System of Measurement and Unitsbieap.gov.in/Pdf/BCMTPaperIII.pdf · Matter: Any substance which occupies space is called matter. ... i.e. it has no size, but has a definite mass

Learning ObjectivesThe branch of science which deals with motion, forces and their effect on

bodies is called mechanics (2) The application of principles of mechanics tocommon engineering problems is known as engineering mechanics.

IntroductionFundamental Units

The physical quantities which do not depend up on other quantities areknown as “Fundamental Quantities” and units for such quantities are known as“Fundamental Units” or “Base Units”. The internationally accepted fundamentalquantities are

(i) Length (ii) Mass and (iii) Time

Derived Units

If the units are expressed in other units which are derived from fundamentalunits are known as “derived units’.

Ex: Units of Area, Velocity, Acceleration, Presure etc.

System of Units

There are four systems of units which are commonly used and universallyrecognised. These are known as

(i) C.G..S. (ii) F.P.S. (iii) M.K.S. and (iv) S.I. Units

1UNIT

System of Measurement andUnits

Building Construction and Maintenance Technician254

C.G.S. Units

In this system, the fundamental units of length, mass and time are centimeter,gram and second.

F.P.S. Units

In this system, the fundamental units of length, mass and time are foot,pound and second respectivley.

M.K.S. Units

In this system, the fundamental units of length, mass and time are metre,kilogram and second respectively.

S.I. Units (International System of Units)

The eleventh general conference of weights and measures (GCWM) hasrecommended a unified, systematically constituted system of fundamental andderived units for international use. In this system the fundamental units are meter(m) kilogram (kg) and second (s) respectively. But there is slight varaition intheir derived units. The following are the derived units.

Force - N (Newton)

Stress (or) Pressure - N / mm2 (or) N/m2

Workdone (in joules) - J =Nm

Power in watts - W

Temperature Degree Kelvin - K

Current (ampere) - A

Units of Physical Quantities

Engineering mechanics and strength of materials are essentially “quantitativesciences”. They involve expressions of quantities. For example.

(i) Height of a building is 12m

(ii) Area of cross section of land is 220 mm2

(iii) Stress in bar is 150 Newtons per mm2.

(iv) Radius of gyration of the section is 12 mm.

In all the above expressions of quantities; we essentially state two items,viz., a number and a known standard of measurement.

Paper - III Engineering Mechanics 255

In the statement “Height of building is 12m, the standard of lengthmeasurement is 1m and the height of the building is 12 times the length of thatstandard. The standard of measurement adopted is known as the “unit” of physicalquantity. Each physical quantity can be expressed in a number of units.

Eg.(1) Length can be measured interms of meters, millimeters, feet, yardsand so on.

(2) The unit of area is the area of a square of side 1m and is stated as 1m2.

(3) Unit of volume is the volume of a cube of side 1m and is stated as 1m3.

(4) Unit of velocity is unit displacement for unit time and stated as 1m/s

(5) Unit of mass density is unit mass per unit volume and is stated as 1 kg/ m3.

(6) Unit of acceleration is unit change of velocity per unit time and is stated as 1m per second square (1m/sec2).

Multiples and Submultiples

Recommended prefixes for the formation of multiples and submultiples ofunits are shown in table 1.

Table 1

S.No. Multiplying factor Prefire Symbol

1. 1012 Tera T

2. 109 Giga G

3. 106 Mega M

4. 103 Kilo K

5. 10-3 Milli m

6. 10-6 Micro

7. 10-9 Nano n

8. 10-12 Pico p

Useful Units

1 KNm = 106 N mm

1 M.Pa = N/mm2


1 G..Pa = 109 Pa = 103 x 106 Pa

= 103 M Pa

= KN / mm2

1 G.Pa = 1 KN/mm2

1 MN/m2 = 1 N/mm2

1 KN/m = 1 N/mm (for u.d.l)

1 tonne = 103 kgs = 103 x 9.81 Newtons

Short Answer Type Questions1. Define a) Base units & b) Derived untis.

2. State the units for the following in S.I.System

a) length b) Mass c) Velocity d) Workdone

3. Write the units in S.I. system for the following quantities

a) Area b) Radius of Gyration c) Sterss d) Volume

4. Mention S.I. units for the following quantities

a) Density b) Acceleration c) Current d) Power

5. List the four systems of measurement commonly used.

6. Mention the base units and derived units used in S.I. Units.

2UNIT

Forces and MomentsLearning Objectives

1. Mechanics is the science that deals with force, and the effect of force onbodies. In short, it deals with the motion of bodies, state of not beingconsidered as a special case of motion.

2. In order to understand the principles of mechanics, it is essential thatforce should be studied in detail and its effects understood.

3. Newton’s first law of motion helps us to define a force as an externalagency which tends to change the state of rest or of uniform motion of abody.

Basic Concepts

Space: Space is the region which extends in all directions and containseverything in it.

Ex: Sun and its Planets, Stars, etc.

Time: Time is a measure of duration between successive events. The unittime is a second which is a fraction (1/86,400) of an average solar day.

Motion: A body is said to be in motion, when it changes its position withrespect to other bodies. The relative change in position is called motion.

Matter: Any substance which occupies space is called matter.

Body: Any matter that is bounded by a closed surface is called a body.


Inertia: A resistance offered by matter to any change in its state of motionis called inertia.

Mass: It is a quantitative measure. It is the resistance offered by anysubstance to the change in its position.

Particle: A particle occupies no space, i.e. it has no size, but has a definitemass concentrated at a point.

Rigid body: A Rigid body is that which does not undergo deformation onthe application of forces.

Definition of Force

According to Newton’s first law of motion, force is defined as follows.

“Force is an action that changes or tends to change the state of rest oruniform motion of a body in a straight line”.

In simple words, the action of one body on any other body can be called aforce. These actions may be of various forms, pull or push on a body, gravitationalforce known as weight of a body, force presented by an elastic spring, forceexerted by a locomotive on the train, resistance offered by the track etc.

“Force is specified with its magnitude, direction and the point of application”.

Unit of Force

The unit of force in the international system is the Newton.

The Newton is that force which when applied to a body having a mass of1kg gives it an acceleration of 1m/s2

Characteristics of a Force

The folowing are the four essential characteristics of a force:

1. Point of application 2. Direction, 3. Magnitude and 4. Sense.

1. Point of Application : It indicates where the force is applied.

2. Direction : The direction of a force is stated with refernce to geographicaldirections (i.e. North, East, South and West) or with any fixed reference line.

3. Magnitude : It is the quantity of force applied and it is expressed inNewtons.

4. Sense: It is represented by placing an arrow head to force line denotingthe nature of force i.e. whether a push or pull.


Fig 2.1 Characteristics of Force

Fig 2.1. Shown a force acting on a body. The point of application is A, andacts in a direction which makes at 400 to the horizontal. It’s magnitude is 400N.

Reference Frame of Axes

We know that space is a region extending in all directions. Position of abody in space can be defined only by referring measurement to a fixed frame ofaxes. The measurement may be linear, angular (or) both.

Let us consider methods to define the position of a particle with respect toreference frame. To determine the position of a particle in space, threemeasurements are necessary to be made. For example the position “P” of aparticle with respect to the reference frame XYZ is defined by three co-ordinatesX, Y, Z as shown is Fig.2.2

Fig.2.2 defining position of “P” in space.

Incase the motion of a particle is taking place in a plane, its position can bedefined by only two measurements. This can be done in the following two ways:

1. Rectangular co-ordinate system.

2. polar co-ordinate system

1. Rectangular Co-ordinate system

In this system, the position of any particle “P” is determined by therectangular co-ordinates (x,y), where “x” is the linear distance of the particle“P” from the origin “o” measured parallel to the X – axis and “Y” is the linear


distance of the particle “p” from the origin “o” measured parallel to the “Y” axis.This is shown is fig 2.3.

2. Polar co-ordinate system

In polar co-ordinate system the position of a particle “P” is defined bywhere “r” is the radial distance from of the particle “p” from the origin “o” and““ is the angle made by the radial line OP with the base line OX as shown infig 2.4.

Fig: 2.4 position of a point “P” in a plane through polar co-ordinate system.

Quantities

A physical quantity is one which can be measured. All physical quantity canbe classified in to two main categories.

1. Scalar quantity 2. Vector quantity.

Scalar Quantity

Scalar quantities are those which can be completely defined by theirmagnitude (or) numerical value only and no direction.

Ex: Mass, Length, Time, Speed etc.,

Fig: 2.3 Defining the position of apoint “P” in a plane throughRectangular Co-ordinates.


Vector Quantity

Vector quantities are those which require not only magnitude but alsodirection (line of action) and sense for completely defining them.

Ex: Displacement, veloncity Force, Momenturn etc.,

1. Representation of a vector:- A vector is represented by directed lineas shown in fig 2.5. It may be noted that the length OA represents the magnitudeof the vector OA

Fig 2.5 Vector OA

The direction of the vector OA is from’o’ (i.e. starting point) to A (i.e,endpoint). It is know as vector P.

2. Unit Vector:- A vector, whose magnitude is unity, is know as unit vector.

3. Equal vectors:- The vectors, which are parallel to each other and havesame direction (i.e. same sense) and equal magnitude are known as equal vectors.

4. Like Vectors:- The vectors, which are parallel to each other and havesame sense but unequal magnitude, are known as like vectors.

5. Addition of vectors:- Consider two vectors PQ and RS, which arerequired to be added as shown in Fig2.6(a)

Fig 2.6 (a) Vector PQ and RS (b) Sum of vectors

Take a point A, and draw line AB parallel and equal in magnitude to thevector PQ to some convenient scale. Through B, draw BC parallel and equal tovector RS to the same scale.

Join AC which will give the required sum of vectors PQ and RS as shownis Fig 2.6 (b).This is the method of addition of vectors.

R

S

PQ

A

C

B


6. Subtraction of Vectors:- Consider two vectors PQ and RS whosedifference is required to be found out as shown in fig2.7(a)

Fig 2.7 (a) Vector PQ and RS (b) Difference of vectors

Take a point A, and draw line AB parallel and equal in magnitude to thevector PQ to some convenient scale. Through B draw BC parallel and equal tothe vector RS, but in opposite direction to that of the vector RS, to the samescale. Join AC which will give the required difference of the vector PQ and RSas shown is fig2.7(b)

System of Forces:- a combination of several forces acting on a body iscalled a “system of forces” (or)”Force system”.

The system of forces can be classified according to the arrangement of thelines of action of the forces of the system. The forces may be classified as

1. Coplanar & Non-Coplanar

2. Concurrent & Non-Concurrent

3. Parallel & Non-Parallel and

4. Collinear Forces.

1. Coplanar&Non Coplanar Forces:- The forces whose lines lie on thesame plane are known as “Coplanar Forces”.If the lines of action of the forcesdo not lie in the same plane, then, the forces are called “Non-Coplanar Forces”.

2. Concurrent, Non-Concurrent Forces:- If the forces acting on a bodymeet at a point, they are called “Concurrent Forces”.

Fig 2.8

P QS

R

A B

C


Forces a , b and c shown in 2.8 are concurrent forces as they are meetingat point “0” whereas forces d , e and f are called “Non-Concurrent Forces”because all the three forces are not meeting at a point.

3. Parallel, Non-Parallel Forces:- The forces whose lines of action areparallel to one another are known as ‘Parallel Forces’. The parallel forces mayfurther be classified into two categories depending upon their directions.

a. Like Parallel Forces

b. Unlike Parallel Forces

a. Like Parallel Forces

The Forces whose lines of action are parallel

to each other and act in the same direction as

shown in fig 2.9 are known as “ Like Parallel Forces”

b. Unlike Parallel Forces

The Forces whose lines of action are parallel

to each other and act is opposite directions as

shown in fig 2.10 are known as “Unlike Parallel Forces.”

4. Collinear Forces:- The forces whose lines of action lie on the sameline are known as collinear forces (fig.2.11)

Fig 2.11 Collinear Forces

Non – parallel forces

The forces whose lines of action are not parallel

to one another are known as “Non – Parallel

forces”. As shown in fig 2.12 forces s , t , u are

Non-parallel forces.

Fig 2.9like parallelforces

Fig 2.10 Unlikeparallel forces

Fig 2.11 Collinear Forces


Resultant of forces at a point

The resultant of a force system can be defined as the simplest – single forcewhich can replace the original system without changing its external effect on arigid body.

Law of Parallelogram of forces

If the forces meet at a point, their resultant may be found by the law ofparallelogram of forces. It states that if two forces acting at a point are such thatthey can be represented in magnitude and direction by the two adjacent sides ofparallelogram, the diagonal of the parallelogram passing through their point ofintersection gives the resultant in magnitude and direction.

Fig 2.13

Consider two forces “P” and “Q” acting a point “O” in the body as shownin fig 2.13 (a). Their combined effect can be found out by constructing aparallelogram using vector “P” and vector “Q” as two adjacent sides of theparallelogram as shown in figure 2.13 (b). the diagonal passing through “O”represents their resultant in magnitude and direction.

Mathematically, magnitude of resultant2R P Q 2PQcos

Where = Angle between two forces P and Q

= Angle between resultant R and force P

= Angle between resultant R and force Q

Triangle law of Forces

It states “If two forces acting at a point berepresented is magnitude and direction by the twosides of a triangle in order, their resultant – may berepresented in magnitude and direction by the thirdside of a triangle taken in opposite order”.

Fig 2.14 Trainglelaw of forces


Lami’s Theorem

It states “If three forces acting at a point are inequilibrium, then, each force is proportional to the sine ofthe angle between the other two forces.”

Mathematically, P Q R

Sin Sin Sin

Polygon law of Forces

It is an extension of triangle law of forces for more than two forces. It states“ If a number of forces acting simultaneously on a particle, be represented inmagnitude and direction, by the sides of a polygon taken in order, then theresultant may be represented, in magnitude and direction by the closing side ofthe polygon taken in opposite order.

Fig 2.16

Resolution of a Force

The process of splitting up the given force into a number of componentswithout changing its effect on the body is called “resolution of forces”.A force isgenerally resolved along two mutually perpenduclar directions.

Fig 2.15

Fig 2.17


Analytical determination of the Resultant of a concurrent coplanar forcesystem

The process of resolution and composition can be very conveniently usedin determining the resultant of a concurrent, coplanar force system.

Fig 2.18

Fig 2.18 (a) shows a concurrent, coplanar force system consisting of fourforces F1, F2, F3 and F4 acting an a particle.

The point of concurrence of the force system is taken as the origin, andrectangular cordinate-axis X and Y are selected as shown. The Forces F1, F2 F3and F4 can now be reserved into rectangular components, which have directionsalong the X and Y axes.

For example the components of force F1 or F1 cos along the X-axis andF1 sin along the Y-axis as shown in fig 2.18(b). The components are takenpositive when they act along the positive directions of X and Y axes.

The calculations are conveniently done in a tabular form as shown intable below.

Force Angle with X-axis X-component ( +) Y-component ( +)

F1 1 1 1F cos 1 1F sinF2 2 2 2F cos 2 2F sin F3 3 3 3F cos 3 3F sin F4 4 4 4F cos 4 4F sin

1 1 2 2 3 3 4 4Fx F cos F cos F cos F cos


1 1 2 2 3 3 4 4Fy F sin F sin F sin F sin

22x yResul tan t(R) F F

The direction of ‘R’ can also be easily determined. If y is makes an angleR with the x-axis, then

yR

x

Ftan

F

(or)y1

Rx

Ftan

F

This method is most general method. It determines the resultant of forcesystems analytically. Using the graphical method of force polygon, the resultantof force can also be determined.

Moment of Force

A force acting on a body not only produces linear motion but also producesrotation when its effect is considered at any other point other than the pointcontact. The effect of rotation is called “Moment of a force”.

Fig 2.19

The moment of a force is equal to the product of the force and theperpendicular distance of the point, about which moment is required, and theline of action of force as shown in fig. 2.19

Mathematically M = pxl

Where p = force acting on the body

l = perpendicular distance of the point and the line of action of the force.


Units of Moment

Moment is the product of force and distance, So if the force is in Newton’sand distance, is in mm, then the unit of moment is Nm. in S.I units it is expressedin KNm.

1KNm = 106 Nm

Types of Moements

The moments are classified in to following two types

1. Clock wise Moments

2. Anti clock wise Moments

1. Clockwise Moment: It is the moment of a force, whose effect is toturn or rotate the body, in the same direction in which the hands of a clockmove.

2. Anticlockwise Moment: it is the moment of a force, whose effect is toturn or rotate the body, in the opposite direction in which the hands of a clockmove.

Generally the sign convention for clockwise moment is taken as positiveand for anticlockwise moment – negative

Couple

Tow equal, opposite and parallel forces having different lines of action forma “Couple”.

Moment of Couple

The moment of a couple is a product of force “P” and the perpendiculardistance (a) between the line of action of two equal and opposite forces asshown in fig. 2.20.

Fig 2.20

Mathematically, moment of a couple,

M = Pxa

Where P = force and

a = perpendicular distance also know as arm of couple


Equilibrium :- a body acted upon by a system of forces is said to be inequilibrium, if it either continues in a state of rest or continues to move is astraight line with uniform velocity.

Conditions of equilibrium:- A body is said to be in equilibrium if it satisfiesthe following conditions.

i) The algebraic sum of all vertical forces is zero i.e V 0 ii) the algebraic sum of all Horizontal forces is zero i.e H 0 iii) the algebraic sum of moments about a point is zero i.e. M 0

V 0

H 0

M 0 Solved ExamplesExample 2.1

A body is acted upon by an upward force of 200N and a horizontal forceof 400N. Find the magnitude and direction of resultant.

Solution

Upward Force P = 200N

Horizontal Force Q = 400N

Resultant

2 2

2 2

R P Q

(200) (400)447.21N

Direction of Resultant

R

1 0R

0 1

P 200tan 0.5Q 400tan (0.5) 26.56

26 33

Resultant is at an angle of 260331 with x-axis i.e. Horizontal axis.


Example 2.2

Two Co-planar forces act at a point with an angle of 600 between them. Ittheir resultant is 120N and one of the forces is 60KN. Calculate the other force.

Solution

Resultant R = 120KN

Force P = 60KN

Angle between two forces θ = 600

Let other force = Q

Resultant =

Example 2.3

Find the Magnitude and Direction of a resultant of two forces 50N and70N acting of a point with an included angle of 500 between them. The force70N being horizontal.

Solution

Horizontal Force P = 70N

Other Force Q = 50N


2 2

2 2 0

2

2

2

2

2

R P Q 2PQCos

120 60 Q 2X60XQCos60

120 3600 Q 60Q

3600 Q 60q 14400Q 60Q 14400 3600 10800Q 60Q 10800 0

-60 60 4(1)( 10800)Q =

2X1-60 3600 43200 -60 216.33

2 278.165KN (taking ve value)

Other Force

∴

∴

∴ 78.165KN


2 2

2 2 0

0

0

1

Magnitudeof Resul tan t R P Q 2PQCos

(70) (50) 2X70X50XCos50109.08N

Let Angle between resul tan t R and force PQSin 50 x Sin50tan

P QCos 70 50 x Cos5050(0.766)tan 0.375

70 50(0.6428)tan (0.375) 20.55

∴

∴

0 0 120 33

Resultant acts at angle of 220331 with force ‘p’ i.e. 70N.

Example 2.4

Determine the magnitude and direction of the resultant of the two forces260N and 180N acting at a point, at mutually perpendicular directions.

Solution

Horizontal Force P = 260N

Vertical Force Q = 180N


2 2

2 2 0

2 2

0

0

1

Magnitudeof Resul tan t R P Q 2PQCos

(260) (180) 2X260X180XCos90

(260) (180) 316.23NLet Angle between resul tan t R and force P

QSin 180 x Sin90 180tan 0.6923P QCos 260 180 x Cos90 260

tan (0.6923) 34

∴

∴0 0 1.69 34 41

Resultant 316.23N acts at an angle of 340411 with Horizontal Force ‘P’i.e. 260N.


Example 2.5

Two forces of magnitude 20 KN and 30 KN act at a point angle of 600

find the magnitude and direction of resultant force.

Solution

P = 20 KN

Q = 30 KN

= 600

Magnitude of the resultant

2 2

2 2 0

R P Q 2PQcos

20 30 2 20 30 cos60

= 43.59 KN

Resultant of the given forces = 43.59 KN

If = angle between resultant ‘R’ and force ‘P’

0

0

30 0.866Qsin 30sin 60tanP Qcos 20 30cos60 20 30 0.5

Direction of resultant – with respect to force ‘p’ = 360 350

Example 2.6

A particle ‘o’ is acted and by the following forces

i) 200 N inclined at 300 to north of east

ii) 250 N to words north

iii) 300 N to words North 450 west

iv) 350 N inclined at 400 to south of west as per the given data, the forcesare acting as shown in fig 2.21

25.98 0.742335

1 0 1tan 0.7423 36 35

Fig 2.21


Solution:

As per the given data, the forces are acting as shown in fig 2.21

Resolving the forces horizontally

Resolving the forces vertically

Magnitude of the resultant


Example 2.7

Find the magnitude and direction of resultant force of following force actingat a point.

a. 80 KN due North

b.20 KN due North – East

c. 40 KN due east

d.60 KN in a direction inclined 300 east to south

e. 70 KN in a direction inclined 600 south of west

0 0 0H 200cos30 300cos 45 350cos 40 250cos90173.21 212.13 268.11 0

307.03N

0 0 0 0V 200cos30 250sin90 300sin 45 350sin 40100 250 212.13 224.97337.16N

2 2

2 2

R H V

307.03 337.16

456 N

R

1R

0

0 1R

V 337.16tanH 307.03

1.0981tan 1.0981

47.6847 40 with x axisasshown in fig 2.22

Fig 2.22

Fig 2.23


Solution

As per the given data the forces are acting as shown in fig 2.23


Resolving the force in vertically

Magnitude of resultant

Direction of resultant RV 3.522tan 0.0495H 71.19

0 1R 20 50 with xaxis

Example 2.8

Determine the resultant magnitude direction ofthe given system of coplanar concurrent.

Solution

The force diagram is drawn as shown in fig2.24 and angles are needed with respect it horizontalaxis i.e., x – axis as shown in fig 2.25

0 0 0 0H 40 20cos 45 60cos30 70cos60 80cos9040 20 0.7071 60 0.866 70 0.5 040 14.142 57.96 3571.102KN

0 0 0 0V 80sin 90 20sin 45 40sin 0 60sin 30 70sin 6080 1 20 0.7071 40 0 60 0.5 70 0.86680 14.142 30 60.623.522KN

2 2 2R H V 71.02 3.522

71.19KN

1 0R

0 1

tan 0.0495 2.834

2 50

Fig 2.24



Resolving the force vertically

Magnitude of Resultant

Direction of Resulatant

0 0 0 0H 30cos 45 40cos30 60cos60 20cos3030 0.7071 40 0.866 60 0.5 20 0.86621.21 34.64 30 17.328.53KN

0 0 0V 30sin 45 40sin 30 60sin 60 20sin 3030 0.7071 40 0.5 60 0.866 20 0.521.21 20 57.96 10

40.75KN

2 2

2 2

R H V

8.53 40.75

41.63KN

R

1R

0

0 1

0 1R

V 40.75tan 4.727H 8.53

tan 4.777

78.1778 10

78 10 with x axis.

Fig 2.25


Example 2.9

Four men pull a tree is the east, south east, south west and North Westdirections with forces 200 N, 300N, 150 N and 350 N respectively. Find theresultant for a ovel its direction.

Fig 2.26

Solution

Force diagram is drawn as shown in fig 2.26

Resolving forces horizontally

Resolving the forces vertically

Magnitude of Resultant

0 0 0 0H 200cos0 300cos 45 150cos 45 350cos 45200 1 300 0.7071 150 0.7071 350 0.7071200 212.13 106.06 247.4858.59 N

0 0 0 0V 200sin 45 150sin 45 300sin 45 200sin 0350 0.7071 150 0.7071 300 0.7071 200 0247.48 106.06 212.13

70.71N

22

2 2

R ( H) V

58.59 70.71

91.83N



= -1.207

Example 2.10

Two Forces of 100N and 80N act at a point making an angle of 600 betweenthem. Determine their resultant.

Solution

Let the two forces to be P and Q ; then P = 100N and Q = 80 N

Angle between the above two forces =600

The resultant force R of the tow forces

Resultant of the two given

24400= 156.2N

Forces = 156.2 N

RV 58.59tanH 70.71

1R

0 0

0 1R

tan 1.207

50.36 50 2150 .21 with x axis

2 2

2 2 0

P Q 2PQcos

100 80 2 100 80cos60

10000 6400 8000

2 2

2 2 0

P Q 2PQ Cos

(100) (80) 2x100x80 Cos60

10000 6400 8000


Short Answer Type Questions1. Define scalar quality and vector quality with one example for each.

2. State the following with neat sketcure

a. Parallelogram law of force

b. Triangle law of force.

3. Define force and mention two characteristics of force.

4. Define the terms

a. Resultant of a force

b. Equilibrium of force

c. Polygon law of forces

5. Define

a. Resultant – of a system of forces

b. Equilibrium of system of forces

6. State the conditions for equilibrium of rigid body

7. Define

a. Co – planar forces

b. Non- coplanar forces.

c. Collinear forces.

Essay Answer Type Questions1. Find the magnitude and direction of a resultant of two forces 60 N and

80 N acting at a point with an included angle 500 between them. The force 80 Nbeing horizontal Ans: R = 127.164 r=28.810

2. Find the resultant of the force as shown in figure

Ans: R = 41.63 r =780.101


3. Determine the magnitude and direction of the resultant of system offorces as shown in figure.

Ans: R = 444.70 r =45.240W

4. Find the magnitude and direction of the resultant of the system of co-planar forces as shown in fig.

Ans:R=31.03KN r=470.401With X-axis

5. Calculate the magnitude and direction of the resultant of a system of co–planar forces given below

i) 200 N inclined 300 to north of the east

ii) 250 N towards north

iii) 300 N north 450 west

iv) 350 N included 400 to south of west.

Ans: R= 456 r= 47.670 with x-axis in II quadrant


6. A telephone pole has five wires radiating from in IInd Floor Quadrantthe top of the pole, producing the following concurrent pulls.

i) 260 N due West

ii) 180 N due East

iii) 170 N at 300 East of North

iv) 240 N due North west

v) 200 N due South west

Find the resultant pull in magnitude and direction

Ans: R=352.87N r= N60.170W

7. Find the magnitude of two forces, such that if they act at right angles,their resultant is but if they act at 600, their resultant is

Ans: 3N and 1N

3UNIT

Centroid & Moment ofInertia

Learning ObjectivesAfter studying this unit, the student will be able to

• Know what is centre of gravity and centroid

• Calculate centroid of geometric sections

Centre of Gravity

Centre of Gravity (or) mass centre of a point in the body where entire mass weight – is assumed to be concentrated. In other words, it is a point in thebody, through which the resultant of the weights of different parts of the body isassumed to be acting. It is generally written as C.G.

Centroid:

The plane figure like triangle, rectangle circle etc have only areas and massis negligible. The centre of area of such plane figures is called ‘Centroid’ (or)“Centre of Area”. It is generally denoted by “G”

Centroidal Axis

The axis which passes through centre of gravity (or)centroid is known as “Centroidal Axis” XX1, YY1, ZZ1

are called Centroidal Axis

Fig 3.1


Axis of Symmetry

Axis of Symmetry is the line dividing the figure into two equal parts likemirror images the centroid always lies on the axis of symmetry.

Fig 3.2

A figure may contain one (or) more axis of symmetry. If there are moreaxis of symmetry the cntroid lies at the intersection of axis of symmetry

Fig 3.3

Position of centroids for Standard Geometric Sections.

S. No Name Shape of figure Position of centroid

1 Rectangle

2 Triangle

At int er sec tions ofDiagonals

Hy31A BH2


Lx2By2

A L X B


3 Parallelogram

4 Circle

5 Semicircle

6 Trapezium (slopingon both sides)

7 Trapezium (One sideis vertical andother side is sloping)


Lx2By2

2a ba b

2a b1 a b

hy ( )3hy ( )3hA (a b)2

2


Dx Radius2Dy Radius2

DA2

2 2a ab bx3 (a b)

hA (a b)2


Centroid of Composite sections

A composite section is a combination of simple regular shapes as rectangle,triangle circle, semi circle etc. For determining the centroid of composite sections,the entire area is divided into two (or) more regular simple shapes, Then theprinciple of moments is applied to determine the centroid.

Centoid of plane figure having hollow Portion

The Centroid of plane figure having hallow portion is determined similar tothe composite sections by applying principle of moments, However the negativesign is taken into consideration of hollow positions which are enclosed in aregular shape.

Sections Symmetrical about both X and Y axes

Sections Symmetrical about – horizontal axis (XX)

Sections Symmetrical about the vertical axes (YY)

Fig 3.4

Fig 3.5

Fig 3.6


Sections un symmetric about – the both axes (X-X,Y-Y)

Methods of determination of centroid

The following three methods are available to locate the cntroid of an area.

1. Analytical method

2. Graphical method

3. Experimental method

Analytical method for location of the centroid

Principle: The sum of the moments of a system of a coplanar forces aboutany point in the plane is equal to the moment of their resultant about the samepoint.

Fig 3.8

Consider a lamina in area “A” divided into number of elementary areas A1,A2, A3, ….etc as shown in fig. 3.8.Let the centroids of these elementary areasbe at a distance of x1, x2, x3….. etc from vertical axis and y1 , y2, y3 from thehorizontal axis.

Let the centroids of the total area “A” is at a distance of x and y fromvertical and horizontal axis respectively. As per the principle of moments, thesum of moments of all the elementary areas about horizontal axis OX is equal tothe moment of the total area about the same horizontal axis i.e OX.

Fig 3.7


1 1A yy

A

where A=A1+A2 +A3 +….

Similarly taking moments of areas about vertical axis i.e. OY

1 1A xx

A

The terms 1 1 1 1A y & A x are know as First movement of areaabout y-axis and x-axis respectively

First moment of area: The First moment of area about a line is the productof area and the perpendicular distance of its centroid from the given line.

Important – Note

1. If the axis passer through the centroid, the moments of areas on oneside of the axis will be equal to the moments of areas on the other side of theaxis.

Example 3.2

Locate the centroids if the trapegezium as shown in figure 3.09

Fig 3.09

Solution

Dived the trapezium into rectangle of size a x h and triangle of base (b-a)and height – “h”

Area of rectangle (1) A1 = a.h

Area of rectangle (2) A2 = 12 (b-a)h

Total area of trapezium hA a b2

Let the centroid of the trapezium be at a distance y above base and xfrom last vertical side. Centroidal distance of rectangle from A i. e ,

1ax2


Centroid distance of triangle from A i. e ; 2b ax a

3

3a b a 2a b3 3

Similarly

Example 3.3

Locate the position of centroid of lamina in fig 3.10

Solution

Y.Y Axis as symmetry centroid lies in this axis divide the section in to asquare and a triangle A1 = 100x100 = 10,000mm

1 1 2 2

1 2

22 2 2

2 2

a 1 2a bah b a hA x A x 2 2 3x hA A a b

2b a 2a bh a

2 3 3a 2ab b 2a abh 3 a ba b2

a ab b3 a b

1 1 2 21 2

1 2

2 2

A y A y h hy y , yA A 2 3

h h h h hah b a ah b a2 2 3 2 3y h ha b a b

2 23ah bh ah 2ah bh h 2a b

3 a b 3 a h 3 a b

a ab b h 2a bcentroid x , y3 a b 3 a b


y1 = 100 50mm2

22

1A 100 60 3000mm2

260Y 100

0

3

120mmabove base

Total area = A1 +A2 = 10,000+3000 = 13000 mm2

Let be centroid distance from base

The centroid of the lamina is 66.15 mm above the base.

Example 3.4

A trapezoidal lamina has uniform batter on both sides. Its top width is 200mm. bottom width is 300mm and height is 600 mm. determine position of centroidfrom base.

Solution

Top width a = 200 mm

Bottom width b = 30 mm

Height h = 600 m

= Position of centroid from the basey

Fig 3.10

y

66.15 mm

500000 + 3,60,000 13,000

8,60,000 13,000

10000(50) + 3000(120) 13,000

A1y1+A2y2 A1+A2

= =

=

=

=

Fig 3.11


Example 3.5

Find the centroid of the following T – section

Solution

Fig 3.12 shows T- Sections Y-Y axis as symmetrical axis. In this axis only.

Taking base X-X axis as reference line.

Dividing the “T” section in to two rectangles areas. (flange +web)

Area of rectangles flange A1 = 100x10 = 1000 mm2

from the bottom ofthe base

Area of rectangle web (2) A2 =140x10 = 1400 mm2

from the bottom of theweb X.X axis

distance of centroid from bottom of web XX i. e.,

y h3

2a+b a+b( )= = 2x200 + 300

200 + 300( )600 3

400 + 300 500

200 ( )=

280 mm=

Fig 3.12

1 1 2 2

1 2

A y A y 1000 145 1400 70yA A 1000 1400

98000 145000 101.25mm2400

y2 = 140 2 = 70 mm

y1 = 10 2

= 145 mm140 +


Example 3.6(imp)

Find the position of the centroid of an I section given.

Top angle : 60 x 20 mm

Web : 20 x 100 mm

Bottom angle : 100x20 mm

Solution

Fig 3.13 shows given I sectrim Y-Y axis as axis of symmetry so centroidlien in this axis only.

we can find

X-X axis is base of the bottom taken as reference line. Dividing I section into three rectangles.

Area of rectangle (1) A1 = 60 x 20 =1200 mm2

from the base of thebottom flange

Area of rectangle (2) A2 = 100 x 20 = 2000 mm2

from the base of the bottom flange


Y3 = = 10 mm from the base of the botttom of the flange

y1 = 20 + 100 + 20 2

130 mm=

y2 = 20 + 100 2

70 mm=

20 2

y


Example 3.7(imp)

A masonry dam of the trapezoidal section with one face is vertical. Topwidth of dam is 3m, bottom width of dam is 6m and height is 6m. Find theposition of centroid.

Solution

(i) Applying for multa

Top width a = 3 mt

Bottom width b = 6 mt

Height of the dam = 6 mt

∴one face is vertical.

Let centroid of the dam be at a distance about baseand turn use vertical face centroid.

IInd method

Trapezium OBCD is divided in to tow simple areas

1. Rectangle OLCD 2. Triangle CLB

x

Fig 3.14

x =a2 + ab + b2

3(a+b)

32 + 3 x 6 + 62

3(3+6)=

9+18+36 27=

= 2.33 mt

y =h3

2a + b a+b( )

63

2x3+6 3+6( )=

12 92 ( )=

= 2.67 mt

Y


There is no axis as axis of symmetry line.

∴ We can find both x and yFinding Vertical face OD as resume line.

For Rectangle OLCD area. A1 = 3x 6 = 18 m2

Xl = = 1.5. from vertical face OD.

For triangle CLB

Area

m

form vertical face.

Finding y Base of the dam OB taken as reference line.

Area of rectangle (1) A1 = 18m2

from the base of the dam

Area of triangle (2) A2 = 9m2

from the baseof the dam

Example 3.8 (imp)

Determine the centroid of the channel section 200 x 100 x 10 mm as shownis fig 3.15

Solution

Fig 3.15 shown the given channel section. X-X axis as axis of symmetryline is this only centroid lies it.

x

16y 3mt2

21A2

3 6 3 2

2

9m

3x 3 4m3

1 1 2 2

1 2

A y A y 18 3 9 2y 2.67mA A 18 9

1 1 2 2

1 2

A x A x 18 1.5 9 4x 2.33A A 18 9


Finding only taking vertical outerface AB as reference line dividing thesection in to three rectangles

Rectangle (1) AreaA1 = 100 10 = 1000 mm2

1100x 50mm

2 from vertical face AB

Area of Rectangle(2) Area A2 = 180 10 = 1800 mm2

Area of rectangle (3) A3 =100 10 = 100mm2

from verticalface AB

From the vertical face AB

Example 3.9

Find the position of centroid for an angle of section from base as shown infig. 3.16

Solution

Fig 3.16 shows given angle section there is no X-X and Y-Y axis are axisof symmetry.

Fig 3.15

23

100x 50mm2

1 1 2 2 3 3

1 2 3

A x A x A xxA A A

1000 50 1800 5 1000 501000 1800 1000

50000 9000 500003800

28.68mm

x2 = = 5mm from vertical face AB10 2


We have to find both x and y

Finding

Vertical face AC as reference line.

Dividing the angle section as two rectangularareas.

Areas of Rectangle 1 A1 = 120 x 20 =2400 mm2

120x 10mm2

from vertical face AC

Area of Rectangle 2 A2 = 100 x 20 = 2000 mm2

from vertical from AC

= 37.27mm from vertical face AC.

Finding

Bottom AB as axis of reference.

1120y 60mm

2 from bottom base AB

from the bottom base AB

= 37.27 from the bottom base AB.

Example 3.10

Determine the centroids of the selection shown in figure 3.17

y

x

1 1 2 2

1 2

A x A x 2400 10 2000 70 24000 140000xA A 2400 2000 4400

2100x 20 70mm

2

220y 10mm2

1 1 2 2

1 2

A y A y 2400 60 2000 10yA A 2400 2000

Fi.g 3.16

120

120

20

20


Solution

Figure 3.17 shows selection has

no X-X and Y-Y axis as axis of

symmetry. So both and can be

determined.

Finding

CD line vertical face taken as axis of reference line.

Dividing given Z selection in to three rectangular areas.

Area of rectangular A1 = 80 x 20 = 1600mm2

180x 40mm2

(from vertical face CD)

Area of rectangle (2) A2 = 220x20 = 4400 mm2

from vertical face CD


from the vertical face CD

= 92.02 mm from the vertical face CD

Finding yBottom base AB as axis of reference

120y 200 210mm2

from bottom base AB

2200y 110mm


from bottom base of AB

x

1 1 2 2 3 3

1 2 3

A x A x A x 1600 40 4400 90 1920 140xA A A 1600 4400 1920

64000 396000 2688007920

380x 100 140mm2

32y 12mm2

x y

220x 80 90mm2

100

200

80

80

20

10020

2 0

Fig. 3.17


= 106.44 mm from bottom base AB

Example 3.11

Find the cenrodidal distance for the built up section shown in figure 3.18

Solution

Figure 3.18 shown Y-Y axis of symmetry

centroid lies in it we can find

Finding

Bottom most layer AB line a as axis of reference.

Built up section has divided in 5 rectangular areas.

Rectangular (1) A1= 100 x 10 = 1000mm2

110y 10 20 150 20 205mm2

Rectangular (2) A2 = 100 x 20 = 2000mm2

220y 10 20 150 190mm2

Rectangular (3) A3 = 150 x 20 = 3000 mm2

3150y 10 20 105mm

2

Rectangular (4) A4 = 20 x 200 = 4000mm2

420y 10 20mm2

Rectangle (5) A5 = 10 x 200 = 2000mm2

510y 5mm2

1 1 2 2 3 3

1 2 3

A y A y A y 1600 210 4400 110 1920 12yA A A 1600 4400 1920

336000 484000 230407920

8430407920

Fig 3.18

y

y


= 82.5mm from bottom base AB

Review QuestionsShort Answer Type Questions

1. Define centre of gravity

2. Determine the center of gravity and centroid

3. Locate the position of centroid of the following figures with a neat sketch

1) rectangle 2) triangle 3) circle 4) Semi circle

4. Find the centroid of triangle of base 80 mm and height 120 mm from thebase and the apex

Essay Answer Type Questions1. A masonry dam is trapezoidal in section with one face vertical. Top

width is 3m and bottom width is 10 m height is 10 m. Find the position ofcentroid axis

Ans. x 3.564m and y = 4.260 m

2. Determine the centre of gravity of I section having the following dimensions

Bottom flange = 300x100mm

Top flange = 150x50mm

Web = 50x400mm

Ans. 198.9mm from bottom flange


3. Find out the centroid of an un equal angle section 100mm x 80mm x20mm

Ans = 25 mm from left face

= 35mm from bottom face

4. Find the centre of gravity of channel section 100 x 50 x 15 mm

Ans = 17.8 mm from outer face of web

5. Find the centroid of the given “T” section

Top flange of 250mmx50mm

Web 50mmx200mm

Ans: y = 169.44mm from bottom of the web.

6. Find the centroid of the section shown in figure

Ans: x = 95.56mm from the left edge

y = 85.55mm from the bottom edge

Fig 3.19

xy

x


Moment of Inertia (M.I)Definition

The product area (A) and perpendicular distance (x) between the point isknow as “the first moment of area (Ax) about the point. If this moment is againmultipled by the distance ‘x’ i.e. Ax.x = Ax2 is called moment of moment of area(or) the second moment of area or simply moment of inertia. Its unit in SI systemis mm4.

Moment of inertia for some regular geometrical sections

Position of centroids for Standard Geometric Sections.

S. No Name Shape of figure MI about MI about XX(Ixx) yy(Iyy)

1 Rectangle

2 Hollow

Rectangle

3 Solid Circular

section

4 4D D64 64

4 Hollow Circular

section

5 Triangle

3 32 2BD DB12 12

db3

12DB3

12-bd3

12BD3

12-

about cg about baseBC

3 3b h b h3 6 1 2

4 4 4 4(D d ) (D d )64 64


Parallel Axis Theorem

It states that if the moment of inertia of a plane area about an axis throughits centre of gravity (IGG), is shown in fig 3.20. Then the moment of inertia of thearea about an axis AB parallel to IGG at a distance of “h” from centre of gravityis given by IAB = IGG + Ah2

Fig 3.20

Where IAB = M.I of the area about an axis AB

IGG = M. I of the area about its C.G

A = Area of the section

h = distance between C.G of the section and the axis AB.

Radius of Gyration

Radius of gyration about a given axis is defined as the effective distancefrom the given axis at which the whole are may be considered to be located withrespect to axis of rotation. It is denoted by “k” or “r”

I = Ak2 (or) Ar2

Where I = moment of inertia

K(or) r = radius of gyration k (or) r =

A= area of cross section

Units for k or r in S.I system is mm

Perpendicular axis theorem

It states that if IXX and IYY be the moment of inertia of plane section abouttwo perpendicular axes meeting at “o” shown in figure 3.21 then, the moment ofinertia IZZ about the axis Z Z which is perpendicular to both XX and YY axises,is given by

IZZ = IXX + IYY

IA


For symmetrical section like circular IXX = IYY

IZZ = IXX + Ixx

IZZ = 2IXX

J = 2 I where “J” is known as polar moment of Inertia

Polar moment of inertia.

Definition :- The moment of inertia of an area( IZZ ) about an axisperpendicular to its plane is called “polar moment of Inertia”. It is denoted by“J”

Solved ProblemsProblem 3.12

Find the moment of inertia of a rectangular

section 400mm wide and 800mm deep about

its base.

Solution

Breadth of bearn B = 400mm

Depth of beam D = 800 mm

M.I . about C.G i.e.

M. I about its base 2ABI I Ah

IAB = 1.706 x 1010 + (400x 800) (400)2

= 1.706 x 1010 + 5.12 x 1010

= 6.826 x 1010 mm4

4 3310 2

XX

400 800BDI 1.706 10 mm12 12

Fig 3.21

Fig 3.22


Problem 3.13

Find the M.I of hollow circular sections whose external diameter is 60mmand internal diameter is 50mm about Centroidal axis

Solutions

External dia D = 60 mm

Internal dia d = 50mm

4 4XX YY

4 4

4

I I D d64

60 5064329.2mm

Moment of inertia about Centroidal axis is = 329.2 mm4

Problem 3.14

Find the moment of inertia of a rectangle 60mm wide and 120mm deepabout Centroidal axis. Find also least radius of gyration.

Solutions

B = 60mm

D = 120mm

M. I about Centroidal axis

Area of rectangle A = BD = 60 x 120 = 7200mm2

Least radius of gyrations k (or) r =

Fig 3.23

IcG A

8.64 x 106

7200=

Fig 3.24


∴ Least radius of gyration = 34.64mm

Problem 3.15

Find the radius of gyration of hollow circular sectors of external diameter300mm and internal dia 200mm.

Solution

External dia D = 300mm

d = 200mm

Area 2 2D d4

2 2 4 2300 200 3.927 10 mm4

=

Radius of gyration

Alternate method

= 90.14mm

Problem 3.16

Find the radius of gyration of a triangle whose base is 40mm and height is60mm about an axis passing through C.G and parallel to base.

Base b = 40mm

H = 60mm

Area = 1 bh2

M.I of triangle about Centroidal axis

8

4

I 3.191 10K 90.14mmA 3.927 10

4 42 2 2 2

2 2

D d D d 300 20064K4 4D d

64

Fig 3.25

Fig 3.26


∴ Radius of gyration

3bh 2 hK36 bh 18

60K 14.14mm18

Problem 3.17

Find the moment of inertia about Centroidal axis of hollow rectangularsections shown in fig 3.27

Solution

B = 200mm

D = 400mm

b = 100mm

d = 200mm

M.I about XX axis for hollow rectangular sections.

= 1000x106 mm4

M. I about Y Y Axis for a hollow rectangular section

= 250 x 106mm4

Problem 3.18

Determine the position of centroid and calculate the moment of inertia aboutits horizontal centroidal axis of a T – beam shown in figure 3.28

Solution

Ixx A

K =

Ixx = 112

[ 200 x 4003 - 100 x 2003]

XXbhI36

Fig 3.27

3 33 3

YYDB db 1I 400 200 200 10012 12 2


Finding Centroid

YY axis is axis of symmetry centroid lies on it.

To Finding y

Take AB line axis of reference.

Dividing T section into two rectangular areas

Area of rectangle (1) A 1 = 300 x 100 = 30000mm2

1100y 200 250mm


Area of rectangle (2) A2 = 200 x 100 = 2000mm2

from bottom base AB

Centroidal distance y from bottom 1 1 2 2

1 2

A y A yA A

30000 250 20000 10030000 20000

= 190mm from bottom base AB.

M. I of a rectangle 1 about centroidal axis

IXX at (1) = IG + Ah12

= 2.5 x 107x108 x 106 = 25 x 106+108 x 106

= 133 x 106 mm4

Fig 3.28

2200y 100mm

2

32

271 1

7 2

300 100 300 100 y y12

2.5 10 300 100 250 190 h y y

2.5 10 300 100 60


M. I of a rectangle 2 of about Centroidal axis

Moment inertia of T – beam about its Centroidal axis

Problem 3.19

An un symmetrical I section has top flange 100x20mm web 100 x 120mmand bottom flange 80x20 mm over all depth is 160mm.

Calculate centroid

Solution

Figure 3.30 Shows given I section.

YY-axis is axis of symmetric line

so centroid lies on it.

Finding yTake line AB, passing through the bottom edge as axis of reference

Divide the section into three rectangular areas.

Area of rectangle (1) A 1 = 100 x 20 = 2000 mm2

120y 20 120 150mm2

from base

2G 2I @ 2 I Ah

32100 200 100 200 90

12

6 6

6 4

I at (1)d I at(2)

133 10 228.7 10361.70 10 mm

7 6

6 6

6 4

6.67 10 162 1066.7 10 162 10228.7 10 mm

since h2 = y- y2

h2 = 190-100 = 90mm



2120y 20

60

280mm from base


320y 10mm2

from the base

1 1 2 2 3 3

1 2 3

A y A y A y 2000 150 1200 80 1600 10yA A A 2000 1200 1600

300000 96000 16000 85.83mm

4800

from the base AB.

Finding M.I of “I” section about X-X axis about centroid

M.I of rectangular (1) about X-X axis

h1 = 150-85.83 = 64.17

M. I of rectangle (2) about a X-axis

h2 = y - y2 = 85.83-50 = 5.83mm

M. I of rectangle (3) about X-X axis

h3 = - y3 = 85.83 - 10 = 75.83 mm

Ixx

1

2xx G 1 1 1 1

3

6 4

I @I I A h h y y

100 20 100 20 64.1712

8.3 10 mm

2

2xx G 2 2

32

6 4

I at 2 I A h

10 120 10 120 5.8312

1.48 10 mm

3

2xx G 3

32

6 4

I at 3 I A h

80 20 80 20 75.8312

9.25 10 mm


Moment of inertia of given I section about X axis

Problem 3.20

Determine the moment of inertia of the un equal angle section of size 150mmx 100mm x 25mm about Centroidal axis.

Solution

Finding centroid

FindingVertical face CD has axis of reference, dividing L section has two

rectangular areas.

Area of rectangle 1 A1= 125 x 25 = 3125 mm2

125x 12.5mm2


Area of rectangle 2 A2 = 100 x 25 = 2500mm2


= 29.17mm from vertical face CD

Finding yBottom base AB has taken as axis of reference

1125y 25 87.5mm

2 from base

225y 12.52

from base

y

xx xx xx6 6 6

6 4

I at1 I at 2 I at 3

8.3 10 1.48 10 9.25 1019.03 10 mm

x

2100x 50mm

2

1 1 2 2

1 2

A x A x 3125 12.5 2500 50xA A 3125 2500

39062.5 1250005625

164062.55625


= 54.17mm from the base AB.

Finding xxI

M. I of rectangle (1) about x-x axis

h1 = y - y1 =54.17-37.5 = 33.33mm

= 7.54 106 mm4

M I of rectangle (2) about x-x axis

h2 = y - y2 =54.17-12.5

= 41.67mm

= 0.13 x 106 + 4.34 x 106

= 4.47 x 106 mm4

Moment Inertia of given angular section about X-X axis

Finding YYI M I of rectangle (1) about Y-Y axis

YYI at 1 = IG1 + AA1h12

= 29.17 - 12.5

y

1

322

xx G 1 1

6 6

25 125I at1 I A h 25 125 33.3312

4.07 10 3.47 10

2

32

xx G 2 2100 25I @.2 I A h 100 25 41.67

12

xx xx6 6

6 4

I at I I at2

7.54 10 4.47 1012.01 10 mm

273437.5 + 31250 5625

=

304687.5 5625

=

y = A1y1 + A2y2 A1 + A2

3125 x 87.5 + 2500 x 12.5 31.25 + 2500=

h1 = x - x1


M.I Rectangular (2) about Y-Y axis

M. I of a given angular section about Y-Y axis

Review QuestionsShort Answer Type Questions

1. Explain a) Parallel axis Theorem

b) Perpendicular axis theorem

2. Define the termsa) Moment of inertia

b) Radius of gylation

3. Find the radius of gyration of circle having diameter “d”

Ans: d4

4. Find the radius of gyration of hollow circular plate of 60mm inner diameterand 100mm outer diameter

(Ans:29.15mm)

2

2YY G 2 2

32

6 6

6 4

I at 2 I A h

25 100 2500 20.8312

2.08 10 1.08 103.16 10 mm

YY YY6 6

6 4

I at I I at2

1.063 10 3.16 104.223 10 mm

32

1 1

32

6 6

6 4

DB A x x12

150 25 3125 29.17 12.512

0.195 10 0.868 101.063 10 mm

h2 = x2 - x = 50 - 29.17 = 20.83


5. Find M.I of a rectangular section 200mm width and 400mm depth aboutthe base

(Ans. 4.267 x 109mm4)

Essay Answer Type Questions1. Find the moment of Inertia of a T Section having flange150mm x 50mm

and web 50 x 150mm about xx and yy-axis through the C. G of thesection.

[ Ans: Ixx = 53.125 x 106 mm4

IYY =15.625 x 106mm4]

2. Determine the moment of Inertia of an unequal angle section of size100mm x 80mm x 20mm about Centroidal axis

[ Ans: Ixx = 2.907 x 106 mm4

IYY =1.627 x 106mm4]

3. Determine the moment of inertia of an I section about XX axis given thattop flange 100mm x 10mm web = 200mm x 10mm different flange160mm x 10mm

[Ans: Ixx = 34.38 x 106 mm4]

4. A built up section is formed by an I section and to flange plates of size280 x 20mm are an each flange find the moment of inertia aboutcentrodial X-X axis as shown in below figure

[Ans: Ixx = 188.22 x 106 mm4]

Key Concepts1. The C.G of a body is the fixed point at which its weight is assumed to be

concentrated.

2. The centroid of a surface is the fixed point at which the area of thesurface is assumed to be concentrated.


3. The centroid of a surface is determined from the equations:

4. The centroid of a composite area is treated by the principle of moments,dividing it into regular simple figures.

5. The M.I of an area about a given axis is the sum of the values of “ax2”where “a” is the area of each element and “x” is the distance of thecentroid of the element from the given axis

2I ax 6. Radius of gylation (Kxx) of an area about given axis is the distance from

the axis at which the area may be assumed concentrated to given the M.I of the area about the given axis

IKA

7. Parallel axis theorem :- if “XX” is an axis is parallel to the centrodal axisC.G of surface of area A and if “d” is the distance between the twoparallel axis.

2CGI I Ad

8. Perpendicular axis theorem: If XX and YY are two perpendicular axisis the plane of the area and ZZ is an axis perpendicular to both of themthrough their intersection.

zz xx YYI I I

9. The M.I about an axis perpendicular to its plane is known as its polarM.I

10. M.I of a built up section = Sum of M.I of all elements of the sectionabout the same axis.

11.M.I of a rectangle bxd about axis through centroid parallel to

side

12. M. I of a triangle ‘bxh’ about axis through centroid parallel to base3bh

36

13. M. I of circle of dia ‘d’ about any diameter 4d

64

1 1 1 1A x A yx and yA A

3bdb12


14. M.I of hollow circular section of diameters “D” and ‘d’ about anydia

15. Polar M.I of a solid shaft of dia ‘d’ about axis 4d32

16. Polar M.I of hollow shaft of dia of diameter ‘D’ and ‘d’

=

4 4D d64

4 4D d32


Learning ObjectivesWhen an engineer under taken the design of a structure, it is essential that

he should have the concept of various forces acting on the structure. The mainobjective of a civil engineer is to design a safe and feasible (most economical)structure. Hence adequate knowledge of the properties of materials and theirbehavior under various external loads is essential. When an external force actson a body, if body tends to under go some deformations. The effect of forces ofbodies are to be studied.

The properties of the materials and their behavior under load are explainedin this chapter a few tests assess the performance of the materials are alsoexplained.

When a body is subjected to a system of external loads, it undergoes de-formation. At that time it offers resistance against the deformation. The internalresistance exerted by the body to resist the applied load or force is termed asstress. In other words it is defined as the force acting on unit area of cross-section.

External force pStressCross sectional area A

Units are 2 2 2N m or kN m or N mm

According to Nature of Stress. It can be classified as

1) Tensile Stress 2) Compressive Stress 3) Shear Stress

4UNIT

Simple Stress and Strains


Tensile Stress: When an external force produces increase in length of thebody then that force is called as tensile force or the body is in tension or pullingforce.

Then the stress developed in the cross section of the body is called tensilestress and is denoted by tf .

Compressive Stress: When an external force causes shortening of thebody then that force is called compressive force or the body is under compres-sion or impushing force.

The stress developed in the body due to compressive force is called ascompressive stress. It is denoted by cf .

Shear Stress: The tangential force acting along the section of the bodythen that force is called shear force.

The stress in the section due to shear force is called as shear stress. It isdenoted by sf .

Due to this force there is no increase or decrease in length. But there ischange in shape.

Strain: It is a measure of the deformation produced by the application ofthe external forces.

This strain is three types.

1) Tensile Strain 2) Compressive Strain 3) Shear Strain

tIncrease in lengthTensile Strain eOriginal length

cDerease in lengthCompressive Strain eOriginal length

Shear Strain se = It is a measure of the angle through which a body isdestorted by the applied force.

dsShear Strain tanL

where = Radian

vChange in volume vVolumetric Strain eOriginal volume v

Strain ‘e’ =Change in dimension Original dimension


Mechanical Properties of Materials

1. Elasticity 2. Plasticity 3. Ductility 4. Brittleness

5. Malleability 6. Stiffness 7. Hardness 8. Toughness

9. Creep 10. Fatigue

1. Elasticity

The property of the material by which a body returns to its original shapeafter the removal of external load is called Elasticity.

If the body regains completely its original shape, then it is said to be per-fectly elastic.

Ex: Rubber, Steel and Mild Steel, Copper, Aluminium

2. Plasticity

It is the property of the material by which the material undergo permanentdeformation. That means it fails to regain its original shape after removal of load.

Ex: Gold, Lead, Copper

3. Ductility

It is the property of material by which the material can be drawn into thinwires after under going a considerable deformation without rupture.

Ex: Mild Steel, For steel, Silver, Aluminium etc.

4. Brittleness

It is the property of material by which it breaks without much deformation.

Ex: Glass, Chinaware, Concrete Rock Materials, Cast Iron etc.

Volumetric Strain

The volumetric strain is the algebric sum of all the linear (or) axial strains ifare the strains in three mutual perpendicular directions, then

Volumetric straing, v x y ze e e e

When a solid cube is subjectred to equal normal forces of the same typeon all forces, wil have x y ze e e equal in value.

Then the volumetric strain equal to three times the linear strain ve 3.e .

ex , ey , ez


5. Malleability

It is the property of material by which it can be beaten or rolled into thinsheets without rupture.

Ex: Copper, Ornamental gold, Ornamental silver, Wrought iron.

6. Stiffness

It is the property of material by which it offers resistance to bending ac-tion.

Stiffness is the load required to be applied on a body to produce unitdeflection and it is denoted by ‘S’ or ‘K’.

Load pStiffness 'S 'Deflection

Ex: Springs

7. Hardness

It is the ability of material to resist impressing scratching or surface abra-sion. It is the relative property of material. Every Material will have its ownhardness number.

Ex: Diamond, Graphtic, Talc, Mild Steel etc.

Diamond is the hardest substance and

Talc is the softest substance in nature.

8. Toughness

It is the property of material which enables it to absorb energy withoutrupture.

Ex: Brass, Mild Steel

9. Creep

It is the property of material by which it develops the slow deformationand strain with time due to constant stress.

Ex: Concrete

10. Fatigue

It is the property of material by which the material with stands to varyingand repeating loads.

Ex: Concrete and Prestressed Concrete.


Stress-Strain Curve of Mild Steel

A = Limit of Proportionality

B= Elastic Limit

C= Upper Yield Point

D = Lower Yiled Point

E = Strain Hardenning Point

F = Ultimate Region

G = Breaking Point

Hooke’s Law: The stress is directly proportional to strain within a elasticlimit i.e. upto proportionality limit. In other words, the ratio of axial stress to thecorresponding axial strain is constant.

Proportionality Limit: This is the point upto which the stress is directlyproportional to strain. Hence upto this limit the stress-strain curve is a straightline.

Elastic Limit: It is the limit upto which the strain produced will dissappearcompletly on the removal of load. It means the body gets the original shapeafter removal of load. But the stress is not proportional to strain between pro-portional limit and elastic limit.

Yiled Limit: i) When tensile load further increases stress reaches yieldstress and material starts yielding. Even for a small increase in stress the in-crease in strain is very large.

Fig 4.1 stress-strain curve of Mild Steel


Stress

Strain curve suddenly falls showing a decrease in stress. The point fromwhere sudden fall to curve occurs is known as upper yield point ‘C’.

The point upto which the fall of the curve occurs is known as lower yieldpoint ‘D’.

Sudden stretching of the material at constant stress from lower yiled point‘D’ to the point ‘E’ is known as Strain Hardening.

The point where the stress is constant from lower yield point is known as“Strain Hardening Point” ‘E’.

Beyond this, the stress increases with the increase of strain.

The portion of the curve beyond strain hardening represents the strain hard-ening range.

Ultimate Point: Ultimate load is defined as maximum load which can beplaced prior to the breaking of specimen. Stress corresponing to ultimate load isknown as ultimate stress.

Breaking load: After reaching ultimate stress, stress-strain curve suddenlyfalls with rapid increase in strain and specimen breaks. The stress correspond-ing to breaking point “G” is known as “Breaking Stress”.

1) Linear Strain or Logitudinal Strain: The deformation or change inlength per unit length in longitudinal direction is known as linear strain or longitu-dinal strain.

Change in length 3lLinear or Longitudinal StrainOriginal length l

2) Lateral Strain : When a material is subjected to uni-axial stress withinelastic limit. It deforms not only longitudinally but also laterally. It tensile force isapplied the linear dimensions increase, whereas lateral dimensions decrease.

If compressive force is applied, the linear dimensions decrease where aslateral dimensions increase.

For rectangular sections

Lateral Strain = Change in lateral dimension Original lateral dimension

Change in width Change in depthOriginal width Original depth


are lateral strains these two equal b nb n

.

For Circular sections

Poisson’s Ratio

The ratio of laternal strain to linear strain. It is denoted by 1 or orm

.Laternal strainPoissons RatioLinear strain

The value of 1m for elastic materials is 0.25 to 0.33.

The value of 1m should not be greater than 0.5.

The value of 1m for steel lies between 0.25 to 0.30.

This poissons ratio is same both in tensile and in compression.

Volumetric Strain: The change in volume of an elastic body due to exter-nal forces per unit original volume is known on volumetric strain and is denotedby ve .

Change in volume vVolumetric StrainOriginal in volume v

Bulk Modulus (K): a) When a body is subjected to uniform direct stressin all the three mutually perpendicular directions, the ratio of volumetric stressto the corresponding volumetric strain is found to be constant. This is calledBulk Modullus ‘K’.

Volumetric StressBulk Modulus 'K 'Volumetric Strain

The volumetric stress may be direct stress.

Relationship between Elastic Constants

Elastic Constants are 1 , E, C, Km .

where 1m = Poissons Ratio

E = Young’s Modulus or Elastic Modulus

C= Modulus of Rigidity


K = Bulk Modulus

Case i): Relationship between 1E, C, and .m1E 2C 1

m

Case ii): Relationship between 1E,K andm .

2E 3K 1m

Case iii): Relationship between E, C, K.9KCE

3K C

Case iv) : Relationship between

1C, K, .m1 3K 2C

m 6K 2C

Solved ProblemsShort Answer Type Questions

1) A mild steel rod of 10mm diameter and 300 mm length elongates 0.18mm under an axial pull of 10kN. Determine the Young’s Modulus of Material?

Ans: Axial load = 10kN = 310 10 N

= 10,000N

Diameter of rod = 10mm

Cross Sectional Area ‘A’ 2d 22 1 10 10

4 7 4

278.54 mm

Stress Load 10,000f

A C.S. Area 78.54

2127.32 N mm

Change in length 3l 0.18Strain eOriginal in length l 300

= 0.0006

2212200 N mm

= 2.12 x 105 N/mm2

Young’s Modulus ‘E’ = 127.320.0006

StressStrain

=


.3

orStress plAE lStrain A ll

2) A wooden size tie 50mm 100mm size is 2 mts long. It is subjected toan axial pull of 20kN. Find out the elongation of tie. If the modulus of Elasticityof wood is 4 21 10 N mm ?

Ans: Axial pull = 320 kN 20 10 N

Cross Sectional Area 2A 50 100 5000 mm

Length of the tie = 2 2 1000 2000 l m mm4 2Young 's Modulus E 1 10 N mm

Elongation 3

4

l 20 10 20003l 0.8 mmAE 5000 1 10

3) A hollow cast iron column carries an axial load of 2000 kN. If the outerdiameter of the column is 30 cm and permissible stress = 28k N cm . Findoutthe thickness of the column?

Ans: Outer diameter D = 30 cm

Inner diameter d =

Axial load = 2000kN

Permissible Stress = 28k N cm

2000. . . .

Load

C S Area C S Area

C.S. Area of hollow section = 2 2( )4

D d

D = External dia d = Internal dia

2 2

200030 , 8(30)

4

But D cm Stressd


2 222 1 200030 d7 4 5

222 900 d 25028

2 22900 d 250 318.1828

2d 900 318.18 581.82

d 581.82 24.12 cm

ButD dD d 2t 2t D d t

2

30.0 24.12 5.88 2.94 cm2 2

Thickness = 29.4 mm

4) A steel rod of 25mm diameter and 600 mm long is subjected to an axialpull of 40,000N find 1) Intensity of stress 2) Elongation of the rod?

Diameter = 25mm

Length = 600mm

Load = 40,000N

Pull = 40, 000N

Young’s Modulus = 5 22 10 N mm

Cross Sectional Area 2 22 1d 25 254 7 4

2491.07 mm

(a) Intensity of Stress 240,000 81.45 N mmA 491.07

(b) Elongation l 40,000 600l 0.224 mm

AE 491.07 2 105


l lE

A l A l

ll Direct for lAE

(5) A short timber post of rectangular section has are side of cross sectionequal to twice the other. When the post is axially loaded with 10kN (Compres-sion). If contracts by 0.0521 mm per meta length. Calculate the cross sectiondimensions of the post if ‘E’ for timber 4 21.5 10 ? N mm

Ans: One side of cross section = t (breadth)

Other side of cross section = 2t (depth)

Cross sectional area A = breadth depth

= 22 2 t t t

Axial load pull = 10kn = 10,000N

Change in length 0.0521l mm

Original length 1 1000 l m mm4 21.5 10 E N mm

0.0521 ll

AE

2 4

10,000 10000.05212d 1.5 10

2 10002d 6397.953.0 0.0521

' Linear StressE Yong s ModulusLinear Strain

. .

LoadC S Area A

Change in lengthOriginal length l

l


d 6397.95 79.98 80mm

b 2d 160 mm

Cross Sectional Dimensions are = 80 160 mm

6) A hollow steel column has to carry an axial load of 3.3MN. The yieldstress of steel is 2282 N mm . Assessing Factor of Safety 2 For steel. Deter--mine the external and internal diameters required for column section. If the ratioof internal dia to external dia is to be 0.5?

Ans: Axial load 63.3 mn 3.3 10 N

Mega = 610 N = Newton

Gega = 910 m= meta

Kilo = 310Paseal = 21 N m

Yield Stress of Steel = 282 2N m

Factor of Safety = 22Yield Stress 282Permissible stress 141 N / mm

Factor of Safety 2

6

2

Load 3.3 10 NCross Sectional Area A NStress 141m

223404.25mmInternal diameter 0.5Outer diameter

d 0.5D

d 0.5 D

2 2D d 23404.254

2 2D (0.5D) 23404.254


20.75 D 23404.254

D 199.32 200 mm

Internal dia d 0.5D 0.5 200

100 mm

Outer dia D= 200 mm

Internal dia d = 100 mm

7) The following observations were made during a tension test on mildsteel bar of 20mm diameter Gange length = 200mm. Extension Gange at load of31.4kN = 0.1mm, Yield load = 88kN, Ultimate load = 132kN, Breaking load= 92kN, Total extension = 54mm, Diameter of rod at failure = 14.2mm, Deter-mine a) Young’s modulus, b) Yield Stress, c) Ultimate Stress, d) BreakiungStress, e) % Elongation, f) % Reduction in Area?

Ans: Cross sectional area 2 22 1A d 20 204 7 4

2314.3 mm

Load = 31.4 kN 31.4 1000 31400 N

2Load 31400Stress on this load 99.9 N mmC.S. Area 314.3

Change in length l 0.1Strain on this load 0.0005Original lengh l 200

1) 5 2Stress 99.9Young 's Modulus 'E ' 1.998 10 N mmStrain 0.0005

= 5 22.0 10 N mm

2) Yield load 88000Yield Stress 279.98 280C.S. Area 314.3

= 2280.0 N mm

2Ultimate Load 1320003) Ultimate Stress 420.0 N mmC.S. Area 314.3


2Breaking load 920004) Breaking Stress 292.7 N mmC.S. Area 314.3

Total Extension 545) % Elongation 100 100 27%Original Length 200

6) Diameter at failure = 14.2 mm.

The area at failure i.e. at breaking stress changes i.e. going to be decrease.

Hence reduction in area = Original Area (i.e.) Area of 1(a ) Cross Sectionat First - Area of Cross Section 2(a ).

2 2 22

22(14.2) (14.2) 158.434 77

a mm

at final or failure2

1 2 314.3 158.43 155.87 a a mm

% Reduction in Area = 1 2 100%4

a a

Re 100 duction in area

Original area

155.87 100 49.6%314.3

8) A copper bar of 20mm diameter and 300mm long registers an alongationof 0.5mm and decrease in diameter of 38.34 10 mm under a direct tensileload of 47.1kN. Determine

1&Em of copper?

Ans: Direct tensile load = 47.1 kN = 47100 N

Diameter of the rod d = 20mm

C.S. Area Sectional Area 22 1 20 20

4 7 4A d

2314.3 N mm

Direct Stress247100' ' 149.85

. . . . 314.3p Loadf N mm

C S Area C S Area

Elongation = 0.5l mm


Linear Strain = 30.5 1.667 10300

l mm

Decrease in dia meter 38.34 10d mm

Change in lateral dimensionFuteral StrainOriginal dimension

Change in diaOriginal dia

348.34 10 4.17 10

20

3 23

149.85 89.9 101.667 10

Linear stressE N mmLinear strain

4

3

4.17 10'1.667 10

Linear stressPoisson s RatioLinear strain

41

3

4.17 10 2.50 101.667 10

2.5 0.2510

2' 89.9Young s Modulus kN mm

Poissons Ratio = 0.25

9) A bar of 40 40 mm cross section is subjected to an axial load of 300N. The contraction was found to be 0.42 mm over a guage length of 170mmwhat will be the change in lateral dimension. If poisson’s ratio is 6.3? Find E= ?Young’s Modulus

Ans: Cross sectional area = 240 40 1600 mm

Axial load p = 300N

Original length 170l mm

Change in length 0.042l mm

Linear strain = 0.042 0.00247170

ll


1' 0.30.00247

Lateral strain Lateral strainPoisson s Ratiom Linear strain

0.3 0.00247Later Strain = 0.00072plChange in length lAE

2300 170 75.89. 1600 0.42plE N mm

A l

10) The following results were obtained from tensil test on mild steel speci-men.

Diameter of Specimen = 50mm

Gange Length = 250mm

Length of Specimen of Failure = 300mm

Extension of Load of 442.5 444 10kN mm

Load of Yield Point = 162.2 kN

Max Load = 250kN

Diameter of Neck = 36mm

Factor of Safety = 3

Calculate

a) E = Young Modulus b) Stress at Yield Point c) Ultimate Stress

d) Working Stress e) % Elongation f) % of Reduction in Area?

Ans: Initial Diameter D = 50mm

Cross Sectional Area 2 22 1 50 504 7 4

d

21964.3 N mm

Original Length 250l mm

Extension of load (42.5kN) is 4444 10 mm

Axial load = 342.5 10 42500 N

Cross Sectional Area 21964 3A mm


242500 21.64. . 1964.3LoadStress N m

C S Area

44444 10 1.776 10

250lStrain

l

4

Stress 21.64Young 's Modulus EStrain 1.776 10

5 21.218 10 N mm

3Load at Yield Po int 162.2 10Yield StressOriginal Area of C.S 1964.3

282.57 N mm

3Ultimate Load 250 10Ultimate StressC.S. Area 1964.3

2127.27 N mm

Ultimate Stress 127.27Factor of SafetyWorking Stress Working Stress

127.27 127.273 ; Working StressWorking Stress 3

242.42 N mm

Change in LengthPercentage of Elongation 100Original Length

(300 250) 100250

50 100 20%250

Length of specimen increase where as C.S. Area decreases. Due to tensiletest.


The length at failure is less than original length.

% Reduction in Area 1 2

1

A A 100A

Diameter at failure is less than original diameter

( 2A = Area at Failure) diameter at neck

= 36 mm

2A = Area at Failure 2 22 1d 36 364 7 4

22 36 3628

21018.28 mm

1 2

1

A A% Re duction in Area 100A

1964.3 1018.28 1001964.3

48.16%

11) A bar of 30mm diameter is subjected to a pull of 60kN. The measuredextension over a guage length of 200mm is 0.9mm and change in dia is found tobe 0.0039mm. Calculate a) E = Young’s Modulus b) Poisson’s Ratioc) Modulus of Rigidity d) Bulk Modulus

Ans: Diameter of Bar = 30mm

Axial pull 3p 60 10 N 60,000 N

Length of bar ‘l’ = 200mm; Change in length 3l=0.9mm

Change in dia 3d = 0.0039mm2 22 1Cross Sectional Area d 30 30

4 7 4

2707.14 mm

2p 60,000Linear Stress 'f ' 84.84 N mmA 707.14


3l 0.9Linear Strain 4.5 10l 200

= 0.0045

d 0.0039Laterial Strain 0.00013d 30

1) Linear Stress1) Young 's Modulus ELinear Strain

3 284.84 18.853 10 N mm

0.0045

1 LateralStrain 0.000132) Poisson 's Ratiom Linear Strain 0.0045

= 0.0288 = 0.029

13) E 2C 1m

318.853 10 2 C(1 0.029)

33 218.853 10C 9.16 10 N mm

2 1.029

2E 2K 1m

3E 18.853 10K

2 3 1 2 0.0293 1m

36.284 10K0.942

3 26.67 10 N mm

12) A bar of 10mm 10mm size 400mm long is subjected to axial pull of12kN. The elongation in length and contraction in lateral dimension is found tobe 0.4mm & 0.0025mm respectively. Determine the

1m , Poissons Ratio, Young

Modullus ‘E’, Modulus of Rigidity ‘c’, Bulk Modulus K of the Material?

Ans: 2L.S. Area 10 10 10mm


Length = 400mm Side a = 10mm3Axial pull 12kN 12 10 12, 000N

Change in Length = 0.4m

Change in dia = 0.0025mm

pl pll EAE Al

2512000 400E 1.2 10 N mm100 0.4

Lateral StrainPoisson 's RatioLinear Strain

l 0.4But Linear Strain 0.001l 400

d 0.0025Lateral Strain 0.00025d 10

Lateral Strain 0.00025Poisson 's Ratio 0.25LinearStrain 0.001

C = Modulus of Rigidity

K = Bulk Modulus

51 E 5.2 10E 2 1 C

1m 2 1 0.252 1m

5

5 5 21.2 10 1.2 10 0.48 10 N mm2 1.25 2.5

52 E 1.2 10E 3K 1 K

2m 3 1 0.253 1m

10 0.01mm1000


5

5 5 21.2 10 1.2K 10 0.8 10 N mm3 1 0.5 1.5

Poisson’s Ratio = 0.25

5 2E 1.2 10 N mm 5 2C 0.48 10 N mm

5 2K 0.8 10 N mm

13) A steel bar of 400 mm length and 50mm 50mm cross sectionaldimensions is subjected to an axial pull of 300kN in the direction of length.Calculate the volumetric strain change volume if = 0.25 .

Ans: Length y steel bar l=400mm

Cross Sectional Area of the Bar = 250 50 2500 mm

Axial Pull = 300kN = 3,00,000N

Poisson’s Ratio = 0.255 2E 2 10 N mm

2p 300 100Longitudinal Stress f 120 N mmA 2500

If K = Bulk Modulus

Then 2E 3K 1m

52 10 3K (1 2 0.25)

5 5 52 10 2 10 2 10K3(1 0.5) 3 0.5 1.5

5 21.33 10 N mm

StressBulk ModulusVolumetric Strain

vStressVolumetric Strain e

K


5

5

120 90.22 101.33 10

vChange in VolumeeOriginal Volume

vChange in Volume e Original Volume

590.2 10 50 50 400C.S. Area Length

590200000 10

3902 mm

Volumetric Strain 5ve 90 10

Change in Volume 3902 mm

14) A steel bar of 240mm length and 20mm in dia meter was stretched by1.2mm under an axial pull of 32kN. Determine Young’s Modulus and ShearModulus take Poisson’s Ratio as 0.25?

Ans: Length of Steel Bar l=2400mm

Diameter d = 20mm

Increase in Length 1.2l mm

Axial Pull = 32kN = 32,000N

E = ? ; C=? ; 1 0.25m

22 1Cross Sectioinal Area 20 207 4

232,000Linear Stress 101.8 N mm314.3

l 1.2Linear Strain 0.0005l 2400

222 20 20 314.3 N mm28


Linear Stress 101 8Young 's Modulus 203600Linear Strain 0.0005

5 22.04 10 N mm

Poisson’s Ratio = 0.25

Shear Modulus = Rigidity Modulus = ‘C’

1E 2C 1m

5 5E 2.04 10 2.04 10C1 2(1 0.25) 2 (1.25)2C 1m

5

5 22.04 10 0.816 10 N mm2.50

5 20.82 10 N mm

15) A rectangular steel bar 60 mm wide and 10 mm thick, 5m long issubjected to an axial pull of 80kN. If the increase in length under the load is1.5mm and decrease in thickness is 0.0014mm. Determine three elastic con-stants of the material and poisson’s ratio. Decrease in width under the load andchange in volume produced?

b = 60mm; t = 10mm; l=5m = 5000mm

Cross Sectional area 2b t 60 10 600 mm

Axial load = 80kN = 80,000N

Increase in Length = 1.5mm

Decrease in Thickness = 0.0014mm

1E, C, K ? ?m

2Load 800001) LinearStress 133.33 N mmC.S. Area 600

4Increase in Length 1.5mm2) Linear Strain 3 10 0.0003Original Length 5000mm

.


5 2Linear Stress 133.333) Young 's Modulus 4.44 10 N mmLinear Strain 0.0003

d 0.00144) Lateral Strain 0.00014d 10

Lateral Strain 0.00014Poisson 's Ratio 0.47Linear Strain 0.0003

If C = Modulus of Rigidity

K = Bulk Modulus

1E 2C 1m

5 5E 4.44 10 4.4 10Modulus of Rigidity 'C '1 2(1 0.46) 2 (1.46)2C 1m

5 21.52 10 N mm

5E 4.44 10Bulk Modulus K2 3(0.08)3 1m

55 24.44 10 18.5 10 N mm

0.24

v

Volumetric Stress 133.33Bulk Modulus KVolumetric Strain e

5v 5

133.33e 7.207 1018.5 10

vChangein VolumeBut eOriginal Volume

v ve Original Volume


5 3v

v

7 207 60 10 5000 10 216.2 mme C.S.Area Length

t 0.0014 bLateral Strain 0.00014t 10 b

b 0.00014 60 0.00084mm

Essay Answer Type Questions1) The following observations were made during a tension test on the mild

steel bar of 20mm diameter of gange length = 300mm.

Extension at a load of 30kN=0.1mm

Yield Point = 80kN

Ultimate Load = 130kN

Total Extension = 50mm

Diameter of Rod at Failure = 14.1mm

Calculate a) Young’s Modulus b) Yield Stress c) Ultimate Stress

d) % of Elongation e) % Reduction in Area?

Ans: 5 2a) 2.56 10 N mm 2b) 254.65 N mm2c) 413.8 N mm d) 16.67% e) 50.3%

2) A hallow pipe metal pupe 2.5m long is subjected to an axial pull of300kN. The piple has an interval diameter 250mm assuming ‘E’ for the metal as

6 20.1 10 N mm . Find the thickness of the pipe. If the elongation of pipe is0.15mm.

Ans: 52.6mm

3) A rectangular bar 50mm wide and 10mm thick is 3000mm long. It issubjected to an axial pull of 50N. If the change in length is 1.5mm and decreasein thickness is 0.0014. Determine the for elastic constants?

Ans: 2a) E 200 kN mm 2b) C 78.125 N mm2c) K 151.52 kN mm

1d) 0.28m


4) A steel bar 2m long, 20mm wide and 10mm thick is subjected to a pullof 20kN in the direction of its length. Find the change in length, breadth andthickness Take 5 2E 1.0 10 N mm and Poisson’s Ratio = 0.3?

Ans: l 1mm; b 0.003mm; t 0.0015mm

5) For a given Material 5 2E 1.0 10 N mm and 5 2C 0.4 10 N mm .Find K and lateral contraction of a round bar of 50mm diameter and 2-5m longwhen stretched 2-5mm

1 0.25m .

Ans: 5 2K 0.67 10 N mm ; d 0.0125mm

6) A bar of 30mm diameter is subjected to a pull of 60kN. The measuredextension an gauge length of 200mm is 0.09 mm and change in diameter is0.0039mm.

Calculate the 1m

Poissons ratio and the values of three elastic constants?

Ans:1a) 0.288m 2b) E 188.67 kN mm

2c) C 73.19 kN mm 2d) K 149.85 kN mm

Key Concepts1. Stress is internal resistance setup by material per unit area to resist

deformation

: PA

2. Strain is deformation per unit length.

3. Hookes law states that stress is proportional to strain within elasticlimits.

4. Young’s modulus is normal stress per unit normal strain within elasticlimits.

If young’s modulus is E, area of cross section A angle length Lload applied on the body “P” the deformation L is the body then

PL PLE or LA L AE

5. Yield point is stress at which strain increases under a steady load.

6. Ultimate strength is stress corresponding to maximum load that can berealised before failure.

2


7. Factor of Safety UltimatestressWorkingstress

8. Shear Stress is stress applied tangalially to a surface.

9. Shear strain is angular strain in radians produced under shear stress.

10. Modules of Rigidify is shear stress per unit shear strain within elasticlimits.

11. Ratio lateral strain to longitudinal strain within elastic limits is poisson’sratio.

12. Bulk modulus is normal stress per unit Volumetric strain within elasticlimits when the body is subjected to three equal mutually perpendicular normalstresses of same kind.

13. The relationship between the three elastic moduli is given by

1 2E Z N 1 3k 1m m

9KN 1Byeliminations3K N m

2

Learning Objectives• We come across several instances of members subjected to compressive

loads. Examples for such loading may be framed structures, rafters in buildings,frames of presses etc.,

ColumnsColumns are the members subjected to direct compressive force. A vertical

member subjected to direct compressive force is called a column or pillar.

Struts:- When a member of a structure is any position and carrying anaxial compressive load is called strut. Strut may be horizontal, inclined, or evenvertical.

Post:- Wooden member carrying compressive force called post.

Stanchion:- a built up rolled steel carrying compressive force is calledstanchion

Boom:- Main compressive member in a jib crane is called a boom.

Short Columns:- In this type of columns, the bucking stresses are verysmall, compared to direct stress or bucking stresses. Fails due to direct stress.

b. Short columns is a column whose slenders ration is less than < 32 orwhose length is <8 times the least latered dimension

c. For mild steel columns slenderness ration is less then 80.

5UNIT

Columns - Struts


Medium columns or Intermediate Columns

Medium columns is a column which fairs either due to direct stress orbuckling stress.

a. For medium columns, the slenderness ration is more than 32 and lessthan 120.

b. For medium columns, their length is more than 8 times but less than 30times. Their least lateral dimmesion.

Long columns

Long Columns is column which fairs primary due to bending stress.

a. Long columns, The direct stress are very small compared to their backingstresses.

b. Long columns is a column whose slenderness raton is greater than 120or whose legth is more then 30 times the lest lateral dimension.

c. For mild Steel Column slender ness ration is >80.

Slenderness Ratio

Slenderness Ratio of a column is defined as the ration of column height toits least radings of gyration

lk

where l = Equivalent length of the column

k = Least radius of gyration.

and IxA

I = Least moment of Inertia

A= Cross sectional area

a. Load carrying capacity of long columns depends on slenderness ratio.

b. For good design the slenderness ration is as small as possible.

Effective length of a Column

The effective length of a given with given end conditions is the length of anequivalent column of the same material and semicross section with hinged endshaving the value of the crippling load equal to that of a given column effectivelength of column mainly depends upon end conditions.


S. No Types and Endless Actual Equivalent conditions crippling length length

1 Both ends hinged 2

2

EIPL

L l = L

2 Both ends fixed 2

2

4 EIPh

LLl2

3 One end fixed other

end free 2

2

2 EIPL

LLl2

4 One end fixed other

end minged. L l = 2L

P = Eulers crippling or buckling load.

E = Young’s modulus

I = Least moment of inertia

l = Equivalent length of columns for given end conditions.

Columns subjected axial load

Equals crippling load or critical load2

cr 2

EIPl

Where EI = Flexural rigidity

l = effective length of columns

equals formula is used for lons columns or l 80k for mild steel columns,

Assumptions made in Euler’s formula

a. The column is initially perfectly straight

b. The Cross section of the column is uniform through out length.

c. We length of the column is very large when compared to its lateraldimension.

L

2

2

EIP4L


d. The Shortening of the columns is neglected

e. the self weight of the columns is ignorable

f. The column fails due to bucking alone.

g. The column material is perfectly elastic, homogeneous is tropic & obeyshook’s law.

h. The pin joints are friction less & fixed ends are rigid.

Rankines formula can be used for both short & long columns the followingvalues are for a columns both eudshinged.

S. NO Material Crushins Stress Fc in N/mm2

1 Wrought Iron 250 1

9000

2 Cast Iron 250 1

1600

3 Mild steel 320 250 1

7500

4 Timber 50 1

750

5 Medium carbon Steel 500 1

5000

The above values for a column both ends hinged.

Bucking loads:- the load at which the column just buckles is called bucklingload or critical load or crippling load and the column is said to have developedan elastic instability.

The value of buckling load is low for long columns and relatively high forshort columns.

Factor of safety :- Factory Safety = crippling load

Safe loadColumn subjected to axial load :-

Euler’s is crippling load = 2

cr 2

EIPl

2

Rakine'scons tan tfcEI

L


EI =flexural rigidity

l = effective length of the column

Euler’s formula is only for long columns l 80K

11.Rankinl’s formula for crippling load :-

cr 2 2c.A crushing loadp

L l1 1K k

fc = Allowable crushing stress

A = Area of cross section of the column

K= least radius of gyration

A = rankine’s constant.

Limitations of Euler’s Formula

1. Euler’s formula is used for long columns & negrech the stress due todirect compressive loads.

2. If slenderness ration is less than 80, the Euler’s formula for mild steelcolumns is not valid.

3. In Euler’s formula that the critical or allowable stress an a columndecreases with the increase in slenderness ratio.

4. Long axially loaded columns tends to defect about the axis of the leastmoment of inertia, the least radius of gyration and it should be usedfor determination the slenderness ration.

Solved Problems1) A mild steel column 5m long and 50mm diameter which is fixed at one

end free at the other end determine the Euler’s crippling load take

E = 2 x 10 N/mm2

Euler’s crippling load = 2

2

EIl

This is condition for its both ends hinged

f

L


But one end fixed other end free

2 5

2

3.14 2 10 Ip

4 l

But I for circular section

= 306919.6mm4

2 5

2

2 5

3.14 2 10 306919.6p

4 5000

3.14 2 306919.6 104 5000 5000

6052.2 N

2. A hollow along tube of 5m long with external dia 40mm and internal dia25mm was to extend 6mm under a tensile load of 60 KN. Find the bucking loadfor the tube when used as strut in both ends pinned. Also find the safe load,taking factor of safty as 4?

Length = 5m

Both ends pinned or hinged l L 5m 5000mm External = 40mm ; internal dia = 25mm

M.I hollow Cross section = 4 4D d64

4 422 1 40 257 64

4

4ITd 22I 5064 1

=> 1 = 2L, p = 2El l2


22 1 256000 335776

7 64

4

22 1 256000 3357767 64

109421.7 mm

Area of hollow section 2 2D d4

2 222 1 40 257 4

222 1600 625 766mm28

l= 6mm; total load = 60KN = 60,000N

Stress 260,000 60,000 78.33N mmc s area 766

Strain 3l 6 stress 78.330.0012 El 5000 strain 0.0012

22

22

3.17 65275 109421.7EIP 2816.9 NL 5000

Safe load = crippling load 2816.9

Fs 4

= 704.2 N

3. A mild steel tube 4m long, 30mm internal dia and 5mm thick is used asstrut with both ends pinned find the Euler’s crippling load take E = 2x 105N/mm2 ?

l = 4 m = 4000mm

Internal dia = 30mm

External dia = 5+30+5

= 40mm1

,


I of hollow circular section 4 4D d64

4 222 1 40 307 64

22 2560000 8100007 64

4

22 10,000 256 817 64

22 1,75,0,000 85937.5mm7 64

Euller’s crippling load 2

2

EIL

Both ends pinned l = L

2 2 5

22

EI 2 10 85937.5L 4000

2 5 22 10 85937.5 2 85937.5

4000 4000 160

P = 10610.6 N

Crippling load = 10610.6N

4. A cast graw hollow column, having 80mm extends dia and 60mm internaldia, is used as column of 3m long using Rakine’s formula, determine cripplingload, when both ends are fixed? Take

fc =500N/mm2 1

1600

D= external dia = 80mm internal dia d = 60mm

2 22 2 22 1A D d 80 604 7 422 226500 360028 28

2800100


= 2200mm2

4 4 4 422 1I 80 60 80 6064 7 64

491.07 4096 1296

4491.07 2800 1374996mm

Least radius of gyration I 1374996KA 2200

K = 25mm

L=3m = 3000mm1

1600 fe = 500N/mm2

Both ends fixed = l 3000L 1500mm2 2

6c

2 2F A 500 2200 1.1 50P 1L 1 1500 11 1 1600K 1600 25

1.1 106 338461.6 N

3.25

= 338.5KN

5. The cross section of mild steel column is hollow rectangular section withthe dimensions 300200mm and vertal thickness 25mm. the length of columnis 3m and both ends are hinged. Farm the safe load is can carry using a Ramkine’sformula:-

Take F-S = 3; Fc = 300N/mm2 1

1500

Feast M.I 3 31 BD bd12

3 31 300 200 250 15012

l

l

6


4

4 4

1 10 2, 40, 000 84375121 1,55625 12968 10 mm

12

L = 3000mm

Both ends hinged 2

fc 'Al LL1 Ak

4 4I 12968 10 12968 10KA GS.area

2

225 0075.92

C. S area = (BD-bd)= 300x200-250x150

= 60 000 -37 , 500 = 22,500mm2

2 2300 22,500 300 22500P

1 1 30001 1K 7500 75.92

300 22,500 5587749.3

1.208

N

= 5587.75 KN

Factory safety =3

Safe load = 5587.75 1862.6KN

3

Short Answer Type Questions1. What is mean of bucking load or criphists load? An what factors does it

depends?

2. Define the terms i) slenderness ratio ii) critical load iii) equivalent length

3. State the different Euler’s formulae for different end conditions ofcolumns?

4. Distinguish between long columns short column of mild steel?

5. Different between a)long columns b) medium columns c) short columns.

C

,

x

x


6. State the assumptions made by Euler’s theory?

7. How do the Euler’s formula for crippling load to the following conditions?i) both ends hinged ii) both ends fixed iii) one end fixed other end free iv) are endfixed and other hinged?

8. What are the limitations of Euler’s buckling theory?

9. A mild steel tube 3m long 30mm internal dia and 4mm thick is used asstrut both ends hinged find the crippling load. Take E = 2105 N/mm2

Ans. Pcr = 13728.98 N

10. Calculate the safe compressive load on a hollow cast iron column withone end fixed and other end hinged of 150 mm external dia and 100mm and 3mlength use angle’s formula with a factor of safety of 3 and E =0.95 x 106 N/mm2

(Safe load = 1384.89)

11. A column of timber section 100 x 200mm is 4.5m long end its bothends are fixed calculate safe load for a column. It can take using Euler’s formula.

Take E = 17.5 x 103N/mm2 & factory safety = 3 ?

(Ans: 189.463 KN = safe load )

12. Determine the section of cast iron hollow cylindrical columns 5m longwith ends firmly build in (both ends field) if is carries axis load of 300 KN. Theration of using factory of safety = 8, take

Ramkin’s constant

Ans D=171.1mm d = 128.3 mm

13. A hollow cast iron column with fixed ends supports axial load of1000KN. If the column is 5m long and has and external dia of 250 mm find thethickness of meta required. Use rankine’s formula taking a constant and assumingthe working stress of 80N/mm2

(Ans d = 215.6mm t = 17.19mm )

11000

Learning Objectives• The graphical statics presents a less tediuos and practical solutions of a

problem in statics by graphical method.

• The accuracy of the graphical solution may not match with that of theanalytical one but is generally sufficient for all practical purposes.

Space Diagram, Bow’s Notation and Vector DiagramThe relative positions of the various vectors acting on a system are

represented, in a figure called the Space Diagram. It is drawn to a linear scale toshow the points of application and the directions of all the vectors. In naming thevectors, a standard practice or notation is used. Bow’s notation is generallyfollowed.

In Bow’s notation, each space on either side of the line of action of eachvector is given a name.

The vector Diagram represents the magnitudes and directions of all thevectors acting on the system. It is drawn to the scale of vectors.

Equilibrant and Resultant of Two Concurrent Forces

These are determined the help of the law of triangle of forces.

Example 7.1 : Determine the equilibrant and hence the resultant of twoforces of 150 N and 250 N acting at a point O if the angle between them is 600.

7UNIT

Graphics Statics


Force Magnitude Inclination with OX

Equilibrant ca 350 N -1420

Resultant ac 320 N 380

1. Draw the space diagram to show the two given forces making 600 witheach other.

2. Name the two forces as per Bow’s notation, the 150 N by AB and the250 N by BC as shown in Fig 7.1. The equilibrate will then berepresented by CA.

3. Select a convenient point (in Fig 7.2) to present the space A. Drawthrough a, ab parallel to the direction of force AB (150 N). Mark thepoint b on ab represents 150 N to the selected scale say 1 mm for 5 N.

4. From b, draw bc parallel to the 250 N force i.e., BC. The length bc isselected such that the magnitude of BC is represented by it to the samescale of 1 mm for 5 N.

5. Join ‘ca’ to get abc, the triangle of forces for the point O. Fig. 7.2 isknown as the Vector Diagram.

6. ‘ca’ represents the magnitude and direction of the equilibrant of the givenforces. Measure its magnitude, Draw a parallel to this direction in thespace diagram, tabulate results and measure the angle made by it withOX.

7. ‘ac’ represents the magnitude and direction of the resultant of the twogiven forces. Measures its magnitude and inclination with OX andtabulate result.

Fig 7.1 Fig 7.2


Note

1. To every space in the space diagram, there will be a corresponding pointin the vector diagram.

2. To every vector in the space diagram, there will be a straight line in thevecot diagram.

3. The equilibrant and the resultant will be collinear, equal and opposite.

4. The vector diagram is a closed figure for a system of forces in equilibrium

Equilibrant and Resultant of more than two Concurent Forces

These are determined by the law of polygon of forces. This is only anextension of the method of triangle of forces.

Example 7.2

Determine the equilibrant and resultant of 4 pulls of 300 N, 600N, 400 Nand 200 N making angles of 300, 1200, 2250 and 3300 respectively with a fixeddirection OX.

Procedure

1. In the space diagram (Fig ) draw the direction OX and the direction ofall the given forces making the stated angles with OX.

2. Name the given forces as AB, BC, CD and DE by using Bow’s notationstarting with 200 N and going clock wise about O. (See Fig ) Let EAbe the equilibrant to he system.


Results

Force Magnitude Inclination with OX

Equilibrant ea 320 N 2980

Resultant ae 320 N 1180

3. Draw ab (Fig ) parallel to the force AB=200 N and to represent itsmagnitude to a scale of 1 mn for 10 N.

4. From ‘b’ draw be parallel to the next force in the order i.e., BC andmark as such that be represents the forces of 300 N to the scale selected.

5. From ‘c’ draw cd parallel and proportional to force CD=600N andform an draw be parallel and proportional to DE =400 N.

6. ‘abcde’ is the vector diagram. Hence by law of polygon of forces ‘ea’represents the equilibrate of the given system of cocurrent forces.

Measure its magnitude to scale (=320 N) and draw a parallel to its directionthrough O in the space diagram. Measure inclination of this line with OX (=2980)and tabulate results.

7. ‘ae’ represents magnitude and direction of the resultant. Measure itsmagnitude and inclination and tabulate results.

Note : The equilibrnat of a system of copanar concurrent forces is also acoplanar force and is concurrent with the system. Hence the resultant passesthrough O, the point of concurrency. Its direction will be parallel to the closingsidce eea of the polygon of forces for the point O.

Reactions of Simply Supported Beams

To find the reactions at the supports of simply supported beams, proceedas follows:

1. Draw space diagram, vector diagram and funicular polygon for all theforces on the beam excepting the reactions.

2. Produce the first ray a/o and the last ray say d/o of the funicular polygonto cut the lines of action of the reactions at the respective supports at p,q respectively. The line pq will be the closing line ofthe funicular polygon.

3. Draw a ray parallel to the closing line ‘pq’ through the pole O of thevector diagram to meet the load line at say e.

4. ‘ea’ will represent the reaction EA and ‘de’ will represent the reactionDE.


Graphical Method of Determing Centroid

As in the analytical method, the composite area is divided into elementaryfigures whose area and centroid can be determined easily. The areas of all suchelements are considered as parallel forces acting in a convenient direction throughthe respective centroid. The line of action of the resultants force is determinedgraphically as per the method. The centoid of the given composite area lies onthe line of action of the resultant force. If there is no symmetry about any axis,the centroid is then located at the intersection of the resultant forces in the twoassumed directions.

Example 4.9 : Determine graphically the centroid of a Tee section 180mm/120mm / 20 mm.

The Tee section is symmetric about the axis of the web. Hence its centroidlies on this axis. The areas of the flange and web will be treated as horizontalforces through their centroids located by intersecting the diagnonals.

Example: 4.10 : Determine graphically the centroid of an unequal angle100 mm x 80 mm x 10 mm.

Diagram is displayed in the next page




Long Answer Type Questions

1. Determine graphically the euilibrant of the forces shown in Fig.

2. Two forces 200 N and 300 N act at an aggle of 1500. Find the magnitudeand direction of the resultnat by graphical method. The 200 N Force is horizontal.

3. Determine the distance of the centroid of the sections shown in Fig.from the bottom most edge and the central vertical axis.

Documents

System of Measurement and Unitsbieap.gov.in/Pdf/BCMTPaperIII.pdf · Matter: Any substance which occupies space is called matter. ... i.e. it has no size, but has a definite mass