When the condition of body is unaffected even though it is acted upon by no. of forces, it is said to be in equilibrium. this PPT is about the general idea about Equilibrium of Forces. I will be uploading the numerical on it soon.

1. Prepared by : Prof. V. V. Nalawade
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4. Definition of Equilibrium
– Stable Equilibrium
– Unstable Equilibrium
– Neutral Equilibrium
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6. Unstable Equilibrium
• A body is said to be in unstable equilibrium if,
– An additional force is set up on slight
displacement which tends to push it away from
the original position of the body.
– It does not return back to its original position
after being slightly displaced by a force.
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7. Neutral Equilibrium
• A body is said to be in neutral
equilibrium if,
–It is initially in a state of static equilibrium.
–No additional force is set up on slight
displacement from initial position.
–It occupies a new position and remains in
static equilibrium .
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8. Equilibrant
• It is the force which,
when applied to a body
acted by the concurrent
force system keeps the
body in equilibrium.
• Equilibrant is always equal
in magnitude and opposite
in direction & collinear to
resultant force.
Fx
Fy
R
E
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9. Analytical conditions of Equilibrium
• We know that magnitude of resultant of
forces is given by,
• Equilibrium means resultant force acting on
the body is ZERO, i.e. R=0
• Therefore, ΣFx = Σfy = 0
• Also the varignon’s theorem of moments
gives the position of the resultant,
• i.e. , ΣM = R* x but as R= 0; M = 0
2 2
x y
R F F
  
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10. Analytical conditions of
Equilibrium
• And hence the analytical conditions of
equilibrium are;
1. ΣFx = 0
2. Σfy = 0
3. ΣM = 0
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11. 1) Equilibrium of collinear force system
ΣF = 0
2) Equilibrium of concurrent force system
ΣFx = 0, ΣFy = 0
3) Equilibrium of parallel force system
ΣF = 0, ΣM = 0
4) Equilibrium of general force system
ΣFx = 0, ΣFy = 0, ΣM =0
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Analytical conditions of Equilibrium
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12. Free Body Diagram (FBD)
• FBD is a sketch of the body showing all active
and reactive forces acting on a body.
• Active forces means applied forces and
weight of the body
• Reactive forces means reactions offered by
floor, wall, strings, cables, supports etc
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13. Importance of FBD
• The sketch of FBD is the key step that
translates a physical problem into a form that
can be analyzed mathematically.
• The unknown forces of equilibrium of each
body can be obtained very easily.
• FBD represents all active and reactive forces.
• All equations of equilibrium can be applied to
each FBD separately.
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14. 1. Draw a neat sketch of the body assuming that all
supports are removed.
2. FBD may consists of an entire assembled structure or
any combination or part of it.
3. Show all the relevant dimensions & angles on the
sketch.
4. Show all the active forces on corresponding point of
application & insert their magnitude & direction if
known.
5. Show all the reactive forces due to each support.
6. If the sense of reaction is unknown, it should be
assumed.
7. Use principle of transmissibility wherever convenient.
Procedure of drawing FBD
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15. • Example 1:
1. Consider an electric lamp suspended by an
electric wire as shown in fig(a),
 it is separated from its surrounding as shown in
fig(b).
 In fig (c), active force (W) & reactive force (T),
acting on a free body are shown. Therefore fig
(C) is the FBD of fig (a).
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16. • Example 2:
1. A sphere suspended by a string but resting
against the wall as shown in fig(a),
 it is separated from its surrounding as shown in
fig(b).
 In fig (c), active force (W) & reactive force (TBC ,
RA), acting on a free body are shown.
(a) (b)
(c)
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17. • Example 3:
1. A sphere resting in a box as shown in fig(a),
 it is separated from its surrounding as shown in
fig(b).
 In fig (c), active force (W) & reactive force (RA &
RB =Reactions at contact points A & B), acting on
a free body are shown.
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18. • Example 4:
1. A block resting on horizontal floor as shown
in fig(a),
 it is separated from its surrounding as shown in
fig(b).
 In fig (c), active force (W) & reactive force (R
=Reaction offered by floor), acting on a free body
are shown.
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19. • Example 5:
1. A ladder having weight W resting against
rough horizontal surface & smooth vertical
wall as shown in fig(a),
 In fig (b), active force (W) & reactive force (RA,
RB & F), acting on a free body are shown.
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21. • Example 8:
1. A ladder having weight W resting against
smooth horizontal surface & smooth vertical
wall as shown in fig(a),
 In fig (b), active force (W) & reactive force (RA,
RB & F), acting on a free body are shown.
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22. Lami’s Theorem
• If three forces acting at a point on a body keep
it at rest, then each force is proportional to
the sine of the angle between other two
forces.
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23. Proof-I
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24. Proof-II
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25. 1. The theorem is applicable only if the body is
in equilibrium.
2. The theorem is not applicable for more or less
than three coplanar concurrent forces.
3. The theorem is applicable only if the nature of
all three forces are same, i.e. all forces are
either push or pull.
4. The theorem is not applicable for parallel or
non-concurrent forces.
5. The theorem is not applicable for non-
coplanar forces.
Limitations of Lami’s Theorem
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26. Problems on Lami’s Theorem
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27. • Problem 1:
 Find the tension in each
rope in fig.
• Solution:
Given Data: 1) Θ = Tan-1 4/3 = 53.13ᵒ
2) w =100 kg =10 x 9.81 = 981N
Step i) Consider the FBD of Point C.
θ1
θ2
θ3
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28. General Calculation :
Angle between W and TBC = 90 – 30 = 60ᵒ
Angle between TBC & TAC = 30 + 90 + (90 - 53.13)
=156.87ᵒ
Angle between TAC & W = 53.13 + 90 =143.13ᵒ
Step ii) Applying Lami’s Theorem,
θ1
θ2
θ3
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29. Types of
Support
Representa
tion
Reaction Reaction Force
Roller Vertical Reaction
Hinged Vertical as well
as Horizontal
Reaction
Fixed Vertical,
Horizontal
Reaction &
Moment
RA
RA
HA
RA
HA
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30. Types of
Support
Representati
on
Allowed Movements
Vertical Horizontal Rotation
Roller NO YES YES
Hinged NO
NO YES
Fixed NO NO NO
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31. Roller Support
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32. Hinge Support
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33. Fixed Support
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34. Load
Point Load
Distributed
Load
Uniformly
Distributed
Load
Uniformly
Varying Load
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38. How to convert UDL in point load
• Intensity of UDL is in w
• We convert it into a point load by following
formula;
 Where, w= intensity of UDL
 l = span of UDL
 And it is acting on a distance
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39. • A UDL is replaced by concentrated point
load.
• The magnitude of this equivalent point load
= the area under loading diagram and it
acts through the centroid.
• i.e. W = w * l 40
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43. How to convert UVL in point load
• Intensity of UVL is in w
• We convert it into a point load by following
formula;
 Where, w= intensity of UVL
 l = span of UVL
 And it is acting on a distance
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44. • A UVL is replaced by concentrated point
load.
• The magnitude of this equivalent point load
= the area under loading diagram and it
acts through the centroid.
• i.e. W = 1/2 * l * w 45
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49. Simply Supported Beam
Simply Supported Beam With overhang
Cantilever Beam
Continuous Beam
Beams linked with internal hinges
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50. • Supported by a hinge at one end and roller at
the other end.
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51. • Beam projected beyond the supports at one
or both the ends.
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52. • A beam fixed at one end and free at other end
is called as cantilever beam.
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53. • A beam which has more than two supports is
called as continuous beam
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54. • When two or
more beams are
connected to
each other by
pin joint and
continuous
beam is
formed. Such a
joint are called
internal hinges.
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56. Problems on Support
reaction of beams
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