Example Design of 20 Story Building ACI

ACI 314 Task Group B/C Draft No. 1

Page 1 of 46

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


Page 2 of 46

TABLE OF CONTENTS

1. Building description……………………………………………………………….. ………..3 1.1 Material……………………………………………………………………………..3 1.2 Design loading………………………………………………………………………3 1.3 Story weight………………………………………………………………. ………..3 1.4 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 12……………6 2.1.2 Design of structural wall (shear wall)…………………………………………6 2.1.3 Design of special moment frame………………………………………………6

3. Lateral Force Analysis:………………………………………………………………………7

3.1 Mapped Spectral Acceleration………………………………………………………7 3.2 Structural System……………………………………………………………………7

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…………………………………....31 6.1 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 4½ inches thick light weight concrete slabs, 12 x 18½ 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 3rd floor (light weight 115 pcf)

fc´ = 5,000 psi below 3rd floor (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 16–20 wi = 2800 kips Floor 9 – 15 wi = 2850 kips Floor 3 – 8 wi = 2900 kips Floor 2 - w2 = 4350 kips Total building weight Σwi = 58,500 kips

1.4 Governing Codes: IBC -2006 ACI 318-05 ASCE 7 -05

1 This 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.


Page 4 of 46

Elevator Opening

Beam

Stair

Girder

Typical Bay

5 Bays @ 28’ 0” = 140’ 0”

5 Bays @ 28’ 0” = 140’ 0”

Fig. 1 - Typical Floor Plan


Page 5 of 46

Roof

20

19

18

17

16

15

14

13

12

11

10

9

8

7

6

5

4

3

2

1

19 Stories @ 13’ 0” = 247’ 0”

23’ 0”

270’ 0”

Fig. 2 - Elevation

Columns


Page 6 of 46

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 Cd and overstrength factor Ω0 Establish mapped maximum considered earthquake spectral response acceleration for short and long

periods Ss and Sl from USGS data base Calculate design spectral response acceleration SDs and SDl Establish a standard response spectrum for design reference Calculate fundamental period Ta using (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 8√fc´ 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, Mn to draw an interaction diagram based on φ = 1 Based on Pu/φ and Mu/φ, 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 reinforcement

requirements

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 318-05

Reference ASCE 7-05 Remarks

3.1 Mapped Spectral Acceleration 11.4.1

Short period Sa = 2.25 From USGS data base One second S1 = 0.75 From USGS data base

Site Class D 11.4.2 Default Site Class Site Coefficent Fa = 1.0 Table 11.4-1

Fv = 1.5 Table 11.4-2 Maximum Considered Earthquake 11.4.3

SMS = Fa Ss = 2.25 (Eq. 11.4-1) SM1 = Fv S1 = 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 data base

For T < T0 Sa = SDS[0.4 + 0.6 T/T0] = (Eq. 11.4-5) T = fundamental period For T0 ≤T ≤ TS Sa = SDS = of structure For TS ≤T ≤ 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 11.4.6 1.5 x Design response

spectrum Occupancy 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 Ωo = 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 ρ = 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. Equivalent Lateral Force Procedure 12.8 Building Height hn = 270 ft Problem statement

Effective Seismic Weight W = 58,500 kip

Calculation of Seismic Response 12.8.1.1 12.8.1.1

Seismic Reponse Coefficient Cs = SDS /[R/I] = 0.214 (Eq. 12.8-2) For T ≤ TL Cs = SD1 /T[R/I] = 0.080 (Eq. 12.8-3) Governs design

> 0.01 (Eq. 12.8-5) For S1 ≥ 0.6 Cs = 0.5 S1/[R/I] = (Eq. 12.8-6)

Building Period 12.8.2.1 Period Parameter Ct = 0.02 Table 12.8-2 Period Parameter x = 0.75 Table 12.8-2

Approx. Fundamental Period T = Ta = Ct hnx = 1.33 sec. (Eq. 12.8-7)

Seismic Base Shear V = Cs W = 4,705 kip (Eq. 12.8-1)

Vertical Distribution of Force 12.8.3

Vertical Distribution Factor Cvx = wx hxk / Σwihi

k (Eq. 12.8-12) =

For T < 0.5 k = 1 For T = 1.33 k = 1.2 Interpolate in between

For T ≥ 2.5 k = 2.5 Story Force Fx = Cvx V

Horizontal Distribution of Force 12.8.4

Vx = i=nxΣFi (Eq. 12.8-13)

Accidental Torsion Mta = 5% 12.8.4.2

Amplification of Mta Ax = [δmax /1.2δavg]2 =

Deflection at center of mass δx = Cd δse/I (Eq. 12.8-15) Period for computing drift δxe Τ = CuTa 12.8.6.2

Cu = Table 12.8-1 P-Δ Effects 12.8.7

Stability Coefficient θ = Px Δ /[Vx hsx Cd] (Eq. 12.8-16) =

θmax = 0.5/ (β Cd) (Eq. 12.8-17)

≤ 0.25


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4.1 Unit Load Typical Floor

Finish floor 2 4½" LW Conc.

Slab 45 Ceiling 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

Level Height to Level x hx

Weight at Level x wx

wx hxk

k=1.2

wx hx

k

Σwihi

Seismic Force

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


Page 10 of 46

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 ΣPD P1 P2 P3 ΣPL

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

Force to

Frame Vs

Int Column

V

Ext Column

V

Int Col M

Ext Col M

Girder M

Girder Shear

Ext col

axial Load

PE

Int col

axial Load

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


Page 12 of 46

Preliminary design of structural wall

Reference ASCE 7-05

ACI 318-05 Remarks

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

fy = 60 ksi Base Shear to structural walls V = 0.85 x 4705 12.2.5.1 At lower story, walls

resist 75 to 95% of story shear = 3,999 kips

Load factor for E = 1.0 Eq (9-5) Factor seismic force ea panel 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 1st iteration = 5,124

Sq in ea pier

Minimum wall length based on Vn = Acv 6 √ fc´ Can increase to 8√fc´ after 1st iteration = 5,124 x 0.424

= 2,174 kips Required shear strength Vu/φ = 1,000 / 0.60 9.3.4 Conservative to

consider shear control = 1,666 kips < Vn Wall reinforcement hw/lw = 270 / 30.5

= 8.9 > 2 αc = 2.0 21.7.4

For #6 @ 12" o.c. ea face ρt = 0.88 / 168 Spcg may be changed after 1st iteration = 0.00524

Vn = Acv (2 √fc´ + ρ t fy) Eq (21-7) = 5,124 x ( 0.141 + 0.314 ) Reg. Wt Conc = 2,335 kips > Vu/φ

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/φ

For #5 @ 12" o.c. ea face h = 14 in Vn = 5,124 x ( 0.120 + 0.221 ) Lt Wt conc.

= 1,751 kips >Vu/φ


= 1,663 kips


= 1,260 kips

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


Page 13 of 46

= 134,978 kip-ft

Mu /φ = 134,978 / 0.65 = 207,658 kip-ft φ may be increased based on εt Mu /φ = 134,978 / 0.90 = 149,975 kip-ft

Min. Ht. Of Boundary element Mu /4Vu = 134,978 / 4000 = 34 ft > lw

Consider building displacement δσε = 0.0015 x 270 T12.12-1 Conservative for dual

system = 0.405 x 12 = 4.9 in

δu = Cd δ 12.12.1 = 5.5 x 4.9 = 26.7 in Δs = 0.025hx = 81 in.

δu/hw = 26.7 / 3240 = 0.008 > 0.007

c = lw ÷ 600(δu/hw) Eq (21-8) = 30.5 / ( 600 x 0.008 ) = 6.2 ft = 74 in.

a = 0.80 x 6.2 R10.2.7 = 4.9 ft

Boundary element

Extend of boundary element c-

0.1lw = 74 - 36.6 = 37.3 < 51" or c/2 = 74 / 2 = 37.0 < 51"

Appro. Tension force T = 134,978 / ( 28.4 - 2.5 ) = 5,209 kip

Less 0.9 D PD = 0.9 x 3,005 = 2,705

Net tensile force due seismic PE = 5,209 - 2,705 = 2,505 kip

Minimum tension reinf. As = PE / φ fy = 2,505 / ( 0.9 x 60 ) = 46.4 sq. in.

Try 36- #11 As = 1.56 x 36 May not be adequate for compression = 56.2 sq. in.

Total factored load to wall Pu = 5,896 x 1.2 Eq.(9-2) + 1,573 x 1.6

Required axial strength = 9,592 kip 1.2D+1.6L Pu/φ = 9,592 / 0.65 = 14,757

Pu = 5,896 x 1.2 Eq (9-5) + 1,573 x 1.0 = 8,648 kip

1.2D+1.0L+1.0E Pu/φ = 8,648 / 0.65 = 13,305 φ may be increased

Pu = 5,896 / 0.9 Eq (9-7) = 6,551 kip

0.9D + 1.0E Pu/φ = 6,551 / 0.65 = 10,079 φ may be increased

Conc Section at Level 1 Ag = 3,060 + 3,696 Ignore L-shape in

prelim design = 6,756 sq. in.


Page 14 of 46

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) + fy Ast at 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/φ = 9,592 / 0.65 9.3.2.2

= 14,757 Nominal Moment Strength

At Pn = 0

Ignore rebar at compression side and wall reinf.

Strain diagram 0.003 εt ε =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 = Σ T = 5,640 kips a = C /( 0.85 fc' b) = 44.2 in. < 51.0 c = 44.2 / 0.80 = 55.3 in. εt = 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 = 119,202

At Pn = 0 + 936 x 23.4 = 21908.8 + 211 x 20.4 = 4309.93

Mn = 145,421 k-ft


Page 15 of 46

Calculate Pb, Mb

at balance strain condition Strain diagram 0.003

0.00207 c εt 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 ε1 = 0.00264 > εy x = 215-25.5 =189.5 At Cs2 ε2 = 0.00212 > εy x = 215-63 =152 in. At Cs3 ε3 = 0.00162 < εy x = 215 -99 =116 in At T1 ε1 = 0.00175 < εy x = 148 -22.5= 125.5 At T2 ε2 = 0.00123 < εy x = 148 - 60 = 88 in At T3 ε3 = 0.00073 < εy 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 = Es εs

Σ C = 18,889 kips T1 = 74.88 x 50.9 = 3,811 fs = Es εs T2 = 15.60 x 35.7 = 557 T3 = 3.52 x 21.1 = 74

Σ 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 = 8697 Cs3 = 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


Page 16 of 46

Calculate Pn, Mn

at 0.005 strain condition

Strain diagram 0.003 0.0050 c Tension control when

εt > 0.0050 εt 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 ε1 = 0.00244 > εy x = 136-25.5 =110.5 At Cs2 ε2 = 0.00161 < εy x = 136-63 =73 in. At Cs3 ε3 = 0.00082 < εy x = 136 -99 =37 in At T1 ε1 = 0.00450 > εy x = 227 -22.5= 204.5 At T2 ε2 = 0.00368 > εy x = 227 - 60 = 167 in At T3 ε3 = 0.00288 > εy 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 = Es εs

Σ C = 14,853 kips T1 = 74.88 x 60.0 = 4,493 fs = Es εs T2 = 15.60 x 60.0 = 936 T3 = 3.52 x 60.0 = 211

Σ 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


Page 17 of 46

Confinement Reinforcement

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

In-plane direction bc = 51.0 - 4.0 = 47.0

fc'/fyt = 5 / 60 = 0.08333 Ash = 0.09sbcfc'/fyt 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'/fyt = 5 / 60 = 0.08333

Ash = 0.09sbcfc'/fyt = 0.195 s

For s = 6 inches Ash = 1.17 Sq. in. # 5 Hoop plus 2 #5 cross ties Ash = 1.24 Sq. in.

Within the 24" of web 21.7.6.5 ρ = 15.60 / 336 = 0.04643

In-plane direction bc = 24.0 - 4.0 = 20.0 fc'/fyt = 5 / 60 = 0.08333

Ash = 0.09sbcfc'/fyt = 0.150 s

For s = 6 inches Ash = 0.90 Sq. in. #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'/fyt = 5 / 60 = 0.08333 Ash = 0.09sbcfc'/fyt

= 0.075 s For s = 6 inches Ash = 0.45 Sq. in.

# 5 Hoop Ash = 0.62 Sq. in.

Development of horizontal wall reinforcement For # 6 bars ld = db (fy ψt ψe λ)/(25√fc') 12.2.2

fc' = 5000 psi = 34 db Straigth development in boundary element = 25.5 in.

For # 5 bars ld = 38 db Straigth development in boundary element fc' = 4000 psi = 23.7 in.


Page 18 of 46

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 Ad2/12 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 Σfc 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 < 0.15fc'

17 1,356 354 1982 14330 0.362 0.395 0.758 16 1,640 433 2401 21064 0.439 0.581 1.020 15 1,925 511 2820 28891 0.515 0.797 1.313

0.15 fc' = 0.600 ksi

May discontinue boundary element at the 18 floor 21.7.6.3

28.00

1.25 3.00 22.00 3.00 1.25

2.50 1.17

PLAN


Page 19 of 46

30.0 14.0

4 spcg @ 12"

48 #11 10 #11 8 #6 51.0 24.0 48.0

DETAIL Confinement not shown for clarity

40,000

P0

Pn 30,000

P (kip)

20,000

Pn PbMb

10,000 Min. eccentricity

εt = 0.005

0 Mn 100,000 200,000

Moment kip-feet

Simple Interaction Diagram


Page 20 of 46

Rf Bar A Bar B

20

19

18

17 # 5 @ 12" EWEF h = 12 "

16

15

14

13

12

11

10

9

8

7 h = 14"

6 # 5 @ 12" EWEF

5

4 Bar B Bar B 3 Bar A Bar A 2

h = 14"

#6 @ 12 EWEF 48 # 11

10 # 11

1 WALL ELEVATION


Page 21 of 46

5 #5 Crossties @ 6" o.c. #5Hoops @ 6" o.c.

Wall Reinf.

30.0 14.0

2 #4 Crossties @ 6" o.c.

ld

48 #11 10 #11 51.0 24.0

PLAN DETAIL BOUNDARY ELEMENT

5. Moment Frame Design 5.1 Two moment frames in each direction Reference

ASCE 7-05

ACI 318-05

Min. Seismic shear to moment frames Vx = 25% x ΣVx 12.2.5.10 Torsion - Accidental ecc = 5% x 140 12.8.4.2

= 7.0 ft Torsion T = 7 Vx

Torsional stiffness J = 4R (70)2

= 19600 R Additional force ΔVx = TcR/J

= 7Vx R x 70 / 19600 R = 0.025 Vx

Force per frame Vx + ΔVx = ( 0.125 + 0.025 )

Vx = 0.150 Vx

Design frame for Fx = 30% Vx Or per frame Fx = 15% Vx


Page 22 of 46

5.2 Seismic Force distribution using Portal Method At 11th Floor ΣV12 = 3521 x 15% = 528 kips

V11 = 216 x 15% = 32 ΣV11 = 3738 x 15% = 561 kips

Exterior Column MA12 = 53 x 6.5 = 343 kip-ft MA11 = 56 x 6.5 = 364 MAB = MA-12+MA-11 = 708 kip-ft

Axial Load PA12 = 274 kips Axial Load PA11 = 325 kips

Interior Column MB12 = 106 x 6.5 = 687 kip-ft MB11 = 112 x 6.5 = 729

MBA =MBC = (MB-12+MB-11) /2 = 708 kip-ft Girder shear VBA =VAB = (MAB+MBA) /28 = 51 kips

At 3rd Floor ΣV4 = 4624 x 15% = 694 kips V3 = 43 x 15% = 6

ΣV3 = 4667 x 15% = 700 kips Exterior Column MA4 = 69 x 6.5 = 451 kip-ft

MA3 = 70 x 6.5 = 455 MAB = MA-12+MA-11 = 906 kip-ft

Axial Load PA4 = 741 kips PA3 = 805 kips Interior Column MB4 = 139 x 6.5 = 902 kip-ft

MB3 = 140 x 6.5 = 910 Axial Load PB4 = 54 kips

MBA =MBC = (MB-12+MB-11) /2 = 906 kip-ft

Girder shear VBA

=VAB = (MAB+MBA) /28 = 65 kips


Page 23 of 46

Remarks Rf Level

70 A B C D E F

3 3

7 14 14 14 14

12th Floor 11th Floor Vu = 528 kips Vu = 561 kips

> 25 % OTMu

= 41,791 kip-ft OTMu = 49,080 kip-ft

Line of symmetry

53 106 106 106 106 53

Above Flr Line Below Flr Line 32

A B C D E F

` 56 112 112 112 112

4th Floor 3rd Floor Vu = 694 kips Vu = 700 kips

OTMu = 108,199 kip-ft OTMu = 117,300 kip-ft

69 139 139 139 139 69

6 A B C D E F

70 140 140 140 140


Page 24 of 46

Loads Dead Load D = 0.09 x 15.2

+ 5.9 x 0.15 = 2.25 k/ft

L = 0.04 x 15.2 = 0.61 k/ft

Load combinations 1.2D = 2.70 k/ft 1.2D +1.6 L = 3.68 k/ft

1.2D +1.0L+1.0E = 3.31 k/ft 0.9D+1.0E = 2.03 k/ft

Fixed end moment FEMTL = wl2/12 = 187.0 k-ft FEMD = = 147.3 k-ft

Member stiffness - Consider column far end fixed

Ic = 0.70Ig = 4.73 ft4

Ig = 0.35Ig = 1.77 ft4 E = 519119.5 Ksf

Kc = 4EIc/L = 754720 Kg = 4EIg/L = 131402.1

DFAB = Ig/Σ(Ic+Ig) = 0.080 DFBA = 0.074

To edge of slab Gravity Load moment distribution

Spandrel wt

Line of symmetry -68

0 86 8.5 -68

A B C D E F

Service Load D.F. Service Load FEM

D -147 147 -147 147 -147 147 -147 147 -147 147

36 Sq column TL -187 187 -187 187 -187 187

27x36 Girder B.J. 15.0 2.9 -2.9 2.9 -2.9 0 0 -12 fc' = 4000 C.O. 0.1 0.6 -0.1 0.1 -0.1 0 0.5 0

B.J. 0 0.0 0.0 0.0 0.0 0

-M -172 191 -190 190 -190 135


Page 25 of 46

5.3 Based on two cycle moment distribution Exterior column MD+L = 86 k-ft

MD = 68 k-ft ML = 18 k-ft ME = 451 k-ft

Interior Column MD+L = 8 k-ft MD = 0 k-ft ML = 8 k-ft ME = 910 k-ft

Girder at ext. support MD+L = -172 k-ft MD = -135 k-ft ML = -36 k-ft ME = 906 k-ft

Girder at int. support MD+L = -190 k-ft MD = -147 k-ft ML = -43 k-ft ME = 906 k-ft

5.4 Column axial load (Between 3rd and 4th Floor) Ext column PD)A4 = 812 kip Above 3rd Flr

PL)A4 = 148 kip PD)A3 = 860 kip Below 3rd Flr PL)A3 = 157 kip

Int Column PD)B4 = 1302 kip Above 3rd Flr PL)B4 = 272 kip PD)B3 = 1379 kip Below 3rd Flr PL)B3 = 289 kip

Frame Girder Design (3rd floor) fc' = 5 ksi fy = 60 ksi

Factored Moment

(1.2D+1.6L) -Mu = -245 k-ft (9-2)

(1.2D+1.0 L+1.0E) -Mu = -1177 k-ft (9-5)

(1.2D+1.0 L-1.0E) -Mu = 635 k-ft

(0.9D+1.0E) +Mu = 773 k-ft (9-7)

ln = 28.0 - 3.0 = 25.0 ft

Aspect ratios bw = 27 in > 10 in. 21.3.1 h = 36 in


Page 26 of 46

ln /d = 8.3 > 4 bw/h = 0.75 > 3

Min. hc = 20 x 1.128 Minimum column width

= 22.6 < 36 in. Eff. d = 36.0 - 3

= 33.0 in Longitudinal reinf. 21.3.2

Min. As = 200bwd/fy = 3.0 Sq. in. Max. As = 0.025bwd = 22.3 Sq. in

Try 6 # 11 top and - a = fy As / 0.85fc'b 4 - # 11 bottom = 60 x 9.36

÷ ( 0.85 x 5 x 27 ) = 4.9 in.

c = a/0.80 = 6.1 -Mn = fy As (d-a/2)

= 60 x 9.36 x ( 33.0 - 2.4 )/ 12

= 1430 k-ft

> Mu/φ = -

1307.38 k-ft φ = 0.90

-εt = 0.003 x 26.9 / 6.1 = 0.013 > 0.005

Similarly +a = 60 x 6.24 ÷ ( 0.85 x 5 x 27 ) = 3.3 in.

+Mn = 60 x 6.24 x ( 33.0 - 1.6 )/ 12

= 979 k-ft

> Mu/φ = 859 k-ft

With 90º std hook ldh = fydb / (65√fc' ) (21-6) = 18 in.

For Straight top bar ldh = 3.25 x 18 21.5.4.2 = 60 in.

For Straignt bott. Bar ldh = 2.5 x 60 = 150 in.


Page 27 of 46

Girder Shear Strength (3rd Floor) 21.3.4 -Mpr = 1752 k-ft Based on 1.25fy +Mpr = 1207 k-ft Based on 1.25fy

wu=1.2D+1.0L +1.0E wuln /2 = 3.31 x 25.0 / 2 = 41.4 kip

Ve = (-Mpr + Mpr)/ln ± wuln/2 = 118.4 ± 41.4 = 160 kips

> ( 160 + 41.4 )/2 > 101 kips

Vc = 0

Consider #4 ties 4"o.c. Vs = Av fy bw/s for 2xh from face of support

= 0.40 x 60 x 27 /4 = 162 kips

Max Vs = 8 √fc' bw d = 504 kips

Vn = Vc + Vs = 0 + 162 = 162 kips ≅ Ve = 160 kips

Beyond 2h from support Vu = 41.4 x 6.5 / 12.5 +( 1177 + 635 ) / 25.0 = 94 kips

Vu / φ = 94 / 0.75 = 125 kips

Vc = 2 √fc' bw d = 126 kips

At 12" o.c. Vs = 54 Vn = 180 kips >Vu / φ

Design Exterior Column (Between 3rd and 4th Floor) fc' = 5 ksi fy = 60 ksi

Factored Moment

(1.2D+1.6L) -Mu = 110 k-ft (9-2)

Pu)A4 = 1211 kip Above 3rd Flr Pu)A3 = 1283 kip Below 3rd Flr

(1.2D+1.0 L+1.0E) -Mu = 550 k-ft (9-5)


(1.2D+1.0 L-1.0E) -Mu = -514 k-ft



Page 28 of 46

(0.9D±1.0E) +Mu = 390 k-ft (9-7)

Pu)A4 = -10 kip Above 3rd Flr Pu)A3 = -75 kip Below 3rd Flr

lu = 13.0 - 3.0 = 10.0 ft

Aspect ratios b = h = 36 in 21.4.1 b/h = 1 > 0.4

Pu > Agfc´/10 = 648 kip 21.4.2

Try 16 # 10 Vert. ρ = 20.32 / 1296 = 0.015679 > 1%

At Pn = 0 a = 680 / 153 9 bars effective = 4.45 in

c = 4.45 / 0.80 = 5.56 in

εt = 0.0030 x 27.44 / 5.56 = 0.0148 > 0.005 Tension control

At Pn = 0 Mnc = 680 x ( 28.0 - 2.2 ) Ave. d = 28.0

= 17,538 kip-in

= 1,462 k-ft

ΣMnc = 2923 k-ft Conservative

ΣMnb = 1430 k-ft See girder abv

6/5ΣMnb = 1716 k-ft

< ΣMnc 21.4.2.2 At Pu/φ = 1863 / 0.65

= 2866 kip

Mnc = 2850 k-ft > Mu/φ = 847 k-ft OK

At Pu/φ = 317 / 0.65 = 487 kip

Mnc = 1650 k-ft > Mu/φ = 791 k-ft OK

Design Interior Column (Between 3rd and 4th Floor) fc' = 5 ksi fy = 60 ksi

Factored Moment

(1.2D+1.6L) -Mu = 14 k-ft (9-2)

Pu)B4 = 1999 kip Above 3rd Flr Pu)B3 = 2118 kip Below 3rd Flr

(1.2D+1.0 L+1.0E) -Mu = 919 k-ft (9-5)


(1.2D+1.0 L-1.0E) -Mu = -902 k-ft


Page 29 of 46


(0.9D±1.0E) +Mu = 910 k-ft (9-7)


lu = 13.0 - 3.0 = 10.0 ft

Aspect ratios b = h = 36 in 21.4.1 b/h = 1 > 0.4

Pu > Agfc´/10 = 648 kip 21.4.2

Try 16 # 10 Vert. ρ = 20.32 / 1296 = 0.015679 < 6% Larger than 1%

At Pn = 0 a = 680 / 153 9 bars effective = 4.45 in

c = 4.45 / 0.80 = 5.56 in

εt = 0.0030 x 27.44 / 5.56 = 0.0148 > 0.005 Tension control

At Pn = 0 Mnc = 680 x ( 28.0 - 2.2 ) Ave. d = 28.0

= 17,538 kip-in

= 1,462 k-ft

Mu/ φ = 910 / 0.9 = 1011 OK

ΣMnc = 2923 k-ft Conservative

ΣMnb = 1430 + 979 = 2409 See girder abv

6/5ΣMnb = 2890 k-ft

< ΣMnc 21.4.2.2 At Pu/φ = 1889 / 0.65

= 2906 kip

Mnc = 2750 k-ft > Mu/φ = 1413 k-ft OK

At Pu/φ = 1118 / 0.65 = 1721 kip

Mnc = 2600 k-ft > Mu/φ = 1400 k-ft OK

Design Column Shear Strength (Between 3rd and 4th Floor) For 36 x 36 column fc' = 5 ksi

fy = 1.25 x 60 = 75 ksi

φ = 1.0

Girders ΣMpr = 1752 + 1207 21.3.4 See Girder abv

= 2959 ft-kip ½ΣMpr = 1480 ft-kip


Page 30 of 46

At Pu / φ = 1782 / 0.65 = 2741 Interaction diagram Column Mpr = 3050 ft-kip

Design for Mpr = 1480 ft-kip R 21.3.4

Probable shear strength Ve = ΣMpr /

lu Consider Mpr top and bottom the same

= 1480 / 10 = 148 kip

From Portal analysis Vu = 139 kip Due to seismic

Vu/ φ = 139 / 0.65 = 213 kip

Vc = 0 Consider ties @ 5.5"o.c. Vs = Av fy bw/s

5 legs = 1.55 x 60 x 36 / s = 582 kips

Max Vs = 8 √fc' bw d

= 672 kips Vn = Vc + Vs

= 0 + 582 = 582 kips

> Vu/ φ = 148 kips OK

Transverse reinforcement

Try #5 ties at s = 5.75 in on center hx = 8 in.

Ach = ( 36 - 3.5 ) 2 = 1056 Sq in

Ag = 1296 Sq in Ash = 0.3 (sbcfc´ /fyt)[(Ag/Ach) - 1] (21-3)

= 1.17 Sq. in. Or Ash = 0.09 sbc fc´ /fyt (21-4)

= 1.55 Sq. in. Say OK Max spacing s0 = 4 + (14 - hx)/3 (21-5)

= 6 in USE: 36 Square Column w/ 16 # 10 Vert.

#5 Hoops plus 3 #5 cross ties @ 5.75" o.c. for 3 feet top and bottom and through joint, balance @ 12" o.c.


Page 31 of 46

6. Preliminary Material Quantities for Superstructure only:

6.1 Typical Shear-wall (4 total) 4.25ft x 2.5ft (typ.) 22ft x 1.17ft (typ.) 48#-11 48 #-11 10-#11 32-#6 10-#11 10-#11 32-#6 10-#11 48#-11 Total weight of longitudinal reinforcement:

• # 11 – 184 * (4 walls) * 270 ft * 5.31 lb/ft/2000: 527 tons

• # 6 – 64 * (4 walls) * 270 ft * 1.50 lb/ft/2000: 52 tons Total weight of transverse reinforcement: Hoops at boundary elements:

• # 5@6” – 26’/ea * (12 elem.) * (270 ft/.5) * 1.04 lb/ft/2000: 88 tons Cross-ties at boundary elements:

• 5-# 5@6” – 2’/ea *5* (12 elem.) * (270 ft/.5) * 1.04 lb/ft/2000: 37 tons


Page 32 of 46

Hoops at wall elements:

• # 5@12” – 24’*(2) * (8 elem.) * (270 ft) * 1.04 lb/ft/2000: 54 tons

• Total weight of reinforcement in shear walls 758 tons

Estimated quantity of concrete:

• Shear walls:

o 84 sq.ft.(270ft)*(4 locations)/27 3,360 cy

6.2 Columns:

Total weight of longitudinal reinforcement:

36 x 36 Col (24 locations)

16 # 10 Vert.

# 11 – 16 * (24) * 270 ft * 5.31 lb/ft/2000: 275 tons

• Total Wt per square foot of total building area – 1033T(2000)/392,000 sq.ft. ~ 6 psf (with .5 psf for miscellaneous steel)

Estimated quantity of concrete:

• Columns:

o 9 sq.ft.(270ft)*(24 locations)/27 ~2,200 cy


Page 33 of 46

6.3 Floor slab:

\ Estimated quantity of reinforcement:

• 4.5” lt. wt. concrete slab (Est. quantity of rebar) 3.5 psf

\ Estimated quantity of concrete:

• slabs:

o 140’x140’x(4.5”/12)*19fl/27 ~5,200 cy


Page 34 of 46

ACI Spring Convention 2007 1Simplified Design of Concrete Structure

Preliminary Design and Economical Impact ofSimplified Design of R/C Structures

Gravity/Lateral Force Resisting System

by Michael Mota and James S. Lai


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Documents

Example Design of 20 Story Building ACI