Chapter 1 STATIC FORCE ANALYSIS - KTU NOTES

Chapter 1

STATIC FORCE ANALYSIS

Introduction to force analysis in mechanisms – Static force analysis

(four bar linkages only) – Graphical methods

Matrix methods – Method of virtual work – Analysis with sliding and pin

friction.

1.1 STATIC FORCE ANALYSIS OF MECHANISMS

For design of machine components, the following forces are considered

Forces - due to - weight of parts

Forces of assembly

Forces from applied loads

Forces of friction

Inertia forces

Spring forces

Impact forces

Forces due to temperature changes

For static force analysis, the inertia forces due to acceleration are

neglected.

For dynamic force analysis, the inertia forces are taken into account.

In most of the cases, machine component weights are small when

compared to other static forces and hence these forces are neglected in static

force analysis.

1.2 STATIC EQUILIBRIUM

A body is in static equilibrium, if it is in rest and tends to remain at rest.

A body is in static equilibrium, if it is in motion and tends to keep

itself in motion.

The above are true according to Newton’s I law.

The state of equilibrium can be changed by application of external

forces (or) moments.

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In a body to be in static equilibrium, the vectorial sum of all the

forces and moments about any point is zero.

Mathematically, Fx 0; Fy 0;

Mz 0 in two dimensional system

Graphically, the force polygon and couple polygon should be closed.

The following are the conditions for static equilibrium1. A body under action of two forces

will be in equilibrium when the

forces F1 and F2 are same in

magnitude and opposite in direction.

2. A body under the action of three

forces will be in equilibrium, if these forces are concurrent forces and

their resultant is zero.

3. A body under four forces will be in equilibrium if the vector sum of

all forces is zero in such a way that resultant of first two forces.

F1 and F2 and remaining two forces F3 and F4 are collinear as shown

in Fig. 1.3.

F1 F2

Fig. 1.1.

F 1

F 2

F 3

F 2 F 3

F1

Fig.1.2. (a) Body under forces Fig.1.2.(b) Force polygon

F3

F4 F 1

F2

F 1F2

F3F4

O 2

O 1

Fig. 1.3 Body in Equilibrium under the Action of Four Forces

1.2 Dynamics of Machinery - www.airwalkbooks.com

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4. A body under several forces will be

in equilibrium, if the vector sum of

all forces is zero and the vector sum

of all couples and moments is zero.

5. A beam under three or more parallel

forces will be in equilibrium if the

algebraic sum of forces and

moments is zero (Fig. 1.4).

F F1 F2 F3 0

MA F3 l F2 h 0

6. A link under the action of two

forces and an applied couple will

be in equilibrium if the forces are

(a) equal in magnitude (b)

parallel in direction and in

opposite sense and (c) the couple

formed by them (by these 2 forces

F1 and F2) should be equal in

magnitude and should act opposite

to the applied torque (Fig. 1.5).

Equilibrium conditions are:

F1 F2

and Couple; T F1 h F2 h

In static force analysis, the force applied by member 1 on member 2

is represented as F12.

F 1F3

F 2

A B

h

Fig. 1.4 Beam in Equilibrium under Parallel Forces

F = F2 1

F = F1 2

h

T

A

B

Static Force Analysis 1.3

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1.3 FREE BODY DIAGRAMA free body diagram is a diagram of the link isolated from the

mechanism on which the forces and moments are shown in action.

Fig. 1.6 shows a slider-crank mechanism (4 bar mechanism). The

piston is subjected to gas force FG. This gas force is transmitted to crank

shaft to deliver power. The free body diagrams of individual links are shown

in Fig. 1.7.

1.3.1 Consider Piston - Link 4

3 forces are acting on piston.

1. Normal reaction force is acting by cylinder on piston. F14

2. Gas force is acting on piston FG

FG

Piston L ink

P

4

11

32

C

OFixed Link

Crank Connecting Rod L ink 3

Lin k

2

Fig. 1.6 A Slider - Crank Mechanism (4 bar M echan ism )

1

F14

F34

FG

Fig. 1.7 (b) Force Polygon

Closes here

L ink 3

Cylinder L ink 1

P istonL ink 4

L ink 4F34

F 14

P iston

Fig. 1.7 (a) Forces on Piston (Link 4)

G as Force

4 F GFG

Connecting

rod


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3. Force by connecting rod is acting on piston. F34, as these 3 forces are

acting on a concurrent point of piston. The concurrent force pass through

this concurrent point. Hence, a polygon drawn by these 3 forces should

be closed for this piston to be in equilibrium.

1.3.2 Consider connecting rod - Link 3Here, the connecting rod is hinged

at two ends and hence it is a two force

system.

2 forces are acting on connecting rod

hinges.

1. F43 Force by piston (4) is acting

on connecting rod (3)

2. F34 Force by connecting rod (3)

is acting on piston (4).

C

P

F34F 23

F32

F43

Link 3Connecting R od

(c ) Forces on Connecting Rod (Link 3)

F12

F 32

hT

O

C

(d) Forces on C rank (L ink 2)

L ink 2 C

rank

Fig. 1.7 Free Body Diagram s of Various L inks of a Slider-Crank M echanism

L ink 4F34

F14

P iston

Fig. 1.7 (a) Forces on Piston (Link 4)

G as Force

4 FG

C

P

F34F23

F32

F43

Link 3

Connecting Rod

Fig. 1 .7 (c) Forces on Connecting Rod (Link 3)


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Both forces are equal in magnitude but opposite in direction.

F34 F43

And force F43 is acting on crank as F32. So F43 F32

1.3.3 Consider Crank - Link 2Since crank shaft is acted by

F32 (ie F43) at C, the fixed end O is

acted upon by F12 in opposite direction

and F32 F12. Hence the crank is in

equilibrium. But these forces F32 and

F12 are equal and opposite in direction

and hence form a couple.

And this couple is balanced (cancelled) by the torque transmitted

by the crank shaft.

Torque F32 h

Note:For a member under the action of 2 forces and applied torque, to

be in equilibrium, conditions are:

The 2 forces should be equal and opposite.

These 2 forces should form a couple which is equal and oppositeto applied torque.

F12

F 32

hT

O

C

Fig. 1.7 (d) Forces on Crank (Link 2)

Link 2

Cra

nk


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1.4 ANALYSIS WITH SLIDING AND PIN FRICTIONFriction in machine is classified as

1. Sliding friction

2. Pin friction (Turning friction)

1.4.1 Sliding FrictionSliding friction is generated when a link

(piston) is sliding on another link (cylinder). (Fig.1.8). In a sliding friction if gas force is acting

towards left, then friction force Rn will be

acting towards right side. Refer Fig. 1.9.

Normal reaction Rn is acting upward.

Vectorial sum of Rn and Rn is F14.

F14 is acting at an angle of .

is called friction angle and

F14 Rn2 Rn

2

and tan Rn

Rn coefficient of friction

Sliding friction is acting in sliding pair.

Cylinder

P iston

Fig. 1.8

2

3

C

A

1 1

P4

FP4 F

R n

F 34

R n

F 14

Fig. 1.9 Free Body Diagram o f Slider and Force Polygon

P4 F

F34

F34

F14

F=

Fo rce po lygo n closes he re

Piston M ov in g le ft Fric tion force in righ t


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1.4.2 Pin Friction (Turning friction)

The turning pair is used to allow

turning (or) revolving motion between

links.

The friction between the pin and

the links for revolving (or) rotary motion

is called turning friction. In this turning

pair, the friction force does not pass

through the pin centre but it is tangent to the friction circle of the Pin. The

line of action of the force is common tangent to the friction circles of two

pins. It is called friction axis.

Problem 1.1: Determine the torque required to be applied at the crank shaftof a slider-crank mechanism to bring it in equilibrium. The slider is subjectedto a horizontal force of 8000 N and a force of magnitude 2000 N is appliedon the connecting rod as shown in Fig. The dimensions of various links areas under: OA 250 mm, AB 750 mm and AC 250 mm, BOA 40

Fig. 1.101

Crank

Lin

k 2

Connecting RodLink 3

Pin

P=2000 N

60o

40oO

A

B

1

LINK 2

LINK 3

FIXED L INK FIXED L INK

3

2

LINK 4(P ISTO N )C

Fig. Slider-Crank Mechanism

11


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Solution: Graphical Method

P=2000 N

F=8000 N

60o

A

B

LIN K 3

3

h=AC COS 30

= 2 16 .5 mm

F43

n

F 43

r

C

30o

Fig.(b) Free body diagram of LINK3

The fo rce F4 3 is reso lved into 2 com ponen ts-

-R ad ia l com p onen t o f F 43 a long the Link 3

-N orm al com pon en t o f F 43 p erp endicular to the L in k 3

Fr43

F n43

30o

30o=2 16.5 m m

750

2 . D raw fre e body d ia gram o f lin k 3 a s sho w n in T he fo rce is b rokenFig. F 43

P x h = F x ABn43

w here h = AC cos =2 50 x cos

Fo rce: = Fn43

2000 x 216 .5 =5 77.35N

( to BC )r

in to tw o com pon en ts - one a lo ng th e link a nd ano the r perpe nd icu lar to theF r43

link .F n43

3 . Tak in g m o m ents abo ut po in t A ,

P=2000 N

60o

40oO

A

B

1

LINK 2

LINK 3

FIXED LINK FIXED LINK

3

2

LINK 4(P ISTON)C

Fig.(a) Configuration d iagram

11

1. Draw configuration d iagram o f the m echanism as shown in F igu re.


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F =8 0 0 0 N

n

F34r

F 14

LIN K 4(P IS TO N )

F ig . (c ) Freeb ody diagram of L INK 4

P = 2 0 00 N

6 0o

4 0 oO

A

B

1

LINK 2

L INK 3

F IX E D L IN K F IX E D L IN K

3

2LIN K 4(P IS TO N )

C

F ig .(a) C o nfiguration diagram 1

1

4 . O u t o f force s a c ting o n lin k 4, fo rce s an d 4 3 a re kn o w n

F F n co m p le te ly

B y m e a sure m e n t,(1 m m = 1 0 0 N )

F 3 4 = 8 3 2 4 N

w ith su itab le s ca le .

Fo rce 1 4 is p e rp e n dicula r to th e s lid in g su r fa ce a n d force (Figure (c)). F

F r3 4 is a lon g th e lin k 3 . C o ns tru c t a fo rce p o lyg o n as sh o w n in F igure (d )

F14

F34

n

F 34

r

F34=8324 N (By m easurement)

F=8000 N

Fig . (d ) Force polygon fo r Link 4

By m easurement = 8843 N . [See ].F23 Figure (e)

- F orces acting on L IN K 4

5. D raw the force po lygon o f link 3 w ith fo rce s P a nd .F , F43 23

F4 3 =8324 N

F2 3 =8843 N

P=2000N

Fig . (e ) Force polygon fo r Link 3


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PRINCIPLE OF VIRTUAL WORKThe principle of virtual (imaginary) work states that ‘the work done

during a virtual displacement from the equilibrium is equal to zero’. Virtual

displacement is defined as an imaginary infinitesimal displacement of the

system. By applying this principle, an entire mechanism can be examined as

a whole and not as individual links.

Consider a slider-crank mechanism (shown in Fig.1.11) acted upon by

The external piston force F

The external crankshaft torque T and

The force at the bearings.

40oO

A

L INK 2

2F

32 =88 43 N

F = = 8843 N 1 2 32

F

T2

h=

190

mm

F ig: (f) F ree B ody D iagram of L ink 2

6 . D raw the free body d iag ram of lin k 2. M easu re d is tance from andh [Figure (f)]

T2 = F32x h = = 1680.2 Nm clock wise8843 x 0 .19

calcu la te .

2

3

A

B4

11

O

Bearing F orces A cting E qualand O pposite Sense

T

x

N orm al R eaction by

C ylinder on Piston

F

Fig. 1.11 Bearing Reaction Force


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If the crank rotates through a small angular displacement , then

corresponding displacement of the piston is x and the various

forces acting on the system are

Bearing reaction at O which performs no work.

Force of cylinder on piston, perpendicular to piston displacement.

ie Normal reaction which produces no work.

Work done by torque T T

Work done by force F F x

We know that workdone is positive if a force acts in the direction of

the displacement and negative if it acts in the opposite direction.

According to the Principle of Virtual work,

W T F x 0

Since virtual displacement takes place simultaneously during the same

interval t,

T ddt

F dxdt

0

T FV 0

where is the angular velocity of the crank and , the linear velocity

of the piston.

T F

V

The negative sign indicates that for equilibrium, T should be applied

in the opposite direction to the angular displacement.

The Problem 1.1 is solved by using principle of virtual work. First

of all, draw the velocity diagram. Refer Fig.(g).

Assume link OB has rad/s

Velocity of A with respect to O

Va radius OA

Va 0.25


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Draw Va 0.25 250 mm oa r to link OA. Because Va is

perpendicular to OA. [Scale: 100mm 0.1 ]

From a, draw line perpendicular to AB (Length not known to

represent Vba (Velocity of b with respect to a).

From O, draw horizontal line to represent velocity of piston Vb.

o

a

b

c

Paralle l to P(2000 N)

V =26 m mc’

V =203.6 m mb

V=250 m

m

a

Paralle l to F(8000 N)

Pe r

pend

icul

ar to

AB

Pe rpendicu la r to P(2000 N )

Perpend icu lar to OA

Fig.(g) VELOCITY DIAGRAM

P=2000 N

60o

40oO

A

B

1

LINK 2

LINK 3

FIXED L INK FIXED L INK

3

2

LINK 4(P ISTO N)C

Fig.(a) Configuration diagram 1

1

c’


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The above two lines intersect at b.

Measure ob 203.6 mm 0.2036 Vb

ab 198 mm 0.198

Mark c on ab in the same ratio as C divides the AB.

ie acab

ACAB

ac

198

250750 ac 66 mm

Join oc. Now resolve oc into 2 components as

Parallel to force P (2000 N) Vc and

Perpendicular to force P (2000 N)

Measure Vc 26 mm 0.026 (Parallel to P 2000 N)

Using principle of virtual work,

T 2000 0.026 8000 0.2036 0

T 2000 0.026 8000 0.2036

52 1629 1680.8 N m

Problem 1.2: A Slider-crank mechanism as shown in Fig is given below.The force acting on slider is 8000 N and coefficient of friction between allthe links is 0.25. Calculate the driving torque if the pin diameters at jointsO, A and B are 40 mm, 40 mm and 20 mm respectively. The dimensions oflinks are:

OA 200 mm ; AB 800 mm and BOA 60

F=8000 NB

FIXED L INK 1

LINK 4 (PIS TO N )

A

O

23

FIXED L INK 1

60O

SliderC onnecting R od Cra

nk

G IVE N FIG


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Solution:

Friction circle radius:

At point O r1 0.25 402

5 mm

At point A r2 0.25 402

5 mm

At point B r3 0.25 202

2.5 mm

Graphical Method

F=8000 NB

FIXED LINK 1

LINK 4(PISTO N)

A

O

23

FIXED LINK 1

60O

F=8000 N

B

A

O

2

3F32

F12 F34

4

F14

(a)

Slider

FrictionC ircle

Friction ax is

Connecting Rod Cra

nk

Fig. Configuration Diagram

1. Draw con figuration d iagram o f s lider-crank mechanism .

= tan -1 (0 .25) = and draw the resu ltan t side thrust

force 14 inclined at friction ang le.F

2. Draw of different radius a t points (5 m m ), (5mm ) and friction circle O A B

(2 .5m m ).

3 . Draw tangents and decide o f links 2 and 3.friction axis

4. Ca lculate fric tion angle


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Note: In actual slider-crank mechanism, the coefficient of friction is very

low. But in this problem, for easy under standing purpose, an imaginary value

of (-Higher value-) is given to draw friction circles easily.

F34

F32F23

F43

3( c)

F F F F34 43 32 23 = = =

6 . D raw the free body d iagram of link 3 as shown in w ith F igure (c)

F=8000 N

F34=7800 N

F14=1050 N

(b)

5. Consider various forces at the slide r and draw the force po lygon w ith su itab le

F F34 14 = 7800 N and = 1050 N

scale (100N = 1mm ) as shown in . F ind the fo rces and byFigure (b) F F34 14

m easurem ent.

A

O

F 32

F12

h = 81.2 m m

A

O

F32

F1 2

T2

(d) (e)

2

7. M ark forces and at the crank as show n in and andF F32 12 Fig. (a) F ig (d) (e).

T2 = F32 x h

= 7800 x81.2

1000

= 633.36 N m (coun ter clock w ise )

8 . M easure d is tance h = 81.2 m m

Torque:


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Problem 1.3: A four-link mechanism with the following dimensions is actedupon by a force 80 N at 140 on the link DC [Fig.(a)].AD 250 mm, AB 250 mm, BC 500 mm, DC 375 mm, DE 175 mm.. Determinethe input torque T on the link AB for the static equilibrium of the mechanismfor the given configuration.

Solution: If the mechanismis in static equilibrium, theneach of its members shouldalso be in equilibriumindividually.

Link 4 is acted upon by

three forces F, F34 and F14.


two forces F23 and F43.


two forces F32 and F12 and a

torque T.

A

E

C

B

D

140 O

120 O

2

1 1

3

4

F=80 N

(a) Given Configuration

C

B

3

F43

F23

(b)

1 . Fo rce on the link is know n (as 80 N ) com pletely. To know the other tw o

F 4

forces ac ting on th is m em ber com pletely, the direc tion o f one m ore force

m ust be know n. To know tha t, the should be cons ide red firs t w h ich is a link 3

tw o -fo rce m em ber.

2 . As the link 3 is a tw o-fo rce m em ber , fo r its equilib rium , and (Fig.(b )) F F23 43

m ust ac t a long Thus , the line of ac tion of is also a long B C . F B C .34


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E

C

D

140O

4

LOA of F34

LOA = LINE OF ACTION

LOA

of F14

F=80 N

( C )To find LO A of F34 and F14

3. .D raw a link Re fer As the

DC

Fig.(c)

fo rce acts through the point F34 C on

the link 4, d raw a lin e parallel to BC

th roug h . Extend fo rce to in tersectC F

O . Now, as the link 4 is a three-force

at mem ber, the third force should F14

pass th rough the intersection O o f F

and as the three forces are to be F34 concurren t fo r equil ibrium of the link

[Fig. (c)].

O

F34 = 42 N

By m

easuremen t F14 = 66 N

F=80 N

SCALE: 1 m m = 1 N

By measurement

(d)

E

C

D

140O

4

LOA of F34

LOA

of F14 F=80 N

( C )To find (LO A) o f F34 and F14

Line Of Action

m agn itudes of F14and can be found out.F34

From fo rce polygon, F34 = 42 N

F F F F34 43 23 32 = = = = 42 N

By draw ing a fo rce polygon for link 4,

(Take scale 1mm = 1N) ( is complete ly known),F


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Matrix Method:First of all, the angular inclinations of the links BC and DC i.e., angles

and are to be determined by drawing the configuration diagram (Fig.(a))

Position vectors:AB 0.25 at 120, BC 0.5 at 17, DC 0.375 at 72, DE 0.175 at 72

The direction of F34 is along BC since it is a two-force member,

F34 F34 at 17

Since the link DC is in static equilibrium, there are no resultant forces

and summation of moments acting on it is zero. Taking moments of the forces

about point D.

MD F4 DE F34 DC 0 ...(i)

Moments are the cross-multiplication of the vector, so it is done in

rectangular coordinates.

F4 80 cos 140 i 80 sin 40 j 61.28 i 51.42 j

A

B

2

T

F12

F32

h=240 mm

(e)

L ink 2 w ill be in equilib rium if is equa l, pa ralle l and opposite to [F ig. (e)] F F12 32

T F h = x = 42 x 0 .240 = 10 N m antic lock w ise32

The inpu t torque has to be equa l and opposite to th is couple, i.e.,

T = 10 Nm (clockwise)

and


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DE 0.175 cos 72 i 0.175 sin 72 j 0.054 i 0.166 j

F34 F34 cos 17 i F34 sin 17 j F34 0.956 i 0.292 j

DC 0.375 cos 72 i 0.375 sin 72 j 0.1159 i 0.357 j

Inserting the values of vectors in equation (i), we get

MD 61.28 i 51.42 j 0.054 i 0.166 j F34 0.956 i 0.292 j

0.1159 i 0.357 j 0

Assembling in matrix form,

i 61.280.054

j

51.420.166

k00

i0.956 F34

0.1159

j0.292 F34

0.357

k00

0

61.28 0.166 51.42 0.054

0.956 F34 0.357 0.292 F34 0.1159 0

12.95 0.307 F34 0 (or) F34 42 N


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Thus, F34 42 at 17

Now,

F32 F23 F43 F34 42 at 180 17 42 cos 197 i 42 sin 197 j

40.16 i 12.28 j

AB 0.25 cos 120 i 0.25 sin 120 j

0.125 i 0.2165 j

T2c F32 AB

T2c

i 40.16 0.125

j

12.280.2165

k00

40.16 0.2165 12.28 0.125 10.22 N m counterclockwise

Thus input torque 10.22 N m clockwise

Principle of Virtual Work

Assume link AB has rad/s

Velocity radius

Vb 0.25

Draw Vb 0.25 25 mm ab r, to link AB.

Refer Fig.(V) [Scale 10 mm 0.1 ]

(Because Vb will be perpendicular to link AB)

Also mark d nearer to a (as both are fixed link (1))

From a draw line, r to link DC from d. (Length not known)

And draw another line, r to link BC from b (Length not known).

Both of the above lines intersect at c.

Locate e in dc in the same ratio as E divides DC in configuration

diagram.


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a,d

b

c

e

Paralle l to F (80 N)

Perpend icular to B

C

Perpendicula r to AB

Perpendicular to DC

e ’ Perpendicu la r to F (80 N )

V’ = 12 .3 mm

e

V = 0.25=25 mm

b

VELOCITY DIAGRAM

A

E

C

B

D

140O

120O

2

1 1

3

4

(a)

F=80 N

17O

O

By m easurem ent

By m easurem ent

MATRIX METHOD

Fig. (V)


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ie dedc

DEDC

de29

175375

de 13.53 mm

Join ae Now resolve ae into 2 components as

– parallel to force F ( 80 N) and

– perpendicular to force F (80 N)

Now measure

Ve 12.3 mm 0.123 (Parallel to F (80 N))

Assume T as counterclockwise , we can apply

Principle of Virtual work

T 80 0.123 0

T 80 0.128 10.24 N m

T 10.24 N m clockwise

Problem 1.4: A four bar mechanism as shown in Fig. is subjected to twoforces, F3 2000 N at 60 from horizontal at mid point of link 3 and

F4 4000 N at 45 from link 4 at mid point of link 4. The dimensions of

links are as under: AB 0.3 m, BC 0.4 m, CD 0.45 m and AD 0.6 mPerform static force analysis and determine resisting torque on link 2.

Solution:This type of problem can be

solved by Principle of superposition. ie,

Net effect is equal to superposition of the

effect of individual loads taken one at a

time.

A

B

C

D

F3

F 4

90o

45o

60o

1

2

3

4

1Given Figure


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A

B

C

D

60O

45O

(L ine o f act ion)

Direc tion o f F43

(Line of a

ction)D

irectio

n of F23

B

C

Both are parallel

F3 = 2000 N

F4 = 4000 N

F3 =

200

0 N

(a)

(b)

e

f

3

4

2

11

Neglecting Force F4

1. Draw the configura tion d iagram of m echan ism as

2.

Consider the effect of force , neglecting force . .F F3 4

shown in Figure. (a ).

3. Draw the free body diagram of link 3 and find the (line of action (LOA)) direction

of forces and F F23 43 [F igure (b)].


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S ca le : 1 00 0 N = 1 0 m m 1 m m = 1 0 0 N

F3 =

20 0

0 N

(20

mm

)

By m

eas urement

73 5 N

F43 =

By measurem en t

1675 N

F23 =

( c )

B y m e a sure m e nt,

F F43 23 = 7 3 5 N a nd = 1 67 5 N

B y u s in g the se d ire c tio n s ,

4 . C o nstru c t fo rce po lyg o n [F ig u re (c )] w ith

s u ita ble sca le for L in k 3 (1 0 00 N = 1 0 m m )

M ea su re th e m ag n itu d e o f fo rce s a n d .F F43 23

w e ca n d raw fo rce po ly go n for L in k (3 )

A

B

F32

F12

T23

h1 = 234 mm

(d)

2

5. Draw free body diagram of link 2 . Measure d istance 1 = 234 mm . .

[Figure.(d)] h

T F h23 23 1 = x = 1675 x 2 341 00 0

= 392 Nm (counterclockwise).

Torque due to fo rce ,F3


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A

B

C

D

60O

45O

F3 = 2000 N

F4 = 4000 N

C

D

45O

F4 = 4000 N

Both are parallel

3

2

4

F23

F43

LO A o f F34

LO A = L INE O F ACTIO N

L OA

of F14

(a )

(e)

4

Repeated For Reference

Neglecting Force F3

6 . Now conside r the e ffec t o f force neg lecting the

F4

e ffec t o f force . F3

7 . Draw the free body d iagram of link 4 and find the

LOA o f fo rces and F F34 14 [F igure.(e)].


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F4

= 40

00 N

(4 0

mm

)

By m

easur ement F 14 = 355 0

F34=1345

(f)

By m easurement

8. C ons truct fo rce polygon for link 4

[Figure.(f)]

F F14 34 = 3550 a nd = 1345 N

w ith su itab le scale (1 m m = 100 N ). M easure the

m agn itud e o f fo rces and . F F14 34

B

C

F23

F43

(g)

3

A

B

C

D

60O

45O

F3 = 2000 N

(a)

e

f

3

4

2

11

F4 = 4000 N

F43

F23

R epea ted For R efe rence

F F F F34 43 23 32= = =


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Principle of Virtual Work

Assume link AB has an instantaneous angular velocity of rad/s

counter clockwise.

Velocity radius

Hence Vb 0.3 [ AB 0.3 m]

By knowing magnitude and direction of Vb and knowing the direction

of velocity Vcb and Vcd, velocity diagram can be drawn.

Draw Vb 0.3 30 mm horizontally as ab. (Perpendicular to link

AB). Refer Fig. (V).

Mark d nearer to a (as both are fixed link 1)

Draw line r to link DC from d and draw another line r to Link

BC from b. Both lines intersect at c.

Now mark point e as mid point of bc and mark f as mid point

of dc.

Join ae.

Then joint df

A

B

2

F23

h2 = 290 m

m

T24

(h)

F12

10. D raw the free body diag ram o f lin k 2

Torque : = x T F h24 32 2

= 1675 x 290100 0

= 486 N m (cou nte rc lockw ise )

Tota l res is ting to rque :

T T T2 23 24 = +

= 392 + 486 = 878 N - m

= 878 N m (cou nte rc lockw ise )

[Figure.(h)]. M easure the d is tance

h2 = 2 90 m m .


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Now resolve ae into 2 components parallel to force F3 Ve and

perpendicular to F3

Measure Ve 0.21 21 mm by measurement

a,d b

c

ef

Pe rpendicu lar to DC

V’ =

21

mm

eV = 0.3 = 30 m mb Perpend icular to AB

Perpen d ic u la r to B

C

P aral

lel t

o F 3

(20 0

0 N

)

Pa r

a lle

l to

F4 (

4 000

N)

V’ =

11

mm

f

f’

e ’

A

B

C

D

6 0O

4 5 O

F3 = 200 0 N

(a)

e

f

3

4

2

11

F4 = 400 0 N

Repeated For Reference

Fig. (V) Velocity Diagram


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Similarly, resolve df into 2 components.

Parallel to force F4 Vf and perpendicular to F4

Measure Vf 0.11 11 mm by measurement. From velocity

diagram

Parallel to F3 Ve 0.21 (ie 21 mm) and

Parallel to F4 Vf 0.11 (ie 11 mm)

Assume T as counterclockwise (positive) and apply principle of virtualwork.

T F3 0.02 F4 0.11 0

T 2000 0.21 4000 0.11 0

T 860 N m

Problem 1.5: For the mechanism shown in Fig., determine the torque onthe link AB for the static equilibrium of the mechanism.

Solution:

If the mechanism is in static equilibrium, each of its members should

also be in equilibrium individually.

Member 4 is acted upon by three forces F1, F34 and F14

Member 3 is acted upon by three forces F2, F23 and F43.

Member 2 is acted upon by two forces F32 and F12 and a torque

T.

Graphical Solution by Superposition Method(Fig. b and c) Neglecting force F2


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22.510

20

30

A

B

C

D50

E

CE=10

CD=30 F1= 40 N

EE F2 = 80 N

50O

B

C

F43

F23

2

3

4

1 1

3

NEGLECTING F2

(a) Given Figure

(b) Equilibrium of link withtwo forces - Neglecting F2

1 . L ink 4 is a three -force m em ber in

C

D

E

F1= 40 N

LO A o f F34

LOA

of F14

4

O

( c) LOA- Line Of Action

w hich on ly one fo rce is known.F1

2 . How ever, the line o f action (LO A) of

F 34 can be obta ined from the

equ ilibr ium of the link 3 w h ich is a

two-force m em ber and is acted upon

by forces and . Thus, lines o fF F23 43

action o f forces and a re alongF F43 34

. If and intersect at , thenBC OF F1 34

l ine o f action of w ill be a long F14 O D

s ince the th ree forces a re to be

concurren t.


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F1= 40 N (40 mm

)

F34=8 NBy measurement

By m

easurement F14=38 N

3 . D raw the force polygon ( is com ple te lyF 1

F F34 14 = 8 N and = 38 N

Also

F F F F34 43 23 32 = = = = 8 N

known) and ob ta in the m agnitudes of forces

F F34 14 and .

4. The direction of is opposite toF32

F23. that of

A

B

2n

F 2 = 8 N3

F 2 = 8 N1

h1 = 13 mm

T1

(e)

5 . Link 2 is sub jec ted to tw o forces

and a

T F h1 32 1 = x = 8 x 13 = 104 N .m m

(clockw ise)

to rque . Fo r equ ilib rium , is equa l, T F1 12

para lle l and oppos ite to .F32


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B

C

E F2 = 80 N

LOA

o f F43

3

LOA

of F23

O

(g)

7. L ink 3 is a th ree-force m em ber in wh ich is com plete ly known , only Fig.(g ). F2

the direction o f is know n (para llel to ) and is com ple te ly unknow n.F F43 23 DC

If the line of F F43 23 action o f and m eet at , then the line o f action of w ill be F2 O

a long as the three O B fo rces are to be concurrent.


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F2 = 80 N

By m

eas urement F 43 = 34 N

By m easurem ent F23 = 73 N

(h)

D raw the fo rce p olygon ( is com ple te ly know n) by tak in g to a suita ble scaleF F2 2

and tw o lines para llel to line s of a ct ion o f an d . M a rk arrow hea ds o n F F F23 43 23

and F43to know th e directions . R e fe r F ig .(h)

F F23 43 = 73 N an d = 34 N

F F 23 32 = = 73 Nand hen ce

Th e direction o f is opp os ite to tha t o f F F32 23 .

A

B

2

h2 =11 .5 mm

F32 = 73 N

F12 = F32 =73 N

T2

(i)

L ink 2 is subjected to tw o forces

( and ) and a To rque F F12 32 T2.

R efer F ig.( i): = = 73 NF F12 32

For equ ilib rium , is equal, para llelF 12

and opposite to .F32

T F h2 32 2= x = 73 x 11 .5 = 839 .5

N .m m c lockw iseTota l to rque

= 104 + 839.5 = 943 .5 N .m m


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Documents

Chapter 1 STATIC FORCE ANALYSIS - KTU NOTES