Stiffness Coefficients for a Flexural Elementwebshaker.ucsd.edu/homework/StiffnessCoefficients.pdf · 2 September 18, 2002 Ahmed Elgamal Stiffness coefficients for a flexural element

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September 18, 2002 Ahmed Elgamal

Stiffness Coefficients for a Flexural Element

Ahmed Elgamal

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Ahmed ElgamalSeptember 18, 2002

Stiffness coefficients for a flexural element (neglecting axial deformations), Appendix 1, Ch. 1 Dynamics of Structures by Chopra.

u1u3u2 u4

4 degrees of freedom

Positive directionsk11k21

k31k41

3


u1 1.0

L

To obtain k coefficients in 1st column of stiffness matrix, move u1 = 1, u2 = u3 = u4 = 0, and find forces and moments needed to maintain this shape.

4


2L6EI

2L6EI

3L12EI

3L12EI

Σ M = 3L12EI - 3L

12EI= 0


k31k41

Note that Σ Forces = 0Σ Moments = 0

3L12EIi.e. remember , and you

can find other forces & moments

These are (see structures textbook)

5


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

44434241

34333231

24232221

14131211

kkkkkkkkkkkkkkkk

k

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

=

6L6L12

12

LEIk 3

, where i is row numberand j is column number

kkij =

6


u21.0

L

u2 = 1, u1 = u3 = u4 = 0

7


2L6EI

2L6EI

3L12EI

3L12EI

u2 = 1


k32k42⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

6L6L1212-

LEIk 3

Σ M = 0, Σ F = 0

8


u3 = 1, u1 = u2 = u4 = 0

u3 = 11.0

L

9


2L6EI

2L6EI

L2EI

L4EI


k33k43⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

L2EIL

4EIL6EIL6EI-

k2

2

Σ M = 0, Σ F = 0

10


u4 = 1, u1 = u2 = u3 = 0

u4 = 1 1.0

L

11


2L6EI

2L6EI

L2EI

L4EI


k34k44⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

L4EIL

2EIL

6EIL6EI-

k2

2

Σ M = 0, Σ F = 0

12


u1u2

ug

m

L

m is lumped at a point & does not contribute in rotation

Example: Water Tank

u2 above was u3 in the earlier section of these notes

13


Example: Water Tank (continued)

u1u2

ug

m

L

k11k21

k12k22

g2

1

2

23

2

1 u0m

uu

L4EI

L6EI

L6EI

L12EI

uu

0m

&&&&

&&⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

“Note Symmetry”Rotational(used to be u3)

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Static Condensation:

Way to solve a smaller system of equations by eliminating degrees of freedom with zero mass.

e.g., in the above, the 2nd equation gives

0uL

4EIuL6EI

212 =+−

or

11122 u2L3u

4L6u

4EIL

L6EIu === -----*

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Example: Water Tank (continued)Substitute * into Equation 1

g1231 umu2L3

L6EI

L12EIum &&&& −=⎟

⎠⎞

⎜⎝⎛ −+

or,

g131 umu2L

18EI24EIum &&&& −=⎟⎠⎞

⎜⎝⎛ −

+

or,

g131 umuL

3EIum &&&& −=⎟⎠⎞

⎜⎝⎛+

Now, solve for u1 and u2 can be evaluated from Equation * above.

Static condensation can be applied to large MDOF systems of equations, the same way as shown above.

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if EIb = 0

Ib

or of column

g131 umuL

3EIum &&&& −=⎟⎠⎞

⎜⎝⎛+

k of water tank as we were given earlier.

17


Mandatory Reading

Example 9.4 page 362-364

Example 9.8 page 368-369

Sample Exercises: 9.5, 9.8, & 9.9

bendingbeam

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(a)

(b)

Reference: Dynamics of Structures, Anil K. Chopra, Prentice Hall, New Jersey, ISBN 0-13-086973-2

Example

19


(c)

(d)


20


(e)

(f)


21


(g)

(h)


22


(i)

(j)


23


,⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

00

mm

M 2

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

=

22

223

8L2L06L2L4L6L6L06L24126L6L1212

LEIk


therefore,

Sample Exercise: For the above cantilever system, write equation of motion and perform static condensation to obtain a 2 DOF system.

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m

L

bendingbeam

3L3EIk = , m

k=ω

beam with EIbeam = 0

⎟⎠⎞

⎜⎝⎛= 3L

3EI2k⇒

Column Stiffness (lateral vibration)

25


u

3L12EI

k =L ⎟⎠⎞

⎜⎝⎛= 3L12EI2k

Case ofEIb = ∞

h1 h2

(rigid roof)

32

2231

11

hI12E

hI12E

k +=

26


h

L = 2h

3c

7h96EIk = if cb EIEI =

beam column

, & EEE bc ==412ρ112ρ

h24EIk 3

c

++

=c

b

4IIρ =

(See example 1.1 in Dynamics of Structures by Chopra)

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u1

θ1θ2

fs

call it u2

call it u3

Obtained by “static condensation” of 3x3 system

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00f

uuu S

3

2

1

force

Use to represent u2 and u3 in terms of u1 & plug back into

Technique can also be used for large systems of equationsand get fs = ku1

(See example 1.1 in Dynamics of Structures by Chopra)

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u1

u2u3

IcIc

Ib

Neglect axial deformationDraft Example

29


L

L

2c

L6EI

( ) ⎟⎠⎞

⎜⎝⎛

3c

L12EI2

2c

L6EI

u1 = 1

u2 = u3 = 0

30


u2 = 1

u1 = u3 = 0

2c

L6EI

L4EIc

L4EIb

L2EIb

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u3 = 1

u1 = u2 = 0

2c

L6EI

L4EIc

L4EIb

L2EIb

( )( ) ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++=

2cb

2bc

2b

2cbc

ccc

3

LII4L2IL6IL2ILII4L6IL6IL6I24I

LEk

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If frame is subjected to lateral force fs

Then (for simplicity, let Ic = Ib = I)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00f

uuu

8L2L6L2L8L6L6L6L24

LEI s

3

2

1

22

223

33


Static condensation:

From 2nd and 3rd equations,

1

1

22

22

3

2 u6L6L

8L2L2L8L

uu

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡−

1

1

22

22

44 u6L6L

8L2L-2L-8L

4L64L1

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−−

=−

1u11

10L6

⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡−

−−

=⎥⎦

⎤⎢⎣

⎡−

acbd

cbad1

dcba 1

Note matrix inverse:

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Substitute into 1st equation

13s13 u10L

168EIfu1036

103624

LEI

==⎥⎦⎤

⎢⎣⎡ −−

or

310L168EIk = (check this result)

35


2L

L

Draft Example 2

u1

u2u3

36


2c

L6EI

( ) ⎟⎠⎞

⎜⎝⎛

3c

L12EI2

2c

L6EI

u1 = 1

u2 = u3 = 0

37


2c

L6EI

L4EIc

2L4EIb

2L2EIb

u2 = 1

u1 = u3 = 0

38


2c

L6EI

L4EIc

2L4EIb

2L2EIb

u3 = 1

u1 = u2 = 0

39


⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛ +=

2c

b2bc

2b

2c

bc

ccc

3

LI2I4LIL6I

LILI2I4L6I

L6IL6I24I

LEk

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00f

uuu

6LL6LL6L6L6L6L24

LEI s

3

2

1

22

223

For simplicity, let Ib = Ic

40


Static condensation:

1

1

22

22

3

2 u6L6L

6LLL6L

uu

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡−

1

1

22

22

44 u6L6L

6LL-L-6L

L36L1

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

−

11 u11

7L6u

11

35L03

⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡−=

41


Substitute in 1st Equation

s13 fu7

367

3624LEI

=⎥⎦⎤

⎢⎣⎡ −−

13s uLEI

796f =

37L96EIk =

or,

or, Same as in Example 1.1, Dynamics of Structures by Chopra

42


Sample Exercise

1.1 Derive stiffness matrix fork

EIc2

EIc1

EIb

h2

h1

L

1.2 For the special case of Ic1 = Ic2 = Ib, h1 = h2 = h and L = 2h,find lateral stiffness k of the frame.

1)

43


Sample Exercise

2) Derive equation of motion for:

m

2m

EIb

EIb

EIc

EIcEIc

EIch

h

2h

use 600 lb/ft

Flexural rigidity of beams and columns

E = 29,000 ksi, Columns W8x24 sections

with Ic = 82.4 in4

h = 12 ftIb = ½ Ic

44


Sample Exercise (Optional)

3) Derive lateral k of system (need to use computer to invert3x3 matrix)

Ib Ib

IcIcIch = 12 ft

24 ft 24 ft

E = 29,000 ksi,

W8x24 sectionsIc = 82.4 in4

Ib = ½ Ic

Documents

Stiffness Coefficients for a Flexural Elementwebshaker.ucsd.edu/homework/StiffnessCoefficients.pdf · 2 September 18, 2002 Ahmed Elgamal Stiffness coefficients for a flexural element