S. K. Ghosh Associates Inc. Palatine, IL and Aliso Viejo, CA

- 1 -

S. K. Ghosh Associates Inc.

Palatine, IL and Aliso Viejo, CA

www.skghoshassociates.com

SEISMIC DESIGN USING STRUCTURAL DYNAMICS

(PART ONE)

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www.skghoshassociates.com 1

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ASCE 7-10 Table 12.6-1 Permitted Analytical Procedures

SeismicDesign

CategoryStructural Characteristics

EquivalentLateralForce

AnalysisSection 12.8

ModalResponseSpectrumAnalysis

Section 12.9

SeismicResponse

HistoryProceduresChapter 16

D, E and F

Structures with no structural irregularities and not exceeding 160 feet in structural height

P P P

Structures exceeding 160 feet in structural height with no structural irregularities and with T < 3.5 Ts

P P P

Structures not exceeding 160 feet in structural height and having only horizontal irregularities of Type 2, 3, 4, or 5 in Table 12.3-1 or vertical irregularities of Type 4, 5a, or 5b in Table 12.3-2

P P P

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ASCE 7-10 Table 12.6-1 Permitted Analytical Procedures



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Design Ground Motion at Site

“Response Spectra”

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1940 El Centro (Imperial Valley, CA Eq.) Accelerogram

N-S component and corresponding derived velocity and displacement plots

-0.4

-0.2

0

0.2

0.4

-12-8-4048

12

-6

-3

0

3

6

g(g

)g

(in./s

ec.)

x g(in

.)

Ground acceleration, g

Ground velocity, g

Ground displacement, xg

0 4010 20 30

Time, sec.



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Computationof Deformation (orDisplacement)ResponseSpectrum

0.4 g

0

-0.4 g

Time, sec.0 10 20 30

-10

0

10

-10

0

10

-10

0

10

Time, sec.0 10 20 30

Def

orm

atio

n, u

, in.

Gro

und

Acc

eler

atio

n,g

(t)

Natural Vibration Period, T, sec.0 1 2 3

5

0

10

20

15

Sd,

in.

Deformation(or Displacement) Response Spectrum

= 2 Percent

El CentroSOOE Component

May 18, 1940

T = 0.5 sec= 0.02

T = 1 sec= 0.02

T = 2 sec= 0.02

xmax = 2.48 in.

xmax = 6.61 in.

xmax = 8.84 in.

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Response Spectra

Sd = umax

Sv = Sd = (2 / T)Sd

Sa = Sv = 2Sd

Sa = (2 / T)Sv =(2 / T)2 Sd



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(a)Deformation (or Displacement)

(b) Pseudo-velocity and (c)pseudo-acceleration

response spectra

El Centro ground motion – S00o Ecomponent. Damping ratio = 2 percent

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Four-Way Logarithmic Plot of Response Spectrum

El Centro ground motion – S00o Ecomponent.Damping ratio = 2 percent



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Response Spectra for Elastic SDOF Systems Subjected to 1940 El Centro Record

N-Scomponent

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Site-Specific Response Spectra



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Smoothed Site-Specific ResponseSpectra

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Applied Technology CouncilRecommendations for Specifying Ground Shaking at a Site



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Normalized Response Spectra Shapes

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Two-Factor Approach to Local Site Response



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1997 UBC Design Response Spectra

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2000, 2003 IBCDesign Spectrum



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ASCE 7-05, -10 Figure 11.4-1: Design Response Spectrum (2006 – 2015 IBC)

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Design Earthquake Ground Motion Prior to IBC

Approximately 90% probability of non-

exceedance in 50 years (approx. 475-year

return period)



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IBC/ASCE 7 Ground Motion Through 2009 IBC/ASCE 7-05

Maximum Considered Earthquake (MCE):

The maximum level of earthquake ground shaking

that is considered reasonable to design buildings to

resist.

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2% probability of exceedance in 50 years

(approx, 2500 yr. return period) where

deterministic approach is not used.




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Deterministic earthquake (in coastal California) - best

estimate of ground motion from maximum magnitude

earthquakes on seismic faults with high probabilities

of occurrence.

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IBC/ASCE 7 Design Ground Motion Through 2009 IBC/ASCE 7-05SS = MCE spectral acceleration in the short period range for Site Class B

S1 = MCE spectral acceleration at 1.0-sec period for Site Class B

SMS= FaSS, MCE spectral acceleration in the short period range adjusted for Site Class effects.

SM1 = FvS1, MCE spectral acceleration at 1.0-sec period adjusted for Site Class effects.

SDS = (2/3) SMS, spectral acceleration in the short period range for the design ground motion

SD1 = (2/3) SM1, spectral acceleration at 1.0-sec period for the design ground motion.



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BREAK!

If you are encountering technical difficulties, please call (847) 991-2700

If you have any questions, please type them in

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Question andAnswer Session





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Seismic Ground Motion UpdatesASCE 7-05/ 2006, 2009 IBC to ASCE 7-10/2012, 2015 IBC

CHANGE 1 Updated source zone models and attenuation relationships

CHANGE 2 Switch to risk-targeted ground motion

CHANGE 3 Switch to maximum-direction ground motion

CHANGE 4 Change in deterministic ground motion

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CHANGE 1:USGS has updated some source zone models, have used Next Generation Attenuation (NGA) relationships, to the exclusion of old relationships, in the western U.S., and have used new attenuation relationships in addition to old ones in the central and eastern U.S.

UPDATED SOURCE ZONES, NEW ATTENUATION RELATIONSHIPS



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CHANGE 2:Uniform-hazard ground motion has now been replaced by risk-targeted ground motion.

This switch from a 2% in 50-year hazard level to a 1% in 50-year collapse risk target has resulted in up to 30% decreases in ground motion in high-hazard areas of the central and eastern U.S. and in coastal Oregon and in some increases in other areas.

UNIFORM HAZARD RISK TARGETED

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CHANGE 3:A switch has been made from “geo-mean” ground motions (square root of the product of ground motions in any two orthogonal directions) to maximum direction ground motions. This has resulted in increases in short-period ground motion by a factor of 1.1 and in long-period ground motion by a factor of 1.3.

GEO-MEAN MAXIMUM DIRECTION



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CHANGE 4:Deterministic ground motions have been changed from 150% of median ground motions to 84th

percentile ground motions, which are 180% of median ground motions.

150% OF MEDIAN 180% OF MEDIAN

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ASCE 7-10 MCER Ground MotionResponse Accelerations11.2 DEFINITIONSRISK-TARGETED MAXIMUM CONSIDERED

EARTHQUAKE (MCER) GROUND MOTIONRESPONSE ACCELERATIONS The most severe earthquake effects considered by this standard determined for the orientation that results in the largest maximum response to horizontal ground motions and with adjustment for targeted risk.



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ASCE 7-10/2012, 2015 IBC Design Ground Motion SS = MCER spectral acceleration in the short period range for Site Class B

S1 = MCER spectral acceleration at 1.0-sec period for Site Class B

SMS= FaSS, MCER spectral acceleration in the short period range adjusted for Site Class effects.

SM1 = FvS1, MCER spectral acceleration at 1.0-sec period adjusted for Site Class effects.

SDS = (2/3) SMS, spectral acceleration in the short period range for the design ground motion

SD1 = (2/3) SM1, spectral acceleration at 1.0-sec period for the design ground motion.

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ASCE 7-10 ELF Design Base Shear



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ASCE 7-10 ELFMinimum Design Base Shear

All Seismic Design Categories…

Vmin = 0.044SDSIeW 0.01W (Eq. 12.8-5)

wherever S1 > 0.6g...

(Eq. 12.8-6)WR

IS.V emin

150

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Mode Superposition Analysis of Earthquake Response

3

2

1

333231

232221

131211

3

2

1

XXX

xxx

3332321313

3232221212

3132121111

XXXxXXXx

XXXxor

orX

Xx 321



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Mode Superposition Analysis of Earthquake Response

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Example

• Given

hs = 10 ft

w = 386.4 kips/floor

E = 4000 ksi

Icol = 4500 in.4 eachcolumn (0.7Ig )



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Determine Mass Matrix

m = w/g = 386.4 / 386.4

= 1.0 kip-sec.2/in.

100010001

m

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Determine Stiffness Matrix

12EI/hs3 = 12 x 4000 x 9000 / (12x10)3 = 250 kips/in.

kij = force corresponding to displacement of coordinate iresulting from a unit displacement of coordinate j 110

121012

250k



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Equations of Motion

Multi-degree-of-freedom system

- mass concentrated at floor levels which are subject to lateral displacements only

p(t)xkxcxm

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Equations of Motion

Free vibration

[C] = [0], {p(t)} = {0}

0xkxm



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Equations of Motion

Motion of a system in free vibration is simple harmonic

tsinAxtcosAx

tsinAx

2

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Equations of Motion

00

2

2

AmkAkAm-



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Find Determinant for Matrix[k] - 2[m]

The period is equal to 2 / :

T1 = 0.893 sec.

T2 = 0.319 sec.

T3 = 0.221 sec.

Setting the determinant of the above matrix equal to zero yields the following frequencies:

2

2

2

2

25025002505002500250500

mk

1 = 7.036 radians/sec.



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Find Mode Shapes

First Mode:

000

03672502500250036750025002500367500

11

21

31

2

2

2

).().(

).(

31 = 1.0, 21 = 0.802, 11 = 0.445



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Second Mode:

000

68519250250025068519500250025068519500

12

22

32

2

2

2

).().(

).(

32 = 1.0, 22 = -0.55, 12 = -1.22

Find Mode Shapes

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33

23

13

Third Mode:

000

49128250250025049128500250025049128500

13

23

33

2

2

2

).().(

).(

33 = 1.0, 23 = -2.25, 13 = 1.802

Find Mode Shapes



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ASCE 7-10 Section 11.4.5General Procedure Design Spectrum

A five percent damped elastic design response spectrum constructed in accordance with Figure 11.4-1, using the values of SDS and SD1 consistent with the specific site.

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ASCE 7-10 Section 11.4.7Site-Specific Ground Motion ProceduresThe site-specific ground motion procedures set forth in Chapter 21 are permitted to be used to determine ground motions for any structure. A site response analysis shall be performed in accordance with Section 21.1 for structures on Site Class F sites, unless the exception to Section 20.3.1 is applicable. For seismically isolated structures and for structures with damping systems on sites with S1 greater than or equal to 0.6, a ground motion hazard analysis shall be performed in accordance with Section 21.2.



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ASCE 7-10 Section 21.2Ground Motion Hazard Analysis

The ground motion hazard analysis shall account for the regional tectonic setting, geology, and seismicity, the expected recurrence rates and maximum magnitudes of earthquakes on known faults and source zones, the characteristics of ground motion attenuation, near source effects, if any, on ground motions, and the effects of subsurface site conditions on ground motions. The characteristics of subsurface site conditions shall be considered either using attenuation relations that represent regional and local geology or in accordance with Section 21.1.

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ASCE 7-10 Section 21.4

Where the site-specific procedure is used to determine the design ground motion, the parameter SDS shall be taken as the spectral acceleration, Sa, obtained from the site-specific spectra at a period of 0.2 sec, except that it shall not be taken less than 90 percent of the peak spectral acceleration, Sa, at any period larger than 0.2 sec. The parameter SD1 shall be taken as the greater of the spectral acceleration, Sa, at a period of 1 sec or two times the spectral acceleration, Sa, at a period of 2 sec. The parameters SMS and SM1 shall be taken as 1.5 times SDS and SD1, respectively. The values so obtained shall not be less than 80 percent of the values determined by the general procedure.



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ASCE 7-10 Section 12.7.3 Mathematical Model

1. A mathematical model of the structure shall beconstructed for the purpose of determining member forces and structure displacements resulting from applied loads and any imposed displacements or P-delta effects. The model shall include the stiffness and strength of elements that are significant to the distribution of forces and deformations in the structure and represent the spatial distribution of mass and stiffness throughout the structure.

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ASCE 7-10 Section 12.7.3Mathematical Model

2. In addition, the model shall comply with the following:

a. Stiffness properties of concrete and masonry elements shall include the effects of cracked sections.

b. For steel moment frame systems, the contribution of panel zone deformations to overall story drift shall be included.



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3. Structures that have horizontal structural irregularity Type 1a, 1b, 4, or 5 of Table 12.3-1 shall be analyzed using a 3-D representation. Where a 3-D model is used, a minimum of three dynamic degrees of freedom consisting of translation in two orthogonal plan directions and rotation about the vertical axis shall be included at each level of the structure. Where the diaphragms have not been classified as rigid or flexible in accordance with Section 12.3.1, the model shall include representation of the diaphragm’s stiffness characteristics and such additional dynamic degrees of freedom as are required to account for the participation of the diaphragm in the structure’s dynamic response.

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3. EXCEPTION: Analysis using a 3-D representation is not required for structures with flexible diaphragms that have Type 4 horizontal structural irregularities.

Type 4 horizontal irregularity: Out-of-plane offset of the LFRS



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12.7.3 Structural Modeling

For horizontal irregularity Type 4 (out-of-plane offsets) in a structure with flexible diaphragms, substantial orthogonal effects are not likely. The eccentricity in the vertical load path will cause a redistribution of seismic design forces from vertical elements at story above to vertical elements essentially in the same direction at story below. The effect on vertical elements in orthogonal direction will be minimal.

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Modal Response Spectrum Analysis

An elastic dynamic analysis of structure utilizing the peak dynamic response of all modes having a significant contribution to total structural response. Peak modal responses are calculated using the ordinates of the appropriate response spectrum curve which correspond to the modal periods. Maximum modal contributions are combined in a statistical manner to obtain an approximate total structural response.



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ASCE 7-10 Section 12.9.4 Design Values (2012, 2015 IBC)

Vt < 0.85 VELF (based on T CuTa)

Must scale up to 0.85VELF

(based on T CuTa)

No scale down, except by R/Ie-factor

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12.9.4.2 Scaling of Drifts

Where the combined response for the modal base shear (Vt) is less than 0.85CsW, where Csis determined in accordance with Eq. 12.8-6, drifts shall be multiplied by 0.85CsW/Vt



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ASCE 7-10 ELF Design Base Shear

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12.9.4.2 Scaling of Drifts

Added in ASCE 7-10 is scaling of drifts where the near-fault minimum base shear equation governs. There is general agreement that this was always the intent and that this requirement is important for long-period structures (for which minimum base shears control).



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Thank You!!

For more information…


Phone: (847) 991-2700 / (949) 215-6560

Email: [email protected]



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SEISMIC DESIGN USING STRUCTURAL DYNAMICS

(PART TWO)


Palatine, IL and Aliso Viejo, CA


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Step 1 – Develop mathematical model of structure... to represent proper spatial distribution of mass and stiffness of structure (ASCE 7-10 Section 12.7.3)

Step 2 – Determine mode shapes { m} and corresponding periods Tm of structure... by eigenvalue analysis



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Step 3 – For each mode m, determine:

Earthquake participation factor:

Modal Mass:1

2

i

imim g

wM

n

i

imim g

wL1

- 68 -


Step 3 – For each mode m, determine (cont’d):

Effective Weight

Participating Mass

gMLW

m

mm

2

WMLPM

m

m2

im

n

iiwW

1Where

displacement at floor level i for mode m

weight at floor level i



- 69 -


Step 4 – Determine number of modes to be considered... to represent at least 90% of participating mass of structure (ASCE 7-10 Section 12.9.1)

PM = (Wm/W) 0.90

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Step 5 – Determine spectral acceleration and seismic design coefficient for each mode

(1) From Design Response Spectrum (ASCE 7-10 Fig. 11.4-1), determine Sam for Tm,



- 71 -


(2) Determine modal seismic design coefficient,

Csm = (Sam x Ie) / R

where Ie = Importance Factor from ASCE 7-10 Table 1.5-2 and R = Response Modification Factor from Table 12.2-1

- 72 -


Step 6 – Determine modal base shears, Vm, and total dynamic base shear, Vt

Vm = (Csm/g) x Wm

Vt = (V12 + V2

2 + ... + Vn2)1/2



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Step 7 – Determine design base shear from Static Force Procedure (ASCE 7-10 Section 12.8) and compare with base shear from dynamic analysis.

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Step 8 – Scale dynamic analysis results (ASCE 7-10 Section 12.9.4)

Scale factor = 0.85(VELF /Vt) 1

VELF = the equivalent lateral force procedure base shear,

computed based on ASCE 7-10 Sections 12.9.4 and

12.8

Vt = the base shear from the required modal

combination



- 75 -


Step 8 (cont’d) – Scale dynamic analysis results (ASCE 7-10 Sec. 12.9.4)

Adjusted Vt = mean root square of adjusted Vm

values

Vt = (V12 + V2

2 + ... + Vn2)1/2

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Step 9 – Distribute base shear for each mode over height of structure

where Fim = lateral force at level ifor mode m

Vm = base shear for mode m

mimi

imiim V

wwF



- 77 -


Step 10 – Perform lateral analysis for each mode... to determine member forces for each mode of vibration being considered (ASCE 7-10 Section 12.9.2)

For rigid diaphragms... include accidental torsion (ASCE 7-10 Section 12.8.4.2) in the distribution of story shears to seismic force-resisting systems.

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Step 11 – Combine dynamic analysis results (moments, shears, axial forces, and displacements) for all considered modes using root mean square combination (SRSS)... to approximate total structural response (ASCE 7-10 Section 12.9.3) or resultant design values



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Example 1

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Modal Response Spectrum Analysis Example

• Given

hs = 10 ft

w = 386.4 kips/floor

E = 4000 ksi

Icol = 4500 in.4 eachcolumn



- 81 -

Find Determinant for Matrix [k] - 2[m]

2

2

2

2

25025002505002500250500

][][ mk

100010001

][m110121012

250][k

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Find Mode Shapes

First Mode:

000

03672502500250036750025002500367500

11

21

31

2

2

2

).().(

).(

31 = 1.0, 21 = 0.802, 11 = 0.445



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Second Mode:

000

68519250250025068519500250025068519500

12

22

32

2

2

2

).().(

).(

32 = 1.0, 22 = -0.55, 12 = -1.22

Find Mode Shapes

- 84 -

33

23

13

Third Mode:

000

49128250250025049128500250025049128500

13

23

33

2

2

2

).().(

).(

33 = 1.0, 23 = -2.25, 13 = 1.802

Find Mode Shapes



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Determine Modal Mass and Participation Factors for Each Mode

= 1.0 kip-sec2/in. ( 11 + 21 + 31)

= 1.0 (0.445 + 0.802 + 1.0) = 2.247 kip-sec2/in.g

wL i

ii

3

11

1

= 1.0 kip-sec2/in. ( 112 + 21

2 + 312)

= 1.0 (0.4452 + 0.8022 + 1.02)

= 1.841 kip-sec2/in.

g

wM i

ii

3

1

21

1

- 86 -

= 1.0 kip-sec2/in. ( 12 + 22 + 32)

= 1.0 (-1.22 - 0.55 + 1.0) = -0.77 kip-sec2/in.g

wL i

ii

3

12

2

= 1.0 kip-sec2/in. ( 122 + 22

2 + 322)

= 1.0 (1.222 + 0.552 + 1.02)


g

wM i

ii

3

1

22

2




- 87 -

= 1.0 kip-sec2/in. ( 12 + 22 + 32)

= 1.0 (1.802 – 2.25 + 1.0) = 0.552 kip-sec2/in.g

wL i

ii

3

13

3

= 1.0 kip-sec2/in. ( 122 + 22

2 + 322)

= 1.0 (1.8022 + 2.252 + 1.02)


g

wM i

ii

3

1

23

3


- 88 -

Determine Effective Weight and Participating Mass for Each Mode

kips451154

kips6512438631095520

kips088243867912770

kips721059secin.

in.seckip4386

84112472

2

3

23

3

2

2

22

2

2

22

1

21

1

.W

...

).(gMLW

...

).(gMLW

....g

MLW

i



- 89 -

9960

011043863

6512

071043863

0882

914043863

721059

33

22

11

.PM

..

.WWPM

..

.WWPM

..

.WWPM


- 90 -


Consideration of Modes 1, 2 and 3 is sufficient per ASCE 7-10 Section 12.9.1.

Indeed, the consideration of just the first mode would have been sufficient (as PM1 0.90).



- 91 -

Determine Seismic Design Coefficient, Csm, for Each Mode

For the example building considered

g.g.

gR

IS

gT.

gT

.g

RTIS

RISC

eDS

mmm

eD

eamsm

1250=8

1×01=

0750=

81×60

==

×=

1

- 92 -

Determine Seismic Design Coefficient, Csm, for Each Mode

Mode 1: T1 = 0.893 sec

Cs1 = (0.075g/0.893) = 0.0840g

Mode 2: T2 = 0.319 sec

Cs2 = (0.075g/0.319) = 0.2351g 0.125gUse 0.125g

Mode 3: T3 = 0.221 sec

Cs3 = (0.075g/0.221) = 0.3394g 0.125gUse 0.125g



- 93 -

Determine Modal Base Shears

Mode 1: V1 = (CS1W1/g) = 0.0840 x 1059.72 = 89.02 kips

Mode 2: V2 = (CS2W2/g) = 0.125 x 82.08 = 10.26 kips

Mode 3: V3 = (CS3W3/g) = 0.125 x 12.65 = 1.58 kips

Vt = [89.022 + 10.262 + 1.582]1/2 = 89.6 kips

msm

m Wg

CV

- 94 -

Design Base Shear

Determine design base shear from static-force procedure and compare with base shear from dynamic analysis

WRIS

WIS

WR

IS

WRT

ISV

e

eDS

eDS

eD

1

1

5.0

01.0W0.044



- 95 -

ASCE 7-10 Section 12.8.2.1Approximate Fundamental Period

Ta = Ct (hn)x

Ct = 0.016 (MRF system)

x = 0.9 (MRF systems)

hn = structural height = 30 ft

Ta = 0.016 x (30)0.9 = 0.342 sec

- 96 -

ASCE 7-10 Section 12.9.4Period of CuTa

T = CuTa = 1.4 x 0.342 = 0.479 sec < T1 = 0.893 sec

Vs = SD1I eW/(RT) = [0.60 x 1 / (8×0.479)]W = 0.157W

< SDSIeW / R = 1.0 × 1 × W / 8 = 0.125W governs

> 0.5S1IeW / R = 0.5x0.6x1xW =0.0375W

0.044SDSIeW = 0.044x1.0x1xW = 0.044W

Use V = 0.125 x 3 x 386.4 = 144.9 kips



- 97 -

ASCE 7-10 Section 12.9.4 Design Base Shear

The base shear, V, using modal analysis should not be less than 85 percent of the calculated base shear using the static procedure based on a period = CuTa

V= 0.85 × 144.9 kips = 123 kips > 89.6 kips

So the modal forces must be scaled up.

Scale factor = 0.85 × 144.9 / 89.6 = 1.375

- 98 -

Scale Up Modal Results

V1 = 1.375 x 89.02 = 122.0 kips

V2 = 1.375 x 10.26 = 14.1 kips

V3 = 1.375 x 1.58 = 2.2 kips

V = [122.02 + 14.12 + 2.22]1/2

= 122.8 kips



- 99 -

Distribute Base Shear for Each Mode over Height of Structure

Lateral force at level i for mode m

mimi

imiim V

wwF

- 100 -


Level,i

Weight, wi

i1 wi i1 Fi1

3 386.4 1 386.4 54.3

2 386.4 0.802 309.9 43.5

1 386.4 0.445 171.9 24.2

= 868.2 122.0

Mode 1 V1 = 122.0 kips



- 101 -

Level,i

Weight, wi

i2 wi i2 Fi2

3 386.4 1 386.4 -18.3

2 386.4 -0.55 -212.5 10.1

1 386.4 -1.22 -471.4 22.3

= -297.5 14.1



- 102 -


Level,i

Weight, wi

i3 wi i3 Fi3

3 386.4 1 386.4 4.0

2 386.4 -2.25 -869.4 -9.0

1 386.4 1.802 696.3 7.2

= 213.3 2.2




- 103 -

Modal Response Spectrum Analysis Example

3 @

10

ft =

30 ft

54.3

43.5

24.2

18.3

10.1

22.3

4.0

9.0

7.2

- 104 -

Example 2



- 105 -

General

A 20-story reinforced concrete building is designed following the requirements of the 2012, 2015 International Building Codes (IBC) and ASCE 7-10.

Building is located on Site Class D.

- 106 -

The modal response spectrum analysis procedure is used as the basis of design.

Comparison with design forces given by the equivalent lateral force procedure is also shown.

General



- 107 -

The building is symmetrical about both principal plan axes. Along each axis, a dual shear wall-frame interactive system is utilized for resistance to lateral forces

General

- 108 -

Typical Floor Plan of Example Building



- 109 -

Elevation of Example Building

- 110 -

Design Criteria

Beams 34 in. x 24 in.

(width = 34 in.)

Interior Columns 40 in. x 40 in.

Exterior Columns 34 in. x 34 in.



- 111 -

Design Criteria

Shear walls

Grade to 9th Floor 16 in. thick

10th Floor to 16th Floor 14 in. thick

17th Floor to Roof 12 in. thick

Shear Wall Boundary Elements 38 in. x 38 in.

- 112 -

Design Criteria

Material Properties

Concrete: fc’ = 4,000 psi

All members are constructed of normal-weight concrete (wc = 150 pcf)

Reinforcement: fy = 60,000 psi



- 113 -

Design Criteria

Service Loads

Partition live load 15 psf

Live Load 80 psf

(2012, 2015 IBC Table 1607.1)

8-in. two-way concrete slab:

(8/12) × 1.0 × 1.0 × 150 = 100 psf

- 114 -

Design Criteria

Service Loads

Superimposed roofing load 10 psf

Penthouse and equipment 200 kips

Live load on roof 20 psf

(2012, 2015 IBC Table 1607.1)

Cladding 8 psf

(Modeled as a line load along perimeter)



- 115 -

Seismic Design Data

It is assumed that, at the site of the structure,

Ss = 1.5g and S1 = 0.6g

Assume Risk Category II (2012, 2015 IBC Table 1604.5)

Seismic Importance Factor, Ie =1.0

(ASCE 7-10 Table 1.5-2)

Use Default Site Class D

- 116 -

Seismic Design Data

Site coefficient Fa = 1.0

(2012, 2015 IBC Table 1613.3.3(1) — ASCE 7-10 Table 11.4-1)

Site coefficient Fv = 1.5

(2012, 2015 IBC Table 1613.3.3(2) — ASCE 7-10 Table 11.4-2)

Soil modified SMS = FaSs = 1.0 x 1.5g = 1.5g

(2012, 2015 IBC Eq. 16-37 — ASCE 7-10 Table 11.4-1)

Soil modified SM1 = FvS1 = 1.5 x 0.6g = 0.9g

(2012, 2015 IBC Eq.16-38 — ASCE 7-10 Table 11.4-2)



- 117 -

Seismic Design Data

Design spectral Response acceleration Parameters (at 5% damping):

SDS = 2/3 SMS = 2/3 x 1.5g = 1.0g

(2012, 2015 IBC Eq. 16-39 — ASCE 7-10 Eq. 11.4-3)

SD1 = 2/3 SM1 = 2/3 x 0.9g = 0.6g

(2012, 2015 IBC Eq. 16-40 — ASCE 7-10 Eq. 11.4-4)

- 118 -

Seismic Design Data

Dual System (RC Shear walls with SMRF)

R = 7, Cd = 5.5 (ASCE 7-10 Table 12.2-1)

where: R and Cd are response modification factor and deflection amplification factor, respectively

Seismic design category:

Based on both SDS (2012, 2015 IBC Table 1613.3.5(1)) and SD1 (2012, 2015 IBC Table 1613.3.5(2)), the Seismic Design Category for the example building is D



- 119 -

Weights on each floor

Floors:

2nd – 20th floors: 100×132.82 / 1000 = 1764 kips

Roof: (100+10)×132.82 / 1000 + 200 = 2140 kips

Beams (portion below floor slab):

150×22.8×(34/12)×[(24-8)/12]×56 / 1000 = 724 kips

- 120 -


Cladding:

2nd floor: 8×15×130×4 / 1000 = 62 kips

3rd – 20th floors: 8×12.5×130×4 / 1000 = 52 kips

Roof: 8×6.25×130×4 / 1000 = 26 kips

Exterior Columns:

2nd floor: 150×15×(34/12)2×20 / 1000 = 361 kips

3rd – 20th floors: 150×12.5×(34/12)2×20 / 1000 = 301 kips

Roof: 150×6.25×(34/12)2×20 / 1000 = 151 kips



- 121 -


Interior Columns:

2nd floor: 150×15×(40/12)2×8 / 1000 = 200 kips

3rd – 20th floors: 150×12.5×(40/12)2×8 / 1000 = 167 kips

Roof: 150×6.25×(40/12)2×8 / 1000 = 83 kips

Boundary Elements:

2nd floor: 150×15×(38/12)2×8 / 1000 = 181 kips

3rd – 20th floors: 150×12.5×(38/12)2×8 / 1000 = 150 kips

Roof: 150×6.25×(38/12)2×8 / 1000 = 75 kips

- 122 -


Shear Walls:2nd floor: 150×15×(16/12)×22.8×4 / 1000 = 274 kips

3rd – 9th floors: 150×12.5×(16/12)×22.8×4 / 1000 = 228 kips

10th floor: 150×[6.25×(16/12)+6.25×(14/12)]×22.8×4 / 1000 = 214 kips

11th – 16th floors: 150×12.5×(14/12)×22.8×4 / 1000 = 200 kips

17th floor: 150×[6.25×(14/12)+6.25×(12/12)]×22.8×4 / 1000 = 185 kips

18th – 20th floors: 150×12.5×(12/12)×22.8×4 / 1000 = 171 kips

Roof: 150×6.25×(12/12)×22.8×4 / 1000 = 86 kips



- 123 -


Total:2nd floor: 1764+724+62+361+200+181+274 = 3566 kips

3rd – 9th floors: 1764+724+52+301+167+150+228 = 3386 kips

10th floor: 1764+724+52+301+167+150+214 = 3372 kips

11th – 16th floors: 1764+724+52+301+167+150+200 = 3358 kips

17th floor: 1764+724+52+301+167+150+185 = 3343 kips

18th – 20th floors: 1764+724+52+301+167+150+171 = 3329 kips

Roof: 1764+724+26+151+83+75+86 = 3285 kips

Total weight of the building: W = Wi = 67,329 kips

- 124 -

Gravity Load Analysis

The service-level axial forces for each shear wall due to dead and live loads are given in Table 1.

Live load reduction factors were used as follows:

Floors (2012, 2015 IBC Section 1607.10.2):

R = 0.08 (A - 150)%

60% (for vertical members)

23.1 (1 + D/L0)% (D is calculated based on floor weight only)



- 125 -


Roof (2012, 2015 IBC Section 1607.12.2):

Lr = L0 R1 R2

R1 = 1.0 for At 200 ft2

= 1.2 – 0.001 At for 200 ft2 At 600 ft2

= 0.6 for At 600 ft2

R2 = 1.0 for flat roof

- 126 -


For shear wall:

At = 26×25 = 1352 ft2 for roof

A = 26×25 = 1352 ft2 for 20th floor

A = 2×1352 ft2 for 19th floor and so on



- 127 -

Table 1. Service-Level Axial Forces Due to DL and LL in Shear wall

FloorLevel

SupportedArea - ft2

DLpsf

Cum.DLkips

LLpsf

Cum.LLkips R-factor RLL

psfRLLkips

Cum.RLLkips

Roof 1352.0 121 163.6 20 27.0 0.600 12.0 16.2 16

20 1352.0 100 298.8 50 94.6 0.600 20.0 27.0 43

19 2704.0 100 434.0 50 162.2 0.600 20.0 54.1 70

18 4056.0 100 569.2 50 229.8 0.600 20.0 81.1 97

17 5408.0 100 704.4 50 297.4 0.600 20.0 108.2 124

16 6760.0 100 839.6 50 365.0 0.600 20.0 135.2 151

15 8112.0 100 974.8 50 432.6 0.600 20.0 162.2 178

14 9464.0 100 1110.0 50 500.2 0.600 20.0 189.3 206

13 10816.0 100 1245.2 50 567.8 0.600 20.0 216.3 233

12 12168.0 100 1380.4 50 635.4 0.600 20.0 243.4 260

11 13520.0 100 1515.6 50 703.0 0.600 20.0 270.4 287

10 14872.0 100 1650.8 50 770.6 0.600 20.0 297.4 314

9 16224.0 100 1786.0 50 838.2 0.600 20.0 324.5 341

8 17576.0 100 1921.2 50 905.8 0.600 20.0 351.5 368

7 18928.0 100 2056.4 50 973.4 0.600 20.0 378.6 395

6 20280.0 100 2191.6 50 1041.0 0.600 20.0 405.6 422

5 21632.0 100 2326.8 50 1108.6 0.600 20.0 432.6 449

4 22984.0 100 2462.0 50 1176.2 0.600 20.0 459.7 476

3 24336.0 100 2597.2 50 1243.8 0.600 20.0 486.7 503

2 25688.0 100 2732.4 50 1311.4 0.600 20.0 513.8 530

- 128 -

BREAK!





- 129 -




- 130 -

Seismic Analysis Procedure Selection

In this example, the equivalent lateral force procedure (ASCE 7-10 Section 12.8) can be used as T < 3.5Ts and it is a regular structure assigned to SDC D.

However, modal response spectrum analysis of ASCE 7-10 Section 12.9 will be illustrated.



- 131 -

Base Shear Using Equivalent Lateral Force Procedure (ASCE 7-10 Section 12.8)

WRT

TIS LeD2

1

ASCE 7-10 Eq. 12.8-1 and12.8-2WR

ISV eDS

ASCE 7-10 Eq. 12.8-4, for T TL

W.WIS. eDS 0100440 ASCE 7-10 Eq. 12.8-5

WR

IS. e150ASCE 7-10 Eq. 12.8-6, for S1 0.6g

- 132 -

Approximate Period(ASCE 7-10 Section 12.8.2.1)Ta = Ct (hn)x (ASCE 7-10 Eq. 12.8-7)

Ct = 0.02 for dual system

x = 0.75 for a dual system

hn = total height = 255 ft

Ta = 0.02 x (255)0.75 = 1.28 sec



- 133 -

Base Shear Based on Approximate Period

kips28867

3296701605050kips67332967010010

kips296232967010104400.044

(governs)kips45092817

329670160

kips96187

329670101

1

1

,...WR

IS.,.W.

,...WIS.

,..WRT

IS

,..WR

IS

e

eDS

eD

eDS

Use V = 4509 kips

- 134 -

Vertical Distribution of Base Shear

20

1i

kii

kxx

Vx

hw

hwC

T = 1.28 seck = 1+ 0.5 (T - 0.5) = 1.39

The distribution of the design base shear along the height of the building is shown in Table 2

VCF Vxx ASCE 7-10 Eq. 12.8-11

ASCE 7-10 Eq. 12.8-12



- 135 -

Table 2. Lateral Forces by Equivalent Lateral Force Procedure

V = 4509 kips T (sec.) = 1.28 k = 1.39

Floor Levelx

Weightwx, kips

Heighthx, ft

wxhxk, ft-kips Lateral Force

Fx, kipsStory Shear

Vx, kips

Roof 3285 255.0 7,271,537 490 49020 3329 242.5 6,871,683 463 95419 3329 230.0 6,384,332 431 138418 3329 217.5 5,907,207 398 178317 3343 205.0 5,463,541 368 215116 3358 192.5 5,028,511 339 249015 3358 180.0 4,580,464 309 279914 3358 167.5 4,144,396 279 307813 3358 155.0 3,720,846 251 332912 3358 142.5 3,310,422 223 355311 3358 130.0 2,913,815 196 374910 3372 117.5 2,542,378 171 39209 3312 105.0 2,135,716 144 40658 3386 92.5 1,830,729 123 41887 3386 80.0 1,496,174 101 42896 3386 67.5 1,181,460 80 43695 3386 55.0 888,773 60 44284 3386 42.5 621,080 42 44703 3386 30.0 382,724 26 44962 3566 17.5 190,548 13 4509

67,329 66,866,338 4509

- 136 -

Lateral AnalysisUsing ETABS computer program, a three-dimensional analysis of the structure was performed under the lateral forces shown in Table 2.

Rigid diaphragm at each level.

Rigid offsets at both ends of each beam.

Accidental torsion (ASCE 7-10 Section 12.8.4.2) accounted for by displacing the center of mass by a distance of 5% of the building dimension perpendicular to the force direction.



- 137 -

Lateral Analysis

Considering cracked section properties (ASCE 7-10 Section 12.7.3):

For columns and shear walls, Ieff = 0.7IgFor beams, Ieff = 0.5Ig

P- effects were considered in the lateral analysis. Please note that this effect can be neglected in many situations as explained in ASCE 7-10 Section 12.8.7.

- 138 -

Table 3. Lateral Displacements of Example Building by Equivalent Lateral Force Procedure

Floor Levelx

Weightwx, kips

Lateral ForceFx, kips

Displacementx, in. wx x

2,kip-in.2 Fx x, kip-in.

Roof 3285 490 8.9 260205 436420 3329 463 8.5 240520 393919 3329 431 8.1 218416 348718 3329 398 7.6 192283 302717 3343 368 7.2 173301 265316 3358 339 6.7 150741 227215 3358 309 6.2 129082 191514 3358 279 5.7 109101 159313 3358 251 5.1 87342 128012 3358 223 4.6 71055 102711 3358 196 4.0 53728 78610 3372 171 3.5 41307 6009 3312 144 3.0 29808 4328 3386 123 2.5 21163 3097 3386 101 2.0 13544 2026 3386 80 1.5 7619 1205 3386 60 1.1 4097 664 3386 42 0.7 1659 293 3386 26 0.4 542 102 3566 13 0.2 143 3

67,329 1,805,654 28,112



- 139 -

Using approximate period (ASCE 7-10 Section 12.8.2.1) results in a conservative design.

A more rational method for computation of period can be used to reduce the design forces.

However, ASCE 7-10 Section 12.8.2 requires the modified period not to exceed CuTa

where: Cu = 1.4 (ASCE 7-10 Table 12.8-1)

Rational Period Computation

- 140 -

The Rayleigh-Ritz procedure is used for rational period determination (based on the values shown in Table 3):

Rational Period Computation

(governs)sec79128141562

1122838665480512

2 2

...sec.

),( / ,,

Fg/wT iiii



- 141 -

Base Shear Based on Refined Period

kips28867

3296701605050kips67332967010010

kips296232967010104400.044

(governs)kips32247917

329670160

kips96187

329670101

1

1

,...WR

IS.,.W.

,...WIS.

,..WRT

IS

,..WR

IS

e

eDS

eD

eDS

Use V = 3224 kips

- 142 -

Table 4. Lateral Forces by Equivalent Lateral Force Procedure – Refined Period

V = 3224 kips T (sec.) = 1.79 k = 1.65

Floor Levelx

Weightwx, kips

Heighthx, ft

wxhxk, ft-kips Lateral Force

Fx, kipsStory Shear

Vx, kips

Roof 3285 255.0 29,874,041 386 38620 3329 242.5 27,871,774 360 74619 3329 230.0 25,547,952 330 107618 3329 217.5 23,304,204 301 137717 3343 205.0 21,231,050 274 165116 3358 192.5 19,229,556 248 189915 3358 180.0 17,218,843 222 212214 3358 167.5 15,296,253 198 231913 3358 155.0 13,464,069 174 249312 3358 142.5 11,724,819 151 264411 3358 130.0 10,081,322 130 277510 3372 117.5 8,572,349 111 28859 3312 105.0 6,997,562 90 29768 3386 92.5 5,807,512 75 30517 3386 80.0 4,573,724 59 31106 3386 67.5 3,458,531 45 31545 3386 55.0 2,469,355 32 31864 3386 42.5 1,615,797 21 32073 3386 30.0 911,069 12 32192 3566 17.5 395,347 5 3224

67,329 249,645,130 3224



- 143 -

Table 5. Lateral Displacements and Drifts of Example Building by Equivalent Force Procedure –Refined Period (in.)

Floor Level xe Cd x Drift,

Roof 5.92 5.50 32.54 1.4520 5.65 5.50 31.09 1.5419 5.37 5.50 29.55 1.6018 5.08 5.50 27.95 1.6817 4.78 5.50 26.27 1.7516 4.46 5.50 24.51 1.8115 4.13 5.50 22.70 1.8914 3.79 5.50 20.82 1.9413 3.43 5.50 18.88 1.9712 3.07 5.50 16.91 1.9911 2.71 5.50 14.91 1.9910 2.35 5.50 12.92 1.969 1.99 5.50 10.96 1.908 1.65 5.50 9.06 1.827 1.32 5.50 7.24 1.716 1.00 5.50 5.52 1.565 0.72 5.50 3.96 1.374 0.47 5.50 2.59 1.143 0.26 5.50 1.46 0.862 0.11 5.50 0.60 0.601 0.00 0.00 0.00 0.00

x = Cd xe/Ie (ASCE 7-10 Eq. 12.8-15) = x,i - x,i-1

- 144 -

Story Drift Limitation

According to ASCE 7-10 Table 12.12-1, the calculated story drifts shall not exceed 0.020 times the story height for this example building

Floor Maximum Allowable Drift Largest Drift (Table 5)

1st 0.02 x 17.5 ft = 4.2 in. > 0.60 in. O.K.

Others 0.02 x 12.5 ft = 3.0 in. > 1.99 in. O.K.



- 145 -

P- Effects (ASCE 7-10 Section 12.8.7)P- effects need not be considered when stability coefficient ( ) is equal to or less than 0.10:

Px = the total vertical design load at and above Level x

= the design story drift

Vx = the seismic shear force acting between Levels x and x 1

hsx = the story height below Level x

Cd = the deflection amplification factor in ASCE 7-10 Table 12.2-1

Ie = importance factor in ASCE 7-10 Table 1.5-2

dsxx

ex

ChVIP

- 146 -

Table 6. Calculation of Stability Coefficient

Max value < 0.1

So, P- effect could have been

neglected

StoryLevel

DLkips

Cum. DLkips

LLpsf

Areasq.ft

Pxkips

Vxkips

hsxft

Drift,in.

Roof 3285 3285 12 16900 3488 386 12.5 1.45 0.016

20 3329 6614 20 33800 7290 746 12.5 1.54 0.018

19 3329 9943 20 50700 10957 1076 12.5 1.60 0.020

18 3329 13272 20 67600 14624 1377 12.5 1.68 0.022

17 3343 16615 20 84500 18305 1651 12.5 1.75 0.024

16 3358 19973 20 101400 22001 1899 12.5 1.81 0.025

15 3358 23331 20 118300 25697 2122 12.5 1.89 0.028

14 3358 26689 20 135200 29393 2319 12.5 1.94 0.030

13 3358 30047 20 152100 33089 2493 12.5 1.97 0.032

12 3358 33405 20 169000 36785 2644 12.5 1.99 0.034

11 3358 36763 20 185900 40481 2775 12.5 1.99 0.035

10 3372 40135 20 202800 44191 2885 12.5 1.96 0.036

9 3312 43447 20 219700 47841 2976 12.5 1.90 0.037

8 3386 46833 20 236600 51565 3051 12.5 1.82 0.037

7 3386 50219 20 253500 55289 3110 12.5 1.71 0.037

6 3386 53605 20 270400 59013 3154 12.5 1.56 0.035

5 3386 56991 20 287300 62737 3186 12.5 1.37 0.033

4 3386 60377 20 304200 66461 3207 12.5 1.14 0.029

3 3386 63763 20 321100 70185 3219 12.5 0.86 0.023

2 3566 67329 20 338000 74089 3224 17.5 0.60 0.012



- 147 -

Modal Analysis

Using ETABS computer program, the first four mode shapes in X-direction are calculated

Self weight is automatically considered by ETABS

The superimposed dead load (SDL) needs to be computed and assigned to each floor level for the purpose of modal analysis

SDL on roof:

(10/1000) + 200/[(5×26 + 34/12)2] = 0.0218 ksf

- 148 -

Mode Period(sec)

Mass Participation (%)

1 2.39 70.6

2 0.62 15.2

3 0.28 6.2

4 0.17 3.1

Modal Analysis



- 149 -

As seen from the Table, consideration of modes 1, 2 and 3 should be adequate for lateral load analysis, as they constitute about 92% (more than 90%) of the participating mass (ASCE 7-10 Section 12.9.1)

Modal Analysis

- 150 -

Base Shear, Vm, For Each Mode Shape Using Modal Response Spectrum Analysis Procedure (ASCE Section 12.9)

The portion of base shear contributed by the mth

mode, Vm, is to be determined from the following equations:

msmm WCV

mmm M/LW 2

n

iimim wL

1

n

iimim wM

1

2



- 151 -

Table 7. Calculation of Lm, Mm for Example Building

Mode 1, T1 = 2.39 secFloorLevel

i

wi i1 wi i1 wi i12

kips kips kipsRoof 3285 1 3285.0 3285.0

20 3329 0.9565 3184.2 3045.619 3329 0.9107 3031.7 2760.918 3329 0.8626 2871.6 2477.017 3343 0.8130 2717.8 2209.516 3358 0.7603 2553.1 1941.115 3358 0.7053 2368.5 1670.614 3358 0.6481 2176.3 1410.413 3358 0.5893 1978.9 1166.212 3358 0.5290 1776.4 939.711 3358 0.4679 1571.3 735.310 3372 0.4061 1369.4 556.19 3312 0.3458 1145.3 396.08 3386 0.2863 969.3 277.57 3386 0.2290 775.4 177.66 3386 0.1756 594.5 104.45 3386 0.1260 426.5 53.74 3386 0.0824 279.2 23.03 3386 0.0466 157.7 7.32 3566 0.0191 68.1 1.3

67,329 33,300 23,238

L1 = 33,300 kips/g

M1 = 23,238 kips/g

- 152 -


i

wi i2 wi i2 wi i22


20 3329 0.8018 2669.2 2140.119 3329 0.5886 1959.6 1153.518 3329 0.3702 1232.5 456.317 3343 0.1518 507.6 77.116 3358 -0.0576 -193.4 11.115 3358 -0.2528 -848.9 214.614 3358 -0.4271 -1434.1 612.513 3358 -0.5729 -1923.9 1102.212 3358 -0.6844 -2298.1 1572.811 3358 -0.7577 -2544.2 1927.710 3372 -0.7906 -2665.8 2107.59 3312 -0.7838 -2596.1 2034.98 3386 -0.7405 -2507.2 1856.57 3386 -0.6649 -2251.4 1497.06 3386 -0.5632 -1907.0 1074.05 3386 -0.4435 -1501.8 666.14 3386 -0.3171 -1073.8 340.53 3386 -0.1945 -658.5 128.02 3566 -0.0890 -317.4 28.2

67,329 -15,068 22,286

L2 = -15,068 kips/g

M2 = 22,286 kips/g

Table 7. (continued)



- 153 -


i

wi i3 wi i3 wi i32


20 3329 0.6325 2105.7 1331.919 3329 0.2366 787.6 186.318 3329 -0.1431 -476.3 68.217 3343 -0.4675 -1562.8 730.616 3358 -0.6927 -2326.0 1611.215 3358 -0.8049 -2702.8 2175.414 3358 -0.7935 -2664.6 2114.313 3358 -0.6618 -2222.3 1470.712 3358 -0.4309 -1446.9 623.511 3358 -0.1341 -450.5 60.410 3372 0.1870 630.5 117.99 3312 0.4789 1586.0 759.58 3386 0.7130 2414.2 1721.47 3386 0.8593 2909.8 2500.56 3386 0.9008 3050.2 2747.65 3386 0.8366 2832.7 2369.84 3386 0.6829 2312.4 1579.23 3386 0.4691 1588.4 745.12 3566 0.2390 852.4 203.7

67,329 10,503 26,402

L3 = 10,503 kips/g

M3 = 26,402 kips/g


- 154 -

where Sam = the modal design spectral response acceleration at period Tm determined from either the general design response spectrum of ASCE 7-10 Section 11.4.5 or a site-specific response spectrum per Section 21.1 or 21.2.

In this example, the general design response spectrum is used.

Modal Seismic Design Coefficients Csm

)I/R(SC

e

amsm



- 155 -


For T >TL Sa = SD1TL/ T2

(For this example, it does not apply. For this 20-story building, TL isgenerally much larger than T.)

For Ts < T < TL Sa = SD1/T

For T0 < T < TS Sa = SDS

For T < T0 Sa = SDS (0.4 + 0.6T/T0)

where TS = SD1/SDS, and T0 = 0.2TS

- 156 -

For example building: Ts = 0.6/1.0 = 0.60 sec

T0 = 0.2×0.6 = 0.12 sec

Mode 1:

T1 = 2.39 sec > 0.60 sec Cs1 = 0.6g/(2.39x(7/1)) = 0.036g

Mode 2:

T2 = 0.62 sec > 0.60 sec Cs2 = 0.6g/(0.62x(7/1)) = 0.138g

Mode 3:

0.12 sec < T3 = 0.28 sec < 0.60 sec

Cs3 = 1.0g/(7/1) = 0.143g




- 157 -

smm

msmmm C

MLCWV

2

Base shear, Vm, for each mode shape

kips597143040226503103

kips1406138028622068152

kips1718036023823300331

2

3

2

2

2

1

g.g,

,: VMode

g.g,

),(: VMode

g.g,

,: VMode

kips229959714061718 222V

SRSS combination (ASCE 7-10 Section 12.9.3):

- 158 -

Participating Mass for the above three modes

.....PM

%.

%.

%.

O.K9029226115970

2610067,32926,402

10,503PM:3Mode

11510067,32922,286

(-15,068)PM:2Mode

90710067,32923,238

33,300PM:1Mode

2

3

2

2

2

1

WMgLPM

m

m2

Therefore, consideration of the first three modes is sufficient.



- 159 -

Scaling of Elastic Response Parameters for DesignASCE 7-10 Section12.9.4.1 stipulates that the base shear using modal analysis must be scaled up when the combined response for the modal base shear Vt isless than 85 percent of the calculated base shear VELFusing the equivalent lateral force procedure.

0.85 × 3224 (VELF from equivalent lateral force procedure) = 2740 kips 2299 kips (Vt from modal response spectrum procedure)

Therefore, the modal forces must be scaled up.

- 160 -

Scaling of Elastic Response Parameters for Design

Scale factor = 0.85VELF / Vt = 1.192

Modified modal base shears:

V1 = 1.192×1718 = 2048 kips

V2 = 1.192×1406 = 1676 kips

V3 = 1.192×597 = 712 kips

kips274071216762048 222V



- 161 -

Distribution of Base Shear

mn

iimi

imiim V

w

wF

1

Lateral force at level i for mode m:

Distribution of modal base shear for each mode is shown in Table 8

- 162 -

Table 8. Distribution of Base Shear for Example Building

Mode 1, T1 = 2.39 sec and V1 = 2048 kips

FloorLevel i

wikips i1

wi i1kips wi i1/ wi i1

Fi1kips

Roof 3285 1 3285.0 0.0986 20220 3329 0.9565 3184.2 0.0956 19619 3329 0.9107 3031.7 0.0910 18618 3329 0.8626 2871.6 0.0862 17717 3343 0.8130 2717.8 0.0816 16716 3358 0.7603 2553.1 0.0767 15715 3358 0.7053 2368.5 0.0711 14614 3358 0.6481 2176.3 0.0654 13413 3358 0.5893 1978.9 0.0594 12212 3358 0.5290 1776.4 0.0533 10911 3358 0.4679 1571.3 0.0472 9710 3372 0.4061 1369.4 0.0411 849 3312 0.3458 1145.3 0.0344 708 3386 0.2863 969.3 0.0291 607 3386 0.2290 775.4 0.0233 486 3386 0.1756 594.5 0.0179 375 3386 0.1260 426.5 0.0128 264 3386 0.0824 279.2 0.0084 173 3386 0.0466 157.7 0.0047 102 3566 0.0191 68.1 0.0020 4

67,329 33,300 1.0000 2048



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Table 8. (continued)Mode 2, T2 = 0.62 sec and V2 = 1676 kips

FloorLevel i

wikips i2


Fi2kips

Roof 3285 1 3285.0 -0.2180 -36520 3329 0.8018 2669.2 -0.1771 -29719 3329 0.5886 1959.6 -0.1300 -21818 3329 0.3702 1232.5 -0.0818 -13717 3343 0.1518 507.6 -0.0337 -5616 3358 -0.0576 -193.4 0.0128 2215 3358 -0.2528 -848.9 0.0563 9414 3358 -0.4271 -1434.1 0.0952 16013 3358 -0.5729 -1923.9 0.1277 21412 3358 -0.6844 -2298.1 0.1525 25611 3358 -0.7577 -2544.2 0.1689 28310 3372 -0.7906 -2665.8 0.1769 2979 3312 -0.7838 -2596.1 0.1723 2898 3386 -0.7405 -2507.2 0.1664 2797 3386 -0.6649 -2251.4 0.1494 2506 3386 -0.5632 -1907.0 0.1266 2125 3386 -0.4435 -1501.8 0.0997 1674 3386 -0.3171 -1073.8 0.0713 1193 3386 -0.1945 -658.5 0.0437 732 3566 -0.0890 -317.4 0.0211 35

67,329 -15,068 1.0000 1676

- 164 -

Table 8. (continued)Mode 3, T3 = 0.28 sec and V3 = 712 kips

FloorLevel i

wikips i3


Fi3kips

Roof 3285 1 3285.0 0.3128 22320 3329 0.6325 2105.7 0.2005 14319 3329 0.2366 787.6 0.0750 5318 3329 -0.1431 -476.3 -0.0454 -3217 3343 -0.4675 -1562.8 -0.1488 -10616 3358 -0.6927 -2326.0 -0.2215 -15815 3358 -0.8049 -2702.8 -0.2573 -18314 3358 -0.7935 -2664.6 -0.2537 -18113 3358 -0.6618 -2222.3 -0.2116 -15112 3358 -0.4309 -1446.9 -0.1378 -9811 3358 -0.1341 -450.5 -0.0429 -3110 3372 0.1870 630.5 0.0600 439 3312 0.4789 1586.0 0.1510 1088 3386 0.7130 2414.2 0.2299 1647 3386 0.8593 2909.8 0.2771 1976 3386 0.9008 3050.2 0.2904 2075 3386 0.8366 2832.7 0.2697 1924 3386 0.6829 2312.4 0.2202 1573 3386 0.4691 1588.4 0.1512 1082 3566 0.2390 852.4 0.0812 58

67,329 10,503 1.0000 712



- 165 -

Lateral AnalysisUsing ETABS computer program, a three-dimensional analysis of the structure was performed under the lateral forces shown in Table 8.

Rigid diaphragm at each level.

Rigid offsets at both ends of each beam.

Accidental torsion (ASCE 7-10 Section 12.8.4.2) accounted for by displacing the center of mass by a distance of 5% of the building dimension perpendicular to the force direction.

- 166 -

Lateral Analysis

Considering cracked section properties (ASCE 7-10 Section 12.7.3):

For columns and shear walls, Ieff = 0.7IgFor beams, Ieff = 0.5Ig

P- effects were considered in the lateral analysis. Please note that this effect can be neglected in many situations as explained in ASCE 7-10 Section 12.8.7.



- 167 -

Table 9. Lateral Displacements and Drifts of Example Building from Dynamic Analysis (in.)

FloorLevel

Mode 1xe1

Mode 2xe2

Mode 3xe3

Resultantxe

Cd xDrift

Roof 3.21 -0.40 0.05 3.23 5.50 17.76 0.81

20 3.07 -0.32 0.03 3.08 5.50 16.95 0.85

19 2.92 -0.23 0.01 2.93 5.50 16.11 0.87

18 2.77 -0.15 -0.01 2.77 5.50 15.23 0.90

17 2.61 -0.06 -0.03 2.61 5.50 14.33 0.93

16 2.44 0.02 -0.04 2.44 5.50 13.40 0.95

15 2.26 0.10 -0.04 2.26 5.50 12.45 0.98

14 2.08 0.17 -0.04 2.09 5.50 11.47 1.00

13 1.89 0.22 -0.03 1.90 5.50 10.46 1.02

12 1.70 0.27 -0.02 1.72 5.50 9.44 1.03

11 1.50 0.30 -0.01 1.53 5.50 8.40 1.0410 1.30 0.31 0.01 1.34 5.50 7.36 1.049 1.11 0.31 0.02 1.15 5.50 6.32 1.03

8 0.92 0.29 0.03 0.96 5.50 5.29 1.00

7 0.73 0.26 0.04 0.78 5.50 4.29 0.97

6 0.56 0.22 0.04 0.60 5.50 3.32 0.89

5 0.40 0.17 0.04 0.44 5.50 2.43 0.81

4 0.27 0.12 0.03 0.29 5.50 1.62 0.69

3 0.15 0.08 0.02 0.17 5.50 0.93 0.53

2 0.06 0.04 0.01 0.07 5.50 0.40 0.40

- 168 -

Story Drift Limitation

According to ASCE 7-10 Table 12.12-1, the calculated story drifts shall not exceed 0.020 times the story height for this example building

Floor Maximum Allowable Drift Largest Drift (Table 5)

1st 0.02 x 17.5 ft = 4.2 in. > 0.40 in. O.K.

Others 0.02 x 12.5 ft = 3.0 in. > 1.04 in. O.K.



- 169 -

Table 10 shows the shear force and bending moment at each floor level of each shear wall (the four shear walls are identical in every respect and are subject to the same forces) due to each considered mode and the resultant load effects

Lateral Analysis

- 170 -

Table 10. Internal Forces in a Shear Wall Due To Lateral Forces Given in Table 8

Story Location ShearV (kips)

MomentM (ft-kips)

Story 20

Mode 1 Top -314 -1,209Bottom -314 -5,129

Mode 2 Top 55 708Bottom 55 1,387

Mode 3 Top 61 -163Bottom 61 596

Resultant Top 324 1,411Bottom 324 5,347

Story 19

Mode 1 Top -94 -6,450Bottom -94 -7,614

Mode 2 Top -165 2,160Bottom -165 97

Mode 3 Top 147 419Bottom 147 2,262


Story 18

Mode 1 Top -25 -8,984Bottom -25 -9,282

Mode 2 Top -267 887Bottom -267 -2,463

Mode 3 Top 176 2,089Bottom 176 4,294




- 171 -



MomentM (ft-kips)

Story 17

Mode 1 Top 54 -10,719Bottom 54 -10,016

Mode 2 Top -339 -1,667Bottom -339 -5,923

Mode 3 Top 166 4,139Bottom 166 6,212


Story 16

Mode 1 Top 115 -11,525Bottom 115 -10,053

Mode 2 Top -379 -5,146Bottom -379 -9,897

Mode 3 Top 124 6,092Bottom 124 7,639


Story 15

Mode 1 Top 182 -11,635Bottom 182 -9,321

Mode 2 Top -378 -9,164Bottom -378 -13,907

Mode 3 Top 52 7,565Bottom 52 8,216


- 172 -



MomentM (ft-kips)

Story 14

Mode 1 Top 255 -10,974Bottom 255 -7,738

Mode 2 Top -355 -13,242Bottom -355 -17,690

Mode 3 Top -29 8,196Bottom -29 7,827


Story 13

Mode 1 Top 312 -9,455Bottom 312 -5,506

Mode 2 Top -296 -17,117Bottom -296 -20,827

Mode 3 Top -111 7,860Bottom -111 6,466


Story 12

Mode 1 Top 367 -7,276Bottom 367 -2,632

Mode 2 Top -214 -20,370Bottom -214 -23,055

Mode 3 Top -180 6,547Bottom -180 4,296




- 173 -



MomentM (ft-kips)

Story 11

Mode 1 Top 419 -4,442Bottom 419 863

Mode 2 Top -113 -22,731Bottom -113 -24,153

Mode 3 Top -224 4,413Bottom -224 1,603


Story 10

Mode 1 Top 466 -968Bottom 466 4,932

Mode 2 Top -1 -23,973Bottom -1 -23,988

Mode 3 Top -237 1,740Bottom -237 -1,223


Story 9

Mode 1 Top 522 3,103Bottom 522 9,712

Mode 2 Top 122 -23,955Bottom 122 -22,426

Mode 3 Top -221 -1,086Bottom -221 -3,850


- 174 -



MomentM (ft-kips)

Story 8

Mode 1 Top 566 7,911Bottom 566 15,078

Mode 2 Top 244 -22,531Bottom 244 -19,473

Mode 3 Top -168 -3,731Bottom -168 -5,839


Story 7

Mode 1 Top 615 13,330Bottom 615 21,113

Mode 2 Top 364 -19,702Bottom 364 -15,138

Mode 3 Top -92 -5,752Bottom -92 -6,902


Story 6

Mode 1 Top 665 19,450Bottom 665 27,854

Mode 2 Top 475 -15,471Bottom 475 -9,516

Mode 3 Top 1 -6,856Bottom 1 -6,846




- 175 -



MomentM (ft-kips)

Story 5

Mode 1 Top 717 26,311Bottom 717 35,363

Mode 2 Top 573 -9,925Bottom 573 -2,736

Mode 3 Top 99 -6,846Bottom 99 -5,610


Story 4

Mode 1 Top 773 33,980Bottom 773 43,721

Mode 2 Top 657 -3,184Bottom 657 5,057

Mode 3 Top 191 -5,651Bottom 191 -3,264


Story 3

Mode 1 Top 832 42,543Bottom 832 53,015

Mode 2 Top 724 4,611Bottom 724 13,693

Mode 3 Top 267 -3,336Bottom 267 5


- 176 -



MomentM (ft-kips)

Story 2

Mode 1 Top 907 52,091Bottom 907 63,487

Mode 2 Top 789 13,296Bottom 789 23,180

Mode 3 Top 327 -81Bottom 327 4,015


Story 1

Mode 1 Top 970 62,875Bottom 970 79,889

Mode 2 Top 808 22,886Bottom 808 37,044

Mode 3 Top 346 3,937Bottom 346 10,004


Because of the location of the shear walls within the plan of the building, the earthquake-induced axial force in each shear wall is equal to zero



- 177 -

Load Combinations For Design

Load combinations used in the strength design method for concrete (considering dead and live loads and earthquake forces only):

(1) U = 1.4D (2012, 2015 IBC Eq. 16-1)

(2) U = 1.2D + 1.6L (2012, 2015 IBC Eq. 16-2)

(3) U = 1.2D + f1L + 1.0E (2012, 2015 IBC Eq. 16-5)

(4) U = 0.9D + 1.0E (2012, 2015 IBC Eq. 16-7)

- 178 -


where:

D = Dead load effect

L = Live load effect

f1 = 0.5 (IBC Section 1605.2)

E = QE + 0.2SDSD when the effects of gravity and seismic loads are additive

(ASCE 7-10 Eq. 12.4-1, 12.4-3, 12.4-4)



- 179 -


E = QE - 0.2SDSD when the effects of gravity and seismic loads are counteractive

(ASCE 7-10 Eq. 12.4-2, 12.4-3, 12.4-4)

QE = Horizontal seismic forces effect

Use redundancy factor, = 1 (the configuration of this dual-system building qualifies for a factor of 1.0, according to ASCE 7-10 Section 12.3.4.2)

- 180 -

Design Loads

Table 11 shows a summary of the axial force, shear force and bending moment at the base of each shear wall



- 181 -

Table 11. Summary of Design Axial Force, Shear Force and Bending Moment for Shear Wall between Grade and Level 2

Loads Load Axial Force(kips)

Shear Force(kips)

BendingMoment(ft-kips)

Dead Load D 5439* 0 0Live Load L 915** 0 0

Lateral Load QE 0 1309 88,626* DL from Table 1 + Self-weight of shear wall + Weight of beams framing into the boundary elements** Reduced LL from Table 1 + 15 psf partition load

Load Combinations1 1.4D 7615 0 0

2 1.2D+1.6L 7991 0 0

3 1.2D+( QE+0.2DSDS)+0.5L 8072 1309 88,626

4 0.9D-( QE+0.2DSDS) 3807 -1309 -88,626

Pu = 8072 kipsVu = 1309 kips

Mu = 88,626 ft-kips

- 182 -






- 183 -

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