Resultant of forces

BASIC CIVIL ENGINEERING
Mechanics of Solids

Mechanics of Solids
PART- II
Mechanics of Deformable
Bodies
PART- I
Mechanics of Rigid
Bodies

COURSE CONTENT IN BRIEF
1. Resultant of concurrent and non-concurrent coplanar forces.
2. Equilibrium of concurrent and non-concurrent coplanar forces.
3. Centroid of plane areas
4. Moment of Inertia of plane areas
5. Kinetics: Newton’s second law, D’Alembert’s principle, Work- Energy,
and Impulse- Momentum principle.
PART I Mechanics of Rigid Bodies
PART II Mechanics of Deformable bodies
6. Simple stresses and strains
7. Statically indeterminate problems and thermal stresses
8. Stresses on inclined planes
9. Stresses due to fluid pressure in thin cylinders

Books for Reference
1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.
2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.
3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition
4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.
5. Machanics of Materials, by E.P.Popov
6. Machanics of Materials, by E J Hearn
7. Strength of materials, by Beer and Johnston
8. Strength of materials, by F L Singer & Andrew Pytel
9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa
10. Strength of Materials, by Ramamruthum
11. Strength of Materials, by S S Bhavikatti

Definition of Mechanics :
In its broadest sense the term ‘Mechanics’ may be defined as
the ‘Science which describes and predicts the conditions of
rest or motion of bodies under the action of forces’.
INTRODUCTION
This Course on Engineering Mechanics comprises of
Mechanics of Rigid bodies and the sub-divisions that come
under it.
PART - I Mechanics of Rigid Bodies

Engineering Mechanics
Mechanics of Solids Mechanics of Fluids
Rigid Bodies Deformable
Bodies
Statics Dynamics
Kinematics Kinetics
Strength of
Materials
Theory of
Elasticity
Theory of Plasticity
Ideal
Fluids
Viscous
Fluids
Compressible
Fluids
Branches of Mechanics

It is defined as a definite amount of matter the parts of which
are fixed in position relative to one another under the
application of load.
Actually solid bodies are never rigid; they deform under the
action of applied forces. In those cases where this deformation
is negligible compared to the size of the body, the body may be
considered to be rigid.
Concept of Rigid Body :

Particle
A body whose dimensions are negligible when compared to the
distances involved in the discussion of its motion is called a
‘Particle’.
For example, while studying the motion of sun and earth, they
are considered as particles since their dimensions are small
when compared with the distance between them.

Force
It is that agent which causes or tends to cause, changes or
tends to change the state of rest or of motion of a mass.
A force is fully defined only when the following four
characteristics are known:
(i) Magnitude
(ii) Direction
(iii) Point of application
(iv) Sense.

Force:
characteristics of the force 100 kN are :
(i) Magnitude = 100 kN
(ii) Direction = at an inclination of 300 to the x-axis
(iii) Point of application = at point A shown
(iv) Sense = towards point A
300
100 kN
A

Scalars and Vectors
A quantity is said to be a ‘scalar’ if it is completely defined by
its magnitude alone.
Example : Length, Area, and Time.
A quantity is said to be a ‘vector’ if it is completely defined
only when its magnitude and direction are specified.
Example : Force, Velocity, and Acceleration.

Principle of Transmissibility : It is stated as follows : ‘The
external effect of a force on a rigid body is the same for all points
of application along its line of action’.
For example, consider the above figure. The motion of the block will be
the same if a force of magnitude P is applied as a push at A or as a pull at
B.
P P
A B
The same is true when the force is applied at a point O.
P P
O

1. RESULTANT OF COPLANAR FORCES
Resultant, R : It is defined as that single force which can
replace a set of forces, in a force system, and cause the
same external effect.
R

=
sameisAparticle,oneffect
321
external
FFFR 
F3
F1
F2
A
A

Parallelogram law of forces : ‘If two forces acting at a point are
represented in magnitude and direction by the two adjacent
sides of a parallelogram, then the resultant of these two forces
is represented in magnitude and direction by the diagonal of
the parallelogram passing through the same point.’
B
C
A
O
P2
P1
R


Contd..
Resultant of two forces acting at a point

In the above figure, P1 and P2, represented by the sides OA and OB have
R as their resultant represented by the diagonal OC of the parallelogram
OACB.
B
C
A
O
P2
P1
R


It can be shown that the magnitude of the resultant is given by:
R = P1
2 + P2
2 + 2P1P2Cos α
Inclination of the resultant w.r.t. the force P1 is given by:
 = tan-1 [( P2 Sin ) / ( P1 + P2 Cos  )]

B
C
A
O
P2
P1
R


2
22
1 PPR 
1
2
tan
P
P


B C
A
O
P2
P1
R
If α = 900 , (two forces acting at a point are at right angle)
Resultant of two forces acting at a point at right angle

Triangle law of forces
‘If two forces acting at a point can be represented both in
magnitude and direction, by the two sides of a triangle taken in
tip to tail order, the third side of the triangle represents both in
magnitude and direction the resultant force F, the sense of the same is
defined by its tail at the tail of the first force and its tip at the tip of
the second force’.

Let F1 and F2 be the two forces acting at a point A and θ is the
included angle.
θ
A
F1
F2
θ
F1
F2
R
‘Arrange the two forces as two sides of a triangle taken in tip to
tail order, the third side of the triangle represents both in magnitude
and direction the resultant force R.
=
the sense of the resultant force is defined by its tail at the tail of the
first force and its tip at the tip of the second force’.



R
F1
F2
θ
A
F1
F2
θ
F1
F2
R
=
(180 -  - ) = θ
)180sin(sinsin
21
 

RFF
where α and β are the angles made by the resultant force
with the force F1 and F2 respectively.

Component of a force, in simple terms, is the effect of a
force in a certain direction. A force can be split into infinite number
of components along infinite directions.
Usually, a force is split into two mutually perpendicular
components, one along the x-direction and the other along y-
direction (generally horizontal and vertical, respectively).
Such components that are mutually perpendicular are called
‘Rectangular Components’.
Component of a force :
The process of obtaining the components of a force is called
‘Resolution of a force’.

Consider a force F making an angle θx with x-axis.
Then the resolved part of the force F along x-axis is given by
Fx = F cos θx
Rectangular component of a force
Fy
x
F
x
F
Fx
The resolved part of the force F along y-axis is given by
Fy = F sin θx
= Fy
x
F
Fx

Let F1 and F2 be the oblique components of a force F. The
components F1 and F2 can be found using the ‘triangle law of
forces’.


F
F1
F2
F1 / Sin  = F2 / Sin  = F / Sin(180 -  - )
Oblique component of a force
The resolved part of the force F along OM and ON can
obtained by using the equation of a triangle.


F
F1
F2
M
O
N

The adjacent diagram gives the sign convention for
force components, i.e., force components that are directed
along positive x-direction are taken +ve for summation along
the x-direction.
Sign Convention for force components:
Also force components that are directed along +ve y-direction are
taken +ve for summation along the y-direction.
+ve
+ve
x
x
y
y

Classification of force system
Force system
Coplanar Forces Non-Coplanar
Forces
Concurrent
Non-concurrent
Concurrent Non-concurrent
A force that can replace a set of forces, in a force system,
and cause the same ‘external effect’ is called the Resultant.
Like parallel Unlike parallel
Like parallel Unlike parallel

Numerical Problems & Solutions(Q1.1)
Resolve the forces shown in figure along x and y
directions.
20 kN
25
0
35 kN
60 kN
3
2

solution:
20 kN
250
35 kN
60 kN
3
2
20 cos θx
60 cos θx
60 sin θx
20 sin θx = 20 sin65
= 60 cos33.7
= 20 cos65
= 60 sin33.7

(Q1.1)
solution:
20 kN
250
35 kN
60 kN
3
2
20 cos θx
60 cos θx
60 sin θx
20 sin θx = 20 sin65
= 60 cos33.7
= 20 cos65
= 60 sin33.7
Force X-comp Y-comp
35kN - 35 0
20kN - 20 cos 65 -20 sin 65
60kN - 60 cos 33.7 + 60 sin 33.7
Answer:

Resolve the forces shown in figure along x and y
directions.
15 kN
15
0
105 kN
75 kN
45 kN
40
0
60 kN
35
0

solution:
15 cos150
45 cos550
45 sin550
60 sin400
60 cos400
15 sin150
15 kN
15
0
105 kN
75 kN
45 kN
40
0
60 kN
35
0
550
(Q1.2)

15 cos150
45 cos550
45 sin550
60 sin400
60 cos400
15 sin150
15 kN
150
105 kN
75 kN
45 kN
40
0
60 kN
350
550
Force X-comp. Y-comp
105 0 +105
75 -75 0
15 + 15 cos15 + 15 sin15
45 - 45 cos55 - 45 sin55
60 + 60 cos40 - 60 sin40
(Q1.2)

Obtain the resultant of the concurrent coplanar forces acting
as shown in figure
15 kN
15
0
105 kN
75 kN
45 kN
40
0
60 kN
35
0

Numerical Problems & Solutions
solution:
15 cos150
45 cos550
45 sin550
60 sin400
60 cos400
15 sin150
15 kN
15
0
105 kN
75 kN
45 kN
40
0
60 kN
35
0
550
(Q1.3)

15 cos150
45 cos550
45 sin550
60 sin400
60 cos400
15 sin150
15 kN
150
105 kN
75 kN
45 kN
40
0
60 kN
350
550
105 0 +105
75 -75 0
15 + 15 cos15 + 15 sin15
45 - 45 cos55 - 45 sin55
60 + 60 cos40 - 60 sin40
------- --------------- ----------------
R ΣFx = ΣFy =
- 40.359 + 33.453
(Q1.3)

∑ Fx = – 75 + 15 cos 15 – 45 cos 55 + 60 Cos 40
= - 40.359 kN = 40.359 kN
∑ Fy = + 105 + 15 Sin 15 – 45 sin 55 – 60 Sin 40
= + 33.453 kN
ΣFx = 40.359kN
ΣFy = 33.453 kN
R
θx
0
1-
69.39
tan;tan
42.52


















y
x
x
x
y
x
yx
F
F
F
F
kNFFR
Answer:
(Q1.3)

Obtain the resultant of the concurrent coplanar forces
acting as shown in figure.
75kN
120
2
3
301
2
25kN
100kN50kN
º
º
(Q1.4)

Solution:
(Q1.4)
75kN
120
2
3
30
1
2
25kN
100kN
50kN
º
º
50 cos θx
= 50 cos 26.3
100 sin θx
= 100 sin 33.7
100 cos θx
= 100 cos 33.7
50 sin θx = 50 sin 26.3
75 cos θx
= 75 cos 30
25 cos θx
= 25 cos 63.43
75 sin θx
= 75 sin 30
25 sin θx
= 25 sin 63.43

Solution:
(Q1.4)
75kN
120
2
3
301
2
25kN
100kN
50kN
º50 cos 26.3 100 sin 33.7
100 cos 33.750 sin 26.3
75 cos 3025 cos 63.43
75 sin 3025 sin 63.43
100 -100 cos33.7 -100 sin33.7
75 -50 cos26.3 +50 sin26.3
15 -25cos 63.43 -25 sin63.43
45 +75 cos30 -75 sin30
------- --------------- ----------------
R ΣFx = ΣFy =
- 74.26 kN -93.17 kN

∑Fx = -50 Cos 26.31- 100 Cos33.69 – 25 Cos 63.43 + 75 Cos 30
74.26kN= -74.26kN =
Contd..
= 93.17kN= -93.17kN
∑FY = 50sin26.31- 100sin 33.69 – 75sin30 – 25sin63.43
(Q1.4)

R = (∑Fx) 2
+ (∑Fy) 2
= 119.14 kN
Θ = tan-1
(∑Fy / ∑Fx ) = 51.44o
∑Fx
∑Fy
R

Answers:
(Q1.4)

A system of concurrent coplanar forces has five forces of which
only four are shown in figure. If the resultant is a force of
magnitude R = 250 N acting rightwards along the horizontal,
find the unknown fifth force.
120N
150N
50N
200N
45º
50°
110º
(Q1.5)

- Assume the fifth force F5 in the first quadrant, at an angle α, as
shown.
The 150 N force makes an angle of 20o
w.r.t. horizontal
R is the resultant of Five forces including F5
150N
50N
200N
120N
45
°
50°
110 º
F5
α
R =250 N
20º
Solution:
(Q1.5)

- Resolve the forces along X & Y axis
150N
50N
200N
120N
45
°
50°
110 º
F5
α
R =250 N
20º
Solution:
(Q1.5)
F5y=F5 sin α
F5x=F5 cos α

Solution:(Q1.5)
150N
50N
200N
120N
45
°
50°
F5
α
R =250 N
20º
F5y=F5 sin α
F5x=F5 cos α
F5 +F5 cosα +F5 sinα
50 -50 cos45 +50 sin45
200 +200cos 50 +200 sin 50
120 0 -120
150 -150 cos20 +150 sin20
------- --------------- ----------------
R ΣFx = ΣFy =
+250 kN 0

∑FX = R = + 250 & ΣFy = 0
ΣFx = + 250 = 200 cos 50 – 150 cos 20 – 50 cos 45 + F5 cos α
 F5 cos α = +297.75 N
because the resultant is acting along x-direction
∑FY = 0 = F5 sin α + 200sin 50 + 150 sin 20 – 120 + 50 sin 45
F5 sin α= -119.87 N
α = 21.90º
119.87N
297.75N
F5 = 320.97N
tan α = F5sin α /F5cos α
=0.402
α = 21.90º
F5= 320.97N
F5cosα =
F5sinα =
Answers
(Q1.5)

A system of concurrent coplanar forces has four forces of which
only three are shown in figure. If the resultant is a force R =
100N acting as indicated, obtain the unknown fourth force.
60
°
45
°
R=100N
50N
25N
70
°
40
°
75N
(Q1.6)

(Q1.6)
60°
45°
R=100N
50N
25N
70°
40°
75N
30°
75 cos70
25 cos30
50 cos45
R cos40
R sin40
75 sin70
25 sin30 F4
α
F4 cos α
F4 sin α
α
 Assume the fourth force (F4) in the first quadrant, at an angle α, as shown.
The 25 N force makes an angle of 30o
w.r.t. horizontal
R is the resultant of Four forces including F4

(Q1.6)
60°
45
°
R=100N
50N
25N
70°
40°
75N
30
°
75 cos70
25 cos30
50 cos45
R cos40
R sin40
75 sin70
25 sin30
F4 +F4 cosα +F4 sinα
50 - 50 cos45 +50 sin45
25 - 25 cos30 +25 sin30
75 +75 cos70 +75 sin70
------- --------------- ----------------
R ΣFx = - R cos40 ΣFy = -R sin40
= -100 cos40kN = - 100 sin40kN
Fx = -Rcos40 = F4cosα + 75cos70 – 50cos45 – 25sin60
F4cosα = - 45.25N
Fx = -Rcos40+ve

Fy = -Rsin40 = F4sinα + 75sin70+25cos60+50sin45
 F4sinα = -182.61N ;
+ve Fy = -Rsin40
α= 76.08º
45.25N
F4=188.13N
182.61N
F4cosα =
F4sinα =
Answers:
tan  = (F4sin /F4cos)
α = 76.08º
& F4 =188.13N
(Q1.6)

The resultant of a system of concurrent coplanar forces is a force
acting vertically upwards. Find the magnitude of the resultant, and the
force F4 acting as shown in figure.
60
°
30
°
15 kN
5 kN
10 kN
70
°
45
°
F4
Contd..
(Q1.7)

F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94
∑Fx = 0
Solution:
+ve
60°
30
°
15 kN
5 kN
10 kN
70°
45°
F4
R
F4 = 21.22kN
Contd..
(Q1.7)

F4cos70 + 10sin60 – 15sin45 + 5sin30 = +R
 +R - 0.342F4 = 0.554
Substituting for F4 , R= +7.81kN
∑Fy = +R
+ve
Solution:
60
°
30
°
15 kN
5 kN
10 kN
70
°
45
°
F4
Fig. 4
R
Answers:
F4 = 21.22 kN
R= +7.81kN
(Q1.7)

Obtain the magnitudes of the forces P and Q if the resultant of the
system shown in figure is zero .
40
°
60
°
P
50N
Q
70
°
45
°
100N
Contd..
(Q1.8)

Contd..
40
°
60°
P
50N
Q
70
°
45
°
100N
For R to be = zero,
∑Fx = 0 and ∑ Fy = 0
∑Fx = 0 :
-Psin45 – Qcos40 + 100cos70 + 50cos60 = 0
0.707P + 0.766Q = 59.2
+ve
(Q1.8)

∑Fy = 0
-Pcos45 + Qsin40 + 100sin70 – 50sin60 = 0
or, -0.707P + 0.642Q = -50.67 ------(b)
+ve
Answers:
Solving (a) & (b)
P = 77.17 N & Q = 6.058N
(Q1.8)

30
°
100N
50N
Forces of magnitude 50N and 100N are the oblique components of a
force F. Obtain the magnitude and direction of the force F.
Refer figure.
Contd..
(Q1.9)

Rotating the axes to have X parallel to 50N,
∑Fx = +50 + 100cos30 = +136.6N
∑ Fy = +100sin30 = +50N
+ve
+ve
30
°
100N
X - AXIS
Y-AXIS
30
°
100N
50N
50N
Contd..
(Q1.9)
Fig. 1.9

F = 145.46N
θ = 20.1º w r t X direction (50N force)
50N
F= (∑Fx)2+(∑Fy)2
= tan-1[(∑Fx)2+(∑Fy)2]
Fig. 6
X - AXISF
θ
Y-AXIS
30
°
100N
X - AXIS
Y-AXIS
30
°
100N
50N
50N
θ
(Q1.9)

Resolve the 3kN force along the directions P and Q. Refer figure.
P
Q
3kN
45
° 60
°
20
°
Contd..
(Q1.10)

3kN
Move the force P parallel to itself to complete a triangle. Using
sine rule,
P/sin45 = Q/sin90 = 3/sin45
Answer :
P = 3kN, and Q = 4.243kN
P
45º
Q
45º
Q
60
°
30
°
X – Axis
P
3kN
45º
(Q1.10)

Resolve the 5kN force along the directions P and Q. Refer Fig. 1.11.
P
Q
3kN
45
° 60
°
20
°
Fig. 1.10 Contd..
(Q1.11)

Q
60
°
20
°
X – Axis
P
5kN
45º
(Q1.11)
Q
60
°
20
°
X – Axis
P
5kN
45º
55
°
Q
60°
X
P
5kN
45º
55°

(Q1.11)
X – Axis
Using sine rule,
P/sin45 = Q/sin80 = 5/sin55
Answer :
P = 4.32 kN, and Q = 6.01 kN
Q
60°
P
45º
55°
800
5N

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 applications.
RESULTANT OF COPLANAR NON CONCURRENT
FORCE SYSTEM
Fig. 1
F2
F1
F3
Fig. 2
F1 F2
F5
F4
F3

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.

MOMENT OF A FORCE ABOUT AN AXIS
Definition: Moment is the
tendency of a force to make a
rigid body to rotate about an
axis.
The applied force can also tend to rotate the body about
an axis in addition to motion. This rotational tendency is
known as moment.
This is a vector quantity
having both magnitude
and direction.

MOMENT OF A FORCE ABOUT AN AXIS
Moment Axis: This is the axis about which rotational
tendency is determined. It is perpendicular to the plane
comprising moment arm and line of action of the force (axis
0-0 in the figure)
Moment Center: This is
the position of axis on co-
planar system. (A).
Moment Arm:
Perpendicular distance
from the line of action of
the force to moment
center. Distance AB = d.

It is computed as the product of the of the force and
the perpendicular distance from the line of action
to the point about which moment is computed.
(Moment center).
Magnitude of moment:
MA = F×d
= Rotation effect because of
the force F, about the point A
(about an axis 0-0)
Unit – kN-m, N-mm etc.

The sense is obtained by ‘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 sense of moment vector’.
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 vice-
versa.
M
M
Sense of moment:

A 100N vertical force is applied to the end of a lever at ‘A’,
which is attached to the shaft at ‘O’ as shown in the figure.
60º
O
A
F=100 N
(Q1.12)
Determine,
1. The moment of 100N force about ‘O’.
2. Magnitude of the horizontal force applied
at ‘A’, which develops same moment
about ‘O’.
3. The smallest force at ‘A’, which develops
same effect about ‘O’.
4. How far from the shaft a 240N vertical
force must act to develop the same effect?

1) Perpendicular distance from the line of
action of force F to the moment center
‘O’ = d
= 240 cos 60º = 120 mm.
Moment about ‘O’ = F × d = 100 × 120
= 12,000 N-mm
(Clockwise)
SOLUTION:
(Q1.1)
60º
O
A
d
F
1. The moment of 100N force about ‘O’.

If force F is acting horizontally then
the perpendicular distance between the line of
action of horizontal force F at A , to moment
center ‘O’
= d = 240 sin 60º = 207.85 mm.
(Q1.12)
Moment about ‘O’ = F × d
= F × 207.85
= 12,000 N-mm (Clockwise)
Therefore,
F = 12,000 / 207.85 = 57.73 N
O
A
60º
d
F
2. Magnitude of the horizontal force applied at ‘A’, which develops same
moment about ‘O’.

3) Solution:
F = M/d
Force is smallest when the perpendicular
distance is maximum so as to produce
same M.
Maximum distance between the point, 0
and the point A is 240 mm.
If the line of action of the force is such that
d = 240 mm
i.e., d = 240 mm.
Therefore, Fmin = 12,000/240 = 50N.
(Q1.12)
O
A
60º
F
3. The smallest force at ‘A’, which develops same effect about ‘O’.

4) Solution:
Distance along x-axis, X = M/F
= 12,000/240
= 50 mm.
Distance along the shaft axis
d = X/cos 60
= 50/cos 60
= 100 mm
O
A
60º
X F
(Q1.12)
4. How far from the shaft a 240N vertical force must act to
develop the same effect?

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Statement: The moment of a force about a moment center or
axis is equal to the algebraic sum of the moments of its
component forces about the same moment center (axis).
Moment of Force P about the
point A,
P x d
θ
P
A
d
Algebraic sum of Moments of
components of the Force P
about the point A,
P cosθ x d1 + P sinθ x d2
=
θ
P
A
P sinθ
P cosθ
d1
d2

VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Q
R
A
O

 p
q
r P
p, r and q are moment arms from ‘O’
of P, R and Q respectively.
,  and  are the inclinations of ‘P’,
‘R’ and ‘Q’ respectively w.r.to X –
axis.

Y
X
Proof (by Scalar Formulation):
Let ‘R’ be the given force.
‘P’ & ‘Q’ are component forces of
‘R’. ‘O’ is the moment center.

We have,
Ry = Py + Qy
R Sin = P Sin + Q Sin  ----(1)
From le AOB, p/AO = Sin 
From le AOC, r/AO = Sin 
From le AOD, q/AO = Sin 
From (1),
 R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)
i.e., R × r = P × p + Q × q
Moment of resultant R about O = algebraic
sum of moments of component forces P &
Q about same moment center ‘O’.
Ry
Py
R
A O
Q
P
p
q r



Y
X
B
C
D
Qy

COUPLE
Two parallel, non collinear (separated by certain
distance) forces that are equal in magnitude and opposite
in direction form ‘couple’.
d
F
F
Hence, couple does not produce any
translation and produces only rotation. M = F x d
The algebraic summation of the
two forces forming couple is zero.

RESOLUTION OF A FORCE INTO A
FORCE-COUPLE SYSTEM
Replace the force F acting at the point A to the point B
F
AB
Apply two equal and opposite forces of same magnitude &
direction as Force F at point B, so that external effect is
unchanged
F
AB
F
F
d

Of these three forces, two forces i.e., one at A and the other
oppositely directed at B form a couple.
Moment of this couple, M = F × d.
Thus, the force F acting at a point such as A in a rigid body can be
moved to any other given point B, by adding a couple M. The
moment of the couple is equal to moment of the force in its
original position about B.
Third force at B is acting in the same direction as that at P.
F
AB
F
F
AB
F
=
M = F x dd

TYPES OF LOADS ON BEAMS
1. Concentrated Loads – This is the load
acting for very small length of the beam.
(also known as point load, Total load W is
acting at one point )
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.
Total intensity, W = w × L
(acts at L/2 from one end of the spread)
W kN
w kN/m
L
W = (w x L) kN
L
L/2

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
spread of the load
W = 1/2× w × L.
(acts at 2×L/3 from ‘Zero’ load end)
w kN/m
L
L
W = ½ × L × w
2/3 ×L
1/3 ×L

A 100N force acts on the corner of a 4m x 3m box as
shown in the Fig. Compute the moment of this force about
A by a) Definition of Moment
b) Resolving the force into components along CA
and CB.
A B
CD
60º
3m
4m
F=50 kN
(Q1.13)

A B
CD
60º
36.87º
60º
d
E
a) By Definition of Moment:
To determine ‘d’:
AC =
CAD = tan-1(3/4) = 36.87º
ECD = 60º
ACE = 60º – 36.87º = 23.13º
From ∆le ACE, d = AC × sin ( ACE)
= 5 × sin 23.13º = 1.96 m.
Moment about A = 50 × 1.96 = 98.20 kNm.
3m
4m
F=50 kN
m534 22
 23.13
(Q1.13)

A B
CD
60º
b) By Components:
Fx = 50 × cos 60 = 25kN.
Fy = 50 × sin 60 = 43.30kN.
+ ΣMA = - Fx × 3 + Fy × 4
= - 25 × 3 + 43.3 × 4
= + 98.20kNm.
3m
4m
F=50kN
Fx
Fy
(Q1.13)

An equilateral triangle of sides 200mm is acted upon by 4
forces as shown in the figure. Determine magnitude and
direction of the resultant and its position from point ‘D’.
30º
80kN
30kN
50kN
60kN
60º
200mm
D
(Q1.14)

Resultant & its inclination:
Resolving forces
+ ΣFx = Rx = +30 + 60 cos30º – 50 cos60º
= +56.96kN.
+ ΣFy = Ry = -80 + 60 sin30º + 50 sin60º
= -6.69kN.
R=
Inclination w.r.to horizontal = θR
= tan-1(Ry/Rx)
= tan-1(6.69/56.96) = 6.7º
30º
80kN
30kN
50kN
60kN
60º
60 cos30
60 sin30
50 sin 60
50 cos 60
200mm
(Q1.14)

b) Position w.r.to D:
Moment of the component forces about D:
+ MD = - 60 × 100 + 80 × 100 = 2000kNmm.
= R × d
where ‘d’ = perpendicular distance from point D to the line
of action of R.
= 2000 × d.
∴ d =2000/57.35 = 34.87mm
(Q1.14)

Find the resultant and its position w.r.to ‘O’ of the non-
concurrent system of forces shown in the figure.
F3=1000N
F2=500N
F1=2500N
F5=2000N
F4=1500N
1m
1m
Ө2Ө4
1
1
Ө5
O
(Q1.15)

A) To find the resultant –
Ө2 = tan-1(1/2) = 26.56°
Ө4 = tan-1(3/2) = 56.31°
Ө5 = tan-1(1/1) = 45°
+ ΣFx = Rx = F2 cosӨ2 +F3
-F4 cosӨ4-F5 cos Ө5
= 500 × cos26.56 + 1000 –
1500 × cos56.31-2000 × cos45
= -799.03N = 799.03N←
+↑ΣFy = Ry= F1+F2 sin Ө2-F4 sin Ө4+F5 sin Ө5
= 2500+500 sin26.56-1500 sin56.31+2000 sin45
=2889.70N ↑
F3=1000N
F2=500N
F1=2500N
F5=2000N
F4=1500N
1m
1m
Ө2Ө4
1
1
Ө5
O
(Q1.15)

∴ Resultant R =
= 2998.14N
ӨR = tan-1 = tan-1(2889.7/799.03) = 74.54°
B) Position of Resultant w.r.to ‘O’:
By Varignon’s theorem, Moment of the resultant about ‘O’
= Algebraic sum of the moments of its components
about ‘O’.
+ Mo =R×d = +2500×2 + 500×sin26.56×5 – 500×
cos26.56×3 - 1000×1- 1500× cos56.31×0
–1500×sin56.3×1+2500× cos45×1-
2500×sin45×0
= 2998.14 × d






x
y
R
R
Rx
Ry
R
ӨR
(Q1.15)

Determine the resultant of three forces acting on a dam
section shown in the figure and locate its intersection with
the base. Check whether the resultant passes through the
middle one-third of the base.
60º
30 kN
120 kN
50 kN
6 m
1 m
2 m
3 m
A B
(Q1.16)

+ ∑Fx = Rx = 50 – 30 × cos30 = 24.02 kN
+ ∑Fy = Ry = -120 – 30 × sin 30= -135 kN
Resultant, R=
60º
30 kN
120 kN
50 kN
6 m
1 m
2 m
3 m 30º
A B
kNRR yx 12.13713502.24 2222

(Q1.16)

θR= tan-1(Ry/Rx) = tan-1(135/24.12) = 79. 91º
Location of the resultant w.r.t. B:
MB= 30×1 + 120 × (6-2) - 50 × 3 = Ry × X
360 = 135× X
Therefore, X = 360/135 = 2.67m from B.
From A, X = 6 –2.67= 3. 33 m.
Middle 1/3rd distance is between 2m and 4m.
2m<3. 33<4m
Hence, the resultant passes through the middle 1/3rd
of the base.
(Q1.16)

A 50 N force is applied to the corner of a plate as
Shown in the fig. Determine an equivalent
force-couple system at A. Also determine an equivalent
system Consisting of a 150 N force at B and another
force at A.
50 N
100 mm
50 mm
30 mm
30º
B
A
(Q1.17)

Force – Couple System at A:
Fx = 50 ×sin 30= 25 N.
Fy = 50 × cos 30= 43.3 N
50 N
100 mm
50 mm
30 mm
30º
50 cos 30
50N
50 sin 30
B
A
60º
(Q1.17)

a) Force – Couple System at A:
Fx = 50 ×sin 30 = 25 N.
Fy = 50 × cos 30 = 43.3 N
These forces can be moved to
A by adding the couple.
Moment of the couple about A
+ ∑MA= Fx×50-Fy×100
= 25×50 - 43.3×100
= -3080 N-mm.
= 3080 N-mm
100 mm
50 mm
30 mm
30º
Fy=50 cos 30
50N
Fx=50 sin 30
B
A
100 mm
50 mm
A
Fy=50 cos 30
Fx=50 sin 30
MA=3080N-mm
(Q1.17)

b) Forces at A and B :
The couple MA is because of two
equal and opposite forces at A
and B.
i.e., MA = 150 × cosθ × 30
= 3080
Therefore, θ = 46.8º.
Resultant force at A:
FX=50×sin30-150×cos46.8
= -77.68N = 77.68N
FY=-50×cos30-150×sin46.8
= -152.65N = 152.65N
100 mm
50 mm
30 mm
B
A
100 mm
50 mm
A
Fy=50 cos 30
Fx=50 sin 30
θ
θ=46.8º
150N
150 N
(Q1.17)

EXERCISE PROBLEMS
Q1.18 A body of negligible weight, subjected to two forces F1=
1200N, and F2=400N acting along the vertical, and the
horizontal respectively, is shown in figure. Find the component
of each force parallel, and perpendicular to the plane.
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
= 1200 N
3
4
Y
F2
F1
= 400 N
1. Resultant of force system

2. Determine the X and Y components of each of the forces shown in
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,
F3X = -306.42 N, F3Y= -257.12N )
30º40º
12
5
300 N
390 N
400 N
X
Y
F1 =
F2 =
F3 =
FIG. 2
FIG.2.
EXERCISE PROBLEMS

600N
200N
800N
20º
40º
30º
FIG. 3
3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3
(Ans: R = 522.67 N, θ = 68.43º)
EXERCISE PROBLEMS

4. A disabled ship is pulled by means of two tug boats as shown in FIG.
4. If the resultant of the two forces T1 and T2 exerted by the ropes
is a 300 N force acting parallel to the X – direction, find :
(a) Force exerted by each of the tug boats knowing α = 30º.
(b) The value of α such that the force of tugboat 2 is minimum,
while that of 1 acts in the same direction.
Find the corresponding force to be exerted by tugboat 2.
( Ans: a. T1= 195.81 N, T2 = 133.94 N
b. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )
T2
R = 300 N
T1
α
20º
FIG. 4
X - direction
EXERCISE PROBLEMS

5. An automobile which is disabled is pulled by two ropes as
shown in Fig. 5. Find the force P and resultant R, such that R is
directed as shown in the figure.
P
Q = 5 kN
R
20º
40º
Fig. 5
(Ans: P = 9.4 kN , R = 12.66 kN)
EXERCISE PROBLEMS

6. A collar, which may slide on a vertical rod, is subjected to three forces
as shown in Fig.6. The direction of the force F may be varied .
Determine the direction of the force F, so that resultant of the three forces is
horizontal, knowing that the magnitude of F is equal to
(a) 2400 N, (b)1400N
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
1200 N
800 N60º
θ
F
ROD
COLLAR
Fig.6
EXERCISE PROBLEMS

7. Determine the angle α and the magnitude of the force Q such that
the resultant of the three forces on the pole is vertically downwards
and of magnitude 12 kN. Refer Fig. 7.
8kN
5kN
Q
30º
α
Fig. 7
(Ans: α = 10.7 º, Q = 9.479 kN )
EXERCISE PROBLEMS

8. Determine the resultant of the parallel coplanar
force system shown in figure.
400 N
1000 N
2000 N
600 N o
60º
60º 30º
10º
(Ans. R=800N towards left, d=627.5mm)
EXERCISE PROBLEMS

9. 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.
10N
40N
a
a
30N
20N
A
B
C
D
(Ans:R=28.28N, θ=45º, x=1.77a)
EXERCISE PROBLEMS

10. 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 2.85m. Determine the magnitude of resultant
and also the distance of the resultant from the left
end.
(Ans: R = 125 N, x = 3.06 m)
EXERCISE PROBLEMS

11. Reduce the given forces into a single force and a
couple at A.
100 N
80 N
1m
1.5m
70.7 kN
200 kN
A
30º45º
30º
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
EXERCISE PROBLEMS

12. Determine the resultant w.r.t. point A.
(Ans: R = 450 kN, X = 7.5 kNm)
EXERCISE PROBLEMS
A
150 N
1.5m3m1.5m
150 Nm
500 N100 N

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