chapter 1 static force analysis - ktu notes
TRANSCRIPT
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
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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)
Static Force Analysis 1.5
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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
Static Force Analysis 1.7
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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.
Static Force Analysis 1.9
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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
Static Force Analysis 1.11
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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’
Static Force Analysis 1.13
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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
Static Force Analysis 1.15
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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.
Link 3 is acted upon by
two forces F23 and F43.
Link 2 is acted upon by
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
Static Force Analysis 1.17
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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
Static Force Analysis 1.19
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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.
Static Force Analysis 1.21
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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
Static Force Analysis 1.23
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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
Static Force Analysis 1.25
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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= = =
Static Force Analysis 1.27
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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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