Tension function

Stress functions are an approach for the analytical solution of boundary value problems of linear elastostatics .

In statics, the local momentum balance is an equation in which only tensions and gravity occur. By expressing the stresses in terms of stress functions that automatically comply with the momentum balance, the solution of a boundary value problem is reduced to finding stress functions that meet the given boundary conditions and the compatibility conditions. The compatibility conditions ensure that a displacement field can be derived from the stresses. An analytical solution often only exists with geometric linearity (small deformations ) and with the assumption of linear elasticity .

These requirements - statics, small deformations and linear elasticity - are given in many applications, especially in the technical field.

history

Chronological sequence in the development of the voltage functions

The history of the stress functions is closely related to the history of the formulation of the compatibility conditions in linear isotropic elasticity. Gustav Robert Kirchhoff derived three of the six compatibility conditions for the distortions (KBV) in 1859 and showed how the displacements can be calculated from the distortions. The approach with stress functions was then devised four years later by George Biddell Airy in 1863. With the Airy stress function named after him today , boundary value problems can be solved in the plane. All six KBV were presented for the first time by Adhémar Jean Claude Barré de Saint-Venant in 1864, but he has not shown that they are sufficient. Stress functions for problems in three dimensions were found by James Clerk Maxwell and Giacinto Morera around 1870 and 1892, respectively. In the meantime, Eugenio Beltrami was able to prove in 1886 that the KBV of St. Venant are actually sufficient. Beltrami found the compatibility conditions for the stresses (KBS) with isotropic elasticity in the absence of gravity in 1892 and Luigi Donati formulated the more general case including gravity in 1894. Nevertheless, this more general equation is referred to as the Beltrami-Michell equation (additionally after John Henry Michell ). Beltrami recognized in 1892 that the stress functions of Airy, Maxwell and Morera that had existed up to that point were special cases of a more general approach. However, Beltrami's solution cannot take a gravitational field into account. In 1953 Hermann Schaefer extended Beltrami's approach to problems with gravity fields. The KBS for transversely isotropic linear elasticity was formulated by Grigore Moisil in 1952.

Shortly

The compatibility conditions for the distortions are

${\displaystyle {\mathfrak {R}}({\boldsymbol {\varepsilon }}):=\sum _{i,j,k,l=1}^{3}\varepsilon _{ij,kl}({\hat {e}}_{l}\times {\hat {e}}_{j})\otimes ({\hat {e}}_{k}\times {\hat {e}}_{i})=\mathbf {0} \,.}$

The vectors form the standard basis belonging to the Cartesian coordinates , " " is the dyadic product and " " the cross product , are the components of the linearized distortion tensor and an index after a comma denotes the derivative according to the corresponding coordinate: ${\displaystyle {\hat {e}}_{1,2,3}}$ ${\displaystyle x_{1,2,3}}$ ${\displaystyle \otimes }$ ${\displaystyle \times }$ ${\displaystyle \varepsilon _{ij}}$ ${\displaystyle {\boldsymbol {\varepsilon }}}$

${\displaystyle (\cdot )_{,k}={\frac {\partial }{\partial x_{k}}}\,.}$

With symmetric arguments , the differential operator delivers divergence-free , symmetric tensors, which also include stress tensors in statics in the absence of gravity. With this differential operator one can find stress tensors satisfying the momentum balances in a simple way : ${\displaystyle {\mathfrak {R}}}$ ${\displaystyle {\boldsymbol {\sigma }}}$

${\displaystyle \mathbf {A} =\mathbf {A} ^{\top }\;\rightarrow \quad {\boldsymbol {\sigma }}={\mathfrak {R}}(\mathbf {A} )={\boldsymbol {\sigma }}^{\top }\;\wedge \;\operatorname {div} ({\boldsymbol {\sigma }})={\vec {0}}\,.}$

The components of the symmetrical argument used are Beltrami's stress functions . In the case of linear isotropic elasticity , the above compatibility condition for the strains in the stresses can be expressed: ${\displaystyle \mathbf {A} }$ ${\displaystyle \sigma _{ij}}$

${\displaystyle \sum _{k=1}^{3}\left(\sigma _{ij,kk}+{\frac {1}{1+\nu }}\sigma _{kk,ij}\right)=0\;,\quad i,j=1,2,3}$

This equation is known as the Beltrami-Michell equation. The material parameter is the Poisson's ratio . ${\displaystyle \nu }$

The solution of a boundary value problem can now be traced back to finding stress functions that result in stresses that comply with the required boundary conditions and the compatibility conditions.

The stress functions found by Airy, Maxwell and Morea fit here as special cases:

author	year	Stress functions	Stress tensor
Airy	1863	${\displaystyle \mathbf {A} ={\begin{pmatrix}0&&\\&0&\\&&\varphi \end{pmatrix}}}$	${\displaystyle {\boldsymbol {\sigma }}={\begin{pmatrix}\varphi _{,yy}&-\varphi _{,xy}&0\\&\varphi _{,xx}&0\\{\textsf {sym}}&&0\end{pmatrix}}}$
Maxwell	1870	${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{1}&&\\&a_{2}&\\&&a_{3}\end{pmatrix}}}$	${\displaystyle {\boldsymbol {\sigma }}={\begin{pmatrix}a_{2,33}+a_{3,22}&-a_{3,12}&-a_{2,13}\\&a_{1,33}+a_{3,11}&-a_{1,23}\\{\textsf {sym}}&&a_{1,22}+a_{2,11}\end{pmatrix}}}$
Morea	1892	${\displaystyle \mathbf {A} ={\begin{pmatrix}0&\omega _{3}&\omega _{2}\\\omega _{3}&0&\omega _{1}\\\omega _{2}&\omega _{1}&0\end{pmatrix}}}$	${\displaystyle {\boldsymbol {\sigma }}={\begin{pmatrix}-2\omega _{1,23}&(\omega _{1,1}+\omega _{2,2}-\omega _{3,3})_{,3}&(\omega _{1,1}-\omega _{2,2}+\omega _{3,3})_{,2}\\&-2\omega _{2,13}&(-\omega _{1,1}+\omega _{2,2}+\omega _{3,3})_{,1}\\{\textsf {sym}}&&-2\omega _{3,12}\end{pmatrix}}}$
Beltrami	1892	${\displaystyle \mathbf {A} ={\begin{pmatrix}A_{11}&A_{12}&A_{13}\\A_{12}&A_{22}&A_{23}\\A_{13}&A_{23}&A_{33}\end{pmatrix}}}$	: ${\displaystyle {\begin{array}{rcl}\sigma _{11}&=&A_{33,22}+A_{22,33}-2A_{23,23}\\\sigma _{22}&=&A_{33,11}+A_{11,33}-2A_{13,13}\\\sigma _{33}&=&A_{22,11}+A_{11,22}-2A_{12,12}\\\sigma _{12}&=&(-A_{12,3}+A_{23,1}+A_{13,2})_{,3}-A_{33,12}\\\sigma _{13}&=&(+A_{12,3}+A_{23,1}-A_{13,2})_{,2}-A_{22,13}\\\sigma _{23}&=&(+A_{12,3}-A_{23,1}+A_{13,2})_{,1}-A_{11,23}\end{array}}}$

definition

The local momentum and angular momentum balance in the absence of gravity are:

${\displaystyle \operatorname {div} ({\boldsymbol {\sigma }})=\mathbf {0} \,,\quad {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{\top }\,.}$

The differential operator “div” gives the divergence of the stress tensor , which due to the angular momentum balance is identical to its transposed one . The stress tensor is therefore symmetrical due to the angular momentum balance. ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle {\boldsymbol {\sigma }}^{\top }}$

If is a tensor field and a differential operator for symmetric arguments, then is ${\displaystyle \mathbf {A} }$ ${\displaystyle {\mathfrak {D}}}$

${\displaystyle {\boldsymbol {\sigma }}={\mathfrak {D}}(\mathbf {A} )}$

a solution to the balance equations if

${\displaystyle \operatorname {div} ({\mathfrak {D}}(\mathbf {A} ))=\mathbf {0} \,,\quad {\mathfrak {D}}(\mathbf {A} )={\mathfrak {D}}(\mathbf {A} )^{\top }}$

is. A field with these properties is called a stress function . ${\displaystyle \mathbf {A} }$

Beltramis tension functions

The differential operator is given

${\displaystyle {\mathfrak {R}}(\mathbf {A} ):=\operatorname {rot(rot} (\mathbf {A} )^{\top }):=-{\hat {e}}_{l}\times \mathbf {A} _{,kl}^{\top }\times {\hat {e}}_{k}=A_{ij,kl}({\hat {e}}_{l}\times {\hat {e}}_{j})\otimes ({\hat {e}}_{k}\times {\hat {e}}_{i})\,.}$

Applied to any symmetric tensor shows: ${\displaystyle \mathbf {A} }$

${\displaystyle {\begin{array}{rcl}{\mathfrak {R}}(\mathbf {A} )^{\top }&=&A_{ij,kl}({\hat {e}}_{k}\times {\hat {e}}_{i}\otimes {\hat {e}}_{l}\times {\hat {e}}_{j}){\stackrel {\begin{array}{c}\scriptscriptstyle i\leftrightharpoons j\\[-1ex]\scriptscriptstyle k\leftrightharpoons l\end{array}}{=}}A_{ji,lk}({\hat {e}}_{l}\times {\hat {e}}_{j}\otimes {\hat {e}}_{k}\times {\hat {e}}_{i})\\&=&A_{ij,kl}({\hat {e}}_{l}\times {\hat {e}}_{j}\otimes {\hat {e}}_{k}\times {\hat {e}}_{i})={\mathfrak {R}}(\mathbf {A} )\\\operatorname {div} ({\mathfrak {R}}(\mathbf {A} ))&=&{\hat {e}}_{m}\cdot A_{ij,klm}({\hat {e}}_{l}\times {\hat {e}}_{j}\otimes {\hat {e}}_{k}\times {\hat {e}}_{i})={\hat {e}}_{j}\cdot ({\hat {e}}_{m}\times {\hat {e}}_{l})A_{ij,klm}{\hat {e}}_{k}\times {\hat {e}}_{i}={\vec {0}}\end{array}}}$

because components with interchanged indices l and m are the same size but have the opposite sign and vanish in the case l = m. So the tensor is a stress function. So in statics in the absence of gravity it delivers ${\displaystyle \mathbf {A} }$

${\displaystyle {\begin{array}{rclcl}{\boldsymbol {\sigma }}&=&{\mathfrak {R}}(\mathbf {A} )={\begin{pmatrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\&\sigma _{22}&\sigma _{23}\\{\textsf {sym}}&&\sigma _{33}\end{pmatrix}}\\\sigma _{11}&=&A_{33,22}+A_{22,33}-2A_{23,23}\\\sigma _{22}&=&A_{33,11}+A_{11,33}-2A_{13,13}\\\sigma _{33}&=&A_{22,11}+A_{11,22}-2A_{12,12}\\\sigma _{12}&=&-A_{12,33}+A_{23,13}+A_{13,23}-A_{33,12}&=&(-A_{12,3}+A_{23,1}+A_{13,2})_{,3}-A_{33,12}\\\sigma _{13}&=&+A_{12,23}+A_{23,12}-A_{13,22}-A_{22,13}&=&(+A_{12,3}+A_{23,1}-A_{13,2})_{,2}-A_{22,13}\\\sigma _{23}&=&+A_{12,13}-A_{23,11}+A_{13,12}-A_{11,23}&=&(+A_{12,3}-A_{23,1}+A_{13,2})_{,1}-A_{11,23}\end{array}}}$

a permissible stress state because it is . The stress tensor still has to meet the compatibility conditions ${\displaystyle \operatorname {div} ({\boldsymbol {\sigma }})={\vec {0}}}$

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

keep so that it is in harmony with a displacement field. The components of the tensor are known as Beltrami's stress functions. Stress functions previously found by other authors turn out to be special cases of Beltrami's solution. ${\displaystyle A_{ij}}$ ${\displaystyle \mathbf {A} }$

Airy's tension function

→ Main article : Airy stress function

The stress function that George Biddell Airy found in 1863 is the special case ${\displaystyle \varphi }$

${\displaystyle \mathbf {A} ={\begin{pmatrix}0&&\\&0&\\&&\varphi \end{pmatrix}}\rightarrow {\boldsymbol {\sigma }}={\begin{pmatrix}\varphi _{,yy}&-\varphi _{,xy}&0\\&\varphi _{,xx}&0\\{\textsf {sym}}&&0\end{pmatrix}}\,.}$

The compatibility condition for homogeneous, isotropic , elastic material can be as follows

${\displaystyle \Delta \Delta \varphi =0}$

write what makes a biharmonic function . ${\displaystyle \varphi }$

Maxwell's stress functions

The stress functions described by Maxwell in 1868 and 1870 are also classified here

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{1}&&\\&a_{2}&\\&&a_{3}\end{pmatrix}}\quad \rightarrow \quad {\boldsymbol {\sigma }}={\begin{pmatrix}a_{2,33}+a_{3,22}&-a_{3,12}&-a_{2,13}\\&a_{1,33}+a_{3,11}&-a_{1,23}\\{\textsf {sym}}&&a_{1,22}+a_{2,11}\end{pmatrix}}}$

a.

Morea's stress functions

In 1892 Morea found stress functions, which here are the special case

${\displaystyle \mathbf {A} ={\begin{pmatrix}0&\omega _{3}&\omega _{2}\\\omega _{3}&0&\omega _{1}\\\omega _{2}&\omega _{1}&0\end{pmatrix}}\quad \rightarrow \quad {\boldsymbol {\sigma }}={\begin{pmatrix}-2\omega _{1,23}&(\omega _{1,1}+\omega _{2,2}-\omega _{3,3})_{,3}&(\omega _{1,1}-\omega _{2,2}+\omega _{3,3})_{,2}\\&-2\omega _{2,13}&(-\omega _{1,1}+\omega _{2,2}+\omega _{3,3})_{,1}\\{\textsf {sym}}&&-2\omega _{3,12}\end{pmatrix}}}$

turn out.

Beltrami-Schäfer tension functions

The Beltrami stress functions above can not represent gravity because of it. The Beltrami-Schäfer solution, ${\displaystyle \operatorname {div} ({\boldsymbol {\sigma }})={\vec {0}}}$

${\displaystyle {\boldsymbol {\sigma }}={\mathfrak {R}}(\mathbf {A} )+{\mathfrak {h}}({\vec {h}})\,,\quad {\mathfrak {h}}({\vec {h}})=\operatorname {grad} ({\vec {h}})+\operatorname {grad} ({\vec {h}})^{\top }-\operatorname {div} ({\vec {h}})\mathbf {I} }$

which Schäfer found in 1953 can also perform boundary value problems with gravity of the form

${\displaystyle {\vec {b}}=-\Delta {\vec {h}}}$

to solve. The tensor A is symmetric as always. Then

${\displaystyle \operatorname {div} ({\boldsymbol {\sigma }})=-{\vec {b}}\quad \wedge \quad {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{\top }}$

for because is ${\displaystyle \operatorname {div(grad} ({\vec {h}})^{\top })=\Delta {\vec {h}}\,,\;\operatorname {div} (p\,\mathbf {I} )=\operatorname {grad} (p)\,,\;\operatorname {grad(div} ({\vec {h}}))=\operatorname {div(grad} ({\vec {h}}))}$

${\displaystyle {\begin{array}{rcl}{\mathfrak {h}}({\vec {h}})^{\top }&=&\operatorname {grad} ({\vec {h}})^{\top }+\operatorname {grad} ({\vec {h}})-\operatorname {div} ({\vec {h}})\mathbf {I} ={\mathfrak {h}}({\vec {h}})\\\operatorname {div} ({\mathfrak {h}}({\vec {h}}))&=&\operatorname {div} [\operatorname {grad} ({\vec {h}})+\operatorname {grad} ({\vec {h}})^{\top }-\operatorname {div} ({\vec {h}})\mathbf {I} ]=\operatorname {div(grad} ({\vec {h}}))+\Delta {\vec {h}}-\operatorname {grad(div} ({\vec {h}}))\\&=&\Delta {\vec {h}}=-{\vec {b}}\end{array}}}$

according to requirements. The tensor A must be chosen so that the compatibility condition

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

and the specified boundary conditions are observed.

Airy's stress function with gravitational field

With Airy's stress function, gravity can also be taken into account in the form : ${\displaystyle {\vec {b}}=-\operatorname {grad} (V)}$

${\displaystyle {\boldsymbol {\sigma }}={\begin{pmatrix}\varphi _{,yy}+V&-\varphi _{,xy}&0\\&\varphi _{,xx}+V&0\\{\textsf {sym}}&&0\end{pmatrix}}}$

This fits into the Beltrami-Schäfer solution with and and a function g to be determined: ${\displaystyle {\vec {h}}=\operatorname {grad} (g)}$ ${\displaystyle V=\operatorname {div} ({\vec {h}})=\Delta g}$

${\displaystyle -{\vec {b}}=\operatorname {grad} (V)=\operatorname {grad} (\Delta g)=\Delta \operatorname {grad} (g)=\Delta {\vec {h}}\,.}$

The compatibility condition can be found here

${\displaystyle \Delta \Delta \varphi =-\kappa \Delta V=-\kappa \Delta \Delta g}$

write in what the material parameter

${\displaystyle \kappa ={\begin{cases}1-\nu &{\textsf {im\ ebenen\ Spannungszustand}}\\{\frac {1-2\nu }{1-\nu }}&{\textsf {im\ ebenen\ Verzerrungszustand}}\end{cases}}}$

reads.

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 these specifications, the stress tensor reads: ${\displaystyle -m}$ ${\displaystyle E}$

${\displaystyle {\begin{array}{rcl}{\boldsymbol {\sigma }}&=&{\begin{pmatrix}-mEz\\&0\\&&0\end{pmatrix}}={\mathfrak {R}}\left({\begin{pmatrix}0\\&-{\frac {mE}{6}}z^{3}\\&&0\end{pmatrix}}\right)\end{array}}\,.}$

The stress function results accordingly to

${\displaystyle \mathbf {A} ={\begin{pmatrix}0\\&-{\frac {mE}{6}}z^{3}\\&&0\end{pmatrix}}\,.}$

The compatibility condition

${\displaystyle \sigma _{ij,kk}+{\frac {1}{1+\nu }}\sigma _{kk,ij}=0\;,\quad i,j=1,2,3}$

is fulfilled because all second derivatives of the normal stress vanish in the x-direction. So there is a displacement field that is compatible with these stresses. With the boundary conditions sketched in the picture, these are shifts

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

In addition to the compatibility condition page, for example , it is shown here that this displacement field is in equilibrium.

See also

• Poisson's equation
• Integrability condition
• Tensoranalysis formula collection
• Formula collection tensor algebra

Footnotes

1. ↑ ME Gurtin (1972), p. 40
2. ↑ ME Gurtin (1972), p. 92
3. ↑ ^{a b} ME Gurtin (1972), p. 54
4. ↑ ^{a b} ME Gurtin (1972), p. 58
5. ↑ Here the rotation of a tensor is defined as. Occasionally it is used in the literature . Then the differential operator reads:
   ${\displaystyle \operatorname {rot} (\mathbf {T} )={\hat {e}}_{k}\times \mathbf {T} _{,k}}$
   
   ${\displaystyle {\tilde {\operatorname {rot} }}(\mathbf {T} )={\hat {e}}_{k}\times \mathbf {T} _{,k}^{\top }}$
   
   ${\displaystyle {\mathfrak {R}}(\mathbf {A} )={\tilde {\operatorname {rot} }}({\tilde {\operatorname {rot} }}(\mathbf {A} ))}$
6. ↑ ME Gurtin (1972), p. 55
7. ^ R. Greve (2003), p. 128 ff

literature

• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .
• Ralf Greve: Continuum Mechanics . Springer, 2003, ISBN 3-540-00760-1 .
• 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 .