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, October 22, 14 2 Strenght of Materials I - DAT

STRESS External and internal forces in a structural member

¢ Consider an element of continuous and cohesive material subjected to a number of externally applied loads.

Wednesday, October 22, 14 Strenght of Materials I - DAT 3

Lecture Notes of Mechanics of Solids, Chapter 2 1

Chapter 2 Stress and Strain 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.

'F'Fn

'Ft

Cross section: A 'A

F1

F2

F3 F5

F4Free Body Diagram

F1

F2

t

n

(a) (b)

Fig. 2.1 External and internal forces in a structural member

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 force in this infinitesimal area is 'F. In fact, 'F is also a distributed force. When 'A is extremely small, we can say that the distributed force 'F is uniform. In other words, if we look at the whole sectioned area, we can say that the entire area A is subject to an infinite number of forces, where each one (of magnitude 'F) acts over a small area of size 'A. Now, we can define stress. Definition: Stress is the intensity of the internal force on a specific plane

passing through a point. Mathematically, stress can be expressed as

AFlim

A ''

Vo' 0

(2.1)

Dividing the magnitude of internal force 'F by the acting area 'A, we obtain the stress. If we let 'A approach zero, we obtain the stress at a point. In general, the stress could vary in the body, which depends on the position that we are concerning. The stress is one of most important concepts that we introduced in mechanics of solids. Normal and Shear Stress As we known, force is a vector that has both magnitude and direction. But in the stress definition, we only consider the magnitude of the force. Obviously, this may easily confuse us. Let’s still take patch 'A as an example. As we can see, force 'F is not perpendicular to the sectioned infinitesimal area 'A. If we only take the magnitude of the force into account,



'F'Fn

'Ft

Cross section: A 'A

F1

F2

F3 F5

F4Free Body Diagram

F1

F2

t

n

(a) (b)




AFlim

A ''

Vo' 0

(2.1)


Definition of Stress

In the previous figures; l  ∆A infinitesimal regions l  ∆F distributed/resultant force on ∆A l  ∆A is extremely small, ∆F is uniform l  Entire area A is subject to an infinite number of

forces, each one acts over a small area ∆A.

Stress is the intensity of the internal force on a specific plane passing

through a point. Wednesday, October 22, 14 Strenght of Materials I - DAT 4

Stress expression

¢ Mathematically, stress can be expressed as;

¢  If we let ∆A approach zero, we obtain the

stress at a point.




'F'Fn

'Ft

Cross section: A 'A

F1

F2

F3 F5

F4Free Body Diagram

F1

F2

t

n

(a) (b)




AFlim

A ''

Vo' 0

(2.1)


Normal and Shear Stress

Normal Stress, σ ¢  Now let’s resolve the

force ∆F 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

Cross section: A 'A

F1

F2

F3 F5

F4Free Body Diagram

F1

F2

t

n

(a) (b)




AFlim

A ''

Vo' 0

(2.1)


¢  Normal stress, σ (sigma) 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 and defined as Negative.



apparently, the stress may not reflect the real mechanical status at this point. In other words, we need to consider both magnitude and direction of the force.

Now let’s resolve the force 'F in normal and tangential direction of the acting area as Fig. 2.1b). The intensity of the force or force per unit area acting normally to section A is called Normal Stress, V (sigma), and it is expressed as:

AF

lim nA '

'V

' 0o (2.2)

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 and defined as Negative.

The intensity or force per unit area acting tangentially to A is called Shear Stress, W (tau), and it is expressed as:

AF

lim tA '

'W

' 0o (2.3)

Average Normal Stress (SI&4th p. 24-31, 3rd p. 27-34) To begin, we only look at beams that carry tensile or compressive loads and which are long and slender. Such beams can then be assumed to carry a constant stress, and Eq. (2.2) can be simplified to:

AF

V (2.4)

We call this either Average Normal Stress or Uniform Uniaxial Stress. 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:

Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=20kPa Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=20MPa Giga Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=20GPa

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:

2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V

Allowable Stress (SI&4th p. 48-49, 3rd p. 51-52) From Example 2.1, we may concern whether or not 80kg would be too heavy, or say 100.6MPa stress would be too high for the wire/cable, from the safety point of view. Indeed, stress is one of most important indicators of structural strength. When the stress (intensity of force) of an element exceeds some level, the structure will fail. For convenience, we usually adopt allowable force or allowable stress to measure the threshold of safety in engineering.

80kgF

a a

Sectiona-a

A

d

FBD

F

Shear Stress, τ ¢ The intensity or force per unit area

acting tangentially to A is called Shear Stress, τ (tau), and it is expressed as:





AF

lim nA '

'V

' 0o (2.2)



AF

lim tA '

'W

' 0o (2.3)


AF

V (2.4)




2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V


80kgF

a a

Sectiona-a

A

d

FBD

F

Average Normal Stress ¢  Let us look at beams that 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:

¢  We call this either Average Normal

Stress or Uniform Uniaxial Stress.





AF

lim nA '

'V

' 0o (2.2)



AF

lim tA '

'W

' 0o (2.3)


AF

V (2.4)




2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V


80kgF

a a

Sectiona-a

A

d

FBD

F


2/31

Strength of Materials

≡ mechanical analysis of deformable solids

Mechanics of materials of structures:

1. Stress analysis → whether or not internal forces (stresses) due to external loading cause failure

STATICS: STRENGTH OF MATERIALS:

force system distributed along a continuous surface 2/31

Strength of Materials

≡ mechanical analysis of deformable solids

Mechanics of materials of structures:

1. Stress analysis → whether or not internal forces (stresses) due to external loading cause failure

STATICS: STRENGTH OF MATERIALS:

force system distributed along a continuous surface

a) Statics b) Strength of materials

Force system distributed along a continuous surface


The Civil and Planning Engineering Education DepartmentFaculty of Engineering – State University of Yogyakarta

• If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment.

• The stress distributions in eccentrically loaded members cannot be uniform or symmetric.

• A uniform distribution of stress in a section infers that the line of action for the resultant of 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.


• The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy

³³ A

ave dAdFAP VV

• The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis.

AP

AF

aveA

''

o'

VV0

lim

• The force intensity on that section is defined as the normal stress.

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



³³ A

ave dAdFAP VV


AP

AF

aveA

''

o'

VV0

lim



Internal forces for an axially loaded member



³³ A

ave dAdFAP VV


AP

AF

aveA

''

o'

VV0

lim



•  The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy

Uniaxial Compressive Test


15/31

Measurement of mechanical (strength) properties

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

15/31

Measurement of mechanical (strength) properties

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

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:


1. Stress and Strain

Theory at a Glance (for IES, GATE, PSU)1.1 Stress ( )When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force per unit area acting on a material or intensity of the forces distributed over a given section is called the stress at a point.

It uses original cross section area of the specimen and also known as engineering stress or conventional stress.

Therefore, PA

P is expressed in Newton (N) and A, original area, in square meters (m2), the stress will be 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)

Let us take an example: A rod 10 mm 10 mm cross-section is carrying an axial tensile load 10 kN. In this rod the tensile stress developed is given by

32

210 10 10 100N/mm 100MPa

10 10 100tP kN NA mm mm mm

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 the shown section is defined as the normal stress.

0lim and avgA

F PA A

Tensile stress ( t)If > 0 the stress is tensile. i.e. The fibres of the component tend to elongate due to the external force. A member subjected to an external force tensile P and tensile stress distribution due to the force is shown in the given figure.

Page 3 of 429

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.





AF

lim nA '

'V

' 0o (2.2)



AF

lim tA '

'W

' 0o (2.3)


AF

V (2.4)




2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V


80kgF

a a

Sectiona-a

A

d

FBD

F

Solution Example 1


¢  The cross sectional area A of the cable and the applied internal force F:




AF

lim nA '

'V

' 0o (2.2)



AF

lim tA '

'W

' 0o (2.3)


AF

V (2.4)




2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V


80kgF

a a

Sectiona-a

A

d

FBD

F




AF

lim nA '

'V

' 0o (2.2)



AF

lim tA '

'W

' 0o (2.3)


AF

V (2.4)




2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V


80kgF

a a

Sectiona-a

A

d

FBD

F




AF

lim nA '

'V

' 0o (2.2)



AF

lim tA '

'W

' 0o (2.3)


AF

V (2.4)




2622

10793.7400315.0

4mdA �u

u SS

N.mgF 7848980 u

so MPa..A

F 6100107937

7846

u

�V


80kgF

a a

Sectiona-a

A

d

FBD

F

Example 2

Wednesday, October 22, 14 Strenght of Materials I - DAT 16 The Civil and Planning Engineering Education DepartmentFaculty of Engineering – State University of Yogyakarta

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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 (SI&4th

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.

Original length

F F

Deformation(elongation)

Original length

F F

Deformation(contraction)

(+) (�)

Fig. 2.2 Deformation due to applied axial forces

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.

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.

A B'SA’

B’

'S’

(a) Before deformed (b) After deformed

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

line AB lies along the n-axis and has an original length of 'S. After deformation, points A

and B are displaced to A’ and B’ and in general the line becomes a curve having a length 'S’

The change in length of the line is therefore 'S-'S’. We consequently define the generalized strain mathematically as

SS'Slim

AB '''H �

o

(BoA along n) (2.8)

Average Normal Strain If the stress in the body is everywhere constant, in other words, the deformation is uniform in

the material (e.g. uniform uniaxial tension or compression) as shown in Fig. 2.2, the strain can

be computed by

LL

LLL

Original

OriginalDeformed 'H �

(2.9)

i.e. the change in length of the body over its original length,



p. 67-68, 3rd



Original length

F F


Original length

F F


(+) (�)





2.3 STRAIN (SI p.68-69; &4th

p. 67-69; 3rd


A B'SA’

B’

'S’


F3

F2

F1

n







SS'Slim

AB '''H �

o

(BoA along n) (2.8)



be computed by

LL

LLL

Original


(2.9)


¢  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.


Strain

Normal Strain ¢ The elongation (+ve) or contraction

(−ve) of a body per unit length is termed Strain.




p. 67-68, 3rd



Original length

F F


Original length

F F


(+) (�)





2.3 STRAIN (SI p.68-69; &4th

p. 67-69; 3rd


A B'SA’

B’

'S’


F3

F2

F1

n







SS'Slim

AB '''H �

o

(BoA along n) (2.8)



be computed by

LL

LLL

Original


(2.9)




p. 67-68, 3rd



Original length

F F


Original length

F F


(+) (�)





2.3 STRAIN (SI p.68-69; &4th

p. 67-69; 3rd


A B'SA’

B’

'S’


F3

F2

F1

n







SS'Slim

AB '''H �

o

(BoA along n) (2.8)



be computed by

LL

LLL

Original


(2.9)


¢ After deformation, the generalized strain mathematically can be expressed as;




p. 67-68, 3rd



Original length

F F


Original length

F F


(+) (�)





2.3 STRAIN (SI p.68-69; &4th

p. 67-69; 3rd


A B'SA’

B’

'S’


F3

F2

F1

n







SS'Slim

AB '''H �

o

(BoA along n) (2.8)



be computed by

LL

LLL

Original


(2.9)


Average Normal Strain

¢  If the stress in the body is everywhere constant, in other words, the deformation is uniform in the material (e.g. uniform uniaxial tension or compression), the strain ε can be computed by;




p. 67-68, 3rd



Original length

F F


Original length

F F


(+) (�)





2.3 STRAIN (SI p.68-69; &4th

p. 67-69; 3rd


A B'SA’

B’

'S’


F3

F2

F1

n







SS'Slim

AB '''H �

o

(BoA along n) (2.8)



be computed by

LL

LLL

Original


(2.9)


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.


Material test & Stress-Strain Diagram



Unit of Strain From Eqs. (2.8) and (2.9), we can notice that the normal strain is a dimensionless quantity since it is a ratio of two lengths. Although this is the case, 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 H is very small, so measurements of strain are in micrometers per meter (Pm/m) or (P/m). Sometimes for experiment work, strain is expressed as a percent, e.g. 0.001m/m = 0.1%. A normal strain of 480Pm for a one-meter length is said:

H= 480u10-6 = 480(Pm/m) = 0.0480% = 480P (micros) = 480Ps (micro strain) Example 2.3: In Example 2.1, if it is measured that the cable was elongated by 1.35 mm due to the weight of the light, what would its strain be?

s.

. PH 9001090051

001350 6 u �

2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI&4th p.83-93; 3rd p.85-95) 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. One of the most important tests to perform in this regard is the tension or compression. To do 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.

F

F

VY

V

H

Elastic Yielding Hardening Necking

StandardSpecimen Yield stress

Ultimatestress Vu

Fracturestress Vf

Plastic behavior

Elasticbehavior

VplProportionallimit

Fig. 2.4 Material test and Stress-Strain Diagram

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 in structural members and mechanical components. Only 10% of engineering work concerns plastic and other nonlinear stage (e.g. metal forming). In this subject, we are only involved in the linear elastic region, in which the relationship between the strain and stress is linear. 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 Eq. (2.10),


Unit of Strain From Eqs. (2.8) and (2.9), we can notice that the normal strain is a dimensionless quantity since it is a ratio of two lengths. Although this is the case, 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 H is very small, so measurements of strain are in micrometers per meter (Pm/m) or (P/m). Sometimes for experiment work, strain is expressed as a percent, e.g. 0.001m/m = 0.1%. A normal strain of 480Pm for a one-meter length is said:

H= 480u10-6 = 480(Pm/m) = 0.0480% = 480P (micros) = 480Ps (micro strain) Example 2.3: In Example 2.1, if it is measured that the cable was elongated by 1.35 mm due to the weight of the light, what would its strain be?

s.

. PH 9001090051

001350 6 u �

2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI&4th p.83-93; 3rd p.85-95) 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. One of the most important tests to perform in this regard is the tension or compression. To do 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.

F

F

VY

V

H

Elastic Yielding Hardening Necking

StandardSpecimen Yield stress

Ultimatestress Vu

Fracturestress Vf

Plastic behavior

Elasticbehavior

VplProportionallimit

Fig. 2.4 Material test and Stress-Strain Diagram

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 in structural members and mechanical components. Only 10% of engineering work concerns plastic and other nonlinear stage (e.g. metal forming). In this subject, we are only involved in the linear elastic region, in which the relationship between the strain and stress is linear. 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 Eq. (2.10),

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;

l  E is Modulus of Elasticity or Young's Modulus with units of N/m2 or Pa (or GPa) . For mild steel it is about 200GPa~210GPa.



HV E (2.10) where E is terms as the Modulus of Elasticity or Young's Modulus with units of N/m2 or Pa. For most of engineering metal material, GPa is used, e.g. mild steel is about 200GPa~210GPa. 2.5 POISSON'S RATIO (SI&4th Ed p104-105; 3rd Ed p107-108) Definition 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 as shown in Fig 2.5(a). Likewise, a compressive force acting on a deformable body cause it to contract in the 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. The strain in axial direction and in lateral/radial direction are respectively

LaxialGH

rr

lateralG

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

Lateral

StrainAxialStrainLateralv

HH

� � (2.11)

Original shape

Defrormed shape

F

F

r GrLL+G

Original shape

Defrormedshape

F

F

L – G

L rGr

(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. Remarks

x The lateral strain is caused only by axial force. No force or stress acts in lateral direction; x Lateral strain is the same in all lateral direction; x Usually 500 .v dd . For most linearly elastic material v=0.3; x Poisson’s ratio is a constant.

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

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.




LaxialGH

rr

lateralG

H


Axial

Lateral


HH

� � (2.11)

Original shape

Defrormed shape

F

F

r GrLL+G

Original shape

Defrormedshape

F

F

L – G

L rGr

(a) (b)





¢  Likewise, a compressive force acting on a deformable body cause it to contract in the direction of force and yet its sides expand laterally




LaxialGH

rr

lateralG

H


Axial

Lateral


HH

� � (2.11)

Original shape

Defrormed shape

F

F

r GrLL+G

Original shape

Defrormedshape

F

F

L – G

L rGr

(a) (b)





¢ When the load P is applied to the bar it changes the bar’s length by δ and its radius by δr. The strain in axial direction and in lateral/radial direction are respectively;




LaxialGH

rr

lateralG

H


Axial

Lateral


HH

� � (2.11)

Original shape

Defrormed shape

F

F

r GrLL+G

Original shape

Defrormedshape

F

F

L – G

L rGr

(a) (b)





¢ Within the elastic range, the ratio of these two strains is a constant, and it is called Poisson’s ratio by v (Nu);

¢ Poisson’s ration is positive, i.e. v≥0.




LaxialGH

rr

lateralG

H


Axial

Lateral


HH

� � (2.11)

Original shape

Defrormed shape

F

F

r GrLL+G

Original shape

Defrormedshape

F

F

L – G

L rGr

(a) (b)





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, October 22, 14 Strenght of Materials I - DAT 31

Strain in Lateral Direction

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



°°°

¯

°°°

®

� �

� �

Evv

Evv

E

xxz

xxy

xx

VHH

VHH

VH

(2.12)

2.6 THERMAL STRAIN (SI&4th Ed p. 148-152; 3rd Ed p. 152-156)

Thermal Deformation When the temperature of a body is changed, its overall size will also change. In other words, temperature change may cause the dimension or shape change in the material. More specially, if the temperature increases, generally a material expands. Whereas if the temperature decreases, the material will contract. It is supposed that this is a common sense for anyone.

L GT

TRef Tref+ 'T VxVx

x

y

z

TRef

Tref+ 'T(a) Thermal deformation (b) Thermal and mechanical deformation

Fig. 2.6 Thermal and mechanical deformation

For the majority of engineering materials this relationship is linear. If we assume that the material is homogeneous and isotropic, from experiment, we can find a linear relation between thermal deformation and temperature change as:

LTT �� 'DG (2.13)

where : D : Coefficient of thermal expansion , units are strain per oC

�� 'T : algebraic change in temperature ( oC ) (increase +; decrease �)

TG : algebraic change in length (“+” = elongation; “�” = contraction)

Thermal Strain

TLT

Thermal '� DG

H (2.14)

Coupled Strain Status If we consider both mechanical strain HV and thermal strain HT in the structure as shown in Fig. 2.6(b), by referring to Eq. (2.12), the total strains in all directions would be computed as:

°°°

¯

°°°

®

�' �

�' �

�' �

ET

ET

ET

xzTz

xyTy

xxTx

VDHHH

VDHHH

VDHHH

V

V

V

(2.15)