Tutorial: 01
Mechanics of solid
Scope & Objective of the course:
The subject of mechanics of solids deals with determination of strength, deformation and
stability of structural and machine elements. The methods are based on
Laws of Newtonian
mechanics, applied to bodies in static equilibrium geometry and experimentation. These laws are
applied to simple situations with engineering judgment to arrive at results of significance to the
designer. At the end of the course the stu
dent will be in a position to design and analyze simple
structural elements, which involve calculation of stresses, strain and deformation. This is an
essential feature in a design process.
What is a Mechanical Engineering?
Mechanical engineering is one
of the largest, broadest, and oldest engineering disciplines.
Mechanical engineers use the principles of energy, materials, and mechanics to design and
manufacture machines and devices of all types. They create the processes and systems that drive
technolo
gy and industry. Mechanical engineering plays a dominant role in enhancing safety,
economic vitality, enjoyment and overall quality of life throughout the world. Mechanical
engineers are concerned with the principles of force, energy and motion. The men an
d women
who work as mechanical engineers are professionals with expert knowledge of the design and
manufacture of mechanical systems and thermal devices and processes. Some examples of
products and processes developed by mechanical engineers include engine
s and control systems
for automobiles and aircraft, electric power generation plants, lifesaving medical devices and
consumer products ranging from air conditioners to personal computers and athletic equipment.
They also design the machines that mass

produ
ce these products. Virtually every aspect of life is
touched by mechanical engineering. If something moves or uses energy, a mechanical engineer
was probably involved in its design or production
.
Mechanics:
Mechanics is the physical science of forces and
motion. It concerned with deformation of
a body acted on by mechanical , thermal or other loads.
During study of mechanics of solid we
mainly concerned with applied mechanics, i.e. applying the principle of mechanics to system of
practical interest in orde
r to :
Understand their behavior.
Develop rational rule for their design.
Solid:
Solid
is one of the three classical states of matter (the others being gas and liquid). It is
characterized by structural rigidity and resistance to changes of shape or
volume.
Here we
mainly concerned with ordinary engineering structural members: rod, beam, shaft, etc.
The mechanical properties of materials describe characteristics such as their strength and
resistance to deformation. For example, steel beams are used i
n construction because of their
high strength, meaning that they neither break nor bend significantly under the applied load.
Mechanical properties include elasticity and plasticity, tensile strength, compressive strength,
shear strength, fracture toughnes
s, ductility (low in brittle materials), and indentation hardness.
So
Solid
mechanics
is the study of the behavior of solid matter under external actions such as
external forces and temperature changes.
Assignment 01
Q. what are the fundamental
Principle
s
of Mechanics?
Forces:
Any action that tends to change the state of rest or change the state of rest of a body to which it is
applied.
There are many kinds of forces, such as gravity force, the force of magnetic attraction,
steam or gas pressure in a cylinder, wind pressure, atmospheric pressure and frictional resistance
between the surfaces.
For the complete definition of a force we mus
t know (a) its magnitude, (b)
its point of application, and (c) its direction
Force is a vector quantity and it makes a interaction between ‘bodies’ or better to say vector
interaction
between the bodies. Force
interaction
may occur either when there is di
rect contact
between systems
(Example spring force)
or occur between
systems
which are physic
all
y
separated (example electric m
agnetic and gravitational forces)
.
Force interaction have two
principle effects, (1) they tend to alter the motion of the systems
involved, and (2) they tend to
deform or distort the shape
of the system.
SI unit of force is
Newton
(N). 1 N is defined as that
force which gives an acceleration of 1 m/s
2
to a mass of 1 kg.
Moment of a Force:
Let
F
be a force vector applied at p and let o be a fixed point in space. The moment or
torque of
F
about the point
o
is defined as the vector cross product r x F, where r is the displacement
vector form
o
to
p
which is shown in fig 1.1 below
Fig. 1.1 The
moment of a force about a point ‘
O
’
The moment itself is a vector quantity. Its direction is perpendicular to the plane containing
position vector
r
and force
F
.
Fig 1.2 Magnitude of cross product r x F is the area of parallelogram
Moment of force F
about
O
i.e.
M = r x F
=
Fr
sin ¢
Where
F
and
r
are the magnitudes of the vectors F and r and ¢ is the angle between r and
F shown in fig 1.2.
M = h
F
whe
re h = r sin¢
i.e. it is the length of the perpendicular dropped from o to AB (line of action of
force).
SI unit of moment is newton

meter (Nm)
.
If two

dimensional structure (X

Y plane)
F= F
x
i + F
y
j & r = x i + y j
M = (x i + y j) x (F
x
i + F
y
j)
= k (
xF
y

y
F
x
)
Or M = k h
F
Where k is unit vector in the z

direction
perpendicular to the plane of X and Y.
Fig. 1.3 Moment about
‘O’
Moment about a line (OQ)
which passes through ‘
O’
The
c
omponent of r x F along
OQ
is called the moment of F about the line or axis
OQ
,
refers
fig.
1.1.
. The magnitude of this component along the line
OQ
is the
projection of the vector M along
OQ
. This is given by the dot product of M and a unit vector in the direction of
OQ
.
M
alog
OQ
= (r x F) cosα
= h
F
cosα
Example 1.1
Determine the moment M about the shaft axis
oo
due to the force P applied to the
crank handle as shown in fig.
below.
Sol.
When several forces F
1
, F
2
, F
3
, …….F
n
act, there total moment about a fixed point o is
defined as the sum
r
1
x F
1
+ r
2
x F
2
+ r
3
x F
3
+ ……….. + r
n
x F
n
=
∑
j
x F
j
where the r
j
are displacement vectors from o to points on the lines of action of the F
j
.
Couple:
when there are two equal and parallel forces F
1
and F
2
which have
opposite sense, such a co
nfigurati
on of forces is called a couple which is shown in
fig. 1.4.
Fig. 1.4 the moment of a couple about the point
O
Sum of moments (M) of F
1
and F
2
about
O
=
r
1
x F
1
+ r
2
x F
2
From fig 1.4, r
1
= r + a
&
F
1
= F
2
= F
,
M = (r
1
+ a) x F
1
+ r
2
x F
2
= r x (F
1
+ F
2
) + a x F
1
Where
r
1
and
r
2
are vectors
to arbitraya point on the lines of
action of
F
1
and
F
2
.
Now F
1
and F
2
are of equle magnitude and opposite sense and therfor cancel when added at the
same point .
F
1
+ F
2
= 0
M = a x F
1
Fig. 1.5 Couple is represented by a moment vector.
M = a x F
= h
F
Where a is the displacement vector going from an arbitrary
point on F
2
to an arbitrary point on
F
1
.
h is the perpendicular distance between the vectors F and
–
F.
Example
1.2
A beam AB of length
L
is supported as snow in fig.
below
and subjected to equal
but opp
os
ite
vertical forces P at its two ends. Find the reactions at the supports C and D.
Sol.
Conditions for Equilibrium
When several forces
F
1
, F
2
, ….F
n
act on a particle
or in the system
so the necessary and
sufficient condition for the
particle
or system
to be in equilibrium are
:
F
1
+ F
2
+ F
3
+ ………….F
n
=
∑
j
= 0
……………………..(Eq
1
)
r
1
x F
1
+ r
2
x F
2
+ r
3
x F
3
+ ……….. + r
n
x F
n
=
∑
j
x F
j
= 0
…..(Eq 2)
Example 1.3
Two
equal cylinder,
each weighing 900N are
placed
in a box as shown
in fig below
.
Neglecting
friction between the cylinder and the box,
estimate
the reactions at A,
B and C.
Sol.
Two

force member
A system is in equilibrium under the action of only two external forces applied at
A
and
B
.
Then
they
must satisfy the conditions:
The
two forces can not have
random
orientation;
they
must be directed along
AB
.
F
A
=

F
B
Three

force member
A system is in
equilibrium under the action of only three external forces applied at A, B,
C. Then it must satisfy the conditions:
The three forces cannot have random orientation.
They must all lie in the plane ABC if the total moment about each of the point
A
,
B
and
C
is to vanish.
They must all intersect in a common point
O.
Concurrent Forces in Sapce & their method of projections
X
i
, Y
i
, Z
i
are the projection of any force F
i
along
th
e rectangular co

ordinate axes x, y, z
.
X
i
= F
i
cosα
i
Y
i
= F
i
cosβ
i
Z
i
= F
i
cos
γ
i
Where α
i
, β
i
, and γ
i
are the angles that the forces makes with the positive direction of the
coordinate axes X, Y and Z respectively.
Let R be the resultant of all forces and α, β, and γ
are the angles that the resultant makes with
(+ve) direction of
the c0

0rdinate axes X, Y and Z respectively.
X
= ∑
X
i
Y
= ∑ Y
i
Z
= ∑
Z
i
So
R = (X
2
+ Y
2
+ Z
2
)
1
\
2
Cos α =
, Cos β =
, Cos γ =
If the system is in equilibrium their resultant must vanish i.e
X = ∑ X
i
= 0
Y = ∑ Y
i
= 0 & so
R = 0
Z = ∑ Z
i
= 0
Example 1.4
Find the resultant force and the angle
α, β, and γ
of the three forces which are
given in the table below.
F
i
Newtons (N)
α
i
β
i
γ
i
F
1
40
56
33
90
F
2
10
46
80
46
F
3
30
77
115
80
Sol.
Couples in Space
The
system of couples
in
space as represented by moment vector M
1
, M
2
,……………M
n
shown
in Fig.
Taking coordinate axes x, y ,z , as shown , and
α
i
,β
i
and γ
i
are the direction angles of the
vector M
i
.
M
x(i)
=M
i
cosα
M
y(i)
=M
i
cosβ
M
z(i)
=M
i
cosγ
P
receding in the
same manner with moment M
1
,M
2
,M
3
,……………M
n
and adding algebraica
lly
all corresponding projections, we ob
tain the three re
ctangular component of
the resultant couple
M
as follows:
M
x
=
∑
x
)
(i)
= M
1
cosα
1
+
M
2
cosα
2
+…………………………………..+ M
n
cosα
n
M
y
=
∑
y
)
(
i
)
=M
1
cosβ
1
+ M
2
cosβ
2
+…………………………………….+ M
n
cosβ
n
M
z
=
∑
z
)
(i)
=
M
1
cos
γ
1
+
M
2
cos
γ
2
+………………………………………..+M
n
cos
γ
n
M=
(
M
x
2
+
M
y
2
+
M
z
2
)
1/2
α,
β and γ are direction of angle
that the normal to th
e plane of the resultant couple M
makes
with the coordinate axes x, y , z respecti
vely.
Cos α =
x
,
cos β =
y
and
cos
γ
=
z
For E
quilibrium
M
x
= 0 ,
M
y
= 0 ,
M
z
= 0
Example
1.5
Five
equal forces P act on the corners of cube with edge of length a as shown in
fig. Find the
equilibrium of this system of forces.
Sol.
Coplanar forces system
( forces in 2D plane)
Taking an arbitrary point in the
plane and an arbitrary orientation of
the xy plane
, the condition for the
vector sum of the external forces to
vanish
(condition of equilibrium)
is
simply
∑
i
=
∑
ix
i + F
iy
j ) = 0
Or
each component of the resultant
force vector must vanish or zero.
∑
ix
= 0
∑
iy
= 0
The condition that the total moment about
O
should vanish may be written
∑
i
x F
i
=
∑
i
i + y
i
j ) x (F
ix
i + F
iy
j )
= k
∑
i
F
iy
–
y
i
F
ix
) = 0
Example 1.6
Using the above method of projections, find the magnitude and direction of
resultant R of the four concurrent forces shown in Fig. below
Sol.
Free body
Diagram
To investigate the equilibrium of a constrained body, we shall always imagine that we remove
the supports and replace them by the reactions which they exert on the body. Thus all the
reactions or forces acting on the body shown by vectors is calle
d a free

body diagram.
Free body diagram
Friction
The surfaces of two bodies are in contact there will be a limited amount of resistance to sliding
between them, which is called friction. The question of friction between clean dry surfaces was
first investigated in a complete manner by coulomb, who publis
hed in 1781 the results of large
number of experiments. For given dry surfaces in contact, the results of these experiments may
be summarized briefly by following law of friction
1.
The total friction that can be developed is independent of magnitude of the
area of
contact.
2.
The total friction that can be developed is proportional to the normal force.
3.
For low velocities of sliding the total friction that can be developed is practically
independent of the velocity, although the experiments shows that the force
necessary to
start sliding is greater than that necessary to maintain sliding.
Merits & demerits
Marits
: friction helps in
walking on a road
motion of locomotive on rails
Transmission of power by belts, gear etc.
The friction between the wheels and t
he road is essential for the car to move forward.
Demerits:
At every joint in a machine, the force of friction arises due to the relative motion between
two parts and hence some energy is wasted in overcoming the friction.
It produces undesirable noise.
It promotes wear and tear between the surfaces.
Because of friction energy is wasted.
The friction classified as:
1.
Friction between unlubricated surfaces:
The friction experienced between two dry and
unlubricated surfaces in contact is known as
dry
or
solid friction
. It is due to the surface
roughness. The dry or solid friction includes the sliding friction and rolling friction as
discussed above.
2.
Friction
Between Lubricated Surfaces:
When lubricant (
i.e.
oil or grease) is applied
between two surfaces in contact, then
the
friction
appears between them
is
called friction
between lubricated surfaces.
Laws of Static Friction
:
Following are the laws of stati
c friction :
1.
The force of friction always acts in a direction, opposite to that in which the body tends to
move.
Friction
Static friction:
it
is the friction,
experienced by
a body, when at
rest
Dynamic friction or
(kinetic friction):
It is the
friction experie
nced by a
body when in motion
and
is less than the static
friction
Sliding friction:
It is the
friction, experienced by
a body, when it
slides
over another
body.
Rolling friction
:
It is
the friction,
experienced
between the surfaces
which has
balls
or
rollers
interposed
between them.
Pivot friction:
It is the
friction, experienced by
a body, due to the
motion of rotation
as
in case of foot step
bearings
2.
The magnitude of the force of friction is exactly equal to the force, which tends the body
to move.
3.
The magnitude of the limiting friction (
F
) bears a constant ratio to the normal reaction
(
R
N
) between the two surfaces. Mathematically
F
/
R
N
= constant
4.
The force of friction is
independent of the area of contact, between the two surfaces.
5.
The force of friction depends upon the roughness of the surfaces.
Laws of Kinetic or Dynamic Friction
:
Following are the laws of kinetic or dynamic friction :
1.
The force of friction always
acts in a direction, opposite to that in which the body is
moving.
2.
The magnitude of the kinetic friction bears a constant ratio to the normal reaction between
the
two surfaces. But this ratio is slightly less than that in case of limiting friction.
3.
For moderate speeds, the force of friction remains constant. But it decreases slightly with
the
increase of speed.
L
imiting friction
This maximum value of frictional force, which comes into play, when a body just
begins to slide
over the surface of the ot
her body, is known as
limiting force of friction
or simply
limiting
friction
. It may be noted that when the applied force is less than the limiting friction, the body
remains at rest, and the friction into play is called
static friction
which may have any
value
between
zero and limiting friction.
Coefficient of Friction:
It is defined as the ratio of the limiting friction (
F
) to the normal
reaction (
R
N
) between the two bodies. It is generally denoted by μ. Mathematically, coefficient of
friction
μ =
F
/
R
N
Limiting Angle of Friction
Consider that a body
A
of weight (
W
) is resting on a horizontal plane
B
, as shown in Fig.
below
.
If a horizontal force
P
is applied to the body, no relative motion will
take place until the applied
force
P
is equal to the force of friction
F
, acting opposite
to the direction of motion. The
magnitude of this
force of friction is
F
= μ.
W
= μ.
R
N
, where
R
N
is the normal reaction.
In the
limiting case, when the motion just beg
ins, the body will be
in equilibrium under the action of the following three forces :
1.
Weight of the body (
W
),
2.
Applied horizontal force (
P
), and
3.
Reaction (
R
) between the body
A
and the plane
B
The reaction
R
must, therefore, be equal and opposite to the resultant of
W
and
P
and will be
inclined at an angle φ to the normal reaction
R
N
. This angle φ is known as the
limiting angle of
friction
. It may be defined as the angle which the resultant reaction
R
makes wi
th the normal
reaction
R
N
. From above Fig., tan φ =
F/R
N
= μ
R
N /
R
N
= μ
Angle of Repose
Consider that a body
A
of weight (
W
) is resting on
an
inclined plane
B
, as shown in
Fig.
If the angle of
inclination α of the plane to the horizontal is such that
the
body begins to move down the plane, then the angle
α is
called the
angle of repose
.
The body will only begin to move
down the plane, when
W
sin α =
F
= μ.
R
N
= μ.
W
cos α
(
R
N
=
W
cos α)
tan α = μ = tan φ or
α = φ
(
μ = tan φ)
fig.
Angle of repose
Minimum Force Required to Slide a Body on a Rough Horizontal Plane
Consider that a body
A
of weight (
W
) is resting on a
horizontal
plane
B
as shown in Fig. Let an effort
P
is
applied at an
angle θ to
the horizontal such that the body
A
just moves. The various forces
acting on the body are shown
in Fig.. Resolving the force
P
into
two components,
i.e.
P
sin θ acting upwards and
P
cos θ acting
horizontally. Now
for the equilibrium of the body
A
,
R
N
+
P
sin θ =
W
or
R
N
=
W
–
P
sin θ ...
(
i
)
and
P
cos θ =
F
= μ.
R
N
...
(
ii
)
(
F
= μ.
R
N
)
fig.
Minimum force required
to slide a body
Substituting the value of
R
N
from equation
(
i
)
,
we have
P
cos θ = μ (
W
–
P
sin θ) = tan φ
(
W
–
P
sin θ)
(
μ = tan φ)
=
sin
φ / cos φ
(
W
–
P
sin θ
)
P
cos θ .cos φ =
W
sin φ
–
P
sin θ.sin φ
P
cos θ.cos φ +
P
sin θ.sin φ =
W
sin φ
P
cos (θ
–
φ) =
W
sin φ
...[
_
cos θ. cos φ + sin θ.sin φ =
cos (θ
–
φ)]
P
=
W
sin φ
/
cos (θ
–
φ)
………(iii)
For
P
to be minimum, cos (θ
–
φ) should be maximum,
i.e.
cos (θ
–
φ) = 1 or θ
–
φ = 0° or θ = φ
In other words, the effort
P
will be minimum, if its inclination with the horizontal is equal to
the
angle of friction.
So
P
min
=
W
sin θ ...[
From equation
(
iii
)
]
Example 10.1.
A body, resting on a rough horizontal plane required a pull of 180 N inclined
at
30
º
to the plane just to move it. It was found that a push of 220 N inclined at 30º to the plane
just
moved the body. Determine the weight of the body and the coefficient of friction.
Solution.
Given : θ = 30º
Let
W
= Weight of the body in
newtons,
R
N
= Normal reaction,
μ = Coefficient of friction, and
F
= Force of friction.
Fig. 1 Fig. 2
First of all, let us consider a pull of 180 N. The force of friction (
F
) acts towar
ds left as shown
in Fig.1
Resolving the forces horizontally,
F
= 180 cos 30º = 180 × 0.866 = 156 N
Now resolving the forces vertically,
R
N
=
W
–
180 sin 30º =
W
–
180 × 0.5 = (
W
–
90) N
We know that
F
= μ.
R
N
or 156 = μ (
W
–
90) ...
(
i
)
Now let us consider a
push of 220 N. The force of friction (
F
) acts towards right as shown in Fig. 2
Resolving the forces horizontally,
F
= 220 cos 30º = 220 × 0.866 = 190.5 N
Now resolving the forces vertically,
R
N
=
W
+ 220 sin 30º =
W
+ 220 × 0.5 = (
W
+ 110) N
We know that
F
= μ.
R
N or 190.5 = μ (
W
+ 110) ...
(
ii
)
From equations
(
i
)
and
(
ii
),
W
= 1000 N, and μ = 0.1714
Ans.
Assignment 01
Q (3.) what do
you mean by plane truss? Write
down their application
s
.
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