approximate analysis of building structures subjected to earthquakes

Budapest University of Technology and Economics

Faculty of Architecture

APPROXIMATE ANALYSIS OF BUILDINGSTRUCTURES SUBJECTED TO EARTHQUAKES

THESIS SUBMITTED TO THE FACULTY OF ARCHITECTURE OF

BUDAPEST UNIVERSITY OF TECHNOLOGY AND ECONOMICS

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS

FOR THE DEGREE OF

DOCTOR OF PHILOSOPHY

Gabriella Potzta

Supervisor:

Laszlo P. Kollar

Budapest, March, 2002

1

Acknowledgments

I wish to sincerely thank the 6 years work of my supervisor Prof. Laszlo P. Kollar. I’m grateful

for his support, all of his help and the knowledge which I owed to him.

Thanks to the chiefs of the departments for providing my Ph.D studies:

Prof. Gyorgy Farkas, Chief of the Department of Structural Engineering,

Prof. Tamas Matuscsak, Chief of the Department of Mechanics and Structures,

Prof. Marta Kurucz, Chief of the Department of Structural Mechanics.

I thank to Prof. Zsolt Gaspar for the possibility to finish my thesis in the Research Group of

Computational Mechanics.

I would like to thank to all of my professors for their help and advices which helped my research

work: Prof. Karoly Zalka, Prof. Endre Dulacska, Prof. Istvan Hegedus, Prof. Lajos Kollar, Prof.

Pal Rozsa.

2

1 Introduction

Analysis of high-rise building structures stiffened by shear walls, trusses, coupled shear walls, and

frames requires time consuming numerical computations. The designer may be well served by

approximate methods, which (i) can be used in the preliminary design when some of the structural

dimensions are not yet known, (ii) can verify the results of the more advanced numerical calculation,

and, last but not least (iii) can shed light on the behavior of the structure which may lead to a

better design.

Our aim is to present an approximate analysis of building structures subjected to earthquakes

which is (i) simple, (ii) robust (i.e. it gives results with acceptable accuracy for structures with

very different characteristics), and (iii) which can also handle the torsional vibration of building

structures.

The building is stiffened by an arbitrary combination of lateral load-resisting subsystems (shear

walls, frames, trusses, coupled shear walls, cores). We consider stories with identical masses,

however the mass at the top floor may be different. The stiffnesses of the structure may vary with

the height. The analysis is based on the continuum method. We developed replacement beams of

building structures, and we solved approximately the spatial vibration problem of the replacement

beam. Simple formulas are given to calculate the periods of vibration and the internal forces of a

building structure subjected to earthquakes.

The utility and accuracy of the method is demonstrated by a numerical examples, in which the

approximate solution is compared to the results of a finite element calculation.

1.1 Continuum method

One of the most widely used approximate calculations is based on the “continuum method” [14],

[33],[34], [40], when the stiffened building structure is replaced by a (continuous) beam.

The simplest replacement beam is a thin-walled beam, characterized by the bending stiffnesses

(D0yy, D0zz, D0yz), the warping stiffness (Dω) and the torsional stiffness (Dt). (When only a

plane problem is considered, torsion is excluded, and the only parameter that plays a role is the

bending stiffness D0 = D0yy in the x− z symmetry plane.) This model is adequate only for solid

and slender shear walls.

When a truss is loaded laterally, it may show, depending on the stiffnesses of the elements,

bending (flexural) deformation, shear deformation, or the mixture of those (Fig.1).

Hence the shear deformation must be included and the replacement continuum is a Timoshenko-

beam [39], characterized by the bending (D0 = D0yy) and the shear stiffnesses (S = Szz) in the

x− z plane. The bending and the shear stiffnesses of typical structures are given e.g. in [39], [34],

[20] and are listed in Table 1.

A wide frame structure can be modelled by a beam which undergoes shear deformation only

and is characterized by the shear stiffness S = Szz.

3

x

z

Figure 1: Flexural deformation, shear deformation, and mixed deformation

D

D

S

0

l

lD

=

D

S

0

a) b)

Figure 2: Replacement beam of a frame (a), the sandwich beam is equivalent to a Timoshenko-

beam supported by a beam with bending deformation only (b)

When a wide frame is braced by solid walls the replacement beam has two stiffnesses: the

shear stiffness (S) due to the frame and the bending stiffness due to the walls (Dl). This model is

referred to a Csonka-beam because P. Csonka developed it for the analysis of building structures

subjected to wind loads [8].

Neither a thin-walled beam, nor a Timoshenko-beam, nor a Csonka-beam is adequate to

characterize a slender frame, or coupled shear walls. The replacement beam can be obtained by

“smearing out” the beams of the frame along the height, and thus we arrive at the model shown

in Fig.2, which is a sandwich beam [33].

The stiffnesses of the replacement sandwich beam are also included in Table 1 ([14], [37], [8],

[4], D0 is the global bending stiffness, Dl is the local bending stiffness, and S is the shear stiffness).

We note that the sandwich beam is the generalization of the previous models, thus we can

derive them from the sandwich beam with the proper choice of the stiffnesses. This is illustrated

in the following table:

4

Stiffnesses of a sandwich beam Choice of the stiffnesses Resulting beam (stiffnesses)

Dl → 0 Timoshenko-beam (D0, S)

D0 → ∞ Csonka beam (Dl, S)

D0, Dl, SD0 → ∞,

Dl → 0

Beam with shear

deformation only (S)

S → ∞ Thin walled beam (D0 +Dl)

S → 0 or D0 → 0 Thin walled beam (Dl)

It is important to note that a sandwich beam with stiffnesses D0, Dl, and S is equivalent to a

Timoshenko-beam (with stiffnesses D0 and S) which is supported laterally by a beam with bending

stiffness Dl (Fig.2). Hence, if we set the stiffness Dl of a sandwich beam equal to zero we obtain a

Timoshenko-beam with stiffnesses D0 and S.

Continuum models were developed by several authors and it was applied successfully for build-

ing structures subjected to wind loads [8], [34], [37], [40],[46], earthquakes [2], [3], [20], [17], [34],

in the dynamic analysis [25], [28], [29], [33], [40], [43], [42], [47], and in the stability analysis [14],

[24], [28], [29], [34], [40], [41], [44], [45].

However, there are two important problems to be solved:

(i) As we stated before the replacement beam of a single lateral load-resisting subsystem (truss,

frame, shear wall etc.) is given in the literature (see Table 1). When there are several parallel lateral

load-resisting subsystems which are connected horizontally along the height the question arises:

how can they be replaced by only one replacement beam? We find answers only for the following

special cases in the literature: (a) When each lateral load-resisting subsystem is a solid wall

(their shear deformation is neglected) the replacement beam is a beam which undergoes bending

deformation only, and its bending stiffness is the sum of the bending stiffnesses of the individual

walls. (b) When there are frames which can be modeled as beams undergo shear deformation only

and solid walls undergo bending deformation only, the replacement beam is a Csonka-beam, which

has two stiffnesses (with the sandwich notationDl and S, while D0 is infinite), the bending stiffness

is the sum of the bending stiffnesses of the walls, while the shear stiffness is the sum of the shear

stiffnesses of the frames. However, when any of the lateral load-resisting subsystem undergoes both

bending and shear deformation (which is the case of trusses, coupled shear walls, tall frames, and

for wide walls) it can be shown that simple summation (S =∑Sk, D0 =

∑D0k, Dl =

∑Dlk)

may result in a structure which is stiffer by orders of magnitudes than the “real” structure. We

will show in Section 4.1 how the replacement stiffnesses of the building should be calculated.

(ii) As an example let us consider a structure the cross section of which is shown in Fig.3.

When the structure is subjected to torsion, in the two parallel trusses both shear and bending

deformations occur. The classical (Vlasov) theory of beams does not include the shear deformation

in torsion with warping and, hence, its application may significantly overestimates the torsional

stiffness of the structure. This problem, for arbitrary arrangements of the walls, will be addressed

5

StructureReplacement continuum

Siffnesses

wall

b

t = thickness

Timoshenko-beam

D0 = EI I = b3t12

S = AGρ = AG

1.2 A = bt

trussesL

Ad

Ac

d

Ac

h

L

Ad

Acd

h

Timoshenko-beam

D0 = 12EAcL

2 D0 = 12EAcL

2

S = 2EhL2Ad

d3 S = 2Eh2d3

L2Ad+ L

4Ab

The shear stiffness of trusses

with other type of bracing can

be found in the literature.

frameAci , Ici

Ibi

li

h

ci

0 1 i n

Sandwich beam

Dl =∑n

i=0EIci

D0 =∑n

i=0EAcic2i

S =(S−1

b + S−1c

)−1

Sb =∑n

i=112EIbi

lih,

Sc =∑n

i=012EIci

h2

coupled

shear

wall

Aci , Ici

h

ci

0 1 i n

si disi+1

Abi , Ibi

Sandwich beam

Dl =∑n

i=0EIci

D0 =∑n

i=0EAcic2i

S =(S−1

b + S−1c

)−1

Sb =∑n

i=1

6EIbi[(di+si)2+(di+si+1)

2]

d3i h

(1+

12ρEIbiGd2

iAbi

)

Sc =∑n

i=012EIci

h2

Table 1: Replacement stiffnesses of different lateral load-resisting subsystems of high-rise buildings

6

trusses

Figure 3: Plan of a building stiffened by three lateral load-resisting subsystems

in Section 4.2.

1.2 Earthquake analysis

During earthquakes the building structure is subjected to dynamic effects. The inertia forces

which arise form the ground motion cause dynamic internal forces. Several methods are available

to perform the earthquake analysis of building structures.

The Time History Analysis [6], [11] follows the “exact” answer of the structure subjected to

known earthquakes. An earthquake can be characterized by the ground motion, ground velocities,

or ground accelerations. Usually the accelerations as the function of the time are recorded. These

records are called accelerograms and can be used to calculate the response of a structure for an

earthquake which happened in the past. For design purposes several recorded accelerograms, or

artificial ones must be used. The answer of the building can be followed numerically by finite

element programs. This method is rather sophisticated, and requires time consuming numerical

calculations.

The Equivalent Lateral Force procedure uses a simple approximation to determine the maximum

base shear force [34], [13]. The dynamic effect of ground motion is replaced by an equivalent static

load distributed along the height, in a prescribed manner. The method is presented in Appendix

A.1. The analysis is simple and fast, however it can be very inaccurate.

The most frequently used method is the Response Modal Analysis (see Appendix A.2) which

combines the advantages of the previous methods. The basic characteristics of the analysis are

the periods of vibration and the mode shapes of the freely vibrating structure. The method does

not follows the exact answer of the structure during the earthquake, only the maximum values

of the modal responses are determined. Average curves, called Design Spectra are worked out in

the design codes for the maximum values of the modal responses as a function of the period of

vibration and the damping ratio. These are then combined to obtain an approximation of the

maximum value of the response.

In the thesis we present an approximate method to determine the circular frequencies, and

the periods of vibration which can be used in the Response Model Analysis (for buildings with

7

M1

! !1 !2

22

21

2

111

!!!+»

D1 D2D1 D2

! !1 !2

22

21

2

111

!!!+»

D1

M2

M1

M2

! !1 !2

D1

D2 D2

D1 = ¥

22

21

2 !!! +»

a) b) c)

Figure 4: Illustration of Dunkerley’s (a), Southwell’s (b) and Foppl’s (c) theorems

identical stories in Section 5.1 and for buildings with different stories in Section 6). We also give

approximate formulas to calculate the internal forces in Section 5.2.

1.3 Approximate calculation of the circular frequency

We use approximate solutions of the continuum with the aid of the following three theorems, which

are illustrated in Fig.4.

Dunkerley’s theorem [5], [12]: The structure contains two sets of masses denoted by M1 and

M2. The circular frequency of the structure can be approximated by 1/ω2 = 1/ω21 + 1/ω2

2, where

ω1 is the circular frequency of the structure if M2 is set equal to zero while ω2 is the circular

frequency if M1 is set equal to zero.

Southwell’s theorem [5], [23]: The structure is characterized by two stiffnesses, denoted by D1

and D2, such that if we set either one of the stiffnesses equal to infinity the structure would become

infinitely rigid. The circular frequency can be approximated by ω2 = ω21 + ω2

2, where ω1 is the

circular frequency of the structure if D2 is set equal to zero while ω2 is the circular frequency if

D1 is set equal to zero.

Foppl’s theorem: Foppl’s theorem was developed for the stability analysis of elastic structures

[38] and is adopted here for the vibration analysis. The structure is characterized by two stiffnesses

denoted by D1 and D2, such that if we set either one of the stiffnesses equal to zero the structure

would become a mechanism and consequently would not be capable to carry any load. According to

Foppl’s theorem the circular frequency of such a structure is approximated by 1/ω2 = 1/ω21+1/ω2

2,

where ω1 is the circular frequency of the structure if D2 is set equal to infinity, while ω2 is the

circular frequency if D1 is set equal to infinity.

The above three approximations give the exact circular frequencies when the two eigenmodes,

which belong to the circular frequencies ω1 and ω2, are identical.

8

1.4 Previous work

Several authors applied continuum models to calculate the most important parameters of earth-

quake analysis (the circular frequency and base internal forces):

Basu [2] and Kollar [17] performed the earthquake analysis by using a Csonka beam (the

compressibility of columns are neglected). They provided design charts to calculate the circular

frequency of lateral vibration, and the base internal forces.

Rosman [29] gives design tables for the circular frequency of uncoupled lateral and torsional

vibration of symmetrical wall-frame structures neglecting the global bending deformation.

Ng and Kuang presented a simple analysis for the coupled lateral-torsional vibration of buildings

braced by frames and solid shear walls, where the global bending deformations were also neglected.

Skattum [33] derived the differential equations of coupled shear walls and sandwich beams. He

determined the exact values of natural frequencies and mode shapes, which results are presented

in a number of figures. He did not give closed solution for the general vibration problem.

Kollar [15] developed a simple approximation for the calculation of the circular frequency,

however his formula has limited application, the accuracy of which were presented in [22].

Stafford Smith [36], and Rutenberg [32] worked out approximate formulas to calculate the

circular frequencies of symmetrical structures, and gave design charts for the base internal forces.

However their solutions are not applicable for rigid frames and braced frames.

Kopecsiri and Kollar solved the vibration problem for any type of single horizontal load-resisting

subsystems [20], [21]. They gave design charts and also approximate formulas to calculate the

circular frequency and base internal forces. They recommended approximation for the circular

frequency of buildings braced by several plane lateral load-resisting subsystems in case of lateral

and lateral-torsional vibration, respectively.

Zalka performed the lateral and the lateral-torsional vibration analysis of sandwich beams [47].

Closed formulas are given using numerical values presented in tables to calculate the circular

frequencies of structures consisting of several horizontal load-resisting subsystems. Zalka showed

throughout numerical examples that his method is usable for practical purposes.

Several methods are available in the literature to calculate the circular frequencies of buildings.

The accuracy and the limits of application of the most robust methods [32], [21] and [47] will be

presented in Section ?? for lateral vibration of sandwich beams. We will compare these solutions

to our approximation.

To calculate the base internal forces from lateral-torsional vibration there is no closed formula

in the literature.

For earthquake analysis of building structures with varying stiffnesses there is no quick approx-

imate method.

9

a) b)

z

y

z

y

Figure 5: Symmetrical (a) and unsymmetrical (b) arrangements of the lateral load-resisting sub-

systems

2 Problem statement

We consider a building structure that consists of an arbitrary combination of lateral load-resisting

subsystems, i.e. shear walls, coupled shear walls, frames, trusses and cores. The arrangement of

the stiffening system is either symmetrical or arbitrary (Fig.5). The stiffnesses of the individual

stories may be different; however, the stiffnesses must not increase with the height. We consider

uniform mass distribution with an additional mass at the top.

Our first aim is to develop a replacement beam model for the building structure and to derive

the stiffnesses of replacement beams. This beam model can be used in the wind, earthquake, or

stability analyses of the building structure.

Our second aim is to apply this replacement beam model in the earthquake analysis and to

develop simple approximate expressions for the calculation of the eigenfrequencies, and of the

seismic forces.

3 Basic assumptions

We assume that the material behaves in a linearly elastic manner.

The building must have at least four stories.

The floors are considered to be rigid in their plane and they transfer only horizontal forces but

no bending or vertical forces to the lateral load-resisting subsystems. In addition, we assume that

the floors connect the stiffening system “continuously”, hence each cross-section of the building

remains undeformed in the horizontal plane during loading.

We will replace the building structure, consisting of discrete elements by a continuous beam,

and we will analyze this continuum instead of the building structure.

Error in the analysis.

We solve the differential equation of the continuum analytically and also with the aid of the

three theorems given in Section 1.3. In the thesis we define the error of the approximation as

10

Er =Aapp −Acont

Acont, (3.1)

where “cont” refers to the exact solution of the continuum while “app” refers to the approximate

solution, and A can be e.g. the circular frequency, the base shear force etc.

The total error of the suggested method comes from three sources (i) each lateral load-resisting

subsystem is replaced by a continuous cantilever beam, (ii) these beams are then replaced by a

single replacement beam, and (iii) the characteristics of the replacement beams are calculated

approximately.

In the numerical examples we verify the approximate method by solving structures with the

aid of a FE method (ETABS) and by our approximate calculation. The error is defined as

Er =Aapp −AETABS

AETABS, (3.2)

where “ETABS” refers to the FE program ETABS [10].

4 Replacement beam

4.1 Plane problem

In this section we consider symmetrical stiffening systems which deform only in the plane of

symmetry (x − z plane). The floors connect the lateral load-resisting subsystems continuously,

hence the horizontal displacements of the lateral load-resisting subsystems are identical.

Our aim is to determine the stiffnesses of the replacement beam of the building structure.

4.1.1 One lateral load-resisting subsystem

As it was stated in the Introduction, the replacement beam of a lateral load-resisting subsystem is

a sandwich beam. The replacement stiffnesses are summarized in Table 1.

For latter use we define the strain energy of a sandwich beam [1]:

U = UT + Ul, (4.3)

where

UT =1

2

∫ (Sγ2 +D0 (χ′)

2)dx, Ul =

1

2

∫Dl (w

′′)2dx. (4.4)

Here UT and Ul are the strain energies of a Timoshenko beam (stiffnesses D0 and S) and of a

beam with bending deformation only (stiffness Dl), respectively. w is the displacement in the x−zplane, χ is the rotation of the cross-section in the x − z plane, and γ is the shear strain. Prime

denotes derivative with respect to x. χ and γ are related to the displacement w by:

11

D

D

S

0

l=

D

D

S

01

1

1

l

D

D

S

0k

lk

k

D

D

S

0n

ln

n

Figure 6: Parallel lateral load-resisting subsystems and the replacement sandwich beam

l

l

l

Figure 7: The load and the deformations of a sandwich beam subjected to a sinusoidal load

w′ = χ+ γ. (4.5)

It can be shown [14] that in a sandwich beam:

γ = −D0

Sχ′′. (4.6)

4.1.2 Several lateral load-resisting subsystems

In this section we consider n lateral load-resisting subsystems (Fig.6) The k-th element has the

stiffnesses D0k, Dlk, and Sk. The stiffnesses of the beam which replaces the n lateral load-resisting

subsystems are denoted by D0, Dl, and S.

We determine the replacement stiffnesses by applying a sinusoidal displacement on the stiffening

system which is caused by a sinusoidal horizontal load (Fig.7).

Then we equate the sum of the strain energies of the individual lateral load-resisting subsystems

to the strain energy of the replacement wall. Hence (see Eq.4.3) we write

1

2

∫ (Sγ2 +D0 (χ′)

2+Dl (w′′)

2)dx =

1

2

∫ n∑

k=1

(Skγ

2k +D0k (χ′

k)2

+Dlk (w′′

k )2)dx, (4.7)

wk, γk, and χk are the displacement, the shear strain, and the rotation of cross-section of the k-th

lateral load-resisting subsystem, and w, γ, and χ are the displacement, the shear strain, and the

12

rotation of cross-section of the replacement beam, respectively. The horizontal displacements of

the lateral load-resisting subsystems are identical, hence we can write:

w1 = w2 = ... = wn = w. (4.8)

We apply the displacements

w = w0 sinπ

lx, χ = χ0 cos

π

lx (4.9)

on the beam. Eqs.(4.5), (4.6), and (4.9) yield

χ =πl

1 + π2

l2D0

S

w0 cosπ

lx, γ =

π2

l2D0

Sχ. (4.10)

By introducing Eqs.(4.8), (4.9) and (4.10) into Eq.(3.181) and performing the integration be-

tween 0 and l we obtain

Dl +D0

1 + π2D0

l2S

=

n∑

k=1

(Dlk +

D0k

1 + π2D0k

l2Sk

). (4.11)

When l is large γ is small (see Eq.4.10) and Eq.(4.11) results in

Dl +D0 =

n∑

k=1

(Dlk +D0k) . (4.12)

As a consequence we may state that for large l the replacement beam is a beam which undergoes

bending deformation only.

The Taylor series expansion of the function 1/(1 + π2D0/

(l2S))

with respect to 1/l2 about

1/l20 is

1

1 + π2D0

l2S

=1

1 + π2D0

l20S

−π2D0

S(1 + π2D0

l20S

)2

(1

l2− 1

l20

)+

(π2D0

S

)2

(1 + π2D0

l20S

)3

(1

l2− 1

l20

)2

− ... =

=

∞∑

i=0

(− π2D0

S

)i

(1 + π2D0

l20S

)i+1

(1

l2− 1

l20

)i

. (4.13)

Introducing Eq.(4.13) into Eq.(4.11) yields

Dl +

∞∑

i=0

D0

(− π2D0

S

)i

(1 + π2D0

l20S

)i+1

(1

l2− 1

l20

)i

=

n∑

k=1

Dlk +

∞∑

i=0

D0k

(− π2D0k

Sk

)i

(1 + π2D0k

l20Sk

)i+1

(1

l2− 1

l20

)i

.

When l is close to l0 the terms multiplied by(

1l2 − 1

l20

)i

(i = 1, 2, ...) vanish and we have

Dl +D0

1 + π2D0

l20S

=

n∑

k=1

Dlk +

D0k

1 + π2D0k

l20Sk

. (4.14)

13

To obtain a good agreement between the replacement beam and the structure the first three

terms in the series are considered. By equating the first term in the series we obtain Eq.(4.14),

while from the second and third terms we have

D0(1 + π2D0

l20S

)2

π2D0

S=

n∑

k=1

D0k(1 + π2D0k

l20Sk

)2

π2D0k

Sk, (4.15)

D0(1 + π2D0

l20S

)3

(π2D0

S

)2

=n∑

k=1

D0k(1 + π2D0k

l20Sk

)3

(π2D0k

Sk

)2

. (4.16)

Eqs.(4.14), (4.15) and (4.16) can be rearranged to yield the replacement stiffnesses of the beam:

S = π2B3

C2, D0 =

1CB2 − 1

l20

C2

B3

, Dl = A− B2

C, (4.17)

where

A =

n∑

k=1

D0k

1 + π2D0k

l20Sk

+Dlk

, (4.18)

B =

n∑

k=1

D0k(1 + π2D0k

l20Sk

)2

π2D0k

Sk,

C =

n∑

k=1

D0k(1 + π2D0k

l20Sk

)3

(π2D0k

Sk

)2

.

The choice of l0 to obtain the “best” replacement stiffnesses will be discussed in Section 4.3.

4.2 Spatial problem

As it was stated in the Introduction, the beam undergoes both shear and bending (flexural) de-

formations in torsion. We adopt here the beam theory given in [18] (which was developed for

composite beams and which is the simplification of Sun and Wu’s theory [48]). Accordingly, the

displacements of the beam are described by the vectors

u =

v

w

ψ

, χ =

χy

χz

ϑB

, (4.19)

while the shear deformations are

γ =

γy

γz

ϑS

. (4.20)

14

90+°»´

90-°»´

#S

#B

Figure 8: Bending (ϑB) and shear (ϑS) deformation in torsion

v and w are the displacements in the y and z directions, ψ is the rotation of the cross section about

the x axis, χy and χz are the rotations of the cross-section about the z and y axes, respectively.

The total twist per unit length (ϑ) contains a shear type (ϑS) and a bending type (ϑB) deformation

(Fig.8):

ϑ = ψ′ = ϑB + ϑS .

(We note that in Vlasov’s theory γy = γz = ϑS = 0 and u′ = χ). The shear deformations

are related to the displacements by (see [18]):

u′ =

v′

w′

ψ′

= χ + γ . (4.21)

The strain energy of this beam is

UT =1

2

∫γT

Syy Syz Syω

Szy Szz Szω

Sωz Sωy Sωω

γ + χ′T

D0zz D0zy D0zω

D0yz D0yy D0yω

D0ωz D0ωy D0ωω

χ′ +Dtϑ

2

dx.

(4.22)

In this equation D0yy (= EIyy), D0yz (= EIyz), D0zz (= EIzz) are the bending stiffnesses,

D0ωω (= EIω) is the warping stiffness. (D0ωz and D0ωy are zero if the coordinate system is

attached to the shear center see Section 4.2.3.) [S] is the shear stiffness matrix and Dt (= GIt) is

the torsional stiffness. Tdenotes the transpose of vector .

The above beam theory is the generalization of the “Timoshenko-beam” theory for spatial

problems. For our case, as in the plane problem, the local stiffnesses must also be included, and

the strain energy becomes

U = UT + Ul, (4.23)

where Ul is the strain energy of a beam which undergoes bending deformation only:

15

z

y

´

³

rk´

®k

rk³

Figure 9: Global coordinates ( y, z) and the local coordinates (η, ζ) attached to the k-th lateral

load-resisting subsystem

Ul =1

2

∫u′′ T

Dlzz Dlzy Dlzω

Dlyz Dlyy Dlyω

Dlωz Dlωy Dlωω

u′′ dx. (4.24)

Here u is the displacement vector (Eq. 4.19), and Dlij are the (local) bending stiffnesses.


First we consider a single lateral load-resisting subsystem (Fig.9).

The stiffness matrices of this bracing element in the η−ζ coordinate system are[Dη−ζ

0

]

k,[Dη−ζ

l

]

k

and[Sη−ζ

]k, and the torsional stiffness is Dη−ζ

tk . The transformation of the stiffnesses into the

y − z coordinate system gives:

[Dl]k = [T ]Tk

[Dη−ζ

l

]

k[T ]k , (4.25)

[D0]k = [T ]Tk

[Dη−ζ

0

]

k[T ]k ,

[S]k = [T ]Tk

[Sη−ζ

]k[T ]k ,

where

[T ]k =

cosαk − sinαk rkη

sinαk cosαk rkζ

0 0 1

, (4.26)

αk is the angle between the axes η and y; rkη and rkς are the distances of the η and ζ axes from

the origin (Fig.9).


To obtain the replacement stiffnesses of a stiffening system containing several lateral load-resisting

subsystems we assume the displacements in the form of

16

u =

v0

w0

ψ0

sin

π

lx, χ =

χy0

χz0

ϑB0

cos

π

lx, (4.27)

and introduce them into the expression of the strain energy of the replacement beam (Eq.4.23)

and into the sum of the strain energies of the individual elements. By equating them we obtain:

1

2

∫ (γT

[S] γ + χ′T [D0] χ′ + u′′T [Dl] u′′ +Dtϑ2)dx (4.28)

=1

2

∫ n∑

k=1

(γkT

[Sk] γk + χk′ T [D0k] χk′ + uk′′ T [Dlk] uk′′ +Dtkϑ2)dx.

Then we follow the same steps as in Section 4.1.2. The algebra is involved but straightforward

and it is presented in Appendix B. The results are

[S] = π2 [B] [C]−1 [B] [C]−1 [B] , (4.29)

[D0] = [B] [C]−1

[B]

[[E] − 1

l20[B]

−1[C]

]−1

,

[Dl] = [A] − [B] [C]−1

[B] ,

Dt =

n∑

k=1

Dtk,

where

A =

n∑

k=1

(([E] +

π2

l20[D0]k [S]

−1k

)−1

[D0]k + [Dl]k

), (4.30)

B =

n∑

k=1

π2

([E] +

π2

l20[D0]k [S]

−1k

)−1

[D0]k [S]−1k

([E] +

π2

l20[D0]k [S]

−1k

)−1

[D0]k ,

C =

n∑

k=1

π4

([E] +

π2

l20[D0]k [S]−1

k

)−1

[D0]k [S]−1k ×

×(

[E] +π2

l20[D0]k [S]−1

k

)−1

[D0]k [S]−1k

([E] +

π2

l20[D0]k [S]−1

k

)−1

[D0]k .

We observe that Eq.(4.29) reduces to Eq.(4.17) when [D0] , [Dl] , and [S] are replaced by D0, Dl

and S.

4.2.3 Location of the replacement beam

The elements of the stiffness matrices [D0] , [Dl] and [S] depends on the choice of the location of

the axis of the replacement beam which passes through the origin of the coordinate system.

17

In the analysis the location of the axis can be chosen arbitrarily. (The stiffnesses, loads,

coordinates of the mass center are influenced by the location, however the eigenfrequency, buckling

loads, internal forces are not.)

There is a special location of the origin called shear center (or more precisely the “bending

deformation shear center” [18]). When the origin of the coordinate system is attached to the

bending deformation shear center the stiffness matrix [D0] simplifies and D0zω and D0yω are

zero. However, this choice of the origin simplifies the analysis only when the shear deformation is

neglected (shear stiffnesses are infinite), because in this case the load applied at the shear center

does not cause the twist of the building.

As a rule, when the beam undergoes both bending and shear deformations there is no such

location and the beam may twist even if the load is applied at the bending deformation shear

center.

However for a given load and boundary conditions we may define a location (which varies with

the height) such a way that the load acting at this location do not cause the twist of the building.

When the building is symmetrical, the load which is applied in the symmetry plane does not

cause twist, and hence it is practical (however not necessary) to place the axis of the beam at the

symmetry plane.

4.3 Practical considerations

In Sections 4.1.2 and 4.2.2 we obtained a replacement beam whose stiffnesses depend on the choice

of l0. When l0 is approximately equal to the variation of the load (see Fig.7) the behavior of

the replacement beam will be very close to the behavior of the structure. Consequently different

replacement beams should be applied for different loading conditions.

The question arises, how should we choose l0 to obtain the “best” replacement beam. We

suggest for a few cases the values l0 which are given in Fig.10.

We note that there are practical cases when the stiffnesses of the replacement beam are not

sensitive to the choice of l0 and hence the same replacement beam can be used for different loading

conditions.

When D0i

Si≈ D0j

Sjor Sil

20 >> D0i Eqs.(4.17) and (4.18) becomes

S =B3

C2,

D0 =B2

C, (4.31)

Dl =n∑

k=1

(D0k +Dlk) −D0,

where

18

Vibration

l0

l0 » 2H

Buckling

l0

l0 » 2H

Lateral load

l0

l0

l0 = 2H

l0

Hl3

20 = Hl

3

20 »

Figure 10: The values of l0 for lateral load, buckling and vibration

B =

n∑

k=1

D20k

Sk, C =

n∑

k=1

D30k

S2k

. (4.32)

Note that these equations are identical to Eqs.(4.17) and (4.18) when l0 → ∞.

When Sil20 << D0i Eqs.(4.17) and (4.18) becomes

S =

n∑

k=1

Sk, D0 = ∞, Dl =

n∑

k=1

Dlk. (4.33)

These simplifications can be carried out also in case of spatial problems: When [D0]i [S]−1i ≈

[D0]j [S]−1j or [S]i l

20 >> [D0]i Eqs.(4.29) and (4.30) become

[S] = [B] [C]−1

[B] [C]−1

[B] ,

[D0] = [B] [C]−1

[B] , (4.34)

[Dl] =

n∑

k=1

([D0]k + [Dl]k) − [D0] ,

where

[B] =

n∑

k=1

[D0]k [S]−1k [D0]k , [C] =

n∑

k=1

[D0]k [S]−1k [D0]k [S]

−1k [D0]k . (4.35)

For [S]i l20 << [D0] Eqs.(4.29) and (4.30) become

19

[S] =

n∑

k=1

[S]k , [D0] = ∞, [Dl] =

n∑

k=1

[Dl]k . (4.36)

4.4 Numerical examples

To demonstrate the utility of the replacement beam we consider two simple examples shown in

Fig. 5.3.

4.4.1 Doubly symmetrical structure

The structure is stiffened by two coupled shear walls, a frame, and two shear walls (Fig.5.3a). We

wish to determine in the x − z symmetry plane (i) the circular frequencies of the structure when

the mass is distributed uniformly along the floors, and (ii) the buckling loads when the load is

applied at the top.

The replacement stiffnesses of the lateral load-resisting subsystems are (see Table 1):

Dfl =

3∑

i=0

EIci = 4 × 1.95 × 1010 × 1.33 × 10−4 = 1.040 × 107 Nm2,

Df0 =

3∑

i=0

EAcic2i = 2 × 1.95 × 1010 × 0.04 × (22 + 62) = 6.240× 1010 Nm2,

Sb =

3∑

i=1

12EIbilih

= 3 × 12 × 1.95 × 1010 × 1.33 × 10−4

4 × 3.05= 7.672 × 106 N, (4.37)

Sc =

3∑

i=0

12EIci

h2= 4 × 12 × 1.95 × 1010 × 1.33 × 10−4

3.052= 1.342× 106 N,

Sf =(S−1

b + S−1c

)−1= 4.881 × 106 N,

Dwl = 0,

Dw0 = EIw = 1.95 × 1010 × 1.07 == 2.087× 1010 Nm2, (4.38)

Sw =GAw

ρ=

1.95×1010

2×1.2 × 0.8

1.2= 5.42 × 109 N,

Dwt = GItw =

1.95 × 1010

2 × 1.2× 1.07 × 10−2 = 1.733 × 108 Nm2,

20

Aw= 0.8 m

2

Itw

= 1.07 m´10-2 4

li= 4 m

0 1 2 n=3

L= 4 m

framez

y

r = 8 m

a = 36 m

b=

18.0

coupled shear walls

frame shear wall

c2=2c1=2

c3=6c0=6

shear wall

r = -8 m

Aci= 0.04 m

2

Ici= 1.33 m´10

-4 4

Ibi= 1.33 m´10

-4 4

Ab i

= 0.08 m2

Ici= 1.33 m´10

-1 4

Ib= 1.07 m´10

-3 4

coupled shear walls

h =

3.0

5 m

2 m

2 m 2 m

Aci= 0.4 m

2

z

y

r = 8 m

a = 36 m

r = -8 m

(a) (b)

(c)

Iw= 1.07 m

4

/mkgm104.12812

1836N/m10.951

kg/m103.058m91.505.330

2722

210

5

´=+

=Q´=

´==´=

mE

MH

Figure 11: Numerical example. (a) symmetrical structure, (b) unsymmetrical structure, (c) lateral

load-resisting subsystems

Dcl =

1∑

i=0

EIci = 2 × 1.95 × 1010 × 1.33 × 10−1 = 5.200 × 109 Nm2,

Dc0 =

1∑

i=0

EAcic2i = 2 × 1.95 × 1010 × 0.4 × 22 = 6.240 × 1010 Nm2,

Sb =6EIb

[(d+ s1)

2+ (d+ s2)

2]

d3h(1 + 12ρEIb

Gd2Ab

) =6 × 1.95 × 1010 × 1.07 × 10−3 × 2 × (2 + 2)2

23 × 3.05(1 + 12×1.2×1.95×1010×1.07×10−3

1.95×1010/2.4×22×0.08

) =(4.39)

= 1.468 × 108 N,

Sc =

1∑

i=0

12EIci

h2= 2 × 12 × 1.95 × 1010 × 1.33 × 10−1

3.052= 6.708× 109 N,

Sc =(S−1

b + S−1c

)−1= 1.436 × 108 N,

21

mode i = 1 i = 2 i = 3

l0i 2H 23H

25H

A[×1011 kNm2

]1.341 0.7018 0.4077

B[×1014 kNm4

]7.669 1.220 0.3119

C[×1017 kNm6

]107.6 2.535 0.3120

D0i

[×1010 kNm2

]9.406 13.29 13.08

Dli

[×1010 kNm2

]7.950 1.149 1.047

Si

[×107 kN

]3.841 27.87 29.06

Table 2: Replacement stiffnesses of the symmetrical structure given in Fig.5.3

deformation 1. mode 2. mode 3. mode

bending (νB1) 3.52 22.03 61.7

shear (νS1) 0.5π 1.5π 2.5π

Table 3: The values of the multiplier, µi for the calculation of the circular frequencies.

mode Approximation ETABS Error [%]

1 0.2607 0.2649 -1.59

2 1.265 1.206 4.89

3 2.690 2.708 -0.66

Table 4: Comparison of the numerical and approximate results for the circular frequencies.

where superscript f , w, and c refers to the frame, to the walls, to the coupled shear walls, respec-

tively.

The structure is symmetrical, thus the replacement stiffnesses of the structure can be calculated

from Eqs.(4.17) and (4.18). We choose l01 = 2H, l02 = 23H , and l03 = 2

5H in the analysis of the

first, second and third mode of vibration, respectively (see Fig.10). The results are given in Table

2.

Using the replacement stiffnesses we approximate the circular frequencies, ωi (in the x − z

symmetry plane) as follows (Section 5.1.1):

ω2i =

1(ωB0

i

)2 +1

(ωS

i

)2

−1

+(ωBl

i

)2

=

(1

µ2Bi

D0i

mH4

+1

µ2Si

Si

mH2

)−1

+ µ2Bi

Dli

mH4, (4.40)

where µBi and µSi for the first three modes are given in Table 3, the mass, M, and the total height

of the building, H are given in Fig.5.3. The approximate value of the circular frequencies (Eq.4.40)

and the results of a finite element calculation (using the ETABS program) are summarized in Table

4. The maximum error is less then 5% (Table 4).

The buckling loads of the structure (when the load is applied at the top) also can be ap-

proximated using the replacement stiffnesses [19]. Buckling loads in the x − z symmetry plane

22

are:

N1cr =

(π2D01

(2H)2

)−1

+1

S1

−1

+π2Dl1

(2H)2 = 3.953 × 104 kN,

N2cr =

(π2D02(23H)2

)−1

+1

S2

−1

+π2Dl2(23H)2 = 1.861× 105 kN, (4.41)

N3cr =

(π2D03(25H)2

)−1

+1

S3

−1

+π2Dl3(25H)2 = 3.004× 105 kN.

These results are identical to the theoretical values calculated by the equations of [14].

We note that simple summation of the stiffnesses would result in the following buckling loads:

N1cr = 4.947 × 104 kN,

N2cr = 2.115 × 105 kN, (4.42)

N3cr = 3.178 × 105 kN,

and hence the maximum error would be 25.14%.

4.4.2 Unsymmetrical structure

The geometrical and material properties of the structure are given in Fig.5.3b. The building is

stiffened by a shear wall, coupled shear walls, and a frame as shown in Fig.5.3b. We wish to

determine the circular frequencies of the first three modes of vibration. The lateral load-resisting

subsystems of the unsymmetrical structure are identical to those of the symmetrical structure

(Section 4.4.1), the replacement stiffnesses are given by Eq.(4.38). The structure has one plane of

symmetry (x− y). The structure vibrates either in the plane of symmetry (Section 4.1) or spatial,

lateral-torsional vibration occurs (Section 4.2).

In the symmetry plane the circular frequencies can be calculate from Eq.(4.40) independently

of the spatial vibration modes. The results for the first three modes are given in Table 5.

To calculate the circular frequencies of the coupled vibration modes first we determine the

replacement stiffness matrices (Section 4.2). The stiffness matrices of the individual lateral load-

resisting subsystems are

23

mode direction Approximation ETABS Error [%]

1 spatial vibration 0.07834 0.07272 7.73

1 x− y plane 0.1095 0.1095 0.00

1 spatial vibration 0.2187 0.2263 -3.36


2 x− y plane 0.6828 0.6813 0.22



3 x− y plane 1.895 1.887 0.24

3 spatial vibration 2.2974 2.314 -0.72

Table 5: Comparison of the numerical and approximate results for the circular frequencies of the

unsymmetrical structure given in Fig.10

[Dη−ζ

0

]

1=

Dc0 0 0

0 0 0

0 0 0

,

[Dη−ζ

l

]

1=

Dcl 0 0

0 0 0

0 0 0

,

[Sη−ζ

]1

=

Sc 0 0

0 ε 0

0 0 ε

,

[Dη−ζ

0

]

2=

Dw0 0 0

0 0 0

0 0 0

,

[Dη−ζ

l

]

2=

Dwl 0 0

0 0 0

0 0 0

,

[Sη−ζ

]2

=

Sw 0 0

0 ε 0

0 0 ε

,(4.43)

[Dη−ζ

0

]

3=

Df0 0 0

0 0 0

0 0 0

,

[Dη−ζ

l

]

3=

Dfl 0 0

0 0 0

0 0 0

,

[Sη−ζ

]3

=

Sf

0 0

0 ε 0

0 0 ε

.

Dt = Dwt

(Because of numerical considerations we replaced the zero elements in the main diagonal of the

shear stiffness matrices by a small element ε (ε > 0)).

The matrices of the transformation from the local (η − ζ) coordinate systems into the y − z

coordinate system are

[T ]1 = [T ]c

=

1 0 −8

0 1 0

0 0 1

, [T ]2 = [T ]

w=

0 −1 0

1 0 0

0 0 1

, [T ]3 = [T ]

f=

1 0 8

0 1 0

0 0 1

.

(4.44)

For l0 = 2H Eq.(4.30) yields:

24

A = 1010 ×

7.362 0 −37.93

0 2.0776 0

−37.93 0 471.1

,

B = 1014 ×

5.556 0 10.89

0 7.865× 10−3 0

10.89 0 356.2

, (4.45)

C = 1018 ×

9.966 0 66.94

0 2.978× 10−5 0

66.94 0 637.8

.

The stiffness matrices of the replacement beam are (Eq.4.29):

D0 = 1011 ×

1.248 0 0

0 0.2087 0

0 0 79.87

kNm2,

Dl = 109 ×

5.210 0 −41.52

0 0 0

−41.52 0 333.5

kNm2, (4.46)

S = 108 ×

1.485 0 −10.99

0 54.17 0

−10.99 0 95.04

kN.

Dt = 1.733 × 108 Nm2.

The circular frequencies of the lateral-torsional vibration modes can be determined as the eigen-

values of the following equation (Section 5.1.1)

[(H4

µ2Bi

[D0]−1

+H2

µ2Si

[S]−1

)−1

+µ2

Bi

H4[Dl] +

µ2Si

H2[G]−ω2

mim [M ]

]

v0m

w0m

ψ0m

= 0, (4.47)

µBi and µSi for the first three modes are given in Table 3, H is the total height of the building

(H = 91.5 m), matrices [M ] and [G] are

[M ] =

1 0 ym

0 1 zm

ym zmΘm + y2

m + z2m

, [G] =

0 0 0

0 0 0

0 0 Dt

, (4.48)

m is the mass per unit height (m = 3.058 × 105 kg/m), and Θ is the polar moment of mass

(per unit height) about the mass center (Fig.5.3), ym, zm are the coordinates of the mass cen-

25

ter. In this example symmetrical mass distribution was considered, thus ym = 0 and zm = 0,

Θ = m(a2 + b2

)/12 = 4.128 × 107 kgm2/m. The approximate values of the first three circular

frequencies, and the results of the ETABS calculation are given in Table 5. The maximum error is

less then 12%.

5 Approximate analysis of building structures with identical

stories subjected to earthquakes

5.1 Circular frequency and period of vibration

In the previous section the replacement beam of building structures were defined. In this section

simple formulas are developed for the calculation of the circular frequencies of a beam with uniform

stiffnesses. We suggest to approximate the circular frequencies of the building by the circular

frequencies of the replacement continuum. We considered uniform mass distribution, however the

mass at the top floor may be different.

We recall that the circular frequency, ω is related to the period of vibration, T and to the

frequency, f by

T =2π

ω, f =

ω

2π. (5.49)

5.1.1 Uniform mass distribution

Plane problem. We consider a building structure with a symmetrical plan which vibrates in

the x−z symmetry plane. The stiffnesses of the replacement beam in the x−z plane are the shear

stiffness (S), the global bending stiffness (D0) and the local bending stiffness (Dl). The circular

frequencies of beams which undergo bending or shear deformations only are summarized in Table

6 [22].

The circular frequency of a sandwich beam with constant mass distribution was determined

in [22]. We obtained the “exact” values using the Rayleigh-Ritz method (Appendix C.1.1). An

approximate expression can be obtained by using Foppl’s and Southwell’s theorems [22]:

ω2mi =

1(ωB0

mi

)2 +1

(ωS

mi

)2

−1

+(ωBl

mi

)2

i = 1, 2, ... (5.50)

where i = 1 belongs to the first mode of vibration, ωB0

mi and ωBl

mi are the circular frequencies of

beams with bending stiffnesses D0 and Dl, respectively; and ωSmi is the circular frequency of a

beam which undergoes shear deformation only and whose shear stiffness is S. From the second row

of Table 6 we have:

26

mass and

boundary

conditions

bending deformation only shear deformation only

m

H

i = 1 i = 3i = 2 i = 1 i = 3i = 2

ωBi = µBi

√D

mH4

µB1 = π

µB2 = 2π

µB3 = 3π

ωSi = µSi

√S

mH2

µS1 = π

µS2 = 2π

µS3 = 3π

m

H

i = 1 i = 3i = 2 i = 1 i = 3i = 2

ωBi = µBi

√D

mH4

µB1 = 3.52

µB2 = 22.03

µB3 = 61.7

ωSi = µSi

√S

mH2

µS1 = 0.5π

µS2 = 1.5π

µS3 = 2.5π

M

H

ωB =√

3√

DMH3 ωS =

√S

MH

Table 6: Circular frequencies of beams capable bending deformations only or shear deformations

only

27

Figure 12: Accuracy of approximate solutions of the circular ferquency in case of lateral vibration

with the aid of (a) Kopecsiri’s formula, (b) the correction of Kopecsiri’s formula, (c) Rutenberg’s

formula, and (d) Zalka’s formula

ωB0

mi = µBi

√D0

mH4, ωBl

mi = µBi

√Dl

mH4, ωS

mi = µSi

√S

mH2, (5.51)

where m is the mass per unit height.

Numerical comparisons with the exact solution for the circular frequencies shows that for i = 1

Eq.(5.50) may underestimate the circular frequency by up to 15 percent (−15% ≤ Er ≤ 0) for

i = 2 may underestimate the circular frequency by up to 11% and overestimate it by up to 9%

(−11% ≤ Er ≤ 9%), while for i = 3 may underestimate the circular frequency by up to 4% and

overestimate it by up to 6% (−4% ≤ Er ≤ 6%). The errors of using Eq.(5.50) for the lowest

circular frequency are illustrated in Fig.12a.

The accuracy of this approximation is suitable for design purposes. However in some special

28

cases the designer would need more accurate results, thus we suggest a correction factor to reduce

the errors of Eq.(5.50). To obtain a more accurate solution we fitted a curve to the values of the

error given in Fig.12a. We derived an approximate formula in the function of the stiffness ratios

α = H√

SDl

and β =√

Dl

D0:

ω2m1 =

1(ωB0

m1

)2 +1

(ωS

m1

)2

−1

+(ωBl

m1

)2

(1 + f) , (5.52)

where

f =2

100(1 + 2.7β)

(1.1

1.1 + e−1.6ξ − e−0.013ξ− 1

), ξ = α (1 + 4β) . (5.53)

Eq.(5.52) may overestimate the lawest circular frequency by up to 2.3% and underestimate it

by up to 4.14% (−4.143% ≤ Er ≤ 2.3%). The errors of Eq.(5.52) compared to the “exact” results

are given in Fig.12b.

Comparison with previous approximate solutions As it was mentioned in Section 1.4

there are several approximate formulas available in the literature for the calculation of the circular

frequency. In this section we present the accuracy of the most robust methods [32], [47], [21].

One lateral load-resisting subsystem. We consider a single lateral load-resisting subsys-

tem replaced by a sandwich beam with stiffnesses D0, Dl, and S. We calculated the lawest circular

frequency with the aid of Rutenberg’s [32], Zalka’s [47] and Kopecsiri’s [21] (Eq.5.50) approxima-

tions, respectively. The errors compared to the “exact” results are given in Fig.?? in the function

of the stiffness ratios α = H√

SDl

and β =√

Dl

D0.

Fig.?? shows that Rutenberg’s method is not applicable for large values of α.

The maximum errors of Zalka’s formula is 16%, however in the region of maximum errors Zalka

suggest to neglect the shear deformation. With this limitation Zalka’ s approximation gives more

accurate results for the practical values of stiffness ratios.

Kopecsiri’s formula can be applied for any value of the stiffness ratios, his approximation may

result 16% error.

Eq.(5.52) is the correction of Kopecsiri’s formula which results less than 5% error for any value

of stiffness ratios (Fig.12).

Several lateral load-resisting subsystems. We consider a building structure consisting of

two different lateral load-resisting subsystems replaced sandwich beams with the stiffnesses D01,

Dl1, S1, and D02, Dl2, S2, respectively. There is two basic concept for estimating the circular

frequency of the whole structure:

Kopecsiri and Zalka approximate the circular frequency as the sum of the circular frequencies

of the individual lateral load-resisting subsystems:

29

ω2 =

(1

(ωB01)2 +

1

(ωS1)2

)−1

+ ς(ωBl1

)2+

(1

(ωB02)2 +

1

(ωS2)2

)−1

+ ς(ωBl2

)2. (5.54)

where ς represents a correction factor given in [47].

If we substitute the following stiffnesses:

Dl1 = 0, Dl2 = 0, S1H2 >> D01, S2H

2 << D02, (5.55)

Eq.(5.54) results in

ω2 =(ωB01

)2+(ωS2

)2, (5.56)

which is equal to the result of Southwell’s theorem applied for a Csonka beam. Eq.(5.56) would

result 16% error.

According to our approximation first the replacement beam of the structure is produced, then

the circular frequency of the replacement beam is calculated. The circular frequency of a sandwich

beam can be determined by Eq.([?]) with an error less than 5%. The accuracy of the replacement

beam model was presented throughout numerical examples in Section 4.4.

Spatial problem. The replacement beam of the stiffening system has the following stiffnesses:

[D0] =

D0zz D0zy D0zω

D0yz D0yy D0yω

D0ωz D0ωy D0ωω

, [Dl] =

Dlzz Dlzy Dlzω

Dlyz Dlyy Dlyω

Dlωz Dlωy Dlωω

(5.57)

[S] =

Syy Syz Syω

Szy Szz Szω

Sωz Sωy Sωω

, Dt,

where [D0] and [Dl] are the matrices of the global and local bending stiffnesses, respectively, [S]

is the shear stiffness matrix and Dt is the torsional stiffness. The calculation of [D0] , [Dl] , and

[S] are given in Section 4.2. (We note that the axis of the replacement beam passes through the

centroid of the coordinate system which may be chosen arbitrarily.)

The circular frequencies for beams where both ends are simply supported and [Dl] is zero were

derived in [18]. Following the same steps as in [18] but taking also [Dl] into account, we obtain

the following eigenvalue problem for the circular frequency, ωmi:

[(H4

µ2Bi

[D0]−1

+H2

µ2Si

[S]−1

)−1

+µ2

Bi

H4[Dl] +

µ2Si

H2[G]−ω2

mim [M ]

]

v0m

w0m

ψ0m

= 0, (5.58)

30

where for simply supported beams µBi and µSi are given in the first row of Table 6; m is the mass

per unit height, matrices [M ] and [G] are

[M ] =

1 0 ym

0 1 zm

ym zmΘm + y2

m + z2m

, [G] =

0 0 0

0 0 0

0 0 Dt

, (5.59)

Θ is the polar moment of mass (per unit height) about the mass center, and ym, zm are the

coordinates of the mass center. The eigenvector of Eq.(5.58) contains the amplitudes of vibration

in the x− z plane (w0m), x− y plane (v0m) and the amplitude of the torsional mode (ψ0m).

For a cantilever beam, we adopt Eq.(5.58) as an approximation, but we introduce values for

µBi and µSi given in the second row of Table 6. We obtain three circular frequencies (and three

corresponding eigenvectors) for each value of i. The accuracy of Eq.(5.58) was also investigated

numerically. We compared the approximate results to the results of the “exact” solution (Appendix

C.1.1). We obtained the same accuracy for ω as for the plane problem.

When both the x− z and the x− y planes are symmetry planes, the building vibrates either in

the x− z plane, or in the x− y plane, or torsionally about the x axis. In this case the off-diagonal

elements of matrices [D0] , [Dl] , and [S] are zero, and, accordingly, Eq.(5.58) gives the following

three sets of circular frequencies

ω2ymi =

(1

µ2Bi

D0zz

mH4

+1

µ2Si

Syy

mH2

)−1

+ µ2Bi

Dlyy

mH4,

vibration in the

x− y plane(5.60)

ω2zmi =

(1

µ2Bi

D0yy

mH4

+1

µ2Si

Szz

mH2

)−1

+ µ2Bi

Dlzz

mH4,

vibration in the

x− z plane

ω2ωmi =

(1

µ2Bi

D0ωω

ΘH4

+1

µ2Si

Sωω

ΘH2

)−1

+ µ2Bi

Dlωω

ΘH4+ µ2

Si

GItΘH2

.torsional

vibration

We observe that the expression for ω2zmi is identical to Eq.(5.50) developed for the plane prob-

lem.

5.1.2 Additional mass at the top

Plane problem. When there is an additional mass (M) at the end of the cantilever (Fig.15),

the lowest circular frequency may be approximated by using Dunkerley’s theorem as:

ω2 =

(1

ω2m1

+1

ω2M

)−1

, (5.61)

where

ω2M '

1(ωB0

M

)2 +1

(ωS

M

)2

−1

+(ωBl

M

)2

, (5.62)

31

and (see Table 6 third row)

ωB0

M =√

3

√D0

MH3, ωBl

M =√

3

√Dl

MH3, ωS

M =

√S

MH. (5.63)

The accuracy of Eq.(5.62) was found to be: −6% ≤ Er ≤ 0.

Spatial problem. In case of spatial vibration the circular frequencies belonging to i = 1 can be

calculated as (see Dunkerley’s theorem)

ω2 =

(1

ω2m1

+1

ω2M

)−1

, (5.64)

where ωM is calculated from the following equation [?]

[(H3

3[D0]

−1+H [S]

−1

)−1

+3

H3[Dl] +

1

H[G]−ω2

miM [Ω]

]

v0M

w0M

ψ0M

= 0, (5.65)

where M is the additional mass at the top and matrix [Ω] is given by

[Ω] =

1 0 yM

0 1 zM

yM zMΘM + y2

M + z2M

, (5.66)

where Θ is the polar moment of mass (at the top) about the mass center, and yM , zM are the

coordinates of the mass center.

Care must be taken in the use of Eq.(5.64). Equation (5.64) may give unacceptable error when

the horizontal distribution of masses at the top is significantly different from the mass distribution

on the other floors. In this case the eigenvectors for the uniform mass distribution and those for

the mass on the top are very different. Hence, the use of Eq.(5.64) is recommended only if the

scalar product of the eigenvectors are close to unity, say v0Mvom + w0Mw0m + ψ0Mψ0m > 0.9.

5.2 Internal forces


In the response modal analysis of buildings subjected to earthquakes an equivalent load is deter-

mined in each mode of vibration (Appendix A.2). For in-plane vibration, when the ground motion

is in the plane of the vibration, the horizontal force is [6]

fi (x) =

H∫0

mφi (x) dx

H∫0

mφ2i (x) dx

mφi (x)SAi (5.67)

32

where SAi is the spectral acceleration (which depends on the period of vibration, damping and the

ground peak acceleration), and φi is the mode shape.

For spatial vibration [6]

fi =

fyi

fzi

fωi

=

H∫0

φi (x)T[M ]dx ι

H∫0

φi (x)T[M ] φi (x) dx

m [M ]

φyi (x)

φzi (x)

φΘni (x)

SAi (5.68)

where fyi and fzi are horizontal forces in the y and z directions, respectively, fωi is the resultant

moment about the x axis, and ι is the influence vector [6], which represents the direction of

excitation. For example if the excitation is in the x−y plane ι = 1, 0, 0 , if the excitation is in the

x−η plane (where η is in the y−z plane, and the angle between y and η is ϑ) ι = cosϑ, sinϑ, 0 ,and for torsional excitation ι = 0, 0, 1 .

Plane problem. In this section we consider symmetrical structures subjected to earthquakes in

the symmetry plane.

Base shear force. The total horizontal load, which is identical to the base shear force, is

obtained by integrating Eq.(5.67) over the height of the building.

For uniform mass distribution integration of Eq.(5.67) gives

Vi =

H∫

0

fi (x) dx =

(H∫0

φi (x) dx

)2

H∫0

φ2i (x) dx

mSAi = νimHSAi, (5.69)

the values of the multiplier, νi are given in Table 7 for the first, second and third periods of

vibration for beams undergo bending deformation (νi = νB1, νi = νB2, νi = νB3) and for beams

undergo shear deformation only (νi = νS1, νi = νS2, νi = νS3).

For a sandwich beam the multiplier lies between these two values i.e. for the first period of

vibration 0.61 < νi < 0.81, and for the second period of vibration 0.09 < νi < 0.188 and for the

third 0.032 < νi < 0.065. To estimate νi for a sandwich beam we adopt the approximate formula

of the circular frequency. Equation (5.50) and (5.51) gives

1 =

(

1

µ2Bi

D0

mH4

+1

µ2Si

SmH2

)−1

+ µ2Bi

Dl

mH4

−1

ω2mi. (5.70)

The value of νi is approximated as

νi =

(

νBi

µ2Bi

D0

mH4

+νSi

µ2Si

SmH2

)−1

+1

νBiµ2

Bi

Dl

mH4

−1

ω2mi, (5.71)

33

mode deformationbase shear

force

shear force

at 3/4H

base

overturning

moment

1. modebending deformation

only (νB1)0.61 0.32 0.45

shear deformation

only (νS1)0.81 0.32 0.52


only (νB2)0.188 0.09 0.039

shear deformation

only (νS2)0.09 0.09 0.018


only (νB3)0.065 0.033 0.00825

shear deformation

only (νS3)0.032 0.033 0.0044

Table 7: The values of the multiplier, νi for the calculation of the internal forces.

where νBi and νSi are given in Table 7. This expression gives the exact value for νi when S → ∞,

or S → 0, or D0 → 0, or D0 → ∞ and Dl → 0. For arbitrary values of D0, Dl and S Eq.(5.71)

was verified numerically. Comparisons with the exact solution of Eq.(5.69) for a sandwich beam

showed that the accuracy of Eq.(5.71) for ν1 is −9% < Er < 0, for ν2 is −37% < Er < 0, while

for ν3 is −27% < Er < 0.

Shear force at height 3/4H. The distribution of the horizontal forces is determined by

the mode shape of the beam (see Eq.5.67). In the first, second, and third modes of vibration the

mode shape, the distribution of horizontal seismic forces and the shear forces along the height are

illustrated in Fig.13. The maximum of the shear force arises at the bottom of the cantilever, but

in the second mode there is another local maximum (Fig.13, middle).

This local maximum for different continuum models and different stiffness distributions was

calculated, and the location of the maximum was found to be between 0.6H and 0.9H . The

maximum can be approximated by evaluating the shear force at height 3/4H :

V3/4i = νimHSAi. (5.72)

The constant for beams with bending deformation or shear deformation only is given in Table

7 (column “shear force at 3/4H”), for the first three modes. For a sandwich beam Eq.(5.72)

νi = νBi = νSi can be used. The maximum error of this approximation for ν1 is −3% < Er < +3%,

for ν2 is −2% < Er < 10%, while for ν3 is −7% < Er < +4% percent.

34

x

H

V1

M1

F (x)1 V (x)1

x

H

V3

F (x)3 V (x)3

V(x)

x

H

V2

F (x)2 V (x)2

M2

M3

Figure 13: Mode shapes of vibration, horizontal seismic force distribution, shear force

(The design value of the shear force can be calculated by combining the modal responses.

There are different combination rules given in design codes. We suggest to calculate the design

value of shear force at the bottom of the cantilever and at height 3/4H. At the top of the cantilever

the shear force is zero. The shear force diagram may be approximated by fitting a second order

parabola to the design value of the shear force at these three point. See Fig.13.)

Base overturning moment. The base overturning moment can be calculated from the

horizontal load Eq.(5.67) as follows

For a uniform mass distribution we have

Mi =

H∫

0

xfi (x) dx =

H∫0

φi (x) dxH∫0

xφi (x) dx

H∫0

φ2i (x) dx

mSAi = νimH2SAi. (5.73)

The multiplier νi is given in Table 7 for a beam undergoes bending deformation and for a

beam undergoes shear deformation only. For a sandwich beam the multiplier lies between these

two values. νi is again calculated from Eq.(5.71) by introducing the values of νi given in the last

column of Table 7. Numerical comparisons showed that Eq.(5.71) gives the value of ν1 with an

error −5% < Er < 1%, while for the second and third modes the results are not acceptable.

35

Vzi

Vyi

V i!

Mzi

Myi

Figure 14: The base shear forces (Vyi, Vzi), the resultant torque (Vωi), and the base overturning

moments (Myi, Mzi)

Spatial problem.

Base shear force. The base shear forces and the resultant torque (Fig.14) can be calculated

from the integration of Eq.(5.68).

Vyi

Vzi

Vωi

=

H∫

0

fyi

fzi

fωi

dx. (5.74)

First we consider the case when the structure consists of lateral load-resisting subsystems which

undergo shear deformation only. In this case the functions in the eigenvector (φyi, φzi, φωi) are

cosines and Eqs.(5.68) and (5.74) give

Vyi

Vzi

Vωi

=

φ0iT[M ] ι

φ0iT [M ] φ0im [M ]Hνi φ0iSAi, (5.75)

where φ0i is the eigenvector of Eq. (5.58)

φ0i =

v0mi

w0mi

ψ0mi

, (5.76)

and (for i = 1) νi = νSi = 0.81 (Table 7). The same equation apply when the lateral load-resisting

subsystems have bending deformation only, but we must substitute νi = νBi = 0.61 into Eq.(5.75).

When the lateral load-resisting subsystems are sandwiches, the horizontal forces can be calcu-

lated by

36

Vyi

Vzi

Vωi

=

φ0T[M ] ι

φ0T

[M ] φ0m [M ]H

νyiv0mi

νziw0mi

νωiψ0mi

SAi, (5.77)

where νyi, νzi, and νωi are multipliers with values between νBi and νSi (i.e. for i = 1 0.61 <

ν1 < 0.81 and for i = 2 0.09 < ν2 < 0.188), see Table 7. To obtain a reasonable estimation for ν

we rearrange the expression obtained for the circular frequencies. By multiplication of Eq.(5.58)

by [M ]−1/mω2

mi we obtain

v0mi

w0mi

ψ0mi

=

(

H4

µ2Bi

[D0]−1

[M ] + H2

µ2Si

[S]−1

[M ])−1

+

+µ2

Bi

H4 [M ]−1

[Dl] +µ2

Si

H2 [M ]−1

[G]

−1

v0mi

w0mi

ψ0mi

ω2

mim. (5.78)

Similarly as it was done for the plane problem (see Eq. 5.71) we introduce ν-s into this equation

and obtain:

νyiv0m

νziw0m

νωiψ0m

=

(

νBiH4

µ2Bi

[D0]−1 [M ] + νSiH

2

µ2Si

[S]−1 [M ])−1

+

+µ2

Bi

νBiH4 [M ]−1

[Dl] +µ2

Si

νSiH2 [M ]−1

[G]

−1

v0mi

w0mi

ψ0mi

ω2

mim, (5.79)

which is an approximate expression for the vector νyiv0mi, νziw0mi, νωiψ0miT . With this ap-

proximation Eq.(5.77) can be directly evaluated for the seismic loads.

(Note that Eqs.(5.77) and (5.79) give the “exact” values of the horizontal forces (Vyi and Vzi)

and the resultant torque (Vωi) when the structure is doubly symmetrical and in one direction it

has shear deformation only, in the other direction it has bending deformation only, and in torsion

it has bending or shear deformation only.)

As a rule, Eqs.(5.77) and (5.79) give only approximate values of the forces and the resultant

moment. The numerical examples showed that the accuracies are the same as for the plane problem.

Base overturning moment. The base overturning moments can be obtained in the same

way as the base shear forces. (Note that we may use the approximation only for the first mode.)

From Eq.(5.68) we arrive at

Mz1

My1

B1

=

H∫

0

x

fy1

fz1

fω1

dx =

φ01T[M ] ι

φ01T[M ] φ01

m [M ]H2

νy1v0m1

νz1w0m1

νω1ψ0m1

SA1 (5.80)

where Mz1 and My1 are the base overturning moments in the x− y and x− z planes. The vector

on the right hand side is approximated by Eq.(5.79) where νB1 = 0.45 and νS1 = 0.52 (see Table

7).

37

DD

S

0

l

H

M

mm

a) b) c)

Figure 15: Sandwich beam (a); constant mass distribution (b); constant mass and additional mass

at the top (c)

We may observe that when there is only bending deformation, νy = νz = νω = 0.45, and when

there is only shear deformation, νy = νz = νω = 0.52.


Plane problem. When there is an additional mass on the top, we have to take the following

additional force and base overturning moment into account

FM = MSA, MM = HMSA.

Spatial problem. In case of spatial vibration when there are additional masses at the top,

concentrated forces and moment about the x axis may arise on the top of the building structure.

The resultant forces (V My and V M

z ) and moment (V Mω ) are [6]

V My

V Mz

V Mω

=

φ0T[M ] ι

φ0T[M ] φ0

M [Ω] φ0SA. (5.81)

The additional base overturning moments have the values:

MMz = HV M

y , MMy = HVM

z .

5.3 Numerical example

In this section a numerical example is presented to demonstrate the accuracy and robustness of

the approximate calculation for building structures with identical stories.

The geometric and material characteristics of the structure are given in Table 8. and Fig. 16.

The approximate calculation was carried out in the following steps:

Step 1. Calculate of the stiffnesses of the lateral load-resisting subsystems [20].

Step 2. Calculate of the replacement stiffnesses of the building by Eq.(4.29).

38

Number of stories 28

Story height, h 2.97 m

Total height, H 83.2 m

Mass / unit height, m 280640 kg/m

Mass moment of inertia / unit height, θ 69001 kgm2/m

Young’s modulus of walls, Ew 1.95 × 107 kN/m2

Shear modulus of walls, Gw 8.125×106 kN/m2

Young’s modulus of beams, Eb 2.3 × 107 kN/m2

Shear modulus of beams, Gb 9.58 × 106 kN/m2

Area of beams, Ab 0.07 m2

Moment of inertia of beams, Ib 5.79 × 10−4 m4

Area of walls , Aw 1.92 m2

Moment of inertia of walls , Iw 8.85 m4

Table 8: Geometric and material characteristics of the shear walls (see Fig. 14)

9.25

s1

= 6

.1

y

z

O

wall

coupled shear walls

11 ´ 3.7

5.55

16.65

1.855.55

k = 2

k = 1

k = 4

k = 3

k=

5

k=

7

k=

6

k=

9

k=

8

s2

= 6

.1d

= 3

.7

Figure 16: Numerical example of Section 5.3

Step 3. Calculate the periods of vibration by Eq.(5.58).

Step 4. Calculate the internal forces: the base shear force (Eq. 5.77) and the base overturning

moment (Eq. 5.80).

The periods of vibration and the internal forces were determined also by a FE code (ETABS).

In the example uniform mass distribution was considered.

Step 1. In the longitudinal, y direction the building structure is stiffened by four shear walls

(k = 1, 2, 3, 4). In the x direction the stiffening system consists three shear walls (k = 5, 6, 7)

39

and two coupled shear walls (k = 8, 9). The replacement stiffnesses of the lateral load-resisting

subsystems are [20]:

Dlk = 0, k = 1, 2, 3, 4, 5, 6, 7,

Dlk = 2EwIw = 3.45 × 108 kNm2, k = 8, 9,

D0k = EwIw = 1.725× 108 kNm2, k = 1, 2, 3, 4, 5, 6, 7,

D0k = 2EwAw (cw)2

= 2.02 × 109 kNm2, k = 8, 9, (5.82)

Sk = 1.2 ×GwAw = 1.87 × 107 kN, k = 1, 2, 3, 4, 5, 6, 7,

Sk =

6EbIb2 (d+ s1)

2

d3h(1 + 12ρEbIb

Gb d2Ab

)

−1

+

(212EwIw

h2

)−1

−1

= 9.99 × 104 kN, k = 8, 9.

Step 2. The stiffnesses in the global coordinate system are:

[Dl]k = Dlk [R]k , [D0]k = D0k [R]k , [S]k = Sk [R]k , (5.83)

where

[R]k =

cos2 αk − cosαk sinαk rk cosαk

− cosαk sinαk sin2 αk −rk sinαk

rk cosαk −rk sinαk r2k

. (5.84)

In the y direction αk = 0 (k = 1, 2, 3, 4), in the x direction αk = π/2 (k = 5, 6, 7, 8, 9), rk is the

distance of the k−th lateral load-resisting subsystem to the origin of the global coordinate system

(Fig 16):

k 1 2 3 4 5 6 7 8 9

rk 1.85 -1.85 1.85 -1.85 -9.25 5.55 1.85 -16.65 -5.55

For l0 = 2H Sil20 >> D0i, thus the replacement stiffness matrices of the structure are (Eqs.

(4.34) and (4.35)):

40

[A] =

16∑

k=1

[D0]k [S]−1k [D0]k =

ε 0 0

0 8.171 × 1016 −9.069× 1017

0 −9.069× 1017 1, 258× 1019

,

[B] =

16∑

k=1

[D0]k [S]−1k [D0]k [S]

−1k [D0]k =

ε 0 0

0 1, 652× 1021 −1.834× 1022

0 −1.834× 1022 2, 544× 1023

,

[D0] = [A] [B]−1

[A] =

ε 0 0

0 4.041 × 1012 −4.486× 1013

0 −4.486× 1013 6.223× 1014

, (5.85)

[S] = [A] [B]−1

[A] [B]−1

[A] =

ε 0 0

0 1.999× 108 −2.219× 109

0 −2.219× 109 3.078 × 1010

,

[Dl] =

16∑

k=1

([D0]k + [Dl]k) − [D0] =

6.876× 1011 0 0

0 1.208× 1012 −70.978× 1012

0 −70.978× 1012 1.293× 1014

.

We note that some of the stiffness matrices are singular because they contain a zero element in

the main diagonal. For numerical purpose we replaced the zero elements in the main diagonal by

a small element ε (ε > 0).

Step 3. The natural frequencies are calculated from Eq.(5.58):

[(H4

µ2Bi

[D0]−1

+H2

µ2Si

[S]−1

)−1

+µ2

Bi

H4[Dl]−ω2

mim [M ]

]

v0m

w0m

ψ0m

= 0, (5.86)

where [M ] is a diagonal matrix:

[M ] =

1 0 0

0 1 0

0 0 θm

=

1 0 0

0 1 0

0 0 159.09

. (5.87)

By substituting µB1 = 3.52, µS1 = 0.5π (Table 6) into Eq.(6.110) we obtain the following three

eigenvalues:

ω2m1 = 1.981, ω2

m2 = 0.635, ω2m3 = 0.352.

The following three eigenvalues are obtained by substituting µB2 = 22.03, µS2 = 1.5π (Table 6).

Equation (6.110) results in:

ω2m4 = 64.61, ω2

m5 = 24.86, ω2m6 = 12.87.

41

Period of vibration, T (sec)

mode direction Approximation ETABS Error [%] Sei

1 lateral (x-z)-torsional 10.59 10.53 0.57 0.0416

2 x-y plane 7.89 8.26 -4.69 0.0751

3 lateral (x-z)-torsional 4.46 4.32 3.24 0.234

4 lateral (x-z)-torsional 1.75 1.86 -5.41 0.889

5 x-y plane 1.26 1.35 -6.67 1.24



8 x-y plane 0.45 0.50 -10.0 2.60


Table 9: Comparison of the numerical and approximate results for the periods of vibration, (T ),

and the spectral accelerations, Sei

The modes 7 to 9 are obtained by introducing µB3 = 61.7, µS3 = 2.5π into Eq.(6.110) which yield:

ω2m7 = 486.02, ω2

m8 = 195.02, ω2m9 = 99.44.

The periods of vibration, T are:

Ti =2π

ωi. (5.88)

The results of the approximate calculation and those of the ETABS code are given in Table 9. The

maximum error is -10%.

Step 4. The seismic forces are determined using the Response Modal Analysis (Appendix A.2).

The calculation was carried out according to Eurocode 8. The spectral accelerations are given in

the function of the period of vibration, [9]:

0 < Ti < TB Sei = agSt

[1 +

Ti(ηβ − 1)

TB

], (5.89)

TB < Ti < TC Sei = agStηβ,

TC < Ti < TD Sei = agStηβ

(TC

Ti

)k1

,

TD < Ti Sei = agStηβ

(TC

TD

)k1(TD

Ti

)k2

.

The parameters in Eq.(5.89) depend on the soil condition, on the location of the structure,

and on the damping ratio. We considered St = 1, ag = 0.08 × 9.81, β = 2.5, TB = 0.15, TC =

0.6, TD = 3, k1 = 1, k2 = 2, η = 1.323.

Substituting the periods of vibration given in Table 9 into Eq.(5.89) the spectral accelerations,

Sei can be determined in each mode of vibration. These are also given in Table 9.

42

mode excitation internal force Approximation ETABS Error [%]

2 (x-y plane) x-y Vy [kN] 1070 998.8 7.13

My [kNm] 65635 61147 7.34

Vz [kN] 2175 2170 0.26

3 ( x-z plane-torsional) x-z Vω [kNm] 22031 22674 -2.84

Mz [kNm] 129450 132772 -2.50

Vz [kN] 223.6 253.7 -11.87

1 ( x-z plane-torsional) x-z Vω [kNm] 3839 3383 8.39

Mz [kNm] 13879 15496 -10.43

5 (x-y plane) x-y Vy [kN] 5423 5232 3.65

My [kNm] 93549 90810 3.02

Vz [kN] 5415.8 5008 8.14


Mz [kNm] 93333 89190 4.64

Vz [kN] 1432 1464 -2.20


Mz [kNm] 24711 25987 -4.91

8 (x-y plane) x-y Vy [kN] 3938 4026 -2.20

My [kNm] 41561 41704 -034

Vz [kN] 2512 2561 -1.90


Mz [kNm] 26543 26812 -1.00

Vz [kN] 1338.8 1336 0.22


Mz [kNm] 14124 14350 -1.57

Table 10: Internal forces of the numerical examle in Section 5.3.

The base shear forces are calculated by Eq.(5.77). The structure has one plane of symmetry

(x − y), thus from the lateral vibration in the x − y plane base shear force arises only in the y

direction. However from the lateral (x − z)- torsional vibration both a shear force and a torque

arise. The base overturning moments are calculated from Eq.(5.80). The internal forces in the

first three modes in case of different earthquake excitations are given in Table 10. Table 10 also

contains the results of the ETABS calculation, and the errors of the approximation.

The design value of the internal forces can be calculated by combining the modal responses [9]:

43

?

Figure 17: Approach: a) structure consisting of discrete elements, b) replacement continuum with

varying stiffnesses, c) replacement continuum with constant stiffnesses

V =9∑

1

√V 2

i =

6786

6659

68358

, M =

9∑

1

√M2

i =

121601

164850

. (5.90)

The ETABS calculation gives:

V =

9∑

1

√V 2

i =

6677

6351

64837

, M =

9∑

1

√M2

i =

118173

165600

. (5.91)

The maximum error is 5.43%. Including 15 modes in the ETABS calculation the design values of

the base shear force are

Vy =

15∑

1

√V 2

y = 7050 kN, (5.92)

Vz =

15∑

1

√V 2

z = 6500 kN,

which result less than 4% error.

6 Approximate analysis of building structures with different

stories subjected to earthquakes

6.1 Approach

The structure consisting of discrete elements is replaced by a continuum, the stiffnesses of which

may vary with the height (Fig.17b). (The calculation of the replacement stiffnesses is discussed in

Section 4.)

44

This continuum is further replaced by one with constant stiffnesses (Fig.17c). Then the circular

frequency (and the internal forces) are determined by the expression given in Section 5.1 (and in

Section 5.2).

According to the above approach we have to determine the stiffnesses of the replacement con-

tinuum with constant stiffnesses. We wish to answer the following question: Can we give a simple

rule how to choose the stiffnesses of the continuum with constant stiffnesses?

We will show that the answer to this question is yes, and we will determine the height where the

stiffnesses of the actual continuum must be evaluated to replace it by one with constant stiffnesses.

The error of the suggested approximate method comes from three sources: First the discrete

structure is replaced by a continuum. Second: the continuum with varying stiffnesses is replaced

by a continuum with constant stiffnesses. Third: the period of vibration of the continuum is

determined by simple approximate formulas.

6.2 Circular frequency and period of vibration


Plane problem. Following the Rayleigh-Ritz method we derived expressions and developed an

algorithm and a computer code for calculating the circular frequencies of Timoshenko-beams and

Csonka-beams with varying stiffnesses. The details of the derivation are given in Appendix C.1.2.

(We note that these expressions and the corresponding computer codes are readily applicable to

beams which undergo bending or shear deformation only.)

The results of the Rayleigh-Ritz method will be referred to as “accurate” solutions of the

continuum. Then we evaluate the stiffness of the beam at ξH (0 < ξ < 1) and determine the

circular frequencies of beams with these uniform stiffnesses by the expressions given in Section 5.1.

We will compare these results to the accurate one.

We determine numerically at which height the stiffnesses of the continuum (with varying stiff-

nesses) should be evaluated to obtain the “best” the replacement beam with uniform stiffnesses.

Beams undergo bending deformation only. The stiffness of a beam, which undergoes

bending deformation only decreases from D to kD (0 < k < 1) from the bottom to the top. Three

different functions of the stiffness were considered: linear (Fig.18a), parabolic with vertical tangent

at the top (Fig.18b) and parabolic with vertical tangent at the bottom (Fig.18c).

The stiffness, D of the replacement beam with constant stiffness was determined at various ξ

and k values, and the circular frequencies was calculated from the following equation (Table 6):

ωB1 = 3.52

√D

mH4, ωB2 = 22.03

√D

mH4, (6.93)

where 1 and 2 refers to the first and second mode of vibration, respectively, m is the mass per unit

height, and H is the total height of the building.

45

x

H

D

kD

kD

D

D

kD

a)

b)

c)

Hx( )[ ]xkD -- 11

( )[ ]211 x-+ kD

( ) ( )[ ]xx kkD ---+ 1211 2

x

H

Hx

x

H

Hx

Figure 18: Bending stiffness distibution: a) linear, b) parabolic with vertical tangent at the top,

c) parabolic with vertical tangent at the bottom

ξ

0.23 -0.60% 1.14% 6.00%

0.24 -0.95% 0.89% 4.63%

0.25 -1.36% 0.63% 3.25%

0.26 -1.83% 0.35% -3.31%

0.27 -2.36% -0.17% -4.07%

Table 11: Error in the first period of vibration, TD1, using replacement constant stiffness for a

B-beam

The maximum errors for different ξ-s are given in Tables 11 and 12.

For practical purposes we suggest ξ = 0.25 for the first mode of vibration and ξ = 0.5 for the

second mode of vibration in case of all these distributions. The maximum error does not exceed

11%.

46

ξ

0.25 22.13% 14.05% 66.21%

0.48 1.69% 3.33% 15.24%

0.50 -2.59% 2.01% 10.81%

0.52 -3.80% -0.77% -8.40%

Table 12: Error in the second period of vibration, TD2, using replacement constant stiffness for a

B-beam

cS

S

S

cS

cS

S

x

H

a)

b)

c)

Hx( )[ ]xcS -- 11

( ) ( )[ ]21211 xx ccS ---+

( )[ ]211 x-+ cS

x

H

Hx

x

H

Hx

Figure 19: Shear stiffness distibution: a) linear, b) parabolic with vertical tangent at the top, c)

parabolic with vertical tangent at the bottom

Beams undergo shear deformation only. The stiffness of a beam, which undergoes shear

deformation only decreases from S to cS (0.1 ≤ c ≤ 1) from the bottom to the top. Three different

functions of the stiffness were considered: linear (Fig.19a), parabolic with vertical tangent at the

top (Fig.19b) and parabolic with vertical tangent at the bottom (Fig.19c).

The stiffness, S of the replacement beam with constant stiffness was determined at various ξ

and c values, and the circular frequencies was calculated from the following equation (Table 6):

47

ξ

0.33 6.93% -4.72% -7.80%

0.35 5.32% -4.05% -5.64%

0.37 3.69% -3.19% -3.08%

0.39 2.03% -2.28% 4.14%

0.42 -2.88% -0.80% 6.27%

Table 13: Error in the first period of vibration, T S1, using replacement constant stiffness for an

S-beam

ξ

0.35 37.65% 27.43% 29.21%

0.60 7.98% 8.83% 9.30%

0.63 -4.97% 5.64% -12.32%

Table 14: Error in the second period of vibration, T S2, using replacement constant stiffness for an

S-beam

ωS1 = 0.5π

√S

mH2, ωS2 = 1.5π

√S

mH2, (6.94)

where 1 and 2 refers to the first and second mode of vibration, respectively, m is the mass per unit

height, and H is the total height of the building.

The maximum errors for different ξ-s are given in Tables 13 and 14.

We obtained the highest errors in the case of the distribution presented in Fig.19c. If the ratio c

is less than 0.1, the error of that distribution can be very high. However, the limit 0.1 is acceptable

for practical purposes.

For practical purposes we suggest ξ = 0.33 for the first mode of vibration and ξ = 0.6 for the

second mode of vibration, in case of all these distributions. Even for this case the maximum error

does not exceed 10%.

(Considering an S-beam the minimum error belongs to ξ = 0.38 as can be seen in Table 13. We

calculated several examples for Timoshenko-beams and for Csonka-beams with different ξ-s. The

suggested value, ξ = 1/3, is a compromise to keep the error of the approximation low for all the

cases and it is also a value easy to remember.)

Timoshenko-beam. We have analyzed a Timoshenko-beam with varying stiffnesses pre-

sented in Fig.20. As it is mentioned in Appendix C.1.2, the flexural stiffness varies linearly for

D0 to kD0, while the inverse of the shear stiffness varies linearly from 1/S to c/S (1 ≤ c ≤ 10).

48

S

c

x

H

D0

kD0

Hx

S

1

( )[ ]xcS

-- 111

( )[ ]xkD -- 110

Figure 20: Stiffness distribution of a Timoshenko-beam

ξD ξS 1.mode 2.mode

0.25 0.30 1.91% -

0.25 0.33 2.63% -

0.25 0.38 7.35% -

0.5 0.6 - 13.96

Table 15: Error in the periods of vibration,T 1 and T 2 using replacement constant stiffness for a

Timoshenko-beam

The Rayleigh-Ritz method yields the circular frequencies as a function of the stiffness ratios c, k

(Appendix C.1.2).

The approximate value of the circular frequencies for constant stiffnesses was determined by

Eq.(5.50) with Dl set equal to 0, which yields

ω2mi =

1(ωB0

mi

)2 +1

(ωS

mi

)2

−1

, i = 1, 2, (6.95a)

where ωB0

mi and ωSmi are given by Eqs.(6.93) and (6.94), respectively (Table 6).

In the calculation of the first period of vibration the stiffnesses were evaluated for the bending

stiffness at the quarter of the height, 14H, and for the shear stiffness at the third of the height, 1

3H .

The maximum differences between the results of the Rayleigh-Ritz method and of Eq.(6.95a) was

found to be only 2.63% (Table 15).

For calculating the second period of vibration we used the stiffnesses evaluated for the bending

stiffness at the half of the height, 12H, and for the shear stiffness at the six tenth of the height,

106 H . The maximum difference between the results of the Rayleigh-Ritz method and Eq.(6.95a) is

less than 14% (Table 15).

Csonka-beam. We have analyzed a Csonka-beam with stiffness distribution presented in

Fig.21. Three different functions of the stiffnesses were considered: linear (Fig.21a), parabolic

with vertical tangent at the bottom (Fig.21b) and parabolic with vertical tangent at the top

(Fig.21c). With the aid of the Rayleigh-Ritz method we can obtain the circular frequencies as a

49

x

H

Dl

dDl

dDl

Dl

Dl

dDl

Hx( ) ( )[ ]xx ddD

l---+ 1211 2

( )[ ]211 x-+ dDl

cS

S

S

cS

cS

S

x

H

a)

b)

c)

Hx( )[ ]xdD

l-- 11( )[ ]xcS -- 11

( ) ( )[ ]21211 xx ccS ---+

( )[ ]211 x-+ cS

x

H

Hx

Figure 21: Stiffness distribution: a) linear, b) parabolic with vertical tangent at the top, c)

parabolic with vertical tangent at the bottom

function of the stiffness ratios d, c (Appendix C.1.2).

The approximate value was determined by Eq.(5.50) with D0 set equal to infinity (D0 = ∞).

It yields:

ω2mi =

(ωS

mi

)2+(ωBl

mi

)2

, i = 1, 2, (6.96)

where ωB0

mi and ωSmi are given by Eqs.(6.93) and (6.94), respectively (Table 6).

The maximum errors between the results of Rayleigh-Ritz method and Eq.(6.96) are compiled

in Table 16.

In the calculation of the first period of vibration the stiffnesses were evaluated for the bending

stiffness at the quarter of the height, 14H, and for the shear stiffness at the third of the height,

13H . The maximum differences between the results of the Rayleigh-Ritz method and Dunkerley’s

theorem was found to be less than 18%.

For calculating the second period of vibration we used the stiffnesses evaluated for the bending

stiffness at the half of the height, 12H, and for the shear stiffness at the six tenth of the height,

106 H . The maximum differences between the results of the Rayleigh-Ritz method and Dunkerley’s

theorem was found to be less than 18%.

50

ξD ξS

0.25 0.33 -16.37% -15.14% -17.89%

0.25 0.35 -16.01% -15.10% -17.12%

0.23 0.35 22.60% - -

Table 16: Error in the first period of vibration,T 1 using replacement constant stiffness for a Csonka-

beam

ξD ξS

0.52 0.60 -13.15% -16.69% -10.52%

0.50 0.60 -12.42% -15.67% -10.12%

Table 17: Error in the second period of vibration,T 2 using replacement constant stiffness for a

Csonka-beam

Sandwich beam. The period of vibration of a sandwich beam with constant stiffnesses can

be determined by Eq.(5.50).

For the calculation of the first period of vibration the flexural stiffnesses, Dl and D0 are evalu-

ated at the quarter of the height, for the shear stiffness, S at the third of the height. In the second

mode Dl and D0 are evaluated at the half of the height, and S at the six tenth of the height.

A computer code for the calculation of the “accurate” period of vibration of sandwiches was

not developed, and hence the accuracy of the above approximation will be investigated numerically

in Section 6.4.

Spatial problem. The spatial vibration problem will be investigated with the aid of a numerical

example in Section 6.4.2.


When there is an additional mass (M) at the end of the cantilever (Fig.15), the lowest circular

frequency may be approximated by using Dunkerley’s theorem as:

ω2 =

(1

ω2m1

+1

ω2M

)−1

, (6.97)

where ω2m1 is given byωM is the circular frequency of a cantilever with a single mass at the top,

the stiffnesses of which varies with the height.

We apply the a similar approximation for a single mass at the top as we applied for uniform

mass distribution (Section 6.2.1). We replace the continuum with varying stiffnesses with a beam

with constant stiffnesses. We suggest to evaluate the replacement constant stiffnesses at different

51

height of the structure. The analysis is derived in [26]. The results are:

D0 = D1/4H0 , Dl = D

1/4Hl , S = S0.64H .

The circular frequency of the replacement beam is (Eq.5.62):

ω2M '

1(ωB0

M

)2 +1

(ωS

M

)2

−1

+(ωBl

M

)2

, (6.98)

where ωB0

M , ωSM , and ωBl

M are given in Table 6. The utility and accuracy of this approximation is

investigated numerically in Section 6.

6.3 Internal forces

We analyzed beams develop bending or shear deformations only, and Csonka beams with the

stiffness distributions given in Figures 18, 19, and 21, respectively. The stiffness distributions are

limited. The ratio of the stiffness at the bottom of the beam and the stiffness at the top of the

beam should not exceed 10. We suggest to approximate the base shear force of the structure with

different stories by the base shear of the replacement beam with constant stiffnesses (Eqs. 5.69

and 5.71). The approximation is acceptable only in the first mode of vibration. The maximum

error for a beam with bending deformation only (ν1 = νB1 = 0.61) is 15.45%, for a beam with

shear deformation only (ν1 = νS1 = 0.81) is 28.08%, and the errors for a Csonka beam (Eq. 5.71)

are under 30%. To get a better approximation we can decrease the ratio of the stiffnesses at the

bottom and at the top. If the ratio is under 5 the maximum error is less than 25%. We produced

more complicated results in the function of the stiffness ratios [27], which gives an error less than

15%, however this results are tedious for practical purposes.

6.4 Numerical examples

In this section numerical examples are presented to demonstrate the accuracy and robustness of

the approximate method for varying stiffnesses.


We present the calculation of the lowest circular frequency of type of lateral load-resisting subsys-

tems.

The approximate calculations was carried out in the following steps:

Step 1. Calculate of the stiffnesses of the building at each level (Table 1).

Step 2. Calculate the replacement stiffnesses of the beam, D, at the quarter of the height and

S at the third of the height.

Step 3. Calculate the period of vibration by Eq.(5.50).

52

H

L1 L2 L3

h

Figure 22: Rigid frame example

The period of vibration of the buildings were determined also by a FE code (ETABS) [10].

Note that in these examples the mass at each level was calculated as mh, while at the very top

either as mh2 or as mh.

Rigid frame The geometrical and material characteristics of the frame are given in Table 18

and Fig. 22. In this example the number of stories were 4, 6, 8, 10, 15 and 21. Here the calculation

for n = 10 is presented.

Step 1. The beams are identical to each other while the columns are different at each level. At

the i−th level the dimensions of the cross-section of the columns are:

a = b =

√(2/3 + i) 3.05 + 15

100, (6.99)

which results in areas and moments of inertia given in Table 18, and illustrated in Fig.23.

Step 2. We calculated the area and the second moment of inertia of columns at each level

(Ac, Ic), and fitted curves to obtain Ac and Ic as a function of the distance from the ground. In

this calculation we assumed that Ac and Ic are values at the third of the heights of the columns

(Fig. 23), then the area and inertia were evaluated at the quarter and third of the total height.

This results in the following area and moment of inertia:

A1/4c =

30.5 × 3/4 + 6

100= 0.379 m2,

I1/4c =

(30.5 × 3/4 + 6

100

)2

/12 = 1.95 × 10−2 m4, (6.100)

I1/3c =

(30.5 × 2/3 + 6

100

)2

/12 = 1.04 × 10−2 m4.

53

Number of storyes n = 4 to 21

Story height, h (m) 3.05

Total height, H (m) nh

Distance between columns, l (m) 5.00

Weight / unit height, mg (kN/m) 588.6

Young’s modulus, E (kN/m2) 2.78×107

Area of beams, Ab (m2) 0.366×0.488 = 0.1786

Moment of inertia of beams, Ib (m4) 3.545×10−3

Area of columns at ith story, Aci (m2) 3.05(n−i+2/3)+1.5n100

Moment of inertia of columns at ith story, Ici (m4) [3.05(n−i+2/3)+1.5n]2

104×12

Table 18: Geometric and material characteristics of the rigid frame example in Section 6.4

1/4H 1/4H1/3H

Ic

1/4Ac

1/4

a) b)

Ic

1/3

Figure 23: Variation of a) the area, b) the moment of inertia of the columns

The replacement stiffnesses are (Table 1)

54

Sc =

4∑

i=0

12EI1/3ci

h2= 4

12 × 2.78 × 107 × 1.04 × 10−2

3.052= 1.49 × 106 kN,

Sb =

3∑

i=1

12EIbilih2

= 312 × 2.78 × 107 × 3.54 × 10−3

5.00 × 3.05= 2.32 × 105 kN,

S =(S−1

b + S−1c

)−1=

(1

14.9 × 105+

1

2.32 × 105

)−1

= 2.01 × 105kN, (6.101)

D0 =4∑

i=0

EA1/4ci c2i = 2 ×

(2.52 + 7.52

)× 2.78 × 107 × 0.379 = 1.32 × 109 kNm2,

Dl =

4∑

i=0

EI1/4ci = 4 × 2.78 × 107 × 1.95 × 10−2 = 1.33 × 106 kNm2.

Step 3. The period of vibration is (Eq.5.50):

ωmi =

1(ωB0

mi

)2 +1

(ωS

mi

)2

−1

+(ωBl

mi

)2

0.5

=

=

(

1H4

3.522

mD0

+1

(0.5π)2 mH2

S

)−1

+H4

3.522

m

Dl

0.5

= (6.102)

=

(

130.54

3.52588.6/9.811.32×109

+1

(0.5π)2 30.52

×588.6/9.812.01×105

)−1

+30.54

3.52

588.6/9.81

1.33 × 106

0.5

= 2.99 1/ sec .

The period of vibration is:

T =2π

ω= 2.098 sec .

The period of vibration of the frame was also calculated by the FE code (ETABS), the result

is

T = 2.000 sec,

which is very close to the approximate value furnished by our method. Results of n = 4, 6, 8, 10, 15

and 21 stories are presented in Table 19.

Coupled shear walls

The geometric and material characteristic of the coupled shear walls are given in Table 20 and

Fig. 24. In this example we wanted to demonstrate the effect of the variation of all the stiffnesses

(S,D0, Dl). To fulfill this purpose the thickness of the wall was chosen for the first nine stories as

w = 465 mm, for the middle nine stories as w = 310 mm and for the top nine stories as w = 155

mm.

55

Number of storiesETABS

mh/2

ETABS

mh

Approximate

method

Error (%)

mh/2

Error (%)

mh

21 4.008 4.122 4.252 -6.09 -3.15

15 2.891 3.006 3.056 -5.71 -1.66

10 2.000 2.122 2.098 -4.89 1.13

8 1.662 1.791 1.733 -4.25 3.24

6 1.452 1.62 1.386 -4.58 14.44

4 1.059 1.24 1.071 -1.13 13.63

Table 19: Period of vibration (T , sec) of the frame given in Fig. 11 calculated by the ETABS code

and by the approximate method. (In the ETABS calculation the concentrated mass on the very

top was either mh/2 or mh, as shown in the table)

H

s sd

h

Figure 24: Coupled shear wall example

Step 1. The areas and the moments of inertia of the beams and columns are given in Table 20,

and their distributions are illustrated in Fig.25.

Step 2. We fitted a curve to obtain Ac and Ic as a function of the distance from the ground.

In the calculation we assumed that Ac and Ic are values at the third of the heights of the columns

(Fig.25), and thee area and inertia were evaluated at the quarter and third of the total height.

This results in:

56

Story 1-9 10-18 19-27

Number of stories 27

Story height, h (m) 2.74

Total height, H (m) 73.98

Weight / unit height, mg (kN/m) 3.21×10−3

Young’s modulus, E (kN/m2) 2.5×107

Shear modulus, G (kN/m2) 1.07×107

Depth of walls, s (m) 3.00

Clear span of openings, d (m) 1.50

Area of beams, Ab (m2) 0.74×0.465 = 0.344 0.74×0.31 = 0.229 0.74×0.155 = 0.115

Moment of inertia of beams, Ib (m4) 15.70×10−3 10.47×10−3 5.23×10−3

Areas of columns, Ac (m2) 3.00×0.456 = 1.368 3.00×0.31 = 0.93 3.00×0.155 = 0.456

Moments of inertia of columns, Ic (m4) 1.046 0.698 0.349

Table 20: Geometric and material characteristics of the coupled shear walls example

1/4H 1/4HA

1/4

c 1/3H I1/4

c

I1/3

c

A1/3

b

1/3H

I1/3

b

1/3H

Figure 25: Variation of the area and inertia of the columns and beams of the coupled shear walls

example

57

A1/4c = 1.20 m2,

I1/4c = 0.900 m4,

I1/3c = 0.81 m4, (6.103)

A1/3b = 0.268 m2,

I1/3b = 12.21 × 10−3 m4.

The replacement stiffnesses are (Table 1):

Sc = 6

2∑

i=0

12EI1/3ci

h2= 6

12 × 25 × 106 × 0.81 × 2

2.742= 3.9 × 108 kN,

Sb = 6

1∑

i=1

6EI1/3bi

[(d+ si)

2+ (d+ si+1)

2]

d3h

(1 +

12ςEI1/3

bi

Gd2A1/3

bi

) = 66 × 25 × 106 × 12.21 × 10−3 (1.5 + 3.00)

2

1.53 × 2.74(1 + 12×1.2×25×106

×12.21×10−3

1.07×107×1.52×0.286

)

= 1.47 × 107 kN, (6.104)

S =(S−1

b + S−1c

)−1=

(1

3.9 × 108+

1

1.47 × 107

)−1

= 1.42 × 107kN,

D0 = 6

2∑

i=0

EA1/4ci c2i = 2 × 6 × 25 × 106 × 4.52 × 1.2 = 1.82 × 109 kNm2,

Dl =

2∑

i=0

EI1/4ci = 6 × 2 × 25 × 106 × 0.9 = 2.7 × 108 kNm2.

Step 3.

The circular frequency is (Eq.5.50):

ωmi =

1(ωB0

mi

)2 +1

(ωS

mi

)2

−1

+(ωBl

mi

)2

0.5

=

=

(

1H4

3.522

mD0

+1

(0.5π)2 mH2

S

)−1

+H4

3.522

m

Dl

0.5

= (6.105)

=

(

173.984

3.523210/9.811.824×109

+1

(0.5π)2 73.982

×3210/9.811.42×107

)−1

+73.984

3.52

3210/9.81

2.7 × 108

0.5

= 1.55 1/ sec .

The period of vibration is:

T =2π

ω= 4.05 sec .

The period of vibration of the structure was also calculated by the FE code (ETABS) and it

gives

58

Total height, H 83.2 m

Mass / unit height, m 280640 kg/m

Mass moment of inertia / unit height, θ 69001 kgm2/m

Young’s modulus of walls, Ew 1.95 × 107 kN/m2

Shear modulus of walls, Gw 8.125 × 106 kN/m2

Young’s modulus of beams, Eb 2.3 × 107 kN/m2

Shear modulus of beams, Gb 9.58 × 106 kN/m2

Story height, h (m) 2.97 m

Area of beams, Ab 0.07 m2

Moment of inertia of beams, Ib 5.79 × 10−4 m4

Area of walls , Aw 1.92 m2

Moment of inertia of walls , Iw 8.85 m4

Coupled shear walls

Story 1-10 11-19 20-28

Depth of walls, s (m) 6.1 4.1 2.1

Clear span of openings, d (m) 3.7 3.7 3.7

Areas of walls, Ac (m2) 1.92 1.44 0.958

Moments of inertia of walls, Ic (m4) 8.85 3.35 0.722

Table 21: Geometric and material characteristics of the numerical example given in Fig.25

T = 3.90 sec,

which is very close to the approximate value provided by our method, the error is 3.86%.


In this section a symmetrical bracing system is considered the stiffnesses of which vary with the

height. We present the approximate calculation of the periods of vibration and the base shear

forces in the first modes of vibration.

The geometric and material characteristics of the structure are given in Table 21. and Fig. 26.

The approximate calculation was carried out in the following steps:

Step 1. Calculate of the stiffnesses of the lateral load-resisting subsystems [20].

Step 2. Calculate of the stiffnesses of the building by Eq.(4.29) at each story.

Step 3. Calculate the replacement stiffnesses of the beam, D, at the quarter of the height and

S at the third of the height.

Step 4. Calculate the periods of vibration by Eq.(5.58).

Step 5. Calculate the base shear forces (Eq. 5.77).

59

4.1y

z

wallcoupled shear walls

11 ´ 3.7

16.65

5.55

4.1

3.7

16.65

2.1y

z

2.1

3.7

6.1y

z

6.1

3.7

1-10 stories

11-19 stories

20-28 stories

Figure 26: Numerical example of Section 6.4.2

The periods of vibration and the internal forces were determined also by a FE code (ETABS).

In the example uniform mass distribution was considered.

Step 1. The lateral load-resisting subsystems at the first ten stories are similar to the structures

of Section 5.3. The replacement stiffnesses are given by Eq.(5.82). The stiffnesses of the coupled

shear walls at stories 11-19 are:

Dl = 2EwIw9−19 = 1.305× 108 kNm2,

D0 = 2EwAw9−19

(cw9−19

)2= 8.53 × 108 kNm2, (6.106)

S =

6EbIb2 (d+ s9−19)

2

d3h(1 + 12ρEbIb

Gb d2Ab

)

−1

+

(212EwIw

9−19

h2

)−1

−1

= 6.33 × 104 kN.

The stiffnesses of the coupled shear walls at stories 20-28 are:

60

Dl = 2EwIw20−28 = 2.817× 107 kNm2,

D0 = 2EwAw20−28

(cw20−28

)2= 6.29 × 108 kNm2, (6.107)

S =

6EbIb2 (d+ s20−28)

2

d3h(1 + 12ρEbIb

Gb d2Ab

)

−1

+

(212EwIw

20−28

h2

)−1

−1

= 3.50 × 104 kN.

Step 2. The stiffness matrices at the different stories are calculated from Eq.(4.29) which results

in:

[D0]1−10 =

0.69 0 0

0 3.51 0

0 0 992.8

× 109 kNm2, [S]1−10 =

74.8 0 0

0 0.203 0

0 0 55.52

× 106 kN,

[Dl]1−10 =

0 0 0

0 1.37 0

0 0 214.6

× 109 kNm2,

[D0]11−19 =

0.69 0 0

0 1.68 0

0 0 472.05

× 109 kNm2, [S]11−19 =

74.8 0 0

0 0.128 0

0 0 35.16

× 106 kN,(6.108)

[Dl]11−19 =

0 0 0

0 0.602 0

0 0 85.21

× 109 kNm2,

[D0]20−28 =

0.345 0 0

0 0.651 0

0 0 171.7

× 109 kNm2, [S]20−28 =

37.4 0 0

0 0.038 0

0 0 9.80

× 106 kN,

[Dl]20−28 =

0 0 0

0 0.711 0

0 0 30.0

× 109 kNm2.

Step 3. We approximate the stiffness distributions by continuous curves as a function of the

distance from the ground. Linear and second order curves are fitted to the elements of the stiffness

matrices at the third of the height of the stories with uniform stiffnesses (Fig.27), and thee re-

placement stiffness matrices are evaluated at the quarter and third of the total height. This results

in:

61

1/3H

S (1,1) S (2,2) S (3,3)

1/4H

Dl (1,1) Dl (2,2) Dl (3,3)

1/4H

D0 (1,1) D0 (2,2) D0 (3,3)

Figure 27: Stiffness distribution in the numerical example of Section 6.4.2

[D0] =

0.675 0 0

0 2.74 0

0 0 771.1

× 109 kNm2, [S] =

73.2 0 0

0 0.159 0

0 0 43.72

× 106 kN,(6.109)

[Dl] =

0 0 0

0 0.972 0

0 0 157

× 109 kNm2.

Step 4. The natural frequencies are calculated from Eq.(5.58):

[(H4

µ2Bi

[D0]−1

+H2

µ2Si

[S]−1

)−1

+µ2

Bi

H4[Dl]−ω2

mim [M ]

]

v0m

w0m

ψ0m

= 0, (6.110)

where [M ] is a diagonal matrix:

62

Period of vibration, T (sec)

mode direction Approximation ETABS Error [%]

1 x-y plane 7.984 8.31 -3.92

2 x-z plane 6.03 6.10 -1.15

3 torsional 5.65 5.98 -5.52

4 x-y plane 1.44 1.51 6.32

5 x-z plane 1.35 1.48 8.78

6 torsional 1.29 1.30 -0.77

Table 22: Comparison of the numerical and approximate results for the periods of vibration, (T )

mode excitation Approximation ETABS Error

1 (x-y plane) x-y 1045.7 981.86 kN 6.50%

2 ( x-z plane) x-z 1904.7 1757.5 kN 8.38%

Table 23: Internal forces of the numerical examle in Section 6.4.2.

[M ] =

1 0 0

0 1 0

0 0 θm

=

1 0 0

0 1 0

0 0 159.09

. (6.111)

By substituting µB1 = 3.52, µS1 = 0.5π (Table 6) into Eq.(6.110) we obtain the first three

eigenvalues.

ω2m1 = 1.235, ω2

m2 = 1.085, ω2m3 = 0.619. (6.112)

The periods of vibration, T are:

T1 =2π

ω1. (6.113)

The results of the approximate calculation and those of the ETABS code are given in Table 22 for

the first six periods of vibration. The maximum error is 8.78%.

Step 4. The seismic forces are determined using the Response Modal Analysis (Appendix

A.2). The calculation was carried out according to Eurocode 8. The spectral accelerations are

calculated from Eq.(5.89). The base shear forces in the first two modes of vibration are calculated

by Eq.(5.77). The results of the approximate calculation and those of the ETABS code are given

in Table 23. The maximum error is 8.38%.

63

7 Discussion

We presented the stiffnesses of the replacement beam of the stiffening system of building structures.

By using an energy approach we derived formulas which show the contribution of the stiffnesses of

the individual lateral load-resisting subsystems to the overall stiffnesses of the structure. We took

the shear deformation not only in the in-plane problem but also in torsion into account.

We worked out a simple approximate method for the calculation of the period of vibrations and

base internal forces of building structures subjected to earthquakes. We considered symmetrical

and unsymmetrical building structures stiffened by shear walls, coupled shear walls, trusses, frames

or cores. We solved the vibration problem in case of lateral, torsional and spatial lateral-torsional

vibration.

We presented a simple approximation to take the effect of varying stiffnesses into account.

Numerical examples were presented to show the utility and usefulness of the replacement beam.

References

[1] H. G. Allen. Analysis and design of structural sandwich panels. Pergamon Press: Oxford,

1969.

[2] A. K. Basu. Seismic design charts for coupled shear walls. J. Struct. Engrg., ASCE 1983;

109(2): 335-352.

[3] A. K. Basu, A. K. Nagpal and S. Kaul. Charts for Seismic Design of Frame-Wall Systems. J.

Struct. Engrg., ASCE 1984; 110(1): 31-46.

[4] H. Beck. Contribution to the analysis of coupled shear walls. Journal of the American Concrete

Institute 1962; 55: 1055-1069.

[5] Bishop, R. E. D and Jonson, D. C, The mechanics of vibration. Cambridge University Press,

London, 1960; 233-239.

[6] A. K. Chopra. Dynamics of Structures: Theory and application to Earthquake Engineering.

Prentice Hall, New Yersey, 1995; 434-438, 468-519.

[7] B. Csak, F. Hunyadi, Gy. Vertes. Foldrengesek hatasa az epıtmenyekre (In hungarian: Earth-

quake effects on buildings). Muszaki Konyvkiado, Budapest, 1981

[8] P. Csonka. Egyszerusıtett eljaras szelerokkel terhelt emeletes keretek szamıtasara (In hungar-

ian: Simplified analysis for the calculation of multistory frames subjected to wind load). MTA

Muszaki Tudomanyok Osztalyanak Kozlemenyei 1965; 35(1-4): 209-229.

[9] Eurocode 8. Design provisions for earthquake resistance of structures.

64

[10] A. Habibullah. ETABS. Three dimensional analysis of building systems. Computers and Struc-

tures Inc., USA, 1995

[11] D. J. Dowrick. Earthquake Resistant Design. John Wiley & Sons, New York, 1987

[12] S. Dunkerley: On the Whirling and Vibration of Shafts. Phil. Trans. Roy. Soc. A. 1894; 185:

279-360.

[13] Gy. Farkas . Magasepıtesi vasbetonszerkezetek, Muegyetemi Kiado, Budapest, 1998

[14] I. Hegedus and L. P. Kollar. Application of the sandwich theory in the stability analysis of

structures. In: Structural stability in engineering practice. Red.: L. Kollar. Spon: London,

etc., 1999.

[15] L. Kollar. A szel dinamiku hatasa magas epıtmenyekre. (In hungarian: The dynamic effect of

wind load on high rise building structures). Muszaki Konyvkiado, Budapest, 1979

[16] L. Kollar. Epıtmenyek meretezese foldrengesre (In hungarian: Earthquake resistant design of

buildings). Tervezesfejlesztesi es Technikai Epıteszeti Intezet, Budapest, 1990

[17] L. P. Kollar. Calculation of plane frames braced by shear walls for seismic load. Acta Technica

Acad. Sci. Hung., Budapest 1991; 104 (1-3): 187-209.

[18] L. P. Kollar. Flexural-torsional vibration of open section composite beams with shear defor-

mation., International Journal of Solids and Structures 2001; 38: 7543-7558.

[19] L. P. Kollar. Flexural-torsional buckling of open section composite beams with shear defor-

mation., International Journal of Solids and Structures 2001; 38: 7525-7541.

[20] A. Kopecsiri and L. P. Kollar. Approximate seismic analysis of building structures by the

continuum method. Acta Technica, Civil Eng. 1999; 108(3-4): 417-446.

[21] A. Kopecsiri and L. P. Kollar. Simple formulas for the analysis of symmetric (plane) bracing

structures subjected to earthquakes. Acta Technica, Civil Eng. 1999; 108(3-4): 447-473.

[22] A. Kopecsiri. Approximate seismic analysis of building structures by the continuum method.

Ph.D thesis, Technical University of Budapest, Budapest, 1997.

[23] H. Lamb and R. V. Southwell. The vibration of a Spinning Disk. Proc. Roy. Soc. A. 1921; 99:

272-80.

[24] Q. S. Li. Sability of tall buildings with shear-wall structures. Engineering Structures. 2000;

23: 1177-1185.

[25] S. C. Ng and J. S. Kuang. Triply coupled vibration of asymmetric wall-frame structures. J.

Struct. Engrg., ASCE 2000; 126(8): 982-987.

65

[26] G. Potzta. Magasepuletek meretezese foldrengesre a kontinuum modszer alapjan (In hun-

garian: Earthquake analysis of high-rise buildings by the continuum method). Diploma

work,Technical University of Budapest, Budapest, 1999

[27] G. Potzta. Epuletek kozelıto vizsgalata foldrengesre (In hungarian: Approximate earthquake

analysis of buildings). Scientific Student Paper, Technical University of Budapest, Budapest,

1999.

[28] R. Rosman. Dynamics and stability of shear wall building structures. Proc. Instn. Civ. Engs

1973; 55: 411-423.

[29] R. Rosman. Stability and dynamics of shear wall frame structures. Build. Sci. 1974; 9: 55-63.

[30] A. Rutenberg. An accurate approximate formula for the natural frequencies of sandwich

beams. Computers & Structures. 1979; 10: 875-878.

[31] P. Rozsa. Linearis algebra es alkalmazasai (In hungarian: Applications of linear algebra).

Budapest Tankonyvkiado, Budapest, 1997.

[32] A. Rutenberg. Approximate natural frequencies for coupled shear walls. Earthquake Engineer-

ing and Structural Dynamics. 1975; 4: 95-100.

[33] S.K. Skattum. Dynamic analysis of coupled shear walls and sandwich beams. PhD thesis,

California Institute of Technology, 1971.

[34] B. Stafford Smith and A. Coull. Tall building structures: Analysis and Design. John Willey

and Sons, Inc., New York etc., 1991.

[35] B. Stafford Smith and E Crowe. Estimating Periods of vibration of Tall Buildings. J. Struct.

Engrg., ASCE 1986; 112(5): 1005-1019.

[36] B. Stafford Smith and Y.-S. Yoon. Estimating seismic base shears of tall wall-frame buildings.

J. Struct. Engrg., ASCE 1991; 117(10): 3026-3041.

[37] L. Szeremi. Stiffening system of multistory buildings by the continuum model. Periodica Po-

litechnika Civ. Engrg 1978; 22(3-4): 205-218.

[38] T. Tarnai. Summation theorems. In: Structural stability in engineering practice. Red.: L.

Kollar. Spon: London, etc., 1999.

[39] S. P. Timoshenko and J. M. Gere, Theory of elastic Stability. 2nd edn. McGraw-Hill: New

York, 1961

[40] K. A. Zalka. Global Structural Analysis of Buildings. E & FN Spon, London, 2000.

[41] K. A. Zalka and G. S. T. Armer. Stability of large structures. Butterworth-Heinemann: Oxford,

1992

66

[42] K. A. Zalka. An analytical procedure for 3-dimensional eigenvalue problems. Building Research

Establishment Note, 1993; 32/93.

[43] K. A. Zalka. Dynamic analysis of core supported buildings. Building Research Establishment

Note. 1994; 127/94.

[44] K. A. Zalka. Equvivalent wall for frameworks for the global stability analysis.Building Research

Establishment Note, 1998; 33/98

[45] K. A. Zalka. Full-height buckling of frameworks with cross-bracing. Proc. Instn. Civ. Engrs.

Struct. & Bldgs. 1999; 134: 181-191.

[46] K. A. Zalka. Stress analysis of buildings under horizontal load. Proc. Instn. Civ. Engrs. Struct.

& Bldgs., 2000; 140: 179-186.

[47] K. A. Zalka. A simplified method for calculation of the natural frequencies of wall-frame

buildings. Engineering Structures, 2001; 23: 1544-1555.

[48] X. Wu and C. T. Sun, Simplified Theory for Composite Thin-Walled Beams. AIAA J., 1992;

30: 2945-2951.

A Earthquake analysis

We present two approximate method to perform the earthquake analysis of multistory building

structures.

A.1 Equivalent Lateral Force Procedure

The Equivalent Lateral Force procedure uses a simple approximation to determine a maximum

base shear force [34], [13]. The dynamic effect of ground motion is replaced by an equivalent static

load distributed along the height,. in a prescribed manner.

The equivalent static load for one concentrated mass is:

Qsi = CsG, (1.114)

where

- G is the weight of the structure,

- Cs is the earthquake parameter:

Cs =ah

gk, (1.115)

- ah the horizontal ground acceleration,

- g acceleration due to gravity,

67

- k depends on the type of the structure [11], [13].

For multistory buildings the equivalent static load is distributed along the height as follows:

Qsi = QsGizi

n∑i=1

Gizi

, (1.116)

where

- Qsi is the equivalent load acting at the height zi,

- zi is the distance of the i-th story from the bottom of the structure.

The method is simple and fast, however it can be very inaccurate.

A.2 Response Modal Analysis

The most frequently used method is the Response Modal Analysis. The method is discussed first

for a structure consisting of a single mass [6], [34]

A.2.1 Single-degree-of-freedom system

The equation of motion of a single-degree-of-freedom system (without damping) is:

M∂2u (t)

∂t2+Ku (t) = p (t) , (1.117)

where

u (t)- is the horizontal displacement of the structure with respect to the ground,

K−is the lateral stiffness of the system,

M - is the mass,

p(t)-external force,

in case of earthquake excitation:

p (t) = −M∂2ug (t)

∂t2, (1.118)

where ug (t)- is the ground motion given as a function of the time (t).

The investigation of the free vibration (p(t) = 0) results the complementary solution of Eq.(1.117):

M∂2uc

∂t2+Kuc = 0. (1.119)

The solution has the following form

uc = Aeiωt, (1.120)

where ω =√

KM is the circular frequency of vibration. The period of vibration of the system is

T =2π

ω. (1.121)

68

Using ω Eq.(1.117) can be written in the following form:

∂2u

∂t2+ ω2u = −∂

2ug

∂t2. (1.122)

For a given ground motion, ug(t) the deformation response, u(t) depends only the circular

frequency ω (or the period of vibration T ) of the system. However for practical purposes we are

interested only the maximum values of the response of the system:

SD = maxt

|u(t)| , (1.123)

SV = maxt

∣∣∣∣du(t)

dt

∣∣∣∣ ,

SA = maxt

∣∣∣∣d2u(t)

dt2

∣∣∣∣ .

For any given earthquake and also for damped systems these values can be calculated by solving

Eq. 1.122. The approximate relation between the responses are:

SV = ωSD, (1.124)

SA = ω2SD.

For different recorded earthquakes the spectral acceleration SA was determined in the design codes,

and average curves called design spectrum are worked out in the function of the period of vibration

and the damping ratio.

In the numerical examples our calculations were carried out according to Eurocode 8. The

spectral accelerations are given in the function of the period of vibration, [9]:

0 < Ti < TB SA = agSt

[1 +

Ti(ηβ − 1)

TB

], (1.125)

TB < Ti < TC SA = agStηβ, (1.126)

TC < Ti < TD SA = agStηβ

(TC

Ti

)k1

,

TD < Ti SA = agStηβ

(TC

TD

)k1(TD

Ti

)k2

.

where

SA− is the ordinate of the elastic response spectrum,

T− is the period of vibration of a linear single degree-of-freedom system,

ag− is the design ground acceleration for the reference return period,

β− is spectral acceleration amplification factor for 5% viscous damping,

TB, TC− are limits of the constant spectral acceleration branch,

TD− is value defining the beginning of the constant displacement range of the spectrum,

69

k1, k2− are exponents which influence the shape of the spectrum for a vibration period greater

than TC , TD

respectively,

S− is soil parameter,

η− is damping correction factor with reference value η = 1 for 5% viscous damping.

The values of the parameters are given in [9].

The equivalent static force acting on the structure:

fS = ku (t) = mω2u (t) ≈ mSA (1.127)

A.2.2 Multi-degree-of-freedom system

The equation of motion of a multi-degree-of-freedom system subjected to earthquake excitation

[6]:

[M ]∂2 u∂t2

+ [K] u = p (t) = − [M ] ι ∂2ug

∂t2, (1.128)

where

[M ]- is the mass matrix,

u- contains the independent displacements,

[K]- is the stiffness matrix,

ι- the influence vector represents the displacements of the masses resulting from static appli-

cation of a unit ground motion.

First we investigate the free vibration of the structure:

[M ]∂2 u∂t2

+ [K] u = 0 . (1.129)

Looking for the solution in the following form

u = φ eiωt,

we obtain the eigenvalue problem:

−ω2 [M ] φ + [K] φ = 0. (1.130)

which results N independent eigenvectors φn and N different ωn, where N is the number of

degrees of freedom of the system.

In Eq.(1.128) the displacement, u can be expressed as the linear superposition of the modal

vectors φ as follows

u (t) =N∑

n=1

φn qn (t) = [Φ] q (t) , (1.131)

70

where [Φ] contains the eigenvectors of the free vibration: We substitute Eq.(1.131) into Eq.(1.128)

and we multiply the equation by ΦT , which results in:

[Φ]T

[M ] [Φ]d2 q (t)dt2

+ [Φ]T

[K] [Φ] q (t) = [Φ]T p (t) . (1.132)

The eigenvectors φn are orthogonal to the [M ] , and [K] matrices, thus [Φ]T

[M ] [Φ] and

[Φ]T

[K] [Φ] are diagonal matrices. We choose the eigenvectors such a way that the following

relationship holds:

[]T

[M ] [] = [E] . (1.133)

It can be shown [6] that the following relationship is satisfied:

[Φ]T [K] [Φ] =⟨ω2

1...ω2n...ω

2N

⟩. (1.134)

Because of the orthogonality the differential equation system (1.132) reduces to N independent

differential equations:

d2qn (t)

dt2+ ω2

nqn (t) = pn (t) , (1.135)

which is equal to the equation of motion of a one-degree-of-freedom system (Eq.1.122), where

pn (t) =φT

n

φTn [M ] φn

p (t) = − φTn [M ] ι

φTn [M ] φn

∂2ug

∂t2= −Γn

∂2ug

∂t2. (1.136)

Eqs.(1.135) and (1.136) result in:

qn(t) = Γnun(t), (1.137)

where un(t) is the displacement function of a one-degree-of-freedom system with eigenfreqvency,

ωn subjected to the ground motion ug (Eq. 1.122). The peak response of the structure in the n

-th mode of vibration (using Eq. 1.123) [6] is

|un (t)|max = φn |qn (t)|max = φn ΓnSDn. (1.138)

The total displacement can be approximated by combining the modal responses. The modal

maximum values will not occur at the same time. However for design purposes superpositions are

recommended in the different codes.

A.2.3 Spatial problem

For unsymmetrical-plane structures the equation of motion has the same form as Eq.(1.128)

[M ]∂2 u∂t2

+ [K] u = p (t) = − [M ] ι ∂2ug

∂t2,

however it has three times more degrees of freedom:

71

[M ]

[M ]

[Θ]

∂2

∂t2

uyuzuΘ

+

[Kyy] [Kyz] [KyΘ]

[Kzy] [Kzz] [KzΘ]

[KΘy] [KΘz] [KΘΘ]

uyuzuΘ

= − [M ] ι ∂

2ug

∂t2,

(1.139)

where the influence vector ι depends on the direction of the earthquake excitation.

If the building is subjected to ground acceleration in the y − z plane:

ι=

100

, (1.140)

in case of ground motion in the x− z plan

ι=

010

, (1.141)

and for torsional excitation

ι=

001

. (1.142)

The method of solving Eq.(1.139) is similar to the method described in the previous section.

A.2.4 Continuum

The equation of motion for a continuous beam is:

m∂2u (x, t)

∂t2+ ku (x, t) = p (t) . (1.143)

The Taylor series expansion of the displacement function is:

u (x, t) =

∞∑

r=1

φr (x) qn (t) . (1.144)

We substitute Eq.(1.144) into Eq.(1.143), and we multiple it by φn. The integration between 0

and H results in:

∞∑

n=1

H∫

0

φnmφr

d2qn (t)

dt2dx+

∞∑

n=1

H∫

0

φnkφrqn (t) dx =

H∫

0

φTnp (t) dx. (1.145)

The eigenvectors φn are orthogonal to each other:

72

H∫

0

φnφrdx = 0. (1.146)

Because of the orthogonality Eq.(1.145) becomes

d2qn (t)

dt2+ ω2

nqn (t) = pn (t) , (1.147)

which is equal to the equation of motion of a one-degree-of-freedom system (Eq. 1.122). pn (t) is

obtained as follows:

pn (t) =

H∫

0

φTnp (t) dx. (1.148)

The solution in the n-th mode can be obtained using the Response Spectrum Analysis as in

section A.2.1. The approximate value of the displacement is the sum of th the peak responses of

the considered modes.

In case of spatial vibration of the continuum the equation of motion

[M ]∂2 u∂t2

+ [K] u = p (t) = − [M ] ι ∂2ug

∂t2, (1.149)

m

m

Θ

∂2

∂t2

uy(x)

uz (x)

uΘ (x)

+

kyy kyz kyΘ

kzy kzz kzΘ

kΘy kΘz kΘ

uy(x)

uz (x)

uΘ (x)

= − [M ] ι ∂

2ug

∂t2. (1.150)

Three independent displacement functions belong to every mode of free vibration. It is denoted

by φn which has the following form:

φn =

φyn

φzn

φΘn

, (1.151)

where φyn, φzn, φΘn are the mode shapes of the vibration along to the y axis, the z axis and the

rotational vibration, respectively. We can use the modal vectors φn and the circular frequency

ωn to solve Eq.(1.150) as in Section A.2.2.

A.2.5 Internal forces

In the Response Modal Analysis equivalent load is determined in each mode of vibration. For

in-plane vibration, when the ground motion is in the plane of the vibration, the horizontal forces

acting on a multistory building are [6]

73

x

H F (x)n

V

M

Figure 28: The horizontal seismic force (Fn(x)), and the internal forces at the bottom of the

canilever (V, M).

fn (t) = [K] un (t) = ω2n [M ] φn qn (t) = ω2

n [M ] φn Γnun (t) , (1.152)

fn = |fn (t)|max = Γn [M ] φn SAn =φT

n [M ] ιφT

n [M ] φn

[M ] φn SAn.

The horizontal force acting on a continuum

fn (x) = Γnmφn (x)SAn =

H∫0

mφn (x) dx

H∫0

mφ2n (x) dx

mφn (x)SAn.

For the spatial vibration of a continuum

fn = Γn [M ] φn SAn =φT

n [M ] ιφT

n [M ] φn

[M ] φn SAn (1.153)

For example in case of earthquake in the y direction

fn =

Vyn

Vzn

Mtn

=

H∫0

mφyn (x) dx

H∫0

(mφ2

yn (x) +mφ2zn (x) + Θφ2

Θn (x))dx

m

φyn (x)

φzn (x)

φΘn (x)

SAn

If the distribution of the horizontal seismic forces, fn is known, the internal forces in the

n-th mode of vibration can be calculated by performing a simple static analysis on the building

(Fig.28).

The total design value can be determined by one of the modal combination rules (given in the

codes), for example by the square-root-of-sum-of-squares rule

74

R '(

N∑

n=1

R2n

)1/2

.

B Replacement stiffnesses for spatial problem

In this section of the final version of the thesis we will give the derivation of Eq.(4.29).

C Solutions of vibration problem of replacement beams

In this section we present an analysis for the free vibration of a cantilever beam with uniform and

varying stiffnesses.

The differential equation of the in-plane free vibration of a sandwich beam with uniform stiff-

nesses is [14], ??:

D0Dl

SwV I

B − (D0 +Dl)wIVB = −mω2w, (3.154)

where m is the mass, D0, Dl and S are the stiffnesses of the sandwich beam, ω is the circular

frequency, and w is the displacement function consisting of two parts[14]:

w = wD + wS . (3.155)

This differential equation has constant coefficients and can be solved analytically ??, when the

stiffnesses vary with the height, the coefficients are not constant, and we may obtain solutions by

using the Rayleigh-Ritz method.

We developed analysis for the free vibration of beams with both uniform and varying stiffnesses

based on the Rayleigh-Ritz method. First the analysis of in-plane and spatial vibration of sandwich

beams with uniform stiffnesses are presented (Section ??), then the in-plane vibration analyses of

Timoshenko-beams and Csonka-beams (Section C.1.2).

On the basis of the presented analyses we developed a computer code written in MATLAB to

determine the eigenfrequencies and internal forces of replacement beams.

C.1 “Exact” solution by the Rayleigh-Ritz method

C.1.1 Uniform stiffnesses

Plane problem. A sandwich beam undergoes both flexural and shear deformations, thus its

mode shape consists of two parts [14]:

w = wD + wS . (3.156)

The relationship between wD and wS is:

75

S w′

S = −(D0 w′′

D)′. (3.157)

In case of constant stiffnesses Eq.(3.157) can be rearranged as:

wS = −D0

Sw′′

D. (3.158)

We approximate wD by an n-th order polynomial:

wD =

n∑

i=0

w0i xi = xT w0 , (3.159)

where w0i -s are yet unknown constants.

This function must satisfy the geometrical boundary conditions, which are at the bottom of

the beam (x = 0):

w = 0, w′

D = 0, and w′ = 0. (3.160)

(Note that the force boundary conditions do not have to be satisfied in the Rayleigh-Ritz

method.)

The remaining constants can be determined from the stationary condition of the potential

energy, which can be written in the following form:

δ (U +K)

δw0i= 0, i = 2, ..., n. (3.161)

where U is the strain energy and K is the kinetic energy [14]:

U =1

2

∫ H

0

S (w′

S)2

+D0( w′′

D)2 +Dl( w′′)2dx, (3.162)

K =1

2

∫ H

0

mω2w2dx, (3.163)

ω is the circular frequency, m is the mass per unit length, D0, Dl are the local and global bending

stiffnesses and S is the shear stiffness of the structure.

After a straightforward algebraic manipulation we obtain from Eqs. (3.156) trough (3.163) the

following equation:

[A] w0 = [B] w0ω, (3.164)

where

76

[A] =1

H2

D20

S

0 0T

0[

(i+2)(i+1)ij(j+2)(j+1)i+j−1 Hi+j+1

]

+

+ (Dl +D0)

4H

2 (i+ 2)Hi+1

T

2 (j + 2)Hj+1[

(i+2)(i+1)(j+2)(j+1)i+j+1 Hi+j+1

]

+ (3.165)

+1

H4

D20Dl

S2

0 0T

0[

(i+2)(i+1)i(i−1)(j−1)j(j+2)(j+1)i+j−3 Hi+j+1

]

−

− 2

H2

D0Dl

S

0

2 (i+ 2) (i+ 1) iHi+1

T

0[

(i+2)(i+1)i(i−1)(j+2)(j+1)i+j+1 Hi+j+1

]

, (3.166)

[B] = m

H5

5

Hi+5

i+5

T

Hj+5

j+5

[Hi+j+5

i+j+5

]

+

1

H4

D20m

S2

0 0T

0[

(i+2)(i+1)(j+2)(j+1)i+j+1 Hi+j+5

]

−

− 2

H2

D0m

S

0

(i+2)(i+1)Hi+5

i+3

T

0[

(i+2)(i+1)i+j+3 Hi+j+5

]

Spatial problem. A sandwich beam has flexural and shear deformations in y direction, in z

direction, and also in torsion, its mode shape consists of two parts:

φ = φD + φS , (3.167)

where φD , and φS contain the following displacement functions:

φD =

vD(x)

wD(x)

ϕD(x)

, φS =

vS(x)

wS(x)

ϕS(x)

.

The relationship between φD and φS is [14]:

[S] φ′S = −([D0] φ′′D)′. (3.168)

In case of constant stiffnesses

φS = [S]−1

[D0] φ′′D (3.169)

We approximate φD by n-th order polynomials:

φD (x) =

vD(x)

wD(x)

ϕD(x)

=

∑ni=0 vi x

i

∑ni=0 wi x

i

∑ni=0 ϕi x

i

= (3.170)

=

1 x x2 ...

0 0 0 0

0 0 0 0

0 0 0 0

1 x x2 ...

0 0 0 0

0 0 0 0

0 0 0 0

1 x x2 ...

v0w0ϕ0

= [X ]

T φ0

77

where φ0 contains yet unknown constants.

The displacement function must satisfy the geometrical boundary conditions, which are at the

bottom of the beam (x = 0):

φ (0) = φD + φS = 0

φ′D = 0

φ′ (0) = φ′D + φ′S = 0 (3.171)

By satisfying these boundary conditions the displacement functions become

φD (x) =

xT 0T 0T

0T xT 0T

0T 0T xT

v0w0ϕ0

= [X ]T φ0 , (3.172)

φS (x) = − [S]−1

[D0]

xTS 0T 0T

0T xTS 0T

0T 0T xTS

v0w0ϕ0

= [X ]

TS φ0 ,

where

xT=

[x2

xi+2

T], (3.173)

xTS =

[0(i+ 2) (i+ i)xi

T].

The derivatives of φD

φ′D = [X ]T1 φ0 ,

φ′′D = [X ]T2 φ0 , (3.174)

φ′′′D = [X ]T3 φ0 ,

φIVD = [X ]

T4 φ0 ,

where matrices [X ]1 , [X ]2 , [X ]3 , [X ]4 have the same form as matrix [X ] , and contain the

derivatives of vector x .The remaining constants can be determined from the stationary condition of the potential

energy, which can be written as

δ (U −K)

δw0i= 0, i = 2, ..., n, (3.175)

where U is the strain energy and K is the kinetic energy [14] (if we neglect GIt)

78

U =1

2

H∫

0

[(φ′S

)T[S] φ′S +

(φ′′D

)T[D0] φ′′D +

(φ′′S + φ′′D

)T[Dl]

(φ′′S + φ′′D

)]dx,

(3.176)

K =1

2

H∫

0

(φS + φD)T ω2 [M ] (φS + φD) dx

Substituting Eq.(3.169) into Eq.(3.175) results in an eigenvalue problem with eigenvector φD

and eigenvalue ω.

[A] φ0 = [B] φ0 ω (3.177)

where

[A] =1

H2[D0] [S]

−1[D0] ⊗

0 0

0 (i+2)(i+1)ij(j+2)(j+1)i+j−1 Hi+j+1

+

+ ([Dl] + [D0]) ⊗

4H 2 (i+ 2)Hi+1

2 (j + 2)Hj+1 (i+2)(i+1)(j+2)(j+1)i+j+1 Hi+j+1

+1

H4[D0] [S]−1 [Dl] [S]−1 [D0] ⊗

0 0

0 (i+2)(i+1)i(i−1)(j−1)j(j+2)(j+1)i+j−3 Hi+j+1

− (3.178)

− 2

H2[Dl] [S]−1 [D0] ⊗

0 2 (i+ 2) (i+ 1) iHi+1

0 (i+2)(i+1)i(i−1)(j+2)(j+1)i+j+1 Hi+j+1

,

[B] = [M ] ⊗

H5

5Hi+5

i+5

Hj+5

j+5Hi+j+5

i+j+5

+

1

H4[D0] [S]−1 [M ] [S]−1 [D0] ⊗

0 0

0 (i+2)(i+1)(j+2)(j+1)i+j+1 Hi+j+5

−

− 2

H2[M ] [S]

−1[D0] ⊗

0 (i+2)(i+1)Hi+5

i+3

0 (i+2)(i+1)i+j+3 Hi+j+5

.

where ⊗ refers to the Kronekker multiplication [31].

C.1.2 Varying stiffnesses

An analysis is presented below to determine the period of in-plane vibration of Timoshenko-beams

and Csonka-beams. The equation of beams capable bending or shear deformation only can be

derived from those of the latter two cases, hence the derivation for those are not presented. The

sandwich beam with varying stiffnesses was investigated numerically. Only lateral vibration was

considered.

Csonka-beam. We approximate the mode shape of a Csonka-beam by an n-th order polynomial:

w =n∑

i=0

wi xi, (3.179)

79

where wi-s are yet unknown constants.

This function must satisfy the geometrical boundary conditions, which are at the bottom of

the beam (x = 0):

w = 0 and w′ = 0. (3.180)

(Note that the force boundary conditions do not have be satisfied in the Rayleigh-Ritz method.)



δ (U −K)

δwi= 0, i = 2, ..., n, (3.181)


U =1

2

∫ H

0

Dl( w′′)2 + S (w′)

2dx, (3.182)

K =1

2

∫ H

0

mω2w2dx. (3.183)

Here ω is the circular frequency, m is the mass per unit length, and stiffnesses Dl and S may vary

with the height.

After a straightforward algebraic manipulation we obtain from Eqs. (3.179) trough (3.183) the

following equation:

[aij ] wj = λ [bij ] wj , λ =m

DlH2ω2, (3.184)

where [aij ] and [bij ] are given in Table 24 for the three different stiffness distributions given in Fig.

29.

Equation (3.184) is a linear eigenvalue problem with eigenvector wj and eigenvalue λ. The

circular frequency and the period of vibration can be calculated from the eigenvalues λ (Eq. 3.184)

as follows :

ω =

√λ

H2

√Dl

m, T =

2π

ω. (3.185)

Timoshenko-beam. A Timoshenko-beam has flexural and shear deformations, its mode shape

consists of two parts:

w = wD + wS . (3.186)

The relationship between wD and wS is [14]:

S w′

S = −(D0 w′′

D)′. (3.187)

80

Stiffness distribution

Csonka-beam ϕ = H√

SDl

Fig. 29aaij = ij

[ϕ2

i+j−1 + ϕ2(c−1)i+j + (i−1)(j−1)

i+j−3 + (d−1)(i−1)(j−1)i+j−2

],

bij = 1i+j−1

Fig. 29caij = ij

[ϕ2

i+j−1 + ϕ2(c−1)i+j+1 + (i−1)(j−1)

i+j−3 + (d−1)(i−1)(j−1)i+j−1

],

bij = 1i+j−1

Fig. 29baij = ij

[ϕ2

i+j−1 + ϕ2(1−c)i+j+1 − 2ϕ2(1−c)

i+j + (i−1)(j−1)i+j−3 + (1−d)(i−1)(j−1)

i+j−1 − 2 1−di+j−2

],

bij = 1i+j−1

Timoshenko-beam α = H√

SDo

Fig. 30

A = ij (i− 1) (j − 1)[

1i+j−3 − (1−k)

i+j−2 + (1−k)2

α2(i+j−3) + (c−1)(1−k)2

α2(i+j−2)

]

B = (1) − ij (i− 1) (j − 1) (1−k)(j−2)α2

[1

i+j−4 − c+k−2i+j−3 − (1−k)(c−1)

i+j−2

]

C = (1) − ij (i− 1) (j − 1) (1−k)(i−2)α2

[1

i+j−4 − c+k−2i+j−3 − (1−k)(c−1)

i+j−2

]

D = ij(i−1)(j−1)(j−2)(i−2)α2

[1

i+j−5 + 2k−c−3i+j−4 + (1−k)(2c+k−3)

i+j−3 + (1−k)2(c−1)i+j−2

]

E =1+ 1

α4 (1−k)2(c−1)2(j−1)(i−1)

i+j+1 + (1−k)2ijα4(i+j−1)

F = ij(j−1)α4(i+j−2) −

α2j(j−1)−ij(j−2)(2−c−k)(1−k)+(1−k)(c−1)j(j−i)(i−1)α4(i+j−1) +

+α2j(j−2)(2−c−k)+i(j−1)(j−2)(c−1)(1−k)2+(1−k)(c−1)(2−c−k)j(i−i)(j−2)α4(i+j)

G = ij(i−1)α4(i+j−2) −

α2i(i−1)−ij(i−2)(2−c−k)(1−k)+(1−k)(c−1)i(i−i)(j−1)α4(i+j−1) +

+α2i(i−2)(2−c−k)+j(i−1)(i−2)(c−1)(1−k)2+(1−k)(c−1)(2−c−k)i(j−i)(i−2)α4(i+j)

H = ij(i−1)(j−1)α4(i+j−3) − (2−c−k)ij[(i−1)(j−2)+(j−1)(i−2)]

α4(i+j−2) +

+ ij(i−2)(j−2)(2−c−k)2−(i−1)(j−1)(1−k)(c−1)[i(j−2)+j(i−2)]α4(i+j−1) +

+ (i−2)(j−2)(c−1)(1−k)(2−c−k)[i(j−1)+j(i−1)]α4(i+j) +

+ (i−2)(j−2)(i−1)(j−1)(c−1)2(1−k)2[i(j−1)+j(i−1)]α4(i+j+1) +

if i = 2, j = 2 : bij = E, aij = A

if i = 2, j 6= 2 : bij = E + F, aij = A+B

if i 6= 2, j = 2 : bij = E +G, aij = A+ C

if i 6= 2, j 6= 2 : bij = E + F +G+H,

aij = A+B + C +D

Table 24: Parameters of Eq. (7) for a Csonka-beam and those of Eq. (17) for a Timoshenko-beam.

We approximate wD by an n-th order polynomial:

wD =

n∑

i=0

wi xi, (3.188)

and wS by an (n− 2)-th order polynomial:

wS =

n−2∑

i=0

zi xi. (3.189)

The geometrical boundary conditions at the bottom of the beam (x = 0) are:

81

H H

H H

x

x

H Hx

Dl

dDl

cS

S

( )úû

ùêë

é --

-+ x

H

dx

H

dD

l

1211 2

2

( )úû

ùêë

é --

-+ x

H

cx

H

cS

1211 2

2

úû

ùêë

é -+ 2

2

11 x

H

dD

l úû

ùêë

é -+ 2

2

11 x

H

cS

S

cSkDl

Dl

Dl

kDl

cS

S

a)

b)

c)

úû

ùêë

é -- x

H

dD

l

11 úû

ùêë

é -- x

H

cS

11

S

Figure 29: Distribution of stiffnesses for a Csonka-beam

S

1

S

c

x

H

D0

kD0

úû

ùêë

é -- x

H

c

S

11

1

úû

ùêë

é -- x

H

kD

110

H

Figure 30: Distribution of stiffness D0 and the inverse of stiffness S for a Timoshenko-beam

wD = 0, w′

D = 0, wS = 0. (3.190)

If the stiffnesses S and D0 are uniform, zi can be determined directly from wi using Eqs.(3.187)

to (3.189). If Eq.(3.187) is satisfied, wD is a polynomial and S is not constant, as a rule, ws must

be a function which is not a polynomial. For the sake of a simple solution we assume that D0 and

the inverse of S vary linearly (Fig.30). By so doing, both wD and wS are polynomials (Eqs.3.188

and 3.189) and zi can be determined as a function of wi from Eq.(3.187). The results are given in

Table 25.



82

i

0z0 = c

1z1 = 2w2

1−kH − 3w3

2z2 = w2

(1−k)(c−1)H2 + 3w3

[3 1−k

H − 2 c−1H

]

i > 2zi = wi

(1−k)(c−1)(i−1)2

H2 + wi+1 (i+ 1)[

(1−k)iH − (c−1)(i−1)

H

]− wi+2 (i+ 1) (i+ 2)

Table 25: Determination of the constants,zi of the shear deformation. c and k are defined in Fig

4.

δ (U −K)

δwi= 0, i = 2, ..., n, (3.191)


U =1

2

∫ H

0

D0( w′′

D)2 + S (w′

S)2dx, (3.192)

K =1

2

∫ H

0

mω2w2dx, (3.193)

ω is the circular frequency and m is the mass per unit length, as before.

Equations (3.188) through (3.193) results in the following algebraic equation:

[aij ] wj = λ [bij ] wj , λ =m

D0H2ω2, (3.194)

where [aij ] and [bij ] are given in Table 24. The circular frequency and the period of vibration can

be calculated from the eigenvalues λ (Eq. 3.194):

ω =

√λ

H2

√D0

m, T =

2π

ω. (3.195)

83

approximate analysis of building structures subjected to earthquakes

Documents

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shear walls

department of mechanics

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simplest replacement

numerical examples