Seismic Analysis Of Structures - Iii 65p5t

Chapters – 5 & 6 Chapter -5

RESPONSE SPECTRUM METHOD OF ANALYSIS T.K. Datta Department Of Civil Engineering, IIT Delhi

Response Spectrum Method Of Analysis

1/1

Introduction Response spectrum method is favoured by earthquake engineering community because of:  It provides a technique for performing an equivalent static lateral load analysis.  It allows a clear understanding of the contributions of different modes of vibration.  It offers a simplified method for finding the design forces for structural for earthquake.

 It is also useful for approximate evaluation of seismic reliability of structures. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/2

Contd…  The concept of equivalent lateral forces for earthquake is a unique concept because it converts a dynamic analysis partly to dynamic & partly to static analysis for finding maximum stresses.  For seismic design, these maximum stresses are of interest, not the time history of stress.  Equivalent lateral force for an earthquake is defined as a set of lateral force which will produce the same peak response as that obtained by dynamic analysis of structures .  The equivalence is restricted to a single mode of vibration. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/3

Contd…  The

response spectrum method of analysis is developed using the following steps.  A modal analysis of the structure is carried out to obtain mode shapes, frequencies & modal participation factors.  Using the acceleration response spectrum, an equivalent static load is derived which will provide the same maximum response as that obtained in each mode of vibration.

 Maximum modal responses are combined to find total maximum response of the structure. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/4

Contd…  The first step is the dynamic analysis while , the second step is a static analysis.  The first two steps do not have approximations, while the third step has some approximations.

 As a result, response spectrum analysis is called an approximate analysis; but applications show that it provides mostly a good estimate of peak responses.  Method is developed for single point, single component excitation for classically damped linear systems. However, with additional approximations it has been extended for multi point-multi component excitations & for nonclassically damped systems. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/5

Development of the method  Equation of motion for MDOF system under single point excitation (5.1) Mx  Cx  Kx   MIx g

 Using modal transformation, uncoupled sets of equations take the form

zi  2ii zi   z  i xg ; i  1  MI i   Mi 2 i i T i T i



m

(5.2)

i is the mode shape; ωi is the natural frequency λ is the more participation factor; ξ is the i i modal damping ratio.

T.K. Datta Department Of Civil Engineering, IIT Delhi


1/6

Contd…  Response of the system in the ith mode is (5.3)

x i = φi z i

 Elastic force on the system in the ith mode

fsi = Kx i = Kφi z i

(5.4)

 As the undamped mode shape i satisfies

Kφi = ωi2Mφi

(5.5)

 Eq 5.4 can be written as

fsi = ωi2Mφi z i

(5.6)

 The maximum elastic force developed in the ith mode fsimax = Mφi ωi2 z imax (5.7) T.K. Datta Department Of Civil Engineering, IIT Delhi


1/7

Contd…  Referring to the development of displacement response spectrum (5.8) zi max  i Sdi i , i  2  Using S a   S d , Eqn 5.7 may be written as

f s i max  i M i S ai  Pei

(5.9)

 Eq 5.4 can be written as

xi max  K 1 f si max  K 1Pei

(5.10)

 Pe i is the equivalent static load for the ith mode of vibration.  Pe i is the static load which produces structural displacements same as the maximum modal displacement. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/8

Contd…  Since both response spectrum & mode shape properties are required in obtaining Pe i , it is known as modal response spectrum analysis.  It is evident from above that both the dynamic & static analyses are involved in the method of analysis as mentioned before.  As the contributions of responses from different modes constitute the total response, the total maximum response is obtained by combining modal quantities.  This combination is done in an approximate manner since actual dynamic analysis is now replaced by partly dynamic & partly static analysis. T.K. Datta Department Of Civil Engineering, IIT Delhi


2/1

Contd… Modal combination rules  Three different types of modal combination rules are popular  ABSSUM  SRSS  CQC  ABSSUM stands for absolute sum of maximum values of responses; If x is the response quantity of interest m x   xi max (5.11) i 1

xi max is the absolute maximum value of response in the ith mode. T.K. Datta Department Of Civil Engineering, IIT Delhi


2/2

Contd…  The combination rule gives an upper bound to the computed values of the total response for two reasons:  It assumes that modal peak responses occur at the same time.  It ignores the algebraic sign of the response.  Actual time history analysis shows modal peaks occur at different times as shown in Fig. 5.1;further time history of the displacement has peak value at some other time.

 Thus, the combination provides a conservative estimate of response. T.K. Datta Department Of Civil Engineering, IIT Delhi


Top floor displacement (m)

2/3

0.4 0.2 0

-0.2 -0.4 0

5 t=6.15

10

15

20

25

30

First generalized displacement (m)

(a) Top storey displacement 0.4 0.2 0 -0.2 -0.4 0

5 t=6.1

10

15 Time (sec)

20

25

30

(b) First generalized displacement

Fig T.K. Datta Department Of Civil Engineering, IIT Delhi

5.1 Response Spectrum Method Of Analysis

Contd…

2/3

Second generalized displacement (m)

0.06

0.04

0.02

0

-0.02

-0.04

-0.06 0

t=2.5

5

10

15

20

25

30

Time (sec)

(c) Second generalized displacement

Fig 5.1 (contd.) T.K. Datta Department Of Civil Engineering, IIT Delhi


2/4

Contd…  SRSS combination rule denotes square root of sum of squares of modal responses  For structures with well separated frequencies, it provides a good estimate of total peak response.

x

m

x i 1

2 i max

(5.12)

 When frequencies are not well separated, some errors are introduced due to the degree of correlation of modal responses which is ignored.

 The CQC rule called complete quadratic combination rule takes care of this correlation. T.K. Datta Department Of Civil Engineering, IIT Delhi


2/5

Contd…  It is used for structures having closely spaced frequencies:

x

m

m

m

2 x  i   ij xi x j i 1

(5.13)

i 1 j 1

 Second term is valid for i of degree of correlation.

j

& includes the effect

 Due to the second term, the peak response may be estimated less than that of SRSS.  Various expressions for  i j are available; here only two are given : T.K. Datta Department Of Civil Engineering, IIT Delhi


2/6

Contd… ij

1     1     4



2

2

ij

2

ij

ij 

(Rosenblueth & Elordy) (5.14)

ij

1       4  1   

8

1  ij

2

3

2

ij

2

2

ij

2

ij

(Der Kiureghian)

(5.15)

2

ij

 Both SRSS & CQC rules for combining peak modal responses are best derived by assuming earthquake as a stochastic process.

 If the ground motion is assumed as a stationary random process, then generalized coordinate in each mode is also a random process & there should exist a cross correlation between generalized coordinates. T.K. Datta Department Of Civil Engineering, IIT Delhi


2/7

Contd…  Because of this,  i j exists between two modal peak responses.

 Both CQC & SRSS rules provide good estimates of peak response for wide band earthquakes with duration much greater than the period of structure.  Because of the underlying principle of random vibration in deriving the combination rules, the peak response would be better termed as mean peak response. Fig 5.2 shows the variation of  i j with frquency ratio.  i j rapidly decreases as frequency ratio increases. T.K. Datta Department Of Civil Engineering, IIT Delhi


2/8

Contd…

Fig 5.2 T.K. Datta Department Of Civil Engineering, IIT Delhi


2/9

Contd…

As both response spectrum & PSDF represent frequency contents of ground motion, a relationship exists between the two.  This relationship is investigated for the smoothed curves of the two. Here a relationship proposed by Kiureghian is presented  2 4   D  ,    S xg              ff      p0   



 2

p0 ( )  T.K. Datta Department Of Civil Engineering, IIT Delhi

2

 2.8  2 ln    2 

(5.16 a)

(5.16 b)


2/10

Contd…

2 -3 PSDF of acceleration sec /rad) (m

Example 5.1 : Compare between PSDFs obtained from the smoothed displacement RSP and FFT of Elcentro record. 0.05

Unsmoothed PSDF from Eqn 5.16a Raw PSDF from fourier spectrum

0.04 0.03 0.02 0.01 00

10

20

30 40 Frequency (rad/sec)

Eqn.5.16a Fourier spectrum of El Centro

0.02

2

-3

60

Unsmoothed

0.025 (m PSDFs of acceleration sec/rad)

50

0.015 0.01 0.005 00

10

20

30

40 50 60 Frequency (rad/sec)

70

80

90

100

5 Point smoothed T.K. Datta Department Of Civil Engineering, IIT Delhi

Fig5.3 Response Spectrum Method Of Analysis

Application to 2D frames  Degree of freedom is sway degree of freedom.  Sway d.o.f are obtained using condensation procedure; during the process, desired response quantities of interest are determined and stored in an array R for unit force applied at each sway d.o.f.  Frequencies & mode shapes are determined using M matrix & condensed K matrix.  For each mode find  (Eq. 5.2) & obtain Pei i (Eq. 5.9) N r W ir  i  N r 1 (5.17) 2 r W     ir r 1



2/11

2/12

Contd…  Obtain R j  RPej ( j  1...r ) ; R j is the modal peak response vector.

 Use either CQC or SRSS rule to find mean peak response. Example 5.2 : Find mean peak values of top displacement, base shear and inter storey drift between 1st & 2nd floors. Solution : ω1 =5.06rad/s; ω2 =12.56rad/s;

ω3 =18.64rad/s; ω 4 = 23.5rad/s φ1T =  -1 -0.871 -0.520 -0.278 ; φ 2T = -1 -0.210 0.911 0.752 φ3T =  -1 0.738 -0.090 -0.347 ; φ T4 = 1 -0.843 0.268 -0.145 T.K. Datta Department Of Civil Engineering, IIT Delhi


2/13

Contd… Table 5.1 Disp (m)

Base shear in of mass (m)

Drift (m)

Approaches

2 modes

all modes

2 modes

all modes

2 modes

all modes

SRSS

0.9171

0.917

1006.558

1006.658

0.221

0.221

CQC

0.9121

0.905

991.172

991.564

0.214

0.214

ABSSUM

0.9621

0.971

1134.546

1152.872

0.228

0.223

Time history

0.8921

0.893

980.098

983.332

0.197

0.198



Application to 3D tall frames  Analysis is performed for ground motion applied to each principal direction separately.  Following steps are adopted:  Assume the floors as rigid diaphragms & find the centre of mass of each floor.

 DYN d.o.f are 2 translations & a rotation; centers of mass may not lie in one vertical (Fig 5.4).  Apply unit load to each dyn d.o.f. one at a time & carry out static analysis to find condensed K matrix & R matrix as for 2D frames.  Repeat the same steps as described for 2D frame T.K. Datta Department Of Civil Engineering, IIT Delhi


3/1

3/2 C.G. of mass line C.G. of mass line

CM1

CM1

L

L CM2 CM2

L

L CM3

CM3

L

L

L

(a)

L

L

xg 



(b)

xg

x

x

Figure 5.4: T.K. Datta Department Of Civil Engineering, IIT Delhi


3/3

Contd… Example 5.3 : Find mean peak values of top floor displacements , torque at the first floor & VX and VY at the base of column A for exercise for problem 3.21. Use digitized values of the response spectrum of El centro earthquake ( Appendix 5A of the book). Solution : ω1 =13.516rad/s; ω2 =15.138rad/s; ω3 = 38.731rad/s; ω4 = 39.633rad/s ; ω5 = 45.952rad/s; ω6 =119.187rad/s

Results are obtained following the steps of section 5.3.4. Results are shown in Table 5.2. T.K. Datta Department Of Civil Engineering, IIT Delhi


3/4

Contd… TABLE 5.2 Torque Approac

displacement (m) (rad)

Vx(N)

Vy(N)

hes (1)

(2)

(3)

SRSS

0.1431

0.0034

0.0020

214547

44081

CQC

0.1325

0.0031

0.0019

207332

43376

0.1216

0.0023

0.0016

198977

41205

Time history

 Results obtained by CQC are closer to those of time history analysis. T.K. Datta Department Of Civil Engineering, IIT Delhi


RSA for multi excitation  Response spectrum method is strictly valid for single point excitation.

 For extending the method for multi excitation, some additional assumptions are required.  Moreover, the extension requires a derivation through random vibration analysis. Therefore, it is not described here; but some features are given below for understanding the extension of the method to multi excitation.  It is assumed that future earthquake is represented by an averaged smooth response spectrum & a PSDF obtained from an ensemble of time histories.



3/5

3/6

Contd…

 Lack of correlation between ground motions at two points is represented by a coherence function.  Peak factors in each mode of vibration and the peak factor for the total response are assumed to be the same.  A relationship like Eqn. 5.16 is established between S d and PSDF.  Mean peak value of any response quantity r consists of two parts: T.K. Datta Department Of Civil Engineering, IIT Delhi


3/7

Contd… • Pseudo static response due displacements of the s

to

the

• Dynamic response of the structure with respect to s. Using normal mode theory, uncoupled dynamic equation of motion is written as: s

zi  2i zi  i2 zi    ki uk ; i  1..m (5.18) k 1

i T MRk ki  T i Mi T.K. Datta Department Of Civil Engineering, IIT Delhi

(5.19)


3/8

Contd…

 If the response of the SDOF oscillator to uk is zki s then zi    ki z ki (5.20) k 1

 Total response is given by s

m

r  t    ak uk  t    i zi  t  k 1

i 1

s

m

s

r  t    ak uk  t    i   ki zki

(5.22)

r  t   a T u  t   T z  t 

(5.23)

k 1



(5.21)

i 1

k 1

φ β and z are vectors of size m x s (for s=3 & m=2)



3/9

Contd…

φβT = 1 β11  1 β21  1 β31  2 β12 2β22  2β32  (5.24a) z T = z11 z21 z31 z12 z22 z32 

 

(5.24b)



 Assuming r t ,u t and z t to be random processes, PSDF of r (t ) is given by:

S rr  a S uua    S zz   a S uz    S zua T

T

T

T

(5.25)

 Performing integration over the frequency range of interest & considering mean peak as peak factor multiplied by standard deviation, expected peak response may be written as: T.K. Datta Department Of Civil Engineering, IIT Delhi


3/10

Contd… E max r  t   = b

T

b +b

uu

b T = a1up1 a 2up2 a3up3

T

uz

φβD + φ

T βD zz

φβD + φ

aSupS 

(5.27a)

T φβD =  φ1β11D11 φ1β 21D21 .... φ1β s1Ds1 ...φ mβ11D1m 

Dij = Di  ω j ,ξ j 

i =1,..,s ; j =1,..,m

 luu , lu z and

lz z

12

b  (5.26)

T βD zu

(5.27b) (5.27c)

are the correlation matrices

whose elements are given by: uiu j

1 = σui σuj

α

 S  ω  dω

-α

uiu j


(5.28)


3/11

Contd… ui zkj

zki zlj

1 = σui σ zkj

1 = σ zki σ zlj

α

* h  j Suiuk  ω dω

(5.29)

-α α

* h h  i j Sukul  ω  dω

(5.30)

-α

Suiuk

coh  i,k  1 21 21 = 2 Sui Suk coh  i,k  = Sug 2 ω ω

(5.31)

Suiuj

coh  i, j  1 21 21 = 4 Sui Suj coh  i, j  = Sug 4 ω ω

(5.32)

1 2 uk

1 2 ul

Sukul = S S coh k,l = coh k,l Sug T.K. Datta Department Of Civil Engineering, IIT Delhi

(5.33)


3/12

Contd…

 For a single train of seismic wave, Dij = Di  ωj ,ξ j  that is displacement response spectrum for a specified ξ ; correlation matrices can be obtained if coh(i, j ) is additionally provided; Su g can be determined from D  ωj ,ξ j  (Eqn 5.6).  If only relative peak displacement is required,third term of Eqn.5.26 is only retained.  Steps for developing the program in MATLAB is given in the book.

Example 5.4 Example 3.8 is solved for El centro earthquake spectrum with time lag of 5s. T.K. Datta Department Of Civil Engineering, IIT Delhi


3/13

Contd… Solution :The quantities required for calculating the expected value are given below: 1 1  1  1 1 1 1 1 φ  ; φ  0.5 1 ; r  3 1 1 1 , 0.5  1       w1  12.24 rad/s ; w2  24.48 rad/s  1111 1 1 1 1 T a   ;    3 1 1 1 2111 0.0259 0.0259 0.0259  T D   0.0129 0.0129 0.0129 T

11 21 1131 12 12 12  22 12  32  21 21 2131 22 12 22  22 22 32 

-0.0015 -0.0015 -0.0015  0.0015 0.0015 0.0015  D11  D21  D31  D(1  12.24)  0.056m D12  D22  D32  D (2  24.48)  0.011m

0 coh  i, j    1   2

1 0

1

2   5   10  1  ; 1  exp  ;   exp 2     2   2  0 



3/14

Contd… uu

uz

zz

 1   0.873  0.765

0.873 1 0.873

0.765  0.873 1 

 0.0382   0.0063   0.0027

0.0061 0.0027 0.0387 0.0063

0.0443 0.0062 0.0068 0.0447

0.0063 0.0387

0.0029 0.0068

 1  0.0008   0.0001   0.0142  0.0007   0.0001

0.0008 1 0.0008 0.0007 0.0142 0.0007

0.0029  0.0068  0.0447 

0.0001  0.0007  1 0.0001 0.0007 0.0142   0.0001 1 0.0007 0.0001  0.0007 0.0007 1 0.0007   0.0142 0.0001 0.0007 1  0.0001 0.0008


0.0142 0.0007 0.0007 0.0142


3/15

Contd…  Mean peak values determined are: (u1 )tot  0.106m ; (u2 )tot  0.099m (u1 ) rel  0.045m ; (u2 ) rel  0.022m

 For perfectly correlated ground motion uu

zz

1   0  0 1 1  1  0 0   0

0 1 0 1 1 1 0 0 0

0 0  uz  null matrix 1  1 0 0 0 1 0 0 0  1 0 0 0  0 1 1 1 0 1 1 1  0 1 1 1 



3/16

Contd…  Mean peak values of relative displacement RSA u1 =0.078m

;

RHA 0.081m

u2 = 0.039m

;

0.041m

 It is seen that’s the results of RHA & RSA match well.

 Another example (example 3.10) is solved for a time lag of a 2.5 sec. Solution is obtained in the same way and results are given in the book. The calculation steps are self evident. T.K. Datta Department Of Civil Engineering, IIT Delhi


4/1

Cascaded analysis  Cascaded analysis is popular for seismic analysis of secondary systems (Fig 5.5). Secondary System

k

..

xf

m Fig 5.5

..

xg Secondary system mounted on a floor of a building frame

c

.. .. .. xa = xf + xg SDOF is to be analyzed for obtaining floor response spectrum

 RSA cannot be directly used for the total system because of degrees of freedom become prohibitively large ; entire system becomes nonclasically damped. T.K. Datta Department Of Civil Engineering, IIT Delhi


4/2

Contd… In the cascaded analysis two systems- primary and secondary are analyzed separately; output of the primary becomes the input for the secondary.  In this context, floor response spectrum of the primary system is a popular concept for cascaded analysis. The absolute acceleration of the floor in the figure is x a  Pseudo acceleration spectrum of an SDOF is obtained for x a ; this spectrum is used for RSA of secondary systems mounted on the floor. T.K. Datta Department Of Civil Engineering, IIT Delhi


4/3

Contd…

Using this spectrum, peak displacement of the secondary system with T=0.811s is 0.8635m.  The time history analysis for the entire system (with C matrix for P-S system) is found as 0.9163m. T.K. Datta Department Of Civil Engineering, IIT Delhi

Displacement (m)

Example 5.6 For example 3.18, find the mean peak displacement of the oscillator for El Centro earthquake.  for secondary system = 0.02 ;  for the main system = 0.05 ;floor displacement spectrum shown in the Fig5.6 is used Solution 1.5

1

0.5

0 0

5

10

15

20

25

30

35

40

Frequency (rad/sec)

Floor displacement response spectrum (Exmp. 5.6) Response Spectrum Method Of Analysis

Approximate modal RSA  For nonclassically damped system, RSA cannot be directly used.  However, an approximate RSA can be performed.  C matrix for the entire system can be obtained (using Rayleigh damping for individual systems & then combining them without coupling )





C1 C  0

0 C2 

matrix is obtained considering all d.o.f. &  T C  becomes non diagonal.

 Ignoring off diagonal , an approximate modal damping is derived & is used for RSA. T.K. Datta Department Of Civil Engineering, IIT Delhi


4/4

4/5

Seismic coefficient method  Seismic coefficient method uses also a set of equivalent lateral loads for seismic analysis of structures & is recommended in all seismic codes along with RSA & RHA.

 For obtaining the equivalent lateral loads, it uses some empirical formulae. The method consists of the following steps: • Using total weight of the structure, base shear is obtained by

Vb  W  Ch

(5.34)

Ch is a period dependent seismic coefficient T.K. Datta Department Of Civil Engineering, IIT Delhi


4/6

Contd… • Base shear is distributed as a set of lateral forces along the height as

Fi  Vb  f (hi )

(5.35)

f (hi ) bears a resemblance with that for the

fundamental mode.

• Static analysis of the structure is carried out with the force Fi (i = 1,2...... n) .  Different codes provide different recommendations for the values /expressions for Ch & f (hi ) . T.K. Datta Department Of Civil Engineering, IIT Delhi


4/7

 Distribution of lateral forces can be written as S

F = ρ × W × φ × a1 1 j j j1 g Fj Wj × φ j1 = ∑ Fj ΣWj × φ j1 Fj = Vb × Fj = Vb Fj = Vb

Wj × φ j1 ΣWj × φ j1

Wj ×h j

(5.37) (5.38) (5.39)

ΣWj ×h j Wj ×h jk ΣWj ×h j

(5.36)

k


(5.40) Response Spectrum Method Of Analysis

4/8

 Computation of base shear is based on first mode. Following basis for the formula can be put forward. Sa

i×) λ V = ΣF =(ΣW × φ × bi ji j ji i g Sa i e Vb i = Wi g

 

Vb ≤ Σ Vb i ≤Σ

S ai

Wie

g S Vb = a1 × W g

(5.41) (5.42) (5.43)

 i = 1to n 


(5.44) (5.45) Response Spectrum Method Of Analysis

5/1

Seismic code provisions  All countries have their own seismic codes.  For seismic analysis, codes prescribe all three methods i.e. RSA ,RHA & seismic coefficient method. Codes specify the following important factors for seismic analysis: • Approximate calculation of time period for seismic coefficient method. • C h Vs T plot. • Effect of soil condition on T.K. Datta Department Of Civil Engineering, IIT Delhi

S A or a g g

& Ch


5/2

Contd… • Seismicity of the region by specifying PGA.

• Reduction factor for obtaining design forces to include ductility in the design. • Importance factor for structure.  Provisions of a few codes regarding the first three are given here for comparison. The codes include: • • • • •

IBC – 2000 NBCC – 1995 EURO CODE – 1995 NZS 4203 – 1992 IS 1893 – 2002



5/3

Contd… IBC – 2000 • C h for class B site,

 1.0 T1  0.4s  Ch   0.4 T1  0.4s T  1

(5.46)

A • for the same site, is given by g  0.4  7.5T n A    1.0 g  0.4  Tn 

0  Tn  0.08s 0.08  Tn  0.4s


(5.47)

Tn  0.4s Response Spectrum Method Of Analysis

5/4

Contd…  T may be computed by  N 2   Wi ui T1  2  i 1N g Fi ui    i 1



     

(5.48)

Fi can have any reasonable distribution.

 Distribution of lateral forces over the height is given by Fi  Vb

W j h kj N

k W h  j j

(5.49)

j 1



5/5

Contd… k={1; 0.5 T +1.5 ; 2 for T 1

1

≤ 0.5s ; 0.5 ≤ T1 ≤ 2.5s; T1 ≥ 2.5s (5.50)

 Distribution of lateral force for nine story frame is shown in Fig5.8 by seismic coefficient method . W 2

9 W

8

W

7

Storey

W W

9@3m

W

6 5

W

4

W

3

W

2 10

T=2sec T=1sec T=0.4sec 2 Storey force

4

Fig5.8 T.K. Datta Department Of Civil Engineering, IIT Delhi


5/6

Contd…  NBCC – 1995 CeU ; Ce = USIF (5.51a) ;(5.51b) • C h is given by Ch = R

A • For U=0.4 ; I=F=1, variations of S & with T g

are given in Fig 5.9.

Seismic response factor S

4.5 4 3.5 3 2.5 2 1.5 10

0.5

Time period (sec)

1

1.5



5/7

Contd…

• For PGV = 0.4ms-1 , A g  1.2 A  =  0.512 g   Tn

is given by

0.03 ≤ Tn ≤ 0.427s Tn > 0.427s

(5.52)

• T may be obtained by   Fu  i 1  T1 = 2π  N  g 1 Fu  i i  N

2 i

1 2

(5.53)

• S and A/g Vs T are compared in Fig 5.10 for v = 0.4ms-1 , I = F = 1; z h = z v (acceleration and velocity related zone) T.K. Datta Department Of Civil Engineering, IIT Delhi


5/8

Contd… 1.4 A/g S

1.2

S or A/g

1

0.8

0.6

0.4

0.2

0 0

0.5

1

1.5

2

2.5

3

3.5

4

Time period (sec)



5/9

Contd… • Distribution of lateral forces is given by Fi =  Vb - Ft 

Wh i i

N

 Wh i=1

0   Ft = 0.07T1Vb  0.25V b 

i

(5.54)

i

T1 ≤ 0.7 s 0.7 < T1 < 3.6 s

(5.55)

T1 ≥ 3.6 s



5/10

Contd…  EURO CODE 8 – 1995 • Base shear coefficient C s is given by C Cs = e • Ce is given by q A   g  1 Ce =   A  Tc  3 g  T    1

(5.56)

0 ≤ T1 ≤ Tc (5.57) T1 ≥ Tc

• Pseudo acceleration in normalized form is given by Eqn 5.58 in which values of Tb,Tc,Td are Tb

Tc

Td

hard

0.1

0.4

3.0

med soft

0.15 0.2

0.6 0.8

3.0 3.0 (A is multiplied by 0.9)



5/11

Contd… • Pseudo acceleration in normalized form ,

is given by Tn  1+1.5 T b   2.5  A =  Tc  ug0  2.5  T   n   Tc Td 2.5  2 T  n

0 ≤ Tn ≤ Tb Tb ≤ Tn ≤ Tc Tc ≤ Tn ≤ Td


(5.58)

Tn ≥ Td


5/12

 Rayleigh's method may be used for calculating T.  Distribution of lateral force is Fi = Vb

Wφ i i1

 Wφ i=1

Fi = Vb

(5.59)

N

i

i1

Wh i i

(5.60)

N

 Wh i=1

i

i

 Variation of ce / u go & A / u go Fig 5.11. T.K. Datta Department Of Civil Engineering, IIT Delhi

are shown in


5/13

Contd… 3

..

A/ug0

..

2.5

Ce/ug0

..

Ce /ug0 or A ../ug0

2

..

1.5

1

0.5

0 0

0.5

1

1.5

2

2.5

3

3.5

4

Time period (sec)

Fig 5.11 T.K. Datta Department Of Civil Engineering, IIT Delhi


6/1

Contd…  NEW ZEALAND CODE ( NZ 4203: 1992) • Seismic coefficient & design response curves are the same. • For serviceability limit, C  T  = Cb  T1,1 RzL s = Cb  0.4,1 RzL s

T1 ≥ 0.45

(5.61a)

T1 ≤ 0.45

(5.61b)

Ls is a limit factor. • For acceleration spectrum, T.K. Datta Department Of Civil Engineering, IIT Delhi

T1 is replaced by T.


6/2

Contd… • Lateral load is multiplied by 0.92. • Fig5.12 shows the plot of cb vs T for   1

• Distribution of forces is the same as Eq.5.60 • Time period may be calculated by using Rayleigh’s method. • Categories 1,2,3 denote soft, medium and hard. • R in Eq 5.61 is risk factor; Z is the zone factor; ls is the limit state factor. T.K. Datta Department Of Civil Engineering, IIT Delhi


6/3

Contd… 1.2 Category 1 Category 2

1

Category 3 0.8

Cb

0.6

0.4

0.2

0 0

0.5

1

1.5

2

2.5

3

3.5

4

Time period (sec)



Contd…

6/4

 IS CODE (1893-2002) • Time period is calculated by empirical formula and distribution of force is given by: Fj = Vb

2 Wh j j N

2 Wh  j j

(5.65)

j=1

• Ce vs T & given by:

Sa vs T are the same; they are g

 1+15T 0 ≤ T ≤ 0.1s Sa  =  2.5 0.1≤ T ≤ 0.4s g  1  0.4 ≤ T ≤ 4.0s  T T.K. Datta Department Of Civil Engineering, IIT Delhi

for hard soil (5.62)


6/5

Contd… Sa g

Sa g

 1+15T  =  2.5  1.36   T  1+15T  =  2.5  1.67   T

0 ≤ T ≤ 0.1s 0.1≤ T ≤ 0.55s

for medium soil

(5.63)

for soft soil

(5.64)

0.55 ≤ T ≤ 4.0s 0 ≤ T ≤ 0.1s 0.1≤ T ≤ 0.67s 0.67 ≤ T ≤ 4.0s

For the three types of soil Sa/g are shown in Fig 5.13

Sesmic zone coefficients decide about the PGA values.



6/6

Contd… 3

Spectral acceleration coefficient (Sa/g)

Hard Soil 2.5

Medium Soil

Soft Soil

2

1.5

1

0.5

0 0

0.5

1

1.5

2

2.5

3

3.5

4

Time period (sec)

Variations of (Sa/g) with time period T Fig 5.13



6/7

Contd… Example 5.7: Seven storey frame shown in Fig 5.14 is analyzed with Concrete density = 24kNm-3 ; E = 2.5×107 kNm-2

Live load = 1.4kNm-1

For mass: 25% for the top three & rest 50% of live load are considered. T1 = 0.753s ; T2 = 0.229s ; T3 = 0.111s R = 3; PGA = 0.4g ; for NBCC, PGA ≈ 0.65g

Solution:  First period of the structure falls in the falling region of the response spectrum curve.  In this region, spectral ordinates are different for different codes. T.K. Datta Department Of Civil Engineering, IIT Delhi


6/8

Contd…

All beams:-23cm  50cm Columns(1,2,3):-55cm  55cm

7@3m

Columns(4-7):-:-45cm  45cm

5m

5m

5m

A Seven storey-building frame for analysis Fig 5.14 T.K. Datta Department Of Civil Engineering, IIT Delhi


6/9

Contd… Table 5.3: Comparison of results obtained by different codes 1st Storey Displacement (mm)

Base shear (KN)

Top Storey Displacement (mm)

Codes SRSS

CQC

SRSS

CQC

SRSS

3

all

3

all

3

all

3

all

IBC

33.51

33.66

33.52

33.68

0.74

0.74

0.74

0.74

NBCC

35.46

35.66

35.46

35.68

0.78

0.78

0.78

NZ 4203

37.18

37.26

37.2

37.29

0.83

0.83

Euro 8

48.34

48.41

48.35

48.42

1.09

Indian

44.19

44.28

44.21

44.29

0.99


3

all

CQC

3

all

10.64 10.64

10.64

10.64

0.78

11.35 11.35

11.35

11.35

0.83

0.83

12.00 12.00

12.00

12.00

1.09

1.09

1.09

15.94 15.94

15.94

15.94

0.99

0.99

0.99

14.45 14.45

14.45

14.45


6/10

Contd… 7

Number of storey

6 5 4 3 2 1

0

2

4

6 8 10 Displacement (mm)

12

IBC NBCC NZ 4203 Euro 8 Indian 14 16

Comparison of displacements obtained by different codes Fig 5.15 T.K. Datta Department Of Civil Engineering, IIT Delhi


Lec-1/74



Chapter - 6

Inelastic Seismic Response of Structures



1/1

Introduction  Under relatively strong earthquakes, structures undergo inelastic deformation due to current seismic design philosophy.  Therefore, structures should have sufficient ductility to deform beyond the yield limit.  For understanding the ductility demand imposed by the earthquake, a study of an SDOF system in inelastic range is of great help. The inelastic excursion takes place when the restoring force in the spring exceeds or equal to the yield limit of the spring. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/2

Contd..  For this, nonlinear time history analysis of SDOF system under earthquake is required; similarly, nonlinear analysis of MDOF system is useful for understanding non-linear behaviour of MDOF system under earthquakes.  Nonlinear analysis is required for other reasons as well such as determination of collapse state, seismic risk analysis and so on.  Finally, for complete understanding of the inelastic behavior of structures, concepts of ductility and inelastic response spectrum are required. The above topics are discussed here. T.K. Datta Department Of Civil Engineering, IIT Delhi


Non linear dynamic analysis

1/3

 If structure have nonlinear either in inertia or in damping or in stiffness or in any form of combination of them, then the equation of motion becomes nonlinear.  More common nonlinearities are stiffness and damping nonlinearities.  In stiffness non linearity, two types of non linearity are encountered : • Geometric • Material (hysteretic type)  Figure 6.1 shows non hysteric type non linearity; loading & unloading path are the same. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/4

Contd.. f Unloading Loading

Dx

x Loading Unloading

Fig.6.1



1/5

Contd.. 

Figure 6.2 shows hysteric type nonlinearity; experimental curves are often idealised as (i) elasto plastic; (ii) bilinear hysteretic ; (iii) general strain hardening f

f f y

fy xy

x

x y

x

Variation of force with displacement under cyclic loading f

f

fy

f y

xy

x

Idealized model of force displacement curve Fig.6.2 T.K. Datta Department Of Civil Engineering, IIT Delhi

x y

x

Idealized model of force displacement curve Response Spectrum Method Of Analysis

1/6

Contd..  Equation of motion for non linear analysis takes the form

M Dx  Ct Dx  Kt Dx  Mr Dxg

(6.1)

C t and K t matrices are constructed for the current time interval.

 Equation of motion for SDOF follows as

mΔx + ct Δx + k t Δx = -mΔxg

(6.2)

 Solution of Eqn. 6.2 is performed in incremental form; the procedure is then extended for MDOF system with additional complexity.  C t and K t should have instantaneous values. T.K. Datta Department Of Civil Engineering, IIT Delhi


1/7

Contd.. 

ct and k t are taken as that at the beginning of the time step; they should be taken as average values.

 Since D x & D x are not known, It requires an iteration.  For sufficiently small D t , iteration may be avoided.

 NewMark’s   Method in incremental form is used for the solution

Δx = Δt x k + δ Δt Δx Δx = Δt x k

Δt   + 2

2

(6.3)

x k + β  Δt  Δx


2

(6.4)


1/8

Contd.. 1

1 1 Δx = Δx xk - xk 2 βΔt 2β β  Δt 

(6.5)

 δ  δ δ Δx = Δx - x k + Δt 1-  x k βΔt β  2β 

(6.6)

kΔx = Δp δ 1 k = kt + ct + m 2 βΔt β  Δt 

(6.7) (6.8 a)

m  m δ   δ   Δp = -mΔx g +  + c t  x k +  + Δt  -1 c t  x k (6.8b)  βΔt β   2β    2β x k+1 = x k + Δx x k+1 = x k + Δx x k+1 = x k + Δx (6.9) T.K. Datta Department Of Civil Engineering, IIT Delhi


1/9

Contd..  For more accurate value of acceleration, it is calculated from Eq. 6.2 at k+1th step.  The solution is valid for non hysteretic non linearity.  For hysteretic type, solution procedure is modified & is first explained for elasto - plastic system.

 Solution becomes more involved because loading and unloading paths are different.  As a result, responses are tracked at every time step of the solution in order to determine loading and unloading of the system and accordingly, modify the value of kt. T.K. Datta Department Of Civil Engineering, IIT Delhi

81 Of Analysis Response Spectrum Method

1/10

Elasto-plastic non linearity  For material elasto plastic behaviour, to be constant.

ct is taken

 k t is taken as k or zero depending upon whether the state is in elastic & plastic state (loading & unloading).  State transition is taken care of by iteration procedure to minimize the unbalanced force; iteration involves the following steps. Elastic to plastic state

 Dx e  ae  Dx 0 T.K. Datta Department Of Civil Engineering, IIT Delhi

(6.10) Response Spectrum Method Of Analysis

1/11

Contd..

ae ) Δp withkt = 0  Use Eq. 6.7, find( Δ x)for(1p Δ x =(Δ x )+ e (Δ x ) p Plastic to plastic state Eq. 6.7 with Kt=0 is used ; transition takes place if x < 0 at the end of the step; computation is then restarted. Plastic to elastic state Transition is defined by x = 0 D x is factored (factor e) such that x = 0 ( Δx ) a is obtained for 1- e  Δp with k t ≠ 0 Δ x =(Δ x )+Factored Δx a T.K. Datta Department Of Civil Engineering, IIT Delhi


2/1

Contd.. Example 6.1 Refer fig. 6.3 ; n  10 rad / s ; find responses at t=1.52 s & 1.64s given responses at t= 1.5s & 1.62s ; m=1kg Solution: x

fx

f x 0.15mg

m

c ..

xg

0.0147m

SDOF system with non-linear spring

x

Force-displacement behaviour of the spring

Fig . 6.3 T.K. Datta Department Of Civil Engineering, IIT Delhi


2/2

t = 1.5s; x = 0.01315m; x = 0.1902m/s; x = 0.46964 m/s 2 ; fx = 1.354 ; c = 0.4Nsm-1; t k =100Nm-1 t k = 10140 Nm-1

Δx g = - 0.00312g Δp = 37.55 N Δx = 0.0037m; Δx = - 0.01ms -1; Δf = k Δx t (f x ) =1.7243N > 0.15mg t+Δt (f x )+ e Δxk = 0.15mg(e = 0.3176) t t T.K. Datta Department Of Civil Engineering, IIT Delhi


2/3

Contd..

kΔx = 1- e  Δp; k = 0; Δx =0.00373m; Δx =- 0.00749ms-1 t 2 x = x + Δx + Δx = 0.01725m x = 0.1827ms -1 1 t+Δt t 2 t+Δt P -c x -f k+1 t k+1 x(k+1) x = = 0.279ms-1 k+1 m At t =1.625s ; x > 0 ; k = 10040; Δp =- 0.4173; Δx =0.000042; Δx = - 0.061; X t + ΔX1 = 0 ; e =- 6.8; Δx = eΔx =- 0.000283; 1 kΔx = 1- e  Δp; k =100; Δx = Δx + Δx =- 4.44×10-5 ; Δx =- 0.061; t 1 2 2 x = x + Δx =0.0298; x = x + Δx = - 0.033 t+Δt t t+Δt t x from Eqn  =3.28; f = f +k Δx = 1.4435N t+Δt  t+Δt xt t 2 T.K. Datta Department Of Civil Engineering, IIT Delhi


2/4

Solution for MDOF System  Sections undergoing yielding are predefined and their force- deformation behaviour are specified as shown in Fig 6.4. 0.5m V 0.5k

0.5k

p1

m

k

x y1

x

x y2

x

x y3

x

k V

m

p2

k m

1.5k

k V p3

1.5k

Fig.6.4

 For the solution of Eqn. 6.1, state of the yield section is examined at each time step. T.K. Datta Department Of Civil Engineering, IIT Delhi


2/5

Contd..

 Depending upon the states of yield sections, stiffness of the are changed & the stiffness matrix for the incremental equation is formed.  If required, iteration is carried out as explained for SDOF.

 Solution for MDOF is an extension of that of SDOF. KΔ x = Δp

(6.11)

δ 1 K = Kt + Ct + M 2 βΔt βΔ t

(6.12a)

M  M δ   δ   Δp =-M r Δ x g +  + C t  x k +  + Δt  -1 C t  x k (6.12b)  β Δt β   2β    2β T.K. Datta Department Of Civil Engineering, IIT Delhi


2/6

Contd.. Example 6.2: Refer to Fig 6.5; K/m = 100; m = 1 kg; find responses at 3.54s. given those at 3.52s. Solution: 1.44977  0.15mg      f = 0.95664  < 0.15mg  k 0.63432  0.15mg     

and

x>0

10260 δ 1 K =K + C + M =  -124 10260 t βΔt t 2 β(Δt)  0 -124 Δx g = 0.5913

sym    10137 

32.6224  M   M   δ  δ Δp = -MIΔx g +  + C x +  + Δt  - 1 C  x = 18.0256   βΔt β t  k  2β  2β  t  k   8.4376  0.0032  Δx = K -1Δp = 0.0018  0.0009  T.K. Datta Department Of Civil Engineering, IIT Delhi


2/7

Contd.. m

3m

k/2

x 3 k/2

m

3m

k/2

x 2 k/2

m

3m

k/2

x 1 k/2

f y

3 storey frame

f y

 0.15m g

xy  0.01475m

x y

Fig.6.5 T.K. Datta Department Of Civil Engineering, IIT Delhi

x

Force displacement curve of the column Response Spectrum Method Of Analysis

2/8

Contd..  0.0032   0.16  K Δx = -0.0014  ; Δf = Δx =  -0.07  2  -0.0009  -0.045  e (0.16)  1.60977   0.15mg  1       f = f + Δf = 0.88664  f + eΔf = f + e (-0.07) = ≤ 0.15mg  k+1 k k k 2   ≤ 0.15mg  0.58932    e (-0.045)    3  e = 0.136; e =1; e =1 1 2 3    e Δx  e Δx 1 1    1 1     Δx e = Δx =  e Δx + e (Δx - Δx )  =  e2 Δx2  1 1 1 1 2 2     e Δx + e (Δx Δx ) + e (Δx Δx ) e Δx   1 2 2 3 3 2   1 1  3 3    0.000435    = -0.000965  -0.001865   

e   1   e 2  =   e   3 

0.1358    0.6893    3.07   



2/9

Contd.. 0.0028  0.00324  0.02009        Δx = 0.0027  ; Δx = Δx + Δx = 0.0018  ; x = x + Δx = 0.00833  1 2 2 k+1 k 0.0026  0.00074  0.0114        0. - 0.0509   δ  δ δ Δx = Δx - x + Δt 1-  x =  -0.0406  ; x = x + Δx = k k k+1 k βΔt β  2β   -0.0524 

 0.1361  0.07    0.0165 

e(0.16)  1.4715   1  0.00  0.0218    Δf = e (-0.07) +  -0.005 =  -0.075  ; f = f + Δf =  0.882  2 k+1 k   -0.005  -0.05   0.584       e (-0.045)   3   -2.289  x = M-1 P - C x - F = -1.7 018  k+1 k+1 t k+1 k+1 -2.2825 







Bidirectional Interaction  Bidirectional interaction assumes importance under: • Analysis for two component earthquake • Torsionally Coupled System  For such cases, elements undergo yielding depending upon the yield criterion used.

 When bidirectional interaction of forces on yielding is considered, yielding of a cross section depends on two forces.  None of them individually reaches yield value; but the section may yield. T.K. Datta Department Of Civil Engineering, IIT Delhi


3/1

3/2

Contd.. If the interaction is ignored, yielding in two directions takes place independently.

 In incremental analysis, the interaction effect is included in the following way.  Refer Fig 6.6; columns translate in X and Y directions with stiffness k exiand keyi  K ex 0 K ex e y    Ke =  0 K ey K ey e x  (6.13a) K ex e y K ey e x K θ   K ex =  K exi ; K ey =  K eyi ; K θ =  K exi e y +  K eyi e x (6.13b) T.K. Datta Department Of Civil Engineering, IIT Delhi


.

3/3

Contd.. D Colm. 2 Colm. 4

CR C.M.

D

X e

y

e x

Y

Colm. 1

Colm. 3



3/4

Contd.. Transient stiffness K t remaining constant over D t is given by

K t = K e -K p

(6.14)

The elements of the modification matrix K p are

B2yi BxiByi B2xi K pxi = ; K pyi = ; K pxyi = K pyxi = Gi Gi Gi

(6.15)

Gi = K exihxi2 +K eyihyi2

(6.16a)

Bxi = K exihxi ; Byi = K eyihyi

(6.16b)

Vyi Vxi hxi = 2 ; hyi = 2 Vpxi Vpyi

(6.16c)



3/5

Contd..  When any of the column is in the full plastic state satisfying yield criterion, k t = 0 .  During incremental solution k t changes as the elements from E-P, P-P, P-E; the change follows E-P properties of the element & yield criterion.

 Yield criterion could be of different form; most popular yield curve is 2

 Vxi   Vyi  i =  +  V   V   pxi   pyi  T.K. Datta Department Of Civil Engineering, IIT Delhi

2

(6.19)


3/6

Contd..  For V pxi  V pyi , curve is circular ; V pxi  V pyi , curve is ellipse; i  1 shows plastic state, i  1 shows elastic state, i  1 is inissible.  If i  1, internal forces of the elements are pulled back to satisfy yield criterion; equilibrium is disturbed, corrected by iteration.  The solution procedure is similar to that for SDOF.  At the beginning of time , check the states of the elements & accordingly the transient stiffness matrix is formed. T.K. Datta Department Of Civil Engineering, IIT Delhi


3/7

Contd.. If any element violates the yield condition at the end of time or es from E-P, then an iteration scheme is used. If it is P-P & i  1 for any element, then an average stiffness predictor- corrector scheme is employed. 1 ' K = The scheme consists of : ta 2 K t0 +K t   DU 1 is obtained with K ta for the time internal Δt & incremental restoring force vector is obtained. DF1  K ta DU1 DFi 1  DFi 

(6.21) force tolerance

(6.22a)

DU i 1  DU i  displacement tolerance (6.22b) T.K. Datta Department Of Civil Engineering, IIT Delhi


3/8

Contd..  After convergence , forces are calculated & yield criterion is checked ; element forces are pulled back if criterion is violated. 1  Fi  Fi (6.23)

i

 With new force vectorK ta is calculated & iteration is continued.  For E-P, extension of SDOF to MDOF is done.  For calculating DU p , the procedure as given in SDOF is adopted. T.K. Datta Department Of Civil Engineering, IIT Delhi


3/9

Contd..  If one or more elements are unloaded from plastic to elastic state, then plastic work increments for the elements are negative Dw pi  Fi DU pi (6.25)

DU pi  DU i  K ei 1DFi

(6.26)

 When unloaded, stiffness within Dt , is taken as elastic. Example 6.3: Consider the 3D frame in Fig 6.8; assume:

D =3.5m; h = 3.5m; M px = M p y = M p  = M0 ; M p B = M p  = 1.5Mo A

D

ko M p C = 2M o ; k x = k y= k A = k o ; k B= k C = 1.5k o ;k C = 2k o ; m = 50 in which mx = my = m = 620kg and  VP A = 152.05 T.K. Datta Department Of Civil Engineering, IIT Delhi


3/10

Contd.. find Initial stiffness & stiffness at t = 1.38s, given that t = 1.36s k D

y 2k

1.5k

3.5m

x

B

3.5m

C

3.5m

3 D frame

Force (N)

A

1.5k

For column A

152.05 N

0.00467m

Displacement (m)

Force-displacement curve of column A T.K. Datta Department Of Civil Engineering, IIT Delhi


3/11

U  0.00336   x    U = 0.00037  y     0.00003    θ k 

U  0.13675  x     U = 0.00345  y      0.00311   θ k  V = 102.83 Ax

   Vx  627.27  VB x = 154.24 F =  Vy  = 70.888  k    773.51 VD x = 158.66   Vθ k  V = 211.54 Cx

U  -0.16679  x     U = -0.11434  y     -0.06153    θ k  V = 10.10 Ay V = 19.56 By V = 15.15 Dy V = 26.08 Cy

; x

gk

= - 0.08613g

Solution: Forces in the columns are pulled back (Eq. 6.23) & displacements at the centre T.K. Datta Department Of Civil Engineering, IIT Delhi


3/12

Contd.. KD 2 2 K ex =  K exi = 6k 0 ; K ey =  K eyi = 6k 0 ; K θ =  = 3k 0  3.5  4 0 54250  186000 K e =  0 186000 54250   54250 54250 1139250  2

2

 Vxi   Vyi  i =  +  V   V   pxi   pyi  A = 0.462 ; B = 0.465; C = 0.491 ; D = 0.488 Kt = Ke 638.6 sym  δ 1  0  ×10 4 K = Kt + Ct + M= K +1 0000M= 638.6 t 2   βΔt β(Δt) 5.425 5.425 1379.76  16282  M  M δ   δ   Δp = -MΔx g +  + C t  Uk +  + Δt  - 1 C t  Uk =  286   βΔt β   2β    2β  612  T.K. Datta Department Of Civil Engineering, IIT Delhi


3/13

Contd..  0.0025   476.1 ΔU = K -1Δp = 0.000001 and ΔF = K t ΔU=  10.2  0.000001 181.2  0.0059    Uk +1 = Uk + ΔU = 0.0004   0.0001   1103.36    Fk +1 = Fk + ΔF =  81.09  ;  954.75     V i =  xi  Vpxi eA =

1

A

2

  Vyi  +    Vpyi  0.824

VAx = 183.89 VBx = 275.84 ; VDx = 275.84 VCx = 367.79

VAy = 13.52 VBy = 20.27 VDy = 20.27 VCy = 27.03

2

  & A = 1.47 ; B = 1.47 ; C = 1.47 ; D =1.47  eB =

1

B

= 0.824

eD =

1

D

= 0.824 e C =

1

C

= 0.824

 With the e values calculated as above, the forces in the columns are pulled back T.K. Datta Department Of Civil Engineering, IIT Delhi


3/14

Contd..  0.254  DU e  0.00122  105 ; e is calculed as 0.00124  16268.57  Dp2  (1  e )Dp   285.76   631.48  ex  ey 

K x K

i i

0.83 0.83 103 in which e = DU xe etc. x   DU x 0.83

 0.29167

i

hxi 

Vxi ; 2 V pxi

hxA hxB hxD hxC

 0.00795  0.0053 ;  0.0053  0.00398

hyi 


Vyi V

2 pyi

;

hyA hyB hyD hyC

   

0.00059 0.00039 0.00039 0.00029


3/15

Contd.. Bxi  K exi hxi ;

BxA



246.56

ByA

 18.12

BxB



246.56

ByB

 18.12

BxD



246.56

ByD

 18.12

BxC



246.56

By xC

 18.12

Gi  K h  K eyi h ; 2 exi xi

K pxi

2 yi

;

Byi  K eyi hyi ;

GA

 1.972

GB

 1.314

GD

 1.314

GC



0.98

Byi2 Bxi2 D2  ; K pyi  ; K p  Gi Gi 2


K

p

; K pxyi  K pyxi 

Bxi Byi Gi


3/16

Contd.. exp 

K K

x

Pyi i

 0.2917; eyp 

Pyi

K K

Pxi

yi

 0.2917

Pxi

sym  1.0    103 K t   16.06 185    0.29 53.96 0  sym  62   1 5  0.16 63.85  K  Kt  Ct  M = K  10000 M = 10  t   Dt  ( Dt ) 2  0 0.54 126.6 

0.002598  0.0000983 0.0026  0.002598  0.0001018       1 DU 2  K Dp2   0.0001 ; DU px    ; DU py    0.002602 0.0000983      0.00    0.002602   0.0001018  T.K. Datta Department Of Civil Engineering, IIT Delhi


3/17

Contd..  K ex  K px DV pi     K pxy VAx VBx

 

151.913 228.043

VDx VCx

 

227.745 303.471

i

 Vxi  V pxi 

   

2

 K pxy   DU px    K ey  K py   i  DU py  i

;

 Vyi  V pyi 

VAy VBy

 

9.78 11.54

VDy VCy

 

11.37 10.98

   

2

 A  1.002; B  1.002; C  1.0;  D  1.00

Because yield condition is practically satisfied, no further iteration is required. T.K. Datta Department Of Civil Engineering, IIT Delhi


Multi Storey Building frames  For 2D frames, inelastic analysis can be done without much complexity.  Potential sections of yielding are identified & elasto–plastic properties of the sections are given.  When IMI = Mp for any cross section, a hinge is considered for subsequent Dt & stiffness matrix of the structure is generated.  If IMI > Mp for any cross section at the end of Dt IMI is set to Mp, the response is evaluated with average of stiffness at t and t  Dt (IMI = Mp ). T.K. Datta Department Of Civil Engineering, IIT Delhi


4/1

4/2

Contd..  At the end of each Dt , velocity is calculated at each potential hinge; if unloading takes place at the end of Dt , then for next Dt , the section behaves elastically. ( Dt ~ small ). Example 6.4 Find the time history of moment at A & the forcedisplacement plot for the frame shown in Fig 6.9 under El centro earthquake; Dt  0.2s ; compare the results for elasto plastic & bilinear back bone curves. • Figs. 6.10 & 6.11 are for the result of elasto -plastic case Figs 6.12 & 6.13 are for the result of bilinear case • Moment in Fig 6.12 does not remain constant over time unlike elasto-plastic case. T.K. Datta Department Of Civil Engineering, IIT Delhi


4/3

Contd.. m

3m

x3

k

k m

x2 3m

k

k m

3m

1.5k

x1

k = 23533 kN/m m = 235.33 

103

1.5k kg

Kd

A

346.23kN Force (kN)

Frame

Ki

0.01471m


 0.1 KK di

Displacement (m)

Force-displacement curve of column Response Spectrum Method Of Analysis

4/4

Contd.. 600000

Moment (N-m)

400000 200000 0 -200000 0

5

10

15

-400000

20

25

30

Time (sec)

-600000 400000

Fig.6.10

300000 200000

Force (N)

100000 0 0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

-100000 -200000 -300000 Displacement (m)

-400000


Fig.6.11 Response Spectrum Method Of Analysis

4/5

800000 600000 400000 200000 0 -200000 0 -400000 -600000 -800000

5

10

15

20

25

30

Time (sec)

500000 400000

Fig.6.12

300000 200000

Shear Force(N)

Moment (N-m)

Contd..

100000 0 -0.005 0 -100000 -200000

0.005

0.01

0.015

0.02

0.025

0.03

Displacement(m)

-300000 -400000 -500000


Fig.6.13 Response Spectrum Method Of Analysis

4/6

Contd..  For nonlinear moment rotation relationship, tangent stiffness matrix for each Dt obtained by considering slope of the curve at the beginning of Dt  If unloading takes place, initial stiffness is considered.  Slopes of backbone curve may be interpolated ; interpolation is used for finding initial stiffness.  If columns are weaker than the beams, then top & bottom sections of the column become potential sections for plastic hinge.  During integration of equation of motion K t is given by K K K (6.27) t


e

p


4/7

Contd..  Non zero elements of Kp are computed using Eqns. 6.15 & 6.16 and are arranged so that they correspond to the degrees of freedom affected by plastification.  The solution procedure remains the same as described before.  If 3D frame is weak beam-strong column system, then problem becomes simple as the beams undergo only one way bending. The analysis procedure remains the same as that of 2D frame. T.K. Datta Department Of Civil Engineering, IIT Delhi


4/8

Contd..  For 2D & 3D frames having weak beam strong column systems, rotational d.o.f are condensed out; this involves some extra computational effort. • Incremental rotations at the member ends are calculated from incremental displacements. • Rotational stiffness of member is modified if plastification/ unloading takes place. • The full stiffness matrix is assembled & rotational d.o.f. are condensed out.  The procedure is illustrated with a frame as shown in the figure (with 2 storey). T.K. Datta Department Of Civil Engineering, IIT Delhi


4/9

Contd..  Elasto-plastic nature of the yield section is shown in Fig 6.16.

 Considering anti-symmetry : M1, M2

Mp1 = Mp2 = Mp3

p



Moment-rotation relationship of elasto-plastic beam fig. 6.16 T.K. Datta Department Of Civil Engineering, IIT Delhi


4/10  k  -k  K= - kl  2  - kl  2

-k 2k kl 2 0

kl 2 kl 2 kl 2  α1 + 0.67  2 kl 2 6

-

-

kl 2

0 kl 2 6 kl 2 2

          α 2 +1.33   

 Δ1  Δ   2    θ1    θ2  

K Δ = K Δ - K ΔθK θ-1K θΔ 1  6 3  α1 + 0.67  -1 Kθ = 2   1 3  α 2 +1.33   kl 

(6.28a)

(6.28b) -1

(6.29a) -1

1  3 1 -1 3  α1 + 0.67  θ=    Δ  1 3  α 2 +1.33   l 1 0   1   1 -1 3k -1 -1 3  α1 + 0.67  KΔ = k    l 1 0 1 3 α +1.33 -1 2       2  T.K. Datta Department Of Civil Engineering, IIT Delhi

(6.29b) -1

1 -1 1 0   

(6.30)


4/11

Contd..  Equation of motion for the frame is given by:

MΔx + CΔx +K Δt Δx = -MIΔx g

(6.31)

C = αK Δ0 + βM

(6.32)

The solution requires K D t to be computed at time t; this requires 1 &  2 to be calculated.  Following steps are used for the calculation x = x + Δx ; x = x + Δx i i-1 i-1 i i-1 i-1 M = M + ΔM ; M = M + ΔM 1i 1i-1 1i-1 2i 2i-1 2i-1 T.K. Datta Department Of Civil Engineering, IIT Delhi

(6.33a) (6.33b)


4/12

Contd..  D 1i 1& D  2 i 1 are obtained using Eqn. 6.29b in which  values are calculated as: r1i-1l r2i-1l α1 = & α2 = 6EIc 6EIc M1i-1 M2i-1 r1i-1 = ; r2i-1 = θ1i-1 θ 2i-1

 D M 1i 1& D M 2 i 1 are then obtained; and hence M 1i & M 2 i  &  are calculated from M 1i 1 2 M and 2 i , K D t is obtained using ( Eq. 6.30).  If Elasto-plastic state is assumed, then 1   2  0 for 1  2   P at the beginning of the time interval; for unloading1 &  2 are obtained by (Eq.6.28a.) T.K. Datta Department Of Civil Engineering, IIT Delhi


4/13

Contd..

Example 6.5: For the frame shown in Fig 6.17, find the stiffness matrix at t = 1.36 s given the response quantities in Table 6.1 θ4

θ1

5 3m

k

6

θ5

k

θ2

3m

k

D2 4 k

θ3

θ6 1

M

k

3 3m

D1

E = 2.48  107 kN/m2 Beam 30  40 cm

50KN-m

D3 2

k

Column 30  50 cm

5m

θY = 0.00109 rad

Frame

θ

Force-displacement curve Fig. 6.17



x

4/14

Contd.. Table 6.1

1

Time x x Step sec m m/s 1.36 0.00293 0.0341

3

1.36

5

θ

θ

θ

M

m/s2 -1.2945

rad 0.00109

rad/s 0.013

rad/s2 -0.452

kNm 50

0.00701 0.0883

-2.8586

0.00095

0.014

-0.297

-23.18

1.36

0.00978 0.1339

-3.4814 x

0.00053 0.009127

-0.098

42.89

2

1.36

0.00293 0.0341

-1.2945

0.00109

0.013

-0.452

-50

4

1.36

0.00701 0.0883

-2.8586

0.00095

0.014

-0.297

23.18

6

1.36

0.00978 0.1339

-3.4814

0.00053 0.009127

-0.098

-42.89

t

x

Table shows that sections 1 & 2 undergo yielding; recognising this, stiffness matrices are given below: T.K. Datta Department Of Civil Engineering, IIT Delhi


4/15

Contd..  1.067  Δ1 -1.067 2.133 Δ sym   2  0  Δ3 -1.067 2.133   0.8 -0.8 0 2.4   θ1  θ2 Κ = 4.83 ×10 4 ×  0.8 0 -0.8 0.8 4   0 0.8 0 0 0.8 3.2   θ3  0.8 θ -0.8 0 0.4 0 0 2.4   4 0 -0.8 0 0.4 0 0.8 4  0.8  θ5  0 0.8 0 0 0 0 0 0.8 3.2  θ6 

ΚΔ

 0.4451 = 4.83 ×10 4 × -0.6177  0.2302


1.276 -1.0552

sym  Δ1 Δ  2 1.811 Δ 3


Push over analysis  Push over analysis is a good nonlinear static (substitute) analysis for the inelastic dynamic analysis.  It provides load Vs deflection curve from rest to ultimate failure.  Load is representative of equivalent static load taken as a mode of the structure & total load is conveniently the base shear.  Deflection may represent any deflection & can be conveniently taken as the top deflection. T.K. Datta Department Of Civil Engineering, IIT Delhi


5/1

5/2

Contd..

 It can be force or displacement control depending upon whether force or displacement is given an increment.  For both , incremental nonlinear static analysis is ‘performed by finding K t matrix at the beginning of each increment.  Displacement controlled pushover analysis is preferred because, the analysis can be carried out up to a desired displacement level.  Following input data are required in addition to the fundamental mode shape(if used). T.K. Datta Department Of Civil Engineering, IIT Delhi


5/3

Contd..  Assumed collapse mechanism  Moment rotation relationship of yielding section.  Limiting displacement.

 Rotational capacity of plastic hinge.  Displacement controlled pushover analysis is carried out in following steps:  Choose suitable D1  Corresponding to D1 , find D 1  D1  r


r


5/4

Contd..  Obtain Dp  D1 ; obtain DVB1  At nth increment, VBn 

 DVB D  D i

1n

1i

 At the end of each increment , moments are checked at all potential locations of plastic hinge.  For this,  n is calculated from condensation relationship.  If | M |  M P , then ordinary hinge is assumed at that section to find K for subsequent increment. T.K. Datta Department Of Civil Engineering, IIT Delhi


5/5

Contd..  Rotations at the hinges are calculated at each step after they are formed.

 If rotational capacity is exceeded in a plastic hinge, rotational hinge failure precedes the mechanism of failure.  VBi V s Δ 1i is traced up to the desired displacement level. Example 6.6  Carry out an equivalent static nonlinear analysis for the frame shown in Fig 6.19. T.K. Datta Department Of Civil Engineering, IIT Delhi


Mmax yy

5/6

Contd.. 3m

3m 3m 3m My

3m

y

3m

c

3m 4m

Frame

Moment rotation curve for beams

4m

Fig.6.19 Cross section

Location

C1 C2 B1 B2

G,1st, 2nd 3rd ,4th, 5th & 6th G,1st, 2nd 3rd ,4th, 5th & 6th

b (mm)

d (mm)

400 400 300 300 400 500 300 300 Table 6.2


(kNm)

(rad)

(rad)

168.9 119.15 205.22 153.88

9.025E-3 0.0133 6.097E-3 8.397E-3

0.0271 0.0399 0.0183 0.0252


5/7

Contd..  Solution is obtained by SAP2000. D (m) 0.110891 0.118891 0.134891 0.142891 0.150891 0.174891 0.190891 0.206891 0.310891 0.318891 0.334891 0.350891 0.518891 0.534891 0.622891 1.448699 1.456699

Base shear (KN) 316.825 317.866 319.457 320.006 320.555 322.201 323.299 324.397 331.498 332.035 333.11 334.185 342.546 343.207 346.843 307.822 308.225

Plastic Hinges at section 1 1,2 1,2,3 1,2,3,4 1,2,3,4,5 1,2,3,4,5,6 1,2,3,4,5,6,7 1,2,3,4,5,6,7,8 1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10,11 1,2,3,4,5,6,7,8,9,10,11,12 1,2,3,4,5,6,7,8,9,10,11,12,13 1,2,3,4,5,6,7,8,9,10,11,12,13,14 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17 Table 6.3



5/8

Contd.. 1 0.9143 400

0.5345 0.3120 0.1988 0.0833

Base shear (kN)

0.7548 300 200 100 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Roof displacement (m)

Fig.6.21 Fig.6.20



5/9

Contd..

15

14 13

10

11 12

7 8 16 2

5 5

9

6 8

17

4

3

15

3

1

3



Ductility & Inelastic spectrum  A structure is designed for a load less than that obtained from seismic coefficient method or RSA (say, for VB / R . ( R  3  4)  The structure will undergo yielding, if it is subjected to the expected design earthquake.  The behavior will depend upon the force deformation characteristics of the sections.  The maximum displacements & deformations of the structure are expected to be greater than the yield displacements. T.K. Datta Department Of Civil Engineering, IIT Delhi


6/1

6/2

Contd..  How much the structure will deform beyond the yield limit depends upon its ductility; ductility factor is defined as μ =

xm xy

(6.35a )

f fo

For explaining ductility , two SDOFs fy are considered with elasto – plastic behavior & the other a corresponding elastic system shown in Fig 6.23.

Elastic Elasto-plastic

Stiffness k

xy xo

xm x

Fig. 6.23 T.K. Datta Department Of Civil Engineering, IIT Delhi


6/3

Contd..  An associated factor, called yield reduction factor, R Y is defined as inverse of f y :

fY x Y fY = = f0 x0

(6.35b)

 R Y  2 means that the strength of the SDOF system is halved compared to the elastic system.

xm xm x y μ = × =μ fy = x0 xy x0 Ry

(6.36)

 With the above definitions, equation of motion of SDOF system becomes: T.K. Datta Department Of Civil Engineering, IIT Delhi


6/4

Contd.. x + 2ξωn x + ω n 2 x y f(x, x) = - xg y

(6.37)

μ + 2ξω n x + ω n x y f(μ,μ) = - ωn 2

f(x, x) ; x(t) = fy

f(x, x) = ay =





fy

m

2

xg ay

(6.38)

μ(t) xy ;

= ω n 2 x 0 fy

depends uponn ,  ,


f y.


6/5

Contd..  Time history analysis shows the following :  For f Y  1 , responses remain within elastic limit & may be more than that for f Y  1.  For f Y  1, two counteracting effects take place (i) decrease of response due to dissipation of energy (ii) increase of response due to decreased equivalent stiffness.  Less the value of f , more is the permanent Y deformation at the end . 



is known if x m for a f Y & x 0 can be calculated.



6/6

Contd..  Effect of time period on  , xm , xo , fY are illustrated in Fig 6.24.

 For long periods, xm  xo  x go & independent of f Y ;   RY .  In velocity sensitive region, x m may be smaller or greater than x o ; not significantly affected by f Y ;  may be smaller or larger than RY .

 In acceleration sensitive region, xm  xo ; increases with decreasing f Y & T; ;   RY for shorter period,  can be very high (strength not very less). T.K. Datta Department Of Civil Engineering, IIT Delhi


6/7

Contd..

f  0.125 y f  0.25 0.1 xxy 0.05 m g0

0.01 0.005

Te =10 Tf =15

T d =3

Tc =0 .5

0.5 1 5 10 T (sec) n

f 0.5 y

0.001 50 100

Normalized peak deformations for elasto-plastic system and elastic system Fig. 6.24 T.K. Datta Department Of Civil Engineering, IIT Delhi

Spectral Regions Vel. Disp. sensitive sensitive 10

fy 0.125 f 0.25 y

5

1

1 f 1 y

0.5 Ta =0 .035

0.05 0.1

x x f 1 m g0 y

5

xm/x0

0.001 0.01

10

0.1 0.05

Tb =0.12 5

0.01 0.005

f  0.5 y

Acc. sensitive

0.1 0.01

0.050.1

0.5

Te =10 Tf =15

1 0.5

Td =3

1 0.5

Tc =0 .5

10 5

Tb =0 .125

10 5

Ta =0.0 35

x0/ xg0 or xm/ xg0

Acc. sensitive

Spectral Regions Vel. Disp. sensitive sensitive

0.5 1 5 10 T (sec) n

0.1 50 100

Ratio of the peak deformations Response Spectrum Method Of Analysis

6/8

Inelastic response spectra  Inelastic response spectrum is plotted for :

Dy = x y

Vy = ωn x y A y = ωn2 x y

(6.39)

 For a fixed value of , and  plots of DY , VY , AY against Tn are the inelastic spectra or ductility spectra & they can be plotted in tripartite plot.  Yield strength of the E-P System. fy = mA y

(6.40)

 Yield strength for a specified  is difficult to obtain; but reverse is possible by interpolation technique. T.K. Datta Department Of Civil Engineering, IIT Delhi


6/9

Contd..  For a given set of Tn &  , obtain response for E-P system for a number of f Y .

 Each solution will give a  ; f o  K xo , x o is maximum displacement of elastic system.  From the set of f & corresponding f Y .

 , find the desired  &

 Using f Y value, find

 for the E-P system.

 Through iterative process the desired f Y and



are obtained .



6/10

Contd..  For different values of Tn , the process is repeated to obtain the ductility spectrum. 1

fy /w =Ay /g

0.8  1

0.6

0.4

1.5

0.2

4

2

8 0 0

0.5

1

1.5

2

2.5

3

Tn (sec)



6/11

Contd..  From the ductility spectrum, yield strength to limit  for a given set of Tn &  can be obtained. 2 x = μ x = μ A ω  Peak deformation m Y Y n.

 1.5

 2 0.37

0.0

 4 5

0.195

 8

0.05 0.01

0.050.1

Te =10 Tf =15

Td =3

Tc =0.5

Tb =0.125

 The plot is shown in Fig. 6.26.

Ta =0.035

0.12 0.1

2

0.5 1 5 10 Tn(sec)

10 20 50 100



Ry

0.5

fy

 If spectrum for   1 is known ,it is possible to plot f vs. Tn for different values of  .

1

1

7/1

Contd..  Above plot for a number of earthquakes are used to obtain idealized forms of f & Tn .  1   -1/2 fy = (2 μ - 1)  -1 μ  

1

Tn < Ta

T < Tn < Tc b

(6.41)

Tn > Tc

 1  1.5  2  4

0.2

0.05 0.01

0.050.1

0.5 1 5 10 Tn (sec)


Tf =33

Tc =1/

Te =10

 8

Tc' Tb =1/8

0.1

Ta =1/33

fy

0.5

50 100

Fig. 6.27 Response Spectrum Method Of Analysis

Construction of the spectra  As RY  1 fY , idealized inelastic design spectrum for a particular  can be constructed from elastic design spectrum.  Inelastic spectra of many earthquakes when smoothed compare well with that obtained as above.  Construction of the spectrum follows the steps below :  Divide constant A-ordinates of segment b  c by RY  2 1 to obtain b'c'. T.K. Datta Department Of Civil Engineering, IIT Delhi


7/2

7/3

 Similarly, divide V ord inates of segments (c  d ) by RY   ; to get c'd ' ; D ordinates of segments (d  e) by RY   to get d 'e'; ordinate f by  to get f ' .

Pseudo-velocity V or Vy (log scale)

Contd..

 f ' & e' ; draw DY  xgo  for Tn  33 s ; take a ' as the same a; a '&b'. 1    Draw AY  xgo for Tn  s 33

.


V

. xgo d

Elastic design c V spectrum V/ 

b

c'

e

d' e'

f

b' Inelastic design spectrum

a a'

f'

Tb=1/8 sec Ta=1/33 sec

Te=10 sec Tf=33 sec

Natural vibration period Tn (sec) (log scale)

Illustration of the Method Fig. 6.28 Response Spectrum Method Of Analysis

7/4

Contd.. Example 6.7 : Construct inelastic   2 design spectrum from the elastic spectrum given in Fig 2.22.

Ac c. (m /se

c

2

Elastic design spectrum

2

)

g

1m

10

1 0.7 0.5 0.4 0.3

Inelastic design spectrum

1m 0.

Pseudo velocity(m/sec)

m 10

10 7 5 4 3

Di sp . (m )

The inelastic design spectrum is drawn & shown in Fig 6.28b. 1g

0.2

g 0.1 0.0 1m

0.1 0.07 0.05 0.04 0.03

1g 0.0

0. 00 1m

1

0.01 0.007 0.005 0.004 0.003

0. 00 01 g

1m 00 0.0

0. 00 1g

0.02

-5

10

m

0.002

Ta 0.001 0.01 0.02

Tc

Tb 0.05

0.1

0.2 0.3 0.50.7 1

Td Te 2

3 4 5 67 10

Tf 20 30 50 70100

Time period (sec) (b)

Inelastic design spectrum for  = 2 Fig. 6.28b



Ductility in multi-storey frames  For an SDOF, inelastic spectrum can provide design yield strength for a given  ; maximum displacement under earthquake is found as  x y  For multi-storey building , it is not possible because  It is difficult to obtain design yield strength of all for a uniform  .  Ductility demands imposed by earthquake on widely differ.  Some studies on multi - storey frames are summarized here to show how ductility demands vary from member to member when designed using elastic spectrum for uniform  . T.K. Datta Department Of Civil Engineering, IIT Delhi


7/5

7/6

Contd..  Shear frames are designed following seismic coefficient method ; VBY is obtained using inelastic spectrum of El centro earthquake for a specified ductility & storey shears are distributed as per code.

 Frames are analysed assuming E-P behaviour of columns for El centro earthquake.  The storey stiffness is determined using seismic coefficient method by assuming storey drifts to be equal. T.K. Datta Department Of Civil Engineering, IIT Delhi


7/7

Contd..  Results show that For taller frames,  are larger in upper & lower stories; decrease in middle storeys. Deviation of storey ductility demands from the design one increases for taller frames. In general  demand is maximum at the first storey & could be 2-3 times the design 

Study shows that increase of base shear by some percentage tends to keep the  demand within a stipulated limit. T.K. Datta Department Of Civil Engineering, IIT Delhi




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