Example314b taskgroupb-c-120730160543-phpapp01

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

Page 3 of 46
1
1. 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

Page 7 of 46
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

Page 8 of 46
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 hn
x
= 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 hx
k
/ 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=
n
xΣFi (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

Page 9 of 46
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 hx
k
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

Page 11 of 46
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
For #
5 @ 12" o.c. ea face h = 12 in
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 dynamic
behaviorRequired 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 εtMu /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
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
0.0050 c Tension control when
εt > 0.0050εt 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
# 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
= 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'/fyt = 5 / 60 = 0.08333
= 0.075 s
# 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 elementfc' = 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-
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-
At Pu/φ = 1118 / 0.65
= 1721 kip
Mnc = 2600
k-
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
diagramColumn 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)
• 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
• 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 of
Simplified Design of R/C Structures
Gravity/Lateral Force Resisting System
by Michael Mota and James S. Lai

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