Cubic anisotropy

The cubic crystal shown is mapped onto itself by two different transformations: rotations by 90 ° around one of the spatial directions shown in black or 120 ° rotations around a spatial diagonal (red). After such rotations, the sample of a cubic anisotropic material shows unchanged behavior.

Cubic crystal structure of the diamond.

The cubic anisotropy , belonging to the crystal system of the same name, is the simplest type of directional dependence of a material. The cubic crystal system is one of the seven crystal systems in crystallography . It includes all point groups that each have a threefold rotational or rotational inversion axis in four different directions . These four threefold axes run in cubic crystals along the four spatial diagonals of the unit cells , the shape of which corresponds to a cube , see picture above. In cubic anisotropic materials, the atoms are arranged in a hexagonal lattice like in the upper picture or like in the diamond , which is the animation below.

Cubic anisotropic materials have the following properties:

1. The force-deformation behavior does not change if the material is rotated by 90 ° around certain mutually perpendicular axes, the orthotropic axes (black in the picture, table salt structure ).
2. In the reference system parallel to these axes, there is no coupling between normal strains and shear distortions.
3. An equally unchanged force-deformation behavior is shown when the material is rotated by 120 ° around certain axes (the spatial diagonals, one of which is red in the picture) or by 180 ° around the orthotropic axes ( diamond structure , see animation on the right).

As a result, the cubic anisotropy has two groups of symmetry.

The special case that a material (on a particle) shows the same force-deformation behavior regardless of the direction of load, is called isotropy . In contrast, the general case that the force-deformation behavior depends on the direction of loading is called anisotropy . The cubic anisotropy is a special case of the orthotropy and contains the isotropy as a special case. A non-isotropic, cubically anisotropic material is not transversely isotropic . (Transverse isotropy is another special case of orthotropy and also contains isotropy as a special case.)

Many metals and their salts are cubically anisotropic, e.g. B. Semiconductor metals , which play an important role in semiconductor technology in electronics , alkali metals and their salts.

Symmetry group

The directional dependence of a material is characterized by the fact that the force-deformation behavior is independent (invariant) with respect to only certain rotations of the material. These rotations form the symmetry group of the cubic anisotropic material. The cubic anisotropy has - as mentioned at the beginning - two groups of symmetry.

One group includes all 90 ° rotations around three specific, mutually perpendicular axes, which are called orthotropic axes. The invariance of these rotations of the material is illustrated by two experiments on a particle: In the first experiment, a certain force is applied to the particle and the resulting deformation is measured. In the second experiment, the material is rotated one after the other about any orthotropic axes - each by 90 °. Then you apply the same force as in the first experiment and measure the deformation again. With cubic anisotropic material, the same deformation will be measured in the second experiment as in the first. Even with non-linear elastic material behavior.

The other symmetry group includes 120 ° rotations around a certain axis, the (111) space diagonal , and certain 180 ° rotations around the orthotropic axes. The invariance of these rotations of the material is shown in the second experiment when the particle is rotated 120 ° around the (111) space diagonal or 180 ° around an orthotropic axis. If you apply the same force as in the first experiment and measure the deformation again, then in the case of cubic anisotropic material you will measure the same deformation in the second experiment as in the first.

The dependence on the rotations of the material can be seen if one does not rotate in multiples of 30 ° in the second experiment. If the special case isotropy is not present, one will now always measure a different deformation than in the first experiment.

Cubic anisotropy in linear elasticity

Are given two tensors second stage and with 3 × 3 coefficients respectively . The most general linear relationship that exists between these coefficients is: ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle {\boldsymbol {\varepsilon }}}$ ${\displaystyle \sigma _{ij}}$ ${\displaystyle \varepsilon _{ij}}$

${\displaystyle \sigma _{ij}=\sum _{k,l=1}^{3}C_{ijkl}\varepsilon _{kl}\,,\quad i,j=1,2,3\,.}$

This contains 81 coefficients with which the nine components are mapped to nine components . In linear elasticity theory , in which there is the symmetric stress tensor and the symmetric strain tensor , the number of independent tensor components is reduced to six, so that only 36 coefficients are independent. In the case of hyperelasticity , the symmetry is still present, so that only 21 coefficients are then independent. This relationship between stresses and distortions can now also be written in Voigt's notation as a matrix equation: ${\displaystyle C_{ijkl}}$ ${\displaystyle \varepsilon _{kl}}$ ${\displaystyle \sigma _{ij}}$ ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle {\boldsymbol {\varepsilon }}}$ ${\displaystyle C_{ijkl}}$ ${\displaystyle C_{ijkl}=C_{klij}}$

${\displaystyle \sigma _{i}^{\text{v}}=\sum _{j=1}^{3}C_{ij}^{\text{v}}\varepsilon _{j}^{\text{v}}\Leftrightarrow {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\\\end{bmatrix}}={\begin{bmatrix}C_{11}^{\text{v}}&C_{12}^{\text{v}}&C_{13}^{\text{v}}&C_{14}^{\text{v}}&C_{15}^{\text{v}}&C_{16}^{\text{v}}\\&C_{22}^{\text{v}}&C_{23}^{\text{v}}&C_{24}^{\text{v}}&C_{25}^{\text{v}}&C_{26}^{\text{v}}\\&&C_{33}^{\text{v}}&C_{34}^{\text{v}}&C_{35}^{\text{v}}&C_{36}^{\text{v}}\\&&&C_{44}^{\text{v}}&C_{45}^{\text{v}}&C_{46}^{\text{v}}\\&{\text{sym.}}&&&C_{55}^{\text{v}}&C_{56}^{\text{v}}\\&&&&&C_{66}^{\text{v}}\end{bmatrix}}\,{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\\\end{bmatrix}}}$

Here the fourfold indexing ijkl with the scheme 11 → 1, 22 → 2, 33 → 3, 23 → 4, 13 → 5, 12 → 6 has been reduced to a clearer double indexing. The stiffness matrix with the 21 independent components represents the elasticity tensor of the material. The superscript v is suppressed in the following, since the number of indices already characterizes Voigt's notation. ${\displaystyle C^{\text{v}}}$ ${\displaystyle C_{ij}^{\text{v}}}$

Law of elasticity and anisotropy factor

A material is cubic anisotropically linearly elastic if an orthonormal basis exists, so that the law of elasticity represented in relation to this basis takes the following form (with only three independent entries):

${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\\\end{bmatrix}}={\begin{bmatrix}C_{11}&C_{12}&C_{12}&0&0&0\\&C_{11}&C_{12}&0&0&0\\&&C_{11}&0&0&0\\&&&C_{44}&0&0\\&{\text{sym.}}&&&C_{44}&0\\&&&&&C_{44}\end{bmatrix}}\,{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\\\end{bmatrix}}}$

In the special case of isotropy , only two of the three entries are independent and with the shear modulus G and the second Lamé constant λ:

${\displaystyle C_{11}=2G+\lambda \,,\quad C_{12}=\lambda \,,\quad C_{44}=G\quad \rightarrow \quad {\frac {2C_{44}}{C_{11}-C_{12}}}={\frac {2G}{2G+\lambda -\lambda }}=1\,.}$

The fraction defines the dimensionless anisotropy factor according to Zener:

${\displaystyle A={\frac {2C_{44}}{C_{11}-C_{12}}}\,.}$

The anisotropy factor is different from one for cubic anisotropy, as the following table shows.

material	${\displaystyle C_{11}}$[ MPa ]	${\displaystyle C_{12}}$ [MPa]	${\displaystyle C_{44}}$ [MPa]	A [-]
Semiconductor (diamond structure)
diamond	1,020,000	250,000	492,000	1.3
silicon	166,000	64,000	80,000	1.6
Germanium	130,000	49,000	67,000	1.7
Alkali metals (body- centered cubic lattice )
lithium	13,500	11,400	8,800	8.4
sodium	7,400	6,200	4,200	7.2
potassium	3,700	3,100	1,900	6.7
Alkali metal chlorides ( table salt structure )
Sodium chloride	48,500	12,500	12,700	0.7
Potassium chloride	40,500	6,600	6,300	0.37
Rubidium chloride	36,300	6,200	4,700	0.31

If the anisotropy factor is less than one, then the crystals are stiffest along the (100) cube edges. If the anisotropy factor is greater than one, then the crystals along the (111) -space diagonal are the most rigid, see also the #example below. You can look up other cubic crystallizing chemical substances in the category: Cubic crystal system .

Material parameters

Cubic anisotropy is a special case of orthotropy , which has nine material parameters for linear elasticity (three moduli of elasticity, three transverse contraction numbers when pulling in the direction of an orthotropic axis, and three shear moduli when shearing in planes perpendicular to the orthotropic axes). The cubic anisotropy has 3 independent parameters:

• a modulus of elasticity,
• a Poisson's ratio and
• a shear module

The dimension of the modulus of elasticity and the modulus of shear is force per area, while the Poisson's ratio is dimensionless. The Poisson's ratio describes how one moves along a direction - e.g. B. the 1-direction - drawn material sample across it - z. B. in 2-direction - contracted. ${\displaystyle E}$ ${\displaystyle G}$ ${\displaystyle \nu }$

The law of elasticity for cubic anisotropic, linear elasticity is:

${\displaystyle {\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\\\end{bmatrix}}=\underbrace {\begin{bmatrix}{\frac {1}{E}}&-{\frac {\nu }{E}}&-{\frac {\nu }{E}}&0&0&0\\&{\frac {1}{E}}&-{\frac {\nu }{E}}&0&0&0\\&&{\frac {1}{E}}&0&0&0\\&&&{\frac {1}{G}}&0&0\\&{\textsf {sym}}&&&{\frac {1}{G}}&0\\&&&&&{\frac {1}{G}}\end{bmatrix}} _{=:S}\,{\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\\\end{bmatrix}}}$

By inverting the compliance matrix S one gets the stiffness matrix C:

${\displaystyle {\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{13}\\\sigma _{12}\\\end{bmatrix}}=\underbrace {\begin{bmatrix}{\frac {1-\nu }{\nu }}\lambda &\lambda &\lambda &0&0&0\\&{\frac {1-\nu }{\nu }}\lambda &\lambda &0&0&0\\&&{\frac {1-\nu }{\nu }}\lambda &\\&&&G&0&0\\&{\text{sym}}&&&G&0\\&&&&&G\end{bmatrix}} _{=:C}\,{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\\\end{bmatrix}}}$

The compliance matrix S and the stiffness matrix C are symmetrical and have values ​​other than zero at the same points. The second Lamé constant hangs over

${\displaystyle \lambda ={\frac {\nu E}{(1+\nu )(1-2\nu )}}}$

with the modulus of elasticity and the Poisson's ratio. The anisotropy factor can be expressed with these material parameters:

${\displaystyle A={\frac {2G(1+\nu )}{E}}={\frac {2\nu G}{(1-2\nu )\lambda }}\,.}$

Isotropy occurs when A = 1 or equivalent

${\displaystyle G={\frac {E}{2(1+\nu )}}\quad {\textsf {oder}}\quad G={\frac {1-2\nu }{2\nu }}\lambda \quad {\textsf {oder}}\quad 2G+\lambda ={\frac {1-\nu }{\nu }}\lambda }$

applies.

Stability criteria

The material parameters cannot be chosen arbitrarily, but must meet certain stability criteria. These follow from the requirement that the stiffness and compliance matrices must be positive and definite . This leads to the conditions:

• All diagonal elements of the stiffness and compliance matrix must be positive (so that the material stretches in the tensile direction when you pull on it and not compresses) and
• the determinant of the stiffness and compliance matrix must be positive (so that it compresses under pressure and does not expand).

If material parameters are identified in a real material that contradict these stability criteria, caution is required. The stability criteria are:

${\displaystyle {\begin{array}{l}E,G>0\\\nu >-1\\1-2\nu >0\end{array}}}$

As the left hand side of the last inequality approaches zero, the material is increasingly resisting hydrostatic compression.

Coordinate-free representation

In the case of orthtropic linear elasticity, the material has no tensile-shear coupling but three preferred directions, the pairwise perpendicular orthotropic axes, in which the material has different properties than perpendicular to it. Because two orthotropic axes already define the third, only two material line elements of length one are required to define the preferred directions . Tensor structural variables are formed from these with the dyadic product " " : ${\displaystyle {\hat {k}}_{1,2}}$ ${\displaystyle {\hat {k}}_{1}\cdot {\hat {k}}_{2}=0}$ ${\displaystyle \otimes }$

${\displaystyle \mathbf {K} _{1}:={\hat {k}}_{1}\otimes {\hat {k}}_{1}\quad {\textsf {und}}\quad \mathbf {K} _{2}:={\hat {k}}_{2}\otimes {\hat {k}}_{2}\,.}$

Further be the third unit vector that complements to a right-handed base and ${\displaystyle {\hat {k}}_{3}:={\hat {k}}_{1}\times {\hat {k}}_{2}}$ ${\displaystyle \{{\hat {k}}_{1},{\hat {k}}_{2}\}}$

${\displaystyle \mathbf {K} _{3}:={\hat {k}}_{3}\otimes {\hat {k}}_{3}=\mathbf {I} -\mathbf {K} _{1}-\mathbf {K} _{2}\,.}$

The tensor I is the unit tensor . With the coefficient and the unit tensor of the fourth level , the constant and symmetrical elasticity tensor of the fourth level is calculated ${\displaystyle h={\frac {1-2\nu }{\nu }}\lambda -2G={\frac {E}{1+\nu }}-2G=2G{\frac {1-A}{A}}}$ ${\displaystyle {\stackrel {4}{\mathbf {I} }}}$

${\displaystyle {\stackrel {4}{\mathbf {C} }}=2G{\stackrel {4}{\mathbf {I} }}+\lambda \mathbf {I} \otimes \mathbf {I} +h(\mathbf {K} _{1}\otimes \mathbf {K} _{1}+\mathbf {K} _{2}\otimes \mathbf {K} _{2}+\mathbf {K} _{3}\otimes \mathbf {K} _{3})\,.}$

Regarding the base , in Voigt's notation : ${\displaystyle \{{\hat {k}}_{1},{\hat {k}}_{2},{\hat {k}}_{3}\}}$

${\displaystyle C^{\text{v}}={\begin{bmatrix}2G+\lambda +h&\lambda &\lambda &0&0&0\\&2G+\lambda +h&\lambda &0&0&0\\&&2G+\lambda +h&0&0&0\\&&&G&0&0\\&{\text{sym.}}&&&G&0\\&&&&&G\end{bmatrix}}={\begin{bmatrix}{\frac {1-\nu }{\nu }}\lambda &\lambda &\lambda &0&0&0\\&{\frac {1-\nu }{\nu }}\lambda &\lambda &0&0&0\\&&{\frac {1-\nu }{\nu }}\lambda &\\&&&G&0&0\\&{\text{sym}}&&&G&0\\&&&&&G\end{bmatrix}}}$

The inverse of the elasticity tensor is:

${\displaystyle {\stackrel {4}{\mathbf {S} }}={\stackrel {4}{\mathbf {C} }}{}^{-1}={\frac {1}{2G}}{\stackrel {4}{\mathbf {I} }}-{\frac {\nu }{E}}\mathbf {I} \otimes \mathbf {I} +\underbrace {\left({\frac {1+\nu }{E}}-{\frac {1}{2G}}\right)} _{*}(\mathbf {K} _{1}\otimes \mathbf {K} _{1}+\mathbf {K} _{2}\otimes \mathbf {K} _{2}+\mathbf {K} _{3}\otimes \mathbf {K} _{3})}$

The prefactor * depends on the anisotropy factor

${\displaystyle {\frac {1+\nu }{E}}-{\frac {1}{2G}}={\frac {A-1}{2G}}}$

together. Regarding the base , in Voigt's notation : ${\displaystyle \{{\hat {k}}_{1},{\hat {k}}_{2},{\hat {k}}_{3}\}}$

${\displaystyle S^{\text{v}}={\begin{bmatrix}S_{11}&-{\frac {\nu }{E}}&-{\frac {\nu }{E}}&0&0&0\\&S_{11}&-{\frac {\nu }{E}}&0&0&0\\&&S_{11}&0&0&0\\&&&{\frac {1}{G}}&0&0\\&{\text{sym.}}&&&{\frac {1}{G}}&0\\&&&&&{\frac {1}{G}}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{E}}&-{\frac {\nu }{E}}&-{\frac {\nu }{E}}&0&0&0\\&{\frac {1}{E}}&-{\frac {\nu }{E}}&0&0&0\\&&{\frac {1}{E}}&0&0&0\\&&&{\frac {1}{G}}&0&0\\&{\text{sym.}}&&&{\frac {1}{G}}&0\\&&&&&{\frac {1}{G}}\end{bmatrix}}}$

With

${\displaystyle S_{11}={\frac {1}{2G}}-{\frac {\nu }{E}}+{\frac {1+\nu }{E}}-{\frac {1}{2G}}={\frac {1}{E}}}$

example

A sample made of silicon and one made of common salt have the material parameters according to the table above

parameter	C 11	λ = C 12	G = C 44	E.	ν	A.
unit	MPa	MPa	MPa	MPa	-	-
silicon	166,000	64,000	80,000	130.383	0.28	1.6
Table salt	48,500	12,500	12,700	43,377	0.20	0.7

These samples are loaded in a Cartesian coordinate system in the x direction, but only small deformations should occur. Then the stand stress tensor and the strain tensor over the Nachgiebigkeitstensor related: . With pure pull in the x-direction : ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle {\boldsymbol {\varepsilon }}}$ ${\displaystyle {\boldsymbol {\varepsilon }}={\stackrel {4}{\mathbf {S} }}:{\boldsymbol {\sigma }}}$ ${\displaystyle {\boldsymbol {\sigma }}=\sigma {\hat {e}}_{x}\otimes {\hat {e}}_{x}}$

${\displaystyle {\boldsymbol {\varepsilon }}=\left[{\frac {1}{2G}}{\stackrel {4}{\mathbf {I} }}-{\frac {\nu }{E}}\mathbf {I} \otimes \mathbf {I} +\left({\frac {1+\nu }{E}}-{\frac {1}{2G}}\right)(\mathbf {K} _{1}\otimes \mathbf {K} _{1}+\mathbf {K} _{2}\otimes \mathbf {K} _{2}+\mathbf {K} _{3}\otimes \mathbf {K} _{3})\right]:\sigma {\hat {e}}_{x}\otimes {\hat {e}}_{x}}$

The Frobenius scalar product “:” formed with the trace operator with the tensor K ₁ extracts the x-component of the basis vector ${\displaystyle k_{1x}}$ ${\displaystyle {\hat {k}}_{1}}$

${\displaystyle \mathbf {K} _{1}:({\hat {e}}_{x}\otimes {\hat {e}}_{x})=({\hat {k}}_{1}\otimes {\hat {k}}_{1}):({\hat {e}}_{x}\otimes {\hat {e}}_{x})=\operatorname {Sp} (({\hat {k}}_{1}\otimes {\hat {k}}_{1})\cdot ({\hat {e}}_{x}\otimes {\hat {e}}_{x}))=({\hat {k}}_{1}\cdot {\hat {e}}_{x})\operatorname {Sp} ({\hat {k}}_{1}\otimes {\hat {e}}_{x})=({\hat {k}}_{1}\cdot {\hat {e}}_{x})^{2}=:k_{1x}^{2}\,,}$

which also applies to the other two tensors K _2,3 . This is how you get the strain tensor

${\displaystyle {\boldsymbol {\varepsilon }}=\sigma \left[{\frac {1}{2G}}{\hat {e}}_{x}\otimes {\hat {e}}_{x}-{\frac {\nu }{E}}\mathbf {I} +\left({\frac {1+\nu }{E}}-{\frac {1}{2G}}\right)\left(k_{1x}^{2}\mathbf {K} _{1}+k_{2x}^{2}\mathbf {K} _{2}+k_{3x}^{2}\mathbf {K} _{3}\right)\right]}$

because the Frobenius inner product of Einheitstensors with the dyad is the track of the dyad, which is equal to one: . The expansion in the x-direction is therefore: ${\displaystyle {\hat {e}}_{x}\otimes {\hat {e}}_{x}}$ ${\displaystyle \mathbf {I} :({\hat {e}}_{x}\otimes {\hat {e}}_{x})=\operatorname {Sp} ({\hat {e}}_{x}\otimes {\hat {e}}_{x})={\hat {e}}_{x}\cdot {\hat {e}}_{x}=1}$

${\displaystyle \varepsilon :={\hat {e}}_{x}\cdot {\boldsymbol {\varepsilon }}\cdot {\hat {e}}_{x}={\boldsymbol {\varepsilon }}:({\hat {e}}_{x}\otimes {\hat {e}}_{x})=\sigma {\biggl [}{\frac {1}{2G}}-{\frac {\nu }{E}}+\left({\frac {1+\nu }{E}}-{\frac {1}{2G}}\right)\underbrace {\left(k_{1x}^{4}+k_{2x}^{4}+k_{3x}^{4}\right)} _{q}{\biggr ]}\,.}$

The elongation is determined by the prefactor

${\displaystyle {\frac {1+\nu }{E}}-{\frac {1}{2G}}={\frac {A-1}{2G}}}$

which is a linear function of the anisotropy factor A. Depending on the sign of A-1, the positive term q, which is dependent on the alignment of the orthotropic axes, is added or subtracted. The first orthotropic axis of the crystal will now be according to the prescription

${\displaystyle {\hat {k}}_{1}=\cos(\alpha ){\hat {e}}_{x}+{\frac {\sin(\alpha )}{\sqrt {2}}}({\hat {e}}_{y}+{\hat {e}}_{z})\quad \rightarrow \quad k_{1x}=\cos(\alpha )}$

oriented with respect to the global base system depending on the angle α. At α = 0 ° the pull is parallel to the (100) direction of the grid, at α≈54.7 ° in the direction of the (111) space diagonal and at α = 90 ° in the (011) direction. The transverse contraction direction, which is determined by the other two axes, is determined by minimizing the deformation energy, which is proportional to the product σ · ε. As a minimum, the factor q must be minimum with positive A-1 and maximum with negative A-1. For table salt, which has a negative A-1, the orientation is ${\displaystyle {\hat {e}}_{x,y,z}}$

Ratio of the measured modulus of elasticity to the material parameter E when pulling in the x-direction on a silicon and a common salt sample depending on the orientation of the sample. ${\displaystyle V_{x}}$

${\displaystyle {\begin{array}{rclcl}{\hat {k}}_{1}^{a}&=&\cos(\alpha ){\hat {e}}_{x}+{\frac {\sin(\alpha )}{\sqrt {2}}}({\hat {e}}_{y}+{\hat {e}}_{z})&\rightarrow &k_{1x}^{a}=\cos(\alpha )\\{\hat {k}}_{2}^{a}&=&{\frac {1}{\sqrt {2}}}({\hat {e}}_{y}-{\hat {e}}_{z})&\rightarrow &k_{2x}^{a}=0\\{\hat {k}}_{3}^{a}&=&-\sin(\alpha ){\hat {e}}_{x}+{\frac {\cos(\alpha )}{\sqrt {2}}}({\hat {e}}_{y}+{\hat {e}}_{z})&\rightarrow &k_{3x}^{a}=-\sin(\alpha )\end{array}}}$

optimal and for silicon where A-1 is positive is the orientation

${\displaystyle {\begin{array}{rclcl}{\hat {k}}_{1}^{b}&=&\cos(\alpha ){\hat {e}}_{x}+{\frac {\sin(\alpha )}{\sqrt {2}}}({\hat {e}}_{y}+{\hat {e}}_{z})&\rightarrow &k_{1x}^{b}=\cos(\alpha )\\{\hat {k}}_{2}^{b}&=&-{\frac {\sin(\alpha )}{\sqrt {2}}}{\hat {e}}_{x}+{\frac {1}{2}}(\cos(\alpha )+1){\hat {e}}_{y}+{\frac {1}{2}}(\cos(\alpha )-1){\hat {e}}_{z}&\rightarrow &k_{2x}^{b}=-{\frac {\sin(\alpha )}{\sqrt {2}}}\\{\hat {k}}_{3}^{b}&=&-{\frac {\sin(\alpha )}{\sqrt {2}}}{\hat {e}}_{x}+{\frac {1}{2}}(\cos(\alpha )-1){\hat {e}}_{y}+{\frac {1}{2}}(\cos(\alpha )+1){\hat {e}}_{z}&\rightarrow &k_{3x}^{b}=-{\frac {\sin(\alpha )}{\sqrt {2}}}\end{array}}}$

optimal.

proof

Only one degree of freedom is available for the orientation of the orthotropic axes , for which the x component k 2x of the second basis vector is selected and which must be adapted so that q becomes extremal. The orthotropic axes to be determined are written with unknowns a and b ${\displaystyle {\hat {k}}_{2,3}}$ ${\displaystyle {\begin{array}{rcl}{\hat {k}}_{2}&=&k_{2x}{\hat {e}}_{x}+(a+b){\hat {e}}_{y}+(a-b){\hat {e}}_{z}\\{\hat {k}}_{3}&=&-{\sqrt {2}}\sin(\alpha )b{\hat {e}}_{x}+\left[{\frac {\sin(\alpha )}{\sqrt {2}}}k_{2x}-\cos(\alpha )(a-b)\right]{\hat {e}}_{y}+\left[\cos(\alpha )(a+b)-{\frac {\sin(\alpha )}{\sqrt {2}}}k_{2x}\right]{\hat {e}}_{z}\,.\end{array}}}$ The basis vector must be perpendicular and have length one: ${\displaystyle {\hat {k}}_{2}}$ ${\displaystyle {\hat {k}}_{1}}$ ${\displaystyle {\begin{array}{rclcl}{\hat {k}}_{1}\cdot {\hat {k}}_{2}&=&0=\cos(\alpha )k_{2x}+{\frac {\sin(\alpha )}{\sqrt {2}}}(a+b+a-b)=\cos(\alpha )k_{2x}+{\sqrt {2}}\sin(\alpha )a&\rightarrow &a=-{\frac {\cos(\alpha )}{{\sqrt {2}}\sin(\alpha )}}k_{2x}\\{\hat {k}}_{2}\cdot {\hat {k}}_{2}&=&1=k_{2x}^{2}+(a+b)^{2}+(a-b)^{2}=k_{2x}^{2}+2{\frac {\cos ^{2}(\alpha )}{2\sin ^{2}(\alpha )}}k_{2x}^{2}+2b^{2}&\rightarrow &b={\frac {1}{\sqrt {2}}}{\sqrt {1-{\frac {k_{2x}^{2}}{\sin ^{2}(\alpha )}}}}\,.\end{array}}}$ The negative value for b leads to an equivalent alignment. The x-component of the third basis vector is accordingly ${\displaystyle k_{3x}=-{\sqrt {2}}\sin(\alpha )b=-{\sqrt {\sin ^{2}(\alpha )-k_{2x}^{2}}}\,.}$ The target variable is calculated using the x components ${\displaystyle q=\cos ^{4}(\alpha )+k_{2x}^{4}+(\sin ^{2}(\alpha )-k_{2x}^{2})^{2}\,,}$ which should be extremal. To determine the extremum, the derivatives are formed: ${\displaystyle {\begin{array}{rcl}{\frac {\mathrm {d} q}{\mathrm {d} k_{2x}}}&=&4k_{2x}^{3}+2(\sin ^{2}(\alpha )-k_{2x}^{2})\cdot (-2k_{2x})=4[2k_{2x}^{2}-\sin ^{2}(\alpha )]k_{2x}{\stackrel {\displaystyle !}{=}}0\\{\frac {\mathrm {d} ^{2}q}{\mathrm {d} k_{2x}^{2}}}&=&24k_{2x}^{2}-4\sin ^{2}(\alpha )\end{array}}}$ In the extremum, the derivative according to k 2x disappears in the first line and the sign of the second derivative in the second line determines the type of extremum: minimum with a positive sign and maximum with a negative one. The one solution k 2x = 0, for which q has a maximum, can be read off directly. Here a = 0 and the alignment is sufficient. The other solution corresponds to a minimum of q and includes , a = cos (α) / 2 and b = 1/2, which leads to the alignment . ${\displaystyle b=1/{\sqrt {2}}}$ ${\displaystyle {\hat {k}}_{1,2,3}^{a}}$ ${\displaystyle k_{2x}=-\sin(\alpha )/{\sqrt {2}}}$ ${\displaystyle {\hat {k}}_{1,2,3}^{b}}$

The modulus of elasticity E _x measured with tension in the x-direction is the ratio σ / ε of the stresses to the elongations in the x-direction. If this E _x = σ / ε is related to the material parameter E, a dimensionless ratio results

${\displaystyle V_{x}:={\frac {\sigma }{E\varepsilon }}={\frac {A}{1+(A-1)[(1+\nu )(k_{1x}^{4}+k_{2x}^{4}+k_{3x}^{4})-\nu ]}}\,.}$

With the selected alignment of the specimen and tension in the x-direction, the factor results as in the picture. At α = 0 ° is in any case and thus . While silicon reacts most stiffly when pulled in the direction of the room diagonal, this is the case with table salt when pulled in the direction of the grid. With isotropic elasticity, the factor would be constantly equal to one regardless of the orientation of the sample, i.e. it would be on a straight line parallel to the abscissa . ${\displaystyle V_{x}}$ ${\displaystyle k_{1x}^{4}+k_{2x}^{4}+k_{3x}^{4}=1}$ ${\displaystyle V_{x}=1}$ ${\displaystyle V_{x}}$

Footnotes

1. ↑ Haupt (2000), p. 363.
2. ↑ Haupt, 2000.
3. ↑ ^{a b} Newnham (2005), p. 111.
4. ↑ H. Altenbach, 2012, p. 331.

literature

• H. Altenbach: Continuum Mechanics: Introduction to the material-independent and material-dependent equations . Springer, 2012, ISBN 3-642-24119-0 .
• J. Betten: Continuum mechanics - elastic and inelastic behavior of isotropic and anisotropic substances . Springer, 2012, ISBN 3-642-62645-9 .
• P. Haupt: Continuum Mechanics and Theory of Materials . Springer, 2000, ISBN 3-540-66114-X .
• RE Newnham: Properties of materials . Oxford University Press, 2005, ISBN 978-0-19-852075-7 .