Stretch tensor

Distance or deformation tensors are unit-free second-level tensors that measure local changes in distance between material elements when a body is deformed. Changes in distance of matter elements correspond to the stretching or compression of the material lines that connect the matter elements under consideration. These changes in the internal arrangement correspond to a change in the external shape of the solid and are visible , for example, as expansion or compression .

The distance sensors are an essential parameter in the description of the kinematics of deformation and in continuum mechanics a number of different distance sensors are defined, which in turn serve to define the strain tensors . In some material models of hyperelasticity , distance sensors are used directly.

Extension of line elements

Stretching and shearing of the tangents (red and blue) on material lines (black) in the course of a deformation

In the quantitative assessment of a deformation of a body, material lines of the body before and after deformation are compared with one another. In practice, strain gauges can be stuck to the body for this purpose. The direction of the strain gage is described mathematically with a material line element in the undeformed initial configuration and in the deformed current configuration , see figure on the right. These line elements are related in a linear approximation via the deformation gradient : ${\displaystyle \mathrm {d} {\vec {X}}}$ ${\displaystyle \mathrm {d} {\vec {x}}}$ ${\displaystyle \mathbf {F} }$

${\displaystyle \mathrm {d} {\vec {x}}=\mathbf {F} \cdot \mathrm {d} {\vec {X}}}$

The stretching of a line element in the direction ${\displaystyle \lambda }$

${\displaystyle {\vec {e}}={\frac {\mathrm {d} {\vec {X}}}{|\mathrm {d} {\vec {X}}|}}}$

is the relationship

${\displaystyle \lambda ={\frac {|\mathrm {d} {\vec {x}}|}{|\mathrm {d} {\vec {X}}|}}={\sqrt {\frac {\mathrm {d} {\vec {x}}\cdot \mathrm {d} {\vec {x}}}{\mathrm {d} {\vec {X}}\cdot \mathrm {d} {\vec {X}}}}}={\sqrt {\frac {(\mathbf {F} \cdot \mathrm {d} {\vec {X}})\cdot (\mathbf {F} \cdot \mathrm {d} {\vec {X}})}{\mathrm {d} {\vec {X}}\cdot \mathrm {d} {\vec {X}}}}}={\sqrt {{\vec {e}}\cdot \mathbf {C} \cdot {\vec {e}}}}}$

The stretch tensor

${\displaystyle \mathbf {C} =\mathbf {F} ^{\top }\cdot \mathbf {F} }$

is called the right Cauchy-Green tensor and is therefore a measure of the stretching of line elements. The superscript " " stands for the transposition . The stretchings are extremal in the direction of the eigenvectors of . The stretching is calculated in the deformed position ${\displaystyle \top }$ ${\displaystyle \mathbf {C} }$

${\displaystyle \lambda ={\sqrt {\frac {\mathrm {d} {\vec {x}}\cdot \mathrm {d} {\vec {x}}}{(\mathbf {F} ^{-1}\cdot \mathrm {d} {\vec {x}})\cdot (\mathbf {F} ^{-1}\cdot \mathrm {d} {\vec {x}})}}}=\left({\frac {\mathrm {d} {\vec {x}}}{|\mathrm {d} {\vec {x}}|}}\cdot \mathbf {c} \cdot {\frac {\mathrm {d} {\vec {x}}}{|\mathrm {d} {\vec {x}}|}}\right)^{-{\frac {1}{2}}}\,.}$

The Cauchy Stretches Ensor

${\displaystyle \mathbf {c} :=\mathbf {F} ^{\top -1}\cdot \mathbf {F} ^{-1}}$

is therefore a spatial measure for the stretching of line elements.

Stretching of normal vectors

Stretching and shearing of the normals (red and blue) on material surfaces (gray) in the course of a deformation

The extension of normal vectors can also be determined with distance sensors . A family of surfaces can be represented by a scalar function

${\displaystyle \Phi ({\vec {x}},t)=C}$

and an area parameter can be defined. ${\displaystyle C}$

The normal vectors to these surfaces are the gradients

${\displaystyle {\vec {n}}:=\operatorname {grad} (\Phi )=\sum _{i=1}^{3}{\frac {\mathrm {d} \Phi }{\mathrm {d} x_{i}}}{\vec {e}}_{i}\,.}$

These are related to the normal in the reference configuration as follows:

${\displaystyle {\begin{aligned}{\vec {N}}:=&\operatorname {GRAD} (\Phi )=\sum _{i=1}^{3}{\frac {\mathrm {d} \Phi }{\mathrm {d} X_{i}}}{\vec {e}}_{i}=\sum _{j,k=1}^{3}{\frac {\mathrm {d} \Phi }{\mathrm {d} x_{k}}}{\frac {\mathrm {d} x_{k}}{\mathrm {d} X_{j}}}{\vec {e}}_{j}\\=&\sum _{i,j,k=1}^{3}{\frac {\mathrm {d} \Phi }{\mathrm {d} x_{k}}}{\vec {e}}_{k}\cdot {\frac {\mathrm {d} x_{i}}{\mathrm {d} X_{j}}}{\vec {e}}_{i}\otimes {\vec {e}}_{j}={\vec {n}}\cdot \mathbf {F} =\mathbf {F} ^{\top }\cdot {\vec {n}}\\\rightarrow {\vec {n}}=&\mathbf {F} ^{\top -1}\cdot {\vec {N}}\end{aligned}}}$

The arithmetic symbol denotes the dyadic product . The stretching of the normal vectors in the deformed and undeformed position in a material point leads to the finger tensor ${\displaystyle \otimes }$ ${\displaystyle {\vec {X}}}$

${\displaystyle {\begin{array}{rcl}\lambda &=&\displaystyle {\frac {|{\vec {n}}|}{|{\vec {N}}|}}={\sqrt {\frac {{\vec {n}}\cdot {\vec {n}}}{{\vec {N}}\cdot {\vec {N}}}}}={\sqrt {\frac {(\mathbf {F} ^{\top -1}\cdot {\vec {N}})\cdot (\mathbf {F} ^{\top -1}\cdot {\vec {N}})}{{\vec {N}}\cdot {\vec {N}}}}}={\sqrt {{\frac {\vec {N}}{|{\vec {N}}|}}\cdot \mathbf {f} \cdot {\frac {\vec {N}}{|{\vec {N}}|}}}}\\\rightarrow \mathbf {f} &=&\mathbf {F} ^{-1}\cdot \mathbf {F} ^{\top -1}=\mathbf {C} ^{-1}\,,\end{array}}}$

which is a measure for the stretching of the material surface normals. The finger tensor operates in the initial configuration.

Its counterpart in the current configuration is the left Cauchy-Green tensor

${\displaystyle \mathbf {b} =\mathbf {F} \cdot \mathbf {F} ^{\top }=\mathbf {c} ^{-1}}$

for the

${\displaystyle \lambda ={\sqrt {\frac {{\vec {n}}\cdot {\vec {n}}}{(\mathbf {F} ^{\top }\cdot {\vec {n}})\cdot (\mathbf {F} ^{\top }\cdot {\vec {n}})}}}=\left({\frac {\vec {n}}{|{\vec {n}}|}}\cdot \mathbf {b} \cdot {\frac {\vec {n}}{|{\vec {n}}|}}\right)^{-{\frac {1}{2}}}}$

can be derived.

Main invariants of the right Cauchy-Green tensor

In the event of a deformation, the material line, surface and volume elements are transformed with the deformation gradient from the initial configuration to the current configuration

${\displaystyle {\begin{array}{rccl}\mathrm {d} {\vec {x}}&=&\mathbf {F} &\cdot \;\mathrm {d} {\vec {X}}\\{\vec {n}}\,\mathrm {d} a&=&\operatorname {cof} (\mathbf {F} )&\cdot \;{\vec {N}}\,\mathrm {d} A\\\mathrm {d} v&=&\operatorname {det} (\mathbf {F} )&\mathrm {d} V\,.\end{array}}}$

The cofactor of the deformation gradient is its transposed adjoint :

${\displaystyle \operatorname {cof} (\mathbf {F} ):=\operatorname {det} (\mathbf {F} )\mathbf {F} ^{\top -1}\,.}$

It turns out that the main invariants of the Cauchy-Green tensor on the right are measures for the change in line, surface and volume elements:

${\displaystyle {\begin{array}{rclclcc}\operatorname {I} _{1}(\mathbf {C} )&=&\operatorname {Sp} (\mathbf {C} )&=&\operatorname {Sp} (\mathbf {F^{\top }\cdot F} )=\mathbf {F} :\mathbf {F} &=&\|\mathbf {F} \|^{2}\\\operatorname {I} _{2}(\mathbf {C} )&=&\operatorname {Sp(cof} (\mathbf {C} ))&=&\operatorname {Sp(cof} (\mathbf {F} )^{\top }\cdot \operatorname {cof} (\mathbf {F} ))&=&\|\operatorname {cof} (\mathbf {F} )\|^{2}\\\operatorname {I} _{3}(\mathbf {C} )&=&\operatorname {det} (\mathbf {C} )&=&\operatorname {det} (\mathbf {F^{\top }\cdot F} )&=&\operatorname {det} (\mathbf {F} )^{2}\end{array}}}$

The Frobenius norm is defined with the Frobenius scalar product ":" of tensors: ${\displaystyle \|(\cdot )\|}$

${\displaystyle \mathbf {A} :\mathbf {B} :=\operatorname {Sp} (\mathbf {A^{\top }\cdot B} )\quad {\text{und}}\quad \|\mathbf {A} \|:={\sqrt {\mathbf {A} :\mathbf {A} }}\,.}$

Physical interpretation

The connection between the right Cauchy-Green tensor and the change in the line, surface and volume elements can be seen macroscopically.

Be

${\displaystyle {\vec {x}}={\vec {\chi }}({\vec {X}},t)}$

the motion function of the particles of a material body. The material coordinates are taken by the particles at a certain time when the body is in its undeformed starting position. The time-dependent vector describes the spatial coordinates that the particles assume during their movement - including deformation - at time t. ${\displaystyle {\vec {X}}}$ ${\displaystyle t_{0}}$ ${\displaystyle {\vec {x}}}$

Lengths of lines

If a material line is marked with the curve parameter in the undeformed initial state , the length of the line results to ${\displaystyle {\vec {X}}(s)}$ ${\displaystyle s\in [0,1]}$

${\displaystyle L=\int _{0}^{1}\left|{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\right|\mathrm {d} s=\int _{0}^{1}{\sqrt {{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}}}\mathrm {d} s=\int _{0}^{1}{\sqrt {{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\cdot \mathbf {I} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}}}\mathrm {d} s\,.}$

Therein is the unit tensor . In the deformed position, this length changes to ${\displaystyle \mathbf {I} }$

${\displaystyle l=\int _{0}^{1}\left|{\frac {\mathrm {d} {\vec {\chi }}({\vec {X}}(s),t)}{\mathrm {d} s}}\right|\mathrm {d} s=\int _{0}^{1}{\sqrt {{\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} s}}\cdot {\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} s}}}}\mathrm {d} s=\int _{0}^{1}{\sqrt {\left(\mathbf {F} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\right)\cdot \left(\mathbf {F} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\right)}}\mathrm {d} s=\int _{0}^{1}{\sqrt {{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\cdot \mathbf {C} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}}}\mathrm {d} s\,.}$

The change in the length of the marked line is determined by the stretch tensor . ${\displaystyle \mathbf {C} }$

Areas

If, in the same way, in the undeformed initial state, a material surface is designated with the surface parameters , the content of the surface results in ${\displaystyle {\vec {X}}(u,v)}$ ${\displaystyle (u,v)\in [0,1]^{2}}$

${\displaystyle F=\int _{0}^{1}\int _{0}^{1}{\biggl |}\overbrace {{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} u}}\times {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} v}}} ^{A{\vec {N}}}{\biggr |}\,\mathrm {d} u\mathrm {d} v=\int _{0}^{1}\int _{0}^{1}|{\vec {N}}|\,\overbrace {A\mathrm {d} u\mathrm {d} v} ^{\mathrm {d} A}=\int _{0}^{1}\int _{0}^{1}{\sqrt {{\vec {N}}\cdot \mathbf {I} \cdot {\vec {N}}}}\,\mathrm {d} A\,.}$

In the deformed position, this area changes to

${\displaystyle {\begin{array}{rcl}f&=&\displaystyle \int _{0}^{1}\int _{0}^{1}{\biggl |}\overbrace {{\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} u}}\times {\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} v}}} ^{a{\vec {n}}}{\biggr |}\,\mathrm {d} u\mathrm {d} v=\int _{0}^{1}\int _{0}^{1}|{\vec {n}}|\,\overbrace {a\mathrm {d} u\mathrm {d} v} ^{\mathrm {d} a}=\int _{0}^{1}\int _{0}^{1}|\operatorname {cof} (\mathbf {F} )\cdot {\vec {N}}|\,\mathrm {d} A\\&=&\displaystyle \int _{0}^{1}\int _{0}^{1}{\sqrt {{\vec {N}}\cdot \operatorname {cof} (\mathbf {F} )^{\top }\cdot \operatorname {cof} (\mathbf {F} )\cdot {\vec {N}}}}\,\mathrm {d} A=\int _{0}^{1}\int _{0}^{1}{\sqrt {{\vec {N}}\cdot \operatorname {cof} (\mathbf {C} )\cdot {\vec {N}}}}\,\mathrm {d} A\end{array}}}$

The change in the content of the marked area is determined by the cofactor of the distance sensor . ${\displaystyle \mathbf {C} }$

Volumes

In the undeformed initial state, a material volume is marked with the location parameters . The volume is then calculated to ${\displaystyle {\vec {X}}(\xi ,\eta ,\zeta )}$ ${\displaystyle (\xi ,\eta ,\zeta )\in [0,1]^{3}}$

${\displaystyle V=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\overbrace {\operatorname {det} {\begin{pmatrix}{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \xi }}&{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \eta }}&{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \zeta }}\end{pmatrix}}\,\mathrm {d} \xi \mathrm {d} \eta \mathrm {d} \zeta } ^{\mathrm {d} V}=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}{\sqrt {\operatorname {det} (\mathbf {I} )}}\,\mathrm {d} V}$

In the deformed position, this volume changes to

${\displaystyle {\begin{array}{rcl}v&=&\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\operatorname {det} {\begin{pmatrix}{\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} \xi }}&{\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} \eta }}&{\frac {\mathrm {d} {\vec {x}}}{\mathrm {d} \zeta }}\end{pmatrix}}\,\mathrm {d} \xi \mathrm {d} \eta \mathrm {d} \zeta =\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\operatorname {det} {\begin{pmatrix}\mathbf {F} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \xi }}&\mathbf {F} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \eta }}&\mathbf {F} \cdot {\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \zeta }}\end{pmatrix}}\,\mathrm {d} \xi \mathrm {d} \eta \mathrm {d} \zeta \\&=&\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\operatorname {det} \left[\mathbf {F} \cdot {\begin{pmatrix}{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \xi }}&{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \eta }}&{\frac {\mathrm {d} {\vec {X}}}{\mathrm {d} \zeta }}\end{pmatrix}}\right]\,\mathrm {d} \xi \mathrm {d} \eta \mathrm {d} \zeta =\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}\operatorname {det} (\mathbf {F} )\,\mathrm {d} V=\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}{\sqrt {\operatorname {det} (\mathbf {C} )}}\,\mathrm {d} V\,,\end{array}}}$

wherein the determinant product theorem has been exploited. The change in volume can thus be expressed with the stretch tensor, as with material lines and surfaces.

Left and right extension tensor and turns

In the case of non-deformation , the distance sensors are equal to the unit tensor and are independent of any rotations of the body that may occur. The reason for this lies in the polar decomposition of the deformation gradient

${\displaystyle \mathbf {F} =\mathbf {R\cdot U} =\mathbf {v\cdot R} \,,}$

which splits the deformation locally into a rotation, mediated by the orthogonal rotation tensor (with and the determinant ), and a pure stretching, mediated by the symmetrical, positive-definite right and left stretch tensor or , respectively . By multiplying the deformation gradient with its transposed one , the rotations and "reverse rotations" cancel each other out: ${\displaystyle \mathbf {R} }$ ${\displaystyle \mathbf {R} ^{\top }=\mathbf {R} ^{-1}}$ ${\displaystyle \operatorname {det} (\mathbf {R} )=+1}$ ${\displaystyle \mathbf {U} }$ ${\displaystyle \mathbf {v} }$ ${\displaystyle \mathbf {R} }$ ${\displaystyle \mathbf {R} ^{\top }=\mathbf {R} ^{-1}}$

${\displaystyle {\begin{array}{ccccccc}\mathbf {F} ^{\top }\cdot \mathbf {F} &=&\mathbf {U\cdot R} ^{\top }\cdot \mathbf {R\cdot U} &=&\mathbf {U\cdot U} &=&\mathbf {C} \\\mathbf {F\cdot F} ^{\top }&=&\mathbf {v\cdot R\cdot R} ^{\top }\cdot \mathbf {v} &=&\mathbf {v\cdot v} &=&\mathbf {b} \end{array}}}$

which of course also applies to the inverse of the right and left Cauchy-Green tensor. The right and left Cauchy-Green tensor and their inverses are therefore unaffected by rotations of the body.

Principal Axis Transformations

The right and left stretch tensor as well as the right and left Cauchy-Green tensor are therefore similar , which is why they have the same eigenvalues and therefore the same main invariants . The eigenvalues ​​of the distance sensors are called main extensions. All distance sensors are symmetrically positive definite and therefore all three eigenvalues ​​are positive and the three eigenvectors are mutually perpendicular (or orthogonalizable) in pairs, so that they form an orthonormal basis . Let be the eigenvectors of , the eigenvectors of and its eigenvalues. Then the major axis transformations are: ${\displaystyle {\vec {N}}_{1,2,3}}$ ${\displaystyle \mathbf {U} }$ ${\displaystyle {\vec {n}}_{1,2,3}}$ ${\displaystyle \mathbf {v} }$ ${\displaystyle \lambda _{1,2,3}}$

${\displaystyle \mathbf {U} =\sum _{i=1}^{3}\lambda _{i}{\vec {N}}_{i}\otimes {\vec {N}}_{i}\qquad {\textsf {und}}\qquad \mathbf {v} =\sum _{i=1}^{3}\lambda _{i}{\vec {n}}_{i}\otimes {\vec {n}}_{i}\,.}$

From it follows: ${\displaystyle \mathbf {v} =\mathbf {R} \cdot \mathbf {U} \cdot \mathbf {R} ^{\top }}$

${\displaystyle \mathbf {R} \cdot {\vec {N}}_{i}={\vec {n}}_{i}\qquad {\text{also}}\qquad \mathbf {R} =\sum _{i=1}^{3}{\vec {n}}_{i}\otimes {\vec {N}}_{i}}$

and further:

${\displaystyle \mathbf {F} =\mathbf {R\cdot U} =\mathbf {R} \cdot \sum _{i=1}^{3}\lambda _{i}{\vec {N}}_{i}\otimes {\vec {N}}_{i}=\sum _{i=1}^{3}\lambda _{i}{\vec {n}}_{i}\otimes {\vec {N}}_{i}}$

${\displaystyle \mathbf {C} =\mathbf {U\cdot U} =\sum _{i=1}^{3}\lambda _{i}^{2}{\vec {N}}_{i}\otimes {\vec {N}}_{i}}$

${\displaystyle \mathbf {f} =\mathbf {C} ^{-1}=\sum _{i=1}^{3}\lambda _{i}^{-2}{\vec {N}}_{i}\otimes {\vec {N}}_{i}}$

${\displaystyle \mathbf {b} =\mathbf {v\cdot v} =\sum _{i=1}^{3}\lambda _{i}^{2}{\vec {n}}_{i}\otimes {\vec {n}}_{i}}$

${\displaystyle \mathbf {c} =\mathbf {b} ^{-1}=\sum _{i=1}^{3}\lambda _{i}^{-2}{\vec {n}}_{i}\otimes {\vec {n}}_{i}}$

Derivation of the elongations

Some material models of hyperelasticity contain functions of the eigenvalues ​​of the left distance tensor and the stresses result from the derivation of these functions according to the left Cauchy-Green tensor . It is therefore worthwhile to provide the derivative of the eigenvalues according to the stretch tensor . It turns out ${\displaystyle \lambda _{i}}$ ${\displaystyle \mathbf {v} }$ ${\displaystyle \mathbf {b} }$ ${\displaystyle \lambda _{i}}$ ${\displaystyle \mathbf {b} }$

${\displaystyle {\frac {\mathrm {d} \lambda _{i}}{\mathrm {d} \mathbf {b} }}={\frac {1}{2\lambda _{i}}}{\hat {n}}_{i}\otimes {\hat {n}}_{i}\,.}$

Is calculated accordingly

${\displaystyle {\frac {\mathrm {d} \lambda _{i}}{\mathrm {d} \mathbf {C} }}={\frac {1}{2\lambda _{i}}}{\hat {N}}_{i}\otimes {\hat {N}}_{i}\,.}$

proof

First the eigenvalues ​​of the left Cauchy-Green tensor are considered. The eigenvalues solve the characteristic polynomial of the left Cauchy-Green tensor: ${\displaystyle \eta _{i}=\lambda _{i}^{2}}$ ${\displaystyle \eta _{i}}$ ${\displaystyle \operatorname {det} (\mathbf {b} -\eta _{i}\operatorname {I} )=-\eta _{i}^{3}+\operatorname {I} _{1}\eta _{i}^{2}-\operatorname {I} _{2}\eta _{i}+\operatorname {I} _{3}=0\,.}$ The coefficients of this polynomial are the three main invariants of the left Cauchy-Green tensor. Implicit differentiation of the characteristic polynomial using the chain rule and the derivatives of the main invariants for symmetric tensors , inserting the formulas of Vieta, yields the main axis transformation of the left Cauchy-Green tensor with its pairwise orthogonal eigenvectors normalized to magnitude one, which agree with the eigenvectors of the left range tensor , and the form of the unit tensor results, for example, for i = 1: For i = 2 and i = 3, a corresponding calculation is made, which is a fixed. The desired derivation of the eigenvalues ​​of the left distance tensor after the left Cauchy-Green tensor is finally determined ${\displaystyle {\frac {\mathrm {d} \operatorname {I} _{1}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\mathbf {I} ,\quad {\frac {\mathrm {d} \operatorname {I} _{2}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\operatorname {I} _{1}\mathbf {I} -\mathbf {b} \quad {\text{und}}\quad {\frac {\mathrm {d} \operatorname {I} _{3}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\operatorname {I} _{3}\mathbf {b} ^{-1}=\mathbf {b\cdot b} -\operatorname {I} _{1}\mathbf {b} +\operatorname {I} _{2}\mathbf {I} }$ ${\displaystyle {\begin{aligned}0=&-3\eta _{i}^{2}{\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}+\eta _{i}^{2}\mathbf {I} +2\operatorname {I} _{1}\eta _{i}{\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}-(\operatorname {I} _{1}\mathbf {I} -\mathbf {b} )\eta _{i}-\operatorname {I} _{2}{\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}+\mathbf {b\cdot b} -\operatorname {I} _{1}\mathbf {b} +\operatorname {I} _{2}\mathbf {I} \\=&-(3\eta _{i}^{2}-2\operatorname {I} _{1}\eta _{i}+\operatorname {I} _{2}){\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}+(\eta _{i}^{2}-\eta _{i}\operatorname {I} _{1}+\operatorname {I} _{2})\mathbf {I} +(\eta _{i}-\operatorname {I} _{1})\mathbf {b} +\mathbf {b\cdot b} \\\rightarrow {\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}=&{\frac {(\eta _{i}^{2}-\eta _{i}\operatorname {I} _{1}+\operatorname {I} _{2})\mathbf {I} +(\eta _{i}-\operatorname {I} _{1})\mathbf {b} +\mathbf {b\cdot b} }{3\eta _{i}^{2}-2\operatorname {I} _{1}\eta _{i}+\operatorname {I} _{2}}}\end{aligned}}}$ ${\displaystyle \operatorname {I} _{1}=\eta _{1}+\eta _{2}+\eta _{3}\quad {\text{und}}\quad \operatorname {I} _{2}=\eta _{1}\eta _{2}+\eta _{2}\eta _{3}+\eta _{3}\eta _{1}\,,}$ ${\displaystyle \mathbf {b} =\eta _{1}{\hat {n}}_{1}\otimes {\hat {n}}_{1}+\eta _{2}{\hat {n}}_{2}\otimes {\hat {n}}_{2}+\eta _{3}{\hat {n}}_{3}\otimes {\hat {n}}_{3}}$ ${\displaystyle {\hat {n}}_{1,2,3}\,,}$ ${\displaystyle \mathbf {I} ={\hat {n}}_{1}\otimes {\hat {n}}_{1}+{\hat {n}}_{2}\otimes {\hat {n}}_{2}+{\hat {n}}_{3}\otimes {\hat {n}}_{3}}$ ${\displaystyle {\begin{aligned}{\frac {\mathrm {d} \eta _{1}}{\mathrm {d} \mathbf {b} }}=&{\frac {\eta _{1}^{2}-\eta _{1}(\eta _{1}+\eta _{2}+\eta _{3})+\eta _{1}\eta _{2}+\eta _{2}\eta _{3}+\eta _{3}\eta _{1}}{3\eta _{1}^{2}-2(\eta _{1}+\eta _{2}+\eta _{3})\eta _{1}+\eta _{1}\eta _{2}+\eta _{2}\eta _{3}+\eta _{3}\eta _{1}}}({\hat {n}}_{1}\otimes {\hat {n}}_{1}+{\hat {n}}_{2}\otimes {\hat {n}}_{2}+{\hat {n}}_{3}\otimes {\hat {n}}_{3})\\&+{\frac {[\eta _{1}-(\eta _{1}+\eta _{2}+\eta _{3})](\eta _{1}{\hat {n}}_{1}\otimes {\hat {n}}_{1}+\eta _{2}{\hat {n}}_{2}\otimes {\hat {n}}_{2}+\eta _{3}{\hat {n}}_{3}\otimes {\hat {n}}_{3})}{3\eta _{1}^{2}-2(\eta _{1}+\eta _{2}+\eta _{3})\eta _{1}+\eta _{1}\eta _{2}+\eta _{2}\eta _{3}+\eta _{3}\eta _{1}}}\\&+{\frac {\eta _{1}^{2}{\hat {n}}_{1}\otimes {\hat {n}}_{1}+\eta _{2}^{2}{\hat {n}}_{2}\otimes {\hat {n}}_{2}+\eta _{3}^{2}{\hat {n}}_{3}\otimes {\hat {n}}_{3}}{3\eta _{1}^{2}-2(\eta _{1}+\eta _{2}+\eta _{3})\eta _{1}+\eta _{1}\eta _{2}+\eta _{2}\eta _{3}+\eta _{3}\eta _{1}}}\\=&{\hat {n}}_{1}\otimes {\hat {n}}_{1}\end{aligned}}}$ ${\displaystyle {\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}={\hat {n}}_{i}\otimes {\hat {n}}_{i}}$ ${\displaystyle {\frac {\mathrm {d} \eta _{i}}{\mathrm {d} \mathbf {b} }}={\frac {\mathrm {d} \lambda _{i}^{2}}{\mathrm {d} \mathbf {b} }}=2\lambda _{i}{\frac {\mathrm {d} \lambda _{i}}{\mathrm {d} \mathbf {b} }}={\hat {n}}_{i}\otimes {\hat {n}}_{i}\quad \rightarrow \quad {\frac {\mathrm {d} \lambda _{i}}{\mathrm {d} \mathbf {b} }}={\frac {1}{2\lambda _{i}}}{\hat {n}}_{i}\otimes {\hat {n}}_{i}}$

example

A square (black) with an inscribed circle (red) is deformed into a rectangle (blue) with an inscribed ellipse (purple)

A square with side length one is stretched to a rectangle with width and height and rotated by an angle , see the illustration on the right. Let the square be positioned at the origin of a Cartesian coordinate system, so that for the points of the square ${\displaystyle b}$ ${\displaystyle h}$ ${\displaystyle \alpha }$

${\displaystyle {\vec {X}}=\left({\begin{array}{c}X\\Y\end{array}}\right)\in {[{0,1}]}^{2}}$

applies. Then is in the deformed state

${\displaystyle {\begin{array}{rcl}{\vec {x}}&=&\left({\begin{array}{c}x\\y\end{array}}\right)=\left({\begin{array}{cc}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{array}}\right)\left({\begin{array}{c}bX\\hY\end{array}}\right)\\&=&\left({\begin{array}{c}bX\cos \alpha -hY\sin \alpha \\bX\sin \alpha +hY\cos \alpha \end{array}}\right)\end{array}}\,.}$

This calculates the deformation gradient and the distance sensors

${\displaystyle {\begin{array}{rcl}\mathbf {F} &=&\left({\begin{array}{cc}b\cos \alpha &-h\sin \alpha \\b\sin \alpha &h\cos \alpha \end{array}}\right)=\left({\begin{array}{cc}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{array}}\right)\left({\begin{array}{cc}b&0\\0&h\end{array}}\right)\\\rightarrow \mathbf {R} &=&\left({\begin{array}{cc}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{array}}\right)\\\rightarrow \mathbf {U} &=&\left({\begin{array}{cc}b&0\\0&h\end{array}}\right),\lambda _{1}=b,{\vec {N}}_{1}=\left({\begin{array}{c}1\\0\end{array}}\right),\lambda _{2}=h,{\vec {N}}_{2}=\left({\begin{array}{c}0\\1\end{array}}\right)\\\rightarrow \mathbf {C} &=&\mathbf {U\cdot U} =\left({\begin{array}{cc}b&0\\0&h\end{array}}\right)\left({\begin{array}{cc}b&0\\0&h\end{array}}\right)=\left({\begin{array}{cc}b^{2}&0\\0&h^{2}\end{array}}\right)=\mathbf {F} ^{\top }\cdot \mathbf {F} \\\rightarrow \mathbf {v} &=&b\mathbf {R} \cdot {\vec {N}}_{1}\otimes \mathbf {R} \cdot {\vec {N}}_{1}+h\mathbf {R} \cdot {\vec {N}}_{2}\otimes \mathbf {R} \cdot {\vec {N}}_{2}\\&=&b\left({\begin{array}{c}\cos \alpha \\\sin \alpha \end{array}}\right)\otimes \left({\begin{array}{c}\cos \alpha \\\sin \alpha \end{array}}\right)+h\left({\begin{array}{c}-\sin \alpha \\\cos \alpha \end{array}}\right)\otimes \left({\begin{array}{c}-\sin \alpha \\\cos \alpha \end{array}}\right)\\&=&\left({\begin{array}{cc}b{\cos }^{2}\alpha +h{\sin }^{2}\alpha &\left(b-h\right)\cos \alpha \sin \alpha \\\left(b-h\right)\sin \alpha \cos \alpha &b{\sin }^{2}\alpha +h{\cos }^{2}\alpha \end{array}}\right)\\&&\rightarrow \lambda _{1}=b,{\vec {n}}_{1}=\left({\begin{array}{c}\cos \alpha \\\sin \alpha \end{array}}\right),\lambda _{2}=h,{\vec {n}}_{2}=\left({\begin{array}{c}-\sin \alpha \\\cos \alpha \end{array}}\right)\\\rightarrow \mathbf {b} &=&\mathbf {v\cdot v} =\left({\begin{array}{cc}b^{2}{\cos }^{2}\alpha +h^{2}{\sin }^{2}\alpha &\left(b^{2}-h^{2}\right)\sin \alpha \cos \alpha \\\left(b^{2}-h^{2}\right)\sin \alpha \cos \alpha &b^{2}{\sin }^{2}\alpha +h^{2}{\cos }^{2}\alpha \end{array}}\right)=\mathbf {F\cdot F} ^{\top }\end{array}}}$

In the middle of the square a straight line of length ½ is marked at an angle to the x-axis, see illustration. In the starting position, the points on the line then have the following coordinates: ${\displaystyle \beta }$ ${\displaystyle s\in [{0,1}]}$

${\displaystyle {\begin{array}{rcl}{\vec {X}}\left(s\right)&=&{\dfrac {1}{2}}\left({\begin{array}{c}1+s\,\cos \beta \\1+s\,\sin \beta \end{array}}\right)\\\rightarrow {\dfrac {\mathrm {d} {\vec {X}}(s)}{\mathrm {d} s}}&=&{\dfrac {1}{2}}\left({\begin{array}{c}\cos \beta \\\sin \beta \end{array}}\right)\rightarrow \left|{\dfrac {\mathrm {d} {\vec {X}}\left(s\right)}{\mathrm {d} s}}\right|={\dfrac {1}{2}}\end{array}}}$

By definition, the length of the line is independent of its direction:

${\displaystyle L(\beta )=\int _{0}^{1}\left|{\dfrac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\right|\mathrm {d} s=\int _{0}^{1}{\dfrac {1}{2}}\mathrm {d} s={\dfrac {1}{2}}\,.}$

In the deformed position the points have the coordinates

${\displaystyle {\begin{array}{rcl}{\vec {x}}(s)&=&\left({\begin{array}{c}x\\y\end{array}}\right)={\dfrac {1}{2}}\left({\begin{array}{c}b(1+s\cos \beta )\cos \alpha -h(1+s\,\sin \beta )\sin \alpha \\b(1+s\,\cos \beta )\sin \alpha +h(1+s\,\sin \beta )\cos \alpha \end{array}}\right)\\\rightarrow {\dfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} s}}&=&{\dfrac {1}{2}}\left({\begin{array}{c}b\cos \alpha \cos \beta -h\sin \alpha \sin \beta \\b\sin \alpha \cos \beta +h\cos \alpha \sin \beta \end{array}}\right)=\left({\begin{array}{cc}b\cos \alpha &-h\sin \alpha \\b\sin \alpha &h\cos \alpha \end{array}}\right){\dfrac {1}{2}}\left({\begin{array}{c}\cos \beta \\\sin \beta \end{array}}\right)=\mathbf {F} \cdot {\dfrac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\\\rightarrow \left|{\dfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} s}}\right|&=&{\dfrac {1}{2}}{\sqrt {b^{2}{\cos }^{2}\beta +h^{2}{\sin }^{2}\beta }}\\&=&{\sqrt {{\dfrac {1}{2}}\left({\begin{array}{c}\cos \beta \\\sin \beta \end{array}}\right)\cdot \left({\begin{array}{cc}b^{2}&0\\0&h^{2}\end{array}}\right){\dfrac {1}{2}}\left({\begin{array}{c}\cos \beta \\\sin \beta \end{array}}\right)}}={\sqrt {{\dfrac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}\cdot \mathbf {C} \cdot {\dfrac {\mathrm {d} {\vec {X}}}{\mathrm {d} s}}}}\end{array}}}$

Length of the lines in the deformed square as a function of the angle . ${\displaystyle l(\beta )}$ ${\displaystyle \beta }$

hence increasing the length of the line

${\displaystyle {\begin{array}{rcl}l(\beta )&=&\displaystyle \int _{0}^{1}\left|{\dfrac {\mathrm {d} {\vec {x}}}{\mathrm {d} s}}\right|\mathrm {d} s=\int _{0}^{1}{\dfrac {1}{2}}{\sqrt {b^{2}{\cos }^{2}\beta +h^{2}{\sin }^{2}\beta }}\mathrm {d} s\\&=&{\dfrac {1}{2}}{\sqrt {b^{2}{\cos }^{2}\beta +h^{2}{\sin }^{2}\beta }}\end{array}}}$

changed. The result is again independent of the angle of rotation . If the square and the rectangle are equal, then ${\displaystyle \alpha }$

${\displaystyle b\cdot h=1\rightarrow h={\dfrac {1}{b}}}$

and the lengths of the deformed line form a curve in a polar diagram as in the figure on the right. There is . ${\displaystyle b=1,5}$

See also

Mechanics:

• Convective coordinates
• Configuration (mechanics)

Mathematics:

• Formula collection tensor algebra
• Tensoranalysis formula collection

Individual evidence

1. ↑ The Fréchet derivative of a scalar function with respect to a tensor is the tensor for which - if it exists - applies: ${\displaystyle f(\mathbf {T} )}$ ${\displaystyle \mathbf {T} }$ ${\displaystyle \mathbf {A} }$

   ${\displaystyle \mathbf {A} :\mathbf {H} =\left.{\frac {\mathrm {d} }{\mathrm {d} s}}f(\mathbf {T} +s\mathbf {H} )\right|_{s=0}=\lim _{s\rightarrow 0}{\frac {f(\mathbf {T} +s\mathbf {H} )-f(\mathbf {T} )}{s}}\quad \forall \;\mathbf {H} }$

   There is and ":" the Frobenius scalar product . Then will too ${\displaystyle s\in \mathbb {R} }$

   ${\displaystyle {\frac {\partial f}{\partial \mathbf {T} }}=\mathbf {A} }$

   written.

literature

• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .
• P. Haupt: Continuum Mechanics and Theory of Materials . Springer, 2010, ISBN 978-3-642-07718-0 .
• A. Bertram: Elasticity and Plasticity of Large Deformations: An Introduction . Springer, 2012, ISBN 978-3-642-24614-2 .