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Page 1: square footings

Foundation Engineering

Foundation EngineeringFoundation Engineering

Prof. Mesut PervizpourProf. Mesut Pervizpour

Structural Design of Shallow Foundationsg

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Spread Footing, Structural Design• Introduction (Types of foundation, Design steps, Footing size, Materials, Loads Reduction factors Cover requirements )Loads, Reduction factors, Cover requirements )

• Structural Design (Two-way shear, One-way shear, Flexure, Development lengths, Dowels, Load transfer)

• Continuous Footing• Rectangular Footing• Examples (square wall rectangular combined eccentricity)Examples (square, wall, rectangular, combined, eccentricity)

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Loads acting on foundationLoads acting on foundation

TT

N l l d

Torsion loadsTT

PP Normal loads(compression-tension, é)

PP

MM Moment loads(in å or ç directions)

MM

VV Shear loads(in å or ç directions)

VV

3Figure 2.1 Coduto

Design LoadsDesign LoadsDead Loads (D)

Allowable Stress Design (ASD)Allowable Stress Design (ASD)Estimates of most critical actual service loads

geotechnical( )

Live Loads (L)

Snow/Rain Loads (S/R)

geotechnicalEx: ANSI/ASCE (Use the largest of design load:)

DD + L + F + H + T + (Lr or S or R)D + L + (L or S or R) + (W or E)Snow/Rain Loads (S/R)

Earth Pressure Loads (H)

Fluid Loads (F) Allowable = ultimate / FS > design load OK

D + L + (Lr or S or R) + (W or E)D + (W or E)

Alternative method for W & E (pg19 Coduto)

( )

Earthquake Loads (E)

Wind Loads (W)

Load & Resistance Factor Design (LRFD)Load & Resistance Factor Design (LRFD)Ultimate strength design, use of factored load (U):

Structural( )

Self-straining Loads (T)

Impact Loads (I)

Load factors: Pu = 1 PD + 2 PL + ….Resistance factor: Pu Pn

nominal load capacity

Ex, ACI 318-05 (Ch9): U = 1.4(D + L)

Stream Flow/Ice Loads (SF/ICE)

Centrifugal/Braking Loads (CF/BF)

U = 1.2(D +F+T )+ 1.6(L+H)+0.5(LR or S or R)U = 1.2D + 1.6(Lr or S or R) + (L or 0.8W)U = 1.2D + 1.6W + L + 0.5(LR or S or R)Or Use ACI 318-05 Appendix C Load factor and

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strength reduction factors BUT …..ANSI/ASCE: U = 1.4D (eqns on pg 23 Coduto)

AASHTO: U = 1.25D + 1.75L (bridges)

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Load & Resistance Factor Design (LRFD)Load & Resistance Factor Design (LRFD)Ultimate strength design, use of factored load (U): Structural

Load factors: Pu = 1 PD + 2 PL + ….Resistance factor: Pu Pn nominal load capacity

ACI 318-05 (Ch9): ACI 318-05 (Appendix C 1999):

U = 1.4(D + L)

U = 1.2(D +F+T )+ 1.6(L+H)+0.5(LR or S or R)

U = 1.2D + 1.6(Lr or S or R) + (L or 0.8W)

U 1 2D 1 6W L 0 5(L S R)

U = 1.4D + 1.7L

U = 0.5(1.4D + 1.4F + 1.7T )

U = 0.9D + 1.4F

U 1 4D 1 7L 1 4FU = 1.2D + 1.6W + L + 0.5(LR or S or R) U = 1.4D + 1.7L + 1.4F

U = 1.4D + 1.7L + 1.7H ….. other

factors:

Fl 0 90

Pu = U = 1.2 D + 1.6 L (ACI)

factors:

Fl 0 90

Flexure = 0.90

Shear = 0.85

Bearing on Concrete = 0.70

Unreinforced footings = 0 65

Note: AASHTO Pu =1.25D + 1.75L

Flexure = 0.90

Shear = 0.75

Bearing on Concrete = 0.65

Unreinforced footings =

If you use ACI 318-05 Appendix C Load factor then use the same strength reduction factors together and do not mix with

Unreinforced footings = 0.65

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Unreinforced footings = factors together and do not mix with Chapter 9 numbers.

Types of Foundations Isolated Spread footingfooting

Combined footing

Property line

C til

Wall (continuous) footingCantilever

(Strap)footing

footing

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Types of Foundations A BCombined footings

Isolated spread footing

Wall footing

Rectangular, PA = PB

Property line

A B

Rectangular, PA < PB

A B

A B

Property line

Mat Footing Pile cap

Rectangular, PA < PB

A B

Property linePiles

Weak soil

7Strap or Cantilever

Bearing stratum

Design Considerations

Footings must be designed to carry the column loads and transmitFootings must be designed to carry the column loads and transmit

them to the soil safely while satisfying code limitations.

1. Area of the footing is based on the soil allowable bearing capacity

2. Footing thickness is based on:

• Two-way shear or punch out (diagonal) shear

• One-way (wide beam) shear

3. Steel reinforcement is based on flexure design at critical section

4. Bearing capacity of columns at their base (load transfer to footing)

5. Dowel requirements

6. Development length of bars

7. Differential settlement

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The area of footing can be determined from the actual external loads such

Footing Size

The area of footing can be determined from the actual external loads such that the allowable soil pressure is not exceeded.

weightselfincludingloadTotalf ifA

pressure soil allowable

ggfootingof Area

WLLDL

Strength design requirements (structural design):

all

f

q

WLLDL Area

f tif

uu

Pq

Strength design requirements (structural design):

the bearing pressure at the base of footing

footing ofareaLLDL Pu 7141 .. ACI 05 - Appendix C

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LLDL 6.12.1 Pu ACI 05 - Chapter 9

Structural DesignMaterial Selection:

Concrete compressive strength (f’c): 2000 – 3000 psi (15 – 20 Mpa)Concrete, compressive strength (f c): 2000 3000 psi (15 20 Mpa)Steel, Grade 40 (fy): 40,000 psi (300 Mpa)

Grade 60 (fy): 60,000 psi (420 Mpa)

Modes of failure: shear and flexureModes of failure: shear and flexure

Design Loads:

Factors:

Factored normal load (ACI) (appendix C), Pu = 1.4 D + 1.7 LFactored normal load (ACI-05) (chapter 9), Pu = 1.2 D + 1.6 L

Factors:Flexure = 0.90 Bearing on concrete = 0.65 (0.70)Shear = 0.75 (0.85) Unreinforced footings = (0.65)

Values in parenthesis are to be used with Appendix C equations.

Minimum cover requirements and standard dimensions:

dEffective depth, d:

d T 3i d 6 i

3 in

db

dT d = T – 3in – db 6 in

Minimum thickness, T:

T 12 in (300 mm)ACI 15.7

10

3 in(70 mm)Flexural Steel

db: nominal diameter of reinforcing steel

T 12 in (300 mm)

Round T to a multiple of 3 in or 100 mm(12, 15, 18 .. in , or 300, 400, 500 mm)

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Reinforced Footing Design:Initial design for shear: find the initial estimate for effective depth, d

No shear reinforcement in footings

Shear resistance provided only by the depth of concrete above steel

One way (beam shear) or Two way (diagonal tension) shear:One-way (beam shear) or Two-way (diagonal tension) shear:

Two-way shear governs when subjected to vertical loads alone for

square footings, Check both two-way and one-way shear for applied

shear/moment loads

Designing for Shear: dOne-way Sheard/2

d/2Two-way Shear

Vuc Vnc

T

Criteria to find “d”

Vuc: factored shear force on critical surface

: resistance factor for shear = 0.75 (App. C=0.85)

Vnc = Vc + Vs dT

dVnc: nominal shear capacity on the critical surface

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Vc: nominal shear load capacity of concreteVs: nominal shear load capacity of reinforcing steel (neglected)

Reinforced Footing Design:Two-way (punching) shear design d/2

d/2Two-way Shear

P

T

d/2Pu

d

T

c1 + d

d/2

c2c2 + d

b0

12c1

b0

Page 7: square footings

Reinforced Footing Design:Two-way (punching) shear design

Vuc Vnc

Assume d.

Determine b0.

b 4( +d) for square columns

1.

2.

b0 = 4(c+d)

b0 = 2(c1+d) +2(c2+d)

for square columns where one side = c

for rectangular columns of gsides c1 and c2.

Shear force Vuc acts at a section that has a length b 4( +d) 2( +d) +2( +d) d d th d

3.

bo

b0 = 4(c+d) or 2(c1+d) +2(c2+d) and a depth d;

the section is subjected to a vertical downward load Pu and vertical upward pressure qu. u p p qu

2

u u uV P q c d

V P q c d c d

for square columns

for rectang lar col mns

c

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u u u 1 2V P q c d c d for rectangular columnsc

Reinforced Footing Design; Vuc Vnc Criteria to find “d”

Two-way (punching) shear design

For “square footings supporting square or circular columns on the interior”, the

nominal two-way shear capacity [ACI 11 12 2 1]:

4. Calculate nominal shear capacity on critical surface

nominal two way shear capacity [ACI 11.12.2.1]:

'4 0 ccnc fdbVV 1

(English Units)Nominal two-way

:b (length of critical shear surface)

'3

10 ccnc fdbVV (SI Units)

shear capacity

:0b (length of critical shear surface)

fc’: 28 – day compressive strength of concrete (psi, MPa)

Vx: in lb or NFor columns with other shapes and located

c, d, b0: in inches or mmFor columns with other shapes and located out of center ACI 11.12.2.1 (next page)

Vuc = Vnc Solve for “d”5. Set

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uc nc

If d is not close to the assumed d, revise your assumptions

Page 8: square footings

G l th i l t h it [ACI 11 12 2 1]

Vuc Vnc Criteria to find “d”Reinforced Footing Design; Two-way (punching) shear designGeneral, the nominal two-way shear capacity [ACI 11.12.2.1]:

Nominal two-way shear capacity (Smallest Value for Vnc calculated below)

(English Units) (SI Units)

'4 0 ccnc fdbVV (English Units)

4

'3

10 ccnc fdbVV

(SI Units)

21 '

42 0 c

ccnc fdbVV

'2

16

10 c

ccnc fdbVV

'c

scnc fdb

b

dVV 0

0

2

'cs

cnc fdbb

dVV 0

0

212

1

Vnc: nominal shear capacity (one critical surface lb/N)

b0: length of critical surface (in/mm)

d: effective depth (in/mm) : long side / short side (c / c ) of column

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p ( )

c: column width (in/mm)c: long side / short side (cl / cs) of column

s: interior = 40, edge = 30, corner = 20 (columns)

Reinforced Footing Design; One-way (wide beam) shear design

Vuc Vnc

F f ti ith b di ti i di ti th iti l ti i l t d

dOne-way Shear

For footings with bending action in one direction the critical section is located at a distance d from face of column.

1 A d1. Assume a d.

2. bo = b.

dT

The ultimate shearing force at section m-m can be calculated

cL b =

3.

d

cLbqV o 22 uu

If no shear reinforcement is to be sed then d

bo

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If no shear reinforcement is to be used, then d can be checked

Page 9: square footings

Vuc Vnc Criteria to find “d”Reinforced Footing Design; One-way (wide beam) shear design

Nominal one-way shear load capacity [ACI 11.3.1.1]:

'2 fdbVV (E li h U it )Nominal

4. Calculate nominal shear capacity on critical surface

2 cocnc fdbVV

'1

fdbVV

(English Units)

(SI Units)

one-way

shear load

capacity6 cocnc fdbVV (SI Units) capacity

Vnc: nominal one-way shear capacity on the critical section (lb/N)

b : length of one way critical shear surface (in/mm)bo: length of one-way critical shear surface (in/mm)

d: effective depth (in/mm)

fc’: 28-day compressive strength of concrete (psi/MPa)

The larger of effective depth d obtained from two way and one way

Vuc = Vnc Solve for “d”5.

If d is not close to the assumed d, revise your assumptions

Set

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The larger of effective depth, d, obtained from two-way and one-way capacity calculations is used for design purposes.

Procedure in determination of effective depth, d

Reinforced Footing Design;

• Assume a trial d (~ column width, note: T multiple of 3 in)

• Compute Vuc and Vnc and check

• Repeat first two steps to find smallest d, satisfying the shear criteria

Vuc Vnc

p p , y g

• Compute footing thickness T (using a db = 1 in)

Actual value of d will be determined from the flexural analysis

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Actual value of db will be determined from the flexural analysis

Page 10: square footings

Critical Sections

Design for Flexure (As)

Reinforced Footing Design; Muc MncFlexural design criteria:

= 0.90 for flexure in RCObtain size of member and amount and location of reinforcing steel.

b 0.85fc’

a/2

dNeutral Axis

c a = 1 c C

a/2jd

= dAs(min)

Grade 40 Neutral Axis(Axis of Zero Strain)

Asf T

d –

a/2

Grade 40As 0.002 AgGrade 60As 0.0018 Ag

Asfy T

2

adfAM ysn

As: steel area: steel ratio

0F

fAabf '850nu MM

'850y

f

fda

b: width ysc fAabf '.850

db

As '59.019.0 cyysuc ffdfAM

Calculate Muc and solve equation for As

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'85.0 cfdb

bf

Mdd

f

bfA

c

uc

y

cs '

353.2

176.1

' 2

Critical Sections

Design for Flexure (As)

Reinforced Footing Design; Muc Mnc

Calculate Muc at the critical section for moment

Location of critical section for bending (l cantilever distance):

c c cc

FlexuralSteel c/4 cp

lB

lB

l = (B c/2) / 2

lB

l = (2B (c + c )) / 4

Concrete Columnl = (B – c) / 2

Masonry Columnl = (B – c/2) / 2

Steel Columnl = (2B – (c + cp )) / 4

for circular/octagonal column use an equivalent area square column

lc Where Muc at the critical section is:

2

2

1lqM uuc

Care to be taken that this is moment per width, for instance for a square footing with width B, the moment is:

20qu

2 uuc2

2

1lBqM uuc

Page 11: square footings

Limiting As:

Reinforced Footing Design;

Minimum steel reinforcement in footing:ACI [10.5.4 and 7.12.2]

Grade 40 steel As 0.0020 Ag

Grade 60 steel As 0.0018 Ag

The minimum steel percentage () required in flexural members is 200/fy with minimum area and maximum spacing of steel bars in the direction of bending shall be as required for shrinkage temperature reinforcement (10.5.4).

Rebar sizing criteria:

• Clear space between bars, the larger of: “db”, or “4/3 times the nominal maximum

aggregate size”

• Center to center spacing of reinforcement less than 3T or 18 in (whichever less)

Rebar sizing criteria:

• Center-to-center spacing of reinforcement less than 3T or 18 in (whichever less)

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Design Data for Steel Reinforcing Bars:

Reinforced Footing Design;

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Page 12: square footings

Reinforced Footing Design;

Development Length of Flexural Steel: ld ld-suppliedCriteria: d d suppliedAvailable (Supplied) development length (ld-supplied):

3 inld-supplied

ld-supplied = l – 3 in (70 mm)

If not satisfied, use smaller db bars

Where l, is the length of critical section for bending (cantilever segment)

lB

Critical section for bending

for bending (cantilever segment)

Transverse Reinforcement Steel Area:

A 2A

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Atr = 2Ab Atr = 4Ab

ACI [12 2 3]

ld ld-suppliedCriteria:

*At l t 2d d t i t l t d

Required Development Length ld for Bars in Tension:

Reinforced Footing Design;

ACI [12.2.3]

fl 3

(English units)

dfl 9

(SI units)

*At least 2db, and concrete spacing at least db

b

trc

y

b

d

dKcf

f

d

l '40

3

b

trc

y

b

d

dKcf

f

d

l '10

9

2 5 b

ns

fAK yttr

tr 1500 ns

fAK yttr

tr 10

2.5

ld: min. required development length (in, mm), at least 300mm (12 in)fyt: transverse reinforcing steel yield strength (psi, MPa)

For spread footings, use of Ktr = 0 is conservative

c: spacing or cover (in, mm), smaller of bar center-to-nearest concrete, 0.5 center-center bar spacing)Atr: total x-sec. area of all transverse reinforcement within spacing s, crossing potential

l f litti th h th d l d i f t (i 2 2) b t k

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plane of splitting through the developed reinforcement (in2, mm2), maybe taken zeros: max. center-to-center spacing of transverse reinforcement within ld (in, mm)n: number of bars or wires being developed along the plane of splitting

Page 13: square footings

ld ld-suppliedCriteria:

Required Development Length ld for Bars in Tension:

Reinforced Footing Design;

List of Modifiers for Development Length Calculations:

Modifier Value

1 3 horizontal (top) reinforcement(reinforcement location)

1.3 horizontal (top) reinforcement within 12 in of concrete

1.0 all other 1.7

(coating factor)

1.5 epoxy coated with spacing < 6db

or cover < 3db

1.2 epoxy coated bars or wires1 0 uncoated bars or wires

(reinforcement factor)

1.0 uncoated bars or wires

0.8 #6 bar and smaller, and deformed wire

1 0 #7 d l

(lightweight concrete factor)

1.0 #7 and larger

1.3 lightweight aggregate concrete1.0 Normal weight concrete

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(lightweight concrete factor) g

Design for Compressive Loads:

Load Transfer from Columns to Footings;

Nominal column bearing capacity

AfP '850 sAfP cnb 185.0Design criterion

nbu PP Pu = factored column load ( 1.2D + 1.6L ACI-05 )

A1 = cross-sectional area of the column = c2

s = (A2/A1)0.5 < 2 if c + 4d < B; otherwise, s = 1s (A2/A1) 2 if c 4d B; otherwise, s 1

A2 = (c + 4d)2

= 0.65

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Note: on column use column fc’, and on footing used footing fc’

Page 14: square footings

Column compressive forces transmitted to footing directly by bearingDesign for Compressive Loads:

Load Transfer from Columns to Footings;

Permitted bearing capacity at base of column:

Column compressive forces transmitted to footing directly by bearingUplift (tension) forces are transferred by developed reinforcing bars

Check the strength of the lower part of the column (only concrete):Permitted bearing capacity at base of column:

where A1 = loaded area (column area)1(0.85 )nb cP f A

A = 0.65A1

450 May be multiplied by for bearing on footing:1

2

A

A

2A2 measuredon this plane where A2 = area of lower base of the largest

21 1

1

(0.85 ) 2 (0.85 )nb c c

AP f A f A

A

1on this plane 2

pyramid cone contained within footing havingside slope 1 vertical to 2 horizontal (or area of portion of supporting footing that is geometrically

27

similar and concentric with column)

If bearing force (Pu) > the smallest of the allowable values carry excess by dowels

• Design for compressive loads

A minimum steel ratio = 0.005 of the column section (Ag) as compared to

Use of Dowels for Connection:

Connections with Superstructure

Wall steel

Design for compressive loads• Design for moment loads• Design for shear loads

( g) p = 0.01 as minimum reinforcement for the column itself. The number of dowel

Wall steel

Lap joint

Concrete ormasonrycolumn

bars needed is four, these may be placed at the four corners of the column. The d l b ll

Doweldowel bars are usually extended into the footing, bent at the ends, and tied to the main footingthe main footing reinforcement. The dowel diameter shall not exceed the diameter of the longitudinal

Use at least 4 dowels with As at least equal to column steel or 0.005 Ag (col area)

gbars in the column by more than 0.15 in.

28

g

Should not be greater than #11 bars (should have 90o hooks at bottom)ld of dowel (ACI 12.3)

Page 15: square footings

Reinforced Footing Design; Required Development Length ld for Bars in Compression:

l lC

absd ll yb

ybab fd

f

fdl 00030020 .

'.

ld ld-suppliedCriteria:

ld: min. required development length (in)lab: basic development length (in)d : nominal bar diameter (in)

cf

db: nominal bar diameter (in) s: Modifier

As-required / As-provided for excess reinforcement0 75 for spirally enclosed reinforcement0.75 for spirally enclosed reinforcement

fc’: 28 day compressive strength of concrete (psi)

SI units:

In column & footing: 0.0750.0043b y

d b y

c

d fl d f

f

SI units:

29

c

Design for Moment Loads:

Load Transfer from Columns to Footings;

Moment in column causes some dowels to act in tension.

Dowels must be embedded at least one development length in to footing.

'c

bdh

f

dl

1200 English

'c

bdh

f

dl

100 SI

lldh: development length for 90o hooks T ldh

70mm

12db

The development length might dictate the footing thickness.

The number and size of dowels should at least be as the steel in the column.

12db

30

Try using a smaller diameter dowel to reduce the development length

Page 16: square footings

Design for Shear Loads:

Load Transfer from Columns to Footings;

Shear (Vu) in column must be transferred to footing by dowels.

Footing and column poured separately (weak shear plane – cold joint)

Minimum required dowel steel area:

uVA

Vu: applied shear load

: 0.75 (0.85) for shear y

s fA ( )

: 0.6 (cold joint), 1.0 (cold joint but roughened)

Ultimate shear load Vu, can not exceed 0.2 f’c Ac.

Ac and f’c are column area and concrete compressive strength.

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Foundation Depth:

Embedment depth D:Dmin

(i )Load P(k)

Dmin

(i )Load

P/b (k/ft)

Square & Rectangular Continuous Footings

Embedment depth, D:

Accommodate footing thickness, T

D should be at least > 300mm (12”)

(in)P(k) (in)P/b (k/ft)0-65

65-140

12

18

0-10

10-20

12

18

Depth of frost penetration

Presence of expansive soil

Potential of scour

140-260

260-420

420-650

24

30

36

20-28

28-36

36-44

24

30

36Dmin

(mm)Load P(kN)

Dmin

(mm)Load

P/b (kN/m)

0-300 300 0-170 300

300-500

500-800

800-1100

400

500

600

170-250

250-330

330-410

400

500

600800 00

1100-1500

1500-2000

2000 2700

600

700

800

900

410-490

490-570

570 650

600

700

800

900

32

2000-2700

2700-3500

900

1000

570-650

650-740

900

1000

Page 17: square footings

Example 1 Square footing:21in x 21in square reinforced concrete column carries a vertical load of 380 k and a vertical live load of 270 k It is to be supported on a square spread footing that will be

ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution

vertical live load of 270 k. It is to be supported on a square spread footing that will be founded on a soil with an allowable bearing pressure of 6500 psf. The groundwater table is well below the bottom of the footing. Determine the required width, B, thickness, T, and effective depth, d.

Solution: Unfactored load: P = k

Depth of embedment for P = 650k D = in

W B2 D B2 3 * 150 450 B2Wf = B2 D c = B2 3 * 150 = 450 B2

Df

a uB

WPq

2 06500

450650000 2

B

uq

WPB f

B 06500 uq Da

B = ft (use in)

Factored load: P = 1 2 (380) + 1 6 (270) = kFactored load: Pu = 1.2 (380) + 1.6 (270) = k

Material properties: Use fc’ = 4000 psi, and fy = 60 k/in2

O l li d l d i ti l ( h li d t) l h k t h

33

Only applied load is vertical (no shear or applied moment) only check two-way shear.

Solution 1 (cont.): Two-way shear: bo = 4(c+d) =

bo =

ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution

bo = 4(c+d)Calculate qu = Pu / A = lbs / 1262

qu = psi

c=21inCalculate shear force acting on bo for a square footing:Vuc = Pu – qu (c + d)2

Vuc =

B = 10.5 ft = 126 in

uc

Vuc =

Two-way Nominal shear capacity for square footing:

'4 0 ccnc fdbVV 00044844 ,ddVnc

2012125221 ddVnc ,, 2759939,15 ddVnc

Set V = V =0.75V and solve for d:

34

Set Vuc Vnc 0.75Vnc and solve for d:

Page 18: square footings

Solution 1 (cont.):

Two-way shear:

815 d2 + 18 291 d 863 304 0 d2 + 22 44 d 1 059 27 0O

863,304 – 2,352d – 56 d2 = 15,939d + 759 d2

ACIACI--05 05 ChCh9 9 Factors SolutionFactors Solution

815 d2 + 18,291 d – 863,304= 0

ina

acbbd

2

27.059,11444.2244.22

2

4 22

d2 + 22.44 d – 1,059.27 = 0Or:

Use d = in T = d + db + cover = + 1 + 3 in

Flexural design: For a concrete column: incB

l 5522

21126

2.

2

2

1lBqM uuc

22

inlbMuc

050,724,92

5.5256126 2

'59.019.0 cyysuc ffdfAM

000,60

5901240006090 sAAM

000,4

,

2412659.0124000,609.0 s

suc AM

3,792.86 As2 - 1,296,000 As + 9,724,050 = 0

Or: 2 785632146934169341

35

Or:As

2 – 341.69 As + 2,563.78 = 02

2

68.72

78.563,21469.34169.341inAs

Solution 1 (cont.): Flexural design:Check limiting As: 0.0018 Ag = 0.0018 * B * T

ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution

g s g

= 0.0018 * * = 6.35 in2 As = 7.68 in2 OK

Use 12 # 8 bars: db = 1in, Ab = 0.79 in2 As = 12 * 0.79 = 9.42 in2 > 7.68in2 OK

61262 Use same reinforcement in

Development length: ld-supplied = l – cover = 52.5 – 3 = 49.5 in

Bar spacing: inn

erBspacing 910

112

6126

1

2.

cov

Use same reinforcement in other direction

trc

y

b

d

Kcf

f

d

l '40

3

525253053.

useKc tr

c = cover + 0.5db = 3 + 0.5 = 3.5 in

bd 2.5

5252531

... usedb

inld

ld

d 462814628462852

1111

0004

00060

40

3...

,min

d d

b 52000440 .,min

ld-supplied = 49.5 in > ld-min = 28.6 in OK Lay steel to 3in to cover.

36

Lay steel to 3in to cover.

Page 19: square footings

Solution 1 (cont.): Column Bearing (Load transfer):

sAfP b 1'850 s = (A /A )0 5 < 2 for c + 4d < B;

ACIACI--05 05 ChCh9 9 Factors SolutionFactors Solution

sAfP cnb 185.0 s = (A2/A1)0.5 < 2 for c + 4d < B;

21 + 4*24 = 117 < 126in

A2 = (c+4d)2 = 13,689 in2

s = (A2/A1)0.5 = (13,689 / 212)1/2 = 5.57 > 2Therefore use s = 2

lbsPnb 80099822210004850 2 ,,,. lbP 22094918009982650

nbu PP 000,888lbsPnb 22094918009982650 ,,,,.

Compression bearing OK

Dowels (column footing):Dowels (column-footing): As-min = 0.005 Ag = 0.005 * 212 = 2.205in2

Use 4 # 7 bars db = 0.875in and Ab = 0.6 in2 As = 0.6*4 = 2.4 in2 > As-min = 2.205 in2

OKDowel development length (in compression):Dowel development length (in compression):

yb

c

ybab fd

f

fdl 00030020 .

'.

000608750

Required ld = 16.60 in

37

infdinl ybab 7515000608750000300003060160004

000608750020 .,....

,

,..

Solution 1 (cont.): Dowel development length (in compression): Required ld = 16.60 in

Available length in footing:

ACIACI--05 Ch9 Factors Solution05 Ch9 Factors Solution

Available length in footing:Footing thickness – cover – 2 (footing bar diameter) – dowel diameterld-available = 28 in – 3 in – 2*1 in – 0.875 in = 22.125 in

l = 22 125” > l = 16 60” OKld-available = 22.125 > ld-required= 16.60 OK

4 # 721 in

Critical section for moment

T = 28 in d = 24 in

B =126 in12 # 8

38

Page 20: square footings

UsingACI – 05 Appendix 5 FactorsFor Solution of Same Examplep

39

Example 1 Square footing:21in2 square reinforced concrete column carries a vertical load of 380 k and a vertical live load of 270 k It is to be supported on a square spread footing that will be founded

ACIACI--05 Appendix C Solution05 Appendix C Solution

live load of 270 k. It is to be supported on a square spread footing that will be founded on a soil with an allowable bearing pressure of 6500 psf. The groundwater table is well below the bottom of the footing. Determine the required width, B, thickness, T, and effective depth, d.

Solution: Unfactored load: P = 380 + 270 = 650 k

Depth of embedment for P = 650k D = 36 in

W B2 D B2 3 * 150 450 B2Wf = B2 D c = B2 3 * 150 = 450 B2

Df

a uB

WPq

2 06500

450650000 2

B

uq

WPB f

B 06500 uq Da

B = 10.36 ft (use 10 ft 6in 126 in)

Factored load: P = 1 4 (380) + 1 7 (270) = 991 kFactored load: Pu = 1.4 (380) + 1.7 (270) = 991 k

Material properties: Use fc’ = 4000 psi, and fy = 60 k/in2

O l li d l d i ti l ( h li d t) l h k t h

40

Only applied load is vertical (no shear or applied moment) only check two-way shear.

Page 21: square footings

Solution 1 (cont.): Two-way shear: bo = 4(c+d) = 4 (21 + d)

bo = 84 + 4d

ACIACI--05 Appendix C Solution05 Appendix C Solution

bo = 4(c+d)Calculate qu = Pu / A = 991,000 lbs / 1262

qu = 62.421 psi

c=21inCalculate shear force acting on bo for a square footing:Vuc = Pu – qu (c + d)2

Vuc = 991,000 – 62.421 (21 + d)2

B = 10.5 ft = 126 in

uc

Vuc = 963,472.3 – 2,621.7d – 62.421 d2

Two-way Nominal shear capacity for square footing:

'4 0 ccnc fdbVV 00044844 ,ddVnc

2012125221 ddVnc ,, 22860206418 ddVnc ..,

Set V = V and solve for d:

41

Set Vuc Vnc and solve for d:

963,472.3 – 2,621.7d – 62.421 d2 = 18,064.2d + 860.2 d2

Solution 1 (cont.):

Two-way shear: 963,472.3 – 2,621.7d – 62.421 d2 = 18,064.2d + 860.2 d2

922 621 d2 + 20 685 9 d 963 472 3 0 d2 + 22 42 d 1 044 28 0O

ACIACI--05 05 Appendix C SolutionAppendix C Solution

922.621 d2 + 20,685.9 d - 963,472.3 = 0

ina

acbbd 9922

2

2804411442224222

2

4 22

..,..

d2 + 22.42 d - 1,044.28 = 0Or:

Use d = 23in T = d + db + cover = 23 + 1 + 3 27 in

Flexural design: For a concrete column: incB

l 5522

21126

2.

2

2

1lBqM uuc

22

inlbM uc

52016839102

55242162126 2

.,,..

'59.019.0 cyysuc ffdfAM

00060

5901230006090,sA

AM

000423126

5901230006090,

.,. ssuc AM

3,792.86 As2 - 1,242,000 As + 10,839,016.52 = 0

Or: 2 748572144632746327

42

Or:As

2 – 327.46 As + 2,857.74 = 02

2

9782

748572144632746327inAs .

.,..

Page 22: square footings

Solution 1 (cont.): Flexural design:Check limiting As: 0.0018 Ag = 0.0018 * B * T

ACIACI--05 Appendix C Solution05 Appendix C Solution

g s g

= 0.0018 * 126 * 27 = 6.124 in2 As = 8.97 in2 OK

Use 12 # 8 bars: db = 1in, Ab = 0.79 in2 As = 12 * 0.79 = 9.42 in2 > 8.97in2 OK

61262 Use same reinforcement in

Development length: ld-supplied = l – cover = 52.5 – 3 = 49.5 in

Bar spacing: inn

erBspacing 910

112

6126

1

2.

cov

Use same reinforcement in other direction

trc

y

b

d

Kcf

f

d

l '40

3

525253053.

useKc tr

c = cover + 0.5db = 3 + 0.5 = 3.5 in

bd 2.5

5252531

... usedb

inld

ld

d 462814628462852

1111

0004

00060

40

3...

,min

d d

b 52000440 .,min

ld-supplied = 49.5 in > ld-min = 28.6 in OK Lay steel to 3in to cover.

43

Lay steel to 3in to cover.

Solution 1 (cont.): Column Bearing (Load transfer):

sAfP b 1'850 s = (A /A )0 5 < 2 for c + 4d < B;

ACIACI--05 05 Appendix C SolutionAppendix C Solution

sAfP cnb 185.0 s = (A2/A1)0.5 < 2 for c + 4d < B;

21 + 4*23 = 113 < 126in

A2 = (c+4d)2 = 12,769 in2

s = (A2/A1)0.5 = (12,769 / 212)1/2 = 5.38 > 2Therefore use s = 2

lbsPnb 80099822210004850 2 ,,,. lbP 22094918009982650

nbu PP 000,991lbsPnb 22094918009982650 ,,,,.

Compression bearing OK

Dowels (column footing):Dowels (column-footing): As-min = 0.005 Ag = 0.005 * 212 = 2.205in2

Use 4 # 7 bars db = 0.875in and Ab = 0.6 in2 As = 0.6*4 = 2.4 in2 > As-min = 2.205 in2

OKDowel development length (in compression):Dowel development length (in compression):

yb

c

ybab fd

f

fdl 00030020 .

'.

000608750

Required ld = 16.60 in

44

infdinl ybab 7515000608750000300003060160004

000608750020 .,....

,

,..

Page 23: square footings

Solution 1 (cont.): Dowel development length (in compression): Required ld = 16.60 in

Available length in footing:

ACIACI--05 Appendix C Solution05 Appendix C Solution

Available length in footing:Footing thickness – cover – 2 (footing bar diameter) – dowel diameterld-available = 27 in – 3 in – 2*1 in – 0.875 in = 21.125 in

l = 21 125” > l = 16 60” OKld-available = 21.125 > ld-required= 16.60 OK

4 # 721 in

Critical section for moment

T = 27 in d = 23 in

B =126 in12 # 8

45