A Quadratic Serendipity Plane Stress Rectangular Element · A Quadratic Serendipity Plane Stress...

Putting It All TogetherMAE 323: Chapter 5

2011 Alex Grishin 1MAE 323 Chapter 5

A Quadratic Serendipity Plane Stress Rectangular Element

•In Chapter 2, we learned two different energy-based methods of:

1. Turning differential equations into integral (or energy)

equations

2. Using this form of the equations to generate discrete

approximations using shape functions

•In Chapter 3, we learned how certain shape functions may be derived

•In Chapter 4, we learned some basic results from elasticity theory.

Namely, the form of the stress equilibrium equation and how stress

relates to strain via some form of Hooke’s Law

•In This chapter, we’d like to put all these ideas together to see how the

finite element method is used a general




•The first thing we’d like to do is create a plane stress element using the

quadratic Serendipity shape functions derived in Chapter 3 over a

rectangular domain. What we mean by this is we’d like to be able to

fully define the discrete algebraic (weak) form of it’s governing

elastostatic equations.

2

2

2

2

1( 1)( 1)( 1)

4

1( 1)( 1)( 1)

4

1( 1)( 1)( 1)

4

1( 1)( 1)( 1)

4( , )

1( 1)( 1)

2

1( 1)( 1)

2

1( 1)(1 )

2

1( 1)( 1)

2

r s r s

r s s r

r s s r

r s s r

r s

r s

r s

r s

r s

− − − + + + − − + + + + −

− − + − −

= − − + −

− − +

− −

N (1)

2

1

rs

3

4

5

67

8




•Formally, the way we’d do this is to start with the differential equation

(from Chapter 4 – remembering that the indices range over spatial

coordinates):

,0

ij j ibσ + =

•Then, using the Galerkin formulation*, we would multiply this with a

trial function. In this context, it would be a vector-valued trial function,

wi

*Alternatively, we could integrate the strain energy density and equate this to the work

done by external nodal forces (i.e. the Rayleigh-Ritz Method)

( ),0

ij j i ib wσ + =

•Then integrate over an element volume

( ),0

ij j i i

V

b w dVσ + =∫




•Now, we won’t go through the complete derivation because it involves

some mathematics most students haven’t seen yet (mostly concepts

from Advanced Calculus). This is only because we are now working in

two spatial dimensions. We will just give the resulting weak form:

ij ij i i i i

V V S

t dV t b w t F w dSσ δε = +∫ ∫ ∫

where: ( ), ,

1

2ij i j j i

w wδε = +(see chapter 4 for the

definition of strain)

•And t is the thru-thickness (normal to the plane) of the domain.

Now replace the stress and strain tensors with their vector

counterparts (as well as the forces), as we saw in Chapter 4, and let’s

assume a unit thickness for t:

T

V V S

dV w wdSδ = +∫ ∫ ∫σ ε b F (2)

Stress

vector

Strain

vector

Body Load

vector

External surface

load vector


2011 Alex Grishin MAE 323 Chapter 5 5

•Now, recall that the Serendipity elements are isoparametric, which

means that if we are going to perform the integrals in (2), we need an

explicit mapping between the isoparametric coordinates and the global

coordinates


rs

x

y

•For the strain matrix, this mapping is supplied

by the Jacobian of (x,y) with respect to (r,s):

x y

r r

x y

s s

∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂

J

such that:i ii

i ii

N NN x y

x xr r r

N NN x y

y ys ss

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= = ∂ ∂∂ ∂ ∂

∂ ∂ ∂ ∂∂

J (3)

1

2

3

4

5

67

8



•If we wanted to convert the equations to the global coordinate system,

we would need the inverse of (3):


where det J is the determinant of J given by :

11

det

i i i

i i i

N N Ny y

x s r r r

N x x N N

y s r s s

−

∂ ∂ ∂∂ ∂ − ∂ ∂ ∂ ∂ ∂

= = ∂ ∂ ∂ ∂ ∂ −

∂ ∂ ∂ ∂ ∂

JJ

det x y y x

r s r s

∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂J

(4)

•Although we could integrate (4) directly, it’s a little inconvenient because it

represents a full coordinate transformation at every point in the integrals

we’re going to perform. Fortunately, we can use a shortcut…



•The shortcut we use will allow us to integrate over the parametric coordinates

(which remember, always range from/to ±1) instead of transforming into global

coordinates. The shortcut is made possible by the concept of substitution of

variables*. Since we are integrating shape functions (or derivatives of shape

functions), and these function are isoparametric, we know that:


Where Nx(r,s), Ny(r,s) are our shape functions for the x and y-directions,

respectively. This is because det J actually represents a differential volume

distortion (a mapping of the differential volume in one coordinate system to

other):

, 1, 1

( , ) ( ( , ), ( , ))det x y

x y

f x y dxdy f N r s N r s drds± ±

=∫∫ ∫∫ J

( , )det J=

( , )

x y

r s

∂

∂

*We’re showing the multivariate version, which is beyond the scope of elementary calculus. See:

http://mathworld.wolfram.com/ChangeofVariablesTheorem.html and:

http://en.wikipedia.org/wiki/Change_of_variables_theorem



•Now, returning to the governing equations:


( )e e

e T e e

SV V

dV w wdSδ = +∫ ∫ ∫σ ε b F

•We are going to discretize this equation with our shape functions. We have

now attached a superscript e to all terms which will be evaluated on an element

basis. Before doing so, we make use of Hooke’s Law for an isotropic material to

covert the stress in the LHS to strain (we want the equation in terms of a single

primary unknown variable. In our case, this will be displacement):

( )e e

ee T e

SV V

dV wdV wdSδ = +∫ ∫ ∫ε C ε b F

•So, we need to write the strain vector in terms of shape functions. You already

got a hint of how we will do in this in Chapter 2. We’re going to write the strain

vector in term of a strain shape function matrix times displacement:

e e e e= ⋅ = ⋅ε B u ∆ d (6)

(5)



•B is the strain shape function matrix, and it is defined by:


•Where, ∆e is a strain operator. In the three dimensions, it is given as:

0 0

0 0

0 0

0

0

0

e

x

y

z

y x

z y

z x

∂ ∂

∂ ∂

∂ ∂

= ∂ ∂

∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∆

e

u

v

w

=

d

e e e= ⋅B ∆ N (7)

e

=

N 0 0

N 0 N 0

0 0 N



•In two dimensions:


u

v

=

d

0

0e

x

y

x y

∂ ∂

∂ = ∂

∂ ∂ ∂ ∂

∆

•Substituting in our shape functions and converting to parametric coordinates:

0

0e

r

s

r s

∂ ∂

∂ = ∂

∂ ∂ ∂ ∂

∆

1

1

( , )

( , )

n

i i

ie

n

i i

i

N r s u

N r s v

=

=

=

∑

∑d

e⋅

= = ⋅

N u N 0 ud

N v 0 N vOr



•Next, re-arrange d for more convenient storage and manipulation. First,

expand it in matrix form:


1

2

3

4

5

6

7

1 2 3 4 5 6 7 8 8

1 2 3 4 5 6 7 8 1

2

3

4

5

6

7

8

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

e

u

u

u

u

u

u

u

N N N N N N N N u

N N N N N N N N v

v

v

v

v

v

v

v

=

d



•Now, rearrange:


1

1

2

2

3

3

4

1 2 3 4 5 6 7 8 4

1 2 3 4 5 6 7 8 5

5

6

6

7

7

8

8

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

e

u

v

u

v

u

v

u

N N N N N N N N v

N N N N N N N N u

v

u

v

u

v

u

v

=

d



•Now, calculate B according to (7):


3 5 6 7 81 2 4

3 5 6 7 81 2 4

3 3 5 5 6 6 7 7 8 81 1 2 2 4 4

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0e

N N N N NN N N

r r r r r r r r

N N N N NN N N

s s s s r s s s

N N N N N N N N N NN N N N N N

r s r s r s r s r r r s r s r s

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

B

•Before continuing, let’s pause and review what we’ve done. We’ve calculated

an element strain shape function matrix, Be according to:

e e e= ⋅B ∆ N

0

0

r

s

r s

∂ ∂

∂ = ⋅ ∂

∂ ∂ ∂ ∂

N 0

0 N



•So, we now have an expression of strain in terms of local parametric shape

functions:


•So, let’s go back to equation (5) and plug in what we’ve got so far:

e e e e e e= ⋅ = ⋅ ⋅ε B u ∆ N u

0

0

r

s

r s

∂ ∂

∂ = ⋅ ⋅ ∂

∂ ∂ ∂ ∂

N 0 u

0 N v

( ) ( ) ( )( ) det det det e e

ee e T e e e e e e

s

SV V

dV dV dS⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ + ⋅ ⋅∫ ∫ ∫B u C B u J b N u J F N u J

( )e e

ee T e

SV V

dV wdV wdSδ = +∫ ∫ ∫ε C ε b F

(8)



•After simplification:


( )( ) det det det e e

ee T e e e e

s

SV V

dV dV dS⋅ ⋅ ⋅ = ⋅ + ⋅∫ ∫ ∫B C B u J b N J F N J

•This is the final set of equations which result in the algebraic system:

e e e= +k u b F

Element

stiffness

Element

displacement

Element body

force

Global external

load vector

•Compare this to the general (but non-parametric) equation offered in Chapter 2

(equation (21)) for the case of no body forces:

T

V

V∂= ⋅ ⋅ =

∂ ∫∆ C ∆ d Fu

(9)



•Let’s focus on the LHS of equation (9). We should remark that all finite

element equations involve a term with this form. In mathematics, it is referred

to as a bilinear form. It always involves an outer product of shape functions and

usually represents the internal energy of the system. In structural mechanics, it

provides us with the stiffness matrix


( )( ) det e

e T e e

V

dV⋅ ⋅ ⋅∫ B C B u J

•We have almost all the ingredients we need now to calculate the stiffness matrix of

a quadratic rectangular Serendipity element for plane stress problems. Equation (1)

provides us with the shape functions, equation (7) provides us with Be , and C for

plane stress is provided from Chapter 4:

2

1 0

1 01

0 0 (1 ) / 2

Eν

νν

ν

= − −

C



•There’s still one thing missing! How do we calculate the integral over the

domain shown below?


( )( ) det e

e T e e

V

dV⋅ ⋅ ⋅∫ B C B u J

•This actually can be done analytically. Either manually or with a Computer Algebra

System (CAS). However, both techniques are too slow in general. What is needed is

a very accurate and robust (easily programmed and widely applicable) method of

doing this – even if it’s still only approximate.

•Historically, the method almost universally adopted is called Gaussian Quadrature,

which tends to give very good results for the integrals of smooth (or piecewise

smooth) functions

rs



•Gaussian Quadrature (sometimes called Legendre-Gauss Quadrature*) works

by sampling the integrand at points prescribed over the domain by the

quadrature rule. These points are then weighted and summed, producing the

an approximation of the integral.


Gaussian Quadrature

( ) ( )1 1

( ) det ( , ) ( , ) det ( , )e

n nT

e T e e e e

i j i j i j i j

i jV

dV r s r s r s w w= =

⋅ ⋅ ⋅ ≈ ⋅ ⋅∑∑∫ B C B u J B C B J

•The two-dimensional quadrature rule is generated by taking the outer product of

one-dimensional rules. Thus, if a three-point rule is used, the one dimension

locations, ri and corresponding weights, wi are found (looked up from a table or

calculated). The two dimensional points and weights are then found by the taking

the outer product of each (thus a three point rule results in nine points in two

dimensions, and 27 points in three dimensions).

*http://mathworld.wolfram.com/Legendre-GaussQuadrature.html


2011 Alex Grishin

r

s

MAE 323 Chapter 5 19

•Quadrature points are usually given on the interval -1<ri<1, and so this is

another convenience provided by the isoparametric coordinates

•Below the coordinates for a two-point rule are shown


Gaussian Quadrature

s

r

1/ 3s =

1/ 3s = −

1/ 3r = − 1/ 3r =


2011 Alex Grishin

rs

MAE 323 Chapter 5 20

•Below are the points for a three-point quadrature


Gaussian Quadrature

s

r

s

s

r=0.0

s

3 / 5s =

3 / 5s = −

3 / 5r =3 / 5r = −



•Below is a table of quadrature points and corresponding weights 2-point and 3-

point quadrature


Gaussian Quadrature

Point Locations Weights

2 -0.5773503 1.0000000

0.5773503 1.0000000

3 -0.7745967 0.5555556

0.0000000 0.8888889

0.7745967 0.5555556

•So, how do we know how many points to use when we integrate using

Gaussian Quadrature? The rules are derived (in one dimension) so as to

integrate all polynomials up to degree 2m-1 exactly, where m is the number of

points used in the quadrature! So, in principle, a 2-point rule would integrate

2nd and 3rd degree functions exactly.

1/ 3−

1/ 3

3 / 5−0

3 / 5

1

1

5 / 9

8 / 9

5 / 9



•Below, two curves: one 2nd degree and one 3rd degree are integrated exactly

with a two-point Gaussian quadrature


Gaussian Quadrature

•So, how do we know how many points to use when integrating elements?

p1

p2

2

1

2 3

2

2 5 3

4 10 2 4

p x x

p x x x

= − +

= − + + −

1

1 1 11

1 1( ) (1) (1) 3 3 6

3 3 3

5 5

3p x dx p p

−

− ≈ + = − + − =

∫

1

2 2 21

1 1 10 10 20( ) (1) (1)

3 3 33 3 3

26

3 3

26

3p x dx p p

−

− − − ≈ + = − − + + =

∫

Exact!

Exact!



•The guideline for exact integration is usually not followed in finite elements.

One reason is that for 2nd and 3rd degree shape functions, the preceding

formula would only be reliable if the sides of the rectangle were straight (if the

mid-side nodes lay on a straight line connecting corner nodes). When this is not

the case, we have a Jacobian with different values at all points within the

domain – this introduces error into the integral. Other reasons have to do with

mesh instabilities (which we’ll discuss later) and matrix assembly efficiency.

•In practice, a two-point quadrature rule is usually used for linear elements,

whereas a three-point rule is frequently used for quadratic elements.


Gaussian Quadrature

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