Deformation (mechanics)

Deformation changes a body's shape or size. Strain quantifies relative changes in lengths and angles, whereas a rigid translation or rotation alone produces no strain. The continuum approximation describes these changes using displacement fields and their spatial variation.

A homogeneous deformation has the same deformation gradient at every material point. On a connected reference domain it is an affine map $y(X)=FX+b$ with constant matrix $F$ and translation $b$. Orientation preservation requires $\det F>0$. Its stretches and local volume ratio are spatially constant. A homogeneous deformation describes the kinematics; a homogeneous hyperelastic material instead describes spatial independence of a constitutive law.

Volume preservation by the elastic accommodation in morphoelasticity. If $\mathbf{F}=\mathbf{AG}$, elastic incompressibility imposes $\det \mathbf{A}=1$, and hence $\det \mathbf{F}=\det \mathbf{G}$. Total volume can increase through growth while the elastic response remains incompressible. A hydrostatic pressure acts as a Lagrange multiplier enforcing this constraint in the Cauchy stress tensor.

For a right stretch tensor $U={\sum }_a{\lambda }_a{n}_a\otimes n_a$, define $E^f=f(U)={\sum }_{a}f({\lambda }_a)n_a\otimes n_a$. A standard normalized choice has $f\in C^1(0,\infty )$, $f(1)=0$, $f^{\prime }(1)=1$ and $f$ strictly increasing; requiring $f^{\prime }>0$ gives a nonsingular stretch-to-strain parameterization. This strain vanishes exactly for a rigid rotation, is independent of superposed spatial rotations, and agrees to first order with the infinitesimal symmetric displacement gradient. Indeed $F=I+H$ gives $U=I+(H+H^T)/2+O(\Vert H{\Vert }^2)$, and Taylor expansion of $f$ proves the last statement.

The Biot strain tensor uses $f(\lambda )=\lambda -1$, so $E^{(1)}=U-I$. Its principal values are extension ratios minus one. It is a spectral strain measure, since $f(1)=0$, $f^{\prime }(1)=1$ and $f^{\prime }>0$. Its reference-volume work conjugate is the symmetric Biot stress.

For $C=F^{T}F$, differentiation of $C^{-1}$ gives ${\dot{E}}^{(-2)}=F^{-1}D{F}^{-T}$. Its reference-power conjugate is $T^{(-2)}=F^T\tau F$. In a convected coordinate basis $g_I=F{e}_I$, these are the covariant components of the Kirchhoff stress tensor. The second Piola-Kirchhoff stress tensor instead gives its contravariant components, since $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$.

With the right Cauchy-Green deformation tensor $C=F^{T}F$, the Green-Lagrange strain tensor measures the change in the material metric. Since ${\dot{E}}^{(2)}=F^{T}DF$, its work-conjugate stress and strain partner is the second Piola-Kirchhoff stress tensor $F^{-1}\tau F^{-T}$. The formula follows by cycling factors in $T:{\dot{E}}^{(2)}$, not by identifying nominal and symmetric material stress.

In plane strain, the deformation map has $x_1=x_1(X_1,X_2)$, $x_2=x_2(X_1,X_2)$ and $x_3=X_3$. The out-of-plane stretch is one and out-of-plane shear vanishes. This kinematic condition need not make the out-of-plane stress vanish.

For an isotropic elastic annulus $a<r<b$ with internal excess pressure $P$ and zero outer gauge pressure, radial equilibrium gives $u=Cr+D/r$. Its stress tensor has ${\tau }_{rr}=2(\lambda +\mu )C-2\mu D/r^2$ and ${\tau }_{\theta \theta }=2(\lambda +\mu )C+2\mu D/r^2$. Enforcing ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$ gives $2\mu D=P{a}^2{b}^2/(b^2-a^2)$, so the maximum difference occurs at the bore and equals the displayed expression. Once an appropriate yield criterion is reached, stress must be related to elastic strain while total strain includes a plastic contribution. The elastic formula alone cannot describe further irreversible deformation.

The displayed scalar stress potential automatically satisfies the two in-plane static momentum equations without body force, because the mixed third derivatives cancel. For homogeneous isotropic linear elasticity, strain compatibility further makes $A$ biharmonic. Since stresses depend only on second derivatives, $A$ is unique only up to $ax+by+c$ on a connected domain. Conversely a function with all second derivatives zero is affine, proving that this is exactly the gauge freedom. It does not license adding an arbitrary biharmonic function, which generally changes stress.

For plane strain in $y>0$ under surface normal traction $-P(x)$ and zero surface shear, use the forward Fourier transform phase $e^{ikx}$. At $k\ne 0$, biharmonicity gives $({\partial }_y^2-k^2{)}^2\hat{A}=0$, whose decaying part is $(C+Dy)e^{-|k|y}$. Boundary traction fixes $C=\hat{P}/k^2$, $D=|k|C$, proving the displayed formula. Constitutive inversion and a decaying displacement mode give $2\mu \hat{v}=[y+2(1-\nu )/|k|]\hat{P}{e}^{-|k|y}$. Multiplication by $-ik$ produces an unambiguous displacement-gradient convolution kernel.

The zero mode is separate: a nonzero resultant load $\hat{P}(0)$ conflicts with an ordinary fully decaying displacement normalization in two dimensions and leads to a logarithmic reference issue. Balanced loading or a distributional/reference-displacement treatment is required before extending the transformed formulas globally. Dividing by $k^2$ and assigning a finite value at $k=0$ is invalid.

Simple shear displaces one coordinate direction in proportion to a transverse coordinate while preserving volume. Its deformation gradient has a single off-diagonal component $\gamma$. Simple shear flow describes the corresponding velocity field rather than the finite deformation.

For the reciprocal-Jacobian quadratic energy under $F=I+e_1\otimes e_2$, $T_R/\mu$ has off-diagonal entries $(1,2)$ and $(2,1)$ equal to one and all others zero. Every unit reference normal in the first two coordinate directions therefore maximizes nominal traction. The Cauchy stress tensor has upper-left block $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$. Its largest absolute eigenvalue is $\phi =(1+\sqrt{5})/2$, with spatial normal proportional to $(\phi ,1,0)$. Pulling this back by $F^T$ gives reference normal proportional to $(1,\phi ,0)$ and plane $x_1+\phi x_2=0$.

For $\gamma >0$, direct multiplication gives $|FN{|}^2-|N{|}^2=\gamma N_2(2N_1+\gamma N_2)$. Thus length-preserving line directions form the union of two planes in direction space: $N_2=0$ and $2N_1+\gamma N_2=0$. Their intersection is the third coordinate axis. This concerns segment length rather than pointwise fixed lines; a segment can move and turn while keeping its length.

The two nontrivial principal stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, with $v_{+}{v}_{-}=1$. The maximizing reference direction is proportional to $(1,v_{+},0)$. Consequently it tends to $(1,1,0)$ as $\gamma$ tends to zero from above. At zero itself all stretches equal one, so every direction maximizes length; the limit selects a direction from a degenerate endpoint. This follows from the largest eigenvector of the right Cauchy-Green deformation tensor.

The principal stretches are the eigenvalues of a stretch tensor, or equivalently the singular values of the deformation gradient. They measure local length ratios along the principal reference directions.

The Jacobian matrix of a deformation map carries an infinitesimal reference line element to its current image. Its determinant $J=\det F$ is the local volume ratio. An orientation-preserving deformation has $J>0$, and incompressibility imposes $J=1$.

Two constant deformation gradients are rank-one connected when their difference is an outer product $a\otimes N$. For a nonzero jump its matrix rank is one; equality $A=B$ is the trivial compatible case. The vector $N$ specifies a reference interface normal, and $a$ is the jump amplitude. The Hadamard jump condition identifies this algebraic condition with continuity of an affine deformation across a planar interface. The normal can be normalized by rescaling $a$ without changing the product.

The unrotated variants differ by $2\delta e_3\otimes e_1$, giving normal $e_1$. The rotation about $e_2$ with $\cos \theta =(1-{\delta }^2)/(1+{\delta }^2)$ and $\sin \theta =2\delta /(1+{\delta }^2)$ gives $U_{+}-\overline{R}{U}_{-}=[2\delta /(1+{\delta }^2)](e_1+\delta e_3)\otimes e_3$, giving the second normal $e_3$. If these are the only two compatible rotations, the only gradients in $SO(3)U_{+}$ compatible with $\overline{R}{U}_{-}$ are $U_{+}$ and $\overline{R}{U}_{+}$. Left multiplication by a rotation preserves rank, so the second classification follows directly from the first.

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.

For a twice continuously differentiable deformation map, each row of its cofactor matrix has zero divergence in reference coordinates. In three dimensions write $(\mathrm{cof}Dy{)}_{iA}=\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{ABC}{y}_{j,B}{y}_{k,C}$. Applying ${\partial }_A$ gives terms in which a symmetric pair of second partial derivatives contracts with an antisymmetric Levi-Civita symbol, so each term vanishes. This identity is the cancellation that makes the Piola-Kirchhoff stress tensor transform force balance correctly between configurations.

An elastic deformation map whose deformation gradient is independent of position at each time. Its spatial material acceleration is $\mathbf{a}=\ddot{F}{F}^{-1}(\mathbf{x}-\mathbf{c})+\ddot{\mathbf{c}}$. Incompressibility requires $\det F=1$.

The factor describing local stress-free growth in the morphoelasticity convention $\mathbf{F}=\mathbf{AG}$. Its determinant is the local grown-volume ratio. For $\mathbf{G}=\mathrm{diag}(1,1,\gamma )$, growth changes only the preferred third-direction length, by factor $\gamma$.

The ratio of current to reference lengths of an infinitesimal material line element. Along an elastic rod, it is $\alpha =ds/dS=|d\mathbf{r}/dS|>0$. A principal stretch is a stretch ratio along a principal direction of the stretch tensor.

For an orientation-preserving deformation gradient $F$, oriented area transforms by its cofactor matrix: $(Fa)\times (Fb)=(\det F)F^{-T}(a\times b)$. Taking the dot product with $Fc$ proves this from the determinant identity for the scalar triple product. Applying it to tangent vectors of a surface proves Nanson's formula. Equating current and reference tractions then gives the material-first nominal stress tensor $P=J{F}^{-1}\sigma$ for symmetric Cauchy stress tensor $\sigma$.

The factor $A$ can encode an internal relaxed configuration, while $F_{\ast }=F{A}^{-1}$ determines the instantaneous elastic energy. Differentiating $F_{\ast }$ at fixed $F$ or fixed $A$ produces the corresponding nominal stress tensor and thermodynamic force conjugate to an internal variable. This is a kinematic decomposition, not an assertion that each factor separately derives from a global deformation map.

For an orientation-preserving invertible deformation gradient, the polar decomposition of an invertible real matrix separates a rotation $R$ from the right stretch tensor $U$ or the left stretch tensor $V$. The rotation is shared by the two factorizations, with $V=RU{R}^T$.

The right Cauchy-Green tensor measures stretch on the reference configuration. Its eigenvalues are the same squared principal stretches as those of the left Cauchy-Green deformation tensor, but its eigenvectors are reference directions.

The left Cauchy-Green tensor measures stretch on the current configuration. It is a positive-definite symmetric matrix for an invertible deformation gradient, with eigenvalues equal to the squares of the principal stretches.

The first principal invariant of the left Cauchy-Green deformation tensor is the sum of the squares of the principal stretches. Since $B=F{F}^T$, it also equals ${\sum }_{i,I}{F}_{iI}^2$. Its derivative with respect to the deformation gradient is $2F$. For a strain energy density $W=w(I_1)$ with incompressibility, the Cauchy stress tensor is $T=2w^{\prime }(I_1)B-pI$.

A deformation map sends each particle's reference configuration position to its current configuration position. Its spatial derivative with respect to the Lagrangian coordinates is the deformation gradient.

A radial deformation map preserving each particle's direction from a common centre. In an orthonormal basis of spherical coordinates, its deformation gradient has principal stretches $f^{\prime }(R),f(R)/R,f(R)/R$. Thus incompressibility gives $f^{\prime }(R)f(R{)}^2=R^2$, and integration gives $f(R{)}^3=R^3+C$.