In continuum mechanics, a compatible deformation (or strain) tensor field in a body is that unique tensor field that is obtained when the body is subjected to a continuous, single-valued, displacement field. Compatibility is the study of the conditions under which such a displacement field can be guaranteed. Compatibility conditions are particular cases of integrability conditions and were first derived for linear elasticity by Barré de Saint-Venant in 1864 and proved rigorously by Beltrami in 1886.[1]
In the continuum description of a solid body we imagine the body to be composed of a set of infinitesimal volumes or material points. Each volume is assumed to be connected to its neighbors without any gaps or overlaps. Certain mathematical conditions have to be satisfied to ensure that gaps/overlaps do not develop when a continuum body is deformed. A body that deforms without developing any gaps/overlaps is called a compatible body. Compatibility conditions are mathematical conditions that determine whether a particular deformation will leave a body in a compatible state.[2]
In the context of infinitesimal strain theory, these conditions are equivalent to stating that the displacements in a body can be obtained by integrating the strains. Such an integration is possible if the Saint-Venant's tensor (or incompatibility tensor)
\boldsymbol{R}(\boldsymbol{\varepsilon})
\boldsymbol{\varepsilon}
\boldsymbol{R}:=\boldsymbol{\nabla} x (\boldsymbol{\nabla} x \boldsymbol{\varepsilon})T=\boldsymbol{0}~.
\boldsymbol{R}:=\boldsymbol{\nabla} x \boldsymbol{F}=\boldsymbol{0}
\boldsymbol{F}
The compatibility conditions in linear elasticity are obtained by observing that there are six strain-displacement relations that are functions of only three unknown displacements. This suggests that the three displacements may be removed from the system of equations without loss of information. The resulting expressions in terms of only the strains provide constraints on the possible forms of a strain field.
For two-dimensional, plane strain problems the strain-displacement relations are
\varepsilon11=\cfrac{\partialu1}{\partialx1}~;~~ \varepsilon12=\cfrac{1}{2}\left[\cfrac{\partialu1
Repeated differentiation of these relations, in order to remove the displacements
u1
u2
\cfrac{\partial2\varepsilon11
The only displacement field that is allowed by a compatible plane strain field is a plane displacement field, i.e.,
u=u(x1,x2)
In three dimensions, in addition to two more equations of the form seen for two dimensions, there arethree more equations of the form
\cfrac{\partial2\varepsilon33
eikr~ejls~\varepsilonij,kl=0
eijk
\boldsymbol{\nabla} x (\boldsymbol{\nabla} x \boldsymbol{\varepsilon})T=\boldsymbol{0}
\boldsymbol{\nabla} x \boldsymbol{\varepsilon}=eijk\varepsilonrj,iek ⊗ er
The second-order tensor
\boldsymbol{R}:=\boldsymbol{\nabla} x (\boldsymbol{\nabla} x \boldsymbol{\varepsilon})T~;~~Rrs:=eikr~ejls~\varepsilonij,kl
For solids in which the deformations are not required to be small, the compatibility conditions take the form
\boldsymbol{\nabla} x \boldsymbol{F}=\boldsymbol{0}
\boldsymbol{F}
eABC~\cfrac{\partialFiB
x=\boldsymbol{\chi}(X,t)
The compatibility condition for the right Cauchy-Green deformation tensor can be expressed as
\gamma | |
R | |
\alpha\beta\rho |
:=
\partial | |
\partialX\rho |
\gamma | |
[\Gamma | |
\alpha\beta |
]-
\partial | |
\partialX\beta |
\gamma | |
[\Gamma | |
\alpha\rho |
]+
\gamma | |
\Gamma | |
\mu\rho |
\mu | |
~\Gamma | |
\alpha\beta |
-
\gamma | |
\Gamma | |
\mu\beta |
\mu | |
~\Gamma | |
\alpha\rho |
=0
k | |
\Gamma | |
ij |
m | |
R | |
ijk |
The problem of compatibility in continuum mechanics involves the determination of allowable single-valued continuous fields on simply connected bodies. More precisely, the problem may be stated in the following manner.[5]
Consider the deformation of a body shown in Figure 1. If we express all vectors in terms of the reference coordinate system
\{(E1,E2,E3),O\}
u=x-X~;~~ui=xi-Xi
\boldsymbol{\nabla}u=
\partialu | |
\partialX |
~;~~ \boldsymbol{\nabla}x=
\partialx | |
\partialX |
What conditions on a given second-order tensor field
\boldsymbol{A}(X)
v(X)
\boldsymbol{\nabla}v=\boldsymbol{A} \equiv vi,j=Aij
For the necessary conditions we assume that the field
v
vi,j=Aij
vi,jk=Aij,k~;~~vi,kj=Aik,j
vi,jk=vi,kj
Aij,k=Aik,j
\boldsymbol{\nabla} x \boldsymbol{A}=\boldsymbol{0}
To prove that this condition is sufficient to guarantee existence of a compatible second-order tensor field, we start with the assumption that a field
\boldsymbol{A}
\boldsymbol{\nabla} x \boldsymbol{A}=\boldsymbol{0}
v
A
B
v(XB)-v(XA)=
XB | |
\int | |
XA |
\boldsymbol{\nabla}v ⋅ ~dX =
XB | |
\int | |
XA |
\boldsymbol{A}(X) ⋅ dX
v
A
B
From Stokes' theorem, the integral of a second order tensor along a closed path is given by
\oint\partial\Omega\boldsymbol{A} ⋅ ds=\int\Omegan ⋅ (\boldsymbol{\nabla} x \boldsymbol{A})~da
\boldsymbol{A}
\oint\partial\Omega\boldsymbol{A} ⋅ ds=0 \implies \intAB\boldsymbol{A} ⋅ dX+\intBA\boldsymbol{A} ⋅ dX=0
v
The compatibility condition for the deformation gradient is obtained directly from the above proof by observing that
\boldsymbol{F}=\cfrac{\partialx
\boldsymbol{F}
\boldsymbol{\nabla} x \boldsymbol{F}=\boldsymbol{0}
The compatibility problem for small strains can be stated as follows.
Given a symmetric second order tensor field
\boldsymbol{\epsilon}
u
\boldsymbol{\epsilon}=
1 | |
2 |
[\boldsymbol{\nabla}u+(\boldsymbol{\nabla}u)T]
Suppose that there exists
u
\boldsymbol{\epsilon}
\boldsymbol{\nabla}u=\boldsymbol{\epsilon}+\boldsymbol{\omega}
\boldsymbol{\omega}:=
1 | |
2 |
[\boldsymbol{\nabla}u-(\boldsymbol{\nabla}u)T]
\boldsymbol{\nabla}\boldsymbol{\omega}\equiv\omegaij,k=
1 | |
2 |
(ui,jk-uj,ik)=
1 | |
2 |
(ui,jk+uk,ji-uj,ik-uk,ji)=\varepsilonik,j-\varepsilonjk,i
\boldsymbol{\omega}
\omegaij,kl=\omegaij,lk
\varepsilonik,jl-\varepsilonjk,il-\varepsilonil,jk+\varepsilonjl,ik=0
\boldsymbol{\nabla} x (\boldsymbol{\nabla} x \boldsymbol{\epsilon})T=\boldsymbol{0}
w
\boldsymbol{\nabla} x \boldsymbol{\epsilon}=\boldsymbol{\nabla}w+\boldsymbol{\nabla}wT
\boldsymbol{\nabla} x (\boldsymbol{\nabla}w+\boldsymbol{\nabla}wT)T =\boldsymbol{0}
Let us now assume that the condition
\boldsymbol{\nabla} x (\boldsymbol{\nabla} x \boldsymbol{\epsilon})T=\boldsymbol{0}
u
\boldsymbol{\omega}
\boldsymbol{\nabla}w
XA
XB
w(XB)-w(XA)=
XB | |
\int | |
XA |
\boldsymbol{\nabla}w ⋅ dX =
XB | |
\int | |
XA |
(\boldsymbol{\nabla} x \boldsymbol{\epsilon}) ⋅ dX
w(XA)
w(X)
XA
Xb
XB | |
\oint | |
XA |
(\boldsymbol{\nabla} x \boldsymbol{\epsilon}) ⋅ dX=\boldsymbol{0}
XB | |
\oint | |
XA |
(\boldsymbol{\nabla} x \boldsymbol{\epsilon}) ⋅ dX=
\int | |
\OmegaAB |
n ⋅ (\boldsymbol{\nabla} x \boldsymbol{\nabla} x \boldsymbol{\epsilon})~da =\boldsymbol{0}
w
\boldsymbol{\omega}
In the next step of the process we will consider the uniqueness of the displacement field
u
u(XB)-u(XA)=
XB | |
\int | |
XA |
\boldsymbol{\nabla}u ⋅ dX =
XB | |
\int | |
XA |
(\boldsymbol{\epsilon}+\boldsymbol{\omega}) ⋅ dX
\boldsymbol{\nabla} x \boldsymbol{\epsilon}=\boldsymbol{\nabla}w=-\boldsymbol{\nabla} x \omega
XB | |
\oint | |
XA |
(\boldsymbol{\epsilon}+\boldsymbol{\omega}) ⋅ dX=
\int | |
\OmegaAB |
n ⋅ (\boldsymbol{\nabla} x \boldsymbol{\epsilon}+\boldsymbol{\nabla} x \boldsymbol{\omega})~da=\boldsymbol{0}
u
u
The compatibility problem for the Right Cauchy-Green deformation field can be posed as follows.
Problem: Let
\boldsymbol{C}(X)
\boldsymbol{C}
x(X)
(1) \left( | \partialx |
\partialX |
\right)T\left(
\partialx | |
\partialX |
\right)=\boldsymbol{C}
Suppose that a field
x(X)
\partialxi | |
\partialX\alpha |
\partialxi | |
\partialX\beta |
=C\alpha\beta
C\alpha\beta=g\alpha\beta
\deltaij~
\partialxi | ~ | |
\partialX\alpha |
\partialxj | |
\partialX\beta |
=g\alpha\beta
Gij=
\partialX\alpha | ~ | |
\partialxi |
\partialX\beta | |
\partialxj |
~g\alpha\beta
Gij
g\alpha\beta
\deltaij=Gij
(x)
k | |
\Gamma | |
ij |
=0
\partial2xm | |
\partialX\alpha\partialX\beta |
=
\partialxm | |
\partialX\mu |
(X)
\mu | |
\Gamma | |
\alpha\beta |
-
\partialxi | ~ | |
\partialX\alpha |
\partialxj | |
\partialX\beta |
(x)
m | |
\Gamma | |
ij |
| |||||||||
\partialX\beta |
=
m | |
F | |
~\mu |
(X)
\mu | |
\Gamma | |
\alpha\beta |
;~~
i | |
F | |
~\alpha |
:=
\partialxi | |
\partialX\alpha |
(X)\Gamma\alpha\beta\gamma=
1 | \left( | |
2 |
\partialg\alpha\gamma | |
\partialX\beta |
+
\partialg\beta\gamma | |
\partialX\alpha |
-
\partialg\alpha\beta | |
\partialX\gamma |
\right)~;~~ (X)
\nu | |
\Gamma | |
\alpha\beta |
=g\nu\gamma(X)\Gamma\alpha\beta\gamma~;~~ g\alpha\beta=C\alpha\beta~;~~g\alpha\beta=C\alpha\beta
(X)
\mu | |
\Gamma | |
\alpha\beta |
=\cfrac{C\mu\gamma
| |||||||||
\partialX\beta |
=
m | |
F | |
~\mu |
~\cfrac{C\mu\gamma
| |||||||||||||
\partialX\beta\partialX\rho |
=
| |||||||||||||
\partialX\rho\partialX\beta |
\implies
| |||||||||
\partialX\rho |
(X)
\mu | |
\Gamma | |
\alpha\beta |
+
m | ||
F | ~ | |
~\mu |
\partial | |
\partialX\rho |
[(X)
\mu | |
\Gamma | |
\alpha\beta |
]=
| |||||||||
\partialX\beta |
(X)
\mu | |
\Gamma | |
\alpha\rho |
+
m | ||
F | ~ | |
~\mu |
\partial | |
\partialX\beta |
[(X)
\mu | |
\Gamma | |
\alpha\rho |
]
m | |
F | |
~\gamma |
(X)
\gamma | |
\Gamma | |
\mu\rho |
(X)
\mu | |
\Gamma | |
\alpha\beta |
+
m | ||
F | ~ | |
~\mu |
\partial | |
\partialX\rho |
[(X)
\mu | |
\Gamma | |
\alpha\beta |
]=
m | |
F | |
~\gamma |
(X)
\gamma | |
\Gamma | |
\mu\beta |
(X)
\mu | |
\Gamma | |
\alpha\rho |
+
m | ||
F | ~ | |
~\mu |
\partial | |
\partialX\beta |
[(X)
\mu | |
\Gamma | |
\alpha\rho |
]
m | |
F | |
~\gamma |
\left((X)
\gamma | |
\Gamma | |
\mu\rho |
(X)
\mu | |
\Gamma | |
\alpha\beta |
+
\partial | |
\partialX\rho |
[(X)
\gamma | |
\Gamma | |
\alpha\beta |
]- (X)
\gamma | |
\Gamma | |
\mu\beta |
(X)
\mu | |
\Gamma | |
\alpha\rho |
-
\partial | |
\partialX\beta |
[(X)
\gamma | |
\Gamma | |
\alpha\rho |
]\right)=0
m | |
F | |
\gamma |
\gamma | |
R | |
\alpha\beta\rho |
:=
\partial | |
\partialX\rho |
[(X)
\gamma | |
\Gamma | |
\alpha\beta |
]-
\partial | |
\partialX\beta |
[(X)
\gamma | |
\Gamma | |
\alpha\rho |
]+ (X)
\gamma | |
\Gamma | |
\mu\rho |
(X)
\mu | |
\Gamma | |
\alpha\beta |
-(X)
\gamma | |
\Gamma | |
\mu\beta |
(X)
\mu | |
\Gamma | |
\alpha\rho |
=0
\boldsymbol{C}
The proof of sufficiency is a bit more involved.[5] [6] We start with the assumption that
\gamma | |
R | |
\alpha\beta\rho |
=0~;~~g\alpha\beta=C\alpha\beta
x
X
\partialxi | |
\partialX\alpha |
\partialxi | |
\partialX\beta |
=C\alpha\beta
| |||||||||
\partialX\beta |
=
i | |
F | |
~\gamma |
~(X)
\gamma | |
\Gamma | |
\alpha\beta |
i | |
F | |
~\alpha |
(X)
\gamma | |
\Gamma | |
\alpha\beta |
=(X)
\gamma | |
\Gamma | |
\beta\alpha |
~;~~
\gamma | |
R | |
\alpha\beta\rho |
=0
i | |
\Gamma | |
jk |
i | |
F | |
~\alpha |
C2
Next consider the system of equations
\partialxi | |
\partialX\alpha |
=
i | |
F | |
~\alpha |
i | |
F | |
~\alpha |
C2
xi(X\alpha)
xi
\det\left| | \partialxi |
\partialX\alpha |
\right|\ne0
\partialxi | |
\partialX\alpha |
~g\alpha\beta~
\partialxj | |
\partialX\beta |
=\deltaij
g\alpha\beta=C\alpha\beta=
\partialxk | ~ | |
\partialX\alpha |
\partialxk | |
\partialX\beta |
\partialx | |
\partialX |
\boldsymbol{C}