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Mechanics of Solids [3 1 0 4]
CIE 101 / 102CIE 101 / 102
First Year B.E. Degree
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Mechanics of Solids
PART- IIPART- I
Mechanics of Deformable
BodiesMechanics 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
and Impulse- Momentum principle.
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
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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
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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
PART - I Mechanics of Rigid Bodies
or motion of bodies under the action of forces’.
This Course on Engineering Mechanics comprises of
Mechanics of Rigid bodies and the sub-divisions that come
under it.
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Engineering Mechanics
Mechanics of Solids Mechanics of Fluids
Rigid Bodies Deformable Ideal Viscous Compressible
Branches of Mechanics
Bodies
Statics Dynamics
Kinematics Kinetics
Strength of
Materials
Theory of
ElasticityTheory of Plasticity
Ideal
Fluids
Viscous
Fluids
Compressible
Fluids
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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.
Concept of Rigid Body :
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.
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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.
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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 fourA force is fully defined only when the following four
characteristics are known:
(i) Magnitude
(ii) Direction
(iii) Point of application
(iv) Sense.
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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(iii) Point of application = at point A shown
(iv) Sense = towards point A
300
100 kN
A
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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.
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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’.
P PA B
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.
The same is true when the force is applied at a point O.
P PO
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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
θθθθ
=
same isA particle,on effect
321
external
FFFR ++=F3
F1
F2
AA
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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
Resultant of two forces acting at a point
is represented in magnitude and direction by the diagonal of
the parallelogram passing through the same point.’
BC
AO
P2
P1
Rα
θ
Contd..
-
In the above figure, P1 and P2, represented by the sides OA and OB have
BC
AO
P2
P1
Rα
θ
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.
It can be shown that the magnitude of the resultant is given by:
R = √P12 + P22 + 2P1P2Cos α
Inclination of the resultant w.r.t. the force P1 is given by:
θ = tan-1 [( P2 Sin α) / ( P1 + P2 Cos α )]
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BC
AO
P2
P1
Rα
θ
If α = 900 , (two forces acting at a point are at right angle)
Resultant of two forces acting at a point at right angle
222
1PPR +=
1
2tanP
P=θ
θ
B C
AO
P2
P1
R
If α = 90 , (two forces acting at a point are at right angle)
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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 intip to tail order, the third side of the triangle represents both in
magnitude and direction the resultant force F, the sense of the same ismagnitude 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’.
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Let F1 and F2 be the two forces acting at a point A and θ is the
included angle.
Triangle law of forces
θ
F1
F
θ
F1R
=
A F2F2
‘Arrange the two forces as two sides of a triangle taken in tip totail 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’.
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Triangle law of forces
θ
A
F1
F2
θ
F1
F2
R=
ββββ
αααα
R
F1
F2
(180 - α - β) = θ
)180sin(sinsin
21
βααβ −−==
RFF
where α and β are the angles made by the resultant force
with the force F1 and F2 respectively.
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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-
Component of a force :
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’.
The process of obtaining the components of a force is called
‘Resolution of a force’.
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Rectangular component of a force
Fyθx
F
θθθθx
F
Fx
= Fyθθθθx
F
Fx
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
The resolved part of the force F along y-axis is given by
Fy = F sin θx
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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’.
ββββF F2
Oblique component of a force
FF2 M
N
ααααF1
F2
F1 / Sin β = F2 / Sin α = F / Sin(180 - α - β)
The resolved part of the force F along OM and ON can
obtained by using the equation of a triangle.
ββββαααα
F1
O
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Sign Convention for force components:
+ve
+vex
x
yy
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.
Also force components that are directed along +ve y-direction are
taken +ve for summation along the y-direction.
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Classification of force system
Force system
Coplanar Forces Non-Coplanar
Forces
ConcurrentNon-concurrent
Concurrent Non-concurrentConcurrent 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
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Numerical Problems & Solutions(Q1.1)
Resolve the forces shown in figure along x and y
directions.
20 kN
250
25
35 kN
60 kN
3
2
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Numerical Problems & Solutions(Q1.1)
solution:
20 kN
250
20 cos θx
20 sin θx = 20 sin65= 20 cos65
35 kN
60 kN
3
2
60 cos θx
60 sin θx
= 60 cos33.7
= 60 sin33.7
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(Q1.1)
solution:
20 kN
250
35 kN
3
2
20 cos θx
60 sin θx
20 sin θx = 20 sin65= 20 cos65
= 60 sin33.7Answer:
60 kN
3
60 cos θx
60 sin θx
= 60 cos33.7
= 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:
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Numerical Problems & Solutions(Q1.2)
Resolve the forces shown in figure along x and y
directions.
15 kN
105 kN
150
75 kN
45 kN
400
60 kN
350
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solution:
15 cos150
15 sin150
15 kN
150
105 kN
(Q1.2)
45 cos550
45 sin550
60 sin400
60 cos400
150
75 kN
45 kN
400
60 kN
350
550
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15 cos150
45 cos550
45 sin550
60 sin400
60 cos400
15 sin150
15 kN
150
105 kN
75 kN
45 kN
400
60 kN
350
550
(Q1.2)
45 sin550
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
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Numerical Problems & Solutions(Q1.3)
Obtain the resultant of the concurrent coplanar forces acting
as shown in figure
15 kN
105 kN
15 kN
150
75 kN
45 kN
400
60 kN35
0
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Numerical Problems & Solutions
solution:
15 cos150
15 sin150
15 kN
150
105 kN
(Q1.3)
45 cos550
45 sin550
60 sin400
60 cos400
150
75 kN
45 kN
400
60 kN
350
550
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Numerical Problems & Solutions
15 cos150
45 cos550
45 sin550
60 sin400
60 cos400
15 sin150
15 kN
150
105 kN
75 kN
45 kN
400
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
(Q1.3)
45 sin550 45 - 45 cos55 - 45 sin55
60 + 60 cos40 - 60 sin40
------- --------------- ----------------
R ΣFx = ΣFy =
- 40.359 + 33.453
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Numerical Problems & Solutions
∑ 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
(Q1.3)
= + 33.453 kN
ΣFx = 40.359kN
ΣFy = 33.453 kN
R
θx
0
1-
69.39
tan ;tan
42.52
=
Σ
Σ=
Σ
Σ=
=Σ+Σ=
θ
θθy
x
x
x
y
x
yx
F
F
F
F
kNFFR
Answer:
-
Obtain the resultant of the concurrent coplanar forces
acting as shown in figure.
120
100kN50kNº
(Q1.4)
75kN
1202
3
3012
25kN
º
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Solution:
(Q1.4)
120
100kN50kN
º 100 sin θx
100 cos θx
= 100 cos 33.750 sin θx = 50 sin 26.3
75kN
1202
3
301
2
25kN
º
º50 cos θx
= 50 cos 26.3
= 100 sin 33.7
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
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Solution:
(Q1.4)
75kN
1202
3
3012
25kN
100kN50kN
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
Force X-comp. Y-comp
100 -100 cos33.7 -100 sin33.7
75 -50 cos26.3 +50 sin26.3
15 -25cos 63.43 -25 sin63.43 75 sin 3025 sin 63.4315 -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 =
∑FY = 50sin26.31- 100sin 33.69 – 75sin30 – 25sin63.43
(Q1.4)
Contd..
= 93.17kN= -93.17kN
-
∑Fx
∑FyR
θ
Answers:
(Q1.4)
R = √(∑Fx) 2 + (∑Fy) 2 = 119.14 kN
Θ = tan-1(∑Fy / ∑Fx ) = 51.44o
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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.
150N200N
(Q1.5)
120N
150N
50N
200N
45º
50°
110º
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�- 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
Solution:
(Q1.5)
150N
50N
200N
120N
45°
50°110 º
F5
α
R =250 N
20º
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�- Resolve the forces along X & Y axis
150N 200N
Solution:
(Q1.5)
F5y=F5 sin α
50N120N
45°
50°110 º
F5
α
R =250 N
20º
F5y=F5 sin α
F5x=F5 cos α
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Solution:(Q1.5)
150N
50N
200N
120N
45°
50°
F5
α
R =250 N
20º
F5y=F5 sin α
F5x=F5 cos α
Force X-comp. Y-comp
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
(Q1.5)
Y 5
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
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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.
75N
(Q1.6)
60°
45°
R=100N
50N
25N
70°
40°
-
(Q1.6)
75N
75 sin70
25 sin30 F4
F4 sin 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
60°
45°
R=100N
50N
25N
70°
40°
30°
75 cos70
25 cos30
50 cos45
R cos40
R sin40
α
F4 cos α
α
-
(Q1.6)
60°
45°
50N
25N
70°
40°
75N
30°
75 cos70
25 cos30
50 cos45
R cos40
R sin40
75 sin70
25 sin30
Force X-comp. Y-comp
F4 +F4 cosα +F4 sinα
50 - 50 cos45 +50 sin45
25 - 25 cos30 +25 sin30
75 +75 cos70 +75 sin70
R=100N
50N50 cos45R sin40------- --------------- ----------------
R ΣFx = - R cos40 ΣFy = -R sin40
= -100 cos40kN = - 100 sin40kN
ΣFx = -Rcos40 = F4cosα + 75cos70 – 50cos45 – 25sin60
F4cosα = - 45.25N
ΣFx = -Rcos40+ve
-
ΣFy = -Rsin40 = F4sinα + 75sin70+25cos60+50sin45
∴ F4sinα = -182.61N ;
+ve ΣFy = -Rsin40
(Q1.6)
α= 76.08º
45.25N
F4=188.13N
182.61N
F4cosα =
F4sinα =
Answers:
tan α = (F4sinα /F4cosα)
α = 76.08º
& F4 =188.13N
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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.
10 kN
70°
F4
(Q1.7)
60°
30°
15 kN
5 kN
70°
45°
Contd..
-
Solution:
60°
30°
10 kN
70°
45°
F4
R
(Q1.7)
F4 sin70 – 10cos 60 – 15cos 45 – 5cos 30 = 0; or, F4sin70 = 19.94
∑Fx = 0+ve
30°
15 kN
5 kN
45°
∴F4 = 21.22kN Contd..
-
Solution:
60°
30°
5 kN
10 kN
70°
45°
F4R
(Q1.7)
F4cos70 + 10sin60 – 15sin45 + 5sin30 = +R
∴ +R - 0.342F4 = 0.554
Substituting for F4 , R= +7.81kN
∑Fy = +R
+ve
15 kN
5 kN
Fig. 4
Answers:
F4 = 21.22 kN
R= +7.81kN
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Obtain the magnitudes of the forces P and Q if the resultant of the
system shown in figure is zero .
40°
Q
70°
100N
(Q1.8)
40°
60°
P
50N
70°
45°
Contd..
-
40°
60°
P
50N
Q
70°
45°
100N
(Q1.8)
Contd..
For R to be = zero,
∑Fx = 0 and ∑ Fy = 0
∑Fx = 0 :
-Psin45 – Qcos40 + 100cos70 + 50cos60 = 0
0.707P + 0.766Q = 59.2
+ve
-
∑Fy = 0
-Pcos45 + Qsin40 + 100sin70 – 50sin60 = 0
or, -0.707P + 0.642Q = -50.67 ------(b)
+ve
(Q1.8)
Answers:
Solving (a) & (b)
P = 77.17 N & Q = 6.058N
-
100N
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.
(Q1.9)
30°
100N
50N
Contd..
-
30°
100N
X - AXIS
Y-AXIS
30°
100N
50N50N
(Q1.9)
Rotating the axes to have X parallel to 50N,
∑Fx = +50 + 100cos30 = +136.6N
∑ Fy = +100sin30 = +50N
+ve
+ve
Contd..
Fig. 1.9
-
Fig. 6
30°
100N
X - AXIS
Y-AXIS
30°
100N
50N
50N
(Q1.9)
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-AXISθ
-
Resolve the 3kN force along the directions P and Q. Refer figure.
Q3kN
(Q1.10)
P
45°60°
20°
Contd..
-
3kN
P
45ºQ
45º
Q
60°
30° X – Axis
3kN
45º
(Q1.10)
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
30° X – Axis
P
-
Resolve the 5kN force along the directions P and Q. Refer Fig. 1.11.
Q3kN
(Q1.11)
P
45°60°
20°
Fig. 1.10 Contd..
-
Q
60°
20°X – AxisP
5kN
45º
(Q1.11)
Q
60°
20° X – Axis
5kN
45ºQ
60°
P
5kN45º
55°
P 20° X – Axis
P55°
60°
X
-
(Q1.11)
X – Axis
Using sine rule,
P/sin45 = Q/sin80 = 5/sin55
Answer :
Q
60°
P
45º
55°
800
5N
X – Axis
P = 4.32 kN, and Q = 6.01 kN
-
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
F2F1
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
The applied force can also tend to rotate the body about
an axis in addition to motion. This rotational tendency is
known as moment.
tendency of a force to make a
rigid body to rotate about an
axis.
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 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×dMA = 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
Sense of moment:
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
-
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.
A
F=100 N
(Q1.12)
Determine,
1. The moment of 100N force about ‘O’.
2. Magnitude of the horizontal force applied
60º
O
F=100 N2. 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
SOLUTION:
(Q1.1)
60º
A
F
1. The moment of 100N force about ‘O’.
= d
= 240 cos 60º = 120 mm.
Moment about ‘O’ = F × d = 100 × 120
= 12,000 N-mm
(Clockwise)
60ºO
d
-
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’
(Q1.12)
d
F
2. Magnitude of the horizontal force applied at ‘A’, which develops same
moment about ‘O’.
center ‘O’
= d = 240 sin 60º = 207.85 mm.
Moment about ‘O’ = F × d
= F × 207.85
= 12,000 N-mm (Clockwise)
Therefore,
F = 12,000 / 207.85 = 57.73 N
OA
60º
-
3) Solution:
F = M/d
Force is smallest when the perpendicular
distance is maximum so as to produce
same M.
(Q1.12)
A
F
3. The smallest force at ‘A’, which develops same effect about ‘O’.
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.
O
60º
F
-
4) Solution:
Distance along x-axis, X = M/F
= 12,000/240
A
(Q1.12)
4. How far from the shaft a 240N vertical force must act to
develop the same effect?
= 12,000/240
= 50 mm.
Distance along the shaft axis
d = X/cos 60
= 50/cos 60
= 100 mm
O
60º
X F
-
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).
θ
P
dθ
PP sinθ
P cosθ
d
Moment of Force P about the
point A,
P x d
θ
A
d
Algebraic sum of Moments of
components of the Force P
about the point A,
P cosθ x d1 + P sinθ x d2
=
A
d1
d2
-
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Y
Proof (by Scalar Formulation):Proof (by Scalar Formulation):
Let ‘R’ be the given force.
‘P’ & ‘Q’ are component forces of ‘R’.
‘O’ is the moment center.
Q
R
AO
ααααβ p
qrP
p, r and q are moment arms from ‘O’
of P, R and Q respectively.
α, β and γ are the inclinations of ‘P’, ‘R’ and ‘Q’ respectively w.r.to X –
axis.
γ
Y
X
-
We have,
Ry = Py + QyR Sinβ = P Sinα + Q Sin γ ----(1)From ∆le AOB, p/AO = Sin αFrom ∆le AOC, r/AO = Sin βFrom ∆le AOD, q/AO = Sin γFrom (1),
∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)
Ry
Py
RQ
P
p
q r
αααα
βγ
Y
B
CD
Qy
∴ 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’.
A O
αααα
X
-
COUPLE
Two parallel, non collinear (separated by certain
distance) forces that are equal in magnitude and opposite
in direction form ‘couple’.
d
F
The algebraic summation of the d
FHence, 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 & Apply two equal and opposite forces of same magnitude &
direction as Force F at point B, so that external effect is
unchangedF
AB
F
Fd
-
Of these three forces, two forces i.e., one at A and the other
F
AB
F
F
AB
F
=
M = F x dd
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.
-
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
W kN
w kN/m
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)
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
w kN/m
L
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)
L
W = ½ × L × w
2/3 ×L1/3 ×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.
(Q1.13)
and CB.
A B
CD60º
3m
4m
F=50 kN
-
A B
CD60º
36.87º
60º
d
a) By Definition of Moment:
To determine ‘d’:
AC =
CAD = tan-1(3/4) = 36.87º
ECD = 60º
3m
4m
F=50 kN
m534 22 =+ 23.13
(Q1.13)
A BdE
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.
-
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
3m
4m
F=50kN
Fx
Fy
(Q1.13)
A B
= - 25 × 3 + 43.3 × 4
= + 98.20kNm.
-
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’.
50kN
(Q1.14)
30º
80kN
30kN
60kN
60º
200mmD
-
Resultant & its inclination:
Resolving forces
+ ΣFx = Rx = +30 + 60 cos30º – 50 cos60º
= +56.96kN.
+ ΣFy = Ry = -80 + 60 sin30º + 50 sin60º
50kN
60kN
60 cos30
60 sin30
50 sin 60
50 cos 60
(Q1.14)
+ 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
30kN60º
200mm
-
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.
∴
(Q1.14)
of action of R.
= 2000 × d.
∴ d =2000/57.35 = 34.87mm
-
Find the resultant and its position w.r.to ‘O’ of the non-
concurrent system of forces shown in the figure.
F2=500NF1=2500N
F5=2000N
Ө2Ө41
(Q1.15)
F3=1000N
F4=1500N1m
1m
Ө2Ө4
1
1
Ө5
O
-
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
F3=1000N
F2=500NF1=2500N
F5=2000N
1m
1m
Ө2Ө4
1
1
Ө5
O
(Q1.15)
+ 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 ↑
F4=1500N1m
-
∴ 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’
x
y
R
R
Rx
RyR
ӨR
(Q1.15)
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
d = 1.43 m from O.
-
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.
(Q1.16)
60º
30 kN
120 kN
50 kN
6 m
1 m2 m
3 m
A B
-
60º
30 kN
120 kN
50 kN
6 m
1 m
2 m3 m 30º
A B
(Q1.16)
+ ∑Fx = Rx = 50 – 30 × cos30 = 24.02 kN
+ ∑Fy = Ry = -120 – 30 × sin 30= -135 kN
Resultant, R=
6 m
kNRR yx 12.13713502.242222 =+=+
-
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.
(Q1.16)
From A, X = 6 –2.67= 3. 33 m.
Middle 1/3rd distance is between 2m and 4m.
2m
-
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.
(Q1.17)
50 N
100 mm
50 mm
30 mm
30º
B
A
-
Force – Couple System at A:
Fx = 50 ×sin 30= 25 N.
Fy = 50 × cos 30= 43.3 N50 N
100 mm
50 mm
30 mm
30º
50N
50 sin 30
B
A
60º
(Q1.17)
50 cos 30
50N
-
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
+ ∑M = F ×50-F ×100
100 mm
50 mm
30 mm
30º
50N
Fx=50 sin 30
B
A
(Q1.17)
+ ∑MA= Fx×50-Fy×100
= 25×50 - 43.3×100
= -3080 N-mm.
= 3080 N-mm
Fy=50 cos 30
50N
100 mm
50 mm
A
Fy=50 cos 30
Fx=50 sin 30
MA=3080N-mm
-
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.
50 mm
30 mmB
A
θ
150 N
(Q1.17)
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
100 mm
50 mm
A
Fy=50 cos 30
Fx=50 sin 30θ=46.8º
150N
-
EXERCISE PROBLEMS
Q1.18 A body of negligible weight, subjected to two forces F1=1200N, and F2=400N acting along the vertical, and thehorizontal respectively, is shown in figure. Find the componentof each force parallel, and perpendicular to the plane.
= 1200 NF
1. Resultant of force system
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
= 1200 N
34
Y
F2
F1
= 400 N
-
2. Determine the X and Y components of each of the forces shown in
12
5
390 N
X
YF2 =
FIG.2.
EXERCISE PROBLEMS1. Resultant of force system
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,
F3X = -306.42 N, F3Y= -257.12N )
30º40º
5
300 N400 N
F1 =F3 =
FIG. 2
-
600N
800N
20º40º
3. Obtain the resultant of the concurrent coplanar forces shown in FIG.3
EXERCISE PROBLEMS1. Resultant of force system
200N
20º40º
30º
FIG. 3
(Ans: R = 522.67 N, θ = 68.43º)
-
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.
EXERCISE PROBLEMS1. Resultant of force system
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
-
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
EXERCISE PROBLEMS1. Resultant of force system
P
Q = 5 kN
R20º
40º
Fig. 5
(Ans: P = 9.4 kN , R = 12.66 kN)
-
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
1200 N
EXERCISE PROBLEMS1. Resultant of force system
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
1200 N
800 N60º
θ
F
ROD
COLLAR
Fig.6
-
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.
8kN5kN
EXERCISE PROBLEMS1. Resultant of force system
Q30º
α
Fig. 7(Ans: α = 10.7 º, Q = 9.479 kN )
-
8. Determine the resultant of the parallel coplanar
force system shown in figure.
1000 N
2000 N
600 N o
60º
60º 30º
10º
EXERCISE PROBLEMS1. Resultant of force system
400 N2000 N 60º10º
(Ans. R=800N towards left, d=627.5mm)
-
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.
EXERCISE PROBLEMS1. Resultant of force system
10N
40N
a
a
30N
20N
AB
CD
(Ans:R=28.28N, θ=45º, x=1.77a)
-
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.
EXERCISE PROBLEMS1. Resultant of force system
(Ans: R = 125 N, x = 3.06 m)
-
11. Reduce the given forces into a single force and a
couple at A.
70.7 kN
EXERCISE PROBLEMS1. Resultant of force system
100 N80 N
1m
1.5m
70.7 kN200 kN
A
30º45º
30º
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
-
12. Determine the resultant w.r.t. point A.
EXERCISE PROBLEMS1. Resultant of force system
150 N
1.5m3m1.5m
150 Nm
(Ans: R = 450 kN, X = 7.5 kNm)
A
1.5m
500 N100 N