" Inertial Reduction in Function of Time "
Research paper. Part One
SMART STRUCTURES
" Inertial Reduction in Function of Time "
Prepared By
Engineer Jaime Alberto Montoya Castaño
National University of Colombia
September 2021
Table of Contents
List of Figures………………………………………………………………………………3
0. Preface …………………………………………………………………………………..4
1. Classic equation of motion for a building subjected to an external force…. 5-10
2. SSI, Soil Structure Interaction equation of motion for a building subjected to an external force...11
3. Inertial Reduction Equation in Function of Time...12-13
List of Figures
Figure 1. (a) Two-story shear frame; (b) Forces acting on the two masses.
Figure 2. Effective earthquake forces.
Figure 3. The “ Bathtub “ Soil-Structure Interaction (SSI) model.
Figure 4. The Electromagnetic Frame.
0. Preface
In light of the dawn of the smart structures concept, where the buildings will have a central brain to control the surrounding ecosystem in a very efficient way, I would like to introduce the research about the new use of the electromagnetic force to help the structures to improve their response under external dynamic forces and earthquakes.
Since we already have the technology to provide a very fast alarm to activate the building protection system under a fast track protocol, we can now use the electromagnetic force to reduce the mass of the building during the earthquake in order to improve the building total response to the event and to protect the lives and assets of the people.
Of course we have been using the electromagnetic force in our buildings in the form of electrical supplies, communications system, HVAC system and fire protection system, but now we are suggesting the use of the force to make the structure more ductile and resilient during an earthquake.
Classic equation of motion for a building subjected to an external force
We first formulate the equations of motion for the simplest possible Multiple Degrees of Freedom MDF system, a highly idealized two-story frame subjected to external forces p1(t) and p2(t) ( Fig. 1). The beams and floor systems are considered rigid in flexure.
Figure 1. (a) Two-story shear frame; (b) Forces acting on the two masses.
We consider the mass as concentrated at the floor levels. We assume that a linear viscous mechanism represents the energy dissipation in the structure.
The number of independent displacements required to define the displaced positions
of all the masses relative to their original equilibrium position is called the number of degrees
of freedom. The two-story frame of Fig. 1, with lumped mass at each floor level,
has two DOFs: the lateral displacements u1 and u2 of the two floors in the direction of the
x-axis.
The forces acting on each floor mass mj are shown in Fig. 1b
By using Newton’s second law of motion then gives for each mass: ( F = ma )
pj − fSj − fDj = mj ¨ uj or mj ¨ uj + fDj + fSj = pj (t) (1.1)
Equation (1.1) contains two equations for j = 1 and 2, and these can be written in matrix
form:
m1 0 Ü1 + fD1 + fS1 = P1(t)
0 m2 Ü2 + fD2 + fS2 = P2 (t) (1.2)
Equation (1.2) can be written compactly as
m¨u + fD + fS = p(t) (1.3)
Assuming linear behavior, the elastic resisting forces fS are next related to the floor
displacements u. For this purpose we introduce the lateral stiffness kj of the j th story; it
relates the story shear Vj to the story deformation or drift, j = uj − uj−1, by
Vj = kj Delta j (1.4)
The story stiffness is the sum of the lateral stiffnesses of all columns in the story. For a
story of height h and a column with modulus E and second moment of area Ic, the lateral
stiffness of a column with clamped ends, implied by the shear-building idealization, is
12E Ic/h3. Thus the story stiffness is
kj = Sum 12E Ic / h3 (1.5)
With the story stiffnesses defined, we can relate the elastic resisting forces fS1 and
fS2 to the floor displacements, u1 and u2. The force fS1 at the first floor is made up of two
contributions: f a S1 from the story above, and f b S1 from the story below. Thus
fS1 = f bS1+ f aS1
which, after substituting Eq. (1.4) (Vj = kj Delta j ) and noting that Delta 1 = u1 and Delta 2 = u2 − u1,
becomes
fS1 = k1u1 + k2(u1 − u2) (1.6a)
The force fS2 at the second floor is
fS2 = k2(u2 − u1) (1.6b)
Observe that f aS1 and fS2 are equal in magnitude and opposite in direction because both
represent the shear in the second story. In matrix form Eqs. (1.6a) and (1.6b) are :
fs1 = k1+k2 -k2 u1
fs2 = -k2 +k2 u2 or fs = ku (1.7)
Thus the elastic resisting force vector fS and the displacement vector u are related through
the stiffness matrix k for the two-story shear building.
The damping forces fD1 and fD2 are next related to the floor velocities ˙ u1 and ˙ u2.
The j th story damping coefficient cj relates the story shear Vj due to damping effects to
the velocity ˙j associated with the story deformation by
Vj = c Delta j (1 .8 )
Therefore, fD1 = c1 ˙ u1 + c2( ˙ u1 − ˙u2) fD2 = c2( ˙ u2 − ˙u1) (1.9)
In matrix form Eq. (1.9) is :
fD1 = c1+c2 -c2 ů1
fD2 = -c2 +c2 ů2 or fd = ců (1.10)
The damping resisting force vector fD and the velocity vector ů are related through the
damping matrix c for the two-story shear building.
We now substitute Eqs. (1.7) and (1.10) into Eq. (1.3) to obtain
m¨u + c˙u + ku = p(t) (1.11)
This matrix equation represents two ordinary differential equations governing the displacements u1(t) and u2(t) of the two-story frame subjected to external dynamic forces p1(t) and p2(t). Each equation contains both unknowns u1 and u2. The two equations are therefore coupled and in their present form must be solved simultaneously.
Now we consider an external force due to a ground motion generated by an earthquake.
Therefore P(t) becomes Üg(t) which represents the ground motion.
The displacement of the ground is denoted by Üg(t) , the total (or absolute) displacement of the mass mj by utj , and the relative displacement between this mass and the ground by uj . At each instant of time these displacements are related by
utj (t) = ug(t) + uj (t) (1.12)
Such equations for all the N masses can be combined in vector form:
ut (t) = ug(t)ɩ + u(t) (1.13)
where ɩ is a vector of order N with each element equal to unity.
However, the inertia forces fI are related to the total accelerations ¨ut of the
masses, therefore
fI = m¨ut (1.14)
thus the equation of motion of a building subjected to an earthquake is:
m¨u + c˙u + ku = −mɩ¨ ug(t) (1.15)
Equation (1.15) contains N differential equations governing the relative displacements
uj (t) of a linearly elastic MDF system subjected to ground acceleration ¨ ug(t). The stiffness
matrix in Eq. (1.15) refers to the horizontal displacements uj and is obtained by the static
condensation method to eliminate the rotational and vertical DOF of the nodes; hence this k is known as the lateral stiffness matrix.
Figure 2. Effective earthquake forces
The ground motion can therefore be replaced by the effective earthquake forces:
peff(t) = −mɩ¨ ug(t) (1.16)
In this general approach the total displacement of each mass is expressed as its displacement usj due to static application of the ground motion plus the dynamic displacement uj relative to the quasi-static displacement:
utj (t) = usj(t) + uj (t) or ut (t) = us(t) + u(t) (1.17)
The quasi-static displacements can be expressed as us(t) = ιug(t), where the influence
vector ι represents the displacements of the masses resulting from static application of a
unit ground displacement; thus Eq. (1.17) becomes
ut (t) = ιug(t) + u(t) (1.18)
The equations of motion are obtained as before, but now the Eq. (1.18) is used
m¨u + c˙u + ku = −mι ¨ ug(t) (1.19)
Now the effective earthquake forces are
peff(t) = −mι ¨ ug(t) (1.20)
The spatial distribution of the effective earthquake forces peff(t) is defined by s =mι. This force distribution can be expanded as a summation of modal inertia force distributions sn:
mɩ = Sum sn = Sum TTmPhi (1.21)
where, TT = Ln/Mn Ln = Phi t m ɩ Mn = Phit m Phin (1.22)
The contribution of the nth mode to mɩ is :
Sn = TTmPhin (1.23)
The contribution of the nth mode to the nodal displacements u(t) is
un(t) = φnqn(t) = TTnφnDn(t) (1.24)
The equivalent static forces associated with the nth-mode response are fn(t) = kun(t).
Where, fn(t) = snAn(t) (1.25)
The nth-mode contribution rn(t) to any response quantity r (t) that can be expressed as a linear combination of the structural displacements u(t) is determined by static analysis
of the structure subjected to external forces fn(t). If r st n denotes the modal static response, the static value (indicated by superscript “st”) of r due to external forces sn,then
rn(t) = rn st An(t) (1.26)
Therefore, the total response is the summary of rn(t):
Sum rn(t) = Sum rn st An(t) (1.27)
2. SSI, Soil Structure Interaction equation of motion for a building subjected to an external force.
We are using the “Bathtub” model in order to couple the structure stiffness matrix with the foundation stiffness matrix. Therefore the above-ground and subterranean portions of the
structure are modeled, horizontal and vertical soil springs are included, the far
end of each spring is fixed against a rigid “bathtub” element, and the horizontal
foundation input motion, uFIM, or free-field motion, Üg(t), is applied to the rigid element.
Figure 3. The “ Bathtub “ SSI model.
By calculating the dynamic stiffness for the group of piles of the foundation, we can obtain the foundation stiffness matrix. Therefore, the equation of motion considering the soil-structure interaction (SSI) effect is:
m¨u + c˙u + (ksu + kfu) = −mι ¨ ug(t) (2.1)
Where ks is the structure stiffness matrix and kf is the foundation stiffness matrix.
By considering the SSI effect , we are expecting a period lengthening for the entire structure, expressed as:
Ť/ t = SQR(1+ks/kx + ksh*h/kyy ) (2.2)
The period lengthening will allow the description of the frequency equation in a more precise way.
3. Inertial Reduction Equation in Function of Time.
We are going to use the electromagnetic force to counterbalance the external force produced by an earthquake in the structure.
By using the Lorenz-Maxwell force equation:
F = q(E+v*B) (3.1)
Where q is the charge, E is the electric field, v is the velocity and B is the magnetic field.
We are going to calculate the total force on the system required to match the external force
Fe =TT mPhinAn generated by the earthquake.
By considering four capacitors in each floor, two capacitors per column and considering the magnetic flux density vector B opposed to the external force ( Figure 4 ) :
B1 = Ž Uo Io / 2 PI r (3.2)
Figure 4. The Electromagnetic Frame.
Where Uo is the permeability of the Supermalloy, Io is the initial current induced into the system and r is the distance between the capacitors and the main wire.
The electromagnetic force in the loop is:
F1 = Integral IdlB1 = Integral ab IdlB +Integral cdIdlB (3.3)
And the total force using four capacitors:
Ft = Ž Uo Io / 2PI ( I1d1/r1 - I1d1/r2 + I2d2/r3 - I2d2/r4 ) (3.4)
Where I1 and I 2 is the current induced in each loop, r1,r2,r3,r4 is the distance between the capacitor and the main wire, d1, d2 is the length of each capacitor. ( Figure 4 )
The total electromagnetic force could be synthesized as follow:
Ft = Ž Uo Io (Ict) / 2PI (3.5)
Where Ict is the topological arrangement of the current induced by the capacitors.
On the other side we have the total force generated by the earthquake:
Ft = TTmPhinAn
Therefore,
TTmPhinAn = Ž Uo Io (Ict) / 2 PI
And the Inertial reduction of the system is expressed by:
mr = Ž Uo Io Ict / 2PITTPhinAn (3.6)
Gama = Ict /TTPhin2PI Is the topological tensor
Therefore, the inertial reduction equation in function of time is the sum of the loop:
mr(t) = Sum ŽU o Io Gama / Sum An (3.7)
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