Compatibility condition

For deformation in the plane, two displacements correspond to three distortion fields and . The displacement field below can be reconstructed from them if the compatibility conditions are met. ${\displaystyle \varepsilon _{11},\varepsilon _{22}}$ ${\displaystyle \varepsilon _{12}}$

Compatibility conditions are in the continuum mechanics , so that conditions which must be met from derivatives can be reconstructed a motion field for the location formed sizes of the motion field. The derived quantities are then compatible with a motion field.

When a body moves through space, deformations occur in the cases that are interesting for continuum mechanics, which can be quantified by the distortions that are calculated from derivatives of the movement field according to location. There are six components of the distortions in the general three-dimensional case. If the three components of the movement in the x, y and z directions are to be reconstructed from them, it is clear that the distortions cannot be independent of one another. In the case of a plane movement, there are three distortion fields and , which correspond to two displacement components in the x and y direction (after renaming according to the scheme 1 → x and 2 → y). The illustration opposite shows such a case. Now the question can be asked whether the movement can be reconstructed from the distortion fields. This can succeed precisely when the distortions comply with the compatibility conditions formulated for them. ${\displaystyle \varepsilon _{11},\varepsilon _{22}}$ ${\displaystyle \varepsilon _{12}}$

By deriving the three components of the movement in the x, y and z directions according to the three position coordinates in the x, y and z directions, a total of nine derivations are created which form the components of the deformation gradient . There are also compatibility conditions for the nine components of the deformation gradient which they must adhere to so that the movement can be restored from them.

The compatibility conditions are used in the theory of stress functions , with the help of which analytical solutions of flat and spatial, linear elastostatics can be calculated, e.g. B. in the Airy stress function .

Movements

In order to describe the movement of a body, a “name” or “label” is first assigned to each particle of the body using the reference configuration. This "name" should be the position here

${\displaystyle {\vec {X}}=\sum _{i=1}^{3}X_{i}{\hat {e}}_{i}\in \mathbb {V} ^{3}}$

of the particle at a given point in time . The numbers are called the material coordinates of the particle and apply in relation to the standard basis of the Euclidean vector space . Mostly it is chosen so that at this point in time the body is undeformed and at rest and movement begins. In the course of its movement through space, each particle travels to its line forward, the movement function ${\displaystyle t_{0}}$ ${\displaystyle X_{1,2,3}\in \mathbb {R} }$ ${\displaystyle {\hat {e}}_{1,2,3}}$ ${\displaystyle \mathbb {V} ^{3}}$ ${\displaystyle t_{0}}$

${\displaystyle {\vec {\chi }}({\vec {X}},t)={\vec {x}}=\sum _{i=1}^{3}x_{i}{\hat {e}}_{i}\in \mathbb {V} ^{3}}$

describes mathematically. With respect to the standard basis, each particle now has spatial coordinates at a time . ${\displaystyle t\geq t_{0}}$ ${\displaystyle x_{1,2,3}\in \mathbb {R} }$

Linearized strain tensor

The linearized strain tensor arises from derivatives of the displacement field . The displacement of a particle is the path it has traveled, mathematically the difference vector between its current position and its position in the initial configuration: ${\displaystyle {\vec {u}}}$

${\displaystyle {\vec {u}}({\vec {X}},t):={\vec {\chi }}({\vec {X}},t)-{\vec {X}}=\sum _{i=1}^{3}u_{i}({\vec {X}},t){\hat {e}}_{i}\,.}$

Often, especially in technical applications, it can be assumed that, firstly, this displacement is small compared to the dimensions of the body and, secondly, the derivatives of the displacements according to location are also small compared to one. Then the material coordinates and the spatial coordinates no longer need to be kept apart and the distortions of the body are measured with the linearized distortion tensor, which the representation ${\displaystyle {\vec {X}}}$ ${\displaystyle {\vec {x}}}$

${\displaystyle {\boldsymbol {\varepsilon }}:={\dfrac {1}{2}}(\operatorname {grad} ({\vec {u}})+\operatorname {grad} ({\vec {u}})^{\top })=\varepsilon _{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\dfrac {1}{2}}(u_{i,j}+u_{j,i}){\hat {e}}_{i}\otimes {\hat {e}}_{j}}$

owns. The operator "grad" calculates the gradient , the superscript symbol stands for the transposition , the arithmetic symbol " " is the dyadic product and Einstein's sum convention was used in the last two equations . Here as in the following, indices occurring twice in a product, i and j above, have to be added up from one to three. Furthermore, an index after a comma is an abbreviation for the derivation according to the coordinate mentioned: ${\displaystyle (\cdot )^{\top }}$ ${\displaystyle \otimes }$

${\displaystyle u_{i,j}:={\dfrac {\partial u_{i}}{\partial x_{j}}}\,.}$

Calculating the rotation of the strain tensor yields:

${\displaystyle {\begin{aligned}\operatorname {rot} ({\boldsymbol {\varepsilon }}):=&{\hat {e}}_{k}\times {\frac {\partial }{\partial x_{k}}}{\boldsymbol {\varepsilon }}={\hat {e}}_{k}\times {\frac {1}{2}}(u_{i,jk}+u_{j,ik})({\hat {e}}_{i}\otimes {\hat {e}}_{j})\\=&{\frac {1}{2}}u_{i,jk}({\hat {e}}_{k}\times {\hat {e}}_{i})\otimes {\hat {e}}_{j}+{\frac {1}{2}}\underbrace {u_{j,ik}({\hat {e}}_{k}\times {\hat {e}}_{i})\otimes {\hat {e}}_{j}} _{=\mathbf {0} }\\\rightarrow \operatorname {rot(rot} ({\boldsymbol {\varepsilon }})^{\top })=&{\hat {e}}_{l}\times {\frac {\partial }{\partial x_{l}}}\left({\frac {1}{2}}u_{i,jk}{\hat {e}}_{j}\otimes ({\hat {e}}_{k}\times {\hat {e}}_{i})\right)={\frac {1}{2}}u_{i,jkl}{\hat {e}}_{l}\times {\hat {e}}_{j}\otimes {\hat {e}}_{k}\times {\hat {e}}_{i}=\mathbf {0} \end{aligned}}}$

The upper right term disappears because components with interchanged indices i and k have the same size but the opposite sign, so that they cancel each other out in the sum, or disappear at i = k, which in the last equation also applies to the indices j and l in applies analogously. The distortions derived from the displacement field therefore satisfy

${\displaystyle {\begin{aligned}\operatorname {rot(rot} ({\boldsymbol {\varepsilon }})^{\top })=\varepsilon _{ij,kl}({\hat {e}}_{l}\times {\hat {e}}_{j}\otimes {\hat {e}}_{k}\times {\hat {e}}_{i})=&\mathbf {0} \\\downarrow &\\2\varepsilon _{12,12}-\varepsilon _{22,11}-\varepsilon _{11,22}=&0\\2\varepsilon _{13,13}-\varepsilon _{33,11}-\varepsilon _{11,33}=&0\\2\varepsilon _{23,23}-\varepsilon _{33,22}-\varepsilon _{22,33}=&0\\\varepsilon _{11,23}+\varepsilon _{23,11}-\varepsilon _{12,13}-\varepsilon _{13,12}=&0\\\varepsilon _{22,13}+\varepsilon _{13,22}-\varepsilon _{12,23}-\varepsilon _{23,12}=&0\\\varepsilon _{12,33}+\varepsilon _{33,12}-\varepsilon _{13,23}-\varepsilon _{23,13}=&0\end{aligned}}}$

These are also the compatibility condition of the distortions, because if these equations are adhered to by a distortion field, then there is a displacement field that causes the given distortions.

Proof 1

The conclusion from the compatibility condition to the existence of the displacement field succeeds with the tensor field , which is trace-free : ${\displaystyle \operatorname {rot(rot} ({\boldsymbol {\varepsilon }})^{\top })=\mathbf {0} }$ ${\displaystyle \operatorname {rot} ({\boldsymbol {\varepsilon }})=\varepsilon _{ij,k}({\hat {e}}_{k}\times {\hat {e}}_{i})\otimes {\hat {e}}_{j}}$ ${\displaystyle \operatorname {Sp} (\operatorname {rot} ({\boldsymbol {\varepsilon }}))=\varepsilon _{ij,k}\,({\hat {e}}_{k}\times {\hat {e}}_{i})\cdot {\hat {e}}_{j}=\varepsilon _{ij,k}\,({\hat {e}}_{i}\times {\hat {e}}_{j})\cdot {\hat {e}}_{k}=0\,,}$ because terms with interchanged indices i and j are the same size, but have opposite signs, so that they cancel each other out in the sum, or they disappear when i = j. After the Poincaré Lemma in the expression a exists obliquely symmetric tensor W whose rotation is: . According to the Poincaré lemma in its expression there is now a vector field for which the following applies: and whose symmetric part is the strain tensor: ${\displaystyle \operatorname {rot} (\mathbf {T} ^{\top })=\mathbf {0} \;{\text{und}}\;\operatorname {Sp} (\mathbf {T} )=0\quad \rightarrow \quad \exists \mathbf {W} \colon \mathbf {T} =\operatorname {rot} (\mathbf {W} )\;{\text{mit}}\;\mathbf {W} ^{\top }=-\mathbf {W} }$ ${\displaystyle \operatorname {rot} ({\boldsymbol {\varepsilon }})}$ ${\displaystyle \operatorname {rot} ({\boldsymbol {\varepsilon }})=\operatorname {rot} (\mathbf {W} )\quad \rightarrow \quad \mathbf {0} =\operatorname {rot} ({\boldsymbol {\varepsilon }})-\operatorname {rot} (\mathbf {W} )=\operatorname {rot} [({\boldsymbol {\varepsilon }}+\mathbf {W} )^{\top }]}$ ${\displaystyle \operatorname {rot} (\mathbf {T} ^{\top })=\mathbf {0} \quad \rightarrow \quad \exists {\vec {u}}\colon \mathbf {T} =\operatorname {grad} ({\vec {u}})}$ ${\displaystyle {\vec {u}}}$ ${\displaystyle {\boldsymbol {\varepsilon }}+\mathbf {W} =\operatorname {grad} ({\vec {u}})}$ ${\displaystyle {\frac {1}{2}}[\operatorname {grad} ({\vec {u}})+\operatorname {grad} ({\vec {u}})^{\top }]={\frac {1}{2}}({\boldsymbol {\varepsilon }}+\mathbf {W} +{\boldsymbol {\varepsilon }}^{\top }+\mathbf {W} ^{\top })={\boldsymbol {\varepsilon }}\,.}$

In plane problems, such as Airy's stress function, where only two coordinates are involved, these compatibility conditions are further reduced to only one of the first three scalar equations.

The compatibility condition can also be written without the rotation:

${\displaystyle {\begin{aligned}\Delta {\boldsymbol {\varepsilon }}+\operatorname {grad(grad(Sp} ({\boldsymbol {\varepsilon }})))-2\operatorname {sym(grad(div} ({\boldsymbol {\varepsilon }})))=&\mathbf {0} \\\leftrightarrow \quad \varepsilon _{ij,kk}+\varepsilon _{kk,ij}-\varepsilon _{ik,jk}-\varepsilon _{jk,ik}=&0\,,\quad i,j=1,2,3\end{aligned}}}$

The operator “Sp” gives the trace of a tensor, “div” is the divergence and “sym” gives the symmetrical part

${\displaystyle \operatorname {sym} (\mathbf {T} )={\frac {1}{2}}(\mathbf {T} +\mathbf {T} ^{\top })\,.}$

Proof 2

The outer tensor product "#" defined as follows is used for the derivation : ${\displaystyle {\begin{aligned}({\vec {a}}\otimes {\vec {g}})\#({\vec {b}}\otimes {\vec {h}}):=&({\vec {a}}\times {\vec {b}})\otimes ({\vec {g}}\times {\vec {h}})\\\mathbf {A} \#\mathbf {B} =&-\operatorname {Sp} (\mathbf {A} )\mathbf {B} ^{\top }-\operatorname {Sp} (\mathbf {B} )\mathbf {A} ^{\top }+\mathbf {A} ^{\top }\cdot \mathbf {B} ^{\top }+\mathbf {B} ^{\top }\cdot \mathbf {A} ^{\top }+[\operatorname {Sp} (\mathbf {A} )\operatorname {Sp} (\mathbf {B} )-\operatorname {Sp} (\mathbf {A} \cdot \mathbf {B} )]\mathbf {I} \,.\end{aligned}}}$ The tensor is the unit tensor . This results in: The operator “ ” is the Laplace operator and “div” is the divergence . The trace from is calculated to with the consequence Therefore disappears exactly when it also disappears: Because if is, then also v = 0 and it follows, conversely , that v = 0 and accordingly applies. So the compatibility of the distortions with a displacement field can also be ensured. ${\displaystyle \mathbf {I} }$ ${\displaystyle {\begin{aligned}{\mathfrak {R}}({\boldsymbol {\varepsilon }}):=&\operatorname {rot(rot} ({\boldsymbol {\varepsilon }})^{\top })=\varepsilon _{ij,kl}({\hat {e}}_{l}\times {\hat {e}}_{j}\otimes {\hat {e}}_{k}\times {\hat {e}}_{i})=\varepsilon _{ij,kl}\overbrace {({\hat {e}}_{l}\otimes {\hat {e}}_{k})} ^{\mathbf {A} }\#\overbrace {({\hat {e}}_{j}\otimes {\hat {e}}_{i})} ^{\mathbf {B} }\\=&\varepsilon _{ij,kl}{\bigl [}-\delta _{lk}({\hat {e}}_{j}\otimes {\hat {e}}_{i})^{\top }-\delta _{ji}({\hat {e}}_{l}\otimes {\hat {e}}_{k})^{\top }+({\hat {e}}_{l}\otimes {\hat {e}}_{k})^{\top }\cdot ({\hat {e}}_{j}\otimes {\hat {e}}_{i})^{\top }\\&+({\hat {e}}_{j}\otimes {\hat {e}}_{i})^{\top }\cdot ({\hat {e}}_{l}\otimes {\hat {e}}_{k})^{\top }+(\delta _{lk}\delta _{ji}-\delta _{kj}\delta _{li})\mathbf {I} {\bigr ]}\\=&-\varepsilon _{ij,kk}({\hat {e}}_{i}\otimes {\hat {e}}_{j})-\varepsilon _{kk,ij}({\hat {e}}_{i}\otimes {\hat {e}}_{j})+\varepsilon _{jk,ik}({\hat {e}}_{i}\otimes {\hat {e}}_{j})+\varepsilon _{ik,jk}({\hat {e}}_{i}\otimes {\hat {e}}_{j})+[\varepsilon _{ii,kk}-\varepsilon _{ij,ij}]\mathbf {I} \\=&-[\underbrace {\Delta {\boldsymbol {\varepsilon }}+\operatorname {grad(grad(Sp} ({\boldsymbol {\varepsilon }})))-2\operatorname {sym(grad(div} ({\boldsymbol {\varepsilon }})))} _{=:{\mathfrak {L}}({\boldsymbol {\varepsilon }})}]+[\underbrace {\Delta \operatorname {Sp} ({\boldsymbol {\varepsilon }})-\operatorname {div(div} ({\boldsymbol {\varepsilon }}))} _{=:v}]\mathbf {I} \\\rightarrow {\mathfrak {R}}({\boldsymbol {\varepsilon }})=&-{\mathfrak {L}}({\boldsymbol {\varepsilon }})+v\mathbf {I} \end{aligned}}}$ ${\displaystyle \Delta }$ ${\displaystyle {\mathfrak {L}}({\boldsymbol {\varepsilon }})}$ ${\displaystyle {\begin{aligned}\operatorname {Sp} ({\mathfrak {L}}({\boldsymbol {\varepsilon }}))=&\operatorname {Sp} (\varepsilon _{ij,kk}{\hat {e}}_{i}\otimes {\hat {e}}_{j}+\varepsilon _{kk,ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}-\varepsilon _{jk,ik}{\hat {e}}_{i}\otimes {\hat {e}}_{j}-\varepsilon _{ik,jk}{\hat {e}}_{i}\otimes {\hat {e}}_{j})\\=&\varepsilon _{ii,kk}+\varepsilon _{kk,ii}-\varepsilon _{ik,ik}-\varepsilon _{ik,ik}=2\Delta \operatorname {Sp} ({\boldsymbol {\varepsilon }})-2\operatorname {div(div} ({\boldsymbol {\varepsilon }}))=2v\end{aligned}}}$ ${\displaystyle \operatorname {Sp} ({\mathfrak {R}}({\boldsymbol {\varepsilon }}))=-2v+3v=v\,.}$ ${\displaystyle {\mathfrak {R}}({\boldsymbol {\varepsilon }})}$ ${\displaystyle {\mathfrak {L}}({\boldsymbol {\varepsilon }})}$ ${\displaystyle {\mathfrak {R}}({\boldsymbol {\varepsilon }})=\mathbf {0} }$ ${\displaystyle {\mathfrak {L}}({\boldsymbol {\varepsilon }})=\mathbf {0} \,.}$ ${\displaystyle {\mathfrak {L}}({\boldsymbol {\varepsilon }})=\mathbf {0} }$ ${\displaystyle {\mathfrak {R}}({\boldsymbol {\varepsilon }})=\mathbf {0} }$ ${\displaystyle {\begin{aligned}{\mathfrak {L}}({\boldsymbol {\varepsilon }})=\Delta {\boldsymbol {\varepsilon }}+\operatorname {grad(grad(Sp} ({\boldsymbol {\varepsilon }})))-2\operatorname {sym(grad(div} ({\boldsymbol {\varepsilon }})))=&\mathbf {0} \\\leftrightarrow \quad \varepsilon _{ij,kk}+\varepsilon _{kk,ij}-\varepsilon _{ik,jk}-\varepsilon _{jk,ik}=&0\,,\quad i,j=1,2,3\end{aligned}}}$

Tensions

In the approach to the equations of motion using stress functions, the stresses are the primary unknowns. If these are found for the given boundary conditions, then it is necessary to reconstruct the motion field from them. This is possible with linear, isotropic elasticity if the stresses in a gravitational field, such as the force of gravity, fulfill the following compatibility conditions formulated for them: ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle {\vec {b}}}$

${\displaystyle {\begin{aligned}\Delta {\boldsymbol {\sigma }}+{\frac {1}{1+\nu }}\operatorname {grad(grad(Sp} ({\boldsymbol {\sigma }})))+{\frac {\nu }{1-\nu }}\operatorname {div} ({\vec {b}})\mathbf {I} +\operatorname {grad} ({\vec {b}})+\operatorname {grad} ({\vec {b}})^{\top }=&\mathbf {0} \\\leftrightarrow \quad \sigma _{ij,kk}+{\frac {1}{1+\nu }}\sigma _{kk,ij}+{\frac {\nu }{1-\nu }}b_{k,k}\delta _{ij}+b_{i,j}+b_{j,i}=&0\;,\quad i,j=1,2,3\end{aligned}}}$

or in the absence of gravity:

${\displaystyle \Delta {\boldsymbol {\sigma }}+{\frac {1}{1+\nu }}\operatorname {grad(grad(Sp} ({\boldsymbol {\sigma }})))=\mathbf {0} \quad \leftrightarrow \quad \sigma _{ij,kk}+{\frac {1}{1+\nu }}\sigma _{kk,ij}=0\;,\quad i,j=1,2,3}$

The symbol is the Kronecker delta and is the Poisson's ratio . ${\displaystyle \delta _{ij}}$ ${\displaystyle \nu }$

Proof 3

The derivation is based on proof 2, which showed that if ${\displaystyle {\mathfrak {L}}({\boldsymbol {\varepsilon }}):=\Delta {\boldsymbol {\varepsilon }}+\operatorname {grad(grad(Sp} ({\boldsymbol {\varepsilon }})))-2\operatorname {sym(grad(div} ({\boldsymbol {\varepsilon }})))}$ disappears, the distortion field is compatible. In statics, the divergence of the stresses is in equilibrium with the specific gravity : The abbreviation follows because of and : With linear, isotropic elasticity, the stress-strain relationship is linear: The material parameter G is the shear modulus . Now the compatibility condition can be expressed with the stresses: Applying the trace to this equation gives and finally leads to the compatibility conditions expressed in the stresses: ${\displaystyle {\boldsymbol {\varepsilon }}}$ ${\displaystyle {\vec {b}}}$ ${\displaystyle \operatorname {div} ({\boldsymbol {\sigma }})+{\vec {b}}={\vec {0}}\quad \leftrightarrow \quad \sigma _{ij,i}+b_{j}=0\,,\quad j=1,2,3\,.}$ ${\displaystyle p:=\operatorname {Sp} ({\boldsymbol {\sigma }})}$ ${\displaystyle \operatorname {div(grad} (p))=\Delta p\,,\;\operatorname {div} (p\,\mathbf {I} )=\operatorname {grad} (p)}$ ${\displaystyle \operatorname {Sp\circ grad} =\operatorname {div} }$ ${\displaystyle {\begin{aligned}{\mathfrak {L}}({\boldsymbol {\sigma }})=&\Delta {\boldsymbol {\sigma }}+\operatorname {grad(grad} (p))+2\operatorname {sym(grad} ({\vec {b}}))\\{\mathfrak {L}}(p\,\mathbf {I} )=&\Delta p\,\mathbf {I} +\operatorname {grad(grad} (p))\,.\end{aligned}}}$ ${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2G}}\left[{\boldsymbol {\sigma }}-{\frac {\nu }{1+\nu }}\operatorname {Sp} ({\boldsymbol {\sigma }})\,\mathbf {I} \right]\quad \leftrightarrow \quad \varepsilon _{ij}={\frac {1}{2G}}\left[\sigma _{ij}-{\frac {\nu }{1+\nu }}\sigma _{kk}\delta _{ij}\right]\,.}$ ${\displaystyle {\mathfrak {L}}(2G{\boldsymbol {\varepsilon }})={\mathfrak {L}}({\boldsymbol {\sigma }})-{\frac {\nu }{1+\nu }}{\mathfrak {L}}(p\,\mathbf {I} )=\Delta {\boldsymbol {\sigma }}+{\frac {1}{1+\nu }}\operatorname {grad(grad} (p))-{\frac {\nu }{1+\nu }}\Delta p\,\mathbf {I} +2\operatorname {sym(grad} ({\vec {b}}))=\mathbf {0} \,.}$ ${\displaystyle \Delta p=-{\frac {1+\nu }{1-\nu }}\operatorname {div} ({\vec {b}})}$ ${\displaystyle \Delta {\boldsymbol {\sigma }}+{\frac {1}{1+\nu }}\operatorname {grad(grad(Sp} ({\boldsymbol {\sigma }})))+{\frac {\nu }{1-\nu }}\operatorname {div} ({\vec {b}})\mathbf {I} +2\operatorname {sym(grad} ({\vec {b}}))=\mathbf {0} \,.}$

These compatibility conditions are called Beltrami-Michell equations.

There are also compatibility conditions for cubically anisotropic (Albrecht 1951) and transversely isotropic (von Moisil 1952) linear elasticity.

Deformation gradient

The components of the deformation gradient are calculated from the derivatives of the motion components according to the material coordinates : ${\displaystyle \chi _{1,2,3}}$ ${\displaystyle X_{1,2,3}}$

${\displaystyle F_{ij}({\vec {X}},t)={\dfrac {\partial \chi _{i}({\vec {X}},t)}{\partial X_{j}}},\quad i,j=1,2,3\,.}$

Now there are components of the deformation gradient that were derived from the three motion functions . ${\displaystyle 3\cdot 3=9}$ ${\displaystyle F_{ij}}$ ${\displaystyle \chi _{1,2,3}}$

Conversely , if the three motion functions are to be obtained from nine components of the deformation gradient , the components of the deformation gradient must adhere to the following compatibility conditions formulated for them: ${\displaystyle F_{ij}({\vec {X}},t)}$ ${\displaystyle \chi _{1,2,3}({\vec {X}},t)}$

${\displaystyle \operatorname {rot} (\mathbf {F} ^{\top }):={\hat {e}}_{k}\times {\frac {\partial F_{ij}}{\partial X_{k}}}{\hat {e}}_{j}\otimes {\hat {e}}_{i}={\begin{pmatrix}{\dfrac {\partial F_{13}}{\partial X_{2}}}-{\dfrac {\partial F_{12}}{\partial X_{3}}}&{\dfrac {\partial F_{23}}{\partial X_{2}}}-{\dfrac {\partial F_{22}}{\partial X_{3}}}&{\dfrac {\partial F_{33}}{\partial X_{2}}}-{\dfrac {\partial F_{32}}{\partial X_{3}}}&\\[2ex]{\dfrac {\partial F_{11}}{\partial X_{3}}}-{\dfrac {\partial F_{13}}{\partial X_{1}}}&{\dfrac {\partial F_{21}}{\partial X_{3}}}-{\dfrac {\partial F_{23}}{\partial X_{1}}}&{\dfrac {\partial F_{31}}{\partial X_{3}}}-{\dfrac {\partial F_{33}}{\partial X_{1}}}&\\[2ex]{\dfrac {\partial F_{12}}{\partial X_{1}}}-{\dfrac {\partial F_{11}}{\partial X_{2}}}&{\dfrac {\partial F_{22}}{\partial X_{1}}}-{\dfrac {\partial F_{21}}{\partial X_{2}}}&{\dfrac {\partial F_{32}}{\partial X_{1}}}-{\dfrac {\partial F_{31}}{\partial X_{2}}}&\\[2ex]\end{pmatrix}}=\mathbf {0} }$

If so, put the Poincaré lemma in the form

${\displaystyle \operatorname {rot} (\mathbf {F} ^{\top })=\mathbf {0} \quad \rightarrow \quad \exists {\vec {\chi }}\colon \mathbf {F} =\operatorname {grad} ({\vec {\chi }})}$

sure that there is a vector field whose gradient is the tensor field . ${\displaystyle {\vec {\chi }}({\vec {X}})}$ ${\displaystyle \mathbf {F} }$

Stretch tensor

The deformation gradient can be due to

${\displaystyle \mathbf {F} ={\dfrac {\partial \chi _{i}}{\partial X_{j}}}{\hat {e}}_{i}\otimes {\hat {e}}_{j}={\dfrac {\partial (\chi _{i}{\hat {e}}_{i})}{\partial X_{j}}}\otimes {\hat {e}}_{j}=:{\vec {g}}_{j}\otimes {\hat {e}}_{j}}$

with the tangent vectors

${\displaystyle {\vec {g}}_{j}:={\dfrac {\partial {\vec {\chi }}({\vec {X}},t)}{\partial X_{j}}}}$

being represented. The components of the right Cauchy-Green tensor are calculated because of ${\displaystyle g_{ik}}$ ${\displaystyle \mathbf {C} =\mathbf {F} ^{\top }\cdot \mathbf {F} }$

${\displaystyle \mathbf {C} =g_{ik}{\hat {e}}_{i}\otimes {\hat {e}}_{k}=\mathbf {F} ^{\top }\cdot \mathbf {F} =\left({\hat {e}}_{i}\otimes {\vec {g}}_{i}\right)\cdot \left({\vec {g}}_{k}\otimes {\hat {e}}_{k}\right)=({\vec {g}}_{i}\cdot {\vec {g}}_{k}){\hat {e}}_{i}\otimes {\hat {e}}_{k}}$

from the scalar products of these tangent vectors:

${\displaystyle g_{ik}:={\vec {g}}_{i}\cdot {\vec {g}}_{k}\,.}$

With the Christmas symbols of the first kind

${\displaystyle \Gamma _{mkl}:={\vec {g}}_{m}\cdot {\dfrac {\partial {\vec {g}}_{k}}{\partial X_{l}}}={\dfrac {1}{2}}\left({\dfrac {\partial g_{km}}{\partial X_{l}}}+{\dfrac {\partial g_{ml}}{\partial X_{k}}}-{\dfrac {\partial g_{lk}}{\partial X_{m}}}\right)=\Gamma _{mlk}}$

it can be shown that given the components of the right Cauchy-Green tensor, the movement can be reconstructed if and only if ${\displaystyle g_{ij}}$

${\displaystyle R_{ijkl}={\dfrac {\partial \Gamma _{ijk}}{\partial X_{l}}}-{\dfrac {\partial \Gamma _{ijl}}{\partial X_{k}}}+g^{pq}(\Gamma _{pik}\Gamma _{qjl}-\Gamma _{pil}\Gamma _{qjk})=0\,,\;{\text{für}}\quad i,j,k,l=1,2,3}$

applies. The components belong to the inverse of the right Cauchy-Green tensor ${\displaystyle g^{pq}}$

${\displaystyle \mathbf {C} ^{-1}=g^{pq}{\hat {e}}_{p}\otimes {\hat {e}}_{q}}$

and are the components of the Riemann-Christoffel curvature tensor . Of the above equations for the 81 components of the Riemann-Christoffel tensor, only six are independent. Because of the linear relationship ${\displaystyle R_{ijkl}}$

${\displaystyle \mathbf {E} ={\dfrac {1}{2}}(\mathbf {C} -\mathbf {I} )}$

between the right Cauchy-Green tensor and the Green-Lagrange strain tensor, there can also be compatibility conditions for the components ${\displaystyle \mathbf {C} }$ ${\displaystyle \mathbf {E} }$

${\displaystyle E_{ij}={\dfrac {1}{2}}(g_{ij}-\delta _{ij})\rightarrow g_{ij}=2E_{ij}+\delta _{ij}}$

of the Green-Lagrange strain tensor, which are, however, much more difficult to solve than in the geometrically linear case, where the linearized strain tensor merges, see above and the following example. ${\displaystyle \mathbf {E} }$ ${\displaystyle {\boldsymbol {\varepsilon }}}$

example

Boundary conditions and deformation (beige) when bending the straight beam (dashed)

On a linearly elastic beam aligned in the x-direction, only a tension proportional to the z-coordinate acts

${\displaystyle \sigma _{xx}={\vec {t}}\cdot {\hat {e}}_{x}=-mEz}$

with proportionality factor and modulus of elasticity of the material of the beam, see figure on the right. According to Hooke's law , the stresses correspond to the elongations ${\displaystyle -m}$ ${\displaystyle E}$

${\displaystyle {\begin{pmatrix}\varepsilon _{xx}\\\varepsilon _{yy}\\\varepsilon _{zz}\end{pmatrix}}={\dfrac {1}{E}}{\begin{pmatrix}1&-\nu &-\nu \\-\nu &1&-\nu \\-\nu &-\nu &1\end{pmatrix}}{\begin{pmatrix}\sigma _{xx}\\\sigma _{yy}\\\sigma _{zz}\end{pmatrix}}={\begin{pmatrix}1&-\nu &-\nu \\-\nu &1&-\nu \\-\nu &-\nu &1\end{pmatrix}}{\begin{pmatrix}-mz\\0\\0\end{pmatrix}}={\begin{pmatrix}-mz\\\nu mz\\\nu mz\end{pmatrix}}}$

because shear stresses are not specified, which is why no shearings occur. The size is the Poisson's ratio of the material of the beam. Because all second derivatives of the strains disappear, the compatibility conditions are met: there is a displacement field that causes the given strains. With the boundary conditions sketched in the picture, these are shifts ${\displaystyle \nu }$

${\displaystyle {\begin{aligned}u=&-mxz\\v=&\nu myz\\w=&{\dfrac {m}{2}}(x^{2}-\nu y^{2}+\nu z^{2})\end{aligned}}}$

because:

${\displaystyle {\begin{array}{lcllcllcl}\varepsilon _{xx}&=&{\dfrac {\partial u}{\partial x}}=-mz,&2\varepsilon _{xy}&=&{\dfrac {\partial u}{\partial y}}+{\dfrac {\partial v}{\partial x}}=0+0&=&0\\[2ex]\varepsilon _{yy}&=&{\dfrac {\partial v}{\partial y}}=\nu mz,&2\varepsilon _{yz}&=&{\dfrac {\partial v}{\partial z}}+{\dfrac {\partial w}{\partial y}}=\nu my-\nu my&=&0\\[2ex]\varepsilon _{zz}&=&{\dfrac {\partial w}{\partial z}}=\nu mz,&2\varepsilon _{xz}&=&{\dfrac {\partial u}{\partial z}}+{\dfrac {\partial w}{\partial x}}=-mx+mx&=&0\end{array}}}$

and

${\displaystyle u(x=0)=0\,,\quad v(x=0,y=0)=0\,,\quad w(x=0,y=0,z=0)=0\,.}$

The example of the stress functions shows that this motion field is also in equilibrium.

See also

• Integrability condition
• Formula collection tensor algebra
• Tensoranalysis formula collection

Footnotes

1. ↑ In the literature there is also the condition , which is not a contradiction in view of the then different definition of rotation ( ). ${\displaystyle \operatorname {{\tilde {rot}}({\tilde {rot}}} ({\boldsymbol {\varepsilon }}))=\mathbf {0} }$ ${\displaystyle \operatorname {\tilde {rot}} \mathbf {T} ={\hat {e}}_{k}\times \mathbf {T} _{,k}^{\top }}$
2. ↑ Beltrami found the compatibility conditions for the stresses with isotropic elasticity in the absence of gravitational acceleration in 1892 and Donati and Michell formulated the more general case including gravitational acceleration in 1894 and 1900, see ME Gurtin (1972), p. 92. Despite this, Donati's work appeared six years earlier as Michells, this more general equation is referred to as the Beltrami-Michell equation.

Individual evidence

1. ↑ ME Gurtin (1972), p. 40
2. ↑ ME Gurtin (1972), p. 92
3. ^ E. Klingbeil: Tensor calculation for engineers . BI Wissenschaftsverlag, 1989, ISBN 3-411-05197-3 , pp. 122 .

literature

• ME Gurtin: The Linear Theory of Elasticity . In: S. Flügge (Ed.): Handbuch der Physik . Volume VI2 / a, Volume Editor C. Truesdell. Springer, 1972, ISBN 3-540-05535-5 .
• P. Haupt: Continuum Mechanics and Theory of Materials . Springer, 2010, ISBN 978-3-642-07718-0 .
• E. Klingbeil: Tensor calculation for engineers . BI Wissenschaftsverlag, 1989, ISBN 3-411-05197-3 .
• Martin H. Sadd: Elasticity - Theory, applications and numerics . Elsevier Butterworth-Heinemann, 2005, ISBN 0-12-605811-3 .
• PK Raschewski: Riemannian geometry and tensor analysis . VEB Deutscher Verlag der Wissenschaft, 1959.
• TY Thomas: Systems of Total Differential Equations Defined over Simply Connected Domains . In: Annals of Mathematics . tape 35 , 1934, pp. 730-734 (English, JSTOR ).