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-v^{2}) 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,
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]
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.
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