nonlinear solid mechanics - university of manchester · nonlinear solid mechanics andrew hazel....

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



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Introduction



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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 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


γij =1

2(Gij − gij) .


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


γij =1

2(Gij − gij) .


R(ξi ) = r(ξi ) + εu(ξi ),

⇒ γij = ε1

2

(gi ·

∂u

∂ξj+ gj ·

∂u

∂ξi

)

The strain tensor


γij =1

2(Gij − gij) .


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 ,

⇒ 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 .

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


RW =

∫∫t·R dS +

∫∫∫ (F− ρR

)·R dV ,


⇒ RW =

∫∫Ti ·R

ni√G

dS +

∫∫∫ (F− ρR

)·R dV ,

where Ti = ti

√(GG ii ) and G = det Gij .

Rate of Work


RW =

∫∫t·R dS +

∫∫∫ (F− ρR

)·R dV ,


⇒ 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


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


RW =

∫∫∫τ ijGj ·

∂R

∂ξidV

Rate of Work


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



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



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



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



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



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






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.






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 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.

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.

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).




u = 0

σxx = 0

σxy = 0

σxx = 0

σxy = 0

u = Key



I Impose a displacement on the top.




u = 0

σxx = 0

σxy = 0

σxx = 0

σxy = 0

σijnj = −pni



I Impose a pressure load on the top.




u = 0

σxx = 0

σxy = 0

σxx = 0

σxy = 0

σijnj = (0,−L)



I Impose a vertical load on the top.




u = 0

ux = 0

σxy = 0

ux = 0

σxy = 0

σijnj = (0,−L)


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

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.

nonlinear solid mechanics - university of manchester · nonlinear solid mechanics andrew hazel....

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