In the case of isotropic linear elasticity, we need only two material parameters to describe the C C C
Lame parameters: λ \lambda λ μ \mu μ
Young's modulus E E E G G G
Young's modulus E E E ν \nu ν
In terms of Lame parameter C C C
C i j k l = λ δ i j δ k l + μ ( δ i k δ j l + δ i l δ j k ) C_{ijkl}=\lambda\delta_{ij}\delta_{kl}+\mu\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right) C ijk l = λ δ ij δ k l + μ ( δ ik δ j l + δ i l δ jk ) The stress-strain relationship is given by following:
σ i j = λ ε k k δ i j + 2 μ ε i j \sigma_{ij}=\lambda\varepsilon_{kk}\delta_{ij}+2\mu\varepsilon_{ij} σ ij = λ ε kk δ ij + 2 μ ε ij or
σ i j = ( 3 λ + 2 μ ) ε k k 3 δ i j + 2 μ dev ( ε i j ) \sigma_{ij}=\left(3\lambda+ 2\mu\right)\frac{\varepsilon_{kk}}{3}\delta_{ij}+2\mu\text{dev}\left(\varepsilon_{ij}\right) σ ij = ( 3 λ + 2 μ ) 3 ε kk δ ij + 2 μ dev ( ε ij ) where dev ( ε ) \text{dev}(\varepsilon) dev ( ε )
dev ( ε ) = ε − 1 3 tr ( ε ) 1 \text{dev}(\varepsilon) = \varepsilon - \frac{1}{3} \text{tr}(\varepsilon) \textbf{1} dev ( ε ) = ε − 3 1 tr ( ε ) 1 The term ( 3 λ + 2 μ ) \left(3\lambda+ 2\mu \right) ( 3 λ + 2 μ )
The Voigt form of the stiffness tensor C C C E E E ν \nu ν
C = E ( 1 + ν ) ( 1 − 2 ν ) [ 1 − ν ν ν ν 1 − ν ν ν ν 1 − ν 1 − 2 ν 2 1 − 2 ν 2 1 − 2 ν 2 ] C = \frac{E}{(1+\nu)(1-2\nu)}\left[\begin{array}{cccccc}
1-\nu & \nu & \nu\\
\nu & 1-\nu & \nu\\
\nu & \nu & 1-\nu\\
& & & \frac{1-2\nu}{2}\\
& & & & \frac{1-2\nu}{2}\\
& & & & & \frac{1-2\nu}{2}
\end{array}\right] C = ( 1 + ν ) ( 1 − 2 ν ) E 1 − ν ν ν ν 1 − ν ν ν ν 1 − ν 2 1 − 2 ν 2 1 − 2 ν 2 1 − 2 ν Similary, the inverse of C C C
C − 1 = 1 E [ 1 − ν − ν − ν 1 − ν − ν − ν 1 2 + 2 ν 2 + 2 ν 2 + 2 ν ] C^{-1} =\frac{1}{E}\left[\begin{array}{cccccc}
1 & -\nu & -\nu\\
-\nu & 1 & -\nu\\
-\nu & -\nu & 1\\
& & & 2+2\nu\\
& & & & 2+2\nu\\
& & & & & 2+2\nu
\end{array}\right] C − 1 = E 1 1 − ν − ν − ν 1 − ν − ν − ν 1 2 + 2 ν 2 + 2 ν 2 + 2 ν Plane stress
In the case of plane-stress, we use the following relationship between stress and strain.
[ σ 11 σ 22 σ 12 ] = E 1 − ν 2 [ 1 ν 0 ν 1 0 0 0 1 − ν 2 ] [ ε 11 ε 22 2 ε 12 ] \begin{bmatrix}\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}
\end{bmatrix}=\frac{E}{1-\nu^{2}}\begin{bmatrix}1 & \nu & 0\\
\nu & 1 & 0\\
0 & 0 & \frac{1-\nu}{2}
\end{bmatrix}\begin{bmatrix}\varepsilon_{11}\\
\varepsilon_{22}\\
2\varepsilon_{12}
\end{bmatrix} σ 11 σ 22 σ 12 = 1 − ν 2 E 1 ν 0 ν 1 0 0 0 2 1 − ν ε 11 ε 22 2 ε 12 [ ε 11 ε 22 2 ε 12 ] = 1 E [ 1 − ν 0 − ν 1 0 0 0 2 + 2 ν ] [ σ 11 σ 22 σ 12 ] \begin{bmatrix}\varepsilon_{11}\\
\varepsilon_{22}\\
2\varepsilon_{12}
\end{bmatrix}=\frac{1}{E}\begin{bmatrix}1 & -\nu & 0\\
-\nu & 1 & 0\\
0 & 0 & 2+2\nu
\end{bmatrix}\begin{bmatrix}\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}
\end{bmatrix} ε 11 ε 22 2 ε 12 = E 1 1 − ν 0 − ν 1 0 0 0 2 + 2 ν σ 11 σ 22 σ 12 Plane strain
Similary, in the case of plane-strain, we use the following relationship between stress and strain.
[ σ 11 σ 22 σ 12 ] = E ( 1 + ν ) ( 1 − 2 ν ) [ 1 − ν ν 0 ν 1 − ν 0 0 0 1 − 2 ν 2 ] [ ε 11 ε 22 2 ε 12 ] \begin{bmatrix}\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}
\end{bmatrix}=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu & \nu & 0\\
\nu & 1-\nu & 0\\
0 & 0 & \frac{1-2\nu}{2}
\end{bmatrix}\begin{bmatrix}\varepsilon_{11}\\
\varepsilon_{22}\\
2\varepsilon_{12}
\end{bmatrix} σ 11 σ 22 σ 12 = ( 1 + ν ) ( 1 − 2 ν ) E 1 − ν ν 0 ν 1 − ν 0 0 0 2 1 − 2 ν ε 11 ε 22 2 ε 12 [ ε 11 ε 22 2 ε 12 ] = 1 + ν E [ 1 − ν − ν 0 − ν 1 − ν 0 0 0 2 ] [ σ 11 σ 22 σ 12 ] \begin{bmatrix}\varepsilon_{11}\\
\varepsilon_{22}\\
2\varepsilon_{12}
\end{bmatrix}=\frac{1+\nu}{E}\begin{bmatrix}1-\nu & -\nu & 0\\
-\nu & 1-\nu & 0\\
0 & 0 & 2
\end{bmatrix}\begin{bmatrix}\sigma_{11}\\
\sigma_{22}\\
\sigma_{12}
\end{bmatrix} ε 11 ε 22 2 ε 12 = E 1 + ν 1 − ν − ν 0 − ν 1 − ν 0 0 0 2 σ 11 σ 22 σ 12