1 CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES FINITE ELEMENT ANALYSIS AND DESIGN Nam-Ho Kim

1

CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES

FINITE ELEMENT ANALYSIS AND DESIGNNam-Ho Kim

2

INTRODUCTION

• We learned Direct Stiffness Method in Chapter 2– Limited to simple elements such as 1D bars

• In Chapter 3, Galerkin Method and Principle of Minimum Potential Energy can be applied to more complex elements

• 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

Interpolationfunction

NodalDOF

Potential of applied loads

Strain energy

U V

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)

q = dv/dx

v(x) L F

x

y

Neutral axis

0

dvu(x,y) u (x) y

dx

20

xx 2

duu d vy

x dx dx

dv

dxq

5

BEAM THEORY cont.

• Euler-Bernoulli Beam Theory cont.– Strain along the beam axis:– Strain xx varies linearly w.r.t. y; Strain yy = 0

– Curvature:– Can assume plane stress in z-dir basically uniaxial

status

• Axial force resultant and bending moment

20

xx 2

duu d vy

x dx dx

0 0du / dx

2 2d v / dx

2

xx xx 0 2

d vE E Ey

dx

2

xx 0 2A A A

22

xx 0 2A A A

d vP dA E dA E ydA

dx

d vM y dA E ydA E y dA

dx

Moment of inertia I(x)

0

2

2

P EA

d vM EI

dx

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

yx

p(x)

F1 F2 F3

C1 C2 C3

y

x

7

GOVERNING EQUATIONS

• Beam equilibrium equations

– Combining three equations together:– Fourth-order differential equation

yy

dVV dx

dx

dMM dx

dx

yVM

dx

p

yy y y

dVf 0 p(x)dx V dx V 0

dx

ydV

p(x)dx

y

dM dxM M dx pdx V dx 0

dx 2

4

4

d vEI p(x)

dx

y

dMV

dx

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

xx 2

d vEy

dx

2

2

d vM EI

dx

xx

M(x)y(x,y)

I

xy

u v v v0

y x x x

0

dvu(x,y) u (x) y

dx

Bending stress

xy

VQ

Ib

9

POTENTIAL ENERGY

• Potential energy• Strain energy

– Strain energy density

– Strain energy per unit length

– Strain energy

U V

2 22 22 2

0 xx xx xx 2 2

1 1 1 d v 1 d vU E( ) E y Ey

2 2 2 dx 2 dx

2 22 22 2

L 0 2 2A A A

1 d v 1 d vU (x) U (x,y,z)dA Ey dA E y dA

2 dx 2 dx

Moment of inertia

22

L 2

1 d vU (x) EI

2 dx

22L L

L 20 0

1 d vU U (x)dx EI dx

2 dx

10

POTENTIAL ENERGY cont.

• Potential energy of applied loads

• Potential energy

– Potential energy is a function of v(x) and slope– The beam is in equilibrium when has its minimum value

CF NNL

ii i i0

i 1 i 1

dv(x )V p(x)v(x)dx Fv(x ) C

dx

CF2 NN2L L

ii i i20 0

i 1 i 1

dv(x )1 d vU V EI dx p(x)v(x)dx Fv(x ) C

2 dx dx

vv*

0v

11

RAYLEIGH-RITZ METHOD

1. Assume a deflection shape

– Unknown coefficients ci and known function fi(x)

– Deflection curve v(x) must satisfy displacement boundary conditions

2. 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 n(c ,c ,...c ) U V

1 2 n

0c c c

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

xv(x) Csin

L

22 2 4L

2 30

1 d v C EIU EI dx

2 dx 4L

L L0

0

0 0

2p LxV p(x)v(x)dx p Csin dx C

L

4

2 03

2p LEIU V C C

4L

44

0 03 5

2p L 4p Ld EIC 0 C

dC 2L EI

13

EXAMPLE – SIMPLY SUPPORTED BEAM cont.

• Exact vs. approximate deflection at the center

• Approximate bending moment and shear force

• Exact solutions

4 40 0

approx exact

p L p LC C

76.5EI 76.8EI

22 20

2 2 3

3 30

y 3 3 2

4p Ld v x xM(x) EI EIC sin sin

dx L L L

4p Ld v x xV (x) EI EIC cos cos

dx L L L

33 40 0 0

20 0

0y 0

p L p L p1v(x) x x x

EI 24 12 24

p L pM(x) x x

2 2p L

V (x) p x2

14

EXAMPLE – SIMPLY SUPPORTED BEAM cont.

• Deflection

• Bendingmoment

• Shear force

0.0

0.2

0.4

0.6

0.8

1.0

0 0.2 0.4 0.6 0.8 1xv(

x)/v

_max

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

Ben

ding

Mom

ent M

(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

She

ar F

orce

V(x

)

V_exact

V_approx

Erro

r increa

ses

15

EXAMPLE – CANTILEVERED BEAM

• Assumed deflection

• Need to satisfy BC

• Strain energy

• Potential of loads

F

C

–p0

E,I,L2 31 2v(x) a bx c x c x

v(0) 0, dv(0) / dx 0 2 31 2v(x) c x c x

L

2

1 2

0

EIU 2c 6c x dx

2

L

1 2 0

0

3 42 3 20 0

1 2

dvV c ,c p v(x)dx Fv(L) C (L)

dx

p L p Lc FL 2CL c FL 3CL

3 4

16

EXAMPLE – CANTILEVERED BEAM cont.

• Derivatives of U:

• PMPE:

• Solve for c1 and c2:

• Deflection curve:

• Exact solution:

L2

1 2 1 21 0

L2 3

1 2 1 22 0

U2EI 2c 6c x dx EI 4Lc 6L c

c

U6EI 2c 6c x xdx EI 6L c 12L c

c

1

2

0c

0c

32 20

1 2

42 3 3 20

1 2

p LEI 4Lc 6L c FL 2CL

3

p LEI 6L c 12L c FL 3CL

4

3 31 2c 23.75 10 , c 8.417 10

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

2 3 41v(x) 5400x 800x 300x

24EI

17


• Deflection

• Bendingmoment

• Shear force

Erro

r increa

ses

0.0

0.0

0.0

0.0

0.0

0 0.2 0.4 0.6 0.8 1xv(

x)/v

_max

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

Ben

ding

Mom

ent M

(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

She

ar F

orce

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 loadF1

F2

C2C1

p(x)

x

19

FINITE ELEMENT INTERPOLATION cont.

• Nodal DOF of beam element– Each node has deflection v and slope q– Positive directions of DOFs– Vector of nodal DOFs

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

• Will write deflection curve v(s) in terms of s

v1v2

q2q1

Lx1

s = 0

x2

s = 1

x

T1 1 2 2{ } {v v } q qq

1x x 1s , ds dx,

L Lds 1

dx 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 30 1 2 3v(s) a a s a s a s

21 2 3

dv dv ds 1(a 2a s 3a s )

dx ds dx Lq

1 1 2 2

dv(0) dv(1)v(0) v v(1) v

dx dx q q

1 0

1 1

2 0 1 2 3

2 1 2 3

v v(0) a

dv 1(0) a

dx Lv v(1) a a a a

dv 1(1) (a 2a 3a )

dx L

q

q

0 1

1 1

2 1 1 2 2

3 1 1 2 2

a v

a L

a 3v 2L 3v L

a 2v L 2v L

q q q q q

21


• Deflection interpolation cont.

• Shape functions

– Hermite polynomials– Interpolation property

2 3 2 3 2 3 2 31 1 2 2v(s) (1 3s 2s )v L(s 2s s ) (3s 2s )v L( s s ) q q

2 31

2 32

2 33

2 34

N (s) 1 3s 2s

N (s) L(s 2s s )

N (s) 3s 2s

N (s) L( s s )

1

11 2 3 4

2

2

v

v(s) [N (s) N (s) N (s) N (s)]v

q q

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 of s 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 22

2

v

d v 1 d v 1[ 6 12s, L( 4 6s), 6 12s, L( 2 6s)]

vdx L ds L

q q 2

2 24 11 4

d v 1{ }

dx L

B q B: strain-displacement vector

2T T

2 24 11 4

d v 1{ }

dx L

Bq

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 EIM(s) EI { }

dx L B q

3

y 3 3

dM d v EIV EI [ 12 6L 12 6L]{ }

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:

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

v1

v2

q 2q 1

1 1 1 11 2 3 42 2 2 2

1 L 1 LN ( ) , N ( ) , N ( ) , N ( )

2 8 2 8

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

2 22 2

v( ) N ( )v N ( ) N ( )v N ( )

v L1 L 1 L0 0 v 0.025

2 8 2 8 2 8

q q

q q

31 2 41 1 2 2

2 2 2 21 1 2 2

dNdN dN dNdv 1 dv 1v v

dx L ds L ds ds ds ds

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

L L

q q

q q

25

EXAMPLE

• A beam finite element with length L

• Calculate v(s)

• Bending moment

3 2

1 1 2 2

L Lv 0, 0, v ,

3EI 2EI q q

1 1 2 1 3 2 4 2v(s) N (s)v N (s) N (s)v N (s) q q

2 3 2 32 2v(s) (3s 2s )v L( s s ) q

2 2

2 22 2 2 2

3 2

2

d v EI d v EIM(s) EI (6 12s)v L( 2 6s)

dx L ds L

EI L L(6 12s) L( 2 6s)

L 3EI 2EI

L(1 s) (L x)

q

L F

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 element’s strain energy

– Strain energy of element (e)

p(x)

F1 F2 F3

C1 C2C3

y

x

F4 F5

C4 C5

1 2 3 45

( )11x

( ) ( )1 22 1x x= ( ) ( )2 3

2 1x x= ( ) ( )3 42 1x x= ( )4

2x

eT 2

e1

NEL NELL x eL L0 x

e 1 e 1

U U (x)dx U (x)dx U

e2

e1

2 22 2x 1e

2 3 2x 0

1 d v EI 1 d vU EI dx ds

2 dx L 2 ds

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 T Te e

2 2 21 4 4 14 1 1 4

d v d v d v{ } { }

ds ds ds

q B B q

e

1 Te e e e e(e) T T3 0

1 EI 1U { } ds { } { } [ ]{ }

2 L 2 q B B q q k q

1e

3 0

6 12s

L( 4 6s)EI[ ] 6 12s L( 4 6s) 6 12s L( 2 6s) ds

L 6 12s

L( 2 6s)

k

28


• Stiffness matrix of a beam element

• Strain energy cont.

– Assembly

2 2

e

3

2 2

12 6L 12 6L

6L 4L 6L 2LEI[ ]

L 12 6L 12 6L

6L 2L 6L 4L

k

Symmetric, positive semi-definite

Proportional to EI

Inversely proportional to L

NEL NEL

e e e(e) T

e 1 e 1

1U U { } [ ]{ }

2

q k q

Ts s s

1U { } [ ]{ }

2 Q K Q

29

EXAMPLE – ASSEMBLY

• Two elements• Global DOFs

F3F2

y

x

1 2 3

2EI EI

2L LT

s 1 1 2 2 3 3{ } {v v v } q q qQ

1 1 2 2

12 2

1 13

22 2

2

v v

3 3L 3 3L v

3L 4L 3L 2LEI[ ]

L 3 3L 3 3L v

3L 2L 3L 4L

q q

q q

k

2 2 3 3

22 2

2 23

32 2

3

v v

12 6L 12 6L v

6L 4L 6L 2LEI[ ]

L 12 6L 12 6L v

6L 2L 6L 4L

q q

q q

k

2 2

s 3 2 2 2

2 2

3 3L 3 3L 0 0

3L 4L 3L 2L 0 0

3 3L 15 3L 12 6LEI[ ]

L 3L 2L 3L 8L 6L 2L

0 0 12 6L 12 6L

0 0 6L 2L 6L 4L

K

30


• Potential energy of applied loads– Concentrated forces and couples

– Distributed load (Work-equivalent nodal forces)

ND

i i i ii 1

V Fv C

q

1

1T

1 1 2 ND s s2

ND

F

C

V v v ...... { } { }F

C

q q

Q F

e2

e1

NEL NELx (e)

xe 1 e 1

V p(x)v(x)dx V

e2

e1

1x(e) (e)

x0

V p(x)v(x)dx L p(s)v(s)ds

1

(e) (e)1 1 1 2 2 3 2 4

0

1 1 1 1(e) (e) (e) (e)

1 1 1 2 2 3 2 4

0 0 0 0

(e) (e) (e) (e)1 1 1 1 2 2 2 2

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

q q

q q

q q

31

EXAMPLE – WORK-EQUIVALENT NODAL FORCES

• Uniformly distributed load

pL/2 pL/2

pL2/12 pL2/12

p

Equivalent

1 1 2 31 10 0

pLF pL N (s)ds pL (1 3s 2s )ds

2

21 12 2 31 20 0

pLC pL N (s)ds pL (s 2s s )ds

12

1 1 2 32 30 0

pLF pL N (s)ds pL (3s 2s )ds

2

21 12 2 3

2 40 0

pLC pL N (s)ds pL ( s s )ds

12

2 2T pL pL pL pL

{ }2 12 2 12

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 (v or q ) or force (F or C) must be

known, not both

– Use standard procedure for assembly, BC, and solution

1 12 2 2

1 13

2 22 2 2

2 2

v F12 6L 12 6L pL / 2

C6L 4L 6L 2L pL / 12EIv FL 12 6L 12 6L pL / 2

C6L 2L 6L 4L pL / 12

q q

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}

T Ts s s s s

1U V { } [ ]{ } { } { }

2 Q K Q Q F

s s s[ ]{ } { }K 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 EIM(s) EI { }

dx L ds L B q

1

3 31

y 3 3 3 32

2

v

dM d v EI d v EIV (s) EI [ 12 6L 12 6L]

vdx dx L ds L

q q

x

My

I

2y

xy 2

1.5V 4y(y) 1

bh h

35

EXAMPLE – CLAMPED-CLAMPED BEAM

• Determine deflection & slope at x = 0.5, 1.0, 1.5 m

• Element stiffness matricesF2 = 240 N

y

x

1 2

1 m

3

1 m

1 1 2 2

1

1(1)

2

2

v v

v12 6 12 6

6 4 6 2[ ] 1000

v12 6 12 6

6 2 6 4

q q

q q

k

2 2 3 3

2

2(2)

3

3

v v

v12 6 12 6

6 4 6 2[ ] 1000

v12 6 12 6

6 2 6 4

q q

q q

k

11

11

2

2

33

33

F12 6 12 6 0 0 v

C6 4 6 2 0 0

24012 6 24 0 12 6 v1000

06 2 0 8 6 2

F0 0 12 6 12 6 v

C0 0 6 2 6 4

q q q

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

v24 0 2401000

0 8 0

q

2

2

v 0.01

0.0

q

12

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

3122 (1)

s

v( ) v N ( ) N ( ) v N ( ) N ( ) 0.01 N ( ) 0.005m

dN1( ) v 0.015rad

L ds

q q

q

2 3 3

32(1)

s 1

v(1) v N (1) 0.01 N (1) 0.01m

dN1(1) v 0.0rad

L ds

q

2 1 1

12(2)

s 0

v(0) v N (0) 0.01 N (0) 0.01m

dN1(0) v 0.0rad

L ds

q

Will this solution be accurate or approximate?

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

1s

2

2

v12 6 12 6

6 4 6 2[ ] 1000

v12 6 12 6

6 2 6 4

q q

K

2 31e

2 311e

0 02 302e

2 32e

F 1/ 2 601 3s 2s

C L / 12 10(s 2s s )Lp L ds p L

F 1/ 2 603s 2s

C L / 12 10( s s )L

L = 1m

38


• FE matrix equation

• Applying BC

• Deflection curve:• Exact solution:

1 1

1 1

2

2

v12 6 12 6 F 60

6 4 6 2 C 101000

v12 6 12 6 60

6 2 6 4 10 50

q q

2

2

v12 6 601000

6 4 60

q

2

2

v 0.01

0.03

q

m

rad

33 4v(s) 0.01N (s) 0.03N (s) 0.01s

4 3 2v(x) 0.005(x 4x x )

39


• Support reaction (From assembled matrix equation)

• Bending moment

• Shear force

2 2 1

2 2 1

1000 12v 6 F 60

1000 6v 2 C 10

q

q 1

1

F 120N

C 10N m

2

1 1 2 22

EIM(s) { }

LEI

( 6 12s)v L( 4 6s) (6 12s)v L( 2 6s)L1000[ 0.01(6 12s) 0.03( 2 6s)]

60s N m

q q

B q

y 1 1 2 23

EIV 12v 6L 12v 6L

L1000[ 12 ( 0.01) 6( 0.03)]

60N

q q

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

q

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

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

1u2u

1v 2v

1q

2q

1u

2u

1v

2v

1q

2q

1u

2u

1v

2v

1q

2q 1

2 3

i i i{u , v , }q

42

PLANE FRAME ELEMENT cont.

• Element-fixed local coordinates• Local DOFs Local forces• Transformation between local and global coord.

1

2f

x

y

Local coordinates

Global coordinates

xy

1u

2u

1v

2v

1q

2q

x y{u,v, }q x y{f , f , c}

x1 x1

y1 y1

1 1

x2 x2

y2 y2

2 2

f fcos sin 0 0 0 0f fsin cos 0 0 0 0

c c0 0 1 0 0 0

f f0 0 0 cos sin 0

f f0 0 0 sin cos 0

0 0 0 0 0 1c c

f f f f f f f f

{ } [ ]{ }f T f

{ } [ ]{ }q T q

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

1 x1

2 x2

u f1 1EAu fL 1 1

y112 2

1 13

2 y22 2

2 2

fv12 6L 12 6L

6L 4L 6L 2L cEIvL 12 6L 12 6L f

6L 2L 6L 4L c

q q

44


• Element matrix equation (local coord.)

• Element matrix equation (global coord.)

• Same procedure for assembly and applying BC

x11 1 1

y12 2 2 2 12 2

2 2 2 2 1 1

1 1 2 x2

2 2 2 2 2 y22 2

2 2 2 2 2 2

fa 0 0 a 0 0 uf0 12a 6La 0 12a 6La v

0 6La 4L a 0 6La 2L a c

a 0 0 a 0 0 u f

0 12a 6La 0 12a 6La v f

0 6La 2L a 0 6La 4L a c

q

q

1

2 3

EAa

LEI

aL

[ ]{ } { }k q f

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

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

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 1

AEP u u

L

2

EIM(s) { }

L B q

y 3

EIV (s) [ 12 6L 12 6L]

L q

y1 1

2 21 1

3y2 2

2 222

V v12 6L 12 6L

M 6L 4L 6L 2LEIV vL 12 6L 12 6L

6L 2L 6L 4LM

q q

Documents

1 CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES FINITE ELEMENT ANALYSIS AND DESIGN Nam-Ho Kim