Vibration of asymmetrical coupled shear walls

Journal of Sound and Vibration (1973) 27(4), 573-581

VIBRATION OF ASYMMETRICAL COUPLED SHEAR WALLS M. PETYT AND W. H. MIRZA Institute 0/ Sound and Vibration Research, University 0/ Southampton, Southampton S09 5NH, England

(Received 18 November 1972, and ill reuisedform 29 January 1973)

Asymmetrical coupled shear walls are assumed to act as thin-walled beams of open cross-section. A finite element displacement model is developed and shown to be accurate and relatively efficient. The importance of including the effect of warping of the crosssection, during torsion, is also investigated.

1. INTRODUCTION

The elevator shaft, service bay and stairwell of a modern high rise building consist of shear walls connected together to form a complex asymmetric shape. Heidebrecht and Raina [1] have analysed such a structure, assuming that it acts as a thin-walled beam of open crosssection. They obtained an exact solution for the frequencies and modes of vibration, for the case of a cross-section with one axis of symmetry. Since their method involves a trial and error numerical process, which is extremely time consuming, they also present an approxlmate Galerkin type solution using the uncoupled mode shapes. Gere and Lin [2] have also presented an approximate solution based upon the Rayleigh-Ritz method. In this paper the finite element displacement method is used to investigate the above problem. However, in this case no axis of symmetry will be assumed. A new beam element with J 2 degrees of freedom per mode is presented.

2. STRAIN AND KINETIC ENERGIES The geometry of the cross-section of a thin-walled open section beam is shown in Figure I. C is the centroid and 0 the shear centre of the cross-section. The axes Cyz are principal axes through the centroid. The displacement of the cross-section is defined by the displacements L', w of the shear centre in directions parallel to the axes Cy, Cz and a rotation about the shear centre. (A list of symbols is given in Appendix I.)

e

o

r

z Figure 1. Cross-section of a thin-walled open section beam. 573

574

M. PETYT AND W . H . MIRZA

The strain energy of the beam is given by [3] U

=

t

IL[

Elu

(02 (oe)2] ox2V)2 + EI~~ (iFoxW)2 + EJ., (iF dx8)2 + GJ ax dx, 2 2

(1)

where x is the axial coordinate along the length of the beam and I zz and In are the second moments of area about Cz and Cy, respectively. J w is the warping constant ofthe cross-section with pole at 0 and J is the Saint-Venant torsion constant. E and G denote the Young's modulus and shear modulus, respectively. The kinetic energy of the beam is given by

T= \

e)2 t, (08)2] + A at dx,

at + (OW at fLu [(Oue)2

(2)

where u is the mass per unit length, I, is the polar moment of area of the cross-section about C and A is the area of the cross-section. V e, We are the displacements of the centroid. These are related to the displacements of the shear centre as follows: We

= W + c~e.

(3)

Substituting equation (3) int o equation (2) gives L

T=tJf1. o

[(OV)2 au se (dW)2 -2c OW 00 +-10 (00)2] -2c --+ dx ot z ot dt ot ~ ot ot A dt '

(4)

where /0 is the polar moment of area of the cross-section about 0.

3. DISPLACEMENT FUNCTIONS

e,

An inspection of equations (I) and (4) shows that v, ou/ox, w, ow/ox, oe/ox are required to be continuous between elements for convergence. This can be achieved by representing v, W, 0 by means of cubic polynomials and taking the above six quantities as nodal degrees of freedom. However, substituting equations (I) and (4) into Hamilton's principle [4] shows that the boundary conditions are as follows. At a clamped boundary

ou/ox =0, W= dW/OX=O, v=

e = de/OX =

(5)

0,

and at a free boundary

a2 v/ox 2 =

d3 u/ox 3

iF w/ox = 0 2

3

w/ox

o()

3

=

0,

=

0,

a0 3

GJ--EJ - = 0 3

ax

tu

ox

'

(6)

In order to satisfy the conditions at the free boundary and also to provide high accuracy, it was decided to represent u, w, 0 by means of seventh-order polynomials. The nodal degrees

575

ASYMMETRICAL SHEAR WALLS

of freedom were taken to be v,

W,

e and their first three derivatives, making a total of twelve

in all. This means that the displacements can be expressed in the form v = N(x)q.,

= N(x)qw, e= N(X)q9'

W

(7)

where N(x) is a row matrix of shape functions. Notice that each component of displacement is represented by the same shape function. This leads to a simplification in determining the element stiffness and mass matrices. The column matrices of nodal degrees of freedom are defined as follows: =

qL'

(v

oV f !

2

0

2 3 aVI+! a VI+! a

v£ a 3 v! v

ax ox 2 ax 2 1+1 ox aw£;)2 WI

03 WI

qw= ( Wf~ ox2 ox3

ox 2

V£+l)

ax3

'

02 Wl+l 03 W I+ 1 ) ~~aT

aW 1+ 1 "'1+1

' (8)

where, i, i + I are the node points at the two ends of the element.

4. STIFFNESS AND MASS MATRICES Substituting the displacement functions (7) into the strain energy expression (1) gives (9) where I

k=

f BI(x) B

2(x)

dx,

1(x)

dx,

o I

I{ =

JBf(x) B

(10)

o

and

(II) primes denoting differentiation with respect to x. When the strain energy is expressed in the form ( 12)

where (13)

it can be seen that the element stiffness matrix is

k=

l

E/ U k

o

0

Ely)'k

o

o

o EJ",k

+ GJk

].

(14)

516

M. PETYT AND W. H. MIRZA

Again, substituting the displacement functions (7) into the kinetic energy expression (4) yields

T=

t(/l 4,·

ID4,·

-

u«, 4,· m46 + /l4", mq", - uc, 4w mqo + /l;O 40 mqo}.

(15)

where

f NT(x) N(x) dx I

m=

(16)

o

and dots denote differentiation with respect to time. When the kinetic energy is expressed in the form (17)

it can be seen that the element mass matrix is

m=

[ pm0

-Pc,m] .

0

-uc,«

(18)

-pel/fit

J.iID

pIa _

-pCl/1it

-m

A

Once the matrices k, k, iii are evaluated, each ofwhich is only of order (8 x 8), the stiffness and mass matrices of the element can be evaluated from equations( 14)and (18). The expressions for these submatrices are given in Appendix II. The formation of the stiffness and mass matrices of the complete structure follows the usual standard procedures [5, 6].

5. APPLICAnONS

The accuracy of the method was first tested by analysing the bending vibrations of a uniform, symmetric cantilever beam. In this case the number of degrees of freedom per node reduce to four. The percentage error in frequency obtained when compared with the exact solution is shown in Table I. For comparison, the corresponding results 0 btained by using the more conventional two degree of freedom per node element [5] are shown in Table 2. It can be seen that, for the same num bel' of total degrees of freedom, the present element gives greater accuracy than the more conventional one. The accuracy of the method was next tested by analysing E-type structure and comparing the results with the exact solution [I]. Details of the structure are given in Figure 2 and Table 3. The structure was analysed by using five finite elements. In the present case the crossTABLE

1

Percentage error in bending frequencies for a uniform cantilever beam-eight degrees of freedom element Mode number No. of elements

No. of d.o.f.

1

4 8

2

+ error < 0·001 %.

A

2

+ 0·004 + +

3

4

5

0·114 12'7 + 0·03 0'084

577

ASYMMETRICAL SHEAR WALLS TABLE

2

Percentage error in bending frequencies for a uniform cantilever beam-four degrees of freedom element Mode number No. of elements

No. of d.o.f,

1 2 3 4

2 4

6 8

A

2 0·49 0·049 0·012 0-004

3

60'3 0·9 21·8 1·3 0'33 0·8 0·12

4

80'5

16'3 1·4

Figure 2. Details of E-type structure.

TABLE 3 Geometric and materialproperties ofan E-type structure

H= 94 in = 2·3876 In b = 40 in = 1·016 m d= 16 in = 0-4064 m h = 0·65 in = 0·0165 m E = 30 x 106 1bjin2 = 20'69 x 1012 N/m 2 v = 0·15 p = 150 lb/ft" = 2·4 Mg/rn"

section ofthe structure exhibits one axis of symmetry. This means that the motion in the direction of this axis is not coupled with the other motions, and can be treated as a simple beam in bending. The finite element solution for the coupled modes of vibration are compared with the exact solution in Table 4. It is clear that the finite element results lie very close to the exact solution, the sixth frequency being within 3 %. The time taken to obtain the first ten frequencies and modes on an r.c.L. 1907 computer was less than 100 seconds. This is in contrast to the high computation time, as referred to in reference [I], that is required for the exact solution. Since the present example

578

M . ?ETYT AND W . H. MIRZA TABLE

4

Frequencies (Hz) ofan E-type structure Mode no.

Predominant motion Torsion Flexure Torsion Torsion Flexure Torsion

1

2 3 4 5 6

Finite element solution 30·23 125·60 188·90 528'00 789'80

Exact solution 30·20 125·00 186·50 52] ,10 784-60 1020'60

1049'70

% error 0·099 0'480 ],287 1·324 0·663 2-849

exhibits one axis of symmetry, the differential equation forthe coupled motion isof order eight. In the general case the differential equation will be of order twelve. The computing time required to obtain the exact solution in this case is expected to be excessive. On the other hand, the finite element model presented will readily solve this case with little increase in computing. The strain energy expression (I) allows for the warping of the cross-section of the beam during torsion through the warping constant, J.,. In order to investigate the importance of this term a channel section structure was analysed for varying wall thicknesses. Details of the structure are given in F igure 3 and Table 5. Figure 4 shows the percentage error obtained when warping is neglected for various thickness-breadth ratios, lilb. It can be seen that neglect ing warping has the effect of reduc ing the frequencies . This reduction is greatest for small values of hlb and reduces as !llh increases. This effect is smallest for the first mode and increases with mode number. The curve for the second mode is not shown since it is a predominantly flexural mode and its frequency is almost unaffected by the warping of the cross-section for all values of hlb.

b/2 ~

T 1 H

/

/

Figure 3. Details of channel section structure.

TABLE 5 Geometrical and material properties of a channel section structure

H - 150 ft = 45'72 m = 10 ft = 3'048 m E = 4'32 x 108 Ib/ft 2 = 20 '69 v = 0·25 p = 150 Ib/ft3 = 2'4 Mg/m 3 b

X

10 9 N/m 2

579

ASYMMETRICAL SHEAR WALLS

-40 - 45 0025 005 0075

010

0125

015

0 20

025

hlb

Figure 4. Effect of warping of the cross-section on channel-section wall frequencies.

6. CONCLUSIONS

A finite element model for thin-walled open sect ion beams has been developed. This has then been used to investigate the vibration characteristics of asymmetrical coupled shear walls . One comparison with an exact solution has shown the model to be accurate and relatively efficient. The importance of including the effect of cross-sectional warping during torsion has also been investigated . In practical building structures the walls are not of constant thickness and so the higher order derivatives will have to be released at the nodes . This feature can be included into general purpose programs. If simpler type elements are preferred, the method proposed in the paper will provide a quick accurate solution for comparison purposes.

REFERENCES 1. A. C. HEIDEBRECHT and R. K. RAINA 1971 Journal 0/ the Engineering Mechanics Division, Proceedings American Society 0/ Civil Engineers 97, 239-252. Frequency analysis of thin-walled

shear walls. 2. J . M . GERE and Y. K. LIN 1958 Journal ofApplied Mechanics 25,373-378. Coupled vibrations of thin-walled beams of open cross-section . 3 . V. Z. VLASOV 1961 Thin-wafted Elastic Beams. Jerusalem, Israel: Israel Program for Scientific

Translation. 4. L. MEIROVITCH 1967 Analytical Methods in Vibrations. New York : The Macmillan Company. 5 . J. S. PRZEMIENIECKI 1968 Theory 0/ Matrix Structural Analysis. New York: McGraw-Hili Book Company, Inc. 6. O. C. ZIENKIEWICZ 1971 The Finite Element Method in Engineering Science. London: McGraw-Hill Book Company, Inc.

580

M. PETYT AND W. H. MIRZA

APPENDIX I LIST OF SYMBOLS

A BI, B 2 Cy , C. E G Ie, /0 I y y , I.. J Jw k

k, k L

I

m iii N q qu, qw, qo D, w x, y, z

e

/.l

cross-sectional area "strain" functions coordinates of centroid relative to shear centre Young's modulus of elasticity shear modulus of elasticity polar moments of area of the cross-section second moments of area of the cross-section Saint-Venant torsion constant warping constant element stiffness matrix submatrices of element stiffness matrix length of beam length of beam finite element element mass matrix submatrix of element mass matrix shape functions element nodal degrees of freedom submatrices of element nodal degree of freedom matrix displacement components of the shear centre Cartesian coordinates rotation of cross-section mass per unit length

APPENDIX II DEFINITION OF ELEMENT SUB MATRICES

280

-II

140

-I II

40 -[2 33 I

I

k=f3

-13 22 280

-II 140

IT[ 40 33

--[2

I

-[3

22

600 -12 77

379 462

-[3

8 231 140

_/4

50 231

-I I

2

99

3465

-fS

40 33 181 -[3 462

- - I --[2 II

380

-[2

77 _181 _ [3

462 5 462[4

Symmetric

4

/6

I

--[3

22

5 4 -I

462

I

5

462

2772

- - /4

15

_ _5_ [5 _ _1_ [6

2772

4620

280

-

11

140

--I II

600

n

l2

40 -j2 33

--{3

379 462

231/

_~[J

~[4

_J..[S

22

231

50

99

4

_2_ /6 3465

581

ASYMMETRICAL SHEAR WALLS

700 429

-

-I

271 858

-[2

E.- p

-[3

_5_ 13 5148

--I

300 1001 123

858

_

I

k =I

73 4 18018 _37 _ Is 101960 23 --[2 858

25 4 18018 271 --I 858

700 429 271 -I 858

--

Symmetric

--I

4004

I --/

6

90090

5 5148

--I

3

97 -p 6006

47 5 4 ---13 - - - I 12012 12012 73 23 _1_/4 ~[J - [2 - _ 15 858 12012 396 720720 5 5 6 73 I - 13 - - - /4 - - - IS - - - I 12012 720720 116424 5148

1042 906 - - I 2574 12012

~ [2 273

274

/2

36036

49 24024 86

--[3

- _ 14

360360

m=1

383 1081080[3 1 /4 100100 26 IS 2162160

123 4004

18018

1143 36036

- - -

1143 36036

- ~ [2

--I

~ [2

36036 I 521 [3 16 1621620 2162160 1042

36036 199 3 - - - 1 144144 -

73 18018

-[3

--I

~r·

_5_[3 5148

3430

300 1001

_ 12

36036

2574 Symmetric

700 429 271 --I 858 23 - 12 858

-

11

__ /4

144144

37 5 - - - I _ 1_ /6

201960

155

36036 ~p

~/4 720720

43 IS 4324320 36036

~p

-~p

274 12

24024

~/4 360360

90090

_

144144

_ 906 I 12012 273

4

521 13 2162 160

_ _1_ 1 _/4 144144 43 IS 4324320 7

12972960

If>

383 [3 1081080 _1_ /4

100100 26 IS 2162 160· 1 1621620

If>