Isolated Footing Design
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Isolated Footing Design(ACI 318-05) Design For Isolated Footing 4 Footing No. -
Group ID -
Length
4
1
172.720 cm
Footing No.
Foundation Geometry Width
172.720 cm
Footing Reinforcement
Thickness
30.480 cm Pedestal Reinforcement
-
Bottom Reinforcement(Mz)
Bottom Reinforcement(Mx)
Top Reinforcement(Mz)
Top Reinforcement(Mx)
Main Steel
Trans Steel
4
#3 @ 17 cm c/c
#3 @ 15 cm c/c
#3 @ 17 cm c/c
#3 @ 17 cm c/c
N/A
N/A
Isolated Footing 4
Input Values Footing Geomtery Design Type : Calculate Dimension Footing Footing Thickness Thickness (Ft) (Ft) : 12.000 12.000 in Footing Footing Length Length - X (Fl) : 40.000 40.000 in Footing Footing Width Width - Z (Fw) (Fw) : 40.00 40.000 0 in Eccentricity Eccentricity along along X (Oxd) : 0.000 0.000 in Eccentricity Eccentricity along Z (Ozd) (Ozd) : 0.000 0.000 in
Column Dimensions Column Column Shape Shape : Rectan Rectangul gular ar Column Length Length - X (Pl) (Pl) : 31.750 31.750 cm Column Width Width - Z (Pw) : 16.662 16.662 cm
Isolated Footing Design
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Pedestal Include Pedestal? Yes Pedestal Shape : Rectangular Pedestal Height (Ph) : 0.000 cm Pedestal Length - X (Pl) : 31.750 cm Pedestal Width - Z (Pw) : 16.662 cm
Design Parameters Concrete and Rebar Properties Unit Weight of Concrete : 150.000 kg/m3 Strength of Concrete : 4.000 ksi Yield Strength of Steel : 60.000 ksi Minimum Bar Size : #3 Maximum Bar Size : #10 Minimum Bar Spacing : 2.000 cm Maximum Bar Spacing : 18.000 cm Pedestal Clear Cover (P, CL) : 3.000 in Footing Clear Cover (F, CL) : 3.000 in
Soil Properties Soil Type : UnDrained Unit Weight : 112.000 lb/ft3 Soil Bearing Capacity : 4.000 kip/ft2 Soil Surcharge : 0.000 kip/in2 Depth of Soil above Footing : 0.000 in Undrained Shear Strength : 0.000 kip/in2
Sliding and Overturning Coefficient of Friction : 0.500 Factor of Safety Against Sliding : 1.500 Factor of Safety Against Overturning : 1.500 ------------------------------------------------------
Design Calculations Footing Size
Initial Length (Lo) = 101.600 cm Initial Width (Wo) = 101.600 cm
Load Combination/s- Service Stress Level Load Combination Number
Load Combination Title
1
CARGA VIVA + CARGA MUERTA
3
COMBINACION 1
101
Load Combination/s- Strength Level Load Combination Number
Load Combination Title
1
CARGA VIVA + CARGA MUERTA
3
COMBINACION 1
Isolated Footing Design
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Applied Loads - Service Stress Level LC
Axial (kgf)
Shear X (kgf)
Shear Z (kgf)
Moment X (kNm)
Moment Z (kNm)
1
11420.946
1835.277
0.000
0.000
0.000
3
16735.719
3401.942
0.000
0.000
-57.252
101
281 56.664
5237.219
0.000
0.000
-57.252
Applied Loads - Strength Level LC
Axial (kgf)
Shear X (kgf)
Shear Z (kgf)
Moment X (kNm)
Moment Z (kNm)
1
11420.946
1835.277
0.000
0.000
0.000
3
16735.719
3401.942
0.000
0.000
-57.252
Reduction of force due to buoyancy = 0.000 kgf Effect due to adhesion = 0.000 kgf Area from initial length and width, A o = L X W = 10322.559 cm 2 o o Min. area required from bearing pressure, A min = P / q = 14442.440 cm 2 max
Note: Amin is an initial estimation. P = Critical Factored Axial Load(without self weight/buoyancy/soil). qmax = Respective Factored Bearing Capacity.
Final Footing Size Length (L2) = 172.720
cm
Governing Load Case :
# 101
Width (W2) = 172.720
cm
Governing Load Case :
# 101
Depth (D2) = 30.480
cm
Governing Load Case :
# 101
Area (A 2) = 29832.196 cm2
Pressures at Four Corners
Pressure at corner 1 (q1)
Pressure at corner 2 (q2)
Pressure at corner 3 (q3)
Pressure at corner 4 (q4)
(kgf/cm2)
(kgf/cm2)
(kgf/cm2)
(kgf/cm2)
(cm2)
1
0.3232
0.4535
0.4535
0.3232
0.000
101
0.0827
1.8141
1.8141
0.0827
0.000
101
0.0827
1.8141
1.8141
0.0827
0.000
1
0.3232
0.4535
0.4535
0.3232
0.000
Load Case
Area of footing in uplift (A u)
If A u is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
Isolated Footing Design
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Pressure at corner 1 (q 1)
Pressure at corner 2 (q 2)
Pressure at corner 3 (q 3)
Pressure at corner 4 (q 4)
Load Case
(kgf/cm2)
(kgf/cm2)
(kgf/cm2)
(kgf/cm2)
1
0.3232
0.4535
0.4535
0.3232
101
0.0827
1.8141
1.8141
0.0827
101
0.0827
1.8141
1.8141
0.0827
1
0.3232
0.4535
0.4535
0.3232
Check for stability against overturning and sliding
Factor of safety against sliding
-
Factor of safety against overturning
Load Case No.
Along XDirection
Along ZDirection
About XDirection
About ZDirection
1
3.156
N/A
N/A
17.885
3
2.484
N/A
N/A
2.123
101
2.701
N/A
N/A
3.287
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction Critical Load Case for Sliding along X-Direction : 3 Governing Disturbing Force : 3401.942 kgf Governing Restoring Force : 8449.695 kgf Minimum Sliding Ratio for the Critical Load Case : 2.484 Critical Load Case for Overturning about X-Direction : 1 Governing Overturning Moment : 0.000 kNm Governing Resisting Moment : 98.109 kNm Minimum Overturning Ratio for the Critical Load Case : N/A
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction Critical Load Case for Sliding along Z-Direction : 1 Governing Disturbing Force : 0.000 kgf Governing Restoring Force : 5792.309 kgf Minimum Sliding Ratio for the Critical Load Case : N/A Critical Load Case for Overturning about Z-Direction : 3 Governing Overturning Moment : -67.421 kNm Governing Resisting Moment : 143.119 kNm Minimum Overturning Ratio for the Critical Load Case : 2.123
Isolated Footing Design
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Shear Calculation Punching Shear Check
Total Footing Depth, D = 30.480cm Calculated Effective Depth, deff =
D - Ccover - 1.0 = 20.320 cm
For rectangular column,
Bcol / Dcol = 1.905
=
1 inch is deducted from total depth to cater bar dia(US Convention).
Effective depth, d eff , increased until 0.75XVc
Punching Shear Force
Punching Shear Force, Vu = 15655.425 kgf, Load Case # 3 From ACI Cl.11.12.2.1, bo for column=
178.105 cm
Equation 11-33, Vc1 =
65966.991 kgf
Equation 11-34, Vc2 =
105625.800 kgf
Equation 11-35, Vc3 =
64370.600 kgf
Punching shear strength, V c =
0.75 X minimum of (Vc1, Vc2, Vc3) =
48277.950 kgf
0.75 X Vc > Vu hence, OK
Along X Direction (Shear Plane Parallel to Global X Axis)
From ACI Cl.11.3.1.1, Vc =
31212.213 kgf
Distance along X to design for shear, Dx =
57.709 cm
Check that 0.75 X V c > Vux where Vux is the shear force for the critical load cases at a distance deff from the face of t he column caused by bending about the X axis. From above calculations,
23409.159 kgf
Isolated Footing Design
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0.75 X Vc = Critical load case for Vux is # 3
576 9. 506
kgf
0.75 X V c > Vux hence, OK
One-Way Shear Check Along Z Direction (Shear Plane Parallel to Global Z Axis)
From ACI Cl.11.3.1.1, V c =
31212.213 kgf
Distance along X to design for shear, Dz =
122.555
cm
Check that 0.75 X V c > V uz where Vuz is the shear force for the critical load cases at a distance d eff from the face of the column caused by bending about the Z axis. From above calculations,
0.75 X Vc =
23409.159 kgf
Critical load case for Vuz is # 3
924 9. 373
kgf
0.75 X V c > Vuz hence, OK
Design for Flexure about Z Axis (For Reinforcement Parallel to X Axis)
Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced Concrete Design (5th ed.) by Salmon and Wang (Ref. 1) Critical Load Case # 3 The strength values of steel and concrete used in the formulae are in ksi Factor
from ACI Cl.10.2.7.3 =
0.850
Isolated Footing Design
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From ACI Cl. 10.3.2,
=
0.02851
From ACI Cl. 10.3.3,
=
0.02138
From ACI Cl. 7.12.2,
=
0.00180
From Ref. 1, Eq. 3.8.4a, constant m =
17.647
Calculate reinforcement ratio
for critical load case
Design for flexure about Z axis is performed at the face of the column at a distance, Dx =
70.485
cm
Ultimate moment,
53.040
kNm
Nominal moment capacity, Mn =
58.933
kNm
Required
0.00203
=
Since
OK
Area of Steel Required, A s =
1.107
in2
Selected bar Size = #3 Minimum spacing allowed (S min) = = 2.000 cm Selected spacing (S) = 15.653 cm Smin <= S <= Smax and selected bar size < selected maximum bar size... The reinforcement is accepted. According to ACI 318-05 Clause No- 10.6.4 Max spacing for Cracking Consideration = 19.050 cm Safe for Cracking Aspect. Based on spacing r einforcement increment; provided reinforcement is #3 @ 15.240 cm o.c.
Required development length for bars =
=30.480 cm
Available development length for bars, DL
62.865
cm
Area of one bar = 0.110
in2
= Try bar size
#3
Number of bars required, N bar =
11
Because the number of bars is rounded up, make sure new r einforcement ratio < ρmax Total reinforcement area, A s_total = deff =
Reinforcement ratio,
Nbar X (Area of one bar) =
1.210 in2
D - Ccover - 0.5 X (dia. of one bar) =
22.384 cm
0.00202
=
From ACI Cl.7.6.1, minimum req'd clear distance between bars, Cd =
max (Diameter of one bar, 1.0, Min. User Spacing) =
Check to see if width is sufficient to accomodate bars
15.653 cm
Isolated Footing Design
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Design for Flexure about X axis (For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced Concrete Design (5th ed.) by Salmon and Wang (Ref. 1) Critical Load Case # 3 The strength values of steel and concrete used in the formulae are in ksi Factor
0.850
from ACI Cl.10.2.7.3 =
From ACI Cl. 10.3.2,
=
0.02851
From ACI Cl. 10.3.3,
=
0.02138
From ACI Cl.7.12.2,
=
0.00180
From Ref. 1, Eq. 3.8.4a, constant m =
17.647
Calculate reinforcement ratio
for critical load case
Design for flexure about X axis is performed at the face of the column at a distance, D z =
78.029
cm
Ultimate moment,
30.281
kNm
Nominal moment capacity, Mn =
33.646
kNm
Required
0.00127
=
Since
OK
Area of Steel Required, A s =
0.933
in2
Selected Bar Size = #3 Minimum spacing allowed (S min) = 2.000 cm Selected spacing (S) = 18.000 cm Smin <= S <= Smax and selected bar size < selected maximum bar size... The reinforcement is accepted. According to ACI 318-05 Clause No- 10.6.4 Max spacing for Cracking Consideration = 19.050 cm Safe for Cracking Aspect. Based on spacing r einforcement increment; provided reinforcement is
Isolated Footing Design
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#3 @ 17.780 cm o.c.
Required development length for bars =
=30.480 cm
Available development length for bars, DL = Try bar size
#3
Area of one bar =
Number of bars required, N bar =
70.409
cm
0.110
in2
9
Because the number of bars is rounded up, make sure new r einforcement ratio < ρmax Total reinforcement area, A s_total = deff =
Reinforcement ratio,
Nbar X (Area of one bar) =
0.990
in2
D - Ccover - 0.5 X (dia. of one bar) =
20.295
cm
0.00182
=
From ACI Cl.7.6.1, minimum req'd clear max (Diameter of one bar, 1.0, Min. 18.000 distance between bars, C d = User Spacing) =
cm
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading. As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depth and surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for M x. Find the area of steel required The strength values of steel and concrete used in the formulae are in ksi Factor
0.850
from ACI Cl.10.2.7.3 =
From ACI Cl. 10.3.2,
=
0.02851
From ACI Cl. 10.3.3,
=
0.02138
From ACI Cl. 7.12.2,
=
0.00180
From Ref. 1, Eq. 3.8.4a, constant m =
17.647
Calculate reinforcement ratio
for critical load case
Isolated Footing Design
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Design for flexure about A axis is performed at the face of the column at a distance, Dx =
78. 02 9 c m
Ultimate moment,
0.236 kNm
Nominal moment capacity, Mn =
0. 26 2 k Nm
=
0.00001
Since
OK
Required
Area of Steel Required, A s =
0.933 in2
Selected bar Size = #3 Minimum spacing allowed (S min) = 2.000 cm Selected spacing (S) = 18.000 cm Smin <= S <= Smax and selected bar size < selected maximum bar size... The reinforcement is accepted. According to ACI 318-05 Clause No- 10.6.4 Max spacing for Cracking Consideration = 19.050 cm Safe for Cracking Aspect. Based on spacing r einforcement increment; provided reinforcement is #3 @ 18 cm o.c.
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift # Calculate the flexural reinforcement for Mz. Find the area of steel required The strength values of steel and concrete used in the formulae are in ksi Factor
0.850
from ACI Cl.10.2.7.3 =
From ACI Cl. 10.3.2,
=
0.02851
From ACI Cl. 10.3.3,
=
0.02138
From ACI Cl.7.12.2,
=
0.00180
From Ref. 1, Eq. 3.8.4a, constant m =
17.647
Calculate reinforcement ratio
for critical load case
Isolated Footing Design
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Design for flexure about A axis is performed at the face of the column at a distance, Dx =
70. 48 5 c m
Ultimate moment,
0.192 kNm
Nominal moment capacity, Mn =
0. 21 4 k Nm
=
0.00001
Since
OK
Required
Area of Steel Required, A s =
0.979 in2
Selected bar Size = #3 Minimum spacing allowed (S min) = 2.000 cm Selected spacing (S) = 18.000 cm Smin <= S <= Smax and selected bar size < selected maximum bar size... The reinforcement is accepted. According to ACI 318-05 Clause No- 10.6.4 Max spacing for Cracking Consideration = 19.050 cm Safe for Cracking Aspect. Based on spacing r einforcement increment; provided reinforcement is #3 @ 18 cm o.c.
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