FEM FOR LINEAR ELASTICITY

FEM FOR LINEAR ELASTICITY

INTRODUCTION TO COMPUTATIONAL MECHANICS OF MATERIALS Civil Engineering, 1st cycle studies, 7th semester

elective subject academic year 2014/2015

Institute L-5, Faculty of Civil Engineering, Cracow University of Technology

Piotr Pluciński Adam Wosatko

Jerzy Pamin

Lecture contents

Equilibrium state of body FEM discretization

FE approximation

Equilibrium of structure discretized using FEs Simplest two-dimensional FE

Types of linear elastic problems Three-dimensional continuum Plane stress

Plane strain

Axi-symmetric problem Numerical examples

Membrane in plane stress conditions Brazilian test – plane strain

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Deformable body

Deformable body V – distances between particles (material points) vary in time

Material point P – geometric point with mass

Y Z

X

S t

P

V

Body and surface forces

Y Z

X

S

n t

u(x , y , z) ρb(x , y , z)

P

P ⁰ V

Vector of body force density [N/m ³ ]

ρ b = ρ



 0 0

−g





Vector of surface force density (traction) [N/m ² ]

t =



 t x

t y

t z





Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Equilibrium state

Y Z

X

S

n t

u(x , y , z) ρb(x , y , z)

P

P ⁰ V

Equilibrium equation for a body:

Z

S

t dS + Z

V

ρ b dV = 0

Static boundary conditions:

t = σ n where σ – stress tensor

(Voigt’s notation) Z

S

σ n dS + Z

V

ρ b dV = 0

Equilibrium equation

Green–Gauss–Ostrogradsky (divergence) theorem:

Z

S

σ n dS = Z

V

L ^T σ dV where L – matrix of differential operators

Navier equations

Z

V

L ^T σ + ρ b
dV = 0 ⇐⇒ L ^T σ + ρ b = 0 ∀P ∈ V

σ ij ,j + ρ b i = 0

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Equilibrium equation

Green–Gauss–Ostrogradsky (divergence) theorem:

Z

S

σ n dS = Z

V

L ^T σ dV where L – matrix of differential operators

Navier equations

Z

V

L ^T σ + ρ b
dV = 0 ⇐⇒ L ^T σ + ρ b = 0 ∀P ∈ V

σ ij ,j + ρ b i = 0

Set of equations for linear elasticity

∀P ∈ V

Equilibrium equations

L ^T σ + ρ b = 0 σ ij ,j + ρ b i = 0

Kinematic equations

Linear relation: small strains

 = L u  ij = ¹ ₂ (u i ,j + u j ,i )

Constitutive equations

Linear relation: Hooke’s law

σ = D σ ij = D ijkl  kl

+ static and/or kinematic boundary conditions

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Equilibrium equation – weak form

We introduce a weighting function w which can be interpreted as kinematically admissible variation of displacement δu (which satisfies kinematic boundary conditions):

Z

V

(δu) ^T L ^T σ + ρb
dV = 0 ∀δu

Weak form

Using of Green–Gauss–Ostrogradsky theorem:

− Z

V

(Lδu) ^T σdV + Z

S

(δu) ^T σndS + Z

V

(δu) ^T ρbdV = 0 Z

V

( L δu) ^T σdV = Z

S

(δu) ^T tdS + Z

V

(δu) ^T ρbdV

virtual work of internal forces = virtual work of external forces

This is the virtual work principle.

Approximation of displacement field u

N ^e – matrix of shape functions, q ^e – vector of displacement DOFs u = N ^e q ^e

N ^e

[3×N] =





N ₁ ^e 0 0 . . . N _w ^e 0 0 0 N ₁ ^e 0 . . . 0 N _w ^e 0 0 0 N ₁ ^e . . . 0 0 N _w ^e



 q ^e

[N×1]

=



 q ^e ₁ . . . q ^e _N





y z

x 2

3 6

8 9

11 14

17

η ζ

ξ

η ζ

ξ i j

k l m

n

o q ^e = I T ^e Q p

I

T ^e – transformation matrix which specifies topology

and directional cosines between local and global coordinate axes

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Equilibrium equation

Application of FE approximation

Equilibrium equation for FE model – weak form

E

X

e=1

Z

V ^e

(L ^e δu ^e ) ^T σ ^e dV ^e − Z

S ^e

(δu ^e ) ^T t ^e dS ^e − Z

V ^e

(δu ^e ) ^T f ^e dV ^e



= 0

where: f ^e = ρb ^e – vector of body forces and P E

e=1 denotes assembly We introduce approximation: δu ^e = N ^e δq ^e

and matrix of derivatives of shape functions: B ^e = L ^e N ^e

E

X

e=1

Z

V ^e

(B ^e δq ^e ) ^T σ ^e dV ^e − Z

S ^e

(N ^e δq ^e ) ^T t ^e dS ^e − Z

V ^e

(N ^e δq ^e ) ^T f ^e dV ^e



= 0

E

X

e=1

(δq ^e ) ^T

Z

V ^e

B ^{e T} σ ^e dV ^e − Z

S ^e

N ^{e T} t ^e dS ^e − Z

V ^e

N ^{e T} f ^e dV ^e



= 0

Equilibrium of discretized structure

We take into account topology and relations between local and global coordinate systems for all FEs

δq ^e = I T ^e δQ

(δQ) ^T

E

X

e=1

I T ^{e T}

Z

V ^e

B ^{e T} σ ^e dV ^e − Z

S ^e

N ^{e T} t ^e dS ^e − Z

V ^e

N ^{e T} f ^e dV ^e



= 0

This equation is satisfied ∀δQ:

E

X

e=1

I T ^{e T}

Z

V ^e

B ^{e T} σ ^e dV ^e − Z

S ^e

N ^{e T} t ^e dS ^e − Z

V ^e

N ^{e T} f ^e dV ^e



= 0

E

X

e=1

I T ^{e T}

Z

V ^e

B ^{e T} σ ^e dV ^e



=

E

X

e=1

I T ^{e T}

Z

S ^e

N ^{e T} t ^e dS ^e + Z

V ^e

N ^{e T} f ^e dV ^e



internal nodal forces = external nodal forces

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Equilibrium of internal and external forces

E

X

e=1

I T ^{e T}

Z

V ^e

B ^{e T} σ ^e dV ^e



=

E

X

e=1

I T ^{e T}

Z

S ^e

N ^{e T} t ^e dS ^e + Z

V ^e

N ^{e T} f ^e dV ^e



internal nodal forces = external nodal forces

We substitute contitutive and kinematic equations

linear elasticity: σ = Dε

linear kinematic relation: ε = Lu

σ ^e = D ^e L ^e u ^e = D ^e L ^e N ^e q ^e = D ^e B ^e T I ^e Q

E

X

e=1

I T ^{e T}

Z

V ^e

B ^{e T} D ^e B ^e dV ^e

 I T ^e Q =

E

X

e=1

I T ^{e T}

Z

S ^e

N ^{e T} t ^e dS ^e + Z

V ^e

N ^{e T} f ^e dV ^e



Element matrices

E

X

e=1

I T ^{e T}

 R

V ^e B ^{e T} D ^e B ^e dV ^e k ^e

 I T ^e Q =

E

X

e=1

I T ^{e T}

 R

S ^e N ^{e T} t ^e dS ^e p ^e _b

+ R

V ^e N ^{e T} f ^e dV ^e p ^e



E

X

e=1

I

T ^{e T} k ^e T I ^e Q =

E

X

e=1

I T ^{e T} p ^e _b +

E

X

e=1

I T ^{e T} p ^e

E

P

e=1

I T ^{e T} k ^e T I ^e

K

Q =

E

P

e=1

I T ^{e T} p ^e _b P b

+

E

P

e=1

I T ^{e T} p ^e

P K Q = P b + P

K Q = F where:

K – global stiffness matrix for the model

Q – global vector of DOFs (nodal displacements) F – right-hand side (nodal force) vector

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Triangular element

u ^e (x , y ) = N ^e (x , y )q ^e

N ^e =

 N _i ^e 0 N _j ^e 0 N _k ^e 0 0 N _i ^e 0 N _j ^e 0 N _k ^e

 , q ^e =















 q 1

q 2

q 3

q 4

q ₅ q ₆

















x ^e y ^e

i

k

j q 1 e

q 2

q ₃ q 4

q 5

q 6

y ^e N i (x ^e , y ^e )

x ^e 1 i

k j

y ^e N j (x ^e , y ^e )

x ^e 1

i k

j y ^e

N k (x ^e , y ^e )

x ^e

1 i

k

j

Quadrilateral element

u ^e (x , y ) = N ^e (x , y )q ^e N ^e =

 N _i ^e 0 N _j ^e 0 N _k ^e 0 N _l ^e 0 0 N _i ^e 0 N _j ^e 0 N _k ^e 0 N _l ^e



q ^e =























 q 1

q 2

q 3

q 4

q 5

q 6

q 7

q 8

























x ^e y ^e

i

k

j l

e q 1

q ₂

q 3

q ₄ q 5

q ₆ q 7

q ₈

y ^e N i (x , y )

x ^e 1

i

j

k

l y ^e

N l (x , y )

x ^e i 1 j

k l y ^e

N j (x , y )

x ^e 1 j i

k

l y ^e

N k (x , y )

x ^e 1

i j

k l

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Three-dimensional continuum

Vector of displacement functions: u = {u(x , y , z), v (x , y , z), w (x , y , z)}

Vector of strain components: ε = {ε _x , ε _y , ε _z , γ _xy , γ _yz , γ _zx } Vector of stress components: σ = {σ x , σ y , σ z , τ xy , τ yz , τ zx }

Constitutive relations for linear elasticity















 σ x

σ y

σ z

τ xy

τ yz

τ zx

















= E n ν

















1 − ν ν ν 0 0 0

ν 1 − ν ν 0 0 0

ν ν 1 − ν 0 0 0

0 0 0 ^1−2ν ₂ 0 0

0 0 0 0 ^1−2ν ₂ 0

0 0 0 0 0 ^1−2ν ₂































 ε x

ε y

ε z

γ xy

γ yz

γ zx

















n ν = (1 + ν)(1 − 2ν)

Plane stress (σ _z = 0)

Vector of displacement functions:

u = {u(x , y ), v (x , y )}

Vector of strain components:

ε = {ε x , ε y , ε z , γ xy }

Vector of stress components:

σ = {σ x , σ y , τ xy }

Constitutive relations for linear elasticity







 σ x

σ y

σ z

τ _xy









= E n ν









1 ν 0 0

ν 1 0 0

0 0 0 0

0 0 0 ^1−ν ₂















 ε x

ε y

ε z

γ _xy







 n _ν = 1 − ν ²

ε z = − _1−ν ^ν (ε

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

x + ε y )

Plane stress (σ _z = 0) – FE

Stiffness matrix for FE

k ^e = Z

A ^e

B ^{e T} D ^e B ^e h ^e dA ^e

Loading vector for FE

p ^e = Z

A ^e

N ^{e T} f ^e h ^e dA ^e

Vector of boundary loading

p ^e _b = Z

Γ ^e

N ^{e T} t ^e h ^e dΓ ^e

A ^e – FE area, h ^e – FE thickness

x ^e y ^e

Γ ^e

A ^e

Plane strain (ε _z = 0)

Vector of displacement functions:

u = {u(x , y ), v (x , y )}

Vector of strain components:

ε = {ε x , ε y , γ xy }

Vector of stress components:

σ = {σ x , σ y , σ z , τ xy }

Constitutive relations for linear elasticity







 σ x

σ y

σ z

τ xy









= E n _ν









1 − ν ν ν 0

ν 1 − ν ν 0

ν ν 1 − ν 0

0 0 0 ^1−2ν ₂















 ε x

ε y

ε z

γ xy









n ν = (1 + ν)(1 − 2ν)

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Axi-symmetric problem

Vector of displacement functions:

u = {u(r , z), v (r , z)}

Vector of strain components:

ε = {ε r , ε θ , ε z , ε rz }

Vector of stress components:

σ = {σ _r , σ _θ , σ _z , σ _rz }

Constitutive relations







 σ _r σ θ

σ z

σ rz









= E n _ν









1 − ν ν ν 0

ν 1 − ν ν 0

ν ν 1 − ν 0

0 0 0 ^1−2ν ₂















 ε _r ε θ

ε z

ε rz









n ν = (1 + ν)(1 − 2ν)

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25 h = 0.2 m

X Y

4 i

5 j

2 k 1

l

Discretization

i j

3 k

1 2

Q 1

Q 2

Q 3

Q 4

Q 5

Q 6

Q 7

Q 8

Q 9

Q 10

elem. no. elem. node nos

1 4 5 2 1

2 5 3 2

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m X

Y Discretization

4

i j

k 1

l 1

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1

Q2 Q3 Q4

Q5

Q6 Q7

Q8 Q9

Q10

Matrix of constitutive relations

D = 18 · 10 ⁶ 1 − 0.25 ²





1 0.25 0

0.25 1 0

0 0 ^1−0.25 ₂



 [kPa]

D =





19.2 4.8 0 4.8 19.2 0

0 0 7.2



 · 10 ⁶ [kPa]

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Shape functions – Element 1

N _i ¹ (x ⁽¹⁾ , y ⁽¹⁾ ) = x ⁽¹⁾ y ⁽¹⁾ − 2x ⁽¹⁾ − 4y ⁽¹⁾ + 8

8 , N _k ¹ (x ⁽¹⁾ , y ⁽¹⁾ ) = x ⁽¹⁾ y ⁽¹⁾ 8

N _j ¹ (x ⁽¹⁾ , y ⁽¹⁾ ) = x ⁽¹⁾ y ⁽¹⁾ − 2x ⁽¹⁾

8 , N _l ¹ (x ⁽¹⁾ , y ⁽¹⁾ ) = x ⁽¹⁾ y ⁽¹⁾ − 4y ⁽¹⁾ 8 N ¹ =

 N _i ¹ 0 N _j ¹ 0 N _k ¹ 0 N _l ¹ 0 0 N _i ¹ 0 N _j ¹ 0 N _k ¹ 0 N _l ¹



Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Matrix K – Element 1

B ¹ (x ⁽¹⁾ , y ⁽¹⁾ ) =











y ⁽¹⁾ 8 − ¹

4 0 ¹ ₄ − ^{y (1)}

8 0 ^{y (1)} ₈ 0 − ^{y (1)}

8 0

0 ^x ₈ ⁽¹⁾ − ¹

2 0 − ^{x (1)}

8 0 ^x ₈ ⁽¹⁾ 0 ¹ ₂ − ^{x (1)}

8 x ⁽¹⁾

8 − ¹ ₂ ^y ₈ ⁽¹⁾ − ₄ ¹ − ^{x (1)} ₈ ¹ ₄ − ^y ₈ ^{(1) x (1)} ₈ ^y ₈ ^{(1) 1} ₂ − ^{x (1)} ₈ − ^{y (1)} ₈











K ¹ = Z ₂

0

Z ₄

0

B ^{1 T} DB ¹ h dx ⁽¹⁾ dy ⁽¹⁾ =























16 6 -1.6 -1.2 -8 -6 -6.4 1.2 6 28 1.2 10.4 -6 -14 -1.2 -24.4 -1.6 1.2 16 -6 -6.4 -1.2 -8 6 -1.2 10.4 -6 28 1.2 -24.4 6 -14 -8 -6 -6.4 1.2 16 6 -1.6 -1.2 -6 -14 -1.2 -24.4 6 28 1.2 10.4 -6.4 -1.2 -8 6 -1.6 1.2 16 -6 1.2 -24.4 6 -14 -1.2 10.4 -6 28























· 10 ⁵

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (2)

Discretization y (2)

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q2 5 = Q3

Q2 6 = Q4 Q2 3 = Q5

Q2 4 = Q6 Q2 1 = Q9

Q2 2 = Q10

Shape functions – Element 2

N _i ² (x ⁽²⁾ , y ⁽²⁾ ) = 2 − y ⁽²⁾

2 , N _k ² (x ⁽²⁾ , y ⁽²⁾ ) = y ⁽²⁾ − x ⁽²⁾ 2 N _j ² (x ⁽²⁾ , y ⁽²⁾ ) = x ⁽²⁾

2 N ² =

 N _i ² 0 N _j ² 0 N _k ² 0 0 N _i ² 0 N _j ² 0 N _k ²



Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (2)

Discretization y (2)

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q2 5 = Q3

Q2 6 = Q4 Q2 3 = Q5

Q2 4 = Q6 Q2 1 = Q9

Q2 2 = Q10

Matrix K – Element 2

B ² (x ⁽²⁾ , y ⁽²⁾ ) =







0 0 ¹ ₂ 0 - ¹ ₂ 0 0 - ¹ ₂ 0 0 0 ¹ ₂ - ¹ ₂ 0 0 ¹ ₂ ¹ ₂ - ¹ ₂







K ² = B ^{2 T} DB ² hA ² =















7.2 0 0 -7.2 -7.2 7.2 0 19.2 -4.8 0 4.8 -19.2 0 -4.8 19.2 0 -19.2 4.8 -7.2 0 0 7.2 7.2 -7.2 -7.2 4.8 -19.2 7.2 26.4 -12 7.2 -19.2 4.8 -7.2 -12 26.4















· 10 ⁵

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m 6 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Vector P _b – Element 1

P ¹ _b = Z

Γ ¹ _ij

N ^1T t b.c. = 0

dΓ + Z

Γ ¹ _jk

N ^1T tdΓ

common edge force balance along line 2-5 t ¹ _jk = −t ² _ki

+ Z

Γ ¹ _kl

N ^1T tdΓ+

Z

Γ ¹ _li

N ^1T tdΓ

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m 6 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Vector P _b – Element 1

Z

Γ ¹ _kl

N ^1T tdΓ = Z 4

0



N ¹ (x ⁽¹⁾ , y ⁽¹⁾ = 2)  ^T

"

0

−3  1 − ^x ₄ ⁽¹⁾ 

− 6 ^x ₄ ⁽¹⁾

# dx ⁽¹⁾

= {0 0 0 0 0 -10 0 -8}

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m 6 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Vector P _b – Element 1

Z

Γ ¹ _li

N ^1T tdΓ = Z 2

0



N ¹ (x ⁽¹⁾ = 0, y ⁽¹⁾ )  T

tdy ⁽¹⁾

= {R ₁ ¹ R ₂ ¹ 0 0 0 0 R ₇ ¹ R ₈ ¹ }

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m 6 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Vector P _b – Element 1

P ¹ _b =























 R ₁ ¹ R ₂ ¹ 0 0 0 -10 R ₇ ¹ R ₈ ¹ -8

























Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m 6 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (2)

Discretization y (2)

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q2 5 = Q3

Q2 6 = Q4 Q2 3 = Q5

Q2 4 = Q6 Q2 1 = Q9

Q2 2 = Q10

Vector P _b – Element 2

P ² _b = Z

Γ ² _ij

N ^2T t w .b. = 0

dΓ + Z

Γ ² _jk

N ^2T tdΓ+

Z

Γ ² _ki

N ^2T tdΓ

common edge force equilibrium along line 2-5 t ¹ _jk = −t ² _ki

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m 6 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (2)

Discretization y (2)

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q2 5 = Q3

Q2 6 = Q4 Q2 3 = Q5

Q2 4 = Q6 Q2 1 = Q9

Q2 2 = Q10

Vector P _b – Element 2

P ² _b = − Z

Γ ² _jk

N ^2T tdΓ =















 0 0 0 -7

0 -6.5

















Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m X

Y Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1

Q2 Q3 Q4

Q5

Q6 Q7

Q8 Q9

Q10

Assembly - stiffness matrix

K =































16 -6 -1.6 1.2 0 0 -6.4 -1.2 -8 6 -6 28 -1.2 10.4 0 0 1.2 -24.4 6 -14 -1.6 -1.2 42.4 -6 -19.2 7.2 -8 -6 -13.6 6 1.2 10.4 -6 54.4 4.8 -7.2 -6 -14 6 -43.6

0 0 -19.2 4.8 19.2 0 0 0 0 -4.8

0 0 7.2 -7.2 0 7.2 0 0 -7.2 0

-6.4 1.2 -8 -6 0 0 16 6 -1.6 -1.2

-1.2 -24.4 -6 -14 0 0 6 28 1.2 10.4 -8 6 -13.6 6 0 -7.2 -1.6 1.2 23.2 -6 6 -14 6 -43.6 -4.8 0 -1.2 10.4 -6 47.2































· 10 ⁵

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m X

Y Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1

Q2 Q3 Q4

Q5

Q6 Q7

Q8 Q9

Q10

Assembly - loading vector

P b =





























 0 -8 0 -16.5 0 -7 0 0 0 0





























 +





























 R ₇ ¹ = R 1

R ₇ ¹ = R 2

0 0 0 0 R ₁ ¹ = R 7

R ₂ ¹ = R 8

0 0































Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m X

Y Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1

Q2 Q3 Q4

Q5

Q6 Q7

Q8 Q9

Q10

FEM system of equations: KQ = P + P _b































16 -6 -1.6 1.2 0 0 -6.4 -1.2 -8 6 -6 28 -1.2 10.4 0 0 1.2 -24.4 6 -14 -1.6 -1.2 42.4 -6 -19.2 7.2 -8 -6 -13.6 6 1.2 10.4 -6 54.4 4.8 -7.2 -6 -14 6 -43.6 0 0 -19.2 4.8 19.2 0 0 0 0 -4.8

0 0 7.2 -7.2 0 7.2 0 0 -7.2 0

-6.4 1.2 -8 -6 0 0 16 6 -1.6 -1.2 -1.2 -24.4 -6 -14 0 0 6 28 1.2 10.4 -8 6 -13.6 6 0 -7.2 -1.6 1.2 23.2 -6 6 -14 6 -43.6 -4.8 0 -1.2 10.4 -6 47.2































· 10 5





























 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q10 Q9































=





























 0 -8 0 -16.5 0 -7 0 0 0 0





























 +





























 R1 R2 0 0 0 0 R7 R8 0 0































Solution:

Q = {0 0 3.881 -11.03 3.949 -19.62 0 0 -3.744 -10.75} · 10 −5 m

R = {-54 16.744 0 0 0 0 54 14.756 0 0} kN

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m X

Y Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1

Q2 Q3 Q4

Q5

Q6 Q7

Q8 Q9

Q10

FEM system of equations: KQ = P + P _b































16 -6 -1.6 1.2 0 0 -6.4 -1.2 -8 6 -6 28 -1.2 10.4 0 0 1.2 -24.4 6 -14 -1.6 -1.2 42.4 -6 -19.2 7.2 -8 -6 -13.6 6 1.2 10.4 -6 54.4 4.8 -7.2 -6 -14 6 -43.6 0 0 -19.2 4.8 19.2 0 0 0 0 -4.8

0 0 7.2 -7.2 0 7.2 0 0 -7.2 0

-6.4 1.2 -8 -6 0 0 16 6 -1.6 -1.2 -1.2 -24.4 -6 -14 0 0 6 28 1.2 10.4 -8 6 -13.6 6 0 -7.2 -1.6 1.2 23.2 -6 6 -14 6 -43.6 -4.8 0 -1.2 10.4 -6 47.2































· 10 5





























 0 0 Q3 Q4 Q5 Q6 0 0 Q10 Q9































=





























 0 -8 0 -16.5 0 -7 0 0 0 0





























 +





























 R1 R2 0 0 0 0 R7 R8 0 0































Solution:

Q = {0 0 3.881 -11.03 3.949 -19.62 0 0 -3.744 -10.75} · 10 −5 m

R = {-54 16.744 0 0 0 0 54 14.756 0 0} kN

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m X

Y Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1

Q2 Q3 Q4

Q5

Q6 Q7

Q8 Q9

Q10

FEM system of equations: KQ = P + P _b































16 -6 -1.6 1.2 0 0 -6.4 -1.2 -8 6 -6 28 -1.2 10.4 0 0 1.2 -24.4 6 -14 -1.6 -1.2 42.4 -6 -19.2 7.2 -8 -6 -13.6 6 1.2 10.4 -6 54.4 4.8 -7.2 -6 -14 6 -43.6 0 0 -19.2 4.8 19.2 0 0 0 0 -4.8

0 0 7.2 -7.2 0 7.2 0 0 -7.2 0

-6.4 1.2 -8 -6 0 0 16 6 -1.6 -1.2 -1.2 -24.4 -6 -14 0 0 6 28 1.2 10.4 -8 6 -13.6 6 0 -7.2 -1.6 1.2 23.2 -6 6 -14 6 -43.6 -4.8 0 -1.2 10.4 -6 47.2































· 10 5





























 0 0 Q3 Q4 Q5 Q6 0 0 Q10 Q9































=





























 0 -8 0 -16.5 0 -7 0 0 0 0





























 +





























 R1 R2 0 0 0 0 R7 R8 0 0































Solution:

Q = {0 0 3.881 -11.03 3.949 -19.62 0 0 -3.744 -10.75} · 10 −5 m

R = {-54 16.744 0 0 0 0 54 14.756 0 0} kN

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (1)

y (1) Discretization

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q1 7 = Q1

Q1 8 = Q2

Q1 5 = Q3 Q1 6 = Q4

Q1 1 = Q7 Q1 2 = Q8

Q1 3 = Q9 Q1 4 = Q10

Return of Element 1

Q ¹ = {0 0 -3.744 -10.75 3.881 -11.03 0 0} · 10 ⁻⁵ ε ¹ = B ¹ Q ¹

ε ¹ =





0.953y − 0.936

−0.034x 0.953x − 0.034y − 2.688



 · 10

−5 , ε ¹ (2, 1) =



 1.708 6.831

−81.600



 · 10

−7

σ ¹ = Dε ¹

σ ¹ =





182.976y − 179.712 − 1.632x 45.744y − 44.928 − 6.528x 68.616x − 2.448y − 193.536



 , σ ¹ (2, 1) =



 0

−12.297

−58.750



 kPa

Membrane in plane stress conditions

Example

3 kN/m

7.5 kN/m

4 m 2 m

2 m

E = 18 GPa ν = 0.25

h = 0.2 m x (2)

Discretization y (2)

4

i j

k 1

l

1 i

j 3 k 2 2

5 Q2 5 = Q3

Q2 6 = Q4 Q2 3 = Q5

Q2 4 = Q6 Q2 1 = Q9

Q2 2 = Q10

Return of Element 2

Q ² = {-3.744 -10.75 3.949 -19.62 0 0} · 10 ⁻⁵ ε ² = B ² Q ²

ε ² =



 3.416 13.660

−48.610



 · 10

−7

σ ² = Dε ²

σ ² =



 0

−24.593

−35.000



 kPa

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Example for plane strain

Brazilian test, quarter of specimen, linear elasticity, four-node elements

Brazilian test, elasticity

Displacements, vertical stress σ _yy and invariant of stress tensor J ₂ ^σ

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Brazilian test, elasticity

Stress σ _yy for coarse and fine mesh

Results depend strongly on mesh density

Brazilian test, perfect plasticity HMH

Final displacements and stress σ yy

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

Brazilian test, perfect plasticity HMH

Why does the load-displacement diagram exhibit hardening?

Brazilian test, perfect plasticity HMH

It is caused by so-called locking in the four-node FE (eight-node element gives correct result)

Project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union

References

N.S. Ottosen, M. Ristinmaa.The Mechanics of Constitutive Modeling., Elsevier, 2005.

O.C. Zienkiewicz, R.L. Taylor, J.Z. Zhu. The Finite Element Method: Its Basis and Fundamentals., VI edition, Elsevier

Butterworth Heineman, 2005.

R. de Borst, L.J. Sluys. Computational Methods in Non-linear

Solid Mechanics. Lecture notes, Delft University of Technology,

CTme5142, Delft, 1999.