Properties of Matter

Course: Diploma

Subject: Applied Science Physics

Unit: II

Chapter: I

ElasticityDeformationAll materials are deformable.

There are two Deformations

1. Elastic deformation

As we have hinted at before, elastic deformation is a deformation which is reversible.

2. plastic deformation

Plastic deformation is characterised by a permanent deformation.

Elastic deformation

As we have hinted at before, elastic deformation is a

deformation which is reversible.

There is no permanent deformation of the substance,

meaning that it will return to its original dimensions.

The stress could be tensile or compressive.

1

Elastic deformationThe particles that make up the substance are each being

displaced a little from their nearest neighbours and then

returning to their original positions.

The particles are, of course, in constant thermal motion

about fixed positions but we shall just deal with the fixed

positions.

Before the particles are put under compression, they will

be in fixed positions where the net forces of attraction and

repulsion are zero, as indicated in the diagram above.

Elastic deformation

The particles that make up the substance are each being

displaced a little from their nearest neighbours and then

returning to their original positions.

If a compressive force is applied from the right in our

diagram, each particle will be displaced one unit to the

left, relative to its nearest neighbour.

F

Elastic deformation

The particles that make up the substance are each being

displaced a little from their nearest neighbours and then

returning to their original positions.

And, of course, a bigger force causes even more

compression.

The change in position of each particle is clearly shown

by the yellow lines.

The compression, Δl, is also shown.

F

2F

Δl

Elastic deformation

The particles that make up the substance are each being

displaced a little from their nearest neighbours and then

returning to their original positions.

And, of course, the process is elastic, so the particles will

always return to their original equilibrium positions.

2

Elastic deformation

The particles that make up the substance are each

being displaced a little from their nearest neighbours

and then returning to their original positions.

If you are confirming elastic behaviour

with a sample, then to see if the sample

returns to its original shape and size when

unstressed.

3

Elastic deformation

Materials which exhibit elastic properties include:

When stressed up to, and

sometimes beyond, the limit

of proportionality, solid

metals will exhibit elastic

behaviour.

Most metals have a fairly

large elastic range. 4

Elastic deformation

Materials which exhibit elastic

properties include:

Thermoplastics are a group of

plastics that can be melted by the

application of heat and then

cooled back to a solid repeatedly.

5

6

Elastic deformation

Materials which exhibit elastic properties include:

Many composites are

manufactured with their

elastic behaviour in mind.

These include carbon fibre

and plywood which have

both been used in the

manufacture of aircraft

fuselages.

7

8

Plastic deformation

Plastic deformation is characterised by a permanent

deformation.

When a sample is unloaded, it

does not return to its original

length or shape.

It is likely to be thinner, if tensile

stresses have been applied and

wider, if compressive stresses

have been applied.

When unloading, the contraction

is likely to be proportional to the

decrease in load.

9

Plastic deformation

Substances that exhibit plastic deformation include:

After showing elastic behaviour

for moderate loads,

thermoplastics then show a

plastic region before they finally

break.

Substantial necking can occur. 10

Plastic deformation

Substances that exhibit plastic deformation

include:

After showing elastic behaviour for

moderate loads, many metals then

show a plastic region before they

finally break.

11

12 13

Plastic deformation

Plastic deformation cannot occur due to molecules

being pulled a “little” way apart so they are still next to

their nearest neighbour.

If that was the case, they would just slip back to their

original positions when the external force (stress) was

removed and the substance would return to its original

shape.

What does happen is that one plane of particles slips

over the surface of another. There is no return from this

new position so the deformation is permanent.

F

Plastic deformationYou will notice that:

•The sample is now longer than it was

•The sample is now thinner that it was

•The positions of many of the particles, relative to

each other, have changed – this creates the

permanent deformation

Plastic deformation

•Many substances exhibit both plastic and elastic

deformation, depending on the stress applied and the

substance used

•On loading (stressing) the material, it will first show

elastic deformation, then possibly plastic deformation

•When unloaded, the contraction will be in proportion

to the change in load

Restoring Force

A variable force that gives rise to

an equilibrium

The restoring force is a function only of

position of the mass or particle. e.g.

spring.

Opposite force

14

Stress-Strain Relationship

EE -- Young’s modulus

GG -- shear modulus

Hooke’s Law:

DIRECT STRESS

When a force is applied to an elastic body, the body deforms. The way in which the body deforms depends upon the type of force applied to it.

Compression force makes the

body shorter.

A tensile force makes the

body longer

15

A

F

Area

ForceStress

2/ mN

Tensile and compressive forces are called

DIRECT FORCES

Stress is the force per unit area upon which

it acts.

….. Unit is

Pascal (Pa) or

Note: Most of engineering fields used k Pa, M

Pa, G Pa.

( Symbol – Sigma )(

L

LStrain

DIRECT STRAIN

In each case, a force F produces a deformation x. In

engineering, we usually change this force into stress and

the deformation into strain and we define these as follows:

Strain is the deformation per unit of the original length.

The symbol = (EPSILON)

Strain has no unit’s since it is a ratio of length to

length. Most engineering materials do not stretch

very mush before they become damages, so strain

values are very small figures. It is quite normal to

change small numbers in to the exponent for 10-6(

micro strain).

Stress- Strain Curve for Mild Steel (Ductile Material)

Strain

Stress

Plastic state

Of material

Elastic State

Of material

Yield stress Point

E = modulus of

elasticity

Ultimate stress point

Breaking stress point

SHEAR STRAIN:

A

B C

D

/2

A

B

C

D

B’C’

D

’/2

B

A

CB” C’’

D

State of simple

Shear on Block

Total change

in corner

angles +/-

Distortion with

side AD fixed

F

Since

is extremely small,

we can assume

BB” = arc with A as centre ,

AB as radius.

So, =BB”/AB=CC”/CD

Elongation of diagonal AC can be nearly taken as

FC”.

Linear strain of diagonal = FC”/AC

= CC”cos 45/CDsec45

B

A

CB” C’’

D

F

= CC”/2CD = (1/2)

but = /N (we know N= / )

so

= /2N ------(8)

Linear strain ‘’is half the shear strain ‘’.

B

A

CB” C’’

D

F

Modulus of Elasticity:

• Stress required to produce a strain of unity.

• i.e. the stress under which the bar would be

stretched to twice its original length . If the

material remains elastic throughout, such

excessive strain.

• Represents slope of stress-strain line OA.

Value of E is same in Tension & Compression.

=E

A

O

• Hooke’s Law:-

Up to elastic limit, Stress is proportional to strain

=E ; where E=Young’s modulus

=F/A and = / L

F/A = E ( / L)

=FL /AE

E

L

L

L

YOUNG’S MODULUS (E):-

-

Young’s Modulus (E) is defined as the

Ratio of Stress () to strain ().

E = /

= -------------(5)LA

FL

BULK MODULUS (K):--

VA

FV

BULK MODULUS (K):--

Where, V/V

Change in volume=

Original volume

Volumetric Strain=

K = -------------

(6) VA

FV

MODULUS OF RIGIDITY (N): OR

MODULUS OF TRANSVERSE ELASTICITY

OR SHEARING MODULUS

Up to the elastic limit,

shear stress () shearing strain()

= N

Expresses relation between shear stress and shear

strain. /=N;

where

Modulus of Rigidity = N = / = F/Aθ -------------(7)

YOUNG’S MODULUS E = FL/A∆L

K = FV/A∆VBULK MODULUS

MODULUS OF RIGIDITYN = F/Aθ

ELASTIC CONSTANTS

-------------(5)

-------------(6)

-------------(7)

LB

DP

P

L+ ∆LB- ∆B

D- ∆D

POISSONS RATIO:- = lateral contraction per

Unit axial elongation, (with in elastic limit)

L(1+)

B(1-)

D(1-)

= (∆B/B)/(∆L/L);

= (∆B/B)/()

So ∆B = B;

New breadth

B - ∆B = B - B

=B(1 - )

Sims . , New depth

D-∆D= D(1- )

for isotropic materials = ¼ for steel = 0.3

Volume of bar before deformation V= L * B*D

new length after deformation L1=L + ∆L = L + L = L (1+ )

new breadth B1= B - ∆B = B - B = B(1 - )

new depth D1= D - ∆D = D - D = D(1 - )

new cross-sectional area = A1= B(1- )*D(1- )= A(1- )2

new volume V1= V - ∆V = L(1+ )* A(1- )2

AL(1+ - 2 )

Since is small

change in volume = ∆V =V1-V = AL (1-2 )

and unit volume change = ∆V/ V = {AL (1-2 )}/AL

∆V/ V = (1-2)

RELATION BETWEEN ELASTIC

CONSTANTS(A) RELATION BETWEEN E and K

Let a cube having a side L be subjected to

three mutually perpendicular stresses of

intensity

By definition of bulk modulus

K= / v

Now v = ∆V/V = /K ------------------( I

)

x

z

y

The total linear strain for each side

v =/E – (μ/E) – (μ/E)

so ∆L / L = =(/E) *(1-2 μ)-----------(ii)

now V=L3 ∆V = 3 L2 ∆L

∆V/V = 3 L2 ∆L/ L3= 3 ∆L/L = 3 (/E) * (1-2 μ) ------------------(iii)

v

v

L B D

L B D

Equating (i) and (iii)

/K = 3( /E)(1-2 μ)

E = 3 K(1-2 μ) ------------------(9)

(B) Relation between E and N

D

B

A

CB

”C’

’

Linear strain of diagonal AC,

= /2 = /2N --------------------( I )

F

A

B C

D

State of simple shear produces tensile

and compressive stresses along

diagonal planes and

=

Strain of diagonal AC, due to these

two mutually perpendicular direct

stresses

= /E - (- μ/E) = (/E)*(1+μ) ---(ii)

But =

so = ( /E)*(1+μ) ------------------(iii)

From equation (i) and (iii)

/2N = ( /E)(1+μ)

OR

E =2N(1+μ)-------(10)

But E = 3 K (1-2μ)------(9)

Eliminating E from --(9) & --(10)

=(3K - 2N) / (6K +2N)-----(11)

Eliminating μ from –(9) & --(10)

E = 9KN / (N+3K) ---------(12)

(C) Relation between E ,K and N:--

E = 3K (1-2μ) --------(9)

E = 9KN / (N+3K) -------(12)

E = 2N(1+μ) -------(10)

Example : 1

A short hollow, cast iron cylinder with wall

thickness of 10 mm is to carry compressive load

of 100 kN. Compute the required outside

diameter `D’ , if the working stress in

compression is 80 N/mm2. (D = 49.8 mm).

Solution: = 80 N/mm2;

F= 100 kN = 100*103 N

A =(/4) *{D2 - (D-20)2}

as = P/A

substituting in above eq. and solving. D = 49.8 mm

D

d

10 mm

Example: 2

A Steel wire hangs vertically under its weight. What is

the greatest length it can have if the allowable tensile

stress t =200 MPa? Density of steel =80 kN/m3.(ans:-

2500 m)

Solution:

t =200 MPa= 200*103

kN/m2 ;

=80 kN/m3.

Wt. of wire P=(/4)*D2*L*

c/s area of wire A=(/4)*D2

t = P/A

solving above eq. L =2500m

L

Example:3 An aluminum bar 1.8 meters long

has a 25 mm square c/s over 0.6 meters of its

length and 25 mm circular c/s over other 1.2

meters . How much will the bar elongate

under a tensile load P=17500 N, if E = 75000

Mpa.

Solution :- ∆L= ∑PL/AE

=17500*600 / (252*75000) +

17500*1200/(0.785*252*75000) =0.794 mm

0.6 m1.2 m

25 mm sq.sect 25 mm cir..sect

17500 N

Example: 4 A prismatic steel bar having cross

sectional area of A=300 mm2 is subjected to axial

load as shown in figure . Find the net increase in

the length of the bar. Assume E = 2 x 10 5 MPa.

( Ans = -0.17mm)

Solution:

∆L = ∑PL/AE =

= 20000*1000/(300*2x10 5)-15000*2000/(300*2 x10 5)

= 0.33 - 0.5 = -0.17 mm (i.e. contraction)

15 kN

1 m 1 m 2 m

20 kN 15 kN

C B A

2020 C00B15 15A

Example 5. A steel wire 10 m long and 2 mm

in diameter is attached to the ceiling and a

200-N weight is attached to the end. What is

the applied stress?

L

L

A

AF

First find area of wire:2 2(0.002 m)

4 4

DA

A = 3.14 x 10-6 m2

F = 200 N

-6 2

200 N

3.14 x 10 m

FStress

A

Stress =6.37 x 107 Pa

Example 6. A 10 m steel wire stretches 3.08

mm due to the 200 N load. What is the

longitudinal strain?

L

L

Given: L = 10 m; L = 3.08 mm

Longitudinal Strain

3.08 x 10-4

0.00308 m

10 m

LSrain

L

Example 7. The elastic limit for steel is 2.48 x

108 Pa. What is the maximum weight that can

be supported without exceeding the elastic

limit?

L

L

A

AF

82.48 x 10 PaF

StressA

Recall: A = 3.14 x 10-6 m2

P= 2.48 x 108 Pa

F = (2.48 x 108 Pa)(3.14 x 10-6

m2)

F = 779 N

Example 8. The ultimate strength for steel is

4.s89 x 108 Pa. What is the maxi- mum weight

that can be supported without breaking the

wire?

L

L

A

AF

84.89 x 10 PaF

StressA

Recall: A = 3.14 x 10-6 m2

P = (4.89 x 108 Pa)

F = (4.89 x 108 Pa)(3.14 x 10-6 m2)

F = 1536 N

Example 9. In our previous example, the stress

applied to the steel wire was 6.37 x 107 Pa and the

strain was 3.08 x 10-4. Find the modulus of elasticity

for steel.

L

L

7

-4

6.37 x 10 Pa

3.08 x 10

StressModulus

Strain

Modulus = 207 x 109 Pa

This longitudinal modulus of elasticity is called

Young’s Modulus and is denoted by the symbol Y.

Example 10: Young’s modulus for

brass is 8.96 x 1011Pa. A 120-N

weight is attached to an 8-m length

of brass wire; find the increase in

length. The diameter is 1.5 mm.

8 m

L

120 N

First find area of wire:

2 2(0.0015 m)

4 4

DA

A = 1.77 x 10-6 m2

or FL FL

Y LA L AY

8 m

L

120 N

Y = 8.96 x 1011 Pa; F = 120 N;

L = 8 m;

A = 1.77 x 10-6 m2

F = 120 N;

L = ?

or FL FL

Y LA L AY

-6 2 11

(120 N)(8.00 m)

(1.77 x 10 m )(8.96 x 10 Pa)

FLL

AY

L = 0.605 mmIncrease in length:

Example 11. A steel stud (S = 8.27 x 1010Pa) 1 cm

in diameter projects 4 cm from the wall. A 36,000

N shearing force is applied to the end. What is the

defection d of the stud?

d

l

F

2 2(0.01 m)

4 4

DA

Area: A = 7.85 x 10-5 m2

; F A F A Fl Fl

S dd l Ad AS

-5 2 10

(36,000 N)(0.04 m)

(7.85 x 10 m )(8.27 x 10 Pa)d

d = 0.222 mm

Example 12. A hydrostatic press contains 5

liters of oil. Find the decrease in volume of

the oil if it is subjected to a pressure of 3000

kPa. (Assume that B = 1700 MPa.)

/

P PVB

V V V

6

9

(3 x 10 Pa)(5 L)

(1.70 x 10 Pa)

PVV

B

V = -8.82 mLDecrease in V;

Stress → Strain curve diagrams

for some materials

16

Breaking stress

breaking stress, is the stress at which a

specimen fails via fracture.

17

Factor of Safety

The ratio of the maximum safe load to the

maximum allowable design load

Magnitude of the factor of safety varies

depending on the loading conditions and

type of forces induced

,

U

P

stress at failuresafety factor N =

stress when loaded

Ultimate stress

Permissible stress

Cranes

Elastic limit =30x107Nm-2

M=104 kg, g =9.8 ms-2

A=???

Bridges

Mountains

Elastic limit =3x108Nm-2

P=hmax ρg= 3x108Nm-2

ρ =3x103 kgm-3

hmax=????

3

34

WL

bd y

Ultimate Strength

The strength of a material is a measure of

the stress that it can take when in use. The

ultimate strength is the measured stress at

failure but this is not normally used for

design because safety factors are required.

The normal way to define a safety factor is :

stressePermissibl

stressUltimate

loadedwhen stress

failureat stress = factorsafety

• Design uncertainties mean we do not push the limit.

• Factor of safety, N

working

y

N

Often N is

between

1.2 and 4

• Ex: Calculate a diameter, d, to ensure that yield does

not occur in the 1045 carbon steel rod below. Use a

factor of safety of 5.

working

y

N

220,000N

d2 / 4

5

DESIGN OR SAFETY FACTORS

18

REFERENCE BOOKS AUTHOR/PUBLICATION

ENGINEERING PHYSICS S S PATEL (ATUL PRAKASHAN)

MODERN ENGINEERING

PHYSICSA S VASUDEVA

ENGINEERING PHYSICS K. RAJGOPALAN

Image reference links

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9. http://postimg.org/image/sbi19315p

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Image reference links

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