example design of 20 story building aci-libre

8/10/2019 Example Design of 20 Story Building ACI-libre

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ACI 314 Task Group B/C Draft No. 1

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Preliminary Design of a 20-story Reinforced Concrete Building

By

Mike Mota, P.E.

Chair Task B-C

Preliminary Design and Economical Impact

Member of ACI and Secretary of Committee 314

Atlantic Regional Manager

CRSI

Jim Lai, S.E.

(Retired)

March 19, 2008


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TABLE OF CONTENTS

1. Building description.. ..3

1.1 Material..3

1.2 Design loading3

1.3 Story weight. ..31.4 Governing codes.3

2. Outline of preliminary design procedure:6

2.1 Loading:..6

2.1.1 Develop seismic loading based on ASCE7-05 Chapter 11 and 126

2.1.2 Design of structural wall (shear wall)6

2.1.3 Design of special moment frame6

3. Lateral Force Analysis:7

3.1 Mapped Spectral Acceleration7

3.2 Structural System7

4. Equivalent Lateral Force Procedure:8

4.1 Unit Loads ..9

4.2 Seismic Story Shear and Building OTM.9

4.3 Preliminary design of structural wall.10

5. Moment Frame Design:..21

5.1 Two moment frames in each direction..22

5.2 Seismic Force distribution using Portal Method...25

5.3 Based on two cycle moment distribution..25

5.4 Column axial load (Between 3rd and 4th Floor)..25

6. Preliminary Material Quantities for Superstructure only....316.1 Shear-walls.31

6.2 Columns.32

6.3 Slabs...32

7. Appendix A: Power-point slides from Atlanta Session..34


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11. Building Description:

20-story office building in Los Angeles, CA has a dual moment resisting frame system of reinforced

concrete structural walls and reinforced concrete moment frames. Typical floor plan and an elevation are

shown in Figures 1 and 2.

The building is square in plan with five 28-ft bays totaling 142 ft 3 inches out to out in each direction.

Story heights are 23 ft from the first to second floors and 13 feet for the remaining 19 stories; the overall

building height is 270 feet.

Typical floor framing consists of 4 inches thick light weight concrete slabs, 12 x 18 beams at 9 ft- 4in

o.c. and 18 x 24 girders; interior columns are 30 inches square for the full height of the building.

Girders at the periphery of the floor are 27 x 36 and columns are 36 inches square for the full height of the

building.

A 28 ft x 84 ft x 13 ft high penthouse with equipment loading at the roof level

A small mezzanine floor at the first story

1.1 Material:

Concrete Strength fc = 4,000 psi above 3rdfloor (light weight 115 pcf)

fc = 5,000 psi below 3rdfloor (normal weight)

Reinforcement - fy= 60,000 psi

1.2 Design Loading:

Partition including miscellaneous dead load = 20 psf Floor Live load = 50 psf (reducible based on tributary area)

1.3 Story weight:

Roof = wrf=2800 kips

Floor 1620 wi= 2800 kips

Floor 9 15 wi= 2850 kips

Floor 3 8 wi= 2900 kips Floor 2 - w2= 4350 kips

Total building weight wi= 58,500 kips

1.4 Governing Codes:

IBC -2006

ACI 318-05 ASCE 7 -05

1This example was originally developed by James S. Lai of Johnson and Nielsen Associates, Structural

Engineers, Los Angeles, CA for BSSC trial design and was published in FEMA 140, Guide to Application of

NEHRP Recommended Provisions in Earthquake-Resistant Building Design, Building Seismic Safety Council,

Washington, D.C. 1990.


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Elevator

Opening

Beam

Stair

Girder

Typical

Bay

5 Bays @ 28 0 = 140 0

5 Bays @ 28 0 = 140 0

Fig. 1 - Typical Floor Plan


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2.0 OUTLINE OF PRELIMINARY DESIGN PROCEDURE:

2.1 LOADING

2.1.1 Develop seismic loading based on ASCE7-05 Chapter 11 and 12.

Establish response modification factor R, deflection amplification factor Cdand overstrength factor

0

Establish mapped maximum considered earthquake spectral response acceleration for short and longperiods Ssand Slfrom USGS data base

Calculate design spectral response acceleration SDsand SDl

Establish a standard response spectrum for design reference Calculate fundamental period Tausing (Eq. 12.8-7)

Calculate seismic response coefficient, Cs

Calculate seismic base shear V

Calculate vertical distribution of story seismic forces

Calculate building overturning Distribute seismic forces to structural walls and building frames accounting for accidental torsion

Approximate building deflection (any suggestions without doing computer run?)

2.1.2 Design of structural wall (shear wall)

Obtain seismic base shear for one wall pier from horizontal distribution

Calculate required seismic shear strength at lower story

Design wall thickness or guess at wall thickness and calculate nominal shear strength base on 8fc Calculate seismic overturning moment by proportion of building overturning or from story force

distribution Calculate gravity loads dead and live with the approximate loading combinations

Base on the calculated seismic OTM, obtain the approximate area of tension reinforcement

Check for requirement of boundary element based on Section 21.7.6 (ACI 318)

Establish P0, Pb, Mb, Mnto draw an interaction diagram based on = 1 Based on Pu/ and Mu/, check that design is within the interaction diagram envelope Check for termination of boundary reinforcement requirement Calculate confinement reinforcement for longitudinal boundary rebar

For upper stories, establish shear strength for reduce wall thicknesses and the minimum reinforcementrequirements

2.1.3 Design of special moment frame

Obtain seismic base shear for one perimeter frame from horizontal distribution (no less than 25% of total

building shear)

Distribute story seismic shear to column based on portal method (or other acceptable method) Calculate seismic axial force and moments in end column and first interior column

Calculate gravity loads dead and live axial loads

Calculate gravity load moments based on approximate coefficients Obtain combined loading combinations for girders and columns

For girder design, calculate minimum required flexural strength and reinforcement

Calculated required shear strength based on probable moment strength of girder, and design shear reinf. For column (design end column and first interior column), design longitudinal reinforcement such that the

column moment strength satisfies equation (21-1.) Calculate probable moment strength of column ends

Calculate required shear strength

Design transverse confinement reinforcement Check joint shear strength requirement


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3. Lateral Force Analysis

(Seismic)

Code: ASCE 7-05 and ACI

318-05Reference

ASCE 7-05 Remark

3.1 Mapped SpectralAcceleration 11.4.1

Short period Sa = 2.25 From USGS da

One second S1 = 0.75 From USGS da

Site Class D 11.4.2 Default Site Cla

Site Coefficent Fa = 1.0 Table 11.4-1

Fv = 1.5 Table 11.4-2

Maximum Considered Earthquake 11.4.3

SMS = FaSs = 2.25 (Eq. 11.4-1)

SM1 = FvS1 = 1.13 (Eq. 11.4-2)

Design Spectral Accel parameter 11.4.4

SDS = 2SMS/3 = 1.50 (Eq. 11.4-3)

SD1 = 2SM1/3 = 0.75 (Eq. 11.4-4)

Design Response Spectrum 11.4.5

T0 = 0.2 SD1/SDS = 0.10 sec

Short period transition period TS = SD1/SDS = 0.50 sec

Long period transition period TL = 12.0 From USGS da

For T < T0 Sa = SDS[0.4 + 0.6 T/T0] = (Eq. 11.4-5) T = fundamenta

For T0T TS Sa = SDS = of structure

For TST TL Sa = SD1/T = 0.563 (Eq. 11.4-6)

For T > TL Sa = SD1 TL/T2 = (Eq. 11.4-7)

MCE Response Spectrum MCE = 1.5 DBS = 0.845 11.4.61.5 x Design res

spectrumOccupancy Category I 11.5.1

Importance Factor I = 1.0 Table 11.5-1

Seismic Design Category 11.6

Based on SDS D SDS 0.50 Table 11.6-1

Based on SD1 D SD1 0.20 Table 11.6-2

3.2 Structural System 12.2

Dual System D3 Table 12.2-1

Response Modification Factor R = 7.0 Table 12.2-1

System overstrength factor o = 2.5 Table 12.2-1

Deflection amplification Factor Cd = 5.5 Table 12.2-1

Height Limit NL Table 12.2-1

Horizontal Structural Irregularity None Table 12.3-1

Vertical Structural Irregularity None Table 12.3-2

Redundancy Factor = 1.0 12.3.4.2

Analysis procedure T < 3.5 Ts = 1.75 Table 12.6-1

USE: Equivalent Static analysis


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4.1 Unit Load

Typical Floor

Finish floor 24" LW Conc.

Slab 45Ceiling 7

Misc 6

Partition 10

Beams 20

Girders 10

Columns 10

Dead Load* 70 90 100 110

Live 50 40 35 30

Total Load 120 130 135 140

* USE same load at roof to allow for equipment wt.

4.2 Seismic Story Shear and Building OTM

LevelHeight to Level

x hx

Weight

at Level

x wx

wxhxk

k=1.2

wxhx

k

wihi

SeismicForce

at

Level x

Story

Shear

Force

OTM

ft kips x 103 Cvx kips kips kip-ft

Roof 270 2,800 2,316 0.099 468

20 257 2,800 2,183 0.094 441 468 6,080

19 244 2,800 2,051 0.088 414 908 17,889

18 231 2,800 1,921 0.082 388 1,323 35,083

17 218 2,800 1,792 0.077 362 1,710 57,319

16 205 2,800 1,664 0.071 336 2,072 84,258

15 192 2,850 1,566 0.067 316 2,408 115,565

14 179 2,850 1,440 0.062 291 2,724 150,983

13 166 2,850 1,315 0.056 266 3,015 190,180

12 153 2,850 1,193 0.051 241 3,281 232,829

11 140 2,850 1,072 0.046 216 3,521 278,607

10 127 2,850 954 0.041 193 3,738 327,200

9 114 2,850 838 0.036 169 3,930 378,296

8 101 2,900 737 0.032 149 4,100 431,590

7 88 2,900 625 0.027 126 4,248 486,820

6 75 2,900 516 0.022 104 4,375 543,690

5 62 2,900 410 0.018 83 4,479 601,913

4 49 2,900 309 0.013 62 4,562 661,214

3 36 2,900 214 0.009 43 4,624 721,327

2 23 4,350 187 0.008 38 4,667 782,002

1 0 4,705 890,218

Total 58,500 23,304 1.000 4,705

Seismic base shear V = 4705 kips


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28.0

1.25 3.00 22.0 3.00 1.25

A A

2.5

P1 P2 P3 h

4.25

A - A

Plan

Elevation

4.3 Preliminary design of structural wall

Dead Load Live Load

Level P1 P2 P3 PD P1 P2 P3 PL

Roof 131 24 65 220 220 41 0 0 41

20 147 56 81 284 504 39 8 31 78

19 147 56 81 284 788 39 8 31 78

18 147 56 81 284 1,072 39 8 31 78

17 147 56 81 284 1,356 39 8 31 78

16 147 56 81 284 1,640 39 8 31 78

15 147 56 81 284 1,925 39 8 31 78

14 147 56 81 284 2,209 39 8 31 78

13 147 56 81 284 2,493 39 8 31 78

12 147 56 81 284 2,777 39 8 31 78

11 147 56 81 284 3,061 39 8 31 78

10 147 62 81 290 3,351 39 8 31 78

9 147 62 81 290 3,641 39 8 31 78

8 147 62 81 290 3,930 39 8 31 78

7 147 62 81 290 4,220 39 8 31 78

6 147 62 81 290 4,510 39 8 31 78

5 147 62 81 290 4,799 39 8 31 78

4 151 66 86 304 5,103 39 8 31 78

3 151 66 86 304 5,407 39 8 31 78

2 227 133 129 489 5,896 59 16 47 122

1

3,005 1,225 1,666 5,896 805 157 612 1,573

Note

1 Wall - Lt Wt above 4th floor

2 Include Mezz. Floor


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Reference table

Perimeter Frame

Based on Portal Method for horizontal force distribution

Level

Forceto

FrameVs

Int

ColumnV

Ext

ColumnV

Int

ColM

Ext

ColM

Girder

M

Girder

Shear

Extcol

axialLoad

PE

Intcol

axialLoad

PE

OTM*0.15/140

Roof 46 3.3

20 70.2 14.0 7.0 91 46 134 9.6 3 5 7

19 136.3 27.3 13.6 177 89 218 15.5 13 11 19

18 198.4 39.7 19.8 258 129 296 21.1 28 15 38

17 256.6 51.3 25.7 334 167 369 26.3 50 20 61

16 310.8 62.2 31.1 404 202 437 31.2 76 24 90

15 361.2 72.2 36.1 470 235 500 35.7 107 28 124

14 408.7 81.7 40.9 531 266 560 40.0 143 32 162

13 452.3 90.5 45.2 588 294 614 43.8 183 35 204

12 492.1 98.4 49.2 640 320 663 47.4 227 38 249

11 528.2 105.6 52.8 687 343 708 50.6 274 41 299

10 560.7 112.1 56.1 729 364 748 53.4 325 43 351

9 589.6 117.9 59.0 766 383 783 55.9 378 46 405

8 614.9 123.0 61.5 799 400 814 58.1 434 48 462

7 637.3 127.5 63.7 828 414 841 60.1 492 49 522

6 656.2 131.2 65.6 853 427 863 61.7 552 51 583

5 671.8 134.4 67.2 873 437 881 63.0 614 52 645

4 684.2 136.8 68.4 890 445 896 64.0 677 53 708

3 693.6 138.7 69.4 902 451 906 64.7 741 54 773

2 700.1 140.0 70.0 910 455 1,267 90.5 805 54 838

1 705.8 141.2 70.6 1,623 812 896 97 954


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Preliminary design of structural wall

Reference

ASCE

7-05

ACI

318-05 Remarks

Material Propoerties fc = 5ksi= 5,000 psi reg wt below 3rd Fl

f = 60 ksiBase Shear to structural walls V = 0.85 x 4705 12.2.5.1 At lower story, wall

resist 75 to 95% ofstory shear= 3,999 kips

Load factor for E = 1.0 Eq (9-5)

Factor seismic force eapanel Vu = 3,999 / 4

1,000 kips

Wall length lw = 30.5 = 366 in

Wall height hw = 270 ft

Consider wall thickness h = 14 in

Gross wall area Acv = 14 x 366

Can increase after 1iteration= 5,124

Sqin ea pier

Minimum wall length based on Vn = Acv 6 fc Can increase to 8fcafter 1

stiteration= 5,124 x 0.424

= 2,174 kips

Required shear strength Vu/ = 1,000 / 0.60 9.3.4 Conservative toconsider shear contr= 1,666 kips < Vn

Wall reinforcement hw/lw = 270 / 30.5

= 8.9 > 2

c = 2.0 21.7.4

For #6 @ 12" o.c. ea face t = 0.88 / 168 Spcg may be chang

after 1st iteration= 0.00524Vn = Acv(2 fc + t f ) Eq (21-7)

= 5,124 x ( 0.141 + 0.314 ) Reg. Wt Conc

= 2,335 kips > Vu/

For #5 @ 12" o.c. ea face h = 14 in

Vn = 5,124 x ( 0.141 + 0.221 ) Reg. Wt Conc

= 1,859 kips >Vu/


Vn = 5,124 x ( 0.120 + 0.221 ) Lt Wt conc.

= 1,751 kips >Vu/


Vn = 4,392 x ( 0.120 + 0.258 ) Lt Wt conc.

= 1,663 kips

For#4 @ 12" o.c. ea face h = 12 in

Vn = 4,392 x ( 0.120 + 0.167 ) Lt Wt conc.

= 1,260 kips

Application of Resultant hx = 0..5 hn = 135 ft Due to dynamicbehaviorRequired moment strength Mu = 1,000 x 135


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Ast = 181.0 + 18.5 = 199.4 in2 Total in wall panel

Average compressive stress Pu/ Ag = 9,592 / 6,756

= 1.4 ksi

< 0.35 fc' = 1.75 ksi

> 0.10 fc' = 0.5 ksi

Nominal axial strength Po = 0.85 fc' (Ag-Ast) + f Astat zero eccentricity = 0.85 x 5.0 x 6,557

+ 60 x 199.4

= 27,865 + 11,966

Po = 39,832 kips

Nominal axial strength Pn = 0.80 Po Eq (10-2)

= 31,865 kips

Pu/ = 9,592 / 0.65 9.3.2.2= 14,757

Nominal Moment Strength

At Pn= 0 Ignore rebar atcompression side

and wall reinf.

Strain diagram 0.003

t =0.011 c

a

Force diagram

T1 T2 T3 Cc

363

T1 = 60 x 74.88 = 4493 48#11 at ends

T2 = 60 x 15.60 = 936 10#11 in web

T3 = 60 x 3.52 = 211 count 8#6 effective

C = T = 5,640 kips

a = C /( 0.85 fc' b) = 44.2 in. < 51.0

c = 44.2 / 0.80 = 55.3 in.

t = 0.003 x 307.7 / 55.3

= 0.017 > 0.005 10.3.4 Tension control

Nominal moment strength Mn = 4,493 x 26.5 = 119,202At Pn= 0 + 936 x 23.4 = 21908.8

+ 211 x 20.4 = 4309.93

Mn = 145,421 k-ft


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Calculate Pb, Mb

at balance strain condition


0.00207 c

t a

Force diagram

Cs3

T1 T2 T3 Cs2 Cs1

Cc2 Cc1

363

c = 363 x 0.003 / 0.0051

= 215 in.

d - c = 148 in.

a = 0.80 x 215 12.2.7.3

= 172 in.At Cs1 1 = 0.00264 > x = 215-25.5 =189.5

At Cs2 2 = 0.00212 > x = 215-63 =152 in.

At Cs3 3 = 0.00162 < x = 215 -99 =116 in

At T1 1 = 0.00175 < x = 148 -22.5= 125.

At T2 2 = 0.00123 < x = 148 - 60 = 88 in

At T3 3 = 0.00073 < x = 148 -96 = 52 in

Compressive force Cc1 = 0.85 fc'b(51) = 6,503

Cc2 = 0.85 fc'b(a-51) = 7,192

Cs1 = 74.88 x 55.8 = 4,175 fs' = fs- 0.85fc'

Cs2 = 15.60 x 55.8 = 870

Cs3 = 3.52 x 42.7 = 150 fs= Ess

C = 18,889 kips

T1 = 74.88 x 50.9 = 3,811 fs= Ess

T2 = 15.60 x 35.7 = 557

T3 = 3.52 x 21.1 = 74

T = 4,442 kips

Pb = 18,889 - 4,442 = 14,447 kips

Moment about C.L of wall Cc1 = 6,503 x 13.1 = 85345.3 k-ft

Cc2 = 7,192 x 6.0 = 42889.9

Cs1 = 4,175 x 13.1 = 54791.1

Cs2 = 870 x 10.0 = 8697Cs3 = 150 x 7.0 = 1051

T1 = 3,811 x 13.1 = 50013.1

T2 = 557 x 10.0 = 5569.59

T3 = 74 x 7.0 = 520

Mb = = 248,878 k-ft


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Calculate Pn, Mn

at 0.005 strain condition


0.0050 c Tension control whe

t> 0.0050t a

Force diagram

Cs3

T1 T2 T3 Cs2 Cs1

Cc2 Cc1

363

c = 363 x 0.003 / 0.0080

= 136 in.

d - c = 227 in.

a = 0.80 x 136

= 109 in.At Cs1 1 = 0.00244 > x = 136-25.5 =110.5

At Cs2 2 = 0.00161 < x = 136-63 =73 in.

At Cs3 3 = 0.00082 < x = 136 -99 =37 in

At T1 1 = 0.00450 > x = 227 -22.5= 204.

At T2 2 = 0.00368 > x = 227 - 60 = 167 i

At T3 3 = 0.00288 > x = 227 -96 = 131 in

Compressive force Cc1 = 0.85 fc'b(51) = 6,503

Cc2 = 0.85 fc'b(a-51) = 3,445

Cs1 = 74.88 x 55.8 = 4,175 fs' = fs- 0.85fc'

Cs2 = 15.60 x 42.5 = 663

Cs3 = 3.52 x 19.5 = 69 fs= Ess

C = 14,853 kips

T1 = 74.88 x 60.0 = 4,493 fs= Ess

T2 = 15.60 x 60.0 = 936

T3 = 3.52 x 60.0 = 211

T = 5,640 kips

Pn = 14,853 - 5,640 = 9,213 kips

Moment about C.L of wall Cc1 = 6,503 x 13.1 = 85345.3 k-ft

Cc2 = 3,445 x 8.6 = 29584.4

Cs1 = 4,175 x 13.1 = 54791.1

Cs2

= 663 x 10.0 = 6628

Cs3 = 69 x 7.0 = 480

T1 = 4,493 x 13.1 = 58968

T2 = 936 x 10.0 = 9360

T3 = 211 x 7.0 = 1478

Mn = = 246,635 k-ft


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Confinement Reinforcement

Reinf. ratio = 74.88 / 1530 = 0.0489 Less than 8%

In-plane direction bc = 51.0 - 4.0 = 47.0

fc'/f t = 5 / 60 = 0.08333

Ash = 0.09sbcfc'/f t Eq. (21-4)

= 0.353 s

For s = 6 inches Ash = 2.12 Sq. in.

# 5 Hoop plus 5 #5 cross ties Ash = 2.17 Sq. in.

=

Out-of-plane direction bc = 30.0 - 4.0 = 26.0

fc'/f t = 5 / 60 = 0.08333

Ash = 0.09sbcfc'/f t

= 0.195 s


# 5 Hoop plus 2 #5 cross ties Ash = 1.24 Sq. in.

Within the 24" of web 21.7.6.5

= 15.60 / 336 = 0.04643

In-plane direction bc = 24.0 - 4.0 = 20.0

fc'/f t = 5 / 60 = 0.08333


= 0.150 s


#5 Hoop plus 2 #4 cross ties Ash = 0.89 Sq. in. # 4 Grade 40

=

Out-of-plane direction bc = 14.0 - 4.0 = 10.0

fc'/f t = 5 / 60 = 0.08333


= 0.075 s


# 5 Hoop Ash = 0.62 Sq. in.

Development of horizontal wall reinforcement

For # 6 bars ld = d (f te)/(25fc') 12.2.2

fc' = 5000 psi = 34 d Straigth developmenin boundary elemen= 25.5 in.

For # 5 bars ld = 38 d Straigth developmenin boundary elemenfc' = 4000 psi = 23.7 in.


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Boundary Element (Cont.)

Reference

ASCE

7-05

ACI

318-05 Remarks

Check when boundary reinforcement may be discontinue

Consider the boundary element size is reduced to 30 x 30 at upper stories

Size Area x Ax2 Ad

2/12

2.5 2.5 6.25 14.0 1225 3

1.0 25.5 25.5 0 0 1382

2.5 2.5 6.25 14.0 1225 3

38.0 2450 1388

I = 2450 + 1388 = 3838 ft4 = 79,590,816

Ag= 38.0 x 144 = 5472 in2

c = 183 in.

Level PD PL Pu Mu Pu/Ag Muc/I fc

kip kip -ft

20 504 119 723 1520 0.132 0.042 0.174

19 788 197 1143 4472 0.209 0.123 0.332

18 1,072 276 1562 8771 0.285 0.242 0.527

example design of 20 story building aci-libre

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