caed lec9frame sep2015

7/21/2019 CAED Lec9frame Sep2015

http://slidepdf.com/reader/full/caed-lec9frame-sep2015 1/28

Frame Element

Internal



Outlines Review

FE Formulation of Frame Structure Element

Example

Internal



Objectives Understand what frame element is and its degree of

freedoms

To formulate the stiffness matrix of frame element To solve a frame problem using the stiffness-

displacement-load matrices approach

Internal



Beam element - review1000 lb/ft

10 ft2.5 ft

1 2 3

2.5 ft

500 lb

Internal



ReviewElement Type DOF K matrix Remark

Spring/ bar U1

Truss U1, U2

Beam U1, θ1

Frame U1, U2, θ1

k k

k k K

22

22

22

22

mlmmlm

lml lml

mlmmlm

lml lml

K

22

22

3

4626

612612

2646

612612

L L L L

L L

L L L L

L L

L EI K

L

EAk

sin

cos

m

l

Internal



Definition: So…what ARE frames??

Internal



Definition: Structural members that may be rigidly connected

with welded joints or bolted joints

Combination of bar (truss) and beam

Internal



X

Y i

j

θ

3 DOF at each node: lateral andlongitudinal displacements, androtation

Combination of bar and beam

Superimpose to get the stiffness matrix

Frame Element

Internal



Revisit: Bar and Beam Elements

22

22

3

4626

612612

2646

612612

L L L L

L L

L L L L

L L

L

EI K

k k

k k K

Internal



Truss Element

Stiffness matrix for members under axial loading

3

2

1

3

2

1

000000000000

0000

000000

000000

0000

j

j

j

i

i

i

L AE

L AE

L AE

L AE

axial

uu

u

u

u

u

K

Internal



Beam Element

Stiffness matrix for lateral displacements and rotations

3

2

1

3

2

1

22

22

3

46026061206120

000000

260460

61206120

000000

j

j

j

i

i

i

xy

uu

u

u

u

u

L L L L L L

L L L L

L L

L

EI K

Internal



Stiffness matrix for Frame

+

=

22

22

3

460260

61206120

000000

260460

61206120000000

L L L L

L L

L L L L

L L

L

EI

000000

000000

0000

000000

0000000000

L AE

L AE

L

AE

L

AE

3

2

1

3

2

1

46

26

612612

2646

612612

22

2323

22

2323

00

00

0000

00

00

0000

j

j

j

i

i

i

L EI L EI L EI L

EI

L

EI

L

EI

L

EI

L

EI

L AE

L AE

L EI

L

EI L EI

L

EI

L EI

L EI

L EI

L EI

L AE

L AE

u

u

u

u

u

u

Internal



Local DOF related toGlobal DOF through the

transformation matrix

In Matrix form: {U}=[T]{u}

Trans form ation Matr ix

X

Y

i

j

θ

Internal



cossin00

sincos00

00cossin

00sincos

T

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

T

In Matrix form: {U}=[T]{u}{F}=[T]{f}

Internal



Stiffness matrix

T eTK TK thatconcludewe

UTk TF

FUTk T

FTTUTk T

FTUTk

f TFanduTU f uk

1

1

11

11

,

From strain energy equation, we can show that;

Internal



Stiffness matrix

100000

0cosθsinθ000

0sinθcosθ000

000100

0000cosθsinθ

0000sinθcosθ

00

00

0000

00

00

0000

100000

0cosθsinθ-000

0sinθcosθ000

000100

0000cosθsinθ-

0000sinθcosθ

K

L

4EI

L

6EI

L

2EI

L

6EI

L

6EI

L

12EI

L

6EI

L

12EI

L

AE

L

AE

L

2EI

L

6EI

L

4EI

L

6EI

L

6EI

L

12EI

L

6EI

L

12EI

L

AE

L

AE

T

e

22

2323

22

2323

Internal



Example 4.5

10 ft

9 ft

800 lb/ft

E=30 x 106 psi

A=7.65 in2

I=204 in4

Determine the deformation of the frame under the

given distributed load.

Internal



Element 1

(1) 21

u22

u21

u23

u12

u11u13

Y

X

L

4EI

L

6EI

L

2EI

L

6EI

L

6EI

L

12EI

L

6EI

L

12EI

L

AE

L

AE

L

2EI

L

6EI

L

4EI

L

6EI

L

6EI

L

12EI

L

6EI

L

12EI

L

AE

L

AE

22

2323

22

2323

00

00

0000

00

00

0000

e

xy K

E =30 x 106 lb/in2

A= 7.65 in2

I = 204 in4

L = 10 ft

Internal



Stiffness matrix with respect to

local C.S

23

22

21

13

12

11

31

2040002550010200025500

255042.50255042.50

001912.5001912.5

1020002550020400025500

25505.420255042.50

005.1912001912.5

10

u

u

u

u

u

u

K xy

Since the local and the global C.S are aligned in the same direction,

[K]1 same as global stiffness matrix.

Internal



Element 2

(2)

3 u32

u31

u33

u22

u21

u23

Y

X

Internal



Local stiffness matrix for element 2

33

32

31

23

22

21

32

2266663148.15-01133333148.150

3148.1558.3-03148.1558.3-0

002125002125

1133333148.1502266663148.150

3148.153.5803148.1558.30

002125-002125

10

u

u

u

uu

u

K xy

Internal



Transformation matrix[T] when Ɵ = 270o

100000

0270cos270sin000

0270sin270cos000

000100

0000270cos270sin

0000270sin270cos

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

T

100000

001000

010000

000100

000001

000010

T

Internal



The stiffness matrix for element 2

becomes,

100000

001000

010000000100

000001

000010

2266663148.15-01133333148.150

3148.1558.3-03148.1558.3-0002125002125

1133333148.1502266663148.150

3148.153.5803148.1558.30

002125-002125

10

100000

001000010000

000100

000001

000010

32

T

xy K

Internal



Perform matrix operation:

33

32

31

23

22

21

32

22666603148.15-11333303148.15

02125002125-0

3148.1503.583148.15058.3-

11333303148.1522666603148.15

021250021250

3148.1503.583148.15058.3

10

u

u

u

u

u

u

K xy

Internal



Assemble the elements

226666015.3148113333015.3148000

021250021250000

15.314803.5815.314803.58000

113333015.3148226666204000255015.3148010200025500

025500255021255.42025505.420

15.314803.5815.3148003.585.1912005.1912

0001020002550020400025500

00025505.42025505.420

000005.1912005.1912

G K

Internal



Nodal loads

Internal



Solve for unknowns

Internal



Recap Frame structure element has 3 DOFs

Stiffness matrix for frame element is

[K]=[T]T

[k][T]

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

T

L4EI

L

6EI

L2EI

L

6EI

L

6EI

L

12EI

L

6EI

L

12EI

LAE

LAE

L

2EI

L

6EI

L

4EI

L

6EI

L

6EI

L

12EI

L

6EI

L

12EI

L

AE

L

AE

22

2323

22

2323

00

00

0000

00

00

0000

e

xy K

caed lec9frame sep2015

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