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2. Strain
CHAPTER OBJECTIVES
   • Define concept of
     normal strain
   • Define concept of
     shear strain
   • Determine normal
     and shear strain in
     engineering
     applications




©2005 Pearson Education South Asia Pte Ltd   1
2. Strain
CHAPTER OUTLINE
  1. Deformation
  2. Strain




©2005 Pearson Education South Asia Pte Ltd   2
2. Strain
 2.1 DEFORMATION
     Deformation
     • Occurs when a force is applied to a body
     • Can be highly visible or practically unnoticeable
     • Can also occur when temperature of a body is
       changed
     • Is not uniform throughout a body’s volume, thus
       change in geometry of any line segment within
       body may vary along its length




©2005 Pearson Education South Asia Pte Ltd                 3
2. Strain
 2.1 DEFORMATION
     To simplify study of deformation
     • Assume lines to be very short and located in
       neighborhood of a point, and
     • Take into account the orientation of the line
       segment at the point




©2005 Pearson Education South Asia Pte Ltd             4
2. Strain
 2.2 STRAIN
     Normal strain
     • Defined as the elongation or contraction of a line
       segment per unit of length
     • Consider line AB in figure below
     • After deformation, Δs changes to Δs’




©2005 Pearson Education South Asia Pte Ltd                  5
2. Strain
 2.2 STRAIN
     Normal strain
     • Defining average normal strain using εavg (epsilon)

                                                        Δs − Δs’
                                                 εavg =
                                                          Δs

     • As Δs → 0, Δs’ → 0

                                                   lim     Δs − Δs’
                                             ε=
                                                B→A along n Δs



©2005 Pearson Education South Asia Pte Ltd                            6
2. Strain
 2.2 STRAIN
     Normal strain
     • If normal strain ε is known, use the equation to
       obtain approx. final length of a short line segment
       in direction of n after deformation.

                                             Δs’ ≈ (1 + ε) Δs

     • Hence, when ε is positive, initial line will elongate,
       if ε is negative, the line contracts



©2005 Pearson Education South Asia Pte Ltd                      7
2. Strain
 2.2 STRAIN
       Units
       • Normal strain is a dimensionless quantity, as
         it’s a ratio of two lengths
       • But common practice to state it in terms of
         meters/meter (m/m)
       • ε is small for most engineering applications, so
         is normally expressed as micrometers per
         meter (μm/m) where 1 μm = 10−6
       • Also expressed as a percentage,
         e.g., 0.001 m/m = 0.1 %



©2005 Pearson Education South Asia Pte Ltd                  8
2. Strain
 2.2 STRAIN
       Shear strain
       • Defined as the change in angle that occurs
         between two line segments that were originally
         perpendicular to one another
       • This angle is denoted by γ (gamma) and
         measured in radians (rad).




©2005 Pearson Education South Asia Pte Ltd                9
2. Strain
 2.2 STRAIN
       Shear strain
       • Consider line segments AB and AC originating
         from same point A in a body, and directed along
         the perpendicular n and t axes
       • After deformation, lines become curves, such that
         angle between them at A is θ’




©2005 Pearson Education South Asia Pte Ltd                   10
2. Strain
 2.2 STRAIN
       Shear strain
       • Hence, shear strain at point A associated with n
         and t axes is
                                                   π      lim
                                             γnt =   −
                                                   2   B→A along n θ’
                                                       C →A along t


       • If θ’ is smaller than π/2, shear strain is positive,
         otherwise, shear strain is negative


©2005 Pearson Education South Asia Pte Ltd                              11
2. Strain
 2.2 STRAIN
   Cartesian strain components
   • Using above definitions of normal and shear strain,
     we show how they describe the deformation of the
     body
   • Divide body into small
     elements with
     undeformed dimensions
     of Δx, Δy and Δz




©2005 Pearson Education South Asia Pte Ltd                 12
2. Strain
 2.2 STRAIN
     Cartesian strain components
     • Since element is very small, deformed shape of
        element is a parallelepiped




     •          Approx. lengths of sides of parallelepiped are
                (1 + εx) Δx                  (1 + εy)Δy   (1 + εz)Δz


©2005 Pearson Education South Asia Pte Ltd                             13
2. Strain
 2.2 STRAIN
     Cartesian strain components
     • Approx. angles between the sides are
         π               π               π
           − γxy           − γyz            − γxz
         2               2               2

     •        Normal strains cause a change in its volume
     •        Shear strains cause a change in its shape
     •        To summarize, state of strain at a point requires
              specifying 3 normal strains; εx, εy, εz and 3 shear
              strains of γxy, γyz, γxz


©2005 Pearson Education South Asia Pte Ltd                          14
2. Strain
 2.2 STRAIN
     Small strain analysis
     • Most engineering design involves applications
       for which only small deformations are allowed
     • We’ll assume that deformations that take place
       within a body are almost infinitesimal, so normal
       strains occurring within material are very small
       compared to 1, i.e., ε << 1.




©2005 Pearson Education South Asia Pte Ltd                 15
2. Strain
 2.2 STRAIN
     Small strain analysis
     • This assumption is widely applied in practical
       engineering problems, and is referred to as
                small strain analysis
     •          E.g., it can be used to approximate sin θ = θ, cos
                θ = θ and tan θ = θ, provided θ is small




©2005 Pearson Education South Asia Pte Ltd                           16
2. Strain
 EXAMPLE 2.1
        Rod below is subjected to temperature increase
        along its axis, creating a normal strain of
        εz = 40(10−3)z1/2, where z is given in meters.
        Determine
        (a) displacement of end B of rod
        due to temperature increase,
        (b) average normal strain in the
        rod.



©2005 Pearson Education South Asia Pte Ltd               17
2. Strain
 EXAMPLE 2.1 (SOLN)
   (a) Since normal strain reported at each point along
       the rod, a differential segment dz, located at
       position z has a deformed length:
                                             dz’ = [1 + 40(10−3)z1/2] dz




©2005 Pearson Education South Asia Pte Ltd                                 18
2. Strain
 EXAMPLE 2.1 (SOLN)
   (a) Sum total of these segments along axis yields
       deformed length of the rod, i.e.,
           0.2 m
              z’ = ∫0 [1 + 40(10−3)z1/2] dz
                 = z + 40(10−3)(⅔ z3/2)|00.2 m
                     = 0.20239 m

              Displacement of end of rod is
              ΔB = 0.20239 m − 0.2 m = 2.39 mm ↓




©2005 Pearson Education South Asia Pte Ltd             19
2. Strain
 EXAMPLE 2.1 (SOLN)
   (b) Assume rod or “line segment” has original
       length of 200 mm and a change in length of
       2.39 mm. Hence,
                       Δs’ − Δs   2.39 mm
                εavg =          =         = 0.0119 mm/mm
                         Δs       200 mm




©2005 Pearson Education South Asia Pte Ltd                 20
2. Strain
 EXAMPLE 2.3
     Plate is deformed as shown in figure. In this
     deformed shape, horizontal lines on the on plate
     remain horizontal and do not change their length.
     Determine
     (a) average normal strain
         along side AB,
     (b) average shear strain
         in the plate relative to
         x and y axes



©2005 Pearson Education South Asia Pte Ltd               21
2. Strain
 EXAMPLE 2.3 (SOLN)
    (a) Line AB, coincident with y axis, becomes line
        AB’ after deformation. Length of line AB’ is
                 AB’ = √ (250 − 2)2 + (3)2 = 248.018 mm




©2005 Pearson Education South Asia Pte Ltd                22
2. Strain
 EXAMPLE 2.3 (SOLN)
    (a) Therefore, average normal strain for AB is,
                          AB’ − AB 248.018 mm − 250 mm
               (εAB)avg =         =
                            AB            250 mm

                                    = −7.93(10−3) mm/mm

               Negative sign means
               strain causes a
               contraction of AB.




©2005 Pearson Education South Asia Pte Ltd                23
2. Strain
 EXAMPLE 2.3 (SOLN)
    (b) Due to displacement of B to B’, angle BAC
        referenced from x, y axes changes to θ’.
        Since γxy = π/2 − θ’, thus
                 γxy =         tan−1         (        3 mm
                                                 250 mm − 2 mm)= 0.0121 rad




©2005 Pearson Education South Asia Pte Ltd                                    24
2. Strain
 CHAPTER REVIEW
        •          Loads cause bodies to deform, thus points in
                   the body will undergo displacements or
                   changes in position
        •          Normal strain is a measure of elongation or
                   contraction of small line segment in the body
        •          Shear strain is a measure of the change in
                   angle that occurs between two small line
                   segments that are originally perpendicular to
                   each other

©2005 Pearson Education South Asia Pte Ltd                         25
2. Strain
 CHAPTER REVIEW
        •          State of strain at a point is described by six
                   strain components:
                        a) Three normal strains: εx, εy, εz
                        b) Three shear strains: γxy, γxz, γyz
                        c) These components depend upon the orientation of
                           the line segments and their location in the body
        •          Strain is a geometrical quantity measured
                   by experimental techniques. Stress in body
                   is then determined from material property
                   relations

©2005 Pearson Education South Asia Pte Ltd                                    26
2. Strain
 CHAPTER REVIEW
        •          Most engineering materials undergo small
                   deformations, so normal strain ε << 1.
                   This assumption of “small strain analysis”
                   allows us to simplify calculations for
                   normal strain, since first-order
                   approximations can be made about their
                   size




©2005 Pearson Education South Asia Pte Ltd                      27

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Files 2 handouts-ch02

  • 1. 2. Strain CHAPTER OBJECTIVES • Define concept of normal strain • Define concept of shear strain • Determine normal and shear strain in engineering applications ©2005 Pearson Education South Asia Pte Ltd 1
  • 2. 2. Strain CHAPTER OUTLINE 1. Deformation 2. Strain ©2005 Pearson Education South Asia Pte Ltd 2
  • 3. 2. Strain 2.1 DEFORMATION Deformation • Occurs when a force is applied to a body • Can be highly visible or practically unnoticeable • Can also occur when temperature of a body is changed • Is not uniform throughout a body’s volume, thus change in geometry of any line segment within body may vary along its length ©2005 Pearson Education South Asia Pte Ltd 3
  • 4. 2. Strain 2.1 DEFORMATION To simplify study of deformation • Assume lines to be very short and located in neighborhood of a point, and • Take into account the orientation of the line segment at the point ©2005 Pearson Education South Asia Pte Ltd 4
  • 5. 2. Strain 2.2 STRAIN Normal strain • Defined as the elongation or contraction of a line segment per unit of length • Consider line AB in figure below • After deformation, Δs changes to Δs’ ©2005 Pearson Education South Asia Pte Ltd 5
  • 6. 2. Strain 2.2 STRAIN Normal strain • Defining average normal strain using εavg (epsilon) Δs − Δs’ εavg = Δs • As Δs → 0, Δs’ → 0 lim Δs − Δs’ ε= B→A along n Δs ©2005 Pearson Education South Asia Pte Ltd 6
  • 7. 2. Strain 2.2 STRAIN Normal strain • If normal strain ε is known, use the equation to obtain approx. final length of a short line segment in direction of n after deformation. Δs’ ≈ (1 + ε) Δs • Hence, when ε is positive, initial line will elongate, if ε is negative, the line contracts ©2005 Pearson Education South Asia Pte Ltd 7
  • 8. 2. Strain 2.2 STRAIN Units • Normal strain is a dimensionless quantity, as it’s a ratio of two lengths • But common practice to state it in terms of meters/meter (m/m) • ε is small for most engineering applications, so is normally expressed as micrometers per meter (μm/m) where 1 μm = 10−6 • Also expressed as a percentage, e.g., 0.001 m/m = 0.1 % ©2005 Pearson Education South Asia Pte Ltd 8
  • 9. 2. Strain 2.2 STRAIN Shear strain • Defined as the change in angle that occurs between two line segments that were originally perpendicular to one another • This angle is denoted by γ (gamma) and measured in radians (rad). ©2005 Pearson Education South Asia Pte Ltd 9
  • 10. 2. Strain 2.2 STRAIN Shear strain • Consider line segments AB and AC originating from same point A in a body, and directed along the perpendicular n and t axes • After deformation, lines become curves, such that angle between them at A is θ’ ©2005 Pearson Education South Asia Pte Ltd 10
  • 11. 2. Strain 2.2 STRAIN Shear strain • Hence, shear strain at point A associated with n and t axes is π lim γnt = − 2 B→A along n θ’ C →A along t • If θ’ is smaller than π/2, shear strain is positive, otherwise, shear strain is negative ©2005 Pearson Education South Asia Pte Ltd 11
  • 12. 2. Strain 2.2 STRAIN Cartesian strain components • Using above definitions of normal and shear strain, we show how they describe the deformation of the body • Divide body into small elements with undeformed dimensions of Δx, Δy and Δz ©2005 Pearson Education South Asia Pte Ltd 12
  • 13. 2. Strain 2.2 STRAIN Cartesian strain components • Since element is very small, deformed shape of element is a parallelepiped • Approx. lengths of sides of parallelepiped are (1 + εx) Δx (1 + εy)Δy (1 + εz)Δz ©2005 Pearson Education South Asia Pte Ltd 13
  • 14. 2. Strain 2.2 STRAIN Cartesian strain components • Approx. angles between the sides are π π π − γxy − γyz − γxz 2 2 2 • Normal strains cause a change in its volume • Shear strains cause a change in its shape • To summarize, state of strain at a point requires specifying 3 normal strains; εx, εy, εz and 3 shear strains of γxy, γyz, γxz ©2005 Pearson Education South Asia Pte Ltd 14
  • 15. 2. Strain 2.2 STRAIN Small strain analysis • Most engineering design involves applications for which only small deformations are allowed • We’ll assume that deformations that take place within a body are almost infinitesimal, so normal strains occurring within material are very small compared to 1, i.e., ε << 1. ©2005 Pearson Education South Asia Pte Ltd 15
  • 16. 2. Strain 2.2 STRAIN Small strain analysis • This assumption is widely applied in practical engineering problems, and is referred to as small strain analysis • E.g., it can be used to approximate sin θ = θ, cos θ = θ and tan θ = θ, provided θ is small ©2005 Pearson Education South Asia Pte Ltd 16
  • 17. 2. Strain EXAMPLE 2.1 Rod below is subjected to temperature increase along its axis, creating a normal strain of εz = 40(10−3)z1/2, where z is given in meters. Determine (a) displacement of end B of rod due to temperature increase, (b) average normal strain in the rod. ©2005 Pearson Education South Asia Pte Ltd 17
  • 18. 2. Strain EXAMPLE 2.1 (SOLN) (a) Since normal strain reported at each point along the rod, a differential segment dz, located at position z has a deformed length: dz’ = [1 + 40(10−3)z1/2] dz ©2005 Pearson Education South Asia Pte Ltd 18
  • 19. 2. Strain EXAMPLE 2.1 (SOLN) (a) Sum total of these segments along axis yields deformed length of the rod, i.e., 0.2 m z’ = ∫0 [1 + 40(10−3)z1/2] dz = z + 40(10−3)(⅔ z3/2)|00.2 m = 0.20239 m Displacement of end of rod is ΔB = 0.20239 m − 0.2 m = 2.39 mm ↓ ©2005 Pearson Education South Asia Pte Ltd 19
  • 20. 2. Strain EXAMPLE 2.1 (SOLN) (b) Assume rod or “line segment” has original length of 200 mm and a change in length of 2.39 mm. Hence, Δs’ − Δs 2.39 mm εavg = = = 0.0119 mm/mm Δs 200 mm ©2005 Pearson Education South Asia Pte Ltd 20
  • 21. 2. Strain EXAMPLE 2.3 Plate is deformed as shown in figure. In this deformed shape, horizontal lines on the on plate remain horizontal and do not change their length. Determine (a) average normal strain along side AB, (b) average shear strain in the plate relative to x and y axes ©2005 Pearson Education South Asia Pte Ltd 21
  • 22. 2. Strain EXAMPLE 2.3 (SOLN) (a) Line AB, coincident with y axis, becomes line AB’ after deformation. Length of line AB’ is AB’ = √ (250 − 2)2 + (3)2 = 248.018 mm ©2005 Pearson Education South Asia Pte Ltd 22
  • 23. 2. Strain EXAMPLE 2.3 (SOLN) (a) Therefore, average normal strain for AB is, AB’ − AB 248.018 mm − 250 mm (εAB)avg = = AB 250 mm = −7.93(10−3) mm/mm Negative sign means strain causes a contraction of AB. ©2005 Pearson Education South Asia Pte Ltd 23
  • 24. 2. Strain EXAMPLE 2.3 (SOLN) (b) Due to displacement of B to B’, angle BAC referenced from x, y axes changes to θ’. Since γxy = π/2 − θ’, thus γxy = tan−1 ( 3 mm 250 mm − 2 mm)= 0.0121 rad ©2005 Pearson Education South Asia Pte Ltd 24
  • 25. 2. Strain CHAPTER REVIEW • Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position • Normal strain is a measure of elongation or contraction of small line segment in the body • Shear strain is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to each other ©2005 Pearson Education South Asia Pte Ltd 25
  • 26. 2. Strain CHAPTER REVIEW • State of strain at a point is described by six strain components: a) Three normal strains: εx, εy, εz b) Three shear strains: γxy, γxz, γyz c) These components depend upon the orientation of the line segments and their location in the body • Strain is a geometrical quantity measured by experimental techniques. Stress in body is then determined from material property relations ©2005 Pearson Education South Asia Pte Ltd 26
  • 27. 2. Strain CHAPTER REVIEW • Most engineering materials undergo small deformations, so normal strain ε << 1. This assumption of “small strain analysis” allows us to simplify calculations for normal strain, since first-order approximations can be made about their size ©2005 Pearson Education South Asia Pte Ltd 27