Applide AScience Physics

1. Properties of Matter
Course: Diploma
Subject: Applied Science Physics
Unit: II
Chapter: I

2. Elasticity
Deformation
All 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.

3. 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

4. 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.
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.

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. Plastic deformation
You 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

17. 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

18. 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

19. Stress-Strain Relationship
 E
E -- Young’s modulus
 G
G -- shear modulus
Hooke’s Law:

20. 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

21. A
F
Area
Force
Stress 
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 )(

22. 
L
L
Strain

  
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).

23. 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

24. 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

25. 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

26.  = 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

27. 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 

28. 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

29. YOUNG’S MODULUS (E):-
-
Young’s Modulus (E) is defined as the
Ratio of Stress () to strain ().
E =  / 
= -------------(5)
LA
FL


30. BULK MODULUS (K):--
VA
FV


31. 


 
BULK MODULUS (K):--
Where, V/V
Change in volume
=
Original volume
Volumetric Strain=
K = -------------
(6) VA
FV


32. 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)

33. YOUNG’S MODULUS E = FL/A∆L
K = FV/A∆VBULK MODULUS
MODULUS OF RIGIDITYN = F/Aθ
ELASTIC CONSTANTS
-------------(5)
-------------(6)
-------------(7)

34. L
B
D
P
P
L+ ∆L
B- ∆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- )

35. 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)

36. 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

37. 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

   
  
  
  

38. Equating (i) and (iii)
/K = 3( /E)(1-2 μ)
E = 3 K(1-2 μ)
------------------(9)

39. (B) Relation between E and N
D
B
A
CB
”
C’
’ 
Linear strain of diagonal AC,
 = /2 = /2N --------------------( I )
F




A
B C
D



40. 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)

41. 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)

42. (C) Relation between E ,K and N:--
E = 3K (1-2μ) --------(9)
E = 9KN / (N+3K) -------(12)
E = 2N(1+μ) -------(10)

43. 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

44. 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

45. 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 m
1.2 m
25 mm sq.sect 25 mm cir..sect
17500 N

46. 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 C00 B
15 15A

47. 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
A
F
First find area of wire:
2 2
(0.002 m)
4 4
D
A
 
 A = 3.14 x 10-6 m2
F = 200 N
-6 2
200 N
3.14 x 10 m
F
Stress
A
 
Stress =6.37 x 107 Pa

48. 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
L
Srain
L

 

49. 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
A
F
8
2.48 x 10 Pa
F
Stress
A
 
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

50. 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
A
F
8
4.89 x 10 Pa
F
Stress
A
 
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

51. 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
Stress
Modulus
Strain
 
Modulus = 207 x 109 Pa
This longitudinal modulus of elasticity is called
Young’s Modulus and is denoted by the symbol Y.

52. 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
D
A
 
 
A = 1.77 x 10-6 m2
or
FL FL
Y L
A L AY
  


53. 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 L
A L AY
  

-6 2 11
(120 N)(8.00 m)
(1.77 x 10 m )(8.96 x 10 Pa)
FL
L
AY
  
L = 0.605 mmIncrease in length:

54. 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
D
A
 
 
Area: A = 7.85 x 10-5 m2
;
F A F A Fl Fl
S d
d 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

55. 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 PV
B
V V V
 
 
 
6
9
(3 x 10 Pa)(5 L)
(1.70 x 10 Pa)
PV
V
B
 
  
V = -8.82 mLDecrease in V;

56. Stress → Strain curve diagrams
for some materials
16

57. Breaking stress
 breaking stress, is the stress at which a
specimen fails via fracture.
17

58. 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 failure
safety factor N =
stress when loaded
Ultimate stress
Permissible stress


 

59.  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
3
4
WL
bd y
 

60. 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
loadedwhenstress
failureatstress
=factorsafety 

61. • 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

62. REFERENCE BOOKS AUTHOR/PUBLICATION
ENGINEERING PHYSICS S S PATEL (ATUL PRAKASHAN)
MODERN ENGINEERING
PHYSICS
A S VASUDEVA
ENGINEERING PHYSICS K. RAJGOPALAN

63. Image reference links
1. http://www.imageupload.co.uk/image/5LHC
2-3. http://postimg.org/image/stmtzsmfx/
4. http://s6.postimg.org/klgb1255p/New_Picture_79.p
ng?noCache=1420609995
5. http://postimg.org/image/klgb1255p
6. http://s6.postimg.org/dtku139ul/New_Picture_86.p
ng
7. http://postimg.org/image/5bgftv9nh/
8. http://postimg.org/image/gxvhxvqn1
9. http://postimg.org/image/sbi19315p
10. http://postimg.org/image/7u6q8akrh
11. http://postimg.org/image/arz3o9319

64. Image reference links
12. http://postimg.org/image/ljwo3cihp
13. http://postimg.org/image/c21ih1m71
14. http://postimg.org/image/gqcwyhlzx
15. http://postimg.org/image/o7m4dpbj1
16. http://postimg.org/image/omxe6pxgd
17. http://postimg.org/image/ehyqh5evx
18. http://postimg.org/image/yzjossqzh
19. http://postimg.org/image/ment34fq5