chapter 7 · 2021. 5. 3. · chapter 7. 2d elements. book chapters [o] v1/ch4 & ch6. institute...

Institute of Structural Engineering

Method of Finite Elements I

Chapter 7

2D Elements

Book Chapters[O] V1/Ch4 & Ch6



Today’s Lecture

• Continuum Elements Plane Stress

Plane Strain

Next Lecture• Structural Elements Plate Elements

Shell Elements

Extending (discretization) to higher Dimensions



In the previous lectures:

The Galerkin method was presented The isoparametric concept was introduced The bar and beam elements were presented with extensions to 3D Some numerical integration methods were presented

In today’s lecture:

– Concepts from previous lectures are combined to formulate elements for the above cases

– Linear 3D elasticity equations are used as a starting point– Plane stress/Plane strain elements are presented as special cases

Today’s Lecture (in more detail)



3D elasticity problem

tn

0Γ

tΓ

uΓ

Ω

Problem variables:

Displacements:

x

y

z

uuu

=

u

x xy xz

xy y yz

xz yz z

ε ε εε ε εε ε ε

=

ε

Strain tensor:




00

t

u

at

t at

at

⋅ = Γ

⋅ = Γ

= Γ

σ n

σ n

u u

tn

0Γ

tΓ

uΓ

Boundary conditions:

ΩV

xF dV

zF dV

yF dV

Traction: Distributed force per unit surface areat −




x

y

z

FFF

=

F

tn

0Γ

tΓ

uΓ

Body Forces:distributed force per unit volume (e.g., weight, inertia, etc)

Ω

NOTE: If the body is accelerating, then the inertia force, is added onto the body force vector

x x

y y

z z

F uF uF u

ρρ ρ

ρ

= − = −

F F u

ρuV

xF dV

zF dV

yF dV



3D elasticity equations Internal stresses

zσ

zyτzxτ

xzτ

xyτ

xσ

yzτyσ

xyτ

z

y

x

σx, σy and σz are normal stresses. τxy, τyz and τxz are the shear stresses.

Notationτxy is the stress on the face perpendicular to the x-axis and points in the positive y direction

Stress Tensor:x xy xz

xy y yz

xz yz z

σ τ ττ σ ττ τ σ

=

σ

*total of 9 stress components of which only 6 are independent since:

xy yx

xz zx

yz zy

τ τ

τ ττ τ

=

==



3D elasticity problem In FE analysis Voigt notation is typically used for strains and stresses:Here we use a variant, that is different to standard Voigt notation, which ordersthe z components of shear strains/stresses last.

x xy xz

xy y yz

xz yz z

σ τ ττ σ ττ τ σ

=

σ

x xy xz

xy y yz

xz yz z

ε ε εε ε εε ε ε

=

ε 222

xx x

yy y

zz z

xy xy

xz xz

yz yz

ε εε εε εε γε γε γ

= =

ε

x

y

z

xy

xz

yz

σσστττ

=

σ

engineering strain



3D elasticity equations

Equilibrium equations:

0

0

0

xyx xzx

xy y yzy

yzxz zz

Fx y z

Fx y z

Fx y z

τσ τ

τ σ τ

ττ σ

∂∂ ∂+ + + =

∂ ∂ ∂∂ ∂ ∂

+ + + =∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

0∇⋅ + =σ F

In vector formComponent wise

where 𝑭𝑭 is the applied body force:

x

y

z

FFF

=

F



3D elasticity equations Strain definition:

( )12

s T= ∇ = ∇ +∇ε u u u

In vector form

Component wise

1 12 2

1 12 2

1 12 2

yx x x z

y y yx z

yxz z z

uu u u ux y x z x

u u uu ux y y z y

uuu u ux z y z z

∂ ∂ ∂ ∂ ∂ + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + ∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂ ∂ + + ∂ ∂ ∂ ∂ ∂

ε

Using Voigt notation

x

y

z

yx

x z

y z

uxuyuz

uuy x

u uz x

u uz y

∂ ∂

∂ ∂

∂ ∂ = ∂ ∂

+ ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂

ε



3D elasticity equations Constitutive equation, using Voigt notation can be conveniently written as:

=σ Eε

1 0 0 01 0 0 0

1 0 0 01 20 0 0 0 0

2(1 )(1 2 )1 20 0 0 0 0

21 20 0 0 0 0

2

x x

y y

z z

xy xy

xz xz

yz yz

E

ν ν νν ν νσ εν ν νσ ε

νσ ετ γν ν

ντ γτ γ

ν

− −

− − = + − − −

Assumption: Linear elastic material (Hooke’s Law) & isotropic

E



3D elasticity equations The weak form of the problem can be obtained using the Galerkin method as:

( ) ( )T T Td d dΩ Ω Γ

Ω = ⋅ Ω+ ⋅ Γ∫ ∫ ∫ε w Eε u w F w t

where is the weight functionx

y

z

www

=

w



Plane stress/strain

In several cases of practical interest:

• The third dimension of the problem is: either very small

or very large but includes no variation in the problem parameters

• The problem equations can be simplified resulting in a 2D problem



Definition of strains in the 2D domain

[O] §4 Fig. 4.3



Definition of stresses in 2D

[O] §4 Fig. 4.4: Definition of stresses σx; σy; σxy and principal stresses σI ; σII in 2D solids


Method of Finite Elements I30-Apr-10

First let’s consider a structure where:• The length in one dimension is

much smaller than the other two

• Loads are applied only within a plane

• The in-plane stresses, strains and displacements are constant in the third dimension

• Normal and shear stresses in the third dimension are negligible

Plane stress assumptions

Such a structure is said to be in a state of plane stress



Plane stress equationsAssuming that stresses along the third dimension are zero:

0),(),(),(

=τ=τ=σ

τ=τ

σ=σσ=σ

yzxzz

xyxy

yy

xx

yxyxyx

0

0

=+∂σ∂

+∂τ∂

=+∂τ∂

+∂σ∂

yyxy

xxyx

Fyx

Fyx

, 00

xz

zx

yy xz

xy yzyx

uuxzu

yuu

y x

εεε εγ ε

∂ ∂=∂ ∂ ∂ = = = ∂ = ∂ ∂ + ∂ ∂

ε

( )

2

1 01 0

110 0

2

1

x x

y y

xy xy

z x y

Eσ ν εσ ν ε

νσ ν γ

νε ε εν

= − −

= − +−

Constitutive equationStrain definition

Equilibrium equations:Assumptions:

How is this derived?



Examples of plane stress problems

Circular Plate UnderEdge Loadings

Thin Plate WithCentral Hole



Examples of plane stress problems

[O] §4 Fig. 4.1



Next we consider a structure where:• The length in one dimension is

much larger than the other two

• Loads are applied only within a plane

• Loads are constant in the third dimension

• Displacements and strains along the third dimension are negligible

Such a structure is said to be in a state of plane strain

Plane strain assumptions



Plane strain equationsNext we assume that strains along the third dimension are zero:

0

0

=+∂σ∂

+∂τ∂

=+∂τ∂

+∂σ∂

yyxy

xxyx

Fyx

Fyx

0, 0

0

x

x zy

y xz

xy yzyx

uxuy

uuy x

ε εε γγ γ

∂ ∂ = ∂ = = = ∂ = ∂ ∂ + ∂ ∂

ε

( )

1 01 0

(1 )(1 2 )1 20 0

2

x x

y y

xy xy

z x y

Eσ ν ν εσ ν ν ε

ν νσ ν γ

σ ν σ σ

− = − + − −

= +

Constitutive equationStrain definition

Equilibrium equations:Assumptions:

( , )( , )

0

x x

y y

z

u u x yu u x yu

==

=



Examples of plane strain

x

y

z

x

y

z

P

Long CylindersUnder Uniform Loading

Semi-Infinite Regions Under Uniform Loadings



Examples of plane strain[O

] §4

Fig

. 4.2



Weak form of the problemThe weak form of the problem is similar to the one used for the 3D problem:

( ) ( )T T Td d dΩ Ω Γ

Ω = ⋅ Ω+ ⋅ Γ∫ ∫ ∫ε w Eε u w F w t

where:

stresses, strains, displacements and constitutive matrices correspond to the plain stress/strain case

integration is carried out over an area instead of a volume



Discretization Several options are available for discretizing the

plane stress/strain problem

One of the most common choices is isoparametricLagrange elements

In the following, we will briefly review some basic properties of isoparametric elements

Some specific, and widely used, elements will be presented for plane stress/strain analysis



1 1( , ) ; ( , )

n n

i i i ii i

x h r s x y h r s y= =

= =∑ ∑

Using the isoparametric concept, geometry can be discretized as:

Isoparametric formulation

and displacements as:

1 1( , ) ; u ( , )

n n

x i xi y i yii i

u h r s u h r s u= =

= =∑ ∑

where are nodal values of the spatial coordinates and displacement components

, , ,i i xi yix y u u



Isoparametric quadrilaterals

1 11 ( , )x y

2 22 ( , )x y

3 33 ( , )x y

4 44 ( , )x y

1 ( 1, 1)− − 2 (1, 1)−

4 ( 1,1)− 3 (1,1)

r

s

x

y

For a 4 noded linear isoparametric quadrilateral (q4), coordinates r and s are defined based on the following transformation:



4 4

1 1( , ) ; ( , ) i i i i

i ix h r s x y h r s y

= =

= =∑ ∑

Then the displacement and geometry approximations specialize to:


4 4

1 1( , ) ; u ( , ) x i xi y i yi

i iu h r s u h r s u

= =

= =∑ ∑

with

1 2

3 4

1 1(1 )(1 ); (1 )(1 );4 41 1(1 )(1 ); (1 )(1 );4 4

h r s h r s

h r s h r s

= − − = + −

= + + = − +




1h 2h

3h 4h



1

1

2

1 2 3 4 2

1 2 3 4 3

3

4

4

0 0 0 00 0 0 0

xyx

h h h h yxh h h h xy

yxy

=

x

x

N

The shape functions can also be written in matrix form as:


1

1

2

21 2 3 4

31 2 3 4

3

4

4

0 0 0 00 0 0 0

x

y

x

x y

y x

y

x

y

uuu

u uh h h hu uh h h h

uuu

=

u N

u

=u N u =x N x



Derivatives with respect to the spatial coordinates can be obtained via use of the Jacobian (Lecture 4):

Shape function derivatives

1

−

∂ ∂= ⇒ =

∂ ∂=

J Γr x x rΓ J

∂ ∂∂ ∂

x yxr r r

x yys s s

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= ∂∂ ∂ ∂

∂ ∂ ∂ ∂ J

r x

∂ ∂∂ ∂

In addition, the infinitesimal surface element can be transformed as:

det( )d dx dy dr dsΩ = = J

https://ethz.ch/content/dam/ethz/special-interest/baug/ibk/structural-mechanics-dam/education/femI/2020/Lectures/Lecture_4_2020.pdf



Using the above transformations, strain components can be obtained, for instance:

4

,1

( , )xx i x xi

i

u h r s ux

ε=

∂= =∂ ∑ , 11 12

( , ) ( , ) ( , )( , ) i i ii x

h r s h r s h r sh r sx r s

∂ ∂ ∂= = +

∂ ∂ ∂Γ Γwith

The above can be written in matrix form as: 1

1

2

21, 2, 3, 4,

3

3

4

4

0 0 0 0

x

y

x

yx x x x x

x

y

x

y

uuuu

h h h huuuu

ε

=




All strain components can be similarly computed and gathered in a matrix using Voigt notation:

1

1

21, 2, 3, 4,

21, 2, 3, 4,

31, 1, 2, 2, 3, 3, 4, 4,

3

4

4

0 0 0 00 0 0 0

x

y

xx x x x

yy y y y

xy x y x y x y x

y

x

y

uuu

h h h hu

h h h hu

h h h h h h h huuu

=

u

B

ε

=ε B u




Next the strains can be substituted in the weak form to obtain the stiffness matrix as:

Stiffness matrix and load vectors

T dΩ

= Ω∫K B EB

In the above, the domain of integration is defined with respect to the xy system and the shape functions with respect to r and s. Therefore a change of variables has to be performed as follows:

1 1

1 1

det( )T Td drdsΩ − −

= Ω =∫ ∫ ∫K B EB B EB J



In a similar manner load vectors can be obtained, for instance due to a body force:

1 1

1 1

det( )T drds− −

= ∫ ∫Ff N F J

For surface tractions, integration has to be carried out along the sides of the quadrilateral, e.g.:

1

1

det( )T dr−

= ∫t s1f N t J

where is the Jacobean determinant of the side where the load is applied

)det(𝐉𝐉𝐬𝐬𝐬𝐬




By reviewing the derived expressions:

1 1

1 1

det( )T drds− −

= ∫ ∫K B EB J

It can be observed that: integration over a general quadrilateral was reduced to

integration over a square the expressions to be integrated are now significantly more

complicated due to the use of the chain rule analytical integration might not be possible

1 1

1 1

det( )T drds− −

= ∫ ∫Ff N F J

1

1

det( )T dr−

= ∫t s1f N t J




To overcome this difficulty, numerical integration is typically employed, thus reducing integrals to sums:

1 1

1 1

det( ) ( , ) ( , )det[ ( , )]T Ti j i j i j i j

i jdrds w w r s r s r s

− −

= =∑∑∫ ∫K B EB J B EB J

where are the weights and coordinates of the Gauss points used

, ,i i iw r s

Notice that the Jacobian is also evaluated at the different Gauss points since it is not constant in general!




Gauss point weights and coordinates are typically precomputed can be found in tables. For instance for a linear quadrilateral:

Numerical integration

1 1

2 2

11, ,3

11,3

w r

w r

= = −

= =

In 1D:

The above can be combined to obtain the 2D coordinates and weights

12

34

1 2

4 3

r

s

x

y



The procedure presented above can be performed for different kinds of elements just by changing the shape functions and Gauss points used, for instance:

8 noded quadratic quadrilateral (q8) Shape functions

Other isoparametric elements

1

2

34

1 2

4 3

r

s

x

y

5

6

7

8

5

6

7

8

( )

( )

( )

( )

( )( )

( )( )

( )( )

( )( )

1

2

3

4

2

5

2

6

2

7

2

8

1 (1 )(1 ) 141 (1 )(1 ) 141 (1 )(1 ) 141 (1 )(1 ) 141 1

21 1

21 1

21 1

2

h r s r s

h r s r s

h r s r s

h r s r s

r sh

r sh

r sh

r sh

= − − − − −

= + − − + −

= + + − + +

= − + − − +

− −=

+ −=

− +=

− −=




8 noded quadratic quadrilateral (q8) Gauss points


12 3

4

1 2

4

3r

s

x

y5 6

7 8

5 6

7 89

9

1 1

2 2

3 3

5 / 9, 0.68 / 9, 0

5 / 9, 0.6,

w rw r

w r

= = −= =

= =

In 1D:

The above can be combined to obtain the 2D coordinates and weights



Two-dimensional Gauss quadrature for rectangular elements

P. W

rigg

ers,

Com

puta

tiona

l Con

tact

Mec

hani

cs (2

006)

https://link.springer.com/book/10.1007/978-3-540-32609-0




Linear or constant strain triangle (CST) Shape functions


1

2

3

1 2

3

r

s

x

y

1

2

3 1

h rh sh r s

=== − −




Linear or constant strain triangle (CST) Gauss points


1

2

3

1 2

3

r

s

x

y

1 1 11, 1/ 3, 1/ 3w r s= = =



Two-dimensional Gauss quadrature for triangular elements

P. W

rigg

ers,

Com

puta

tiona

l Con

tact

Mec

hani

cs (2

006)

https://link.springer.com/book/10.1007/978-3-540-32609-0

chapter 7 · 2021. 5. 3. · chapter 7. 2d elements. book chapters [o] v1/ch4 & ch6. institute...

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