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EQUILIBRIUM OF NON-CONCURRENT
COPLANAR FORCE SYSTEM
When a body is in equilibrium, it has neither translatory
nor rotatory motion in any direction. Thus the resultant
force R and the resultant couple M are both zero, and we
have the equilibrium equations for two dimensional forcesystem
Fx = 0;
Fy = 0 Eq(1)
M = 0
These requirements are both necessary and sufficient
conditions for equilibrium.
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Supports: A structure is subjected to external forces
and transfers these forces through the supports on to thefoundation. Therefore the support reactions and the
external forces together keep the structure in equilibrium.
There are different types of supports. Some of them are
a) Roller Support b) Hinged or pinned support c) Fixed
or built in support
Some supports are shown in the figure along with the
reactions that can be mobilised.
Types of supports
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Types of Supports Action on body
(a) Flexible cable ,belt ,chain,
rope
BODY
BODY
T
Force exerted by cable is
always a tension away from
the body in the direction of
cable
(b) Smooth surfacesContact forces are normal to
the surfaces
FF
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(c) Roller
supportContact force is normal to thesurface on which the roller moves.
The reaction will always be
perpendicular to the plane of the
roller . Roller support will offeronly one independent reaction
component.(Whose direction is
known.)
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( d )pinned Support / hinged support
This support does not allow any translatory movement of
the rigid body. There will be two independent reaction
components at the support. The resultant reaction can be
resolved into two mutually perpendicular components.
Or it can be shown as resultant reaction inclined at an
angle with respect to a reference direction.
R vR
Rh
θ
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R v
MR H
(e) Fixed or Built-in Support
M
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(contd .)
This type of support not only prevents the translatory
movement of the rigid body, but also the rotation of the
rigid body. Hence there will be 3 independent reactioncomponents of forces. Hence there will be 3 unknown
components of forces, two mutually perpendicular
reactive force component and a reactive moment asshown in the figure.
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TYPES OF BEAMS
A member which is subjected to predominantly transverse loads
and supported in such a way that rigid body motion is preventedis known as beam. It is classified based on the support
conditions. A beam generally supported by a hinge or roller at
the ends having one span(distance between the support) is
called as simply supported beam. A beam which is fixed at oneend and free at another end is called as a cantilever beam.
(a) Simply supported beam
span
MA
VA
B
span
AHA
(b) Cantilever beam
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If one end or both ends of the beam project beyond the supportit is known as overhanging beam.
A cantilever with a simple support anywhere along its length is
a propped cantilever.
(c) Overhanging beam
(right overhang)
MA
VA
B
span
A
HA
(d) Propped Cantilever
beam
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A beam which is fixed at both ends is called a fixed beam.A beam with more than one span is called continuous beam.
VC
(f) Two Span continuous beam
VA VB
HA HB
MA
VA
span
VB
(e) Fixed beam
MB
HA
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Statical ly determinate beam and statical ly indeterminatebeam :
Using the equations of equilibrium given in EQ(1) ,if all the
reaction components can be found out, then the beam is a
statical ly determinate beam ,and if all the reaction
components can not be found out using equations of
equilibrium only, then the beam is a statical ly indeterminate
beam.
In the above fig (a),(b)and ( c ) are statically determinate
beams ,where as (d),(e) and ( f) are statically Indeterminate
beams .
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If the number of reaction components is more than thenumber of non-trivial equilibrium equations available then
such a beam is a statically indeterminate beam.
If the number of reaction components is equal to the
number of non-trivial equilibrium equations available then
such a beam is a statically determinate beam
If the number of reaction components is less than the
number of non-trivial equilibrium equations available then
such a beam is an unstable beam.
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Determination of Beam reactions
Since three equilibrium equations are available, for a planar
structure a maximum of three unknown independent reactioncomponents can be determined using these equations.
Step I: Draw the free body diagram of the structure showing
the given loadings and the reactions at the supports.
Step 2: Apply the equations Fx = 0, Fy = 0, M = 0.
Assuming some directions and senses for unknown forces
and moments.
Step 3: solve for unknown reactions. If any of them is positive,
it is along the sense initially assumed while drawing the FBD.
If it is negative, it is opposite to the initially assumed sense
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4kN/m
12kN/m
(1)Find the reactions at A,B,C and D for the beam loadedas shown in the figure(Ans.R A=R B =34kN;R C=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
Problems for practice
12kN/m
4kN/m
20 kN
30kN
1m 2m 1m 1m 2m 1m 1m 2m
AB
C
34
40kNm
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(2)A uniform bar AB of weight 50N shown in the figuresupports a load of 200N at its end. Determine the
tension developed in the string and the force supported
by the pin at B.(Ans. T=529.12N;R B=807.15N, θB=64.6˚)
2.5m2.5m200N
2 . 5 m
A
B
60˚
string
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(3)Find the position of the hinged support (x),such that thereactions developed at the supports of the beam are equal..
(Ans.x=2m.)
2.0m 1.4m1.0m 3.0m0.6
15kN18kN/m
10kN/m
x
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(4)A right angled bar ABC hinged at A as shown in figcarries two loads W and 2W applied at B &C .Neglecting
self weight of the bar find the angle made by AB with
vertical(Ans:θ =18.44˚)
0.5L2W
θ
B
L m
WC
A
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