Engineering Mechanics for First Year B.E. Degree Students

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Oct 29, 2013 (3 years and 11 months ago)

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Engineering Mechanics

for

First Year B.E. Degree Students

COURSE

CONTENT IN BRIEF

1.
Introduction.

2.
Resultant of concurrent and non
-
concurrent coplanar forces.

3.
Equilibrium of concurrent and non
-
concurrent coplanar forces.

4.
Analysis of plane trusses.

5.
Friction.

6.
Centroid and Moment of Inertia.

7.
Resultant and Equilibrium of concurrent non
-
coplanar forces.

8.
Rectilinear and Projectile motion.

9.
Newton’s second law, D’Alembert’s principle, banking and super
elevation.

10.
Work, Energy, and Power.

11.
Impulse
-

Momentum principle.




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.



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’.


CHAPTER


I

INTRODUCTION

This Course on Engineering Mechanics comprises of
Mechanics of Rigid bodies and the sub
-
divisions that come
under it.

1

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

2

Fundamental Concepts and Axioms

Rigid body :




It

is

defined

as

a

definite

amount

of

matter

the

parts

of

which

are

fixed

in

position

relative

to

one

another
.

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
.


3

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
.


4

Space

The

concept

of

space

is

associated

with

the

notion

of

the

position

of

a

point,

defined

using

a

frame

of

reference,

with

respect

to

which

the

position

of

the

point

is

fixed

through

three

measures

specific

to

the

frame

of

reference
.

These

three

measures

are

known

as

the

co
-
ordinates

of

the

point,

in

that

particular

frame

of

reference
.

x

y

z

5

Mass

:


It

is

a

measure

of

the

quantity

of

matter

contained

in

a

body
.

It

may

also

be

treated

as

a

measure

of

inertia,

or

resistance

to

change

the

state

of

rest,

or

of

uniform

motion

along

a

straight

line,

of

a

body
.

Two

bodies

of

the

same

mass

will

be

attracted

by

the

earth

in

the

same

manner
.


Continuum

:




A

particle

can

be

divided

into

molecules,

atoms,

etc
.

It

is

not

feasible

to

solve

any

engineering

problem

by

treating

a

body

as

a

conglomeration

of

such

discrete

particles
.

A

body

is

assumed

to

be

made

up

of

a

continuous

distribution

of

matter
.

This

concept

is

called

‘Continuum’
.

6

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

and

(iv)

Line

of

action
.


7

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
.



Whereas

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
.

8

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.

(
More detailed discussion on Resultant will follow in Chapter 2

)

Like

parallel

Unlike parallel

Like

parallel

Unlike parallel

9




Axioms of Mechanics

(1)
Parallelogram

law

of

forces

:

It

is

stated

as

follows

:

‘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

P
2

P
1

R





10

Contd..

In

the

above

figure,

P
1

and

P
2
,

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

P
2

P
1

R





It can be shown that the magnitude of the resultant is given by:

R =

P
1
2

+ P
2
2

+ 2P
1
P
2
Cos
α

Inclination of the resultant w.r.t. the force P
1

is given by:



= tan
-
1

[( P
2

Sin

) / ( P
1

+ P
2

Cos


)]

11

Contd..

(
2
)


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

A

B

P

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

P

O

12




(3) Newton’s Laws of motion:



(i)

First Law :


If

the

resultant

force

acting

on

a

particle

is

zero,

the

particle

will

remain

at

rest

(if

originally

at

rest)

or

will

move

with

constant

speed

in

a

straight

line

(if

originally

in

uniform

motion)
.





(ii) Second Law :



If

the

resultant

force

acting

on

a

particle

is

not

zero,

the

particle

will

have

an

acceleration

proportional

to

the

magnitude

of

the

resultant

and

in

the

direction

of

this

resultant

i
.
e
.
,

F

α

a

,or

F

=

m
.
a

,

where

F,

m,

and

a,

respectively

represent

the

resultant

force,

mass,

and

acceleration

of

the

particle
.





(iii) Third law:



The

forces

of

action

and

reaction

between

bodies

in

contact

have

the

same

magnitude,

same

line

of

action,

and

opposite

sense
.

13

Note :


1
.

‘Axioms’

are

nothing

but

principles

or

postulates

that

are

self



evident

facts

which

cannot

be

proved

mathematically

but

can

only

be

verified

experimentally

and/or

demonstrated

to

be

true
.

2
.

The

three

basic

quantities

of

mechanics

are

length,

time,

and

force
.

Throughout

this

Course

we

adopt

SI

units

and

therefore

they

are

expressed

in

meters,

seconds,

and

Newtons,

written

as

m,

s,

and

N

respectively
.




3.

The ‘external effect’ of a force on a body is manifest in a change in
the state of inertia of the body. While the ‘internal effect’ of a force on a
body is in the form of deformation.

14

RESULTANT OF CONCURRENT COPLANAR FORCES

CHAPTER



2

Y
-
Direction

X
-
Direction

F
3

F
1

R


F
x


F
y



F
2


In the above diagram F
1
, F
2
, F
3

form a system of concurrent
coplanar forces. If R is the resultant of the force system, then its
magnitude and direction are given by:

Composition of forces and Resolution of force


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
.

15

Contd..

(i)


Magnitude, R =


(

F
x
)
2

+
(

F
y
)
2






(ii)

Direction,

θ

=

tan


1
(

F
y

/


F
x
)

,

where
:


ΣF
x

=

Algebraic

summation

of

x
-
components

of

all

individual

forces
.



ΣF
y

=

Algebraic

summation

of

y
-
components

of

all

individual

forces
.



θ

=

Angle

measured

to

the

resultant

w
.
r
.
t
.

x
-
direction
.



The

process

of

obtaining

the

resultant

of

a

given

force

system

is

called

‘Composition

of

forces’
.


Note
:

The

orientation

of

x
-
y

frame

of

reference

is

arbitrary
.

It

may

be

chosen

to

suit

a

particular

problem
.

16

Contd..



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

:

F
y

Fig. 1

F
x



F

F
y



F

F
x

Fig. 2

Fig. 3

F

F
y

F
x







The

process

of

obtaining

the

components

of

a

force

is

called

‘Resolution

of

a

force’
.


17


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:

+ve

+ve

x

x

y

y

Also force components that are directed along +ve y
-
direction
are taken +ve for summation along the y
-
direction.

When

the

components

of

a

force

are

not

mutually

perpendicular

they

are

called

‘Oblique

Components’
.

Consider

the

following

case
.


Oblique Components of a force
:

18

Contd..

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’
.





F

F
1

F
2

Let

F
1

and

F
2

be

the

oblique

components

of

a

force

F
.

The

components

F
1

and

F
2

can

be

found

using

the

‘triangle

law

of

forces’,

which

states

as

follows
:

‘If

two

forces

acting

at

a

point

can

be

represented

both

in

magnitude

and

direction,

by

the

two

sides

of

a

triangle






F

F
1

F
2

F
1

/ Sin


=
F
2
/ Sin


=
F / Sin(180
-



-


)



19

Contd..

Numerical

Problems & Solutions


∑ Fx = + 15 Cos 15


75


45 Sin 35


+ 60 Cos 40





∑ Fy = + 15 Sin 15 + 105


45 Cos 35




60 Sin 40


= + 33.453 kN

15

kN

15
0

105

kN

75 kN

45

kN

40
0

60 kN

35
0

Fig.1A

(1A)

+
ve

=
-

40.359 kN


=
40.359 kN


Obtain the resultant of the concurrent coplanar forces acting


as shown in Fig. 1A.


Solution
:

+ve

20

Contd..

R =

(
∑Fx )
2

+ (
∑Fy)
2

=

(
-

40.359)
2

+ (33.453)
2


Θ

= tan
-
1
(
∑Fy/
∑Fx)

Magnitude,R = 52.42 kN

Inclination,
Θ

= 39.69
o

15

kN

15
0

105 kN

75 kN

45 kN

40
0

60 kN

35
0

Fig.1A

(1A)

∑Fx

Θ

Answer
:

(w.r.t. X


direction)

∑Fy

R

21

Contd..

(1B)

50 kN

2

3

100 kN

α

26.31
o

75 kN

30
o

1

2

25kN

Obtain the resultant of the

concurrent coplanar forces acting

as shown in Fig. 1B.

∑Fx =
-
50 Cos 26.31
-

100 Cos33.69


25 Cos 63.43 + 75 Cos 30


+ ve

74.26kN


-
74.26kN =

75kN

120

2

3

30

1

2

Fig
.
1B

25kN

100kN

50kN

º

º

α

= tan
-
1
(2/3)=33.69


Solution:

β

º

β
= tan
-
1
(2/1)


=


63.43
º

22

Contd..




(1B)

50 kN

2

3

100 kN

α

26.31
o

75 kN

30
o

1

2

β

25kN

=
-
93.17kN

=

93.17kN

23


F
Y
= 50sin26.31
-

100sin 33.69


75sin30


25sin63.43

ve

+

Contd..

Contd..

R =

(
∑Fx)
2
+ (∑Fy)
2

= 119.14 kN

Θ

= tan
-
1
(∑Fy /
∑Fx

)

= 51.44
o

∑Fx


Fy

R

Θ

100

kN

50kN

26.31
o

75 kN

30
o

(1B)

25kN

33.69
º

63.43
º

Answers:

Contd..

24

-

Assume the fifth force F
5

in the
first quadrant, at an angle
α
, as
shown.

-

The 150 N force makes an angle
of 20
o
w.r.t. horizontal

(2)


150N

50N

200N

120N

45
°

50
°

110
º

F
5

α

R =250 N

20
º

A

system

of

concurrent

coplanar

forces

has


five

forces

of

which

only

four

are

shown

in


Fig
.
2
.

If

the

resultant

is

a

force

of

magnitude


R

=

250

N

acting

rightwards

along

the


horizontal,

find

the

unknown

fifth

force
.

Fig. 2

120N

150N

50N

200N

45
º

50
°

110
º

Solution:

25

Contd..

200 cos 50


150 cos 20


50

cos 45 +F
5
cos
α

= 250.



F
5

cos
α

= +297.75 N



+ve


F
X

= R

150N

50N

200N

120N

45
°

50
°

110
º

F
5

α

R =250 N

20
º

26

Contd..

Contd..


∑F
Y

= 0.

F
5

sin
α

+ 200sin 50 + 150 sin 20




120 + 50 sin 45 = 0

F
5

sin
α
=
-
119.87N = 119.87N

α

= 21.90
º

119.87N

297.75N

F
5
= 320.97N

tan
α

= F
5
sin

α

/F
5
cos
α




=0.402





α

= 21.90
º


F
5
= 320.97N

F
5
cos
α

=


F
5
sin
α

=

+ve

Answers

Contd..

27

150N

50N

200N

120N

45
°

50
°

110
º

F
5

α

R =250 N

20
º

(3) A system of concurrent coplanar forces has four forces
of which only three are shown in Fig.3. If the resultant is a
force R = 100N acting as indicated, obtain the unknown
fourth force.


Fig. 3

60
°

45
°

R=100N

50N

25N

70
°

40
°

75N

28

Contd..

-

Assume the fourth force F
4

in the 1
st

quadrant, making an
angle
α

as shown


α

60
°

45
°

R=100N

50N

25N

70
°

40
°

75N

F
4

F
4
cos
α

+
75cos70


50cos45


25sin60 =
-
100cos40

Or, F
4
cos
α

=
-

45.25N ; or, F
4
cos
α

= 45.25N



Fx =
-
Rcos40

+ve

29

Contd..

60
°

45
°

R=100N

50N

25N

70
°

40
°

75N

α

F
4



Fy =
-
Rsin40


F
4
sin
α

+ 75sin70+25cos60+50sin45 =
-

100sin40



F
4
sin
α

=
-
182.61N ; or, F
4
sin
α

= 182.61N

+ve

30

Contd..

Contd..

α
=
76.08
º

45.25N

F
4
=188.13N

182.61N

F
4
cos
α

=

F
4
sin
α

=

Answers:





= tan
-
1
(F
4
sin


/F
4
cos

)




= 76.08
º

& F
4
=188.13N

31

Contd..


.


(4) The resultant of a system of concurrent coplanar forces is a force
acting vertically upwards. Find the magnitude of the resultant, and the
force F
4

acting as shown in Fig. 4.


60
°

30
°

15 kN

5 kN

10 kN

70
°

45
°

F
4

Fig. 4

32

Contd..



F
4

sin70


10cos 60


15cos 45


5cos 30 = 0; or, F
4
sin70 = 19.94

∑Fx = 0

60
°

30
°

15 kN

5 kN

10 kN

70
°

45
°

F
4

Fig. 4

Solution:

+ve

R

33


F
4
= 21.22kN

Contd..

Contd..

F
4
cos70 + 10sin60


15sin45 + 5sin30 = +R




+R
-

0.342F
4

= 0.554

Substituting for F
4

,

R= +7.81kN


Fy = +R


+ve

Solution:

60
°

30
°

15 kN

5 kN

10 kN

70
°

45
°

F
4

Fig. 4

R

Answers:

F
4
= 21.22 kN

R= +7.81kN

34

Contd..


(5) Obtain the magnitudes of the forces P and Q if the resultant of the
system shown in Fig. 5 is zero

.

40
°

60
°

P

50N

Q

70
°

45
°

Fig. 5

100N

35

Contd..

Contd..

40
°

60
°

P

50N

Q

70
°

45
°

Fig
. 5

100N

36

Contd..

For R to be = zero,

∑F
x
= 0 and ∑ F
y

= 0

∑Fx = 0 :


-
Psin45


Qcos40 + 100cos70 + 50cos60 = 0

Or, 0.707P + 0.766Q = 59.2

+ve

40
°

60
°

P

50N

Q

70
°

45
°

Fig
. 5

100N


∑Fy = 0

-
Pcos45 + Qsin40 + 100sin70


50sin60 = 0


or,
-
0.707P + 0.642Q =
-
50.67

+ve

Answers:

(b)

Solving (a) & (b)

P = 77.17 N & Q = 6.058N

37

Contd..

30
°

100N

50N

Fig
.
6

(6) Forces of magnitude 50N and 100N are the oblique components of
a force F. Obtain the magnitude and direction of the force F. Refer
Fig.6.

38

Contd..

Rotating the axes to have X parallel to 50N,


Fx = +50 + 100cos30

=
+136.6N




Fy = +100sin30 = +50N


+ve

+ve

30
°

100N

(6)

X
-

AXIS

Y
-
AXIS



30
°

100N

50N

Fig
.
6

50N

39

Contd..

Contd..


F = 145.46N

θ

= 20.1
º

w r t X direction (50N force)

50N


F=

(∑Fx)
2
+(

Fy)
2

= tan
-
1
[(∑Fx)
2
+(∑Fy)
2
]

Fig
.
6

X
-

AXIS

F

θ

Y
-
AXIS

30
°

100N

(6)

X
-

AXIS

Y
-
AXIS

30
°

100N

50N

50N

θ

40

Contd..

(7)

Resolve the 3kN force along the directions P and Q. Refer Fig. 7.

P

Q

3kN

45
°

60
°

30
°

Fig. 7

41

Contd..

3kN

Fig. 7

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


45
º

Q

45º

Q

60
°

30
°


X


Axis


P

3kN

45
º

42

Contd..

EXERCISE

PROBLEMS

1
.

A

body

of

negligible

weight,

subjected

to

two

forces

F
1
=

1200
N,

and

F
2
=
400
N

acting

along

the

vertical,

and

the

horizontal

respectively,

is

shown

in

Fig
.
1
.

Find

the

component

of

each

force

parallel,

and

perpendicular

to

the

plane
.


Ans : F
1X

=
-
720 N, F
1Y

=
-
960N, F
2X

= 320N, F
2Y

=
-
240N



FIG. 1

=

1200

N

3

4

Y

F
2

F
1

=

400

N

43

2. Determine the X and Y components of each of the forces shown in

(Ans : F
1X

= 259.81 N, F
1Y
=
-
150 N, F
2X
=
-
150N, F
2Y
= 360 N,


F
3X
=
-
306.42 N, F
3Y
=
-
257.12N )




30
º

40
º

12

5

300 N

390 N

400 N

X

Y

F
1

=

F
2

=

F
3

=

FIG. 2

FIG.2.

44

45

600
N

200
N

800
N

20
º

40
º

30
º

FIG
.

3

3
.

Obtain

the

resultant

of

the

concurrent

coplanar

forces

shown

in

FIG
.
3

(Ans
:

R

=

522
.
67

N,

θ

=

68
.
43
º)

4
.

A

disabled

ship

is

pulled

by

means

of

two

tug

boats

as

shown

in

FIG
.

4
.

If

the

resultant

of

the

two

forces

T
1

and

T
2

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

=

30
º
.

(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
.

T
1
=

195
.
81

N,

T
2

=

133
.
94

N



b
.

α

=

70
º
,

T
1

=

281
.
91

N,

T
2
(min)

=

102
.
61

N

)



T
2

R

=

300

N

T
1

α

20
º

FIG
.

4

X
-

direction

46

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

20
º

40
º

FIG
.

5

(Ans
:

P

=

9
.
4

kN

,

R

=

12
.
66

kN)

47

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)
1400
N

(

Ans
:

a
.

θ

=

41
.
81
º

;


b
.

The

resultant

cannot

be

horizontal
.
)

1200 N

800 N

60
º

θ

F

ROD

COLLAR

FIG.6

48

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

30
º

α

Fig. 7

(Ans:
α

= 10.7
º, Q = 9.479 kN

)

49