Dynamic behavior of ancient frames consisting of classical ... · Arch Appl Mech (2016)...

Arch Appl Mech (2016) 86:2033–2052DOI 10.1007/s00419-016-1170-4

ORIGINAL

George T. Michaltsos · Ioannis G. Raftoyiannis

Dynamic behavior of ancient frames consisting of classicalcolumns with architrave

Received: 30 October 2015 / Accepted: 7 August 2016 / Published online: 1 September 2016© Springer-Verlag Berlin Heidelberg 2016

Abstract This paper deals with the dynamic behavior of ancient frames consisting of two classical columnswith an architrave resting on them. This free-to-rock plane frame is studied through a simple analytical processafter some generally accepted simplifications. Furthermore, an analytical approach method for the solution ofthe fully nonlinear system of equations of motion (including the vertical component of earthquake excitation)is proposed, while the corresponding static and dynamic conditions against rocking and sliding are studied indetail. Finally, some characteristic examples are presented and useful conclusions are drawn.

Keywords Seismic response · Classical columns · Ancient frames · Columns with architrave

1 Introduction

A problem that has attracted the attention of several investigators during the last three decades is the studyof the dynamic stability of blocks against sliding, rocking or sliding–rocking, under the action of strongground motion. This type of problem usually appears during the seismic excitation of tanks, heavy machines,monuments, and mainly, of ancient structures, classical temples or monuments that are found mostly in Greeceand Southern Italy.

A complete list and discussion on relative studies performed up to 1980 have been presented by Ishiyama[1], Spanos and Koh [2] and Shi et al. [3]. Lately, a large number of studies have been published using eitheranalytical [4–15] or experimental methods [16–19], or through the use of regular or modified FEM [20–23].A detailed list of recent relative studies has been presented by Konstandinidis and Makris [11].

Although the aforementioned problem seems easy, an analytical treatment is very difficult because of itshighly nonlinear nature since the equations of motion change when the angle of rotation reverses sign and alsobecause at the instant of the sign’s change an impact takes place. In a recent publication of the authors [24],the problem of the stability of a single multidrum column has been studied in detail.

The present work deals with the dynamic stability of an ancient plane frame consisting of two multidrumancient columns with an architrave resting on them due to seismic excitation. Bibliography is rather pooron this problem. The system is considered free to rock and is studied through a simple analytical processafter some generally accepted simplifications. Furthermore, an analytical approach method for the solution ofthe fully nonlinear equation of motion (including the vertical earthquake excitation) is presented, while thecorresponding static and dynamic conditions against rocking and sliding are studied in detail.

Energy dissipation will be due to sliding or impact. It should be noted at this stage that the main goal of thepresent work is to evaluate the stability for the frame against rocking (overturning) or sliding and not to studythe motion of such frames (that can be easily done by just employing the equations derived herein). Thus, the

G. T. Michaltsos · I. G. Raftoyiannis (B)Department of Civil Engineering, National Technical University of Athens, 15780 Athens, GreeceE-mail: [email protected]

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2034 G. T. Michaltsos, I. G. Raftoyiannis

Fig. 1 Ancient frame and two characteristic possible modes of failure

Fig. 2 A column part with height hi

frame is considered to lose its stability at the initial phase of seismic action prior to impact. On the other hand,the sliding surfaces are already smooth enough, and hence, neither the coefficient of friction nor the contactsurfaces do increase in time due to wear.

In this work, only the most unfavorable modes of failure are presented and studied in detail in order toascertain the dynamic stability of this structural system while the rest possible modes are proved to be lessimportant. Finally, some representative examples are presented and useful conclusions are drawn.

2 Basic considerations and kinematic relations

Let us consider the ancient plane frame of Fig. 1, consisting of two classical columns with an architrave restingon them. For a strong ground motion, the frame may either collapse due to excessive rocking or sliding orit may return to its original equilibrium state. Two characteristic possible modes of failure are schematicallyshown in Fig. 1.

For the geometry of a column part with height hi and top width bo and bottom width bi (see Fig. 2), thefollowing geometric relations hold:

AE = bo + bi2

(1a)

Dynamic behavior of ancient frames 2035

2Ri =√h2i +

(bo + bi

2

)2

(1b)

2Ri =√h2i +

(bi2

)2

(1c)

where over-lined symbols refer to the midpoint of the top surface of each drum. Moreover, the angles ai andai are defined as follows:

tan ai = AE

hi= b0 + bi

2hi(1d)

tan ai = bi2hi

(1e)

For a more clear presentation and in order to facilitate the analysis since the equations of motion change whenthe angle of rotation reverses sign, the positive and negative rotation of the blocks will be treated separately inthe following sections.

2.1 Case a: positive seismic motion

In this case, the ground is moving to the right causing a counter clockwise rotation of the block. Angles θ2 andϕ are expressed with respect to angle θ1, while the horizontal displacements u1, u2, u3 and the vertical onesυ1, υ2, υ3 are determined as follows (Fig. 3):

Projection on the horizontal axis gives:

u1 = u2 = u3 (2a)

On the other hand, it is υ1 − υ3 = L · sin ϕ, and since ϕ is very small, one can write:

ϕ = υ1 − υ3

L(2b)

Fig. 3 Deformed state of a possible mode of failure (positive seismic motion)


Taking into account that counter clockwise rotations are positive, one has:

u1 = bi − 2Ri sin(ai + θ1)

u2 = b j − 2R j sin(a j + θ2)

u3 = b j

2− 2R j sin(a j + θ2)

with: θ1 < 0, and θ2 < 0 (2c)

Easily, one can also find:

υ1 = 2Ri [cos(ai + θ1) − cos ai ]

υ2 = 2R j[cos(a j + θ2) − cos a j

]υ3 = 2R j

[cos(a j + θ2) − cos a j

](2d)

Introducing Eqs. (2c) into (2a) and taking into account that the angles ai , a j , θ1, θ2 are small, one may write:bi − 2Ri (ai + θ1) = b j − 2R j (a j + θ2), or:

θ2 = Aiθ1 + Bi

where: Ai = Ri

R j, Bi = (b j − bi ) − 2R ja j + 2Riai

2R j(2e)

2.2 Case b: negative seismic motion

In this case, the ground is moving to the left causing a clockwise rotation of the block. Through a similarprocess, one can determine the following relations (Fig. 4):

u1 = u2 (2a′)

ϕ = υ2 − υ3

L(2b′)

Fig. 4 Deformed state of a possible mode of failure (negative seismic motion)


u1 = bi − 2Ri sin(ai − θ1) (2c′)u2 = b j − 2R j sin(a j − θ2)

u3 = bi2

− 2Ri sin(ai − θ2)

with: θ1 > 0, and θ2 > 0

and similarly:

υ1 = 2Ri [cos(ai − θ1) − cos ai ] (2d′)υ2 = 2R j

[cos(a j − θ2) − cos a j

]υ3 = 2R j [cos(ai − θ2) − cos ai ]

θ2 = A jθ1 + Bj (2e′)

where: A j = Ri

R j, Bj = (bi − b j ) − 2Riai + 2R ja j

2R j

3 The complete equations of motion

Assuming a positive earthquake action for both fx and fy , the drums will initially rotate by a negative angle(θ < 0) and if overturning does not occur, they will reverse to a positive rotation and this back-and-forthmotion will continue until stable equilibrium.

The Lagrange’s equation for rigid body motion of the above system shown in Fig. 5 is [25]:

d

dt

[∂K

∂θ1

]− ∂K

∂θ1+ ∂�

∂θ1= 0 (3a)

where K is the total kinetic energy and � the potential of the external forces. Equation (3a) holds in the limitstates associated with overturning and sliding.

Fig. 5 Deformed state of a possible mode of failure for positive earthquake action


By virtue of relations (2), the displacements of the block gravity centers can be determined as follows:

uGi = fx − u12

= fx − bi2

+ Ri sin(ai + θ1)

uGj = fx − u22

= fx − b j

2+ R j sin(a j + θ2)

ua = fx − u1 = fx − bi + 2Ri sin(ai + θ1)

υGi = fy + υ1

2= fy + Ri [cos(ai + θ1) − cos ai ]

υGj = fy + υ2

2= fy + R j


]υa = fy + υ1 + υ3

2= fy + Ri [cos(ai + θ1) − cos ai ] + R j


](3b)

From the above equations, one obtains:

uGi = fx + Ri θ1 cos(ai + θ1)

uG j = fx + R j θ2 cos(a j + θ2)

ua = fx + 2Ri θ1 cos(ai + θ1)

υGi = f y − Ri θ1 sin(ai + θ1)

υGj = f y − R j θ2 sin(a j + θ2)

υa = f y − Ri θ1 sin(ai + θ1) − R j θ2 sin(a j + θ2) (3c)

The total kinetic energy of block i will be:

Ki = 1

2JGi θ

21 + 1

2mi (u

2Gi + υ2

Gi )

= 1

2JGi θ

21 + 1

2mi

[f 2x + R2

i θ21 cos

2(ai + θ1) + 2 fx θ1Ri cos(ai + θ1)]

+ 1

2mi

[f 2y + R2

i θ21 sin

2(ai + θ1) − 2 f y θ1Ri sin(ai + θ1)]

which finally concludes to the following expression:

Ki = 1

2Joi θ

21 + mi Ri fx θ1 cos(ai + θ1) − mi Ri fy θ1 sin(ai + θ1) + mi

f 2x + f 2y2

(4a)

Similarly, for block j , one obtains:

K j = 1

2Joj θ

22 + m j R j fx θ2 cos(a j + θ2) − m j R j fy θ2 sin(a j + θ2) + m j

f 2x + f 2y2

(4b)

and for block a:

Ka = maf 2x + f 2y

2+ maR2

i θ21

2+ maRi fx θ1 cos(ai + θ1)

−ma fy[Ri θ1 sin(ai + θ1) + R j θ2 sin(a j + θ2)

](4c)

In the above expressions, the higher-order terms can be neglected due to their small contribution, while onecan set:

Joi = JGi + mi R2i

Joj = JGj + m j R2j (4d)

Finally, the potential of the external forces is:

� = migυGi + m j gυGj + magυa (4e)


Taking into account Eq. (2e), one can write:

∂Ki

∂θ1= Joi θ1 + mi Ri fx cos(ai + θ1) − mi Ri fy sin(ai + θ1)

∂K j

∂θ1= Joj A

2i θ1 + m j R j Ai fx cos(a j + θ2) − m j R j Ai f y sin(a j + θ2)

∂Ka

∂θ1= maR

2i θ1 + maRi fx cos(ai + θ1) − ma fy

[Ri sin(ai + θ1) + R j Ai sin(a j + θ2)

](5a)

and after differentiation with respect to time:

d

dt

(∂Ki

∂θ1

)= Joi θ1 − mi Ri fx θ1 sin(ai + θ1) + mi Ri fx cos(ai + θ1)

−mi Ri fy θ1 cos(ai + θ1) − mi Ri fy sin(ai + θ1)

d

dt

(∂K j

∂θ1

)= Joi A

2i θ1 − m j R j A

2i fx θ1 sin(a j + θ2) + m j R j Ai fx cos(a j + θ2)

−m j R j A2i f y θ1 cos(a j + θ2) − m j R j Ai f y sin(ai + θ1)

d

dt

(∂Ka

∂θ1

)= maR

2i θ1 − maRi fx θ1 sin(ai + θ1) − ma fy

[Ri θ1 cos(ai + θ1) + R j A

2i θ1 cos(a j + θ2)

]+maRi fx cos(ai + θ1) − ma fy


](5b)

On the other hand, one can determine the following terms:

∂Ki

∂θ1= −mi Ri fx θ1 sin(ai + θ1) − mi Ri fy θ1 cos(ai + θ1)

∂K j

∂θ1= −m j R j Ai fx θ2 sin(a j + θ2) − m j R j Ai f y θ2 cos(a j + θ2)

∂Ka

∂θ1= −maRi fx θ1 sin(ai + θ1) − ma fy

[Ri θ1 cos(ai + θ1) + R j Ai θ2 cos(a j + θ2)

](5c)

and finally:

∂�

∂θ1= −migRi sin(ai + θ1) − m j gR j Ai sin(a j + θ2) − mag


](5d)

Thus, the equation of motion Eq. (3a) can be readily formulated.

4 Linear approximation

As already mentioned above, for small angles θ and a, one can write:

cos(a + θ) ∼= 1, sin(a + θ) ∼= a + θ, for θ < 0

cos(a − θ) ∼= 1, sin(a − θ) ∼= a − θ, for θ > 0 (6)

therefore, for θ1 < 0 and taking into account Eqs. (5), (3a) becomes:

I θ1 + fx (mi Ri + m j R j Ai + maRi )

− f y[mi Ri (ai + θ1) + m j R j Ai (a j + θ2) + maRi (ai + θ1) + ma R j (a j + θ2)

]−g

[mi Ri (ai + θ1) + m j R j Ai (a j + θ2) + maRi (ai + θ1) + ma R j (a j + θ2)

] = 0

where: I = Joi + Joj + maR2i , θ2 = Aiθ1 + Bi , and Ai , Bi from Eq. (2e) (7a)


Equation (7a) can be written as follows:

I θ1 + fx (mi Ri + m j R j Ai + maRi )

− ( f y + g)(mi Ri + m j R j A

2i + maRi + ma R j A

2i

)θ1

− ( f y + g)[mi Riai + m j R j Ai (a j + Bi ) + maRiai + ma R j Ai (a j + Bi )

] = 0 (7b)

For θ1 > 0, one can find through a similar way:

I θ1 + fx (mi Ri + m j R j A j + maRi )

+ ( f y + g)(mi Ri + m j R j A

2j + ma Ri + maR j A

2j

)θ1

− ( f y + g)[mi Riai + m j R j A j (a j − Bj ) + ma Riai + maR j A j (a j − Bj )

] = 0

and A j , Bj from Eq. (2e)’ (7c)

Thus, a linearized equation of motion Eq. (7c) is obtained.

5 Solution of the equations of motion

Equation (7) can be rewritten in the following form:

I θ1 − (f y + g

)(�1θ1 + �2) = −�3 fx (8a)

where g is the acceleration of gravity and the terms �i (i = 1, 2, 3) are:

�1 = (mi Ri + m j R j A

2i + maRi + ma R j A

2i

)for θ1 < 0

�1 = −(mi Ri + m j R j A

2j + ma Ri + maR j A

2j

)for θ1 > 0

�2 = mi Riai + m j R j Ai (a j + Bi ) + maRiai + ma R j Ai (a j + Bi ) for θ1 < 0

�2 = mi Riai + m j R j A j (a j − Bj ) + ma Riai + maR j A j (a j − Bj ) for θ1 > 0

�3 = mi Ri + m j R j Ai + maRi for θ1 < 0

�3 = mi Ri + m j R j A j + maRi for θ1 > 0 (8b)

Equation (8a) can be also written as follows:

I θ1 − �1( f y + g)(θ1 + �2

�1

)= −�3 fx and finally:

θ1 − ω2o

g( f y + g)

(θ1 + �2

�1

)= −�3

Ifx

where: ω2o = g�1

I(9)

5.1 The free vibrating frame

When the system is free to rotate aboutO orO ′ (Fig. 5), it vibrates freely following the formof the correspondingmode according to the following equation:

θ − ω2oθ = ω2

o�, with � = �2/�1 (10a)

and ωo, from eq (9). The solution of Eq. (10a) is given by

θ = c1 Sinh ωot + c2 Cosh ωot − � (10b)

With initial conditions θ(0) = θo, θ (0) = θo, Eq. (10b) becomes:

θ = θo

ωoSinh ωot + (θo + �) Cosh ωot − � (10c)


Fig. 6 Free vibrating frame

Let us assume next that the frame is released from position (1) as shown in Fig. 6, with initial angle θo andinitial angular velocity θo = 0, and oscillates without impact (no energy loss). Equation (10c) for t = T/4gives: (θo + �) Cosh ωoT

4 − � = 0 or finally:

T = 4

ωo· Cosh−1

(�

� + θo

)(11)

from which the period of the free vibrating frame without energy loss can be obtained (a case physically notaccepted). This period depends of course on the initial angle θo through a nonlinear relation.

5.2 The forced vibrating frame

5.2.1 Vertical ground motion is ignored

Ignoring the vertical component of ground motion, Eq. (9) becomes:

θh − ω2oθh = ω2

o� − �3

I· fx (12a)

The general solution of the above equation is given by:

θh(t) = c1 Sinh ωot + c2 Cosh ωot − � + Fx (t) (12b)

where Fx (t) is a particular solution of Eq. (12a). In order to address the complicated expression of fx (t), andto determine Fx (t), one may employ the Lagrange’s method or the so-called method of variation of parameters(see application in Sect. 6.2.1).

With time conditions θh(to) = θo, and θh(to) = θo, Eq. (12b) concludes to the following:

θh(t) ={

− [θo − Fx (to)] sinhωoto +[θo − Fx (to)

]Cosh ωoto

ωo

}Sinh ωot

+{[θo − Fx (to)] Cosh ωoto −

[θo − Fx (to)

]sinhωoto

ωo

}Cosh ωot − � + Fx (t) (12c)

If the above conditions are the initial conditions, i.e., for to = 0, one obtains:

θh(t) =[θo − Fx (0)

]ωo

Sinh ωot + [θo − Fx (0)] Cosh ωot − � + Fx (t) (12d)


5.2.2 The complete equation of motion including vertical ground motion

Taking into account the vertical ground motion fy , one has:

θ − ω2oθ = ω2

o� +(

ω2o

g�k f − �3

I

)· fx + ω2

o

gk f fxθ

with: k f = f y/ fx (13a)

In order to solve nonlinear Eq. (13a), one can apply an approach method proposed by Kounadis [26], withproven accuracy and rapid convergence, especially if one uses as first approach a function close to the finalsolution of the equation.

Ignoring the nonlinear term ω2og k f fxθ , one gets the equation:

θ − ω2oθ = ω2

o� +(

ω2o

g�k f − �3

I

)· fx (13b)

which has the form of Eq. (12a) and it can be solved in a similar way.Introducing, as a first approach, the solution θh of Eq. (13b) into the last nonlinear term of Eq. (13a), one

obtains:

θ − ω2oθ = ω2

o� +(

ω2o

g�k f − �3

I

)· fx + ω2

o

gk f fxθ (13c)

This last can be solved in a similar way as Eq. (12a).

6 Rocking and sliding conditions

There are two types of conditions that determine the stability of the system, the static and the dynamicones. Static conditions are those necessary for the lift-up of the frame, which conclude to a relation givingthe minimum fx required for the frame to start rocking. Dynamic conditions are those necessary to causeoverturning of the frame at time to without sliding. Numerical solutions for the determination of these limitstates (rocking and sliding) are presented in the works of Baggio and Trovaluisci [27] and Ferris and Tin-Loi[28]. In the present work, each joint has different characteristics, i.e., contact area and normal force, and hence,it is assumed that rocking and sliding will not occur simultaneously.

6.1 The static conditions

Let us consider the frame of Fig. 7, which is composed by the movable parts i and j of the columns and thearchitrave a, being in the critical mode of motion.

6.1.1 Rocking condition

Taking equilibrium of moments about the point O , one obtains:

mi fxhi2

− mi (g − f y)bi2

+ ma fx

(hi + ha

2

)− ma(g − f y)

L + bi2

+m j fx

(hi − h j

2

)− m j (g − f y)

(L + bi

2

)= 0

which finally gives:

fxg

≤ mi (hi + k f bi ) + ma[2hi + ha + k f (L + bi )

] + m j[2hi − h j + k f (2L + bi )

]mi + ma(L + bi ) + m j (2L + bi )

(14a)


Fig. 7 Static conditions—equilibrium of overturning forces

6.1.2 Sliding condition

The sliding condition is Fh ≤ μFv , where μ is the required friction coefficient, Fh is the resultant of thehorizontal forces and Fv is the resultant of the vertical forces. This gives:

mi fx + ma fx + m j fx ≤ μ[mi (g − f y) + ma(g − f y) + m j (g − f y)

]which concludes to the following expression:

μ ≥ fxg − k f fx

(14b)

where k f is a coefficient determined from the masses mi , m j and ma in Eq. (14b).

6.2 The dynamic conditions

If the rocking condition Eq. (14a) is not valid, the motion of a slender frame is initiated by acceleration fx .

6.2.1 The overturning condition

This inequality specifies the acceleration fx as a function of g, required for the frame to start tilting. Since theframe starts to move, overturning depends not only on the magnitude of fx but also on the duration of motion.

Applying the Lagrange’s method, in order to determine a particular solution, one obtains the followingsystem:

c′1(t)Sinh ωot + c′

2(t)Cosh ωot = 0

c′1(t)Cosh ωot + c′

2(t)Sinh ωot = ω2o� − �3

Ifx

which has the solution:

c′1(t) = 1

ωo

(ω2o� − �3

Ifx

)Cosh ωot

c′2(t) = − 1

ωo

(ω2o� − �3

Ifx

)Sinh ωot


and then one obtains:

c1(t) = �Sinh ωot − �3

ωo I

∫fx Cosh ωotdt

c2(t) = −� Cosh ωot + �3

ωo I

∫fxSinh ωotdt

Therefore, the searching particular solution is given by the following relation:

Fx (t) = −� + �3

ωo I

[−Sinh ωot

∫fx Cosh ωotdt + Cosh ωot

∫fxSinh ωotdt

](15a)

From the above, one can get:

Fx (t) = �3

I

[− Cosh ωot

∫fx Cosh ωotdt + Sinh ωot

∫fxSinh ωotdt

](15b)

For initial conditions θo = θo = 0, the solution of Eq. (12d) becomes:

θ(t) = − Fx (0)

ωoSinh ωot − Fx (0) Cosh ωot + Fx (t)

with: Fx (0) = −� + �3

ωo I

∫fxSinh ωotdt

∣∣∣∣t=0

Fx (0) = −�3

I

∫fx Cosh ωotdt

∣∣∣∣t=0

(15c)

Considering in addition the amplitude pg of the acceleration fx , the following relation is also valid:

fx = pg · ¨fx (15d)

Let us assume that at time t = to the frame under the earthquake’s action takes its critical position, i.e., theframe gets an angle θ(to) = −a and a speed θ (to) = 0, while it stops rotating counterclockwise and it startsto rotate clockwise.

In this case, Eq. (12d) becomes:

θ(to) = − Fx (0)

ωoSinh ωoto + [−a − Fx (0)] Cosh ωoto + Fx (to) or

− Fx (0)

ωoSinh ωoto − Fx (0) Cosh ωoto + Fx (to) + a − a · Cosh ωoto = 0 (15e)

The above, because of Eqs. (15a), (15b), (15c) and (15d), becomes:

−�3 pgωo I

Sinh ωoto

∫¨fx Cosh ωotdt

∣∣∣∣t=0

+ �3 pgωo I

Cosh ωoto

∫¨fxSinhωotdt

∣∣∣∣t=0

+ �3 pgωo I

Sinhωoto

∫¨fx Cosh ωotdt

∣∣∣∣t=to

− �3 pgωo I

Cosh ωoto

∫¨fxSinh ωotdt

∣∣∣∣t=to

+ a · (1 − Cosh ωoto) − � = 0

which concludes to the following condition:

pgg

= ωo I · [� − a · (1 − Cosh ωoto)]

g · �3 · D

where: D = Sinh ωoto

(∫¨fx Cosh ωotdt

∣∣∣∣t=to

−∫

¨fx Cosh ωotdt

∣∣∣∣t=0

)

− Cosh ωoto

(∫¨fxSinh ωotdt

∣∣∣∣t=to

−∫

¨fxSinh ωotdt

∣∣∣∣t=0

)(15f)


Fig. 8 Usual studied cases

The above equation specifies the acceleration’s amplitude as a fraction of g, required for the overturning ofthe frame.

Usually, four cases are studied (Fig. 8) in practice. For these four cases, one can search for the relationbetween pg and to.

a) The step pulse.

In this case, it is fx = pg and¨fx = 1. From the second of Eq. (15f), one finds:

D = −1 − Cosh ωotoωo

(16a)

b) The half-sine wave pulse.

In this case, it is fx = pg sin�t and ¨fx = sin�t with � = πto.

From the second of Eq. (15f), one can determine D = ωo sin�to−�·Sinh ωoto�2+ω2

o, but at t = to it will be

sin�to = 0, and finally:

D = � · Sinh ωoto�2 + ω2

o(16b)

c) The quarter-cosine wave pulse.

In this case, it is fx = pg cos�t and ¨fx = cos�t with � = π2to

.


Fig. 9 Sliding and forces

From the second of Eq. (15f), one can determine D = ωo(cos�to− Cosh ωoto)�2+ω2

o, but at t = to it will be

cos�to = 0 and finally:

D = ωo · Cosh ωoto�2 + ω2

o(16c)

d) The damped half-sine wave pulse.

In this case, it is: fx = pge−βt sin�t , and ¨fx = e−βt sin�t with � = πto.

At t = to, it will be sin�to = 0, cos�to = 1, and therefore, one can finally determine:

D = −−2β�ωoe−βto + 2β�ωo · Cosh ωoto − �(β2 + �2 + ω2o)Sinh ωoto

β4 + 2β2(�2 − ω2o) + (�2 + ω2

o)2 (16d)

6.2.2 The sliding condition

One can search for the condition against sliding at time t = 0, prior to rocking.Due to rotations θ1 and θ2, one has the following displacements (see Fig. 9):

υi (t) = Riθ1(t)

υ j (t) = R jθ2(t) (17a)

Taking into account Eq. (2d), one can set for the inertia force components:

Phi (0) = mi fx (0) + mi Ri θ1(0) cos ai = mi fx (0) + mihi2

θ1(0)

Phj (0) = m j fx (0) + m j R j θ2(0) cos a j = m j fx (0) + m jh j

2Ai θ1(0)

Pha(0) = ma fx (0)

Pυi (0) = mig − mik f fx (0) − mi Ri θ1(0) sin ai = mig − mik f fx (0) − mibi2

θ1(0)

Pυ j (0) = m jg − m jk f fx (0) − m j R j θ2(0) sin a j = m j g − m jk f fx (0) − m jb j

2Ai θ1(0)

Pυa(0) = ma[g − k f fx (0)

](17b)


Equilibrium of moments about point O (see Fig. 9) gives

Phi (0) · hi2

+ Phj (0) ·(hi − h j

2

)+ Pha(0) ·

(hi + ha

2

)

−Pυi (0) · bi2

− Pυ j (0) ·(L + bi

2

)− Pυa(0) · L + bi

2+ Joi θ1(0) + Joj Ai θ1(0) = 0 (17c)

Introducing the force components Ph , Pυ from Eq. (17b) into Eq. (17c), one finally obtains:

θ1(0) =mi

hi2 + m j

(hi − h j

2

)+ ma

L+bi2 + mik f

bi2 + m jk f

(L + bi

2

)+ mak f

L+bi2

Q· fx (0)

−mi + m j + ma

Q· g

with: Q = mi

(hi2

)2

+ m j Aih j

2

(hi − h j

2

)− m j

(bi2

)2

− m j Aib j

2

(L + bi

2

)+ Joi + Joj

(17d)

In order to avoid sliding at t = 0, the following condition must be fulfilled: μ ≥∑

Ph(0)∑Pυ(0) , or because of

Eq. (17b):

μ ≥ Phi (0) + Phj (0) + Pha(0)

Pυi (0) + Pυ j (0) + Pυa(0)(17e)

where θ1(0) required to compute the force components Ph , Pυ is taken from Eq. (17d).

7 Numerical examples and discussion

As stressed above, the main goal of the present work is to evaluate the dynamic stability of the frame againstrocking (overturning) or sliding and not to study the motion of the frame. From most available accelerograms,it becomes clear that the observed maximum instantaneous periods are smaller than 0.70 s either for usual orfor extreme ground motions. Therefore, the corresponding instantaneous frequency is higher than ∼9.00 s−1.These values T ≤ 0.70 and � ≥ 9.00 will be considered next as the critical ones.

From the most representative accelerograms, one can see that an earthquake with amplitude pg = 6 m/s2

is a rather unfavorable well-defined earthquake for analysis of the temples in Greece and Southern Italy.In the following, the behavior of the columns of temples of Zeus and Apollo under the action of a sinusoidal

type earthquake without damping and with strong damping with coefficient β = 1, amplitude pg = 6 m/s2,� = 8s−1 or T = 0.785s and coefficients k f = 0, 0.2, 0.5 and 0.8 will be studied.

In Fig. 10, one can see the two representative cases of ancient frames studied herein.Each frame consists from two columns of height 6m, at distance 2.70m between each other, with an

architrave resting on them with dimensions 2.70 × 0.85 × 1.20 m.The difference between the above two cases is that the columns of the first case consist of two drums

(blocks) while the columns of the second case of three drums (see Fig. 10).

7.1 Columns with two drums

Let us consider now the ancient frames shown in Fig. 10a, with columns consisting of two drums. The studiedpossible modes of instability are shown in Fig. 11. Note that the architrave rotates and translates in mode 1due to its geometry and contact points (see also Figs. 5 and 7). Note also that modes 3 and 4 are not similarsince the architrave rotates and translates in a different way for the same as above reasons.

From the frame data and the corresponding equations of motion, Table 1 is formulated.In Table 2, one can see a list with the parameters required to study the frame’s stability.Applying the relations given in paragraph 6, one can determine the results shown in Table 3. In this table,

the minimum values of fx/g andμ required for loss of stability are listed for each one of the considered modesshown in Fig. 11.


Fig. 10 Two representative column cases: a two drums and b three drums

Fig. 11 Possible modes of instability for two-drum columns

Table 1 Geometrical and mass inertia data of the frames with two drums

bo bi hi mi Ri Ri ai ai acr IGi ei ISi

One drum 0.90 1.00 3 575 1.573 1.521 0.307 0.165 0.327 474 1.473 1722Two drums 0.90 1.10 6 1280 3.041 3.013 0.165 0.091 0.189 3945 2.900 14715

Table 2 Parameters required for stability study of the frames with two drums

1 (2–2) 2 (1–1) 3 (1–2) 4 (2–1)

Ai 1 1 0.517 1.934Bi 0 0 0.011 −0.021A j 1 1 0.517 1.394Bj 0 0 −0.011 0.021I 36552 5391 18305 23784�1 12517 4244 4782 12544�2 1890 1133 1139 1882�3 10145 3046 4147 8065� 0.151 0.267 0.238 0.150ω2o 3.424 7.871 2.612 5.274

Table 3 Static and dynamic stability factors for the frames with two drums

Modes Static conditions Dynamic conditions

Rocking fx/g > Sliding μ > Rocking fx/g > Sliding μ >

1 (2–2) 2.016 0.60 1.447 1.18672 (1–1) 1.272 0.60 1.624 0.98983 (1–2) 0.635 0.60 2.683 0.83864 (2–1) 2.854 0.60 1.325 0.7470


Fig. 12 Mode 3: a influence of β on pg = fx/g and b influence of � on pg = fx/g

Fig. 13 Possible modes of instability for three-drum columns

In order to determine the static conditions, the following values of the ground motion characteristics havebeen considered: fx = 6 m/s2, � = 8 s−1 and β = 0.05.

From Table 3, one can draw the following conclusions:Static conditions are fulfilled. The ratio fx/g represents the horizontal mass inertia required for rocking.

Mode 3 is the most unfavorable against rocking, but the corresponding dynamic condition for rocking is notfulfilled for this mode. As long as the frame starts rocking (because of the static condition), dynamic slidingwill prevail (because of the dynamic condition) since μ > 0.8386 is required, while the usual values for μare: 0.75–0.85.

In Fig. 12, one can see the influence of the parameters β and � on the dynamic condition of rocking. Fromthe above plots, one can see that the frequency� has a strong influence of on the dynamic condition of rockingwhile the influence of damping β is not significant.


Table 4 Geometrical and mass inertia data of the frames with three drums

bo bi hi mi Ri Ri ai ai acr IGi ei ISi

One drum 0.90 0.97 2 371 1.104 1.029 0.437 0.238 0.456 151 0.987 512Two drums 0.90 1.04 4 799 2.058 2.017 0.238 0.129 0.260 1127 1.952 4173Three drums 0.90 1.10 6 1280 3.041 3.013 0.165 0.091 0.189 3945 2.900 14715

Table 5 Parameters required for stability study of the frames with three drums

1 (3–3) 2 (2–2) 3 (1–1) 4 (1–3) 5 (1–2) 6 (2–3) 7 (3–1) 8 (2–1) 9 (3–2)

Ai 1 1 1 0.3629 0.5364 0.6765 2.7741 1.8642 1.4787Bi 0 0 0 0.0156 0.0134 0.0065 −0.0430 −0.0250 −0.0096A j 1 1 1 0.3629 0.5364 0.6765 2.7742 1.8642 1.4787Bj 0 0 0 −0.0156 −0.0134 −0.0065 0.0430 0.0250 0.0096I 36843 11667 1984 16250 5645 22286 22575 8020 26237�1 12614 6489 2498 3015 2618 7005 13595 6843 12219�2 1918 1371 930 910 902 1393 1928 1413 1884�3 10242 4901 1688 2702 2161 5920 7444 4028 8750� 0.152 0.211 0.372 0.302 0.344 0.199 0.142 0.206 0.154ω2o 3.424 5.512 12.590 1.855 4.638 3.143 6.022 8.533 4.657

Table 6 Static and dynamic stability factors for the frames with three drums

Modes Static conditions Dynamic conditions

Rocking fx/g > Sliding μ > Rocking fx/g > Sliding μ >

1 (3–3) 2.016 0.60 1.448 1.1882 (2–2) 1.528 0.60 1.516 1.0823 (1–1) 0.994 0.60 1.889 0.8044 (1–3) 0.197 0.60 4.435 0.7595 (1–2) 0.580 0.60 2.632 0.7346 (2–3) 1.077 0.60 1.967 0.9687 (3–1) 3.191 0.60 1.378 1.2298 (2–1) 2.054 0.60 1.482 1.1429 (3–2) 2.545 0.60 1.321 1.244

7.2 Columns with three drums

Let us consider now the ancient frame shown in Fig. 10b, with columns consisting of three drums. The possiblemodes of instability studied are shown in Fig. 13. Note that the architrave rotates and translates in a differentway in modes 1, 4 and 7, 5 and 8, 6 and 9 for the same as in Fig. 11 reasons.

From the frame data and the corresponding equations of motion, Table 4 is formulated.In Table 5, one can see a list with the parameters required for the study of the frame’s stability.Applying the relations given in paragraph 6, one can determine Table 6 showing the minimum values of

fx/g andμ required for stability loss for each of the considered modes presented in Fig. 13 with fx = 6 m/s2,� = 8 s−1 and β = 0.05 for the static conditions.

From Table 6, one can draw the following conclusions:Static conditions are fulfilled. Mode 4 is the most unfavorable against rocking, but the dynamic condition

for this mode is not fulfilled. The frame will suffer dynamic sliding since μ > 0.759 is required, while theusual values for μ are: 0.75–0.85. Modes 5 and 3 are also unfavorable but not likely to occur since mode 4will prevail.

In Fig. 14, one can see the influence of the parameters β and � on the dynamic condition of rocking. Fromthe above plots, one can verify the negligible influence of damping β and the strong influence of the frequency� on the dynamic condition of rocking as in the previous case of Fig. 12.

8 Conclusions

From the ancient frame models studied above, one can draw the following conclusions:


Fig. 14 Mode 4: a influence of β on pg = fx/g and b influence of � on pg = fx/g

• Ancient frames consisting of two columns with an architrave resting on them are more stable againstoverturning compared to single columns.

• The frames with columns containing two drums are more stable than the ones containing three or moredrums due to the reduced friction required for sliding.

• The most unfavorable mechanisms of failure are the ones involving sliding.• Even if the static conditions are not fulfilled, the corresponding dynamic conditions show that the framesare stable against overturning.

• The most likely-to-slide surfaces are not the ones located at the base of the frame but those located atintermediate levels due to the reduced friction required for sliding.

• The smoother the contact surfaces between drums are, the much higher is the danger of column failure dueto drum sliding since the required friction is smaller.

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