1. ENGINEERING
MECHANICS MADE
SIMPLE
COMPILED
BY
Prof.S.RAJENDIRAN,
Ashoka Institution

2. R2
= d2
+g2
= (m+n)2
+g2
= m2
+n2
+2mn+g2
= F12
+ 2F1F2cosά+F22
-n2
+n2
hence R = F12
+ F2
+2F1F2cosά
Pl note that if F1=F2= R to find angle between F1andF2 apply triangular
formula
i.e. =
Parallelogram law
If two forces acts on a particle ,are represented as
adjacent sides of the parallelogram, then the resultant is the
diagonal of the parallelogram passing through the
point of application of the forces
Proof
Force F1andF2 acts at point o(fig)
Let the resultant is R
let us extend force vector as shown
And all construction can be as per fig
R = F12
+ F2
- 2F1F2cosά
F2
F1
R
m n
d= (m+n)
g
ά ά
F2
o
Then We get

3. Equilibrium
Is a force which brings the body to state of equilibrium. It is equal in
magnitude but opposite in direction to the resultant force.
Centre of Gravity
Centre of gravity can be defined as a point through which resultant force
of gravity of the body acts.
Centroid
The centre point obtained by dividing first moment of line segment by the
total length of the line or first moment of area about reference axis with a
total area or first moment of volume about reference axis by total volume
is termed as centroid.
Moment of inertia of area.
Second moment of area is sum of area times the square of the distance
from the axis is considered, which is the plane of the area. This is also
termed as second moment of area or it is moment of a moment of an
area.

4. Polar Moment
It is second moment of area about an axis normal to the area is
termed as polar moment of area.
I.e. IZZ =IXX+IYY
Moment
Moment of a force is a measure of its tendency to cause a body
to rotate about a specific point or axis. This is different from the
tendency for a body to move, or translate, in the direction of the
force. In order for a moment to develop, the force must act upon
the body in such a manner that the body would begin to twist. This
occurs every time a force is applied so that it does not pass
through the centroid of the body. A moment is due to a force not
having an equal and opposite force directly along it’s line of action.

5. F
d
Where
‘d' is perpendicular distance between force ‘F’ and to the end of cantilever beam
‘F’ Force act on the diameter of the beam where beam cross section is circular
M = Moment
Moment of force M = F X d
Z
Y XF
d
O
Moment of force “F” about “O” is given by M = F X d
Where “d” is perpendicular distance between force “F” and
point “O”
M = Moment, (force F creates rotary motion about O)
Example.1
Example.2
Moment

6. F
F
d
Couple
• Two parallel forces equal in magnitude but opposite in direction separated by
particular distance acts on a body form a Couple
• System of forces whose total of force F = ∑ Fi is zero.
• Couple is a pure moment. Please note that moment of a couple is independent
of reference point.
• Couple is a system of forces with a resultant moment but no resultant force.
• Magnitude of couple is F X d (which is also called moment of couple if d
perpendicular distance between forces)
• Where d is the perpendicular distance between forces of F
• Since the forces are equal and oppositely directed, the resultant force is zero.
But the displacement of the force couple create a couple moment

7. Torque
F
R
Torque = F X R
Where
“F” is a force
“R” is a radius
It is true that force F act radially as shown. Torque is a resultant of
couple. Torque is a moment of force. Force applied on diameter like
moment will not make it rotate whereas force applied at radius as shown
in above fig will make it rotate.
The forces have a turning effect or moment called a torque about an
axis which is normal to the plane of force.
Cantiliver beam

8. force
A force is a vector quantity that, when applied to
some rigid body, has a tendency to produce
translation (movement in a straight line) or
translation and rotation of body. When problems
are given, a force may also be referred to as a
load or weight.
Characteristics of force are the magnitude,
direction(orientation) and point of application.

9. Scalar and vector quantityScalar and vector quantity
Scalar Quantity has magnitude only (not
direction) and can be indicated by a point on a
scale. Examples are temperature, mass, time
and dollars.
Vector Quantities have magnitude and
direction. Examples are wind velocity, distance
between to points on a map and forces.

10. Collinear
Collinear : If several forces lie along the same
line-of –action, they are said to be collinear.
Coplanar When all forces acting on a body are
in the same plane, the forces are coplanar.

11. Introduction - DefinitionIntroduction - Definition
Concurrent Force Systems:
A concurrent force system contains forces
whose lines-of action meet at some one point.
Forces may be compressive (pushing)

12. Lami’s theorem
Law of sines
Law of cosines
a b c
Sin A sin B sin C
Law of sines

13. Moment of a couple
F
F
D
d
Couple is formed when two forces equal in magnitude act at a distance of D in opposite direction
Magnitude of couple is FD where as moment of couple is Fd

14. Varignon’s principle (Theorem)
θ
θ1
θ2 F1
F
F2
d1
d
d2
Fx1
Fx2
Fx
O
A
x
Y
θ
θ1
θ2
Proof
By principle of resolving forces about X axis we know that
Fx = Fx1 + Fx2
Multiplying OA the above equation is equal to
OAFx = OAFx1 + OAFx2 ------------------- (Eq 1)
We also know that Fx is equal to F cos θ, similarly Fx1 is equal to F1 cos θ1
and Fx2 is equal to F2 cos θ2
therefore rewriting Eq 1
OAFCos θ = OA F1 cos θ1 + OA F2 cos θ2 -------------- (Eq 2)
Statement
Varignon’s theorem states that the moment
of a force F about an axis is equal to the
sum of moments of its compliments (F1 &
F2) about the same axis.
i.e, Fd = F1d1 + F2d2

15. We know that
θ1
θ1
O x
Y
A
d1= OA Cos θ1
F1
OA Cos θ1 = d1 from above fig
Substituting the above d, d1 and d2 values the Eq 2 can be rewritten
as
Fd = F1d1 + F2d2
Hence Varinon’s theorems is proved
O
A
x
Y
θ2
d2 =
θ2
F2
OA Cos θ2 = d2 from the above figOA Cos θ = d from above fig
θ
O
A
F
θ
Y
x
d = OA Cos θ
OA Cos θ2

17. Principle of transmissibility
AB is line of action
Looking at the above fig we know that when an object acted upon by
a force F, when line of action (AB) remains the same whereas changing
the point of application of force without changing the magnitude of
force. The condition of the object remains same i.e. motion, or
equilibrium.
A
B
Object
A
B
Object
Point of application
Point of application
=
F
F

18. Definition
When an object subjected to rotation and translation motion,
then at an instant a point exists (which may inside the object or out
side the object) which is the plane of other points may looked as
pure rotation about this point and that point has the velocity is zero,
which is called instantaneous centre. Axis passing through and
directed at right angles to the plane of motion is called
instantaneous axis of rotation
Points to remember
• The instantaneous center is not a fixed point its location keeps on
changing at every instant and path traced or locus by it is called
centrode
• The instantaneous center may within the body or outside the body if
it is within the body then it is called body centrode if it is outside the
body it is called space centrode
• The velocity at instantaneous center is zero
Instantaneous center (Instantaneous centre of rotation)

20. The centre point obtained by dividing first moment of area about the reference axis
by the total area termed as centre of area.
The centre point obtained by dividing first moment of volume about the reference
axis by the total volume termed as centre of volume.
Volume centre x = ∑ΔVi
Xi
V
Volume centre y = ∑ΔVi
Yi
V
Centroid or Centre of Area or area center
Centroid x = ∑ΔAi
Xi
A
Centroid y = ∑ΔAi
Yi
A
Centre of Volume or volume center
Volume centre
Centroid
WHERE V=∑ΔVi
WHERE A=∑ΔAi

21. • A differential strip parallel to the x axis is chosen
for dA.
dyldAdAydIx == 2
• For similar
triangles,
dy
h
yh
bdA
h
yh
bl
h
yh
b
l −
=
−
=
−
=
• Integrating dIx from y = 0 to y = h,
( )
h
hh
x
yy
h
h
b
dyyhy
h
b
dy
h
yh
bydAyI
0
43
0
32
0
22
43






−=
−=
−
== ∫∫∫
12
3
bh
I x=
Determine the moment of inertia of a triangle with respect to its base.

22. • An annular differential area element is
chosen,
( ) ∫∫∫ ===
==
rr
OO
O
duuduuudJJ
duudAdAudJ
0
3
0
2
2
22
2
ππ
π
4
2
rJO
π
=
• From symmetry, Ix = Iy,
xxyxO IrIIIJ 2
2
2 4
==+=
π
4
4
rII xdiameter
π
==
Determine the centroidal polar moment of inertia of a circular area by direct integration.

23. Centroid Location
Symmetrical Objects
Centroid location is determined by an
object’s line of symmetry.
Centroid is located on the
line of symmetry.
When an object has multiple lines of symmetry, its centroid is
located at the intersection of the lines of symmetry.

24. H
B
Centroid Location
The centroid of a triangle is located at a
distance of 1/3 its height and 1/3 its base.

25. Centroid Location
The centroid of a ½ circle or semi-circle is
located at a distance of 4*R/3π away from the
axis on its line of symmetry
•
=
π
4 R
3
•
=
π
4 2in.
3
0.849 in. = 0.8in
.849in.

26. Centroid Location Equations
Complex Shapes
i i
i
y A
y=
A
∑
∑
i i
i
x A
x=
A
∑
∑ i i
i
z A
z=
A
∑
∑

27. 9 - 27
• Second moments or moments of inertia of an area with
respect to the x and y axes,
∫∫ == dAxIdAyI yx
22
• Evaluation of the integrals is simplified by choosing
dΑ to be a thin strip parallel to one of the coordinate
axes.
• For a rectangular area,
3
3
1
0
22
bhbdyydAyI
h
x === ∫∫
• The formula for rectangular areas may also be applied to
strips parallel to the axes,
dxyxdAxdIdxydI yx
223
3
1 ===
Second moments or moments of inertia of an area with respect to the x and y axes

28. Moment of inertia off regular shape

29. Polar Moment of Inertia
• The polar moment of inertia is an important parameter in
problems involving torsion of cylindrical shafts and rotations of
slabs.
∫= dArJ 2
0
• The polar moment of inertia is related to the rectangular
moments of inertia,
( )
xy II
dAydAxdAyxdArJ
+=
+=+== ∫∫∫∫
22222
0

30. Radius of Gyration of an Area
• Consider area A with moment of inertia Ix. Imagine
that the area is concentrated in a thin strip parallel
to the x axis with equivalent Ix.
A
I
kAkI x
xxx == 2
kx = radius of gyration with respect to the x
axis
• Similarly,
A
J
kAkJ
A
I
kAkI
O
OOO
y
yyy
==
==
2
2
222
yxO kkk +=

31. Parallel Axis Theorem
• Consider moment of inertia I of an area A with respect to
the axis AA’
∫= dAyI 2
• The axis BB’ passes through the area centroid and is called
a centroidal axis.
( )
∫∫∫
∫∫
+′+′=
+′==
dAddAyddAy
dAdydAyI
22
22
2
2
AdII += parallel axis theorem

32. Parallel Axis Theorem
• Moment of inertia IT of a circular area with respect to a
tangent to the circle,
( )
4
4
5
224
4
12
r
rrrAdIIT
π
ππ
=
+=+=
• Moment of inertia of a triangle with respect to a centroidal axis,
( )
3
36
1
2
3
1
2
13
12
12
2
bh
hbhbhAdII
AdII
AABB
BBAA
=
−=−=
+=
′′
′′

33. Principal Axes and Principal Moments of Inertia
Given
∫
∫∫
=
==
dAxyI
dAxIdAyI
xy
yx
22
we wish to determine moments and product
of inertia with respect to new axes x’ and y’.
θθ
θθ
θθ
2cos2sin
2
2sin2cos
22
2sin2cos
22
xy
yx
yx
xy
yxyx
y
xy
yxyx
x
I
II
I
I
IIII
I
I
IIII
I
+
−
=
+
−
−
+
=
−
−
+
+
=
′′
′
′
• The equations for Ix’ and Ix’y’ are the parametric
equations for a circle,
( )
2
222
22
xy
yxyx
ave
yxavex
I
II
R
II
I
RIII
+




 −
=
+
=
=+− ′′′
• The equations for Iy’ and Ix’y’ lead to the same circle.
θθ
θθ
sincos
sincos
xyy
yxx
−=′
+=′Note:

34. Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips






−===






−==





−=
b
x
hyyxx
dx
b
x
hdxydA
b
x
hy
elel 1
11
2
1
2
1
Integrating dIx from x = 0 to x = b,
( )
bb
b
elelxyxy
b
x
b
xx
hdx
b
x
b
xx
h
dx
b
x
hxdAyxdII
0
2
4322
0
2
32
2
0
2
2
2
1
83422
1






+−=








+−=






−===
∫
∫∫∫
22
24
1 hbIxy =

35. 9 - 35
Apply the parallel axis theorem to evaluate the product of inertia with respect to the
centroidal axes.
hybx 3
1
3
1 ==
With the results from part a,
( )( )( )bhhbhbI
AyxII
yx
yxxy
2
1
3
1
3
122
24
1 −=
+=
′′′′
′′′′
22
72
1 hbI yx −=′′′′

36. 9 - 36
Moments of Inertia of Common Geometric Shapes

38. Mt
ObjectF1
F2
F4F5
F3
C.G=
Before D'Alembert's principle
When an object subjected to
multiple forces (F1, F2,…..etc)
then all forces reduces to one
resultant = R and one net torque =
Mt
which produces translation and
rotary motion.
As per D'Alembert's principle
R – ma = 0
Mt
– Iά = 0
D'Alembert's principle
D'Alembert's principles introduces that
body will be in dynamically equilibrium such that
inertia force –ma acts opposite to the translation
motion produced by resultant and – Iά inertia force
due to rotation in the opposite direction of net
torque Mt
Mt
– Iά
– ma
R
C.G
= 0
The force system consisting of external forces and inertia force can be
considered to keep the object in equilibrium. Since the resultant force externally
acting on the object is not zero, the object is said to be in dynamic equilibrium. This
known as D'Alembert's principle.

39. Proof
Let us consider a disc rotates with a constant angular velocity on the surface
of the floor as shown in the fig below which will have translation velocity shown
in Fig 1 and rotation velocity shown in Fig 2. Superimposing Fig 1 & 2 we get Fig 3,
where at point M the velocity equal to zero since translation velocity cancels
rotation velocity.
From the fig 1, 2 & 3 the
instantaneous centre is well
defined where M is the
instantaneous centre and at that
point velocity is zero
M
V
Translation velocity
Fig 1
V
V
M
V
Rotation velocity
Fig 2
M
2V
Fig 3
Rotation velocity Translation velocitycancels
-V
+V
Velocity at point M = 0

40. Vω
R
O
Object
V1
V2
A
B
O
Object
Different methods of locating instantaneous centre
Since V = ωR we can find out R = V/ ω on the
velocity line make a perpendicular line as shown
above and mark the distance = R then O is the
instantaneous centre
Similarly, when an object subjected by rotary and
translation motion if we know the velocity of points A
& B as shown above (which are not parallel) are drawn
and we can draw perpendicular lines from A & B as
shown above (angular velocity is perpendicular to
velocity) both the lines will intersect at point O. which
is called instantaneous centre
Let us assume that an object subjected to rotary and
translation motion and if we know the angular
acceleration equal to ω and translation velocity equal
to V at an instant shown below.
II Method
I Method

41. III Method
From the above fig we know that an object subjected to rotary and
translation motion point A1 occupies the position of A2 and point B1
occupies the position of B2. Join A1 A2 and B1 B2 draw a perpendicular
bisector for both the lines which will intersect at point O then point o is called
instantaneous centre.
Centrode
The locus of the instantaneous centre is called centrode. If it lies inside the
object, then it is called body centrode, if it lies out side the object it is space
centrode.
O
A1
B1
A2
B2
Object

43. Area Theorem
If a line or profile rotates about an axis then
the surface area formed by the line about the
axis is equal to product of the line length and
distance traveled by the C.G of the line.
Let us consider that cone of height h and
base radius R is being formed by the rotation of
the line AB (slant length square root of h2
+ R2
))
about the axis MN .
Then surface area of the cone as per Pappus
theorem is AB multiplied by distance traveled
by the C.G
M
N
B
A
R
R/2
R/2
C.G
h
Slant length AB=√ h2 + R2
Distance traveled by C.G = 2π R/2 = πR
Therefore surface area = πR √ h2 + R2
Theorems of Pappus - Guldinus.

44. M
N
B
A
R
C.G
R/3
h
R/3
Pappus Theorems on Volume
If a cross section of a solid revolves
about an axis then, volume formed by
generation of the cross section is
equivalent to product of the area of
cross section of area and distance
traveled by the C.G while cross section
revolves
Let us consider the cone cross section
is a right angle triangle shown below
rotates about axis MN then volume of the
cone generated equal to product of
cross section area of triangle (R h/2) and
distance made by the C.G equal to2πR/3
Then Volume of cone = π R2
h
3

45. Resolution of force to couple and a force
F
F
F F
F
Fig A Fig B Fig C
= =
Object
M M
N
Let us assume that a single force F act on an object (at point M) as shown in Fig A.
And Let us also introduce equivalent opposite force at point N (shown in Fig B).
By introduction of the forces at N the condition of the object remains same (since
we have introduced force equal in magnitude but opposite in direction at point N and
net force at Point N is zero)
Now if we look at Fig c. we were able to understand that the right side force at N
and left side force at M forms a couple. The net result is a force and a couple as
shown in Fig C. Thus the single force is resolved into a couple and a force.

46. Centre of Area or area center
The centre point obtained by dividing first moment of area about the reference
axis by the total area termed as centre of area.
Centre of mass or mass center
The centre point obtained by dividing first moment of masses about the
reference axis by the total mass termed as centre of mass.
Centre of mass, Centre of volume, Centre of area
Xi, Yi, are the coordinates of ΔMi/ΔAi/ΔVi
ΔMi/ΔAi/ΔVi
Object
Xi
Yi
Mass Centre
mass centre x = ∑ΔMi
Xi
M
mass centre y = ∑ΔMi
Yi
M
WHERE M=∑ΔMi

47. Drawing a FBD, steps
1). Decide which body to analyze.
2). Separate this body from everything else and sketch the contour,
3). Draw all applied forces (weight).
4). Draw all reactions.
5). Include any necessary dimensions and coordinate axis.
If you don't know a direction assume a direction and let the sign of the
answer tell you if the direction is correct or not.
Rules:
The magnitude and direction of known forces should be clearly indicated
(usually applied forces)
Indicate the direction of the force exerted on the body, not the force exerted
by the body.
Unknown forces are usually the reactions (constraining forces).
Reactions are supports and connections in 2-D.
Free - body diagrams

49. Friction
Dry Friction
(Coulomb Friction)
Fluid Friction
W
N
F
Fig 1
P
Friction
• Occurs un-lubricated surfaces
are in contact during sliding
• Friction force always oppose
theθ
• sliding motion
• Occurs when the adjacent layers
in a fluid (liquid, gas) are moving
at different velocities
• This motion causes friction
between fluid elements
• Depends on the relative velocity
between layers
• No relative velocity –no fluid
friction
• Depends on the viscosity of fluid
– (measure of resistance to
shearing action between the
fluid layers)
From the above fig it is seen that an object having weight W rest on a floor is
pushed by a force F. the contact area of the floor and the area of the object which
in contact with the floor experience in frictional force (F) which is against the
motion of the object.
θ

50. Limiting Friction
The maximum value of frictional force which comes into play, when the motion is impending,
is known as limiting friction. In this case, applied force is less than the friction force.
Static Friction
The frictional force for which the body at rest is called static friction.
Dynamic Friction
If the value of applied force is exceeding the limiting friction, the body starts moving over
the floor and the frictional resistance are experienced by the body is known as dynamic
friction.
Sliding Friction
It is the friction experienced by a body when it slides over the other body
Rolling Friction
It is the friction experienced by the body when it rolls over a surface.
Coefficient of friction
The limiting friction bears a constant ratio to the normal reaction between the two forces is
called coefficient of friction. = F/N
Where,
F – Limiting friction
N – Normal reaction
Which is otherwise Coefficient friction denoted as μ,
Hence, μ = F/N or = tan θ

51. Laws of Coloumb Friction
• The force of friction always acts in a direction opposite to that in which the
body tends to move.
• Till the limiting value is reached, the magnitude of friction is exactly equal to the
force which tends to move the body.
• The magnitude of the limiting friction bears a constant ratio to the normal
reaction between the two surfaces
• The force of friction depends upon the roughness/smoothness of the surfaces
• The force of friction is independent of the area of contact between the two
surfaces
• After the body starts moving, the dynamic friction comes into play, the
magnitude of which is less than that of limiting friction and it bears a constant
ratio with normal force. This ratio is called coefficient of dynamic friction.
P
F
F limit
Static
state
Dynamic State

52. The maximum inclination of the plane on which a body free from external
applied force can repose (sleep) is called angle of repose.
θ
N
F
θ
W
From the above fig we conclude that
N = W cos θ and F= W sin θ
Since,
F/N = W sin θ / W Cos θ
F/N = tan θ = μ
F/N = tan ά (ά – angle of friction)
θ = ά
The value of angle of repose same as the value of limiting friction
Angle of repose

53. Cone of Friction
W
F
P
ά ά
N
F
R
When a body having impending motion in the direction of P, the frictional force will
be the limiting friction and the resultant reaction R will make limiting friction angle with
the normal as shown in the fig if body is having impending motion with some other
direction, again the resultant reaction makes limiting frictional angle ά with the normal
in that direction. Thus, when the direction of force P is gradually changed through 3600
the resultant R generates right circular cone with the semicentral angle equal to ά.
If the resultant reaction is on the surface of this inverted right circular cone whose
semi central angle in limiting frictional angle ά, the motion of the body is impending. If
the resultant is within the cone the body is stationary. This inverted cone with semi
central angle, equal to limiting frictional angle ά, is called cone of friction.

54. Impulse momentum = Final Momentum – Initial Momentum
We know R is the resultant force acting on a body of mass m, then from Newton’s
second law,
R=ma
But acceleration a = dv/dt
R=m dv/dt
Rdt = mdv
∫ Rdt = ∫ mdv
If initial velocity is u and after time interval t it becomes v, then
t v
∫ Rdt = m [v] = mv - mu
0 u
Impulse Momentum – mv – mu
The impulse momentum principle can be stated as the component of the resultant
linear impulse along any direction is equal to the component of momentum in that
direction.
Impulse Momentum

55. If mv- mu is equal to zero then, momentum is conserved
The principle of conservation of momentum may be stated as the momentum is
conserved in a system in which resultant force is zero. In other words in a system
the resultant force is zero, initial momentum will remain equal to the final
momentum.
Coefficient of restitution may be defined as the ratio between relative velocity of
separation and relative velocity of approach along the line of impact, when two
bodies collide. v2-v1
u1-u2
Conservation momentum
Coefficient of restitution
e =

56. The displacement of the particle in rotation is measured in terms of angular
displacement θ, where θ is in radins. Thus when a particle moves from position A to
B, the displacement is θ as shown in the Fig. 1. This displacement has the direction
– clockwise or counter clockwise.
S
B
A
θ Reference Axis
r
The distance traveled by a particle A to B is S which is equal to r θ
S = r θ
ds/dt = r dθ/dt
v = r ω, (since dθ/ dt = ω)
dv/dt = r dω/dt
at
= r ά (where dω/dt = ά)
ά = at
/r
we know that an
= v2
/r = (r ω )2
/r = r ω2
Relation between ω, v, r and ά

57. Span
A B
Supports
Beam
Span
A B
Supports
Beam
OverhangingOverhanging
Pl note that If there is no inclined or horizontal load on the beam, then the reaction at the
hinged support will be vertical, otherwise the reaction will be inclined to the vertical.
Beams are classified on the basis of supports.
Simply supported beam
When supports are provided at both ends of the beam is called simply supported beam.
When the beam is projecting beyond the supports is called overhanging beam
Beams

58. Cantilever beam, Fixed beam and Continuous Beam
Fig A Fig B Fig C
When one end is fixed other end is free it is called cantilever beam (Fig A)
When both ends are fixed it is called fixed beam (Fig B)
When beam is supported more than two positions it is called Continuous Beam
(Fig C)