Table of Contents
Project 2 Final Report

Project 2 — David Wagner 2007/07/04 18:28

Sketch of project.

Structural Configuration

The plan and elevation with overall dimensions provided by the architect has been simplified to what is presented here. For example, the 23”x12” rectangular structural columns used here are inscribed within the 26” circular columns planned, the diameter having been determined by dividing the 15' bay spacing by the golden ratio four times.1)

Member Dimensions

  • Roof Slab: 121½' x 36' x 8”
  • Beams: 36' x 18” wide x 10” web height
  • Structural Columns: 23” x 12” inscribed within 26” diameter circular architectural columns.
  • Footing: 6' x 6' x 2'

Although the beam, columns, and footings of the two end frames may require less reinforcement than interior members, the full amount is specified to allow for future expansion of this building.

Construction Sequence

The construction sequence is standard for reinforced concrete structures.

  1. After grading the site, frame and pour the column and wall footings.
  2. Frame and pour the columns.
  3. Install insulation and vapor barrier, and fill to new grade.
  4. Frame and pour the roof beams and roof slab.
  5. Frame and pour the floor slab.
  6. Install exterior walkways after erecting exterior walls.

Materials

The structural materials needed are widely available standard construction materials.

  • Soil fill is 120 lb/cf.
  • The structural steel is grade 60.
  • All structural concrete has f’c=4,000 psi. Reinforced concrete density is 150 lb/cf.
  • The sod roof assembly includes vegetation and soil cover, lightweight concrete to taper the roof to slope laterally 2% from the center, waterproofing, and provisions for drainage. The entire assembly weighs 110 psf.
  • Exterior walkways are permeable concrete lattice, 120 lb/cf.

Loading

All loads act vertically downward except wind loads, which act transversely on the longitudinal face of the building.

Dead Loads

  • Slab Weight: 150 lb/cf x 8/12 = 100 psf
  • Sod Roof: 110 psf
  • Plaster Ceiling Finish: 5 psf
  • Mechanical and Electrical: 4 psf
  • Beam Web Weight: 150 lb/cf x 18/12 x 10/12 = 187.5 lb/ft
  • Column Weight: 150 lb/cf x 242/12 x ¼π * (26/12)² = 11,150 lb

Live Loads

In addition to the standard live and wind loads, when it rains the soil of the sod roof can become saturated equivalent to a uniform distribution of 6” of water.

  • Roof: 10 psf
  • Rain: 62 pcf x 0.5 = 31 psf
  • Wind: 26.9⇒ psf

Transverse Wind Load

Longitudinal wind loads are borne by shear walls not considered as part of this design. For simplicity, roof loads are not considered and the transverse wind load is approximated by a single uniform load acting on the 18' of exposed wall surface.

V=90 mph, I=1.00, K_z(20') = 0.90, K_{zt}=1, K_d = 1

q_z=0.00256 K_z K_{zt} K_d V^2 I = 0.00256*K_z*K_{zt}*K_d*V^2*I = 0.00256*K_z*1*1*90^2 * 1.00 =20.736 K_z

p=qGC_p - q_h(GC_{pi}) = 0.85 q C_p pm (18.9*0.18) = 0.85 q C_p pm 3.4

z (ft) K_z q_z (psf) Windward Wall p (psf)
0-15 0.85 17.6 11.968 ± 3.4
20 0.90 18.7 12.7 ± 3.4
h=21 ft 0.91 q_h=18.9 12.9 ± 3.4

The weighted average over the 18' exposed height of the windward wall is 12.1 ± 3.4

Leeward Wall (L/B=36/120=0.3→C_p=-0.5) p = -8.0 ± 3.4

Total wW,max = 12.1+3.4 + 8.0+3.4 = 26.9⇒ psf

Slab Shear and Bending Envelopes

The one-way roof slabs span 15'=180” center-to-center and have a clear span of 13.5'=162”. Longitudinal wind loads are borne by shear walls and not considered to effect the slab.

Total wD: 219 psf, wL = 41 psf, l = 15 ft = 180 in, ln = 13.5 ft = 162 in

  • (9-1) U = 1.4D → w_u=1.4*219=306.6 psf
  • (9-3) U = 1.2D + 1.6L → w_u=1.2*219 + 1.6*41 =328.4 psf Controls

The maximum positive moment is near the middle of each end span. The maximum negative moment is at the exterior faces of the interior columns supporting the end spans. The maximum shear is at the face of the supporting beam.

  • M_{u+}={w_u {l_n}^2}/14={0.3284*{13.5}^2}/144.275 kip-ft = 51.3 kip-in
  • M_{u-}=-{w_u {l_n}^2}/10=-{0.3284*{13.5}^2}/10=-5.985 kip-ft = -71.8 kip-in
  • V_u = {w_u l}/2={0.3284*13.5}/22.217 kip

Frame Shear and Bending Envelopes

Each beam and pair of supporting columns form a simple symmetrical frames with small cantilevered overhangs on either side. Assuming the base of the frame is pinned results in the most conservative maximum moment calculations.

The total dead loads on the beam are 219 psf + 187.5 lb/ft. the total live loads on the beam are 41 psf down and 26.9 psf lateral wind load.

wW,max = 26.9⇒ psf

w_D=219*15 + 187.5=3.473 kip/ft

w_L=41*15 =0.615 kip/ft

w_W=26.9*18=0.484⇒ kip/ft

  • (9-1) U = 1.4D → w_u=1.4*3.473=4.862 kip/ft
  • (9-3) U = 1.2D + 1.6L + 0.8W → w_u=1.2*3.473 + 1.6*0.615 + 0.8*0.484right =5.152 kip/ft down, and 0.387⇒ kip/ft laterally. Controls
  • (9-6) U= 0.9D + 1.6W = 0.9*3.473 + 1.6*0.484=3.126 kip/ft down, and 0.774⇒ kip/ft laterally.

Since this building is much wider laterally (36') than it is tall (18'), by inspection (9-3) will produce the greatest shear and bending moment in the beam.

Ignoring more complicated wind effects, the negative moment on the beam at the exterior face of the column is simply the reaction to the cantilevered load, a much smaller value than the moment on the interior face.

M_{u-,ext}={w_u l^2}/2={5.152*4.25^2}/2 =46.53 kip-ft = 558.3 kip-in

h=20.25 ft, L=26 ft

k= {I_2 h}/{I_1 L} = {18*18^3*20.25*12}/{12*23^3*26*12}=0.5600

H={w L^2}/{4h(2k+3)}= {5.152*26^2}/{4*20.25*(2*0.5600+3)}=10.44 kip

M=-Hh=10.44*20.25=-211.4 kip-ft

H_b={{wh}/{8}} {{6+5k}/{2k+3}}={{0.387*20.25}/{8}}*{{6+5*0.5600}/{2*0.5600+3}}=0.9796*{8.8/4.12}=2.092

H_a=wh-H_b = 0.387*20.25-2.092=5.745

M_1=H_a h -0.5wh^2=5.745*20.25-0.5*0.387*20.25^2=36.98 kip-ft

M_2=-H_b h = -2.092*20.25=-42.36 kip-ft

M_{u-,int}=-211.4-42.4=-253.8 kip-ft = -3045 kip-in

V_{u,col}=H+H_b=10.4+2.1=12.5 kip

The maximum positive bending moment occurs near the center of the span. Consider a section from the face in which the positive moment is decreased by wind loads to a distance x from this face.

M(x)=-211.4+37.0 +{{37.0+42.4}/26}x -{{5.152*26}/2}x -{5.152x^2}/2 = -2.576x^2+63.9x-174.4

M prime(x)=V_u(x)=-5.152x + 63.9 = 0 right x=12.4

M_{u+}=-2.576x^2+63.9x-174.4 = -2.576*12.4x^2+63.9*12.4-174.4 = -396.1+792.4-174.4=221.9 kip-ft = 2663 kip-in

Bending Envelope (kip-ft)

JpGraph Function Plot

Shear Envelope (kip)

JpGraph Function Plot

Dead and live loads are transfered into the columns, and the wind load causes a relatively small axial reaction.

  • P_D=15*{36/2}*219 + {36/2}*187.5 + 11150=59130+3375+11150=74 kip
  • P_L=15*{36/2}*41=11 kip
  • P_W=15*18*26.9*{9/{2*26}}=1.2 kip
  • P_u=1.2*74+1.6*11+0.8*1.2=88.8+17.6+1=107 kip

The axial load may be insufficient to ensure column slenderness so a pinnacle tops each column. These pinnacles may be of any durable material so long as each weighs at least

The maximum bending moment in the column occurs at the top and is the difference between the interior and exterior face negative bending moments.

M_{u,col}=M_{u-,int}-M_{u-,ext}=3045-558=2487 kip-in

Slab Design

t_{min}=180/24=7.5, say t=8 in

d≈8-0.75-0.5=6.75

A_{s,min}=0.0018 A_g = 0.0018*12*8=0.1728 in²/ft

Find grade 60 steel for the top to handle the larger negative moment.

A_{s,top} approx {71.8}/{0.9*60*0.9*6.75} approx0.2189 in²/ft

Try #4 bars (the smallest permissible on top), designing a 1' strip of slab as a 1' wide beam, b=bw=12”.

s={12*A_{bar}}/{A_s} = {12*0.20}/{0.2189} =10.96→ #4@10”

A_{s,top}={12*0.20}/10 =0.24 in²/ft

d=d_t=8-0.75-{4/16}=7 in

a={0.24*60}/{0.85*4*12}=0.3529 in

The β1 of 4000 psi concrete is 0.85.

epsilon_t = 0.003{{beta_1 d_t - a}/a} = 0.003*{{0.85*7 - 0.3529}/0.3529}=0.04758 > 0.005 → φ=0.9

phi M_n = 0.9*0.24*60*(7-{0.3529/2})=88.43 kip-in > 71.8✔

A_{s,bot} approx {51.3}/{0.9*60*0.9*6.75} approx0.1564 in²/ft < 0.1728 ∴ Use As,min

s={12*A_{bar}}/{A_s} = {12*0.20}/{0.1728} =13.89 → #4@12”

A_{s,bot}={12*0.20}/12 =0.20 in²/ft

d=d_t=8-0.75-{4/16}=7 in

a={0.2*60}/{0.85*4*12}=0.2941 in

epsilon_t = 0.003{{beta_1 d_t - a}/a} = 0.003*{{0.85*7 - 0.2941}/0.2941}=0.05769 > 0.005 → φ=0.9

Longitudinal (121½')Lateral (36')
Slab Top Mat #4 @10” #4 @12”
Slab Bottom Mat #4 @12” #4 @12”

Though all the mid spans and the edge ends require less reinforcement in the top mat, it doesn't make sense to add unnecessary complexity to change the bar spacing at most 2” just for this. To simplify construction further, two kinds of welded wire mesh providing equivalent steel areas can be used for top and bottom mats in place of the rebar specified here.

If bars are used, the 121½' lengths can be made from a pair of 64' lengths, with one bar of each pair cut more-or-less randomly within its middle third and the resulting three pieces lap-spliced.

Beam Design

The beams are T-beams, their flanges integral with the slab they support.

T-beam Web Reinforcement

Start with positive bending moment (web steel).

h=18 in, bw=18 in

d approx h-2.5=18-2.5=15.5

b_e=min[26*12/4,18+16*10,15*12]=min[78,178,180]=78 in

A_s approx {2899}/{0.9*60*0.9*15.5} approx3.848 in

Try As= 4#9 = 4.00 in², bw >= 12.2 in✔

d=d_t=18-1.5-0.5-{9/16}=15.4375 in

A_{s,min}=max[{3*sqrt{4000}*18*15.4375}/{60000}, {200*18*15.4375}/{60000}] = max[0.8787,0.9263]=0.9263 < 4.00✔

a={4.00*60}/{0.85*4*78}=0.9050

epsilon_t=0.003*{{0.85*15.4375-0.905}/{0.905}}=0.0405 > 0.005 → φ=0.9

A_{s,max}={0.85*4*0.85*78*3*15.4375}/{60*8}=21.75 > 4.00✔

phi M_n = 0.9*0.85*4*0.905*78*(15.4375-{0.905/2})=3237 kip-in > 2662✔

20' bars are sufficient to develop across the positive moments shown in the bending diagram.

T-Beam Web Reinforcement3#9, 20'

Shear Reinforcement

#3 U stirrups: Av=0.22 in²

V_c=2 b_w d sqrt{f prime_c}=2*18*15.4375*sqrt{4000}=35.150 kip

4 b_w d sqrt{f prime_c}=70.3 kip

8 b_w d sqrt{f prime_c}=140.6 kip

phi V_c=0.75*35.15=26.36 kip; {phi V_c}/2=0.75*35.15/2=13.18 kip

V_{u,cant}={w_u l}={5.152*4.25} =21.9 kip

{phi V_c}/2 < 21.9 kip < phi V_c

s_1={A_v f_y}/{50 b_w}={0.22*60000}/{50*18}=14.67

s_2={A_v f_y}/{0.75 b_w sqrt{f prime_c}}={0.22*60000}/{0.75*18*sqrt{4000}}=15.46

s_3=d/2=15.4375/2=7.719 in Controls Minimum spacing

s_4=24 in

Cantilever #3 U-stirrups: 1@2”, 7@7”

d_{V,crit}={h_{col}/{2*12}}+{d/12} = {23/{2*12}}+15.4375/12=0.958+1.286=2.245 ft

V_u(2.245)=-5.152x + 63.90=-5.152*2.245 + 63.90=52.3 kip

V_u(26-2.245)=-5.152x + 63.90=-5.152*23.76 + 63.90=58.5 kip Controls

V_n=V_u / phi = 58.5/0.75=78.00 kip

V_s=V_n-V_c = 78.00-35.15=42.85 kip

2 b_w d sqrt{f prime_c} < V_s < 4 b_w d sqrt{f prime_c} → 35.15 < 42.85 < 70.3

s_5={A_v f_y d}/{V_s} = {0.22*60*15.4375}/{42.85}=4.756 in Controls initial spacing

Spacing may increase to 7” at Vu ≥ φVc

-5.152(26-x) + 63.90= -13.2=63.9-134+5.152x right x=11.044 ft = 132.5”

Distance 0 to 27” 27” to 133” 133” to 156”
Vu 58.5 58.5-26.426.4-13.213.2-0
Minimum Spacing 4” 7” 7” none
T-beam Stirrups: #3U 1@3”, 6@4”, 16 @7”

T-beam Flange Reinforcement

d approx h-2.5=18-2.5=15.5

A_s approx {3045}/{0.9*60*0.9*15.5} approx4.042 in

With an effective flange width of 78”, these can be spread out quite a bit. Try 6#8 = 4.74 in²

d=18-1.5-0.5-{8/16}=15.5 in

a={4.74*60}/{0.85*4*18}=4.647

phi M_n = 0.9*4.74*60*(15.5-{4.647/2})=3373 > 3045✔

Top Bar Length

c+K_{tr}/d_b} = {2.5+0}/1 = 2.5

l_d = {d_b 3 f_y psi_t psi_e psi_s lambda}/{40 sqrt{f prime_c} {c+K_{tr}/d_b}}= {1* 3 *60000 *1.3* 1.0* 1.0*1.0}/{40*sqrt{4000} *2.5}}=37 in

Length Needed: 5' cantilever - 3” cover + 4' inside span + 37” development length ≈ 12'

T-beam Flange Reinforcement6#8, 12' (2 sets each beam)

Column Design

e = 2487/107=23 in > 18 in → Use a different inscribed rectangle or the full circular section.

Try using an inscribed 23” x 12” rectangle. The diagonal is sqrt{23^2 + 12^2}=25.9 in < 26 in✔

{K L}/r = {1.0*(21*12-18)}/{0.3*23}=234/{0.3*23}33.91

34- {12 M_1}/M_2 = 34-{12*1}/{2251} =33.99

{K L}/r <= 34- {12 M_1}/M_2 → Short Column

e/h= {M_u}/{P_u h}= 2487/{107*23}=1.0

K_n= {P_u}/{phi f prime_c A_g} = {107}/{0.9*4*23*12}=0.1077

R_n = {P_u e}/{phi f prime_c A_g h} = 0.1077 * 1.0 =0.1077

gamma = 16/23=0.6957

rho_g approx0.01

A_s=rho_g A_g=0.01*23*12=2.76 in²

Use 4#8 : 3.16 in²

rho_g = 3.16/{23*12}=0.01145

epsilon_t > 0.005 → φ=0.9

Column Reinforcement4#8, 20'9” long, 15” 90º Hook on top

(Development length calculations are the same as the footing dowel calculations below.)

Placing the centers of the bars on the corners of a 16” x 10” rectangle provides sufficient cover: {26/2}- {sqrt{16^2 + 10^2}}/2 - {8 / {2*8} }=3.066 in > 3 in✔

Column Ties

Vu,col = 11.561 kip

V_c >= 2 sqrt{f prime_c} b_w d = 2* sqrt{4000} * 23*12=34.91 > 11.561 → No additional shear reinforcement necessary. Use standard #3 column ties.

s=min[48d_t , 16 d_b , b]=min[48*{3/8},16*{8/8},12]=min[18,16,12]=12”

Column Ties12”x23” 5#4@4” at the top, then #3@12”

Footing Design

The circular column is treated as a square of same area (ACI 318-05 15.3). First, determine net pressure on the bottom of the footing.

h_{col}=sqrt{0.25 pi d^2}=sqrt{0.25*pi*26^2}=23 in

h_{col} <= h <= 1.6h_{col} right 23 <= h <= 37 Try h = 24 in

P_{foot}=0.150*2=0.300 ksf

P_{soil}=0.120*3=0.360 ksf

P_{floor}=0.150*{4/12}+0.075=0.125 ksf

P_{net}=4-0.3-0.36-0.125=3.215 ksf

Estimate dimensions.

A_{req}={P_D + P_L}/P_{net} = {74+11}/3.215=26.439 sf

L=B=6 ft = 72 in A=LB=36 sf

P_{u,net}={1.2*74+1.6*11}/36=2.956 ksf

d approx 24-3-1=20 in

B=h_{col}+d=23+20=43 in = 3.583 ft

V_u = P_{u,net}(A-B^2)=3.215*(36-3.583^2)=74.47 kip

beta_c=1, alpha_s=40, b_0=4B=172

V_c=min[2+4/{beta_c} , {alpha_s d}/{b_0} +2, 4 ]sqrt{f prime_c} b_0 d= min[2+4/{1} , {40*20}/{172} +2, 4 ]*sqrt{4000}*172*20= min[6 , 6.651, 4 ]*217.6=870.3 kip

phi V_n=phi V_c=0.75*870.3=652.7 kip > V_u✔

G= {L/2} -{h_{col}/2}-d = {6/2} - {23/12*2} - {20/12}=3-0.9583-1.667=0.375 ft

V_u = P_{u,net} B G = 2.956*6*0.375=6.651 kip

phi V_c = phi 2 srt{f prime_c} L d = 0.75*2*sqrt{4000}*72*20=136.6 kip > V_u✔

Footing Dimensions6' x 6' x 2'

Determine reinforcement.

F={L/2}-{h_{col}/2}={72/2}-{23/2}=24.5 in = 2.042 ft

M_u={P_{u,net} B F^2}/2 = {2.956*6*2.042^2}/2 =36.97 ft-kip = 443.6 in-kip

A_s approx M_u/{phi A_s f_y (0.95d)} = 443.6/{0.9*60*0.95*20}=0.4323 in²

A_{s,min}=0.0018 B h=0.0018*72*24=3.11 in² Controls → φ=0.9

Try 6#7 = 3.60 in²

d=24-3-{7/8}=20.13

a={A_s f_y}/{0.85 f prime_c B} = {3.6*60}/{0.85*4*72}=0.8824 in

phi M_n = phi A_s f_y (d - {a/2}) = 0.9*3.6*60*(20.13-{0.8824/2})=3828 in-kip > 443.6✔

Footing Reinforcement6#7, 5½' (x2)

Place reinforcement uniformly across the bottom in both directions.

Check development length.

Length Available={{72-23}/2} -3=21.5 in

c=min[3+{7/16},11/2,]= min[3.437,5.5] =3.437

{c+k_{tr}}/d_b = {3.437+0}/0.4375 =7.857 > 2.5 → Use 2.5

l_d = {d_b 3 fy psi_t psi_e psi_s lambda}/{40 sqrt{f prime_c} ({{c+k_{tr}}/d_b}) }= {0.4375* 3 *60000*1.0*1.0*1.0*1.0}/{40*sqrt{4000}*2.5 }=12.45 in < 21.5✔

Check bearing strength.

sqrt{A_2/A_1}=sqrt{{72^2}/{23^2}}=9.8 > 2.0 → use 2.0

phi P_{n,foot}=phi 0.85 f prime_c A_1 sqrt{A_2/A_1}=0.65*0.85*4*23^2*2.0=2338 kip

phi P_{n,col}=phi 0.85 f prime_c A_1 = 0.65*0.85*4*23^2=1169 kip

P_u={1.2*74+1.6*11}=106.4 → Minimum dowels

A_{s,min}=0.005*23^2=2.645 in²

Use 4#8 = 3.16 in²

l_{d8}=max[{0.02 d_b f_y}/{sqrt{f prime_c}} , 0.0003 f_y d_b ,8]= max[{0.02*1*60000}/{sqrt{4000}} , 0.0003*60000*1,8]= max[18.97,18,8]=18.97 in

Available length in footing: 24-3-1-1-0.75=18.25 in < 18.97 → use #4 ties @4” around the dowels in the footing.

l_{d8}=18.97*0.75=14.23 < 18.25✔

Lap splice length: max[0.0005 f_y d_b,12]=max[0.0005*60000*1,12]=30

L=30+15=45”

Footing Dowels4#8, 45” long, 15” 90º Hook

Dowels are tied by 4#4 ties and the dowels extend at least 15” into footing and least 30” into the column.

Quantities and Costs

Number1818991
FootingsColumnsBeam WebBeam FlangeSlab
Length ft 6 1.92 36 121.5
Width ft 6 1.92 1.5 36Quantity Cost
Height ft 2 19.5 0.83 0.67TotalUnitTotal
Volume cy 48 47.76 15 108 218.76 $68.50 $14,984.85
#7, 5.5' 12 216 $11.22 $2,423.52
#8, 45”, 15” 90 Hook 4 72 $18.57 $1,337.22
#4 12”x23” Ties 4 5 162 $14.12 $2,287.33
#3 12”x23” Ties 18 324 $12.32 $3,991.25
#8, 20'9”, 15” 90 Hook 4 72 $63.96 $4,605.30
#3 18”x18” U Stirrups 31 279 $11.75 $3,279.55
#9, 20' 3 27 $68.00 $1,836.00
#8, 12' 12 108 $32.04 $3,460.32
#4, 63' 158 158 $42.75 $6,754.82
#4, 35.5' 242 242 $23.71 $5,738.79
$50,698.95

Final Drawings

Illustration showing placement of reinforcement. Illustration showing placement of reinforcement.

1) Φ = ½√5 + ½ ≈ 1.618; 15'Φ-4 ≈ 2.189' ≈ 26”

Discussion


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