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Unit-III
ELEMENTS OF CIVIL ENGINEERING AND
ENGINEERING MECHANICS
by
Prof. Karisiddappa, MCE, Hassan
COMPOSITION OF FORCES: The reduction of a given system of forces
to the simplest system that will be its equivalent is called the problem of
composition of forces.
• RESULTANT FORCE: It is possible to find a single force which
will have the same effect as that of a number of forces acting on a
body. Such a single force is called resultant force.
• The process of finding out the resultant force is called composition of
forces.
COMPOSITION OF CO-PLANAR CONCURRENT FORCE SYSTEM
COMPOSITION OF TWO FORCES: It is possible to reduce a given
system of forces i.e., two forces to the simplest system as its equivalent
(resultant force) with the help of parallelogram law of forces.
• LAW OF PARALLELOGRAM OF FORCES:
If two forces, which act at a point be represented in magnitude and
direction by the two adjacent sides of a parallelogram drawn from one of its
angular points, their resultant is represented by the diagonal of the
parallelogram passing through that angular point, in magnitude and
direction.
RR
B C
A O
α
R
θ
αcos2 21
2
2
2
1 FFFFR ++=
RR
B C
A O
α
α
PROOF:
R
θ
D
Consider two forces F1 and F2 acting at point O as shown in
figure. Let α be the angle between the two forces.
Complete the parallelogram ACBO .Drop perpendicular CD to
OA produced. Let R be the resultant force of forces and
.Let θ be the inclination of the resultant force with the
line of action of the force.
From triangle OCD,
+=
+=
=
++=
++=
+++=
++=
====
++=
+=
−
αα
θ
αα
θ
θ
α
α
ααα
αα
αα
cos
sintan
cos
sintan
tan
cos2
cos2
sincoscos2
)sin()cos(
,sin,cos,
)(
21
21
21
2
21
2
2
2
1
2
221
2
1
2
22
2
22
221
2
1
2
2
2
2
21
2
221
222
222
FF
F
FF
F
OD
CD
FFFFR
FFFFR
FFFFFR
FFFR
ROCFCDFADFOA
CDADOAOC
CDODOC
21
0
21
0
21
0
,180
,0
,90
FFR
FFR
FFR
−==
+==
+==
α
α
αIF IF IF
1F
1F
1F
2F
2F
2FR
• TRIANGLE LAW OF FORCES: If two forces acting simultaneously on a body are
represented by the sides of a triangle taken in order,
their resultant is represented by the closing side of
the triangle taken in the opposite order.
O
1F
2F
1F
2F
O A
B
R
• POLYGON LAW OF FORCES: If a number of concurrent forces acting simultaneously on a body ,are
represented in magnitude and direction by the sides of a polygon,
taken in order , then the resultant is represented in magnitude and
direction by the closing side of the polygon, taken in opposite order.
O
1F
2F
3F
4F
O
1F 2F
3F
4F
R
D
A
B
C
1R
2R
COMPOSITON OF FORCES BY
RESOLUTION(Principle of resolved parts)
• The components of each force in the system in two mutually
perpendicular directions are found.
• Then, the components in each direction are algebraically
added to obtain the two components.
• These two component forces which are mutually
perpendicular are combined to obtain the resultant force.
1θ
4θ
2θ
3θ
X
Y
1F2F
3F
4F
Algebraic sum of the components of forces in X
direction
44332211 coscoscoscos θθθθ FFFFFx +−−=∑
Algebraic sum of the components of forces in Y
f direction
Now the system of forces is equal to two
mutually perpendicular forces ,
44332211 sinsinsinsin θθθθ FFFFFy −−+=∑
∑∑ YX FF &
=
+=
∑∑
∑ ∑
−
X
Y
YX
F
F
FFR
1
22
tanθ
NUMERICAL PROBLEMS
1. Determine the magnitude and direction of the resultant of the
two forces of magnitude 12 N and 9 N acting at a point ,if the
angle between the two forces is
GIVEN:
NF 121 = NF 92 = 030=α
0
0
01
21
21
022
21
2
2
2
1
81.12
30cos912
30sin9tan
cos
sintan
3.20
30cos9122912
cos2
=
+=
+=
=
×××++=
++=
−
−
θ
θ
αα
θ
α
FF
F
NR
R
FFFFR
2.Find the magnitude of two equal forces acting at a point with an
angle of 600 between them, if the resultant is equal to N330 GIVEN:
3.The resultant of two forces when they act at right angles is 10 N .Whereas, when they act at a angle of 600 , the resultant is N. Determine the magnitude of the two forces.
Let F1 and F2 be the two forces,
Given – when α =900
R = 10N
When α =600
R = N
We have,
When α =900
Squaring both sides 100= F12 + F2
2 (1)
When α =600
0
21
2
2
2
1 60cos2148 FFFF ++=
5.02 21
2
2
2
1 ×++ FFFF
sayFFF ,21 ==
060,330 == αNR
NF
FR
FFFR
FFFFR
FFFFR
30
3
60cos2
cos2
222
022
21
2
2
2
1
=
=
++=
××++=
++= α
148
148
αcos2 21
2
2
2
1 FFFFR ++=
2
2
2
110 FF +=
squaring both sides
148 = F1
2 + F2
2 (2)
substituting (1) in (2)
148 = 100+F1F2
F1F2 = 48 (3)
squaring equation (3),we get
F12 + F2
2 = 48
2 (4)
From (1) F22
= 100 – F1
2 (5)
Subtracting (5) in (4)
( )
( )( )
NFNF
F
F
F
F
FF
FF
6&8
64
1450
19650
504850
48100
48100
21
2
1
2
1
22
1
2222
1
22
1
4
1
22
1
2
1
==
=
=−
=−
+−=−
−=−
=−
5.The 26 KN force is the resultant of two forces. One of the force is as
shown in figure .Determine the other force.
4.Find the magnitude and direction of the resultant
force for the system of concurrent forces shown
below.
N20
N25
N30
N35
030045
040
X
Y
∑∑
−=
−−=
NF
F
X
X
70.30
40cos3545cos3030cos20 000
∑∑
=
−++=
NF
F
Y
Y
72.33
40sin3545sin302530sin20 000
( ) ( )NR
R
FFRYX
60.45
72.3370.3022
22
=
+−=
+= ∑ ∑
0
1
1
68.47
70.30
72.33tan
tan
=
=
=
−
−
∑∑
θ
θ
θX
Y
F
F
∑ XF
∑ YF
R
θ
0
y 26kN
12 5 10kN 3
4
X
0
Let F be magnitude of unscnorm force with Fx and Fy as its
components in x and y directions.
Component of R in x directions 13 θ1 12
Rx = 26 x cos θ1
= 26 x 5/13 = 10kN 5
Component of R in y direction
Ry = 26 x sin θ1 = 26 x 12/13 = 24kN
Component F and 10kN in X direction
= Fx +10 cos θ2 5
= Fx + 10x 4/5 = Fx +8 θ2 3
4
Component of F and 10kN in y direction
= Fx + 10 x Sin θ2 = Fy + 10 x 3/5
= Fy + 6
Using R/x = /Fx
10 = Fx +8
24 = Fy + 6
Fx = 2kN, Fy = 18kN
But F = √Fx2+Fy
2 = √2
2 + 18
2
F = 18.11kN
θ2 = tan -1
(Fy /Fx) = tan -1
(18/2) = 83.660
θ2 = 83.660 ( inclination of F w.r.t x – axis)
6.Three forces act at a point in a plate as shown in figure. If the
resultant of these forces is vertical, find the resultant force and
angle α..
100N
160N α. 120 N α
0
Since the resultant force is vertical, algebraic sum of horizontal components of these must
be equal to zero.
160 cos α – 120 – 100 sin α = 0
120 + 100 sin α = 160 cos α
6 + 5 sin α = 8cos α
Squaring both the sides
(6+5 sine α )2 = (8 cos α )
2
36 + 60 sin α + 25 sin2α = 64 (1-sin
2 α)
25 sin2 α +64 sin
2 α + 60sin α = 64-36
89 sin2 α + 60 sin α = 28
Sin2 α + 0.674 sin α =0.315
(sin α + 0.337)2
= 0.315 + 0.3372
= 0.428
sin α + 0.337 = √0.428 = 0.654
sin α = 0.654 – 0.337 = 0.317
α = sin-1
(0.317) = 18.50
Resultant force R = Σ Fy
= 160 sin α + 100 cos α
= 160sin 18.50 + 100 cos 18.50
R = 145.60 N
7.ABCDE is a regular hexagon. Forces 90 N,P,Q,240 N and 180 N act along
AB,CA,AD,AE and FA respectively as shown in the figure. Find the forces
P and Q such that the resultant force is zero.
C D
B
P Q E 90N 300 300 240N
300
60
0 30
0
A 180N F X
Since the resultant force is equal to zero, Σ Fx = 0 and Σ Fy = 0
Σ Fx = -180 +240 cos 300 + Q cos 60
0 – p
cos 90
0 + 90 cos 120
0 = 0
-180 + 207.85 + 0.5 Q – 45 =0
0.5Q = 17.15
Q = 34.308N
Σ Fy = 180 sin00+240
sin30
0 + Q sin60
0 – P + 90 sin120
0 = 0
120 + 34.308 x sin 600 – P + 90 sin 120
0 = 0
P = 227.654 N
Moment of force F about O= F x a
= AB x OC
= twice the area of triangle OAB
Thus moment of F about O= 2 x Area of triangle OAB
COMPOSITION OF COPLANAR NON-
CONCURRENT FORCE SYSTEM
MOMENT OF A FORCE: Moment is
defined as the product of the magnitude of the force and
perpendicular distance of the point from the line of
action of the force.
GEOMETRICAL REPRESENTATION OF MOMENT
Consider a force F represented ,in magnitude and direction
by the line AB. Let O be a point about which the moment
of the force F is required. Let OC be the perpendicular
drawn. Join OA and OB
c
FA
B
O
a
VARIGNON’S PRINCIPLE OF MOMENTS:
If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum
of the moments of all the forces about any point is equal to the moment of their resultant
force about the same point.
PROOF:
For example, consider only two forces F1 and F2
represented in magnitude and direction by AB and AC as shown in figure below.
Let O be the point, about which the moments are taken. Construct the parallelogram
ABCD and complete the construction as shown in fig.
By the parallelogram law of forces, the diagonal AD represents, in magnitude and
direction, the resultant of two forces F1 and F2, let R be the resultant force.
By geometrical representation of moments
the moment of force about O=2 Area of triangle AOB
the moment of force about O=2 Area of triangle AOC
the moment of force about O=2 Area of triangle AOD
But,
Area of triangle AOD=Area of triangle AOC + Area of triangle ACD
Also, Area of triangle ACD=Area of triangle ADB=Area of triangle AOB
Area of triangle AOD=Area of triangle AOC + Area of triangle AOB
Multiplying throughout by 2, we obtain
2 Area of triangle AOD =2 Area of triangle AOC+2 Area of triangle AOB
i.e., Moment of force R about O=Moment of force F1 about O + Moment of force F2
about O
Similarly, this principle can be extended for any number of forces.
CO
BA
D
1F
2FR
NUMERICAL PROBLEMS
1.For the non-concurrent coplanar system shown in fig below,
determine the magnitude , direction and position of the resultant
force with reference to A.
R
N35
N25
N50
N20
A
B
C
D
( )
∑∑
−=−−=
→=−=
NF
NF
Y
X
853550
52025
( )↓= N85
( ) NFFR YX 15.858552222 =−+=+= ∑∑
∑ XF
∑ YF
R
θmd
dR
525.215.85
75140
325435
=+
=
×+×=×
or
mx
x
53.285
75140
32543585
=+
=
×+×=×
∑ XF
∑ YF
R
θ
2.Determine the resultant of the force system acting on the plate
as
shown in figure given below wirh respect to AB and AD.
10N 5N
60
0 D 10Nm 30
0 C
3m
A 4m B
14.14N 1 1 20 N
Σ Fx = 5cos300 + 10cos600 + 14.14cos450 = 19.33N Σ Fy = 5sin300 - 10sin600 + 14.14sin450 = -16.16N R = √( Σ Fx2 + Σ Fy2) = 25.2N θ= Tan-1(Σ Fx/ Σ Fy) θ= Tan-1(16.16/19.33) = 39.890 D C y A x θ B R 16.16N Tracing moments of forces about A and applying varignon’s principle of moments we get +16.16X = 20x4 + 5cos300x3-5sin300x4 + 10 + 10cos600x3
Θ 19.33N 19.33N
X = 107.99/16.16 = 6.683m Also tan39.89 = y/6.83 y = 5.586m.
3.The system of forces acting on a crank is shown in figure
below. Determine the magnitude , direction and the point of application of the resultant force.
500 N 150 700N 150
600 600 150mm 150 mm 150 Cos600=75mm Σ Fx = 500cos600 – 700 = 450N Σ Fy = 500sin600 = -26.33N R = √( Σ Fx2 + Σ Fy2) = √(-450)2 + (-2633)2 R = 267.19N (Magnitude) Σ Fx
θ θ= Tan-1(Σ Fx/ Σ Fy) = Tan-1(2633/450) R Σ Fy θ= 80.300 (Direction) ΣFx Θ x R ΣFy
Tracing moments of forces about O and applying varignon’s principle of moments we get -2633x x= -500x sin600x300-1000x150+1200x150cos600 -700x300sin600 X = -371769.15/-2633 X = 141.20mm from O towards left (position).
4.For the system of parallel forces shown below, determine the magnitude of the resultant
and also its position from A .
100N 200N 50N 400N
R
A B C D 1m 1.5m 1m
X
Σ Fy = +100 -200 -50 +400 = +250N ie. R = Σ Fy =250N ( ) Since Σ Fx = 0 Taking moments of forces about A and applying varignon’s principle of moments -250 x = -400 x 3.5 + 50 x 2.5 + 200 x 1 – 100 x 0 X = -1075/ -250 = 4.3m
5.The three like parallel forces 100 N,F and 300 N are acting as shown in figure below. If
the resultant R=600 N and is acting at a distance of 4.5 m from A ,find the magnitude of
force F and position of F with respect to A.
100N F 600 N 300N
A B C D 4.5m 2.5m
X
Let x be the distance from A to the point of application of force F
Here R = Σ Fy 600=100+F+300 F = 200 N
Taking moments of forces about A and applying varignon’s principle of moments, we get 600 x 4.5 = 300 x 7 + F x 200 x = 600 x 4.5 -300 x 7
X = 600/200 = 3m from A
6.A beam is subjected to forces as shown in the figure given below. Find the magnitude ,
direction and the position of the resultant force.
17kN 10kN 20kN 10kN 5kN
α
θ 4kN
A B C D E 2m 3m 2m 1m
Given tan θ = 15/8 sin θ = 15/ 17 cos θ = 8/17 tan α = 3/4 sin α = 3/5 cos α = 4/5
Σ Fx = 4 +5 cos α – 17 cos θ = 4+5 x 4/5 – 17 x 8/17 Σ Fx = 0 Σ Fy = 5 sin α -10 +20 – 10 + 17 sin θ = 5 x 3/5 -10+20 – 10 + 17 x 15/17
Σ Fy = 18 kN ( )
Resultant force R = √ 2Fx 2 + Σ Fy2 = √ 0+182 R = 18 kN ( ) Let x = distance from A to the point of application R Taking moments of forces about A and applying Varignon’s theorem of
moments -18 x = -5 x sin α x 8 +10 x 7 -20 x 5 + 10 x 2
= -3 x 8 +10 x7 – 20 x 5 + 10 x 2
X = -34/-18 = 1.89m from A (towards left)