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Chapter 1
Introduction to mechanical
metallurgy
Subjects of interest
               • Introduction to mechanical metallurgy
               • Strength of materials – Basic assumptions
               • Elastic and plastic behaviour
               • Average stress and strain
               • Tensile deformation of ductile metals
               • Ductile vs brittle behaviour
               • What constitute failure?
               • Concept of stress and the types of stresses
               • Units of stress and other quantities


 Suranaree University of Technology     Tapany Udomphol        May-Aug 2007
Objectives


     • This chapter provides a background of continuum
     description of stress and strain and extends it to the defect
     mechanisms of flow and fracture of metals.
     • Elastic and plastic behaviours of metals are highlighted
     and factors influencing failure in metals are finally
     addressed.




Suranaree University of Technology   Tapany Udomphol                 May-Aug 2007
Introduction
Mechanical metallurgy : Response of metals to forces or loads.



Mechanical assessment of                  Forming of metals into
Materials                                 useful shapes
• Structural materials                    • Forging, rolling, extrusion.
• Machine, aircraft, ship, car etc        • drawing, machining, etc


    We need to know                         We need to know conditions
    limiting values of which                of load and temperature to
    materials in service can                minimise the forces that are
    withstand without                       needed to deform metal
    failure.                                without failure.


 Suranaree University of Technology   Tapany Udomphol                  May-Aug 2007
Strength of materials
     Strength of materials deals with relationships between;
     • internal resisting forces
     • deformation
     • external loads
     , which act on some part of a body (member) in equilibrium.


• In equilibrium condition, if there are the
                                                               Internal force   External force
external forces acting on the member,
there will be the internal forces resisting                    ∫ σdA                        P
the action of the external loads.

• The internal resisting forces are usually expressed by the
stress acting over a certain area, so that the internal force is the
integral of the stress times the differential area over which it acts.

                                       P = ∫ σdA               …Eq.1

  Suranaree University of Technology         Tapany Udomphol                               May-Aug 2007
Assumptions in strength of materials
The body (member) is
                                                                Remark:
• Continuous:             No voids or empty spaces.             Anisotropic is when
• Homogeneous: Has identical properties at all                  the body has
               points.                                          property that varies
                                                                with orientation.
• Isotropic:              Has similar properties
                          in all directions or orientation.


Macroscopic scale, engineering materials such
as steel, cast iron, aluminium seems to be
continuous, homogeneous and isotropic.

Microscopic scale, metals are made up of an
aggregate of crystal grains having different
properties in different crystallographic directions.
   However, these crystal grains are very small,
   and therefore the properties are homogenous
                                                                     Microstructure of
   in the macroscopic scale.                                         low carbon steel
   Suranaree University of Technology         Tapany Udomphol                May-Aug 2007
Elastic and plastic behaviour
• All solid materials can be deformed                     stress           Maximum
when subjected to external load.                                           tensile strength
                                                                 Elastic
• In elastic region, stress is proportional to                   limit                        Fracture
strain. This follows Hook’s law up to the
elastic limit . The material now has elastic
behaviour.                                                                    σmax

• At the elastic limit, when the load is               Elastic Plastic
removed, the material will change back to
its original shape.
                                                                                         strain
• Beyond the elastic limit, material
                                                                  Load and extension curve of
permanently deformed or the material has                         under uniaxial tensile loading.
undergone plastic deformation.

                       Note: elastic deformations in metals are relatively small
                       in comparison to plastic deformations.

     Suranaree University of Technology        Tapany Udomphol                            May-Aug 2007
Average stress and strain
    No load applied                             Load applied          Lo+ δ
                           Lo                                          Lo
                                                                                      P




• A uniform cylindrical bar which is           The external load P is balanced by
subjected to an axial tensile load P.          the internal resisting force, giving
The original gauge length Lo has               the equilibrium equation;
undergone a slight increase in                  Internal           External
length to Lo+ δ with a slight                                                 P = ∫ σdA
                                                force                 force

decrease in diameter.                           ∫σdA                    P


• The average elastic strain e is               If the stress σ is uniform over the area
the ratio of change in length to the            A, σ = constant, then
original length.
                                                                  P = σ ∫ dA = σA
                 δ∆L L − Lo
…Eq.2      e=   =   =                                                         P
              Lo Lo    Lo                    The average stress σ =                 …Eq.3
                                                                              A
        Suranaree University of Technology      Tapany Udomphol                     May-Aug 2007
Average stress and strain
           • In general, the stress is not uniform over the area A
               average stress.
           • On a microscopic scale, metals consist of more
           than one phase and therefore give rise to
           nonuniformity of stress.
                                                       stress           Maximum
                                                                        tensile strength
• Below the elastic limit, Hook’s law                         Elastic
can be applied, so that the average                           limit                        Fracture
stress is proportional to the average
strain,
                        σ               …Eq.4                              σmax
                          =E
                        ε
 The constant E is the modulus of
 elasticity or Young’s modulus.                                                       strain
                                                                Load and extention curve of
                                                               under uniaxial tensile loading
   Suranaree University of Technology       Tapany Udomphol                            May-Aug 2007
Tensile deformation of ductile metal
• In tension test, specimen is subjected to axial tensile load
until fracture. Load and extension are measured and expressed
as stress and strain, see Fig.
• The initial linear portion of the curve (OA) is the elastic region,
following Hook’s law.                                                                       D
• Point A is the elastic limit (the greatest stress that the metal
                                                                                        B
can withstand without undergone permanent or plastic                    Elastic
                                                                                   A’
                                                                                  A
deformation. A’ is the proportional limit where the curve                 limit
deviates from linearity.
• The slope of the linear portion is the modulus of elasticity.
• Point B is the yield strength, defined as the stress which will
produce a small amount of strain equal to 0.002 (OC).                             C
• As the plastic deformation
increases, the metal
becomes stronger (strain
hardening) until reaching the
maximum load, giving
ultimate tensile strength D.
• Beyond D, metal necks
(reduce in x-section). Load
needed to continue
deformation drops off till
failure.
        Suranaree University of Technology           Tapany Udomphol                        May-Aug 2007
Ductile and brittle behaviour
• Completely brittle materials i.e.,
ceramics would fracture almost at
the elastic limit.




                                             stress




                                                                     stress
• Brittle metals such as cast iron
show small amounts of plasticity
before failure.
                                                      strain                  strain
• For engineering materials,                    Completely brittle     Slightly brittle with
                                                                       small amount of
adequate ductility is important                                        ductility
because it allows the materials to
redistribute localised stresses.

      • Fracture behaviour of metal (ductile or brittle) also depends on
      some conditions, i.e., temperature, tension or compression, state
      of stresses, strain rate and embrittling agent.


   Suranaree University of Technology   Tapany Udomphol                             May-Aug 2007
What constitute fracture?
  Three general ways that cause failures in structural
  members and machine elements:

1) Excessive elastic deformation




                                                           stress
2) Yielding or excessive plastic deformation                                  Plastic
                                                                              energy
                                                                    Elastic
                                                                    energy
3) Fracture
                                                                        strain




     Types of Load/forces                                 Good structural
                                                          components in
     Nature of Materials
                                                          service without
     Structural design                                    failure

   Suranaree University of Technology   Tapany Udomphol                                 May-Aug 2007
Excessive elastic deformation

  • Failure due to excessive elastic deformation are controlled by
  the modulus of elasticity not the strength of the materials.
  Two general types of excessive elastic
     deformation:

  1) Excessive elastic deformation under




                                                        stress
     condition of stable equilibrium.
                                                                 Elastic
  EX: too much deflection in a shaft can cause                   energy
  wear in bearing and damage other parts.                           strain


  2) Excessive elastic deformation under
     condition of unstable equilibrium.

   EX: A sudden deflection or buckling of a slender column.


 Suranaree University of Technology   Tapany Udomphol                        May-Aug 2007
Excessive plastic deformation

  • Excessive plastic deformation occurs when the elastic
  limit is exceeded     yielding.
  • Yielding produce permanent change of shape and can
  cause fracture.




                                                           stress
  EX: When component changed in shape,
  it cannot function properly any longer.                                     Plastic
                                                                    Elastic   energy
                                                                    energy
• Failure by excessive plastic                                          strain
deformation is controlled by the yield
strength of the materials. Even in more
complex loading conditions, yield strength
is still a significant parameter.

                                                    Large deformation that cause
                                                                failure in a pipe
 Suranaree University of Technology   Tapany Udomphol                                   May-Aug 2007
Fracture
             Metals fail by fracture in three general ways:

             1) Sudden brittle fracture
             Ex : some ductile metal such as plain carbon steel
                will undergo ductile to brittle transition with
                decreasing temperature, increasing rate of loading
                and triaxial state of stresses.

             2) Fatigue, or progressive fracture
                Most fracture in machine parts are due to fatigue
                (subjected to alternating stresses). Failure is due
                to localised tensile stress at spot or notch or stress
                concentration.

             3) Delayed fracture
             Ex: Stress-rupture failure, which occurs when a
                metal has been statistically loaded at an elevated
                temperature.
 Suranaree University of Technology     Tapany Udomphol                  May-Aug 2007
Working stress

 To prevent structural members or machine elements from failure,
such members should be used under a stress level that is lower than
its yield stress σo. This stress level is called the working stress σw.
 According to American Society of Mechanical Engineering
 (ASME), the working stress σw may be considered as either the
 yield strength σo or the tensile strength σu divided by a number
 called safety factor.

                                      σo                          σu
        …Eq.5               σw =           or         σw =             …Eq.6
                                      No                          Nu
        Where σw                  = working stress
                     σo           = yield strength
                     σu           = tensile strength
                     No           = safety factor based on yield strength
                     Nu           = safety factor based on tensile strength
  Safety factor depends on loading/service conditions, consequences, etc.
 Suranaree University of Technology             Tapany Udomphol                May-Aug 2007
Concept of stress and the type of
stresses

 Definition: stress is force per unit area.

 There are two kinds of external forces which may act on a body;
 1) Surface forces : forces distributed over the surface of the body,
    i.e., hydrostatic pressure


 2) Body forces : forces distributed over the volume of the body,
    i.e., gravitational force, magnetic force, centrifugal force, thermal
    stress.




Suranaree University of Technology   Tapany Udomphol               May-Aug 2007
Stress at a point
  • Consider a body having forces F1…F4
  acting on it, see Fig (a). The body is cut by a
  plane passing through point O. If one half is
  removed and replaced by an equivalent force
  F, acting on the x-sectional area A to remain
  the static equilibrium, see Fig (b).
  • We can resolve F into component normal to
  the plane Fn and tangential to the plane Fs.                        (a) Equilibrium of
                                                                      an arbitrary body.
  • The concept of stress at a point is shrinking
  the area A into infinitesimal dimensions.


                     Fn                            Fs
…Eq.7    σ = lim              and        τ = lim        …Eq.8
              ∆A→0    A                     ∆A→0   A

        Note: σ and τ depend on the orientation of the plane            (b) Force acting
        passing through P and will vary from point to point.                   on parts.
    Suranaree University of Technology              Tapany Udomphol            May-Aug 2007
The total stress can be resolved into two components;
            1) Normal stress σ perpendicular to A.
            2) Shear stress τ lying in the plane of the area.

  • The force F makes an angle θ with the normal z to the plane x-y of
  the area A.
  • The plane containing the normal z and F intersects the plane A along
  a dash line that makes an angle φ with the y axis.

The normal stress                         F                 The shear stress in the plane
is given by                        σ=       cos θ
              …Eq.9                       A                 acting along OC has the
                                                            magnitude.
                                                                                 F
                          F                                                 τ = sin θ
                                                                  …Eq.10         A

                                                                         F
                                                       x direction τ =     sin θ sin φ …Eq.11
                                                                         A
                                                                         F
                                                       y direction τ =     sin θ cos φ …Eq.12
      Resolutions of total stress into its                               A
      components.
     Suranaree University of Technology             Tapany Udomphol                 May-Aug 2007
Concept of strain and types of
 strain
 The average linear strain was defined as the ratio of the
 change in length to the original length.
            δ∆L L − Lo                       Where e = average linear strain
      e=   =   =
         Lo Lo    Lo                               δ = deformation.
                                       …Eq.2

Strain at a point is the ratio of the deformation to the gauge
length as the gauge length              0.

True strain or natural strain is the strain as the change in
linear dimension divided by the instantaneous value of the
dimension.
                        Lf
                           dL      Lf
                     ε=∫      = ln                       …Eq.13
                        Lo
                            L      Lo

  Suranaree University of Technology           Tapany Udomphol             May-Aug 2007
Shear strain

          • Shear strain is the angular change in a right angle.
          • The angle at A, which is originally 90o, is decreased by
          a small amount θ when the shear stress is applied.
          • The shear strain γ will be given by

                a
                                                      a
                θ                                  γ = = tan θ = θ
      h                                               h

                                                           …Eq.14
            A

             Shear strain


Suranaree University of Technology    Tapany Udomphol                  May-Aug 2007
Units of stress and other quantities

         • Following SI unit

Length                               metre (m)
Mass                                 kilogram (kg)        Frequency    (s-1, Hz)
Time                                 second (s)           Force        Newton (N)
Electric current                     ampere (A)           Stress       (N.m-2, Pa)
Temperature                          Kelvin (K)           Strain       dimensionless
Amount of substance                  mole (mol)
Luminous intensity                   candela (cd)




       Suranaree University of Technology            Tapany Udomphol               May-Aug 2007
Example: The shear stress required to nucleate a grain boundary
       crack in high-temperature deformation has been estimated to be
                                                                 1
                                                3πγ b G        2
                                            τ =
                                                8(1 − ν ) L 
                                                             
                                                            

       Where γb is the grain boundary surface energy ~ 2 J.m-2, G is shear
       modulus, 75 GPa, L is the grain boundary sliding distance, assume =
       grain diameter 0.01 mm, and the Poisson’s ratio ν = 0.3.

Checking the unit
                     1
    Nm N         1 2                                       1
    2 × 2                   3π × 2 × 75 × 10       9
                                                           2
         m  =N  = N τ =
                2 2
     m                                                      = 15.89 × 10 7 N .m −2
τ =           4
              m 
                              8(1 − 0.3) × 10 − 2 × 10 −3 
                                                          
      m          m 2
                       τ = 158.9 MN .m −2 = 158.9 MPa
          



       Suranaree University of Technology             Tapany Udomphol     May-Aug 2007
References

     • Dieter, G.E., Mechanical metallurgy, 1988, SI metric edition,
     McGraw-Hill, ISBN 0-07-100406-8.
     • Sanford, R.J., Principles of fracture mechanics, 2003,
     Prentice Hall, ISBN 0-13-192992-1.
     • www.indiamart.com.
     • www.technion.ac.il/.../ Figures/Light-steel.jpg
     • www.uvi.edu




 Suranaree University of Technology     Tapany Udomphol                May-Aug 2007

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01 introduction to_mechanical_metallurgy

  • 1. Chapter 1 Introduction to mechanical metallurgy Subjects of interest • Introduction to mechanical metallurgy • Strength of materials – Basic assumptions • Elastic and plastic behaviour • Average stress and strain • Tensile deformation of ductile metals • Ductile vs brittle behaviour • What constitute failure? • Concept of stress and the types of stresses • Units of stress and other quantities Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 2. Objectives • This chapter provides a background of continuum description of stress and strain and extends it to the defect mechanisms of flow and fracture of metals. • Elastic and plastic behaviours of metals are highlighted and factors influencing failure in metals are finally addressed. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 3. Introduction Mechanical metallurgy : Response of metals to forces or loads. Mechanical assessment of Forming of metals into Materials useful shapes • Structural materials • Forging, rolling, extrusion. • Machine, aircraft, ship, car etc • drawing, machining, etc We need to know We need to know conditions limiting values of which of load and temperature to materials in service can minimise the forces that are withstand without needed to deform metal failure. without failure. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 4. Strength of materials Strength of materials deals with relationships between; • internal resisting forces • deformation • external loads , which act on some part of a body (member) in equilibrium. • In equilibrium condition, if there are the Internal force External force external forces acting on the member, there will be the internal forces resisting ∫ σdA P the action of the external loads. • The internal resisting forces are usually expressed by the stress acting over a certain area, so that the internal force is the integral of the stress times the differential area over which it acts. P = ∫ σdA …Eq.1 Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 5. Assumptions in strength of materials The body (member) is Remark: • Continuous: No voids or empty spaces. Anisotropic is when • Homogeneous: Has identical properties at all the body has points. property that varies with orientation. • Isotropic: Has similar properties in all directions or orientation. Macroscopic scale, engineering materials such as steel, cast iron, aluminium seems to be continuous, homogeneous and isotropic. Microscopic scale, metals are made up of an aggregate of crystal grains having different properties in different crystallographic directions. However, these crystal grains are very small, and therefore the properties are homogenous Microstructure of in the macroscopic scale. low carbon steel Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 6. Elastic and plastic behaviour • All solid materials can be deformed stress Maximum when subjected to external load. tensile strength Elastic • In elastic region, stress is proportional to limit Fracture strain. This follows Hook’s law up to the elastic limit . The material now has elastic behaviour. σmax • At the elastic limit, when the load is Elastic Plastic removed, the material will change back to its original shape. strain • Beyond the elastic limit, material Load and extension curve of permanently deformed or the material has under uniaxial tensile loading. undergone plastic deformation. Note: elastic deformations in metals are relatively small in comparison to plastic deformations. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 7. Average stress and strain No load applied Load applied Lo+ δ Lo Lo P • A uniform cylindrical bar which is The external load P is balanced by subjected to an axial tensile load P. the internal resisting force, giving The original gauge length Lo has the equilibrium equation; undergone a slight increase in Internal External length to Lo+ δ with a slight P = ∫ σdA force force decrease in diameter. ∫σdA P • The average elastic strain e is If the stress σ is uniform over the area the ratio of change in length to the A, σ = constant, then original length. P = σ ∫ dA = σA δ∆L L − Lo …Eq.2 e= = = P Lo Lo Lo The average stress σ = …Eq.3 A Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 8. Average stress and strain • In general, the stress is not uniform over the area A average stress. • On a microscopic scale, metals consist of more than one phase and therefore give rise to nonuniformity of stress. stress Maximum tensile strength • Below the elastic limit, Hook’s law Elastic can be applied, so that the average limit Fracture stress is proportional to the average strain, σ …Eq.4 σmax =E ε The constant E is the modulus of elasticity or Young’s modulus. strain Load and extention curve of under uniaxial tensile loading Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 9. Tensile deformation of ductile metal • In tension test, specimen is subjected to axial tensile load until fracture. Load and extension are measured and expressed as stress and strain, see Fig. • The initial linear portion of the curve (OA) is the elastic region, following Hook’s law. D • Point A is the elastic limit (the greatest stress that the metal B can withstand without undergone permanent or plastic Elastic A’ A deformation. A’ is the proportional limit where the curve limit deviates from linearity. • The slope of the linear portion is the modulus of elasticity. • Point B is the yield strength, defined as the stress which will produce a small amount of strain equal to 0.002 (OC). C • As the plastic deformation increases, the metal becomes stronger (strain hardening) until reaching the maximum load, giving ultimate tensile strength D. • Beyond D, metal necks (reduce in x-section). Load needed to continue deformation drops off till failure. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 10. Ductile and brittle behaviour • Completely brittle materials i.e., ceramics would fracture almost at the elastic limit. stress stress • Brittle metals such as cast iron show small amounts of plasticity before failure. strain strain • For engineering materials, Completely brittle Slightly brittle with small amount of adequate ductility is important ductility because it allows the materials to redistribute localised stresses. • Fracture behaviour of metal (ductile or brittle) also depends on some conditions, i.e., temperature, tension or compression, state of stresses, strain rate and embrittling agent. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 11. What constitute fracture? Three general ways that cause failures in structural members and machine elements: 1) Excessive elastic deformation stress 2) Yielding or excessive plastic deformation Plastic energy Elastic energy 3) Fracture strain Types of Load/forces Good structural components in Nature of Materials service without Structural design failure Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 12. Excessive elastic deformation • Failure due to excessive elastic deformation are controlled by the modulus of elasticity not the strength of the materials. Two general types of excessive elastic deformation: 1) Excessive elastic deformation under stress condition of stable equilibrium. Elastic EX: too much deflection in a shaft can cause energy wear in bearing and damage other parts. strain 2) Excessive elastic deformation under condition of unstable equilibrium. EX: A sudden deflection or buckling of a slender column. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 13. Excessive plastic deformation • Excessive plastic deformation occurs when the elastic limit is exceeded yielding. • Yielding produce permanent change of shape and can cause fracture. stress EX: When component changed in shape, it cannot function properly any longer. Plastic Elastic energy energy • Failure by excessive plastic strain deformation is controlled by the yield strength of the materials. Even in more complex loading conditions, yield strength is still a significant parameter. Large deformation that cause failure in a pipe Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 14. Fracture Metals fail by fracture in three general ways: 1) Sudden brittle fracture Ex : some ductile metal such as plain carbon steel will undergo ductile to brittle transition with decreasing temperature, increasing rate of loading and triaxial state of stresses. 2) Fatigue, or progressive fracture Most fracture in machine parts are due to fatigue (subjected to alternating stresses). Failure is due to localised tensile stress at spot or notch or stress concentration. 3) Delayed fracture Ex: Stress-rupture failure, which occurs when a metal has been statistically loaded at an elevated temperature. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 15. Working stress To prevent structural members or machine elements from failure, such members should be used under a stress level that is lower than its yield stress σo. This stress level is called the working stress σw. According to American Society of Mechanical Engineering (ASME), the working stress σw may be considered as either the yield strength σo or the tensile strength σu divided by a number called safety factor. σo σu …Eq.5 σw = or σw = …Eq.6 No Nu Where σw = working stress σo = yield strength σu = tensile strength No = safety factor based on yield strength Nu = safety factor based on tensile strength Safety factor depends on loading/service conditions, consequences, etc. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 16. Concept of stress and the type of stresses Definition: stress is force per unit area. There are two kinds of external forces which may act on a body; 1) Surface forces : forces distributed over the surface of the body, i.e., hydrostatic pressure 2) Body forces : forces distributed over the volume of the body, i.e., gravitational force, magnetic force, centrifugal force, thermal stress. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 17. Stress at a point • Consider a body having forces F1…F4 acting on it, see Fig (a). The body is cut by a plane passing through point O. If one half is removed and replaced by an equivalent force F, acting on the x-sectional area A to remain the static equilibrium, see Fig (b). • We can resolve F into component normal to the plane Fn and tangential to the plane Fs. (a) Equilibrium of an arbitrary body. • The concept of stress at a point is shrinking the area A into infinitesimal dimensions. Fn Fs …Eq.7 σ = lim and τ = lim …Eq.8 ∆A→0 A ∆A→0 A Note: σ and τ depend on the orientation of the plane (b) Force acting passing through P and will vary from point to point. on parts. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 18. The total stress can be resolved into two components; 1) Normal stress σ perpendicular to A. 2) Shear stress τ lying in the plane of the area. • The force F makes an angle θ with the normal z to the plane x-y of the area A. • The plane containing the normal z and F intersects the plane A along a dash line that makes an angle φ with the y axis. The normal stress F The shear stress in the plane is given by σ= cos θ …Eq.9 A acting along OC has the magnitude. F F τ = sin θ …Eq.10 A F x direction τ = sin θ sin φ …Eq.11 A F y direction τ = sin θ cos φ …Eq.12 Resolutions of total stress into its A components. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 19. Concept of strain and types of strain The average linear strain was defined as the ratio of the change in length to the original length. δ∆L L − Lo Where e = average linear strain e= = = Lo Lo Lo δ = deformation. …Eq.2 Strain at a point is the ratio of the deformation to the gauge length as the gauge length 0. True strain or natural strain is the strain as the change in linear dimension divided by the instantaneous value of the dimension. Lf dL Lf ε=∫ = ln …Eq.13 Lo L Lo Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 20. Shear strain • Shear strain is the angular change in a right angle. • The angle at A, which is originally 90o, is decreased by a small amount θ when the shear stress is applied. • The shear strain γ will be given by a a θ γ = = tan θ = θ h h …Eq.14 A Shear strain Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 21. Units of stress and other quantities • Following SI unit Length metre (m) Mass kilogram (kg) Frequency (s-1, Hz) Time second (s) Force Newton (N) Electric current ampere (A) Stress (N.m-2, Pa) Temperature Kelvin (K) Strain dimensionless Amount of substance mole (mol) Luminous intensity candela (cd) Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 22. Example: The shear stress required to nucleate a grain boundary crack in high-temperature deformation has been estimated to be 1  3πγ b G  2 τ =  8(1 − ν ) L     Where γb is the grain boundary surface energy ~ 2 J.m-2, G is shear modulus, 75 GPa, L is the grain boundary sliding distance, assume = grain diameter 0.01 mm, and the Poisson’s ratio ν = 0.3. Checking the unit 1  Nm N  1 2 1  2 × 2  3π × 2 × 75 × 10 9 2 m  =N  = N τ = 2 2 m  = 15.89 × 10 7 N .m −2 τ =  4 m   8(1 − 0.3) × 10 − 2 × 10 −3     m    m 2   τ = 158.9 MN .m −2 = 158.9 MPa   Suranaree University of Technology Tapany Udomphol May-Aug 2007
  • 23. References • Dieter, G.E., Mechanical metallurgy, 1988, SI metric edition, McGraw-Hill, ISBN 0-07-100406-8. • Sanford, R.J., Principles of fracture mechanics, 2003, Prentice Hall, ISBN 0-13-192992-1. • www.indiamart.com. • www.technion.ac.il/.../ Figures/Light-steel.jpg • www.uvi.edu Suranaree University of Technology Tapany Udomphol May-Aug 2007