Appendices

1531 1 1 1 ELASTIC SOIL-STRUCTURE- INTERACTION (SSI) The purpose of this chapter is to develop some necessary structural dynamics for a single degree of freedom structure mounted on a rigid foundation block. The eventual aim of the chapter is to use the response spectrum method to estimate the actions applied to the foundation. The main parts of the chapter follow the approach of John Wolf in his 1985 book “Dynamic Soil-Structure Interaction”. After some preliminaries we will follow Wolf in developing the response of a single degree of freedom structure with a massless foundation. In this we will evaluate the steady state response to sinusoidal excitation at various frequencies. Next we extend the Wolf material to evaluate the steady state response of a simple structure on a foundation with mass and rotational inertia, so the system now has three degrees of freedom – lateral displacement of the structural mass, lateral displacement of the foundation mass and rotation of the foundation mass. Once again we evaluate the steady state response. These steady state studies provide useful insight into the behaviour of the system and the relative importance of various parameters, particularly frequencies. However, since an earthquake does not apply a steady state response we need another approach to estimate the maximum foundation actions which occur during an earthquake. For this the response spectrum method is applied. In its simplest form, it has the above-ground structure represented with a single equivalent mass; this can be extended easily to systems with more degrees of freedom, that is buildings represented with discrete masses for each above- ground storey. The final part of the chapter develops these tools. Design of Earthquake Resistant Foundations 154 u h h e h e u s M e u =u s +u h +h e Figure 11.1 Single degree of freedom (SDOF) structure – foundation model All the analysis discussed in this chapter is for elastic response – elastic soil behaviour as well as elastic structural behaviour. As far as the foundation soil is concerned this is represented as pseudo-elastic and a soil modulus much less than that associated with truly elastic soil behavior is used to approximate the decrease in stiffness of the soil during earthquake shaking. 11.1 SINGLE DEGREE OF FREEDOM MODEL As explained in Chapter 1 much of our work will be concerned with the simple model shown in Figure 11.1 (a repeat of Figure 1.1). The first part of the chapter assumes that the mass of the foundation is zero, so the only mass in the system represents the structure above the foundation. Note the notation for the height of the mass and the mass – h e and M e for effective height and effective mass, ie the effective mass may include a contribution from the foundation as well the elevated mass. Next the foundation mass (and rotational inertia) is included, but the structural part is still considered as a single degree of freedom (that is there is only one structural mass). 11.2 RESPONSE SPECTRUM METHOD Equations 10.2 to 10.9 give the steady state response of an SDOF system to sinusoidal excitation. As stated above they are of use in design of foundations for vibrating machines. What we need is a method for handling the earthquake excitation of an SDOF structure which is not a steady state response. The key to this is the response spectrum. Recalling that each point on the response spectrum gives the maximum response of a SDOF oscillator having a period corresponding to the abscissa of the point in question, it is apparent that the spectrum can be used to estimate the maximum response of a given SDOF system. In Chapter 10 the amplitude of the sinusoidal forcing function is denoted by Q o . The response spectrum gives the required information as follows: 2 v d a a ( , T) ( , T) ( , T) ( T) ( , T) S S S = M ; = = S max earthquake f , ç ç ç ç ç e e (11.1) Chapter 11: Elastic structure-foundation-soil interaction 155 where: f max earthquake (ç,T) is the maximum base shear experienced in an earthquake, T period of the structure (fixed base) of the foundation structure system ç damping ratio M is the SDOF mass S a (ç,T) is the sprectral acceleration from the response spectrum, S v (ç,T) is the spectral velocity, and S d (ç,T) is the relative spectral displacement between the structure and the surrounding ground. Since the response spectrum method will be used for evaluating the response of the SDOF model we need to have a way of estimating S a etc when the value of ç is other than that given in the available response spectrum. A number of equations for making this modification can be found in the literature; some comparisons are given by Priestley et al (2007). The following equation by Kawashima and Aizawa (1986) achieves this relative to ç = 0.05 (5%) (a frequently used damping value associated with code-based design spectra): ( ) ( ) a a S , T 1.5 0.5 S 0.05, T 40 1 ç ¦ ¹ = + ´ ` ç + ¹ ) (11.2) The equation below is taken from the 2003 edition of Eurocode EC8 (CEN (2003), and also changes the spectral values relative to those for a spectrum with ç e = 0.05 (5%). ( ) ( ) ç ¦ ¹ = ´ ` + ç ¹ ) 0.5 a e a e S , T 0.10 S 0.05, T 0.05 (11.3) 11.3 KINEMATIC AND INERTIAL SOIL-STRUCTURE INTERACTION Before we go much further we need to consider how the building foundation interacts with the earthquake motion. There are two facets to this. First there is the question of how the geometry of the foundation interacts with the incoming seismic waves. Note that the incoming waves have already been modified by any site effects; what we are now considering is how these waves are influenced by the foundation independent of the structure attached. The second is how the motions are affected by the inertia of the structure as it responds to the earthquake motion. These two effects are known as kinematic and inertial interaction respectively. The concepts are illustrated in Figure 11.2. In this chapter it is inertial interaction that is under consideration. Design of Earthquake Resistant Foundations 156 Figure 11.2 Kinematic and Inertial soil-structure interaction, illustrating how the analysis is broken down into two stages. (after Mylonakis et al 2006). Table 11.1 Average reduction factors (± one standard deviation) for shear wave velocity and shear modulus within 20 m of the ground surface (EC8 Table 4.1). PGA (g) V s /V s, max G/G max 0.10 0.20 0.30 0.9(±0.07) 0.7(±0.15) 0.6(±.......) 0.80(±0.10) 0.50(±0.20) 0.35(± ) Table 11.2 Effective shear moduli and shear wave velocity (NEHRP 1997, Table 4-3) PGA (g) Vs/Vs, max G/Gmax 0.10 0.70 0.71 0.54 0.50 0.20 Chapter 11: Elastic structure-foundation-soil interaction 157 11.4 CHOICE OF SOIL STIFFNESS VALUES FOR ESTIMATION OF EARTHQUAKE RESPONSE Next we have to decide what values to use for the soil stiffness. Perhaps we need to refer back to the various levels of design analysis mentioned in Chapter 1. What we have in mind here is the so-called Level 2 analysis, that is an analysis aimed at simplicity rather than rigour and which uses elastic methods but alters the elastic modulus of the soil to account for a representative level of strain. One such suggestion is given in EC8 and reproduced in Table 11.1. Another can be found in the NEHRP document on seismic retrofit published in 1999, it is given in Table 10.2. An alternative approach, in use by the writer for some time, is to investigate the response of the foundation with the small strain shear modulus of the soil and then with the shear modulus reduced by a factor of 4. This is seen to bracket the values in the above EC8 table and gives a rapid assessment of the significance of nonlinear soil behaviour. 11.5 SINGLE DEGREE OF FREEDOM STRUCTURE-FOUNDATION MODEL Figure 10.1 is intended to be a simple conceptual model of a vibrating system, so there is only one spring and one damper. We have pointed out how it could be used for to represent a vibrating foundation on the ground surface. An example of such an application would be the foundation for a turbine, air compressor or some other type of machine that produces vibration. However, the system we are interested in at this point, shown Figure 11.3, continues to have only one mass but has several springs and several dampers – stiffness and damping for the superstructure, horizontal stiffness and damping for the foundation, and rotational stiffness and damping for the foundation. What we need to do is derive the equivalent of equation 10.1 in which the stiffness term includes contributions from the three distinct springs in Figure 11.3 and the damping term includes contributions from the three damping elements. Because Figure 11.3 considers the vibration of only one mass it is still an SDOF system. The total displacement of the mass is given by: = + + |+ t g h e u u u h u (11.4a) where: u g , u h , u and | 1 , the various displacement parameters, are defined in Figure 11.2. 1 The notation | for foundation rotation could be confused with the usual symbol for the angle of shearing resistance for soil. The context should usually make things clear. But also the notation for angle of shearing resistance is |' given that this is an effective stress parameter. Design of Earthquake Resistant Foundations 158 u h h e h e u s M e u =u s +u h +h e K s K h K Figure 11.3 Definition of various terms used in the SDOF structure-foundation model. What we are likely to want more frequently is the relative displacement of the mass with respect to the initial position of the foundation: = + |+ relative h e u u h u (11.4b) We will calculate all three displacement components on the right hand side of equation (11.4b). 11.6 DAMPING AND DYNAMIC RESPONSE What is now needed is the value for the damping coefficient. Firstly we have to consider what is meant by damping for the behaviour of an elastic system. If the soil is modelled as an elastic half space and an object vibrates at, or close to, the surface then energy is radiated away to infinity. It is the function of damping in the above equations to model this transfer of energy away from the source of vibration. It is thus referred to as radiation damping (or sometimes as geometric damping). In addition there is internal dissipation of energy within the soil (an inelastic process), this is referred to as material damping (Figures 3.25 and 4.42). In a real foundation situation both these damping mechanisms occur. The material damping is shear strain dependent. The process of determining radiation damping is based on recognising that equations 10.2, 10.11 and 10.13 on a simple model of the response of an elastic soil deposit to surface excitation. What is needed to understand the behaviour of the system is a rigorous continuum solution for the problem. This has been obtained and compared with the simple SDOF approach above. The details are discussed by Richart, Hall and Woods (1970). It appears from such work that the damping depends on the mode of vibration and the geometry of the foundation. Chapter 11: Elastic structure-foundation-soil interaction 159 So for a given mode of vibration of a shallow foundation there is more than one mechanism of damping. We need a way of combining these; this will be discussed below. 11.6.1 Natural frequencies and damping parameters From Figure 11.3 we have a number of natural frequencies and damping parameters: (a) Superstructure 2 2 s s s e K K C M ç e e = = (11.5) (b) Foundation horizontal 2 2 h h h h h e h K K C M ç e e = = (11.6) (c) Foundation rotational 2 2 2 e K K C M h | | | | | | ç e e = = (11.7) The damping terms in the above equations represent the elastic radiation of energy away from the foundation. Matching the analytical steady state response of shallow foundation vibrating at the surface of an elastic half space with the solution from equation 10.1 it becomes apparent the radiation damping is a function of frequency (more details of this are given in Chapter 10). We can now consider how the foundation will produce dynamic resistance. Consider horizontal displacement only for a shallow foundation; for dynamic loading the dynamic reaction consists of two components: = +  h h h h h H K u C u (11.8a) Note (i) that the reaction force is related to the relative displacement between the foundation and soil (as shown in Figure 11.3), and (ii) that the second term increases as the deformation velocity increases. One the other hand material damping is known to a very good approximation to be frequency independent (cf Chaps. 3 and 4). Thus our two damping mechanisms are in some ways incompatible. Even so they are expected to be acting at the same time. That is energy is being radiated from the foundation and energy is being dissipated by hysteresis in the soil adjacent to the foundation. The equivalent viscous damping ratio for soil is usually related to the shear strain amplitude (ie it is not a material constant cf. chaps 3 & 4). We assume a value and make an addition to the damping parameter above as follows: ç = + + = + ç + ç e e   g h h h h h h h h h h g h 2 K 2K H K u (C )u K u ( )u (11.8b) Design of Earthquake Resistant Foundations 160 where: ç g is the hysteretic damping of the soil expressed as an equivalent viscous damping ratio. (Note the difference in notation: ç for hysteretic damping and ç for radiation damping.) Similarly if the foundation is subject to moment loading: | | | | | | | , = |+ + | = |+ ç + ç | e e   g g 2 K 2K Mom K (C ) K ( ) (11.9) 11.6.2 Harmonic response and complex representations of displacement, velocity and acceleration Returning to the differential equation 10.1, we will consider the solution of this second order ordinary differential equation. Firstly let’s transform the equation by dividing through by M e : | e e   2 n o u + 2 u + u = f sin( t ) (11.10) where: e n is the natural frequency (for the particular mode under consideration), | is C/M e = 2e n ç, and f o is Q o /M e . Following Taylor (2005), it turns out to be helpful to consider the excitation as a complex quantity, rather than sinusoidal as in equation (11.10); that is substitute: e = + = O i t u(t ) x(t ) iy(t ) e (11.11) where i = \-1. Equation 11.10 then becomes: e | e   2 i t n o u + 2 u + u = f e (11.12) This is helpful as it enables us to handle both excitation in terms of sine and cosine waves, and also because of the properties of exponential function e i e t . Recalling Euler’s formula and its derivatives: i t i t i t 2 i t 2 i t 2 2 2 e cos( t ) i sin( t ) d e i e sin( t ) i cos( t ) dt d e e cos( t ) i sin( t ) dt e e e e e = e + e = e = ÷e e + e e = ÷e = ÷e e ÷ e e (11.13) Chapter 11: Elastic structure-foundation-soil interaction 161 The second and third of these illustrate how differentiation of a complex exponential is simply a matter of multiplying successively by ie. Using these relations, equation 11.12 becomes: e ÷e |e e 2 2 i t n o ( + 2i + )u(t ) = f e (11.14) Note that we have substituted a solution into the equation, so 11.14 is no longer a second order differential equation but an algebraic equation. From here we conclude that the complex solution to 11.14 is of the form: e = O i t u(t ) e (11.15) From comparison of equations 11.14 and 11.15 it is apparent that: O = e ÷e + |e o 2 2 n f 2i (11.16) 11.7 THE CORRESPONDENCE PRINCIPLE Wolf explains briefly the correspondence principle on pages 15 and 16 of his book. It is also introduced in Flugge (1967). It is a very useful concept in considering the dynamics of foundations on and within elastic media. The idea is that if we have a solution for the static response of a system then we can write the solution for dynamic behaviour simply by replacing the static stiffness by a complex quantity called the impedance. Returning to and reorganising the equation above using the Ae i e t : ( ) = + = + ç  h h h h h h h h H K u C u K 1 2 i u (11.17) The term ( ) + ç h h K 1 2 i is known as the impedance. As explained above we convert from a static situation to a dynamic on simply be exchanging the stiffness terms for the impedances. 11.8 LAGRANGE’S EQUATIONS Lagrange’s equations are differential equations of motion expressed in terms of generalised coordinates. They are developed in detail in most texts on theoretical mechanics; two examples with accessible presentations are Thomson (1988) and Taylor (2005). The attraction of Lagrange’s equations is that they provide a reliable method of deriving equations of motion, particularly for systems involving both linear and rotational motion. By generalised coordinates is meant a set of coordinates pairs of which are “linked” – such as force and acceleration, moment and curvature etc. The equations are based on writing down expressions for the kinetic and the Design of Earthquake Resistant Foundations 162 potential energies of the system. The form of the equations for a system with N generalised coordinates is: i i i d T T U . . . 0 i 1, 2, N dt q q q | | c c c ÷ + = = | c c c \ .  (11.18) where: T is the kinetic energy of the system and U is the potential energy, and q is the usual symbol for the generalized coordinates. 11.9 WOLF’S SDOF MODEL Wolf (1985) covers elastic methods for calculating soil structure interaction. In this section we will follow the development he presents in his section 3.4 of his book. The details of Wolf’s model and the associated coordinates are shown in Figure 11.3. 11.9.1 Equations of motion First we need the expressions for the kinetic and potential energy, so that we can derive the equations of motion using Lagrange’s equations. Since the system has only one mass all the kinetic energy is located there. Referring to Figure 3, we can write: = + + + | = + + + | + + + |+ |+ |+                     2 e g h s e 2 2 2 2 2 e g h s e g h g s g e h e s e h s 1 T M (u u u h ) 2 1 M (u u u h 2u u 2u u 2u h 2u h 2u h 2u u ) 2 (11.19) Note that the kinetic energy given above depends on the total movement of the mass, that is it includes the movement, or more correctly, the velocity of the ground on which the structure is sitting. Note also, that the kinetic energy contains no terms in the displacement components directly, so the cT/cq i terms won’t appear in our equations of motion. The expression for the stored elastic energy in the system at the instant the velocities are zero is: | = + + | 2 2 2 h h s s 1 1 1 U K u K u K 2 2 2 (11.20) Now we obtain equations for each of the displacement coordinates u o , u s and |. u s : Chapter 11: Elastic structure-foundation-soil interaction 163 + + + | + = + + | + = ÷         e h g s e s s e h s e s s e g d M (u u u h ) K u 0 dt which becomes: M (u u h ) K u M u u o : + + + | + = + + | + = ÷         e h g s e h h e h s e h h e g d M (u u u h ) K u 0 dt which becomes: M (u u h ) K u M u |: | | | + + + | + | = + + | + = ÷         2 e e h e g e s e 2 e h e s e e e g d M (h u h u h u h ) K 0 dt which becomes: M (hu h u h ) K u M h u We assume that we need the steady state solution to these equations for harmonic input only, Oe i e t . This might seem like a very severe limitation, but recall that we are considering only an elastic system so superposition applies. (Once we have the solution of a simple harmonic excitation we could calculate what will happen for quite complex inputs using Fourier series to reduce the input to a collection of harmonic excitations of varying frequencies and amplitudes.) If the steady state response is adequate then we can reformat these equations as we did for equation11.14. We derived the above equations of motion for an undamped system. For the next stage of the development we replace the stiffness with the impedance and hence include the damping in the equations (this assumes that the forms of the stiffness and damping matrices for the system are the same). Substitution of a solution of the form of 11.15 gives: ÷ e + + | + + ç = e 2 2 e s h e s s s e g M (u u h ) K (1 2 i )u M u (11.21) ÷ e + + | + + ç + ç = e 2 2 e s h e h h g s e g M (u u h ) K (1 2 i 2 i )u M u (11.22) | | ÷ e + + | + + ç + ç | = e 2 2 e s h e g e e e g e K M h (u u h ) (1 2 i 2 )h M h u h (11.23) Now dividing 11.21 and 11.22 by Me 2 and 11.23 by Mhe 2 , and noting the definitions of the square of the natural frequencies in equations, 11.4, 11.5 and 11.6, we obtain: Design of Earthquake Resistant Foundations 164 | | ( e ¦ ¹ ¦ ¹ + ç ÷ ÷ ÷ ( e ¦ ¦ ¦ ¦ ( ¦ ¦ ¦ ¦ e ( ¦ ¦ ¦ ¦ ÷ + ç + ç ÷ ÷ = ´ ` ´ ` ( e ¦ ¦ ¦ ¦ ( ¦ ¦ ¦ ¦ e ( ÷ ÷ + ç + ç ÷ ¦ ¦ ¦ ¦ ( ¹ ) ¹ ) e ¸ ¸ | 2 s s 2 2 h h g g 2 2 g e 2 s h 1 (1 2 i ) 1 1 1 1 1 (1 2 i 2 i ) 1 1 u 1 1 (1 2 i 2 i ) 1 1 u u h (11.24) These are the same as Wolf’s equations (3.49). 11.9.2 Reduction to SDOF form Equations 11.24 show how the three deformation components relate. However, despite the three equations, there is only one dynamic degree of freedom, so there should only be one equation of motion. We can use two of equations 11.24 to express, say, u h and h e | in terms of u s and then substitute these into the remaining to obtain just one equation of motion. Subtracting the second of 11.24 from the first, we obtain: e + ç = e + ç + ç 2 s s h s 2 h h g 1 2 i u u 1 2 i 2 i (11.25) | | e + ç | = e + ç + ç 2 s s e s 2 g 1 2 i h u 1 2 i 2 i (11.26) Now substitute equations 11.25 and 11.26 into the first of equations11.24: | | | | + ç + ç e e e e + ç ÷ ÷ ÷ = | | e e + ç + ç e + ç + ç e \ . 2 2 2 2 s s s s g 2 2 2 2 s h h g g s 1 2 i 1 2 i 1 2 i u u 1 2 i 2 i 1 2 i 2 i (11.27) This equation shows how the various stiffness and damping components combine to determine one displacement component of the model in Figure 11.3 (the displacement of the structural column). Once u s is available then u h and | can be obtained from equations 11.25 and 11.26 respectively. 11.9.3 Equivalent fixed base SDOF model Now equation 11.27 gives the response of the model shown in Figure 11.3 to harmonic ground excitation. However, if we look at Figure 1.2, which illustrates the model used for the response spectrum calculation, we note that this is a simple fixed base structure. Thus one wonders if equation 11.27 can be transformed to produce an equivalent fixed base SDOF model. If this was available then one could apply the response spectrum method to estimate the maximum displacements during an earthquake time history. Chapter 11: Elastic structure-foundation-soil interaction 165 Continuing to follow Wolf (1985), what we seek is an equation of the following form: ÷ e + e + = e    2 2 e s e g ( M i C K)u M u (11.28) where: e =   2 e K M (11.29) 2K C ç = e    (11.30) So reorganising 11.28 we obtain: 2 2 s g 2 2 (1 2 i )u u e e + ç ÷ = e e     (11.31) Our requirement, then, is to assign values to the parameters in 11.31 which have a tilde such that the structural deformation is essentially the same as given by equation11.27. Note that the mass is the same in the two systems. First we need the equivalent natural period of the system. For an undamped system the steady state response is infinite at the natural period. Thus in equation 11.27 we set all the damping terms to zero and the coefficient of u s on the left hand side to zero. This yields: 2 2 2 2 s h 1 1 1 1 | = + + e e e e  (11.32) Substituting the stiffness expressions from the above equations, 11.32 reduces to: 2 2 s 2 s s e h K K h 1 K K | e e = + +  (11.33) Since the bottom line of equation 11.33 will always be greater than unity the natural frequency of the equivalent SDOF model will always smaller than the fixed-base frequency of the structure, e s . Next we need to find the equivalent damping for the system. This requires comparison of equations 11.27 and11.31. Looking first at 11.27 we need to remove the complex terms from the bottom side of several of the expressions. The standard way of converting a complex denominator to a real quantity is the multiply the expression by unity in the form of the complex conjugate in both numerator and denominator. For example taking the fourth term on the left hand side of 11.27 and ignoring products of damping ratios as these will be small: Design of Earthquake Resistant Foundations 166 h g h g s s s h g h g h g h g h g 1 2 i 2 i 1 2 i 2 i 2 i 1 2 i 1 2 i 2 i 2 i 1 2 i 2 i 1 2 i 2 i 1 2 i 2 i 2 i 2 i ÷ ç ÷ ç ÷ ç ÷ ç + ç + ç = = + ç ÷ ç ÷ ç + ç + ç ÷ ç ÷ ç + ç + ç ÷ ç ÷ ç Using this technique we can rewrite equation 11.27 as: ( ) 2 2 2 2 s s h g s g s g 2 2 2 2 s h s 1 2 i (1 2 i 2 i 2 i ) 1 2 i 2 i 2 i u u | | | | e e e e + ç ÷ ÷ + ç ÷ ç ÷ ç ÷ + ç ÷ ç ÷ ç = | | e e e e \ . (11.34) The next step is to substitute 11.32 into11.31: 2 2 2 2 s g 2 2 2 2 s h (1 2 i )u u | e e e e + ç ÷ ÷ ÷ = e e e e    (11.35) Now we find the equivalent system damping by equating the left hand sides of 11.34 and11.35. They have the same real parts so we obtain the equivalent system damping as: 2 2 2 2 2 2 s h g 2 2 2 2 2 2 h h h (1 ) | | | | | | e e e e e e ç = ç ÷ ÷ + ç + ç + ç + | | e e e e e e \ .  (11.36) Wolf suggests that this be evaluated at resonance and then this value applied right across the frequency range. Substituting 11.32 into 11.36 gives: 2 2 2 2 s h g 2 2 2 2 s h s 1 | | | | e e e e ç = ç + ç + ç + ç ÷ | e e e e \ .      (11.37) Finally we obtain the equivalent ground motion by comparing the right hand sides of 11.27 and 11.31, from which we conclude that the effective seismic input to our equivalent fixed base structure will always be less than the actual seismic motion (this is reminiscent of the information obtained from the measured performance of actual structures in Chapter 2 (Figures 2.2, 2.4, 2.6 and 2.8): 2 g g 2 s u u e = e   (11.38) In summary then, we started with the structure-foundation model shown in Figure 11.3 and have reduced it to an equivalent fixed base single degree of freedom model. Equation 11.32 shows how the various stiffness components combine to give the natural frequency of the equivalent system, 11.37shows how the damping parameters are combined and 11.38 shows how the actual ground motion is modified to become the input for the SDOF model. The solution to equation 11.31 gives the structural displacement. Equations 11.25 and 11.26 give the other deformation components in terms of the Chapter 11: Elastic structure-foundation-soil interaction 167 structural displacement. We can simplify these equations by eliminating the complex quantities from the denominators: ( ) 2 s h s h g s 2 h u 1 2 i 2 i 2 i u e = + ç ÷ ç ÷ ç e (11.39) ( ) 2 s e s g s 2 h 1 2 i 2 i 2 i u | | e | = + ç ÷ ç ÷ ç e (11.40) The total displacement of the mass relative to the free field is given by: ( ) 2 h s h e s s g s 2 2 2 2 2 s h 2 1 1 1 2 u u h 2 i i i u | | | | ç | | ç + + | = e + ÷ ç ÷ç ÷ ÷ | | | e e e e e \ . \ .   (11.41) The shear force and moment sustained by the foundation are: = = s s e s s H K u Mom h K u (11.42) 11.9.4 Solution of the above equations The solution 11.15 transforms the second order differential equation 11.12 to the algebraic equation 11.14 making the evaluation of the displacements simple at the expense, although modest, of dealing with complex quantities. Rearranging 11.31 to obtain u s and making the denominator real gives: 2 2 2 2 g g s 2 2 2 2 2 2 2 2 2 2 2 g 2 2 2 2 2 (1 2 i ) u u u (1 2 i ) (1 2 i ) (1 2 i ) (1 2 i ) u 1 4 e ÷ ç ÷ e e e = = e e e e e + ç ÷ + ç ÷ ÷ ç ÷ e e e e ÷ ç ÷ e e = e ¦ ¹ ( e ¦ ¦ ÷ + ç ´ ` ( e ¸ ¸ ¦ ¦ ¹ )                   (11.43) Next we need the magnitude (which is real) of u s which is obtained by taking the square root of the product of the above complex quantity with its complex conjugate, and using equation 11.38 the actual ground displacement is inserted: 2 2 g 2 g 2 s s 2 2 2 2 2 2 2 2 u u u 1 4 1 4 e e e e = = ( ( e e ÷ + ç ÷ + ç ( ( e e ¸ ¸ ¸ ¸       (11.44) Equation 11.44 gives the bending of the structural column shown in Figure 11.3; the other two components are given in equations 11.25 and 11.26. Design of Earthquake Resistant Foundations 168 11.9.5 Dimensionless parameters The solution presented in the above equations can be expressed in terms of the following dimensionless parameters. - The ratio of the stiffness of the structure to the stiffness of the soil: s e s h s V e = (11.45) where: V s is the shear wave velocity of the soil. - The slenderness ratio: e h h a = (11.46) where: a is the radius of the foundation. - The mass ratio: = µ e 3 M m a (11.47) - Poisson’s ratio for the soil and the hysteretic damping ratios of the structure and soil. 11.9.6 Example calculation Example 11.1 As an example of the application of the ideas outlined in the above section consider a water tank holding 50 m 3 of water at a height of 20 m above a clay foundation. The shallow foundation is circular and, for the purposes of this example, rests on the ground surface. The fixed base natural period of the structure is assumed to be 1 second and the equivalent viscous damping ratio is 3.5%. The soil on which the foundation sits is saturated clay with an undrained shear strength of 50 kPa and the shear wave velocity of the clay is 118 m/sec. The calculations for this example can be found in Appendix 11.1. As a first step we consider the fixed base response of the system; that is we assume that the foundation is fixed so that there is no compliance from the material below. This means that K h and K | are infinite, so the bottom line of equation 11.33 is unity, and the thus the natural frequency of the system is that of the water tank/tower alone. Figure 11.4 shows the calculated steady state response of the fixed base system to sinusoidal excitation over a range of frequencies. The peak amplitude is controlled by the system damping of 3.5%. Allowing for the effect of the soil beneath the foundation it is first necessary to find a size of foundation that satisfies the ultimate limit state requirements. For the Chapter 11: Elastic structure-foundation-soil interaction 169 purposes of the example calculation a horizontal acceleration of 0.2g is assumed to be applied to the water tank. Also the bearing strength is evaluated using the partial factor approach with a factor on the undrained shear strength of 1.4 (that is the shear strength used in the calculations, from equation 6.4, is 50/1.4 = 36 kPa), a foundation of radius 4.6 m is required. The foundation disk is assumed to be weightless for this example (foundation mass is included in Example 11.2) Three sets of calculations are done including the effect of the compliance of the soil beneath the foundation; one using the small strain shear modulus (shear wave velocity 118 m/sec.) and the other following reduction suggested in Table 11.1which reduces the “shear wave velocity” to 83 m/sec. The response for these cases is plotted in Figure 11.4. It is apparent that elastic soil-structure-interaction, V s = 118 m/sec, has only a small effect on the response. Introduction of the 50% reduction in shear modulus following Table 11.1, V s = 83 m/sec, demonstrates further slight changes from the reduced soil stiffness. Comparing these three curves in Figure 11.4 one could hardly say that, for the example structure considered here, that soil-structure interaction is significant. To achieve a significant change as a consequence of soil structure interaction a substantial reduction in the soil modulus is needed. The final plot in the figure shows what happens if the shear wave velocity used to estimate the modulus of the soil is reduced to 37 m/sec (that is G/G max = 0.1), only then do we see a significant change, but a shear wave velocity of 37 m/sec is an extremely low value, so it seems that for this structure elastic soil-structure interaction effects are not significant. Three dimensionless parameters are defined in equations 11.45, 46 and 47. Two of them are not affected by changes in the soil stiffness. That given by equation 11.45 has a value of 1.1 when the shear wave velocity is 118 m/sec, 1.5 when it is 83 m/sec, and 3.4 when it is 37 m/sec. The calculations on which Figure 11.4 is based assume that there is full contact between the underside of the foundation and the clay below. An approximate way of checking if this might be true would be to estimate the vertical settlement of the foundation under the static vertical load and compare this with the uplift at the edge of the foundation during the maximum actions. That is we need the vertical settlement under the weight of the tank and the rotation of the foundation under a moment of 0.2xweightxheight. These are estimated as part of the calculations in Appendix 11.1 from which it is apparent that once the horizontal acceleration gets beyond about 0.15 g then there will be some loss of contact between the foundation and the underlying soil. Recently there has been completed a thorough investigation of the response of buildings during recent earthquakes in California. Some results of the study are plotted in Figure 11.6. In this the authors compared the peak free-field acceleration with the peak acceleration recorded at the base of the structure. The data in Figure 11.5 indicate that soil-structure interaction effects are not significant for a large number of buildings in the study. Design of Earthquake Resistant Foundations 170 0 0.5 1 1.5 2 0 5 10 15 Frequency ratio A m p l i f i c a t i o n 0 Fixed base structure (period 1 second) V s =118 m/sec V s =83 m/sec V s =37 m/sec Figure 11.4 Steady state response to sinusoidal excitation of an elevated water tank with shallow foundation conditions ranging from fixed base to very low soil shear wave velocity. Figure 11.5 Comparison of free-field and foundation level structural motions: (a) peak acceleration data, (b) 5% damped spectral acceleration at the natural period of the structure-foundation system (after Stewart et al 1999). Chapter 11: Elastic structure-foundation-soil interaction 171 u h h e h e u s M u =u s +u h +h e M f , I f K h , K K s Figure 11.6 Three degree of freedom structure-foundation-soil model 11.10 INCLUSION OF FOUNDATION MASS AND INERTIA We now extend the above to cover the model a foundation mass in addition to the structural mass as shown in Figure 11.6. This will now have three degrees of freedom because the foundation block has both mass and inertia. The expressions for the kinetic and the potential energy are now: ( ) ( ) = + + |+ + + + |       2 2 2 g h e s f g h f 1 1 1 T M u u h u M u u I 2 2 2 2 2 2 s s h h 1 1 1 U K u K u K 2 2 2 | = + + | Applying Lagrange’s equations to the above two equations gives the following coupled equations of motion for the system shown in Figure 11.6 when there is no damping present: | ¦ ¹ ( ( ¦ ¹ ¦ ¹ ¦ ¦ ¦ ¦ ¦ ¦ ( ( + + = ÷ + ´ ` ´ ` ´ ` ( ( ¦ ¦ ¦ ¦ ¦ ¦ ( ( + | | ¸ ¸ ¹ ) ¹ ) ¸ ¸ ¹ )     e s s s f e o f o g f 2 e e e f e M M Mh u k 0 0 u M M M M Mh u 0 K 0 u u M M Mh Mh Mh I 0 0 K Mh (11.48) Adding damping by using the impedance we can transform these equations as above by substituting u = Ce i e t and simplifying. The following three equations are obtained: Design of Earthquake Resistant Foundations 172 ( ) 2 s s h e s s g u u h 1 2i u u e | | ÷ ÷ ÷ |+ + ç = | e \ . (a) ( ) 2 h s h h h e g f f M M u u 1 2i u h u M M M M e | | ÷ ÷ + + ç ÷ | = | + e + \ . (b) ( ) | | e | | | || | ÷ ÷ ÷ + |+ + + ç | = | | | e \ . \ . \ . 2 f f s h e e g 2 2 e e I I u u 1 h 1 1 2 i h u Mh Mh (c) In matrix form the above three equations become: ( ) ( ) ( ) | ¢ ( ¦ ¹ e ¦ ¦ | | ( + ç ÷ ÷ ÷ ´ ` | e ( \ . ¦ ¦ ¹ ) ( ¦ ¹ ¦ ¹ ¦ ¹ ( e ¦ ¦ ¦ ¦ ¦ ¦ | | ÷ + ç ÷ ÷ = ´ ` ´ ` ´ ` | ( + e + \ . ¦ ¦ ¹ ) ¦ ¦ ¦ ¦ ( | ¹ ) ¹ ) ( ¦ ¹ e | | | | ¦ ¦ ( ÷ ÷ + + ç ÷ ´ ` | | ( e \ . \ . ¦ ¦ ¹ ) ¸ ¸ 2 s s s 2 h h h g f f e 2 f 2 e 1 2i 1 1 1 u 1 M M 1 2i 1 u u 1 M M M M h 1 I 1 1 1 1 2i 1 Mh (11.49) Equations (11.48) are three coupled second order differential equations. In investigating the steady state response to an Oe i e t input, these become in equation (11.49), three simultaneous algebraic equations. From these equations we can obtain the steady state displacement amplitudes of three degrees of freedom. These solutions, as functions of the excitation frequency e, are given by: ( ) ( ) ( ) ( ) ( ) + e ÷ e e = O e s 1 DC D B u Det (11.50) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) e = O + e ÷ + e ÷ + e e ÷ e ÷ e + + e ÷ e h u 1 DC D E C DE ED A C D A D C D D EA 1 Det (11.51) ( ) ( ) ( ) ( ) | e e e = O e e h A B Det (11.52) where: Det(e) is the value of the determinant of the 3x3 matrix on the left hand side of equation (11.49), A(e) is e s 2 (1+2iç s )/e 2 B(e) is e h 2 (1 + 2iç h + 2i, g )/e 2 Chapter 11: Elastic structure-foundation-soil interaction 173 C(e) is e | 2 (1 + 2iç | + 2i, g )/e 2 D is 1 + I f /Mh e 2 and E is M/(M + M f ) 11.10.1 More dimensionless parameters Introduction of the foundation block means that more dimensionless parameters are now required to specify our system. - The ratio of the stiffness of the structure to the mass of the foundation block: M/M f - The moment of inertia of the foundation block with respect to the rotational inertia of the structural mass about the base of the foundation: I f /Mh 2 - When the foundation is embedded, the depth ratio for the foundation block: D f /B 11.10.2 Further calculations with the water tower structure Example 11.2 We now return to the water tower example but now with foundation mass. Assuming a concrete foundation 1 m thick, a foundation of radius 3 m satisfies the ULS requirement for the foundation (the smaller foundation is a consequence of the of the foundation weight improving the moment capacity of the system), but we will keep to the 4.6 m radius foundation used in Example 11.1 to aid comparison. The calculations for this example are given in Appendix 11.2 and the results are: 0 1 2 0 2 4 6 8 h|M usM uhM T M 50 = M f 172.838 = V s 37.268 = a 4.6 = e s 6.283 = e h 18.838 = e | 7.867 = e eq 4.905 = Design of Earthquake Resistant Foundations 174 Figure 11.7 Steady state response to sinusoidal excitation of an elevated water tank on a shallow foundation with mass. The soil has a very low soil shear wave velocity to emphasise SSI effects. Figure 11.7 shows that because the foundation has mass then there are two more degrees of freedom – horizontal displacement and rocking of the foundation. In the plot components of the displacement of the tower mass are plotted separately. It is clear that the horizontal displacement mode contributes very little to the displacement. The fixed base period of this system is the same as in Example 11.1at 1 second. The peak response at 1.28 seconds is a consequence of the period lengthening due to the soil compliance. 11.11 MAXIMUM EARTHQUAKE FOUNDATION DISPLACEMENTS The material presented in sections 11.7 and 11.8 provides a helpful framework for coming to an understanding of the significance of various parameters on the steady state response. However, it does not provide an estimate of the earthquake induced foundation actions and displacements. As explained in Chapter 1 we will use the response spectrum method coupled with modal analysis to find these. Given here is a very brief statement of the modal analysis procedure, basically in the form of a series of steps. It is assumed that a computer mathematics package is available for the calculations. The steps explained below are illustrated with a numerical example in Appendices 11.3 and 11.4. A more thorough explanation of the modal analysis procedure is given in texts such as Chopra (1995) or Thomson (1988). In the development below we follow Butterworth (2007). 11.11.1 Calculation steps in modal analysis We start with equation 11.48 rewritten to include damping terms: | | ¦ ¹ ¦ ¹ ( ( ¦ ¦ ¦ ¦ ( ( + + + ´ ` ´ ` ( ( ¦ ¦ ¦ ¦ ( ( + | | ¸ ¸ ¸ ¸ ¹ ) ¹ ) ( ¦ ¹ ¦ ¹ ¦ ¦ ¦ ¦ ( = ÷ + ´ ` ´ ` ( ¦ ¦ ¦ ¦ ( | ¹ ) ¹ ) ¸ ¸        e s s s f e h h h 2 e e e f s s f h g f e M M Mh u C 0 0 u M M M Mh u 0 C 0 u Mh Mh Mh I 0 0 C k 0 0 u M 0 K 0 u u M M 0 0 K Mh (11.53) The operations for modal analysis are done using a computer mathematics package, what is presented here are the steps involved. Step (i) Find the modal frequencies of the system Chapter 11: Elastic structure-foundation-soil interaction 175 Since we have three degrees of freedom there will be three modal frequencies. The mathematical operations involved here are done by requesting eigenvalues. These are obtained by operating only on the stiffness and mass matrices, so at this stage the terms in the damping matrix are not involved in the calculations. We are assuming here that damping does not affect the modal frequencies. Reference to Figure 10.2 shows that for a single degree of freedom system this is reasonably true for damping coefficients up to about 20%. Step (ii) Obtain the mode shapes These are known as eigenvectors and are obtained by calling the appropriate function in the computer mathematics package being used. Step (iii) Normalise the mode shapes (not essential) Although not essential, this step is frequently done as it aids visualising the mode shapes; it has no effect on subsequent operations. This is usually done by dividing the components of each mode shape vector through by the component of the vector having the largest absolute magnitude. This means that the largest value in the eigen vector becomes unity. Combining these eienvectors into a 3x3 matrix gives us what is called the modal matrix. In what follows we will denote the modal matrix as u 2 , and the mode shape vectors (one for each mode) as ¢ i where, in the example we are discussing i ranges between 1 and 3. Step (iv) Plot the mode shapes (not essential) Now we have completed the preliminary calculations. The above discussion assumes that a computer mathematics package is available. However, since there are only three degrees of freedom, all the calculations to this point could be done by hand. In what follows the calculations could also be done by hand for a three degree of freedom system, but it is so much more convenient to have a computer method of doing the computations. We have seen that there three distinct modal periods. This suggests that when the structure-foundation system is subject to an excitation containing many frequencies, then all three modes are likely to contribute to the overall response of the system. This is where, the so-called, normal mode superposition comes into play. What we have calculated so far is the modal periods and mode shapes. Now we have to find how these modes interact and contribute to the deformations of the system and in generating the foundation actions. What we have to do is solve equations 11.53 for some specified input. However, these are a set of three coupled second order differential equations. One wonders if there is a simpler way; the answer is yes. Using some clever operations in linear algebra we can transform the coupled equations into three independent 2 Not to be confused with the strength reduction factor of the LRFD approach to ultimate limit state calculations discussed Chapter 6. Presumably context will eliminate any confusion. Design of Earthquake Resistant Foundations 176 equations which can be solved independently of each other. What is involved here is diagonalising the 3x3 matrices in equations 11.53. We then need to find the modal participation factors which tell us how the final displaced shape is made up of the summation of contributions from each mode. The concept here is that a given displacement component is obtained by summing the contributions from each of the modes: 1 1 2 2 3 3 u(t ) Y (t ) Y (t ) Y (t ) Y(t ) = ¢ + ¢ + ¢ = u   (11.54) where: u(t )  is the vector of the displacement components (system solution) Y 1 (t) etc are the modal coordinates (calculation details below) and u is the modal matrix and ¢ 1 etc individual mode shape vectors. Equation 11.54 tells is that if we have the modal coordinate amplitudes we can get the displacement of the system as we already have the modal matrix, u. The assumption underlying equation 11.54 is that the normal coordinates have no coupling (in other words they are orthogonal to use the mathematical terminology). Step (v) Reduce equations 11.53 to three uncoupled differential equations. A short hand form of equation 11.53 is: u u u + + =   M C K F (11.55) Comparison with equation 11.53 indicates the notation in equation 11.55. Now we can substitute equation 11.54 into 11.55 and obtain: Y Y Y u + u + u =   M C K F (11.56) These are still three coupled equations. We can now reduce them to independent equations by multiplying through by the transpose of u: u u u u u + u u + u u = u   T T T T M C F (11.57) As this represents three independent equations we can separate them: i i i Y Y Y T T T T i i i i i i i i F ¢ ¢ + ¢ ¢ + ¢ ¢ = ¢   M C K (11.58) Since there are three modes i ranges from 1 to 3. The factors in front of the Y terms the above equations are the generalized mass for mode i, the generalized damping, the generalized stiffness, and, on the right hand side, the generalized force. If we divide through by the generalized mass, we obtain: Chapter 11: Elastic structure-foundation-soil interaction 177 T i i i i i i i i C K Y Y Y M M ¢ + + =   i i F M (11.59) Using the definitions given in equations 10.2 and 10.4 this becomes: 2 i i i i i i Y 2 Y Y ¢ + , e + e =   T i i i F M (11.60) We now solve each of these three independent equations to obtain the vector of modal displacements. From there we obtain the actual displacements by applying equation11.54. The right hand sides of equations 11.59 and 11.60 are in effect an acceleration. The response spectrum gives separate accelerations for each modal frequency. This gives rise the modal participation factors defined as: T T i i i i T i i F F P M ¢ ¢ = = ¢ ¢ i M (11.61) 11.11.2 Response spectrum estimates of maximum displacements For each modal frequency (or period) and damping value we obtain the peak accelerations from the response spectrum and from there obtain the peak modal accelerations: 1 max 1 A 1 1 2 max 2 A 2 2 3 max 3 A 3 3 Y PS ( , ) Y P S ( , ) Y P S ( , ) = , e = , e = , e    (11.62) These can be converted to modal displacements using: 1 A 1 1 1 max 2 1 2 A 2 2 2 max 2 1 3 A 3 3 3 max 2 3 PS ( , ) Y P S ( , ) Y P S ( , ) Y , e = e , e = e , e = e (11.63) The above are the modal displacements, we convert to the actual displacement components for each mode as follows: Design of Earthquake Resistant Foundations 178 1 i Ai 2 i i max i 2 1 3 i max u PS u Y u ( ( = ¢ = ¢ ( e ( ¸ ¸ (11.64) We calculate a vector of maximum displacements for each mode. We obtain our “best estimate” of the actual maximum for each mode from the usual “square root of the sum of the squares”: 2 2 2 i max i mode 1 i mode 2 i mode 3 u u u u = + + (11.65) 11.11.3 Foundation actions Having obtained an estimate of the maximum displacement for each mode we use these displacements and the appropriate stiffness values to obtain the foundation actions: | = = = = | base s max s side wall h f 1 base h f foundation H u K H u K H u K Mom K (11.66) 11.11.4 Further calculations with the water tower structure Example 11.3 This example returns to Example 11.2 and uses the modal analysis procedure to calculated the modal periods. The calculations for this example are given in Appendix 11.3. It is found that the modal periods are the same as those The modal analysis procedure is used to get the modal periods shown in Figure 11.7, namely: 0.13, 0.31, and 1.28 seconds. 11.11.5 Multistorey building example using mode superposition Example 11.4 This example uses the modal analysis procedure to evaluate the response of a simple 12 storey building founded on a single storey basement. Soil structure interaction effects are included so the total number of modes for the structure- foundation system is 14 – one for each floor and horizontal translation and rotation of the foundation. Note that the size of the basement, which has a plan area substantially larger than that of the 12 storey tower, is required so that the foundation satisfies the LRFD requirement. Chapter 11: Elastic structure-foundation-soil interaction 179 The calculations for this example are given in Appendix 11.4. 11.12 SUMMARY REFERENCES Butterworth, J. W. (2007) Structural Dynamics Notes, pers. Comm. Chopra, A. (1995) Dynamics of Structures, Prentice Hall, New Jersey. Flugge, W. (1967) Viscoelasticity, Blaisdell, Waltham. Gazetas, G. (1991) “Foundation Vibrations”, Chapter 15 of Foundation Engineering Handbook, edited by Hsai-Yang Fang, Van Nostrand, 2nd edition, pp. 553-593. Richart, F. E., Hall, J. R. and Woods, R. D. (1970) Vibrations of Soils and Foundations, Prentice-Hall, New Jersey Thomson, W. T. (1988) Theory of Vibration with Applications, 3 rd edition, Unwin Hyman, London. Taylor, J. R. (2005) Classical Mechanics, University Science Books, Sausalito Wolf, J. P. (1985) Dynamic Soil-Structure Interaction, Prentice-Hall, Englewood Cliffs, New Jersey. APPENDIX 11-1 Soil-structure interaction calculations for Fig. 11.4 File: Appendix 11-1 28/08/2012 Verification of Fig. 3-22 in Wolf (1985). ORIGIN 1 := RIGID CIRCULAR FOUNDATION AT GROUND SURFACE (no embedment) 2 2 2 2 2 2 1 4 g s u u e e e , e = ( ÷ + ( ¸ ¸     Response ratio from Equation 11.44 in notes. First fixed base structure: ζ s 0.035 := Structural damping M 50 := Structural mass (tonnes) h 20 := Height of column (m) T 1.0 := Natural period (secs) (assumed) ω s 2 π · T := ω s 6.283 = Natural frequency (radians/sec) i 1 100 .. := ω i ω s 0.025 · i · := Ratio i ω i ω s := This gives the ratio of the steady state displacement amplitude of the mass to that the ground displacement amplitude. Response i Ratio i ( ) 2 1 Ratio i ( ) 2 ÷ ¸ ( ¸ 2 4 ζ s 2 · + := 0 1 2 3 0 5 10 15 Response Ratio Size foundation to provide the required bearing strength on clay with s u = 50 kPa s u 50 := Clay (kPa) ρ 1.8 := Soil density (tonnes/m^3) grav 9.81 := Vert M grav · := Vert 490.5 = Weight of structure (kN) Hor 0.2Vert · := Hor 98.1 = Assumed horizontal accn 0.2g Mom 0.2Vert · h · := Mom 1.962 10 3 × = a 4.6 := Trial radius for the foundation size (m) e Mom Vert := e 4 = ( ) 2 2 1 2 025 025 1 2 1 1 1 1 1 1 . . cos : / A r B L L L A and B A where e r ç ç ç ç ç ç ç ç ç ç ÷ ÷ ( ' ' ' = ÷ ÷ = = ¸ ¸ + | | | | + ÷ ' ' = = | | ÷ + \ . \ . = ξ e a := ξ 0.87 = Area 2 a 2 · acos ξ ( ) ξ 1 ξ 2 ÷ · ÷ ( ) · := Area 3.685 = Ld Area 1 ξ + 1 ξ ÷ | \ | | . 0.25 · := Bd Area Ld := Ld 3.735 = Bd 0.987 = q a Vert Area := q a 133.117 = q u_EC7 5.14 s u 1.4 · 1 0.2 Bd Ld · + | \ | | . · 0.5 · 1 1 Hor BdLd · s u 1.4 · ÷ | \ | | | . 0.5 + ¸ ( ( ( ( ¸ · := q u_EC7 145.389 = Note partial factor of 1.4 applied to the undrained shear strength - hence the label EC7 Q u_EC7 q u_EC7 Area · := Q u_EC7 535.717 = Vert 490.5 = Thus a surface circular foundation 4.6 m radius is OK, 4.5m is not. Now look at SSI effects with small strain shear modulus: k s ω s 2 M · := k s 1.974 10 3 × = Tower stiffness (kN/m) s u 50 := Stiff clay (kPa) ρ 1.8 := Soil density (tonnes/m^3) G 500 s u · 1.0 · := G 2.5 10 4 × = Small strain shear modulus (kPa) V s G ρ := V s 117.851 = Shear wave velocity (m/sec) a 4.6 := Foundation radius (m) q a M 9.81 · π a 2 · := q a 7.379 = 6 s u · q a 40.658 = Static bearing strength easily OK k h 8 G · a · 2 0.5 ÷ := k h 6.133 10 5 × = Horizontal stiffness of foundation (kN/m) ω h k h M := ω h 110.755 = Rotational stiffness of foundation (kNm/radian) k ϕ 8 G · a 3 · 3 1 0.5 ÷ ( ) · := k ϕ 1.298 10 7 × = ω ϕ k ϕ M h 2 · := ω ϕ 25.474 = ω s 6.283 = Equivalent natural frequency (radians/sec) (Reduction from e s mainly a consequence of h.) Equation 11.33 in notes. ω eq ω s 2 1 k s k h + k s h 2 · k ϕ + := ω eq 6.091 = Following Wolf, we will calculate the damping values at the system natural frequency. c h 4.6ρ · V s · a 2 · 2 0.5 ÷ := ζ h c h ω eq · 2 k h · := ζ h 0.068 = Horizontal damping ratio ρ 1.8 = c ϕ 0.4ρ · V s · a 4 · 1 0.5 ÷ := ζ ϕ c ϕ ω eq · 2 k ϕ · := ζ ϕ 0.018 = Rocking damping ratio ζ g 0.05 := Soil material damping ratio ζ eq ω eq 2 ω s 2 ζ s · 1 ω eq 2 ω s 2 ÷ | \ | | | . ζ g · + ω eq 2 ω h 2 ζ h · + ω eq 2 ω ϕ 2 ζ ϕ · + := Equation 11.37 in the notes. ζ eq 0.037 = i 1 110 .. := ωSSI i ω eq 0.05 · i · := RatioSSI i ωSSI i ω eq := ω i ω s 0.05 · i · := Ratio i ω i ω s := ResponseSSI1 i ωSSI i ω s | \ | | | . 2 1 ωSSI i ω eq | \ | | | . 2 ÷ ¸ ( ( ( ¸ 2 4 ζ eq 2 · + := Response i Ratio i ( ) 2 1 Ratio i ( ) 2 ÷ ¸ ( ¸ 2 4 ζ s 2 · + := 0 1 2 0 5 10 15 Response ResponseSSI1 RatioSSI Ratio ω eq ω s · , The above plot shows that when a foundation that satisfies bearing strength requirements and the small strain stiffness of the soil is used for the SSI calculations, the effect of the soil flexibility is negligible. Now repeat the above calculations reducing the soil modulus to 10% of the small strain value. k s ω s 2 M · := k s 1.974 10 3 × = Tower stiffness (kN/m) s u 50 := Stiff clay (kPa) ρ 1.8 := Soil density (tonnes/m^3) G 500 s u · 0.1 · := G 2.5 10 3 × = Small strain shear modulus (kPa) V s G ρ := V s 37.268 = Shear wave velocity (m/sec) a 4.6 = Foundation radius (m) q a M 9.81 · π a 2 · := q a 7.379 = 6 s u · q a 40.658 = Static bearing strength easily OK k h 8 G · a · 2 0.5 ÷ := k h 6.133 10 4 × = Horizontal stiffness of foundation (kN/m) ω h k h M := ω h 35.024 = Rotational stiffness of foundation (kNm/radian) k ϕ 8 G · a 3 · 3 1 0.5 ÷ ( ) · := k ϕ 1.298 10 6 × = ω ϕ k ϕ M h 2 · := ω ϕ 8.055 = ω s 6.283 = Equivalent natural frequency (radians/sec) (Reduction from e s mainly a consequence of h.) Equation 11.33 in notes. ω eq ω s 2 1 k s k h + k s h 2 · k ϕ + := ω eq 4.905 = Following Wolf, we will calculate the damping values at the system natural frequency. c h 4.6ρ · V s · a 2 · 2 0.5 ÷ := ζ h c h ω eq · 2 k h · := ζ h 0.174 = Horizontal damping ratio c ϕ 0.4ρ · V s · a 4 · 1 0.5 ÷ := ζ ϕ c ϕ ω eq · 2 k ϕ · := ζ ϕ 0.045 = Rocking damping ratio ζ g 0.05 := Soil material damping ratio ζ eq ω eq 2 ω s 2 ζ s · 1 ω eq 2 ω s 2 ÷ | \ | | | . ζ g · + ω eq 2 ω h 2 ζ h · + ω eq 2 ω ϕ 2 ζ ϕ · + := Equation 11.37 in notes. ζ eq 0.061 = i 1 110 .. := ωSSI i ω eq 0.05 · i · := RatioSSI i ωSSI i ω eq := ResponseSSI2 i RatioSSI i ( ) 2 1 RatioSSI i ( ) 2 ÷ ¸ ( ¸ 2 4 ζ eq 2 · + := 0 1 2 0 5 10 15 Frequency ratio A m p l i f i c a t i o n So we need a very severe reduction in soil modulus to observe a significant effect from the compliance of the soil beneath the foundation. Wolf's dimensionless parameters: Parameter from equation 11.45 Sbar ω s h · V s := Sbar 3.372 = hbar h a := hbar 4.348 = Parameter from equation 11.46 mbar M ρ a 3 · := mbar 0.285 = Parameter from equation 11.46 Only the first of the above three depends on the soil stiffness. When the small strain shear modulus is used Sbar =1.1. When we use Table 11.1which, for an acceleration of 0.2g, gives a reduction factor of 0.5, Sbar =1.5. If we try extreme softening of the soil and use a reduction factor of 0.1, then Sbar =3.4. Appendix 11-2 Tower structure with foundation mass 28/08/2012 File: Solution of equation 11.49 23/08/2011   ( ) ( ) ( ) | ¢ ( ¦ ¹ e ¦ ¦ | | ( + ç ÷ ÷ ÷ ´ ` | e ( \ . ¦ ¦ ¹ ) ( ¦ ¹ ¦ ¹ ¦ ¹ ( e ¦ ¦ ¦ ¦ ¦ ¦ | | ÷ + ç ÷ ÷ = ´ ` ´ ` ´ ` | ( + e + \ . ¦ ¦ ¹ ) ¦ ¦ ¦ ¦ ( | ¹ ) ¹ ) ( ¦ ¹ e | | | | ¦ ¦ ( ÷ ÷ + + ç ÷ ´ ` | | ( e \ . \ .¦ ¦ ¹ ) ¸ ¸ 2 s s s 2 h h h g f f e 2 f 2 e 1 2i 1 1 1 u 1 M M 1 2i 1 u u 1 M M M M h 1 I 1 1 1 1 2i 1 Mh These solutions, as functions of the excitation frequency e, are given by: ( ) ( ) ( ) ( ) ( ) + e ÷ e e = O e s 1 DC D B u Det (11.50) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) e = O + e ÷ + e ÷ + e e ÷ e ÷ e + + e ÷ e h u 1 DC D E C DE ED A C D A D C D D EA 1 Det (11.51) ( ) ( ) ( ) ( ) | e e e = O e e h A B Det (11.52) where: Det(e) is the value of the determinant of the 3x3 matrix on the left hand side of equation (11.49), A(e) is e s 2 (1+2iç s )/e 2 B(e) is e h 2 (1 + 2iç h + 2i, g )/e 2 C(e) is e| 2 (1 + 2iç| + 2i, g )/e 2 D is 1 + I f /Mh e 2 and E is M/(M + M f ) ORIGIN 1 := ξ s 0.035 := Structural damping M 50 := Structural mass (tonnes) h 20 := Height of column (m) Ts 1.0 := Structure natural period (secs) (assumed) ω s 2 π · Ts := ω s 6.283 = Natural frequency (radians/sec) Thickness of foundation (m) D f 1 := a 4.6 := Radius of surface foundation (m) M f π a 2 · D f · 2.6 · := M f 172.838 = Mass of foundation (tonnes) j 1 3000 .. := ω j ω s 200 j · := Ratio j ω j ω s := T j 2 π · ω j := max ω ( ) 94.248 = min T ( ) 0.067 = I f 1 4 M f · a 2 · 1 3 M f · D f 2 · + := I f 971.925 = Foundation moment of inertia (tonne m 2 ) k s ω s 2 M · := k s 1.974 10 3 × = Tower stiffness (kN/m) s u 50 := Stiff clay (kPa) ρ 1.8 := Soil density (tonnes/m^3) Ψ 0.1 := Modifier for the soil stiffness G 500 s u · Ψ · := G 2.5 10 3 × = Small strain shear modulus (kPa) V s G ρ := V s 37.268 = Shear wave velocity (m/sec) q a M 9.81 · π a 2 · := q a 7.379 = 6 s u · q a 40.658 = Static bearing strength easily OK k h 8 G · a · 2 0.5 ÷ := k h 6.133 10 4 × = Horizontal stiffness of foundation (kN/m) ω h k h M f := ω h 18.838 = Th 2 π · ω h := Th 0.334 = Rotational stiffness of foundation (kNm/radian) k ϕ 8 G · a 3 · 3 1 0.5 ÷ ( ) · := k ϕ 1.298 10 6 × = ω ϕ k ϕ M h 2 · I f + := ω ϕ 7.867 = Tϕ 2 π · ω ϕ := Tϕ 0.799 = ω s 6.283 = For comparison purposes the equivalent natural frequency for an SDOF system (radians/sec). Equation 11.33 in notes. ω eq ω s 2 1 k s k h + k s h 2 · k ϕ + := ω eq 4.905 = Teq 2 π · ω eq := Teq 1.281 = Following Wolf, we will calculate the damping values at the system natural frequency. c h 4.6ρ · V s · a 2 · 2 0.5 ÷ := ξ h c h ω eq · 2 k h · := ξ h 0.174 = Horizontal damping ratio c ϕ 0.4ρ · V s · a 4 · 1 0.5 ÷ := ξ ϕ c ϕ ω eq · 2 k ϕ · := ξ ϕ 0.045 = Rocking damping ratio ζ g 0.05 := Soil material damping ratio ξ eq ω eq 2 ω s 2 ξ s · 1 ω eq 2 ω s 2 ÷ | \ | | | . ζ g · + ω eq 2 ω h 2 ξ h · + ω eq 2 ω ϕ 2 ξ ϕ · + := ξ eq 0.07 = Equation 11.37 in notes. Solution of equation 11.49. (Recall that this is an algebraic equation rather than the coupled second order differential equations 11.48. Considering the steady state response to the excitation u = Ce iωt enables this simplification.) i 1 ÷ := MAT j ω s ω j | \ | | | . 2 1 2 i · ξ s · + ( ) · 1 ÷ M ÷ M M f + 1 ÷ 1 ÷ ω h ω j | \ | | | . 2 1 2 i · ξ h · + ( ) · 1 ÷ 1 ÷ 1 ÷ M ÷ M M f + ω ϕ ω j | \ | | | . 2 1 2 i · ξ ϕ · + ( ) · 1 ÷ ¸ ( ( ( ¸ 1 I f M h 2 · + | \ | | | . · ¸ ( ( ( ( ( ( ( ( ( ( ( ¸ := DET j MAT j := AA j ω s ω j | \ | | | . 2 1 2 i · ξ s · + ( ) · := BB j ω h ω j | \ | | | . 2 1 2 i · ξ h · + ( ) · := CC j ω ϕ ω j | \ | | | . 2 1 2 i · ξ ϕ · + ( ) · := DD 1 I f M h 2 · + := EE M M M f + := hϕ j AA j BB j · DET j ÷ ÷÷÷÷ := hϕM j Rehϕ j ( ) 2 Imhϕ j ( ) 2 + := us j 1 DD CC j · + DD ÷ ( ) BB j · DET j := usM j Reus j ( ) 2 Imus j ( ) 2 + := uh j 1 DD CC j · + DD ÷ ( ) EE CC j DD · EE · + EE DD · ÷ AA j CC j · DD · + AA j DD · ÷ CC j DD · ÷ DD + EE AA j · + 1 ÷ ( · DET j := uhM j Reuh j ( ) 2 Imuh j ( ) 2 + := 0 2 4 6 8 0 2 4 6 8 hϕM usM uhM Ratio ω eq ω s · Ratio ω eq ω s · , Ratio ω eq ω s · , M 50 = M f 172.838 = V s 37.268 = a 4.6 = ω s 6.283 = ω h 18.838 = ω ϕ 7.867 = ω eq 4.905 = 0 1 2 0 2 4 6 8 hϕM usM uhM T M 50 = M f 172.838 = V s 37.268 = a 4.6 = ω s 6.283 = ω h 18.838 = ω ϕ 7.867 = ω eq 4.905 = These plots show that because there are three degrees of freedom the system has three distinct periods. Note also that the lateral displacement of the tower mass is dominated by the contributions from foundation rotation and flexing of the tower, lateral foundation displacement is small. Finally, observe that the three periods of the response above (0.13, 0.31 and 1.28 seconds) the greatest response occurs at the equivalent period of the structure foundation system calculated above. Size foundation to provide the required bearing strength on clay with s u = 50 kPa and a acceleration amplitude of 0.2g. s u 50 := Clay (kPa) ρ 1.8 := Soil density (tonnes/m^3) grav 9.81 := a 3.0 := Trial radius for the foundation size (m) M f 172.838 = Vert M M f + ( ) grav · := Vert 2.186 10 3 × = Total weight of structure (kN) Hor 0.2Vert · := Hor 437.208 = Mom 0.2M · grav · h · := Mom 1.962 10 3 × = Assumed horizontal accn 0.2g e Mom Vert := e 0.898 = For a circular foundation the effective area is given by: ( ) 2 2 1 2 025 025 1 2 1 1 1 1 1 1 . . cos : / A r B L L L A and B A where e r ç ç ç ç ç ç ç ç ç ç ÷ ÷ ( ' ' ' = ÷ ÷ = = ¸ ¸ + | | | | + ÷ ' ' = = | | ÷ + \ . \ . = ξ e a := ξ 0.299 = Area 2 a 2 · acos ξ ( ) ξ 1 ξ 2 ÷ · ÷ ( ) · := Area 17.667 = Ld Area 1 ξ + 1 ξ ÷ | \ | | . 0.25 · := Bd Area Ld := Ld 4.905 = Bd 3.602 = q a Vert Area := q a 123.735 = q u_EC7 5.14 s u 1.4 · 1 0.2 Bd Ld · + | \ | | . · 0.5 · 1 1 Hor BdLd · s u 1.4 · ÷ | \ | | | . 0.5 + ¸ ( ( ( ( ¸ · := q u_EC7 163.603 = Note partial factor of 1.4 applied to the undrained shear strength - hence the label EC7 Q u_EC7 q u_EC7 Area · := Q u_EC7 2.89 10 3 × = Vert 2.186 10 3 × = Thus a surface circular foundation 3.0 m radius is OK, 2.9m is not. a 3 = This foundation size is smaller than that required in Example 11-1. There the required foundation radius was 4.6m. The difference between the two is this in this case the mass the foundation slab enhances the bearing strength reducing the eccentricity. The calculations in the first part of this example were done with the 4.6m radius foundation to make comparison with Example 11-1 more direct. Appendix 11-3 Modal analysis of Tower structure 28/08/2012 Developed from File: One storey basement.mcd 15, 17/06/02, 15/10/02, 6/11/02, 7/11/02, 9/11/02, 11/11/02 Circular foundation with mass at the ground surface supporting tank structure from Example 11-1. g 9.81 := Gravitational acceleration (m/sec 2 ) ORIGIN 1 := Soil properties: ν 0.5 := Poisson's ratio s u 50 := Stiff clay (kPa) ρ 1.8 := Soil density (tonnes/m^3) Ψ 0.1 := Modifier for the soil stiffness G 500 s u · Ψ · := G 2.5 10 3 × = Small strain shear modulus (kPa) V s G ρ := V s 37.268 = Shear wave velocity (m/sec) Structure properties ξ s 0.035 := Structural damping h 20 := Tower height (m) M 50 := Mass (tonnes) a 4.6 := Foundation radius (m) D f 1.0 := Thicknss of foundation slab (m) M f π a 2 · D f · 2.6 · := M f 172.838 = Foundation mass (tonnes) J f 1 4 M f · a 2 · 1 3 M f · D f 2 · + := J f 971.925 = Foundation moment of inertia(tonne m 2 ) Stiffness values: Tower natural period (sec) and frequency T 1 1 := ω s 2 π · T 1 · := ω s 6.283 = k s ω s 2 M · := k s 1.974 10 3 × = Tower stiffness (kN/m) Foundation lateral stiffness k h 8 G · a · 2 ν ÷ := k h 6.133 10 4 × = Horizontal stiffness of foundation (kN/m) ω h k h M f := ω h 18.838 = Th 2 π · ω h := Th 0.334 = Rotational stiffness of foundation (kNm/radian) k ϕ 8 G · a 3 · 3 1 ν ÷ ( ) · := k ϕ 1.298 10 6 × = ω ϕ k ϕ M h 2 · J f + := ω ϕ 7.867 = Tϕ 2 π · ω ϕ := Tϕ 0.799 = ω s 6.283 = For comparison purposes the equivalent natural frequency for an SDOF system (radians/sec). Equation 11.33 in notes. ω eq ω s 2 1 k s k h + k s h 2 · k ϕ + := ω eq 4.905 = Teq 2 π · ω eq := Teq 1.281 = In this document we will find the modal periods only, so we do not need the damping values. Equations of motion and modal periods K θ k ϕ := K o1 k h := K o1 6.133 10 4 × = Lateral stiffness for EQ parallel to side K o K o1 := Thus parallel case or diagonal with no lift off M o M f := Mass of the foundation J o J f := m M := J 0 := Single storey structure above the foundation K1 K o k s + k s h · k s ÷ k s h · K θ k s h 2 · + k s ÷ h · k s ÷ k s ÷ h · k s | \ | | | | | . := M1 M o 0 0 0 J o J + 0 0 0 m | \ | | | | . := fsq1 genvals K1 M1 , ( ) := f1 fsq1 := T1 2 π · f1 := Note that these modal periods are the same as those obtained for Example 11-2. f1 46.563 19.013 4.886 | \ | | | . = T1 0.135 0.33 1.286 | \ | | | . = sort T1 ( ) 0.135 0.33 1.286 | \ | | | . = Appendix 11-4 Modal analysis of a 12 storey building with a one storey basement File: Example 11-4 Modal analysis of a twelve storey building plus single storey basement and including elastic SSI. 15, 17/06/02, 15/10/02, 6/11/02, 9/11/02, 12/11/02, 28/08/2012 g 9.81 := Gravitational acceleration (m/sec 2 ) Soil properties: β 250 := Stiffness multiplier s u 50 := Undrained shear strength of clay (kPa) γ 18 := Unit weight of soil (kN/m 3 ) G β s u · := Shear modulus (kPa) V s Gg · γ := V s 82.538 = Shear waave velocity (m/sec) ν 0.5 := Poisson's ratio Structure properties s 8.0 := Column spacing (m) L s 14 + := Length of foundation slab (m) (Additional length beyond tower needed for bearing strength.) B L := Width of foundation slab (m) h 3.5 := Storey height (m) m s s · 7.2 · 9.81 := m 46.972 = Floor mass (tonnes) (plus DL and LL - 7.5kPa) n 12 := Number of floors D f 4.5 := Depth of foundation (m) d D f := Depth of sidewall contact (m) M f D f L 2 · 0.5 · := M f 1.089 10 3 × = Foundation mass (tonnes) A f L 2 := Foundation base area (m 2 ) A w 2L D f · := Sidewall area of foundation (m 2 ) J f M f 12 L 2 D f 2 + | \ | . · := J f 4.576 10 4 × = Moment of inertia of foundation block (tonne.m 2 ) J fl m 12 s 2 0.5 2 + ( ) · := J fl 251.498 = Moment of inertia of floor slab Stiffness values: Structure stiffness - need to set k so that 1st mode period with a rigid foundation is given by: 0.11H 0.75 (H height of building - m) T 1 0.11 n h · ( ) 0.75 · := T 1 1.815 = k 35500 := Lateral stiffness of 4 columns (kN/m) Calculated to get the natu period correct. Foundation lateral stiffness Kh GL · 2 ν ÷ := Is 2 2.5 A f L 2 | \ | | | . 0.85 · + := Id 1 0.15 2 D f B · | \ | | . 0.5 · + := Iw 1 0.52 8 D f 2 · A w B L 2 · · | \ | | | . 0.4 · + := Kh KhIs · Id · Iw · := Kh 1.208 10 6 × = Foundation rotational stiffness Second moment of area of foundation (m 4 ) Ibx B L 3 · 12 := Krx GIbx 0.75 · 1 ν ÷ := Krx 4.129 10 7 × = Trx 1 2.52 d B | \ | | . · 1 2 d B · | \ | | . D f d | \ | | . 0.2 · B L | \ | | . 0.5 · + ¸ ( ( ( ¸ · + := Sry 3 L B | \ | | . 0.15 · := Sry 3 = KR Krx Sry · Trx · := KR 2.138 10 8 × = (kNm/rad.) Equations of motion and modal periods K o Kh := K θ KR := M o M f := J o J f := m m := J J fl := Twelve storey structure above the foundation. One degree of freedom for each storey plus horizontal displacement and rotation of the foundation. i 0 13 .. := j 0 13 .. := K i j , 0 := M i j , 0 := K12 K 0 0 , K o k + ÷ K 0 1 , k h · ÷ K 0 2 , k ÷ ÷ K 1 0 , k h · ÷ K 1 1 , K θ 12k · h 2 · + ÷ K 1 13 , k ÷ h · ÷ K 2 0 , k ÷ ÷ K 2 2 , 2 k · ÷ K 2 3 , k ÷ ÷ K l l 1 ÷ , k ÷ ÷ K l l , 2 k · ÷ K l l 1 + , k ÷ ÷ l 3 12 .. e for K 13 12 , k ÷ ÷ K 13 13 , k ÷ K 13 1 , k ÷ h · ÷ K := M12 M i i , m ÷ i 0 13 .. e for M 1 1 , J o 12J · + ÷ M 0 0 , M o ÷ M := fsq12 genvals K12 M12 , ( ) := f12 fsq12 := T12 2 π f12 · := Gets eigenvalues (ie modal frequencies) f12 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 67.027 54.55 53.26 51.133 48.205 44.528 40.181 35.478 33.345 29.258 23.319 3.409 10.265 16.936 = T12 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 0.094 0.115 0.118 0.123 0.13 0.141 0.156 0.177 0.188 0.215 0.269 1.843 0.612 0.371 = sort T12 ( ) 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 0.094 0.115 0.118 0.123 0.13 0.141 0.156 0.177 0.188 0.215 0.269 0.371 0.612 1.843 = Now get eigen vectors to eventually determine the mode shapes: PHI12 genvecs K12 M12 , ( ) := I 1 0 1 1 1 1 1 1 1 1 1 1 1 1 ( ) T := P12 PHI12 1 ÷ I · := P12 T 0 1 2 3 4 5 0 -4 1.88·10 -3 -6.207·10 0.014 -0.024 0.041 ... = Get the absolute maximum for each modal vector: MaxPHI j max PHI12 j ( ) ( ) 2 ¸ ( ¸ := Normalise modal vectors: PHI12 i j , PHI12 i j , MaxPHI j := 0 13 i P12 i ¿ = 3.861 ÷ 10 3 ÷ × = PHI12 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 0.034 -3 -4.513·10 -3 9.399·10 -0.015 0.024 -0.039 1 -4 -2.027·10 -4 -3.965·10 -4 -4.776·10 -4 -5.932·10 -4 -5.794·10 -3 -9.296·10 0.253 -0.49 0.696 -0.858 1 -3 2.53·10 -0.486 0.85 -1 0.898 -0.585 -4 -6.832·10 0.688 -1 0.764 -0.108 -0.635 -4 1.651·10 -0.848 0.903 -0.115 -0.783 0.981 -5 3.207·10 0.954 -0.584 -0.596 0.949 0.024 -4 -2.916·10 -1 0.12 0.985 -0.237 -0.996 -3 1.118·10 0.984 0.373 -0.841 -0.694 0.597 -3 -4.119·10 -0.906 -0.774 0.243 0.983 0.623 0.015 0.771 0.984 0.487 -0.362 -0.986 -0.056 -0.588 -0.952 -0.953 -0.594 -3 -8.769·10 0.204 0.368 0.684 0.905 1 0.991 -0.749 -0.125 -0.248 -0.367 -0.481 ... = Modal participation factors: P12 PHI12 1 ÷ I · := P12 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 -4 1.88·10 -3 -6.207·10 0.014 -0.024 0.041 -0.074 0.172 -0.674 0.867 0.491 0.325 -1.286 0.465 -0.314 = Storey 0 1 2 3 4 5 6 7 8 9 10 11 12 | \ | | | | | | | | | | | | | | | | | . := 0 13 i P12 i ¿ = 3.861 ÷ 10 3 ÷ × = Set up normalised mode shapes for plotting by removing the rotational components - row 1 Plot stack submatrix PHI12 0 , 0 , 0 , 13 , ( ) submatrix PHI12 2 , 13 , 0 , 13 , ( ) , ( ) := Set up normalised mode shapes for plotting by removing the rotational components - row 1 Plot stack submatrix PHI12 0 , 0 , 0 , 13 , ( ) submatrix PHI12 2 , 13 , 0 , 13 , ( ) , ( ) := Plot 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 11 12 0.034 -3 -4.513·10 -3 9.399·10 -0.015 0.024 -0.039 -3 -9.296·10 0.253 -0.49 0.696 -0.858 1 -3 2.53·10 -0.486 0.85 -1 0.898 -0.585 -4 -6.832·10 0.688 -1 0.764 -0.108 -0.635 -4 1.651·10 -0.848 0.903 -0.115 -0.783 0.981 -5 3.207·10 0.954 -0.584 -0.596 0.949 0.024 -4 -2.916·10 -1 0.12 0.985 -0.237 -0.996 -3 1.118·10 0.984 0.373 -0.841 -0.694 0.597 -3 -4.119·10 -0.906 -0.774 0.243 0.983 0.623 0.015 0.771 0.984 0.487 -0.362 -0.986 -0.056 -0.588 -0.952 -0.953 -0.594 -3 -8.769·10 0.204 0.368 0.684 0.905 1 0.991 -0.749 -0.125 -0.248 -0.367 -0.481 ... = 2 ÷ 1 ÷ 0 1 2 0 3 6 9 12 Storey Storey Storey Storey Storey Storey Plot 11 ( ) Plot 13 ( ) , Plot 12 ( ) , Plot 0 ( ) , Plot 1 ( ) , Plot 2 ( ) , Plot of some of the mode shapes. Red - st mode, dotted green - 2nd mode, dotted blue - 3rd mode. Spectral accelerations: From Table 3.1 in DR 1170.4/PPC3 C(T) =C h (T)ZRN(T,D) where: Z is the hazard factor - take as 0.4 (central part of NZ) R is the return period factor - take as 1.0 (500 year return period) N(T,D) is the near fault factor - take as 1.0 For ductile structures ductility and the so-called Structural performance factor reduce the design actions below those for elastic response. The equation is now: C(T) =C h (T)ZRN(T,D)Sp/µ where: Sp is the Structural Performance Factor µ is the ductility We will use the following: Z 0.40 := Sp 0.67 := μ 3 := The values of Ch(T) are given in Table 3.1 of the draft standard and in equation form in the Commentary. There are three categories of site condition: rock, shallow soil, deep soil and soft soil. Rocka TT ( ) EQ i 1.00 TT i 0 = if 1.00 1.35 TT i 0.1 | \ | | . · + ¸ ( ( ¸ 0 TT i < 0.1 s if 2.35 0.1 TT i < 0.3 s if 1.60 0.5 TT i | \ | | . 0.75 · ¸ ( ( ( ¸ 0.3 TT i < 1.5 s if 1.05 TT i 1.5 TT i < 3 s if 3.15 TT i ( ) 2 TT i 3 > if ÷ i 0 length TT ( ) 1 ÷ .. e for EQ := Shallowa TT ( ) EQ i 1.33 TT i 0 = if 1.33 1.60 TT i 0.1 | \ | | . · + ¸ ( ( ¸ 0 TT i < 0.1 s if 2.93 0.1 TT i < 0.3 s if 2.0 0.5 TT i | \ | | . 0.75 · ¸ ( ( ( ¸ 0.3 TT i < 1.5 s if 1.32 TT i 1.5 TT i < 3 s if 3.96 TT i ( ) 2 TT i 3 > if ÷ i 0 length TT ( ) 1 ÷ .. e for EQ := Deepa TT ( ) EQ i 1.12 TT i 0 = if 1.12 1.88 TT i 0.1 | \ | | . · + ¸ ( ( ¸ 0 TT i < 0.1 s if 3.00 0.1 TT i < 0.56 s if 2.4 0.75 TT i | \ | | . 0.75 · ¸ ( ( ( ¸ 0.56 TT i < 1.5 s if 2.14 TT i 1.5 TT i < 3 s if 6.42 TT i ( ) 2 TT i 3 > if ÷ i 0 length TT ( ) 1 ÷ .. e for EQ := VSofta TT ( ) EQ i 1.12 TT i 0 = if 1.12 1.88 TT i 0.1 | \ | | . · + ¸ ( ( ¸ 0 TT i < 0.1 s if 3.00 0.1 TT i < 1.0 s if 3.0 1.0 TT i | \ | | . 0.75 · ¸ ( ( ( ¸ 1.0 TT i < 1.5 s if 3.32 TT i 1.5 TT i < 3 s if 9.96 TT i ( ) 2 TT i 3 > if ÷ i 0 length TT ( ) 1 ÷ .. e for EQ := T12 sort T12 ( ) := Arrange the modal periods in order Shallow elastic site condition: Spect_acc Z Shallowa T12 ( ) · g · := 0 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 1.2 Spect_acc g T12 Plot of the spectral accelerations at the various modal periods Eliminate 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ( ) T := P12 P12 T diag Eliminate ( ) · := P12 P12 T := Modal acceleration amplitudes Yacc P12Spect_acc · ( ) ÷ ÷÷÷÷÷÷ := Modal displacement amplitudes Ymax Yacc fsq12 | \ | | . ÷ ÷÷÷ := Displacements in each mode u PHI12diag Ymax ( ) · := Max probable displacements (SRSS basis) umax u u T · := j 0 13 .. := umax2 j umax j j , := Only need diagonal components of umax umax umax2 ÷ ÷÷÷ := u 0 u u 1 u 2 umax 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 -3 8.68·10 -4 6.16·10 0.14 0.272 0.401 0.523 0.638 0.742 0.835 0.915 0.981 1.032 1.068 1.087 = Ymax 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 -7 4.647·10 -5 -2.398·10 -5 5.499·10 -4 -1.063·10 -4 2.051·10 -4 -4.289·10 -3 1.226·10 -3 -6.156·10 -3 8.965·10 -3 6.588·10 -3 6.871·10 -1.086 0.03 -3 -3.078·10 = Weight M f n m · + ( ) g · := Weight 1.621 10 4 × = Bottom storey shear and moment (J ohn Butterworth) F12 M12 PHI12 · diag Yacc ( ) · := Storey forces (all modes) I2 I := I2 0 0 := Contrived to sum storey forces excluding foundation level V I2 T F12 · := Bottom storey shears for all modes V1 V V T · := SRSS combination for probable maximum V1 4.742 10 3 × = The answer! V1 n m · g · 0.858 = F x y , ( ) if y 2 < 0 , y 1 ÷ , ( ) h · := I3 matrix 1 14 , F , ( ) := Mom I3F12 · := M1 MomMom T · := M1 1.313 10 5 × = Moment at base of 1st storey F x y , ( ) if y 3 < 0 , y 2 ÷ , ( ) h · := I4 matrix 1 14 , F , ( ) := Mom I4F12 · := M2 MomMom T · := M2 1.149 10 5 × = Moment at base of 2nd storey Foundation shear Found_sh K o umax 0 · := Found_sh 1.048 10 4 × = Found_sh Weight 0.647 = Foundation moment Found_mom K θ umax 1 · := Found_mom 1.317 10 5 × = Found_mom Found_sh n · h · 0.299 = Foundation bearing strength Need to iterate until the foundation plan dimension satifies LRFD - this is the reason that the dimensions of the basement are larger than the tower. FOUNDATION LOADS (Raft foundation, diagonal actions): V Weight := Vertical load, kN H Found_sh 2 := H 7.412 10 3 × = Horizontal shear, kN 2 s u · B · D f · 9.9 10 3 × = H 0.0 := Assume all the shear is taken in passive pressure. Mx Found_mom 2 := Moment about the x (B) axis, kNm My Found_mom 2 := Moment about the y (L) axis, kNm EFFECTIVE DIMENSIONS: ex My V := ex 5.744 = Bd B 2 ex · ÷ := Bd 10.511 = ey Mx V := ey 5.744 = Ld L 2 ey · ÷ := Ld 10.511 = Need to check that Bd <Ld, if not switch: LT Ld := BT Bd := Bd if LT BT > BT , LT , ( ) := Ld if BT LT > BT , LT , ( ) := Ld 10.511 = Bd 10.511 = APPLIED STRESSES: Surcharge pressure: q γ D f · := q 81 = (kPa) Gross applied pressure: qa V Ld Bd · := qa 146.741 = (kPa) SHAPE FACTORS: λcs 1 0.2 Bd Ld · + := λcs 1.2 = λqs 1 := DEPTH FACTORS: λcd if D f Bd < 1 0.4 D f Bd | \ | | . · + , 1 0.4atan D f Bd | \ | | . · + , ¸ ( ( ¸ := λcd 1.171 = λqd 1.0 := λ cd 1.0 := Depth factor set to zero because of the passive pressure mobilisation INCLINATION FACTORS: H set to zero on the assumption that it is taken by passive pressire λci 0.5 1 1 H Ld Bd · s u · ÷ + | \ | | . · := λci 1 = λci λci 2 := Horizontal shear in two directions λqi λci := COHESION COMPONENT OF GROSS ULTIMATE BEARING PRESSURE: quc s u 5.14 · λcs · λcd · λci · := quc 361.212 = SURCHARGE COMPONENT OF GROSS ULTIMATE BEARING PRESSURE: quq q λqs · λqd · λqi · := quq 81 = GROSS ULTIMATE BEARING PRESSURE: qu quc quq + := qu 442.212 = Φ 0.5 := Strength reduction factor for bearing Φ qu · 221.106 = qa 146.741 = OK (could have used net bearing pressure and perhaps achieved a smaller foundation) Summary max T12 ( ) 1.843 = Found_sh 1.048 10 4 × = V1 4.742 10 3 × = V s 82.538 = Found_mom 1.317 10 5 × = M1 1.313 10 5 × = s u 50 =