1 ! +KKGK 4%

8# A vertical double acting steam engine has a cylinder ;&& mm

diameter and 85& mm stro*e and runs at 3&& rpm# .he reciprocating

part has a mass of 335 *g and the piston rod is 5& mm diameter# .heconnecting rod is 0#3 m long# ?hen the cran* has turned through 0357

from the top dead centre! the steam pressure above the piston is

;&*,:m3 and below the piston is 0#5 *,:m3# 1alculate the effective

turning moment on the cran* shaft#

Dept.of Mechanical Engg. AAMEC Page


10/48


11/48


)olution

! /L +G6E ! +GGC/ rad6s@

0# Net load on the gudgeon !in

F#F I'= P F

( ) 1C/% GKL

pDK

L

F//

' =××=×=

+==

n

cos/AcosAr mamF

/0 0 0 I

( )

+−=

nr x

/sin cos+

/

θ θ

n ! l 6r ! /6C ! K

( )

×

+−×=× −−K/

sin cos++C++

/11 θ θ

( )

+−=

G

sin cos+C+

/θ

θ

( )

+

−=

G

sin

G

cos+GC+

/θ θ

( ) ( ) θ θ /sin cos+GC6+G +−=×

θ θ /

sincosGGE+ +−=

θ θ /

cos+cosGGE+ −+−= ( )θ θ // cos+sin −= E+coscosGN

/ =−−− θ θ

KcosGcos/ =−+ θ θ

)ol(ing the abo(e quadratic equation

cos ! G1

! cos#+*G1 ! 11+

+== ncos/A

cosAr mamF/

0 0 0 I

+××=

K

cosEEcos11C+GGC/+ /

FI ! +E+ %

F#FF I'8 = +E+#1C/= !+GKN %

3# Thrust in the connecting rod

cos5

FF 8> =

nθ φ sin sin = K

11sin = KCKKE= +1E+=

( )+1E+sin +−=φ G/=

cos5

FF 8> =

cosG/

+GKN= %+GEE1=

;# 'eaction between the !iston and cylinder

F % ! F8 tan !+GKN tanG/ϕ ! /CK %

8# >ngine speed at which the above values will become @ero

F p ! F' # FI

! F' # FI &hen Fp $ &

FI $ FC

p D ×=

+ //+0

Kn

cos/AcosAr m

π



12/48


GKK

cosEEcos11C+

//+ ××=

+××

π

1C/K /+ =

KGNKK61C//+ ==

rad6sE/1+ ="uivalent Dynamical )ystem

In order to determine the motion of a rigid body, under the action of

e3ternal forces, it is usually con(enient to replace the rigid body by two

masses placed at a fi3ed distance apart, in such a way that,

+ The sum of their masses is equal to the total mass of the body

/ The centre of gra(ity of the two masses coincides with that of the

body and

1 The sum of mass moment of inertia of the masses about their centre

of gra(ity is equal to the mass moment of inertia of the body

&hen these three conditions are satisfied, then it is said to be an

e&uialent dynamical system#


13/48


( )

( ) /

/+

/+/+!

l l

l l l l =

+

+

/

/+ ! l l =

This equation gi(es the essential condition of placing the two masses,

so that the system becomes dynamical equi(alent

,ote

&hen the radius of gyration * + is not *nown, then the position of the

second mass may be obtained by considering the body as a compound

pendulum

'ength of the simple pendulum which gi(es the same frequency as the

rigid body *ie compound pendulum is

( )

h

h!

#

// +

= O *0eplacing h by l +

( )

+

/+

/

l

l ! #

+=

&e 4now that

l $%l & = '! ( &

/++

/+/+ l l

l

l l l # +=

+=

This means that the second mass is situated at the centre of

oscillation or percussion of the body, which is at a distance of

l & = '! ( & )l $

%# The connecting rod of a gasoline engine is ** mm long between its

centers# +t has a mass of ," kg and mass moment of inertia of

-***kgmm. # +ts centre of graity is at .** mm from its small end centre#

/etermine the dynamical e&uialent twomass system of the connecting rod if one of the masses is located at the small end centre #

+iven

l ! 1 mm@

m ! +C 4g@

I ! 4g#mm/@

l+ ! / mm

)olution

Mass moment of inertia *I, ( )/

! " * =( ) /+C ! ×=

( ) KEE+C6/ ==!

""! E/+=

&e 4now that

For a dynamical equi(alent system,

l $%l & = *4 - /

l & = *4 - / )l $ ! KEE6/ ! /11 mm

m+ ! Mass placed at the small end centre, and

m/ ! Mass placed at a distance l / from the centre of gra(ity -

/+

/+

mm

l l

l

+=

11//

+C11/

+×

= !g +=



14/48


/+

+/

mm

l l

l

+=

11//

/+C

+×

= !g G1+K=

-# 0 connecting rod is sus!ended from a !oint ." mm aboe the centre of

small end1 and %"* mm aboe its centre of graity1 its mass being -#"kg#

2hen !ermitted to oscillate1 the time !eriod is found to be ,#3-seconds#

Find the dynamical e&uialent system constituted of two masses1 one of

which is located at the small end centre#

+iven

h ! EC mm ! EC m@

l+ ! EC " /C ! E/C mm ! E/C m@

m ! 1C 4g@

t p ! +G s

)olutionFor a compound pendulum, time period of oscillation *t p,

( )

gh

h! t p

//

/ +

= π

( )

gh

h! //

/G+ +

= π

)quaring both sides

( )

1GE

K//CGGC

/ += !

( ) ( ) K//C1GEGGC/ −×=! !+K/C m/

"! 1+K/C ==

For a dynamically equi(alent system

*4 - / ! l $%l &

l / ! *4 - /6l $ !+K/C6E/C! //Gm

/+

/+

mm

l l

l

+=

//GE/C

C1//G

+×= !g +=

/+

+/

mm

l l

l

+=

//GE/C

C1E/C

+×

= !g C/=

3# The following data relate to a connecting rod of a reci!rocating

engine:

4ass 5 "" kg6 /istance between bearing centers 5 3"* mm6 /iameter of

small end bearing 5 -" mm6 /iameter of big end bearing 5 ,** mm6Time of oscillation when the connecting rod is sus!ended from small

end 5 ,#3 s6 Time of oscillation when the connecting rod is sus!ended

from big end 5 ,#%3 s# /etermine:

,# The radius of gyration of the rod about an a7is !assing through the

centre of graity and !er!endicular to the !lane of oscillation6

.# The moment of inertia of the rod about the same a7is6 and

# the dynamically e&uialent system for the connecting rod1 constituted

of two masses1 one of which is situated at the small end centre#

+iven m ! CC 4g

l ! GC mm ! GC m

d+ ! C mm ! C m

d/ ! + mm ! + m

t p+ ! +G1 s



15/48


t p/ ! +EG s

)olution

First of all, let us find the lengths of the equi(alent simple pendulum when

suspended

(a) From the to! of small end bearing6 and

(b) From the to! of big end bearing

&e 4now that for a simple pendulum

g

#t p

++ /π =

g

#t p +/

+

/=

π

... (Squaring both sides)

/

+

+/

= π

pt g #

/

/

G1+G+N = π "G1/=

)imilarly,/

/

//

=

π

pt g #

/

/

EG+G+N

=

π "=

0# 'adius of gyration of the rod about an a7is !assing through the centre

of graity and !er!endicular to the !lane of oscillation:

Bqui(alent length of simple pendulum ( )

h

h! #

// +=

( ) ,*// h #hh #h! −=−=

&hen the rod is suspended from the top of small end bearing,

( ) ,* +++/

h #h! −= *i

and when the rod is suspended from the top of big end bearing

( ) ,* ////

h #h! −= *ii

Also, from the geometry of the Fig

//

/+

/+

d

l

d

hh ++=+

/

+GC

/

C++=

h+=h/ !N1C m

h/ !N1C # h+

From equations *i and *ii,

( ) ,* ///+++ h #hh #h −=−

)ubstituting the (alue of h/ from equation *iii,

( ) ,,N1C*,*N1C*G1/ ++++ hhhh −−−=−

( ) ( ) /++/++ +C+//1G1/ hhhh −+−=−/111K1 + =h

"h EC1K16/11+ == %ow from equation *i,

( ) ,ECG1/*EC/ −=! ++G1=

( ) "! 1K1++G1 ==

3# =oment of inertia of the rod

( )/

! " * = ( )//

m#4gC+E1K1CC ==

;# Dynamically e"uivalent system for the rod

)ince one of the masses *m+ is situated at the centre of small end

bearing, therefore its distance from the centre of gra(ity, -, is



16/48


"hl E+/C1CEC,/6C*++ =−=−=

m/ ! Magnitude of the second mass, and

l / ! Distance of the second mass from the centre of gra(ity, -, towards big

end bearing

For a dynamically equi(alent system

*4 - / ! l $%l &

l & = *4 - / )l $ =+%$$-)+%.$&/= +%$0-"

/+

/+

mm

l l

l

+=

//GE+/C

CC+N1

+×

= !g +G+1=

1orrection 1ouple to be applied to ma*e .wo-=ass )ystem

Dynamically >"uivalent

The torque required to accelerate the body α α /,* ! " * T == ############## *i

Torque required to accelerate the two#mass system placed arbitrarily

α α /+++ ,* ! " * T == 111111111111111111 'ii(

Difference between the torques required accelerating the two#mass system

and the torque required to accelerate the rigid body

α α //++P

,*,* ! "! "T T T −=−= α //

+ ,*,* ! ! " −= 111111 'iv(

The difference of the torques TP is 4nown as correction couple# .his

couple must be applied! when the masses are placed arbitrarily to ma4e

the system dynamical equi(alent

*4 - / ! l $%l &

*4 + / ! l $%l -


17/48


the line of centers# The radius of gyration about an a7is through the

C#9# !er!endicular to the !lane of rotation is ,,* mm# Find the

e&uialent dynamical system if only one of the masses is located at

gudgeon !in#

+f the connecting rod is re!laced by two masses1 one at the gudgeon

!in and the other at the crank !in and the angular acceleration of therod is . *** rads. clockwise1 determine the correction cou!le a!!lied to

the system to reduce it to a dynamically e&uialent system #

+iven

m ! / 4g@

l ! /C mm ! /C m@

l $ ! + mm ! +m@

4 - ! ++ mm ! ++ m @

H ! /1 rad6s/

)olution

>"uivalent dynamical system

It is gi(en that one of the masses is located at the gudgeon pin

'et the other mass be located at a distance l/ from the centre of gra(ity

For an e"uivalent dynamical system

*4 - / ! l $%l &

l & = *4 - / )l $ !++

/6+! +/+m

m+ ! Mass placed at the gudgeon pin,

m/ ! Mass placed at a distance l / from


18/48


.urning moment diagram The turning moment diagram *also 4nown as

cran4# effort diagram is the graphical representation of the turning

moment or cran4#effort for (arious positions of the cran4

.urning =oment Diagram for a )ingle 1ylinder Double Acting )team

>ngine:

Turning moment r n

F T P ×

−+=

θ

θ θ

//sin/

/sinsin

.urning =oment Diagram for a Four )tro*e 1ycle Internal

1ombustion >ngine:

.urning =oment Diagram for a =ulti-cylinder >ngine:



19/48


Ma3Fluctuation of Bnergy Q B!Ma3imum energy " Minimum energy


20/48


nergy )tored in a Flywheel

Mean 4inetic energy of the flywheel, /

/

+ω * 4 = //

/

+ω "! =

Ma3imum fluctuation of energy R 4 ! Ma3imum .B " Minimum .B

//

/+ I

/

+ I

/

+ B −=∆

( )///++ I/

+ −=

( ) ( )[ ]/+/+ I/

+ −+=∆ 4

( )/+ I −= ω

+=

/

/+ ω ω ω

−=ω

ω /+/ I

−=ω

ω /+/ I

sC /I ω =

−=

ω

ω ω /+ sC

sC //m4 ω = [ ]/"! * =

RB ! / 4 C )

=/

/

+

ω * 4

The radius of gyration *! may be ta4en equal to the mean radius of

the rim * 5, because the thic4ness of rim is (ery small as compared to the

diameter of rim Therefore, substituting ! ! 5

s s C "5C 4 ////m4 ω ω ==∆

sC "v 4 /=∆ [ ] 5v ω =

Dept.of Mechanical Engg. AAMEC Page 2!


21/48


,ote

)ince ! / π 3 6E, therefore equation may be written as

−×=∆

E

%/

E

%/

E

%/I B /+

π π π ( )/+

/

I

1E

K 3 3 3 −××= π

( )/+/

/

m4 N

3 3 3 −××= π

sC 3 ×××=

///

m4 N

π

−=

3

3 3 C s

/+

,,# The mass of flywheel of an engine is %#" tonnes and the radius of

gyration is ,#3 metres# +t is found from the turning moment diagram that

the fluctuation of energy is "% kNm# +f the mean s!eed of the engine is

,.* r#!#m#1 find the ma7imum and minimum s!eeds#

+ivenMass m ! EC t ! EC 4g@

0adius of gyration 4 ! +G m@

Fluctuation of energyR B ! CE 4%#m ! CE +1 %#m@

)peed % ! +/ rpm

)olution

%+ and %/ ! Ma3imum and minimum speeds respecti(ely

Fluctuation of energy sC 3 "! 4 //

/

N×=∆

π

,*N

+CE /+/

/

1 3 3 3 "! −×=× π

,*+/G+EN

/+/

/

3 3 −×××= π

,*/+CCE /+ 3 3 −= *+

Mean speed , /

/+ 3 3 3 +

=

/+/I /+

3 3 +=

/+//+ ×=+ 3 3 /K/+ =+ 3 3 */

)ol(ing equations (,) and (.)1

%+ ! +/+ rpm

%/! ++N rpm



22/48


03# .he flywheel of a steam engine has a radius of gyration of 0 m and

mass 35&& *g# .he starting tor"ue of the steam engine is 05&& ,-m and

may be assumed constant# Determine

i# .he angular acceleration of the flywheel! and

ii .he *inetic energy of the flywheel after 0&

seconds from the start#+iven

* $ 0 m4

m $ 35&& *g4

. $ 05&& ,-m

)olution

0# Angular acceleration of the flywheel

H ! Angular acceleration of the flywheel

Mass moment of inertia of the flywheel

I = "%! & ! 2500 × 12 = 2500 kg-m2

)tarting torque of the engine T ! I H

+C ! *%6

+C! /C H

H ! +C 6 /C !E rad 6s/

;inetic energy of the flywheel:

Assuming uniform acceleration

+ ! Angular speed at rest !

/ ! Angular speed after + seconds,

t ! Time in seconds

&e 4now that

/ ! + = H t ! = E +! E rad 6s

.inetic energy of the flywheel ( ) ///

+ω * ×= ( ) /E/C

/

+×=

m# %KC=

0;# A hori@ontal cross compound steam engine develops ;&& *?at9&

r#p#m# .he coefficient of fluctuation of energy as found from the

turning moment diagram is to be and the fluctuation of speed is to

be *ept within E of the mean speed# Find the weight of theflywheel re"uired! if the radius of gyration is 3 metres#

+iven

8 ! 1 4& ! 1 +1 &@

% ! N rpm@


23/48


Ma3imum fluctuation of energy RB ! &or4 done per cycle < B

! / +1 + ! / +1 %#m

m ! Mass of the flywheel

Ma3imum fluctuation of energy RB $m#* 3#3#1)

/ +1 ! m // *NK/E/ +

/! 1CCK mm !/ +161CCK ! CE1 4g

,


24/48


,"# The turning moment diagram for a multi cylinder engine has been

drawn to a scale , mm5 %** Nm ertically and , mm5 = hori>ontally#

The interce!ted areas between the out!ut tor&ue cure and the meanresistance line1 taken in order from one end1 are as follows:

? ".1 @ ,. $m#*

3

#

3#

1) CKK ! m *C/ *E/GK/ 1

m ! CKK 6 /NE ! +G1 4g



25/48


,%# 0 shaft fitted with a flywheel rotates at ."* r#!#m# and dries a

machine# The tor&ue of machine aries in a cyclic manner oer a !eriod

of reolutions# The tor&ue rises from -"* Nm to *** Nm uniformlyduring ,. reolution and remains constant for the following reolution#

+t then falls uniformly to -"* Nm during the ne7t ,. reolution and

remains constant for one reolution1 the cycle being re!eated thereafter#

/etermine the !ower re&uired driing the machine and !ercentage

fluctuation in s!eed1 if the driing tor&ue a!!lied to the shaft is constant

and the mass of the flywheel is "** kg with radius of gyration of %** mm

+iven

% ! /C rpm

m ! C 4g @4 ! E mm ! E m

)olution

! /L /C6E ! /E/ rad6s

The turning moment diagram for the complete cycle is shown in Fig

Torque required for one complete cycle! Area of figure A2


26/48


8

89

A

#9 =

C1

+GC1

−−

=π

#9

'M ! CL

From similar triangles


27/48


T/

!+C %#m

/

! G

wrt D< !+G=G !/E

wrt ID<

&or4 done per cycle ! Area of triangle A- = Area of triangle -')

××+

××= #9 : A8O

/

+

/

+

××+

××= +C

/+/

/+ π π

!+C L %#m ############ *i

&or4 done per cycle ! Tmean / L %#m ######## *ii

From equations *i and *ii,

Tmean ! +C L 6 / L ! GC %#m

From similar triangles A


28/48


O8 A8

A4 C4 ×= O8

A8

48 A8×

−=

N

K

/

GC/ π ×

−= rad

K

π =

KN

K π π θ −=C rad

1E

π =

π

π +G

1E

×= 1C=

Again from similar triangles ABD and A2-,

A8

A4

8

4D =

8 A8

A4 4D ×= ,* O8O

A8

48 A8−×

−=

−×

−=

N

K

/

GC/ π π

rad 4DN

G/ π =

N

G/

N

K π π θ += D rad

N

GE π = π

π +G

N

GE ×= ! +1E

,3# 0 three cylinder single acting engine has its cranks set e&ually at

,.*= and it runs at %** r#!#m# The tor&uecrank angle diagram for each

cycle is a triangle for the !ower stroke with a ma7imum tor&ue of 8* Nm

at %*=from dead centre of corres!onding crank# The tor&ue on the

return stroke is sensibly >ero# /etermine

i# Power deelo!ed#

ii% Coefficient of fluctuation of s!eed1 if the mass of the flywheel is

,.kg and has a radius of gyration of 3* mm1

iii# Coefficient of fluctuation of energy1 andi# 4a7imum angular acceleration of the flywheel#

+iven

% ! E rpm

Tma3 ! N %#m@

m ! +/ 4g@

4 ! G mm ! G m

)olution

8ower de(eloped ! wor4 done6cycle ! Area of three triangles

N/

+1 ×××= π m# %K/K=

eangle6cycl


29/48


( )KCCEE/

+−××=

π ! CGN %#m !a

a/!Area of triangle 2b<P

/

+;; 8C ××=

O*2< ! E !L61 rad

( )CEN1/

+ −××= π m# %G++=

ECK1a aaa ====

%ow, let the total energy at A ! B, then referring to Fig

Bnergy at 2 ! B " CGN

Bnergy at < ! B " CGN = ++G ! B = CGN

Bnergy at D ! B = CGN " ++G ! B " CGN

Bnergy at B ! B " CGN = ++G ! B = CGN

Bnergy at - ! B = CGN " ++G ! B " CGN

Bnergy at V ! B " CGN = ++G ! B = CGN

Bnergy at ! B = CGN " CGN ! B ! Bnergy at A

Ma3imum energy! B = CGN

Minimum energy! B " CGN

=a


30/48


It is assumed to be triangular during compression and e3pansion stro4es,

neglecting the suction and e3haust stro4es

n = 3)& ! 1 6 / ! +C&or4 done6cycle ! P .+ ) n ! / +1 E 6 +C ! G %#m OO *i

eangle6cycl


31/48


Assuming the resisting tor"ue to be uniform! find the mass of the rim

of a flywheel re"uired to *eep the speed between 3&3 and 09B r#p#m#

.he mean radius of the rim is 0#3m#

+iven

)uction stro4e a$ = +%/ $+ 2- "& >

B3haust stro4e a = +%./ $+ 2- "&>

+ m/of area !1 M%#m

3 $ = &+& r%p%">

3 & = $0 r%p%"%>

5 = $%& " )olution

The turning moment cran4 angle diagram for a four stro4e engine is shown

in Figure

The areas below the Yero line of pressure are ta4en as negati(e while the

areas abo(e the Yero line of pressure are ta4en as positi(e

%et area ! a1 " *a+=a/=aK

( ) ( ) ( ) ( )( )1111 +EC+++KC+GE −−−− ×+×+×−×= ! K+#1m/

Bnergy scale is + m

/

! 1 M%#m! 1 +

E

%#m %et wor4 done per cycle ! K + "1 1 +E ! +/ +1 %#m (i)

&or4 done per cycle! T "ean ? (ii)

From equations *i( and 'ii(,

T "ean x ? = $& $+-

T "ean = $& $+- ) ? = 0// 31"

&or4 done during e3pansion stro4e! a- 4nergB scale

! EG + "1 1 +E

! /K +1 %#m (iii)

&or4 done during e3pansion stro4e ! Area of triangle A8C

! X 8C A

! X L A ! +C+ A

*iv(

From equations (iii) and (i)1

A = &+% $+- )$%/@$ = $& 0/ 31"



32/48


B3cess torque,

T excess = AF ! A 2 F ! +/ NGC " NCC = $& +-+ 31"

)imilar triangles AD4 and A8C,

A

AF

8C

D4 =

8C A

AF D4 ×= π ×=+/NGC

+/1 radN/=

Ma3imum fluctuation of energy

R 4 =Area o E AD4 AF D4 ××=/

++/1N/

/

+××= = $@ 31

"

4ass of the rim of a flywheel

/

/+ 3 3 3 +

=/

+NG// += =&++ rp"

Ma3imum fluctuation of energy *R 4 (, H $m#k . #D.#C E

( )/+/

/

N+KKK 3 3 3 "5 −×=

π

( ) ( )+NG////+N

+KKK/

/

−××××= "π

4g+1G+1E+/6+KKK =="

30# .he turning moment curve for an engine is represented by the

e"uation!

. $ (3& &&& ' 95&& sin 3% 5&& cos 3 ,-m! where is the angle

moved by the cran* from inner dead centre# If the resisting tor"ue isconstant! find

i# Power developed by the engine4

ii =oment of inertia of flywheel in *g-m 3! if the

total fluctuation of speed is not e


33/48


The turning moment diagram for one stro4e*ie Valf re(olution of the

cran4shaft is shown in Fig

)ince at points 2and D, the torque e3erted on the cran4shaft is equal to the

mean resisting torque on the flywheel

Therefore, T ! Tmean

/ = NC sin /" C cos / ! / NC sin / ! C cos /

tan /! sin /6cos / ! C6NC ! E

∴ /! 1+J or

! +CCJ

∴2! +CCJ and

D! NJ = +CCJ ! +CCJ

Ma3imum fluctuation of energy, ( )∫ =∆ D

8

4

θ

θ

θ dT#T mean

( )∫ +=∆C+C

C+C

d/#Ccos/#NCsin// θ θ θ 4

C+C

C+C/

Csin/#

/

NCcos/

−=

θ θ m# %++G=

Ma3imum fluctuation of energyH $m#k . #D.#C E

++G ! I *+GGC/ +

++G ! 1CC I

I ! ++G61CC ! 1+/+ 4g#m/

Angular acceleration of the flywheel

'et H! Angular acceleration of the flywheel,

and

! Angle turned by the cran4 from inner

dead centre ! KCJ *-i(en

The angular acceleration in the flywheel is produced by the e3cess torque o(er the mean torque

B3cess torque at any instant "eanexcess T T T −=

/#Ccos/#NCsin// θ θ +=θ θ Ccos/#NCsin/=

B3cess torque at KCJ

! NC sin NJ " C cos NJ ! NC %#m

B3cess torque!IH ! 1+/+ H *ii

From equations *iand *ii

H! NC61+/+ ! 1KK rad 6s/

33# A certain machine re"uires a tor"ue of (5&&& ' 5&& sin ,-m to

drive it! where is the angle of rotation of shaft measured from

certain datum# .he machine is directly coupled to an engine which

produces a tor"ue of (5&&& ' 6&& sin 3,-m# .he flywheel and the



34/48


other rotating parts attached to the engine has a mass of 5&& *g at a

radius of gyration of m# If the mean speed is 05& r#p#m#! find

i# .he fluctuation of energy!

ii .he total percentage fluctuation of speed! and

iii .he ma


35/48


Differentiating this e3pression with respect to and equating to Yero for

ma3imum or minimum (alues

( ) sinC/sinE =−∴ θ θ θ d

d

cosC/cos+/ =− θ θ

cosC/cos+/ =− θ θ *cos/!/cos/#+

+/cosCcos/K/ =−− θ θ

( )

/K/

/K+/K/CCcos

×××+±

=θ KG

11KC ±= !G+G *or #E+K

! 1CJ or +/EJ

)ubstituting ! 1CJ in equation (i)1 we hae ma7imum tor&ue

T "ax = .++ sin @+ 2 /++ sin -/

T "ax = &@@ 31"

)ubstituting !+/EJ in equation (i) we hae minimum tor&ue

T "in = .++ sin &//%& 2 /++ sin $&@%. ! " NE %#m

Ma3imum acceleration *

T ma3ma3 =α /

ma3

"!

T =

*" * = "%! & (

( ) /KC

/

×= /rad6sKE1=

Minimum acceleration *or ma3imum retardation, *

T minmin =α /

min

"!

T =

*" *

= "%! & (

( ) /KC

NE

×=

/rad6s/+/=

3;# .he e"uation of the turning moment curve of a three cran*

engine is (5&&& ' 05&& sin ; ,-m! where is the cran* angle in

radians# .he moment of inertia of the flywheel is 0&&& *g-m3

and themean speed is ;&& r#p#m# 1alculate

i# Power of the engine! and

ii .he ma

3 = -++ r%p%"% or

! / L 16E ! 1+K/ rad 6s)olution

0# Power of the engine

&or4 done per re(olution ∫ +=π

θ θ

/

,d+Csin1*C

π θ

θ

/

1

+Ccos1C

−= ! + L %#m

∴ Mean resisting torqueπ

π

π /

+

/

6==

rev


36/48


)ince the resisting torque is constant, therefore the torque e3erted on the

shaft is equal to the mean resisting torque on the flywheel

T = T "ean

C = +C sin 1 ! C

+C sin 1 ! or sin 1 ! 1 ! J or +GJ

! J or EJ

Ma3imum fluctuation of energy, ( )∫ =∆E

mean dT#T θ 4

( )∫ +=E

dC#+Csin1C θ θ ∫ =E

d+Csin1 θ θ

E

1

+Ccos1

−=

θ

R B!+%#m

Ma3imum fluctuation of energy H $m#k . #D.#C E

+ = $+++ '-$%&(& C

+ ! NG /+E C

C = $+++ ) 0@ &$. = +%++$ or +%$H

(ii) 2hen resisting tor&ue is ("*** ? %** sin ) Nm

The turning moment diagram is shown in Fig,

)ince at points 8 and C, the torIJe e3erted on the shaft is equal to the

mean resisting torque on the flywheel

Therefore

C = +C sin 1 ! C = E sin

/C sin 1 ! sin /C *1 sin " K sin1 !sin

*" sin 1 ! 1 sin " K sin1

1 " K sin/ ! K

*Di(iding by /C sin

K

K1sin

/ −=θ !EC

)in!GE/

! C1J or +/E1J

2 ! C1J, and


37/48


[ ]∫ −=∆1+/E

C1

dEsin+Csin1 θ θ θ 4

1+/E

C1

cosE1

1cos+C

+−

= θ θ

! " +ECE %#mMa3imum fluctuation of energy , H $m#k . #D.#C E

+ECE = $+++ '-$%&(& C

+ECE! NG /+E C

C = $./. ) 0@ &$. = +%++ $. or +%$.H

38# .he turning moment diagram for a multi-cylinder engine has been

drawn to a scale of 0 mm to 5&& ,-m tor"ue and 0 mm to 67 of cran*

displacement# .he intercepted areas between output tor"ue curve and

mean resistance line ta*en in order from one end! in s"# mm are

% ;&! ' 80&! % 3B&! ' ;3&! % ;;&! ' 35&! % ;6&! ' 3B&! % 36& s"#

mm! when the engine is running at B&& r#p#m#.he engine has a stro*e of ;&& mm and the fluctuation of

speed is not to e


38/48


Turning moment scale is

+ mm ! C %#m and

cran4 angle scale is+ mm ! EJ! L 61 rad

+ mm/ on the turning moment diagram

! C L 6 1 ! C/1 %#m

Bnergy at A = 4, then reerring to Fig

Bnergy at 8 = 4 2 -+ (=inimum energy

Bnergy at C = 4 2 -+ G $+ ! 4 G -+

Bnergy at D = 4 G -+ 2 &+ ! 4 G $++

Bnergy at 4 = 4 G $++ G -&+ ! 4 G &+ (=a

D = & " or

5 = $ " >

L = @&++ !g)"-

)olution

= & ? + ).+ = % rad)s

)ince the fluctuation of speed is S / of mean speed, therefore total

fluctuation of speed,

+ " / ! K

+ " / ! K


39/48


Ma3imum fluctuation of energy

R 4 =


40/48


Bnergy at 8 = 4 G $.+

Bnergy at C = 4 G $.+ 2 $@& = 4 2 $&

Bnergy at D = 4 2 $& G $. = 4 G $/.

Bnergy at 4 = 4 G $/. 2 $0$ = 4 2 -/ (=inimum energy

Bnergy at F = 4 2 -/ G $0@ = 4 G $.& (=a

a& = &$ $+2/ "&>

a- = / $+2/ "&>

a = $+2/ "&>

Dept.of Mechanical Engg. AAMEC Page 4!


41/48


3 & = 0 r%p%"%>

3 $ = $+& r%p%"%>

L = $/+ !g)"->

N = @%/ 9Pa = @%/ $+. 3)"&

olJtion:

The tJrning "o"ent1cran! angle diagra" or a oJr stro!e engine is shown in Fig%

The areas ;elow the ero line o pressJre are ta!en as negative while the

areas a;ove the ero line o pressJre are ta!en as positive

3et area = a- 2 'a$ G a& G a (

= / $+/ 2 '/ $+/ G &$ $+ 2/ G $+ 2/ (

= /$ $+ 2/ "&

$"& = $ 931"= $ $+. 31" o wor!

Thereore

3et wor! done per cBcle= /$ $+ 2/ $ $+. = @$+ 31"%%%(i)


42/48


/

/+ 3 3 C s−

=/

NG+/ −= = +%+

9axi"J" lJctJation o energB 'E4(

E4 = *%&%C

$+ $.+ = * '$+%@(& +%+

$+$.+ = %-/ *

* = $+$.+ ) %-/ = &-$@ !g1"&

Ei>e of flywheel

7oop stress N=L % v&

@%/ $+. = $/+ v&

N/G+C

+CE

/ =×

=v

m6s 11=vv = ?D3).+

D = v .+)?3 = -+%- .+)? $++

D= /%@. "

9axi"J" lJctJation o energB, E4 = "%v& C

$+ $.+ = " '-+%-(& +%+

$+$.+ = -.%@& "

" = $+ $.+)-.%@& = &@.%@ !g

" = MolJ"e densitB g

A D ρ

π ××= g

t ; D ρ

π ×××=

G+N

+GKGEC/E

K×××××= t t π

&@.%@ = /0%-$+t &

t & =&@.%@) '/0%-$+ ( =%..$+1

t=+%+&$."=&$%.""

; = t = &$%. = .% ""

.3# 0n Gtto cycle engine deelo!s "* k2 at ,"* r#!#m# with -" e7!losions

!er minute# The change of s!eed from the commencement to the end of

!ower stroke must not e7ceed *#"B of mean on either side# Find the

mean diameter of the flywheel and a suitable rim crosssection haing

width four times the de!th so that the hoo! stress does not e7ceed <

4Pa# 0ssume that the flywheel stores ,%," times the energy stored by

the rim and the work done during !ower stroke is ,#


43/48


ince the explosions per "inJte are eIJal to 3)&, thereore, the engine is a

oJr stro!e cBcle engine%


44/48


@-$ = ?D A L

@-$ = ? - t & @&++

@-$ = &@$ .0 t &

t & = -$) &@$ .0 = +%+&

t = +%$@" = $@+ ""

; = t = $@+ = .+ ""Flywheel in Punching Press:

.8# Punching !ress is drien by a constant tor&ue electric motor# The

!ress is !unching !ress is drien by a constant tor&ue electric motor# The

!ress is !roided with a flywheel that rotates at ma7imum s!eed of .."

r#!#m# The radius of gyration of the flywheel is *#" m# The !ress !unches

-.* holes !er hour6 each !unching o!eration takes . second and

re&uires ," kNm of energy# Find the !ower of the motor and the

minimum mass of the flywheel if s!eed of the same is not to fall below

.** r# !# m

9ien:

3 =&&/ r%p%" >

! = +%/ " >

7ole pJnched = @&+ per hr>

4= $/ !31"= $/ $+- 31" >

3=&++ r%p%"%

Eolution:

Power o the "otor

Total energB reIJired per second = 4nergB reIJired ) hole 3o% o holes )

s

=$/ $+- @&+)-.++ = -+++ 31")s

Power o the "otor = -+++


45/48


Power of the motor re&uired

heared area, A =? d t = ? - -& =-&+ ""&

ince the energB reIJired to pJnch a hole is @ 31")""& o sheared area,

thereore total energB reIJired per hole,

4 $ = @ -&+ = &.@+ 31"

Also the ti"e reIJired to pJnch a hole is $+ second, thereore energBreIJired or pJnching wor! per second

= &. @+)$+ = &.@ 31")s

Power o the "otor reIJired = &.@

! = +%. " >

3 $ = -++ r%p%"%

Eolution

$ = &? -++).+ = -$%& rad)s

E!eed of the flywheel immediately after rieting

&= AngJlar speed o the lBwheel i""ediatelB ater riveting

4nergB sJpplied ;B the "otor, 4 & = -!< = -+++ < = -+++31")s

8Jt energB a;sor;ed dJring one riveting operation which ta!es $ second,

4 $ = $+ +++ 31"

∴ 4nergB to ;e sJpplied ;B the lBwheel or each riveting operation per

second or the "axi"J" lJctJation o energB

E4 = 4 $ S 4 & = $+ +++ S -+++ = @+++ 31"

9axi"J" lJctJation o energB 'E4(,

( ) ( )[ ]///+//

+ω ω −××=∆ "! 4

( ) ( )[ ]/

/

//

K/1+E+C/

+ ω −××=

( ) ///NGF/FFIII ω −=& = 0@%& 2 '@+++) &@( = @&

= &.%0 rad)s


D i f M hi U i I F A l i


46/48


Corresponding speed in r%p%"%

3 & = &.%0 .+) & ? = &/@%. r%p%"

Number of riets that can be closed !er minute

ince the energB a;sor;ed ;B each riveting operation which ta!es $

second is $+ +++ 31",

Thereore, nJ";er o rivets that can ;e closed per "inJte

E+

/ ×= 4

4 E

+

1×= =$

.# 0 !unching !ress is re&uired to !unch

4nergB reIJired = . 31")""&>

Ti"e = $)$+ s = +%$ s>

3 $ = $.+ r%p%"%>

3 & = $+ r%p%"%>

! = $"

Eolution:

heared area per hole = ? d %t = ? + $/ = $/ ""&

4nergB reIJired to pJnch a hole 4 $ = . $/ = $$ -$+ 31"

4nergB reIJired or pJnching wor! per second=4nergB reIJired per hole3o% o holes per second

= $$ -$+ -+).+ = /.// 31")s

ince the pJnching ta!es $)$+ o a second,

Thereore, energB sJpplied ;B the "otor in $)$+ second,

4 & = /.// $)$+ = /./%/ 31"

4nergB to ;e sJpplied ;B the lBwheel dJring pJnching a hole or "axi"J"

lJctJation o energB o the lBwheel

E4 = 4 $ S 4 & = $$ -$+ S /./%/ = $+ @%/ 31"

9ean speed o the lBwheel

/

/+ 3 3 3 +

=/

+K+E += =$/+ rp"

9axi"J" lJctJation o energB 'E4(

( )/+/

/

N 3 3 3 "! 4 −××=∆

π

( )+K+E+C+N

C+KK //

−×××= "π

$+ @%/ = --"

" = $+@%/ ) -- = -&@ !g

# 0 !unching machine makes ." working strokes !er minute and is

ca!able of !unching ." mm diameter holes in ,3 mm thick steel !lates

haing ultimate shear strength ** 4Pa# The !unching o!eration takes

!lace during ,,*th of a reolution of the crankshaft#

stimate the !ower needed for the driing motor1 assuming a

mechanical efficiency of 8" !ercent# /etermine suitable dimensions for

the rim crosssection of the flywheel1 haing width e&ual to twice

thickness# The flywheel is to reole at 8 times the s!eed of the

crankshaft# The !ermissible coefficient of fluctuation of s!eed is *#,#

The flywheel is to be made of cast iron haing a working stress

(tensile) of % 4Pa and density of -."* kgm # The diameter of the

flywheel must not e7ceed ,#< m owing to s!ace restrictions# The hub and

the s!okes may be assumed to !roide "B of the rotational inertia of thewheel#

9ien :

n = &/>

d $ = &/ "" = +%+&/ ">


D i f M hi U it I F A l i


47/48


t $ = $ "" = +%+$ " >

= -++ 9Pa= -++ $+. 3)"& >

= 0/H = +%0/ >

C = +%$>

N = . 9Pa = . $+. 3)"&>

L = @&/+ !g)"

-

D = $% " or>

5 = +%@ "

Eolution:

Power needed for the driing motor

Area o plate sheared, A s = ?d $ t $ = ? +%+&/ +%+$ = $$ $+S. "&

9axi"J" shearing orce reIJired or pJnching

J s s A F τ ×=

= $$ $+S. -++ $+. = & &++3

4nergB reIJired per stro!e= Average shear orce Thic!ness o plate

+/+ t F s ××= +GK/K//

+ ××== -$@% 31"

4nergB reIJired per "in=4nergB)stro!e 3o% o wor!ing stro!es)"in

= -$@% &/ = 0/ /+ 31"

Power needed or the driving "otor "η ×

=E

min perrequiredBnergy

NCE

NCKC

×=

= $.@/


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DOM unit-I