Project : Date : 23-Apr-2020
Footing No : Code : ACI 318-08
Given Data
Materials informations Exterior Columnn Interior Columnn Soil informations
Fcu footing = 30 Mpa D.L. = 650.00 KN D.L. = 920.00 KN qall = 150.0 KN/m²
Fcu column = 30 Mpa L.L. = 400.00 KN L.L. = 580.00 KN γs = 18.0 KN/m³
Fy = 420 Mpa Lx₁ = 0.50 m Lx₂ = 0.70 m Df = 1.50 m
Cc = 70 mm Ly₁ = 0.25 m Ly₂ = 0.30 m h = 700 mm
S = 4.00 m ex = 0.30 m
Design Procedure
Determination of Resultant force location
Pw₁ = P D.L. + P L.L. = 1050.0 KN
Pw₂ = P D.L. + P L.L. = 1500.0 KN
R = Pw₁ + Pw₂ = 2550.0 KN
XR (Location of Resultant force from exterior column center) = Pw₂ S / R = 2.35 m
Determination of Footing dimensions
γa = ( γc + γs )/2 = 21.5 KN/m³
qna = qall - γa Df = 117.8 KN/m²
Areq = R / qna = 21.66 m² 15.2.2
L = 2 ( XR + Lx₁ /2 + ex ) = 5.85 m
B = A / L = 3.75 m
Af = LB = 21.94 m²
qact = R / Af = 116.2 KN/m² < qna ….. safe
Assume footing thickness ( h ) = 700 mm
Then effective footing depth ( d ) = h - Cc - Ø = 619 mm
Determination of Ultimate design forces
Pu₁ = 1.2 P D.L. + 1.6 P L.L. = 1420.0 KN Eq. (9-2)
Pu₂ = 1.2 P D.L. + 1.6 P L.L. = 2032.0 KN
qs = (Pu₁ + Pu₂) / Af = 157.4 KN/m² 15.2.1
Wu = (Pu₁ + Pu₂) / L = 590.1 KN/m
Z (Point of zero shear from footing edge) = Pu₁ / Wu = 2.41 m
X₁ (outer face of exterior column to footing edge) = 0.30 m
X₂ (outer face of interior column to footing edge) = 0.95 m
V₁ = Wu X₁ = 177.0 KN
V₂ = V₁ + Wu Lx₁/2 = 324.5 KN
V₃ = V₂ - Pu₁ = -1095.5 KN
V₄ = V₃ + Wu Lx₁/2 = -947.9 KN
V₅ = V₄ + Wu (S - Lx₁ /2 - Lx₂ /2) = 1058.4 KN
V₆ = V₅ + Wu Lx₂ /2 = 1264.9 KN
V₇ = V₆ - Pu₂ = -767.1 KN
V₈ = V₇ + Wu Lx₂/2 = -560.6 KN
Vmax (at distance d from column face) = 693.1 KN
M₁ = Wu X₁²/2 = 26.6 KN.m.
M₂ = Wu (X₁+Lx₁/2)²/2 = 89.3 KN.m.
M₃ = Wu (X₁+Lx₁)²/2 - Pu₁ (Lx₁/2) = -166.2 KN.m.
M₄ = Wu (L - X₂ - Lx₂)²/2 - Pu₁ (S - Lx₂/2) = 21.6 KN.m.
M₅ = Wu (L - X₂ - 0.5 Lx₂)²/2 - Pu₁ S = 428.1 KN.m.
M₆ = Wu (L - X₂)²/2 - Pu₁ (S + 0.5 Lx₂) - Pu₂ (0.5 Lx₂) = 195.8 KN.m.
Mmax (at distance Z from footing edge) = Wu Z²/2 - Pu₁ (Z - X₁ - Lx₁/2) = -927.6 KN.m.
Pu₁ = 1420.00 KN. Pu₂ = 2032.00 KN.
Wu = 590.09 KN/m
1264.9 KN.
324.5 KN.
-767.1 KN.
-1095.5 KN.
-927.6 KN.m.
89.3 KN.m.
428.1 KN.m.
89.3 KN.m.
428.1 KN.m.
Check of wide beam action
f = 0.75 9.3.2.3
Vu max = 693.1 KN
fVn = f 0.17 Bd= 1449.9 KN > Vu max ….. safe Eq. (11-3)
f 'c
Check of two way action
For Exterior column
bo = 2 (Lx₁ + X + d/2) + (Ly₁ + d) = 3.09 m
β = Lx₁ / Ly₁ = 2.00
αs = 30 Exterior column
0.17 (1 +2/β) = 0.34 Eq. (11-31)
fVn
bo d
= the minimum of [
0.083 (αs d / bo) + 2
0.33
= 0.499
= 0.33 (governs)
Eq. (11-32)
Eq. (11-33)
f 'c
Vup₁ = Pu₁ - qs (X₁ + Lx₁ + d/2) (Ly₁ + d) = 1268.3 KN
fVn = f * 0.33 bo d = 2317.7 KN > Vup₁ ….. safe
f 'c
For Interior column
bo = 2 (Lx₂ + Ly₂ + 2d) = 4.48 m
β = Lx₂ / Ly₂ = 2.33
αs = 40 Interior column
0.17 (1 +2/β) = 0.316 Eq. (11-31)
fVn
bo d
= the minimum of [
0.083 (αs d / bo) + 2
0.33
= 0.459
= 0.33 (governs)
Eq. (11-32)
Eq. (11-33)
f 'c
Vup₂ = Pu₂ - qs (Lx₂ + d) (Ly + d) = 1841.3 KN
fVn = f * 0.316 bo d = 3214.0 KN > Vup₂ ….. safe
f 'c
Determination of footing reinforcement
Assume tension controlled section ( f = 0.9 ) 9.3.2.1 , 10.3.4
β₁ = 0.85
rmin = 0.0018 7.12.2.1
0.85 f'c 600
rmax = 0.75
fy
β₁ [ 600 + fy ] = 0.018214
Longitudinal top reinforcement
Mu = 927.6 KN.m. 15.4.2
Mu
Rn = = 0.717282
f B d²
0.85 f'c
rreq =
fy [ 1 1
2 Rn ] = 0.001739
0 . 85 f ' c
rgross = rreq ( d / h ) = 0.001538 ρ < ρ min……take ρ = ρ min 7.12.2.1
As = r B d = 4178.3 mm²
Use 11 Ø22 (As = 4179.3 mm² )
Bars spacing = 359 mm < 450 mm…...safe 10.5.4
Longitudinal bottom reinforcement
Mu = 428.1 KN.m. 15.4.2
Mu
Rn = = 0.331065
Rn = = 0.331065
f B d²
0.85 f'c
rreq =
fy [ 1 1
2 Rn ] = 0.000795
0 . 85 f ' c
rgross = rreq ( d / h ) = 0.000703 ρ < ρ min……take ρ = ρ min 7.12.2.1
As = r B d = 4178.3 mm²
Use 21 Ø16 (As = 4220.2 mm² )
Bars spacing = 180 mm < 450 mm…...safe 10.5.4
Transverse direction reinforcement
Exerior colunm hidden beam
b₁ = X₁ + Lx₁ + d/2 = 1.11 m
Y₁ = (B - Ly₁) / 2 = 1.75 m
W₁ = Pu₁ / B = 378.7 KN/m
M₁ = W₁ Y₁² /2 = 579.8 KN.m.
M₁
Rn = = 1.515486
f b₁ d²
0.85 f'c
rreq =
fy [ 1 1
2 Rn ] = 0.003753
0 . 85 f ' c
rgross = rreq ( d / h ) = 0.003319 ρ max > ρ > ρ min…...safe 7.12.2.1
As = r b₁ d = 2577.7 mm²
Use 7 Ø22 (As = 2659.6 mm² )
Bars spacing = 181 mm < 450 mm…...safe 10.5.4
Inerior colunm hidden beam
b₂ = Lx₂ + d = 1.32 m
Y₂ = (B - Ly₂) / 2 = 1.73 m
W₂ = Pu₂ / B = 541.9 KN/m
M₂ = W₂ Y₂²/2 =806.2 KN.m.
M₂
Rn = = 1.772442
f b₂ d²
0.85 f'c
rreq =
fy [ 1 1
2 Rn ] = 0.004421
0 . 85 f ' c
rgross = rreq ( d / h ) = 0.00391 ρ max > ρ > ρ min…...safe 7.12.2.1
As = r b₂ d = 3609.8 mm²
Use 10 Ø22 (As = 3799.4 mm² )
Bars spacing = 144 mm < 450 mm…...safe 10.5.4
Secondary reinforcement
As = 0.2 As = 222.8 mm²
Use 5 Ø12 (As = 565.2 mm² )
Bars spacing = 200 mm < 450 mm…...safe 10.5.4
Check development of reinforcement
Longitudinal top reinforcement
fy ψt ψe ψs
Ld = [ 1.1 λ *
(Cb + Ktr)/db ]db Eq. (12-1)
f 'c
clear cover = 70.0 mm (governs) 12.2.3
Cb = the minimum of [ one half bar spacing = 179.4 mm
Ktr = 0
(Cb + Ktr) / db = 3 > 2.5 take it = 2.5
ψt = 1.3 ( top bars ) 12.2.4 ( a )
note ψt ψe = 1.3 < 1.7
ψe = 1 ( uncoated bars ) 12.2.4 ( b )
ψs = 1 ( Larger than No 22 bars ) 12.2.4 ( c )
λ = 1 ( normal weight concrete ) 12.2.4 ( d )
fy ψt ψe ψs
Ld = [ 1.1 λ *
(Cb + Ktr)/db ] db = 891.6 mm Eq. (12-1)
f 'c
Lavailable = Z - Cc = 2336.4 mm »» reinforcement can be fully developed
Longitudinal bottom reinforcement
fy ψt ψe ψs
Ld = [ 1.1 λ *
(Cb + Ktr)/db ] db Eq. (12-1)
f 'c
clear cover = 70.0 mm (governs) 12.2.3
Cb = the minimum of [ one half bar spacing = 89.9 mm
Ktr = 0
(Cb + Ktr) / db = 4 > 2.5 take it = 2.5
ψt = 1 ( bottom bars ) 12.2.4 ( a )
note ψt ψe = 1 < 1.7
ψe = 1 ( uncoated bars ) 12.2.4 ( b )
ψs = 0.8 ( smaller than No. 19 bars) 12.2.4 ( c )
λ = 1 ( normal weight concrete ) 12.2.4 ( d )
fy ψt ψe ψs
Ld = [ 1.1 λ *
(Cb + Ktr)/db ] db = 399.0 mm Eq. (12-1)
f 'c
Lavailable = X₂ - Cc = 880.0 mm »» reinforcement can be fully developed
Check tensile strain (εt)
As fy 1755323
a= = = 22.9 mm
0.85 f'c B 76500
a
C= = 27.0 mm
β₁
0.003
εt = ( C ) d - 0.003 = 0.0658 > 0.005 section is tension controlled 10.3.4
Forces transfer at column footing interface
Exterior column
Bearing strengrh of footing
f = 0.65 9.3.2.4
fPnb = f ( 0.85 f'cf A₁ ) A2 10.14.1
A1
A₁ = a b = 125000.0 mm²
A₂ = 6820000.0 mm²
A 2 = 7.39 >2 take it = 2
A1
fPnb = 3315.0 KN
Bearing strengrh of column
fPnb = f ( 0.85 f'cc A₁ )
fPnb = 1657.5 KN
fPnb = least of fPnb footing and fPnb column
fPnb = 1657.5 KN > Pu₁……Use As minimum
Determination of column dowels
true As = ( Pu - fPnb ) / f fy = -870.0 mm² 15.8.1.2
Asmin = 0.005 A₁ = 625.0 mm² 15.8.2.1
As req. = 625.0 mm²
Use 4 Ø16 (As = 803.8 mm² )
As required 12.3.3 ( a )
Kr = = 0.778
As provided
Development of column dowels
In footing
Ldc = 0.24 fy / λ f ' c db = 329.2 mm (governs) 12.3.2
Ldc minmum = 0.043 fy db = 289.0 mm
L req = Kr Ldc = 256.0 mm 12.3.3 ( a )
Lavailable = h - Cc - Ø - Øb₁ - Ødowel = 576.0 mm > Ldc……safe
In column
dowels compression splice
0 Ls = 0.071 fy db = 477.1 mm > 300mm……… safe 12.16.1
1 Ls = (0.13 fy - 24 ) db = 489.6 mm > 300mm……… safe
1 Ls = 1.33Ls = 634.6 mm 12.16.1
column bars development
Ldc = 0.24 fy / λ f ' c db = 329.2 mm (governs) 12.3.2
Ldc minmum = 0.043 fy db = 289.0 mm
L req = the larger of dowels splice and column bars development = 477.1 mm
Interior column
Bearing strengrh of footing
f = 0.65 9.3.2.4
fPnb = f ( 0.85 f'cf A₁ ) A2 10.14.1
A1
A₁ = a b = 210000.0 mm²
A₂ = 9455000.0 mm²
A 2 = 6.7 >2 take it = 2
A1
fPnb = 5569.2 KN
Bearing strengrh of column
fPnb = f ( 0.85 f'cc A₁ )
fPnb = 2784.6 KN
fPnb = least of fPnb footing and fPnb column
fPnb = 2784.6 KN > Pu₂……Use As minimum
Determination of column dowels
true As = ( Pu - fPnb ) / f fy = -2756.8 mm² 15.8.1.2
Asmin = 0.005 A₁ = 1050.0 mm² 15.8.2.1
As req. = 1050.0 mm²
Use 6 Ø16 (As = 1205.8 mm² )
As required 12.3.3 ( a )
Kr = = 0.871
As provided
Development of column dowels
In footing
Ldc = 0.24 fy / λ f ' c db = 329.2 mm (governs) 12.3.2
Ldc minmum = 0.043 fy db = 289.0 mm
L req = Kr Ldc = 286.7 mm 12.3.3 ( a )
Lavailable = h - Cc - Ø - Øb₂ - Ødowel = 576.0 mm > Ldc……safe
In column
dowels compression splice
0 Ls = 0.071 fy db = 477.1 mm > 300mm……… safe 12.16.1
1 Ls = (0.13 fy - 24 ) db = 489.6 mm > 300mm……… safe
1 Ls = 1.33Ls = 634.6 mm 12.16.1
column bars development
Ldc = 0.24 fy / λ f ' c db = 329.2 mm (governs) 12.3.2
f 'c
Ldc minmum = 0.043 fy db = 289.0 mm
L req = the larger of dowels splice and column bars development = 477.1 mm
FINAL RESULTS
Dimensions Long Reinforcement Transverse Reinforcement
L (m) B (m) h (m) Long Top Long Bottom Column₁ Column₂
5.85 m 3.75 m 0.70 m 11 Ø22 21 Ø16 7 Ø22 10 Ø22
Check of wide beam shear Safe
Check of two way shear (punching) Safe
Check of reinforcement development Safe
Check of dowels development Safe
