Material time derivatives
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Material time derivatives are needed for many updated Lagrangian formulations
of finite element analysis.
Recall that the motion can be expressed as
x = φ ( X , t ) or X = φ − 1 ( x , t ) {\displaystyle \mathbf {x} ={\boldsymbol {\varphi }}(\mathbf {X} ,t)\quad {\text{or}}\qquad \mathbf {X} ={\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t)} If we keep X {\displaystyle \mathbf {X} } fixed, then the velocity is given by
V ( X , t ) = ∂ φ ∂ t ( X , t ) {\displaystyle \mathbf {V} (\mathbf {X} ,t)={\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}(\mathbf {X} ,t)} This is the material time derivative expressed in terms of X {\displaystyle \mathbf {X} } .
The spatial version of the velocity is
v ( x , t ) = V ( φ − 1 ( x , t ) , t ) {\displaystyle \mathbf {v} (\mathbf {x} ,t)=\mathbf {V} ({\boldsymbol {\varphi }}^{-1}(\mathbf {x} ,t),t)} We will use the symbol v {\displaystyle \mathbf {v} } for velocity from now on by slightly abusing the
notation.
We usually think of quantities such as velocity and acceleration as spatial
quantities which are functions of x {\displaystyle \mathbf {x} } (rather than material quantities which
are functions of X {\displaystyle \mathbf {X} } ).
Given the spatial velocity v ( x , t ) {\displaystyle \mathbf {v} (\mathbf {x} ,t)} , if we want to find the acceleration
we will have to consider the fact that x ≡ x ( X , t ) {\displaystyle \mathbf {x} \equiv \mathbf {x} (\mathbf {X} ,t)} , i.e., the
position also changes with time. We do this by using the chain rule. Thus
D v ( x , t ) D t = a ( x , t ) = ∂ v ( x , t ) ∂ t + ∂ v ( x , t ) ∂ x ⋅ ∂ φ ( X , t ) ∂ t = ∂ v ∂ t + ∇ v ⋅ V = ∂ v ∂ t + ∇ v ⋅ v {\displaystyle {\cfrac {D\mathbf {v} (\mathbf {x} ,t)}{Dt}}=\mathbf {a} (\mathbf {x} ,t)={\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial t}}+{\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial \mathbf {x} }}\cdot {\frac {\partial {\boldsymbol {\varphi }}(\mathbf {X} ,t)}{\partial t}}={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {V} ={\frac {\partial \mathbf {v} }{\partial t}}+{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} } Such a derivative is called the material time derivative expressed
in terms of x {\displaystyle \mathbf {x} } . The second term in the expression is called the
convective derivative ..
Velocity gradient
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Let the velocity be expressed in spatial form , i.e., v ( x , t ) {\displaystyle \mathbf {v} (\mathbf {x} ,t)} .
The spatial velocity gradient tensor is given by
l := ∂ v ( x , t ) ∂ x = ∇ v {\displaystyle {\boldsymbol {l}}:={\frac {\partial \mathbf {v} (\mathbf {x} ,t)}{\partial \mathbf {x} }}={\boldsymbol {\nabla }}\mathbf {v} } The velocity gradient l {\displaystyle {\boldsymbol {l}}} is a second order tensor which can expressed as
l = l i j e i ⊗ e j = ∂ v i ∂ x j e i ⊗ e j {\displaystyle {\boldsymbol {l}}=l_{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}={\frac {\partial v_{i}}{\partial x_{j}}}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}} The velocity gradient is a measure of the relative velocity of two points in the current configuration.
Time derivative of the deformation gradient
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Recall that the deformation gradient is given by
F = ∂ φ ∂ X {\displaystyle {\boldsymbol {F}}={\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}} The time derivative of F {\displaystyle {\boldsymbol {F}}} (keeping X {\displaystyle \mathbf {X} } fixed) is
F ˙ = ∂ ∂ t ( ∂ φ ∂ X ) = ∂ ∂ X ( ∂ φ ∂ t ) = ∂ v ∂ X = ∇ ∘ v {\displaystyle {\dot {\boldsymbol {F}}}={\frac {\partial }{\partial t}}~\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}\right)={\frac {\partial }{\partial \mathbf {X} }}~\left({\frac {\partial {\boldsymbol {\varphi }}}{\partial t}}\right)={\frac {\partial \mathbf {v} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{\circ }\mathbf {v} } Using the chain rule
F ˙ = ∂ v ∂ x ⋅ ∂ x ∂ X = ∂ v ∂ x ⋅ ∂ φ ∂ X = l ⋅ F {\displaystyle {\dot {\boldsymbol {F}}}={\frac {\partial \mathbf {v} }{\partial \mathbf {x} }}\cdot {\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\frac {\partial \mathbf {v} }{\partial \mathbf {x} }}\cdot {\frac {\partial {\boldsymbol {\varphi }}}{\partial \mathbf {X} }}={\boldsymbol {l}}\cdot {\boldsymbol {F}}} Form this we get the important relation
l = F ˙ ⋅ F − 1 . {\displaystyle {\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}~.} Time derivative of strain
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Let d X 1 {\displaystyle d\mathbf {X} _{1}} and d X 2 {\displaystyle d\mathbf {X} _{2}} be two infinitesimal material line segments in a
body. Then
d x 1 = F ⋅ d X 1 ; d x 2 = F ⋅ d X 2 {\displaystyle d\mathbf {x} _{1}={\boldsymbol {F}}\cdot d\mathbf {X} _{1}~;~~d\mathbf {x} _{2}={\boldsymbol {F}}\cdot d\mathbf {X} _{2}} Hence,
d x 1 ⋅ d x 2 = ( F ⋅ d X 1 ) ⋅ ( F ⋅ d X 2 ) = d X 1 ⋅ ( F T ⋅ F ) ⋅ d X 2 = d X 1 ⋅ C ⋅ d X 2 = d X 1 ⋅ ( 2 E + 1 ) ⋅ d X 2 {\displaystyle d\mathbf {x} _{1}\cdot d\mathbf {x} _{2}=({\boldsymbol {F}}\cdot d\mathbf {X} _{1})\cdot ({\boldsymbol {F}}\cdot d\mathbf {X} _{2})=d\mathbf {X} _{1}\cdot ({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})\cdot d\mathbf {X} _{2}=d\mathbf {X} _{1}\cdot {\boldsymbol {C}}\cdot d\mathbf {X} _{2}=d\mathbf {X} _{1}\cdot (2~{\boldsymbol {E}}+{\boldsymbol {\mathit {1}}})\cdot d\mathbf {X} _{2}} Taking the derivative with respect to t {\displaystyle t} gives us
∂ ∂ t ( d x 1 ⋅ d x 2 ) = d X 1 ⋅ ∂ C ∂ t ⋅ d X 2 = 2 d X 1 ⋅ ∂ E ∂ t ⋅ d X 2 {\displaystyle {\frac {\partial }{\partial t}}(d\mathbf {x} _{1}\cdot d\mathbf {x} _{2})=d\mathbf {X} _{1}\cdot {\frac {\partial {\boldsymbol {C}}}{\partial t}}\cdot d\mathbf {X} _{2}=2~d\mathbf {X} _{1}\cdot {\frac {\partial {\boldsymbol {E}}}{\partial t}}\cdot d\mathbf {X} _{2}} The material strain rate tensor is defined as
E ˙ = ∂ E ∂ t = 1 2 ∂ C ∂ t = 1 2 C ˙ {\displaystyle {\dot {\boldsymbol {E}}}={\frac {\partial {\boldsymbol {E}}}{\partial t}}={\frac {1}{2}}~{\frac {\partial {\boldsymbol {C}}}{\partial t}}={\frac {1}{2}}~{\dot {\boldsymbol {C}}}} Clearly,
E ˙ = 1 2 ∂ ∂ t ( F T ⋅ F ) = 1 2 ( F ˙ T ⋅ F + F T ⋅ F ˙ ) . {\displaystyle {\dot {\boldsymbol {E}}}={\frac {1}{2}}~{\frac {\partial }{\partial t}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}})={\frac {1}{2}}~({\dot {\boldsymbol {F}}}^{T}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})~.} Also,
1 2 ∂ ∂ t ( d x 1 ⋅ d x 2 ) = d X 1 ⋅ E ˙ ⋅ d X 2 = ( F − 1 ⋅ d x 1 ) ⋅ E ˙ ⋅ ( F − 1 ⋅ d x 2 ) = d x 1 ⋅ ( F − T ⋅ E ˙ ⋅ F − 1 ) ⋅ d x 2 {\displaystyle {\frac {1}{2}}~{\frac {\partial }{\partial t}}(d\mathbf {x} _{1}\cdot d\mathbf {x} _{2})=d\mathbf {X} _{1}\cdot {\dot {\boldsymbol {E}}}\cdot d\mathbf {X} _{2}=({\boldsymbol {F}}^{-1}\cdot d\mathbf {x} _{1})\cdot {\dot {\boldsymbol {E}}}\cdot ({\boldsymbol {F}}^{-1}\cdot d\mathbf {x} _{2})=d\mathbf {x} _{1}\cdot ({\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1})\cdot d\mathbf {x} _{2}} The spatial rate of deformation tensor or stretching tensor is
defined as
d = F − T ⋅ E ˙ ⋅ F − 1 = 1 2 F − T ⋅ C ˙ ⋅ F − 1 {\displaystyle {\boldsymbol {d}}={\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1}={\frac {1}{2}}~{\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {C}}}\cdot {\boldsymbol {F}}^{-1}} In fact, we can show that d {\displaystyle {\boldsymbol {d}}} is the symmetric part of the velocity
gradient, i.e.,
d = 1 2 ( l + l T ) {\displaystyle {\boldsymbol {d}}={\frac {1}{2}}~({\boldsymbol {l}}+{\boldsymbol {l}}^{T})} For rigid body motions we get d = 0 {\displaystyle {\boldsymbol {d}}={\boldsymbol {\mathit {0}}}} .
Lie derivatives
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Most of the operations above can be interpreted as push-forward
and pull-back operations. Also, time derivatives of these tensors can
be interpreted as Lie derivatives .
Recall that the push-forward of the strain tensor from the material
configuration to the spatial configuration is given by
e = ϕ ∗ [ E ] = F − T ⋅ E ⋅ F − 1 {\displaystyle {\boldsymbol {e}}=\phi _{*}[{\boldsymbol {E}}]={\boldsymbol {F}}^{-T}\cdot {\boldsymbol {E}}\cdot {\boldsymbol {F}}^{-1}} The pull-back of the spatial strain tensor to the material configuration
is given by
E = ϕ ∗ [ e ] = F T ⋅ e ⋅ F {\displaystyle {\boldsymbol {E}}=\phi ^{*}[{\boldsymbol {e}}]={\boldsymbol {F}}^{T}\cdot {\boldsymbol {e}}\cdot {\boldsymbol {F}}} Therefore, the rate of deformation tensor is a push-forward of the
material strain rate tensor, i.e.,
d = F − T ⋅ E ˙ ⋅ F − 1 = ϕ ∗ [ E ˙ ] {\displaystyle {\boldsymbol {d}}={\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {E}}}\cdot {\boldsymbol {F}}^{-1}=\phi _{*}[{\dot {\boldsymbol {E}}}]} Similarly, the material strain rate tensor is a pull-back of the rate
of deformation tensor to the material configuration, i.e.,
E ˙ = F T ⋅ d ⋅ F = ϕ ∗ [ d ] {\displaystyle {\dot {\boldsymbol {E}}}={\boldsymbol {F}}^{T}\cdot {\boldsymbol {d}}\cdot {\boldsymbol {F}}=\phi ^{*}[{\boldsymbol {d}}]} Now,
E = ϕ ∗ [ e ] ⟹ E ˙ = ∂ ∂ t ( ϕ ∗ [ e ] ) {\displaystyle {\boldsymbol {E}}=\phi ^{*}[{\boldsymbol {e}}]\quad \implies \quad {\dot {\boldsymbol {E}}}={\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {e}}]\right)} Also,
d = ϕ ∗ [ E ˙ ] = ϕ ∗ [ ∂ ∂ t ( ϕ ∗ [ e ] ) ] {\displaystyle {\boldsymbol {d}}=\phi _{*}[{\dot {\boldsymbol {E}}}]=\phi _{*}\left[{\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {e}}]\right)\right]} Therefore the rate of deformation tensor can be obtained by first pulling
back e {\displaystyle {\boldsymbol {e}}} to the reference configuration, taking a material time
derivative in that configuration, and then pushing forward the result to
the current configuration.
Such an operation is called a Lie derivative . In general, the Lie
derivative of a spatial tensor g {\displaystyle \mathbf {g} } is defined as
L ϕ [ g ] := ϕ ∗ [ ∂ ∂ t ( ϕ ∗ [ g ] ) ] . {\displaystyle {\mathcal {L}}_{\phi }[{\boldsymbol {g}}]:=\phi _{*}\left[{\frac {\partial }{\partial t}}\left(\phi ^{*}[{\boldsymbol {g}}]\right)\right]~.} Spin tensor
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The velocity gradient tensor can be additively decomposed into a symmetric
part and a skew part:
l = 1 2 ( l + l T ) + 1 2 ( l − l T ) = d + w {\displaystyle {\boldsymbol {l}}={\frac {1}{2}}~({\boldsymbol {l}}+{\boldsymbol {l}}^{T})+{\frac {1}{2}}({\boldsymbol {l}}-{\boldsymbol {l}}^{T})={\boldsymbol {d}}+{\boldsymbol {w}}} We have seen that d {\displaystyle {\boldsymbol {d}}} is the rate of deformation tensor. The quantity
w {\displaystyle {\boldsymbol {w}}} is called the spin tensor .
Note that d {\displaystyle {\boldsymbol {d}}} is symmetric while w {\displaystyle {\boldsymbol {w}}} is skew symmetric, i.e.,
d = d T ; w = − w T . {\displaystyle {\boldsymbol {d}}={\boldsymbol {d}}^{T}~;~~{\boldsymbol {w}}=-{\boldsymbol {w}}^{T}~.} So see why w {\displaystyle {\boldsymbol {w}}} is called a "spin", recall that
l = F ˙ ⋅ F − 1 {\displaystyle {\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}} Therefore,
w = 1 2 ( F ˙ ⋅ F − 1 − F − T ⋅ F ˙ T ) {\displaystyle {\boldsymbol {w}}={\frac {1}{2}}({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}-{\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {F}}}^{T})} Also,
F = R ⋅ U ⟹ F ˙ = R ˙ ⋅ U + R ⋅ U ˙ {\displaystyle {\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\quad \implies \quad {\dot {\boldsymbol {F}}}={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {U}}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}} Therefore,
F ˙ ⋅ F − 1 = ( R ˙ ⋅ U + R ⋅ U ˙ ) ⋅ ( U − 1 ⋅ R T ) = R ˙ ⋅ R T + R ⋅ U ˙ ⋅ U − 1 ⋅ R T {\displaystyle {\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}=({\dot {\boldsymbol {R}}}\cdot {\boldsymbol {U}}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}})\cdot ({\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T})={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T}} and
F − T ⋅ F ˙ T = ( R ⋅ U − 1 ) ⋅ ( U ⋅ R ˙ T + U ˙ ⋅ R T ) = R ⋅ R ˙ T + R ⋅ U − 1 ⋅ U ˙ ⋅ R T {\displaystyle {\boldsymbol {F}}^{-T}\cdot {\dot {\boldsymbol {F}}}^{T}=({\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1})\cdot ({\boldsymbol {U}}\cdot {\dot {\boldsymbol {R}}}^{T}+{\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T})={\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}+{\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T}} So we have
w = 1 2 ( R ˙ ⋅ R T + R ⋅ U ˙ ⋅ U − 1 ⋅ R T − R ⋅ R ˙ T − R ⋅ U − 1 ⋅ U ˙ ⋅ R T ) {\displaystyle {\boldsymbol {w}}={\frac {1}{2}}~({\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}\cdot {\boldsymbol {R}}^{T}-{\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}-{\boldsymbol {R}}\cdot {\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}}\cdot {\boldsymbol {R}}^{T})} Now
R ⋅ R T = 1 ⟹ R ˙ ⋅ R T + R ⋅ R ˙ T = 0 {\displaystyle {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}={\boldsymbol {\mathit {1}}}\quad \implies \quad {\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\boldsymbol {R}}\cdot {\dot {\boldsymbol {R}}}^{T}={\boldsymbol {\mathit {0}}}} Therefore
w = R ˙ ⋅ R T + 1 2 R ⋅ ( U ˙ ⋅ U − 1 − U − 1 ⋅ U ˙ ) ⋅ R T {\displaystyle {\boldsymbol {w}}={\dot {\boldsymbol {R}}}\cdot {\boldsymbol {R}}^{T}+{\frac {1}{2}}~{\boldsymbol {R}}\cdot ({\dot {\boldsymbol {U}}}\cdot {\boldsymbol {U}}^{-1}-{\boldsymbol {U}}^{-1}\cdot {\dot {\boldsymbol {U}}})\cdot {\boldsymbol {R}}^{T}} The second term above is invariant for rigid body motions and zero for an
uniaxial stretch. Hence, we are left with just a rotation term. This is
why the quantity w {\displaystyle {\boldsymbol {w}}} is called a spin.
The spin tensor is a skew-symmetric tensor and has an associated axial vector
ω {\displaystyle {\boldsymbol {\omega }}} (also called the angular velocity vector) whose components are
given by
ω = [ w 1 w 2 w 3 ] {\displaystyle {\boldsymbol {\omega }}={\begin{bmatrix}w_{1}\\w_{2}\\w_{3}\end{bmatrix}}} where
w = [ 0 − w 3 w 2 w 3 0 − w 1 − w 2 w 1 0 ] {\displaystyle \mathbf {w} ={\begin{bmatrix}0&-w_{3}&w_{2}\\w_{3}&0&-w_{1}\\-w_{2}&w_{1}&0\end{bmatrix}}} The spin tensor and its associated axial vector appear in a number of modern
numerical algorithms.
Rate of change of volume
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Recall that
d v = J d V where J = det F {\displaystyle dv=J~dV\qquad {\text{where}}~J=\det {\boldsymbol {F}}} Therefore, taking the material time derivative of d v {\displaystyle dv} (keeping X {\displaystyle \mathbf {X} } fixed),
we have
d d t ( d v ) = J ˙ d V = J ˙ J d v {\displaystyle {\cfrac {d}{dt}}(dv)={\dot {J}}~dV={\cfrac {\dot {J}}{J}}~dv} At this stage we invoke the following result from tensor calculus:
If A {\displaystyle {\boldsymbol {A}}} is an invertible tensor which depends on t {\displaystyle t} then
d d t ( det A ) = ( det A ) tr ( d A d t ⋅ A − 1 ) {\displaystyle {\cfrac {d}{dt}}(\det {\boldsymbol {A}})=(\det {\boldsymbol {A}})~{\text{tr}}\left({\cfrac {d{\boldsymbol {A}}}{dt}}\cdot {\boldsymbol {A}}^{-1}\right)}
In the case where A = F , J = det F {\displaystyle {\boldsymbol {A}}={\boldsymbol {F}},~J=\det {\boldsymbol {F}}} we have
d d t ( J ) = J tr ( F ˙ ⋅ F − 1 ) {\displaystyle {\cfrac {d}{dt}}(J)=J~{\text{tr}}\left({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}\right)} or,
J ˙ = J tr ( l ) = J tr ( d ) {\displaystyle {\dot {J}}=J~{\text{tr}}({\boldsymbol {l}})=J~{\text{tr}}(\mathbf {d} )} Therefore,
d d t ( d v ) = tr ( d ) d v {\displaystyle {\cfrac {d}{dt}}(dv)={\text{tr}}(\mathbf {d} )~dv} Alternatively, we can also write
J ˙ = 1 2 J C − 1 : C ˙ {\displaystyle {\dot {J}}={\frac {1}{2}}~J~{\boldsymbol {C}}^{-1}:{\dot {\boldsymbol {C}}}} These relations are of immense use in numerical algorithms - particularly
those which involved incompressible behavior, i.e., when J ˙ = 0 {\displaystyle {\dot {J}}=0} .