CHAPTER – II
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RESULTANT OF COPLANAR NON CONCURRENT
FORCE SYSTEM
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Coplanar Non-concurrent Force System:
This is the force system in which lines of action of
individual forces lie in the same plane but act at different
points of application.
F1
F3
Fig. 1
F2
F2
F1
F5
F3
F4
Fig. 2
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TYPES
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1. Parallel Force System – Lines of action of individual
forces are parallel to each other.
2. Non-Parallel Force System – Lines of action of the
forces are not parallel to each other.
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MOMENT OF A FORCE ABOUT AN AXIS
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The applied force can also tend to rotate the body about
an axis, in addition to causing translatory motion. This
rotational tendency is known as moment.
Definition: Moment is the tendency of a force to make a
rigid body rotate about an axis.
This is a vector quantity.
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Moment Axis: This is the axis about which
rotational tendency is determined. It is
perpendicular to the plane comprising the
moment arm and line of action of the force.
Moment Center (B): This is the point at
A
d
which the moment axis intersects the plane of
the coplanar system.
F
Moment Arm: The perpendicular distance
from the moment center to the line of action
of the force.
Distance AB = d.
B
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EXAMPLE FOR MOMENT
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Consider the example of a pipe wrench.
The force applied which is
perpendicular to the handle of the
wrench tends to rotate the pipe about
its vertical axis. The magnitude of
this tendency depends both on the
magnitude of the force and the
moment arm ‘d’.
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MAGNITUDE OF MOMENT
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It is computed as the product of the
the perpendicular distance to the line
of action of the force from the
moment center and the magnitude of
the force.
MA = d×F
Unit – Unit of Force × Unit of distance
kN-m, N-mm etc.
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B
d
F
A
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SENSE OF THE MOMENT
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The sense is obtained by the ‘Right Hand Thumb’ rule.
‘If the fingers of the right hand are curled in the
direction of rotational tendency of the body, the extended
thumb represents the +ve sense of the moment vector’.
M
M
For the purpose of additions, the moment direction may
be considered by using a suitable sign convention such as +ve
for counterclockwise and –ve for clockwise rotations or viceversa.
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VARIGNON’S THEOREM
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(PRINCIPLE OF MOMENTS)
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Statement: The moment of a force about a moment center
(axis) is equal to the algebraic sum of the moments of the
component forces about the same moment center (axis).
Proof (by Scalar Formulation):
Let ‘R’ be the given force.
‘P’ & ‘Q’ be the component forces
of ‘R’.
‘O’ be the moment center.
p, r, and q be the moment arms of
P, R, and Q respectively from ‘O’.
, , and  be the inclinations of
‘P’, ‘R’, and ‘Q’ respectively w.r.t.
the X – axis.
Y
Ry
Q
R
Qy
Py

A
q r
 
P
p
O
X
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Y
We have,
Ry = Py + Qy
Ry
R Sin = P Sin + Q Sin  ----(1)
Q
From le AOB, p/AO = Sin 
Qy
r

From le AOC, r/AO = Sin 
Py

From le AOD, q/AO = Sin 

From (1),
A
 R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)
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i.e., R × r = P × p + Q × q
Moment of the resultant R about ‘O’ = algebraic
sum of moments of the component forces P & Q
about same moment center ‘O’.
R
q
p
O
P
X
VARIGNON’S THEOREM – PROOF BY VECTOR
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FORMULATION
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Consider three forces F1, F2, and F3
concurrent at point ‘A’ as shown in fig.
Let r be the position vector of ‘A’ w.r.t
‘O’. The sum of the moments about ‘O’
for these three forces by cross-product is,
Mo = ∑(r×F) = (r×F1) + (r×F2) + (r×F3).
By the property of cross product,
Mo = r × (F1+F2+F3)
=r×R
where, R is the resultant of the three
forces.
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APPLICATIONS OF VARIGNON’S
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THEOREM
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1. It simplifies the computation of moments by
judiciously selecting the moment center. The
moment can be determined by resolving a force into X
& Y components, because finding x & y distances in
many circumstances may be easier than finding the
perpendicular distance (d) from the moment center to
the line of action.
2. Location of resultant - location of line of action of the
resultant in the case of non-concurrent force systems,
is an additional information required, which can be
worked out using Varignon’s theorem.
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COUPLE
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Two parallel, non collinear forces (separated by a
certain distance) that are equal in magnitude and opposite
in direction form a ‘couple’.
The algebraic summation of the
F
two forces forming the couple is zero.
Hence, a couple does not produce any
translation, but produces only rotation.
d
F
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Moment of a Couple: Consider two equal and opposite
forces separated by a distance ‘d’. Let ‘O’ be the moment
center at a distance ‘a’ from one
F
of the forces. The sum of moments
d
of two forces about the point ‘O’ is,
a
+ ∑ Mo = -F × ( a + d) + F × a = -F× d
F
O
Thus, the moment of the couple about ‘O’ is independent
of the location, as it is independent of the distance ‘a’.
The moment of a couple about any point is a
constant and is equal to the product of one of the forces
and the perpendicular distance between them.
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RESOLUTION OF A FORCE INTO A
FORCE-COUPLE SYSTEM
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F
F
F
F
d
Q
P
Q
=
=
M=F × d
P
F
Fig. (a)
Fig. (b)
Fig. (c)
A given force F applied at a point can be replaced
by an equal force applied at another point Q, together with
a couple which is equivalent to the original system.
Two equal and opposite forces of same magnitude F
and parallel to the force F at P are introduced at Q.
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Q
=
P
Fig. (a)
F
F
F
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M=F × d
d
=
F
Q
P
F
Fig. (b)
Fig. (c)
Of these three forces, two forces i.e., one at P and the
other oppositely directed at Q form a couple.
Moment of this couple, M = F × d.
The third force at Q is acting in the same direction as that at P.
The system in Fig. (c) is equivalent to the system in Fig. (a).
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Thus, the force F acting at a point such as P on a rigid
body can be moved to any other given point Q, by adding a
couple M. The moment of the couple is equal to moment of
the force in its original position about Q.
RESULTANT OF COPLANAR NON-CONCURRENT
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FORCE
SYSTEMS
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The resultant of coplanar, non-concurrent force
systems is the one which will produce same rotational and
translational effect as that of the given system. It may be a
force and a moment or a pure moment.
Let F1, F2 and F3 constitute
a non concurrent system as
shown in the fig.
‘O’ – be any convenient reference
point in the plane.
F1
O
F3
F2
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F1
F1
F2
∑Mo
O
Fig. A
F3
R
O
Fig. B
F3
F2
∑Mo
O
Fig. C
Each force can be replaced by a force of the same magnitude
and acting in the same direction at ‘O’ and a moment about
‘O’ as in Fig B. The non concurrent forces can be combined
as in the case of concurrent system to get the resultant force
R. Thus, the resultant of the system is equal to a force R at
‘O’ and a moment ‘∑Mo’ as shown in fig.C.
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R
R
∑Mo
d
O
O
Fig. C
 Ry
force RRx and
tan  
Fig. D
The single
the moment ‘∑Mo’ shown in the fig.C
can be replaced by a single force R acting at a distance ‘d’
Mo
from ‘O’, which gives the same effect. Thus,
resultant can
the
d
R
be reduced to a single force acting at a certain distance from
‘O’. Mathematically,
R  Rx 2  Ry 2
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X and Y intercepts of Resultant:
In some problems, it may be required to determine
distances of the resultant R along x-axis and y-axis i.e., X
and Y intercepts. Let ‘d’ be the perpendicular distance of the
resultant from ‘O’ as shown in the fig.
Let Rx=∑Fx and Ry=∑Fy be the
components of the resultant in X and Y Y-axis
directions.
A
By Varignon’s theorem,
Rx
Rx
R×d= ∑Mo
Y
d
R
At B, ∑Mo = Rx×0 + Ry×X
X-axis
O
X
Ry
B
Ry
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Therefore,
X= ∑Mo/Ry
Similarly, at A,
Ry×0 + Rx×Y = ∑Mo
Therefore,
Y= ∑Mo/Rx
Y-axis
A
Y
Rx
Rx
d
R
X-axis
B
O
X
Ry
Ry
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TYPES OF LOADS ON BEAMS
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1. Concentrated Loads – This is the load acting for a very
small length of the beam.
2. Uniformly distributed load – This is the load acting for a
considerable length of the beam with same intensity of W
kn/m throughout its spread.
W kN/m
Total intensity = W × L
(acts at L/2 from one end of the spread)
L
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3. Uniformly varying load – This load acts for a considerable
length of the beam with intensity varying linearly from ‘0’
at one end to W kN/m to the other representing a triangular
distribution. Total intensity of load = area of triangular
W kN/m
spread of the load = 1/2× W × L. (acts at 2×L/3 from
‘Zero’ load end)
L
PROBLEMS FOR PRACTICE
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1. Determine the resultant of the parallel coplanar
force system shown in fig. Take radius of the
circle=1860mm
(Ans. R=2000N towards left, d=626.9mm)
600 N
1000 N
60º
2000 N
10º
o
30º
60º
400 N
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2. Four forces of magnitudes 10N, 20N, 30N and 40N
acting respectively along the four sides of a square
ABCD as shown in the figure. Determine the
magnitude, direction and position of resultant w.r.t. A.
(Ans:R=28.28N, θ=45º, X=1.77a)
20N
D
C
30N
a
A
a
40N
B
10N
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3. Four parallel forces of magnitudes 100N, 150N, 25N
and 200N acting at left end, 0.9m, 2.1m and
2.85m respectively from the left end of a horizontal
bar of length 2.85m. Determine the magnitude of
resultant and also the distance of the resultant from
the left end.
(Ans:R=125N, X=3.06m)
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4. Reduce the given forces into a single force and a
couple at A.
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
70.7kN
200kN
45º
30º
1.5m
A
1m
30º
100KN
80KN
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5. Determine the resultant w.r.t. point A.
(Ans:R=450kN, X=7.5kNm)
150Nm
150N
1.5m
3m
1.5m
A
100N
500N