1. 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.

2. B C
RR
a
R
q
O A
2 cos a 1 2
R = F 2
+ F 2
+ F F
1 2

3. B C
RR
a
O A
α
PROOF:
R
q
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 q be the inclination of the resultant force with the
line of action of the force.

4. From triangle OCD,
= +
2 2 2
OC OD CD
= + +
2 2 2
( )
OC OA AD CD
= = = =
a a
, cos , sin ,
OA F AD F CD F OC R
= + +
a a
( cos ) ( sin )
R F F F
= + + +
a a a
2 cos cos sin
R F F F F F
= + +
R F F F F
2 cos
= + +
R F F F F
CD
OD
F
2
F F
1 2
F
1 2
sin
+
q
tan
=
+
=
=
−
a
a
q
a
a
q
a
a
sin
cos
tan
cos
tan
2 cos
1 2
1 2
2
2
2
1
2
1 2 2
2
1
2
2 2
2
2 2
1 2 2
2
1
2
2
2
2
1 2
2
1 2 2
F F

5. = = +
R F F
= = +
R F F
= = −
1 2
90 ,
0
1 2
0
1 2
0
0 ,
180 ,
R F F
IF a
a
a
IF
IF
1 F
1 F
1 F
2 F
2 F
2 F
R
• 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
1 F
2 F
1 F
2 F
O A
B
R

6. • 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
1 F
2 F
3 F
4 F
O
3 F
4 F
1 F 2 F
R
D
A
B
C
1 R
2 R
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.

7. 1 q
4 q
2 q
3 q
X
Y
1 F
2 F
3 F
4 F
Algebraic sum of the components of forces in X
direction
1 1 2 2 3 3 4 4 F F cosq F cosq F cosq F cosq x = − − +
Algebraic sum of the components of forces in Y
f direction
1 1 2 2 3 3 4 4 F F sinq F sinq F sinq F sinq y = + − −
Now the system of forces is equal to two
mutually perpendicular forces ,
X Y F F
R F F
X Y
= +
=
−
F
Y
F
X
1
2 2
q tan

8. 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:
F 12N 1 = F 9N 2 = 0 a = 30
= + +
R F F F F
= + + × × ×
2 2 0
12 9 2 12 9 cos30
20.3
−
F
1 2
0
0
a
0
1
sin
1 2
1 2
2
2
2
1
12.81
9sin 30
12 9cos30
tan
cos
tan
2 cos
=
=
+
=
+
=
−
R
q
q
a
q
a
F F
R N

9. 2.Find the magnitude of two equal forces acting at a point with an
angle of 600 between them, if the resultant is equal to30 3 N
GIVEN:
= + + a
2 cos
2
2
2
1
R F F F F
1 2
= + + × ×
2 2 0
R F F F F
= + +
R F F F
3
=
R F
30
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,
2
1 R = F + F + F F
When α =900
Squaring both sides 100= F1
2 + F2
2 (1)
When α =600
0
148 = F 2
+ F 2
+ 2F F cos 60
1 2
1 2
2
1 F + F + F F ×
2 0.5 1 2
2
2
F F F, say 1 2 = =
0 R = 30 3N,a = 60
F N
2 cos60
2 2 2
=
148
148
2 cosa 1 2
2
2
2
2
2
1 10 = F + F

10. squaring both sides
148 = F1
2 + F2
2 (2)
substituting (1) in (2)
148 = 100+F1F2
F1F2 = 48 (3)
squaring equation (3),we get
F1
2 + F2
2 = 482 (4)
From (1) F2
2 = 100 – F1
2 (5)
Subtracting (5) in (4)

11. ( )
− =
2 2
1
100 48
2
1
F F
− = −
2 2
1
100 48
4
1
F F
( )
( )
2 2 2 2
1
F
− = − +
2 2
1
F
2
1
F
2
1
F
50 48 50
− =
50 196
− =
50 14
64
=
= =
8 6
F N F N
1 2
4.Find the magnitude and direction of the resultant
force for the system of concurrent forces shown
below.
= − −
F
X
20cos30 30cos45 35cos400 0 0
= + + −
F
Y
20sin 30 25 30sin 45 35sin 400 0 0
= +
2 2
R F FX Y
R
30.70 33.72 2 2
= − +
45.60
=
−
1
F
Y
F
33.72
tan
q
q
5.The 26 KN force is the resultant of two forces. One of the force is as
shown in figure .Determine the other force.
20N
25N
30N
35N
0 30
0 45
0 40
X
Y
= −
F N
X
30.70
=
F N
Y
33.72
( ) ( )
R N
0
1
47.68
30.70
tan
=
=
=
−
q
X
X F
Y F
R
q
0

12. 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 q1 12
Rx = 26 x cos q1
= 26 x 5/13 = 10kN 5
Component of R in y direction
Ry = 26 x sin q1 = 26 x 12/13 = 24kN
Component F and 10kN in X direction
= Fx +10 cos q2 5
= Fx + 10x 4/5 = Fx +8 q2 3
4
Component of F and 10kN in y direction
= Fx + 10 x Sin q2 = Fy + 10 x 3/5
= Fy + 6
Using R/x = /Fx
10 = Fx +8
24 = Fy + 6
Fx = 2kN, Fy = 18kN
But F = ÖFx2+Fy2 = Ö22 + 182
F = 18.11kN
q2 = tan -1 (Fy /Fx) = tan -1 (18/2) = 83.660
q2 = 83.660 ( inclination of F w.r.t x – axis)

13. 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 ..
100N
160N
.
120 N
0
Since the resultant force is vertical, algebraic sum of horizontal components of these must
be equal to zero.
160 cos – 120 – 100 sin = 0
120 + 100 sin = 160 cos
6 + 5 sin = 8cos
Squaring both the sides
(6+5 sine )2 = (8 cos )2
36 + 60 sin + 25 sin2 = 64 (1-sin2 )
25 sin2 +64 sin2 + 60sin = 64-36
89 sin2 + 60 sin = 28
Sin2 + 0.674 sin =0.315
(sin + 0.337)2 = 0.315 + 0.3372
= 0.428
sin + 0.337 = Ö0.428 = 0.654
sin = 0.654 – 0.337 = 0.317
= sin-1 (0.317) = 18.50
Resultant force R = Fy
= 160 sin + 100 cos
= 160sin 18.50 + 100 cos 18.50

14. 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
600 300
A 180N F X
Since the resultant force is equal to zero, Fx = 0 and Fy = 0
Fx = -180 +240 cos 300 + Q cos 600 – p cos 900 + 90 cos 1200 = 0
-180 + 207.85 + 0.5 Q – 45 =0
0.5Q = 17.15
Q = 34.308N
Fy = 180 sin00+240 sin300 + Q sin600 – P + 90 sin1200 = 0
120 + 34.308 x sin 600 – P + 90 sin 1200 = 0
P = 227.654 N

15. 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
Moment of force F about O= F x a
A
= AB x OC
F
c
B
O
a
= twice the area of triangle OAB
Thus moment of F about O= 2 x Area of triangle OAB

16. 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.
C
O
B
A
D
1 F
2 F
R
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.

17. 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
25N
35N
50N
20N
A
B
C
D
( )
= − = ®
25 20 5
F N
= − − = −
X
F N
Y
50 35 85
= 85N(¯)
R F F ( ) N X Y 5 85 85.15 = 2 + 2 = 2 + − 2 =
X F
Y F
R
× = × + ×
R d
35 4 25 3
2.525
140 75
d m q
85.15
=
+
=
or
× = × + ×
85 x
35 4 25 3
2.53
140 75
x m
85
=
+
=

18. 2.Determine the resultant of the force system acting on the plate
X F
Y F
R
q
as
shown in figure given below wirh respect to AB and AD.
10N 5N
600 D 10Nm 300
C
3m
A 4m B
14.14N 1
1 20 N
Fx = 5cos300 + 10cos600 + 14.14cos450
= 19.33N
Fy = 5sin300 - 10sin600 + 14.14sin450
= -16.16N
R = ( Fx2 + Fy2) = 25.2N
= Tan-1( Fx/ Fy)
= Tan-1(16.16/19.33) = 39.890
D C
y
19.33N
19.33N
A x 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

19. 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
Fx = 500cos600 – 700
= 450N
Fy = 500sin600
= -26.33N
R = ( Fx2 + Fy2) = (-450)2 + (-2633)2
R = 267.19N (Magnitude) Fx
= Tan-1( Fx/ Fy)
= Tan-1(2633/450) R Fy
= 80.300 (Direction)
Fx
x
R Fy
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).

20. 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
Fy = +100 -200 -50 +400
= +250N
ie. R = Fy =250N ( ) Since 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 = 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

21. 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
4kN
A B C D E
2m 3m 2m 1m
Given tan = 15/8 sin = 15/ 17 cos = 8/17
tan = 3/4 sin = 3/5 cos = 4/5
Fx = 4 +5 cos – 17 cos
= 4+5 x 4/5 – 17 x 8/17
Fx = 0
Fy = 5 sin -10 +20 – 10 + 17 sin
= 5 x 3/5 -10+20 – 10 + 17 x 15/17
Fy = 18 kN ( )
Resultant force R = Ö 2Fx 2 + Fy2 = Ö 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 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)