NATURE OF DYNAMIC LOADING IN STRUCTURES DUE TO EARTHQUAKES
RESPONSE OF SINGLE DEGREE OF FREEDOM LINEAR STRUCTURES: Single degree of freedom systems involves simple structures like water tanks,
pergolas etc that can be easily idealized as a single mass concentrated at the top of a mass less stiffness system (column).
The response involves determination of displacements in the structure and then the member forces and moments from the displaced configuration of the structure. In case of single storey systems, it is considered as a single DOF system by taking
in to account, only the horizontal DOF and eliminating the other DOFs by condensation methods in which the rotational and the vertical components are condensed out.
Generalized Single Degree Of System (Idealization And Equilibrium formulation):
Idealisation:
Mathematical equilibrium formulation:
Forming the dynamic equilibrium of the structure shown above, acted by force P(t), spring force F(s) and damping force F(d), the equation can be given as:
Mǘ + Ců + Ku = P(t)
Where ǘ is represents the acceleration and ů represents velocity. Equation formulation for earthquake excitation:
In case of earthquake excitation, the external force acting on the structure is not considered, but the ground motions during the earthquake are considered as base motions and are converted to equivalent excitations as shown below:
Here the total displacement (ut) of the structure can be given as the sum of relative
(ur) and the base displacement (ub).
i.e. ut = ur + ub --- 1
The equilibrium equation can be given as: Mut’’ + Cur’ + kur = 0 --- 2
Substituting 1 in 2 gives M (ur’’ + ub’’) + Cu’ + Ku = 0
=> Mur’’ + Cur’ + kur = - Mub’’ --- 3
(u’’ and u’ represents the time based differentials of the time function)
Equation 3 is similar to normal dynamic equilibrium equation based on D’Alembert’s principle in which the excitation component P(t) is replaced by the component - Mub’’. Hence the earthquake excited SDOF system can be analysed by
solving the dynamic equilibrium equation obtained after the above said replacement of the excitation function.
Solution Of Equation For Single DOF System:
The dynamic differential equation for SDOF system subjected to earthquake (Given by equation 3) can be solved by any of the following methods based on the nature of the equation:
1) Classical method 2) Duhamel’s integral 3) Transform methods & 4) Numerical methods
In case of linear systems the response of the system is separately determined for the dynamic excitations and for the forces acting before the commencement of the dynamic excitations (static forces) and the results are added, where as such a separation of the problem is not possible in case of non-linear systems.
The classical method of solution comprises of determining the complementary function and the particular integral and is used in case of such simple linear differential equations that can be easily solved by this method. The solution of a freely vibrating system can be given as:
A SIN (שt) + B COS (שt) = 0,
which gives the complementary function. Te particular integral depends on the excitation force. The boundary conditions are used to determine the constants A and B.
If the function is of a sort that can be integrated easily, the Duhamel’s integral can be used in which the function is assumed to act a sequence of infinitesimally small impulses. The displacement solution using Duhamel’s integral method can be given as:
U(t) = 1/(m ש) ∫ (p(τ) sin [ש (t – τ)] dτ) (limits 0 to τ)
Where ש represents the natural frequency, and τ, the time instant considered.
In case of transform methods either the Fourier or Laplace transform is used, in which the given differential equation in terms of variable t is converted into an algebraic function in terms of iש so that the operations on the equation can be performed easily,
after which inverse transformation is performed to get the solution.
The solution of the Fourier transform, F[P(t)] = ∫ e –i שt p(t) dt, can be given as,
U = (ao) / (2k) + Σ (an cos (nωt) + bn sin (nωt)) / (k) (0 to n)
Where,
ao = (2/τ) ∫ f(t) dt (limits 0 to τ)
an = (2/τ) ∫ f(t) cos (nωt) dt (limits 0 to τ)
bn = (2/τ) ∫ f(t) sin (nωt) dt (limits 0 to τ)
Where τ is a time instant considered.
The numerical methods are involved if the excitations are too complex to be integrated or in case of non linear systems which can be performed either as numerical solution for duhamel’s integral or using finite difference or finite element approach. Displacement Solution For Nonlinear Systems:
In case of code based design for earthquakes the structure is allowed to deform in the inelastic range and provisions are given for required ductility in the structure. Thus the study of the structure in inelastic range is important in case of earthquake excitation. The force deformation behavior can be as such taken as variations given by hysteresis curves or can be idealized as elasto-plastic systems. Here up to linear range the initial stiffness value has to be used after which the a trace has to made on the history of loading at each time instant and based on the load deformation curve the stiffness at each infinitesimal time instant is determined as the tangent modulus of the curve at that instant. Hence the stiffness can be given as a function of force and velocity.
Various numerical time stepping procedures can be used for the solution of the above problem. The different procedures are:
1) Those based on interpolation of excitation 2) Based on finite difference expressions and
3) Based on assumed variation of acceleration
In these methods, the time domain is divided in to a lot of infinitesimal time instants Δt, and for each variation the variation in response Δu is determined and is added to the displacement u up to the time considered for getting the response at time t + Δt. The force value is linearly interpolated between the times t and t + Δt, the stiffness values are taken as per the load deformation variation and a constant damping value as per the linear system is considered.
Determination of other response components – Forces and Rotations.
After determination of displacement component, the other responses can be determined by either finding the equivalent static force to bring about the given deformed shape and then analyzing the structure for the lateral forces acting, or, by calculating the forces directly from the deformation knowing the stiffness of the system. The former method is usually adopted.
Example:
For the above system the deflection at the top end of the two columns is the same, and equal to U which is obtained from the normal procedures of SDOF analysis by considering the mass of the system to be lumped at the center of the beam and considering only the horizontal Degree of Freedom.
Knowing the end displacement of both the columns, the story shears of the columns are determined as (k1*U) & (k2*U) respectively. After determination of story shear the story moment is calculated at the end of each column by multiplying the story shear of the column with corresponding length of the column.
Thus the story moments at the column ends would be (k1*U*L1) and (k2*U*L2) respectively.
Solution For Multi Degree Of Freedom System: Problem Formulation:
In the same way as that of a SDOF system the equilibrium equation of a multi DOF system can be given as
M u’’ + C u’ + K u = 0
And for earthquake excitation, the equation can be given as M ur’’ + C ur’ + K ur = - ι Mb’’
Where M is the mass matrix of the structure elements of which are obtained as the mass required at each i th DOF to counteract an unit acceleration at the j th DOF, K is the stiffness matrix and C is the damping coefficient matrix obtained as the damping required at each i th DOF to counteract an unit velocity at the j th DOF.
The total deformation response can be given as a sum of the rigid body deformation of the structure to the earthquake excitation and the deformation of the structure due to its flexibility. ‘ι’ represents the rigid body deformation matrix of each DOF and will be unity if all the DOFS of the structure undergoes the same rigid body deformation.
The general solution can be given as [ K – Mω]Φ = 0
where Φ represents the modal matrix of the structure and from the above equation the mode shapes and natural frequencies are determined.
Orthogonality of the modes and its implications:
The orthogonality conditions of the modes can be given as, ΦnT k Φr = 0 and
ΦnT m Φr = 0. This implies that, if a modal mass or a modal stiffness matrix M = ΦT m Φ
or K = ΦT k Φ respectively are formed, they would be diagonal matrices. Further it can be
implied that [(fI)nT ur = 0] ((fI)n = inertia force in nth DOF, ur = rth mode displacement)
and [(fs)nT ur = 0] where fs represents the static force, thus proving that work done by nth
mode forces in going through rth mode displacements (where r not = n) is zero. Thus it implies that the forces due to displacements in any mode will not affect the displacements in any other mode.
Solution Of Differential Equation:
The solution can be obtained either by classical methods or by modal methods, the former being used very rarely. In case of modal methods, due to the modal
orthogonality, the solution is determined separately for each mode and then the solutions obtained for different modes are combined together based on the modal contribution factors using one of different combination rules such as SRSS, CQC or Absolute sum rules.
Modal analysis for MDOF systems:
The classical solution for the linear differential equation Mu’’ + Ku = p(t) (1) for MDOF system will not be efficient for systems with more DOFS, nor is it feasible for systems excited by other types of forces. Consequently, it is advantageous to transform these equations to modal coordinates.
The system displacement can be expanded in terms of modal contributions and the dynamic response of a system can be expressed as
U(t) = Σ Φr qr (t) (r = 1 to n) = Φ.q(t) (2)
Substituting 2 in 1 and pre multiplying each term by ΦT and applying orthogonal
relation, thereby eliminating all terms of the summation except when r = n, the relation reduces to
Mnq’’(t) + K qn(t) = pn(t)
Where M, K and pn are the modal matrices formed by pre and post multiplication by the
modal matrix.
From the above equation ‘q’ is determined knowing the mass, stiffness matrices, time periods and the modal matrix from which the solution is determined as explained in the model problem.
Model problem for Modal analysis of MDOF system:
Prob: A three-story frame shown in the figure is subjected to an excitation F cosωt at the top story level. Determine the response at the top level on the consideration of
1. First mode only 2. First two modes 3. All the three modes
Given Data: Ω = 0, 0.5 ω1 & 1.3 ω2 k1 = k2 = 160 kN / mm, k3 = 240 kN / mm m = 20000 kg Solution: [k] = 8 X 107 (Taking 80 outside)
[m] = 2 X 104 (from the figure)
From the given mass and stiffness matrices the natural periods and frequencies can be determined as:
ω2 = (1924.733, 14437.2, 27889)T, => ω = (43.87, 120.15, 167)T
Φ =
Modal mass matrix Mr = ΦT m Φ =
Modal force matrix Fr = ΦT f(t) = F cosωt (11 1)T
Displacement at the top story level =
u(t) = 1.0 * F cosωt / ((20000 * 1.6896) (1924.733 - Ω2)) +
1.0 * F cosωt / ((20000 * 2.9667) (14437.2 - Ω2)) +
Consideration of 1st term alone implies 1st mode shape alone, and as per the number of
mode shapes to be used corresponding number of terms are utilized and solution is attained for different Ω values.
From the results it can be implied that the initial fewer modes will have more effect than the higher modes on the result and hence, it would be sufficient to consider only few initial modes in case of multi story buildings.
The forces and moments, from the displacement result, can be obtained either by equivalent static method or by determining the storey shears of each story from the corresponding inter story drifts in case of shear frames, by multiplying them with the corresponding story stiffness values and for moment frames, from the condensation equations or by reverse engineering.