Simple Stress and Strain

Simple Stress and Strain
Lecturer;
Dr. Dawood S. Atrushi

Content
¢ Stress
¢ Strain (Deformation)
¢ Stress-Strain relationship
¢ Hook’s law
¢ Poisson’s ration
¢ Thermal or Temperature stress and
strain
Wednesday, 2 Strenght of Materials I - DAT October 22, 14

STRESS
Chapter Chapter 2 2 Stress Stress and and Strain
Strain
External and internal forces in a structural member
2.0 INTRODUCTION
In the first Chapter we discuss the equations of statics, and how to determine the ground
reaction for any structure. The method can also be used to determine the internal loads carried
by the members or parts of a body. We now need to define how these internal loads are
distributed and carried by the material and the deformation they create.
2.1 STRESS (SI&4th p. 22-23, 3rd p. 22-23)
Consider an element of continuous (no voids) and cohesive (no cracks, breaks and defects)
material subjected to a number of externally applied loads as shown in Fig. 2.1a). It is
supposed that the member is in equilibrium.
2.0 INTRODUCTION
In the first Chapter we discuss the equations of statics, and how to determine the ground
reaction for any structure. The method can also be used to determine the internal loads carried
by the members or parts of a body. We now need to define how these internal loads are
distributed ¢ Consider and carried by the material an element and the deformation of they continuous
create.
2.1 STRESS and (SI&cohesive 4th p. 22-23, 3rd p. 22-material 23)
subjected to a
Consider number an element of continuous of externally (no voids) and cohesive (no cracks, breaks material subjected to a number of externally applied applied loads as shown loads.
and defects)
in Fig. 2.1a). It is
supposed that the member is in equilibrium.
Free Body Diagram
Free Body Diagram
'F
n
'Fn
'F
'Ft
F4
F4
F3 F5
'Fn
F3 F5
Cross section: A 'A
Cross section: A 'A
F1
F2
t
t
F1
F2
n
F1
F2
(a) (b)
(a) (b)
Fig. 2.1 External and internal forces in a structural member
'Ft
F2
F1
If we now cut this body, the applied forces can be thought of as being distributed over the cut
area A as in Fig. 2.1b). Now if we look at infinitesimal regions 'A, we assume the resultant
If we now cut this body, the applied forces can be thought of as being distributed over the cut
area A as in Fig. 2.1b). Now if we look at infinitesimal regions 'A, we assume the resultant

Definition of Stress
In the previous figures;
l ΔA infinitesimal regions
l ΔF distributed/resultant force on ΔA
l ΔA is extremely small, ΔF is uniform
l Entire area A is subject to an infinite number of
forces, each one acts over a small area ΔA.
Stress is the intensity of the internal
force on a specific plane passing
through a point.

look at the whole sectioned area, we can say number of forces, we Stress can define expression
where each one (of magnitude stress.
Definition: Stress is the intensity of the ¢ Mathematically, passing stress through can be
a point.
Mathematically, expressed as;
stress can be expressed as
lim F
A '
A
'
V
' o0
Dividing the magnitude of internal force 'F by let ¢ If we let ΔA approach zero, we obtain the
stress at a point.
'A approach zero, we obtain the stress at body, which depends on the position that important concepts that we introduced in mechanics 5 Strenght of Materials I - DAT Wednesday, October 22, 14
Normal and Shear Stress

and carried by the material and the deformation they create.
STRESS (SI&4th p. 22-23, 3rd p. 22-23)
Normal and Shear Stress
an element of continuous (no voids) and cohesive (no cracks, breaks and defects)
subjected to a number of externally applied loads as shown in Fig. 2.1a). It is
that the member is in equilibrium.
Normal Stress, σ
¢ Now let’s resolve the
force ΔF in normal and
tangential direction of
the acting area. The
intensity of the force or
force per unit area
acting normally to
section A is called
Normal Stress
'F
'Fn
'Ft
F4
Cross section: A 'A
F1
F2
F3 F5
Free Body Diagram
F1
F2
t
n
(a) (b)
cut this body, the applied forces can be thought of as being distributed over the cut

we need to consider both magnitude and direction Now let’s resolve the force 'F in normal and tangential 2.1b). The intensity of the force or force per unit Normal Stress, V (sigma), and it is expressed as:
¢ Normal stress, σ (sigma) is expressed
as:
F
'
lim n
A '
V
' o0
A
If this stress “pulls” on the area it is referred as “pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expressed as:
¢ If this stress “pulls” on the area it is
referred as Tensile Stress and defined
as Positive.
¢ If it “pushes” on the area it is called
Compressive Stress Negative.
'
F
and defined as
lim t
A '
W
' o0
A
7 Strenght of Materials I - DAT Wednesday, October 22, 14
Average Normal Stress (SI&4th p. 24-31, 3rd p. 27-

Now let’s resolve the force 'F in normal and 2.1b). The intensity of the force or force per unit Normal Stress, V (sigma), and it is expressed as:
F
'
lim n
A '
V
A
If this stress “pulls” on the area it is referred “pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expressed as:
Shear Stress, τ
' o0
¢ The intensity or force per unit area
acting tangentially to A is called Shear
Stress, τ (tau), and it is expressed
as:
F
'
lim t
A '
W
' o0
A
Average Normal Stress (SI&4th p. 24-31, 3rd p. To begin, we only look at beams that carry tensile and slender. Such beams can then be assumed Wednesday, 8 Strenght of Materials I - DAT October 22, 14

“pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expressed as:
Average Normal Stress
F
'
lim t
A '
W
Average Normal Stress (SI&4th p. 24-31, 3rd p. 27-To begin, we only look at beams that carry tensile and slender. Such beams can then be assumed to simplified to:
¢ Let us look at beams ' o0
that A
carry tensile or
compressive loads and which are long
and slender. Such beams can then be
assumed to carry a constant stress, and
the Equation can be simplified to:
V F A
We call this either Average Normal Stress or Uniform Units of Stress
The units in the SI system is the Newton per square In engineering, Pa seems too small, so we usually Kilo Pascal KPa (=Pau103) ¢ We call this either Average Normal
Stress or Uniform Uniaxial Stress.

mechanical analysis of deformable solids
force system
distributed along a
continuous surface ≡ Mechanics of materials of structures:
Mechanics of materials of structures:
1. Stress analysis → whether or not internal forces
1. Stress analysis → whether or not internal forces
(stresses) due to external STATICS: STRENGTH OF MATERIALS:
(stresses) due to external loading cause a) Statics b) Strength of materials
STATICS: STRENGTH OF MATERIALS:
Force system distributed
along a continuous surface
force system
distributed along a
continuous surface

• The loaded perpendicular • The the • The resultant of the internal forces for an axially
loaded member is normal to a section cut
perpendicular to the member axis.
• The force intensity on that section is defined as
the normal stress.
• A uniform distribution of stress in a section
infers that the line of action for the resultant the internal forces passes through the centroid
of the section.
• • A uniform distribution of stress is only
possible if the concentrated loads on the end
sections of two-force members are applied at
the section centroids. This is referred to as
centric loading.
• If a two-force member is eccentrically loaded,
then the resultant of the The Civil and Planning Faculty stress of Engineering distribution Engineering Education – State University in of • The normal stress
at a particular
'
F
point may not be
equal to the
average stress but
the resultant of the
stress distribution
must satisfy
V V
P V aveA dF V dA
P
The equal stress • The indeterminate, alone.
Internal forces for an
axially loaded member
• The normal stress at a particular point may not be
equal to the average stress but the resultant of the
stress distribution must satisfy
³ ³
A
A
A
ave
A

'

' o
0
lim
• The detailed distribution of stress is statically
indeterminate, i.e., can not be found from statics
alone.

Measurement of mechanical (strength) properties
Uniaxial Compressive
Test
Measurement of mechanical (strength) properties
Uniaxial compressive test of a concrete block:
Uniaxial compressive test of a concrete block:
http://www.muszakiak.com/diagnosztika-es-anyagvizsgalat/beton-szilardsagi-vizsgalatok.html
http://metrumkft.hu/termekek/betonszilardsagvizsgalo
F
F
http://www.muszakiak.com/diagnosztika-es-anyagvizsgalat/beton-szilardsagi-vizsgalatok.html
F
F

Units of Stress
¢ The units in the SI system is the
Newton per square meter or Pascal,
i.e. : Pa = N/m2.
¢ In engineering, Pa seems too small, so
we usually use:
Stress (ı)
material is subjected to an external force, a resisting force is set up within the component.
internal resistance force per unit area acting on a material or intensity of the forces distributed
given section is called the stress at a point.
It uses original cross section area of the specimen and also known as engineering stress conventional stress.
Therefore, P
A
T
P is expressed in Newton (N) and A, original area, in square meters (m2), the stress ǔ will expresses in N/ m2. This unit is called Pascal (Pa).
As Pascal is a small quantity, in practice, multiples of this unit is used.
1 kPa = 103 Pa = 103 N/ m2 (kPa = Kilo Pascal)
1 MPa = 106 Pa = 106 N/ m2 = 1 N/mm2 (MPa = Mega Pascal)
1 GPa = 109 Pa = 109 N/ m2 (GPa = Giga Pascal)
take an example: A rod 10 mm q10 mm cross-section is carrying an axial tensile load this rod the tensile stress developed is given by
3
10 10 10 P kN q
N

Example 1
An 80 kg lamp is
supported by a
single electrical
copper cable of
diameter d = 3.15
mm. What is the
stress carried by
the cable.
A d 0.00315
u m
F 784
100 6
V
a a
80kg
F
F
Section
a-a
A
d
FBD
Stress
the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m2.
engineering, Pa seems too small, so we usually use:
Pascal KPa (=Pau103) e.g. 20,000Pa=20kPa
Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=20MPa
Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=20GPa
2.1: An 80 kg lamp is supported by a single electrical
of diameter d = 3.15 mm. What is the stress carried
determine the stress in the wire/cable as Eq. (2.4), we need
sectional area A of the cable and the applied internal
6 2
2 2
7.793 10
4
S S
4
u

F mg 80u9.8 784N
. MPa
A .
7 793 10
6
u
Stress (SI4th p. 48-49, 3rd p. 51-52)
Example 2.1, we may concern whether or not 80kg would be too heavy, or say
stress would be too high for the wire/cable, from the safety point of view. Indeed,
of most important indicators of structural strength. When the stress (intensity of

Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=Giga Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=Example 2.1: An 80 kg lamp is supported by a single electrical
copper cable of diameter d = 3.15 mm. What is the stress carried
by the cable.
To determine the stress in the wire/cable as Eq. (2.4), we need
the area A of the cable and the applied internal
force Pascal MPa (=Pau106) e.g. 20,000,000Pa=Giga Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=Example 2.1: An 80 kg lamp is supported by a single electrical
copper cable of diameter d = 3.15 mm. What is the stress carried
by the cable.
To determine the stress in the wire/cable as Eq. (2.4), we need
the cross sectional area A of the cable and the applied internal
force F:
Allowable Stress (SI4th p. 48-49, 3rd p. 51-52)
From Example 2.1, we may concern whether or not 80kg 100.6MPa stress would be too high for the wire/cable, from the stress is one of most important indicators of structural strength. Mega In engineering, Pa seems too small, so we usually use:
From Example 2.1, we may concern whether or not 80kg 100.6MPa stress would be too high for the wire/cable, from Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=Solution
Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=Example Giga 1
Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=Example 2.1: An 80 kg lamp is supported by a single electrical
copper cable The cross of diameter sectional d = 3.15 area mm. A What of the
is the stress carried
¢ by the cable.
To determine cable the and stress the in applied the wire/internal
cable as Eq. (2.4), we need
the cross force sectional F:
area A of the cable and the applied internal
force F:
A d u m
6 2
2 2
7.793 10
0.00315
2 2
0.00315
2 2
A d 0.00315
4
u m
S S

S S
4
u
4
4
S
u

u

F mg 80u9.8 784N
7.793 u 10
m
7.793 10
4
4
F mg 80u9.8 784N
F mg 80u9.8 784N
F 784
100 6
F 784
100 6
so . MPa
V
6 2
so . MPa
V
F A 7 .
793 784
10
100 6
6
u
6 2
so . MPa
V
A .
6
7 793 10
6
u
A .
7 793 u
10
From 15 Example 2.1, Strenght we may of Materials concern I - DAT whether Wednesday, or October not 22, 80kg 14
would 100.6MPa would be too high for the wire/cable, from the safety stress is one of most important indicators of structural strength. When

Review of Static
Example 2
• 7KHVWUXFWXUHLVGHVLJQHG
WRVXSSRUWDN1 ORDG
• 7KHVWUXFWXUHFRQVLVWVRID
ERRPDQGURGMRLQHGE
SLQV]HURPRPHQW
FRQQHFWLRQV

DWWKH
MXQFWLRQVDQGVXSSRUWV
• 3HUIRUPDVWDWLFDQDOVLV
WRGHWHUPLQHWKHLQWHUQDO
IRUFHLQHDFKVWUXFWXUDO
PHPEHUDQGWKHUHDFWLRQ
IRUFHVDWWKHVXSSRUWV
The Civil and Planning Engineering Education Department
Faculty of Engineering – State University of Yogyakarta

Deformation
¢ Whenever a force is applied to a body,
its shape and size will change. These
changes are referred as deformations.
¢ Positive (elongation) or negative
(contraction) deformation;
2.2 DEFORMATION (SI4th p. 67-68, 3rd p. 70)
Whenever a force is applied to a body, its shape and size will change. These changes referred as deformations. These deformations can be thought of being either positive
(elongation) or negative (contraction) in sign as shown in Fig. 2.2.
DEFORMATION (SI4th p. 67-68, 3rd p. 70)
Whenever a force is applied to a body, its shape and size will change. These changes are
referred as deformations. These deformations can be thought of being either positive
elongation) or negative (contraction) in sign as shown in Fig. 2.2.
F F
F
F F
(+) ()
(+) ()
Original length
Deformation
(elongation)
F Deformation
Deformation
(contraction)
(contraction)
Original length
Deformation
(elongation)
Original length
Fig. 2.2 Deformation due to applied axial forces
Original length
It is however very hard to make a relative comparison between bodies or structures of different
however very hard to make a relative comparison between bodies or structures of different

¢ It is however very hard to make a
relative comparison between bodies
or structures of different size and
length as their individual deformations
will be different. This requires the
development of the concept of Strain,
which relates the body’s deformation
to its initial length.

F F
(+) ()
Original length
Deformation
(elongation)
F F
Deformation
(contraction)
Deformation
(contraction)
Original length
Original length
Strain
Deformation
(elongation)
Normal Strain
¢ The elongation (+ve) or contraction
(−ve) of a body per unit length is
termed Strain.
Original length
size and length as their individual deformations will be different. This requires the development
of the concept of Strain, which relates the body’s deformation to its initial length.
2.3 STRAIN (SI p.68-69; 4th p. 67-69; 3rd p.71-72)
Normal Strain
The elongation (+ve) or contraction (ve) of a body per unit length is termed Strain.
size and length as their individual deformations will be different. This requires the development
of the concept of Strain, which relates the body’s deformation to its initial length.
2.3 STRAIN (SI p.68-69; 4th p. 67-69; 3rd p.71-72)
Normal Strain
The elongation (+ve) or contraction (ve) of a body per unit length is termed Strain.
n
A 'S B A’
A 'S B A’
B’
F1
'S’
B’
'S’
F2
(a) Before deformed (b) After deformed
F3
(a) Before deformed (b) After deformed
Fig. 2.3 Generalized deformation due to applied forces
F3
F2
F1
n
Fig. 2.3 Generalized deformation due to applied forces
Let’s take the arbitrarily shaped body in Fig. 2.3 as an example. Consider the infinitesimal
line segment AB that is contained within the undeformed body as shown in Fig. 2.3(a). The
Let’s take the arbitrarily shaped body in Fig. 2.3 as an example. Consider the infinitesimal

(a) Before deformed Fig. 2.3 Generalized deformation due Let’s take the arbitrarily shaped body in Fig. 2.3 as an line segment AB that is contained within the undeformed line AB lies along the n-axis and has an original length ¢ After deformation, the generalized
and B are strain displaced mathematically to A’ and B’ can and be
in general the line The change expressed in length as;
of the line is therefore 'S-'S’. We strain mathematically as
'
'
lim S' S
B A '
H

(BoA along n) o
S
Average Normal Strain
If the stress in the body is everywhere constant, in other the material (e.g. uniform uniaxial tension or compression) be computed 20 by
Strenght of Materials I - DAT Wednesday, October 22, 14
L
L

and B are displaced to A’ and B’ and in general the The change in length of the line is therefore 'S-'S’. strain Average mathematically Normal as
Strain
'
'
lim S' S
B A '
If the H
stress
in the body (Bis oeverywhere
A along n) ¢ o
S
constant, in other words, the
Average deformation Normal Strain
is uniform in the material
If the stress (e.g. in uniform the body uniaxial is everywhere tension or
constant, in other the material compression), (e.g. uniform the uniaxial strain ε tension can be
or compression) be computed computed by
by;
L

Deformed Original '
L
i.e. the change in length of the body over its original L L
H
L
Original

Unit of Strain
¢ The normal strain is a dimensionless quantity
since it is a ratio of two lengths, but it is common in
practice to state it in terms of a ratio of length
units. i.e. meters per meter (m/m)
¢ Usually, for most engineering applications ε is very
small, so measurements of strain are in
micrometers per meter (μm/m) or (μ/m).
¢ Sometimes for experiment work, strain is
expressed as a percent, e.g. 0.001m/m = 0.1%. A
normal strain of 480μm for a one-meter length is
said: ε= 480×10-6 = 480(μm/m) = 0.0480% = 480μ
(micros) = 480μs (micro strain)

Example 3
¢ If it is measured that the a cable was
elongated by 1.35 mm due to the
weight of the light, what would its
strain be?

Stress-Strain Relationship
Material Test and Stress-Strain Diagram
¢ The material strength depends on its
ability to sustain a load without undue
deformation or failure.
¢ The property is inherent in the
material itself and must be determined
by experiment.
¢ The tests are performed in universal
test machine.

2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI4th p.83-85-95)
Material Test and Stress-Strain Diagram
The material strength depends on its ability to sustain a load without undue deformation failure. The property is inherent in the material itself and must be determined by experiment.
One of the most important tests to perform in this regard is the tension or compression. so, a bunch of standard specimen is made. The test is performed in universal test machine.
Shown in Fig. 2.4 is the specimen and test result of Stress-Strain Diagram.
2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI4th p.83-93; p.85-95)
Material Test Material and Stress-Strain The material strength depends test Diagram
on its
ability to sustain a load without undue deformation failure. The property is inherent in the material itself and must be determined by experiment.
One of the most Stress-important Strain tests to perform Diagram
in this regard is the tension or compression. To so, a bunch of standard specimen is made. The test is performed test machine.
Shown in Fig. 2.4 is the specimen and test result of Stress-Strain F
Standard
Specimen Yield stress
F
V
V
Elastic Yielding Hardening VY
VY
Ultimate
stress Vu
Fracture
stress Vf
Plastic behavior
Elastic
behavior
ProportionalVpl
limit
Fig. 2.4 Material test and Stress-Strain Diagram
H
Elastic Yielding Hardening Necking
Standard
Specimen Yield stress
Ultimate
stress Vu
Fracture
stress Vf
Plastic behavior
Elastic
behavior
ProportionalVpl
limit
Fig. 2.4 Material test and Stress-Strain Diagram
F
F
H
The Stress-Strain diagram consists of 4 stages during the whole process, elastic, yielding,
hardening and necking stages respectively. From yielding stage, some permanent plastic
deformation occurs. About 90% of engineering problems only concern the elastic deformation
The Stress-Strain diagram consists of 4 stages during the whole process, elastic, yielding,
hardening and necking stages respectively. From yielding stage, some permanent

Hooke’s Law
¢ The stress-strain linear relationship was
discovered by Robert Hook in 1676 and is
known as Hooke's law.
¢ It is mathematically represented by;
V EH where E is terms as the Modulus of Elasticity For l E most is Modulus of engineering of Elasticity or metal Young's
Modulus with units of N/m2 or Pa (material, or
GPa GPa) . For mild steel it is about
2.5 200GPa~POISSON'S 210GPa.
RATIO (SI4th Ed Definition

When the load P is applied to the bar it changes strain in axial direction and in lateral/radial direction axial Poisson’s Ratio
¢ When a
deformable body is
stretched by a
tensile force, not
only does it
elongate but it also
contract laterally,
i.e. it would
contract in other
two dimensions.
v Lateral Strain
H
G
L
H
In early 1800s, French scientist Poisson realized two strains is a constant. We called the constant Lateral
Axial Strain
H
Axial
Original shape
Defrormed shape
F
F
r Gr
L
L+G
(a) Fig. 2.5 Relationship of the axial The negative sign is used here since longitudinal contraction (negative strain), vice versa. So Poisson’s

direction of force and yet its sides expand laterally as in Fig. 2.5(b).
When the load P is applied to the bar it changes the bar’s length by G and its radius by Gr
G
H
¢ Likewise, a
compressive
force acting on a
deformable body
cause it to
contract in the
direction of force
and yet its sides
expand laterally
r
H
Lateral
v Lateral Strain
(2.11)
L
Gr
. The
strain in axial direction and in lateral/radial direction are respectively
axial L
r
lateral
G
H
In early 1800s, French scientist Poisson realized that within elastic range the ratio of these
two strains is a constant. We called the constant as Poisson’s ratio by v (Nu)
Axial
Axial Strain
H
Original shape
Defrormed shape
F
F
r Gr
L
L+G
Original shape
Defrormed
shape
F
F
L – G
r
(a) (b)
Fig. 2.5 Relationship of the axial strain with the lateral strain
The negative sign is used here since longitudinal elongation (positive strain) cause lateral
contraction (negative strain), vice versa. So Poisson’s ration is positive, i.e. vt0.

terms as the Modulus of Elasticity or Young's Modulus with units engineering metal material, GPa is used, e.g. mild steel is about 200GPa~POISSON'S RATIO (SI4th Ed p104-105; 3rd Ed p107-108)
deformable body ¢ is When stretched the by load a tensile P is applied force, not to only the bar
does it elongate laterally, i.e. it would it changes contract the in bar’s other length two dimensions by δ and
as shown compressive force its radius acting by on δr. a The deformable strain in body axial
cause it to force and yet its sides expand laterally as in Fig. 2.5(b).
load P is applied to direction the and in lateral/radial direction
direction and are in lateral/respectively;
bar it changes the bar’s length by G and its radius radial direction are respectively
G
H
axial L
G
r
r
H
lateral
1800s, French scientist Poisson realized that within elastic range the a constant. We called the constant as Poisson’s ratio by v (Nu)
H
Lateral
Lateral Strain
H

Likewise, a compressive force acting on a deformable direction of force and yet its sides expand laterally as in When the load P is applied to the bar it changes the bar’s strain in axial direction and in lateral/radial direction are G
H axial
¢ Within the elastic range, the ratio of
these two strains is a constant, L
and it
is called Poisson’s ratio by v (Nu);
v Lateral Strain
H
¢ Poisson’s ration is positive, i.e. v≥0.
H lateral
In early 1800s, French scientist Poisson realized that two strains is a constant. We called the constant as Poisson’s Lateral
Axial Strain
H
Axial
Original shape
Defrormed shape
F
F

Remarks
¢ The lateral strain is caused only by
axial force. No force or stress acts in
lateral direction;
¢ Lateral strain is the same in all lateral
direction;
¢ Usually 0 ≤ v ≤ 0.5 . For most linearly
elastic material v=0.3;
¢ Poisson’s ratio is a constant. Wednesday, 31 Strenght of Materials I - DAT October 22, 14

Strain in Lateral Direction
¢ For bars subjected to a tensile stress
σx, the strains in the y and z planes
are:

H
°°°
V
H H
®
°°°
H H
¯

x
E
v v

x
y x
V
v v

E
V
E
x
x
z x
2.6 THERMAL STRAIN (SI4th Ed p. 148-152; 3rd 32 Strenght of Materials I - DAT Wednesday, October 22, 14
Thermal Deformation
When the temperature of a body is changed, its overall

Simple Stress and Strain

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Ähnlich wie Simple Stress and Strain

Ähnlich wie Simple Stress and Strain (20)

Mehr von Msheer Bargaray

Mehr von Msheer Bargaray (11)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Simple Stress and Strain