Geometric linearization

In geometric linearization , the kinematic equations of continuum mechanics are linearized with respect to the displacements . Displacements are the distances covered by its particles when a body moves. Expansions occur when neighboring particles have very different displacements, which is why the geometric linearization includes a linearization with regard to the expansions. Due to the geometric linearization, the equations of continuum mechanics for solid bodies experience a considerable simplification, which is permissible if the displacements are small compared to a characteristic dimension of the body and the strains are small compared to one. Then we speak of small displacements or deformations in contrast to large or finite displacements or deformations. In many applications in the technical area, small shifts are assumed or have to be kept small for safety reasons.

From the geometric linearization to distinguish the physical linearization, the material models or other physical non-linearities such as body contact is concerned. In physically linear systems, the equations of continuum mechanics are linear functions of the displacements after geometric linearization . In this case, the displacements cannot have an effect on the rigidity of a body, as is the case with buckling and buckling . Rotations of more than one degree or expansions of more than 3–8% are not correctly mapped geometrically linear. Therefore, the geometric linearization may only be carried out if the aforementioned effects are insignificant.

The geometric linearization is used, because this allows the equations of continuum mechanics in the Lagrangian description z. B. in the displacement method , simplify considerably. The strength of materials used, the geometric linearization into disrepair. In physically linear systems, geometric linearization enables the application of the Airy stress function or modal analysis .

definition

Displacement and its gradient

→ Main article : Displacement gradient

The displacement is the difference vector between the current position of a particle and its starting position:

${\displaystyle {\vec {u}}={\vec {\chi }}({\vec {X}},t)-{\vec {X}}}$.

In it is

• ${\displaystyle {\vec {X}}=\sum _{i=1}^{3}X_{i}{\vec {e}}_{i}=X{\vec {e}}_{1}+Y{\vec {e}}_{2}+Z{\vec {e}}_{3}}$the position of a particle with material coordinates X _1,2,3 or X , Y and Z with respect to the standard base at a certain time in the undeformed starting position of the body, ${\displaystyle {\vec {e}}_{1,2,3}}$ ${\displaystyle t_{0}}$
• ${\displaystyle {\vec {\chi }}({\vec {X}},t)}$the motion function , which indicates the current position of the particle at the time in space, and ${\displaystyle t\geq t_{0}}$
• ${\displaystyle {\vec {u}}({\vec {X}},t)=\sum _{i=1}^{3}u_{i}{\vec {e}}_{i}=u{\vec {e}}_{1}+v{\vec {e}}_{2}+w{\vec {e}}_{3}}$the displacement with components u _1,2,3 or u , v and w .

The displacement gradient is formed with the dyadic product of the derivatives of the displacements according to the material coordinates: ${\displaystyle \otimes }$

${\displaystyle \mathbf {H} ={\frac {\mathrm {d} u_{i}}{\mathrm {d} X_{j}}}{\vec {e}}_{i}\otimes {\vec {e}}_{j}={\begin{pmatrix}{\frac {\mathrm {d} u}{\mathrm {d} X}}&{\frac {\mathrm {d} u}{\mathrm {d} Y}}&{\frac {\mathrm {d} u}{\mathrm {d} Z}}\\{\frac {\mathrm {d} v}{\mathrm {d} X}}&{\frac {\mathrm {d} v}{\mathrm {d} Y}}&{\frac {\mathrm {d} v}{\mathrm {d} Z}}\\{\frac {\mathrm {d} w}{\mathrm {d} X}}&{\frac {\mathrm {d} w}{\mathrm {d} Y}}&{\frac {\mathrm {d} w}{\mathrm {d} Z}}\end{pmatrix}}}$.

Geometric linearization

The geometric linearization refers to the kinematic equations which in continuum mechanics primarily define the deformation gradient and the strain tensors as functions of the displacement gradient . Let be a scalar or tensor valued function of the displacement gradient , e.g. B. the deformation gradient or one of its main invariants . The geometrically linearized function is then obtained by neglecting all terms that contain the Frobenius norm of the displacement gradient in a higher order than one. Sei and the Landau symbol for terms that are at least square in and can be neglected. Then it is expressed mathematically: ${\displaystyle f(\mathbf {H} )}$ ${\displaystyle \mathbf {H} }$ ${\displaystyle f_{t}(\mathbf {H} )}$ ${\displaystyle \parallel \mathbf {H} \parallel }$ ${\displaystyle \delta :=\parallel \mathbf {H} \parallel }$ ${\displaystyle {\mathcal {O}}(\delta ^{2})}$ ${\displaystyle \delta }$

${\displaystyle f(\mathbf {H} )=f_{t}(\mathbf {H} )+{\mathcal {O}}(\delta ^{2})}$.

With the formalism of linearization , the geometrically linearized function with the Gâteaux differential

${\displaystyle \mathrm {D} f(\mathbf {H} )[\mathbf {X} ]:=\left.{\frac {\mathrm {d} }{\mathrm {d} s}}f(\mathbf {H} +s\mathbf {X} )\right|_{s=0}=\lim _{s\rightarrow 0}{\frac {f(\mathbf {H} +s\mathbf {X} )-f(\mathbf {H} )}{s}}}$

can be defined as follows:

${\displaystyle f_{t}(\mathbf {H} ):=f(\mathbf {0} )+\mathrm {D} f(\mathbf {0} )[\mathbf {H} ]=f(\mathbf {0} )+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}f(s\mathbf {H} )\right|_{s=0}=f(\mathbf {0} )+\lim _{s\rightarrow 0}{\frac {f(s\mathbf {H} )-f(\mathbf {0} )}{s}}}$

The function approximates the function f linearly, namely at the point . In principle, however, linearization can also be carried out at any other point. ${\displaystyle f_{t}}$ ${\displaystyle \mathbf {H} =\mathbf {0} }$

A direct dependence on the displacements can occur in the boundary conditions of a continuum mechanical boundary value problem . Let be a displacement- dependent boundary condition , e.g. B. a predetermined displacement or rotation in the storage of the body. Then this is analogous to the function f using ${\displaystyle g({\vec {u}})}$

${\displaystyle g_{t}({\vec {u}}):=g({\vec {0}})+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}g(s{\vec {u}})\right|_{s=0}=g({\vec {0}})+\lim _{s\rightarrow 0}{\frac {g(s{\vec {u}})-g({\vec {0}})}{s}}.}$

geometrically linearized at the point . ${\displaystyle {\vec {u}}={\vec {0}}}$

Geometric linearization after pre-deformation

A geometric linearization away from the origin occurs in practice when a body that has already been subjected to finite (not small) deformations and / or rotations, e.g. B. after a forming process, a modal analysis should be carried out. This is a linear procedure that makes a linearization of the model equations absolutely necessary and therefore implies a geometric linearization. The linearized terms are then calculated according to

${\displaystyle f_{t}(\mathbf {H} )=f(\mathbf {H} _{0})+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}f(\mathbf {H} _{0}+s\Delta \mathbf {H} )\right|_{s=0}=f(\mathbf {H} _{0})+\lim _{s\rightarrow 0}{\frac {f(\mathbf {H} _{0}+s\Delta \mathbf {H} )-f(\mathbf {H} _{0})}{s}}.}$

This contains the operating point at which the linearization is carried out and a small ( ) deviation from the operating point. The linearization of the boundary conditions takes place in an analogous way at the operating point : ${\displaystyle \mathbf {H} _{0}}$ ${\displaystyle \Delta \mathbf {H} =\mathbf {H-H} _{0}}$ ${\displaystyle \parallel \Delta \mathbf {H} \parallel \ll 1}$ ${\displaystyle {\vec {u}}_{0}}$

${\displaystyle g_{t}({\vec {u}})=g({\vec {u}}_{0})+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}g({\vec {u}}_{0}+s\Delta {\vec {u}})\right|_{s=0}=g({\vec {u}}_{0})+\lim _{s\rightarrow 0}{\frac {g({\vec {u}}_{0}+s\Delta {\vec {u}})-g({\vec {u}}_{0})}{s}}.}$

Non-linear geometric effects

Fig. 1: Deformation of a pipe (from gray to yellow): a) Deformation of the cross section changes the stiffness (red) b) Deflection changes the stiffness c) Deformation changes the direction of force (blue), indicated in green: linear calculation

Fig. 2: When buckling, the stiffness is reduced by a sideways deviation of the bending line (black, thin) from the rod or beam axis (black, bold). The evasion is further intensified by the load (red).

Fig. 3: Tank under hydrostatic pressure (white), a) geometrically linear, b) geometrically non-linear

The above figures show examples where significant geometric non-linearity occurs. The effects of geometric nonlinearity can be divided into two categories:

1. Large stretches: The change in shape caused by the stretching affects the stiffness of the body or the external forces. The change in the pipe cross-section (a, red) that occurs in Fig. 1 is an example of this and reduces the bending stiffness of the pipe there. Fig. 3 shows in the right part of the picture how large expansions can also cause considerable rotations.
2. Large deflections: These can affect the stiffness of the body or external forces. Fig. 1 shows how the deformation of the pipe causes the point of application of force to migrate, increasing the bending moment (b) and changing the direction of the load (c, blue). Large twists without stretching also fall into this category.

The mentioned effects of the geometric non-linearity can no longer be mapped after the linearization: A reaction of the deformation on the stiffness of the body or the external forces is neglected. Therefore, the pipe in Fig. 1 reacts much more rigidly with a linear calculation (green) than with a geometrically non-linear calculation (yellow).

The examples in Figures 2 and 3 above are examined in more detail below.

Buckling of the straight bar

→ Main article : Buckling

The buckling of the straight bar is an effect of the geometric non-linearity due to large deflections, see Fig. 2 above. As long as the load remains below a critical load, the structure can withstand this. Above the critical load, there is positive feedback between the increase in load and the decrease in stiffness due to the deflection, which can lead to the dramatic failure of the structure.

Container under pressure

Hydrostatic pressure load acting on a container wall, as indicated in white in Fig. 3, is the typical case for a load following deformation, because pressure always acts perpendicularly on surfaces. The figure shows a container which is rotatably but immovably mounted at the bottom and at the top edge (c) and is exposed to a hydrostatic pressure p ₀ , as it develops when the container is filled with a liquid.

In a geometrically linear calculation (a, on the left in the figure), the horizontal pressure component decreases linearly over the height of the container wall. Without taking into account the twisting of the container wall , see section #Following load below, the pressure acts on it in the horizontal direction (white) and the wall deforms into a blue curve. With linear consideration of the rotation, the purple drawn load p _gl and the red curve of the wall line, which is almost congruent with the blue curve, are created. Although the vertical pressure component is significantly larger here than in the geometrically non-linear calculation (pink), the deformation of the wall is smaller because the vertical component is more easily tolerated by the stiffness of the vertical wall calculated in the undeformed state.

The right part of the picture (b) shows the result of a geometrically non-linear calculation, where the pressure p _gnl (pink) always acts perpendicularly on the container _wall . On the one hand, the bulging reduces the supporting effect of the wall in the vertical direction. On the other hand, the expansion of the container wall in the vertical direction is significant because the membrane forces it causes counteract the pressure in the wall, an effect that the geometrically linear calculation does not reflect, as will be explained in the following section.

Analytical considerations

In this section, examples of movements are examined by means of which the effects of geometric linearization are macroscopically visible and analytically verifiable. In order to bring out the effects, large movements scaled with a parameter are considered, in which the application of geometric linearization is not appropriate. With a small value of close to 0.01, the deformation would be in a range where geometric linearization would be permissible. ${\displaystyle \alpha }$ ${\displaystyle \alpha }$

rotation

V. Mises equivalent stress in a linear elastic steel block due to engineering strains with a rigid body rotation of 5 °

In the case of a pure rotation of a body in the xy-plane around a fixed point in space p , the motion function and displacement lie

${\displaystyle {\vec {\chi }}({\vec {X}})={\vec {p}}+{\begin{pmatrix}\cos(\alpha )&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )\end{pmatrix}}({\vec {X}}-{\vec {p}})\rightarrow {\vec {u}}={\begin{pmatrix}\cos(\alpha )-1&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )-1\end{pmatrix}}({\vec {X}}-{\vec {p}})}$

in front. The matrix is ​​the displacement gradient

${\displaystyle \mathbf {H} :={\begin{pmatrix}{\frac {\partial u}{\partial X}}&{\frac {\partial u}{\partial Y}}\\{\frac {\partial v}{\partial X}}&{\frac {\partial v}{\partial Y}}\end{pmatrix}}={\begin{pmatrix}\cos(\alpha )-1&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )-1\end{pmatrix}}}$.

From this the relationship that is also valid for large rotations can be derived

${\displaystyle {\frac {\partial u}{\partial Y}}=-{\frac {\partial v}{\partial X}}}$

read off. The symmetrical part of the gradient are the engineering strains:

${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}(\mathbf {H} +\mathbf {H} ^{\mathrm {T} })={\begin{pmatrix}\cos(\alpha )-1&0\\0&\cos(\alpha )-1\end{pmatrix}}}$,

which at small angles of rotation - and only then - approximately disappear. The animation shows the v. Mises equivalent stress in a linear elastic steel block ( E = 200,000 MPa, G = 77,000 MPa) due to engineering strains with a rigid body rotation of a maximum of 5 °. The stresses calculated with the erroneous assumption of small strains are above the yield strength of many steels.

Distortion-free twist stretching

Fig. 4: a : Rotation and stretching of a rectangle (ABCD, blue) to another rectangle (ACC'D ', red). b : Deflection v of a linear, pliable structure perpendicular to its alignment and restoring force F, which is a geometrically non-linear effect.

The restoring force F of a linear, limp structure (rope, framework or membrane) deflected perpendicular to its alignment by v in Fig. 4b is a geometrically non-linear effect and cannot be reproduced in geometrically linear calculations. The following analytical consideration shows why.

With the rotational stretching of a rectangle shown in Fig. 4 a , the displacement in the x-direction depends only on the y-coordinate and the displacement in the y-direction only on the x-coordinate of the displaced point: ${\displaystyle u}$ ${\displaystyle v}$

${\displaystyle {\begin{pmatrix}u\\v\end{pmatrix}}=\tan(\alpha ){\begin{pmatrix}-Y\\X\end{pmatrix}}.}$

Formation of the gradient provides the displacement gradient:

${\displaystyle \mathbf {H} :={\begin{pmatrix}{\frac {\partial u}{\partial X}}&{\frac {\partial u}{\partial Y}}\\{\frac {\partial v}{\partial X}}&{\frac {\partial v}{\partial Y}}\end{pmatrix}}=\tan(\alpha ){\begin{pmatrix}0&-1\\1&0\end{pmatrix}}}$

and the linearized strain tensor

${\displaystyle {\boldsymbol {\varepsilon }}={\frac {1}{2}}(\mathbf {H+H} ^{\mathrm {T} })={\begin{pmatrix}0&0\\0&0\end{pmatrix}},}$

which is the symmetrical part of the displacement gradient and which disappears here.

In the rotated rectangle, paths occur in a geometrically linear approximation

${\displaystyle {\frac {\partial u}{\partial Y}}=-{\frac {\partial v}{\partial X}}.}$

no shear distortion, which is not surprising because this identity is also present with a pure rotation, see #rotation above.

It is noteworthy, however, that the displacement of point B to C perpendicular to the line (AB) does not cause any stretching of the line (AB) in a geometrically linear approximation. The deformation gradient , which is the sum of the displacement gradient and the unit tensor I , and its polar decomposition

${\displaystyle {\begin{array}{rcl}\mathbf {F} &=&\mathbf {H+I} ={\begin{pmatrix}1&-\tan(\alpha )\\\tan(\alpha )&1\end{pmatrix}}=\mathbf {R} \cdot \mathbf {U} \\\mathbf {R} &=&{\begin{pmatrix}\cos(\alpha )&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )\end{pmatrix}}\\\mathbf {U} &=&{\frac {1}{\cos(\alpha )}}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}\end{array}}}$

a rotation and a rotation-free extension show the cause of this behavior. The normal stretching in the x and y directions ${\displaystyle \mathbf {R} }$ ${\displaystyle \mathbf {U} }$

${\displaystyle U_{xx}=U_{yy}={\frac {1}{\cos(\alpha )}}=1+{\mathcal {O}}(\alpha ^{2})}$

are equal to one in a linear approximation, because the angle is of the same order as the yx component of the displacement gradient: ${\displaystyle \alpha }$

${\displaystyle 0<\alpha ^{2}\leq \tan ^{2}(\alpha )=\left({\frac {\partial v}{\partial X}}\right)^{2}={\mathcal {O}}(\delta ^{2})}$.

In the absence of a stretch but also the normal strains that would cause the body normal stresses and in Fig. 4 disappear b the restoring force would cause.

Shear distortion-free shear

Fig. 5: Rotation and shear of a rectangle (ABCD, width b, height h, blue) to form a parallelogram (AB'CD ', red)

Given is a rectangle with width b and height h , which, as in Fig. 5, lies at the origin and is aligned parallel to the coordinate axes (blue). With the displacement field

${\displaystyle {\vec {u}}={\begin{pmatrix}u\\v\end{pmatrix}}=\alpha {\begin{pmatrix}X-{\frac {b}{h}}Y\\{\frac {b}{h}}X-{\frac {b^{2}}{h^{2}}}Y\end{pmatrix}}\quad {\text{und}}\quad \alpha =50\,\%}$

the rectangle is deformed to the parallelogram shown in red. The displacement gradient and linearized strain tensor are:

${\displaystyle \mathbf {H} :={\begin{pmatrix}{\frac {\partial u}{\partial X}}&{\frac {\partial u}{\partial Y}}\\{\frac {\partial v}{\partial X}}&{\frac {\partial v}{\partial Y}}\end{pmatrix}}=\alpha {\begin{pmatrix}1&-{\frac {b}{h}}\\{\frac {b}{h}}&-{\frac {b^{2}}{h^{2}}}\end{pmatrix}}\quad \rightarrow \;{\boldsymbol {\varepsilon }}=\alpha {\begin{pmatrix}1&0\\0&-{\frac {b^{2}}{h^{2}}}\end{pmatrix}}}$

When the rectangle is sheared, the shear distortions disappear because of the identity that is also present with rotations

${\displaystyle {\frac {\partial u}{\partial Y}}=-{\frac {\partial v}{\partial X}}}$,

see #rotation above. A geometrically non-linear calculation results in the Green-Lagrange shear distortion

${\displaystyle E_{xy}=-\alpha ^{2}{\frac {b}{h}}\left(1+{\frac {b^{2}}{h^{2}}}\right)}$

which are omitted because of the linearized strain tensor. The shear stiffness of a component can thus be underestimated with geometrically linear calculations and large deformations. ${\displaystyle \alpha ^{2}=(\partial u/\partial X)^{2}={\mathcal {O}}(\delta ^{2})}$

Follow-up load

Fig. 6: A force F acts vertically on a pivoted wall

On a vertically standing, rotatable wall (length L, blue in Fig. 6), a force F ₀ acts vertically at point X so that the wall rotates through an angle and the force application point shifts around the vector (red): ${\displaystyle \alpha }$ ${\displaystyle {\vec {u}}\;{\textsf {nach}}\;{\vec {x}}}$

${\displaystyle {\vec {X}}=L{\begin{pmatrix}0\\1\end{pmatrix}}\;,\quad {\vec {x}}=L{\begin{pmatrix}\sin(\alpha )\\\cos(\alpha )\end{pmatrix}}\;\rightarrow {\vec {u}}=L{\begin{pmatrix}\sin(\alpha )\\\cos(\alpha )-1\end{pmatrix}}}$

The force follows the rotation, making it a function of the angle:

${\displaystyle {\vec {F}}=F{\begin{pmatrix}\cos(\alpha )\\-\sin(\alpha )\end{pmatrix}}}$

In a geometrically linear approximation:

${\displaystyle {\begin{array}{rcl}{\vec {u}}_{t}&=&L\left.{\begin{pmatrix}\sin(\alpha )\\\cos(\alpha )-1\end{pmatrix}}\right|_{\alpha =0}+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}L{\begin{pmatrix}\sin(s\alpha )\\\cos(s\alpha )-1\end{pmatrix}}\right|_{s=0}=L{\begin{pmatrix}\alpha \\0\end{pmatrix}}\\{\vec {F}}_{t}&=&F\left.{\begin{pmatrix}\cos(\alpha )\\-\sin(\alpha )\end{pmatrix}}\right|_{\alpha =0}+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}F{\begin{pmatrix}\cos(s\alpha )\\-\sin(s\alpha )\end{pmatrix}}\right|_{s=0}=F{\begin{pmatrix}1\\-\alpha \end{pmatrix}}\end{array}}}$.

The force application point shifts in a geometrically linear approximation exclusively in the x direction and the y component of the force changes linearly with the angle of rotation. This effect was taken into account when calculating the container under hydrostatic pressure in Fig. 3, which leads to the load according to the purple arrows and the deformation according to the red curve in the left part of the picture (a).

Conclusion

The geometric linearization simplifies the continuum mechanical calculations considerably, because the primary unknown - the shift - only occurs linearly in the distortions. This simplification is appropriate and permissible for small displacements and distortions. In the case of large displacements, geometrically non-linear effects occur which are decisive for the calculated results and, if not observed, can lead to dramatic misjudgments.

Examples

Scalar function

The determinant of the deformation gradient should be linearized geometrically. The deformation gradient is the sum of the displacement gradient and the unit tensor . With the characteristic polynomial the following is calculated: ${\displaystyle \operatorname {det} (\mathbf {F} )}$

${\displaystyle {\begin{array}{rcl}\operatorname {det} _{t}(\mathbf {F} =\mathbf {I} +\mathbf {H} )&=&\operatorname {det} (\mathbf {I} )+\mathrm {D} \operatorname {det} (\mathbf {0} )[\mathbf {H} ]\\&=&1+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}\operatorname {det} (\mathbf {I} +s\mathbf {H} )\right|_{s=0}\\&=&1+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}\left[s^{3}\operatorname {det} (\mathbf {H} +{\frac {1}{s}}\mathbf {I} )\right]\right|_{s=0}\\&=&1+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}\left[s^{3}\left({\frac {1}{s^{3}}}+{\frac {1}{s^{2}}}\operatorname {Sp} (\mathbf {H} )+{\frac {1}{s}}\operatorname {I} _{2}(\mathbf {H} )+\operatorname {det} (\mathbf {H} )\right)\right]\right|_{s=0}\\&=&1+\left.{\frac {\mathrm {d} }{\mathrm {d} s}}\left[1+s\operatorname {Sp} (\mathbf {H} )+s^{2}\operatorname {I} _{2}(\mathbf {H} )+s^{3}\operatorname {det} (\mathbf {H} )\right]\right|_{s=0}\\&=&1+\left.(\operatorname {Sp} (\mathbf {H} )+2s\operatorname {I} _{2}(\mathbf {H} )+3s^{2}\operatorname {det} (\mathbf {H} ))\right|_{s=0}\\&=&1+\operatorname {Sp} (\mathbf {H} ).\end{array}}}$

The function is the second main invariant . So in the geometrically linear case ${\displaystyle \operatorname {I} _{2}}$

${\displaystyle \operatorname {det} (\mathbf {F} )\approx 1+\operatorname {Sp} (\mathbf {H} )}$.

Tensor function

The inverse of the deformation gradient should be geometrically linearized. Out ${\displaystyle \mathbf {F} ^{-1}(\mathbf {H} ):=(\mathbf {I} +\mathbf {H} )^{-1}}$

${\displaystyle (\mathbf {I+H} )\cdot (\mathbf {I-H} )=\mathbf {I+H-H-H\cdot H} =\mathbf {I} +{\mathcal {O}}(\delta ^{2})}$

the following results in a linear approximation:

${\displaystyle \mathbf {F} ^{-1}\approx \mathbf {I-H} }$.

This can be confirmed with the differential calculus as follows. From the (vanishing) Gâteaux differential of the unit tensor, the product rule is used to determine the differential of the inverse deformation gradient:

${\displaystyle {\begin{array}{rcl}\mathrm {D} \mathbf {I} (\mathbf {0} )[\mathbf {H} ]&=&\mathrm {D} (\mathbf {F\cdot F} ^{-1})(\mathbf {0} )[\mathbf {H} ]\\&=&\underbrace {\mathrm {D} \mathbf {F} (\mathbf {0} )[\mathbf {H} ]} _{\mathbf {H} }\cdot \underbrace {\mathbf {F} ^{-1}(\mathbf {0} )} _{\mathbf {I} }+\underbrace {\mathbf {F} (\mathbf {0} )} _{\mathbf {I} }\cdot \mathrm {D} \mathbf {F} ^{-1}(\mathbf {0} )[\mathbf {H} ]=\mathbf {0} \\\rightarrow \mathrm {D} \mathbf {F} ^{-1}(\mathbf {0} )[\mathbf {H} ]&=&-\mathbf {H} .\end{array}}}$

With this differential, the geometrically linear inverse of the deformation gradient is:

${\displaystyle \mathbf {F} _{t}^{-1}(\mathbf {H} )=\underbrace {\mathbf {F} ^{-1}(\mathbf {0} )} _{\mathbf {I} }+\underbrace {\mathrm {D} \mathbf {F} ^{-1}(\mathbf {0} )[\mathbf {H} ]} _{-\mathbf {H} }=\mathbf {I} -\mathbf {H} .}$

See also

• Structural analysis # theory I., II. Or III. order
• P-delta effect
• Formula collection tensor algebra
• Tensoranalysis formula collection

literature

• H. Altenbach: Continuum Mechanics . Springer, 2012, ISBN 978-3-642-24118-5 .
• P. Haupt: Continuum Mechanics and Theory of Materials . Springer, 2000, ISBN 3-540-66114-X .