fness Mat rix f or the E u ler Beam Theory te el ement analysis –“ very ” ba si cs )

fness Mat rix f or the E u ler Beam Theory te el ement analysis –“ very ” ba si cs )

Lecture material for Topology Optimization Design #1

Contents

Euler Beam Theory

1

Stiffness Matrix for Euler Beam

2

Assembly of Stiffness Matrices

3

4

Boundary Conditions

5

Euler Bea m Theory

To set up the Euler Beam Theory, need to know ƒ Ki ne ma tic s ƒ Constitutive ƒ Equilibrium ƒ Re su lt an ts

dd w EI p dx dx ⎡⎤ = ⎢⎥ ⎣⎦

p

Kinematics (Euler Beam Theory)

Kir ch h off Assum p tio n s for Normals ƒ rem ain straight (do not bend) ƒ rem ain uns tretched (keep the s am e length) ƒ rem ain nor m al (a lways 90

to n eutra l p lan e)

θ− yv uy

χ = ⋅ (1) du dx ε =

( , ) (2) d xy y dx χ ε =⋅

and

( 3 ) dv

dx χθ =− =−

Fig. 1

Constitutive and Equilibr ium (Euler Beam T h eory)

By 1-D Hooke’s Equation(,)(,) (4)xyExyσε=⋅ Constitutive

Equilibrium

dx p

VdV+

MdM+ 0 (5)y dVFpdx =⇒=−

∑

0 (6)z dMMVdx =⇒=

∑

Fig. 2

Resultants (Euler Beam Theo ry)

y (, ) xy σ () ( , ) ( 7 ) Mx y x y d y d z σ =⋅ ⋅ ⋅ ∫∫

x y (, )

xy σ () ( , ) ( 9 )

Vx x y d y d z σ =⋅ ⋅ ∫∫

Shear 2Iydydz=⋅⋅

∫∫

(8) My

I σ =

Fig. 3

Fig. 4

Derivation (Euler Beam Theory)

2

2 (10) dMpdx =− Combine (5) and (6)

(7)Æ(10)

()

2 (11) dydydzpdx σ⋅⋅⋅=−

∫∫

(4)Æ(11)

()

2 (12) dEydydzpdx ε⋅⋅⋅=−

∫∫

(2)Æ(12) 22

2 (13) ddEydydzpdxdx χ⎛⎞⋅⋅=−⎜⎟⎝⎠

∫∫

(3)Æ(13) 222

22 (14) ddvEydydzpdxdx ⎛⎞⋅⋅=⎜⎟⎝⎠

∫∫

Recall

2Iydydz=⋅⋅

∫∫

( 1 5)

⎡⎤ = ⎢⎥ ⎣⎦

Stiffness Matr ix for Euler Beam Theory

DOF of a Beam (2D)

xy 2

u

v

1

u

v

1

θ

θ

1x f 1y f

1 M2x f 2y f

2 M

An Euler beam has 3 DOFsin each node to make 6 DOFsin 2D,corresponding with 6 external loads. A stiffness matrix which shows the relations between these DOFsshould be defined. Fig. 5a

Fig. 5b

l

Stiffness Matrix for E u ler Beam

From (8) 2

2

( 1 6) ME d v EM E I Iy R d x σ χ == = ⇒ = −

The bending moment in Fig. 5 will be11

(17 )

MM f x = −

(17)Æ(16) 2

112

= ( 1 8 )

dv EI f x M dx −

Integrating (18) twice 2

112

2

111

32

1112

2

( 1 9) 62

y

dv EI f x M dx dv x EI f M x C dx

xx EIv f M C x C =− =− +

=− + +

Stiffness Matrix for E u ler Beam

2

2111

32

21111

( 2 1a ) 2

( 2 2b) 62

llEIvfMEIlEIv

θθ

θ =− +

=− + +

1122

at 0 a nd (20a )

at a nd ( 20b) dv xv v dx dv xl v v dx θ θ == =

== =

(20)Æ(19)

Solving for 11

and

() ()

112122

12 12 6 6 (23a )

6 6 4 2 (23b)

θθ θθ =− + +

=− + +

Stiffness Matrix for E u ler Beam

11

2211

322

2222

12 6 1 2 6

64 6 2 ( 2 6) 12 6 1 2 6

62 6 4

yy

fv ll

M ll l l EI

fv ll l

M ll l l θ θ − ⎧⎫ ⎧ ⎫ ⎡⎤ ⎪⎪ ⎪ ⎪ ⎢⎥ − ⎪⎪ ⎪ ⎪ ⎢⎥ = ⎨⎬ ⎨ ⎬ ⎢⎥−− − ⎪⎪ ⎪ ⎪ ⎢⎥ ⎪⎪ ⎪ ⎪ − ⎣⎦ ⎩⎭ ⎩ ⎭

(24)Æ(23)

In matrix form,

() ()

212122 121266 (25a)

6624 (25b) y EIfvvlllEIMvvlll θθ

θθ =−+−−

=−++

Stiffness Matrix for E u ler Beam

()

( 2 7a )

EA fu u l =− For axi a l f o rce/ Disp,

ix form,

11

22

11 ( 2 8) 11

xx

f u EA

f u l − ⎧⎫ ⎧⎫ ⎡⎤ = ⎨⎬ ⎨ ⎬ ⎢⎥−⎣⎦ ⎩⎭ ⎩⎭ ()

and (27b)

EA fu u l =− +

,

f u E A

ere 1111

122221

22122221

2112

222222

22222222

00 0 0

01 2 6 0 1 2 6

06 4 0 6 2

00 0 0

01 2 6 0 1 2 6

06 2 0 6 4

x

y

e

x

y

fC C u

fC C L C C L v

MC L C L C L C L

fC C u

fC C L C C L v

MC L C L C L C L θ

θ − ⎧⎫ ⎡ ⎤ ⎧ ⎫ ⎪⎪ ⎢ ⎥ ⎪ ⎪ − ⎪⎪ ⎢ ⎥ ⎪ ⎪ ⎪⎪ ⎢ ⎥ ⎪ ⎪ − ⎪⎪ ⎪ ⎪ ⇒= ⎢⎥ ⎨⎬ ⎨ ⎬ −⎢⎥ ⎪⎪ ⎪ ⎪ ⎢⎥ ⎪⎪ ⎪ −− − ⎢⎥ ⎪⎪ ⎪ − ⎢⎥ ⎪⎪ ⎪ ⎩⎭ ⎣ ⎦ ⎩ ⎭ ( 2 9)

⎪ ⎪ ⎪

123

and A EE I CC L L ==

Stiffness Matrix for E u ler Beam

Transformation Matrix

Since not every beam is parallel to the global coordinates, the stiffness matrix in (29) should be modified according to its inclination angle,

α

Transformation Matrix

thLet's define displacement and force vector of the elementin the global coordinate as as and, and local ones as and. Also, global stiffness matrix and local one as and . Then, eeee e euFuFKK the following relationship holds:

( 3 0b)

Ku = f (30 a)

ee

u= T u

f= T f

T

(31a ) and (31b)

= K T u = Tf T K Tu f

T

(30a)Æ(30b)

Thus

T

(32)

= TK T K

Transformation Matrix

co s si n 0 0 0 0

sin cos 0 0 0 0

00 1 0 0 0 (33) 00 0 co s si n 0

00 0 si n co s 0

00 0 0 0 1

α α αα

αα αα ⎡⎤ ⎢⎥− ⎢⎥ ⎢⎥ = ⎢⎥ ⎢⎥ ⎢⎥ − ⎢⎥ ⎣⎦ T

⇔

TT = I T = T

Using this, every beam element can be analyzed in global coordinate.

node1node2node3 Beam 1

Beam 2 node1node2

node1node2

Assembly of Stiffness Matrice s

{}

{ }

{ }

{ }

{}

{ }

{ }

How can this be done?

{}

Assemble

{}

{ }

,,

{ }

{ }

,,

{ }

{ }

Assembly of Stiffness Matrice s

ty:

: 11122233

3 uvuvuv θθθ ⎧⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪=⎨⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎭ u

The displacement of the

node is

element of the displacement

vector thi

{ }

uuu

uu

⎛⎞ = ⎜⎟ == ⎜⎟ ⎜⎟ = ⎝⎠

Same for and jjii vθ

11

212

33 1

12

2 xx

xxx

xx ff

fff

ff

⎛⎞ = ⎜⎟ += ⎜⎟ ⎜⎟⎜⎟ = ⎝⎠

Same for and i jjxi fM <Detailed Procedure>

Assembly of Stiffness Matrice s

11

2221

222221/212

2222222222 0000

260126

4062 0000

260126

2064 Cu

CLCCLv

LCLCLCL

Cu

CLCCLv

LCLCLCL π θ

θ −⎤⎧⎫⎥⎪⎪−⎥⎪⎪⎥⎪⎪−⎪⎪⎨⎬⎪⎪−−−⎪−⎪⎦⎩⎭ ⎪⎪⎪⎪ 2211

22222222222222003311

33222222332222 0000

01260126

064062 0000

01260126

062064 xy

T

x

y fuCC

fvCCLCCL

MCLCLCLCL

fuCC

fvCCLCCL

MCLCLCLCL θ

θ −⎧⎫⎧⎫⎡⎤⎪⎪⎪⎪⎢⎥−⎪⎪⎪⎪⎢⎥⎪⎪⎪⎪⎢⎥−⎪⎪⎪⎪=⎢⎥⎨⎬⎨⎬−⎢⎥⎪⎪⎪⎪⎢⎥⎪⎪⎪⎪−−−⎢⎥⎪⎪⎪⎪−⎢⎥⎪⎪⎪⎪⎣⎦⎩⎭⎩⎭ TT Beam 2

13

2223

222223/214

2224

222224 0000

01260126

064062 0000

01260126

062064 Cu

CLCCLv

LCLCLCL

Cu

CLCCLv

LCLCLCL π θ

θ −⎤⎧⎫⎥⎪⎪−⎥⎪⎪⎥⎪⎪−⎪⎪⎨⎬⎪⎪−−−⎪−⎪⎦⎩⎭ ⎪⎪⎪⎪ 2112

222222

22222222/4/44114

42222422422224 0000

01260126

064062 0000

01260126

062064 xy

T

xy fCCu

fCCLCCLv

MCLCLCLCL

fCCu

fCCLCCLv

MCLCLCLCL ππ θ

θ −⎧⎫⎡⎤⎧⎫⎪⎪⎢⎥⎪⎪−⎪⎪⎢⎥⎪⎪⎪⎪⎢⎥⎪⎪−⎪⎪⎪⎪=⎢⎥⎨⎬⎨⎬−⎢⎥⎪⎪⎪⎢⎥⎪⎪⎪−−−⎢⎥⎪⎪⎪−⎢⎥⎪⎪⎪⎩⎭⎣⎦⎩⎭ TT ⎪⎪⎪⎪ Beam 4 ent Stiffness Matrices and vectors

Assembly of Stiffness Matrice s

11

1

2

2

2 uvuv θθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪=⎨⎬⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢ •••••••••••••••••••••••••••••••••••• ⎥⎪⎪⎣⎦⎩⎭ 11

22

22

33

3 uvuv θθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪=⎨⎬⎢ •••••••••••••••••••••••••••• ⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪ ••••⎪⎣⎦⎩ •••⎭ • 2 3333

44

4 uvuv θθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪=⎨⎬⎢ •••••••••••••••••••••••••••• ⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪ ••••⎪⎣⎦⎩ •••⎭ • 3Ku

22

24

44

4 uvuv θθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪=⎨⎬⎢⎥⎪⎪⎢ ••••••••••••••••••••••••••••••• ⎥⎪⎪⎢•• ⎥⎪⎪⎢⎥⎪⎪⎣⎦⎩⎭ ••• 4Ku

Each element stiffness matrix will be added to the global stiffness matrix withrespect to the displacement DOFsit has, as seen in (34).

Assembly of Stiffness Matrice s

11

12

22

33

34

44 xyxyxyxy ffMffMffMffM •••••••••••• ••••••• ••••••••••••••••••••••••••••••• ••••••• •••••••••••••••••••••• ⎧⎫⎪⎪⎪⎪

•••• ⎪⎪⎪⎪⎪⎪⎪⎪

•••••••••• ⎪⎪⎪⎪

•••• •

• •

• =⎨⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪••• ••••••••• ⎪⎪ ••⎪⎪⎪• ••

••⎪⎪ ••• •

⎩ •

⎭ •••• 11122233344

4 (34) uvuvuvuv θθθθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎨⎬⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥••••••• ⎪⎪⎢⎥⎪⎪⎢⎥⎣⎦ ••••⎩ •••⎭ •

Boundary Conditions

23

4 000000000 xy ffM •••••••• ••••••••••••••••••••••••••••••••• ••••••••••••••••••••••••••••••••••••••• ⎧⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

••• ••••• ••

•••••• ⎪⎪⎪⎪⎪⎪=⎨⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ ••••••••••

••••••••••• ••••⎪⎪⎩⎭•• •• ••••••••••••••••••• 11122233344

4 uvuvuvuv θθθθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎨⎬⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎣⎦⎩⎭ Applying forces

4

M

3y

f

, known, applied forces)M=

Boundary Conditions

Fixed DOFs

The restricted DOF in a structure will result in a value “0”in the displacement vector. This will cause the stiffness matrix to be reduced. 22333444 0000 uuvuv θθθ ⎧⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎭

Boundary Conditions

23

4 000000000 xy ffM •••••••• ••••••••••••••••••••••••••••••••• ••••••••••••••••••••••••••••••••••••••• ⎧⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

••• ••••• ••

•••••• ⎪⎪⎪⎪⎪⎪=⎨⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ ••••••••••

••••••••••• ••••⎪⎪⎩⎭•• •• ••••••••••••••••••• 2233344

4 0000 uuvuv θθθ ⎧⎫⎡⎤⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎨⎬⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎪⎪⎢⎥⎣⎦⎩⎭

Because there are 8 unknowns, only 8 equations are needed. Therefore,the underlined rows and columns do not need to be considered. The result becomes Eq. (35) in the next page.

Boundary Conditions

22

2

3

33

3

44

44

0 0

0

0

0

fu

u

fv

u v

M θ θ

θ ••• • •• ••• • •• ••• • •• ••• • •• ••• • •• ••• • ⎧ ⎫ ⎡⎤ ⎪ ⎪ ⎢⎥ ⎪ ⎪ ⎢⎥ ⎪ ⎪ ⎢⎥ ⎪ ⎪ ⎢⎥ ⎪ ⎪ ⎢⎥ = ⎨ ⎬ ⎢⎥ ⎪ ⎪ ⎢ •• • • • •• • • • •

⎥ ⎪ ⎪ ⎢⎥ ⎪ ⎪ ⎢⎥ •••• •• • • • •• • • • •• • • • •• • • •

•• • • •

• ⎪⎪ ⎪ ⎢⎥ ⎪ ⎪ ⎢⎥⎣⎦ •• •• • • • •• • • ⎭ • •

⎭ •

• (3 5 )

By solving , we can obtain the displacements at nodes. 1−

u= K f