introduction to structural mechanics book

NATIONALDIPLOMAIN

BUILDINGTECHNOLOGY

UNESCONIGERIATECHNICAL&VOCATIONALEDUCATIONREVITALISATIONPROJECTPHASEII

YEARISEMESTERII

THEORY

Version1:December2008

INTRODUCTIONTOSTUCTURALMECHANICS

COURSECODE:BLD108

BLD 108 INTRODUCTION TO STRUCTURAL MECHANICS

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Table of Contents

WEEK 1: NEWTONS LAWS OF MOTION (1.1) Newtons Laws of Motion and Their Application ...................................

(1.1) First Law .......................................................................................

(1.1) Application...........................................................................

(1.1) Second Law ...................................................................................

(1.1) Application...........................................................................

(1.1) Momentum ...........................................................................

(1.1) Change in Momentum .........................................................

(1.1) Rate of Change of Momentum .............................................

(1.1) Representation of the Second Law ......................................

(1.1) Third Law .....................................................................................

(1.1) Application...........................................................................

Exercises .....................................................................................................................

WEEK 2: NEWTONS LAWS OF MOTION (1.1) Example 1 ........................................................................................................

(1.1) Example 2 ........................................................................................................

(1.1) Example 3 ........................................................................................................

(1.1) Example 4 ........................................................................................................

Exercises .....................................................................................................................

WEEK 3: IMPULSE AND MOMENTUM (1.2) Impulse and Momentum ..................................................................................

(1.2) Impulse...................................................................................................

Law of Conservation of Momentum .........................................................................

(1.2) Example 1 ........................................................................................................

(1.2) Example 2 ........................................................................................................

(1.2) Example 3 ..............................................................................................................


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(1.2) Example 4 ...............................................................................................................

Exercises ..........................................................................................................................

WEEK 4: KINETIC ENERGY (1.3) Kinetic Energy ..................................................................................................

Energy of a Pendulum ......................................................................................

(1.3) Example 2 .........................................................................................................

(1.3) Example 3 .........................................................................................................

(1.3) Example 4 .........................................................................................................

Exercises ....................................................................................................................

WEEK 5: KINEMATICS OF POINTS (1.3) Kinematics ........................................................................................................

Motion ..............................................................................................................

Types of Motion ...................................................................................

Random Motion .......................................................................

Translational Motion ................................................................

Rotational Motion ....................................................................

Oscillatory Motion ...................................................................

Motion in a Straight Line .................................................................................

Equations of Uniformly accelerated Motion ....................................................

(1.4) Example 1 .........................................................................................................

(1.4) Example 2 .........................................................................................................

Exercises ....................................................................................................................

WEEK 6: RESOLUTION AND COMPOSITION OF VECTORS (1.5) Resolution and Composition of Vectors ...........................................................

(1.6) Relative Velocity and Acceleration ..................................................................

(1.7) Representation of Vectors ................................................................................

Unit Vector...........................................................................................


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Equal Vectors .......................................................................................

Like Vectors .........................................................................................

Addition of Vectors..............................................................................

Parallelogram Law of Vectors .............................................................

Triangle Law of Vectors ......................................................................

Polygon Law of Vectors ......................................................................

(1.5) Example 1 .........................................................................................................

(1.6) Example 2 .........................................................................................................

(1.7) Example 2 .........................................................................................................

Exercises ....................................................................................................................

WEEK 7: LOAD AND TENSION AND COMPRESSION (2.1) Load ..................................................................................................................

(2.2) Tension and Compression ................................................................................

(2.2) Tension ..................................................................................................

(2.2) Compression ..........................................................................................

(2.2) Example 1 .........................................................................................................

(2.2) Example 2 .........................................................................................................

(2.2) Example 3 .........................................................................................................

Exercises ....................................................................................................................

WEEK 8: RESOLUTION AND COMPOSITION OF VECTORS (2.3) Stress and Strain ...............................................................................................

(2.3) Stress ........................................................................................................

(2.3) Strain ........................................................................................................

(2.4) Hookes Law .....................................................................................................

(2.5) Modulus of Elasticity ........................................................................................

(2.3) Example 1 .........................................................................................................

(2.3) Example 2 .........................................................................................................

(2.3) Example 3 .........................................................................................................


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Exercises ....................................................................................................................

WEEK 9: RELATIONSHIP BETWEEN STRESS AND STRAIN IN TENSION

(2.6) Relationship Between Stress and Strain in Tension .........................................

(2.6) Example 1 .................................................................................................

(2.7) Limit of Proportionality ....................................................................................

(2.7) Elastic Limit .....................................................................................................

(2.7) Yield Point........................................................................................................

(2.7) Ductility ............................................................................................................

(2.7) Brittleness .........................................................................................................

(2.7) Permanent Set ...................................................................................................

Exercises ....................................................................................................................

WEEK 10: SHEAR STRESS, SHEAR STRAIN , MODULUS OF RIGIDITY AND STRAIN ENERGY

(2.8) Shear Stress .......................................................................................................

(2.8) Shear Strain .......................................................................................................

(2.8) Modulus of Rigidity ..........................................................................................

(2.8) Strain Energy ....................................................................................................

(2.8) Example 1 .........................................................................................................

(2.8) Example 2 .........................................................................................................

Exercises ....................................................................................................................

WEEK 11: ANALYSIS OF A COMPOSITE BODY (2.9) Analysis of a Composite Body .........................................................................

(2.9) Example 1 .........................................................................................................

(2.9) Example 2 .........................................................................................................

(2.9) Example 3 .........................................................................................................

Exercises ....................................................................................................................


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WEEK 12: EQUILIBRIUM OF COPLANAR FORCES (3.1) Relationship Between Stress and Strain in Tension .........................................

(3.1) Coplanar Forces........................................................................................

(3.1) Concurrent Coplanar Forces .....................................................................

(3.2) Polygon of Forces .............................................................................................

Exercises ....................................................................................................................

WEEK 13: ANALYSIS AND RESOLUTION OF FORCES (3.3) Example 1 .........................................................................................................

(3.3) Example 2 .........................................................................................................

Exercises ....................................................................................................................

WEEK 14: RELATIONSHIP BETWEEN STRESS AND STRAIN IN TENSION

(2.6) Relationship Between Stress and Strain in Tension .........................................

(2.6) Example 1 .................................................................................................

(2.7) Limit of Proportionality ....................................................................................

(2.7) Elastic Limit .....................................................................................................

(2.7) Yield Point........................................................................................................

(2.7) Ductility ............................................................................................................

(2.7) Brittleness .........................................................................................................

(2.7) Permanent Set ...................................................................................................

Exercises....................................................................................................................................

WEEK 15: REVISION Example 1 ...................................................................................................................

Example 2 ...................................................................................................................

Example 3 ...................................................................................................................

Example 4 ...................................................................................................................

Example 5 ...................................................................................................................


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Example 6 ...................................................................................................................

Example 7 ...................................................................................................................


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1.1 NEWTONS LAWS OF MOTION AND THEIR APPLICATION

There are three Newtons Laws of motion:

First Law: A body will continue to remain in its state of rest or if in a state of uniform

motion in a straight line, will continue to maintain its state of motion unless an external

force acts on the body which will alter that state of rest or of the uniform motion.

Application

The tendency of bodies to remain at rest allows the production of Maps of regions

and Layouts of towns and cities. Nigeria has always remained in its present position

so are other countries of the world. States within Nigeria and their towns will

continue to remain where they have always been.

Figure 1.1.1 Map of the World.


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Figure 1.1.2 Layout of a City

Object placed on a position will continue to maintain its position. Anytime the object

is needed, it will be found on that particular position. If the object was not found

where it was placed, somebody or

something altered the objects position.

Figure 1.1.3 Objects placed on a position.

A person riding an escalator feels at rest and will continue to move with uniform

motion in a straight line.

Figure 1.1.4 People on escalator


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Passengers in airplane feels impact only when the plane is taking-off or landing or

where there is a storm, when the plane reached its altitude and a uniform motion is

maintained, the passengers feel at rest and go on with their activities.

Figure 1.1.5 Passengers in airplane.

Second Law: The rate of change of momentum of a body is directly proportional to the

applied force and is acting in the direction in which the force acts.

MOMENTUM

Momentum of a body is defined as the product of the mass (m) of the body and the

velocity (v) the body is moving with. That is:

Momentum = mass x velocity (kgm/s)

CHANGE IN MOMENTUM

Mass of a body is the same everywhere on the earth; therefore, momentum change

of a body will only be as a result of change in the velocity of the movement.

If the first (initial) velocity is represented by u, the momentum of the body is mu.

If the second (final) velocity is represented by v, the momentum is mv.

Change in momentum = mv mu = m (v u).


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RATE OF CHANGE OF MOMENTUM

The rate of change of momentum of a body is the ratio of the change in momentum of the

body to the time taken for the change to occur. If the time taken is represented by t;

Rate of change of momentum = m (v u) / t.

REPRESENTATION OF THE SECOND LAW

If the applied force is represented by F, the second law is represented as;

F t

u) - (v m

F = k t

u) - (v m

The constant k is determined to be = 1. So that, in reality the constant of proportionality

is the mass (m) of the body.

F = t

u) - (v m

And t

u) - (v = acceleration (a)

Therefore, F = ma

A body moving with uniform motion has velocity v = u, so that,

Acceleration (a) = tu -u = 0.

The applied force,

F = m x 0 = 0

This implies that a body moving with uniform motion has no force acting on it.


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As an illustration of the definition, suppose a boy or girl on a bicycle, total mass 50 kg,

has a velocity of 1m/s and that, pedaling faster for 5 s, the velocity increases to 3 m/s. Then,

since rate of change means change per second,

Average force = ( )s

kg5

m/s 1 x 50 - 3 x 50 Time

Change Momentum =

= 20 kg m/s2.

1 kg m/s2 is the force acting on a mass of 1 kg which gives it an acceleration of 1 m/s2. By

definition, this force is the newton. Hence the force is 20 newtons. If the cyclist pedals harder, so

that his velocity increases from 1 m/s to 3 m/s in 2 s, the momentum change per second is greater

than before. The force is now given by

N 50 2

1 x 50 - 3 x 50 =

Suppose the cyclist, moving with a momentum of 150 units, applies the brakes slowly and comes

to rest steadily in 6s. Then the momentum change/s = 150/6 = 25 newtons. If the cyclist pulls up

sharply and comes to rest in s, the retarding force = 150/ = 300 newtons. The cyclist suffers

a more violent force in this case. For this reason, a person landing on his feet from a height is

well-advised to flex the knees, so as to provide a less rapid momentum change on coming to rest.

Application

In design of building subjected to dynamic vibrations such as earthquake resisting

structures, the Newtons second law of motion is applied.

Third Law: To every action there is equal and opposite reaction. If a body (first object)

exerts a force (action) on another body (second object), an equal but opposite force

(reaction) is exerted on the body (first object) by the other body (second object).


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Application

Rocket Propulsion: The motion of a rocket is much like the motion of a balloon

losing air. When the balloon is sealed, the air inside pushes on the entire interior

surface of the balloon with equal force. If there is an opening in the balloons surface,

the air pressure becomes unbalanced, and the escaping air becomes a backward

movement balanced by the forward movement of the balloon.

Rockets produce the force that moves them forward by burning their fuel inside a

chamber in the rocket and then expelling the hot exhaust that results. Rockets carry

their own fuel and the oxygen used for burning their fuel. In liquid-fueled rockets, the

fuel and oxygen-bearing substance (called the oxidizer) are in separate compartments.

The fuel is mixed with the oxygen and ignited inside a combustion chamber. The

rocket, like the balloon, has an opening called a nozzle from which the exhaust gases

exit. A rocket nozzle is a cup-shaped device that flares out smoothly like a funnel

inside the end of the rocket. The nozzle directs the rocket exhaust and causes it to

come out faster, increasing the thrust and efficiency of the rocket.

Figure 1.6 Rocket Propulsion


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Exercises

1. Answer the following questions:

a. Two persons were inside a car, the driver who wore his seat belt and a passenger next

to him, who did not wear his seat belt. The car was running at a very high speed, then

suddenly the break was applied; the passenger was thrown out through the wind

screen as result while the driver remained in his seat. Applying the Newtons laws o f

motion explain what happened to the two persons inside the car .

b. Sand falls gently at a constant rate of 50 g/s onto a horizontal belt moving steadily at

40 cm/s. Find the force on the belt in newtons exerted by the sand.

2. Apply the Newtons laws of motion and explain the transition of the forces and acceleration

of a man travelling in an elevator


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EXAMPLE 1

A body of mass 2 kg undergoes a constant horizontal acceleration of 5 m/s2. Calculate the

resultant horizontal force acting on the body. What will be the resultant force on the body when

it moves with a uniform velocity of 10m/s?

Solution:

Data Given

Mass of body (m) = 2 kg

Acceleration (a) = 5 m/s2

Required: (i) Resultant Horizontal force (f) acting on the body

(ii) Resultant Force (f) if body moves with uniform velocity of 10 m/s

Resultant Horizontal Force

From Newtons Second Law, F = ma

Mass=2kg


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F = 2 x 5 = 10N

Therefore, the resultant horizontal force on the body = 10N

Resultant Horizontal Force moving with Uniform Velocity

If the velocity is uniform,

Initial velocity (U) = Final velocity (V) so that

Acceleration (a) = 0

F = 2 x 0 = 0

Therefore, the resultant horizontal force on the body = 0 N.

EXAMPLE 2

A car of mass 600 kg, moving with a forward acceleration of 5 m/s2 is acted upon by a constant

resistive force of 1000 N. Calculate the force exerted from the engine to maintain this forward

acceleration.

Solution:

Data Given


Acceleration (a) = 5 m/s2

Resistive force (f0) = 1000 N

Required: Force exerted from engine of car (f)


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Let F be the force of the engine, Resultant force acting on the car will be equal to,

F 1000.

F 1000 = ma.

F 1000 = 600 x 5

F 1000 = 3000

F = 3000 + 1000 = 4000N

Therefore, F = 4000N = 4KN

Force exerted from the engine = 4KN.

EXAMPLE 3

A bullet of mass 0.045 kg is fired from a gun of mass 9 kg, the bullet moving with an initial

velocity of 200 m/s. Find the initial backward velocity of the gun.

F0 = 1000N

Mass = 600kg

Mass = 9kg Mass = 0.045kg

u = 200 m/s


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Solution:

Data Given

Mass of bullet (mb) = 0.045 kg

Mass of gun (mg) = 9 kg

Initial velocity of bullet (ub) = 200 m/s2

Required: Initial velocity of gun (ug)

Let Fg stands for the force action of the gun, mg the mass of the gun and ag the acceleration of the gun and,

Fb, mb and ab represent that of the bullet.

Fb = mb ab

ab = (200 0) / t = 200/t

Fb = 0.045kg x 200/t

Fb = 9/t

Fg = mg ag

Fg = 9(v/t)

According to Newtons third law, Fb = Fg

9/t = 9v/t

v = 1 m/s

Therefore the initial backward velocity of the gun = 1m/s.


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EXAMPLE 4

A body of mass 5 kg is to be given an acceleration of 20 m/s2. Calculate the force required when

the acceleration is vertically upwards. Take g = 10 m/s2.

Solution:

Data Given

Mass of body m = 5 kg

Needed acceleration (a) = 20m/s2.

Acceleration due to gravity (g) = 10m/s2.

Let F be the force required, expressed in newtons and applied vertically upwards in the direction

of acceleration.

The weight of the body mg acts downwards. Therefore, the resultant force on the body upwards

is:

F mg

Therefore, the equation of motion is F mg = ma

We have

F 5 x 10 = 5 x 20

F = 50 + 100

F = 150 N


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Exercises

3. A car of mass 500kg accelerates at 2m/s2. What is the force acting on the car?

4. Find the acceleration of an object of mass 200g when a force of 0.12N acts on it. If the object

is accelerated from rest, find the distance travelled in min.

5. A rocket expels gas at the rate of 0.5 kg/s. If the force produced by the rocket is 100 newtons,

what is the velocity with which the gas is expelled?


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1.2 IMPULSE AND MOMENTUM

IMPULSE: Impulse of a body is defined as the product of force (F) that produced momentum

change in the body and the time taken for the change to occur. That is,

Impulse = force x time (Ns)

As earlier defined,

Momentum = mass x velocity (kgm/s)

Recall:

From Newtons second law:

t

u) - v( m =F

u) - m(v t =F

The expression on the left hand side (Ft) defines impulse of a body.

Where as the expression on the right hand side (m (v u)) defines the change of momentum of a

body.

From the above equation, it can be seen that impulse of a body is the same as the

momentum change of the body.

Therefore, it could be stated that Ns is the same as kgm/s.

In other words, impulse of a body is as a result of the change in momentum of the body.


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There are many examples of momentum change caused by forces of action and reaction. The

reaction force was put to a very useful purpose by Whittle when he invented the jet aeroplane.

Here a jet of very hot gases issues from behind the aeroplane; the speed and mass per second of

the gas are so high that considerable momentum is imparted to the stream of gas (figure 1.2.1

(a)). An equal and opposite momentum is imparted to the aeroplane, which undergoes a forward

thrust. The same principle is used in rockets for launching satellites. Jets of hot gas issue

downwards; an equal opposite momentum is given to the rocket, which then moves upwards.

Sprinklers on lawns throw out water in one direction, thus giving it momentum, and themselves

rotate in the opposite direction owing to the equal and opposite momentum (figure 1.2.1. (b)).

LAW OF CONSERVATION OF MOMENTUM

Newtons second and third laws enable us to formulate an important conservation law known as

the law of conservation of momentum.

(a) (b)

Figure 1.2.1 Forces of Reaction


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It can also be stated as follows:

By closed or isolated system, it is meant that the system where no external force act.

Let u1 and u2 and v1 and v2 be the initial and final velocities of two colliding bodies of masses m1

and m2. The conservation law can be stated as

m1u1 + m2u2 = m1v1 + m2v2

EXAMPLE 1

A body of mass 5kg moving with a speed of 30m/s is suddenly hit by another body moving in

the same direction, thereby changing the speed of the former body to 60m/s. What is the impulse

received by the first body?

Solution:

Data Given


Initial velocity of body (u) = 30 m/s

Final velocity of body (v) = 60 m/s

In a system of colliding objects the total momentum is conserved, provided there is no net external force acting on the system.

The total momentum of an isolated or closed system of colliding bodies remains constant.

Thus if two or more bodies collide in a closed system, the total momentum before collision is equal to the total momentum after collision.


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Impulse = Change in momentum

= mv mu

= 5(60 30)

= 5 x 30

Therefore, impulse received by the first body = 150Ns.

EXAMPLE 2

A player hits a ball of mass 0.3kg which was moving eastwards with a velocity of 10m/s, causing

it now to move with a velocity 15m/s westwards. The force of the blow acts on the ball for 0.01s.

Calculate the average force exerted on the ball by the player.

Solution:

Data Given

Mass of ball (m) = 0.3kg

Initial velocity (u) = -10 m/s (eastwards assumed negative direction)

Final velocity (v) = 15 m/s (westwards assumed positive direction)

Time force acts on ball (t) = 0.01 s

Mass = 5kg

u=30m/s v=60m/s

Mass = 5kg


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Required: Calculate average force (f).

m = 0.3kg

u = -10m/s

v = 15m/s

t = 0.01s

a = (v u)/t

a = (15 - -10)/ 0.01

a = 2500m/s

F = ma = 0.3 x 2500

F = 750N

Therefore, the average force exerted on the ball by the player = 750N (westwards)

EXAMPLE 3

Object A of mass 20kg moving with a velocity of 3m/s makes an head-on collision with object

B, mass 10kg moving with a velocity of 2m/s in the opposite direction. If A and B stick together

after the collision, calculate their common velocity v in the direction of A.

Solution:

Data Given

Mass of object A (mA) = 20kg


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Mass of object B (mB) = 10kg

Initial velocity of object A (uA) = 3m/s

Initial velocity of object B (uB) = 2m/s

Required: Common velocity (v)

Using mAuA + mBuB = v (mA + mB)

We have

20 x 3 + 10 x (-2) = v (20 + 10)

60 20 = 30v

v = sm

34

Therefore their common velocity = sm

34

A B A

B

uA = 3 uB = 2 m/s v


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Exercises

1. Two identical free-running trolleys are on a smooth horizontal runway. One trolley is at rest

and the other approaches it at a constant speed of 20 cm/s. After the impact the two trolleys

together move at 10 cm/s. Explain this result, stating the physical principal involved.

2. A ball of mass 100 g falls from a height of 3 m on to a horizontal surface and rebounds to a

height of 2 m. Calculate the change in momentum of the ball.

3. Define the impulse of a force. If a force of 6.0 N acts on a body for five seconds, what is the

change in momentum?

4. Explain why the velocity of a recoiling gun is lower than that of the ejected bullet.


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1.3 KINETIC ENERGY

Kinetic energy is the energy possessed by an object, resulting from the motion of that object.

Kinetic energy is energy due to motion; in contrast, potential energy is the energy of a stationary

object due to level or position. The magnitude of the kinetic energy depends on both the mass

and the velocity of the object according to the equation

Ek = mv2

Where m is the mass of the object and

v2 is its velocity multiplied by itself.

A fast-moving cricket ball which crashes into a window pane hurls splinters of glass through big

distances. A moving train coming to rest at a station may compress powerful springs at the

buffers. Thus, since it can do work, these examples show that a moving object has energy. It is

this energy that is called kinetic energy.

Figure 1.31 Energy of a Pendulum


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Energy of a Pendulum

A moving pendulum changes potential energy into kinetic energy and back again. When

the bob (weight on the end of string) is first released, it has potential energy due to its height, but

no kinetic energy since it is not yet moving. As the bob accelerates downward, potential energy

is traded for kinetic. At the bottom of its swing, the bob has no potential energy since it cannot

fall any further. The bob is moving quickly at this point since all of its former potential energy

has been transformed into kinetic energy.

EXAMPLE 1

A girl of mass 30kg is running with a speed of 4m/s. What is her kinetic energy?

Solution:

Given Data

Mass of girl (m) = 30kg

Speed of movement (v) = 4m/s.

Mass = 30kg

V = 4 m/s


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Required: Kinetic energy.

Ek = mv2

Ek = x 30 x 42

Ek = 240 J

Therefore, the kinetic energy of the boy = 240J

EXAMPLE 2

A bullet of mass 40g is moving with a speed of 216km/h. Calculate its kinetic energy.

Solution:

Given Data

Mass of bullet (m) = 40g

Speed of movement (v) = 216km/h.

Required: Kinetic energy.

First convert all units to SI units.

Mass = 40g

V = 216 km/h


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m = 40g = 40/100kg = 0.04kg

216km/h = 60 x 60

1000 x 216 m/s = 60m/s

Ek = x 0.04 x 602

Ek = 72 J

Therefore, the kinetic energy of the bullet is 72 J

EXAMPLE 3

A ball of mass 2kg falls from rest from a height of 200m. Calculate its kinetic energy after

falling a distance of 50m. (Neglect air resistance and take g = 10m/s2). See figure below.

Solution:

Given Data

Mass of ball (m) = 2kg

Height above ground (h) = 100m


Required: Kinetic energy after falling a distance of 50m.

Figure Example 3

150m

50m


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v2 = 2 x 10 x 50

v2 = 1000

Ek = mv2 = x 2 x 1000

Ek = 1000 J = 1 kJ

Therefore, the total kinetic energy at 50m free fall = 1 kJ

EXAMPLE 4

A body of mass 2kg falls from rest through a height of 20m and comes to rest having

penetrated a distance of 0.5m into sandy ground. Calculate the average force exerted by the

sand in bringing the body to rest. (Take g = 10m/s2).

Solution:

Given Data

Mass of body (m) = 2kg

Height above ground (h) = 20m.

Distance penetrated into ground (s) = 0.5m


Required: Average force exerted by sand in bringing body to rest.

If u is the velocity on reaching the ground, the conservation of energy gives:

mu2 = mgh


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= 2 x 10 x 20

= 400J

This kinetic energy is converted into work which is the done in penetrating a distance s = 0.5m

with an average force,

= F x s = 400

F x 0.5 = 400

F = 800N

EXAMPLE 5

A stationary sledge of mass 4 kg is acted on by a force of 8 N for 5 seconds. Calculate the energy

of the sledge.

Solution:

Given Data

Force (F) = 8 N.

Time (t) = 5 s.

Mass (m) = 4 kg.

Initial velocity (u) = 0 (stationary).

Required: Energy.

Acceleration (a) = Force / Mass

a = 8/4

a = 2 m/s2.

Velocity (v) in 5s = u + at


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v = 2 x 5

v = 10 m/s

K.E. = mv2.

K.E. = x 4 x 102

Kinetic energy = 200 J.

Exercises

1) A body of mass 0.5 kg, and initially at rest, is subjected to a force of 2 N for 1 second.

Calculate:

a) The change in momentum of the body during the second.

b) The change in kinetic energy of the body during the second.

2) Niagara falls are 50 m high. Calculate the potential energy of 5 kg of water at the top,

relative to the bottom. What is the kinetic energy of this water just before it reaches the

bottom, and what happens to the energy after the water reaches the bottom?

3) Calculate the kinetic energy of a trolley of mass 80 kg moving with a velocity of 6 m/s.

4) A ball of mass 100 g falls from a height of 3 m on to a horizontal surface and rebounds to

a height of 2 m. Calculate the change:

a) In momentum.

b) In kinetic energy.


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1.4 KINEMATICS OF POINTS

Kinematics is a branch of physics that deals with the study of motion of a body or system

without reference to force and mass.

Motion involves a change of position of a body over a period of time.

Types of Motion

There are four types of motion:

1. Random Motion This motion has no pattern. It occurred in a disorder fashion, as in the

case of gaseous particles or a feather falling from the sky.

2. Translational Motion Any movement upwards, downwards or any other direction that

involves no rotation is referred to as a translational motion.

3. Rotational Motion This involves motion about a circular path about a center or an

axis, such as the motion of a spinning wheel or a rotating fan.

4. Oscillatory Motion This is a to and fro motion, as in the case of a swinging pendulum.

Motion in a Straight Line

Four parameters are required to describe motion in a straight line. These are: distance or

displacement, speed or velocity, acceleration and time.

Displacement (s) is defined as the distance travelled in a specified direction. The SI unit

of displacement is m.


Page 36 of 122

Speed is defined as the rate of change of distance.

Velocity (u or v) is defined is defined as the rate of displacement or the rate of change of

distance in a specified direction. The SI unit of velocity is m/s. If the rate of change of

displacement is constant, the velocity is said to be uniform.

Acceleration (a) is defined as the rate of change of velocity. The SI unit of acceleration is

m/s2. If the rate of change of velocity is constant the acceleration is said to be constant.

EQUATIONS OF UNIFORMLY ACCELERATED MOTION

1. V = U + at

2. S = Ut + at2

3. V2 = U2 2aS

Where V = final velocity,

U = initial velocity,

a = acceleration

t = time taken

S = distance covered

EXAMPLE 1

A train slows from 108km/h with a uniform retardation of 5m/s2. How long will it take to reach

18km/h, and what is the distance covered?

Solution:

U = 108km/h = sm /60 x 60

10000 x 108 = 30m/s


Page 37 of 122

V = 18km/h = sm /60 x 60

1000 x 18 = 5m/s

a = -5m/s2

Time taken

V = U + at

t = (V U)/a

t = 5-30 - 5

t = 5s

Distance covered

V2 = U2 2aS

S = (V2 U2)/2a

S = 5- x 2

30 - 5 22

S = 87.5m

EXAMPLE 2

A car starts from rest and accelerates uniformly until it reaches a velocity of 30m/s after 5s. It

maintained the velocity reached for 15s and is then brought to rest in 10s with a uniform

retardation. Determine:

a) The acceleration of the car,

b) The retardation,


Page 38 of 122

c) The distance covered after 5s,

d) The total distance covered.

Solution:

The movement of the car can be broken into three stages:

First Stage

u = 0

v = 30m/s

t = 5s

From v = u + at we have

a = (v u) / t

= (30 0) / 5

a = 6m/s2.

Acceleration = 6m/s2.

From v2 = u2 2as we have,

s = (v2 u2) / 2a

s = (302 02) / 2 x 6

s = 75m

Distance covered = 75m

Second Stage

Acceleration (a) is constant = 0


Page 39 of 122

u = 30m/s

t = 15s

From s = ut + at2 we have,

s = 30 x 15

s = 450m.

Third Stage

u = 30m/s

v = 0

t = 10s

From v = u + at we have,

a = (v u) / t

a = (0 30) / 10

a = - 3m/s2.

Retardation = 3m/s2.

From v2 = u2 2as we have,

s = (v2 u2) / 2a

s = (02 302) / 2 x -3

s = 150m

a) Acceleration (a) = 6m/s2 (From first stage)

b) Retardation = 3m/s2. (From third stage)


Page 40 of 122

c) The distance covered after 5s = 75m (From first stage)

d) The total distance covered =

Distance covered in first stage + distance covered in second stage + distance covered in

third stage

= 75m + 450m + 150m

= 675m

Therefore, Total distance covered = 675m.

Using Velocity Time graph

30 m / s

0 5 s 20 s 30 sTIME TAKEN

VELO

CI T

Y

1 st S

tage

2nd

Stag

e

3rd

Sta

ge

a) Acceleration (a) is the slope of the curve at first stage of movement

a = 5

30 0 - 50 - 30 =

a = 6m/s2.

b) Retardation (-a) is the slope of the curve at the third stage of movement


Page 41 of 122

a = 1030-

20 - 3030 - 0 =

a = - 3m/s2.

Retardation = 3m/s2.

c) The distance covered after 5s is the area of the graph at the first stage.

= x 5 x 30

= 75m

Distance covered after 5s = 75m

d) The total distance covered is the area under the graph

= (15 + 30) x 30

675m

The total distance covered = 675m


Page 42 of 122

Exercises

1) A train starts from rest from a station and travels with uniform acceleration 0.5 m/s2 for 20

seconds. It travels with uniform velocity for another 30 seconds, the brakes are then applied

so that a uniform retardation is obtained and the train comes to rest in further 10 seconds.

Sketch the velocity-time graph of this motion. Using your graph, calculate the total distance

travelled by the train.

2) A motor car is uniformly retarded and brought to rest from a velocity of 36 km/m in 5

seconds. Find the retardation and the distance covered during this period.

3) A body is dropped from rest at a height of 80 m. How long does it take to reach the ground?

(g = 10 m/2, ignore air resistance).

4) An object moves in a straight line, starting from rest. There are two stages in the journey:

a) It gains speed uniformly for 20 seconds and attains a speed of 8 m/s.

b) It continues at this speed for further 1.5 seconds.

Draw a sketch-graph of speed against time.

Find:

a) The acceleration in stage (a)

b) The acceleration in stage (b)

c) The total distance moved during stages (a) and (b)


Page 43 of 122

1.5 RESOLUTION AND COMPOSITION OF VECTORS

Essential to a study of structural behavior is knowing the net result of the interaction of several

vector forces acting on a body. This interaction can be studies in terms of the laws of vector

addition. These laws and fundamental postulates are base on experimental observation.

Historically, the first method of adding vector quantities was based on the parallelogram law. In

terms of force vectors, this proposition states that when the lines of action of two forces intersect,

there is a single force or resultant, exactly equivalent to these two forces, which can be

represented by the diagonal of the parallelogram formed by using the force vectors as sides of the

parallelogram [see figure 1.5.1].

In general, a resultant force is the simplest force system to which a more complex set of forces

may be reduced and still produce the same effect on the body acted upon.

A graphic technique for finding the resultant force of several force vectors whose lines of action

intersect is illustrated in figure 2. The individual vectors, drawn to scale, are jointed in tip to tail

fashion. The order of combination is not important. Unless the resultant force is zero, the force

F1

F2

(a)

F1

F2

R

(b)

Figure 1.5.1


Page 44 of 122

polygon thus formed does not form a closed figure. The closure line is identical to the resultant

force of the several individual vectors (i.e. the resultant is that vector which would extend from

the tail of the first vector to the tip of the last vector in the group). The resultant closes the force

polygon. This general technique follows from the parallelogram law.

A process that follows directly from the fundamental proposition of the parallelogram of forces

is that of breaking up a single force into two or more separate forces that form a force system

equivalent to the initial force. This process is usually referred to as resolving a force into

components. The number of components that a single force can be resolved into is limitless.

In structural analysis it is often most convenient to resolve a force into rectangular, or Cartesian,

components. By utilizing right angles, components can be found by using simple trigonometric

functions. When a force F is resolved into components on the x and y axes, the components

become

Fx = Fcos and

F1 F3

F2

(a)

R

F1

F3

F2

(b)

F1

F3

F2 R

(c)

Figure 2


Page 45 of 122

Fy = Fsin

The process can be reversed if Fx and Fy are given and it is desired to know the resultant force F:

2y

2x F F +=F And

X

Y

FF 1-tan =

1.6 RELATIVE VELOCITY AND ACCELERATION

Relative velocity compares the velocity of one moving object in relation to the velocity of

another moving object, so that their velocities combined can result in higher or lower velocity

value. Relative velocity changes the velocity of a moving object by comparing it with another

moving object

On the other hand, acceleration results in a change in the velocity of a moving object, by the

action of the object itself.

In Comparison, both relative velocity and acceleration results in changes in the values of

velocities, but while relative velocity results in change in velocity by comparing with another

velocity, acceleration result in change in velocity by the action of the body itself.

1.7 REPRESENTATION OF VECTORS

A vector can be represented by a line in a diagram, say OA in figure 3. The magnitude of the

vector is represented by the length of the line. We can refer to the vector as OA, meaning that the

vector has magnitude equivalent to the length OA and acts from O to A. A vector of 40 units

may be represented on paper by any convenient length, say 5cm, so that 1cm represents 8 units.

The direction of the vector OA is represented by the angle which it makes with a given line,

say OX. The sense of the vector, that is form O to A rather than form A to O, is shown on the

diagram by an arrow.


Page 46 of 122

UNIT VECTOR: A vector, whose magnitude is unity, is known as a unit vector.

EQUAL VECTORS: The vectors, which are parallel to each other and have the same direction

(i.e. same sense) and equal magnitude, are known as equal vectors.

LIKE VECTORS: The vectors, which are parallel to each other and have sense but unequal

magnitude, are known as like vectors.

ADDITION OF VECTORS: Consider two vectors PQ and PS, which are required to be added

as shown in figure 1.7.2 (a). Take a point A, and draw line AB parallel and equal in magnitude to

the vector PQ to some convenient scale. Through B, draw BC parallel and equal to vector RS to

the same scale. Join AC which will give the required sum of vectors PQ and RS as shown in

figure 1.7.2 (b).

This method of adding the two vectors is called the Triangle Law of Addition of Vectors.

Similarly, if more than two vectors are to be added, the same may done first by adding the third

vector to the resultant of the first two and so on. This method of adding more than two vectors is

called Polygon Law of Addition of Vectors.

O

A

X

40 units

5cm

Figure 1.7.1

Figure 1.7.1 (a) Figure 1.7.1 (b)


Page 47 of 122

PARALLELOGRAM LAW OF VECTORS: It states, If two vectors acting simultaneously

on a particle be represented, in magnitude and direction, by the two adjacent sides of a

parallelogram, their resultant may be represented, in magnitude and direction, by the diagonal of

the parallelogram passing through the point of their intersection.

TRIANGLE LAW OF VECTORS: It states, If two vectors acting simultaneously on a

particle be represented in magnitude and direction, by the two sides of a triangle taken in order,

their resultant may be represented, in magnitude and direction, by the third side of the triangle

taken in opposite order.

POLYGON LAW OF VECTORS: It states, If a number of forces acting simultaneously on a

particle be represented in magnitude and direction by the sides of a polygon taken in order, their

resultant may be represented, in magnitude and direction, by the closing side of the polygon

taken in opposite order.

EXAMPLE 1

A foolish boy removes a nail from a vertical wall by pulling on a string attached to the nail in a

direction 300 to the wall. If the tension in the string is 10 N, calculate the magnitude of the force

which is not effective in removing the nail and the effective force used in pulling out the nail. Is

it possible to remove the nail this way if the minimum force required is 7 N? Why is the foolish?

Solution:

300

Nail

T = 10 N

10 sin 300

10 cos 300


Page 48 of 122

From the diagram, the component of the force parallel to the wall = 10 cos300 = N 35 ; this

force is not effective.

The component perpendicular to the wall is the one which is effective in removing the nail. It has

a value of 10 sin 300 = 5 N.

The minimum force required is 7 N. Since this value is less than 7 N he cannot remove the nail

this way as long as the tension remains 10 N and the angle of inclination to the wall is less than

300.

The boy is foolish because if he had applied the of 10 N perpendicular to the wall the nail would

be removed easily.

EXAMPLE 2

A car travelling on a straight road at 100 km/h passes a bus going in the same direction at 60

km/h; calculate:

a) The velocity of the car relative to the bus.

b) The velocity of the car relative to the bus if the bus is now moving in opposite direction

to the car with the same velocity.

Solution:

(a) The velocity of the car relative to the bus =

(-60 + 100) km/h

= 40 km/h.

(b) The velocity of the car relative to the bus =

(60 + 100) km/h

= 160 km/h.


Page 49 of 122

EXAMPLE 3

a) A man travels 7.0 km due north then 10.0 km east. Find the resultant displacement.

b) If he travels 7.0 km 300 east of north then 10.0 km east, find the resultant displacement.

c) If he travels 7.0 km 400 west of north then 10.0 km east, find the resultant displacement.

Solution:

a)

R = 22 10 7 +

R = 12.4 km

tan = 7/10 = 0.7

= 350.

R = 12.4 km, N 550E

b)

N

Resultant

= 12.2 km

350

550

7.0 km

10.0 km

7.0 km

10.0 km

300 Resultant

= 14.8 km

N


Page 50 of 122

R2 = 72 + 102 - 2 x 7 x 10 x cos (30 + 90)

R2 = 149 + 70

R2 = 219

R = 14.8 km

sin 7

120 sin8.14 =

14.8

7sin120 sin =

= 240

R = 14.8 km N 660E

c)

R2 = 72 + 102 2 x 7 x 10 x cos 50

R2 = 149 90

R2 = 59

R = 7.7 km

10.0 km N

Resultant

= 7.7 km 7.0 km 400


Page 51 of 122

sin 7

50 sin7.7 =

7.7

7sin50 sin =

= 440

R = 7.7 km, N 440E

Exercises

1. Find the resultant of the system of forces given in the figure.

450

6 N

4 N

2 2 N


Page 52 of 122

2.1 LOAD

Load could be defined as anything that has the tendency of generating internal stresses in a

structure. The weights of objects supported on other object constitute loads on the supporting

object. If the supporting object in turn is supported by another object, then its weight and the

weights of all objects its supports constitute load on the object that support them all.

Structures are designed and constructed to support weight of other objects. The effect of winds

on a structure also constitutes a load.

The loads imposed on a building are classified as either dead or live.

Dead loads include the weight of the building itself and all major items of fixed

equipment. Dead loads always act directly downward, act constantly, and are additive

from the top of the building down.

Live loads include wind pressure, seismic forces, vibrations caused by machinery,

movable furniture, stored goods and equipment, occupants, and forces caused by

temperature changes. Live loads are temporary and can produce pulsing, vibratory, or

impact stresses.

In general, the design of a building must accommodate all possible dead and live loads to

prevent the building from settling or collapsing and to prevent any permanent distortion,

excessive motion, discomfort to occupants, or rupture at any point.


Page 53 of 122

The plate and its content

constitute load on table.

Likewise the legs of the table

exert load action on the floor

or surface on which it is

placed.

Figure 1: Load on a table

The weight of the head constitute load on the arms that

supports it.

The weights of the arms and head they support constitute

load on the arm of chair that supports them.

The weight of the person on the surface of the chair

constitute load on the chair.

The toes of the person constitutes load on the surface they

were placed.

The legs of the chair exert load action on the surface that supports them.

Figure 2: Load on a chair


Page 54 of 122

Figure 3: Tension and Compression

2.2 TENSION AND COMPRESSION

Tension

A body is said to be in tension if the loads acting on it tends to stretch or elongate it.

Compression

A body is said to be in compression if the loads acting on it tends to shorten or crush it.

The figure illustrated the action of

tension and compression on a

prismatic bar.

The first diagram shows the body in

its original state before subjected to

loading.

The second diagram shows the body

under tension. The body elongates

under the action of tension.

The third diagram shows the body under compression. The body shortens under the

action of compression.

An example of tension is provided by the rope attached to a crane hook, and of compression the

leg of a table. In each case the load consist of two equal and opposite forces acting in line and

tending to fracture the member. The forces on the crane rope are the load being raised at the

other end and the pull of the winding gear at the other end, and on the table leg a portion of the

table weight on top and the reaction of the ground underneath.


Page 55 of 122

EXAMPLE 1

A prismatic bar of circular section is loaded by tensile forces P = 85kN figure EX.1. The bar has

length L = 3.0m and diameter d = 30mm. It is made of aluminium with modulus of elasticity E =

70kN/mm2 and Poissons ratio v = 31 . Calculate the elongation and the decrease in diameter

d of the bar.

Solution:

The longitudinal stress in the bar can be obtained from the equation

AP =

= 4

3010 x 852

3

= 120.3N/mm2.

The axial strain may be obtained from Hookes Law,

= E

= 310 x 703.120

P = 85 kN

P = 85 kN L = 30m

d = 30mm

Figure EX. 1


Page 56 of 122

= 0.001718

The total elongation is

= L

= 0.001718 x 3 x 103

= 5.15mm

The lateral strain is obtained from Poissons Ratio

lateral = -v

= - x 0.001718

lateral = -0.0005726

The decrease in diameter equals numerically the lateral strain times the diameter

d = lateral x d

0.0005726 x 30mm

d = 0.0172mm

EXAMPLE 2

A steel rod 1 m long and 20 mm x 20 mm in cross-section is subjected to a tensile force of 40

kN. Determine the elongation of the rod, if modulus of elasticity for the rod material is 200

kN/mm2.


Page 57 of 122

Solution:

Given Data

Length (L) = 1 m = 1 x 103 mm;

Cross-sectional area (A) = 20 x 20 mm2. = 400mm2.

Tensile force (P) = 40 kN = 40 x 103 N.

Modulus of Elasticity (E) for the material = 200 kN/mm2. = 200 x 103 N/mm2.

Strain () = ( )(E) Modulus ' sYoung

Stress

= AEP

= LL

Therefore, L = L

L = AEPL

= ( ) ( )( )333

10 x 20 x 40010 x 1 x 10 x 40

L = 0.5 mm.

EXAMPLE 3

A load of 5 kN is to be raised with the help of a steel wire. Find the minimum diameter of the

steel, if the stress is not to exceed 100 N/mm2.


Page 58 of 122

Solution:

Given Data

Load (P) = 5 kN = 5 x 103 N.

Stress () = 100 N/mm2.

Let d = Diameter of the wire in mm.

We know that stress in the steel wire (),

AP 100 =

23

2

3 10 x 6.366 (d) x

4

10 x 5 100d

==

100

10 x 6.336 d3

2 = = 63.66

d = 7.98 mm say 8 mm.


Page 59 of 122

Exercises

1) A hollow steel tube 3.5 m long has external diameter of 120 mm. In order to determine the

internal diameter, the tube was subjected to a tensile load of 400 kN and extension was

measured to be 2 mm. If the modulus of elasticity for the tube material is 200 kN/mm2,

determine the internal diameter of tube.

2) A straight bar of 500 mm length has its cross-sectional area of 500 mm2. Find the magnitude

of the compressive load under which it

3) A 25 mm square-cross-section bar of length 300mm carries an axial compressive load of 50

kN. Determine the stress set up in the bar and its change of length when the load is applied.

Take E for bar as 200 kN/mm2.

4) A steel tube, 25 mm outside diameter and 12 mm inside diameter, carries an axial tensile load

of 40 kN. What will be the stress in the bar? What further increase in load is possible if the

stress in the bar is limited to 225 MN/m2?


Page 60 of 122

2.3 STRESS AND STRAIN

Stress

Stress is defined as the internal resistance developed in a body when subject to loading.

Whenever a body is loaded, the fibers or particles that make up the body will tend to reorient

themselves in such a way that equilibrium is maintained with the applied load. These movements

will tend to distort the original shape of the body. The degree to which this distortion is noticed

depends on the strength of the material that makes up the body and the value of the applied load.

This internal resistance that tends to distort the body under load is what is referred to as Stress.

Stresses which are normal to the plane on which they act are called direct stresses, and are either

tensile or compressive.

Normal stress is represented by the symbol, And the S.I. unit of stress is, N/m2

Figure 1: Stresses in a material


Page 61 of 122

Across any section such as XX of the member represented in figure 1, the total force carried

must equal the load P. This is distributed among the internal forces of cohesion, which are called

stresses.

If the member is imagined cut through the section XX (figure 1), each portion is in equilibrium

under the action of the external load P and the stresses at XX.

The force transmitted across any section, divided by the area of that section, is called the

intensity of stress or more simply the stress (). If it is assumed that the load is uniformly

distributed over the section, then,

= AP

Where A is the cross-sectional area.

Strain

Strain is defined as the measure of deformation produced in a member subjected to loading. As

mentioned earlier, internal resistance of bodies under load will tend deform the material as

equilibrium is being maintained, the measure of this deformation is what is referred to as strain.

Direct stresses produce direct strain. Direct strain is represented by the symbol,

Direct strain () is defined as the ratio

(L)length Original

)(length in Change L

= LL

Where L is the change in length and

L is the original length.


Page 62 of 122

2.4 HOOKES LAW

Hookes law states that within the limit of proportionality strain is proportional to the stress

producing it.

Hookes law can be represented as,

Constant )( )( Stress =

Strain

It may be noted that Hookes Law equally holds good for tension as well as compression.

2.5 MODULUS OF ELASTICITY (Youngs Modulus)

Within the limit where Hookes law is obeyed, the ratio between stress and strain defines the

Modulus of Elasticity. The symbol for Modulus of Elasticity is E and its S.I. unit is N/m2.

That is,

= E

E is a constant of proportionality known as modulus of elasticity or Youngs Modulus

.Numerically, it is that value of tensile stress, which when applied to a uniform bar will increase

its length to double the original length if the material of the bar could remain perefectly elastic

throughout such an excessive strain.

EXAMPLE 1

A tie bar on a vertical pressing machine is 2m long and 4cm diameter. What is the stress and

extension under a load of 100kN? Take E as 205 kN/mm2.


Page 63 of 122

Solution:

4

40100

AP 2

3

==

The stress = 79.6 n/mm2.

Extension = E

LAPL

E =

= 4

d x 205

10 x 2 x 1002

3

mm

The extension = 0.78mm

EXAMPLE 2

A prismatic bar with rectangular cross section (20 x 40mm) and length L = 2.8m is subjected to

an axial tensile force of 70kN (figure Ex.2). The measured elongation of the bar is = 1.2mm.

Calculate the tensile stress and strain in the bar.

Solution:

Assuming that the axial forces at the centroids of the end cross sections, the tensile stress in the

bar can be calculated as thus:

40 x 20

70 AP ==

2.8m

20mm

40mm

Figure EX. 2


Page 64 of 122

Therefore, the tensile stress in the bar = 87.5kN/mm2.

The tensile strain can be calculated as thus:

310 x 2.81.2

L ==

Therefore, the tensile strain in the bar = 429 x 10-6

EXAMPLE 3

Determine the stress in each section of the bar shown in Figure EX. 3 when subjected to an axial

tensile load of 20kN. The central section is 30mm square cross-section; the other portions are of

circular section, their diameters being indicated. What will be the total extension of the bar? For

the bar material E = 210 kN/mm2.

Solution:

Stress = AP =

AreaLoad

Figure EX.3

250mm 100mm 400mm

20mm30mm

15mm

20kN 20kN


Page 65 of 122

Stress in section (1) =

4)10 x (20

10 x 2023-

3

= 6-3

10 x 400 x 10 x 80

Stress in section (1) = 63.66 MN/m2.

Stress in section (2) = 6-3

10 x 30 x 3010 x 20


Stress in section (3) =

4)10 x (15

10 x 2023-

3

= 6-3

10 x 225 x 10 x 80


Now extension of a bar can always be written in terms of the stress in the bar since,

L

=E

= EL

Extension of section (1) = 63.66 x 106 x 9-3

10 x 21010 x 250


Page 66 of 122

= 75.8 x 10-6 m


10 x 21010 x 100

= 10.6 x 10-6 m


10 x 21010 x 400

= 215.6 x 10-6 m

Therefore, total extension = (75.8 + 10.6 + 215.6)10-6 m = 0.302mm


Page 67 of 122

Exercises

1) A brass tube 5 cm diameter, 4 cm bore, and 30 cm long is compressed between two end

washers by a load of 25 kN, and the reduction in length measure is 0.2 mm. Assuming

Hookes Law to apply, calculate Youngs modulus.

2) A hollow cylinder 4 m long has outside and inside diameter of 75 mm and 60 mm

respectively. Find the stress and the deformation of the cylinder, when it is carrying an axial

tensile load of 50 kN. Take E = 100 kN/mm2.

3) A steel wire 1 mm diameter is freely hanging under its own weight. If the extension of the

wire should not exceed 2.5 mm, what should be its maximum length? Take E for the wire

material as 200 kN/mm2 and its specific weight as 78.5 kN/m3.

4) A brass rod 1.5 m long and 20 mm diameter was found to deform 1.9 mm under a tensile

load of 40 kN. Calculate the modulus of elasticity of the rod.


Page 68 of 122

2.6 RELATIONSHIP BETWEEN STRESS AND STRAIN IN TENSION

The relationship between Stress and Strain in tension can be best described by carrying out an

experiment.

The figure above shows a graph of a test carried out on a mild-steel bar.

The test is carried out on a bar of uniform cross-section, usually circular, in a testing machine

which indicates the tensile load being applied. For the very small strains involved in the early

part of the test, the elongation of a measured length (called the gauge length) is recorded by an

extensometer or strain gauge".

The load is increased gradually,

At first the elongation, and hence the strain, is proportional to the load (and hence the

stress). This relation (i.e. Hookes law) holds up to a value of the stress known as the

limit of proportionality. This point is shown as A in figure 1.

Figure 1: Tensile Test


Page 69 of 122

Hookes law ceases to be obeyed beyond this point although the material may still be

in the elastic state, in the sense if the load were removed, the strain would also return

to zero. This point is shown as B in figure 1.

If the material is stressed beyond this point, some plastic deformation will occur, i.e.

strain which is not recoverable if the load is removed. At this point, the material is

said to be permanently set.

The next important occurrence is the yield point C, at which the material shows an

appreciable strain even without further increase in load. With mild steel careful

testing will reveal a drop in load immediately yielding commences, so that there are

two values, known as upper and lower yield points.

After yielding has taken place, further straining can only be achieved by increasing

the load, the stress-strain curve continuing to rise up to the point D. The strain in the

region from C to D is in region of 100 times that from O to C, and is partly elastic by

mainly plastic.

At this state (D) the bar begins to form a local neck, the load falling off from the

maximum until fracture at E.

Although in design the material will only be used in the range O-A, it is useful to

examine the other properties obtained from the test.

The maximum or ultimate tensile stress is calculated by dividing the load at D by the original

cross-sectional area. Here it should be pointed that the true stress occurring in the necked portion

is much higher than this, and in fact reaches its greatest value at the breaking point, but it is the

stress which a member can stand distributed over its original area which interests the designer.


Page 70 of 122

EXAMPLE 1

The following results were obtained in a tensile test on a mild steel specimen of original diameter

2 cm. and gauge length 4 cm. At the limit of proportionality the load was 80 kN and the

extension 0.048 mm. The specimen yielded at a load of 85 kN, and the maximum load withstood

was 150 kN.

When the two parts were fitted together after being broken, the length between gauge points was

found to be 5.56 cm, and the diameter at the neck was 1.58 cm.

Calculate the Youngs modulus and the stress at the limit of proportionality, the yield stress, and

ultimate tensile stress; also the percentage elongation and contraction.

Solution:

Young modulus

Hookes law is obeyed up to the limit of proportionality.

E = =

AxPL

P = 80 kN = 80000 kN

L = 4 cm = 40 mm

x = 0.048 mm

A = 4

2d = 4202

A = 100 mm2

E = 0.048 x 100

40 x 80000

E = 213000 N/mm2


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Stress at Limit of Proportionality

Aality)proportion oflimit P(at =

10080000 =

= 255 N/mm2

Yield Stress

Apoint) yieldP(at =

10085000 =

= 271 N/mm2

Ultimate Tensile Stress

Aload) P(maximum =

100150000 =

= 478 N/mm2

Percentage Elongation


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Percentage elongation = 100 x Lengthin IncreasengthOriginalLe

0/0

= 100 x 4

4 - 5.56

Percentage elongation = 390/0

Percentage Contraction

Percentage contraction = Area

Areain DecreaseOriginal

=

2

22

21.58 - 2 x

4 x 1000/0

Percentage contraction = 380/0

2.7.1 Limit of Proportionality: - This limit defines the maximum load a material can

support so that the ratio between stress and strain is constant, that is, maximum limit that

the strain in a material will remain a linear function of the stress. Within the limit of

proportionality, Hookes law is obeyed.

2.7.2 Elastic Limit: - The maximum stress that can be applied to a material without the

material becoming permanently deformed. Any additional loading beyond this limit,

irrecoverable deformation (strain) will remain in the material even if the entire load is

removed.

2.7.3 Yield Point: - The point where there is a considerable increase in strain without

corresponding increase in stress. At yield point, the material will exhibit appreciable

deformation with no increase in load.


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2.7.4 Ductility: - This is the ability of material to exhibit appreciable deformation

before failing. Ductile material will show remarkable deformation under load before

fracture.

2.7.5 Brittleness: - This is the ability of material to fail at a very low value of strain.

Brittle materials will fail under load without any noticeable deformation.

2.7.6 Permanent Set: - Permanent set is said to occur in a material when permanent

deformation remains in the material even when the loads that caused the deformation are

entirely removed.

Exercises

1) Mild steel bar deforms continuously under load at a constant load at certain point during a

tensile test. The bar continued to support extra load even after being highly deformed.

Explain the phenomenon involve in this transformation.

2) A material is loaded until a point is reached where deformation in the material is only

partially recovered when the load is removed. What is this transformation referred to and is

there tendency of recovering the deformation through other means?


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2.8 SHEAR STRESS, SHEAR STRAIN, MODULUS OF RIGIDITY AND STRAIN ENERGY

SHEAR STRESS

Shear stresses are stresses that act tangentially or parallel to the surface of the material, unlike normal stresses that always act normal to the surface of the material. Instead of tending to elongate or shorten a material, shear stresses tend to distort the shape of the material in an effort to create a slip zone inside the material, so that one surface will tend to slide over the other.

The symbol used for shear stress is , and its S.I. unit is N/m2.

The average shear stress is obtained by dividing the total shear force V by the area A over which it acts.

AV =

Figure 1a: A body under the action of Shear

X X

V

V

Figure 1b: section X -

V

V

X X


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SHEAR STRAIN

Under the action of shear stresses, a body deforms resulting in shear strains. In order to visualize these strains, it should be noted first that shear stresses have no tendency to elongate or to shorten the element in the x, y and z directions; in other words, the lengths of the sides of the element do not change. Instead, the shear stresses produce a change in the shape of the element, as shown in figure 2b below. The original element is deformed into an oblique parallepiped, and the front face pqrs of the element becomes a rhomboid.

The angles between the faces at points q and s, which were 2 (900) before deformation, are

reduced by a small angle to (2 - ). At the same time, the angles at p and r are increased to (

2

+ ). The angle is a measure of the distortion, or change in shape, of the element and is called the shear strain. The unit of shear strain is the radian.

MODULUS OF RIGIDITY

For many materials, the initial part of the shear stress-strain diagram is a straight line, just as in tension. For this linearly elastic region, the shear stress and stress strain are directly proportional, and we have the following equation for Hookes law in shear:

G = G is the modulus of rigidity

In other words, the modulus of rigidity G, of a material is defined as,

p q

r s Figure2a

q p

s r

Figure2b


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=G

STRAIN ENERGY

The strain energy (U) of a material subjected to loading is defined as the work done by the load in straining the material. When an elastic material deformed under load, if the load is removed, the deformation will disappear if its elastic limit is not exceeded. Obviously the deformation is as a result of the applied load, but the restoration is as a result of energy stored inside the material. This stored energy is what is referred to as the strain energy.

When a tensile or compressive load P is applied to a bar there is a change in length x which, for an elastic material, is proportional to the load (Figure 3). The strain energy (U) of the bar is defined as the work done by the load in straining it.

For gradually applied or static load the work done is represented by the shaded area in figure 3, giving

U = Px --------------------------------------- (1)

To express the strain energy in terms of the stress and dimensions, for a bar of uniform section A and length L, substitute P = A and x = L/E in equation (1), giving

U = x A x L/E

U = AL x 2

2

E ------------------------------------ (2)

P

x Extension

Load

Figure 3


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But AL is the volume of the bar; hence equation (2) can be stated: The strain energy per

unit volume (usually called resilience) in simple tension or compression is 2

2

E

EXAMPLE 1:

The coupling shown in figure EX.1 is constructed from steel of rectangular cross-section and is designed to transmit a tensile force of 50 kN. If the bolt is of 15 mm diameter calculate:

a) The shear stress in the bolt;

b) The direct stress in the plate;

c) The direct stress in the forked end of the coupling

Solution:

(a) The bolt is subjected to double shear, tending to shear it as shown in figure EX.1 (b). There is thus twice the area of the bolt resisting the shear.

Shear stress in bolt = 2AP

= ( )23- 3

10 x 15 x 24x 10 x 50

Figure EX. 1

25 kN

25 kN

50 kN

(b) (a)

50 kN 50 kN


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= ( )23-3

10 x 1510 x 100

= 141.5 MN/m2.

(b) The will be subjected to a direct tensile stress given by

AP=

= 6-3

10 x 6 x 5010 x 50

= 166.7 MN/m2.

(c) The force in the coupling is shared by the forked end pieces, each being subjected to a direct stress

AP=

= 6-3

10 x 6 x 5010 x 25

= 83.3 MN/m2.

EXAMPLE 2:

An axial pull of 20 kN is applied on a steel rod 2.5 m long and 1000 mm2 in cross-section. Calculate the strain energy, which can be absorbed in the rod. Take E = 200 kN/mm2.

Solution:

Given Data

Axial pull on the rod (P) = 20 kN = 20 x 103 N.

Length of rod (L) = 2.5 m = 2.5 x 103 mm;

Cross-sectional area of rod (A) = 1000 mm2.

Modulus of elasticity (E) = 200kN/mm2 = 200 x 103 N/mm2.


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= AP

= 1000

10 x 20 3

= 440 N/mm2.

Volume of the rod = L x A

= (2.5 x 103) x 1000

= 2.5 x 106 mm3.

Therefore, Strain energy which can be absorbed in the rod,

U = V x 2

2

E

= ( ) ( ) mm-N 10 x 2.5 x )10 x (200 x 2

40 63

2

U = 10 x 103 N-mm = 10 kN-mm

Exercises

1) A rod 1 m long is 10 cm2 in area for a portion of its length and 5cm2 in area for the remainder. The strain energy of this stepped bar is 40% of that of a bar 10 cm2 in area in 1 m long under the same maximum stress. What is the length of the portion 10 cm2 in area?

2) Compare the strain energies of two bars of the same material and length and carrying the same gradually applied compressive load if one is 25mm diameter throughout and the other is turned down to 20 mm diameter over half of its length, the remainder being 25 mm diameter.

3) The vertical load P acting on the wheel of a travelling crane is 53 kN. What is the average shear stress average in the 38 mm axle?


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2.9 ANALYSIS OF A COMPOSITE BODY

A composite body is made from two or more materials. Example is reinforced concrete, where

steel and concrete are used together to support load.

Analysis of composite body is quite different in the sense that the different materials have

different load carrying capa

introduction to structural mechanics book

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