arce504 fem-----------++++

California Polytechnic State University Department of Architectural Engineering FEM for Building Structures, Winter Quarter 2010 Instructor: Ansgar Neuenhofer

C:\calpoly\arce504\Handouts\Winter2010\HandoutsChapters1-3.doc 4/29/2010

- 1 -

Review: Illustration of structure stiffness matrix assembly process (ARCE 306) Assume, each element has 2 nodes and 1 degree of freedom at each node, the rotation. Each quartet of crosses represents the 2x2 element stiffness matrix of a member and is placed into the 6x6 structure stiffness matrix at the appropriate location. When more than one element is connected to a node, we obtain the corresponding diagonal ele-ment of the stiffness matrix by adding the stiffness of all elements connected to that node.

1

1

2

2

3

3

4

4

5

5

6

6

1

2

3

4

5

6

x

xx

x

x

xx

x

x

xx

x

x

xx

x

x

xx

x

x

x x

x

x

xx

x

x

xx

x

x

xx

x



- 2 -

Statics and PVF for simple 3-D problem (loads act perpendicular to plane of structure)

Problem: Find the displacement delta for the given loading. Solution: The internal forces in the structure are the shear forceV , the bending momentM and the torsional moment TM . The two moment diagrams are shown above. We ignore shear deformation, such that the virtual force integral becomes

( )4 41 1

d d

1 1 1 140 k-ft 8 ft 8 ft 30k-ft 6 ft 6 ft 1 40k-ft 8 ft 6 ft

3 329,000ksi 150 in /144 11,500ksi 300 in /144

0.0402 ft 0.080 ft 0.120 ft 1.44 in( )

T TMM x M M xEI GJ

ans

= + = + +

= + = =

4

Given:

29,000 ksi

=150 in (tubular section)

0.3

E

I

=

=

( ) ( )4 4

29, 000ksi11,150 ksi

2 1 2 1 0.3

2 2 150 in 300 in

EG

J I

= = =+ += = =



- 3 -

1 INTRODUCTION

1.1 Key idea (finite vs. infinitesimal) the key idea is the exact same as in ARCE 306, that is subdivide the structure into elements and nodes,

displacements and rotations (if they exist) are the unknowns, results in huge systems of algebraic equations assemble element stiffness matrices in the exact same manner as learned in ARCE 306 what is different is that we use elements other than frame elements (e.g. membrane, plate or shell elements) the finite element method generally provides an approximate solution whose accuracy increases as more

element, and hence unknowns, are used. the idea of subdividing a structure into many elements would have languished had there not been the

spectacular development in digital data processing.

1.2 Hand vs. computer analysis of frame structure

1.2.1 Hand analysis semicircular 2-pinned arch Find , ,N V M @ locationP

( )

cos 30 sin 30

sin 30 cos 30

1 cos 30 sin 30

N A B

V A B

M A r B r

= += +=

D D

D D

D D

Note: structure indeterminate to degree one use three equilibrium equations and one compatibility equation to solve for reactions, then cut free body

and determine , ,N V M process is difficult to automate, what for example if we had 3 pinned arch => statically determinate => four

EEQ, zero compatibility equation



- 4 -

1.2.2 Computer analysis subdivide structure into nodes and elements (probably straight), node-element concept, internal forces are , ,N V M

: vector of element nodal displacements

: vector of element nodal forces

: element stiffness matrix

: (assume linear structural behavior)=

q

Q

K

Q Kq

The above equation is a force-displacement relation in matrix form, which forms the basis of electronic structural analysis. In order to get K , we need (1) stress-strain relation (constitutive relation) (2) strain-displacement relation (kinematic relation) (3) Equilibrium Those three fundamental relations of structural analysis are part of the force method, the slope-deflection method, the solution to the differential equation or any conventional method of structural analysis (see ARCE 302, ARCE 306), i.e. we can use any of these methods to derive K . ALTERNATIVE: energy method (e.g. Rayleigh-Ritz method of ARCE 501)

1 1,q Q

2 2,q Q

3 3,q Q

4 4,q Q

5 5,q Q

6 6,q Q

2 nodes, 3 dofs per node

i

i

i

j

j

j

u

v

u

v

=

q

node-element concept

linear connection between nodes



- 5 -

1.3 Membrane structure

it is problematic to model shear wall with frame elements use area elements (or membrane elements), e.g. 4 node-element, 2 dofs per node there are no nodal rotations (why?) there are no internal forces are , ,N V M but stresses , ,xx yy xy

: element nodal displacements

: element nodal forces

: element stiffness matrix

: (assume linear structural behavior)

(8x1) = (8x8)x(8x1)

=

q

Q

K

Q Kq

1 1,q Q

2 2,q Q

3 3,q Q4 4,q Q

5 5,q Q6 6,q Q

7 7,q Q

8 8,q Q

shear wall

node-element concept



- 6 -

Again, the above equation is a force-displacement relation at the element level in matrix form.

To get that relation, need (as before for frame elements)

(1) stress-strain relation (constitutive relation) (2) strain-displacement relation (kinematic relation) (3) Equilibrium

However, force method, slope-deflection method, differential equation do not work for this type of element only option: energy method

Memorize: energy principle for frame element: one option among many energy principle for other types of elements (membrane, slab, shell, volume): only option

when energy principle is used to derive a matrix force-displacement relation that strategy is called finite element method (FEM)

when other procedures are used to derive a matrix force-displacement relation that strategy is called (direct stiffness method, matrix structural analysis, ARCE 306)

1.4 Historical Background The finite element method is of dual nature. It was developed for stress analysis but soon researchers recognized the mathematical foundation of the method and the finite element method was generalized for application in other fields. Important people: Zienkiewicz (University of Swansea, Wales, UK, General mechanics, 1960s), Clough (UC Berkeley, Structural engineering 1960s), Argyris (University of Stuttgart, Germany, Aerospace applications, 1950s) Some software packages: General purpose ABAQUS ANSYS NASTRAN LS-DYNA Structural engineering ETABS, SAP, RISA, LARSA and many more



- 7 -

1.5 Fields of application of FEM FEM is used in all areas of engineering and more and more so in medicine and even in the food industry, as illustrated below.



- 8 -

Cost of computation 1984: 15,000,000 $ per GFLOP 2007: 0.42 $ per GFLOP GFLOP = 1 Billion floating point operations per second => each year cost of computation has gone down by more than 50%

1 12007-1984 23

6 60.42 0.42= =0.47

1510 1510



- 9 -

2 REVIEW OF RAYLEIGH-RITZ METHOD AND TRANSITION TO FEM

2.1 Key idea Assume displaced shape (trial functions) and use minimum potential energy to find the best solution among

the trial functions Mathematically this step consists of finding the constantsc of the trial functions trial functions must satisfy geometric boundary conditions if the set of trial functions contains the exact solution, then the method will find that solution

Example: Cantilever with uniformly distributed load Problem: Use the Rayleigh-Ritz method to find the deflected shape and the moment diagram of the above cantilever beam. Use both a quadratic and a cubic trial function. Side-trip: Beam kinematics The concept of plane section remain plane leads to ( , ) ( ) ( )x y y x y x = =

If the cross section is assumed to remain perpendicular to the beam axis (i.e. if we ignore shear deformation), we have

N( , ) = ( )v

x y y v x

=

P

L

( )v x

x

EI

0v =

v

pure bending

pure shear

(cross section perpendicular to beam axis)v =

(cross section not perpendicular to beam axis)v

v = =

v =

: curvature

: angle of rotation of cross section

: angle of rotation of beam axisv



- 10 -

2.2 Quadratic function

2.2.1 Trial function and kinematics 2( )

( ) 2

( ) 2 ( ) curvature

v x c x

v x c x

v x c x

= = = = =

(2.1)

2.2.2 Hookes law and minimum of potential energy

N0

2 2

0

2 2

2

2

1( ) ( )d ( )

2

1( )d

2

14

2

4 04

( )4

( ) ( )2 2

e

i

i e

L

W

W

L

W W W

x M x x Pv L

EI x x PcL

EI c L PcL

W PLEIcL PL c

c EIPL

v x xEI

PL PLM x EIv x EI

EI

=

=

= +

= +

= + = = =

= = =

(2.2)

2.3 Cubic function

2.3.1 Trial function and kinematics 2 3

1 2

21 2

1 2

( )

( ) 2 3

( ) 2 6 ( ) curvature

v x c x c x

v x c x c x

v x c c x x

= + = + = + = =

(2.3)



- 11 -

2.3.2 Hookes law and minimum of potential energy

( )

( )( )

( )

( )

2 31 2

2

1

3

2

2

0

2

1 2

0

2 2 21 1 2 2

0

2 2 2 31 1 2 2

2 21 2

1 1 1

21

2 2 2

( )

1( )d

2

12 6 d

2

14 24 36 d

2

2 6 6

4 6 0

6 12

e

e

e

L

i

L

L

i e

i e

W Pv L P c L c L

WPL

c

WPL

c

W EI x x

EI c c x x

EI c c c x c x x

EI c L c c L c L

W WWc L c L PLEI

c c c

W WWEI c L

c c c

= = + = =

=

= +

= + +

= + +

= = + + = = = +

( )

( )( )

( ) ( )

3 32

2 21

1 22 3 32

2 3 2 3

33 3

0

4 6 1 1,

2 66 12

1 1( ) 3

2 6 6

( ) 3 exact6 3

( ) ( ) 6 66

c L PL

L L c PL PL PEI c cc EI EIL L PL

PL P Pv x x x Lx x

EI EI EIP PL

v L L LEI EI

PM x EIv x EI L x P L x

EI

+ =

= = = = + = +

= + = =

= = + =

(2.4)

Note than by using the chain rule we can simplify the calculation (we avoid calculating the square of the curvature)

( )( )

21 2 1 2

1 1 10 0 0

2 31 2 1 2

2 2 20 0 0

( ) ( )2( ) d ( ) d 2 6 2d 4 6

2

( ) ( )2( ) d ( ) d 2 6 6 d 6 12

2

L L Li

L L Li

W x xEI x x EI x x c c x x c L c L

c c c

W x xEI x x EI x x c c x x x c L c L

c c c

= = = + = + = = = + = +

(2.5)



- 12 -

2.3.3 Illustration of potential energy

Figure 2.1: Potential energy as a function of constants 1c and 2c (cubic trial function).

2.4 Numerical results

Figure 2.2: Normalized deflected shape for quadratic and cubic trial functions.

2.5 Summary We have seen that we obtain a more accurate solution is to increase the degree of the polynomial of the trial func-tions. We have a second option: That is decrease the size of the element for which each trial functions is valid, that is subdivide the beam into several elements and assume independent trial functions for each of those elements. In the RR method with free constants c it is difficult to enforce the boundary conditions between adjacent elements. A na-tural way to enforce the boundary conditions is to express the trial function in terms of nodal parameters (displace-ment and rotation) rather the free constants as in the Rayleigh-Ritz method. This is the key idea of the finite element method. We will demonstrate this concept in the next chapter using beam elements, the simplest family of finite ele-ments.

0 0.2 0.4 0.6 0.8 11

0.8

0.6

0.4

0.2

0

Nor

mal

ized

Def

lect

ion

/x Lexact( )v

v L

3

exact( ) 3PL

v LEI

=

quadratic

cubic

(exact)

1c2c

W

2c

1c



- 13 -

3 BEAM FINITE ELEMENTS

3.1 Introduction

Figure 3.1: Sample beam.

In this chapter we demonstrate the basics of the finite element method using the simple beam above. The analytical solution for the deflected shape ( )v x and moment diagram ( )M x is given by

( )( )

32 4

2

( ) 7 10 3360

( ) 16

wL x xv x

EI LwLx

M x

= + =

= (3.1)

In what follows we use the finite element method to analyze the beam numerically. Using different beam elements we show that the finite element method generally yields approximate results only. The accuracy of the approximate solution depends both on the number of elements used (the density of the mesh) and on the properties of the elements (the degree of the trial function).

3.2 Conventional four degree-of-freedom beam element

Figure 3.2: Beam element with four degrees of freedom.

3.2.1 Shape functions The four nodal parameters allow us to interpolate the deflection by a cubic polynomial. As in the Rayleigh-Ritz method, we start by using free constants c .

( )2 3 2 3

1 2 3 4

22 3 4

( ) 1

1( ) 2 3

xv c c c c

L

v c c cL

= + + + = = = + +

c (3.2)

Next, we use the four boundary conditions

EI

Liv

i j

jv

EILx

p( )v x



- 14 -

(0)

(0)

(1)

(1)

i

i

j

j

v v

v

v v

v

= =

= =

(3.3)

to express the nodal parameters q in terms of the free constants c

1

2

3

4

1 0 0 0

10 0 0

1 1 1 1

1 2 30

i

i

j

j

v c

cLv c

c

L L L

=

(3.4)

or

=q Gc (3.5) with

i

i

j

j

v

v

=

q (3.6)

We substitute the inverse of this relation

1

2 1

3

4

1 0 0 0

0 0 0or

3 2 3

2 2

i

i

j

j

vc

Lc

c vL Lc L L

= =

c G q

into Eq. (3.2) and obtain

2 3 1( ) 1 ( )v = = G q N q (3.7) with

2 3 2 3 2 3 3 2( ) =[1 3 2 ( 2 ) 3 2 ( )]x

L LL

+ + =N (3.8)

where N is a 4x1 row vector of shape functions. Note that the concept of shape functions for beam elements is to fit a curve between two points at which both the ordinate ( 1 3,q q ) and the slope ( 2 4,q q ) are known (four data items define a cubic polynomial).



- 15 -

Figure 3.3: Shape functions for beam element with four degrees of freedom.

3.2.2 Strain-displacement transformation vector In structural analysis of beams, we work with a generalized strain, the curvature. In linear structural analysis (small displacement theory) and ignoring shear deformation, the curvature ( )x is the second derivative of the deflected shape ( )v x , thus

2 2 2

2 2 2 2

d ( ) d 1 d( ) ( ) ( ) ( )

d d dv xx x L L = = = = =N q N q B q (3.9)

The 4x1 vector

( ) ( ) ( ) ( )2

2 2 2 2

1 d 1 1 1 1( ) ( ) 6 12 4 6 6 12 2 6

dL L L L L

= = + + + B N (3.10)

is commonly referred to as strain-displacement transformation vector since it relates the strain (curvature) and nodal displacements q .

3.2.3 Hookes law The law of linear elasticity (HOOKE) relates the curvature, the generalized strain for beam elements, to the bending moment (the corresponding generalized stress)

( ) ( ) ( )M EI EI = = B q (3.11)

3.2.4 Minimum potential energy The internal energy stored in a beam element (neglecting shear deformation) is

0 0

0 0

2

0 0

0

1 1d d

2 2

1 1d d d

2 2

1 1d ( ) ( ) d

2 2

1( ) ( ) d

2

L L

i

L L

L L

LT T

W V E V

yE y V yE y A x

EI x x x EI x

x x EI x

= =

= =

= =

=

B q B q

q B B q

(3.12)

0 0.2 0.4 0.6 0.8 10.2

0

0.2

0.4

0.6

0.8

1

1.2

Def

lect

ion

4

1

3

2

N

N

N

N



- 16 -

Note that

( )y y = (3.13) and

2dy A I= (3.14) The external work done on a beam element is the work done by the nodal forces along the nodal displacements and the work done by external element loading along the element displacements (if element loading is present).

0

0

0

( ) ( )d

( ) ( )d

( ) ( )d

LT

e

LT

LT T T

T T

W v x p x x

x p x x

x p x x

= +

= +

= +

= +

q Q

q Q N q

q Q q N

q Q q F

(3.15)

with

0

( ) ( )dL

T x p x x= F N (3.16) The vector F contains forces applied to the nodes that are equivalent (in an energy sense) to the element loading. It is thus often referred to as vector of equivalent nodal forces.

Note that the potential energy is now a function of the nodal displacements q , in the Rayleigh-Ritz method it is a function of the free constants c . Taking derivative with respect to q yields

1

2

0

( ) ( )L

T

n

Wq

WW q x EI x

Wq

= =

B B q Q Fq #

( ) ( )0

3 2 3 2

2

3 2

d

12 6 12 6

4 6 2

12 6

4symm.

LT x EI x x

L L L L

L L LEI

L L

L

=

=

K B B

(3.17)



- 17 -

( ) ( ) ( )

( )

1

11 1 1 1 1

0 01 1

22 2 3

0 0

133 30

( ) ( )d ( ) ( ) d

1 16 12 6 12 d 6 12 d

121 16 12

12 3

L

k B x EI B x x B EI B L

EIEI L

L L L

EIEI

L L

= =

= =

= =

(3.18)

3.2.5 Equivalent nodal forces As an example, we calculate the first component of the vector of equivalent nodal forces F due to a trapezoidal load with intensities ip and jp .

Figure 3.4: Beam element with trapezoidal loading.

( ) ( )

( ) ( )( )( )

1

1 1

0

12 3

0

12 3 3 4

0

( ) ( ) d

1 3 2 d

1 3 2 3 2 d

3 2 1 3 21

3 4 2 4 57 320 20

i j i

i j i

i j i

i j

F N p L

p p p L

p p p L

p p p L

p p L

=

= + +

= + + + = + + + = +

(3.19)

The complete vector of equivalent nodal forces for the four-degree-of-freedom beam element is

( )

( )

21 9

3 2

9 2160

2 3

i j

i j

i j

i j

p p

L p pLp p

L p p

+ = +

F (3.20)

ipjp

L1F

2F

3F

4F



- 18 -

3.3 Higher order beam element (three nodes, five degrees of freedom)

Figure 3.5: Beam element with five degrees of freedom.

3.3.1 Shape functions We derive the shape functions using the procedure of the previous section.

( )2 3 4

1 2 3 4 5

2 32 3 4 5

1

2

3

4

( )

1( ) 2 3 4

(0)

(0)

(0.5)

(1)

(1)

1 0 0 0 0

10 0 0 0

1 1 1 11

2 4 8 161 1 1 1 1

1 2 3 40

i

i

k

j

j

i

i

k

j

j

v x c c c c c

v x c c c cL

v v

v

v v

v v

v

v c

L c

v c

v c

L L L L

= + + + + = + + +

= =

==

= =

5

1

2

13

4

5

or

1 0 0 0 0

0 0 0 0

11 4 16 5

18 5 32 14 3

8 2 16 8 2

i

i

k

j

j

c

vcLc

c vL Lc vL Lc L L

=

= =

q Gc

c G q (3.21)

( )

( )

2 3 4

2 3 4

2 3 4

2 3 4

2 3 4

1 11 18 8

4 5 2

16 32 16

5 14 8

3 2

T

L

L

+ + + = + +

N (3.22)

EI

/2L

iv

i j

jv/2L

kv



- 19 -

Figure 3.6: Shape functions for beam element with five degrees of freedom.

3.3.2 Strain-displacement transformation vector

( )

( )

2

2

22

2

2

22 108 96

8 30 241

32 192 192

10 84 96

2 18 24

T

L

L

L

+ + + = + +

B (3.23)

3.3.3 Element stiffness matrix

2 2

30

( ) ( )d5

LT

L L

L L L LEI

x EI x x LL

= =K B B

316 94 -512 196 -34 36 -128 34 -6 1024 -512 128 symm. 316

2

L

L

-94 36

(3.24)

3.3.4 Equivalent nodal forces

The 5x1vector F of equivalent nodal forces for the five-degree-of-freedom beam element is

13

16 1660

13

i j

i

i j

i j

j

p p

LpL p p

p p

Lp

=

F (3.25)

0 0.2 0.4 0.6 0.8 10.2

0

0.2

0.4

0.6

0.8

1

1.2

Def

lect

ion

3

1

5

2

4

N

N

N

N

N



- 20 -

3.4 Higher order beam element (three nodes, six degrees of freedom)

Figure 3.7: Beam element with six degrees of freedom.

3.4.1 Shape functions

( )2 3 4 5

1 2 3 4 5 6

2 3 42 3 4 5 6

( )

1( ) 2 3 4 5

(0)

(0)

(0.5)

(0.5)

(1)

(1)

1 0 0 0 0 0

10 0 0 0 0

1 1 1 1 11

2 4 8 16 321 1 3 1 5

04 2 16

1 1 1 1 1

i

i

k

k

j

j

i

i

k

k

j

j

v x c c c c c c

v x c c c c cL

v v

v

v v

v

v v

v

vL

v

v L L L L L

= + + + + + = + + + +

= =

= =

= =

=

1

2

3

4

5

6

1

2

3

4

5

6

or

1

1 2 3 4 50

1 0 0 0 0 0

0 0 0 0 0

23 6 16 8 7

66 13 32 32 34 5

68 12 16 40 52 8

24 4 0

c

c

c

c

c

c

L L L L L

cLc

L L Lc

c L L Lc L L Lc

L

=

=

q Gc

1

16 24 4

i

i

k

k

j

j

v

v

v

L L

=

c G q

(3.26)

EI

/2L

iv

i j

jv/2L

kvk



- 21 -

( )

( )

( )

2 3 4 5

2 3 4 5

2 3 4

2 3 4 5

2 3 4 5

2 3 4 5

1 23 66 68 24

6 13 12 4

16 32 16

8 32 40 16

7 34 52 24

5 8 4

T

L

L

L

+ + + + + = + + + + +

N (3.27)

Figure 3.8: Shape functions for beam element with six degrees of freedom.

3.4.2 Strain-displacement transformation vector

( )

( )

( )

2 3

2 3

2

2 2 3

2 3

2 3

46 396 816 480

12 78 144 80

32 192 1921

16 192 480 320

14 204 624 480

2 30 96 80

T

L

L L

L

+ + + + + + = + + + + +

B (3.28)

0 0.2 0.4 0.6 0.8 10.2

0

0.2

0.4

0.6

0.8

1

1.2

Def

lect

ion

4

1

6

2

5

3

N

N

N

N

N

N



- 22 -

3.4.3 Element stiffness matrix

0

2 2 2

3

( ) ( )d

35

LT x EI x x

L L L

L L L L L

LEI

L

=

=

K B B5092 1138 -3584 1920 -1508 242 332 -896 320 -242 38 7168 0 -3583 896 2 2

2

L L L

L

L

symm. 1280 -1920 320 5092 -1138 332

(3.29)

3.4.4 Equivalent nodal forces

The 6x1vector F of equivalent nodal forces for the six-degree-of-freedom beam element is

( )

( )

( )

79 19

5 2

112 112

420 8 8

19 79

2 5

i j

i j

i j

i j

i j

i j

p p

L p p

p pL

L p p

p p

L p p

+ = +

F (3.30)



- 23 -

3.5 Numerical results In this section we summarize the results of the finite element analysis of the beam with trapezoidal load. We use the three beam elements derived before and subdivide the beam inton elements of equal length.

3.5.1 Conventional four degree-of-freedom beam element

Figure 3.9: Normalized deflected shape and moment diagram for four-degree-of-freedom element for different number of elements used.

.

0 0.2 0.4 0.6 0.8 17

6

5

4

3

2

1

0

Def

lect

ion

0 0.2 0.4 0.6 0.8 180

70

60

50

40

30

20

10

0

Mom

ent

2

4

1

exact

n

n

n

===

4

8

1

ct

2

exa

n

n

n

n

=

=

==

4

1000EI

pL

2

1000M

pL

xL

=

xL

=



- 24 -

3.5.2 Five degree-of-freedom beam element

Figure 3.10: Normalized deflected shape and moment diagram for five-degree-of-freedom element for different number of elements used.

0 0.2 0.4 0.6 0.8 17

6

5

4

3

2

1

0

Def

lect

ion ct

1

exa

n =

4

1000EI

pL

0 0.2 0.4 0.6 0.8 170

60

50

40

30

20

10

0

10

Mom

ent 2

4

1

exact

n

n

n

===

2

1000M

pL

xL

=

xL

=



- 25 -

3.5.3 Six degree-of-freedom beam element

Figure 3.11: Normalized deflected shape and moment diagram for six-degree-of-freedom element.

3.6 Summary The finite element method modifies the classical Rayleigh-Ritz method in the sense that the former uses trial func-tion that only apply over a certain domain, the element. If the number of elements is large enough we obtain satis-factory solutions even with low order trial functions. For example, we obtain accurate results for the moment dia-gram using the beam element with four degrees of freedom (cubic shape (trial) function) if we subdivide the beam into eight elements (magenta curve in Figure 3.9). Using the beam element with five degrees of freedom (fourth or-der shape functions) we can achieve comparable level of accuracy with only two elements (see green curve in Figure 3.10). Again (already addressed in HW 2 solution) we point to the important phenomenon that the internal forces as derivatives of the displacements are always less accurate than the displacements.

0 0.2 0.4 0.6 0.8 17

6

5

4

3

2

1

0

Def

lect

ion

0 0.2 0.4 0.6 0.8 170

60

50

40

30

20

10

0

Mom

ent

1(exact)n =

4

1000EI

pL

2

1000M

pL

1(exact)n =

xL

=

xL

=


C:\calpoly\arce504\Handouts\Winter2010\HandoutsChapter4.doc 4/29/2010 - -

26

4 Gauss Integration

4.1 General Remarks In finite element analysis, the element stiffness matrix is obtained by evaluating an integral. For line element (also termed 1-D elements, like rod, truss, beam, frame), we integrate over the length of the element, since the integration over the cross section has already been carried out by using cross section properties like the areaA and the second moment of inertia I . For 2D elements, the integral is over the area of the element, for volume elements we integrate over the volume. In finite element analysis, the integrand usually becomes quite complicated and it is not possible to solve the integrals in closed form. We have to use numerical procedures in these situations. In finite element calcula-tions, simple numerical integration schemes such as the trapezoidal rule or Simpsons formula do not work very well. The Gauss numerical integration scheme or Gauss quadrature, has becomes the standard tool to calculate element stiffness matrices. We derive the Gauss formula for one-dimensional integrals in the following section and then easily extend it to two and three-dimensional integrals.

4.2 Gauss numerical integration in one dimension As in other numerical integration schemes, the basic idea is to represent the integral in the following form

1

11

( )dg ( )n

i ii

I f g f g w=

= = (4.1) in which ig are then sample points at which we evaluate the integrand. These location are commonly referred to as Gauss points. The coefficients iw are the corresponding weights. Thus, to calculate I , we evaluate the integrand at each of several Gauss points ig (the sample points) to obtain ordinates ( )if g . We multiply each ( )if g by an appro-priate weight and add.

4.2.1 Two Gauss points 1 2

1 1 2 211

( )dg ( ) ( ) ( )i ii

I f g f g w f g w f g w=

= = + (4.2) We determine the four unknowns (two Gauss points and two weights) by requiring that the formula give the exact result if the integrand is equal to the first four terms in a polynomial, hence

1 1

1 2

1 11 1

1 1 2 2

1 11 1

2 2 2 21 1 2 2

1 11 1

3 3 3 31 1 2 2

1 1

( ) 1 ( )dg 1dg 2 2

( ) ( )dg dg 0 0

2 2( ) ( )dg dg

3 3

( ) ( )dg dg 0 0

f g f g w w

f g g f g g g w g w

f g g f g g g w g w

f g g f g g g w g w

= = = + =

= = = + =

= = = + =

= = = + =

(4.3)

Solving the previous four simultaneous nonlinear equations, we obtain

1 2 1 21 1

13 3

w w g g= = = = (4.4)

Hence 1 2

11

1 1( )d ( ) ( ) 1 ( ) 1

3 3i iiI f g g f g w f f

== = + (4.5)

Using Gauss integration with two points gives us the exact result up to a polynomial of order three. For other func-tions the results will be approximate.



27

4.2.2 Three Gauss points 1 3

1 1 2 2 3 311

( )d ( ) ( ) ( ) ( )i ii

I f g g f g w f g w f g w f g w=

= = + + (4.6) We determine the four unknowns by requiring that the formula give the exact result if the integrand is equal to the first four terms in a polynomial, hence

1 1

1 2 3

1 11 1

1 1 2 2 2 3

1 11 1

2 2 2 2 21 1 2 2 3 3

1 11 1

3 3 3 3 31 1 2 2 3 3

1 11

4 4

1

( ) 1 ( )dg 1dg 2 2

( ) ( )dg dg 0 0

2 2( ) ( )dg dg

3 3

( ) ( )dg dg 0 0

( ) ( )dg dg

f g f g w w w

f g g f g g g w g w g w



f g g f g g

= = = + + =

= = = + + =

= = = + + =

= = = + + =

= =

14 4 41 1 2 2 3 3

11 1

5 5 5 5 51 1 2 2 3 3

1 1

2 25 5

( ) ( )dg dg 0 0

g w g w g w


= + + =

= = = + + =

(4.7)

Solving the previous five simultaneous equations, we obtain

1 2 3

1 2 3

5 8 59 9 9

3 30

5 5

w w w

g g g

= = =

= = = (4.8)

Hence 1 3

11

3 5 8 3 5( )d ( ) ( ) (0) ( )

5 9 9 5 9i iiI f x x f g w f f f

== = + + (4.9)

Using Gauss integration with three points gives us the exact result up to a polynomial of order five. For other func-tions the results will be approximate.

The table below lists the GAUSS points and corresponding weights for 1,2, 3, 4n = . See any book on numerical analysis for more extensive tabulations.



28

#of GAUSS points weights 1

0 2.000000000

20.577350269 1.0000000000.577350269 1.000000000

30.774596669 0.5555555550.000000000 0.8888888890.774596669 0.555555555

40.861136311 0.3478548450.339981043 0.6

wn

n

n

n

=

=

=

= 52145154

0.861136311 0.3478548450.339981043 0.652145154

(4.10)

4.2.3 Example We calculate the integral

11 d

1I x x= +

(4.11)

The analytical solution is

( ) ( )13 3

2 21

1 2 21 d 1 1 1 0 1.88562

3 31I x x x

= + = + = + =

(4.12)

A numerical solution using a single GAUSS points yields

1 = 1 d 0 1 2 2.00000

1I x x+ + =

(4.13)

Using two GAUSS points gives

( )1

1

= 1 d 0.57735 1 1 0.57735 1 1.90604I x x

+ + + + = (4.14) The result is more accurate than that obtained with one GAUSS point. If we use three GAUSS points the result is

1 = 1 d 0.774596 1 0.55556 0 1 0.88889 0.774596 1 0.55556 1.89273

1I x x+ + + + + + =

(4.15)

The result is more accurate than that obtained with two GAUSS points. As shown before, a numerical integration with n GAUSS points gives the exact solution if the integrand is a polynomial of degree 2 1n or smaller. Consequently, if the integrand is a polynomial, we can always find a number of GAUSS points, such that the integral is evaluated exactly. If ( )f g is not a polynomial, as is the case in the example above, the GAUSS integration yields an approximate solution whose accuracy increases as more GAUSS points are used.



29

4.2.4 Integration limits other than -1/1 In order to use the Gauss integration procedure for integration limits other than -1 and 1, we have to use coordinate transformation as illustrated below by an example.

( )

6

2

6 1

2 1

say sin d

(2) 1 2

(6) 1 6

12

2

12 2 4 d 2d

2

sin d sin 2 4 2d

I x x

y ax b

y a b

y a b

a

b

y x x y x y

x x y y

=

= += = += = +== = = + =

= +

4.3 Gauss numerical integration in two dimensions When the region of integration is a 2x2 square, the one-dimensional Gauss formula easily extends to two dimensions.

1 1

1 2 1 2 1 21 11 1

( , )d ( , )m n

i ji i

I f g g g dg g g w w= =

= = (4.16)

Usually the number of integration points along the two directions is the same such thatm n= .

4.3.1 Example Evaluate the integral

( )1 1 2 21 1

+ d dI x y xy x y

= We find an exact solution by using two Gauss points in both directions.

( ) ( )1 1 2 22 2 2 21 2 1 21 11 1

2 2

2 2

2 2

+ d d +

1 1 1 11 1

3 3 3 31 1 1 1

1 13 3 3 3

1 1 13 3 3

i j i j i ji j

I x y xy x y g g g g w w= =

= = = + + + + +

2 2

11 1

31 1 1 1

1 13 3 3 3

1 1 1 1 1 1 1 1 49 3 9 3 9 3 9 3 9

+ +

= + + + + + =

(4.17)

Analytical integration gives

( )1 1 1 11 1 12 2 3 2 2 2 31 1 11 1 1 1

1 1 2 2 4+ d d + d +0 d

3 2 3 9 9I x y xy x y x y x y y y y y

= = = = = (4.18)


C:\calpoly\arce504\Handouts\Winter2010\HandoutsChapter5.doc 4/29/2010

30

5 Introduction to plane stress and strain

5.1 General stress/strain state

Figure 5-1: General state of stress.

5.2 Plane stress/strain state For finite element analysis we need to express the stresses in terms of the strains. However, as a starting point, it is easier to write strains as a function of stress.

( )( )( )

( )

( )

( )

1

1

1

2 11

2 11

2 11

xx xx yy zz

yy yy xx zz

zz zz xx yy

xy xy xy

xz xz xz

yz yz yz

E

E

E

G E

G E

G E

=

=

= += =+= =+= =

(5.1)

zz

x y

z

axial stres

shear stres

s

s==

yyxx

zx zy

yx

yz

xy

xz

x

y

z

t

yy

xx

xy

yx

all stresses uniformly

distributed across

thickness

xy yx

t

=



31

5.2.1 Plane stress If the body (plate, slab, diaphragm) is thin, it is reasonable to assume that the following stress components are zero.

0zz xz yz = = = Hence

( )( )( )

( )

1

1

1

2 1

0

1

xx xx yy

yy yy xx

xy xy x

x

y

zz x yy

E

E

E

G E

=

=

+= =

= (5.2)

or in matrix notation

( )

1 01

1 0

0 0 2 1

xx xx

yy yy

xy xyE

= + (5.3)

The inverse of this relation (stress as a function of strain) is

( )2

1 0

1 01

10 0 1

2

xx xx

yy yy

xy xy

E

=

(5.4)

5.2.2 Plane strain (rare in buildings) If the body is very thick, it is reasonable to assume that the following strain components are zero.

0zz xz yz = = = (5.5) Hence

( )( )( )

( )

1

1

2 11

10zz zz yy xx zz x

xx xx yy zz

yy yy xx zz

xy xy

x y

xy

yE

E

E

G E

= =

=

=

+=

=

= + (5.6)

or in matrix notation

( )( )

( )

2

2

1 1 01

1 1 0

0 0 2 1

xx xx

yy yy

xy xyE

+ = + +

(5.7)

The inverse of this relation (stress as a function of strain) is



32

( )( )( )

( )

1 01

11 0

1 1 2 11 2

0 02 1

xx xx

yy yy

xy xy

E

= +

(5.8)

Figure 5-2: Illustration of plane strain and plane stress ( 1 2 3, , are , ,x x x x y z ).

5.2.3 Summary of stress-strain relation in vector/matrix notation

( )( )

2

vector of strain

vector of stress

1 0

1 0 Elasticity (constitutive) matrix for plane stress1

10 0

2

1 0

1 01 1 2

1 20 0

2

T

xx yy xy

T

xx yy xy

E

E

= =

= = +

d

D

C

C Elasticity (constitutive) matrix for plane strain

(5.9)

5.2.4 In-plane vs. out-of-plane loading

plane strain

plane stress

in-plane loading (membrane action) out-of-plane loading (plate action)



33

5.3 Comparison Beam-General 2D Structural Element

Figure 5-3: Comparison of Beam and General 2D Structural Element.

x

y

xxxx

d

d

LLxx

MyI

=

BeamGeneral 2-D element

dx

V

Md

MM x

x+

dV

V xx

+

( )v x ( )w x

EI

dx

dy dxxxx xx + xx

dyyyy yy

+

yy

dxyxy xx

+

xy

dxyxy yy

+

xy

X

Y

depth d



34

5.4 Deriving differential equation

5.4.1 Equilibrium of differential element

Figure 5-4: Membrane stresses acting on differential element.

Writing equilibrium in the two directions gives

0 d d d d d d d d 0

0 d d d d d d d d 0

xyxxx xx xx xy xy

yy xyy yy yy xy xy

F t y x t y t x y t x X t x yx y

F t x y t x t y x t y Y t x yy x

= = + + + + + = = = + + + + + =

Deviding by d dx y t gives

0

0

xyxxx

yy xyy

F Xx y

F Yy x

= = + + = = + +

(5.10)

Statics yields two equations with three unknowns => statically indeterminate ,X Y (force per volume) are so called body forces (Self weight, inertia)

5.4.2 Strain-displacement relation and compatibility condition

xx

yy

xy

uxvyu vy x

= = = +

(5.11)

Eq. (5.11) involves three strain components but only two displacements, which means the three strain components can't be independent. We can establish the relation between the strain components (the compatibility equation) by taking derivative of the preceding strain-displacement relation which gives

dx

dy dxxxx xx

+ xx

dyyyy yy

+

yy

dxyxy xx

+

xy

dxyxy yy

+

xy

X

Y



35

2 3

2 2

2 3

2 2

2 3 3

2 2

xx

yy

xy

u

y x y

v

x x y

u vx y x y x y

= = = +

(5.12)

Hence 22 3 3

2 2 2 2

2 3 3

2 2

yx

xy

u v

y x x y x y

u vx y x y x y

+ = + = +

(5.13)

or 2 22

2 2yy xyxx

x yy x

+ = (5.14)

We now have two equilibrium equations, one compatibility equation and three stress-strain relations (HOOKE), that is six equations for the six unknowns ,, , , ,xx yy xy xx yy xy . This allows us to write a set of three simultaneous second order differential equations, either for the three strain components or the three stress components.

5.4.3 Airy-Stress Function and Differential Equation

Airy (1863) simplified the problem of finding the differential equation by introducing a stress functionF from which we can obtain the stress components by differentiation.

2 2 2

2 2 ( )yy xx xyF F F

Xy Yxx yx y

= = = + (5.15)

We can easily check that the stress function satisfies the equilibrium conditions

0

0

xyxxx

y xyy

F Xx yy

F Yy x

= = + + = = + +

(5.16)

from before 3 3

2 2

3 3

2 2

0 0

0 0

xyxxx

yy xyy

F FF X X X ok

x y x y x y

F FF Y Y Y ok

y x x y x y

= = + + = + = = = + + = + =

(5.17)

Differentiating Hooke's law (strain as a function of stress)

( )( )

( )

1

1

1 2 1

xx xx yy

yy yy xx

xy xy xy

E

E

G E

=

= += =

(5.18)

yields



36

( )

22 2

2 2 2

2 2 2

2 2 2

2 2

1

1

2 1

yyxx xx

yy yy xx

xy xy

Ey y y

Ex x x

x y E x y

= =

+=

(5.19)

Substituting Eq. (5.17) into the compatibility equation (5.14) gives

( )2 2 22 22 2 2 2

1 2 10yy yy xyxx xx

E E x yy y x x

+ + + =

(5.20)

which we can write as (using Airy's function)

( )4 4 4 4 4

4 2 2 4 2 2 2 22 1 0F F F F Fy x y x x y x y

+ + + = (5.21) or

4 4 4

4 2 2 42 0 (4th order linear, partial differential equation) F F Fx x y y

+ + = The challenge in the pre-finite element age was to solve this differential equation for different loading and support conditions. After obtaining a solution forF , the stresses are given by Airy's function. The strains can then be derived using Hooke's law and finallyu andv by integrating the strains.


C:\calpoly\arce504\Handouts\Winter2010\HandoutsChapter6-7.doc 4/29/2010 - -

37

6 The 4-node membrane element

6.1 Shape functions Membrane elements have two degrees of freedom per node, the two translational displacements. We use membrane elements to model structures whose out-of-plane stiffness is negligible, e.g. shear walls. Contrary to conventional beam elements, which respond in a state of one-dimensional stress and strain, membrane elements are either in a state of plane stress or plane strain. The simplest membrane element is the four-node rectangular element of side lengths 2a and 2b with 4x2=8 degrees of freedom. The vector of nodal displacements is

Figure 6.1: Four-node rectangular membrane element.

1 1 2 4

1 3 5 7 1 2 3 4

2 4 6 8 1 2 3 4

T

T T

uT T

v

u v u v

q q q q u u u u

q q q q v v v v

= = = = =

q

q

q

(6.1)

The calculations for this element are easier if we select a local coordinate system with the origin at the center. We can relate the global coordinates x andy to the local as given above. As for the beam element before, we first derive the shape functions , 1...4iN i = to interpolate the displacements inside the element in terms of the nodal displacements. The difference is that we now have displacements along both coordinate directions, that is -u andv displacements which are a function of both x andy .

4

1

4

1

( , ) ( , ) ( , )

( , ) ( , ) ( , )

u i ii

v i ii

u N u

v N v

=

=

= =

= =

N q

N q

(6.2)

or

1 2 3 4

1 2 3 4

( , ) ( , ) 0 ( , ) 0 ( , ) 0 ( , ) 0

( , ) 0 ( , ) 0 ( , ) 0 ( , ) 0 ( , )

u N N N N

v N N N N

= q (6.3)

Note that we need only four shape functions, since the shape functions for the -u and v displacements are identical. In deriving the shape functions we start with the general expression for the deflected shape (the same we did for the beam element). The four nodal values allow us to write a bilinear polynomial

1 2 3 4

5 6 7 8

( , ) 1

( , ) 1

u

v

u c c c c

v c c c c

= + + + = = + + + =

c

c

(6.4)

Using the boundary conditions, we can write

a a

b

b

1 1,q Q2 2,q Q

3 3,q Q

4 4,q Q

5 5,q Q

6 6,q Q

7 7,q Q

8 8,q Q

1 2

34

, ,x u

, ,y v,

1 1

1 1

x ya b

a x a

b y b

= =



38

1

2

3

4

( 1, 1)

(1, 1)

(1,1)

( 1,1)

u u

u u

u u

u u

= =

= =

(6.5)

u u=q G c (6.6) with

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

=

G (6.7)

The inverse of this relation is 1

u u=c G q (6.8)

with

1

1 1 1 1 1 1 1 11 1 1 1 14 1 1 1 1

= G (6.9)

Substituting Eq. (6.8) into Eq. (6.4) yields the desired shape functions

1( , ) 1 = N G (6.10) with

( )( ) ( )( )

( )( ) ( )( )

1 2

3 4

1 1 1 1 1 1

4 4

1 1 1 1 1 1

4 4

N N

N N

= = +

= + + = + (6.11)

or in short

( )( )1 1 1 1...44i i i

N i = + + = (6.12)

where i and i are the and coordinates of the four element nodes.

Element node 1 -1 -1 2 1 -1 3 1 1 4 -1 1 We verify that the shape function for node 1 evaluated at the four nodes follows the definition of a shape function.



39

( )( )

N ( )( )

N ( )( )

N ( )( )

N ( )( )

1

1, 1

1, 1

1, 1

1, 1

1 1 1

4

11 1 1 1 1

41

1 1 1 1 041

1 1 1 1 041

1 1 1 1 04

N

ok

ok

ok

ok

= =

= =

= =

= =

=

= + + =

= + =

= =

= + =

Figure 6.2: Plot of shape function 3( , )N for four-node membrane element. Note that for illustrational purposes we plot the value of the shape function along the vertical axis (in the 3-D drawing above), the shape function, of course, interpolates the horizontal displacement. It is interesting to ask whether these shape functions guarantee inter-element compatibility. In particular, if two elements are compatible at common nodes, are they compatible between the nodes. The answer is yes since the variation of the displacement along an element edge is linear. A linear function is completely defined in terms of two parameters, the displacements at the two nodes.

6.2 Strain-displacement relation (Kinematic relation) In a membrane problem we have three strain components, the two axial strains xx and yy and the shear strain xy . From mechanics of materials we recall the strain-displacement relation for the membrane problem (two-dimensional state of strain)

; ;xx yy xyu v u vx y y x

= = = + (6.13)

Figure 6.3: Strain-displacement relation.



40

We collect the strain components in a vector T

xx yy x y = d (6.14) Using the strain-displacement relation and the shape functions we can now interpolate the strain from the nodal di-splacement by taking the first derivatives of the shape functions with respect to x and y . We can write

xx

yy

xy

= = Bqd (6.15)

with the strain-displacement matrix

1 4

1 4

1 1 4 4

0 0

( , ) 0 0

N Nx x

N N

y yN N N N

y x y x

=

B

"

"

"

(6.16)

The matrix B is a 3x8-matrix since the four-node membrane element has three strain components and eight degrees of freedom. Recall that for the conventional beam element with four degrees-of-freedom B was a 1x4 vector containing the second derivative of the shape functions, because the beam element has only one (generalized) strain, the curvature (the second derivative of the displacement), and four degrees of freedom. Taking the derivative of the shape functions we obtain

1 1 1 10 0 0 0

1 1 1 1 10 0 0 0

41 1 1 1 1 1 1 1

a a a a

b b b b

b a b a b a b a

+ + + + = + + + +

B (6.17)

Note that

( , ) ( , ) ( , ) ( , )1 1x a y b

= = N N N N (6.18)

Because of the bilinear form of the displacements the axial strain xx is constant in the x -direction and varies linearly in they -direction, the axial strain yy is constant in the y -direction and varies linearly in thex direction, and the shear strain xy varies linearly in both directions (see Eq. 6.19).

2 4

7 8

3 4 6 8

( , )( , )

( , )( , )

( , ) ( , )( , )

xx

yy

xy

u x yx y c c y

xv x y

x y c c xy

u x y v x yx y c c x c c y

y x

= = += = + = + = + + +

(6.19)

Figure 6.4: Qualitative variation of strains , ,xx yy xy in 4-node element.



41

6.3 Stress-strain relation (constitutive relation)

6.3.1 Plane strain For a membrane problem, the stress components are

xx

yy

xy

= d (6.20)

and the stress-strain relation for a linear isotropic material is

=D C d (6.21) for the plane stress condition the elasticity matrix is

2

1 0

1 01

10 0

2

E

=

C (6.22)

whereE is the modulus of elasticity and is Poissons ratio.

6.4 Equilibrium As for the beam element, we use the principle of minimum potential energy to formulate equilibrium. The internal work is work done by the stresses on the corresponding strains integrated over the volume of the element. The membrane element has uniform thickness t such that the integral reduces to an integral over the area of the element.

1( , ) ( , ) ( , ) ( , ) ( , ) ( , )d

2

1 1( , ) ( , )d ( , ) ( , ) d

2 2

( , ) ( , ) d ( , ) ( , )d

i xx xx yy yy xy xy

V

T T T

V V

T Ti

V V

W x y x y x y x y x y x y V

x y x y V x y x y V

Wx y x y V x y x y V

= + +

= =

= ==

d D q B CB q

B CB q B CB qq

Kq

(6.23)

with

N

N

NN

d

d d

d d

1 1

1 1

( , ) ( , ) d

( , ) ( , ) d

( , ) ( , )d d

( , ) ( , )d d

T

V t A

T

x yA

a bT

a ba b

T

x y x y V

t x y x y A

t x y x y x y

ab t

=

=

=

=

K B CB

B CB

B CB

B CB

(6.24)



42

The integral in the preceding equation is usually evaluated numerically (see GAUSS integration). Since the integrand is a quadratic polynomial (the linear B - matrix is squared), two Gauss points in each direction are more than sufficient for exact integration.

1 1

1 12 2

1 1

( , ) ( , )d d

( , ) ( , )

T

Ti j i j i j

i j

a b t

t w w

= =

=

=

K B CB

B CB (6.25)

The element stiffness matrices K are assembled using the same procedure we learned for other elements

After solving the system of equations

=Kq F (6.26) at the structure level, we return to the element level and calculate any element response we are interested in (strain, stress, principal stress, principal directions etc). For example, we obtain the element stresses by first extracting the element displacements q from the structure displacements and then using

, , , = = B qD( ) C d( ) C ( ) (6.27) When we use the matrix method to analyze frame structure, we obtain the exact solution to the mechanical problem. We satisfy all three fundamental equations exactly, the stress-strain relation, the strain-displacement and the equili-brium conditions. When using the matrix method for other structural elements (membrane, plate, shell and solid ele-ments) this is no longer the case. The results are only approximate and the degree of accuracy depends on (1) the number of elements and (2) the type of element used.



43

6.5 Example

Figure 6.5: Cantilever for beam theory vs. four-node finite element comparison.

In this example, we use the four-node membrane element to model a simple cantilever and compare the results to those of conventional beam theory.

3

beam theory

maxmax,beam theory

2 3

maxmax,beam theory 2

320 k-ft

30 ksf11 2 ft

61k

1.5 1.5 0.75 ksf1 2 ft

PLv

EIMS

V Q VI t A

=

= = =

= = = =

\

Figure 6.6:

(a) Deflection normalized with respect to result from beam theory for various numbers of elements used. (b) Axial stress at top of section for various numbers of elements used. (c) Contour plot of axial stress for analysis with four elements.

0 5 10 15 20

0

0.2

0.4

0.6

0.8

1

Def

lect

ion

Elements in

longitudinal direction

8

2

4

1

16

32

beam theory/v v0 5 10 15 200

5

10

15

20

25

30Ax

ial S

tress

ft

[ksf ]xx

(a)(b)

[ksf ]xx

(c)

1kP =

1 ftt =v

, 0E =



44

Figure 6.7:

(a) Shear stress at neutral axis for various numbers of elements used. (b) Contour plot of shear stress for analysis with four elements.

Figure 6.8: Contour plot of axial and shear stresses for analysis with 32 elements.

[ksf]xy =

ft

Elements in longitudinal direction,, ,

,18

1 26432

[ksf]xy =

[ksf ]xx [ksf]xy =



45

6.6 Element characteristics (Shortcomings)

Figure 6.9: Shortcoming of four-node element (too stiff in bending).

from Robert D. Cook, Finite Element Modeling for Stress Analysis, John Wiley & Sons (1995).

As seen on the previous page, the shear stresses calculated with the conventional four-node element are erratic and completely useless. The reason for this behavior is that the element cannot model a state of pure bending (despite its ability to represent a strain xx that varies linearly withy ), i.e. it generates shear stresses even though shear stresses do not exist. This is the well-known phenomenon of shear locking. Consider the figure above. We know from beam theory that xy is absent, that plane sections remain plane and that the deformed top and bottom edges are circular arcs. Hence a "correct" element should deform as shown in shown in Figure 6.9(b). The four-node element in pure bending, however, deflects as shown in Figure 6.9(c). Sides rotate as shown by the dashed lines but top and bottom edges remain straight. Therefore right angles are not preserved and as a consequence shear strain appears everywhere in the element except along they axis (the element center). We can see the same result by revisiting the general form of the trial functions and the derivatives that gives us the strain components

1 2 3 4

5 6 7 8

2 4

7 8

3 4 6 8

( , )

( , )

xx

yy

xy

u c c c c

v c c c c

c c

c c

c c c c

= + + += + + += += += + + +

(6.28)

The constant 4c needs to be non-zero such that xx varies linearly withy , but 4c also appears in the expression for xy . Consequently, an element that bends also develops shear strains, which makes the element too stiff in bending since it resists an applied moment by spurious shear stresses.



46

7 The 4-node membrane element with incompatible modes

7.1 General discussion The key defect of the four-node rectangular element is its over-stiffness in bending. A remedy for this problem is to expand the shape functions of the element by two terms that describe a state of constant curvature (displacement quadratic). The extra terms allow the edges of the element to become curved.

( ) ( )( ) ( )

42 2

1 21

42 2

3 41

( , ) ( , ) 1 1

( , ) ( , ) 1 1

ii

ii

u u c c

v u c c

=

=

= + +

= + +

N

N (7.1)

The shear strain in the element us 4 4

2 31 1

( , ) ( , )1 1 1 1 2 2xy i i

i i

v u v uv u c c

x y a b a b b a

= = = + = + = + N N (7.2)

For pure bending the negative terms are equal in magnitude to positive terms produced by the summations (shape functions of the original element) This element is usually the default option for a four-node rectangular element in commercial software. The degrees of freedom 1 4-c c are internal degrees of freedom. They are not connected to corresponding degrees of freedom in adja-cent elements such that overlaps or gaps between adjacent elements occur. The internal degrees-of-freedom can be eliminated at the element level by static condensation and thus do not show up in the structure stiffness matrix. Because of the gaps or overlaps that develop, the shape functions associated with degrees of freedom 1 4-c c are in-compatible. However, one can prove mathematically, that the gap converges to zero as the size of the elements approach zero (i.e. with increasing number of elements). Figure 7.1: Incompatible displacement modes for the four-node element

, ,x u , ,x u

, ,x u, ,x u

, ,y v

, ,y v, ,y v

22(1 )u c =

21(1 )u c =

23(1 )v c = 24(1 )v c =

, ,y v



47

Figure 7.2: Four-node element with incompatible modes (a) Tip deflection normalized with respect to result from beam theory for various numbers of elements used. (b) Axial stress at top of section for various numbers of elements used. (c) Contour plot of axial stress for analysis with four elements. Figure 7.3: Four-node element with incompatible modes (a) Shear stress at neutral axis for various numbers of elements used (two elements across depth). (b) Contour plot of shear stress for analysis with four elements (eight elements across depth).

exact/v v

Elements in


2

1

4

8

16

ft

[ksf ]xx

[ksf ]xx

[ksf]xy =

[ksf]xy =

Elements in


2

1

4

8

16

2 elements across depth



48

7.2 Results for beam bending (conventional four-node element and four-node element with incompatible modes)

Figure 7.4: (a) Axial stresses xx and (b) shear stresses xy using conventional four-node element, four elements along length, one along depth (values from beam theory (values from beam theory are max max7.5 ksf, 0.75 ksf = = ). .

Figure 7.5: (a) Axial stresses xx and (b) shear stresses xy using conventional four-node element, 16 elements along length, one along depth (values from beam theory (values from beam theory are max max7.5 ksf, 0.75 ksf = = ). .

20 ft

2f

t

1.5

1

0.5

0

0.5

1

1.5

1.5

1

0.5

0

0.5

1

1.5

3000ksi 1ft(thickness) 0.1k/ftE t p= = =

(a)

(b)

20 ft

2f

t

6

4

2

0

2

4

6

1.5

1

0.5

0

0.5

1

1.5


(a)

(b)



49

Figure 7.6: (a) Axial stresses xx and (b) shear stresses xy using four-node element with incompatible modes, four elements along length, one along depth (values from beam theory are max max7.5 ksf, 0.75 ksf = = ).

Figure 7.7: (a) Axial stresses xx and (b) shear stresses xy using four-node element with incompatible modes, ten elements along length, four along depth (values from beam theory are max max7.5 ksf, 0.75 ksf = = ).

0.3

0.2

0.1

0

0.1

0.2

0.3

20 ft

2f

t


element with incompatible modes

6

4

2

0

2

4

6

(a)

(b)

0.6

0.4

0.2

0

0.2

0.4

0.6

20 ft

2f

t

6

4

2

0

2

4

6


element with incompatible modes

(a)

(b)



50

HW#9: Frame/wall Behavior-Overturning resistance

0.250 =

1.00(reference) =

34C

M PH=

32C

M PH=

no frame action

some frame action

0GI =

GI =

P GI

GI

GI

CI

CI

CIH

H

H

CI

CI

CI



51

0.00154(41% flexure, 22% shear, 37% axial) =

0.0278 =

14C

M PH=

380C

M PH=

significant frame

action

extreme frame

action

GI =

GI =

P GI

GI

GI

CI

CI

CIH

H

H

CI

CI

CI


C:\calpoly\arce504\Handouts\Winter2010\HandoutsChapter8-9.doc 4/29/2010

52

8 Mapped elements using isoparametric transformation

8.1 Mathematical Formulation We use the so-called isoparametric formulation to transform elements of non-rectangular shape or curved sides to the formulation of a unit square element in local coordinates und . In local coordinates, element sides are always defined by 1, 1 = = . The physical coordinates of a point ,x y within an element are defined as

4

1

( , ) ( , ) ( , ) ii

x N x =

= = N x (8.1) und

4

1

( , ) ( , ) ( , ) ii

y N y =

= = N y (8.2) in which ix and iy are the coordinates of the corner nodes and the four interpolation functions iN (or shape func-tions) are the same as those used for the interpolation of the displacements ( , )u and ( , )v within the element. The name isoparametric thus derives from using the same functions to interpolate both the coordinates ( , )x and ( , )y and the displacementsu andv .

Figure 8.1: (a) Four-node plane isoparametric element in xy space. (b) Plane isoparametric element in space. from Robert D. Cook, Finite Element Modeling for Stress Analysis, John Wiley & Sons (1995). Shape functions iN and their derivatives are defined in local coordinates und . In order to calculate the strain, however, we need the derivatives of the shape functions with respect to the global coordinates x andy . For rectan-gular elements we can easily calculate the global derivative by dividing by the element dimensionsa andb (see B -matrix in Eq. 6.17). Also when integrating over the element, we need to express the area element d dx y as a function of d d , Again, for rectangular elements, we simply have

d d d dx y ab =

(8.3)

Consider the derivative of a shape function iN with respect to the local coordinates and . Using the chain rule gives



53

i i i

i i i

N N Nx yx y

N N Nx yx y

= + = +

(8.4)

or more general

i i i

i ii

N x y N Nx xN NN x yy y

= =

J

(8.5)

The inverse of this relation is

1

ii

i i

NNxN Ny

=

J

(8.6)

Since the coordinatesx andy are an explicit function of and , we can calculate the 2x2 Jacobian matrix 1 2

1 1

2 21 2

i ii i

i ii i

N N N Nx yx y

x yN N N Nx y

= =

J

# #

(8.7)

As always, we obtain the element stiffness matrix as an integral over the element 1 1 2 2

1 11 1

det d d ( , ) ( , ) detT T T i j i j i ji j

t t w w = =

= = K B C B J B C B J where det J is the determinant of the Jacobian matrix. The integration is carried out numerically using two Gauss points in each direction. The Jacobian matrix J of the transformation is familiar in elementary calculus. For example, when an integral is transformed from Cartesian coordinates into polar coordinates,d dx y is replaced by dr dr , in which J r= .

Figure 8.2: Mesh of quadrilateral elements of general shape.



54

8.2 Some remarks on element shapes When using mapped elements, we must do so in a way that no element has singular or near-singular Jacobian ma-trix. This is because we need to calculate the inverse of the Jacobian matrix and no inverse matrix exists if the ma-trix is singular and numerical instability may occur if the matrix is near singular (ill-conditioned). Numerical stabi-lity is guarantied if elements have compact and regular shape. The ideal triangular element is an equilateral triangle, the best rectangle is a square. from Robert D. Cook, Finite Element Modeling for Stress Analysis, John Wiley & Sons (1995). from M. Asghar Bhatti, Fundamental Finite Element Analysis and Applications, John Wiley & Sons (2005).

Figure 8.3: Guidelines for mapped elements shapes.



55

If the determinant of the Jacobian matrix is zero or near zero anywhere in the range of 1 1, 1 1 < < < < , the analysis becomes numerically unstable.

Figure 8.4: Contour plots of determinant of Jacobian matrix for a good and bad element.

det( )J

det( )J

range of integration

1 1

1 1

<

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