engineering mechanics · 2018-07-06 · block, screw jack, calculation of mechanical advantage,...

FY CIVIL Second Sem

Sanjay Ghodawat Polytechnic, Atigre 1

ENGINEERING MECHANICS

1. SIMPLE MACHINES 20 Marks

Specific Objectives:

Calculate velocity ratio for given machine.

Find Efficiency of given machine.

Contents:

1.1 Definitions: …………………………………………..... (06 Marks)

Simple machine, compound machine , load , effort , mechanical advantage , velocity

ratio , input of a machine ,output of a machine efficiency of a machine , ideal

machine, ideal effort and ideal load, load lost in friction, effort lost in friction.

1.2 Analysis: …………………………………………..... (04 Marks)

Law of machine, maximum mechanical advantage and maximum efficiency of a

machine, reversibility of a machine, condition for reversibility of a machine, self locking

machine. Simple numerical problems.

1.3 Velocity Ratio for Simple Machines: …………….... (10 Marks)

Simple axle and wheel, differential axle and wheel, Weston‟s differential pulley block,

single purchase crab, double purchase crab, worm and worm wheel, geared pulley

block, screw jack, calculation of mechanical advantage, efficiency, identification of

type such as reversible or not etc.

Man invented various types of machines for his easy work.

Sometimes, one person cannot do heavy work, but with the help of

machine, the same work can be easily done.

To change the tyre of a car, number of person will be required. But

with the help of a “Jack”, the same work can be done by a single man.

Therefore, jack acts as a machine by which the load of a car can be

lifted by applying very small force as compared to the load of car.

Simple Machine or Lifting Machine:

A machine a device by which heavy load can be lifted by

applying less effort as compared to the load.

e.g. Heavy load of car can be lifted with the help of simple screw

jack by applying small force.

FY CIVIL Second Sem


Compound Machine:

Compound machine is a device which may consists of number of

simple machines. A compound machine may also be defined as a

machine which has multiple mechanisms for the same purpose.

Compound machines do heavy work with less efforts and greater

speed.

e.g. In a crane, one mechanism (gears) are used to drive the rope

drum and other mechanism (pulleys) are used to lift the load. Thus, a

crane consists of two simple machines or mechanisms i.e. gears and

pulleys. Hence, it is a compound machine.

Effort:

It may be defined as, the force which is applied so as to

overcome the resistance or to lift the load.

It is denoted by „P‟.

Magnitude of effort (P) is small as compared to the load (W).

Load:

The weight to be lifted or the resistive force to be overcome with the

help of a machine is called as load (W).

Velocity Ratio (V.R.):

It is defined as the ration of distance traveled by the effort (P) to

the distance traveled by the load (W)

loadby travelledDistance

effortby travelledDistance V.R.

Velocity ratio will be always more than one and for a given

machine, it remains constant.

Mechanical Advantage:

It is defined as the ratio of load to be lifted to the effort applied.

FY CIVIL Second Sem


P

W

(P)Effort

(W) Load M.A.

Input:

The amount of work done by the effort is called as input and is

equal to the product of effort and distance travelled by it.

Input = P x X, where, P – Effort and X – distance

travelled by the effort

Output:

The amount of work done by the load is called as output and is

equal to the product of load and distance travelled by it.

Output = W x Y where, W – Load and Y – distance

travelled by the load

Efficiency:

The ratio of output to input is called as efficiency of machine and

it is denoted by Greek letter eta (η)

Generally, efficiency is expressed in percentage

% 100xInput

Outputη

It is always less than 100 because of friction, therefore output <

input.

But Output = W.Y and Input = P.X

% 100xXxP

YxW100x

Input

Outputη

% )Y

XV.R.and

P

WM.A.(Since100x

V.R.

M.A.100x

Y

XP

W

η

FY CIVIL Second Sem


Therefore, efficiency of a machine is also defined as the ratio of

mechanical advantage (M.A.) to the velocity ratio (V.R.). It is also

expressed in percentage.

% 100xV.R.

M.A.η

It is always less than 100 because of friction, therefore M.A. < V.R.

Actual Machine:

The machine whose efficiency is always less than 100 % due to

frictional resistance offered by the different moving component parts of

the machine is called as actual machine.

For such machines, η < 100 % and hence M.A. < V.R.

Ideal Machine:

The machine whose efficiency is 100 % and in which friction is

totally absent or zero, is called as ideal machine.

For ideal machines, η = 100 % and hence M.A. = V.R.

Ideal Effort (Pi):

The effort which is required to lift the load when there is no friction

is called as an ideal effort (Pi)

Ideal Effort Pi = V.R.

W

Where, Pi = Ideal Effort, W = Load to be lifted, V.R. = Velocity Ratio

Ideal Load (Wi):

The load which can be lifted by an effort (P), when there is no

friction, is called as an ideal load (Wi)

Ideal Load Wi = P x V.R.

Where, P = Effort applied, Wi = Ideal Load, V.R. = Velocity Ratio

Lever Arm:

FY CIVIL Second Sem


A rigid bar which is provided in machines so as to apply the effort

(P) is called as lever arm or handle.

LAW OF MACHINE:

The equation which gives the relation between load lifted and

load applied in the form of a slope and intercept of a straight line is

called as Law of a machine i.e. P = mW + C

Where, P = effort applied, W = load lifted, m = slope of the line and

C = y – intercept of the straight line.

To draw the graph of Load (W) V/s Effort (P), effort is applied on a

machine and the corresponding values of the loads are noted down.

The graph of Load (W) V/s Effort (P) is drawn by taking load (W) on the x-

axis and the effort (P) on the y-axis as shown in the figure.

12

12

XX

YYθtanm

It has been observed that,

the graph of load v/s effort is a

straight line cuts the Y-axis giving

the intercept „C‟ which indicates

the effort lost on friction, when no

load is applied.

Is must be noted that, if the

machine is an ideal machine, the

straight line of the graph will pass

through the origin.

Comparing to the equation of straight line i.e. y = mX + C, we get

P = mW + C

Where, P =Effort applied, W = Load applied, m = slope of the line

and C = Y-intercept of the line.

Effort

(P)

Load (W)

C

P = mw + C

(X2, Y2)

(X1, Y1)

O

O

FY CIVIL Second Sem


Note:

If we know the law of machine i.e. values of „m‟ and „C‟ are

known, then for a given load effort can be found out or for a given effort

corresponding value of the given load can be found out. The law of

machine also indicates the friction in the machine and maximum M.A.

Maximum Mechanical Advantage (Max. M.A.):

We know that,

m

1M.A.Max.

maximumbewillM.A.,W

CratiotheNeglecting

smallverybewillW

Cratiothemore,isW''ifequationabovetheIn

W

Cm

1M.A.

getweW''byrDenominato&numeratortheDividing

CmW

WM.A.

CmWPBut,

P

WM.A.

Maximum Efficiency:

The ratio of maximum M.A. to the V.R. is called as maximum

efficiency.

It is also expressed in percentage as

)m

1M.A.MaxSince(100x

V.R.

1x

m

1100x

V.R.

M.A.MaxηMaximum%

FY CIVIL Second Sem


Reversible Machine:

When a machine is capable of doing some work in the reverse

direction even on removal of effort, it is called as reversible machine.

Condition for Reversible Machine:

The efficiency of the machine should be more than 50%.

Irreversible Machine / Non-reversible Machine / Self Locking Machine:

When a machine is not capable of doing some work in the reverse

direction even on removal of effort, it is called as irreversible machine or

non-reversible machine or self locking machine.

Condition for Irreversible Machine:

The efficiency of the machine should be less than 50%.

Friction in Machines in terms of Effort and Load:

In any machine, there are number of parts which are in contact

with each other in their relative motion. Hence, there is always a

frictional resistance and due to which the machine is unable to produce

100 % efficiency.

Let, P = Actual Effort, Pf = Effort Lost in friction, Pi = Ideal Effort

Effort Lost in friction (Pf) = Actual Effort (P) – Ideal Effort (Pi)

)V.R.

WP(Since

V.R.

WPPPP iif

Let, W = Actual load lifted, Wf = Load Lost in friction, Wi = Ideal

Load

Load Lost in friction (Wf) = Ideal Load (Wi) – Actual load lifted

(W)

V.R.)xPW(SinceWV.R.)x(PWWW iif

FY CIVIL Second Sem


GRAPHS:

1. Load v/s Effort:

The graph of load against

effort is a straight line, cuts the y-axis

giving the intercept „C‟ which

represents the effort lost in friction at

zero load.

12

12

XX

YYθtanm

2. Load v/s Percentage Efficiency (%η):

The graph of load v/s % efficiency is

a curve as shown in the above figure. As

load increases, percentage efficiency

also increases and therefore gives rise to a

smooth curve gradually increasing and

becomes more or less parallel to x-axis.

3. Load v/s Mechanical Advantage:

The graph of load v/s Mechanical

Advantage is a curve as shown in the

above figure. As load increases,

mechanical advantage also increases

and therefore gives rise to a smooth

gradually increasing curve.

Effort

(P)

Load (W)

C

P = mw + C

(X2, Y2)

(X1, Y1)

O

O

X

Y

O

Load (W)

% n

X

Y

O

Load (W)

M.A.

Max. M.A. = 1/m

FY CIVIL Second Sem


4. Load v/s Ideal Effort (Pi):

The graph of load v/s ideal effort is a straight line passing through

origin as shown in the above figure.

5. Load v/s Effort lost in friction (Pf):

The graph of load against effort lost in friction is a straight line as

shown in the figure.

X

Y

O

Load (W)

Ideal

Effort

(Pi)

Effort

lost in

friction

(Pf)

Load (W)

O

FY CIVIL Second Sem


STUDY OF SOME SIMPLE MACHNES:

We know that, velocity ratio of a machine is the ratio of distance

travelled by the effort to the distance travelled by the load. It is

observed that, the distance travelled by the effort is greater than the

distance moved by the load. Experimentally, it has been found that, the

velocity ratio remains constant for all loads. The velocity ratio changes

from machine to machine but remains constant for a given machine.

1. Simple Wheel and Axle:

In simple wheel and axle, effort wheel and axle are rigidly

connected to each other and mounted on a shaft. A string is wound

round the axle so as to lift the load (W) another string is wound round the

effort wheel in opposite direction so as to apply the effort (P) as shown in

the figure.

Let, W = Load lifted, P = Effort Applied D = Diameter of the

effort wheel d = diameter of the load axle.

When, the

effort wheel

completes

one revolution,

the effort

moves through

a distance

equal to the

circumference

of the effort

wheel (πD)

and simultaneously the load moves up through a distance equal to the

circumference of the load axle (πd)

d

D

dπ

Dπ

LoadthebytravelledDistance

EffortthebytravelledDistanceV.R.

FY CIVIL Second Sem


2. Differential Axle and

Wheel:

Differential axle and

wheel is a further

modification and

improvement over the

simple axle and wheel.

It consists of a load

axle made up of bigger

axle of diameter d1 and

smaller axle of diameter

d2 rigidly connected to each other and an effort wheel of diameter „D‟.

Since, the load axle is made up of two axles of different diameters; it is

called as a differential axle. Differential axle and effort wheel are

mounted on the same shaft which is supported on ball bearings as

shown in the figure.

A string is wound round the effort wheel so as to apply the

effort „P‟. Another string is wound round the bigger axle further passing

over the pulley carrying the load „W‟ attached to the snatch block. The

same string is further wound round the smaller axle in the opposite

direction to that of bigger axle. The winding of string on effort wheel and

smaller axle is done in the same direction; the string unwinds from the

effort wheel & smaller axle and winds over the bigger axle

simultaneously, when the effort „P‟ is applied.

When effort wheel complete one revolution, the differential axle

also completes one revolution.

Distance travelled by the effort = πD

Length of the string wound over the bigger axle = πd1

Length of the string unwound over the smaller axle = πd2

Total winding over the bigger axle = πd1 - πd2 = π(d1 – d2)

FY CIVIL Second Sem


But, the load „W‟ is lifted through half of the total winding because

snatch block with a movable pulley supports the load „W‟

)dd(2

π)dd(π

2

1winding)total(

2

1loadthebytravelledDistance 2121

We know that,

21

21

dd

D2

)dd(2

π

Dπ

loadthebytravelledDistance

effortthebytravelledDistanceV.R.

Where, D = Diameter of effort wheel, d1 = diameter of bigger axle,

d2 = diameter of smaller axle.

If radius of effort wheel, bigger axle and smaller axle are given

then,

21 RR

R2V.R.

Where, R = Diameter of effort wheel, R1 = diameter of bigger axle,

R2 = diameter of smaller axle.

FY CIVIL Second Sem


3. A Simple Screw Jack:

A screw jack is commonly used for lifting and supporting the

heavy load. A very small effort can be applied at the end of the lever or

handle or tommy bar for lifting the heavy loads. This effort is very small as

compared to the load to be lifted. As jack has a simple mechanism, it is

most commonly used in repair work of vehicles.

When the effort is applied to the handle or lever arm to complete

one revolution then load is lifted through one pitch of the screw (p),

therefore the distance moved by the load is equal to the pitch of the

screw and the distance moved by the effort is equal to 2πL

Let, L = length of the handle or lever arm and p = pitch of the thread or

screw, then

p

L2π



If the effort wheel is used at the place of handle or lever arm for

applying the effort, then

p

Dπ



Where, D = diameter of the effort wheel and p = pitch of thread or

screw.

4. Weston’s Differential Pulley Block:

It consists of upper and lower

block. Upper block is having bigger

pulley and smaller pulley of different

diameter mounted on same common

axle and that of lower block is having a

single pulley. The weight „W‟ is attached

to the lower block. Upper block is fixed

and lower block is movable.

FY CIVIL Second Sem


An endless chain passes over the bigger and pulley and single

pulley in lower block as shown in the figure.

When there is one complete revolution of the bigger pulley,

distance moved by the effort is equal to the circumference of the

bigger pulley (πD). When the bigger pulley completes one revolution,

the smaller pulley also completes one revolution because both pulleys

are fixed on the same axle. Therefore, for one complete revolution, the

length of the chain wound round the bigger pulley is equal to the

circumference of the bigger pulley (πD) and at the same time, the

length of the chain unwound from the smaller pulley is equal to the

circumference of the smaller pulley (πd).

Net winding = πD – πd

As the load (W) attached to the lower block, is equally distributed

between the two parts of the chain, distance moved by load is equal to

the half of the net winding.

Distance moved by the load = ½ ( πD – πd )

dD

D2V.R.

d)πDπ(2

1

Dπ



Where, D = diameter of bigger or upper pulley, d = diameter of

smaller or lower pulley

If radii of bigger and smaller pulley are given, then

rR

R2V.R.

Where, R = radius of bigger or upper pulley, r = radius of smaller or

lower pulley

If number of teeth or cogs of the bigger and smaller pulley are

given, then

FY CIVIL Second Sem


21

1

TT

T2V.R.

Where, T1 = number of teeth or cog on bigger or upper pulley, T2 =

number of teeth or cog on smaller or lower pulley

5. Worm and Worm

Wheel:

This machine is

made of toothed wheel

known as worm wheel

and a square threaded

screw known as worm.

Worm and worm wheel is

geared with each other.

The effort wheel is

attached to the worm. A

load drum is centrally

fixed to the worm wheel as shown in the figure. The effort can be either

by wheel or handle. The worm may be single threaded or multi-

threaded.

Let, w = load lifted, T = number of teeth on the worm wheel,

P = effort applied, r = radius of load drum, R = radius of effort

wheel.

Consider the worm is single threaded.

For one complete revolution of effort wheel, distance moved by

the effort P = 2 π R and load drum performs (1/T) revolution.

Distance moved by the load = T

rπ2

We know that,

FY CIVIL Second Sem


r

TRV.R.

r

TR

T

rπ2

Rπ2



If handle is used in place of effort wheel,

r

TLV.R. Where, L = length of the handle

In general, if worm is „n‟ threaded,

rn

TL

rn

TRV.R. Where, n = number of thread on worm

If the worm is double threaded,

r2

TL

r2

TRV.R. since, n = 2

6. Single Gear Crab or Single Purchase Winch Crab:

This machine consists of mainly the larger gear wheel known as

Spur or main gear and smaller gear known as pinion. Spur or main gear is

mounted rigidly on the load drum or load axle and the spur is geared to

the pinion which is further mounted rigidly on the shaft to which the

effort wheel or handle is attached.

FY CIVIL Second Sem


A rope or string is wound round the load drum so as to lift the load

„W‟ and an another string is wound round the wheel so as to apply the

effort „P‟ as shown in the figure.

Let, W = Load lifted, P = Effort applied, D = diameter of effort

wheel, T1 = number of teeth on spur or main gear, T2 =

number of teeth on pinion d = diameter of load drum or load

axle.

For one complete revolution of effort wheel, distance moved by

the effort = πD and the pinion also completes one revolution, that time

spur performs 1

2

T

T revolutions and as load drum and spur being rigidly

connected, the load drum also performs 1

2

T

T revolutions.

Distance moved by the load = 1

2

T

T x circumference of load drum

= 1

2

T

Tx πd

We know that,

1

2

1

2

T

Tx

d

DV.R.

dxπT

T

Dπ



If handle of length „L‟ is used in place of effort wheel,

1

2

T

Tx

d

L2V.R. where, L =length of the handle.

7. Double Gear Crab or Double Purchase Winch Crab:

This machine consists of two larger gear wheels A and C called as

spurs or main gears and smaller gear wheels B and D called as pinions as

FY CIVIL Second Sem


shown in the figure. A load drum or load axle is rigidly connected to the

spur A, which is further geared with pinion B. Pinion B and spur C are

rigidly mounted on same shaft called as intermediate shaft or axle and

spur C is further geared with pinion D mounted on the shaft. So called as

effort axle to which the effort wheel is attached.

Let, W = Load lifted, P = Effort applied, D = diameter of effort wheel,

d = diameter of load drum or load axle.

T1 = number of teeth on spur A, T2 = number of teeth on

pinion B mounted on intermediate shaft, T3 = number of teeth

on spur C mounted on intermediate shaft,

T4 = number of teeth on pinion D mounted on effort axle

For one complete revolution of effort wheel,

Distance moved by effort = πD

The pinion D on the effort axle also makes one revolution and

therefore spur C and pinion B on the intermediate shaft performs

3

4

T

Trevolution.

FY CIVIL Second Sem


That time, spur A and load drum performs same revolutions as

1

2

3

4

T

Tx

T

T because both are rigidly connected to each other.

Distance moved by the load= 1

2

3

4

T

Tx

T

T revolutions x

circumference of load drum = 1

2

3

4

T

Tx

T

T x πd

We know that,

4

3

2

1

1

2

3

4

T

Tx

T

Tx

d

DV.R.

dπxT

Tx

T

T

Dπ



If handle of length „L‟ is used in place of effort wheel,

4

3

2

1

T

Tx

T

Tx

d

L2V.R. where, L =length of the handle.

8. Geared Pulley Block:

It consists of effort wheel or cog wheel A and a small gear wheel

called as pinion B mounted on

the same shaft. It also consists

of a load wheel C and a spur

wheel D, mounted on the same

shaft. The spur wheel D is

geared with the pinion wheel B.

An endless chain passes

over effort wheel from which

effort „P‟ is applied. The load

„W‟ is attached to a chain

which passes over the load

wheel C as shown in the figure.

Let,

T1 = number of cogs on the effort wheel A,

FY CIVIL Second Sem


T2 = number of teeth on pinion wheel B,

T3 = number of teeth on spur wheel or main gear D,

T4 = number of cogs on the load wheel C

For one complete revolution of effort wheel A (cg wheel), we get

Distance moved by effort = T1

Pinion wheel B also completes one revolution and this causes the

spur wheel D to rotate through 3

2

T

Trevolutions.

Distance moved by the load = 3

2

T

Tx T4

We know that,

4

3

2

1

3

2

T

Tx

T

TV.R.

TxT

T

T



4

1

If the diameters of effort wheel A and load wheel C are given, then

4

3

T

Tx

d

DV.R.

Where, D = diameter of effort wheel A, d = diameter of

load wheel C

T3 = number of teeth on spur wheel B,

T4 = number of teeth on pinion wheel D

FY CIVIL Second Sem


2. FORCE SYSTEMS 12 Marks


Define related terms in mechanics.

Calculate Components of forces.

Contents:

2.1 Fundamentals and Force Systems:……………………..... (06 Marks)

Definitions of mechanics, Engineering mechanics, statics, dynamics, Kinetics,

Kinematics, rigid body, classification of force system according to plane coplanar and

non coplanar, sub classification of coplanar force system-collinear , concurrent, non

concurrent, parallel, like parallel, unlike parallel, general etc. Definition of a force, S.I.

unit of a force, representation of a force

by vector and by Bow‟s notation method. Characteristics of a force, effects of a force,

principle of transmissibility.

2.2 Resolution of a force and Moment of a force: ………..... (04 Marks)

Definition, Method of resolution, along mutually perpendicular direction and along two

given direction. Definition of moment, S. I. unit, classification of moments, sign

convention, law of moments Varignon‟s theorem of moment and it‟s use, definition of

couple, S.I. unit, properties of couple with example.

Necessity of Knowledge of Engineering Mechanics:

In technical field, everywhere we are coming across different

forces and we have to study their effects. Engineering mechanics deals

with the study of forces and their effects. The knowledge of mechanics is

required in the subjects like strength of materials, theory of machines,

design of machines, in future. Hence for every engineer it is necessary to

have knowledge of Engineering Mechanic, so that he can develop

some technologies in any field of Engineering.

Rigid body:

The body, which do not undergo any change in its dimensions

even after application of the force.

No body is perfectly rigid. Everybody undergoes negligible

deformation, but as the deformation is negligible, the body can be

considered as rigid body. In Engineering Mechanics all the bodies are

FY CIVIL Second Sem


assumed as rigid bodies so that it becomes convenient and easy to

solve problems.

Particle:

The point mass (definitely small mass) which possesses negligible

size, shape mass and volume.

A particle is a very small amount of matter which may be

assumed to occupy a single point in space. Practically, any object

having very small dimensions as compared to its range of motion can be

called as particle, e.g. stars, planets, rockets etc

Body:

An object having definite mass, occupying definite space and

possessing properties like mass, volume etc.

Elastic Body:

It is defined as a body which regains the original size and shape

after removal of entire external load causing deformation.

Scalar Quantity:

It is defined as a physical quantity which requires only magnitude

and no direction.

e.g. Time, mass, Speed, Power, Volume etc.

Vector Quantity:

It is defined as a physical quantity which requires both magnitude

and direction. e.g. Force, Velocity, Acceleration, Displacement etc.

TYPES OF UNITS:

Basic Units:

These are the units which do not depend on any other units for

their measurements.

Length (l), mass (m) and Time (t) are said to be basic units.

Derived Units:

FY CIVIL Second Sem


These are the units which depend on other units/ basic units for

their measurements.

e.g. Velocity =Displacement / Time, Force = mass x acceleration

SYSTEMS OF UNITS:

There are four systems of units which are commonly used

M. K. S. Unit:

In this system basic units are measured as follows:

Length is measured in mm/m, Mass is measured in Kg and Time is

measured in sec.

C. G. S. Unit:


Length is measured in cm, Mass is measured in gram (g) and Time

is measured in sec.

F. P. S. Unit:


Length is measured in foot (ft), Mass is measured in Pounds and

Time is measured in sec.

S. I. Unit:

This system is universally accepted and being used in majority of

countries in the world. In this system basic units are measured as follows:

Length is measured in m, Mass is measured in Kg and Time is

measured in sec.

Engineering Mechanics or Applied Mechanics:

It is the branch of science, which deals with study of system of

forces and their effects on bodies in motion or at rest.

FY CIVIL Second Sem


Thus, the subject Engineering Mechanics covers not only the study

of forces but also it covers the effects of the forces on the bodies. The

body under consideration may be in motion or at rest. Depending on

the state of the body (in motion or at rest) Applied Mechanics is further

divided as:

Statics:

It is defined as the branch of Engineering Mechanics, which deals

with the study of system of forces and their effects on bodies at rest.

Dynamics:

It is defined as the branch of Engineering Mechanics, which deals

with the study of system of forces and their effects on bodies in motion.

The motion of a body will depend on its mass and some external

force acting on it. Depending on this, Dynamics is further divided as:

Kinematics:

It is defined as the branch of Dynamics, which deals with the study

of system of forces and their effects on bodies in motion, without

considering the mass of the body and the force causing the motion.

Kinetics:

It is defined as the branch of Dynamics, which deals with the study

of system of forces and their effects on bodies in motion, considering the

mass of the body and the force causing the motion.

FORCE:

It is an external agency, which changes or tends to change the

state of the body i.e. force will tend to change the static state of a body

to dynamic or vice versa.

Force is a vector quantity, having „N‟ (Newton) as S.I. unit.

1 KN (kilo Newton) = 1000 N

1 MN (Megs Newton) = 106 N

1 GN (Giga Newton) = 109 N

Unit Newton force:

FY CIVIL Second Sem


It is the force, which will produce an acceleration of 1 m/s2 to an

object having mass 1 kg.

Representation of Force by Vector:

A force can be represented

by a vector, graphically by drawing

a straight line parallel to the line of

action of the force, to a suitable

scale, indicating magnitude of

force and direction by an arrow head.

A force in the figure is represented by a vector of length 5 cm

(scale 1 cm = 5 N) by drawing a line parallel to the given force and

arrowhead indicates the direction of the force.

Graphical Representation of a Force:

A force can be represented graphically by drawing a straight line

to a suitable scale and parallel to the line of action of the given force

and an arrowhead indicates the direction.

Bow’s Notation:

Any force systems, when represented by a drawing, some spaces

are formed around each force. These spaces are named by capital

letters such as A, B, C …. in order. This method of putting capital letters

for each space (on either side of a force) in order is called as Bow‟s

Notation.

Bow‟s notations are used to represent any force in the graphical

solution of the problem.

F1 F2

B

AB

a

b

25 N

5 cm

FY CIVIL Second Sem


C

F3 A

D F4

In the above figure, force „F1‟ lies in between the spaces B and C,

so it can be represented as force „bc‟. The force „F4‟ lies in between the

spaces D and A, hence it can be represented as force „da‟.

Space Diagram:

It is a diagram in which all the forces of a system are drawn in the

space to a suitable scale and spaces so formed are named by Bow‟s

notations.

F2 F1

B

C θ3 θ2 A

θ1

F3 θ4

D F4

A, B, C, D --- Bow‟s Notations

Θ1, θ2, θ3, θ4 --- angles accurately measured

In case of non-concurrent forces, a suitable scale is also selected

to represent the distance between the forces

F2

C

F1 F2 F3 F4 F3

A B C D

2 m 3 m 3.5 m D 4 m

B

FY CIVIL Second Sem


F1

F4 4 m A

Vector Diagram:

The diagram in which, the forces are taken to a suitable scale and

drawn parallel to their respective lines of action of forcers drawn in the

space diagram by maintaining the same order as it was maintained in

the space diagram, is called as vector diagram.

F1=100 N

F2=80 N

F4=60 N A

BDC

F3=40 N

a b

R

a'

c

d

Space Diagram Vector

Diagram

Characteristics of a force:

1. Magnitude: The quantity of force is known as its magnitude. It is

expressed in terms of N or kN.

2. Direction: The line of action along which force acts is called as

direction. Generally it is the angle made by the force with

reference line or horizontal.

FY CIVIL Second Sem


3. Sense of Nature: The sense is indicated by an arrowhead from

which the sense like pull / push can be easily understood.

4. Point of Application: The exact point at which the force acts is

known as point of application.

Effects of a Force:

Whenever a force acts on a body, the following may be the

effects of that force on the body:

1. It may change the state of the body i.e. from ret to motion or

motion to rest.

2. It may accelerate or retard the motion of the body.

3. It may change the size and shape of the body.

4. It may turn or rotate the body.

5. It may keep the body in equilibrium (rest).

System of Forces:

When two or more forces act on a body, it is the system of forces.

FY CIVIL Second Sem


There are two main systems of forces: Coplanar and Non-coplanar

forces.

Coplanar Forces:

If lines of action (the line along which the force acts) of all forces

lie in a single plane, the force system is called as coplanar force system.

(Refer fig. 1)

F1

F3

F2 Fig: 1

Fig. 2

Systems of Forces

Coplanar Forces Non-coplanar Forces

Collinear Concurrent Non-concurrent Parallel

Forces Forces Forces Forces

Like Parallel Unlike Parallel

Forces Forces

FY CIVIL Second Sem


F2

Non-Coplanar Forces:

If lines of action (the line along which the force acts) of all forces

do not lie in a single plane, the force system is called as non-coplanar

force system. (Refer fig.2)

Types of Coplanar Forces:

Collinear Force System:

If all the forces are acting along the same straight line, the force

system is called as collinear force system. (Refer fig. 3)

Concurrent forces:

If lines of action of all forces are passing

through a common point, the force system is

called as a concurrent force system. (Refer

fig. 4)

Non-concurrent forces:

F1 F2 F3 F4

Fig. 3

F1 F2

F3

F4 F5

Fig. 4

F1 F2

F3

Fig

F5 Fig. 5 F4

FY CIVIL Second Sem


If lines of action of all forces are not passing through a common

point, the force system is called as non-concurrent force system. (Refer

fig. 5)

Parallel Forces:

If the lines of action of all forces are parallel to each other, the

force system is called as non-concurrent force system. A parallel force

system is also a non-concurrent force system.

Like Parallel Forces:

If the lines of action of all forces are parallel to each other and the

all forces are acting in same direction, the force system is called as like

parallel force system. (Refer fig. 6)

Unlike Parallel Forces: If the lines of action of all forces are parallel to

each other but all forces are not acting in same direction, the force

system is called as unlike parallel force system. (Refer fig. 7)

Principle of Transmissibility of Force:

“ If a force acts at a point on a rigid body, it is assumed to act at

any other point on the line of action of the force within the same body”.

In fig (a), as per law of transmissibility of forces, the force „F‟ is

assumed to act at any other point on the line AB (without changing the

magnitude).

F1 F2 F3 F1 F2

F3

Fig. 6 Fig. 7

A B A B A B

Fig. (a) Fig. (b) Fig. (c)

FY CIVIL Second Sem


Limitation of Principle of Transmissibility of Force:

If two equal, opposite and collinear forces are acting on the

body, and if the forces are transmitted, then the combined effect is

different than the earlier.

In fig. (b), initially, the forces are trying to create the tension, but

after transmitting these forces, they try to create the compression (fig.

(c)). Hence, this law is not applicable to equal, opposite and collinear

forces.

Resolution of a Force:

The method of splitting a given force into its components, without

changing its effect on the body.

Generally, it is convenient to resolve the force along x and y axis.

These components along x and y axis are known as Orthogonal

Components/ rectangular components.

FFy

FxO

Y

X

-VE

-VE +VE

+VE

From figure,

θCosF.FxF

FxθCos

θSinF.FyF

FyθSin

θ

FY CIVIL Second Sem


Sign Convention:

Along x-axis: Towards right - +ve

Towards left - -ve

Along y-axis: Upward - +ve

Downwards - -ve

A force can also be resolved along the two directions which are

not at right angle to each other.

F1O

F2

2

FF2

1 O F1A

F2 F

B CF1

Let, F1 and F2 be the components of a force „F‟ along the axes 1

and 2 as shown in figure at an angles α and β with the direction of force

„F‟.

Complete the parallelogram OACB.

In triangle OAC, applying sine rule, we get

βα180Sin

F

αSin

F

βSin

F 21

βαSin

F

αSin

F

βSin

F 21

βαSin

F.SinF

βαSin

βF.SinF 21

Moment of a Force:

A force when act on a body produces turning or rotational effect

is called as a moment of a force.

β

α

β

α (α+β) 180-(α+β)

FY CIVIL Second Sem


Moment (M) of a force is equal to the product of the force (F) and

the perpendicular distance between the line of action of the force and

the reference point (x). FxM

F

Line of action of force

x

A

If the force is passing through the reference point, then the

Moment (M) of a force is zero because there is no perpendicular

distance between the force and the point.

Classification of Moments:

When any force acts on a body, the body may be rotated in

clockwise or anticlockwise direction, depending on how the force is

acting on the body.

If the body is rotated in clockwise direction, the moment of the

force is clockwise moment. (Refer fig. 1).

O

d

F

d

O

FFig. 1

FY CIVIL Second Sem


If the body is rotated in anti-clockwise direction, the moment of

the force is anticlockwise moment. (Refer fig. 2)

O

d

F

d

O

Fig. 2

F

VARIGNON’S THEOREM OF MOMENTS:

“When the number of forces are acting on a body, then the algebraic

sum of moments of all the forces about any point is equal to the

moment of their resultant about the same point”. RAFA MM

Consider, a rectangle ABCD, acted upon by three forces as

shown in the figure. Let, „R‟ be the resultant of the given force system.

Let „x‟ be the perpendicular distance between the resultant „R‟ and the

point „A‟.

Now, taking moments of all the forces about point „A‟,

(considering clockwise moments +ve and anti-clockwise moments –ve)

we get

332211FA x.Fx.Fx.FM

Now, moment of resultant about point „A‟, x.RMRA

FY CIVIL Second Sem


Now, according to Varignon‟s theorem of moments,

RAFA MM

R

x.Fx.Fx.Fx

x.Rx.Fx.Fx.F

332211

332211

F2

F1R

F3

X1

X2X3

X

B

CD

A

SIGN CONVENTION FOR MOMENTS

+VE MOMENT(CLOCKWISE)

-VE MOMENT(ANTI-CLOCKWISE)

Thus, using Varignon‟s theorem of moments, we can find out the

position of the resultant w.r.t. to the reference point.

Couple:

Two equal, opposite and parallel forces acting on a body forms a

couple.

As the two forces are equal and opposite, their resultant is zero,

but they will produce rotational effect i.e. they will create the moment.

FY CIVIL Second Sem


a

P

P

O A B

Properties of a Couple:

1. Couple can be replaced by a single force.

2. Couple can not produce translatory motion (straight line motion)

but it can produce rotation.

3. Moment of a couple = force x lever arm.

4. The moment of a couple is independent of the position of moment

centre and it remains constant.

e.g. from figure, ΣMA = P x a

ΣMO = (P x OB) – ( P x OA) = P ( OB – OA)

ΣMO = P x a

Important Formulae: Fx = X – component of a force = F cosθ

Fy = Y – component of a force = Fsin θ

θ = Angle made by F with horizontal

FY CIVIL Second Sem


3. COMPOSTION OF FORCE 20 Marks


Calculate resultant analytically for given force system.

Calculate resultant graphically.

Contents:

3.1 Analytical method: ……………………………………… (10

Marks)

Definition of Resultant force, methods of composition of forces, Law of

parallelogram of forces, Algebraic method for determination of resultant for

concurrent and non concurrent, parallel coplanar force system.

3.2 Graphical method: …………………………………………... (10 Marks)

Space diagram, vector diagram, polar diagram, and funicular polygon.

Resultant of concurrent and parallel force system only.

Resultant:

It is the single force, which will produce the same effect (in

magnitude and direction) as it is produced by number of forces acting

together.

Thus, the resultant is a single force, which will give the same result

as it is produced by the number of forces acting on a body. We can

replace all the forces by the resultant. In other words, we can say that,

the resultant is the combined effect of all the force acting on the body

(in magnitude and direction).

Composition of Forces:

The process of determining the resultant of number of forces

acting simultaneously on the body is known as the composition of the

forces.

In short, the method to find the resultant of the given force system

is called as composition of forces.

FY CIVIL Second Sem


It is the method of reducing the given force system to its

equivalent simplest system of single force or couple. Combining the

forces of any given system is known as composition of forces.

Methods of Composition of Forces:

The following are the two methods used to find the resultant of a

given force system:

1. Analytical Methods:

(a) Trigonometric Method (Law of parallelogram of forces)

(b) Algebraic Method (Method of Resolution)

2. Graphical Methods:

(a) Triangle Law of Forces

(b) Polygon Law of Forces

Parallelogram Law of Forces:

“If two coplanar concurrent forces acting at and away from the

point and are represented by the two adjacent sides of a parallelogram

in magnitude and direction, then the resultant is given by the diagonal

of the parallelogram passing through the same common point”.

P = 100 NA

Q = 80 N

R

P B

Q

D C

θ θ

FY CIVIL Second Sem


Analytical Solution for Parallelogram Law of Forces:

Consider, two forces „P‟ and „Q‟ acting at and away from point

„A‟ as shown in figure.

Let, the forces P and Q are represented by the two adjacent sides

of a parallelogram AD and AB respectively as shown in fig. Let, θ be the

angle between the force P and Q. Extend line AB and drop

perpendicular from point C on the extended line AB to meet at point E.

BPA E

Q R

D C

Consider Right angle triangle ACE,

AC2 = AE2 + CE2 =(AB + BE) 2 + CE2

= AB2 + BE2 + 2.AB.BE + CE2 = AB2 + BE2 + CE2 + 2.AB.BE

……1

Consider right angle triangle BCE,

BC2 = BE2 + CE2 and BE = BC.Cos θ

……2

Putting BC2 = BE2 + CE2 in equation (1), we get

AC2 = AB2 + BC2 + 2.AB.BE

……3

Putting BE = BC. Cos θ in equation (3),

θ

α

FY CIVIL Second Sem


AC2 = AB2 + BC2 + 2.AB. BC. Cos θ

But, AB = P, BC = Q and AC = R

θ2.P.Q.CosQPR22

In triangle ACE,

BEAB

CE

AE

CEtan

But, CE = BC. Sin θ

θ.CosQP

θSinQ.αtan

STEPS TO FIND THE RESULTANT:

1. Resolve all the forces horizontally and find the algebraic sum of

all the horizontal components (Σ Fx). Here, Σ Fx represents the horizontal

component of the resultant.

2. Resolve all the forces vertically and find the algebraic sum of all

the vertical components (Σ Fy). Here, Σ Fy represents the vertical

component of the resultant.

3. Resultant „R‟ is given by the equation, 22R FyFx

4. Let, α be the angle made by the resultant „R‟ with the

horizontal, then Fx

Fytan

RESULTANT OF NON-CONCURRENT FORCE SYSTEM

STEPS TO FIND RESULTANT OF NON-CONCURRENT FORCES:

1. Resolve all the forces horizontally and find the algebraic sum of

all the horizontal components

(Σ Fx). Here, Σ Fx represents the horizontal component of the

resultant.

FY CIVIL Second Sem


2. Resolve all the forces vertically and find the algebraic sum of

all the vertical components (Σ Fy).

Here, Σ Fy represents the vertical component of the resultant.

3. Resultant „R‟ is given by the equation, 22R FyFx

4. Let, α be the angle made by the resultant „R‟ with the

horizontal, then Fx

Fytan

5. Select some reference point (say point A) and take moments of

all the forces about that point using

correct sign conventions.

6. Find algebraic sum of moments of all the forces about the

reference point i.e. ( FAM )

7. Let, „x‟ be the perpendicular distance between the resultant

and the reference point. Take moment of

the resultant about the reference point (MRA).

8. Apply Varignon‟s theorem of moments to find the position (i.e.

distance x) of the resultant about the

reference point.

Note: 1. If FAM is + ve, then the resultant will produce clockwise

moment about the reference point.

2. If FAM is - ve, then the resultant will produce anticlockwise

moment about the reference point.

FY CIVIL Second Sem


1.6 COMPOSITION OF FORCES

I. ANALYTICAL METHODS

Type 1: Examples on parallelogram law of forces.

Type 1.1: Examples on resultant of two equal forces.

Type 1.2: Examples on resultant of two mutually perpendicular forces.

Type 1.3: Examples on maximum and minimum values of resultant.(when

θ=0or 180)

Type 1.4: General examples on parallelogram law of forces.

Type 1.5: Examples on calculating unknown forces or angles.

Type 2: Examples on resultant of concurrent forces.

Type 2.1: Examples on pull type concurrent forces.

R2 = P2 + Q2 + 2PQ Cos Ɵ OR R = 2Pcos (Ɵ/2)

R = √(P2 + Q2) ……… Magnitude of resultant

α = tan _1 (p/Q) ……. Direction of resultant

when Ɵ =O : R = P + Q and when Ɵ =18O : R = P - Q

R =√( P2 + Q2 + 2PQ Cos Ɵ) …….. Magnitude of resultant

tan α = Q sin Ɵ / (P + Q cos Ɵ) ……. Direction of resultant

FY CIVIL Second Sem


Type 2.2: Examples on pull and push type concurrent forces.

2. Type 2.3: Examples on E, N, W, S, NE, NW, SE, SW and other

directions.

FY CIVIL Second Sem


Type 2.4: Examples on pentagon and hexagon.

Type 2.5: Given R, to find out unknown force and its inclination.

Type 2.6: Miscellaneous examples.

Type 3: Examples on resultant of non-concurrent forces.

Type 3.1: Examples on resultant of non-concurrent forces acting along

the sides of hexagon and pentagon.


the sides a square or rectangle.


the sides a triangle.

Type 3.4: Examples on resultant of non-concurrent forces acting on bent

up bar, straight bar and vertical pole.

Type 4: Resultant of parallel forces.

Type 4.1: Resultant of two like and unlike parallel forces.

Type 3: Examples on resultant of non-concurrent forces.

Department of Civil Engineering

Sanjay Ghodawat Polytechnic, Atigre Page 1

For Details contact:

Mr. V.S. kumbhar

HOD

Civil Department,

Sanjay Ghodawat Polytechnic, Atigre.

Mob. No.: 7798306363

Ph. No.: 0230 -246312

Email ID: [email protected]

mailto:[email protected]

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