3D stress -strain relationship

  1. εHi,

    Based on that hoary schematic cube representation in 3D for stress-strain relationship. Stress tensor {σxx, σyy, σzz, σxy, σyz, σzx} and strain tensors {εxx, εyy, εzz, εxy, εyz, εzx} can be written interchangably. Let's suppose that σzz=σzx=σzy=0 then the reamining terms are written in matrix form then how that E/(1-v2) coefficient is emerging.

    At first sight, is seems that its very simple to express it by multipliers but couldn't figured out .

    Any help please,
     
  2. jcsd
  3. You are missing the off diagonal relationships in your tensor.
     
  4. The relationships betwenn stress and strain are

    [tex]\begin{array}{l}
    {\varepsilon _x} = \frac{1}{E}\left[ {{\sigma _x} - \nu ({\sigma _y} + {\sigma _z}} \right] \\
    {\varepsilon _y} = \frac{1}{E}\left[ {{\sigma _y} - \nu ({\sigma _x} + {\sigma _z}} \right] \\
    {\varepsilon _z} = \frac{1}{E}\left[ {{\sigma _z} - \nu ({\sigma _x} + {\sigma _y}} \right] \\
    {\gamma _{xy}} = \frac{1}{G}{\tau _{xy}} \\
    {\gamma _{yz}} = \frac{1}{G}{\tau _{yz}} \\
    {\gamma _{xz}} = \frac{1}{G}{\tau _{xz}} \\
    \end{array}[/tex]

    Written in matrix form this becomes

    [tex]\varepsilon = C\sigma [/tex]

    where C is the matrix

    [tex]C = \frac{1}{E}\left[ {\begin{array}{*{20}{c}}
    1 & { - v} & { - v} & 0 & 0 & 0 \\
    { - v} & 1 & { - v} & 0 & 0 & 0 \\
    { - v} & { - v} & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & {\frac{E}{G}} & 0 & 0 \\
    0 & 0 & 0 & 0 & {\frac{E}{G}} & 0 \\
    0 & 0 & 0 & 0 & 0 & {\frac{E}{G}} \\
    \end{array}} \right][/tex]
     
  5. you do this by solving (inverting) my matrix equation

    ε=Cσ

    this is possible and leads to

    [tex]\begin{array}{l}
    {\sigma _x} = A\left[ {{\varepsilon _x} + B\left( {{\varepsilon _y} + {\varepsilon _z}} \right)} \right] \\
    {\sigma _y} = A\left[ {{\varepsilon _y} + B\left( {{\varepsilon _x} + {\varepsilon _z}} \right)} \right] \\
    {\sigma _z} = A\left[ {{\varepsilon _z} + B\left( {{\varepsilon _x} + {\varepsilon _y}} \right)} \right] \\
    {\tau _{xy}} = G{\gamma _{xy}} \\
    {\tau _{yz}} = G{\gamma _{yz}} \\
    {\tau _{xz}} = G{\gamma _{xz}} \\
    \end{array}[/tex]

    Also

    [tex]\begin{array}{l}
    G = \frac{E}{{2\left( {1 + \nu } \right)}} \\
    A = E\frac{{\left( {1 - \nu } \right)}}{{\left( {1 + \nu } \right)\left( {1 - 2\nu } \right)}} \\
    B = \frac{\nu }{{\left( {1 - v} \right)}} \\
    \end{array}[/tex]

    As regards to the LaTex this site uses for formulae, I am seriously deficient in latex lore so I use MathType and copy/paste.
    I wouldn't recommend MT, however as it is too expensive for what it is and does not allow the inclusion of images.
     
  6. There seems to be a missing post, however to complete a matrix expression for stress in terms of strain is

    [tex]\sigma = D\varepsilon [/tex]

    where D is the matrix

    [tex]D = A\left[ {\begin{array}{*{20}{c}}
    1 & B & B & 0 & 0 & 0 \\
    B & 1 & B & 0 & 0 & 0 \\
    B & B & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & {\frac{G}{A}} & 0 & 0 \\
    0 & 0 & 0 & 0 & {\frac{G}{A}} & 0 \\
    0 & 0 & 0 & 0 & 0 & {\frac{G}{A}} \\
    \end{array}} \right][/tex]

    A and B have the same meaning as before
     
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