Cauchy-Euler's laws of motion

The Cauchy-Euler's laws of motion by Augustin-Louis Cauchy and Leonhard Euler are the local forms of the momentum and angular momentum balance in continuum mechanics . They are equations of motion which, if they are fulfilled in every point of a body , ensure that the movement of the body as a whole - including deformations - obeys the momentum or angular momentum balance.

The first Cauchy-Euler law of motion corresponds to the momentum balance and, in the geometrically linear case, reads at a material point on the body:

${\displaystyle \rho {\vec {a}}=\rho {\vec {k}}+\operatorname {div} \;{\boldsymbol {\sigma }}\,.}$

Here ρ is the density , the acceleration of the material point, the acceleration due to gravity , the Cauchy's stress tensor and div is the divergence operator . The specific change in momentum is determined by the specific gravity and the drive by the voltages. All of the variables in the equation are generally dependent on both location and time. ${\displaystyle {\vec {a}}}$ ${\displaystyle {\vec {k}}}$ ${\displaystyle {\boldsymbol {\sigma }}}$

The second Cauchy-Euler law of motion corresponds to the local angular momentum balance , which is reduced to the requirement for the symmetry of the Cauchy stress tensor:

${\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{\top }\,.}$

The superscript "┬" marks the transposition . The symmetry means that the shear stresses in a plane are just as great as in planes perpendicular to it.

In the case of large displacements, both laws of motion can be formulated in a Lagrangian material or Eulerian perspective spatially . The structure of the equations is retained, but there are modifications in the dependencies or in the stress tensor. For a clarification of terms, it is recommended to read the article on continuum mechanics. The operators and calculation rules used are listed in the formulas for tensor algebra and tensor analysis .

The Cauchy-Euler's laws of motion are the basis for the Euler's equations in fluid mechanics , the Navier-Stokes equations and the Navier-Cauchy equations . One of the basic equations of the displacement method in the finite element method is the principle of d'Alembert in the Lagrangian version , which is a statement equivalent to the first Cauchy-Euler's law.

First Cauchy-Euler's law of motion

The first Cauchy-Euler law of motion follows from the second Newton's law formulated by Isaac Newton in 1687 and named after him , which corresponds to the momentum balance , according to which the change in momentum over time is equal to the external forces acting on a body:

${\displaystyle {\frac {\mathrm {d} {\vec {p}}}{\mathrm {d} t}}={\vec {F}}_{v}+{\vec {F}}_{a}\,.}$

The vector represents the pulse represents the time change from volume distributed and superficial forces introduced or obtained. In which continuum mechanics idealizes the body as a point set, the above equation becomes an integral equation in which the specific momentum, the specific acceleration due to gravity and the forces acting on the surface are integrated over the volume or over the surface. In the case of small deformations, the first Cauchy-Euler law of motion can be derived from the volume element. ${\displaystyle {\vec {p}}}$ ${\displaystyle {\vec {F}}_{v}}$ ${\displaystyle {\vec {F}}_{a}}$

Momentum balance on the volume element

Tensions on a cut-free pane element

The two-dimensional case in a plane stress state is easier to illustrate and should therefore be put in front. For this purpose, a flat disk of thickness h is considered, which is loaded by forces acting in the plane, see upper part of the figure. A rectangular piece (yellow) is mentally cut out of this disk, parallel to the edges of which a Cartesian coordinate system is defined in which it has the width d x and height d y . According to the cutting principle, cutting stresses occur at the cut surfaces, which take the place of the cut-away part. In the case of an (infinitesimally) small disk element, the cutting stresses can be assumed to be constant over the area. The cutting stresses occur on the surface with the normal in the x-direction and operate accordingly on the surface with the normal in the y-direction. In the component , the first index relates to the surface normal and the second index to the direction of action. According to the assumption there are no stresses perpendicular to the plane of the pane. On the surfaces whose normals point in a positive coordinate direction , there is a positive cutting edge and the stresses act in a positive direction. On the surfaces whose normals point in the negative coordinate direction, the negative edge of the cut is and the stresses act in the negative direction, see picture. ${\displaystyle {\vec {t}}_{x}=\sigma _{xx}{\hat {e}}_{x}+\sigma _{xy}{\hat {e}}_{y}}$ ${\displaystyle {\vec {t}}_{y}=\sigma _{yx}{\hat {e}}_{x}+\sigma _{yy}{\hat {e}}_{y}}$ ${\displaystyle \sigma _{ij}}$

Newton's second law states that the stresses acting on the disk element - multiplied by their effective area - accelerate the disk element. This is done on the disk element, taking into account the gravitational acceleration in the x and y directions

${\displaystyle {\begin{array}{rcl}\mathrm {d} m\,a_{x}&=&\mathrm {d} m\,k_{x}+\sigma _{xx}(x+\mathrm {d} x,y)h\,\mathrm {d} y+\sigma _{yx}(x,y+\mathrm {d} y)h\,\mathrm {d} x-\sigma _{xx}(x,y)h\,\mathrm {d} y-\sigma _{yx}(x,y)h\,\mathrm {d} x\\\mathrm {d} m\,a_{y}&=&\mathrm {d} m\,k_{y}+\sigma _{xy}(x+\mathrm {d} x,y)h\,\mathrm {d} y+\sigma _{yy}(x,y+\mathrm {d} y)h\,\mathrm {d} x-\sigma _{xy}(x,y)h\,\mathrm {d} y-\sigma _{yy}(x,y)h\,\mathrm {d} x\,.\end{array}}}$

The mass of the disk element results from the density ρ of the material and the volume . Division by this volume provides the limit value and the local momentum balance in the x or y direction: ${\displaystyle \mathrm {d} m=\rho h\mathrm {d} x\mathrm {d} y}$ ${\displaystyle h\mathrm {d} x\mathrm {d} y}$ ${\displaystyle \mathrm {d} x\rightarrow 0}$ ${\displaystyle \mathrm {d} y\rightarrow 0}$

${\displaystyle \left.{\begin{array}{rcl}\rho a_{x}&=&\rho k_{x}+{\frac {\mathrm {d} \sigma _{xx}}{\mathrm {d} x}}+{\frac {\mathrm {d} \sigma _{yx}}{\mathrm {d} y}}\\\rho a_{y}&=&\rho k_{y}+{\frac {\mathrm {d} \sigma _{xy}}{\mathrm {d} x}}+{\frac {\mathrm {d} \sigma _{yy}}{\mathrm {d} y}}\end{array}}\right\}\leftrightarrow \quad \rho a_{i}=\rho k_{i}+\sum _{j=1}^{2}{\frac {\mathrm {d} \sigma _{ji}}{\mathrm {d} x_{j}}}\,,\quad i=1,2\,.}$

if - as usual - the coordinates are numbered consecutively according to the scheme x → 1, y → 2, z → 3. The same differential equations result analogously in three dimensions, only one to three is added:

${\displaystyle \rho a_{i}=\rho k_{i}+\sum _{j=1}^{3}{\frac {\mathrm {d} \sigma _{ji}}{\mathrm {d} x_{j}}}\,,\quad i=1,2,3\,.}$

Multiplying these equations with the basis vector of the standard basis and adding the resulting three equations leads to the vector equation ${\displaystyle {\hat {e}}_{i}}$

${\displaystyle {\begin{array}{cccccccl}\underbrace {\sum _{i=1}^{3}\rho a_{i}{\hat {e}}_{i}} &=&\underbrace {\sum _{i=1}^{3}\rho k_{i}{\hat {e}}_{i}} &+&\underbrace {\sum _{k=1}^{3}{\hat {e}}_{k}{\frac {\mathrm {d} }{\mathrm {d} x_{k}}}} &\cdot &\underbrace {\sum _{i,j=1}^{3}\sigma _{ji}{\hat {e}}_{j}\otimes {\hat {e}}_{i}} \\\rho {\vec {a}}&=&\rho {\vec {k}}&+&\nabla &\cdot &{\boldsymbol {\sigma }}&=\;\rho {\vec {k}}+\operatorname {div} \;{\boldsymbol {\sigma }}\,.\end{array}}}$

The Nabla operator " " provides the divergence div of the Cauchy stress tensor in the scalar product , which is a sum of dyads that are formed with the dyadic product " " of the basis vectors and its components . The vector equation is the coordinate-free version of the local momentum balance that applies in any coordinate system. ${\displaystyle \nabla }$ ${\displaystyle {\boldsymbol {\sigma }}}$ ${\displaystyle \otimes }$ ${\displaystyle \sigma _{ji}}$

The intersectional stress vector

${\displaystyle {\vec {t}}_{x}=\sigma _{xx}{\hat {e}}_{x}+\sigma _{xy}{\hat {e}}_{y}+\sigma _{xz}{\hat {e}}_{z}={\boldsymbol {\sigma }}^{\top }\cdot {\hat {e}}_{x}={\hat {e}}_{x}\cdot {\boldsymbol {\sigma }}}$

in the intersection with the normal vector in the x direction is entered line by line in the Cauchy stress tensor, which also applies accordingly to intersection stress vectors in the y and z directions.

Momentum balance in Lagrangian version

In the Lagrangian version , the global momentum balance is

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{V}\rho _{0}({\vec {X}})\,{\dot {\vec {\chi }}}({\vec {X}},t)\,\mathrm {d} V=\int _{V}\rho _{0}({\vec {X}})\,{\vec {k}}_{0}({\vec {X}},t)\,\mathrm {d} V+\int _{A}{\vec {t}}_{0}({\vec {X}},t)\,\mathrm {d} A\,.}$

uses the physical quantities assigned to the material points (particles), see momentum balance . The particles are identified by their material coordinates in the volume V of the body at a specified time t ₀ in the reference state . The density ρ ₀ assigned to a material point is not a function of time due to the mass balance . The patch point is here and below for the substantial derivative , so for the time derivative while holding the particles, which is also the operator defined: ${\displaystyle {\vec {X}}\in V}$ ${\displaystyle {\tfrac {\mathrm {D} }{\mathrm {D} t}}}$

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}(\cdot ):=\left.{\frac {\mathrm {d} }{\mathrm {d} t}}(\cdot )\right|_{{\vec {X}}\;{\text{fest}}}\,.}$

In the integral above, the substantial derivation should mean that the integration area is treated as material in the time derivation, i.e. moves with the body without new particles being added to the area or being removed.

Because the reference volume V does not depend on time, the time derivative of the integral can be shifted in the integrand:

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{V}\rho _{0}({\vec {X}})\,{\dot {\vec {\chi }}}({\vec {X}},t)\,\mathrm {d} V=\int _{V}\rho _{0}({\vec {X}})\,{\ddot {\vec {\chi }}}({\vec {X}},t)\,\mathrm {d} V\,.}$

The externally engaging, surface distributed forces (voltages) are connected to the Nominalspannungstensor N transformed normal vectors on the surface A of the body: . The surface integral of these surface tensions is converted into a volume integral using the Gaussian integral theorem : ${\displaystyle {\vec {t}}_{0}}$ ${\displaystyle {\vec {N}}}$ ${\displaystyle {\vec {t}}_{0}=\mathbf {N} ^{\top }\cdot {\vec {N}}}$

${\displaystyle \int _{A}{\vec {t}}_{0}\,\mathrm {d} A=\int _{A}\mathbf {N} ^{\top }\cdot {\vec {N}}\,\mathrm {d} A=\int _{V}\operatorname {DIV} (\mathbf {N} )\,\mathrm {d} V\,.}$

The divergence operator DIV is capitalized here because it contains the material derivatives according to the material coordinates and is to be distinguished from the spatial operator div, which carries out the spatial derivatives according to the spatial coordinates and which is required in Euler's version. With the results available, the momentum balance can be expressed as a vanishing volume integral: ${\displaystyle {\vec {X}}}$ ${\displaystyle {\vec {x}}}$

${\displaystyle \int _{V}(\rho _{0}{\ddot {\vec {\chi }}}-\rho _{0}{\vec {k}}_{0}-\operatorname {DIV} \mathbf {N} )\,\mathrm {d} V={\vec {0}}\,.}$

This equation applies to every body and every part of it, so that - assuming the integrand is continuous - to the first Cauchy-Euler law of motion in the Lagrangian version

${\displaystyle \rho _{0}({\vec {X}})\,{\ddot {\vec {\chi }}}({\vec {X}},t)=\rho _{0}({\vec {X}})\,{\vec {k}}_{0}({\vec {X}},t)+\operatorname {DIV} \mathbf {N} ({\vec {X}},t)}$

can be closed. The occurrence of the material coordinates and the nominal stress tensor N instead of the Cauchy stress tensor takes into account the change in shape of the part of the body cut out when looking at the volume element in the case of large deformations. In the case of small displacements, there is no need to distinguish between the material and spatial coordinates, which results in the law of motion specified at the beginning. ${\displaystyle \mathbf {N} \approx {\boldsymbol {\sigma }}}$

Impulse balance in Eulerian version

In the Eulerian version , the global momentum balance is

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}\rho ({\vec {x}},t){\vec {v}}({\vec {x}},t)\,\mathrm {d} v=\int _{v}\rho ({\vec {x}},t){\vec {k}}({\vec {x}},t)\,\mathrm {d} v+\int _{a}{\vec {t}}({\vec {x}},t)\,\mathrm {d} a}$

The spatial points are identified by their spatial coordinates in the current volume v at time t, see momentum balance . In contrast to the Lagrangian version, the integration limits as surfaces of the body are dependent on time, which must be taken into account when calculating the change in momentum. According to Reynolds' transport theorem: ${\displaystyle {\vec {x}}\in v}$

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}\rho ({\vec {x}},t){\vec {v}}({\vec {x}},t)\,\mathrm {d} v=\int _{v}\left[{\frac {\mathrm {D} }{\mathrm {D} t}}(\rho {\vec {v}})+\operatorname {div} ({\vec {v}})\rho {\vec {v}}\right]\,\mathrm {d} v=\int _{v}(\underbrace {{\dot {\rho }}{\vec {v}}+\operatorname {div} ({\vec {v}})\rho {\vec {v}}} _{={\vec {0}}}+\rho {\dot {\vec {v}}})\,\mathrm {d} v=\int _{v}\rho {\dot {\vec {v}}}\,\mathrm {d} v\,.}$

The added point stands for the substantial derivation and the term in brackets does not contribute anything in the Eulerian version due to the local mass balance . As in the Lagrangian version, the surface integral of the external stresses is converted into a volume integral with the Gaussian integral theorem: ${\displaystyle {\dot {\rho }}+\rho \operatorname {div} {\vec {v}}=0}$

${\displaystyle \int _{a}{\vec {t}}\,\mathrm {d} a=\int _{a}{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a=\int _{v}\operatorname {div} ({\boldsymbol {\sigma }})\,\mathrm {d} v\,.}$

With the results available, the momentum balance can be expressed as a vanishing volume integral:

${\displaystyle \int _{v}(\rho {\dot {\vec {v}}}-\rho {\vec {k}}-\operatorname {div} {\boldsymbol {\sigma }})\,\mathrm {d} v={\vec {0}}\,.}$

This equation is valid for every volume, so that - assuming continuity of the integrand - the first Cauchy-Euler law of motion in Euler's version

${\displaystyle \rho ({\vec {x}},t)\,{\dot {\vec {v}}}({\vec {x}},t)=\rho ({\vec {x}},t)\,{\vec {k}}({\vec {x}},t)+\operatorname {div} {\boldsymbol {\sigma }}({\vec {x}},t)}$

can be derived. Here is the substantial time derivative of the velocity with a captured particle that is at the location at time t and has the velocity to be formed: ${\displaystyle {\vec {X}}}$ ${\displaystyle {\vec {x}}}$ ${\displaystyle {\vec {v}}({\vec {x}},t)={\dot {\vec {\chi }}}({\vec {X}},t)}$

${\displaystyle {\dot {\vec {v}}}({\vec {x}},t):={\frac {\mathrm {D} }{\mathrm {D} t}}{\vec {v}}({\vec {\chi }}({\vec {X}},t),t):=\left.{\frac {\mathrm {d} }{\mathrm {d} t}}{\vec {v}}({\vec {\chi }}({\vec {X}},t),t)\right|_{{\vec {X}}\;{\text{fest}}}={\frac {\partial }{\partial {\vec {x}}}}{\vec {v}}({\vec {x}},t)\cdot {\dot {\vec {\chi }}}({\vec {X}},t)+{\frac {\partial }{\partial t}}{\vec {v}}({\vec {x}},t)=\operatorname {grad} ({\vec {v}})\cdot {\vec {v}}+{\frac {\partial {\vec {v}}}{\partial t}}}$

The spatial operator grad calculates the spatial velocity gradient with derivatives according to the spatial coordinates x _1,2,3 . The convective component in the substantial acceleration takes into account the flow of the material through the volume v recorded at the top of the volume element when looking at large displacements. In the case of small displacements, this quadratic convective component can be neglected, so that with ${\displaystyle \operatorname {grad} ({\vec {v}})\cdot {\vec {v}}}$

${\displaystyle {\vec {a}}={\dot {\vec {v}}}={\frac {\partial {\vec {v}}}{\partial t}}}$

the law of motion given at the beginning arises.

Influence of jumps in the momentum balance

A jump point on the surface a _s separates two areas of space v ⁺ and v ^-

The required local continuity of the integrands is violated under real conditions if, for example, density jumps at material boundaries or shock waves occur. Such flat jumps can, however, be taken into account if the surface itself is locally continuously differentiable and thus has a normal vector in each of its points . The surface - hereinafter called the jump point - does not have to be a material surface, so it can move with a different speed than the mass itself. This surface divides the mass into two pieces v ⁺ and v ^- and it is agreed that the normal vector the jump point a _s in the direction of the jump point speed and the volume v ⁺ wise, see figure on the right. ${\displaystyle {\vec {v}}_{s}}$

Then the Reynolds transport theorem with a jump point reads:

${\displaystyle {\begin{aligned}{\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}\rho {\vec {v}}\,\mathrm {d} v=&\int _{v}{\frac {\partial (\rho {\vec {v}})}{\partial t}}\,\mathrm {d} v+\int _{a}\rho {\vec {v}}\;({\vec {v}}\cdot \,\mathrm {d} {\vec {a}})+\int _{a_{s}}[[\rho {\vec {v}}\;(({\vec {v}}_{s}-{\vec {v}})\cdot {\vec {n}})]]\mathrm {d} a_{s}\\=&\int _{v}\rho {\dot {\vec {v}}}\,\mathrm {d} v+\int _{a_{s}}[[\rho {\vec {v}}\;(({\vec {v}}_{s}-{\vec {v}})\cdot {\vec {n}})]]\mathrm {d} a_{s}\end{aligned}}}$

The second term with the jump bracket [...] is added. The integrals of the external forces are calculated separately for the volumes v ⁺ and v ^- :

${\displaystyle {\begin{aligned}\left(\int _{v}\rho {\vec {k}}\;\mathrm {d} v+\int _{a}{\vec {t}}\,\mathrm {d} a\right)^{+}=&\int _{v^{+}}\rho {\vec {k}}\;\mathrm {d} v^{+}+\int _{a^{+}}{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a^{+}-\int _{a_{s}}{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a_{s}\\\left(\int _{v}\rho {\vec {k}}\;\mathrm {d} v+\int _{a}{\vec {t}}\,\mathrm {d} a\right)^{-}=&\int _{v^{-}}\rho {\vec {k}}\;\mathrm {d} v^{-}+\int _{a^{-}}{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a^{-}+\int _{a_{s}}{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a_{s}\,.\end{aligned}}}$

The normal should always be directed outwards and therefore enter the jump point once with a positive and once with a negative sign. The union of the surfaces a ⁺ and a ^- results in the surface a of the entire volume v, to whose surface the inner surface a _s does not belong. The sum of the three equations shows after transformations as they were already given above

${\displaystyle \int _{v}(\rho {\dot {\vec {v}}}-\rho {\vec {k}}-\operatorname {div} {\boldsymbol {\sigma }})\,\mathrm {d} v=\int _{a_{s}}[[\rho {\vec {v}}\;(({\vec {v}}-{\vec {v}}_{s})\cdot {\vec {n}})-{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}]]\mathrm {d} a_{s}\,.}$

Beyond the jump point the right side disappears and the local momentum balance without a jump point follows. At the (areal) jump point dv = 0 and the left side can be neglected, so that if the integrand is continuous with the jump bracket in the area

${\displaystyle [[\rho {\vec {v}}\;(({\vec {v}}_{s}-{\vec {v}})\cdot {\vec {n}})+{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}]]=[[\rho {\vec {v}}\otimes ({\vec {v}}_{s}-{\vec {v}})+{\boldsymbol {\sigma }}^{\top }]]\cdot {\vec {n}}={\vec {0}}}$

can be derived. If the crack is a material surface, such as between two adjoining materials of different densities, then it is and it follows: ${\displaystyle {\vec {v}}_{s}={\vec {v}}}$

${\displaystyle [[{\boldsymbol {\sigma }}^{\top }]]\cdot {\vec {n}}={\vec {0}}\quad \Leftrightarrow \quad {{\boldsymbol {\sigma }}^{+}}^{\top }\cdot {\vec {n}}={\vec {t}}^{+}={{\boldsymbol {\sigma }}^{-}}^{\top }\cdot {\vec {n}}={\vec {t}}^{-}\,.}$

The cutting stresses on both sides of a material crack must be the same.

Second Cauchy-Euler's law of motion

The second Cauchy-Euler law of motion follows from the 1754 by Leonhard Euler established and named after him Euler's angular momentum , by which the temporal variation of the angular momentum is equal to the externally acting torque is: ${\displaystyle {\vec {L}}}$

${\displaystyle {\dot {\vec {L}}}={\vec {M}}_{v}+{\vec {M}}_{a}\,.}$

The vector stands for the torque emanating from volume-distributed forces and the vector for the moment introduced on the surface. ${\displaystyle {\vec {M}}_{v}}$ ${\displaystyle {\vec {M}}_{a}}$

Angular momentum balance on the volume element

Cutting stresses on a cube-shaped part of the body

A loaded body is considered, from which a cube-shaped partial body (yellow in the picture) is cut out, which has the edge length 2L and in the center of gravity a Cartesian coordinate system is placed parallel to the cube edges. According to the cutting principle, cutting stresses arise on the cube surfaces , which take the place of the cut-away part of the body and which, according to Cauchy's fundamental theorem, are the normal vectors to the cutting surface, transformed with Cauchy's stress tensor. In the case of an infinitesimally small cube, the sectional stresses can be assumed to be constant over the surface and integrated into a resultant, which load the cube in the center of the surface for reasons of symmetry. The following applies to the moments acting in the middle of the cube: ${\displaystyle {\vec {t}}}$

• From the center of gravity of the cube, the vector points to the center of the cutting surface on the positive cutting edge with the normal in the + x direction and the cutting stress acts there on the surface 4L². ${\displaystyle L{\hat {e}}_{x}}$ ${\displaystyle {\vec {t}}_{+x}={\hat {e}}_{x}\cdot {\boldsymbol {\sigma }}_{+x}}$
• The moment the average voltage at the positive section of the banks is thus: . The arithmetic symbol "×" forms the cross product . ${\displaystyle {\vec {M}}_{+x}=4L^{2}(L{\hat {e}}_{x})\times ({\hat {e}}_{x}\cdot {\boldsymbol {\sigma }}_{+x})}$
• On the negative cutting bank is the lever arm and the average voltage operates on the same surface 4L²: . ${\displaystyle -L{\hat {e}}_{x}}$ ${\displaystyle {\vec {t}}_{-x}=-{\hat {e}}_{x}\cdot {\boldsymbol {\sigma }}_{-x}}$ ${\displaystyle {\vec {M}}_{-x}=4L^{2}(-L{\hat {e}}_{x})\times (-{\hat {e}}_{x}\cdot {\boldsymbol {\sigma }}_{-x})}$
• The moments of the cutting stress add up . ${\displaystyle {\vec {M}}_{x}=4L^{3}{\hat {e}}_{x}\times [{\hat {e}}_{x}\cdot ({\boldsymbol {\sigma }}_{+x}+{\boldsymbol {\sigma }}_{-x})]}$
• In the other two directions in space results in accordance with and . ${\displaystyle {\vec {M}}_{y}=4L^{3}{\hat {e}}_{y}\times [{\hat {e}}_{y}\cdot ({\boldsymbol {\sigma }}_{+y}+{\boldsymbol {\sigma }}_{-y})]}$ ${\displaystyle {\vec {M}}_{z}=4L^{3}{\hat {e}}_{z}\times [{\hat {e}}_{z}\cdot ({\boldsymbol {\sigma }}_{+z}+{\boldsymbol {\sigma }}_{-z})]}$
• In the infinitesimally small cube, a location-independent density ρ and a location-independent gravity field can be assumed, which therefore does not cause a moment in the center of the cube.

The constant density has the inertia tensor result is proportional to the unit tensor I , and the angular acceleration transformed into the variation of the angular momentum: . According to Euler's equations of gyro theory , this rate is equal to the sum of the moments: ${\displaystyle {\boldsymbol {\Theta }}={\frac {16}{3}}L^{5}\rho \mathbf {I} }$ ${\displaystyle {\dot {\vec {\omega }}}}$ ${\displaystyle {\boldsymbol {\Theta }}\cdot {\dot {\vec {\omega }}}={\dot {\vec {L}}}}$

${\displaystyle {\boldsymbol {\Theta }}\cdot {\dot {\vec {\omega }}}={\frac {16}{3}}L^{5}\rho {\dot {\vec {\omega }}}=4L^{3}\sum _{k=1}^{3}{\hat {e}}_{k}\times [{\hat {e}}_{k}\cdot ({\boldsymbol {\sigma }}_{+k}+{\boldsymbol {\sigma }}_{-k})]\,.}$

Division by the volume L³ leads to L → 0 in the limit value and thus ${\displaystyle {\boldsymbol {\sigma }}_{+k}\approx {\boldsymbol {\sigma }}_{-k}\approx {\boldsymbol {\sigma }}}$

${\displaystyle {\begin{aligned}{\vec {0}}=&\sum _{k=1}^{3}{\hat {e}}_{k}\times ({\hat {e}}_{k}\cdot {\boldsymbol {\sigma }})=\sum _{k=1}^{3}{\hat {e}}_{k}\times \left({\hat {e}}_{k}\cdot \sum _{i,j=1}^{3}\sigma _{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}\right)=\sum _{i=1}^{3}{\hat {e}}_{i}\times \left(\sum _{j=1}^{3}\sigma _{ij}{\hat {e}}_{j}\right)\\=&\sum _{i,j=1}^{3}\sigma _{ij}{\hat {e}}_{i}\times {\hat {e}}_{j}=(\sigma _{12}-\sigma _{21}){\hat {e}}_{3}+(\sigma _{23}-\sigma _{32}){\hat {e}}_{1}+(\sigma _{31}-\sigma _{13}){\hat {e}}_{2}\,.\end{aligned}}}$

The sum can be expressed without coordinates using the scalar cross product " " of tensors: ${\displaystyle \cdot \!\times }$

${\displaystyle \sum _{i,j=1}^{3}\sigma _{ij}{\hat {e}}_{i}\times {\hat {e}}_{j}=\mathbf {I} \cdot \!\!\times \sum _{i,j=1}^{3}\sigma _{ij}{\hat {e}}_{i}\otimes {\hat {e}}_{j}=\mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}\,.}$

In the case of a non-cube-shaped cuboid with different dimensions in the x, y and z directions, the result is that the symmetry of the Cauchy stress tensor follows: ${\displaystyle \mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}={\vec {0}}}$

${\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{\top }\,.}$

This tensor equation is the coordinate-free version of the local angular momentum balance, which is valid in any coordinate system.

Angular momentum balance in Lagrangian version

The law of angular momentum reads in global Lagrangian formulation:

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{V}({\vec {\chi }}({\vec {X}},t)-{\vec {c}})\times \rho _{0}({\vec {X}}){\dot {\vec {\chi }}}({\vec {X}},t)\,\mathrm {d} V=\int _{V}({\vec {\chi }}({\vec {X}},t)-{\vec {c}})\times \rho _{0}({\vec {X}}){\vec {k}}_{0}({\vec {X}},t)\,\mathrm {d} V+\int _{A}({\vec {\chi }}({\vec {X}},t)-{\vec {c}})\times {\vec {t}}_{0}({\vec {X}},t)\,\mathrm {d} A}$

The vector is an arbitrary, temporally fixed position vector, see angular momentum balance . As with the momentum balance, the time derivative of the first integral can be shifted in the integrand: ${\displaystyle {\vec {c}}}$

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{V}({\vec {\chi }}-{\vec {c}})\times \rho _{0}{\dot {\vec {\chi }}}\,\mathrm {d} V=\int _{V}[\underbrace {{\dot {\vec {\chi }}}\times \rho _{0}{\dot {\vec {\chi }}}} _{={\vec {0}}}+({\vec {\chi }}-{\vec {c}})\times \rho _{0}{\ddot {\vec {\chi }}}]\,\mathrm {d} V=\int _{V}({\vec {\chi }}-{\vec {c}})\times \rho _{0}{\ddot {\vec {\chi }}}\,\mathrm {d} V}$

The argument list of the functions has been omitted for the sake of clarity. As before, the surface integral is rewritten into a volume integral using the Gaussian integral theorem:

${\displaystyle {\begin{aligned}\int _{A}({\vec {\chi }}-{\vec {c}})\times {\vec {t}}_{0}\,\mathrm {d} A=&\int _{A}({\vec {\chi }}-{\vec {c}})\times (\mathbf {N} ^{\top }\cdot {\vec {N}})\,\mathrm {d} A=-\int _{A}{\vec {N}}\cdot \mathbf {N} \times ({\vec {\chi }}-{\vec {c}})\,\mathrm {d} A=-\int _{A}[\mathbf {N} \times ({\vec {\chi }}-{\vec {c}})]^{\top }\cdot {\vec {N}}\,\mathrm {d} A\\=&-\int _{V}\operatorname {DIV} [\mathbf {N} \times ({\vec {\chi }}-{\vec {c}})]\,\mathrm {d} V=-\int _{V}[\operatorname {DIV} (\mathbf {N} )\times ({\vec {\chi }}-{\vec {c}})-\operatorname {GRAD} ({\vec {\chi }}-{\vec {c}})\cdot \!\!\times \mathbf {N} ]\,\mathrm {d} V\\=&\int _{V}[({\vec {\chi }}-{\vec {c}})\times \operatorname {DIV} (\mathbf {N} )+\mathbf {F} \cdot \!\!\times \mathbf {N} ]\,\mathrm {d} V\end{aligned}}}$

The product rule and the definition of the deformation gradient were used here. The operator GRAD forms the material gradient with derivatives according to the material coordinates X _1,2,3 , which is why the operator grad is capitalized here in contrast to the spatial gradient. ${\displaystyle \operatorname {DIV} (\mathbf {T} \times {\vec {f}})=\operatorname {DIV} (\mathbf {T} )\times {\vec {f}}-\operatorname {GRAD} ({\vec {f}})\cdot \!\!\times \mathbf {T} }$ ${\displaystyle \mathbf {F} =\operatorname {GRAD} {\vec {\chi }}}$

With the results available, the angular momentum balance can be expressed as a vanishing volume integral:

${\displaystyle \int _{V}[({\vec {\chi }}-{\vec {c}})\times (\underbrace {\rho _{0}{\ddot {\vec {\chi }}}-\rho _{0}{\vec {k}}_{0}-\operatorname {DIV} \mathbf {N} } _{={\vec {0}}})-\mathbf {F} \cdot \!\!\times \mathbf {N} ]\,\mathrm {d} V=-\int _{V}\mathbf {F} \cdot \!\!\times \mathbf {N} \,\mathrm {d} V={\vec {0}}\,.}$

The term in brackets does not contribute anything because of the local momentum balance. The last integral is valid for any part of the body, so that with a continuous integrand and - as with the derivation of the volume element - the symmetry of can be derived. The local angular momentum balance in the Lagrangian version is therefore reduced to the requirement ${\displaystyle \mathbf {F} \cdot \!\!\times \mathbf {N} =\mathbf {I} \cdot \!\!\times (\mathbf {F\cdot N} )={\vec {0}}}$ ${\displaystyle \mathbf {F\cdot N} }$

${\displaystyle \mathbf {F\cdot N} =(\mathbf {F\cdot N} )^{\top }=\mathbf {N^{\top }\cdot F^{\top }} \,.}$

Multiplication of left with and right with gives the same meaning: ${\displaystyle \mathbf {F} ^{-1}}$ ${\displaystyle \mathbf {F} ^{\top -1}}$

${\displaystyle \mathbf {F} ^{-1}\cdot \mathbf {F\cdot N\cdot F} ^{\top -1}=\mathbf {F} ^{-1}\cdot \mathbf {N^{\top }\cdot F^{\top }\cdot F} ^{\top -1}\;\leftrightarrow \;\mathbf {N\cdot F} ^{\top -1}=(\mathbf {N\cdot F} ^{\top -1})^{\top }\,.}$

The tensor is the second Piola-Kirchhoff stress tensor whose symmetry is according to ${\displaystyle {\tilde {\mathbf {T} }}:=\mathbf {N\cdot F} ^{\top -1}}$

${\displaystyle {\tilde {\mathbf {T} }}={\tilde {\mathbf {T} }}^{\top }}$

ensures compliance with the angular momentum balance. For small displacements of the second piola-Kirchhoff and Cauchy stress tensor approximately tune agree: . ${\displaystyle {\tilde {\mathbf {T} }}\approx {\boldsymbol {\sigma }}}$

Angular momentum balance in Eulerian version

In global Euler's formulation, the angular momentum theorem reads:

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}({\vec {x}}-{\vec {c}})\times \rho ({\vec {x}},t){\vec {v}}({\vec {x}},t)\,\mathrm {d} v=\int _{v}({\vec {x}}-{\vec {c}})\times \rho ({\vec {x}},t){\vec {k}}({\vec {x}},t)\,\mathrm {d} v+\int _{a}({\vec {x}}-{\vec {c}})\times {\vec {t}}({\vec {x}},t)\,\mathrm {d} a\,.}$

As in the Lagrangian version, is an arbitrary, temporally fixed position vector, see angular momentum balance . The spatial coordinates represent integration variables which therefore do not depend on time. As with the momentum balance, the first integral is calculated using Reynolds' transport theorem: ${\displaystyle {\vec {c}}}$ ${\displaystyle {\vec {x}}}$

${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}({\vec {x}}-{\vec {c}})\times \rho {\vec {v}}\,\mathrm {d} v&=&\displaystyle \int _{v}\left\{{\frac {\mathrm {D} }{\mathrm {D} t}}[({\vec {x}}-{\vec {c}})\times \rho {\vec {v}}]+\operatorname {div} ({\vec {v}})({\vec {x}}-{\vec {c}})\times \rho {\vec {v}}\right\}\,\mathrm {d} v\\&=&\displaystyle \int _{v}({\vec {x}}-{\vec {c}})\times [\underbrace {{\dot {\rho }}\,{\vec {v}}+\operatorname {div} ({\vec {v}})\rho \,{\vec {v}}} _{={\vec {0}}}+\rho \,{\dot {\vec {v}}}]\,\mathrm {d} v=\int _{v}({\vec {x}}-{\vec {c}})\times \rho \,{\dot {\vec {v}}}\,\mathrm {d} v\,.\end{array}}}$

The term in brackets does not contribute anything due to the mass balance. The surface integral in the angular momentum balance is rewritten into a volume integral analogously to the Lagrangian version with the Gaussian integral theorem:

${\displaystyle {\begin{array}{rcl}\displaystyle \int _{a}({\vec {x}}-{\vec {c}})\times {\vec {t}}\,\mathrm {d} a&=&\displaystyle \int _{a}({\vec {x}}-{\vec {c}})\times ({\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}})\,\mathrm {d} a=-\int _{v}[\operatorname {div} ({\boldsymbol {\sigma }})\times ({\vec {x}}-{\vec {c}})-\operatorname {grad} ({\vec {x}}-{\vec {c}})\cdot \!\!\times {\boldsymbol {\sigma }}]\,\mathrm {d} v\\&=&\displaystyle \int _{v}[({\vec {x}}-{\vec {c}})\times \operatorname {div} ({\boldsymbol {\sigma }})+\mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}]\,\mathrm {d} v\end{array}}}$

Deviating from the Lagrangian version, the Cauchy stress tensor takes the place of the nominal stress tensor and, because of the unit tensor, the place of the deformation gradient. ${\displaystyle \operatorname {grad} {\vec {x}}=\mathbf {I} }$

With the results available, the angular momentum balance can be expressed as a vanishing volume integral:

${\displaystyle \int _{v}[({\vec {x}}-{\vec {c}})\times (\underbrace {\rho {\dot {\vec {v}}}-\rho {\vec {k}}-\operatorname {div} {\boldsymbol {\sigma }}} _{={\vec {0}}})-\mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}]\,\mathrm {d} v=-\int _{v}\mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}\,\mathrm {d} v={\vec {0}}\,.}$

The term in parentheses does not contribute because of the local momentum balance and the last integral is valid for any volume, so that with a continuous integrand it can be concluded. Analogous to the Lagrangian version, the angular momentum balance in the Eulerian version is reduced to the requirement for the symmetry of the Cauchy's stress tensor: ${\displaystyle \mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}={\vec {0}}}$

${\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}^{\top }\,.}$

Influence of jump points in the angular momentum balance

Analogous to Cauchy-Euler's first law of motion, the Reynolds transport theorem with a jump point reads here:

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}({\vec {x}}-{\vec {c}})\times \rho {\vec {v}}\,\mathrm {d} v=\int _{v}({\vec {x}}-{\vec {c}})\times \rho {\dot {\vec {v}}}\,\mathrm {d} v+\int _{a_{s}}[[({\vec {x}}-{\vec {c}})\times \rho {\vec {v}}\;(({\vec {v}}_{s}-{\vec {v}})\cdot {\vec {n}})]]\mathrm {d} a_{s}}$

The second term with the jump bracket [...] is added. The integrals of the external forces are calculated separately for v ⁺ and v ^- :

${\displaystyle {\begin{aligned}\left(\int _{v}({\vec {x}}-{\vec {c}})\times \rho {\vec {k}}\;\mathrm {d} v+\int _{a}({\vec {x}}-{\vec {c}})\times {\vec {t}}\,\mathrm {d} a\right)^{+}=&\int _{v^{+}}({\vec {x}}-{\vec {c}})\times \rho {\vec {k}}\;\mathrm {d} v^{+}+\int _{a^{+}}({\vec {x}}-{\vec {c}})\times {\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a^{+}\\&-\int _{a_{s}}({\vec {x}}-{\vec {c}})\times {\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a_{s}\\\left(\int _{v}({\vec {x}}-{\vec {c}})\times \rho {\vec {k}}\;\mathrm {d} v+\int _{a}({\vec {x}}-{\vec {c}})\times {\vec {t}}\,\mathrm {d} a\right)^{-}=&\int _{v^{-}}({\vec {x}}-{\vec {c}})\times \rho {\vec {k}}\;\mathrm {d} v^{-}+\int _{a^{-}}({\vec {x}}-{\vec {c}})\times {\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a^{-}\\&+\int _{a_{s}}({\vec {x}}-{\vec {c}})\times {\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a_{s}\,.\end{aligned}}}$

The sum of the three equations shows after transformations as they were already given above

${\displaystyle \int _{v}[({\vec {x}}-{\vec {c}})\times (\rho {\dot {\vec {v}}}-\rho {\vec {k}}-\operatorname {div} {\boldsymbol {\sigma }})-\mathbf {I} \cdot \!\!\times {\boldsymbol {\sigma }}]\,\mathrm {d} v=\int _{a_{s}}[[({\vec {x}}-{\vec {c}})\times \rho {\vec {v}}\;(({\vec {v}}-{\vec {v}}_{s})\cdot {\vec {n}})-({\vec {x}}-{\vec {c}})\times {\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}]]\mathrm {d} a_{s}\,.}$

Beyond the jump point the right-hand side disappears and the symmetry of the stress tensor follows as above. At the (areal) jump point dv = 0 and the left side can be neglected, so that if the integrand is continuous with the jump bracket in the area

${\displaystyle ({\vec {x}}-{\vec {c}})\times [[\rho {\vec {v}}\;(({\vec {v}}_{s}-{\vec {v}})\cdot {\vec {n}})+{\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}]]=({\vec {x}}-{\vec {c}})\times [[\rho {\vec {v}}\otimes ({\vec {v}}_{s}-{\vec {v}})+{\boldsymbol {\sigma }}^{\top }]]\cdot {\vec {n}}={\vec {0}}}$

can be derived, which is fulfilled identically in the first Cauchy-Euler's law of motion due to the jump condition.

Conclusions from the laws of motion

Further, material-independent equations equivalent to principles can be deduced from the laws of motion. The first Cauchy-Euler law of motion reads:

${\displaystyle {\begin{array}{llcl}{\text{materiell:}}&\rho _{0}\,{\ddot {\vec {\chi }}}&=&\rho _{0}\,{\vec {k}}_{0}+\operatorname {DIV} \mathbf {N} \\{\text{räumlich:}}&\rho \,{\dot {\vec {v}}}&=&\rho \,{\vec {k}}+\operatorname {div} {\boldsymbol {\sigma }}\,.\end{array}}}$

These equations are scalar multiplied by a vector field and integrated over the volume of the body. It arises: ${\displaystyle {\vec {q}}}$

${\displaystyle {\begin{array}{llcl}{\text{materiell:}}&\int _{V}\rho _{0}\,{\ddot {\vec {\chi }}}\cdot {\vec {q}}\,\mathrm {d} V+\int _{V}{\tilde {\mathbf {T} }}:\operatorname {sym} (\mathbf {F} ^{\top }\cdot \operatorname {GRAD} {\vec {q}})\,\mathrm {d} V&=&\int _{V}\rho _{0}\,{\vec {k}}_{0}\cdot {\vec {q}}\,\mathrm {d} V+\int _{A}{\vec {t}}_{0}\cdot {\vec {q}}\,\mathrm {d} A\\{\text{räumlich:}}&\int _{v}\rho \,{\dot {\vec {v}}}\cdot {\vec {q}}\,\mathrm {d} v+\int _{v}{\boldsymbol {\sigma }}:\operatorname {sym(grad} ({\vec {q}}))\,\mathrm {d} v&=&\int _{v}\rho \,{\vec {k}}\cdot {\vec {q}}\,\mathrm {d} v+\int _{a}{\vec {t}}\cdot {\vec {q}}\,\mathrm {d} a\,.\end{array}}}$

Different statements result depending on the vector field . ${\displaystyle {\vec {q}}}$

proof

Scalar multiplication of Cauchy-Euler's first law of motion with the vector field and integration over the volume of the body yields: ${\displaystyle {\vec {q}}}$ ${\displaystyle {\begin{array}{llcl}{\text{materiell:}}&\int _{V}\rho _{0}\,{\ddot {\vec {\chi }}}\cdot {\vec {q}}\,\mathrm {d} V&=&\int _{V}\rho _{0}\,{\vec {k}}_{0}\cdot {\vec {q}}\,\mathrm {d} V+\int _{V}\operatorname {DIV} (\mathbf {N} )\cdot {\vec {q}}\,\mathrm {d} V\\{\text{räumlich:}}&\int _{v}\rho \,{\dot {\vec {v}}}\cdot {\vec {q}}\,\mathrm {d} v&=&\int _{v}\rho \,{\vec {k}}\cdot {\vec {q}}\,\mathrm {d} v+\int _{v}\operatorname {div} ({\boldsymbol {\sigma }})\cdot {\vec {q}}\,\mathrm {d} v\,.\end{array}}}$ The last term on the right is transformed with the product rule: ${\displaystyle {\begin{array}{rlcl}{\text{materiell:}}&\operatorname {DIV} (\mathbf {N} \cdot {\vec {q}})&=&\operatorname {DIV} (\mathbf {N} )\cdot {\vec {q}}+\mathbf {N} ^{\top }:\operatorname {GRAD} {\vec {q}}\\\rightarrow &\operatorname {DIV} (\mathbf {N} )\cdot {\vec {q}}&=&\operatorname {DIV} (\mathbf {N} \cdot {\vec {q}})-\mathbf {N} ^{\top }:\operatorname {GRAD} {\vec {q}}\\{\text{räumlich:}}&\operatorname {div} ({\boldsymbol {\sigma }}\cdot {\vec {q}})&=&\operatorname {div} ({\boldsymbol {\sigma }})\cdot {\vec {q}}+{\boldsymbol {\sigma }}^{\top }:\operatorname {grad} {\vec {q}}\\\rightarrow &\operatorname {div} ({\boldsymbol {\sigma }})\cdot {\vec {q}}&=&\operatorname {div} ({\boldsymbol {\sigma }}\cdot {\vec {q}})-{\boldsymbol {\sigma }}:\operatorname {grad} {\vec {q}}\,.\end{array}}}$ The deformation gradient is also multiplied in the material form: ${\displaystyle \mathbf {N} ^{\top }:\operatorname {GRAD} {\vec {q}}=\operatorname {Sp} (\mathbf {N} \cdot \operatorname {GRAD} {\vec {q}})=\operatorname {Sp} (\underbrace {\mathbf {N} \cdot \mathbf {F} ^{\top -1}} _{={\tilde {\mathbf {T} }}}\cdot \mathbf {F} ^{\top }\cdot \operatorname {GRAD} {\vec {q}})={\tilde {\mathbf {T} }}:\operatorname {sym} (\mathbf {F} ^{\top }\cdot \operatorname {GRAD} {\vec {q}})\,.}$ In the last step, the fact that in the scalar product with a symmetric tensor only the symmetric components sym () contribute something, which is also used in the spatial formulation: The volume integral of the divergence term is converted into a surface integral with the Gaussian integral theorem: ${\displaystyle {\tilde {\mathbf {T} }}}$ ${\displaystyle {\boldsymbol {\sigma }}:\operatorname {grad} {\vec {q}}={\boldsymbol {\sigma }}:\operatorname {sym(grad} ({\vec {q}}))\,.}$ ${\displaystyle {\begin{aligned}{\text{materiell:}}&\int _{V}\operatorname {DIV} (\mathbf {N} \cdot {\vec {q}})\,\mathrm {d} V=\int _{A}(\mathbf {N} \cdot {\vec {q}})\cdot {\vec {N}}\,\mathrm {d} A=\int _{A}{\vec {q}}\cdot \mathbf {N} ^{\top }\cdot {\vec {N}}\,\mathrm {d} A=\int _{A}{\vec {q}}\cdot {\vec {t}}_{0}\,\mathrm {d} A\\{\text{räumlich:}}&\int _{v}\operatorname {div} ({\boldsymbol {\sigma }}\cdot {\vec {q}})\,\mathrm {d} v=\int _{a}({\boldsymbol {\sigma }}\cdot {\vec {q}})\cdot {\vec {n}}\,\mathrm {d} a=\int _{a}{\vec {q}}\cdot {\boldsymbol {\sigma }}^{\top }\cdot {\vec {n}}\,\mathrm {d} a=\int _{a}{\vec {q}}\cdot {\vec {t}}\,\mathrm {d} a\,.\end{aligned}}}$ Combining these results results in the given equations.

Principle of d'Alembert

→ Main article : d'Alembert's principle

The principle of d'Alembert is of fundamental importance for the solution of initial boundary value problems in continuum mechanics, especially the displacement method in the finite element method . For the vector field are virtual shifts used, the bias field independent imaginary largely arbitrary, differential displacements and which are compatible with the geometric bonds of the body. The virtual displacements must disappear wherever displacement boundary conditions of the body are given. Let be the part of the surface of the body on which displacement boundary conditions are explained. For a vector field of the virtual displacements is then ${\displaystyle {\vec {q}}}$ ${\displaystyle \delta {\vec {u}}}$ ${\displaystyle {\vec {u}}}$ ${\displaystyle {A}^{u}}$ ${\displaystyle A}$ ${\displaystyle \delta {\vec {u}}_{0}}$

${\displaystyle {\begin{aligned}{\text{materiell}}\quad &\delta {\vec {u}}_{0}({\vec {X}})={\vec {0}}\quad {\text{für alle}}\;{\vec {X}}{\in }{A}^{u}\\{\text{bzw. räumlich}}\quad &\delta {\vec {u}}({\vec {x}})={\vec {0}}\quad {\text{für alle}}\;{\vec {x}}{\in }{a}^{u}\end{aligned}}}$

to promote. In no surface tension can then be specified. That is why it describes the part of the surface of the body on which surface tensions (can) act, which is also defined in the spatial formulation. This is how: ${\displaystyle {A}^{u}}$ ${\displaystyle {A}^{\sigma }=A\setminus A^{u}}$

${\displaystyle {\begin{array}{llcll}{\text{materiell:}}&\displaystyle \int _{V}\rho _{0}\,{\ddot {\vec {\chi }}}\cdot \delta {\vec {u}}_{0}\,\mathrm {d} V+\int _{V}{\tilde {\mathbf {T} }}:\delta \mathbf {E} \,\mathrm {d} V&=&\displaystyle \int _{V}\rho _{0}\,{\vec {k}}_{0}\cdot \delta {\vec {u}}_{0}\,\mathrm {d} V+\int _{A^{\sigma }}{\vec {t}}_{0}\cdot \delta {\vec {u}}_{0}\,\mathrm {d} A&{\text{für alle}}\;\delta {\vec {u}}_{0}\in {\mathcal {V}}_{0}\\{\text{räumlich:}}&\displaystyle \int _{v}\rho \,{\dot {\vec {v}}}\cdot \delta {\vec {u}}\,\mathrm {d} v+\int _{v}{\boldsymbol {\sigma }}:\delta {\boldsymbol {\varepsilon }}\,\mathrm {d} v&=&\displaystyle \int _{v}\rho \,{\vec {k}}\cdot \delta {\vec {u}}\,\mathrm {d} v+\int _{a^{\sigma }}{\vec {t}}\cdot \delta {\vec {u}}\,\mathrm {d} a&{\text{für alle}}\;\delta {\vec {u}}\in {\mathcal {V}}\,.\end{array}}}$

The set contains all permissible, material or spatial, virtual displacement fields. On the left side is the virtual work of the inertial forces and the virtual deformation work and on the right side the virtual work of the external forces (volume and surface distribution.) ${\displaystyle {\mathcal {V}}_{(0)}}$

In the physical version, the virtual distortions stand for the variation of the Green-Lagrangian strain tensor: ${\displaystyle \delta \mathbf {E} }$

${\displaystyle \delta \mathbf {E} :={\frac {1}{2}}(\mathbf {F} ^{\top }\cdot \delta \mathbf {F} +\delta \mathbf {F} ^{\top }\cdot \mathbf {F} )=\operatorname {sym} (\mathbf {F} ^{\top }\cdot \delta \mathbf {F} )\,.}$

This is where the virtual deformation gradient is. In the spatial version, the virtual strain tensor is formed from the virtual displacement gradient : ${\displaystyle \delta \mathbf {F} :=\operatorname {GRAD} \delta {\vec {u}}}$ ${\displaystyle \delta {\boldsymbol {\varepsilon }}}$ ${\displaystyle \delta \mathbf {H} :=\operatorname {grad} \delta {\vec {u}}}$

${\displaystyle \delta {\boldsymbol {\varepsilon }}:=\operatorname {sym\,grad} \delta {\vec {u}}={\frac {1}{2}}(\delta \mathbf {H} +\delta \mathbf {H} ^{\top })\,.}$

Mechanical energy balance

If the velocity field is used, the balance of mechanical energy follows: ${\displaystyle {\vec {q}}}$

${\displaystyle {\begin{array}{llcl}{\text{materiell:}}&{\frac {\mathrm {D} }{\mathrm {D} t}}\int _{V}{\frac {\rho _{0}}{2}}\,{\dot {\vec {\chi }}}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} V&=&\int _{V}\rho _{0}\,{\vec {k}}_{0}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} V+\int _{A}{\vec {t}}_{0}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} A-\int _{V}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}\,\mathrm {d} V\\{\text{räumlich:}}&{\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}{\frac {\rho }{2}}{\vec {v}}\cdot {\vec {v}}\,\mathrm {d} v&=&\int _{v}\rho \,{\vec {k}}\cdot {\vec {v}}\,\mathrm {d} v+\int _{a}{\vec {t}}\cdot {\vec {v}}\,\mathrm {d} a-\int _{v}{\boldsymbol {\sigma }}:\mathbf {d} \,\mathrm {d} v\,.\end{array}}}$

On the left side is the change in kinetic energy over time and on the right side is the power of the external forces (volume and area distributed) minus the deformation power. This set is also called the working set .

proof

The time derivative of the kinetic energy is equal to the power of the inertial forces ${\displaystyle {\begin{array}{llcl}{\text{materiell:}}&{\frac {\mathrm {D} }{\mathrm {D} t}}\int _{V}{\frac {\rho _{0}}{2}}\,{\dot {\vec {\chi }}}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} V&=&\int _{V}\rho _{0}\,{\ddot {\vec {\chi }}}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} V\\{\text{räumlich:}}&{\frac {\mathrm {D} }{\mathrm {D} t}}\int _{v}{\frac {\rho }{2}}{\vec {v}}\cdot {\vec {v}}\,\mathrm {d} v&=&\int _{v}\left[{\frac {\mathrm {D} }{\mathrm {D} t}}\left({\frac {\rho }{2}}{\vec {v}}\cdot {\vec {v}}\right)+\operatorname {div} ({\vec {v}}){\frac {\rho }{2}}{\vec {v}}\cdot {\vec {v}}\right]=\int _{v}\left[{\frac {\dot {\rho }}{2}}{\vec {v}}\cdot {\vec {v}}+\rho {\vec {v}}\cdot {\dot {\vec {v}}}+\operatorname {div} ({\vec {v}}){\frac {\rho }{2}}{\vec {v}}\cdot {\vec {v}}\right]\\&&=&\int _{v}{\Bigl [}(\underbrace {{\dot {\rho }}+\rho \operatorname {div} ({\vec {v}})} _{=0}){\frac {1}{2}}{\vec {v}}\cdot {\vec {v}}+\rho {\dot {\vec {v}}}\cdot {\vec {v}}{\Bigr ]}\,\mathrm {d} v=\int _{v}\rho {\dot {\vec {v}}}\cdot {\vec {v}}\,\mathrm {d} v\,,\end{array}}}$ which establishes the first terms on the left. Reynolds' transport theorem and the mass balance were used in the spatial formulation. The material gradient of the velocity is the time derivative of the deformation gradient and the symmetrical part of the spatial velocity gradient is the distortion velocity tensor d . The deformation performance is thus written: ${\displaystyle \operatorname {GRAD} {\vec {v}}={\dot {\mathbf {F} }}}$ ${\displaystyle \mathbf {l} :=\operatorname {grad} {\vec {v}}}$ ${\displaystyle {\begin{aligned}{\text{materiell}}:&\int _{V}{\tilde {\mathbf {T} }}:\operatorname {sym} (\mathbf {F} ^{\top }\cdot {\dot {\mathbf {F} }})=:\int _{V}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}\,\mathrm {d} V\\{\text{räumlich}}:&\int _{v}{\boldsymbol {\sigma }}:\operatorname {sym} (\mathbf {l} )\,\mathrm {d} v=\int _{v}{\boldsymbol {\sigma }}:\mathbf {d} \,\mathrm {d} v\end{aligned}}}$ The performance of the external forces results from the replacement of the vector by the velocity vector . ${\displaystyle {\vec {q}}}$

Conservation of energy law

In a conservative system there is a scalar-valued function W _a , the potential energy , its negative time derivative according to

${\displaystyle -{\frac {\mathrm {D} }{\mathrm {D} t}}W_{a}:=\int _{V}\rho _{0}\,{\vec {k}}_{0}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} V+\int _{A}{\vec {t}}_{0}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} A=\int _{v}\rho \,{\vec {k}}\cdot {\vec {v}}\,\mathrm {d} v+\int _{a}{\vec {t}}\cdot {\vec {v}}\,\mathrm {d} a}$

is the power of the external forces, and a deformation energy W _i , its time derivative

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}W_{i}:=\int _{V}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}\,\mathrm {d} V=\int _{v}{\boldsymbol {\sigma }}:\mathbf {d} \,\mathrm {d} v}$

is the deformation power. With the abbreviation

${\displaystyle K:=\int _{V}{\frac {\rho _{0}}{2}}\,{\dot {\vec {\chi }}}\cdot {\dot {\vec {\chi }}}\,\mathrm {d} V=\int _{v}{\frac {\rho }{2}}{\vec {v}}\cdot {\vec {v}}\,\mathrm {d} v}$

for the kinetic energy the balance of the mechanical energy is written:

${\displaystyle {\frac {\mathrm {D} }{\mathrm {D} t}}K+{\frac {\mathrm {D} }{\mathrm {D} t}}W_{i}+{\frac {\mathrm {D} }{\mathrm {D} t}}W_{a}={\frac {\mathrm {D} }{\mathrm {D} t}}(K+W_{i}+W_{a})=0\quad \Leftrightarrow K+W_{i}+W_{a}=E\,.}$

The total mechanical energy E, consisting of the kinetic energy, the deformation energy and the potential energy, is therefore constant over time in a conservative system, which is known as the law of conservation of energy .

Clapeyron's Theorem

If the displacement field is used for small deformations, linear elasticity and in the static case for the vector field , then the displacement gradient is and so that all terms that are higher than one can be neglected. It follows: ${\displaystyle {\vec {q}}}$ ${\displaystyle {\vec {u}}({\vec {X}},t):={\vec {\chi }}({\vec {X}},t)-{\vec {X}}}$ ${\displaystyle \operatorname {grad} {\vec {q}}=\operatorname {grad} {\vec {u}}=\mathbf {H} }$ ${\displaystyle \|\mathbf {H} \|\ll 1\,,}$ ${\displaystyle \|\mathbf {H} \|}$

${\displaystyle {\begin{aligned}\operatorname {GRAD} {\vec {u}}=&\mathbf {H} =\operatorname {GRAD} {\vec {\chi }}-\mathbf {I} =\mathbf {F-I} \\\rightarrow \quad \operatorname {sym} (\mathbf {F} ^{\top }\cdot \operatorname {GRAD} {\vec {u}})=&\operatorname {sym} (\mathbf {(I+H^{\top })\cdot H} )\approx \operatorname {sym} (\mathbf {H} )={\frac {1}{2}}(\mathbf {H+H^{\top }} )={\boldsymbol {\varepsilon }}\,.\end{aligned}}}$

The symmetrical part of the displacement gradient is the linearized strain tensor . The second Piola-Kirchhoff stress tensor changes into the Cauchy stress tensor in the case of small deformations, and the result is the working set

${\displaystyle \int _{v}{\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}\,\mathrm {d} v=\int _{v}\rho \,{\vec {k}}\cdot {\vec {u}}\,\mathrm {d} v+\int _{a}{\vec {t}}\cdot {\vec {u}}\,\mathrm {d} a\,.}$

The integrand on the left is twice the strain energy and the result is Clapeyron's theorem ${\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\varepsilon }}=2w}$

${\displaystyle 2\int _{v}w\,\mathrm {d} v=\int _{v}\rho \,{\vec {k}}\cdot {\vec {u}}\,\mathrm {d} v+\int _{a}{\vec {t}}\cdot {\vec {u}}\,\mathrm {d} a\,.}$

Footnotes

1. ↑ In the literature, the first Piola-Kirchhoff stress tensor and a different definition of the divergence of a tensor are used: Then is . ${\displaystyle \mathbf {P} =\mathbf {N} ^{\top }}$
   ${\displaystyle \int _{A}{\vec {t}}_{0}\,\mathrm {d} A=\int _{A}\mathbf {P} \cdot {\vec {N}}\,\mathrm {d} A=\int _{V}\operatorname {\tilde {DIV}} (\mathbf {P} )\,\mathrm {d} V\,.}$
   ${\displaystyle \operatorname {\tilde {DIV}} \mathbf {P} =\operatorname {DIV} (\mathbf {P} ^{\top })}$
2. ↑ The Fréchet derivative of a function according to is the limited linear operator which - if it exists - corresponds to the Gâteaux differential in all directions , so it holds. In it is scalar, vector or tensor valued but and similar. Then it is also written. ${\displaystyle f}$ ${\displaystyle x}$ ${\displaystyle {\mathcal {A}}}$ ${\displaystyle h}$
   ${\displaystyle {\mathcal {A}}(h)=\left.{\frac {\mathrm {d} }{\mathrm {d} s}}f(x+sh)\right|_{s=0}=\lim _{s\rightarrow 0}{\frac {f(x+sh)-f(x)}{s}}\quad {\text{für alle}}\;h}$
   ${\displaystyle s\in \mathbb {R} \,,f,x\,{\textsf {und}}\,h}$ ${\displaystyle x}$ ${\displaystyle h}$
   ${\displaystyle {\mathcal {A}}={\frac {\partial f}{\partial x}}}$
   
3. ↑ The scalar cross product of tensors is defined with vectors and the dyadic product " " via The scalar cross product of the unit tensor with a dyad exchanges the dyadic product for the cross product: ${\displaystyle {\vec {a}},\,{\vec {b}},\,{\vec {c}},\,{\vec {d}}}$ ${\displaystyle \otimes }$
   ${\displaystyle ({\vec {a}}\otimes {\vec {b}})\cdot \!\!\times ({\vec {c}}\otimes {\vec {d}}):=({\vec {b}}\cdot {\vec {c}}){\vec {a}}\times {\vec {d}}\,.}$ ${\displaystyle \mathbf {I} \cdot \!\!\times ({\vec {c}}\otimes {\vec {d}})=\sum _{i=1}^{3}({\hat {e}}_{i}\otimes {\hat {e}}_{i})\cdot \!\!\times ({\vec {c}}\otimes {\vec {d}})=\sum _{i=1}^{3}({\hat {e}}_{i}\cdot {\vec {c}}){\hat {e}}_{i}\times {\vec {d}}={\vec {c}}\times {\vec {d}}}$

Individual evidence

1. ^ WH Müller: Forays through the continuum theory . Springer, 2011, ISBN 978-3-642-19869-4 , pp. 72 .
2. ^ 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 , pp. 60 .
3. ^ Martin H. Sadd: Elasticity - Theory, applications and numerics . Elsevier Butterworth-Heinemann, 2005, ISBN 0-12-605811-3 , pp. 110 .

literature

• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .
• 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 .