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Behavior Under Axial

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					                              3/6/2009




Behavior under Axial Forces

       Iswandi Imran, PhD




   Observed Response of Bar
      Encased in Concrete




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                                                                3/6/2009




                       Load Sharing between
                     Concrete and Reinforcement




        Member Subjected to Axial
                Forces



        Compatibility                      Equilibrium
             Δ
ε        =
                                       ∫ fdA = N
    c
             L
ε   s    = ε   c
                                       A

ε        = ε         + Δε        Ac f c + As f s + Ap f p = N
    p            c          p




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Calculation of Strain
    Difference




Time Dependent Effect

   ε c = ε cf + ε sh + ε cth
   ε s = ε sf + ε sth
   ε p = ε pf + ε pth




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        Creep and Relaxation

                                              Eci
                           Ec ,eff =
                                          1 + Φ (t , ti )
                                                     f ci
                          ε cf (t , ti ) =
                                                    Ec ,eff
                                               fp
                           E p ,eff =                Ep
                                            f pi




Stress-Strain Relationship
           ⎡ 2ε cf ⎛ ε cf ⎞ 2 ⎤
f c = f c' ⎢      −⎜
                   ⎜      ⎟ ⎥ if ε cf < ε cr
                          ⎟
           ⎢ ε co ⎝ ε co ⎠ ⎥
           ⎣                  ⎦
             ⎡    f cr         ⎤
f c = α1'α 2 ⎢                 ⎥ if ε cf > ε cr
             ⎢1 + 500ε cf
             ⎣                 ⎥
                               ⎦

f s = Esε sf if ε sf < ε sy
f s = f sy
         y   if ε sff > ε sy
                           y

              ⎡              0.975            ⎤
f p = E pε pf ⎢0.025 +                        ⎥ ≤ 1860 MPa ⇒ Low − relaxation
              ⎢
              ⎣           (
                       1 + (118ε pf )10
                                         0.10
                                              ⎥
                                              ⎦)
              ⎡              0.970            ⎤
f p = E pε pf ⎢0.030 +                        ⎥ ≤ 1860 MPa ⇒ Stress − Re lieved
              ⎢
              ⎣           (
                       1 + (121ε pf )       )
                                      6 0.167
                                              ⎥
                                              ⎦




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                  Linear Elastic Response
Equilibrium                                Strain Relationship

N = Ac f c + As f s + A p f p               ε c = ε cf + ε sh + ε cth
Compatibility                               ε s = ε sf + ε sth
εs = εc
                                            ε p = ε pf + ε pth
ε p = ε c + Δε p
                                           N − ε c Relationship
Constitutive Relationship
f c = Ec ε cf
f s = E s ε sff                                           N − N0
                                                   εc =
f p = E p ε pf                                            Ec Atrans
                                   Es     Ep
                   Atrans = Ac +      As + Ap
                                   Ec     Ec
                   No = ApEpΔε p −(AcEcεsh + AcEcεcth + As Esεsth + ApEpε pth)




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