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Nonlinear Solid Mechanics
Andrew Hazel
Introduction
I Typically, want to determine the response of a solid body toan applied load.
I If a solid body is not rigid, then it can deform.
Introduction
I Typically, want to determine the response of a solid body toan applied load.
I If a solid body is not rigid, then it can deform.
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Introduction
I Typically, want to determine the response of a solid body toan applied load.
I If a solid body is not rigid, then it can deform.
@@��
I Question: How do we measure the (finite) deformation of adeformable body?
Lagrangian coordinates
I We parametrise the position vector to any material pointwithin the undeformed body, r by Lagrangian coordinates, ξi .
Tangent vector
gi = ∂r∂ξi
Lagrangian coordinates
I As the body deforms, the Lagrangian coordinates remain“attached” to the same material points in the body.
Tangent vector
gi = ∂r∂ξi
Deformed position, R(ξi )
Tangent vector
Gi = ∂R∂ξi
The metric tensor
I A vector line segment dr = gi dξi has length (squared)
(ds)2 = dr·dr = gi dξi ·gj dξj = gij dξi dξj ,
wheregij = gi ·gj ,
is called the (covariant) metric tensor.
I gij expresses information about the length of material lines inthe undeformed body.
I If the global and Lagrangian coordinates coincide gij = δij .
The metric tensor
I A vector line segment dr = gi dξi has length (squared)
(ds)2 = dr·dr = gi dξi ·gj dξj = gij dξi dξj ,
wheregij = gi ·gj ,
is called the (covariant) metric tensor.
I gij expresses information about the length of material lines inthe undeformed body.
I If the global and Lagrangian coordinates coincide gij = δij .
The strain tensor
I An objective measure of the deformation (strain) is given bythe Green strain tensor
γij =1
2(Gij − gij) .
I gij is the metric tensor of the undeformed configuration.
I Gij is the metric tensor of the deformed configuration.
I Decompose the deformed position into
R(ξi ) = r(ξi ) + εu(ξi ),
The strain tensor
I An objective measure of the deformation (strain) is given bythe Green strain tensor
γij =1
2(Gij − gij) .
I Decompose the deformed position into
R(ξi ) = r(ξi ) + εu(ξi ),
where u(ξi ) is the displacement field and ε� 1, then
Gi = gi+ε∂u
∂ξi⇒ Gij = gij+ε
(gi ·
∂u
∂ξj+ gj ·
∂u
∂ξi
)+O(ε2).
The strain tensor
I An objective measure of the deformation (strain) is given bythe Green strain tensor
γij =1
2(Gij − gij) .
I Decompose the deformed position into
R(ξi ) = r(ξi ) + εu(ξi ),
⇒ γij = ε1
2
(gi ·
∂u
∂ξj+ gj ·
∂u
∂ξi
)
The strain tensor
I An objective measure of the deformation (strain) is given bythe Green strain tensor
γij =1
2(Gij − gij) .
I Decompose the deformed position into
R(ξi ) = r(ξi ) + εu(ξi ),
⇒ γij = ε1
2
(gi ·
∂u
∂ξj+ gj ·
∂u
∂ξi
)I If ξi are chosen to be global Cartesian coordinates
γij ≈ ε1
2
(∂ui
∂xj+∂uj
∂xi
), the infinitessimal strain tensor.
Example
I Consider the simple radial expansion of a unit cube0 ≤ ξi = x i ≤ 1, with deformed position given by
R = 2 r = 2 x.
I Lagrangian coordinates are global Cartesian coordinates so
gi =∂ x
∂x i= ei (a unit vector) ⇒ gij = δij .
I The deformed metric tensor is
Gi =∂ 2x
∂x i= 2ei ⇒ Gij = 4δij .
I Hence the strain tensor is
γij =3
2δij ,
I Note that the infinitessimal strain tensor (with ε = 1) wouldgive γij ≈ δij
An aside: Non-Cartesian tensors
I Starting from the position vector as a function of theLagrangian coordinates r(ξi ), we found the tangent vectors
gi =∂r
∂ξi.
I For a general set of coordinates, ξi , these vectors are notnecessarily orthonormal,
gi ·gj 6=
{1 i = j
0 otherwise
I We define another set of vectors gj so that
gi ·gj = δji =
{1 i = j
0 otherwise
I The “up-down” index notation is used to simplify notation.
An aside: Non-Cartesian tensors
I We decompose r into global Cartesian base vectors
r =∑k
rkek ,
where ek is a unit vector in the k-th global Cartesian direction.
I Hence,
gi =∂rk
∂ξiek ,
I It follows that
gj =∂ξj
∂rnen,
... from which we deduce that
gi = gij gj and gj = g ji gi , where g ji = gj ·gi .
Forces and loads
I A deformable body is typically subject to surface loads (ortractions), t, and body forces, F.
I The stress vector t on a surface ∆S within the strained bodyis defined by
t = lim∆S→0
∆F
∆S,
where ∆F is the (statically equivalent) force acting on thesurface.
I t represents the force per unit area exerted by the materiallocated to one side of the surface on that to the other.
The stress tensor
I We consider a force balance on an infinitessimal tetrahedronin the deformed configuration.
I Three faces are aligned with the planes ξi = const andspanned by the other two covariant (lower index) tangentvectors Gj .
I The vector representation of the face ξi = const is
Gi∆Si
2√
G ii,
where Gi/√
(G ii ) is a unit vector normal to the face and thearea of the face is ∆Si/2.
I Note that we have had to use the contravariant (upper index)vector to ensure orthogonality.
The stress tensor
I Vector representation of the remaining face is n∆S/2 and so
n∆S =∑
i
Gi∆Si√(G ii )
.
I Decomposing the normal n = niGi , then
ni
√(G ii )∆S = ∆Si .
I If the tetrahedron is in equilibrium then
t∆S =∑
i
ti∆Si ,
The stress tensor
I Vector representation of the remaining face is n∆S/2 and so
n∆S =∑
i
Gi∆Si√(G ii )
.
I Decomposing the normal n = niGi , then
ni
√(G ii )∆S = ∆Si .
I If the tetrahedron is in equilibrium then
t∆S =∑
i
ti∆Si ,
The stress tensor
I Vector representation of the remaining face is n∆S/2 and so
n∆S =∑
i
Gi∆Si√(G ii )
.
I Decomposing the normal n = niGi , then
ni
√(G ii )∆S = ∆Si .
I If the tetrahedron is in equilibrium then
t∆S =∑
i
ti∆Si ,
⇒ t =∑
i
ni
√(G ii ) ti ,
The stress tensor
t =∑
i
ni
√(G ii )ti ,
I t is invariant if n remains the same (Cauchy’s principle).
I However, ni are components of a covariant vector so ti
√(G ii )
must be contravariant.
I In other words
ti
√(G ii ) = τ ijGj .
I The quantity τ ij is called the stress tensor.
I Physical components of the stress tensor are obtained byexpressing the stress vectors in terms of unit tangent vectors
ti =∑
j
σ(ij)Gj/√
(Gjj) ⇒ σ(ij) =√
(Gjj)/(G ii ) τ ij .
The stress tensor
t =∑
i
ni
√(G ii )ti ,
I t is invariant if n remains the same (Cauchy’s principle).
I However, ni are components of a covariant vector so ti
√(G ii )
must be contravariant.
I In other words
ti
√(G ii ) = τ ijGj .
I The quantity τ ij is called the stress tensor.
I Physical components of the stress tensor are obtained byexpressing the stress vectors in terms of unit tangent vectors
ti =∑
j
σ(ij)Gj/√
(Gjj) ⇒ σ(ij) =√
(Gjj)/(G ii ) τ ij .
Rate of Work
I In the strained body, the total rate of work is
RW =
∫∫t·R dS +
∫∫∫ (F− ρR
)·R dV ,
where the R is the velocity of the material and R is its theacceleration.
Rate of Work
I In the strained body, the total rate of work is
RW =
∫∫t·R dS +
∫∫∫ (F− ρR
)·R dV ,
where the R is the velocity of the material and R is its theacceleration.
⇒ RW =
∫∫Ti ·R
ni√G
dS +
∫∫∫ (F− ρR
)·R dV ,
where Ti = ti
√(GG ii ) and G = det Gij .
Rate of Work
I In the strained body, the total rate of work is
RW =
∫∫t·R dS +
∫∫∫ (F− ρR
)·R dV ,
where the R is the velocity of the material and R is its theacceleration.
⇒ RW =
∫∫Ti ·R
ni√G
dS +
∫∫∫ (F− ρR
)·R dV ,
where Ti = ti
√(GG ii ) and G = det Gij .
I Now use the divergence theorem (see Green & Zerna)∫∫aini dS =
∫∫∫1√G
∂
∂ξi(ai√
G ) dV .
Rate of Work
I The rate of work becomes
RW =
∫∫∫1√G
∂
∂ξi
(Ti · R
)+(
F− ρR)
·R dV ,
Rate of Work
I The rate of work becomes
RW =
∫∫∫1√G
∂
∂ξi
(Ti · R
)+(
F− ρR)
·R dV ,
=
∫∫∫1√G
Ti ·∂R
∂ξi+
1√G
[∂Ti
∂ξi+√
G(
F− ρR)]
·R dV ,
Rate of Work
I The rate of work becomes
RW =
∫∫∫τ ijGj ·
∂R
∂ξidV
Rate of Work
I The rate of work becomes
RW =
∫∫∫τ ijGj ·
∂R
∂ξidV
I By symmetry of the stress tensor
RW =1
2
∫∫∫τ ij
(Gj ·
∂R
∂ξi+ Gi ·
∂R
∂ξj
)dV
Rate of Work
I The rate of work becomes
I By symmetry of the stress tensor
RW =1
2
∫∫∫τ ij
(Gj ·
∂R
∂ξi+ Gi ·
∂R
∂ξj
)dV
I But recall
γij =1
2(Gij − gij) ,
so
γij =1
2
∂
∂t(Gi ·Gj) =
1
2
(Gi ·
∂R
∂ξj+∂R
∂ξi·Gj
).
I Hence
RW =
∫∫∫τ ij γij dV .
Rate of Work
I The rate of work becomes
I By symmetry of the stress tensor
RW =1
2
∫∫∫τ ij
(Gj ·
∂R
∂ξi+ Gi ·
∂R
∂ξj
)dV
I But recall
γij =1
2(Gij − gij) ,
so
γij =1
2
∂
∂t(Gi ·Gj) =
1
2
(Gi ·
∂R
∂ξj+∂R
∂ξi·Gj
).
I Hence
RW =
∫∫∫τ ij γij dV .
I τ ij and γij are a work conjugate pair.
The principle of virtual diplacements
I Consider a deformable body that is load by a surface tractionT and a body force F
I The body is subject to a virtual displacement δR
The principle of virtual diplacements
I Consider a deformable body that is load by a surface tractionT and a body force F
I The body is subject to a virtual displacement δR
I The virtual work induced by the displacement is∫∫∫ (F− ρ∂
2R
∂t2
)·δR dV +
∫∫T·δR dS
The principle of virtual diplacements
I Consider a deformable body that is load by a surface tractionT and a body force F
I The body is subject to a virtual displacement δRI The virtual work induced by the displacement is∫∫∫ (
F− ρ∂2R
∂t2
)·δR dV +
∫∫T·δR dS
which must be balanced by a change of internal energy∫∫∫τ ijδγij dV
The principle of virtual diplacements
I Consider a deformable body that is load by a surface tractionT and a body force F
I The governing variational principle is∫∫∫τ ijδγij dV−
∫∫∫ (F− ρ∂
2R
∂t2
)·δR dV−
∫∫T·δR dS = 0.
The principle of virtual displacements
I The variation of the strain is given by
δγij =1
2
(Gi ·δ
∂R
∂ξj+ δ
∂R
∂ξi·Gj
).
I Using the symmetry of the stress tensor we can write thevariational principle as∫∫∫
τ ij ∂R
∂ξi·δ ∂R
∂ξj−(
F− ρ∂2R
∂t2
)·δR dV −
∫∫T·δR dS = 0.
Working with the variational principle
I The integral is over the deformed domain, which is unknown.
I It is much more convenient to integrate over the undeformeddomain.
I We already know the mapping from undeformed to deformeddomain R(ξi ).
I The Jacobian of the mapping is√
G/g .
I Hence dV =√
G/g dV0 and
Working with the variational principle
I The integral is over the deformed domain, which is unknown.
I It is much more convenient to integrate over the undeformeddomain.
I We already know the mapping from undeformed to deformeddomain R(ξi ).
I The Jacobian of the mapping is√
G/g .
I Hence dV =√
G/g dV0 and∫∫∫ √G
gτ ij ∂R
∂ξi·δ ∂R
∂ξj−
√G
g
(F− ρ∂
2R
∂t2
)·δR dV0−
∫∫T·δR dS = 0.
Working with the variational principle
I The integral is over the deformed domain, which is unknown.
I It is much more convenient to integrate over the undeformeddomain.
I We already know the mapping from undeformed to deformeddomain R(ξi ).
I The Jacobian of the mapping is√
G/g .
I Hence dV =√
G/g dV0 and∫∫∫σij ∂R
∂ξi·δ ∂R
∂ξj−(
f − ρ0∂2R
∂t2
)·δR dV0 −
∫∫T·δR dS = 0.
where σij =√
G/g τ ij is the second Piola–Kirchhoff stress tensor,f is the body force per unit undeformed volume,ρ0 is the undeformed density.
Finite element approximation of the Largange coordinates
I The undeformed position is given by
r(ξi ) = rk(ξi )ek .
I If there are no special symmetries, choose Lagrangiancoordinates as the global Cartesian coordinates.
I Then components of undeformed position are r i (ξi ) = ξi .
I The tangent vectors are Cartesian base vectors
gi = ei
I Undeformed metric tensor is Kronecker delta gij = δij .
I Approximate the Lagrangian coordinates by a finite elementbasis
ξi =∑
l
ξilψl ,
ξil is the i-th Lagrangian coordinate at the l-th node.
Finite element approximation of the Largange coordinates
I The undeformed position is given by
r(ξi ) = rk(ξi )ek .
I If there are no special symmetries, choose Lagrangiancoordinates as the global Cartesian coordinates.
I Then components of undeformed position are r i (ξi ) = ξi .
I The tangent vectors are Cartesian base vectors
gi = ei
I Undeformed metric tensor is Kronecker delta gij = δij .
I Approximate the Lagrangian coordinates by a finite elementbasis
ξi =∑
l
ξilψl ,
ξil is the i-th Lagrangian coordinate at the l-th node.
Finite element approximation of the Largange coordinates
I The undeformed position is given by
r(ξi ) = rk(ξi )ek .
I If there are no special symmetries, choose Lagrangiancoordinates as the global Cartesian coordinates.
I Then components of undeformed position are r i (ξi ) = ξi .
I The tangent vectors are Cartesian base vectors
gi = ei
I Undeformed metric tensor is Kronecker delta gij = δij .
I Approximate the Lagrangian coordinates by a finite elementbasis
ξi =∑
l
ξilψl ,
ξil is the i-th Lagrangian coordinate at the l-th node.
Finite element approximation of the Largange coordinates
I The undeformed position is given by
r(ξi ) = rk(ξi )ek .
I If there are no special symmetries, choose Lagrangiancoordinates as the global Cartesian coordinates.
I Then components of undeformed position are r i (ξi ) = ξi .
I The tangent vectors are Cartesian base vectors
gi = ei
I Undeformed metric tensor is Kronecker delta gij = δij .
I Approximate the Lagrangian coordinates by a finite elementbasis
ξi =∑
l
ξilψl ,
ξil is the i-th Lagrangian coordinate at the l-th node.
Finite element approximation of the Largange coordinates
I The undeformed position is given by
r(ξi ) = rk(ξi )ek .
I If there are no special symmetries, choose Lagrangiancoordinates as the global Cartesian coordinates.
I Then components of undeformed position are r i (ξi ) = ξi .
I The tangent vectors are Cartesian base vectors
gi = ei
I Undeformed metric tensor is Kronecker delta gij = δij .
I Approximate the Lagrangian coordinates by a finite elementbasis
ξi =∑
l
ξilψl ,
ξil is the i-th Lagrangian coordinate at the l-th node.
Finite element approximation of the variational principle
I Use the same basis function for the unkown positions(isoparametric mapping)
Rk =∑
l
Rkl ψl .
I The basis functions are fixed so
δR =∑
l
δRkl ψlek and δ
∂R
∂ξj=∑
l
δRkl
∂ψl
∂ξjek .
I Thus the principle of virtual displacements becomes∑l
∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ∂2Rk
∂t2
)ψl
]δRk
l dξ1 dξ2 dξ3
−∫∫
TkψlδRkl dS = 0.
Finite element approximation of the variational principle
I Use the same basis function for the unkown positions(isoparametric mapping)
Rk =∑
l
Rkl ψl .
I The basis functions are fixed so
δR =∑
l
δRkl ψlek and δ
∂R
∂ξj=∑
l
δRkl
∂ψl
∂ξjek .
I Thus the principle of virtual displacements becomes∑l
∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ∂2Rk
∂t2
)ψl
]δRk
l dξ1 dξ2 dξ3
−∫∫
TkψlδRkl dS = 0.
Finite element approximation of the variational principle
I The discrete variations may be taken outside the integrals
∑l
{∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
−∫∫
[Tkψl ] dS
}δ Rk
l = 0.
I The variations of the nodes are independent, so the terms inbraces give one discrete equation for each nodal unknown.∫∫∫ [
σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
−∫∫
[Tkψl ] dS = 0.
I These may be assembled in an element-by-element manner.
Finite element approximation of the variational principle
I The discrete variations may be taken outside the integrals
∑l
{∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
−∫∫
[Tkψl ] dS
}δ Rk
l = 0.
I The variations of the nodes are independent, so the terms inbraces give one discrete equation for each nodal unknown.∫∫∫ [
σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
−∫∫
[Tkψl ] dS = 0.
I These may be assembled in an element-by-element manner.
Summary of the method
I Divide the undeformed domain into elements.
I For each element compute the contribution to the discretevolume residual
Rkl =
∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
I Note that the integral is over the Lagrangian coordinates(undeformed domain).
I Loop over the surfaces to add any tractions∫∫[Tkψl ] dS .
I Note that this integral is over the deformed surface
I Assemble the contributions into a global residuals vector.
I Compute the Jacobian (by finite differences if necessary).
I Solve the linear system.
Summary of the method
I Divide the undeformed domain into elements.
I For each element compute the contribution to the discretevolume residual
Rkl =
∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
I Note that the integral is over the Lagrangian coordinates(undeformed domain).
I Loop over the surfaces to add any tractions∫∫[Tkψl ] dS .
I Note that this integral is over the deformed surface
I Assemble the contributions into a global residuals vector.
I Compute the Jacobian (by finite differences if necessary).
I Solve the linear system.
Summary of the method
I Divide the undeformed domain into elements.
I For each element compute the contribution to the discretevolume residual
Rkl =
∫∫∫ [σij ∂Rk
∂ξi∂ψl
∂ξj−(
f k − ρ0∂2Rk
∂t2
)ψl
]dξ1 dξ2 dξ3
I Note that the integral is over the Lagrangian coordinates(undeformed domain).
I Loop over the surfaces to add any tractions∫∫[Tkψl ] dS .
I Note that this integral is over the deformed surface
I Assemble the contributions into a global residuals vector.
I Compute the Jacobian (by finite differences if necessary).
I Solve the linear system.
Constitutive Laws
I Assembling the residuals requires knowledge of the stresstensor.
I For an elastic material, the stress depends only on the currentstate of strain
σij(γjk).
I The specific relationship between stress and strain is known asa constitutive law.
I Given a constitutive law and a compressible material then wesimply compute
σij
(1
2(Gij − gij)
),
at all the integration points within the element.
I What about incompressible materials?
Incompressible Solid Mechanics
I If a solid material is incompressible, its volume cannot change
det Gij = det gij (1)
I Enforce the condition (1) by a Lagrange multiplier that playsthe role of a pressure so that
σij = −p G ij + σij(γkl),
where σij is the deviatoric part of the stress tensor.
I The resulting linear system is again of saddle-point type andso the combinations of pressure/velocity finite elementsappropriate for Navier–Stokes equations can be reused.
I If material is “nearly” incompressible it is also advantageousto use a mixed formulation.
Incompressible Solid Mechanics
I If a solid material is incompressible, its volume cannot change
det Gij = det gij (1)
I Enforce the condition (1) by a Lagrange multiplier that playsthe role of a pressure so that
σij = −p G ij + σij(γkl),
where σij is the deviatoric part of the stress tensor.
I The resulting linear system is again of saddle-point type andso the combinations of pressure/velocity finite elementsappropriate for Navier–Stokes equations can be reused.
I If material is “nearly” incompressible it is also advantageousto use a mixed formulation.
Incompressible Solid Mechanics
I If a solid material is incompressible, its volume cannot change
det Gij = det gij (1)
I Enforce the condition (1) by a Lagrange multiplier that playsthe role of a pressure so that
σij = −p G ij + σij(γkl),
where σij is the deviatoric part of the stress tensor.
I The resulting linear system is again of saddle-point type andso the combinations of pressure/velocity finite elementsappropriate for Navier–Stokes equations can be reused.
I If material is “nearly” incompressible it is also advantageousto use a mixed formulation.
Example problem: Compression of a block
I A square block of material is compressed.
I What are appropriate boundary conditions?
u = 0
σxx = 0
σxy = 0
σxx = 0
σxy = 0
I Fix the base vertically (and horizontally?)
I Leave the sides traction free (do nothing).
Example problem: Compression of a block
I A square block of material is compressed.
I What are appropriate boundary conditions?
u = 0
σxx = 0
σxy = 0
σxx = 0
σxy = 0
u = Key
I Fix the base vertically (and horizontally?)
I Leave the sides traction free (do nothing).
I Impose a displacement on the top.
Example problem: Compression of a block
I A square block of material is compressed.
I What are appropriate boundary conditions?
u = 0
σxx = 0
σxy = 0
σxx = 0
σxy = 0
σijnj = −pni
I Fix the base vertically (and horizontally?)
I Leave the sides traction free (do nothing).
I Impose a pressure load on the top.
Example problem: Compression of a block
I A square block of material is compressed.
I What are appropriate boundary conditions?
u = 0
σxx = 0
σxy = 0
σxx = 0
σxy = 0
σijnj = (0,−L)
I Fix the base vertically (and horizontally?)
I Leave the sides traction free (do nothing).
I Impose a vertical load on the top.
Example problem: Compression of a block
I A square block of material is compressed.
I What are appropriate boundary conditions?
u = 0
ux = 0
σxy = 0
ux = 0
σxy = 0
σijnj = (0,−L)
I Fix the base vertically (and horizontally?)
I Constrain sides horizontally
I Impose a vertical load on the top.
I Incompressible material?
Example problem: Compression of block
I Incompressible Mooney-Rivlin material
I Loaded on top by P cos(x)N
Example problem: Compression of block
I Incompressible Mooney-Rivlin material
I Loaded on top by P cos(x)N
Example problem: Compression of block
I Incompressible Mooney-Rivlin material
I Loaded on top by P cos(x)N
Example problem: Compression of block
I Incompressible Mooney-Rivlin material
I Loaded on top by P cos(x)N
Example problem: Compression of block
I Incompressible Mooney-Rivlin material
I Loaded on top by P cos(x)N
Example problem: Compression of block
I Incompressible Mooney-Rivlin material
I Loaded on top by P cos(x)N
ux = 0, σxy = 0
uy = 0, σxy = 0
σxx = 0
σxy = 0
Summary of the method
I Choose type of element (pick an LBB stable one ifincompressible problem).
I Choose a timestepper (must compute second derivatives).
I Generate mesh in undeformed solid domain.I Specify boundary and initial conditions
I For Dirichlet conditions, replace the discrete weak form.I For traction conditions, assemble the surface integral.
I Loop over elements and assemble the global residuals andJacobian matrix for each time step
I Solve the (non)linear residual equations using Newton’smethod.
I Repeat for as many timesteps as desired.