3d finite element method
TRANSCRIPT
Finite element method
U. Saravanan
Associate ProfessorDepartment of Civil Engineering
IIT Madras
April 2014
U. Saravanan (IIT Madras) Finite element method April 2014 1 / 81
Outline
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 2 / 81
Outline
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 2 / 81
Finite element formulation
Solution Procedure
Applied forces Motion / displacement
Stress Deformation gradient or strain
Constitutive relation
Equilibrium equations
Strain displacement relation
U. Saravanan (IIT Madras) Finite element method April 2014 3 / 81
Finite element formulation
Basic equations in mechanics
Strain-displacement relation
h = grad(u), ε =1
2[h + ht ] (1)
Constitutive relation - Isotropic Hooke’s law:
σ = λtr(ε)1 + 2µε = Dε. (2)
where
D =
λ+ 2µ λ λ 0 0 0λ λ+ 2µ λ 0 0 0λ λ λ+ 2µ 0 0 00 0 0 2µ 0 00 0 0 0 2µ 00 0 0 0 0 2µ
, (3)
U. Saravanan (IIT Madras) Finite element method April 2014 4 / 81
Finite element formulation
Basic equations in mechanics
Strain-displacement relation
h = grad(u), ε =1
2[h + ht ] (1)
Constitutive relation - Isotropic Hooke’s law:
σ = λtr(ε)1 + 2µε = Dε. (2)
where
D =
λ+ 2µ λ λ 0 0 0λ λ+ 2µ λ 0 0 0λ λ λ+ 2µ 0 0 00 0 0 2µ 0 00 0 0 0 2µ 00 0 0 0 0 2µ
, (3)
U. Saravanan (IIT Madras) Finite element method April 2014 4 / 81
Finite element formulation
Basic equations in mechanics
Equilibrium Equations - Spatial form
div(σ) + ρb = ρa. (4)
U. Saravanan (IIT Madras) Finite element method April 2014 5 / 81
Finite element formulation
Initial-Boundary value problem - Spatial formulation
To find a differentiable displacement field, u so that:
div(σ) + ρb = ρa,u(x, t) |t=0 = uo ,
DuDt |t=0 = vo ,
, on At (5)
u = u, on ∂Aut , (6)
t(n) = σn = t, on ∂Aσt . (7)
where ∂Aut and ∂Aσ
t denote the boundary of the body where displacementand traction are specified respectively and is such that
∂Aut ∩ ∂Aσ
t = ∅, and ∂Aut ∪ ∂Aσ
t = ∂At (8)
U. Saravanan (IIT Madras) Finite element method April 2014 6 / 81
Finite element formulation
Displacement field
u = υ0(x) +n∑
i=1
ci (t)υi (x) (9)
where, υi ’s are specified a priori and ci ’s are to be determined.
Then, the displacement gradient is:
h = grad(υ0) +n∑
i=1
cigrad(υi ) (10)
U. Saravanan (IIT Madras) Finite element method April 2014 7 / 81
Finite element formulation
Desirable properties of υ0
υ0 = υ0(x)
Is a differentiable function
Ensures that the displacement boundary condition is met, i.e.,
υ0(x) = u, ∀ x ∈ ∂Aut . (11)
U. Saravanan (IIT Madras) Finite element method April 2014 8 / 81
Finite element formulation
Desirable properties of υ0
υ0 = υ0(x)
Is a differentiable function
Ensures that the displacement boundary condition is met, i.e.,
υ0(x) = u, ∀ x ∈ ∂Aut . (11)
U. Saravanan (IIT Madras) Finite element method April 2014 8 / 81
Finite element formulation
Desirable properties of υi , i 6= 0
Basis Functions:
υi = υi (x)
Satisfy homogeneous displacement boundary condition
υi (x) = o, ∀ x ∈ ∂Aut , (12)
υi ’s to be differentiable functions
The set {υi} to be complete.
The set {υi} to be linearly independent.
U. Saravanan (IIT Madras) Finite element method April 2014 9 / 81
Finite element formulation
Desirable properties of υi , i 6= 0
Basis Functions:
υi = υi (x)
Satisfy homogeneous displacement boundary condition
υi (x) = o, ∀ x ∈ ∂Aut , (12)
υi ’s to be differentiable functions
The set {υi} to be complete.
The set {υi} to be linearly independent.
U. Saravanan (IIT Madras) Finite element method April 2014 9 / 81
Finite element formulation
Desirable properties of υi , i 6= 0
Basis Functions:
υi = υi (x)
Satisfy homogeneous displacement boundary condition
υi (x) = o, ∀ x ∈ ∂Aut , (12)
υi ’s to be differentiable functions
The set {υi} to be complete.
The set {υi} to be linearly independent.
U. Saravanan (IIT Madras) Finite element method April 2014 9 / 81
Finite element formulation
Desirable properties of υi , i 6= 0
Basis Functions:
υi = υi (x)
Satisfy homogeneous displacement boundary condition
υi (x) = o, ∀ x ∈ ∂Aut , (12)
υi ’s to be differentiable functions
The set {υi} to be complete.
The set {υi} to be linearly independent.
U. Saravanan (IIT Madras) Finite element method April 2014 9 / 81
Finite element formulation
Desirable properties of υi , i 6= 0
Basis Functions:
υi = υi (x)
Satisfy homogeneous displacement boundary condition
υi (x) = o, ∀ x ∈ ∂Aut , (12)
υi ’s to be differentiable functions
The set {υi} to be complete.
The set {υi} to be linearly independent.
U. Saravanan (IIT Madras) Finite element method April 2014 9 / 81
Finite element formulation
Weakened Equilibrium equation - Spatial form
Let the weight function, φ = φ(x), then∫At
[div(σ) + ρb− ρa] · φdv = 0 (13)
Recollecting:
div(Atu) = div(A) · u + A · grad(u), and
∫At
div(u)dv =
∫∂At
u · nds
(14)
∫At
[div(σtφ)− σ · grad(φ) + ρb · φ− ρa · φ
]dv = 0 (15)
U. Saravanan (IIT Madras) Finite element method April 2014 10 / 81
Finite element formulation
Weakened Equilibrium equation
Hence weakened equilibrium equation in spatial form
∫At
[σ · grad(φ) + ρa · φ] dv =
∫At
ρb · φdv +
∫∂At
t(n) · φds (16)
∫At
[σ · εφ + ρa · φ
]dv =
∫At
ρb · φdv +
∫∂At
t(n) · φds (17)
U. Saravanan (IIT Madras) Finite element method April 2014 11 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Weighted residual methods
How do we find the unknown ci ’s?
Such that the weakened equilibrium equations are satisfied
Galerkin method:
φ = υi , for i = {1, . . . , n}
Petrov-Galerkin method:
φ = $i , for i = {1, . . . , n}, $i 6= υi
Properties of $i
Be a differentiable function
Satisfy homogeneous boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 12 / 81
Finite element formulation
Why these restrictions on the basis and weight functions?
We require υ0 to satisfy the prescribed boundary conditions and υi i∈ {1, . . . , n} to satisfy the homogeneous boundary conditions so thatthe value of ci ’s are independent of the boundary condition
The restriction on weight functions is so that the following term can becomputed independent of ci ’s∫
∂At
t(n) · φds =
∫∂Au
t
t(n) · φds +
∫∂Aσ
t
t(n) · φds =
∫∂Aσ
t
t · φds, (18)
by virtue of the required form of the boundary condition and therequirement on the weight function, φ
U. Saravanan (IIT Madras) Finite element method April 2014 13 / 81
Finite element formulation
Why these restrictions on the basis and weight functions?
We require υ0 to satisfy the prescribed boundary conditions and υi i∈ {1, . . . , n} to satisfy the homogeneous boundary conditions so thatthe value of ci ’s are independent of the boundary condition
The restriction on weight functions is so that the following term can becomputed independent of ci ’s∫
∂At
t(n) · φds =
∫∂Au
t
t(n) · φds +
∫∂Aσ
t
t(n) · φds =
∫∂Aσ
t
t · φds, (18)
by virtue of the required form of the boundary condition and therequirement on the weight function, φ
U. Saravanan (IIT Madras) Finite element method April 2014 13 / 81
Finite element formulation Six steps in finite element analysis
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 14 / 81
Finite element formulation Six steps in finite element analysis
Introduction
What does the finite element method offer?
The finite element method provides a systematic way of generatingthe basis functions
The finite element analysis of a typical problem involves six stepswhich we shall list next
U. Saravanan (IIT Madras) Finite element method April 2014 15 / 81
Finite element formulation Six steps in finite element analysis
Introduction
What does the finite element method offer?
The finite element method provides a systematic way of generatingthe basis functions
The finite element analysis of a typical problem involves six stepswhich we shall list next
U. Saravanan (IIT Madras) Finite element method April 2014 15 / 81
Finite element formulation Six steps in finite element analysis
Introduction
What does the finite element method offer?
The finite element method provides a systematic way of generatingthe basis functions
The finite element analysis of a typical problem involves six stepswhich we shall list next
U. Saravanan (IIT Madras) Finite element method April 2014 15 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 1: Representation of the given domain into a collection ofpreselected finite elements
a. Construct the finite element mesh of preselected elements
b. Number the nodes and elements
c. Generate the geometric properties needed for the problem
U. Saravanan (IIT Madras) Finite element method April 2014 16 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 1: Representation of the given domain into a collection ofpreselected finite elements
a. Construct the finite element mesh of preselected elements
b. Number the nodes and elements
c. Generate the geometric properties needed for the problem
U. Saravanan (IIT Madras) Finite element method April 2014 16 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 1: Representation of the given domain into a collection ofpreselected finite elements
a. Construct the finite element mesh of preselected elements
b. Number the nodes and elements
c. Generate the geometric properties needed for the problem
U. Saravanan (IIT Madras) Finite element method April 2014 16 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 1: Representation of the given domain into a collection ofpreselected finite elements
a. Construct the finite element mesh of preselected elements
b. Number the nodes and elements
c. Generate the geometric properties needed for the problem
U. Saravanan (IIT Madras) Finite element method April 2014 16 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 2: Derivation of element equations for all typical elements in themesh
a. Consider the weighted residual formulation of the balance of linearmomentum equation over the typical element
b. Assume that the displacement is of the form,
u =n∑
i=1
uiυi ,
and substitute it into step 2a to obtain element equation of the form
[Ke ]{ue} = {Fe} (19)
c. Derive element interpolation functions, υi and compute the elementmatrices.
U. Saravanan (IIT Madras) Finite element method April 2014 17 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 2: Derivation of element equations for all typical elements in themesh
a. Consider the weighted residual formulation of the balance of linearmomentum equation over the typical element
b. Assume that the displacement is of the form,
u =n∑
i=1
uiυi ,
and substitute it into step 2a to obtain element equation of the form
[Ke ]{ue} = {Fe} (19)
c. Derive element interpolation functions, υi and compute the elementmatrices.
U. Saravanan (IIT Madras) Finite element method April 2014 17 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 2: Derivation of element equations for all typical elements in themesh
a. Consider the weighted residual formulation of the balance of linearmomentum equation over the typical element
b. Assume that the displacement is of the form,
u =n∑
i=1
uiυi ,
and substitute it into step 2a to obtain element equation of the form
[Ke ]{ue} = {Fe} (19)
c. Derive element interpolation functions, υi and compute the elementmatrices.
U. Saravanan (IIT Madras) Finite element method April 2014 17 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 2: Derivation of element equations for all typical elements in themesh
a. Consider the weighted residual formulation of the balance of linearmomentum equation over the typical element
b. Assume that the displacement is of the form,
u =n∑
i=1
uiυi ,
and substitute it into step 2a to obtain element equation of the form
[Ke ]{ue} = {Fe} (19)
c. Derive element interpolation functions, υi and compute the elementmatrices.
U. Saravanan (IIT Madras) Finite element method April 2014 17 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 3: Assembly of element equations to obtain the equations of thewhole problem
a. Identify the inter-element continuity conditions on the displacementby relating element nodes to global nodes
b. Assemble element equations using 3a. and equilibrium equations.
Step - 4: Imposition of the boundary conditions of the problem
a. Identify the specified global displacement degrees of freedom
b. Identify the specified traction
U. Saravanan (IIT Madras) Finite element method April 2014 18 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 3: Assembly of element equations to obtain the equations of thewhole problem
a. Identify the inter-element continuity conditions on the displacementby relating element nodes to global nodes
b. Assemble element equations using 3a. and equilibrium equations.
Step - 4: Imposition of the boundary conditions of the problem
a. Identify the specified global displacement degrees of freedom
b. Identify the specified traction
U. Saravanan (IIT Madras) Finite element method April 2014 18 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 3: Assembly of element equations to obtain the equations of thewhole problem
a. Identify the inter-element continuity conditions on the displacementby relating element nodes to global nodes
b. Assemble element equations using 3a. and equilibrium equations.
Step - 4: Imposition of the boundary conditions of the problem
a. Identify the specified global displacement degrees of freedom
b. Identify the specified traction
U. Saravanan (IIT Madras) Finite element method April 2014 18 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 3: Assembly of element equations to obtain the equations of thewhole problem
a. Identify the inter-element continuity conditions on the displacementby relating element nodes to global nodes
b. Assemble element equations using 3a. and equilibrium equations.
Step - 4: Imposition of the boundary conditions of the problem
a. Identify the specified global displacement degrees of freedom
b. Identify the specified traction
U. Saravanan (IIT Madras) Finite element method April 2014 18 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 3: Assembly of element equations to obtain the equations of thewhole problem
a. Identify the inter-element continuity conditions on the displacementby relating element nodes to global nodes
b. Assemble element equations using 3a. and equilibrium equations.
Step - 4: Imposition of the boundary conditions of the problem
a. Identify the specified global displacement degrees of freedom
b. Identify the specified traction
U. Saravanan (IIT Madras) Finite element method April 2014 18 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 3: Assembly of element equations to obtain the equations of thewhole problem
a. Identify the inter-element continuity conditions on the displacementby relating element nodes to global nodes
b. Assemble element equations using 3a. and equilibrium equations.
Step - 4: Imposition of the boundary conditions of the problem
a. Identify the specified global displacement degrees of freedom
b. Identify the specified traction
U. Saravanan (IIT Madras) Finite element method April 2014 18 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 5: Solution of the assembled equations
Step - 6: Postprocessing of the results
a. Compute the gradient of the determined displacement field so thatthe stresses can be computed
b. Represent the results in a tabular and/or graphical form.
U. Saravanan (IIT Madras) Finite element method April 2014 19 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 5: Solution of the assembled equations
Step - 6: Postprocessing of the results
a. Compute the gradient of the determined displacement field so thatthe stresses can be computed
b. Represent the results in a tabular and/or graphical form.
U. Saravanan (IIT Madras) Finite element method April 2014 19 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 5: Solution of the assembled equations
Step - 6: Postprocessing of the results
a. Compute the gradient of the determined displacement field so thatthe stresses can be computed
b. Represent the results in a tabular and/or graphical form.
U. Saravanan (IIT Madras) Finite element method April 2014 19 / 81
Finite element formulation Six steps in finite element analysis
Steps involved in the finite element analysis
Step - 5: Solution of the assembled equations
Step - 6: Postprocessing of the results
a. Compute the gradient of the determined displacement field so thatthe stresses can be computed
b. Represent the results in a tabular and/or graphical form.
U. Saravanan (IIT Madras) Finite element method April 2014 19 / 81
Finite element formulation for linearized elasticity
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 20 / 81
Finite element formulation for linearized elasticity Decomposition of the domain - Selection of element
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 21 / 81
Finite element formulation for linearized elasticity Decomposition of the domain - Selection of element
Elements
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
At the end of thisstep, the (x , y , z)coordinates ofthe respectivenodes as well asthe connectivitymatrix containingthe elementnumber and thenodes thatconstitute thiselement areassumed to beavailable.
U. Saravanan (IIT Madras) Finite element method April 2014 22 / 81
Finite element formulation for linearized elasticity Decomposition of the domain - Selection of element
Elements
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
At the end of thisstep, the (x , y , z)coordinates ofthe respectivenodes as well asthe connectivitymatrix containingthe elementnumber and thenodes thatconstitute thiselement areassumed to beavailable.
U. Saravanan (IIT Madras) Finite element method April 2014 22 / 81
Finite element formulation for linearized elasticity Decomposition of the domain - Selection of element
Elements
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
At the end of thisstep, the (x , y , z)coordinates ofthe respectivenodes as well asthe connectivitymatrix containingthe elementnumber and thenodes thatconstitute thiselement areassumed to beavailable.
U. Saravanan (IIT Madras) Finite element method April 2014 22 / 81
Finite element formulation for linearized elasticity Decomposition of the domain - Selection of element
Elements
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
At the end of thisstep, the (x , y , z)coordinates ofthe respectivenodes as well asthe connectivitymatrix containingthe elementnumber and thenodes thatconstitute thiselement areassumed to beavailable.
U. Saravanan (IIT Madras) Finite element method April 2014 22 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 23 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Degrees of freedom
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
{d} =
u1
v1
w1...u4
v4
w4
(20)
U. Saravanan (IIT Madras) Finite element method April 2014 24 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Degrees of freedom
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
{d} =
u1
v1
w1...u4
v4
w4
(20)
U. Saravanan (IIT Madras) Finite element method April 2014 24 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Degrees of freedom
ex
ey
ez
4
1
2
3
Figure: Tetrahedral solid element
{d} =
u1
v1
w1...u4
v4
w4
(20)
U. Saravanan (IIT Madras) Finite element method April 2014 24 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Interpolation functions
Since there are 12 degrees of freedom, there should be 12 interpolationfunctions, υi , one corresponding to each degree of freedom, so that
u = uex + vey + wez =12∑i=1
diυi (21)
U. Saravanan (IIT Madras) Finite element method April 2014 25 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Interpolation functions
To maintain the meaning of the degrees of freedom, we choose theinterpolation functions as
υ1 = N1(x , y , z)ex , υ2 = N1(x , y , z)ey , υ3 = N1(x , y , z)ez
υ4 = N2(x , y , z)ex , υ5 = N2(x , y , z)ey , υ6 = N2(x , y , z)ez
υ7 = N3(x , y , z)ex , υ8 = N3(x , y , z)ey , υ9 = N3(x , y , z)ez
υ10 = N4(x , y , z)ex , υ11 = N4(x , y , z)ey , υ12 = N4(x , y , z)ez
where Ni ’s are the Lagrange interpolation functions and are such that
Ni (xj , yj , zj) = δij , (22)
where (xj , yj , zj) are the coordinates of the four node points of thetetrahedron.
U. Saravanan (IIT Madras) Finite element method April 2014 26 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Interpolation functions
To maintain the meaning of the degrees of freedom, we choose theinterpolation functions as
υ1 = N1(x , y , z)ex , υ2 = N1(x , y , z)ey , υ3 = N1(x , y , z)ez
υ4 = N2(x , y , z)ex , υ5 = N2(x , y , z)ey , υ6 = N2(x , y , z)ez
υ7 = N3(x , y , z)ex , υ8 = N3(x , y , z)ey , υ9 = N3(x , y , z)ez
υ10 = N4(x , y , z)ex , υ11 = N4(x , y , z)ey , υ12 = N4(x , y , z)ez
where Ni ’s are the Lagrange interpolation functions and are such that
Ni (xj , yj , zj) = δij , (22)
where (xj , yj , zj) are the coordinates of the four node points of thetetrahedron.
U. Saravanan (IIT Madras) Finite element method April 2014 26 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of linear interpolation function
Assuming, the interpolation function is linear along each edge and on eachplane of the tetrahedron,
Ni = bi + cix + diy + eiz (23)
where bi , ci , di and ei are constants determined from the condition thatNi (xj , yj , zj) = δij , where (xj , yj , zj) are the coordinates of the four nodepoints of the tetrahedron.
Then, the requirement: N1(xj , yj , zj) = δ1j , can be written as:1 x1 y1 z1
1 x2 y2 z2
1 x3 y3 z3
1 x4 y4 z4
b1
c1
d1
e1
=
1000
(24)
U. Saravanan (IIT Madras) Finite element method April 2014 27 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of linear interpolation function
Assuming, the interpolation function is linear along each edge and on eachplane of the tetrahedron,
Ni = bi + cix + diy + eiz (23)
where bi , ci , di and ei are constants determined from the condition thatNi (xj , yj , zj) = δij , where (xj , yj , zj) are the coordinates of the four nodepoints of the tetrahedron.
Then, the requirement: N1(xj , yj , zj) = δ1j , can be written as:1 x1 y1 z1
1 x2 y2 z2
1 x3 y3 z3
1 x4 y4 z4
b1
c1
d1
e1
=
1000
(24)
U. Saravanan (IIT Madras) Finite element method April 2014 27 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of linear interpolation function
Solving the above equation we obtain:
b1 =1
∆det
x2 y2 z2
x3 y3 z3
x4 y4 z4
, c1 = − 1
∆det
1 y2 z2
1 y3 z3
1 y4 z4
d1 =
1
∆det
1 x2 z2
1 x3 z3
1 x4 z4
, e1 = − 1
∆det
1 x2 y2
1 x3 y3
1 x4 y4
Where
∆ = det
1 x1 y1 z1
1 x2 y2 z2
1 x3 y3 z3
1 x4 y4 z4
(25)
U. Saravanan (IIT Madras) Finite element method April 2014 28 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of linear interpolation function
Solving the above equation we obtain:
b1 =1
∆det
x2 y2 z2
x3 y3 z3
x4 y4 z4
, c1 = − 1
∆det
1 y2 z2
1 y3 z3
1 y4 z4
d1 =
1
∆det
1 x2 z2
1 x3 z3
1 x4 z4
, e1 = − 1
∆det
1 x2 y2
1 x3 y3
1 x4 y4
Where
∆ = det
1 x1 y1 z1
1 x2 y2 z2
1 x3 y3 z3
1 x4 y4 z4
(25)
U. Saravanan (IIT Madras) Finite element method April 2014 28 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Representation for the displacement field
Hence, the displacement field over the element can be represented inmatrix form as:
u =
uvw
= (NFEM){d} (26)
Where
(NFEM) =
N1 0 0 N2 0 0 N3 0 0 N4 0 00 N1 0 0 N2 0 0 N3 0 0 N4 00 0 N1 0 0 N2 0 0 N3 0 0 N4
(27)
U. Saravanan (IIT Madras) Finite element method April 2014 29 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Representation for the displacement field
Hence, the displacement field over the element can be represented inmatrix form as:
u =
uvw
= (NFEM){d} (26)
Where
(NFEM) =
N1 0 0 N2 0 0 N3 0 0 N4 0 00 N1 0 0 N2 0 0 N3 0 0 N4 00 0 N1 0 0 N2 0 0 N3 0 0 N4
(27)
U. Saravanan (IIT Madras) Finite element method April 2014 29 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Infinitesimal strain tensor
A representation of the infinitesimal strain tensor
ε =
(ε)xx(ε)yy(ε)zz
2(ε)xy2(ε)yz2(ε)xz
=
∂u∂x∂v∂y∂w∂z
∂u∂y + ∂v
∂x∂v∂z + ∂w
∂y∂w∂x + ∂u
∂z
(28)
The strain in the element can be written as
{ε} = (BFEM){d} (29)
U. Saravanan (IIT Madras) Finite element method April 2014 30 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Infinitesimal strain tensor
A representation of the infinitesimal strain tensor
ε =
(ε)xx(ε)yy(ε)zz
2(ε)xy2(ε)yz2(ε)xz
=
∂u∂x∂v∂y∂w∂z
∂u∂y + ∂v
∂x∂v∂z + ∂w
∂y∂w∂x + ∂u
∂z
(28)
The strain in the element can be written as
{ε} = (BFEM){d} (29)
U. Saravanan (IIT Madras) Finite element method April 2014 30 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Definition of (BFEM)
(BFEM) =N1,x 0 0 N2,x 0 0 N3,x 0 0 N4,x 0 0
0 N1,y 0 0 N2,y 0 0 N3,y 0 0 N4,y 00 0 N1,z 0 0 N2,z 0 0 N3,z 0 0 N4,z
N1,y N1,x 0 N2,y N2,x 0 N3,y N3,x 0 N4,y N4,x 00 N1,z N1,y 0 N2,z N2,y 0 N3,z N3,y 0 N4,z N4,y
N1,z 0 N1,x N2,z 0 N2,x N3,z 0 N3,x N4,z 0 N4,x
(30)
U. Saravanan (IIT Madras) Finite element method April 2014 31 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Solution Procedure
Applied forces Motion / displacement
Stress Deformation gradient or strain
Constitutive relation
Equilibrium equations
Strain displacement relation
U. Saravanan (IIT Madras) Finite element method April 2014 32 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Evaluation of Cauchy stress tensor
Constitutive relation: Hooke’s Law
σ = Dε
A representation for Cauchy stress tensor
{σ} =
σxxσyyσzzσxyσyzσxz
(31)
Now, the Hooke’s law can be written as
{σ} = (DFEM){ε} (32)
U. Saravanan (IIT Madras) Finite element method April 2014 33 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Evaluation of Cauchy stress tensor
Constitutive relation: Hooke’s Law
σ = Dε
A representation for Cauchy stress tensor
{σ} =
σxxσyyσzzσxyσyzσxz
(31)
Now, the Hooke’s law can be written as
{σ} = (DFEM){ε} (32)
U. Saravanan (IIT Madras) Finite element method April 2014 33 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Evaluation of Cauchy stress tensor
Constitutive relation: Hooke’s Law
σ = Dε
A representation for Cauchy stress tensor
{σ} =
σxxσyyσzzσxyσyzσxz
(31)
Now, the Hooke’s law can be written as
{σ} = (DFEM){ε} (32)
U. Saravanan (IIT Madras) Finite element method April 2014 33 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Evaluation of Cauchy stress tensor
For isotropic materials
(DFEM) =E
(1 + ν)(1− 2ν)
1− ν ν ν 0 0 0ν 1− ν ν 0 0 0ν ν 1− ν 0 0 00 0 0 1− 2ν 0 00 0 0 0 1− 2ν 00 0 0 0 0 1− 2ν
(33)
Substituting for the strain in the constitutive relation we obtain
{σ} = (DFEM)(BFEM){d} (34)
U. Saravanan (IIT Madras) Finite element method April 2014 34 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Evaluation of Cauchy stress tensor
For isotropic materials
(DFEM) =E
(1 + ν)(1− 2ν)
1− ν ν ν 0 0 0ν 1− ν ν 0 0 0ν ν 1− ν 0 0 00 0 0 1− 2ν 0 00 0 0 0 1− 2ν 00 0 0 0 0 1− 2ν
(33)
Substituting for the strain in the constitutive relation we obtain
{σ} = (DFEM)(BFEM){d} (34)
U. Saravanan (IIT Madras) Finite element method April 2014 34 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Solution Procedure
Applied forces Motion / displacement
Stress Deformation gradient or strain
Constitutive relation
Equilibrium equations
Strain displacement relation
U. Saravanan (IIT Madras) Finite element method April 2014 35 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of elemental equations
Weakened equilibrium equation∫At
[σ · εφ + ρa · φ
]dv =
∫At
ρb · φdv +
∫∂At
t(n) · φds (35)
Defining
ευi =1
2
[grad(υi ) + grad t(υi )
](36)
Identifying the weight function, φ with υi∫At
[σ · ευi + ρa · υi ] dv =
∫At
ρb · υidv +
∫∂At
t(n) · υids (37)
for i ∈ {1, 2, . . . , 12}
U. Saravanan (IIT Madras) Finite element method April 2014 36 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of elemental equations
Weakened equilibrium equation∫At
[σ · εφ + ρa · φ
]dv =
∫At
ρb · φdv +
∫∂At
t(n) · φds (35)
Defining
ευi =1
2
[grad(υi ) + grad t(υi )
](36)
Identifying the weight function, φ with υi∫At
[σ · ευi + ρa · υi ] dv =
∫At
ρb · υidv +
∫∂At
t(n) · υids (37)
for i ∈ {1, 2, . . . , 12}
U. Saravanan (IIT Madras) Finite element method April 2014 36 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of elemental equations
Weakened equilibrium equation∫At
[σ · εφ + ρa · φ
]dv =
∫At
ρb · φdv +
∫∂At
t(n) · φds (35)
Defining
ευi =1
2
[grad(υi ) + grad t(υi )
](36)
Identifying the weight function, φ with υi∫At
[σ · ευi + ρa · υi ] dv =
∫At
ρb · υidv +
∫∂At
t(n) · υids (37)
for i ∈ {1, 2, . . . , 12}
U. Saravanan (IIT Madras) Finite element method April 2014 36 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Computation of acceleration
Acceleration:
a =∂2u
∂t2, (38)
When the developed displacement gradient, spatial velocities and itsgradient are very small
U. Saravanan (IIT Madras) Finite element method April 2014 37 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Computation of acceleration
Assuming small deformations
a ≈ ∂2u
∂t2=
[4∑
i=1
Nid2uidt2
]ex +
[4∑
i=1
Nid2vidt2
]ey +
[4∑
i=1
Nid2wi
dt2
]ez
= (NFEM){d} (39)
Where
{d} =
d2u1dt2
d2v1dt2
...d2v4dt2
d2w4dt2
(40)
U. Saravanan (IIT Madras) Finite element method April 2014 38 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Computation of acceleration
Assuming small deformations
a ≈ ∂2u
∂t2=
[4∑
i=1
Nid2uidt2
]ex +
[4∑
i=1
Nid2vidt2
]ey +
[4∑
i=1
Nid2wi
dt2
]ez
= (NFEM){d} (39)
Where
{d} =
d2u1dt2
d2v1dt2
...d2v4dt2
d2w4dt2
(40)
U. Saravanan (IIT Madras) Finite element method April 2014 38 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Notation and Identities
Defining
{di} as a vector with only the i th element of the array in d i being 1 andall the other elements are 0
Then
υi = (NFEM){di} (41)
ευi = (BFEM){di} (42)
U. Saravanan (IIT Madras) Finite element method April 2014 39 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Notation and Identities
Defining
{di} as a vector with only the i th element of the array in d i being 1 andall the other elements are 0
Then
υi = (NFEM){di} (41)
ευi = (BFEM){di} (42)
U. Saravanan (IIT Madras) Finite element method April 2014 39 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of elemental equations
Substituting the expressions we obtained for the Cauchy stress andacceleration in the weakened spatial form of the equilibrium equations andusing the above identities we obtain
∫At
[(DFEM)(BFEM){d} · (BFEM){di}+ ρ(NFEM){d} · (NFEM){di}
]dv
=
∫At
ρb · (NFEM){di}dv +
∫∂At
t(n) · (NFEM){di}ds
where the integration is performed over the volume or the surface of thetetrahedron under investigation respectively
The above equation holds for di , i ∈ {1, 2, . . . , 12}Hence, it is equivalent to requiring that this hold for an arbitraryvector da (Why?)
U. Saravanan (IIT Madras) Finite element method April 2014 40 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of elemental equations
Substituting the expressions we obtained for the Cauchy stress andacceleration in the weakened spatial form of the equilibrium equations andusing the above identities we obtain
∫At
[(DFEM)(BFEM){d} · (BFEM){di}+ ρ(NFEM){d} · (NFEM){di}
]dv
=
∫At
ρb · (NFEM){di}dv +
∫∂At
t(n) · (NFEM){di}ds
where the integration is performed over the volume or the surface of thetetrahedron under investigation respectively
The above equation holds for di , i ∈ {1, 2, . . . , 12}
Hence, it is equivalent to requiring that this hold for an arbitraryvector da (Why?)
U. Saravanan (IIT Madras) Finite element method April 2014 40 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Formulation of elemental equations
Substituting the expressions we obtained for the Cauchy stress andacceleration in the weakened spatial form of the equilibrium equations andusing the above identities we obtain
∫At
[(DFEM)(BFEM){d} · (BFEM){di}+ ρ(NFEM){d} · (NFEM){di}
]dv
=
∫At
ρb · (NFEM){di}dv +
∫∂At
t(n) · (NFEM){di}ds
where the integration is performed over the volume or the surface of thetetrahedron under investigation respectively
The above equation holds for di , i ∈ {1, 2, . . . , 12}Hence, it is equivalent to requiring that this hold for an arbitraryvector da (Why?)
U. Saravanan (IIT Madras) Finite element method April 2014 40 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Elemental equation
The above equation can be written as
(Ke){d}+ (Me){d} = {feb}+ {fes } (43)
Where
(Ke) =
∫At
(BFEM)t(DFEM)(BFEM)dv , Element stiffness matrix
(Me) =
∫At
ρ(NFEM)t(NFEM)dv , Element mass matrix
{feb} =
∫At
ρ(NFEM)tbdv , Element body force vector
{fes } =
∫∂At
(NFEM)tt(n)ds, Element surface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 41 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Elemental equation
The above equation can be written as
(Ke){d}+ (Me){d} = {feb}+ {fes } (43)
Where
(Ke) =
∫At
(BFEM)t(DFEM)(BFEM)dv , Element stiffness matrix
(Me) =
∫At
ρ(NFEM)t(NFEM)dv , Element mass matrix
{feb} =
∫At
ρ(NFEM)tbdv , Element body force vector
{fes } =
∫∂At
(NFEM)tt(n)ds, Element surface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 41 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Points to be understood
While the stiffness, mass matrices and body force vector could becomputed, surface traction could not be computed at this stage
However, if part or the entire boundary of the element underconsideration happened to coincide with the boundary of the bodywhere the traction is prescribed, then this term can be computed
But to maintain generality, the surface traction vector is assumed tobe unknown and hence at this stage there are 2n unknowns - ndisplacement degrees of freedom and n components of the surfacetraction vector, where n is 12 for the tetrahedral element underconsideration
U. Saravanan (IIT Madras) Finite element method April 2014 42 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Points to be understood
While the stiffness, mass matrices and body force vector could becomputed, surface traction could not be computed at this stage
However, if part or the entire boundary of the element underconsideration happened to coincide with the boundary of the bodywhere the traction is prescribed, then this term can be computed
But to maintain generality, the surface traction vector is assumed tobe unknown and hence at this stage there are 2n unknowns - ndisplacement degrees of freedom and n components of the surfacetraction vector, where n is 12 for the tetrahedral element underconsideration
U. Saravanan (IIT Madras) Finite element method April 2014 42 / 81
Finite element formulation for linearized elasticity Derivation of elemental equations
Points to be understood
While the stiffness, mass matrices and body force vector could becomputed, surface traction could not be computed at this stage
However, if part or the entire boundary of the element underconsideration happened to coincide with the boundary of the bodywhere the traction is prescribed, then this term can be computed
But to maintain generality, the surface traction vector is assumed tobe unknown and hence at this stage there are 2n unknowns - ndisplacement degrees of freedom and n components of the surfacetraction vector, where n is 12 for the tetrahedral element underconsideration
U. Saravanan (IIT Madras) Finite element method April 2014 42 / 81
Finite element formulation for linearized elasticity Assembly of element equations
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 43 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Introduction
While deriving the element equations, we isolated a typical element(say, the kth) from the mesh and developed its finite element model
To solve the full problem, we must sum the contributions from eachof the elements that make up the domain of the problem
Why sum?
Because
u =N∑i=1
Uiν i (44)
U. Saravanan (IIT Madras) Finite element method April 2014 44 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Introduction
While deriving the element equations, we isolated a typical element(say, the kth) from the mesh and developed its finite element model
To solve the full problem, we must sum the contributions from eachof the elements that make up the domain of the problem
Why sum?
Because
u =N∑i=1
Uiν i (44)
U. Saravanan (IIT Madras) Finite element method April 2014 44 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Introduction
While deriving the element equations, we isolated a typical element(say, the kth) from the mesh and developed its finite element model
To solve the full problem, we must sum the contributions from eachof the elements that make up the domain of the problem
Why sum?
Because
u =N∑i=1
Uiν i (44)
U. Saravanan (IIT Madras) Finite element method April 2014 44 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Introduction
While deriving the element equations, we isolated a typical element(say, the kth) from the mesh and developed its finite element model
To solve the full problem, we must sum the contributions from eachof the elements that make up the domain of the problem
Why sum?
Because
u =N∑i=1
Uiν i (44)
U. Saravanan (IIT Madras) Finite element method April 2014 44 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Concepts used in assembling the equations
The assembly of the finite element equations is based on:
1. Continuity of the displacement field - Related to the single valuednature of the displacement
2. Balance of the surface force field - Refers to the requirement that t(n)
+ t(−n) = o
Imposing the continuity of the displacement field at the nodescommon to the elements amounts to the correspondence between thelocal and global nodal values
U. Saravanan (IIT Madras) Finite element method April 2014 45 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Concepts used in assembling the equations
The assembly of the finite element equations is based on:
1. Continuity of the displacement field - Related to the single valuednature of the displacement
2. Balance of the surface force field - Refers to the requirement that t(n)
+ t(−n) = o
Imposing the continuity of the displacement field at the nodescommon to the elements amounts to the correspondence between thelocal and global nodal values
U. Saravanan (IIT Madras) Finite element method April 2014 45 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Concepts used in assembling the equations
The assembly of the finite element equations is based on:
1. Continuity of the displacement field - Related to the single valuednature of the displacement
2. Balance of the surface force field - Refers to the requirement that t(n)
+ t(−n) = o
Imposing the continuity of the displacement field at the nodescommon to the elements amounts to the correspondence between thelocal and global nodal values
U. Saravanan (IIT Madras) Finite element method April 2014 45 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Concepts used in assembling the equations
The assembly of the finite element equations is based on:
1. Continuity of the displacement field - Related to the single valuednature of the displacement
2. Balance of the surface force field - Refers to the requirement that t(n)
+ t(−n) = o
Imposing the continuity of the displacement field at the nodescommon to the elements amounts to the correspondence between thelocal and global nodal values
U. Saravanan (IIT Madras) Finite element method April 2014 45 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Illustrative mesh
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Global degrees offreedom
{UFEM} =
U1
V1
W1...U5
V5
W5
(45)
U. Saravanan (IIT Madras) Finite element method April 2014 46 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Illustrative mesh
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Global degrees offreedom
{UFEM} =
U1
V1
W1...U5
V5
W5
(45)
U. Saravanan (IIT Madras) Finite element method April 2014 46 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Illustrative mesh
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Global degrees offreedom
{UFEM} =
U1
V1
W1...U5
V5
W5
(45)
U. Saravanan (IIT Madras) Finite element method April 2014 46 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Illustrative mesh
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Global degrees offreedom
{UFEM} =
U1
V1
W1...U5
V5
W5
(45)
U. Saravanan (IIT Madras) Finite element method April 2014 46 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Establishing the relationship between global and localdegrees of freedom
For the finite element mesh being studied the relation between the localand global degrees of freedom is:
u11 = U1, v1
1 = V1, w11 = W1, u1
2 = U2, v12 = V2, w1
2 = W2
u13 = U3, v1
3 = V3, w13 = W3, u1
4 = U4, v14 = V4, w1
4 = W4
u21 = U1, v2
1 = V1, w21 = W1, u2
2 = U4, v22 = V4, w2
2 = W4
u23 = U3, v2
3 = V3, w23 = W3, u2
4 = U5, v24 = V5, w2
4 = W5
This transformation from the local degrees of freedom to the global isaccomplished by a boolean transformation matrix, TFEM , so that
{d} = (TFEM){UFEM}. (46)
U. Saravanan (IIT Madras) Finite element method April 2014 47 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Establishing the relationship between global and localdegrees of freedom
For the finite element mesh being studied the relation between the localand global degrees of freedom is:
u11 = U1, v1
1 = V1, w11 = W1, u1
2 = U2, v12 = V2, w1
2 = W2
u13 = U3, v1
3 = V3, w13 = W3, u1
4 = U4, v14 = V4, w1
4 = W4
u21 = U1, v2
1 = V1, w21 = W1, u2
2 = U4, v22 = V4, w2
2 = W4
u23 = U3, v2
3 = V3, w23 = W3, u2
4 = U5, v24 = V5, w2
4 = W5
This transformation from the local degrees of freedom to the global isaccomplished by a boolean transformation matrix, TFEM , so that
{d} = (TFEM){UFEM}. (46)
U. Saravanan (IIT Madras) Finite element method April 2014 47 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Illustrative transformation matrix, (TFEM)
For the element 1
(T1FEM) =
1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 0 0 0 0 00 0 0 0 0 0 1 0 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 0 0 0 0 00 0 0 0 0 0 0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 0 0 0 0 1 0 0 0 00 0 0 0 0 0 0 0 0 0 0 1 0 0 0
, (47)
U. Saravanan (IIT Madras) Finite element method April 2014 48 / 81
Finite element formulation for linearized elasticity Assembly of element equations
An important identity
0 =
∫At
(DFEM)(BFEM)(TeFEM){UFEM} · (BFEM)(Te
FEM){UiFEM}dv
+
∫At
ρ(NFEM)(TeFEM){UFEM} · (NFEM)(Te
FEM){UiFEM}dv
−∫At
ρb · (NFEM)(TeFEM){Ui
FEM}dv
−∫∂At
t(n) · (NFEM)(TeFEM){Ui
FEM}ds
where {UiFEM} is such that the equation
di = (TeFEM){Ui
FEM}
holds.
U. Saravanan (IIT Madras) Finite element method April 2014 49 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Global matrices and force vector
The global stiffness and mass matrix are obtained from
(K) =k∑
e=1
(TeFEM)t(Ke)(Te
FEM), (M) =k∑
e=1
(TeFEM)t(Me)(Te
FEM),
(48)
The global force vector is obtained from
{FFEM} =k∑
e=1
(TeFEM)t [{feb}+ {fes }], (49)
U. Saravanan (IIT Madras) Finite element method April 2014 50 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Global matrices and force vector
The global stiffness and mass matrix are obtained from
(K) =k∑
e=1
(TeFEM)t(Ke)(Te
FEM), (M) =k∑
e=1
(TeFEM)t(Me)(Te
FEM),
(48)
The global force vector is obtained from
{FFEM} =k∑
e=1
(TeFEM)t [{feb}+ {fes }], (49)
U. Saravanan (IIT Madras) Finite element method April 2014 50 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
1. (TeFEM) could be obtained from the connectivity matrix
2. Even though the above formulation provides an algorithm for addingthe different element matrices and vectors, it is computationallyexpensive
3. In practice, the global matrices are directly updated withoutassembling the element matrices.
4. The continuity of the displacement at the inter element nodesguarantees the continuity of the displacement along the entire interelement boundary
U. Saravanan (IIT Madras) Finite element method April 2014 51 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
1. (TeFEM) could be obtained from the connectivity matrix
2. Even though the above formulation provides an algorithm for addingthe different element matrices and vectors, it is computationallyexpensive
3. In practice, the global matrices are directly updated withoutassembling the element matrices.
4. The continuity of the displacement at the inter element nodesguarantees the continuity of the displacement along the entire interelement boundary
U. Saravanan (IIT Madras) Finite element method April 2014 51 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
1. (TeFEM) could be obtained from the connectivity matrix
2. Even though the above formulation provides an algorithm for addingthe different element matrices and vectors, it is computationallyexpensive
3. In practice, the global matrices are directly updated withoutassembling the element matrices.
4. The continuity of the displacement at the inter element nodesguarantees the continuity of the displacement along the entire interelement boundary
U. Saravanan (IIT Madras) Finite element method April 2014 51 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
1. (TeFEM) could be obtained from the connectivity matrix
2. Even though the above formulation provides an algorithm for addingthe different element matrices and vectors, it is computationallyexpensive
3. In practice, the global matrices are directly updated withoutassembling the element matrices.
4. The continuity of the displacement at the inter element nodesguarantees the continuity of the displacement along the entire interelement boundary
U. Saravanan (IIT Madras) Finite element method April 2014 51 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Balance of surface forces
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
We require
(t1(n))(1,3,4) = −(t2
(n))(1,3,2), (50)
where the subscript (1, 3, 4) (or (1, 3, 2))denotes the local node numbers thatform the vertices of the surface beingconsidered
In the finite element method, we impose the above relation in a weighted-integralsense: ∫
∂A1134
(t1(n)) · υ
1i ds +
∫∂A2
132
(t2(n)) · υ
2j ds = 0, (51)
for (i , j) ∈ {(1, 1), (2, 2), (3, 3), (7, 7), (8, 8), (9, 9), (10, 4), (11, 5), (12, 6)}, wherethe choice of i and j correspond to the basis functions associated with eachcommon node in elements 1 and 2
U. Saravanan (IIT Madras) Finite element method April 2014 52 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Balance of surface forces
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
We require
(t1(n))(1,3,4) = −(t2
(n))(1,3,2), (50)
where the subscript (1, 3, 4) (or (1, 3, 2))denotes the local node numbers thatform the vertices of the surface beingconsidered
In the finite element method, we impose the above relation in a weighted-integralsense: ∫
∂A1134
(t1(n)) · υ
1i ds +
∫∂A2
132
(t2(n)) · υ
2j ds = 0, (51)
for (i , j) ∈ {(1, 1), (2, 2), (3, 3), (7, 7), (8, 8), (9, 9), (10, 4), (11, 5), (12, 6)}, wherethe choice of i and j correspond to the basis functions associated with eachcommon node in elements 1 and 2
U. Saravanan (IIT Madras) Finite element method April 2014 52 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Balance of surface forces
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
We require
(t1(n))(1,3,4) = −(t2
(n))(1,3,2), (50)
where the subscript (1, 3, 4) (or (1, 3, 2))denotes the local node numbers thatform the vertices of the surface beingconsidered
In the finite element method, we impose the above relation in a weighted-integralsense: ∫
∂A1134
(t1(n)) · υ
1i ds +
∫∂A2
132
(t2(n)) · υ
2j ds = 0, (51)
for (i , j) ∈ {(1, 1), (2, 2), (3, 3), (7, 7), (8, 8), (9, 9), (10, 4), (11, 5), (12, 6)}, wherethe choice of i and j correspond to the basis functions associated with eachcommon node in elements 1 and 2
U. Saravanan (IIT Madras) Finite element method April 2014 52 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Balance of surface forces
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
We require
(t1(n))(1,3,4) = −(t2
(n))(1,3,2), (50)
where the subscript (1, 3, 4) (or (1, 3, 2))denotes the local node numbers thatform the vertices of the surface beingconsidered
In the finite element method, we impose the above relation in a weighted-integralsense: ∫
∂A1134
(t1(n)) · υ
1i ds +
∫∂A2
132
(t2(n)) · υ
2j ds = 0, (51)
for (i , j) ∈ {(1, 1), (2, 2), (3, 3), (7, 7), (8, 8), (9, 9), (10, 4), (11, 5), (12, 6)}, wherethe choice of i and j correspond to the basis functions associated with eachcommon node in elements 1 and 2
U. Saravanan (IIT Madras) Finite element method April 2014 52 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Balance of surface forces
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
We require
(t1(n))(1,3,4) = −(t2
(n))(1,3,2), (50)
where the subscript (1, 3, 4) (or (1, 3, 2))denotes the local node numbers thatform the vertices of the surface beingconsidered
In the finite element method, we impose the above relation in a weighted-integralsense: ∫
∂A1134
(t1(n)) · υ
1i ds +
∫∂A2
132
(t2(n)) · υ
2j ds = 0, (51)
for (i , j) ∈ {(1, 1), (2, 2), (3, 3), (7, 7), (8, 8), (9, 9), (10, 4), (11, 5), (12, 6)}, wherethe choice of i and j correspond to the basis functions associated with eachcommon node in elements 1 and 2
U. Saravanan (IIT Madras) Finite element method April 2014 52 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Why balance of surface forces?
Required from the physics of the problem - But what then is theequilibrium equation?
Required to reduce the number of unknowns in surface force vector
Required to enforce the traction boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 53 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Why balance of surface forces?
Required from the physics of the problem - But what then is theequilibrium equation?
Required to reduce the number of unknowns in surface force vector
Required to enforce the traction boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 53 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Why balance of surface forces?
Required from the physics of the problem - But what then is theequilibrium equation?
Required to reduce the number of unknowns in surface force vector
Required to enforce the traction boundary condition
U. Saravanan (IIT Madras) Finite element method April 2014 53 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Alternate notation
Let the surface with vertices 1− 2− 3 be called surface 1, vertices1− 2− 4 surface 2, vertices 1− 3− 4 surface 3 and vertices 2− 3− 4surface 4 and then let
F eiJ =
∫surfaceJ
(te(n)) · υei ds (52)
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Now previous equation can bewritten succinctly as
F 1i3 + F 2
j1 = 0 (53)
for (i , j) ∈{(1, 1), (2, 2), (3, 3), (7, 7), (8, 8),(9, 9), (10, 4), (11, 5), (12, 6)}
U. Saravanan (IIT Madras) Finite element method April 2014 54 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Alternate notation
Let the surface with vertices 1− 2− 3 be called surface 1, vertices1− 2− 4 surface 2, vertices 1− 3− 4 surface 3 and vertices 2− 3− 4surface 4 and then let
F eiJ =
∫surfaceJ
(te(n)) · υei ds (52)
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Now previous equation can bewritten succinctly as
F 1i3 + F 2
j1 = 0 (53)
for (i , j) ∈{(1, 1), (2, 2), (3, 3), (7, 7), (8, 8),(9, 9), (10, 4), (11, 5), (12, 6)}
U. Saravanan (IIT Madras) Finite element method April 2014 54 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Alternate notation
Let the surface with vertices 1− 2− 3 be called surface 1, vertices1− 2− 4 surface 2, vertices 1− 3− 4 surface 3 and vertices 2− 3− 4surface 4 and then let
F eiJ =
∫surfaceJ
(te(n)) · υei ds (52)
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Now previous equation can bewritten succinctly as
F 1i3 + F 2
j1 = 0 (53)
for (i , j) ∈{(1, 1), (2, 2), (3, 3), (7, 7), (8, 8),(9, 9), (10, 4), (11, 5), (12, 6)}
U. Saravanan (IIT Madras) Finite element method April 2014 54 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Alternate notation
Let the surface with vertices 1− 2− 3 be called surface 1, vertices1− 2− 4 surface 2, vertices 1− 3− 4 surface 3 and vertices 2− 3− 4surface 4 and then let
F eiJ =
∫surfaceJ
(te(n)) · υei ds (52)
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Now previous equation can bewritten succinctly as
F 1i3 + F 2
j1 = 0 (53)
for (i , j) ∈{(1, 1), (2, 2), (3, 3), (7, 7), (8, 8),(9, 9), (10, 4), (11, 5), (12, 6)}
U. Saravanan (IIT Madras) Finite element method April 2014 54 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Alternate notation
Let the surface with vertices 1− 2− 3 be called surface 1, vertices1− 2− 4 surface 2, vertices 1− 3− 4 surface 3 and vertices 2− 3− 4surface 4 and then let
F eiJ =
∫surfaceJ
(te(n)) · υei ds (52)
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Now previous equation can bewritten succinctly as
F 1i3 + F 2
j1 = 0 (53)
for (i , j) ∈{(1, 1), (2, 2), (3, 3), (7, 7), (8, 8),(9, 9), (10, 4), (11, 5), (12, 6)}
U. Saravanan (IIT Madras) Finite element method April 2014 54 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
For the tetrahedron element being studied
{f1s }1 =
∫surface1
(t1(n1))xN1ds +
∫surface2
(t1(n2))xN1ds
+
∫surface3
(t1(n3))xN1ds +
∫surface4
(t1(n4))xN1ds, (54)
where ni denotes the outward normal to the i th surface
Hence
{fes }i = F ei1 + F e
i2 + F ei3 + F e
i4 (55)
where {fes }i denotes the i th component in the surface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 55 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
For the tetrahedron element being studied
{f1s }1 =
∫surface1
(t1(n1))xN1ds +
∫surface2
(t1(n2))xN1ds
+
∫surface3
(t1(n3))xN1ds +
∫surface4
(t1(n4))xN1ds, (54)
where ni denotes the outward normal to the i th surface
Hence
{fes }i = F ei1 + F e
i2 + F ei3 + F e
i4 (55)
where {fes }i denotes the i th component in the surface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 55 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
ex
ey
ez
4
1
2
3
{f1s }1 is not the net force acting
on the surface of thetetrahedron along the exdirection
The 12 components of {f1s }i
correspond to the weightedaverage of 3 components of thetraction vector with the 4 shapefunctions over the surface of thetetrahedron
U. Saravanan (IIT Madras) Finite element method April 2014 56 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
ex
ey
ez
4
1
2
3
{f1s }1 is not the net force acting
on the surface of thetetrahedron along the exdirection
The 12 components of {f1s }i
correspond to the weightedaverage of 3 components of thetraction vector with the 4 shapefunctions over the surface of thetetrahedron
U. Saravanan (IIT Madras) Finite element method April 2014 56 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
ex
ey
ez
4
1
2
3
{f1s }1 is not the net force acting
on the surface of thetetrahedron along the exdirection
The 12 components of {f1s }i
correspond to the weightedaverage of 3 components of thetraction vector with the 4 shapefunctions over the surface of thetetrahedron
U. Saravanan (IIT Madras) Finite element method April 2014 56 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
ex
ey
ez
4
1
2
3
{f1s }1 is not the net force acting
on the surface of thetetrahedron along the exdirection
The 12 components of {f1s }i
correspond to the weightedaverage of 3 components of thetraction vector with the 4 shapefunctions over the surface of thetetrahedron
U. Saravanan (IIT Madras) Finite element method April 2014 56 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Points to understand
ex
ey
ez
4
1
2
3
{f1s }1 is not the net force acting
on the surface of thetetrahedron along the exdirection
The 12 components of {f1s }i
correspond to the weightedaverage of 3 components of thetraction vector with the 4 shapefunctions over the surface of thetetrahedron
U. Saravanan (IIT Madras) Finite element method April 2014 56 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Illustrative mesh
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
U. Saravanan (IIT Madras) Finite element method April 2014 57 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Surface traction vector for the mesh being studied
{Fs} =
(f 1s )1 + (f 2
s )1
(f 1s )2 + (f 2
s )2
(f 1s )3 + (f 2
s )3
(f 1s )4
(f 1s )5
(f 1s )6
(f 1s )7 + (f 2
s )7
(f 1s )8 + (f 2
s )8
(f 1s )9 + (f 2
s )9
(f 1s )10 + (f 2
s )4
(f 1s )11 + (f 2
s )5
(f 1s )12 + (f 2
s )6
(f 2s )10
(f 2s )11
(f 2s )12
=
F 111 + F 1
12 + F 113 + F 2
11 + F 114 + F 2
12 + F 213 + F 2
14
F 121 + F 1
22 + F 123 + F 2
21 + F 124 + F 2
22 + F 223 + F 2
24
F 131 + F 1
32 + F 133 + F 2
31 + F 134 + F 2
32 + F 233 + F 2
34
F 141 + F 1
42 + F 143 + F 1
44
F 151 + F 1
52 + F 153 + F 1
54
F 161 + F 1
62 + F 163 + F 1
64
F 171 + F 1
72 + F 173 + F 2
71 + F 174 + F 2
72 + F 273 + F 2
74
F 181 + F 1
82 + F 183 + F 2
81 + F 184 + F 2
82 + F 283 + F 2
84
F 191 + F 1
92 + F 193 + F 2
91 + F 194 + F 2
92 + F 293 + F 2
94
F 1101 + F 1
102 + F 1103 + F 2
41 + F 1104 + F 2
42 + F 243 + F 2
44
F 1111 + F 1
112 + F 1113 + F 2
51 + F 1114 + F 2
52 + F 253 + F 2
54
F 1121 + F 1
122 + F 1123 + F 2
61 + F 1124 + F 2
62 + F 263 + F 2
64
F 2101 + F 2
102 + F 2103 + F 2
104
F 2111 + F 2
112 + F 2113 + F 2
114
F 2121 + F 2
122 + F 2123 + F 2
124
U. Saravanan (IIT Madras) Finite element method April 2014 58 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Surface traction vector for the mesh being studied
It then follows from the balance of surface traction that only theunderlined terms are zero
The remaining terms of each equation will either be known becauset(n) is known on the boundary or will remain unknown because thedisplacement is specified on the boundary
U. Saravanan (IIT Madras) Finite element method April 2014 59 / 81
Finite element formulation for linearized elasticity Assembly of element equations
Surface traction vector for the mesh being studied
It then follows from the balance of surface traction that only theunderlined terms are zero
The remaining terms of each equation will either be known becauset(n) is known on the boundary or will remain unknown because thedisplacement is specified on the boundary
U. Saravanan (IIT Madras) Finite element method April 2014 59 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 60 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Illustrative boundary conditions
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Let us assume that a uniformpressure, px acts along the exdirection on the surface withvertices 1− 3− 5
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
All the remaining surfaces arefree of traction
U. Saravanan (IIT Madras) Finite element method April 2014 61 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Illustrative boundary conditions
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Let us assume that a uniformpressure, px acts along the exdirection on the surface withvertices 1− 3− 5
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
All the remaining surfaces arefree of traction
U. Saravanan (IIT Madras) Finite element method April 2014 61 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Illustrative boundary conditions
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Let us assume that a uniformpressure, px acts along the exdirection on the surface withvertices 1− 3− 5
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
All the remaining surfaces arefree of traction
U. Saravanan (IIT Madras) Finite element method April 2014 61 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Illustrative boundary conditions
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Let us assume that a uniformpressure, px acts along the exdirection on the surface withvertices 1− 3− 5
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
All the remaining surfaces arefree of traction
U. Saravanan (IIT Madras) Finite element method April 2014 61 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Illustrative boundary conditions
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Let us assume that a uniformpressure, px acts along the exdirection on the surface withvertices 1− 3− 5
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
All the remaining surfaces arefree of traction
U. Saravanan (IIT Madras) Finite element method April 2014 61 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Illustrative boundary conditions
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
Let us assume that a uniformpressure, px acts along the exdirection on the surface withvertices 1− 3− 5
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
All the remaining surfaces arefree of traction
U. Saravanan (IIT Madras) Finite element method April 2014 61 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
Surface traction free condition implies
F 1i2 = F 1
i4 = F 2i2 = F 2
i4 = 0, ∀ i ∈ {1, 2, . . . , 12} (56)
Traction on surface 1− 3− 5 yields
F 2i3 =
∫surface3
pxex · υ2i ds (57)
Traction on surface 1− 3− 5 acts only along ex , hence
F 2i3 = 0, ∀ i ∈ {2, 3, 4, 5, 6, 8, 9, 11, 12} (58)
F 2i3 = px
∫surface3
Nids = pxS135
3, ∀ i ∈ {1, 7, 10} (59)
where S135 denotes the surface area of the triangular surface with vertices1− 3− 5
U. Saravanan (IIT Madras) Finite element method April 2014 62 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
Surface traction free condition implies
F 1i2 = F 1
i4 = F 2i2 = F 2
i4 = 0, ∀ i ∈ {1, 2, . . . , 12} (56)
Traction on surface 1− 3− 5 yields
F 2i3 =
∫surface3
pxex · υ2i ds (57)
Traction on surface 1− 3− 5 acts only along ex , hence
F 2i3 = 0, ∀ i ∈ {2, 3, 4, 5, 6, 8, 9, 11, 12} (58)
F 2i3 = px
∫surface3
Nids = pxS135
3, ∀ i ∈ {1, 7, 10} (59)
where S135 denotes the surface area of the triangular surface with vertices1− 3− 5
U. Saravanan (IIT Madras) Finite element method April 2014 62 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
Surface traction free condition implies
F 1i2 = F 1
i4 = F 2i2 = F 2
i4 = 0, ∀ i ∈ {1, 2, . . . , 12} (56)
Traction on surface 1− 3− 5 yields
F 2i3 =
∫surface3
pxex · υ2i ds (57)
Traction on surface 1− 3− 5 acts only along ex , hence
F 2i3 = 0, ∀ i ∈ {2, 3, 4, 5, 6, 8, 9, 11, 12} (58)
F 2i3 = px
∫surface3
Nids = pxS135
3, ∀ i ∈ {1, 7, 10} (59)
where S135 denotes the surface area of the triangular surface with vertices1− 3− 5
U. Saravanan (IIT Madras) Finite element method April 2014 62 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
ex
ey
ez
4
1
2
3
For the assumed form for Ni , the value ofNi on the plane opposite to the i th node iszero
Hence, F 243 = 0 because N2 is zero on the
surface with vertices 1− 3− 4.
For similar reasons
F 1i1 = F 2
i1 = 0, ∀ i ∈ {10, 11, 12}F 1i3 = 0, ∀ i ∈ {4, 5, 6}
U. Saravanan (IIT Madras) Finite element method April 2014 63 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
ex
ey
ez
4
1
2
3
For the assumed form for Ni , the value ofNi on the plane opposite to the i th node iszero
Hence, F 243 = 0 because N2 is zero on the
surface with vertices 1− 3− 4.
For similar reasons
F 1i1 = F 2
i1 = 0, ∀ i ∈ {10, 11, 12}F 1i3 = 0, ∀ i ∈ {4, 5, 6}
U. Saravanan (IIT Madras) Finite element method April 2014 63 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
ex
ey
ez
4
1
2
3
For the assumed form for Ni , the value ofNi on the plane opposite to the i th node iszero
Hence, F 243 = 0 because N2 is zero on the
surface with vertices 1− 3− 4.
For similar reasons
F 1i1 = F 2
i1 = 0, ∀ i ∈ {10, 11, 12}F 1i3 = 0, ∀ i ∈ {4, 5, 6}
U. Saravanan (IIT Madras) Finite element method April 2014 63 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
ex
ey
ez
4
1
2
3 For the assumed form for Ni , the value ofNi on the plane opposite to the i th node iszero
Hence, F 243 = 0 because N2 is zero on the
surface with vertices 1− 3− 4.
For similar reasons
F 1i1 = F 2
i1 = 0, ∀ i ∈ {10, 11, 12}F 1i3 = 0, ∀ i ∈ {4, 5, 6}
U. Saravanan (IIT Madras) Finite element method April 2014 63 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
ex
ey
ez
4
1
2
3 For the assumed form for Ni , the value ofNi on the plane opposite to the i th node iszero
Hence, F 243 = 0 because N2 is zero on the
surface with vertices 1− 3− 4.
For similar reasons
F 1i1 = F 2
i1 = 0, ∀ i ∈ {10, 11, 12}F 1i3 = 0, ∀ i ∈ {4, 5, 6}
U. Saravanan (IIT Madras) Finite element method April 2014 63 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Inference from the traction boundary condition
ex
ey
ez
4
1
2
3 For the assumed form for Ni , the value ofNi on the plane opposite to the i th node iszero
Hence, F 243 = 0 because N2 is zero on the
surface with vertices 1− 3− 4.
For similar reasons
F 1i1 = F 2
i1 = 0, ∀ i ∈ {10, 11, 12}F 1i3 = 0, ∀ i ∈ {4, 5, 6}
U. Saravanan (IIT Madras) Finite element method April 2014 63 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Surface traction vector for the mesh being studied
{Fs} =
(f 1s )1 + (f 2
s )1
(f 1s )2 + (f 2
s )2
(f 1s )3 + (f 2
s )3
(f 1s )4
(f 1s )5
(f 1s )6
(f 1s )7 + (f 2
s )7
(f 1s )8 + (f 2
s )8
(f 1s )9 + (f 2
s )9
(f 1s )10 + (f 2
s )4
(f 1s )11 + (f 2
s )5
(f 1s )12 + (f 2
s )6
(f 2s )10
(f 2s )11
(f 2s )12
=
F 111 + F 1
12 + F 113 + F 2
11 + F 114 + F 2
12 + F 213 + F 2
14
F 121 + F 1
22 + F 123 + F 2
21 + F 124 + F 2
22 + F 223 + F 2
24
F 131 + F 1
32 + F 133 + F 2
31 + F 134 + F 2
32 + F 233 + F 2
34
F 141 + F 1
42 + F 143 + F 1
44
F 151 + F 1
52 + F 153 + F 1
54
F 161 + F 1
62 + F 163 + F 1
64
F 171 + F 1
72 + F 173 + F 2
71 + F 174 + F 2
72 + F 273 + F 2
74
F 181 + F 1
82 + F 183 + F 2
81 + F 184 + F 2
82 + F 283 + F 2
84
F 191 + F 1
92 + F 193 + F 2
91 + F 194 + F 2
92 + F 293 + F 2
94
F 1101 + F 1
102 + F 1103 + F 2
41 + F 1104 + F 2
42 + F 243 + F 2
44
F 1111 + F 1
112 + F 1113 + F 2
51 + F 1114 + F 2
52 + F 253 + F 2
54
F 1121 + F 1
122 + F 1123 + F 2
61 + F 1124 + F 2
62 + F 263 + F 2
64
F 2101 + F 2
102 + F 2103 + F 2
104
F 2111 + F 2
112 + F 2113 + F 2
114
F 2121 + F 2
122 + F 2123 + F 2
124
U. Saravanan (IIT Madras) Finite element method April 2014 64 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Surface traction vector for the mesh being studied afterapplication of boundary condition
{Fs} =
F 111 + px
S135
3F 1
21
F 131
F 141
F 151
F 161
F 171 + px
S135
3F 1
81
F 191
000
pxS135
300
There are 9 unknowns in thesurface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 65 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Surface traction vector for the mesh being studied afterapplication of boundary condition
{Fs} =
F 111 + px
S135
3F 1
21
F 131
F 141
F 151
F 161
F 171 + px
S135
3F 1
81
F 191
000
pxS135
300
There are 9 unknowns in thesurface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 65 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Surface traction vector for the mesh being studied afterapplication of boundary condition
{Fs} =
F 111 + px
S135
3F 1
21
F 131
F 141
F 151
F 161
F 171 + px
S135
3F 1
81
F 191
000
pxS135
300
There are 9 unknowns in thesurface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 65 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Surface traction vector for the mesh being studied afterapplication of boundary condition
{Fs} =
F 111 + px
S135
3F 1
21
F 131
F 141
F 151
F 161
F 171 + px
S135
3F 1
81
F 191
000
pxS135
300
There are 9 unknowns in thesurface traction vector
U. Saravanan (IIT Madras) Finite element method April 2014 65 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Displacement boundary condition
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
The non-zero components of{UFEM} is also same as that of{UFEM}There are 6 unknowndisplacement dof
{UFEM} =
U1
V1
W1
U2
V2
W2
U3
V3
W3
U4
V4
W4
U5
V5
W5
=
000000000U4
V4
W4
U5
V5
W5
U. Saravanan (IIT Madras) Finite element method April 2014 66 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Displacement boundary condition
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
The non-zero components of{UFEM} is also same as that of{UFEM}There are 6 unknowndisplacement dof
{UFEM} =
U1
V1
W1
U2
V2
W2
U3
V3
W3
U4
V4
W4
U5
V5
W5
=
000000000U4
V4
W4
U5
V5
W5
U. Saravanan (IIT Madras) Finite element method April 2014 66 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Displacement boundary condition
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
The non-zero components of{UFEM} is also same as that of{UFEM}There are 6 unknowndisplacement dof
{UFEM} =
U1
V1
W1
U2
V2
W2
U3
V3
W3
U4
V4
W4
U5
V5
W5
=
000000000U4
V4
W4
U5
V5
W5
U. Saravanan (IIT Madras) Finite element method April 2014 66 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Displacement boundary condition
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
The non-zero components of{UFEM} is also same as that of{UFEM}
There are 6 unknowndisplacement dof
{UFEM} =
U1
V1
W1
U2
V2
W2
U3
V3
W3
U4
V4
W4
U5
V5
W5
=
000000000U4
V4
W4
U5
V5
W5
U. Saravanan (IIT Madras) Finite element method April 2014 66 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Displacement boundary condition
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
The non-zero components of{UFEM} is also same as that of{UFEM}There are 6 unknowndisplacement dof
{UFEM} =
U1
V1
W1
U2
V2
W2
U3
V3
W3
U4
V4
W4
U5
V5
W5
=
000000000U4
V4
W4
U5
V5
W5
U. Saravanan (IIT Madras) Finite element method April 2014 66 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Displacement boundary condition
ex
ey
ez
4
1
2
3 5 I II
2 4
1
3I
1
3II
2 4
4
5
13
(a) (b)
The surface with vertices1− 2− 3 is fixed in all threedirections for all times
The non-zero components of{UFEM} is also same as that of{UFEM}There are 6 unknowndisplacement dof
{UFEM} =
U1
V1
W1
U2
V2
W2
U3
V3
W3
U4
V4
W4
U5
V5
W5
=
000000000U4
V4
W4
U5
V5
W5
U. Saravanan (IIT Madras) Finite element method April 2014 66 / 81
Finite element formulation for linearized elasticity Imposition of the boundary condition
Finite element governing equation
The 15 unknowns - 9 in surface traction vector and 6 in displacementvector has to be determined from 15 linear equations
(K){UFEM}+ (M){UFEM} = {Fs}+ {Fb} = {FFEM} (60)
U. Saravanan (IIT Madras) Finite element method April 2014 67 / 81
Finite element formulation for linearized elasticity Solution procedure
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 68 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Reformulation of the finite element governing equation
((K11) (K12)(K21) (K22)
){{U1}{U2}
}=
{{F1}{F2}
}, (61)
Where
{U1} is the column of knowndisplacement variables
{U2} is the column of unknowndisplacement variables
{F1} is the column of unknownsurface traction vector
{F2} is the column of knownsurface traction vector
(Kij) are matrices obtained from partitioning the global stiffnessmatrix, depending on the number of known and unknowndisplacement variables
U. Saravanan (IIT Madras) Finite element method April 2014 69 / 81
Finite element formulation for linearized elasticity Solution procedure
Determination of unknown displacement and tractionvector
Writing the above equation as two matrix equations
(K11){U1}+ (K12){U2} = {F1} (62)
(K21){U1}+ (K22){U2} = {F2} (63)
Solving for {U2} we obtain
{U2} = (K22)−1[{F2} − (K21){U1}] (64)
Then {F1} is computed from
{F1} = (K11){U1}+ (K12)(K22)−1[{F2} − (K21){U1}] (65)
U. Saravanan (IIT Madras) Finite element method April 2014 70 / 81
Finite element formulation for linearized elasticity Solution procedure
Determination of unknown displacement and tractionvector
Writing the above equation as two matrix equations
(K11){U1}+ (K12){U2} = {F1} (62)
(K21){U1}+ (K22){U2} = {F2} (63)
Solving for {U2} we obtain
{U2} = (K22)−1[{F2} − (K21){U1}] (64)
Then {F1} is computed from
{F1} = (K11){U1}+ (K12)(K22)−1[{F2} − (K21){U1}] (65)
U. Saravanan (IIT Madras) Finite element method April 2014 70 / 81
Finite element formulation for linearized elasticity Solution procedure
Determination of unknown displacement and tractionvector
Writing the above equation as two matrix equations
(K11){U1}+ (K12){U2} = {F1} (62)
(K21){U1}+ (K22){U2} = {F2} (63)
Solving for {U2} we obtain
{U2} = (K22)−1[{F2} − (K21){U1}] (64)
Then {F1} is computed from
{F1} = (K11){U1}+ (K12)(K22)−1[{F2} − (K21){U1}] (65)
U. Saravanan (IIT Madras) Finite element method April 2014 70 / 81
Finite element formulation for linearized elasticity Post processing of the results
1 Finite element formulationSix steps in finite element analysis
2 Finite element formulation for linearized elasticityDecomposition of the domain - Selection of elementDerivation of elemental equationsAssembly of element equationsImposition of the boundary conditionSolution procedurePost processing of the resultsRemarks
U. Saravanan (IIT Madras) Finite element method April 2014 71 / 81
Finite element formulation for linearized elasticity Post processing of the results
Computing the displacement field
u =
U1 =
∑ni=1 d
1i υ
1i
U2 =∑n
i=1 d2i υ
2i
...Ue =
∑ni=1 d
ei υ
ei
, (66)
where e is the number of elements in the mesh and n is the number ofdegrees of freedom. Depending on the value of x, the correspondingelement in equation (66) is used
U. Saravanan (IIT Madras) Finite element method April 2014 72 / 81
Finite element formulation for linearized elasticity Post processing of the results
Computing the displacement gradient
grad(u) =
grad(U1) =
∑ni=1 d
1i grad(υ1
i )grad(U2) =
∑ni=1 d
2i grad(υ2
i )...
grad(Ue) =∑n
i=1 dei grad(υe
i )
(67)
Note that the derivative calculated from different elements meeting at anode is always discontinuous in all approximations in which only thefunction values are interpolated, unless the approximate solution coincideswith the actual solution
U. Saravanan (IIT Madras) Finite element method April 2014 73 / 81
Finite element formulation for linearized elasticity Post processing of the results
Computing the surface traction
Two ways to compute surface traction
1. Determined from elemental equation:
{fs}equil = (Ke){d}+ (Me){d} − {Fb} (68)
2. Determined from the definition:
{fes }def =
∫∂At
t(n) · υids (69)
where t(n) is computed from its definition, t(n) = σn, where theCauchy stress is determined from Hooke’s law and strain deducedfrom the computed gradient of the displacement
U. Saravanan (IIT Madras) Finite element method April 2014 74 / 81
Finite element formulation for linearized elasticity Post processing of the results
Computing the surface traction
Two ways to compute surface traction
1. Determined from elemental equation:
{fs}equil = (Ke){d}+ (Me){d} − {Fb} (68)
2. Determined from the definition:
{fes }def =
∫∂At
t(n) · υids (69)
where t(n) is computed from its definition, t(n) = σn, where theCauchy stress is determined from Hooke’s law and strain deducedfrom the computed gradient of the displacement
U. Saravanan (IIT Madras) Finite element method April 2014 74 / 81
Finite element formulation for linearized elasticity Post processing of the results
Points to understand
Since {fes }defi are calculated using the approximate displacement field,
they are not as accurate as {fes }equili
{fes }defi 6= {fes }equili
Because of computational aspects, in finite element computer codes{fes }defi are calculated instead of {fes }
equili
However, the difference between {fes }defi and {fes }equili decreases as
the number of elements is increased or the degree of the interpolationis increased, hopefully!
U. Saravanan (IIT Madras) Finite element method April 2014 75 / 81
Finite element formulation for linearized elasticity Post processing of the results
Points to understand
Since {fes }defi are calculated using the approximate displacement field,
they are not as accurate as {fes }equili
{fes }defi 6= {fes }equili
Because of computational aspects, in finite element computer codes{fes }defi are calculated instead of {fes }
equili
However, the difference between {fes }defi and {fes }equili decreases as
the number of elements is increased or the degree of the interpolationis increased, hopefully!
U. Saravanan (IIT Madras) Finite element method April 2014 75 / 81
Finite element formulation for linearized elasticity Post processing of the results
Points to understand
Since {fes }defi are calculated using the approximate displacement field,
they are not as accurate as {fes }equili
{fes }defi 6= {fes }equili
Because of computational aspects, in finite element computer codes{fes }defi are calculated instead of {fes }
equili
However, the difference between {fes }defi and {fes }equili decreases as
the number of elements is increased or the degree of the interpolationis increased, hopefully!
U. Saravanan (IIT Madras) Finite element method April 2014 75 / 81
Finite element formulation for linearized elasticity Post processing of the results
Points to understand
Since {fes }defi are calculated using the approximate displacement field,
they are not as accurate as {fes }equili
{fes }defi 6= {fes }equili
Because of computational aspects, in finite element computer codes{fes }defi are calculated instead of {fes }
equili
However, the difference between {fes }defi and {fes }equili decreases as
the number of elements is increased or the degree of the interpolationis increased, hopefully!
U. Saravanan (IIT Madras) Finite element method April 2014 75 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 1: Determining {fes }i for point forces
If a point force, Fo acts at a point, say xo , then
t(n)(x) = Foδ(x− xo), (70)
where the Dirac delta function δ(·) is defined by∫ ∞−∞
F (x)δ(x− xo)dv = F (xo). (71)
Hence, when traction vector is given by (70)
{fes }i =
∫∂At
t(n) · υids = Fo · υi (xo). (72)
U. Saravanan (IIT Madras) Finite element method April 2014 76 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 1: Determining {fes }i for point forces
If a point force, Fo acts at a point, say xo , then
t(n)(x) = Foδ(x− xo), (70)
where the Dirac delta function δ(·) is defined by∫ ∞−∞
F (x)δ(x− xo)dv = F (xo). (71)
Hence, when traction vector is given by (70)
{fes }i =
∫∂At
t(n) · υids = Fo · υi (xo). (72)
U. Saravanan (IIT Madras) Finite element method April 2014 76 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 2: Sources of error in the finite element solution
1. Domain approximation error, which is due to the approximation of thedomain
2. Computational errors, which are due to inexact evaluation of thecoefficients of the element stiffness matrix and force vector or areintroduced owing to the finite arithmetic in a computer.
3. Approximation errors, which is due to approximation of the solutionsby piecewise polynomials.
The main source of error in a boundary value problem is the thirderror.
U. Saravanan (IIT Madras) Finite element method April 2014 77 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 2: Sources of error in the finite element solution
1. Domain approximation error, which is due to the approximation of thedomain
2. Computational errors, which are due to inexact evaluation of thecoefficients of the element stiffness matrix and force vector or areintroduced owing to the finite arithmetic in a computer.
3. Approximation errors, which is due to approximation of the solutionsby piecewise polynomials.
The main source of error in a boundary value problem is the thirderror.
U. Saravanan (IIT Madras) Finite element method April 2014 77 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 2: Sources of error in the finite element solution
1. Domain approximation error, which is due to the approximation of thedomain
2. Computational errors, which are due to inexact evaluation of thecoefficients of the element stiffness matrix and force vector or areintroduced owing to the finite arithmetic in a computer.
3. Approximation errors, which is due to approximation of the solutionsby piecewise polynomials.
The main source of error in a boundary value problem is the thirderror.
U. Saravanan (IIT Madras) Finite element method April 2014 77 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 2: Sources of error in the finite element solution
1. Domain approximation error, which is due to the approximation of thedomain
2. Computational errors, which are due to inexact evaluation of thecoefficients of the element stiffness matrix and force vector or areintroduced owing to the finite arithmetic in a computer.
3. Approximation errors, which is due to approximation of the solutionsby piecewise polynomials.
The main source of error in a boundary value problem is the thirderror.
U. Saravanan (IIT Madras) Finite element method April 2014 77 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 3: Symmetry of the stiffness and mass matrix
The element matrices - stiffness and mass matrix - derived fromGalerkin method is symmetric, irrespective of the type of elementused or the order of the polynomial used to approximate the solution
Because of the symmetry of the element matrices, the assembledglobal matrix will also be symmetric
The symmetry of the global and elemental matrices depends on thedifferential equation, the weighted residual formulation andnumbering of the finite element equations.
U. Saravanan (IIT Madras) Finite element method April 2014 78 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 3: Symmetry of the stiffness and mass matrix
The element matrices - stiffness and mass matrix - derived fromGalerkin method is symmetric, irrespective of the type of elementused or the order of the polynomial used to approximate the solution
Because of the symmetry of the element matrices, the assembledglobal matrix will also be symmetric
The symmetry of the global and elemental matrices depends on thedifferential equation, the weighted residual formulation andnumbering of the finite element equations.
U. Saravanan (IIT Madras) Finite element method April 2014 78 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 3: Symmetry of the stiffness and mass matrix
The element matrices - stiffness and mass matrix - derived fromGalerkin method is symmetric, irrespective of the type of elementused or the order of the polynomial used to approximate the solution
Because of the symmetry of the element matrices, the assembledglobal matrix will also be symmetric
The symmetry of the global and elemental matrices depends on thedifferential equation, the weighted residual formulation andnumbering of the finite element equations.
U. Saravanan (IIT Madras) Finite element method April 2014 78 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 4: Sparseness of the stiffness and mass matrix
Since, Kij = Mij = 0 if global nodes i and j do not belong to thesame element, the global matrices are banded, i.e., all componentsbeyond a certain distance from the diagonal are zero
A banded matrix is a sparse matrix
Thus, the sparseness of the matrix is a result of the finite elementinterpolation functions, which have non-zero values only over anelement of the domain.
U. Saravanan (IIT Madras) Finite element method April 2014 79 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 4: Sparseness of the stiffness and mass matrix
Since, Kij = Mij = 0 if global nodes i and j do not belong to thesame element, the global matrices are banded, i.e., all componentsbeyond a certain distance from the diagonal are zero
A banded matrix is a sparse matrix
Thus, the sparseness of the matrix is a result of the finite elementinterpolation functions, which have non-zero values only over anelement of the domain.
U. Saravanan (IIT Madras) Finite element method April 2014 79 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 4: Sparseness of the stiffness and mass matrix
Since, Kij = Mij = 0 if global nodes i and j do not belong to thesame element, the global matrices are banded, i.e., all componentsbeyond a certain distance from the diagonal are zero
A banded matrix is a sparse matrix
Thus, the sparseness of the matrix is a result of the finite elementinterpolation functions, which have non-zero values only over anelement of the domain.
U. Saravanan (IIT Madras) Finite element method April 2014 79 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 5: Estimate of stress
The balance of the surface forces at the inter element surfaces isexpressed in a weighted integral sense, is an approximation over theactual requirement
Even this condition will not be satisfied by the finite element solution
Since, {fes }defi 6= {fes }equili , finite element solution also only ensures
that (F eiJ)equil + (F e+1
kL )equil = 0 and not (F eiJ)def + (F e+1
kL )def = 0.
Thus, the conditions on the traction are only satisfied approximatelyand consequently the estimate of the stresses by the finite elementmethod is also poor in comparison to the estimate of the displacement
U. Saravanan (IIT Madras) Finite element method April 2014 80 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 5: Estimate of stress
The balance of the surface forces at the inter element surfaces isexpressed in a weighted integral sense, is an approximation over theactual requirement
Even this condition will not be satisfied by the finite element solution
Since, {fes }defi 6= {fes }equili , finite element solution also only ensures
that (F eiJ)equil + (F e+1
kL )equil = 0 and not (F eiJ)def + (F e+1
kL )def = 0.
Thus, the conditions on the traction are only satisfied approximatelyand consequently the estimate of the stresses by the finite elementmethod is also poor in comparison to the estimate of the displacement
U. Saravanan (IIT Madras) Finite element method April 2014 80 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 5: Estimate of stress
The balance of the surface forces at the inter element surfaces isexpressed in a weighted integral sense, is an approximation over theactual requirement
Even this condition will not be satisfied by the finite element solution
Since, {fes }defi 6= {fes }equili , finite element solution also only ensures
that (F eiJ)equil + (F e+1
kL )equil = 0 and not (F eiJ)def + (F e+1
kL )def = 0.
Thus, the conditions on the traction are only satisfied approximatelyand consequently the estimate of the stresses by the finite elementmethod is also poor in comparison to the estimate of the displacement
U. Saravanan (IIT Madras) Finite element method April 2014 80 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 5: Estimate of stress
The balance of the surface forces at the inter element surfaces isexpressed in a weighted integral sense, is an approximation over theactual requirement
Even this condition will not be satisfied by the finite element solution
Since, {fes }defi 6= {fes }equili , finite element solution also only ensures
that (F eiJ)equil + (F e+1
kL )equil = 0 and not (F eiJ)def + (F e+1
kL )def = 0.
Thus, the conditions on the traction are only satisfied approximatelyand consequently the estimate of the stresses by the finite elementmethod is also poor in comparison to the estimate of the displacement
U. Saravanan (IIT Madras) Finite element method April 2014 80 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 6: Estimate of the displacement field
The displacement field obtained from finite element method satisfiesthe equilibrium equation only in an approximate sense, since theprocedure is based on the weak form of the differential equation.
The assumed displacement field, in many cases, does not contain theactual displacement field and hence there will be an error in theestimated displacements
U. Saravanan (IIT Madras) Finite element method April 2014 81 / 81
Finite element formulation for linearized elasticity Remarks
Remark - 6: Estimate of the displacement field
The displacement field obtained from finite element method satisfiesthe equilibrium equation only in an approximate sense, since theprocedure is based on the weak form of the differential equation.
The assumed displacement field, in many cases, does not contain theactual displacement field and hence there will be an error in theestimated displacements
U. Saravanan (IIT Madras) Finite element method April 2014 81 / 81