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Common to All Branches

1 & 2 Semester

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Elements of Civil Engineering & Engineering Mechanics

[10CIV13]

Physics Cycle

2010 Scheme

Branch Name: Common to all branches SEM: 1/2 University: VTU Syllabus: 2010

Table of Contents:

Elements of Civil Engg. & Engg. Mechanics (10CIV13):

Sl. No. Units 1 (a) Introduction(Building the Future)

(b) Roads (c) Bridges & Dams

2 Engineering Mechanics 3 (a) Composition of Forces

(b) Coplanar Non-Concurrent Force Systems

4 Centroid 5 Equilibrium of Forces 6 Supports & Beams 7 Friction 8 Moment of Inertia

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Unit-I

CIVIL ENGINEERING – BUILDING THE FUTURE By

Prof. P. Nanjundaswamy, SJCE, Mysore

Civil engineers have one of the world's most important jobs: they build our quality of life.

With creativity and technical skill, civil engineers plan, design, construct and operate the

facilities essential to modern life, ranging from bridges and highway systems to water

treatment plants and energy efficient buildings. Civil engineers are problem solvers,

meeting the challenges of pollution, traffic congestion, drinking water and energy needs,

urban development and community planning.

During the past century, clean water supplies have extended general life expectancies.

Transportation systems serve as an economic and social engine. New bridges, blending

strength and beauty, speed transport and bring communities closer together. Public and

private construction, for which engineers provide the essential underpinnings of design

and project oversight, produces hundreds of thousands of jobs and drives community

development. From the functional and beautiful Golden Gate Bridge in the United

States, Petronas Towers in Malaysia, and Pont du Gard in France to the largely hidden

water supply and sanitary sewer systems, civil engineers have made their mark in many

aspects of the daily life of essentially everyone around the globe.

The American Society of Civil Engineers defines civil engineering as “…the profession

in which a knowledge of the mathematical and physical sciences gained by study,

experience, and practice is applied with judgment to develop ways to utilize,

economically, the materials and forces of nature for the progressive well-being of

humanity in creating, improving and protecting the environment, in providing facilities for

community living, industry and transportation, and in providing structures for the use of

humanity.”

Entrusted by society to create a sustainable world and enhance the global quality of life,

civil engineers serve competently, collaboratively, and ethically as master:

• Planners, designers, constructors, and operators of society’s economic and

social engine – the built environment

• Stewards of the natural environment and its resources

• Innovators and integrators of ideas and technology across the public, private,

and academic sectors

• Managers of risk and uncertainty caused by natural events, accidents, and other

threats and

• Leaders in discussions and decisions shaping public environmental and

infrastructure policy.

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world. Whatever area you choose, be it design, construction, research, planning,

teaching or management, civil engineering offers you a wide range of career choices.

And there's no limit to the personal satisfaction you will feel from helping to make our

world a better place to live.

Civil engineering is an umbrella field comprised of many related specialties. The

following figure shows the broad categories of fields under civil engineering.

Building materials technology deals with proper use of desired material for construction

economically and safely. Brick, tiles, soil, cement, stone, sand, steel, aggregates, glass,

wood, plastics etc. include construction materials. Some are natural and many are

manmade. The mechanical properties of these materials shall be sufficient to avoid

failure and excessive deformation and provide durability. The chemical properties shall

be to maintain good environment.

Structural engineers face the challenge of designing structures that support their own

weight and the loads they carry, and that resist extreme forces from wind, earthquakes,

bombings, temperature and others. Bridges, buildings, amusement park rides and many

other kinds of projects are included within this speciality. Structural engineers develop

appropriate combinations of steel, concrete, timber, plastic and new exotic materials.

They also plan and design, and visit project sites to make sure work is done properly.



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The skills of environmental engineers have become increasingly important as we

protect our fragile resources. Environmental engineers translate physical, chemical and

biological processes into systems to destroy toxic substances, remove pollutants from

water, reduce nonhazardous solid waste volumes, eliminate contaminants from the air

and develop groundwater supplies. Environmental engineers are called upon to resolve

the problems of providing safe drinking water, cleaning up contaminated sites with

hazardous materials, disposing of wastewater and managing solid wastes.

Geotechnical engineering is required in all aspects of civil engineering because most

projects are supported by the ground. A geotechnical engineer may develop projects

below the ground, such as tunnels, foundations and offshore platforms. They analyse

the properties of soil and rock that support and affect the behaviour of these structures.

They evaluate potential settlements of buildings, the stability of slopes and fills, the

seepage of ground water and the effects of earthquakes. They investigate rocks and

soils at a project site and determine the best way to support a structure in the ground.

They also take part in the design and construction of dams, embankments and retaining

walls.

Water is essential to our lives, and water resources engineers deal with the physical

control of water. They work with others to prevent floods, supply water for cities,

industry and agriculture, to protect beaches or to manage and redirect rivers. They

design, construct and maintain hydroelectric power facilities, canals, dams, pipelines,

pumping stations, locks, seaport facilities or even waterslides.

The quality of a community is directly related to the quality of its transportation system.

Transportation engineers work to move people, goods and materials safely and

efficiently. They find ways to meet our ever-increasing travel needs on land, air and sea.

They design, construct and maintain all types of transportation facilities, including

airports, highways, railroads, mass transit systems and ports. An important part of

transportation engineering is upgrading our transportation capability by improving traffic

control and mass transit systems, and by introducing highspeed trains, people movers

and other intermodal transportation methods.

The construction phase of a project represents the first tangible result of a design. Using

technical and management skills, construction engineers turn designs into reality on

time and within budget. They apply their knowledge of construction methods and

equipment, along with the principles of financing, planning and managing, to turn the

designs of other engineers into successful facilities.

Planners are concerned with the full development of a community. They analyse a

variety of information to co-ordinate projects, such as projecting street patterns,

identifying park and recreation areas, and determining areas for industrial and

residential growth. They employ their technical and people skills to co-ordinate with

other authorities to integrate freeways, airports and other related facilities.



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Infrastructure

It is the framework of supporting system consisting of transportation, energy,

communication, lifeline facilities, irrigation facilities, etc., for the economic development

of a country by the growth of industrial and agricultural fields. Economic infrastructure

contributes directly to the economic development of the country while social

infrastructure like education & training, social welfare, housing, water supply, etc., will

have indirect influence on the economic development. Urban growth only can lead to

population drift from rural sectors leading to explosion in population in cities and

inadequate development of villages and improper care for agricultural sector. Use of

infrastructural facility only by upper class leads to imbalance. Demands for sustainable

energy, fresh water, clean air, and safe waste disposal drive global infrastructure

development.

The infrastructural development has the following major impacts on a country

• Increase in food production

• Protection from drought, famine, flood

• Healthy and comfortable housing facility

• Safe domestic and industrial water supply

• Safe and scientific waste disposal

• Improvement in communication and transportation

• Generation of electricity from, nuclear, hydel, thermal, solar or wind energy

• Improved, wealth, prosperity, standard of living

• Overall growth of a nation

Large-scale budget allocation for infrastructure leads to agricultural and industrial

developments. It provides employment, eradicates poverty and enhances per capita

income.

Role of Civil engineers in Infrastructural development are

• Construction of roads, railway, ports, harbors and airports

• Construction of dams and proper utilization of water resources

• Construction of Housing, commercial and industrial complexes

• Maintenance of facilities

• Rebuilding, Rehabilitation, Retrofitting and Repair

Concluding Remarks



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� Civil engineers served, serving and will serve as master builders, environmental

stewards, innovators and integrators, managers of risk and uncertainty, and

leaders in shaping public policy.

� Civil Engineering is about community service, development, and improvement

� In essence, Civil Engineering may be regarded as the profession that makes the

world a more agreeable place to live

--------------------------



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Roads

Transportation is a non-separable part of any society and is responsible for the

development of civilizations. It meets travel requirement of people and transport

requirement of goods and it is one of the key infrastructures of a country & considered a

mark of its progress.

The roles of transportation in society are:

� Advancement of community

� Economic prosperity and general development of a country

� Strategic movement in emergency

� Safety, Pollution, Energy consumption

� Other impacts

Roadways or Highways are one of the primary modes of transportation. Roads provide

best bet for achieving inclusive growth of our society than any other modes of transport.

Following are the characteristics of roadways

� Maximum flexibility for travel

� Route, Direction, Time and Speed

� Safety decreases

� Door to door service

� Feeder system for other modes

� Used by various types of vehicles

� For short distance travel – saves time

� Requires relatively small investment

India has the second largest road network in the world, next only to USA. However,

large stretches of our roads still suffer from deficiencies in road geometry and riding

quality resulting in hazardous conditions and poor road safety. Civil engineers face the

challenge of designing safe highways and at the same time improving the operational

speeds of the vehicles to reduce the travel time.

Classification of Roads

Based on road pavement

� Paved roads

� Unpaved roads

Based on use during different seasons

� All-weather roads

� Fair-weather roads

Based on traffic volume



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� Heavy

� Medium and

� Light traffic

Based on tonnage

� Class I, II etc. or Class A, B etc.

Based on location and function

Non-Urban Roads – as per Nagpur Road Plan

� National Highways (NH)

� State Highways (SH)

� Major District Roads (MDR)

� Other District Roads (ODR)

� Village Roads (VR)

Non-Urban Roads – as per third road development plan

� Primary system – Expressways and NH

� Secondary system – SH and MDR

� Tertiary system or rural roads – ODR and VR

Urban Roads

� Arterial roads

� Sub-arterial roads

� Collector streets

� Local streets

Components of a Road

Typical section of a roadway

A roadway consists of Geometric Elements and Structural Elements. The geometric

elements are the visible elements across the roadway while the various layers in the

carriage way constitute the structural elements. The geometric elements include Cross

section Elements, Sight distance considerations, Horizontal and Vertical alignment

details, and Intersection elements. The structural elements consist of typical layers of

varying thicknesses and materials. The common layers in a roadway are: Soil

Subgrade, Sub-base course, Base course and Surface course.

--------------------------



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A bridge is a structure built to span a gorge, valley, road, railroad track, river, body of

water, or any other physical obstacle. A bridge is designed for trains, pedestrian or road

traffic, or pipeline or waterway for water transport or barge traffic. A road-rail bridge

carries both road and rail traffic.

Types of Bridges

Based on Action

• Beam bridges

• Cantilever bridges

• Arch bridges

• Suspension bridges

• Cable-stayed bridges

• Truss bridges

Based on Material used

• Concrete Bridge

• Steel Bridge

• Timber Bridge

• Composite Bridge

Based on purpose

• Road Bridge

• Rail Bridge

• Rail & Road Bridge

• Pedestrian Bridge

• Aqueduct



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Basic Bridge Types



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Culverts

Culverts are smaller bridges, normally with one span built across small streams, drains

or sewer carrying road on top

Bridge Components

• Foundation

• Abutment

• Bridge Pier

• Bearing

• Deck Slab

• Roadway

• Railing

References

1) ASCE (2007), The Vision for Civil Engineering in 2025, American Society of Civil

Engineers

2) Syed Shakeeb Ur Rahman and Madhava Rao V (2006), Elements of Civil

Engineering and Engineering Mechanics, Sanguine Technical Publishers.

--------------------------



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Unit-II

ENGINEERING MECHANICS By Prof. V. Madhava Rao, SJCE, Mysore

MECHANICS

It’s a branch of science, which deals with the action of forces on bodies at rest or

in motion.

ENGINEERING MECHANICS

It deals with the principles of mechanics as applied to the problems in

engineering.

BASIC CONCEPTS

1. Matter: Anything which has mass and requires space to occupy is called matter.

2. Mass: It is a measure of quality of matter contained by the body.

SI unit: Kg.

3. Volume: It is a measure of space occupied by the body.

Unit: m3

Note: Liter � Unit of volume

1000 liters = 1 m3

TMS – Thousand million cubic feet.

� 109 ft3

� 10009 ft x 10000 ft x 1000 ft

4. State of rest and motion: State of rest and state of motion are relative and depend

on the frame of reference. A body is said to be in a state of rest w.r.t. a frame of

reference if the position of the body w.r.t. that frame of reference is not changing

with time. A body is laid to be in a state of motion w.r.t. a frame of reference if the

position of the body w.r.t. that frame of reference is changing with time.

5. Scalar and Vector Quantities: Quantities which require only magnitude to

represent them are called scalar quantities.

Eg: Mass, Time interval.

Quantitites which require both magnitude and direction to represent them are called

vector quantities.

Eg: Force, Velocity, etc.



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A

B

Distance

Displacement

6. Displacement and distance travelled: The total linear movement made by a body

to change its position from one point to another is called distance travelled by the

body. It is a scalar quantity.

Unit: Meter (m)

mm – Millimeter � 10-3m

km – Kilo Meter � 103m

The total linear movement made by a

body to change its position from one

point to another moving along a particular

direction is called displacement.

Displacement is a vector quantity.

Unit: Meter (m).

7. Speed and Velocity: The distance travelled in a unit time is speed.

Unit: m/s � ms-1

The displacement in unit time is called velocity.

Unit: m/s � ms-1

8. Uniform motion and non-uniform motion: If the velocity of the moving body

remains constant then the motion is said to be uniform. If the velocity is changing

with time, the motion is laid to be non-uniform.

9. Acceleration and retardation: The time rate of change of velocity is called

acceleration.

If the velocity is increasing with time then acceleration is positive. If the velocity is

decreasing with time then acceleration is negative. Negative acceleration is called

retardation or deceleration.

Unit: m/s2 � ms-2

10. Momentum: It is the capacity of a moving body to impart motion to other bodies.

Momentum of a moving body is given by the product of mass and velocity of the

moving body.

Momentum = Mass X Velocity

Unit: kg m/s or kg ms-1.

11. Newton’s I Law of Motion: “Everybody continues to be in its state of rest or uniform

motion unless compelled by an external agency”.



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F u

m m

Time interval = t

V

12. Inertia: It is the inherent property of a body by virtue of which it can retain its state of

rest or uniform motion unless compelled by an external agency.

13. Force: It is an external agency, which overcomes or tends to overcome the inertia of

a body. It is a vector quantity.

14. Elements of a force: There are four elements:

a. Magnitude

b. Direction

c. Line of action

d. Point of action or application

15. Newton’s II Law of motion: “The rate of change of moment of a body is directly

proportional to the magnitude of the force applied and takes place in the direction of

the force applied”.

Explanation:

Initial momentum = mu

Final momentum = mv

Change in momentum over a time interval ‘t’ = mv – mu

Rate of change of momentum = t

mumv −

According to Newton’s II law,

maKF

maF

t

uvmF

t

mumvF

=

α

−α

−α

In SI, unit force is defined as that force which acts on a body of unit mass producing

unit acceleration.

i.e., F = 1 when m = 1 and a = 1

then 1 = k . 1 . 1

∴ k = 1

F = ma

Unit of force: newton (N) is the unit of force. One newton is that force which acts on

a body of mass 1 kg producing an acceleration of 1 m/s2.



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kN – Kilo newton – 103N

MN – Mega newton – 106N

GN – Giga newton – 109N

16. Newton’s III law of motion: “For every action there is equal and opposite reaction”.

17. Branches of Mechanics:

Statics: Statics deals with the action of forces on bodies at rest or in equilibrium.

Dynamics: Dynamics deals with the action of forces on bodies in motion.

Kinematics: It deals with the study of geometry of motion without considering the

cause of motion.

Kinetics: Kinetics deals with a study of motion considering the course of motion.

18. Rigid body: The concept of rigid body is purely theoretical or imaginary. A rigid body

is said to undergo, no deformation under the action of any external agency such as

force and moments.

In other words relative positions of the modules of a rigid body are fixed in space.

19. Particle: Concept of particle is purely theoretical or imaginary. A particle is said to

have mass but requires no space to occupy. In other words, a particle is a point

mass.

Fluid

Dynamics

Mechanics

Solid Mech. Fluid Mech.

Rigid

Body

Mech.

Mech. of

Deformable

Bodies

Fluid

Statics

Fluid

Kinematics

Statics Dynamics

Kinematics Kinetics



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The concept of particle cannot be used if the shape and size of the body is

influencing the motion.

Eg: i) Motion of a swimmer.

ii) Motion of a body along a curved path.

20. Continuum: The concept of continuum is purely theoretical or imaginary.

Continuum is said to be made up of infinite number of molecules packed in such a

way that, there is no gap between the molecules so that property functions remain

same at all the points.

21. Point force: The concept of point force in purely theoretical or imaginary, here the

force is assumed to be acting at a point or over infinity small area.

22. Principle physical independence of forces:

Action of forces on bodies are independent, in other words the action of forces on a

body is not influenced by the action of any other force on the body.

23. Principle of superposition of forces:

F1 a1

F1 a2

F1 a1, a2

F2 a1, a2

F2

F1

M

M

M

M



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Net effect of forces applied in any sequence on a body is given by the algebraic sum

of effect of individual forces on the body.

24. Principle of transmissibility of forces:

The point of application of a force on a rigid body can be changed along the same

line of action maintaining the same magnitude and direction without affecting the

effect of the force on the body.

Limitation of principle of transmissibility: Principle of transmissibility can be used

only for rigid bodies and cannot be used for deformable bodies.

25. Assumptions made in Engineering Mechanics

i) All bodies are rigid.

ii) Particle concept can be used wherever applicable.

iii) Principle of physical independence of forces is valid.

iv) Principle of superposition of forces is valid.

v) Principle of transmissibility of forces is valid.

F A

M

Rigid body

M

Fa =

B F M

Rigid body

M

Fa =Line of action

Line of action

F1 a1

F1 a2

F1 (a1+a2)

F2 (a2+a1)

F2

F1

M

M

M

M



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SYSTEM OF FORCES

A group or set of forces is called system of forces.

Types:

1. Coplanar force system:

If the lines of action of forces forming the system lie in the same plane, then the

system is said to be coplanar.

2. Non-coplanar forces:

If the lines of action of forces forming the system do not lie in the same plane then

the system is said to be non-coplanar.

Note: Our study is restricted to coplanar forces.

3. Collinear force system:

If the forces forming the system have common line of action then the system is said

to be collinear.

A

F1

B

F2 F3

C

F1

F2 F3 F4 F6

F5

F1

F2 F3



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4. Concurrent force system:

If the line of action of forces forming the system pass through a common point (point

of concurrence) then the system is said to be concurrent.

5. Non-concurrent force system:

If the lines of action of forces forming the system do not pass through a common

point, then the system is said to be non-concurrent.

6. Parallel force system:

It is a particular case of non-concurrent force system in which the line of action of

forces forming the system are parallel.

RESOLUTION OF A FORCE

F2F1 F3 F4

F2

F1 F3

F4

Like Unlike

F2

F1

F3

F4

F4

F2

F3

A

F1

x x

D

C B

0

F2

F1

F3

F4 O

F2

F1

F3

F4

O

Y



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The force F is producing, simultaneous x displacement and y-displacement. The

part of the force F which is producing x displacement is called x component or

horizontal component of the force F (Fx). The part of the force F which produces y –

displacement is called y component of the force or vertical component of force F (Fy).

The technique of finding a component of a force along any direction is called

resolution of force. The component of a force along any direction is called the resolved

component. The components of a force determined along two mutually perpendicular

direction are called rectangular components.

To resolve a force along any direction

OA represents the force F both in magnitude and direction ‘θ’ is the acute angle

mode by the force w.r.t. x direction.

We have,

OA

OACos 1

=θ

F

FCos x

=θ

θ= CosFFx (�)

OA

AASin 1

=θ

OA

AASin 2

=θ

Y

X

A A2

A1

Fy

FX

F

θ

Y Displacement

X Displacement

X

F



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+

–

+

–

F

FSin

y=θ

θ= SinFFy (�)

X component of a force is given by the product of magnitude of the force and

cosine of acute angle made by the force w.r.t. x-direction.

Y component of a force is given by the product of magnitude of the force and

sine of acute angle made by the force w.r.t. x-direction.

Note:

1. Sign convention for the direction of components.

2. θ = 0

OCosFFx =

= F

OSinFFy =

= O

The horizontal component or ‘X’ component of a force acting along x direction is the

force itself. Whereas, its vertical component or y-component is zero.

3. 90CosFFx =

= O

90SinFFy =

= F

‘x’ component of a force acting along Y direction is zero. Whereas, its ‘y’ component

is equal to itself.



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100 kN

30o

20 kN

30o

4

3

200 N

4. If a force is inclined at 45o w.r.t. x axis or y axis then its x component will be equal to

y component (Fx = Fy).

Problems

1. Find X and Y components of forces in the following cases.

a)

Fx = + 100 Cos 30

= + 86 . 60 kN

= 86 . 60 kN (�)

Fy = + 100 Sin 30

= + 50 . 00 kN

= 50 . 00 kN (�)

b)

Fx = + 20 Cos 70

= + 6 . 840 kN

= 6 . 840 kN (�)

Fy = + 20 Sin 70

= + 18 . 79 kN

= 18 . 79 kN (�)

c)

Method-I

4

3tan =θ

o87.36=θ

Fx = - 200 Cos 36.87o

= – 160 N

= 160 N ()



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Fy = - 200 Cos 36.87o

= – 120 N

= 120 N (�)

Method-II

8.05

4Cos ==θ

6.05

3Sin ==θ

Fx = - 200 Cos θ

= – 200 x 0.8

= – 160 N

= 160 N ()

Fy = – 200 Sin θ

= – 200 x 0.6

= – 120 N

= 120 N (�)



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RESULTANT FORCE OF A SYSTEM OF FORCES

The resultant of a system of forces is a single calculated force which is capable

of producing the same effect as that of system of forces on the body. It is the vector

sum of forces of the system.

COMPOSITION OF FORCES

The technique of finding the resultant of forces is called composition of forces.

MOMENT OF A FORCE

It is the capacity of a force to produce rotator motion. In other words moment of a

force is its rotating capacity.

Based on the direction of rotation produced moment of a force can be classified

into

a) Clockwise moment

b) Anticlockwise moment / counter clockwise moment.

F A

Moment of F about A in

clockwise

F A

Moment of F about A in

anticlockwise

α

R

F1

F2

F3

F4

α



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X

F = 10 kN

2 m A

X

F = 10 kN

2 mA

X

F = 20 kN

3 m

A

Calculation of Moment of a Force about a Point

Moment of a force about any point is given by the product of magnitude of force

and perpendicular distance between the line of action of a force and the point about

which moment is considered.

Sign Convention for Moment of a Force

Clockwise moment positive and anticlockwise moment negative.

1) Find moment of force ‘F’ about ‘A’ in the following cases.

a)

b)

c)

MA = FL

Unit: Nm

X

F



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50 kN

2 m

A

X

F = 10 kN

2 m A

X 4 m B

F = 20 kN

3 m

B

A 4 m

d)

2) Find moment of the force about A and B in the following

a)

b)



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Unit-III

ELEMENTS OF CIVIL ENGINEERING AND

ENGINEERING MECHANICS

by

Prof. Karisiddappa, MCE, Hassan

COMPOSITION OF FORCES: The reduction of a given system of forces

to the simplest system that will be its equivalent is called the problem of

composition of forces.

• RESULTANT FORCE: It is possible to find a single force which

will have the same effect as that of a number of forces acting on a

body. Such a single force is called resultant force.

• The process of finding out the resultant force is called composition of

forces.

COMPOSITION OF CO-PLANAR CONCURRENT FORCE SYSTEM

COMPOSITION OF TWO FORCES: It is possible to reduce a given

system of forces i.e., two forces to the simplest system as its equivalent

(resultant force) with the help of parallelogram law of forces.

• LAW OF PARALLELOGRAM OF FORCES:

If two forces, which act at a point be represented in magnitude and

direction by the two adjacent sides of a parallelogram drawn from one of its

angular points, their resultant is represented by the diagonal of the

parallelogram passing through that angular point, in magnitude and

direction.



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RR

B C

A O

α

R

θ

αcos221

2

2

2

1FFFFR ++=



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RR

B C

A O

α

α

PROOF:

R

θ

D

Consider two forces F1 and F2 acting at point O as shown in

figure. Let α be the angle between the two forces.

Complete the parallelogram ACBO .Drop perpendicular CD to

OA produced. Let R be the resultant force of forces and

.Let θ be the inclination of the resultant force with the

line of action of the force.



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From triangle OCD,

+

=

+

=

=

++=

++=

+++=

++=

====

++=

+=

−

α

αθ

α

αθ

θ

α

α

ααα

αα

αα

cos

sin

tan

cos

sin

tan

tan

cos2

cos2

sincoscos2

)sin()cos(

,sin,cos,

)(

21

21

21

2

21

2

2

2

1

2

221

2

1

2

22

2

22

221

2

1

2

2

2

2

21

2

221

222

222

FF

F

FF

F

OD

CD

FFFFR

FFFFR

FFFFFR

FFFR

ROCFCDFADFOA

CDADOAOC

CDODOC



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21

0

21

0

21

0

,180

,0

,90

FFR

FFR

FFR

−==

+==

+==

α

α

αIF

IF

IF 1

F

1F

1F

2F

2F

2F

R

• TRIANGLE LAW OF FORCES: If two forces acting simultaneously on a body are

represented by the sides of a triangle taken in order,

their resultant is represented by the closing side of

the triangle taken in the opposite order.

O

1F

2F

1F

2F

O A

B

R



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• POLYGON LAW OF FORCES:If a number of concurrent forces acting simultaneously on a body ,are

represented in magnitude and direction by the sides of a polygon,

taken in order , then the resultant is represented in magnitude and

direction by the closing side of the polygon, taken in opposite order.

O

1F

2F

3F

4F

O

1F

2F

3F

4F

R

D

A

B

C

1R

2R

COMPOSITON OF FORCES BY

RESOLUTION(Principle of resolved parts)

• The components of each force in the system in two mutually

perpendicular directions are found.

• Then, the components in each direction are algebraically

added to obtain the two components.

• These two component forces which are mutually

perpendicular are combined to obtain the resultant force.



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1θ

4θ

2θ

3θ

X

Y

1F

2F

3F

4F

Algebraic sum of the components of forces in X

direction

44332211coscoscoscos θθθθ FFFFFx +−−=∑

Algebraic sum of the components of forces in Y

f direction

Now the system of forces is equal to two

mutually perpendicular forces ,

44332211sinsinsinsin θθθθ FFFFFy −−+=∑

∑∑ YX FF &

=

+=

∑∑

∑ ∑

−

X

Y

YX

F

F

FFR

1

22

tanθ



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NUMERICAL PROBLEMS

1. Determine the magnitude and direction of the resultant of the

two forces of magnitude 12 N and 9 N acting at a point ,if the

angle between the two forces is

GIVEN:

NF 121

= NF 92

=0

30=α

0

0

0

1

21

21

022

21

2

2

2

1

81.12

30cos912

30sin9

tan

cos

sin

tan

3.20

30cos9122912

cos2

=

+

=

+

=

=

×××++=

++=

−

−

θ

θ

α

αθ

α

FF

F

NR

R

FFFFR



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2.Find the magnitude of two equal forces acting at a point with an

angle of 600 between them, if the resultant is equal to N330

GIVEN:

3.The resultant of two forces when they act at right angles is 10 N.Whereas, when they act at a angle of 600 , the resultant is N. Determine the magnitude of the two forces.

Let F1 and F2 be the two forces,

Given – when α =900

R = 10N

When α =600

R = N

We have,

When α =900

Squaring both sides 100= F12 + F2

2 (1)

When α =600

0

21

2

2

2

160cos2148 FFFF ++=

5.0221

2

2

2

1×++ FFFF

sayFFF ,21

==

0

60,330 == αNR

NF

FR

FFFR

FFFFR

FFFFR

30

3

60cos2

cos2

222

022

21

2

2

2

1

=

=

++=

××++=

++= α

148

148

αcos221

2

2

2

1FFFFR ++=

2

2

2

110 FF +=



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squaring both sides

148 = F12 + F2

2 (2)

substituting (1) in (2)

148 = 100+F1F2

F1F2 = 48 (3)

squaring equation (3),we get

F12 + F2

2 = 48

2(4)

From (1) F22

= 100 – F1

2(5)

Subtracting (5) in (4)



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( )

( )

( )

NFNF

F

F

F

F

FF

FF

6&8

64

1450

19650

504850

48100

48100

21

2

1

2

1

22

1

222

2

1

22

1

4

1

22

1

2

1

==

=

=−

=−

+−=−

−=−

=−

5.The 26 KN force is the resultant of two forces. One of the force is as

shown in figure .Determine the other force.

4.Find the magnitude and direction of the resultant

force for the system of concurrent forces shown

below.

N20

N25

N30

N35

0

30

0

45

0

40

X

Y

∑∑

−=

−−=

NF

F

X

X

70.30

40cos3545cos3030cos20000

∑∑

=

−++=

NF

F

Y

Y

72.33

40sin3545sin302530sin20000

( ) ( )

NR

R

FFRYX

60.45

72.3370.3022

22

=

+−=

+= ∑ ∑

0

1

1

68.47

70.30

72.33

tan

tan

=

=

=

−

−

∑∑

θ

θ

θ

X

Y

F

F

∑ XF

∑ YF

R

θ

0



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y 26kN

12

5 10kN

3

4

X

0

Let F be magnitude of unscnorm force with Fx and Fy as its

components in x and y directions.

Component of R in x directions 13 θ1 12

Rx = 26 x cos θ1

= 26 x 5/13 = 10kN 5

Component of R in y direction

Ry = 26 x sin θ1 = 26 x 12/13 = 24kN

Component F and 10kN in X direction

= Fx +10 cos θ2 5

= Fx + 10x 4/5 = Fx +8 θ2 3

4

Component of F and 10kN in y direction

= Fx + 10 x Sin θ2 = Fy + 10 x 3/5

= Fy + 6

Using R/x = /Fx

10 = Fx +8

24 = Fy + 6

Fx = 2kN, Fy = 18kN

But F = √Fx2+Fy

2 = √2

2 + 18

2

F = 18.11kN

θ2 = tan -1

(Fy /Fx) = tan -1

(18/2) = 83.660

θ2 = 83.660 ( inclination of F w.r.t x – axis)



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6.Three forces act at a point in a plate as shown in figure. If the

resultant of these forces is vertical, find the resultant force and

angle α..

100N

160N

α.

120 N α

0

Since the resultant force is vertical, algebraic sum of horizontal components of these must

be equal to zero.

160 cos α – 120 – 100 sin α = 0

120 + 100 sin α = 160 cos α

6 + 5 sin α = 8cos α

Squaring both the sides

(6+5 sine α )2

= (8 cos α )2

36 + 60 sin α + 25 sin2

α = 64 (1-sin2

α)

25 sin2

α +64 sin2

α + 60sin α = 64-36

89 sin2

α + 60 sin α = 28

Sin2

α + 0.674 sin α =0.315

(sin α + 0.337)2

= 0.315 + 0.3372

= 0.428

sin α + 0.337 = √0.428 = 0.654

sin α = 0.654 – 0.337 = 0.317

α = sin-1

(0.317) = 18.50

Resultant force R = Σ Fy

= 160 sin α + 100 cos α

= 160sin 18.50 + 100 cos 18.50



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R = 145.60 N

7.ABCDE is a regular hexagon. Forces 90 N,P,Q,240 N and 180 N act along

AB,CA,AD,AE and FA respectively as shown in the figure. Find the forces

P and Q such that the resultant force is zero.

C D

B

P Q E

90N 300

300

240N

300

600

300

A 180N F X

Since the resultant force is equal to zero, Σ Fx = 0 and Σ Fy = 0

Σ Fx = -180 +240 cos 300

+ Q cos 600

– p

cos 900

+ 90 cos 1200

= 0

-180 + 207.85 + 0.5 Q – 45 =0

0.5Q = 17.15

Q = 34.308N

Σ Fy = 180 sin00

+240

sin300

+ Q sin600

– P + 90 sin1200

= 0

120 + 34.308 x sin 600

– P + 90 sin 1200

= 0

P = 227.654 N



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Moment of force F about O= F x a

= AB x OC

= twice the area of triangle OAB

Thus moment of F about O= 2 x Area of triangle OAB

COMPOSITION OF COPLANAR NON-

CONCURRENT FORCE SYSTEM

MOMENT OF A FORCE: Moment is

defined as the product of the magnitude of the force and

perpendicular distance of the point from the line of

action of the force.

GEOMETRICAL REPRESENTATION OF MOMENT

Consider a force F represented ,in magnitude and direction

by the line AB. Let O be a point about which the moment

of the force F is required. Let OC be the perpendicular

drawn. Join OA and OB

c

FA

B

O

a



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VARIGNON’S PRINCIPLE OF MOMENTS:

If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum

of the moments of all the forces about any point is equal to the moment of their resultant

force about the same point.

PROOF:

For example, consider only two forces F1 and F2

represented in magnitude and direction by AB and AC as shown in figure below.

Let O be the point, about which the moments are taken. Construct the parallelogram

ABCD and complete the construction as shown in fig.

By the parallelogram law of forces, the diagonal AD represents, in magnitude and

direction, the resultant of two forces F1 and F2, let R be the resultant force.

By geometrical representation of moments

the moment of force about O=2 Area of triangle AOB

the moment of force about O=2 Area of triangle AOC

the moment of force about O=2 Area of triangle AOD

But,

Area of triangle AOD=Area of triangle AOC + Area of triangle ACD

Also, Area of triangle ACD=Area of triangle ADB=Area of triangle AOB

Area of triangle AOD=Area of triangle AOC + Area of triangle AOB

Multiplying throughout by 2, we obtain

2 Area of triangle AOD =2 Area of triangle AOC+2 Area of triangle AOB

i.e., Moment of force R about O=Moment of force F1 about O + Moment of force F2

about O

Similarly, this principle can be extended for any number of forces.

CO

BA

D

1F

2F

R



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NUMERICAL PROBLEMS

1.For the non-concurrent coplanar system shown in fig below,

determine the magnitude , direction and position of the resultant

force with reference to A.

R

N35

N25

N50

N20

A

B

C

D

( )

∑∑

−=−−=

→=−=

NF

NF

Y

X

853550

52025

( )↓= N85

( ) NFFR YX 15.858552222

=−+=+= ∑∑

∑ XF

∑ YF

R

θmd

dR

525.2

15.85

75140

325435

=+

=

×+×=×

or

mx

x

53.2

85

75140

32543585

=+

=

×+×=×



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∑ XF

∑ YF

R

θ

2.Determine the resultant of the force system acting on the plate

as

shown in figure given below wirh respect to AB and AD.

10N 5N

600 D

10Nm 300

C

3m

A 4m B

14.14N 1

1 20 N

Σ Fx = 5cos300 + 10cos600 + 14.14cos450

= 19.33N Σ Fy = 5sin300 - 10sin600 + 14.14sin450

= -16.16N R = √( Σ Fx2 + Σ Fy2) = 25.2N

θ= Tan-1(Σ Fx/ Σ Fy) θ= Tan-1(16.16/19.33) = 39.890

D C

y

A x θ B

R 16.16N

Tracing moments of forces about A and applying varignon’s principle of moments we get

+16.16X = 20x4 + 5cos300x3-5sin300x4 + 10 + 10cos600x3

Θ 19.33N

19.33N



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X = 107.99/16.16 = 6.683m

Also tan39.89 = y/6.83 y = 5.586m.

3.The system of forces acting on a crank is shown in figurebelow. Determine the magnitude , direction and the point of application of the resultant force.

500 N 150 700N 150

600 600

150mm 150 mm 150 Cos600=75mm

Σ Fx = 500cos600 – 700 = 450N

Σ Fy = 500sin600

= -26.33N R = √( Σ Fx2 + Σ Fy2) = √(-450)2 + (-2633)2

R = 267.19N (Magnitude) Σ Fx

θ θ= Tan-1(Σ Fx/ Σ Fy) = Tan-1(2633/450) R Σ Fy

θ= 80.300 (Direction)

ΣFx

Θ x

R ΣFy

Tracing moments of forces about O and applying varignon’s principle of moments we get -2633x x= -500x sin600x300-1000x150+1200x150cos600 -700x300sin600

X = -371769.15/-2633 X = 141.20mm from O towards left (position).



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4.For the system of parallel forces shown below, determine the magnitude of the resultant

and also its position from A .

100N 200N 50N 400N

R

A B C D

1m 1.5m 1m

X

Σ Fy = +100 -200 -50 +400 = +250N

ie. R = Σ Fy =250N ( ) Since Σ Fx = 0

Taking moments of forces about A and applying varignon’s principle of moments

-250 x = -400 x 3.5 + 50 x 2.5 + 200 x 1 – 100 x 0

X = -1075/ -250 = 4.3m

5.The three like parallel forces 100 N,F and 300 N are acting as shown in figure below. If

the resultant R=600 N and is acting at a distance of 4.5 m from A ,find the magnitude of

force F and position of F with respect to A.

100N F 600 N 300N

A B C D

4.5m 2.5m

X

Let x be the distance from A to the point of application of force F

Here R = Σ Fy 600=100+F+300 F = 200 N

Taking moments of forces about A and applying varignon’s principle of moments, we get

600 x 4.5 = 300 x 7 + F x 200 x = 600 x 4.5 -300 x 7



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X = 600/200 = 3m from A

6.A beam is subjected to forces as shown in the figure given below. Find the magnitude ,

direction and the position of the resultant force.

17kN 10kN 20kN 10kN 5kN

α

θ 4kN

A B C D E

2m 3m 2m 1m

Given tan θ = 15/8 sin θ = 15/ 17 cos θ = 8/17

tan α = 3/4 sin α = 3/5 cos α = 4/5

Σ Fx = 4 +5 cos α – 17 cos θ= 4+5 x 4/5 – 17 x 8/17

Σ Fx = 0

Σ Fy = 5 sin α -10 +20 – 10 + 17 sin θ = 5 x 3/5 -10+20 – 10 + 17 x 15/17

Σ Fy = 18 kN ( )

Resultant force R = √ 2Fx 2 + Σ Fy2 = √ 0+182

R = 18 kN ( )

Let x = distance from A to the point of application R

Taking moments of forces about A and applying Varignon’s theorem of moments -18 x = -5 x sin α x 8 +10 x 7 -20 x 5 + 10 x 2

= -3 x 8 +10 x7 – 20 x 5 + 10 x 2

X = -34/-18 = 1.89m from A (towards left)



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Overview of System of forces It is well known that a system of coplanar forces can occur in different configurations some of the possibilities are

• Coplanar, Collinear, Concurrent

• Coplanar and Concurrent

• Coplanar and Non Concurrent

To determine the resultant of any system of forces we adopt the principle of Resolution

and Composition.

The following figures depict the principles involved.

.Composition of system of forces

)(1

22

tan

)()(

i

i

ii

x

y

R

yx

f

f

ffR

∑

∑

∑∑

−

=

+=

α

COPLANAR NON-CONCURRENT FORCE SYSTEMS By

Prof. G. Ravi



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Equilibrium: Equilibrium is the status of the body when it is subjected to a system of forces. We know that for a system of forces acting on a body the resultant can be determined. By Newton’s 2nd Law of Motion the body then should move in the direction of the resultant with someacceleration. If the resultant force is equal to zero it implies that the net effect of the system of forces is zero this represents the state of equilibrium. For a system of coplanar concurrent forces for the resultant to be zero, hence

Equilibriant : Equilbriant is a single force which when added to a system of forces brings the status of equilibrium . Hence this force is of the same magnitude as the resultant but opposite in sense. This is depicted in Fig 4.

Free Body Diagram: Free body diagram is nothing but a sketch which shows the various forces acting on the body. The forces acting on the body could be in form of weight, reactive forces contact forces etc. An example for Free Body Diagram is shown below.

0f

0f

i

i

y

x

=

=

∑∑



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Equilibrium of 3 Forces: When a set of three forces constituting coplanar concurrent system act on a body Lami’s theorem can be made use of for examining the status of equilibrium. This is depicted in the following figure.

Example 1 : A spherical ball of weight 75N is attached to a string and is suspended from the ceiling. Compute tension in the string if a horizontal force F is applied to the ball. Compute the angle of the string with the vertical and also tension in the string if F =150N

γβα Sin

F

Sin

F

Sin

F 321==

150cos0cos150

0cos

0

=

=−

=−

=∑

θ

θ

θ

T

T

Tf

fix



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NTNT

T

TT

FBD

BCAB

AB

ABBC

75,275

075sin

0f0cos

0fB of

1

y

1

x

i

i

==

=−

=

=−

=

∑

∑

θ

θ

NW

WT

NT

TT

FBD

c

cCD

CD

CDBC

57.1280sin

0f85.148

0cos

0fC of

2

y

2

x

i

i

=

=−

=

=

=+−

=

∑

∑

θ

θ

Example 2: A string or cable is hung from a horizontal ceiling from two points A and D. The string AD, at two points B and C weights are hung. At B, which is 0.6 m from a weight of 75 N is hung. C, which is 0.35 m from D, a weight of wc is hung. Compute wc such that the string portion BC is horizontal.



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• LAMI’S Theorem

Example 4: Three smooth circular cylinders are placed in an arrangement as shown. Two cylinders are of radius 052mm and weight 445 N are kept on a horizontal surface. The centers of these cylinders are tied by a string which is 406 mm long. On these two cylinders, third cylinder of weight 890N and of same diameter is kept. Find the force S in the string and also forces at points of contact.

• LAMI’S Theorem

NN

NF

Sin

NR

Sin

F

Sin

R

ooo

76.10159.63

)3290()32180(90120

=

=

+

=

−

=

N 598 F598N F

A of

BA

AC

=

=

FBD

NR

NF

f

f

FBD

D

BC

y

x

i

i

8905.399

0

0B of

=

=

=

=

∑∑

Example 3: A block of weight 120N is kept on a smooth inclined plane. The plane makes an angle of 320 with horizontal and a force F allied parallel to inclined plane. Compute F and also normal reaction.



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Transformation of force to a force couple system:

It is well known that moment of a force represents its rotatary effect about an axis or a point. This concept is used in determining the resultant for a system of coplanar non-concurrent forces. For ay given force it is possible to determine an equivalent force – couple system. This concept is shown in Fig below.

Resultant for a coplanar non-concurrent system:

By using the principles of resolution composition & moment it is possible to determine

analytically the resultant for coplanar non-concurrent system of forces.

The procedure is as follows:

1. Select a Suitable Cartesian System for the given problem.

2. Resolve the forces in the Cartesian System

3. Compute ∑ fxi and ∑fyi

4. Compute the moments of resolved components about any point taken as the moment

centre O. Hence find ∑ M0

22

+

= ∑∑i

yfi

xfR

=

∑

∑

ixf

iyf

R tan 1-α



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5. Compute moment arm

6. Also compute x- intercept as

7. And Y intercept as

Example 1: Compute the resultant for the system of forces shown in Fig 2 and hence compute the Equilibriant.

R

Md

o

R

∑=

∑∑

=

ix

o

Rf

MX

∑∑

=

ix

o

Rf

My

KN 28.8

60 cos 32 - 44.8 o

=

=∑ ixf

KNM

M

f

oo

o

yi

34.62 )3(60sin32)4(60cos32)3(4.14

49.83 KN 44.6 R

KN 34.11 -

60sin 32- 14.4 - 8

oR

o

−=

−+−=+

=

=

=

=

∑

∑

ς

α

m 164.28.28

34.62y

m 827.1 11.3434.62 x

m 396.164.4434.62d

R

R

R

==

==

==



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Example 2: Find the Equilibriant for the rigid bar shown in Fig 3 when it is subjected to forces.

• Resultant and Equilibriant

Equilibrium: The concept of equilibrium is the same as explained earlier. For a system of

Coplanar Non concurrent forces for the status of equilibrium the equations to be satisfied are

The above principles are used in solving the following examples.

;90

516

0

o

R

y

x

KNf

f

i

i

=

−=

=

∑∑

α

KNM 1462- )4(344)2(172)1(430

=

−+−=+∑ AMς

;0 ;0 ;0 === ∑∑∑ oyx Mffii



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Example 3: A bar AB of length 3.6 m and of negligible weight is acted upon by a vertical force F1 = 336kN and a horizontal force F2 = 168kN shown in Fig 4. The ends of the bar are in contact with a smooth vertical wall and smooth incline. Find the equilibrium position of the bar by computing the angle θ.

• Eq. 1 gives HA=420 KN

• Beams – Laterally loaded bending

• Supports – Hinge, Roller, Fixed

• Equilibrium Concept for support reactions

• Equations are

o87.362.1

9.0tan

=

=

α

α

;420013.53sin

0)1..(..........013.53cos

0

1

2

KNR

FR

f

RFH

f

B

o

B

y

o

BA

x

i

i

=

=−

=

=−−

=

∑

∑

o

A

B

H

M

3.280.538 tan

0 cos 705.6 sin 1310.4-0)sin (1.2 168 -)cos1.2(336)sin6.3(

;0

=

=

=+

=+−

=+∑

θ

θ

θθ

θθθ

ς

;0 ;0 ;0 === ∑∑∑ oyx Mffii



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SUPPORT REACTIONS IN BEAMS: Beams are structural members which are generally

horizontal. They are subjected to lateral forces which act orthogonal to the length of the member.

There are various types of mechanisms used for supporting the beams. At these supports the

reactive forces are developed which are determined by using the concept of equilibrium. The

different types of supports are depicted in the table below.

SUPPORT REACTION NO.OF REACTIONS

ROLLER

(1)

HINGE

(2)

FIXED

(3)

VA



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TYPES OF LOADS ACTING ON BEAMS: There are various types of forces or loads which act on beams. They are (a) Concentrated or point load (b) Uniformly distributed load (UDL) (c) Uniformly varying load (UVL) (d) Arbitrary distributed load. The methodology of converting UDL, UVL to equivalent point load is shown in the Fig below. Some example problems of determining support reactions in beams are illustrated next.



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Example 4: Determine the support reactions for the beam shown in Fig 7 at A and B.


;0

;0

;0

=

=

=

∑∑∑

o

y

x

M

f

f

i

i

KNV

KNV

V

M

KNVV

VV

A

B

B

A

BA

BA

7.233.43

0)10()9(32)5(25)2(100

;670322510

=

=

=+−−−

=+

=+

=+−−−

∑ς

;35V45V

0)8(V)7(40)2(400 80VV

0V40-40-;0

0H ;0

A

B

B

BA

B

A

KN

KN

M

Vf

f

A

Ay

x

i

i

=

=

=+−−=

=+

=+=

==

∑∑

ς



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Review • Coplanar system of Forces.• Concurrent, Non Concurrent.• Resultant, Equilibrium.• Concept of Equilibrium.• Examples.• Analysis of Trusses

KNH

H

f

A

A

xi

32.17032.17

;0

=

=−

=∑

KN 10 KN; 450)11(10)9(15)8()6(2025210

055

010152010

0

==

=−−+−+×−

=+

=+

=+−−−−

=

∑

∑

AB

B

A

BA

BA

y

VV

V

M

VV

VV

fi

ς

;62.4;24.9;12

0)6(20)10(;0

20866.0

030cos20;0

5.0030sin

;00

KNH

KNR

KNV

V

M

RV

RVf

RH

RH

f

A

B

A

A

B

BA

o

BAy

BA

BA

x

i

i

=

=

=−

=+−

=+

=+

=+−=

=

=−

=

∑

∑

∑

ς

Example 6: Determine the support reactions for the beam shown in Fig 9 at A and B. Compiled by www.studyeasy.in


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ANALYSIS OF PLANE TRUSSES: Trusses are special structures which are formed by joining different members. Trusses are used as part of roofing systems in industrial buildings, factories workshops etc. Prominent features of trusses are

• Trusses are articulated Structures.• The basic Geometry used in a truss is a triangle.• Every member is pin connected at ends.• Trusses carry loads only at joints. Joints are junctions where members meet.• Self weight is neglected.• The forces in various members of the truss are axial in nature.

A typical figure of a plane truss and the scheme by which truss configuration is arrived at is shown by the following figures.

Plane Trusses

Truss configuration

•• A truss is said to be perfect if m= 2 j – 3 where m � Members; j � Joints•



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1. Method of Joints.2. Method of Sections.

These two methods of analysis are illustrated by the following examples

Example 1: • Analyse the truss shown in Figure and hence

compute member forces• Step 1: Draw FBD• Step 2: Compute support Reactions (HA, VA,

VB).• Draw FBD’s of Joints to compute member

forces.

• ∑fxi=0

• ∑fyi=0

• HA= - 10 KN

• VA+VB =27.32

• ζ + ∑MA = 0

• -17.32(3) - 10(3) - 10(2.25) + 6VB=0

• VB = 17.41 KN; VA= 9.91 KN

• FBD of joint A

• ∑fxi=0

• -10+PAC cos θ + PAD = 0

• ∑fyi=0; VA + PACsin θ =0

• PAC =-16.52 KN

• PAD=23.21 KN

• ∑fxi=0

• -PAD + PDB = 0

;87.36325.2 tan

o

AD

CD

=

==

θ

θ

Analysis of Trusses: Analysis of trusses would imply determining forces in various members. These forces will be in the form of Axial Tension (or) Compression. The Equilibrium concept is made use of for analyzing the trusses. The two methods of analysis are



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• PDB = 23.21 KN

• ∑fyi=0

• -10+PCD = 0

• PCD = 10 KN

• ∑fxi=0

• -PBD – PBC cos θ =0

• PBC = -29.02 KN

• ∑fyi=0

• VB +PBC sin θ = 0

• 17.41 – 29.02 sin θ = 0

•

Sl.No Member Force Nature

1 AC 16.52 C

2 AD 23.21 T

3 CB 29.02 C

4 CD 10 T

5 DB 23.21 T



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• ∑fxi=0

• HA-10+10=0; HA = 0

• ∑fyi=0

• VA+ VB – 20= 0

• VA+ VB= 20

• ζ + ∑MA = 0

• 10(4)-20(3)+10(4)+VE(6)=0

• VE = 10 KN;

• VA =10 KN;

• Symmetrical

o Geometry ;

o Loads

• ∑fxi = 0

• PAC=0

• ∑fyi = 0

• PAB + 10 =0

• PAB = - 10KN

• tan θ = 4/3

• θ=53.13o

• ∑fxi = 0

• -10 + PBD+PBC cos θ =0

• PBD +0.6PBC =10

• ∑fyi = 0

• -PBA− PBC sin θ =0

• -(-10)-0.8 PBC = 0

• PBC= 12.5 KN

• PBD =2.5 KN

Example 2 : Analyse the truss shown in figure and hence compute member forces. Compiled by www.studyeasy.in


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• ∑fxi = 0

• -PDF – PDB = 0

• PDF = -2.5 KN

• ∑fyi = 0

• PDC=0

• Symmetrical

Sl.No Member Force Nature

1 AB, EF 10 KN C

2 AC, CE 0 -

3 BC, FC 12.5 KN T

4 BD, FD 2.5 KN T

5 DC 0 -

Example 3: Analyse the truss shown in figure and hence compute member forces.

• Isosceles triangle;

• CD = DB = a

• ∑fxi = 0 HA = 0

• ∑fyi = 0

• VA+VB = 5

• + ∑MA=0

• -5(2a)+VB(3a) = 0

• VB = 3.33 KN; VA = 1.67 KN

ac

dCaooo

2

;60sin90sin30sin

=

==



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• ∑fxi = 0

• PAC cos 300 + PAD = 0

• ∑fyi = 0

• 1.67+PAC sin 300 = 0

• PAC = -3.34 KN

• PAD = 2.89 KN

• ∑fxi = 0

• -PDC cos 600 -2.89 +PDB = 0

• ∑fyi = 0

• PDC sin 600 – 5 = 0

• PDC = 5.77 KN

• PDB = 5.77 KN

• ∑fxi = 0 – PBC cos 300 –5.77 = 0

• PBC = -6.66 KN

•



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Sl. No Member Force Nature

1 AC -3.34 KN C

2 AD 2.89 KN T

3 BC 6.66 KN C

4 BD 5.77 KN T

5 CD 5.77 KN T

• Method of Sections: Another method of analysis of trusses is method of sections wherein

which the concept of equilibrium of a system of coplanar non concurrent forces is made use

of. The concept of free body diagram is an important part in this method. This method will be

very useful when only few member forces are required. The equation of moment equilibrium

becomes an important tool in this method. The method is illustrated in following figure.

PROCEDURE FOR METHOD OF SECTIONS

• Step 1: Compute support reactions (if need be).

• Step 2: Place the section to cut not more than three members.

• Step 3: Write FBD, unknown forces away from section(T).

• Step 4: Use equilibrium concept to get member forces

This procedure is used for analyzing some examples as shown below.



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• tan θ = ¾; sin θ = 0.6; cos θ = 0.8

• ∑MF = 0

• - 20(3)+PEC(4) = 0

• PEC = 15 KN (T)

• ∑fxi = 0; 20-PFC cos θ = 0

• PFC = 25 KN (T);

• ∑fyi = 0; -PEC - PFC sin θ – PFD = 0;

• PFD = - 30 KN

• = 30 KN (C)

Example 4 : Compute the forces in members EC, FC and FD of the truss shown in figure.



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Example 5 : Compute the forces in members BE, BD and CD of truss shown in Figure.

•

• ζ + ∑MB =0

• -20(3)-PCD(BC) = 0

• PCD = -34.64 KN = 34.64 KN (C)

• ∑fxi=0

• - PCD –PBD cos300 - PBE cos300 =0

• PBD+PBE=40

• ∑fyi=0

• PBE − PBD=80

• Solve to get PBE = 60 KN; PBD = -20=20 KN (C)



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• Support reactions

• ∑fxi=0; HA + 24 =0

• HA = -24 KN

• ∑fyi=0; VA + VB = 40+31+40=111 KN

• ζ + ∑MA=0

• -40(3.6)-31(2)(3.6)-40(3)3.6-24(2.7)+4(3.6)VB = 0

• VB = 60 KN; VA = 51 KN

• ζ + ∑MC=0

• -VA (3.6)- PBD(2.7) = 0

• PBD= - 68 KN;

• = 68 KN (c)

• ζ + ∑MD=0

• - VA(2)(3.6)+2.7HA+40(3.6)+PCE(2.7)=0

• PCE=106.67 KN (T)

• ∑fyi=0; 51 -40 + PCD sin θ =0; PCD = -

18.33 KN=18.33 KN(C).

Example 6: Compute the forces in members BD, CD and CE of the truss shown in figure.Compiled by www.studyeasy.in


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Coplanar Non-concurrent Force System:This is the force system in which lines of action of

individual forces lie in the same plane but act at different points of application.

CHAPTER – IIRESULTANT OF COPLANAR NON CONCURRENT

FORCE SYSTEM

Fig. 1

F2F1

F3

Fig. 2

F1 F2

F5F4

F3

1Compiled by www.studyeasy.in


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TYPES

1. Parallel Force System – Lines of action of individual forces are parallel to each other.

2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.



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MOMENT OF A FORCE ABOUT AN AXIS

The applied force can also tend to rotate the body aboutan axis, in addition to causing translatory motion. This

rotational tendency is known as moment.

Definition: Moment is the tendency of a force to make arigid body rotate about an axis.

This is a vector quantity.



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Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising the moment arm and line of action of the force.

Moment Center (B): This is the point at which the moment axis intersects the plane of the coplanar system. Moment Arm: The perpendicular distance from the moment center to the line of action of the force.

Distance AB = d.

BAd

F



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Consider the example of a pipe wrench.

The force applied which is

perpendicular to the handle of the

wrench tends to rotate the pipe about

its vertical axis. The magnitude of

this tendency depends both on the

magnitude of the force and the

moment arm ‘d’.

EXAMPLE FOR MOMENT5Compiled by www.studyeasy.in


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MAGNITUDE OF MOMENT

B

F

Ad

It is computed as the product of the the perpendicular distance to the line of action of the force from the moment center and the magnitude of the force.

MA = d×F

Unit – Unit of Force × Unit of distancekN-m, N-mm etc.



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SENSE OF THE MOMENT

The sense is obtained by the ‘Right Hand Thumb’ rule.

‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the +ve sense of the moment vector’.

M M

For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or vice-versa.



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VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)

Statement: The moment of a force about a moment center (axis) is equal to the algebraic sum of the moments of the component forces about the same moment center (axis).

Proof (by Scalar Formulation):

Ry

Py

Q R

A Oαβ pq r P

Qy

Let ‘R’ be the given force.‘P’ & ‘Q’ be the component forces of ‘R’. ‘O’ be the moment center. p, r, and q be the moment arms of P, R, and Q respectively from ‘O’.α, β, and γ be the inclinations of ‘P’, ‘R’, and ‘Q’ respectively w.r.t. the X – axis.

γ

Y

X

8 Compiled by www.studyeasy.in


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We have,Ry = Py + QyR Sinβ = P Sinα + Q Sin γ ----(1)From ∆le AOB, p/AO = Sin αFrom ∆le AOC, r/AO = Sin βFrom ∆le AOD, q/AO = Sin γFrom (1),∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)

i.e., R × r = P × p + Q × q

Moment of the resultant R about ‘O’ = algebraic sum of moments of the component forces P & Q about same moment center ‘O’.

Ry

Py

R

A O

Qy

Q

Pp

qr

αβγ

Y

X



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VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION

Consider three forces F1, F2, and F3 concurrent at point ‘A’ as shown in fig. Let r be the position vector of ‘A’ w.r.t ‘O’. The sum of the moments about ‘O’ for these three forces by cross-product is,Mo = ∑(r×F) = (r×F1) + (r×F2) + (r×F3).By the property of cross product,Mo = r × (F1+F2+F3)

= r × Rwhere, R is the resultant of the three forces.

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1. It simplifies the computation of moments by judiciously selecting the moment center. Themoment can be determined by resolving a force into X & Y components, because finding x & y distances in many circumstances may be easier than finding the perpendicular distance (d) from the moment center to the line of action.

2. Location of resultant - location of line of action of the resultant in the case of non-concurrent force systems, is an additional information required, which can be worked out using Varignon’s theorem.

APPLICATIONS OF VARIGNON’S THEOREM



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COUPLE

Two parallel, non collinear forces (separated by a certain distance) that are equal in magnitude and opposite in direction form a ‘couple’.

The algebraic summation of the two forces forming the couple is zero. Hence, a couple does not produce any translation, but produces only rotation. d

F

F



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Moment of a Couple: Consider two equal and opposite forces separated by a distance ‘d’. Let ‘O’ be the moment center at a distance ‘a’ from oneof the forces. The sum of moments of two forces about the point ‘O’ is,+ ∑ Mo = -F × ( a + d) + F × a = -F× d

Thus, the moment of the couple about ‘O’ is independent of the location, as it is independent of the distance ‘a’.

The moment of a couple about any point is a constant and is equal to the product of one of the forces and the perpendicular distance between them.

O

da

F

F



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RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM

P

F

=P

FF

Q

F

=Q M=F × d

d

A given force F applied at a point can be replaced by an equal force applied at another point Q, together with a couple which is equivalent to the original system.

Two equal and opposite forces of same magnitude F and parallel to the force F at P are introduced at Q.

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

F

F



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P

F

=P

F

Q

F

= Q

M=F × dd

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

F

F

Of these three forces, two forces i.e., one at P and the other oppositely directed at Q form a couple.

Moment of this couple, M = F × d.

The third force at Q is acting in the same direction as that at P.

The system in Fig. (c) is equivalent to the system in Fig. (a).



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Thus, the force F acting at a point such as P on a rigid body can be moved to any other given point Q, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about Q.



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The resultant of coplanar, non-concurrent force systems is the one which will produce same rotational and translational effect as that of the given system. It may be a force and a moment or a pure moment.

RESULTANT OF COPLANAR NON-CONCURRENTFORCE SYSTEMS

Let F1, F2 and F3 constitutea non concurrent system as shown in the fig.‘O’ – be any convenient referencepoint in the plane.

F1

F3

F2O

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Each force can be replaced by a force of the same magnitude and acting in the same direction at ‘O’ and a moment about ‘O’ as in Fig B. The non concurrent forces can be combined as in the case of concurrent system to get the resultant force R. Thus, the resultant of the system is equal to a force R at ‘O’ and a moment ‘∑Mo’ as shown in fig.C.

F1

F2

O O

∑MoO∑Mo

F3

R

Fig. A Fig. B Fig. C

18

F1

F2

F3



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The single force R and the moment ‘∑Mo’ shown in the fig.C can be replaced by a single force R acting at a distance ‘d’ from ‘O’, which gives the same effect. Thus, the resultant can be reduced to a single force acting at a certain distance from ‘O’. Mathematically,

O∑Mo

R

Fig. C

O

R

Fig. D

d

∑∑=

RxRy

θtan

22 RyRxR +=

RMo

d ∑=



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X and Y intercepts of Resultant:In some problems, it may be required to determine

distances of the resultant R along x-axis and y-axis i.e., X and Y intercepts. Let ‘d’ be the perpendicular distance of the resultant from ‘O’ as shown in the fig.

Let Rx=∑Fx and Ry=∑Fy be thecomponents of the resultant in X and Ydirections.

By Varignon’s theorem,R×d= ∑Mo

At B, ∑Mo = Rx×0 + Ry×X RY

XO

d

Y-axis

X-axis

Rx

Ry

Rx

Ry

A

B



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Therefore,X= ∑Mo/Ry

Similarly, at A, Ry×0 + Rx×Y = ∑Mo

Therefore,Y= ∑Mo/Rx

RY

XO

d

Y-axis

X-axis

Rx

Ry

Rx

Ry

A

B



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TYPES OF LOADS ON BEAMS

1. Concentrated Loads – This is the load acting for a very small length of the beam.

2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of W kn/m throughout its spread.

Total intensity = W × L (acts at L/2 from one end of the spread)

3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to W kN/m to the other representing a triangular distribution. Total intensity of load = area of triangular spread of the load = 1/2× W × L. (acts at 2×L/3 from ‘Zero’ load end)

W kN/m

L

W kN/m

L



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PROBLEMS FOR PRACTICE

1. Determine the resultant of the parallel coplanarforce system shown in fig. Take radius of the circle=1860mm

(Ans. R=2000N towards left, d=626.9mm)

400 N

1000 N

2000 N

600 N

o

60º

60º30º

10º



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2. Four forces of magnitudes 10N, 20N, 30N and 40Nacting respectively along the four sides of a squareABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.

(Ans:R=28.28N, θ=45º, X=1.77a)

10N

40N

a

a

30N

20N

AB

CD



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3. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontalbar of length 2.85m. Determine the magnitude of resultant and also the distance of the resultant from the left end.

(Ans:R=125N, X=3.06m)



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4. Reduce the given forces into a single force and a couple at A.(Ans:F=320kN, θ=14.48º, M=284.8kNm)

100KN80KN

1m

1.5m

70.7kN200kN

A

30º45º

30º



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5. Determine the resultant w.r.t. point A.(Ans:R=450kN, X=7.5kNm)

A

150N

1.5m3m1.5m

150Nm

500N100N



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CENTROID

CENTRE OF GRAVITY

Centre of gravity : of a body is the point at which the whole weight of the body may be assumed to be concentrated.

It is represented by CG. or simply G or C.

A body is having only one center of gravity for all positions of the body.

1

Contd.



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CENTRE OF GRAVITY

Consider a three dimensional body of any size and shape, having a mass m.

If we suspend the body as shown in figure, from any point such as A, the body will be in equilibrium under the action of the tension in the cord and the resultant W of the gravitational forces acting on all particles of the body.

2

Contd.



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Resultant W is collinear with the Cord

Assume that we mark its position by drilling a hypothetical hole of negligible size along its line of action

Cord

Resultant

CENTRE OF GRAVITY

3

Contd.



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To determine mathematically the location of the centre of gravity of any body,

Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero.

we apply the principle of moments to the parallel system of gravitational forces.

CENTRE OF GRAVITY

6

Contd.



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We repeat the experiment by suspending the body from other points such as B and C, and in each instant we mark the line of action of the resultant force.

For all practical purposes these lines of action will be concurrent at a single point G, which is called the centre of gravity of the body.

CENTRE OF GRAVITY

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w

A

A

w

B

AG

B

A

w

AB

CGB

A

C

C

B

Example:CENTRE OF GRAVITY

5



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if, we apply principle of moments, (Varignon’s Theorem) about y-axis, for example,

The moment of the resultant gravitational force W, about any axis

=the algebraic sum of themoments about the sameaxis of the gravitationalforces dW acting on allinfinitesimal elements ofthe body.

∫ ×dWxWhere W =

∫ dW

=⋅Wx

The moment of the resultant about y-axis =

The sum of moments of its components about y-axis

CENTRE OF GRAVITY

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where = x- coordinate of centre of gravity

x

x

x

W

dWxx ∫ ⋅=

Similarly, y and z coordinates of the centre of gravity are

W

dWyy ∫ ⋅=

W

dWzz ∫ ⋅=and ----(1)

CENTRE OF GRAVITY

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x

W

dWxx ∫ ⋅=

W

dWyy ∫ ⋅=

W

dWzz ∫ ⋅=

With the substitution of W= m g and dW = g dm

m

dmxx ∫ ⋅=

m

dmyy ∫ ⋅=

m

dmzz ∫ ⋅=

----(1)

----(2),,

,,

(if ‘g’ is assumed constant for all particles, then )

the expression for the coordinates of centre of gravity become

CENTRE OF MASS

9

Contd.



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∫∫

⋅

⋅⋅=

dV

dVxx

ρ

ρ

∫∫

⋅

⋅⋅=

dV

dVyy

ρ

ρ

∫∫

⋅

⋅⋅=

dV

dVzz

ρ

ρand ----(3)

If ρ is not constant throughout the body, then we may write the expression as

,

CENTRE OF MASS

The density ρ of a body is mass per unit volume. Thus, the mass of a differential element of volume dV becomes dm = ρ dV .

10

Contd.



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m

dmxx ∫ ⋅=

m

dmyy ∫ ⋅=

m

dmzz ∫ ⋅= ----(2),,

This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel.

CENTRE OF MASS

Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass.

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When the density ρ of a body is uniform throughout, it will be a constant factor in both the numerators and denominators of equation (3) and will therefore cancel.The remaining expression defines a purely geometrical property of the body.

∫∫

⋅

⋅⋅=

dV

dVxx

ρ

ρ

∫∫

⋅

⋅⋅=

dV

dVyy

ρ

ρ

∫∫

⋅

⋅⋅=

dV

dVzz

ρ

ρand, ----(3)

CENTROID

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When speaking of an actual physical body, we use the

term “centre of mass”.

Calculation of centroid falls within three distinct

categories, depending on whether we can model the

shape of the body involved as a line, an area or a

volume.

The term centroid is used when the calculation concerns

a geometrical shape only.

13

Contd.



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LINES: for a slender rod or a wire of length L, cross-sectional area A, and density ρ, the body approximates a line segment, and dm = ρA dL. If ρ and A are constant over the length of the rod, the coordinates of the centre of mass also becomes the coordinates of the centroid, C of the line segment, which may be written as

L

dLxx ∫ ⋅=

L

dLyy ∫ ⋅=

L

dLzz ∫ ⋅=

The centroid “C” of the line segment,

,,

14

Contd.



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AREAS: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t dA.

If ρ and t are constant over entire area, the coordinates of the ‘centre of mass’ also becomes the coordinates of the centroid, C of the surface area and which may be written as

A

dAxx ∫ ⋅=

A

dAyy ∫ ⋅=

A

dAzz ∫ ⋅=

The centroid “C” of the Area segment,

,,15

Contd.



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VOLUMES: for a general body of volume V and density ρ,the element has a mass dm = ρ dV .

If the density is constant the coordinates of the centre of mass also becomes the coordinates of the centroid, C of the volume and which may be written as

V

dVxx ∫ ⋅=

V

dVyy ∫ ⋅=

V

dVzz ∫ ⋅=

The centroid “C” of the Volume segment,

, ,

16



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Centroid of Simple figures: using method of moment ( First moment of area)

Centroid of an area may or may not lie on the area in question.

It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.

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Moment of Total area ‘A’ about y-axis

=Algebraic Sum of moment of elemental ‘dA’ about the same axis

where (A = a1 + a2 + a3 + a4 + ……..+ an)

(A) x = (a1) x1 + (a2) x2 + (a3) x3 + ……….+(an) xn

= First moment of area

The coordinates of the centroid of the surface area about any axis can be calculated by using the equn.

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If an area has an axis of symmetry, then the centroid must lie on that axis.If an area has two axes of symmetry, then the centroid must lie at the point of intersection of these axes.

AXIS of SYMMETRY:

It is an axis w.r.t. which for an elementary area on one side of the axis , there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other)

19 Contd.



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For example:

The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle. xx

dada

da × x = da × x

Moment of areas,da about y-axis cancel each other

da × x + da × x = 020

Contd.



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AXIS of SYMMETYRY

‘C’ must lie at the intersectionof the axes of symmetry

‘C’ must lie on the axis

of symmetry

‘C’ must lie on the axis of symmetry

21



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EXERCISE PROBLEMS

Locate the centroid of the shaded area shown

Problem No.1:50

40

10

10

Ans: x=12.5, y=17.5

2222



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Locate the centroid of the shaded area shownProblem No.2:

300

300

5001000

mm

1000 mm

500

r=600D=600

Ans: x=474mm, y=474mm23

EXERCISE PROBLEMS

23



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Locate the centroid of the shaded area w.r.t. to the axes shown

Problem No.3:

x-axis

y-axis90 20

20

60120 r=40

Ans: x=34.4, y=40.324

EXERCISE PROBLEMS

24



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Problem No.4:

250 mm2010

380

10200 mm x-axis

y-axis

Ans: x= -5mm, y=282mm

10

25

EXERCISE PROBLEMS

25



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Problem No.5

3050

4020 20

40

x

y

r=20

Ans:x =38.94, y=31.46

30

26

EXERCISE PROBLEMS

26



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Problem No.6

2.4 m

1.0

1.5

1.5

1.0 1.0

r=0.6

x

y

Ans: x=0.817, y=0.2427

EXERCISE PROBLEMS

27



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Problem No.7Locate the centroid of the shaded area w.r.t. to the axes shown

Ans: x= -30.43, y= +9.5828

EXERCISE PROBLEMSCompiled by www.studyeasy.in


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Problem No.8Locate the centroid of the shaded area.

Ans: x= 0, y= 67.22(about base)

20

29



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Problem No.9Locate the centroid of the shaded area w.r.t. to the base line.

Ans: x=5.9, y= 8.17

2

30



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Ans: x=21.11, y= 21.1131



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Ans: x= y= 22.2232



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FORCES IN SPACE(Noncoplanar System of Forces)

FS - 1



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Forces in spaceA Force in space: A Force is said to be in space if its line of

action makes an angle α, β and γwith respect to rectangular co-ordinate axes X, Y and Z respectively as shown the Fig. 1.

F Fig. 1. A Force in space

FS - 2 Compiled by www.studyeasy.in


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Forces in space

Noncoplanar system of forces (Forces in Space) and Their Classifications

System of forces which do not lie in a single plane is called noncoplanar system of forces(Forces in space ). A typical noncoplanar system of forces (forces in space) is shown in the Fig. 2. below

Fig. 2 Forces in space (noncoplanar system of forces)

FS - 3Compiled by www.studyeasy.in


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Forces in space

Noncoplanar system of forces(Forces in space) can be broadly classified into three categories. They are

1. Concurrent noncoplanar system of forces

2. Nonconcurrent noncoplanar system of forces

3. Noncoplanar parallel system of forces



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Forces in space

1. Concurrent noncoplanar system of forces: Forces which meet at a point with their lines of action do not lie in a plane are called “Concurrent noncoplanar system of forces”. A typical system of Concurrent noncoplanar system of forces is shown in the Fig.3.

Fig. 3. Concurrent noncoplanar system of forces



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Forces in space2. Nonconcurrent noncoplanar system of forces: Forces

which do not meet at a point and their lines of action do not lie in a plane, such forces are called “Nonconcurrent noncoplanar system of forces”. A typical system of nonconcurrent noncoplanar system of forces is shown in the Fig.4.

Fig. 4. Nonconcurrent noncoplanar system of forces



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Forces in space

3. Noncoplanar parallel system of forces: If lines of action of all the forces in a system are parallel and they do not lie in a plane such a system is called Non-coplanar parallel system of forces. If all the forces are pointing in one direction then they are called Like parallel forces otherwise they are called unlike parallel forces as shown in the Fig.5.

Fig. 5 Noncoplanar parallel system of forces



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Forces in space

Fig. 6. Resolving a force in space into rectangular components

Rectangular components of a force in space



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Rectangular components of a force in spaceIn the Fig.6(a) a force F is acting at the origin O of the system of rectangular coordinate axes X,Y,Z. Consider OBAC plane passing through the force F. This plane makes an angle φ with respect to XOY plane. Force F makes an angle θy with respect to Y-axis.

Forces in space

Fig.6(a)



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In the Fig.6(b), the force F is resolved in the vertical (Y- axis) and horizontal direction (X – axis) as

Fy = F Cosθy and Fh = F Sinθy respectively.

Forces in space

Fig.6(b)



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In the Fig 6(c) the horizontal component Fh is again resolved in the X and Z axes directions. These components are

Fx = Fh cosφ = F sinθy cosφFz = Fh Sinφ = F sinθy sinφ

Forces in space

Fig.6(c)



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Now applying Pythagorean theorem to the triangles OAB and OCDF2 = (OA)2 = OB2 + BA2 = Fy2 +Fh2 ----------------(1)Fh2 = OC2 = OD2 + DC2 = Fx2 +Fz2 ----------------(2)Substituting equation (2) into the equation (1), we getF2 = Fx2 +Fy

2 + Fz2

F = √ Fx2 + Fy2 + Fz2 ----------------(3)

Forces in space FS - 12Compiled by www.studyeasy.in


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Forces in spaceThe relationship existing between the force F and its three components Fx,

Fy, Fz is more easily visualized if a box having Fx, Fy, Fz for edges is drawn as shown below. The force F is then represented by the original OA of this

box.

Fig. 7 Relationship between Force F and its components Fx, Fy and Fz



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From the above Figure (Fig. 7)

Fx = F Cos θx, Fy = F Cosθy, Fz = F Cosθz ------------(4)

Where θx, θy, θz are the angles formed by the force F with X, Y, Z axes respectively.Fx,Fy,Fz are the rectangular components of the force F in the directions of X, Y, Z

axes respectively.

Cos θx = Fx/F; Cosθy = Fy/F; Cosθz = Fz/F

Substituting equation (4) into the equation (3), we get

F = √ Fx2 + Fy2 + Fz2

F = √ F2Cos2θx + F2Cos2θy + F2Cos2θzF2 = F2 ( Cos2θx + Cos2θy + Cos2θz )

1 = Cos2θx + Cos2θy + Cos2θz -------------(5)



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From the adjacent Fig. 8.

dx= d Cos θx, dy = d Cosθy,dz = d Cosθz

----(6)d = √ dx2 + dy2 + dz2

---(7)Dividing member by

member the relations (4) and (6), we obtain

Fx /dx = Fy/dy = Fz/dz = F/d ----------------------------(8)

Forces in spaceForce Defined by its magnitude and two points on its line of action

Fig 8



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Forces in space

Resultant of concurrent forces in Space:-Resolve all the forces into their rectangular components in X, Y and Z axes directions. Adding algebraically all the horizontal components in the x direction gives

Rx = ∑ Fx, Similarly adding algebraically all the components in y and z directions yield the following relations

Ry = ∑ Fy,Rz = ∑ Fz

Thus magnitude of resultant

R = √ Rx2 + Ry2 + Rz2

Angles θx, θy, θz resultant forms with the axes of coordinates are obtained by

RRCos

RR

CosRRCos z

zy

yx

x === θθθ ;;



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Problems:(1) A tower guy wire is anchored by means of a bolt at A is shown in the following

Figure. The tension in the wire is 6000N. Determine (a) The components Fx, Fy, Fz of the forces acting on the bolt. (b) The angles θx, θy, θz defining the direction of the force.



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Solution: (a) Here dx = 50m, dy = 200m, dz = -100m

Total distance A to B

d = √ dx2 + dy2 + dz2

= √ (50)2 + (200)2 + (-100)2

= 229.13 m

Using the equation (8) Fx /dx = Fy/dy = Fz/dz = F/d∴Fx = dx . (F/d) = (50 x 6000)/ 229.13 = 1309.3 N

Fy = dy . (F/d) = (200 x 6000)/ 229.13 = 5237.20 N

Fz = dz . (F/d) = (-100 x 6000)/ 229.13 = -2618.6 N



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(b) Directions of the force:

Cos θx = dx /d , θx = Cos–1 (50/229.13) = 77.4°

θy = Cos–1 (dy /d) = Cos–1 (200/229.13) = 29.2°

θz = Cos–1 (dz /d) = Cos–1 (-100/229.13) = 115.88°



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Problem(2) Determine (a) the x , y and z components of the 250 N force acting as shown below

(b) the angles θx, θy, θz that the force forms with the coordinate axes.



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Fh = 250 Cos60 = 125 N Fy = 250 Sin60 = 216.5 NFx = 125 Cos25 = 113.29 N Fz = 125Sin25 = -52.83 N

θx = Cos–1 ( Fx /F) = Cos–1 (113.29/250) = 63°θy = Cos–1 ( Fy /F) = Cos–1 (216.5/250) = 30°θz = Cos–1 ( Fz /F) = Cos–1 (-52.83/250) = 102.11o

∴Components of 250 N in the x, y, z axes directions are

Fx = 113.29N Fy = 216.5 N Fz = -52.83N



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Problem 3. Find the resultant of the system of forces as shown below



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Solution

Components of Force F1 = 3000 N:

Fy1 = 3000 x Cos40o = 2298.13 NFh1 = 3000 x Sin40o = 1928.36 NFx1 = 1928.36 x Cos30o =1670 NFz1 = 1928.36 x Sin30o = 964.18 N



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Components of Force F2 = 2000 N:

Fy2 = 2000 x Cos20 = 1879.39 NFh2 = 2000 x Sin20 = 684.04 NFx2 = 684.04 x Sin35° = 392.35 NFz2 = 684.04 x Cos35° = 560.33N



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Rx = ∑ Fx = Fx1 + Fx2 = 1670 – 392.35 = 1277.65Ry = ∑ Fy = Fy1 + Fy2 = 2298.13 + 1879.39 = 4177.52 NRz = ∑ Fz = Fz1 + Fz2 = 964.18 + 560.33 = 1524.51 N

Resultant R = √Rx2 + Ry2 +Rz2

= √1277.652 +4177.522 +1524.512

= 4626.9 NIts inclinations with respect to x, y and z axes are calculated as

θx = Cos–1 (1277.65 /4626.9) = 73° 58' 13.1" θy = Cos–1 (4177.52/4626.9) = 25° 27' θz = Cos–1 (1526.51/4626.9) = 70o45’36”



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Problem 4. In the Fig shown below, the forces in the cables AB and AC are 100 kN and 150 kN respectively. At the joint ‘A’ loading is as shown in the Fig. Find the resultant of system of forces in space and its inclination with rectangular coordinates x,y and z axes.



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Solution: Force in the cable AB = 100 kN

Force in the cable AC=150 kN

For the cable ABdx = -20 mdy = 15mdz = 5mdAB = √ dx2 +dy2 + dz2

= √ (-20)2 +(15)2 + (5)2 = √ 650 = 25.5 m

For the cable ACdx = -20 mdy = 25mdz = -10mdAc = √(-20)2 +(25)2 + (-10)2 = √1125 = 33.54 m

FS - 27Forces in spaceCompiled by www.studyeasy.in


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For the cable AB (1)

Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = F/d

Fx1/-20 = Fy1/15 = Fz1/5 = 100/25.5

∴Fx1 = -78.41 kN Fy1 = 58.82 kNFz1 = 19.61 kN

For the Cable AC (2)

Fx2/-20 = Fy2/25 = Fz2/-10 = 150/33.54

Fx2 = - 89.45 kNFy2 = 111.81 kNFz2 = - 44.72 kN



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Component of force 60 kN

Fx3 = 60 x Cos(70°) = 20.52 kNFy3 = 60 x Cos(30o) = 51.96 kNFz3 = 60 x Cos(111°23)= - 21.88 kN

Component of the force 50KN

Fx4 = 50 kNFy4 = 0Fz4 = 0



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Algebraic summation of Rectangular Components in X, Y and Z axes directions yield: Rx = Σ Fx = Fx1 + Fx2 + Fx3 + Fx4

= -78.43 - 89.45 + 20.52 + 50 = -97.36 kN

Ry = Σ Fy = Fy1 + Fy2 + Fy3 + Fy4= 58.82 + 111.81 + 51.96 + 0 = 222.59 kN

Rz = Σ Fz = Fz1 + Fz2 + Fz3 + Fz4= 19.61 – 44.72 – 21.88 + 0 = -46.99 kN

∴ Resultant R = √ Rx2 + Ry2 + Rz2

= √ (-97.36)2 + (222.59)2 + (-46.99)2

= 247.45 kN



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Inclinations of the resultant with X,Y and Z axes:θx = Cos-1(Rx / R) = Cos-1 (-97.36/247.45) = 113° 10’θy = Cos-1(Ry / R) = Cos-1 (222.59/247.45) = 25° 54’θz = Cos-1(Rz / R) = Cos-1 (-46.99/247.45) = 100° 56’ 47”

Check:

Cos2θx + Cos2θy + Cos2θz = 1Cos2(113°10’) + Cos2(25°54’) + Cos2(100°56°) = 1

1 = 1 Hence OK



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Problem 5. a) Forces F1, F2, and F3 pass through the origin and points whose coordinates are given. Determine the resultant of the system of forces.

F1 = 20 kN, (3,-2,1)F2 = 35 kN, (-2,4,0)F3= 25 kN, (1,2,-3)



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Forces in space

Solution:



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Force F1 = 20 kNd = √ 32 + (-2)2 + 12 = √ 14 = 3.74Cos θx1 = dx / d = 3/3.74 = 0.802Cos θy1 = -2/3.74 = -0.535; Cos θz1 = 1/3.74 = 0.267

Force F2 = 35 kNd = √ (-2)2 + 42 +0 = √20 = 4.47Cos θx2 = -2/4.47 = -0.45Cos θy2 = 4/4.47 = 0.9; Cos θz2 = 0

Force F3 = 25 kNd = √ (1)2 + 22 +(-3)2 = √14 = 3.74Cos θx3 = 1/3.74 = 0.267Cos θy3 = 2/3.74 = 0.535; Cos θz3 = -3/3.74 = -0.802



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Summation of the rectangular components in X, Y and Z axes directions yield:

Rx = ΣFx = 20 x 0.802 + 35 x (-0.45) +25 x 0.267 = 6.965 kNRy = Σ Fy = 20 x (-0.535) + 35 x 0.9 + 25 x 0.535 = 34.175 kNRz = Σ Fz = 20 x 0.267 + 35 x 0 + 25 x (-0.802) = -14.71 kN

Resultant R = √ Rx2 + Ry2 + Rz2

= √ 6.9652 + 34.1752 + (-14.71)2

= 37.85 kN

Inclination of the resultant R with respect to X, Y and Z axes are calculated as

θx = Cos–1(Rx / R) = Cos-1(6.965/37.85 ) = 79.4°θy = Cos–1(Ry /R) = Cos-1(34.175/37.85) = 25.46°θz = Cos–1(Rz /R) = Cos-1(-14.71/37.85) = 112.87°



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Forces in space

Equilibrium of Concurrent non-coplanar system of forces:When a rigid body subjected to concurrent noncoplanar system of forces F1, F2…. ..FN as shown in the Fig. given below, is in equilibrium, then algebraic summation of all the components of the forces in three mutually perpendicular directions must be equal to zero.

Fig. A rigid body subjected to concurrent noncoplanar system of forces

i.e. ∑ Fx = 0

∑ Fy = 0

∑ Fz = 0 (1)

Above equations represent the static conditions of equilibrium for concurrent noncoplanar system of forces



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Problem (1) Find the forces in the rods AB , AC and AO subjected to loading as shown below

Forces in space FS - 37



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Solution:

For the cable AB:

dx1 = 0 - 4 = - 4 dy1= 8 - 0 = 8dz1 = 15 – 0 =15

d1 = √ (-4)2 +(8)2 +(15)2 = 17.46 m

Fx1/-4 = Fy1/8 = Fz1/15 = FAB/17.46

Fx1 = -0.23FAB, Fy1 = 0.46FAB, Fz1 = 0.86FAB



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For the Cable AC:dx2 = 0 - 4 = -4mdy2 = 8 - 0 = 8mdz2 = -20 – 0 = -20m

d2 = √ (-4)2 +(8)2 +(-20)2 = √480 = 21.91mFx2/-4 = Fy2/8 = Fz2/-20 = FAC/21.91Fx2 = -0.18FAC, Fy2 = 0.365FAc, Fz2 = -0.91FAc

For the Force 120 N:Fx3 = 120 N, Fy3 = 0, Fz3 = 0

For the force Fx4 = 300 N:Fx4 = 0, Fy4 = -300N, Fz4 = 0



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Conditions of Equilibrium∑ Fx = 0, ∑ Fy = 0, ∑ Fz = 0

Considering ∑ Fx = 0Fx1 + Fx2 + Fx3 + Fx4 + Fx5= 0-FAD-0.23F – 0.18FAC + 120 + 0 = 0FAD + 0.23FAB + 0.183FAC = 120 ----------------------(1)

∑ Fy = 00.46FAB + 0.365FAC + 0 +0 – 300 = 00.46FAB + 0.365FAC = 300 ------------------------------(2)

∑ Fz = 00.86 FAB – 0.91 FAC + 0 = 0 -----------------------(3)



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Solving Equations (1), (2) and (3), we get,FAB = 372.675 N; FAC = 352.25 N ; FAO = - 30.18 N

Force in the rod AB , FAB = 372.675 N (Tensile)Force in the rod AC, FAC = 352.25 N (Tensile)Force in the rod AO, FAO = 30.18 N (Compressive)



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2) Three cables are connected at D and support, the 400 kN load as shown in the Fig given below. Determine the tensions in each cable



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Let FDA, FDB and FDC are the forces in the cables AD, BD, and CD respectively.For Cable DA:dx1 = (0 – 3) = -3mdy1 = (6 - 4) =2 mdz1 = (6 – 0 = 6 mdAD = √ dx2 + dy2 + dz2

dAD = √ (-3)2 + (2)2 + (6)2 = √9 + 4 + 36 = 7 mFx1/dx1 = Fy1/dy1 = Fz1/dz1 = FDA/dDA

Fx1/(-3) = Fy1/2 = Fz1/6 = FDA/7∴ Fx1 = -0.43FDA, Fy1 =0.286FDA, Fz1 = 0.857FDA



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For Cable DB:dx2= (0 – 3) = -3mdy2 = (6 - 4) =2 mdz2 = (-6 – 0 = -6 mdBD = d2 = √(-3)2 + (2)2 + (-6)2 = √ 49 = 7 mFx2/dx2 = Fy2/dy2 = Fz2/dz2 = FDB/dDB

Fx2/(-3) = Fy2/2 = Fz2/-6 = FDB/7

∴Fx2 = -0.43FDB, Fy2 =0.286FDB, Fz2 = -0.857FDB



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For Cable DC:dx3 (0 – 3) = -3mdy3= (0- 4) = -4mdz3= (0 0 = 0m

dDC= d3= √ (-3)2 + (-4) + (0)2

= √ 9 + 16= 5m

Fx3/dx3 = Fy3/dy3 = Fz3/dz3 = FDC/d DCFx3/(-3) = Fy3/-4 = Fz3/0 = FDC/5

∴Fx3 = -0.6FDC, Fy3 =-0.8FDC, Fz3 = 0



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For the force 400KNdx4 = 12mdy4= -4mdz4= 3md4= d400 = √ (12)2 + (4) + (3)2

= √144 +16 + 9= √69 = 13m

Fx4/dx4 = Fy4/dy4 = Fz4/dz4 = F400/d4

Fx4/(12) = Fy4/-4 = Fz4/3 = 400/13

∴Fx4 = 369.23 kN, Fy4 = -123.08 kN, Fz4 = +92.31 kN



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For Equilibrium, the algebraic summation of resolved component in a particular direction is equal to zero.i.e. ∑ Fx = 0 -------------(1)

∑ Fy = 0 ------------- (2)∑ Fz = 0 --------------(3)

(1) Σ Fx = Fx1 + Fx2 + Fx3 + Fx4 = 0+ 0.43FDA + 0.43FDB + 0.6FDC = 369.23 ---------(1)+ 0.286FDA + 0.286FDB – 0.8FDC = 123.08 ------(2)+ 0.857FDA - 0.857FDB + 0 = 92.31 ------------(3)



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Solving equations (1), (2) and (3), we get

FDB = 304.1kN

FDA = 411.8 kN

FDC = 102 kN



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Forces in spacePractice Questions1. A tower guy wire is anchored by means of a bolt at A as shown below.

The tension in the wire is 2500 N. Determine (a) the components Fx, Fyand Fz of the force acting on the bolt, (b) the angles θx, θy, θz defining the direction of the force

(Ans: Fx= -1060 N , Fy = 2120 N, Fz= + 795 N Θx = 115.1 o ; Θy= 32.0 o ; Θz= 71.5 o )



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Forces in space

Practice Questions2. Determine (a) x, y and z the components of the force 500 N in the below Figure. (b) the angles θx, θy, θz that the force forms with the coordinate axes

(Ans: Fx= + 278 N , Fy = + 383 N, Fz= + 160.7 N

Θx = 56.2 o ; Θy= 40.0 o ; Θz= 71.3 o )



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Forces in space

Practice Questions3. In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in the cable AB is 10 kN, determine the components of the force exerted by the cable AB on the truck

(Ans: Fx= -6.30 kN , Fy = 6.06 kN, Fz= + 4.85 kN)



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Forces in space

Practice Questions4. A 200 kg cylinder is hung by means of two cables AB and AC, which are attached to the top of a vertical wall. A horizontal force P perpendicular to the wall holds the cylinder in the position shown. Determine the magnitude of the force P and the tension in each cable

(Ans: P = 235 N , TAB = 1401 N, TAC= + 1236 N )



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Forces in space

Practice Questions5. Three cables are connected at A, where the forces P and Q are applied as shown. Determine the tension in each cable when P = 0 and Q = 7.28 kN

(Ans: TAB= 2.88 kN , TAC = 5.76 kN, TAD= 3.6 kN)



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Forces in space6. A container of weight w = 400 N is supported by cables AB and AC which

are tied to ring A. Knowing that Q = 0, determine (a) the magnitude of the force P which must be applied to the ring to maintain the container in the position shown in figure below, (b) the corresponding the values of the tension in cables AB and AC

(Ans: P = 138 N , TAB = 270N, TAC = 196N )

Practice Questions



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Forces in space7. A container supported by three cables as shown below. Determine the

weight of the container, knowing that the tension in the cable AB is 4 kN

Ans: 9.32 kN

Practice Questions



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Forces in space8. Determine the resultant of the two forces shown below.

Practice Questions

(Ans: R = 498 N , Θx = 68.9 o ; Θy= 26.3 o ; Θz= 75.1 o )



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Forces in space9. A container of weight W = 1165 N is supported by three cables as shown

below. Determine the tension in each cable.

Ans: TAB = 500 NTAC = 459 NTAD = 516 N

Practice Questions



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EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM

When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction. Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system

∑ Fx = 0; ∑ Fy = 0 Eq(1)

∑M = 0These requirements are both necessary and sufficient conditions for equilibrium.



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Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium.

There are different types of supports. Some of them are a) Roller Support b) Hinged or pinned support c) Fixed or built in supportSome supports are shown in the figure along with the

reactions that can be mobilised.

Types of supports



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Types of Supports Action on body

(a) Flexible cable ,belt ,chain, rope

BODYBODY

T

Force exerted by cable is always a tension away from the body in the direction of cable

(b) Smooth surfacesContact forces are normal to the surfaces

FF

3



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(c) Roller support Contact force is normal to the

surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component.(Whose direction is known.)



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( d )pinned Support / hinged support

This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.

RvR

Rhθ



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Rv

MRH

(e) Fixed or Built-in Support

M



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(contd .)This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.



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TYPES OF BEAMSA member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span(distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam.

(a) Simply supported beam

span

MA

VA

B

span

AHA

(b) Cantilever beam



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If one end or both ends of the beam project beyond the support it is known as overhanging beam.

A cantilever with a simple support anywhere along its length is a propped cantilever.

(c) Overhanging beam (right overhang)

MA

VA

B

span

A

HA

(d) Propped Cantileverbeam



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A beam which is fixed at both ends is called a fixed beam.

A beam with more than one span is called continuous beam.

VC

(f) Two Span continuous beam

VA VB

HA HBMA

VA

spanVB

(e) Fixed beam

MB

HA



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Statically determinate beam and statically indeterminate beam:Using the equations of equilibrium given in EQ(1) ,if all the reaction components can be found out, then the beam is a statically determinate beam ,and if all the reaction components can not be found out using equations of equilibrium only, then the beam is a statically indeterminate beam.In the above fig (a),(b)and ( c ) are statically determinate beams ,where as (d),(e) and ( f) are statically Indeterminate beams .



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If the number of reaction components is more than the number of non-trivial equilibrium equations available then such a beam is a statically indeterminate beam.

If the number of reaction components is equal to the number of non-trivial equilibrium equations available then such a beam is a statically determinate beam

If the number of reaction components is less than the number of non-trivial equilibrium equations available then such a beam is an unstable beam.



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Determination of Beam reactionsSince three equilibrium equations are available, for a planar structure a maximum of three unknown independent reaction components can be determined using these equations.

Step I: Draw the free body diagram of the structure showing the given loadings and the reactions at the supports.

Step 2: Apply the equations ∑ Fx = 0, ∑ Fy = 0, ∑M = 0. Assuming some directions and senses for unknown forces and moments.Step 3: solve for unknown reactions. If any of them is positive, it is along the sense initially assumed while drawing the FBD. If it is negative, it is opposite to the initially assumed sense



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4kN/m

12kN/m

(1)Find the reactions at A,B,C and D for the beam loaded as shown in the figure(Ans.RA=RB =34kN;RC=28.84kN;

MC=-140kNm ; θC=-33.69 ˚ )

Problems for practice

12kN/m

4kN/m

20 kN

30kN

1m 2m 1m 1m 2m 1m 1m 2m

A BC

34

40kNm



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(2)A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B.(Ans. T=529.12N;RB=807.15N, θB=64.6˚)

2.5m 2.5m200N

2.5m

A

B

60˚

string



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(3)Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal..(Ans.x=2m.)

2.0m 1.4m1.0m 3.0m0.6

15kN18kN/m

10kN/m

x



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(4)A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical(Ans:θ =18.44˚)

0.5L 2W

θ

B

L m

WC

A



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FRICTION

Friction is defined as the contact resistance exerted byone body upon another body when one body moves ortends to move past another body. This force which opposesthe movement or tendency of movement is known asfrictional resistance or friction. Friction is due to theresistance offered by minute projections at the contactsurfaces. Hence friction is the retarding force, alwaysopposite to the direction of motion. Friction has bothadvantages & disadvantages.

Disadvantages ---- Power loss, wear and tear etc.

Advantages ---- Brakes, traction for vehicles etc.

Friction 1



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Frictional resistance is dependent on the amount of wedgingaction between the hills and vales of contact surfaces. Thewedging action is dependent on the normal reaction N.

N

F (Friction)

P

W

Hills & Vales Magnified Surface

Friction 2



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Frictional resistance has the remarkable property ofadjusting itself in magnitude of force producing or tendingto produce the motion so that the motion is prevented.

When P = 0, F = 0 block under equilibrium

When P increases, F also increases proportionately tomaintain equilibrium. However there is a limit beyondwhich the magnitude of this friction cannot increase.

Friction 3



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When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional resistance is called the static friction.

Further if P is increased, the value of F decreases rapidlyand then remains fairly a constant thereafter. However athigh speeds it tends to decrease. This frictional resistanceexperienced by the body while in motion is known asDynamic friction OR Kinetic Friction.

Friction 4



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Rolling friction friction experiencedby a body when it rolls over a surface.

Dynamic Friction

Sliding friction friction experienced when a body slides over another surface.

Friction 5

5



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Where Fmax = Limiting Friction

N= Normal Reaction between the contact surfaces

µ =Coefficient of friction

Note : Static friction varies from zero to a maximum value. Dynamicfriction is fairly a constant.

∴ µ = Fmax

N

Fmax

(limiting friction)

P

W

φ

N R

F α N

Fmax = µN

Friction 6



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Angle of FrictionThe angle between N & Rdepends on the value of F. Thisangle θ, between the resultantR and the normal reaction N istermed as angle of friction. AsF increases, θ also increasesand will reach to a maximumvalue of φ when F is Fmax(limiting friction)

i.e. tanφ = (Fmax )/N = µ

Angle φ is known as Angle of limiting Friction.

Fmax

(limiting friction)

P

W

φ

N R

Friction 7



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Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending.

Angle of repose

When granular material is heaped, there exists a limit for theinclination of the surface. Beyond that angle, the grains startrolling down. This limiting angle upto which the grainsrepose (sleep) is called the angle of repose of the granularmaterial.

Friction 8Compiled by www.studyeasy.in


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Significance of Angle of repose:

The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose.

Angle of repose is numerically equal to Angle of limiting friction

Friction 9



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Laws of dry friction

1. The magnitude of limiting friction bears a constant ratioto the normal reaction between the two surfaces.(Experimentally proved)

2. The force of friction is independent of the area ofcontact

between the two surfaces.

3. For low velocities the total amount of friction that canbe developed is practically independent of velocity.It is less than the frictional force correspondingto impending motion.

Friction 10



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The transmission of power by means of belts or rope drivesis possible only because of friction between the wheels andthe belt. Tension in the belt is more on the side it is pulledand less on the other side. Accordingly they are called astight side and slack side.

PullW

T2 (Tight side)T1 (Slack side)

β

FRICTION IN BELT/ ROPE DRIVESFriction 11



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P w T2T1

β

A load W is being pulled by a force P over a fixed drum. Let the force on tight side be T2 and on slack side be T1. (T2>T1 because of frictional force between drum and the rope). Let β be the angle of contact in radians between rope and the drum. Consider an elemental length of rope as shown. Let T be the force on slack side and T+dTon tight side. There will be normal reaction N on the rope in the radial direction and frictional force F= µN in the tangential direction.

FN

dβT+dT

TF

dβ/2

RELATIONSHIP BETWEEN TIGHTSIDE AND SLACKSIDE FORCES IN A ROPEFriction 12



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Σ Forces in radial direction = 0N-T Sin dβ/2 – (T+dT)Sin dβ/2 = 0 { Sin dβ/2 = dβ/2 as dβ is small }∴ N-T dβ/2- (T+dT) dβ/2 = 0 i.e. N = ( T+dT/2) dβ ------(1)We know that F = µN ∴ F = µ ( T+dT/2) dβ-----(2)

Σ Forces in tangential direction = 0(T+dT) Cos dβ/2 = F + T Cos dβ/2 { Cos dβ/2 = 1 as dβ is small }∴ T + dT = F + T i.e. dT = F------(3)From (2) & (3) dT = µ ( T+dT/2) dβNeglecting small quantity of higher order, dT = µT dβ

dT/T = µ dβ



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Integrating both sides,T2 β

∫ dT/T = ∫ µ dβT

10

T2 β(log T) = µ(β)

T1

0

Log (T2/T1) = µβ

∴ T2/ T1 = eµβ where T2

= Force on tightside

T1

= Force on slackside

β = Angle of contact in radians



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1 ) For the block shown in fig., determine the smallestforce P required

a) to start the block up the planeb) to prevent the block moving down the plane.

Take μ = 0.20[Ans.: (a) Pmin = 59.2N (b) Pmin = 23.7N θ = 11.3o]

EXERCISE PROBLEMSFriction15

25°

θ

P

100N



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2) A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If μ between the block and the plane is 0.35, determine the unknown force Pfor impending motion

(a) to the right(b) to the left

[Ans.: (a) P = 132.8N (b) P = 1252N]

Friction 16

2000NP

800N30°



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3) Determine value of angle θ to cause the motion of500N block to impend down the plane, if μ for allcontact surfaces is 0.30.

[Ans.: θ = 28.4°]

Friction 17

500N

200N

θ = ?



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4) In Figure, μ between rope and the fixed drum andbetween all contact surfaces is 0.20. Determine theminimum weight W to prevent the downward motionof the 1000N body.

[Ans. : T1 = 0.76W, T2 = 1.424W, W = 253N]

Friction 18

1000NW

34



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5) A horizontal bar 10m long and of negligible weightrests on rough inclines as shown in fig. If angle offriction is 15o, how close to B may the 200N force beapplied before the motion impends.

[Ans.: x = 3.5m]

Friction 19

100N 200N

2 mX = ?

A B

30° 60°



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6) Determine the vertical force P required to drive thewedge B downwards in the arrangements shown in fig.Angle of friction for all contact surfaces is 12o.

[Ans.: P = 328.42N]

Friction 20

A

B1600N

P

20°



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7) Determine the force P which is necessary to start thewedge to raise the block A weighing 1000N. Selfweight of the wedge may be ignored. Take angle offriction, φ = 15o for all contact surfaces.

[Ans.: P = 1192N]

Friction 21

20°

A

Pwedge



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8) A ladder of weight 200N, 6m long is supported as shownin fig. If μ between the floor and the ladder is 0.5 &between the wall and the ladder is 0.25 and it supports avertical load of 1000N, determinea) the least value of α at which the ladder may be placedwithout slippingb) the reactions at A & B[Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]

Friction 22

α

1000N

5m

A

B

22



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9) An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. μ between the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor?

[Ans.: FA = 52N]

Friction 23

Smooth wall

12m

5mA

B

23



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10)A ladder of length 5m weighing 500N is placed at 45o

against a vertical wall. μ between the ladder and thewall is 0.20 & between ladder and ground is 0.50. If aman weighing 600N ascends the ladder, how high willhe be when the ladder just slips. If a boy now stands onthe bottom rung of the ladder, what must be his leastweight so that the man can go to the top of the ladder.

[Ans.: (a) x = 2.92m (b) Wboy = 458N]

Friction 24



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MOMENT OF INERTIAMoment of Inertia:

The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis.

Ixx = ∫dA. y2

Iyy = ∫dA. x2

x

y

x

ydA

T-1Compiled by www.studyeasy.in


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It is also called second moment of area because first moment of elemental area is dA.y and dA.x; and if it is again multiplied by the distance,we get second moment of elemental area as (dA.y)y and (dA.x)x.



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Polar moment of Inertia (Perpendicular Axes theorem)

The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy

O

y

x

r

z x

Y



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Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.

PERPENDICULAR AXIS THEOREMT-4Compiled by www.studyeasy.in


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Parallel Axis Theorem

y_d

x x

x0 x0

dA

y´*G



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Ixx = ∫dA. y2 _

= ∫dA (d +y')2

_ _= ∫dA (d2+ y'2 + 2dy')

_= ∫dA. d2 + ∫dAy΄2 + ∫ 2d.dAy'

_d2 ∫dA = A.(d)2

∫dA. y'2 = Ix0x0_

2d ∫ dAy’ = 0

( since Ist moment of area about centroidal axis = 0)_

∴Ix x = Ix0 x0+Ad2



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Hence, moment of inertia of any area about an axis xx isequal to the M.I. about parallel centroidal axis plus theproduct of the total area and square of the distancebetween the two axes.

Radius of Gyration

It is the perpendicular distance at which the whole area may be assumed to be concentrated, yielding the same second moment of the area above the axis under consideration.



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Iyy = A.ryy2

Ixx = A.rxx2

∴ryy = √ Iyy/AAnd rxx = √ Ixx /A

y

y

A

x x

rxx and ryy are called the radii of gyration

ryy



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MOMENT OF INERTIA BY DIRECT INTEGRATION

d

dy

x0

xx

d/2x0

y

b

M.I. about its horizontal centroidal axis :

G.

RECTANGLE :

IXoXo = -d/2 ∫ +d/2

dAy2

=-d/2∫+d/2

(b.dy)y2

= bd3/12

About its base

IXX=IXoXo +A(d)2

Where d = d/2, the distance between axes xx and xoxo

=bd3/12+(bd)(d/2)2

=bd3/12+bd3/4=bd3/3



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h

x0

xh/3

x0

b

xy

(h-y)

dy

(2) TRIANGLE :

(a) M.I. about its base : Ixx = ∫ dA.y2 = ∫ (x.dy)y2

From similar triangles b/h = x/(h-y)

∴ x = b . (h-y)/hh

Ixx = ∫ (b . (h-y)y2.dy)/h0

= b[ h (y3/3) – y4/4 ]/h= bh3/12



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(b) Moment of inertia about its centroidal axis:_

Ixx = Ix0x0+ Ad2

_Ix0x0

= Ixx – Ad2

= bh3/12 – bh/2 . (h/3)2 = bh3/36



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Ix0x0= ∫ dA . y2

R 2π= ∫ ∫ (x.dθ.dr) r2Sin2θ

0 0

R 2π=∫ ∫ r3.dr Sin2θ dθ

0 0

R 2π=∫ r3 dr ∫ {(1- Cos2θ)/2} dθ

0 R

0 2π

=[r4/4] [θ/2 – Sin2θ/4] 0 0

= R4/4[π - 0] = πR4/4

IXoXo = π R4/4 = πD4/64

xx

x0x0Rθ

dθ

y=rSinθ

3. CIRCULAR AREA:

r



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Ixx = ∫ dA . y2

R π= ∫ ∫ (r.dθ.dr) r2Sin2θ

0R

0π

=∫ r3.dr ∫ Sin2θ dθ0 0

R π=∫ ∫ r3 dr (1- Cos2θ)/2) dθ

0 0 π

=[R4/4] [θ/2 – Sin2θ/4]0

= R4/4[π/2 - 0] = πR4/8

4R/3π

y0

y0

xxx0

x0

4. SEMI CIRCULAR AREA:

R



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About horizontal centroidal axis:

Ixx = Ix0x0+ A(d)2

Ix0x0= Ixx – A(d)2

= π R4/8 πR2/2 . (4R/3π)2

Ix0x0= 0.11R4



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QUARTER CIRCLE:

Ixx = Iyy

R π/2Ixx = ∫ ∫ (r.dθ.dr). r2Sin2θ

0 0

R π/2=∫ r3.dr ∫ Sin2θ dθ

0 0

R π/2 =∫ r3 dr ∫ (1- Cos2θ)/2) dθ

0 0 π/2

=[R4/4] [θ/2 – (Sin2 θ)/4] 0

= R4 (π/16 – 0) = πR4/16

x x

x0 x0

y

y y0

y0

4R/3π

4R/3π



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Moment of inertia about Centroidal axis,_

Ix0x0= Ixx - Ad2

= πR4/16 - πR2. (0. 424R)2

= 0.055R4

The following table indicates the final values of M.I. about X and Y axes for different geometrical figures.



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Sl.No Figure I x0-x0I y0-y0

I xx I yy

1bd3/12 - bd3/3 -

2bh3/36 - bh3/12 -

3πR4/4 πR4/4 - -

40.11R4 πR4/8 πR4/8 -

50.055R4 0.055R4 πR4/16 πR4/16

b

dx0

x

x0

xd/2

b

h

xxx0

x0h/3

x0x0

y0

y0

OR

y0

y0xxx0

x0

4R/3π

x0

y y0

4R/3π

4R/3π

Y

Y Xo



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EXERCISE PROBLEMS ON M.I.

Q.1. Determine the moment of inertia about the centroidal axes.

100mm

20

30mm

30mm

30mm

[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4

Iyy = 1.855 x 106mm4]

EP-1Compiled by www.studyeasy.in


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Q.2. Determine second moment of area about the centroidal horizontal and vertical axes.

[Ans: X = 99.7mm from A, Y = 265 mm

Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]

200mm

200

300mm

300mm

900mm



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Q.3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.

[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4

rxx = 62.66mm, ryy = 45.63mm]

60

100mm

200mm

140mm

20

20



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Q.4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.

[Ans: X = 83.1mm

Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]

60

60 60

X X2020



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Q.5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.

[Ans: d/2 = 223.9mm d=447.8mm]

200mm

600mm

d400mm

200mm

200mm

200mm



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Q.6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each.

Properties of I section are

Ixx = 7983.9 x 104mm4

Iyy = 2011.7 x 104mm4

[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]4000mm

2500mm

160mm

160mmCross sectional area=6971mm2



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Q.7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.


Properties of ISA

Cross sectional area = 4400mm2

Ixx = Iyy ;Cxx = Cyy =18.5mm

18.5mm18.5mm

20mm

200mm



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Q.8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section.

[Ans: d = 183.1mm]

Properties of ISMC300

C/S Area = 4564mm2

Ixx = 6362.6 x 104mm4

Iyy = 310.8 x 104mm4

Cyy = 23.6mmX X

Y

Y

Lacing

d

380mm

23.6mm



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Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.


160mm

40mm

40mm

40mm



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