CHAP 4 FINITE ELEMENT ANALYSIS OF BEAMS AND FRAMES INTRODUCTION

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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

N

x

q

Beam deflection Interpolation function Nodal DOF Potential of applied loads Strain energy

U V

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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

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) = _dv/_dx v(_x) L F x y Neutral axis 0

( , )

( )

dv

u x y

u x

y

dx

0 2 2 xx

du

u

d v

y

x

dx

dv dx

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5

BEAM THEORY

cont

.

• Euler-Bernoulli Beam Theory cont.

– Strain along the beam axis:

– Strain _xxvaries 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

2 0 2 xx

du

u

d v

y

x

dx

0

du dx

0

/

2 2

/

d v dx

2 0 2 xx xx

d v

E

Ey

dx

2 0 2 2 2 0 2 xx A A A xx A A A

d v

P

dA E

ydA

dx

d v

M

y

dA

E

ydA E

y dA

dx

Moment of inertia I(x) 0 2 2

P

EA

d v

M

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 v

M

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

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7

GOVERNING EQUATIONS

• Beam equilibrium equations

– Combining three equations together:

– Fourth-order differential equation

y y dV V dx dx y y y dM M dx dx y V

M

dx

p

0 ( )

y

0

y y y

dV

f

p x dx

V

dx

V

dx

dV

y

p x

( )

dx

0

2

y

dM

dx

M

dx

pdx

V dx

dx

4 4

( )

d v

EI

p x

dx

y

dM

V

dx

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 2 xx

d v

Ey

dx

M

EI

d v

2₂

dx

( )

( , )

xx

M x y

x y

I

0

xy

u

v

y

x

( , )

0

( )

dv

u x y

u x

y

dx

Bending stress xy

VQ

Ib

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9

POTENTIAL ENERGY

• Potential energy

• Strain energy

– Strain energy density

– Strain energy per unit length

– Strain energy U V 2 2 2 2 2 2 0 2 2

1

1 (

)

2

xx xx

2

xx

2

2 d v

d v

U

E

y

Ey

dx

2 2 2 2 2 2 0 2 2

1

1 ( )

( , , )

2

L A A A

d v

U x

U x y z dA

Ey

dA

E

y dA

dx

Moment of inertia 2 2 2

1 ( )

2

L

d v

U x

EI

dx

2 2 2 0 0

1 ( )

2

L L L

d v

U

U x dx

EI

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

0 1 1

( )

( ) ( )

( )

C F N N L i i i i i i

dv x

V

p x v x dx

Fv x

C

dx

2 2 2 0 0 1 1

( )

1 ( ) ( )

( )

2

C F N N L L i i i i i i

dv x

d v

U V

EI

dx

p x v x dx

Fv x

C

dx

v v*

0 v

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11

RAYLEIGH-RITZ METHOD

1. Assume a deflection shape

– Unknown coefficients c_iand known function f_i(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 n}

( )

v x

c f x

1 2

( , ,... )

c c

c

_n

U V

EXAMPLE

–

SIMPLY SUPPORTED BEAM

• Assumed deflection curve

• Strain energy

• Potential energy of applied loads (no reaction forces)

• Potential energy

• PMPE:

p

₀

E,I,L

( ) sin x v x C L

2 2 2 4 2 3 0

1

2

4

L

d v

C EI

U

EI

dx

L

0 0 0 0

2 ( ) ( )

sin

L L

p L

x

V

p x v x dx

p C

dx

C

L

4 2 0 3

2

4 p L

EI

U V

C

L

4 4 0 0 3 5

2

4

0

2 p L

p L

d

EI

C

dC

L

EI

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13

EXAMPLE

–

SIMPLY SUPPORTED BEAM

cont

.

• Exact vs. approximate solutions

• Approximate bending moment and shear force

• Exact solutions

4 4 0 0 approx exact

76.5

76.8 p L

p L

C

EI

2 2 2 0 2 2 3 3 3 0 3 3 2

4 ( )

sin

4 ( )

cos

y

p L

d v

x

M x

EI

EIC

dx

L

p L

d v

x

V x

EI

EIC

dx

L

3 3 4 0 0 0 2 0 0 0 0

1 ( )

24

12

24 ( )

2

2 ( )

2

y

p L

p

v x

x

EI

p L

p

M x

x

p L

V 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 x 0.6 0.8 1 v( x) /v _ m a x 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 B endi ng Moment 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 x 0.6 0.8 1 S hear F o rc e V (x ) V_exact V_approx Error increases

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15

EXAMPLE

–

CANTILEVERED BEAM

• Assumed deflection

• Need to satisfy BC

• Strain energy

• Potential of loads

F C –p₀ E,I,L 2 3 1 2

( )

v x

a bx c x

c x

(0) 0,

(0) /

0 v

dv

dx

v x

( )

c x

₁ 2

c x

₂ 3

2 1 2 0

2

6

2

L

EI

U

c

c x dx

1 2

0 0 3 4 2 3 2 0 0 1 2

,

( )

2

3

4

L

dv

V c c

p v x dx Fv L

C

L

dx

p L

c

FL

CL

c

FL

CL

EXAMPLE

–

CANTILEVERED BEAM

cont

.

• Derivatives of

U

:

• PMPE:

• Solve for c

₁

and c

₂

:

• Deflection curve:

• Exact solution:

2 1 2 1 2 1 0 2 3 1 2 1 2 2 0

2

6

4

6

2

6

12

L L

U

EI

c

c x dx EI

Lc

L c

c

U

EI

c

c x xdx EI

L c

c

1 2

0

0 c

c

3 2 0 2 1 2 4 2 3 0 3 2 1 2

4

6

2

3

6

12

3

4 p L

EI

Lc

L c

FL

CL

p L

EI

L c

FL

CL

3 3 1

23.75 10 ,

2

8.417 10

c

3 2 3

( ) 10

23.75

8.417 v x

x

2 3 4

1 ( )

5400

800

300

24 v x

x

EI

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17

EXAMPLE

–

CANTILEVERED BEAM

cont

.

• Deflection

• Bending

moment

• Shear force

Error increases 0.0 0.0 0.0 0.0 0.0 0 0.2 0.4 x 0.6 0.8 1 v( x) /v _ m a x v-exact v-approx. -100.00 0.00 100.00 200.00 300.00 400.00 500.00 0 0.2 0.4 x 0.6 0.8 1 B endi ng Moment M( x ) M_exact M_approx 200.0 300.0 400.0 500.0 600.0 0 0.2 0.4 x 0.6 0.8 1 S hear F o rc e 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

F

₁

F

₂

C

₂

C

₁

p

(

x

)

x

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19

FINITE ELEMENT INTERPOLATION

cont

.

• Nodal DOF of beam element

– Each node has deflection v and slope

– Positive directions of DOFs

– Vector of nodal DOFs

• Scaling parameter s

– Length Lof the beam is scaled to 1 using scaling parameter s

• Will write deflection curve

v

(

s

) in terms of

s

v

₁

v

₂

₁

L

x

₁

s

= 0

x

₂

s

= 1

x

1 1 2 2

{ } {

q

v

}

T 1

_,

1 _,

1 ,

x x

s

ds

dx

L

ds

dx Lds

dx

L

FINITE ELEMENT INTERPOLATION

cont

.

• 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 ds

a

a s

dx

ds dx

L

1 1 2 2

(0)

(1)

(0)

dv

(1)

dv

v

dx

1 0 1 1 2 0 1 2 3 2 1 2 3

(0)

1 (0)

(1)

1 (1)

(

2 3 )

v

a

dv

a

dx

L

v

a

dv

a

dx

L

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

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21

FINITE ELEMENT INTERPOLATION

cont

.

• 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 v

L s

s

2 3 1 2 3 2 2 3 3 2 3 4

( ) 1 3

2 ( )

(

2 )

( ) 3

2 ( )

(

)

N s

s

N s

L s

s

N s

s

N s

L s

s

1 1 1 2 3 4 2 2

( ) [

( )

( )]

v

v s

N s

v

! !

_{" #}

! !

$ %

( )

{ }

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

FINITE ELEMENT INTERPOLATION

cont

.

• 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 2 1 2 2 2 2 2 2

1

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

v

d v

s L

s

s L

s

v

dx

L ds

L

! !

_{" #}

! !

$ %

2 2 2 4 1 1 4

1 { }

d v

dx

L

B

q

B: strain-displacement vector 2 2 2 4 1 1 4

1 {

}

T T

d v

dx

L

q

B

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23

FINITE ELEMENT INTERPOLATION

cont

.

• 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

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}

q

T L v₁ v2 ₂ 1 1 1 1 1 1 2 2 2 3 2 4 2

1

1 ( )

,

( )

,

( )

,

( )

2

8

2

8 L

L

N

1 1 1 1 1 1 1 2 1 3 2 4 2 2 2 2 2 2 2 2 2 2

( )

1

0

0.025

2

8

2

8

2

8 v

N

v

N

v

N

L

v

L

v

3 1 2 4 1 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 dN dv dv dN dN dN v v dx L ds L ds ds ds ds v s s s s v s s s s L L

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25

EXAMPLE

• A beam finite element with length L

• Calculate v(s)

• Bending moment

3 2 1

0,

1

0,

2

,

2

3

2 L

L

v

EI

1 1 2 1 3 2 4 2

( )

v s

N s v

N s

N s v

N s

2 3 2 3 2 2

( ) (3

2 )

(

)

v s

s

s v

L s

s

&

'

2 2 2 2 2 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

s

L

s

L

EI

L

s

L x

(

)

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 _C 2 C3 y x F₄ F₅ C₄ C₅ 1 2 3 4 5 1 1 x x₂1 x₁2 2 2 1 x x₂2 x₁3 3 3 1 x x₂3 x₁4 4 4 1 x x₂4 2 1 0 1 1

( )

e T e NEL NEL L x _e L _x L e e

U

U x dx

U

2 1 2 2 2 ₁ 2 2 3 ₀ 2

1

2

e e x e x

d v

EI

d v

U

EI

dx

ds

dx

L

ds

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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

2 2 2 2 2 2 2 1 4 _{4 1} _{1 4} 4 1

{

e

}

T T

{

e

}

d v

ds

q

B

B q

1 ( ) 3 ₀

1

1 {

}

{

}

{

} [

]{

}

2

e T e e e e e e T

EI

T

U

ds

L

(

)

q

B

q

k

q

&

'

1 3 ₀ 6 12 ( 4 6 ) [ ] 6 12 ( 4 6 ) 6 12 ( 2 6 ) 6 12 ( 2 6 ) e s L s EI s L s s L s ds s L L s ( )

k

FE EQUATION FOR BEAM

cont

.

• 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 L EI L L L L L L L ( ) k

Symmetric, positive semi-definite Proportional to EI Inversely proportional to L ( ) 1 1

1 {

} [

]{

}

2

NEL NEL e e e e T e e

U

q

k

q

1 { } [

]{ }

2

T s s s

U

Q

K Q

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29

EXAMPLE

–

ASSEMBLY

• Two elements

• Global DOFs

F₃ F₂ y x 1 2 3 2EI EI 2L L

{ }

Q

s T

{

v

₁

v

₂

v

₃

}

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 L EI L L L v L L L L ( ) 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 L EI L L L v L L L L ( ) k 2 2 2 2 2 3 2 2

3

0

3

4

3

2

0

3

15

3

12

6 [

]

3

2

3

8

6

2

0

12

6

12

6

0

6

2

6

4

s

L

EI

L

(

)

K

30

FE EQUATION FOR BEAM

cont

.

• Potential energy of applied loads

– Concentrated forces and couples

– Distributed load (Work-equivalent nodal forces)

1 ND i i i i i

V

Fv

C

1 1 1 1 2

...

2

{

} { }

T ND s s ND

F

C

V

v

F

C

!

"

#

!

$

%

Q

F

2 1 ( ) 1 1

( ) ( )

e e NEL _x NEL e x e e

V

p x v x dx

V

2 1 1 ( ) ( ) 0

( ) ( )

e e x e e x

V

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

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31

EXAMPLE

–

WORK-EQUIVALENT NODAL FORCES

• Uniformly distributed load

pL/2 pL/2 pL2_/12 _pL2_/12 p Equivalent 1 1 2 3 1 ₀ 1

( )

₀

(1 3

2 )

2 pL

F

pL N s ds

pL

s

s ds

2 1 ₂ 1 ₂ ₃ 1 ₀ 2

( )

₀

(

2 )

12 pL

C

pL N s ds

pL

s

s ds

1 1 ₂ ₃ 2 ₀ 3

( )

₀

(3

2 )

2 pL

F

pL N s ds

pL

s

s ds

2 1 ₂ 1 ₂ ₃ 2 ₀ 4

( )

₀

(

)

12 pL

C

pL N s ds

pL

s

s ds

2 2

{ }

2

12

2

12

T

pL

"

#

$

%

F

FE EQUATION FOR BEAM

cont

.

• 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 ) or force (For C) 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

F

L

pL

C

L

pL

EI

v

F

L

pL

L

C

L

pL

(

)

! !

_{! ! !}

!

_{! ! !}

_{" # "}

_{# " #}

_{! ! !}

$ %

$

%

$ %

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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 s

U V

Q

K Q

Q

F

[K Qs]{ s} { } Fs [ ]{ } { }K Q F [K_s] 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 3 1 3 3 3 3 2 2

( )

[ 12

6

12 6 ]

y

v

dM

d v

EI d v

EI

V s

EI

L

v

dx

L ds

L

! !

_{" #}

! !

$ %

x

My

I

2 2

1.5 ₄

( )

y

1

xy

V

y

bh

h

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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 ( ) 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 v v v ( ) k 1 1 1 1 2 2 3 3 3 3 12 6 12 6 0 0 6 4 6 2 0 0 240 12 6 24 0 12 6 1000 0 6 2 0 8 6 2 0 0 12 6 12 6 0 0 6 2 6 4 F v C v F v C ( ) ! ! ! ! ! ! ! ! ! ! ! ! " # " # ! ! ! ! ! ! ! ! ! ! ! ! $ % $ %

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

240 1000

0

8

0 v

(

)

" # "

#

$ %

$

%

2 2

0.01

0.0 v

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

( )

( ) 0.01

( ) 0.005m

1 ( )

0.015rad

s

v

v N

N

v N

N

dN

v

L

ds

2 3 3 3 2 (1) 1

(1)

(1) 0.01

(1) 0.01m

1 (1)

0.0 rad

s

v

v N

N

dN

v

L

ds

2 1 1 1 2 (2) 0

(0)

(0) 0.01

(0) 0.01m

1 (0)

0.0 rad

s

v

v N

N

dN

v

L

ds

(19)

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

4

6

2 [

] 1000

12

6

12

6

2

6

4

s

v

(

)

K

2 3 1 2 3 1 1 0 ₀ 2 3 0 2 2 3 2

1/ 2

60 1 3

2 /12

10 (

2 )

1/ 2

60

3

2 /12

10 (

)

e e e e

F

s

C

s

s L

L

p L

ds

p L

F

s

C

s

s L

L

!

! !

!

! !

!

"

#

"

#

"

# "

#

!

! !

!

_$

! !

_{% $}

!

_%

$

%

$

%

L= 1m 38

EXAMPLE

–

CANTILEVERED BEAM

cont

.

• FE matrix equation

• Applying BC

• Deflection curve:

• Exact solution:

1 1 1 1 2 2

12

6

12

6

60

6

4

6

2

10 1000

12

6

12

6

60

6

2

6

4 10 50

v

F

C

v

(

)

! !

_{! ! !}

!

_!

_{" # "}

_#

_{! ! !}

_!

_{! ! !}

_!

$ %

$

%

2 2

12

6

60 1000

6

4

60 v

(

)

" # "

#

$ %

$

%

2 2

0.01

0.03 v

m

rad

3 3 4

( )

0.01 ( ) 0.03

( )

0.01 v s

N s

s

4 3 2

( ) 0.005(

4 )

v x

x

(20)

39

EXAMPLE

–

CANTILEVERED BEAM

cont

.

• 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

1 1

120 N

10 N m

F

C

&

'

2 1 1 2 2 2

( )

{ }

( 6 12 )

( 4 6 )

(6 12 )

( 2 6 )

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

60 N m

EI

M s

L

EI

s v

L

s

s v

L

s

L

s

B q

&

1 1 2 2

'

3

12

6

12

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

60 N

y

EI

V

v

L

v

L

EXAMPLE

–

CANTILEVERED BEAM

cont

.

• Comparisons

-0.010 -0.008 -0.006 -0.004 -0.002 0.000 0 0.2 0.4 x 0.6 0.8 1 v FEM Exact -0.030 -0.025 -0.020 -0.015 -0.010 -0.005 0.000 0 0.2 0.4 x 0.6 0.8 1 FEM Exact -60 -50 -40 -30 -20 -10 0 10 0 0.2 0.4 x 0.6 0.8 1 M FEM Exact -120 -100 -80 -60 -40 -20 0 0 0.2 0.4 x 0.6 0.8 1 Vy FEM Exact Deflection Slope

(21)

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 1 u 2 u 1 v v2 11 22 1 u 2 u 1 v 2 v 11 22 1 u 2 u 1 v 2 v 11 22 1 2 3

{ , , }

u v

i i

i 42

PLANE FRAME ELEMENT

cont

.

• Element-fixed local coordinates

• Local DOFs

Local forces

• Transformation between local and global coord.

1 2 * x y Local coordinates Global coordinates

x

y

1

u

2

u

1

v

2

v

11 22

x y

{ , , }

u v

{ ,

f

x

f c

y

, }

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 0 0 0 0 0 0 1 x x y y x x y y f f f f c c f f f f c c * * * * * * * * ( ) ! ! ! ! ! ! ! ! ! ! ! ! " # " # ! ! ! ! ! ! ! ! ! ! ! ! $ % $ %

{ } [ ]{ }

f

T f

{ } [ ]{ }q T q

(22)

43

PLANE FRAME ELEMENT

cont

.

• 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 1 2 2

1

x x

f

u

EA

f

u

L

(

)

" # "

#

$ % $

%

1 1 2 2 1 1 3 2 2 2 2 2 2

12

6

12

6

4

6

2

12

6

12

6

2

6

4

y y

v

L

f

L

c

EI

v

L

f

L

c

(

)

!

! !

_{! ! !}

_!

_{" # "}

_#

_{! ! ! !}

$ % $ %

PLANE FRAME ELEMENT

cont

.

• Element matrix equation (local coord.)

• Element matrix equation (global coord.)

• Same procedure for assembly and applying BC

1 1 1 1 2 2 2 2 1 1 2 2 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

0

12

6

0

12

6

0

6

4

0

6

2

0

12

6

0

12

6

0

6

2

0

6

4

x y x y

a

u

f

a

La

a

La

v

f

La

L a

La

L a

c

a

u

f

a

La

a

La

v

f

La

L a

La

L a

c

(

)

! ! !

!

! ! !

!

! ! !

!

" # "

#

! ! !

!

! ! !

!

! ! !

!

$ %

$

%

1 2 3

EA

a

L

EI

a

L

[ ]{ } { }

k

q

f

[ ][ ]{ } [ ]{ }

k T q

T f

[ ] [ ][ ]{ } { }

T k T q

T

f

[ ]{ } { }

k q

f

[ ] [ ] [ ][ ]

k

T k T

T

(23)

45

PLANE FRAME ELEMENT

cont

.

• Calculation of element forces

– Element forces can only be calculated in the local coordinate

– Extract element DOFs {q} from the global DOFs {Q_s}

– 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

AE

P

u

L

2

( )

EI

{ }

M s

L

B q

+ ,

3

( )

[ 12

6

12 6 ]

y

EI

V s

L

q

1 1 2 2 1 1 3 2 2 2 2 2 2

12

6

12

6

4

6

2

12

6

12

6

2

6

4

y y

V

L

v

M

EI

L

V

L

v

M

L

(

)

!

! !

!

! !

"

#

" #

!

! !

!

! !

$

%

$ %