Combined Footings FPS System

References: ACI Code / Structural Concrete, Theory and Design (Copyright 1998), By M. Nadim Hassoun

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FoundationsBefore starting with the foundations design, a soil investigation has to be performed to check thesoil characteristics and propertires. The site soil must be checked if its capable to hold the loadtransferred to it through the footings (Shear test, Ultimate Bearing Capacity "qu", etc...).The Bearing Capacity is required for designing the foundation. If the soil investigation indicatespoor ground conditions, (Such as the bearing capacity is less than the allowable), A GroundImprovement technique must be performed to treat and improve the soil that support the footingand make it safe to build with.

Common Solutions & Ground Improvement Techniques:

Grouting, Dewatering, Soil Reinforcement (Geograde, Geogrids, etc...), Piles,Dynamic Compaction, Vibroflotation, Vibro-Compaction, Snad Drains, Soil Replacement.

Most Common Types of Soil Prone to Improvement TechniquesCollapsible SoilExpansive SoilSanitary LandfillsKarastic Profile

Designing NotesIt is preferable to check the sizes or dimensions of each footing before starting designing them.This must be done to choose the required and preferred system of footing (Single, Combined, etc.).A fast calculation can be performed using the following formula:Service Load / StressWhere the Service load can be taken = 1.5 of the Ultimate Load that is being transferred fromThe Columns. And the Stress is the one that is provided from the soil tests (i.e. 3 Kg/cm²)

ExamplePu = 250 Ton Area of footing = (250,000/1.5) / 3 = 55555.55 cm² qu = 3 kg/cm² or A = 5.55 m²

Grouting

Geogrids

Piles

Dynamic Compaction Vibro-Compaction

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Project NameSubject Design of Combined Footing

YOPUR LOGOElement Number F1Done ByDate 1/23/2015 6:16

DESIGN OF COMBINED FOOTING (ACI Strength Design Method)

fc' = 4 Ksify = 60 Ksi

Concrete Density = 150 16" 20"Weight of Soil = 100 16" I II 20"

Column Dimentions:16 x 16 in20 x 20 in

Column Spaces Center-Center = 16 ft 2' 16' 4'Space from Property line to center of 22'

2 ft Property LineResultant

Loads Carried By Columns:D.L. = 180 Kips 2' 9'L.L. = 120 Kips

D.L. = 250 KipsL.L. = 140 Kips

Allowable Soil Pressure = 5 KsfBottom of footing below the final grade = 5 ft

Resultant of the column loads location

x = 9The distance of the resultant from the properly line is 11 ft

The length of the footing = 22 ft

Area of the footingAssume the footing total depth = 36 in Assume Cover = 4.5 in

d = 31.5 inTotal actual (working) loads = 690 Kips

Net upward pressure = 4350 psfRequired Area = 158.6207

Width of footing = 7.210031 ft Use 7.3 ftThen, Area of the footing = 160.6

Ultimate upward pressure using ultimate loads456 Kips588 Kips Net ultimate upward pressure qu = 6.501 Ksf

qu = 47.4573Factored shearing force

63.2764 Kips329.4472 Kips150.2815 Kips358.6231 Kips

x = 6.942582 ft

lb/ft³lb/ft³

Column IColumn II

Column I

Column I

Column II

Taking moments about the center of the exterior column I:ft from Column I

ft²

ft²

Pu1 (Column I) =Pu2 (Column II) =

K/ft (per foot length of footing)

Vu (at outer face column I ) =Vu (at interior face column I ) =Vu (at outer face column II ) =

Vu (at interior face column II ) =

Point of zero shear, x: distance between interior faces of columns I and II is:

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Factored moment42.18427 K.ft237.9456 K.ft

The maximum moment occures at zero shear:1278.7691290.685 1290.685 K.ft

Checking the depth for one-way shearMaximum shear occures at a distance "d" from the

31.5 in 16" 20"7.3' 16" I II 20"

Vu1 = 234.0476 Kipsd = 24.84971 in

The effective depth provided = 31.5 inWhich is > 24.85"

Thus, Footing is adequate 2' 16' 4'22'

Checking the depth for two-way shear 358.6231 KFor the interior column,

(c + d) = 4.291667 ft 63.2764 Kbo = 17.16667 ft 6.94' 7.56'

The shear at a section d/2 from all sides of the column:Vu2 = 468.262 Kips

d = 10.57092 in 150.2815The effective depth provided = 31.5 in 329.4472 K Kips

Which is > 10.57" 1290.685 K.ftThus, Footing is adequate 1.33'

Mu (max) 3.17'Checking the depth for moment & Determiningthe required reinforcement in the long direction

Maximum bending moment = 1290.685 K.ft 42.18427 K.ftRu = 178.1871 psi 237.9456

0.003609 0.003333 ρ > ρ(min) O.K. K.ft

As = 9.958058Min. As (Shrinkage) = 5.67648 9.958058

Insert bar No. = 9 Concrete Cover = 6 in Then bar diameter = 1.128 in

Use 10 of Bar No. 9 Spacing = 9.066667 in

The bars are extended between the columns at the top of the footing with a concrete cover 3 in. Placeminimum reinforcement at the bottom of the projecting ends of the footing beyond the columns to take careof the positive moments. Extend the bars a development length ld beyond the side of the column.The minimum shrinkage reinforcement As = 5.676 in2.

Insert bar No. = 8 Concrete Cover = 6 in Then bar diameter = 1 in

Use 8 of Bar No. 8 Spacing = 11.65714 inThe development length required for the main top bars is 1.3 ld = 70.2

Mu1 (at outer face column I) =Mu2 (at outer face column II) =

Maximum Mu (Calculated from column I side) = K.ft Then:Maximum Mu (Calculated from column II side) = K.ft Mu (max) =

interior face of column II d =

ρ = ρ(min) =

in²in² Hence, As = in² Controls

in Beyond the point of maximum

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YOPUR LOGOElement Number F1Done ByDate 1/23/2015 6:16moment. Development length provided for both columns are adequate.

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Reinforcement in short directionThe reinforcement under each column is to be placed within a maximum bandwidth equal to the column widthplus twice the effective depth "d" of the footing.

16"x16" 20"x20"Bandwidth = 6.6 ft 7.3'

Net upward pressure in the short direction under I62.46575 K/ft

Distance from the free end to the face of the column= 2.983333 ft

277.9813 K.ft 6.6' 7'Ru = 42.44734 psi 22'

0.000839As = 2.091901 Min. As (Shrinkage) = 5.13216

Hence, As (Control) = 5.13216



Bandwidth = 7 ft80.54795 K/ft

Distance to face of column = 2.816667 ft319.518 K.ft

Ru = 47.97346 psi0.000949

As = 2.509985 Min. As (Shrinkage) = 5.4432Hence, As (Control) = 5.4432



2.983' 16" 2.983' 2.817' 20" 2.817'

8 # 8 7 # 8 8 # 8 7 # 83' 3'

7.3' 7.3'

ShrinkageSteel

1.333' 16" 20" 3.167'

7 # 8 8 # 8 10 # 9 bars 8 # 8 7 # 8 3'

a. Reinforcement under the exterior column I

IIUnder column I =

Mu (at face of column I) =

ρ =in² in²

in²

b. Reinforcement under the interior column II

Net upward pressure in the short direction under under column I =

Mu (at face of column II in short direction) =

ρ =in² in²

in²

Exterior Footing I Interior Footing II

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22'

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Price: 10 US $

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Combined Footings FPS System