fem beams.pdf

8/22/2019 fem beams.pdf

1/23

1

CHAP 4 FINITE ELEMENT ANALYSIS OF

BEAMS AND FRAMES

2

INTRODUCTION

We learned Direct Stiffness Method in Chapter 2

Limited to simple elements such as 1D bars

we will learn Energy Method to build beam finite element

Structure is in equilibrium when the potential energy is minimum

Potential energy: Sum of strain energy and potential of

applied loads

Interpolation scheme:

( ) ( ) { }v x x N q

Beam

deflection

Interpolation

function

Nodal

DOF

Potential of

applied loadsStrain energy

U V3


2/23

3

BEAM THEORY

Euler-Bernoulli Beam Theory

can carry the transverse load

slope can change along the span (x-axis)

Cross-section is symmetric w.r.t. xy-plane

The y-axis passes through the centroid Loads are applied in xy-plane (plane of loading)

L F

x

y

F

Plane of loading

y

z

Neutral axis

A

4

BEAM THEORY cont.

Euler-Bernoulli Beam Theory cont. Plane sections normal to the beam axis remain plane and normal to

the axis after deformation (no shear stress)

Transverse deflection (deflection curve) is function of x only: v(x)

Displacement in x-dir is function of x and y: u(x, y)

y

y(dv/dx)

T= dv/dx

v(x)L

F

x

y

Neutral axis

0( , ) ( )dv

u x y u x ydx

2

0

2xx

duu d vy

x dx dxHw w

dv

dxT


3/23

5

BEAM THEORY cont.

Euler-Bernoulli Beam Theory cont.

Strain along the beam axis:

Strain Hxxvaries linearly w.r.t. y; Strain Hyy = 0

Curvature:

Can assume plane stress in z-dir basically uniaxial status

Axial force resultant and bending moment

2

0

2xx

duu d vy

x dx dxHw w0 0 /du dxH

2 2/d v dx

2

0 2xx xx

d vE E Ey

dxV H H

2

0 2

22

0 2

xx

A A A

xxA A A

d vP dA E dA E ydA

dx

d v

M y dA E ydA E y dAdx

V H

V H

Moment of inertia I(x)

0

2

2

P EA

d vM EI

dx

H

EA: axial rigidity

EI: flexural rigidity

6

BEAM THEORY cont.

Beam constitutive relation We assume P= 0 (We will consider non-zero P in the frame element)

Moment-curvature relation:

Sign convention

Positive directions for applied loads

2

2

d vM EI

dx Moment and curvature is linearly dependent

+P +P

+M+M+Vy

+Vy

y

x

p(x)

F1 F2 F3

C1 C2 C3

y

x


4/23

7

GOVERNING EQUATIONS

Beam equilibrium equations

Combining three equations together:

Fourth-order differential equation

y

y

dVV dx

dx

dMM dxdxy

VM

dx

p

0 ( ) 0y

y y y

dVf p x dx V dx V

dx

( )y

dVp x

dx

02

ydM dxM M dx pdx V dxdx

4

4( )

d vEI p x

dx

ydMVdx

8

STRESS AND STRAIN

Bending stress

This is only non-zero stress component for Euler-Bernoulli beam

Transverse shear strain

Euler beam predicts zero shear strain (approximation)

Traditional beam theory says the transverse shear stress is

However, this shear stress is in general small compared to

the bending stress

2

2xx

d vEy

dxV

2

2

d vM EI

dx

( )( , )xx

M x yx y

IV

0xyu v v v

y x x xJw w w w w w w w 0

( , ) ( )dv

u x y u x ydx

Bending stress

xy

VQ

IbW


5/23

9

POTENTIAL ENERGY

Potential energy

Strain energy

Strain energy density

Strain energy per unit length

Strain energy

U V3

2 22 2

2 2

0 2 2

1 1 1 1( )

2 2 2 2xx xx xx

d v d vU E E y Ey

dx dx

V H H

2 22 2

2 2

0 2 2

1 1( ) ( , , )

2 2L

A A A

d v d vU x U x y z dA Ey dA E y dA

dx dx

Moment of

inertia

22

2

1( )

2L

d vU x EI

dx

22

20 0

1( )

2

L L

L

d vU U x dx EI dx

dx

10

POTENTIAL ENERGY cont.

Potential energy of applied loads

Potential energy

Potential energy is a function ofv(x) and slope

The beam is in equilibrium when 3 has its minimum value

01 1

( )( ) ( ) ( )

CF NNLi

i i i

i i

dv xV p x v x dx F v x C

dx

22

20 01 1

( )1( ) ( ) ( )

2

CF NNL Li

i i i

i i

dv xd vU V EI dx p x v x dx F v x C

dx dx

3

3

vv*

0v

w3w


6/23

11

RAYLEIGH-RITZ METHOD

1. Assume a deflection shape

Unknown coefficients ciand known function fi(x)

Deflection curve v(x) must satisfy displacement boundary conditions2. Obtain potential energy as function of coefficients

3. Apply the principle of minimum potential energy to determine

the coefficients

1 1 2 2( ) ( ) ( )..... ( )n nv x c f x c f x c f x

1 2( , ,... )nc c c U V 3

12

EXAMPLE SIMPLY SUPPORTED BEAM

Assumed deflection curve

Strain energy

Potential energy of applied loads (no reaction forces)

Potential energy

PMPE:

p0

E,I,L

( ) sinx

v x CL

S

22 2 4

2 30

1

2 4

L d v C EI U EI dx

dx L

S

00

0 0

2( ) ( ) sin

L Lp Lx

V p x v x dx p C dx C L

S

S

42 0

3

2

4

p LEIU V C C

L

S

S3

44

0 0

3 5

2 40

2

p L p Ld EIC C

dC L EI

S

S S

3


7/23

13

EXAMPLE SIMPLY SUPPORTED BEAM cont.

Exact vs. approximate solutions

Approximate bending moment and shear force

Exact solutions

4 4

0 0approx exact

76.5 76.8

p L p LC C

EI EI

22 2

0

2 2 3

3 3

0

3 3 2

4( ) sin sin

4( ) cos cos

y

p Ld v x xM x EI EIC

dx L L L

p Ld v x xV x EI EIC

dx L L L

S S S

S

S S S

S

33 40 0 0

20 0

00

1( )

24 12 24

( )2 2

( )2

y

p L p L pv x x x x

EI

p L pM x x x

p LV x p x

14

EXAMPLE SIMPLY SUPPORTED BEAM cont.

Deflection

Bending

moment

Shear force

0.0

0.2

0.4

0.6

0.8

1.0

0 0.2 0.4 0.6 0.8 1x

v(x)/v_

ma

v-exact

v-approx.

-0.14

-0.12

-0.10

-0.08

-0.06

-0.04

-0.02

0.00

0 0.2 0.4 0.6 0.8 1

x

BendingM

omentM(x)

M_exact

M_approx

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0 0.2 0.4 0.6 0.8 1x

ShearForceV(x)

V_exact

V_approx

Errorinc

reases


8/23

15

EXAMPLE CANTILEVERED BEAM

Assumed deflection

Need to satisfy BC

Strain energy

Potential of loads

F

C

p0

E,I,L2 3

1 2( )v x a bx c x c x

(0) 0, (0) / 0v dv dx 2 3

1 2( )v x c x c x

2

1 2

0

2 62

LEI

U c c x dx

1 2 00

3 42 3 20 0

1 2

, ( ) ( ) ( )

2 33 4

Ldv

V c c p v x dx Fv L C Ldx

p L p Lc FL CL c FL CL

16

EXAMPLE CANTILEVERED BEAM cont.

Derivatives ofU:

PMPE:

Solve for c1 and c2:

Deflection curve:

Exact solution:

2

1 2 1 2

1 0

2 3

1 2 1 2

2 0

2 2 6 4 6

6 2 6 6 12

L

L

U EI c c x dx EI Lc L cc

UEI c c x xdx EI L c L c

c

w w

w w

1

2

0

0

c

c

w3w

w3w

32 20

1 2

42 3 3 20

1 2

4 6 23

6 12 34

p LEI Lc L c FL CL

p L

EI L c L c FL CL

3 3

1 223.75 10 , 8.417 10c c u u

3 2 3( ) 10 23.75 8.417v x x x

2 3 41

( ) 5400 800 30024

v x x x xEI


9/23

17


Deflection

Bending

moment

Shear force

Errorincreases

0.0

0.0

0.0

0.0

0.0

0 0.2 0.4 0.6 0.8 1x

v(x)/v_

ma

v-exact

v-approx.

-100.00

0.00

100.00

200.00

300.00

400.00

500.00

0 0.2 0.4 0.6 0.8 1x

BendingMomentM(x) M_exact

M_approx

200.0

300.0

400.0

500.0

600.0

0 0.2 0.4 0.6 0.8 1x

ShearForce

V(x)

V_exact

V_approx

18

FINITE ELEMENT INTERPOLATION

Rayleigh-Ritz method approximate solution in the entire beam Difficult to find approx solution that satisfies displacement BC

Finite element approximates solution in an element

Make it easy to satisfy displacement BC using interpolation technique

Beam element

Divide the beam using a set of elements

Elements are connected to other elements at nodes

Concentrated forces and couples can only be applied at nodes Consider two-node bean element

Positive directions for forces and couples

Constant or linearly

distributed load

F1 F2

C2C1

p(x)

x


10/23

19

FINITE ELEMENT INTERPOLATION cont.

Nodal DOF of beam element

Each node has deflection vand slope T

Positive directions of DOFs

Vector of nodal DOFs

Scaling parameter s Length L of the beam is scaled to 1 using scaling parameters

Will write deflection curve v(s) in terms ofs

v1 v2

T2T1

Lx1

s = 0

x2

s = 1

x

1 1 2 2{ } { }Tv vT Tq

1 1, ,

1,

x xs ds dx

L L

dsdx Lds

dx L

20


Deflection interpolation Interpolate the deflection v(s) in terms of four nodal DOFs

Use cubic function:

Relation to the slope:

Apply four conditions:

Express four coefficients in terms of nodal DOFs

2 3

0 1 2 3( )v s a a s a s a s

2

1 2 3

1( 2 3 )

dv dv dsa a s a s

dx ds dx LT

1 1 2 2

(0) (1)(0) (1)

dv dvv v v v

dx dxT T

1 0

1 1

2 0 1 2 3

2 1 2 3

(0)

1(0)

(1)

1(1) ( 2 3 )

v v a

dva

dx L

v v a a a a

dva a a

dx L

T

T

0 1

1 1

2 1 1 2 2

3 1 1 2 2

3 2 3

2 2

a v

a L

a v L v L

a v L v L

T

T T

T T


11/23

21


Deflection interpolation cont.

Shape functions

Hermite polynomials

Interpolation property

2 3 2 3 2 3 2 3

1 1 2 2( ) (1 3 2 ) ( 2 ) (3 2 ) ( )v s s s v L s s s s s v L s sT T

2 3

1

2 3

2

2 3

3

2 34

( ) 1 3 2

( ) ( 2 )

( ) 3 2

( ) ( )

N s s s

N s L s s s

N s s s

N s L s s

1

1

1 2 3 4

2

2

( ) [ ( ) ( ) ( ) ( )]

v

v s N s N s N s N s v

T

T

-

( ) { }v s N q

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0

N1 N3

N2/L

N4/L

22


Properties of interpolation Deflection is a cubic polynomial (discuss accuracy and limitation)

Interpolation is valid within an element, not outside of the element

Adjacent elements have continuous deflection and slope

Approximation of curvature

Curvature is second derivative and related to strain and stress

B is linear function ofs and, thus, the strain and stress

Alternative expression:

If the given problem is linearly varying curvature, the approximation is

accurate; if higher-order variation of curvature, then it is approximate

1

2 21

2 2 2 2

2

2

1 1[ 6 12 , ( 4 6 ), 6 12 , ( 2 6 )]

v

d v d vs L s s L s

vdx L ds L

T

T

-

2

2 24 11 4

1{ }

d v

dx L uu B q B: strain-displacement vector

2

2 24 11 4

1{ }T T

d v

dx L uu Bq


12/23

23


Approximation of bending moment and shear force

Stress is proportional to M(s); M(s) is linear; stress is linear, too

Maximum stress always occurs at the node

Bending moment and shear force are not continuous between adjacent

elements

2

2 2( ) { }

d v EI M s EI

dx L B q

3

3 3

[ 12 6 12 6 ]{ }y

dM d v EI V EI L L

dx dx L q

Linear

Constant

24

EXAMPLE INTERPOLATION

Cantilevered beam Given nodal DOFs

Deflection and slope at x = 0.5L

Parameter s = 0.5 at x = 0.5L

Shape functions:

Deflection at s = 0.5:

Slope at s = 0.5:

{ } {0, 0, 0.1, 0.2}T qL

v1

v2

T2T1

1 1 1 11 2 3 42 2 2 2

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

2 8 2 8

L LN N N N

1 1 1 1 11 1 2 1 3 2 4 22 2 2 2 2

2 22 2

( ) ( ) ( ) ( ) ( )

1 10 0 0.025

2 8 2 8 2 8

v N v N N v N

L L v Lv

T T

TT

u u u u

31 2 41 1 2 2

2 2 2 2

1 1 2 2

1 1

1 1

( 6 6 ) 1 4 3 (6 6 ) 2 3 0.1

dNdv dv dN dN dN v v

dx L ds L ds ds ds ds

v s s s s v s s s sL L

T T

T T


13/23

25

EXAMPLE

A beam finite element with length L

Calculate v(s)

Bending moment

3 2

1 1 2 20, 0, ,

3 2

L Lv v

EI EIT T

1 1 2 1 3 2 4 2( ) ( ) ( ) ( ) ( )v s N s v N s N s v N sT T

2 3 2 3

2 2( ) (3 2 ) ( )v s s s v L s s T

> @2 2

2 22 2 2 2

3 2

2

( ) (6 12 ) ( 2 6 )

(6 12 ) ( 2 6 )3 2

(1 ) ( )

d v EI d v EI M s EI s v L s

dx L ds L

EI L Ls L sL EI EI

L s L x

T

LF

Bending moment cause by unit force at the tip

26

FINITE ELEMENT EQUATION FOR BEAM

Finite element equation using PMPE A beam is divided by NEL elements with constant sections

Strain energy

Sum of each elements strain energy

Strain energy of element (e)

p(x)

F1 F2 F3

C1 C2 C3

y

x

F4 F5

C4 C5

1 2 3 45

1

1x1 2

2 1x x2

1

2 3

2 1x x3

1

3 4

2 1x x4

1

4

2x

2

10 1 1

( ) ( )e

T

e

NEL NELL x e

L Lx

e e

U U x dx U x dx U

2

1

2 22 21

2 3 20

1 1

2 2

e

e

xe

x

d v EI d vU EI dx ds

dx L ds


14/23

27

FE EQUATION FOR BEAM cont.

Strain energy cont.

Approximate curvature in terms of nodal DOFs

Approximate element strain energy in terms of nodal DOFs

Stiffness matrix of a beam element

22 2 2

2 2 21 4 4 14 1 1 4

{ } { }T Te ed v d v d v

ds ds ds u uu u

q B B q

1( )3 0

1 1{ } { } { } [ ]{ }

2 2

eTe e e e ee T TEIU ds

L

q B B q q k q

> @1

30

6 12

( 4 6 )[ ] 6 12 ( 4 6 ) 6 12 ( 2 6 )

6 12( 2 6 )

e

s

L sEIs L s s L s ds

sLL s

k

28


Stiffness matrix of a beam element

Strain energy cont.

Assembly

2 2

3

2 2

12 6 12 6

6 4 6 2[ ]

12 6 12 6

6 2 6 4

e

L L

L L L LEI

L LL

L L L L

k

Symmetric, positive semi-definite

Proportional to EI

Inversely proportional to L

( )

1 1

1{ } [ ]{ }

2

NEL NELe e ee T

e e

U U

q k q

1{ } [ ]{ }

2

T

s s sU Q K Q


15/23

29

EXAMPLE ASSEMBLY

Two elements

Global DOFs

F3F2

y

x

1 23

2EI EI

2L L 1 1 2 2 3 3{ } { }T

s v v vT T TQ

1 1 2 2

1

2 2

1 1

3

2

2 2

2

3 3 3 3

3 4 3 2[ ]

3 3 3 3

3 2 3 4

v v

L L v

L L L LEI

L LL v

L L L L

T T

T

T

k

2 2 3 3

2

2 2

2 2

3

3

2 2

3

12 6 12 6

6 4 6 2[ ]

12 6 12 6

6 2 6 4

v v

L L v

L L L LEI

L LL v

L L L L

T T

T

T

k

2 2

2 2 23

2 2

3 3 3 3 0 0

3 4 3 2 0 0

3 3 15 3 12 6[ ]3 2 3 8 6 2

0 0 12 6 12 6

0 0 6 2 6 4

s

L L

L L L L

L L LEIL L L L L LL

L L

L L L L

K

30


Potential energy of applied loads Concentrated forces and couples

Distributed load (Work-equivalent nodal forces)

1

ND

i i i i

i

V F v C T

1

1

1 1 2 2...... { } { }T

ND s s

ND

F

C

V v v F

C

T T

-

Q F

2

1

( )

1 1

( ) ( )e

e

NEL NELxe

xe e

V p x v x dx V

2

1

1

( ) ( )

0

( ) ( ) ( ) ( )e

e

xe e

xV p x v x dx L p s v s ds

1

( ) ( )

1 1 1 2 2 3 2 4

0

1 1 1 1

( ) ( ) ( ) ( )

1 1 1 2 2 3 2 4

0 0 0 0

( ) ( ) ( ) ( )

1 1 1 1 2 2 2 2

( )

( ) ( ) ( ) ( )

e e

e e e e

e e e e

V L p s v N N v N N ds

v L p s N ds L p s N ds v L p s N ds L p s N ds

v F C v F C

T T

T T

T T


16/23

31

EXAMPLE WORK-EQUIVALENT NODAL FORCES

Uniformly distributed load

pL/2 pL/2

pL2/12 pL2/12

p

Equivalent

1 12 3

1 10 0

( ) (1 3 2 )2

pLF pL N s ds pL s s ds

21 1

2 2 3

1 20 0

( ) ( 2 )

12

pLC pL N s ds pL s s s ds

1 12 3

2 30 0

( ) (3 2 )2

pLF pL N s ds pL s s ds

21 1

2 2 3

2 40 0

( ) ( )12

pLC pL N s ds pL s s ds

2 2

{ }2 12 2 12

T pL pL pL pL-

F

32


Finite element equation for beam

One beam element has four variables

When there is no distributed load, p = 0

Applying boundary conditions is identical to truss element

At each DOF, either displacement (vorT) or force (ForC) must be

known, not both

Use standard procedure for assembly, BC, and solution

1 1

2 2 2

1 1

3

2 2

2 2 2

2 2

12 6 12 6 / 2

6 4 6 2 /12

12 6 12 6 / 2

6 2 6 4 /12

v FL L pL

CL L L L pLEI

v FL L pLL

CL L L L pL

T

T

- - -


17/23

33

PRINCIPLE OF MINIMUM POTENTIAL ENERGY

Potential energy (quadratic form)

PMPE

Potential energy has its minimum when

Applying BC

The same procedure with truss elements (striking-the-rows and

striking-he-columns)

Solve for unknown nodal DOFs {Q}

1{ } [ ]{ } { } { }

2

T T

s s s s sU V3 Q K Q Q F

[ ]{ } { }s s sK Q F

[ ]{ } { }K Q F

[Ks] is symmetric & PSD

[K] is symmetric & PD

34

BENDING MOMENT & SHEAR FORCE

Bending moment

Linearly varying along the beam span

Shear force

Constant

When true moment is not linear and true shear is not constant, many

elements should be used to approximate it

Bending stress

Shear stress for rectangular section

2 2

2 2 2 2( ) { }

d v EI d v EI M s EI

dx L ds L B q

1

3 31

3 3 3 3

2

2

( ) [ 12 6 12 6 ]y

v

dM d v EI d v EI V s EI L L

vdx dx L ds L

T

T

-

x

My

IV

2

2

1.5 4( ) 1

y

xy

V yy

bh hW


18/23

35

EXAMPLE CLAMPED-CLAMPED BEAM

Determine deflection &

slope at x = 0.5, 1.0, 1.5 m

Element stiffness matrices

F2 = 240 N

y

x

1 2

1 m

3

1 m

1 1 2 2

1

1(1)

2

2

12 6 12 6

6 4 6 2[ ] 1000

12 6 12 6

6 2 6 4

v v

v

v

T T

T

T

k

2 2 3 3

2

2(2 )

3

3

12 6 12 6

6 4 6 2[ ] 1000

12 6 12 6

6 2 6 4

v vv

v

T T

T

T

k

11

11

2

2

33

33

12 6 12 6 0 0

6 4 6 2 0 0

24012 6 24 0 12 6100006 2 0 8 6 2

0 0 12 6 12 6

0 0 6 2 6 4

Fv

C

v

Fv

C

T

T

T

- -

36

EXAMPLE CLAMPED-CLAMPED BEAM cont.

Applying BC

At x = 0.5 s = 0.5 and use element 1

At x = 1.0 either s = 1 (element 1) or s = 0 (element 2)

2

2

24 0 2401000

0 8 0

v

T

- -

2

2

0.01

0.0

v

T

12

1 1 1 1 1 11 1 1 2 2 3 2 4 32 2 2 2 2 2

3122 (1)

( ) ( ) ( ) ( ) ( ) 0.01 ( ) 0.005m

1( ) 0.015rads

v v N N v N N N

dNvL ds

T T

T

u

2 3 3

32(1)

1

(1) (1) 0.01 (1) 0.01m

1(1) 0.0 rad

s

v v N N

dNv

L dsT

u

2 1 1

12(2 )

0

(0) (0) 0.01 (0) 0.01m

1(0) 0.0 rad

s

v v N N

dNv

L dsT

u

Will this solution be accurate or approximate?


19/23

37

EXAMPLE CANTILEVERED BEAM

One beam element

No assembly required

Element stiffness

Work-equivalent nodal forces

C=50 N-m

p0 = 120 N/m

EI= 1000 N-m2

1

1

2

2

12 6 12 6

6 4 6 2[ ] 1000

12 6 12 6

6 2 6 4

s

v

v

T

T

K

2 31

2 311

0 02 302

2 32

1/ 2 601 3 2

/12 10( 2 )

1/ 2 603 2

/12 10( )

e

e

e

e

F s s

C Ls s s Lp L ds p L

F s s

C Ls s L

- - - -

L = 1m

38


FE matrix equation

Applying BC

Deflection curve:

Exact solution:

1 1

1 1

2

2

12 6 12 6 60

6 4 6 2 101000

12 6 12 6 60

6 2 6 4 10 50

v F

C

v

T

T

- -

2

2

12 6 6010006 4 60

vT - -

2

2

0.010.03

vT

m

rad

3

3 4( ) 0.01 ( ) 0.03 ( ) 0.01v s N s N s s 4 3 2( ) 0.005( 4 )v x x x x


20/23

39


Support reaction (From assembled matrix equation)

Bending moment

Shear force

2 2 1

2 2 1

1000 12 6 60

1000 6 2 10

v F

v C

T

T

1

1

120N

10N m

F

C

> @

2

1 1 2 22

( ) { }

( 6 12 ) ( 4 6 ) (6 12 ) ( 2 6 )

1000[ 0.01(6 12 ) 0.03( 2 6 )]

60 N m

EIM sL

EIs v L s s v L s

L

s s

s

T T

B q

> @1 1 2 23 12 6 12 6

1000[ 12 ( 0.01) 6( 0.03)]

60 N

y EIV v L v LL

T T

u

40


Comparisons

-0.010

-0.008

-0.006

-0.004

-0.002

0.000

0 0.2 0.4 0.6 0.8 1x

v

FEM

Exact

-0.030

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0 0.2 0.4 0.6 0.8 1x

T

FEM

Exact

-60

-50

-40

-30

-20

-10

0

10

0 0.2 0.4 0.6 0.8 1x

M

FEM

Exact

-120

-100

-80

-60

-40

-20

0

0 0.2 0.4 0.6 0.8 1x

Vy

FEM

Exact

Deflection Slope

Bending moment Shear force


21/23

41

PLANE FRAME ELEMENT

Beam

Vertical deflection and slope. No axial deformation

Frame structure

Can carry axial force, transverse shear force, and bending moment

(Beam + Truss) Assumption

Axial and bending effects

are uncoupled

Reasonable when deformation

is small

3 DOFs per node

Need coordinate transfor-

mation like plane truss

F

p

1

2 3

4

1u

2u

1v 2v

11

22

1u

2u

1v

2v

11

22

1u

2u

1v

2v

11

221

2 3

{ , , }i i iu v T

42

PLANE FRAME ELEMENT cont.

Element-fixed local coordinates Local DOFs Local forces

Transformation between local and global coord.

1

2I

x

y

Local coordinates

Global coordinates

xy

1u

2u

1v

2v

11

2

x y{ , , }u v T { , , }x yf f c

1 1

1 1

1 1

2 2

2 2

2 2

cos sin 0 0 0 0

sin cos 0 0 0 0

0 0 1 0 0 0

0 0 0 cos sin 0

0 0 0 sin cos 00 0 0 0 0 1

x x

y y

x x

y y

f f

f f

c c

f f

f fc c

I I

I I

I I

I I

- -

{ } [ ]{ }f T f

{ } [ ]{ }q T q


22/23

43


Axial deformation (in local coord.)

Beam bending

Basically, it is equivalent to overlapping a beam with a bar

A frame element has 6 DOFs

11

22

1 1

1 1

x

x

fuEA

fuL

- -

1 1

2 2

1 1

3

2 2

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

y

y

vL L f

L L L L cEI

vL LL f

L L L L c

T

T

- -

44


Element matrix equation (local coord.)

Element matrix equation (global coord.)

Same procedure for assembly and applying BC

1 1 11

2 2 2 2 11

2 2

2 2 2 2 11

1 1 22

2 2 2 2 22

2 2

2 2 2 2 22

0 0 0 0

0 12 6 0 12 6

0 6 4 0 6 2

0 0 0 0

0 12 6 0 12 6

0 6 2 0 6 4

x

y

x

y

a a fu

a La a La fv

La L a La L a c

a a fu

a La a La fv

La L a La L a c

T

T

- -

1

2 3

EAa

L

EIa

L

[ ]{ } { }k fq

[ ][ ]{ } [ ]{ }k T q T f [ ] [ ][ ]{ } { }T T k T q f [ ]{ } { }k q f

[ ] [ ] [ ][ ]Tk T k T


23/23

45


Calculation of element forces

Element forces can only be calculated in the local coordinate

Extract element DOFs {q} from the global DOFs {Qs}

Transform the element DOFs to the local coordinate

Then, use 1D bar and beam formulas for element forces

Axial force

Bending moment

Shear force

Other method:

{ } [ ]{ }q T q

2 1AE

P u uL

2( ) { }

EIM s

L B q

^ `3

( ) [ 12 6 12 6 ]yEI

V s L LL q

1 1

2 2

1 1

3

2 2

2 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

y

y

V vL L

M L L L LEI

V vL LL

M L L L L

T

T

- -

fem beams.pdf

Documents

Transcript of fem beams.pdf