3/11/2014

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H24SFEFinite Element Formulation of Bar and Beam Elements

Finite Element formulationstiffness matrices of bar element and a beam element.

The direct formulation is presented and then compared with that following the Finite Element method.

Examples to demonstrate the use of the Finite Element Method for trusses and frames.

Contents

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Capital letters are used to represent the local coordinates while small letters are used for the global coordinates

The direct formulation results from solving the differential equation of a bar element

Bar Element

EAXF

EdXdu XXX

X)(

===

EAXF

dXdu XX )(

=

210 and sBC'

XLXXXXXuuuu ==

==

Bar Element

constant a is where,

)( substitute and )( Integrate

1

1

CCXEAF

u

FXFEA

XFdXdu

XX

XXXX

+

=

==

sBC'for Substitute

*

)0(*

11

2

2

11

1

10

XX

X

XLXX

XX

X

XXX

uLEAF

u

uu

uCCEAF

u

uu

+

=

=

=+

=

=

=

=

)( 211 XXX uuLEAF =

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Bar Element

)(

)(in substitute

:mequilibriu From

212

211

12

XXX

XXX

XX

uuLEAF

uuL

EAF

FF

=

=

=

=

2

1

2

1

1111

X

X

X

X

u

u

LEA

FF

element.bar a ofmatrix stiffness local theis 1111

LEA

Using the Finite Element formulation

Bar Element

freedom of degrees ofnumber the toequal unknowns ofnumber with polynomial a using written becan endleft its from distance aat nt displaceme the

and freedom of degrees Two

21

21

XAAu

Xuu

X

XX

+=

210 and :sBC' theusing

freedom of degrees two theof in terms written becan unknowns twoThe

XLXXXXXuuuu ==

==

TXXX

XX

XXXX

uuuLX

LXH

uHu

XL

uuuu

][ and ]1[

where,

or

)(

21

121

==

=

+=

v

v

3/11/2014

4

Bar Element

]11[1]11[ where,

:follows as calculated becan strain The

=

=

===

LLLB

uBudXdH

dXdu

XXX

Xvv

TXXX

XX

uuuLX

LXH

uHu

][ and ]1[ 21==

=

v

v

=

=

=

=

1111]11[

11

)(

02

0

LEAdX

LEAK

EC

AdXCBBK

LL

LTL

Transform the local stiffness matrices from local to global coordinates The relationship between the local displacements and the global

displacements are given by

Matrix T transform the global displacements to local displacements.

Bar Element

sincosand

sincos

222

111

yxX

yxX

uuu

uuu

+=

+=

[ ]

=

=

2

2

1

1

2

1

2

2

1

1

2

1or

sincos0000sincos

formmatrix In

y

x

y

x

X

X

y

x

y

x

X

X

u

u

u

u

Tu

u

u

u

u

u

u

u

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The relationship between the global forces and the local forces are given by

Bar Element

sin,cos

,sin,cos

22

22

11

11

Xy

Xx

Xy

Xx

FFFF

FFFF

=

=

=

=

=

=

2

1

2

2

1

1

2

1

2

2

1

1

or

sin0cos0

0sin0cos

:formmatrix In

X

XT

y

x

y

x

X

X

y

x

y

x

FF

T

FFFF

FF

FFFF

Now the relationship between the global forces and the global displacements can be written as:

Bar Element

[ ]

=

2

1

2

2

1

1

X

XT

y

x

y

x

FF

T

FFFF

[ ]

=

2

2

1

1

2

1

y

x

y

x

X

X

u

u

u

u

Tu

u

=

2

2

1

1

2

2

1

1

sincos0000sincos

1111

sin0cos0

0sin0cos

y

x

y

x

y

x

y

x

u

u

u

u

LEA

FFFF

[ ]

=

2

1

2

1

X

XL

X

X

u

uK

FF

[ ] [ ] [ ] [ ][ ]GLTG uTKTF =

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Bar Element

=

2

2

1

1

2

2

1

1

sincos0000sincos

1111

sin0cos0

0sin0cos

y

x

y

x

y

x

y

x

u

u

u

u

LEA

FFFF

=

2

2

1

1

22

22

22

22

2

2

1

1

y

x

y

x

y

x

y

x

u

u

u

u

scsscs

csccsc

scsscs

csccsc

LEA

FFFF

cos and sin where, == cs

Example 1

Bar Element

N2m3

x

y

m4

N2x

y

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We first need to calculate the member stiffness matrices in global coordinates

Bar Element - Example

N2x

y

[ ]

=

22

22

22

22

scsscs

csccsc

scsscs

csccsc

LEAK G

For Element 1 (nodes 1-2) =0

Bar Element Example (contd)

N2x

y

x

y

X

Y

4321

00000333.00333.000000333.00333.04 3 2 1

1

= EAK G

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For Element 2 (nodes 2-3) =2.2143

Bar Element Example (contd)

N2x

y

6543

128.0096.0128.0096.0096.0072.0096.0072.0128.0096.0128.0096.0

096.0072.0096.0072.06 5 4 3

2

= EAK G

x

y

X

Y

The global stiffness matrix of the whole structure:

Bar Element Example (contd)

6543

128.0096.0128.0096.0096.0072.0096.0072.0128.0096.0128.0096.0

096.0072.0096.0072.06 5 4 3

2

= EAK G

654321

128.0096.0128.0096.000096.0072.0096.0072.000128.0096.0128.0096.000

096.0072.00960.0405.00333.0000000000333.00333.06 5 4 3 2 1

= EAK G

4321

00000333.00333.000000333.00333.04 3 2 1

1

= EAK G

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The global force-displacement relationship reads:

Bar Element Example (contd)

=

3

3

2

2

1

1

3

3

2

2

1

1

128.0096.0128.0096.000096.0072.0096.0072.000128.0096.0128.0096.000

096.0072.00960.0405.00333.0000000000333.00333.0

y

x

y

x

y

x

y

x

y

x

y

x

u

u

u

u

u

u

EA

FFFFFF

2 and 0and

0 :sBC'

22

3311

==

====

yx

yxyx

FF

uuu u

Bar Element Example (contd)

2 and 0and

0 :sBC'

22

3311

==

====

yx

yxyx

FF

uuu u

N2x

y

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Bar Element Example (contd)

=

3

3

2

2

1

1

3

3

2

2

1

1

128.0096.0128.0096.000096.0072.0096.0072.000128.0096.0128.0096.000

096.0072.00960.0405.00333.0000000000333.00333.0

y

x

y

x

y

x

y

x

y

x

y

x

u

u

u

u

u

u

EA

FFFFFF

0 :sBC' 3311 ==== yxyx uuu u

Bar Element Example (contd)

=

3

3

2

2

1

1

3

3

2

2

1

1

128.0096.0128.0096.000096.0072.0096.0072.000128.0096.0128.0096.000

096.0072.00960.0405.00333.0000000000333.00333.0

y

x

y

x

y

x

y

x

y

x

y

x

u

u

u

u

u

u

EA

FFFFFF

2 and 0128.0096.00960.0405.0

22

2

2

2

2

==

=

yx

y

x

y

x

FF

u

uEA

FF

0 :sBC' 3311 ==== yxyx uuu u

=

=

195.41

20

128.0096.00960.0405.01

1

2

2

EAEAuu

y

x

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Beam element subjected to end forces and moments The direct formulation results from solving the differential equation of a

beam element (also known as Euler-Bernoulli beam equation)

Beam Element

XFMXMdX

udEI YY 1122

)( +==

22

10

10

and

and

sBC'

==

==

=

=

=

=

LX

YYLXY

X

YYXY

dXdu

uu

dXdu

uu

Beam Element

22

10

10

and

and

sBC'

==

==

=

=

=

=

LX

YYLXY

X

YYXY

dXdu

uu

dXdu

uu

11

3

1

2

1

10

21

3

1

2

1

1

2

11

1110

1

2

11

62

62

:Integrate

2

C

2

:Integrate

YYY

YXY

YY

YY

X

Y

YY

EIuXEIXFXMEIu

uu

CXEIXFXMEIu

EIXFXMdXduEI

EIdXdu

CXFXMdXduEI

++=

=

++=

+=

==

++=

=

=

XFMXMdX

udEI YY 1122

)( +==

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Beam Element

22

10

10

and

and

sBC'

===

===

=

=

=

=

LX

YYLXY

X

YYXY

dXdu

uu

dXdu

uu

11

3

1

2

12

2

2

112

2

62

2

:at sBC' with Subsitute

YYY

YLXY

Y

LX

Y

EIuLEILFLMEIu

uu

EILFLMEI

dXdu

LX

++=

=

+=

=

=

=

=

Beam Element

22122121

221223131

11

3

1

2

12

2

112

2466

661212

:Rearrange62

2

LEI

LEI

uLEI

uLEIM

LEI

LEI

uLEI

uLEIF

EIuLEILFLMEIu

EILFLMEI

YY

YYY

YYY

Y

+++=

=

++=

+=

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The vertical equilibrium of the beam in the Y directions and taking moments around node 2 result in the following two equations:

Beam Element

LFMM

FF

Y

YY

112

12

and=

=

The end forces (including moments) can be written in terms of the end degrees of freedom:

Beam Element

=

2

2

1

1

matrixstiffnesslocalThe

22

2323

22

2323

2

2

1

1

4626

612612

2646

612612

Y

Y

Y

Y

u

u

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

MFMF

444444 3444444 21

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Using the Finite Element formulation

Beam Element

34

2321

2211

freedom of degrees ofnumber the toequal unknowns ofnumber with polynomial a using written becan endleft its from distance aat nt displaceme the

and , , freedom of degreesFour

xAxAxAAu

Xuu

Y

YY

+++=

=

4

3

2

1

2

32

2

2

1

1

32101

00100001

:conditionsboundary four thengSubstituti

AAAA

LLLLLu

u

Y

Y

2432 32

from calculated becan X distance aat slope) (therotation of angle The

xAxAAdXduY

==

Beam Element

=

=

=

2

2

1

1

2323

22

4

3

2

1

4

3

2

1

2

32

2

2

1

1

1212

132300100001

32101

00100001

Y

Y

Y

Y

u

u

LLLL

LLLLAAAA

A

AAAA

LLLLLu

u

r

EICdxdEIMuB

dxd

CB

CBdXBKL

T

===

=

and

:equations following thefrom calculated are and matrices thewhere,

from calculated is beam a ofmatrix stiffness The0

r

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Beam Element

[ ]AxxAAdxd

xAxAAr

620062

32 teDeffrentia

43

2432

==

=

[ ]

=

=

=

2

2

1

1

2323

22

2

2

1

1

2323

22

4

3

2

1

1212

132300100001

6200

1212

132300100001

But

Y

Y

Y

Y

u

u

LLLL

LLLLx

dxd

u

u

LLLL

LLLLAAAA

Ar

Beam Element

[ ]

+

+

+

=

=

2

2

1

1

232232

2

2

1

1

2323

22

6212664126

1212

132300100001

6200

Y

Y

Y

Y

u

u

Lx

LLx

LLx

LLx

Ldxd

u

u

LLLL

LLLLx

dxd

+

+

+

= 2322326212664126

Lx

LLx

LLx

LLx

LB

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Beam Element

+

+

+

=

==

232232

0

6212664126

and

Lx

LLx

LLx

LLx

LB

EICCBdXBKL

T

+

+

+

+

+

+

=L

L dXLX

LLX

LLX

LLX

LEI

LX

L

LX

L

LX

L

LX

L

K0

232232

2

32

2

32

6212664126

62

126

64

126

Beam Element

+

+

+

+

+

+

=

LL dX

LX

LLX

LLX

LLX

LEI

LX

L

LX

L

LX

L

LX

L

K0

232232

2

32

2

32

6212664126

62

126

64

126

=

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

K L

4626

612612

2646

612612

22

2323

22

2323

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The stiffness matrix of a beam element subjected to both bending, shear and axial forces is assembled from the stiffness matrices:

Beam Element

=

=

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

K

LEAK

L

L

4626

612612

2646

612612and

1111

22

2323

22

2323

The global force-displacement relationship reads:

Beam Element

=

2

2

2

1

1

1

22

2323

22

2323

2

2

2

1

1

1

460260

61206120

0000

260460

61206120

0000

Y

X

Y

X

Y

X

Y

X

u

u

u

u

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEA

LEA

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEA

LEA

MFFMFF

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Beam Element

22

222

222

11

111

111

sincossincos

sincossincos

:rotations and ntsdisplaceme endfor used, are ipsrelationsh following thes,coordinate global toscoordinate local from transformTo

=

=

+=

=

=

+=

xyY

yxX

xyY

yxX

uuu

uuu

uuu

uuu

Beam Element

TKTK

FTF

cs

sc

cs

sc

T

uTu

LTG

LTG

GL

=

=

=

=

:bygiven ismatrix stiffness global eFinally th

:bygiven are forces local and global ebetween th iprelationsh The1000000000000000010000000000

vr

rr

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Example 2

The 2D frame shown has two elements with the same axial stiffness (EA) and bending stiffness (EI), calculate the displacements of all nodes.

Examples

N 20

m2

m3

Local stiffness matrix for a beam Element

Examples (contd)

=

LLLL

LLLL

LIA

LIA

LLLL

LLLL

LIA

LIA

EIK L

460260

61206120

001001

260460

61206120

001001

22

2323

22

2323

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For Element 1 (nodes 1-2)

Examples (contd)

654321

25.1015.105.15.105.15.10

002500250.15.1025.105.15.105.15.10

002500256 5 4 3 2 1

1

= EIK Gx

y

X

Y

pi =

TKTK LTG =

For Element 2 (nodes 2-3)

Examples (contd)

987654

33.1067.067.0067.0067.160067.16067.0044.067.0044.067.0067.033.1067.0067.160067.16067.0044.067.0044.0

9 8 7 6 5 4

2

= EIK G

23pi =

TKTK LTG =

y

X

Y

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The global force-displacement relationship reads:

Examples (contd)

++

++

+

=

3

3

3

2

2

2

1

1

1

3

3

3

2

2

2

1

1

1

33.1067.067.0067.0000067.160067.16000067.0044.067.0044.000067.0067.033.1205.167.0015.10067.16005.167.165.105.15.1067.0044.067.00044.0250025

0000.15.1025.100005.15.105.15.1000000250025

y

x

y

x

y

x

y

x

y

x

y

x

u

u

u

u

u

u

EI

MFFMFFMFF

0 ,0 ,0 0 ,20 ,0

and0 :sBC'

222

111

333

===

===

===

MFF

MFF

u u

yx

yx

yx

N 20

m2

m3

Examples (contd)

++

++

+

=

3

3

3

2

2

2

1

1

1

3

3

3

2

2

2

1

1

1

33.1067.067.0067.0000067.160067.16000067.0044.067.0044.000067.0067.033.1205.167.0015.10067.16005.167.165.105.15.1067.0044.067.00044.0250025

0000.15.1025.100005.15.105.15.1000000250025

y

x

y

x

y

x

y

x

y

x

y

x

u

u

u

u

u

u

EI

MFFMFFMFF

0 :sBC' 333 === yx u u

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Applying the same procedure followed in example 1, the global displacements and forces of the frame are given by

Examples (contd)

[ ]TGEI

u 0001202.11801605.2941801 =

[ ]TGF 402000000200 =

Examples (contd)

e with valuload ddistributeuniformly a from resulting forces end thecalculate element, beam aFor

==L

TL

TL dxHdxHF00

Example 3

===L

T

LT

LTL dx

x

x

x

LLLL

LLLLdxHdxHF

03

2

2323

2200

1

1212

132300100001

=

12

2

12

2

2

2

L

L

L

L

F L

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Examples (contd)

[ ]

[ ]

1212

132300100001

1

1212

132300100001

but 1

2

2

1

1

2323

2232

2

2

1

1

2323

22

4

3

2

1

4

3

2

1

32

34

2321

=

=

=

+++=

Y

Y

Y

Y

Y

Y

Y

u

u

LLLL

LLLLxxxu

u

u

LLLL

LLLLAAAA

AAAA

xxxu

xAxAxAAu

Examples (contd)

[ ]

=

=

3

2

2323

22

2

2

1

1

2323

2232

1

1212

132300100001

1212

132300100001

1

x

x

x

LLLL

LLLLH

u

u

LLLL

LLLLxxxu

T

T

Y

Y

Y

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24

Example 4

Examples (contd)

Starting from the stiffness matrix subjected to both bending-shear and axial forces(shown below), calculate the stiffness matrix of an element with zero moment on theleft hand side and then calculate the stiffness matrix when the right hand sidemoment is equal to zero.

Examples (contd)Separate out the influence of the rotational DoF at node 1:

If M1 = 0, then: Then the remaining 5 equations are:

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Examples (contd)Express in terms of the other degrees of freedom and sub back in to the system ofequations

1

Stiffness matrix for anelement with zero moment onthe left hand side

Examples (contd)Repeating this process with zero moment on the right hand side yields,

Confirm this result yourself.

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Examples (contd)

Example 6

a) Using the Finite Element formulation of a beam element, show that the equivalent force vector of the load shown in the figure is given by

TL

LLxx

LLxx

LLxx

LLxLxWF ])(,)32(,)(,))(2([ 21

21

31

21

2

211

3

211 +

=

Examples (contd)

b) Using the results in (a) calculate the vertical displacement and rotation of nodes 1, 2, and 3 of the continuous beam shown

The beam has uniform cross-section with bending stiffness EI kN/m2.m4

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27

Hints to solve this example

Examples (contd)

=

+=

LT

L

Ts

TvL

dxxfxHF

dssfHdvHF

0

)()(

}][{}][{ vv

Examples (contd)

How do we describe a single point load as a continuous function?

We utilise the helpful properties of the Delta Dirac function,

)()( 1xxWxF = The function (x - a) equals zero for all values EXCEPT when x = a, so if we say..,

we now have a continuous function that is zero at all locations except at x = x1,where the function equals W.

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Examples (contd)

We will also need to know what the integral of the above function looks like, anymath text will tell us that

So the integral between the limits -infinity and +infinity, is simply the value of thefunction at a, or x1 in our case. (The use of infinity is mathematically correcthowever practically speaking were only integrating between 0 and L)

)()()( afdxaxxf =

Examples (contd)

=L

TL dxxfxHF

0

)()(

)()( 1xxWxf =

( )( )dxxxWx

x

x

LLLL

LLLLF

L

T

L 10

3

2

2323

22

1

1212

132300100001

=

( )( )( )( )

dx

xxWxxxWxxxxW

xxW

LLLL

LLLLF

L

T

L

=

0

13

12

1

1

2323

22

1212

132300100001

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29

Examples (contd)

Applying what we know about the Delta Dirac function:

=

31

21

1

2323

22

1212

132300100001

WxWxWxW

LLLL

LLLLF

T

L

After multiplying out and simplifying, we have:

Note the transpose

TL

LLxx

LLxx

LxLx

LxLLxWF ])(,)32(,)(,))(2([ 21

21

31

21

2

211

3

211 +

=

Examples (contd)

Part (b): Using the previous result, calculate displacements and rotations at all nodes

Constructing the element stiffness matrices for both elements

=

2.0000 1.5000 1.0000 1.5000- 1.5000 1.5000 1.5000 1.5000- 1.0000 1.5000 2.0000 1.5000- 1.5000- 1.5000- 1.5000- 1.5000

1 EIKG

=

1.0000 0.3750 0.5000 0.3750- 0.3750 0.1875 0.3750 0.1875- 0.5000 0.3750 1.0000 0.3750- 0.3750- 0.1875- 0.3750- 0.1875

2 EIKG

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30

Examples (contd)Construct the global stiffness matrix

=

1.0000 0.3750 0.5000 0.3750- 0 0 0.3750 0.1875 0.3750 0.1875- 0 0 0.5000 0.3750 3.0000 1.1250 1.0000 1.5000- 0.3750- 0.1875- 1.1250 1.6875 1.5000 1.5000-

0 0 1.0000 1.5000 2.0000 1.5000- 0 0 1.5000- 1.5000- 1.5000- 1.5000

EIK G

Examples (contd)

We need to account for the linear spring at node 2

The spring influences the vertical deflection at node 2, so we can easily see whereit needs to fit into our global stiffness matrix

=

3

3

2

2

1

1

3

3

2

2

1

1

1.0000 0.3750 0.5000 0.3750- 0 0 0.3750 0.1875 0.3750 0.1875- 0 0 0.5000 0.3750 3.0000 1.1250 1.0000 1.5000- 0.3750- 0.1875- 1.1250 1.6875 1.5000 1.5000-

0 0 1.0000 1.5000 2.0000 1.5000- 0 0 1.5000- 1.5000- 1.5000- 1.5000

y

y

y

y

y

y

u

u

u

EI

MFMFMF

+ Kspring = 0.25 EI

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31

Examples (contd)

=

3

3

2

2

1

1

3

3

2

2

1

1

1.0000 0.3750 0.5000 0.3750- 0 0 0.3750 0.1875 0.3750 0.1875- 0 0 0.5000 0.3750 3.0000 1.1250 1.0000 1.5000- 0.3750- 0.1875- 1.1250 1.9375 1.5000 1.5000-

0 0 1.0000 1.5000 2.0000 1.5000- 0 0 1.5000- 1.5000- 1.5000- 1.5000

y

y

y

y

y

y

u

u

u

EI

MFMFMF

Examples (contd)Now simply set up the force vector

The equivalent force vector for a UDL is

And weve also determined that the equivalent force vector for the point load

where = -2 kN/m and L = 2 m

=

+

=

=

88

88

)(

)32(

)(

))(2(

21

21

31

21

2

211

3

211

3

3

2

2

LLxx

LLxx

LxLx

LxLLx

W

MFMF

Fy

y

=

=

=

0.6667- 2.0000- 0.6667 2.0000-

12

2

12

2

2

2

2

2

1

1

L

L

L

L

MFMF

Fy

y

where W = 16 kN and L = 4 m and x1 = 2 m

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32

Examples (contd)

Therefore the global force vector is:

=

+

=

=

88

333.710667.0

2

88

88

00

00667.02

667.02

3

3

2

2

1

1

MFMFMF

F

y

y

y

G

=

3

3

2

2

1

1

1.0000 0.3750 0.5000 0.3750- 0 0 0.3750 0.1875 0.3750 0.1875- 0 0 0.5000 0.3750 3.0000 1.1250 1.0000 1.5000- 0.3750- 0.1875- 1.1250 1.9375 1.5000 1.5000-

0 0 1.0000 1.5000 2.0000 1.5000- 0 0 1.5000- 1.5000- 1.5000- 1.5000

000.8000.8

333.7000.10

667.0000.2

y

y

y

u

u

u

EI

Examples (contd)

=

3

3

2

2

1

1

1.0000 0.3750 0.5000 0.3750- 0 0 0.3750 0.1875 0.3750 0.1875- 0 0 0.5000 0.3750 3.0000 1.1250 1.0000 1.5000- 0.3750- 0.1875- 1.1250 1.9375 1.5000 1.5000-

0 0 1.0000 1.5000 2.0000 1.5000- 0 0 1.5000- 1.5000- 1.5000- 1.5000

000.8000.8

333.7000.10

667.0000.2

y

y

y

u

u

u

EI

Account for known displacements at supports:

=

2

2

1

000.31250.1000.1125.19375.1500.1000.1500.1000.2

333.7000.10

667.0

yuEI

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33

Examples (contd)

Solve for unknown displacements:

=

0909.52727.15

2424.91

2

2

1

EIu y

=

=

00

0909.52727.15

2424.90

1

3

3

2

2

1

1

EIu

u

u

u

y

y

y

G