CEE570 / CSE 551 Class...

Previous class:

One dimensional elements (formulations and general procedure);

Review of CEE360/CEE 470 (Structural Analysis)

This class: One Dimensional Elements (cont’d)

I-Clickers

Bar and beam elements of arbitrary orientation

Assembly of elements

Review of CEE360/CEE 470 (Structural Analysis)

Remarks on modeling

Textbook sections

Chapter 2 (2.1 to 2.7, 2.9, 2.12)

CEE570 / CSE 551 Class #4

1

2

Bar / beam element

2

2v

)( 2u

1

1v

)( 1u

Direct approach to assemble stiffness

matrix: give unit d.o.f, calculate

corresponding loads at each node

2

2

1

1

2

2

1

1

44434241

34333231

24232221

14131211

M

F

M

F

v

v

kkkk

kkkk

kkkk

kkkk

Review of previous class

3

Review of Beam Theories

Beam theory planar cross-section assumption

____________ beam – neglect interlayer shear

deformation

n

n

_____________beam – consider interlayer shear

t

Bernoulli-Euler

Timoshenko

4

1D element of arbitrary orientation

Describes element behavior

eeefuK

Defined under base vector coordinate

F1

'

2u

'

1u

F2

x2v

Fy2

2u

1v

1u

Fx1

Fx1

Fy1

Local coord. Global coord.

fdK fdK

xx

5


F1

'

2u

'

1u

F2

x

Local coord.

2v

F2y

2u

1v

1u

F2x

F1x

F1y

Global coord.

fdK fdK x x

'

2

'

1

u

ud

2

1

F

Ff

2

2

1

1

v

u

v

u

d

2

2

1

1

y

x

y

x

F

F

F

F

f

22K 44K

6


2

2

1

1

'

2

'

1

00

00

v

u

v

u

sc

sc

u

udTd

Find relationship ,

then compute K }{}{ ff }{}{ dd

cosc

sins

Transformation matrix, T

fTffTf T

1

2

3

42222444

TKTKT

fdK fdK

7


Find relationship ,

then compute K }{}{ ff }{}{ dd

Direct approach: obtain the same result

22

22

22

22

44

scsscs

csccsc

scsscs

csccsc

kK

2v

F2y

2u

1v

1u

F2x

F1x

F1y

Global coord.

fdK x

Bar element in 3D space:

111

111

000

000

nml

nmlT

k = EA/L

8


Bar element in 3D space:

111

111

000

000

nml

nmlT

Beam element in 3D space:

Direction cosines of local x’ w.r.t. global x,y,z

A0000A0000A0000A

T1212

333

222

111

nml

nml

nml

A

9


Coordinate transformation from local coordinate (x’,y’,z’) to

global coordinate (x,y,z)

Rotational transformation does not alter intrinsic element properties

(just change of base)

See Prof. Paulino’s CEE471 (former cee361) web page:

http://cee.uiuc.edu/paulino/cee361

A vector V can be expressed in terms of components uvw in global system xyz

or in terms of components u’v’w’ in local system x’y’z’

10

• Element-level force-displacement relations

eeefuK

Structure nodal equilibrium & displacement

compatibility (assembly and BC)

sssfuK

• Solve system of linear equations

• Compute gradients (stress, strain)

sssfKu \

u~,

Recall: General Computational Procedure

11

Assembly of elements (general concept)

Discretization Assembly

eeefuK sss

fuK

Each structural node must be in equilibrium;

Assembly of elements and loads produces a

set of equations stating “equilibrium” (F=0)

Structural loads come from:

• Element deformation

• Initial stress

• External loads (distributed and concentrated)

12


Structural loads applied on nodes include:

• __________________

• ________________ (d.o.f=0)

• __________________

node element

Load applied by element to

structure nodes

dK rdK

Load applied by element to

maintain node equilibrium

dKr

er

P

Structural nodes in equilibrium:

0Prr

elsels N

iie

N

i

i

11

Element deformation

Structural support

Direct external loads

13


Structural nodes in equilibrium:

0Prr

elsels N

iie

N

i

i

11

dKr

Internal

force

Reaction

force External

force

RDK Global equilibrium:

elsN

i

i

1

kK

elsN

iie

1

rPR

- {Internal force} =

{reaction force} +

{external force}

14

Applying boundary condition and solving

4

6

fx1=H1

fy1=V1

0

fy2=10

0

0

0

0

0

0

0

fy6=V6

k11

k21

k31

…

…

…

…

…

…

…

…

kn1

1

2

3

5

P=10

V6

H1

V1

unknown

kn1

kn2

kn3

…

…

…

…

…

…

…

…

knn

u1=0

v1=0

u2

v2

u3

v3

u4

v4

u5

v5

u6

v6=0

…

…

…

…

…

…

…

…

…

…

…

…

k13

k23

k33

…

…

…

…

…

…

…

…

kn3

known

unknown

known

k12

k22

k32

…

…

…

…

…

…

…

…

kn2

• Without applying BCs, the stiffness matrix is

________.

• At the dofs where displacements are known,

reactions are not known and vice-versa.

• The system can be reduced by removing the

highlited rows and columns (see below)

singular

15

Interpretation

r

f

r

f

rrrf

frff

f

f

u

u

KK

KK

rfrffff uKfuK

rrrfrfr uKuKf

• Partitioning the system

corresponding to free dofs

and restrained (prescribed)

dofs (denoted by f and r

subscripts)

• Solve for unknown (free)

displacements by these

equations:

(all right-hand side terms

are known)

• Calculate unknown

reactions by these

equations:

(all right-hand side terms

are known)

16

Assembly and Structural Node Numbers

Structural nodes in equilibrium

1 1

2 2

1 2

3

1

2 3

1 4

3 2

Local numbering Structural numbering

3

2

1

987

654

321

11

d

d

d

aaa

aaa

aaa

dk

3

2

1

987

654

321

22

d

d

d

bbb

bbb

bbb

dk

17

Assembly and Structural Node Numbers

1 1

2 2

1 2

3

1

2 3

1 4

3 2

Local numbering Structural numbering

3928173

3625142

3322111

dadadar

dadadar

dadadar

+ =

Contribution to nodal force in

local numbering by element1:

294817

264514

234211

2

4

1

DaDaDar

DaDaDar

DaDaDar

s

s

s


structural numbering by element1:

18

Assembly and Structural Node Numbers 3

1

2

1 4

3 2

(Element [K])

1

2

1 2

3

1

2

Local numbering

+ =

294817

264514

234211

2

4

1

DaDaDar

DaDaDar

DaDaDar

s

s

s


structural numbering by element 1:

4

3

2

1

564

897

231

0

0000

0

0

4

3

2

1

D

D

D

D

aaa

aaa

aaa

r

r

r

r

s

s

s

s

987

654

321

1

aaa

aaa

aaa

k

s1k (structural) HW (optional): Repeat the procedure for element #2

19

Assembly and Structural Node Numbers 3

1

2

1 4

3 2

(Element [K])

1

2

1 2

3

1

2

Local numbering

+ =

987

654

321

1

aaa

aaa

aaa

k

123

456

789

564

897

231

21

0

0

0

0000

0

0000

0

0

bbb

bbb

bbb

aaa

aaa

aaa

sskkK

987

654

321

2

bbb

bbb

bbb

k

20

Remarks on Modeling

Choice of appropriate elements

Truss use bar element; not beam element

Inappropriate refinement only adds trouble

Limit of “exact solution” of beam element

Explore sections 2.12, 2.13

A B C D E F G H

Solution of FEM Linear Systems

Relationship between structural node numbering

and available solution schemes

Direct versus Iterative Solvers

Next class

21

CEE570 / CSE 551 Class...

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