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Isolated Footing Design(ACI 318-11)

Footing No. Group ID Foundation Geometry
- - Length Width Thickness
1 1 3.40m 3.40m 0.45m

Footing No. Footing Reinforcement Pedestal Reinforcement

- Bottom Reinforcement(Mz) Bottom Reinforcement(Mx) Top Reinforcement(Mz) Top Reinforcement(Mx) Main Steel Trans Steel
1 30 - ϕ16 30 - ϕ16 N/A N/A 12 - #16 #10 @ 200 mm

Isolated Footing 1
2.04 m

1.3 m 1.75 m

0.45 m

Elevation

X 1.7 m

0.5 m
3.4 m

0.5 m
1.7 m

3.4 m

Plan

Input Values

Footing Geomtery

Calculate
Design Type :
Dimension
Footing Thickness (Ft) : 450.00 mm
Footing Length - X (Fl) : 3400.00 mm
Footing Width - Z (Fw) : 3400.00 mm
Eccentricity along X (Oxd) : 0.00 mm
Eccentricity along Z (Ozd) : 0.00 mm

Column Dimensions

Column Shape : Rectangular

Column Length - X (Dcol) : 0.50 m
Column Width - Z (Bcol) : 0.50 m

Pedestal

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Include Pedestal : Yes

Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.30 m
Pedestal Length - X (Pl) : 0.50 m
Pedestal Width - Z (Pw) : 0.50 m

Design Parameters

Concrete and Rebar Properties

Unit Weight of Concrete : 24.00 kN/m3

Strength of Concrete : 35.00 N/mm2
Yield Strength of Steel : 420.00 N/mm2
Minimum Bar Size : ϕ16
Maximum Bar Size : ϕ20
Top Footing Minimum Bar Size : ϕ16
Top Footing Maximum Bar Size : ϕ20
Pedestal Minimum Bar Size : ϕ16
Pedestal Maximum Bar Size : ϕ16
Minimum Bar Spacing : 116.00 mm
Maximum Bar Spacing : 116.00 mm
Pedestal Clear Cover (P, CL) : 75.00 mm
Bottom Footing Clear Cover (F, CL) : 50.00 mm

Soil Properties

Soil Type : Cohesive Soil

Unit Weight : 22.00 kN/m3

Soil Bearing Capacity : 100.00 kN/m2

Soil Bearing Capacity Type : Gross Bearing Capacity

Soil Surcharge : 0.00 kN/m2

Depth of Soil above Footing : 2050.00 mm

Type of Depth : Fixed Bottom

Cohesion : 0.00 kN/m2

Bearing Capacity Input Method : Fixed Bearing Capacity

Sliding and Overturning

Coefficient of Friction : 0.40

Factor of Safety Against Sliding : 1.40
Factor of Safety Against Overturning : 1.40

Global Settings

Top Reinforcement Option : Calculate only when foundation is subjected to uplift forces
Concrete Design Option : Net Pressure(Gross Pressure - Self Weight Pressure)
Top Reinforcement Factor : 1.00

------------------------------------------------------

Design Calculations

Footing Size

Initial Length (Lo) = 3.40 m

Initial Width (Wo) = 3.40 m

Load Combinations

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Load Combination/s- Service Stress Level

Load Load Soil Self
Combination Load Combination Title Combination Bearing Weight
Number Factor Factor Factor
101 1.000 x DL 1.00 1.00 1.00
161 0.600 x DL 1.00 1.00 1.00

Load Combination/s- Strength Level

Load Load Soil Self
Combination Load Combination Title Combination Bearing Weight
Number Factor Factor Factor
201 1.400 x DL 1.00 1.00 1.00
202 1.200 x DL 1.00 1.00 1.00
226 0.900 x DL 1.00 1.00 1.00

Applied Loads on Top of Pedestal

Before consideration of self weight and load safety factor table

Moments are about the center of footing / pile cap (does not include moments caused by lateral loads)

Applied Loads - Service Stress Level

Axial Shear X Shear Z Moment X Moment Z
LC
(kN) (kN) (kN) (kNm) (kNm)
101 5.00 0.00 2.90 0.00 0.00
161 3.00 0.00 1.74 0.00 0.00

Applied Loads - Strength Level

Axial Shear X Shear Z Moment X Moment Z
LC
(kN) (kN) (kN) (kNm) (kNm)
201 7.00 0.00 4.06 0.00 0.00
202 6.00 0.00 3.48 0.00 0.00
226 4.50 0.00 2.61 0.00 0.00

Reduction of force due to buoyancy = 0.00 kN

Effect due to adhesion = 0.00 kN

2
Area from initial length and width, Ao = Lo X W o = 11.56 m

2
Min. area required from bearing pressure, Amin = 6.50 m

Note: Amin is an initial estimation considering self-weight, axial load and moment against factored bearing capacity.

Final Footing Size

Length (L2) = 3.40 m Governing Load Case : # 101

Width (W2) = 3.40 m Governing Load Case : # 101
Depth (D2) = 0.45 m
Depth is governed by Ultimate Load Case
(Service check is performed with footing thickness requirements from concrete check)

Area (A2) = 11.56 m2

Final Pedestal Height = 2.05 m
Final Soil Height = 2.05 m
Foundation Self Weight = 137.14 kN
Soil Weight On Top Of Footing = 510.06 kN

Pressures at 4 Corners
Please note that pressures values displayed in tables below are calculated after dividing by soil bearing factor

Pressure at Pressure at Pressure at Pressure at Area of footing

Load Case / top left top right bottom bottom left in uplift (Au)
Combination corner corner right corner corner 2
(kN/m2) (kN/m2) (kN/m2) (kN/m2) (m )

161 55.5822 55.5822 56.9103 56.9103 0.00

161 55.5822 55.5822 56.9103 56.9103 0.00

101 55.3125 55.3125 57.5260 57.5260 0.00

101 55.3125 55.3125 57.5260 57.5260 0.00

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Pressure at Pressure at Pressure at Pressure at Area of footing

Load Case / top left top right bottom bottom left in uplift (Au)
Combination corner corner right corner corner 2
(kN/m2) (kN/m2) (kN/m2) (kN/m2) (m )

If Au 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 four Corners

Pressure at
Pressure at top Pressure at top bottom right Pressure at
Load Case / left corner right corner corner bottom left corner
Combination (kN/m2) (kN/m2) (kN/m2) (kN/m2)

161 55.5822 55.5822 56.9103 56.9103

161 55.5822 55.5822 56.9103 56.9103

101 55.3125 55.3125 57.5260 57.5260

101 55.3125 55.3125 57.5260 57.5260

Stability Check

2.04 m OTM

Sliding Force 1.3 m 1.75 m

Frictional Force 0.45 m

Factor of safety against

- Factor of safety against sliding
overturning

Load
Along X- Along Z- Required About X- About Z- Required
Case Resultant
Direction Direction FOS Direction Direction FOS
No.

101 N/A 89.96 89.96 1.40 152.93 N/A 1.40

161 N/A 149.47 149.47 1.40 254.10 N/A 1.40

Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction

Critical Load Case for Sliding along X-Direction : N/A

Governing Disturbing Force : N/A
Governing Restoring Force : N/A
Minimum Sliding Ratio for the Critical Load Case : N/A
Critical Load Case for Overturning about X-Direction : 101
Governing Overturning Moment : 7.25 kNm
Governing Resisting Moment : 1108.73 kNm
Minimum Overturning Ratio for the Critical Load Case : 152.93

Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction

Critical Load Case for Sliding along Z-Direction : 101

Governing Disturbing Force : 2.90 kN
Governing Restoring Force : 260.88 kN
Minimum Sliding Ratio for the Critical Load Case : 89.96
Critical Load Case for Overturning about Z-Direction : N/A
Governing Overturning Moment : N/A
Governing Resisting Moment : N/A

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Minimum Overturning Ratio for the Critical Load Case : N/A

Critical Load Case And The Governing Factor Of Safety For Sliding Along Resultant Direction
Critical Load Case for Sliding along Resultant Direction : 101
Governing Disturbing Force : 2.90 kN
Governing Restoring Force : 260.88 kN
Minimum Sliding Ratio for the Critical Load Case : 89.96

Compression Development Length Check

Development length required based on ACI 318 12.3

= 0.29m

Footing depth available

D - Clear Cover = 0.40m
D > Lu Footing depth is OK for column reinforcement compressive development check

Shear Calculation

Punching Shear Check

X 1.7 m

Z
0.19 m
1.7 m

Plan
Total Footing Depth, D = 0.45m
Calculated Effective Depth, d = D - Ccover - 1 * db = 0.38 m
For rectangular column, = Bcol / Dcol = 1.00

Effective depth, d, increased until 0.75XVc Punching Shear Force

Punching Shear Force, Vu = -14.78kN, Load Case # 226

From ACI Cl.11.11.2, bo for column= = 3.54 m

Equation 11-31, Vc1 = = 4097.06 kN

Equation 11-32, Vc2 = = 4229.96 kN

Equation 11-33, Vc3 = = 2651.20 kN

Punching shear strength, Vc = 0.75 X minimum of (Vc1, Vc2, Vc3) = 1988.40 kN

0.75 X Vc > Vu hence, OK

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One-Way Shear Along X

(Shear Plane Parallel to Global X Axis)

X 1.7 m

Z 1.07 m

1.7 m 1.07 m

Plan
From ACI Cl.11.2.1.1, Vc = = 1285.73 kN

Distance along Z to design for shear, Dx = = 1.07 m

Check that 0.75 X Vc > Vux where Vux is the shear force for the critical load cases at a distance d from the face of the column caused by bending about the X axis.

From above calculations, 0.75 X Vc = 964.29 kN

Critical load case for Vux is # 201 = 2.52 kN

0.75 X Vc > Vux hence, OK

One-Way Shear Along Z

(Shear Plane Parallel to Global Z Axis)

X 1.7 m

1.06 m 1.06 m
1.7 m

Plan
From ACI Cl.11.2.1.1, Vc = = 1340.44 kN

Distance along X to design for shear, Dz = = 1.06 m

Check that 0.75 X Vc > Vuz where Vuz is the shear force for the critical load cases at a distance d from the face of the column caused by bending about the Z axis.

From above calculations, 0.75 X Vc = 1005.33 kN

Critical load case for Vuz is # 201 = 0.00 kN

0.75 X Vc > Vuz hence, OK

Flexure About Z-Axis

Design For Bottom Reinforcement Parallel to X Axis

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30 - ϕ16

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 # 201

The strength values of steel and concrete used in the formulae are in ksi

Bars parallel to X Direction are placed at bottom

Effective Depth d = 0.39 m
Factor from ACI Cl.10.2.7.3 =
= 0.80

From Appendix B 8.4.2, = = 0.03317

From ACI Cl. 10.3.3, = = 0.02488

From ACI Cl. 7.12.2, = = 0.00180

From Ref.1, Eq. 3.8.4a, constant m = = 14.12

Calculate reinforcement ratio for critical load case

Design for flexure about Z axis is performed at the face

= 1.45 m
of the column at a distance, Dx =

Ultimate moment = = 0.00 kNm

Nominal moment capacity, Mn = = 0.00 kNm

(Based on effective depth) Required = = 0.00000

(Based on gross depth) x d / Depth = 0.00000

Since ρ < ρmin, select ρ= ρmin ρmin Governs
Area of Steel Required, As = = 2754.01 mm2

Selected bar Size = 16

Minimum spacing allowed (Smin) = 116.00mm
Selected spacing (S) = 116.00mm
Smin<= S <= Smax and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318 Clause No- 10.6.4 1st
Max spacing for Cracking Consideration = 254.89mm

Safe for Cracking Aspect.

Based on spacing reinforcement increment; provided reinforcement is

#16 @ 115mm o.c.

Required development length for bars = = 0.55 m

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Available development length for bars,DL = = 1.40 m

Try bar size # 16 Area of one bar = 201.06 mm2

Number of bars required, Nbar = = 30

Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax

Total reinforcement area, As_total = Nbar X (Area of one bar) = 6031.92 mm2
d= D - Ccover - 0.5 X (dia. of one bar) = 0.39 m

Reinforcement ratio, = = 0.00453

From ACI Cl.7.6.1, minimum req'd clear distance between bars

Cd = max (Diameter of one bar, 1.0" (25.4mm), Min. User Spacing) = 116.00mm
Provided Steel Area / Required Steel Area = 2.19

Flexure About X-Axis

Design For Bottom Reinforcement Parallel to Z Axis

30 - ϕ16

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 # 201

The strength values of steel and concrete used in the formulae are in ksi

Bars parallel to X Direction are placed at bottom

Effective Depth d = = 0.38 m
Factor from ACI Cl.10.2.7.3 =
= 0.80

From Appendix B 8.4.2, = = 0.03317

From ACI Cl. 10.3.3, = = 0.02488

From ACI Cl.7.12.2, = = 0.00180

From Ref.1, Eq. 3.8.4a, constant m = = 14.12

Calculate reinforcement ratio for critical load case

Design for flexure about X axis is performed at

= 1.45 m
the face of the column at a distance, Dz =

Ultimate moment = = 2.64 kNm

Nominal moment capacity, Mn = = 2.94 kNm

(Based on effective depth) Required = = 0.00001

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(Based on gross depth) x d / Depth = = 0.00001

Since ρ < ρmin, select ρ= ρmin ρmin Governs

Area of Steel Required, As = 2754.01 mm2

Selected Bar Size = #16

Minimum spacing allowed (Smin) = 116.00mm
Selected spacing (S) = 116.00mm
Smin<= S <= Smax and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318 Clause No- 10.6.4 3rd
Max spacing for Cracking Consideration = 254.89mm

Safe for Cracking Aspect.

Based on spacing reinforcement increment; provided reinforcement is

#16 @ 115mm o.c.

Required development length for bars

= 0.55 m
=

Available development length for bars,

= 1.40 m
DL =
Try bar size # 16 Area of one bar = 201.06 mm2

Number of bars required, Nbar= = 30

Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax

Total reinforcement area, As_total = Nbar X (Area of one bar) = 6031.92 mm2
d= D - Ccover - 1.5 X (dia. of one bar) = 0.38 m

Reinforcement ratio, = 0.00472

From ACI Cl.7.6.1, minimum req'd clear distance between bars

Cd = max (Diameter of one bar, 1.0" (25.4mm), Min. User Spacing) = 116.00mm
Provided Steel Area / Required Steel Area = 2.19

Pedestal Design
Pedestal Reinforcement Arrangement : 4 Faces

Calculation Of Shear Capacity

#10 @ 200 mm #16 - 12

Design Passed in Shear

Critical Load Case For Shear Along X = 226

Critical Load Case For Shear Along Z = 201
Shear along X = 0.00 kN
Shear along Z = 4.06 kN
(ϕ = 0.75, ref ACI 318 Clause No 9.3.2.3)

Shear Capacities
ϕVcX = 150.68 kN

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(ϕVcX ref eqn ACI 318-11 Eqn 11.4 and Clause no 11.2.1.2)

ϕVsX = 98.96 kN

(ϕVsX ref eqn ACI 318-11 Eqn 11.15 and Clause No 11.4.7.2)

ϕVcZ = 150.79 kN

(ϕVcZ ref eqn ACI 318-11 Eqn 11.4 Clause no 11.2.1.2)

ϕVsZ = 98.96 kN

(ϕVsZ ref eqn ACI 318-11 Eqn 11.15 and Clause No 11.4.7.2)

Tie bar for shear X = 10

Tie bar for shear Z = 10
Tie bar provided = 10
Legs for resisting FX = 2
Legs for resisting FZ = 2
(Spacing Calculation ref Clause no 11.4.5.1 and 11.4.6.3 and Eqn 11.13)
Tie spacing required to resist FX = 0.20 m
Tie spacing required to resist FZ = 0.20 m
Spacing provided = 0.20 m

Note- Actual stirrup provided is dependent upon design requirement and bar binding and positioning. So Stirrup provided in detail drawing may vary from designed stirrup
requirements.

Critical Load Case Based on

201
Main Reinforcement Design :

Check for Pedestal Strength Against Axial Load

Factored Axial Load acting on Pedestal = -29.69 kN

Gross Section Area = 249999.99 sq.mm

(Area Calculated Considering Minimum Reinforcement)
Required Area of steel to resist Axial Force = 2000.00 sq.mm
Provided Steel Area = 2412.77 sq.mm
Provided Steel Area > 2000.00sq.mm,Hence Safe

Strength and Moment Along Reinforcement in X direction

Axial Load [kN]

1000
Factored Moment: 12.8
0 Factored Axial Load: 29.69
0 90 180 270 360 450 549.166377
540
Tension Control Zone
Moment - X [kNm]
Axial Capacity

Note: Factored pedestal design loads (inclusive of Φ = 0.00 based on ACI 318 Chapter 9)

Factored Moment = 12.80 kNm

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Factored Axial Load = -29.69 kN

Bar size = = 16 mm
Number of Bars = 12
Required Steel Area = 1999.9999 sq.mm
Provided Steel Area = 2412.7692 sq.mm

Balanced Neutral Axis Depth = = 0.2453 m

Actual Neutral Axis Depth (Xb) = = 0.0703 m

Strength and Moment from Concrete

Cc = = 833.16 kN

Mc = = 184.95 kNm

Calculate strength and moment from one bar.

Distance between extreme fiber and bar (db) = 0.09 m

Strain in bar, = = -0.0008

Maximum Strain, = = 0.0021

as

= -159026.64 kN/m2

= 0.0021

as

= 0.00 kN/m2

Tension Resisted by Bars = = -31.97 kN

Moment Resisted by Bars = = -5.15 kNm

Total Bar Capacity,Cs= (Bar Stress-Concrete Stress)X Bar Area = -803.47 kN

Capacity of Column = Capacity of

effective concrete+ Capacity of steel = 29.69 kN
= Cc + Cs
Total Moment Capacity provided by (Individual Bar Capacity X Moment Arm) = 33.79 kNm
bars,Ms =
Total Moment, Mox = Mc + Ms = 218.75 kNm

Check for uni-axial bending, = 0.06

Design Moment Mnx = 12.80 kNm

Total Moment Mox = 218.75 kNm
if Mnx or Mnz = 0, then a = 1.0
otherwise, a = 1.5

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30/11/2017
X 1.7 m
2.04 m
Z

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0.5 m 1.3 m 1.75 m
3.4 m

0.5 m
1.7 m

3.4 m
0.45 m
Plan

Print Calculation Sheet

1.7 m
Elevation
BOTTOM 30 - ϕ16
Dimensions (m) Reinforcement

TOP N/A
L B D AstX(T) AstZ(T) AstX(B) AstZ(B)
Foundation
1.7 m

3.4 m 3.4 m 0.45 m N/A N/A 30 - ϕ16 30 - ϕ16

l b h Main Reinforcement Trans Reinforcement
TOP N/A Pedestal
0.5 m 0.5 m 1.3 m #16 - 12 #10 @ 200 mm
BOTTOM 30 - ϕ16
Plan