# Structural Design of Footing

**Critical section for moment**

**Distribution of Flexural Reinforcement**

**Distribution of Flexural Reinforcement**

**SHEAR IN FOOTINGS**

**Punching Shear**

Beam Shear

### One-Way Shear/ Beam Shear

Vu ≤ ∅ Vn

≤∅(2√(fc) bw d)

**Two way shear/ Punching shear**

Vu ≤ =minimum

(2+( 4)/( β_c ) )=√(f_c) b d

((a_{s}d)/( b) +2) √(fc) bd

4 √(fc) bd

βc = ratio of long side to short side of the column, concentrated load or reaction area

αs = 40 for interior columns

= 30 for edge columns

= 20 for corner columns

bo = perimeter of critical section

### Area of Footing

Determine the base area Af required for a square spread footing with the following design conditions:

Service dead load = 350 kips

Service live load = 275 kips

Service surcharge = 100 psf

Assume average weight of soil and concrete above footing base= 130 pcf

Permissible soil pressure = 4.5 ksf

The base area of the footing is determined using service (unfactored) loads with the net permissible soil pressure.

Total weight of surcharge = (0.130 × 5) + 0.100 = 0.750 ksf

Net permissible soil pressure = 4.5 - 0.75 = 3.75 ksf

Required base area of footing:

A_{f} = (350 + 275)/ 3.75

= 167 ft2

Use a 13 × 13 ft square footing (A_{f} = 169 ft^{2})

### Factored loads and soil reaction

To proportion the footing for strength (thickness and required reinforcement)

Pu = 1.4 (350) + 1.7 (275) = 957.5 kips

q_{s} =P_{u}/A_{f}= 957.5/169= 5.70 ksf

### Thickness of Footing

fc′ = 3000 psi

Pu = 957.5 kips

qs = 5.70 ksf

1. Beam Shear / One way Shear

V_{n} = q_{s} x tributary area

B_{w }13 ft = 156 in.

Assumed footing thickness,

T = 33 in.

Avg. effective thickness,

d = 28 in.

tributary area = 13 (6.0 - 2.33) = 47.7 ft^{2 }

V_{n} = 5.7 x 47.7 = 272 kips

V_{n} =

= 0.85 x 156 x28/1000

= 407 kips > V_{n }O.K.

2. Punching shear/ Two way Shear

Vn = qs x tributary area

tributary area =[(13x13)- ( (30+28)(12+28))/144 ]=152.9 ft2

Vn = 5.70 x 152.9 = 872 kips

(v)/(√(fc) bd)=minimum

2 +( 4)/(βc )=3.6 (governs)

( a_{s}d)/( b0)+2=7.7

4

### 2. Punching shear/ Two way Shear

b0 = 2 (30 + 28) + 2 (12 + 28) = 96 in.

βc = ( 30)/12 = 2.5

( b_0 )/12d = ( 196)/28 = 7

as = 40 for interior column

2. Punching shear/ Two way Shear

∅Vc = 0.85x 3.6√3000 x 196 x28/1000

= 920 kips > 872 kips O.K.

**Footing Reinforcement**

fc’ = 3000 psi

fy = 60,000 psi

Pu = 957.5 kips

qs = 5.70 ksf

1. Critical section for moment is at the face of column

Mu = 5.70 × 13 × 6^{2}/2 = 1334 ft.-kips

Mu = 5.70 × 6^{2}/2 = 102.6 ft.-kips/ft.

2. Compute A_{s} (in^{2}/ft.) required

**Footing Reinforcement**

As = ( M )/(φƒy(d-( a)/2))= ( 102.6*12 )/(0.9*60* (28-( 3)/2))=0.86 in2

** **

Mu = 5.70 × 6^{2}/2 = 102.6 ft.-kips/ft.

3. Check for compression depth (a)

as = ( A_s )/(0.85*fc)= ( 0.86*60 )/(0.85*3*12)=1.7

As = ( M )/(φƒy(d-( a)/2))= ( 102.6*12 )/(0.9*60* (28-( 1.7)/2))=0.84 in2

4. Check minimum reinforcement

r_{min}= 0.0018 for Grade 60 rebar

= 0.0018*12*33=0.72 in^{2}/ft. < As

5. Spacing of longitudinal rebar

s= (0.5/0.84)*12= f20 @ 7” c/c

As= 0.84 in2/ft.

As,min= 0.72 in2/ft.

s=f20 @ 7” c/c

( Reinforcing in band width )/(Total Reinforcing in short direction )

= ( 2 )/(β +1)= γ ACI: Eqn. 15-1

Band width for steel in the short direction for

rectangular isolated footings.

ACI 318-08 / BNBC 2017

Column = 18”x18”

DL= 185 kip

LL = 150 kip

f_{y}= 60,000 psi

f’_{c}= 4000 psi

q_{a}= 4000 psf

D_{f}= 5 ft.

Pu= 1.2*DL+1.6*LL

f ( torsion and shear) = 0.75

### Footing Area and Dimension

1. Assume thickness T= 24” and d= 19.5”

qe = 4000psf – (( 24 in)/(12 in/ft))(150 pcf)- ((36 in)/(12 in/ft)) ( 100 pcf)= 3400 psf

Area required = (185 k+150 k )/(3.4 ksf)=98.5 ft^2

qu ((1.2)(185 k)+ (1.6)(150 k))/〖98.0 ft〗^2 =4.71 ksf

Use 7 ft.× 14 ft.= 98.0 ft2

### Thickness of Rectangular Footing

2. Checking Thickness for One-Way Shear

b = 7 ft

Vu1 (7.0 ft) ( 4.625 ft) (4.70 ksf) = 152.49 k

D = ( 152,490 lb )/((0.75)(1.0)(2√(4000 psi))(84 in.))=19.14 in.,h=d+4.5 in.=23.64 in.

Use T= 24”

3. Checking Thickness for Two-Way Shear

18”+19.5”=37.5”=3.125ft.

3. Checking Thickness for Two-Way Shear

b0 = (4) (37.5 in.) = 150in

Vu2 = [98.0 ft^2-(3.125 ft^2 )](4.71 ksf)=415.58 k

d = (415.580 ld )/((0.75)(1.0)(4√(4000 psi))(150 in.))=14.60 in. < 19.5 in

d = (415.580 ld )/((0.75)(( 40 in x 19.5 in)/(150 in.) + 2)(√(4000 psi))(150 in.)) = 8.11 in < 19.5 in

Does not govern

### Rebar of Rectangular Footing

4. Design of Longitudinal steel

Lever arm = ( 14 ft )/2- ( 9 in )/(12in/ft )=6.25 ft

Mu = (6.25 ft) (7.0 ft) (4.71 ksf) (( 6.25 ft)/2) = 643.9 ft-k

Mu = 6.25 × 4.71 x 6.25/2 = 92 ft.-kips/ft.

Compute As (in2/ft.) required

As = ( M )/(φƒy(d-( a)/2))= ( 92*12 )/(0.9*60* (19.5-( 4)/2))=1.17 in2

Mu = 92 ft.-kips/ft.

Check for compression depth (a)

a = ( A_s f_y )/(0.85*f_c^' )= ( 0.23*60 )/(0.85*4*12)=0.4

As = ( M )/(φƒy(d-( a)/2))= ( 92*12 )/(0.9*60* (19.5-( 0.4)/2))=0.2 in2

Check minimum reinforcement

r_{min}= 0.0018 for Grade 60 rebar

= 0.0018*12*24=0.52 in^{2}/ft. > As

r_{min}= 0.0033

= 0.0033*12*19.5=0.77 in^{2}/ft. > As

Spacing

s= (0.32/0.52)*12= f16 @ 7” c/c

Total bars = 14*12/7+1=25

Total bars = 14*12/7+1=25

( Reinforcing in band width )/(Total Reinforcing in short direction )= ( 2 )/(2+1)= ( 2 )/3

2/3*25=16 nos. shall be placed in band width

Temperature and shrinkage rebar governed, so this type distribution is not required

However, it is wise to provide flexure minimum in the middle band

### Combined Footing

· Footings support more than one column

· Necessary when isolated individual footings would run into each other

· Where a column is close to a property line

### Combined Footing Example

qa = 5 ksf

f’c = 3 ksi

fy = 60 ksi

Df = 6 ft

Solution

Assume 27-in. Footing (d=22.5 in.)

qe = 5000 psf- (( 27 in.)/2) (150 pcf) – (( 45 in )/(12 in/ft)) (100 pcf) = 4287 psf

Area required = ((120 k+100 k)+ (200 k+150 k) )/(4287 ksf )=132.96 ft2

Locate Center of Gravity of Column Service Loads

X from c.g. of left column = ((200 k+150 k)+ (12ft) )/((120 k+100 k)+ (200 k+150 k))= 7.37 ft

Distance from property line to c.g. = 0.75 ft + 7.37 ft = 8.12 ft

Length of fooing = (2 x 8.12 ft) = 16.24 ft, 16 ft 3 in.

Required footing width = ( Area required )/(length )= ( 132.96 ft^2)/(16.25 ft)=8.18 ft

1.Find cg of two column loads

2.Determine length of footing so that cg of footing and load are at approximately same location

3.Determine width of footing for required area

Use 16-ft 3-in. x 8-ft-3-in. footing (A = 134 ft2).

qu ((1.2)(320 k) +(1.6) (250 k ))/(134 ft^2 ) = 5085 ksf

Bearing pressure applied on the footing by soil

Shear and moment diagrams for combined footing

** **

**Depth Required for One-Way Shear **

The largest shear force is 271.1 k at the left face of the right column. At a distance d to the left of this location, the value of shear is

Vu1 271.1 k – 48.26 klf (( 22.5 in )/(12 in/ft)) = 180.61 k

d = ( v_u1)/(∅2x√(f_c^' ) b)= (415.580 ld )/(0.75 (2) (1) √3000 psi (8.25 in)(12 in/ft.))= 22.2 in < 22.5 in.

Depth Required for Two-Way Shear

Vu2 at right column = 480 k – ((42.5 in)/(12.in/ft))^2 = 406.6 k

d = (406,600 ld )/((0.75) [(4) (10) √3000 psi] (4 x 42.5 in))

= 14.56 in. < 22.5 in. ok

Vu2 at leftt column = 304 k – (( 29.25 in.x 40.5 in.)/(144 in^2/ft^2 )) = 255.9 k

d = (406,600 ld )/((0.75) [(4) (1.0) √3000 psi] (2 x 29.25 in +40.5 in.))

= 15.73 in. < 22.5 in. ok

Design of Longitudinal Steel

Mu = −729.5 ft-k

As = ( M )/(φƒy(d-( a)/2))= ( 729.5*12 )/(0.9*60* (22.5 - ( 3)/2))=7.72 in2

a = ( A_s f_y )/(0.85*f_c^' )= (7.72 * 60 )/(0.85 * 3 *8.25*12)=2

As = ( M )/(φƒy(d-( a)/2))= ( 729.5*12 )/(0.9*60* (22.5 - ( 2)/2))=7.5 in2

Design of Longitudinal Steel

As,min= 0.0033*b*d

As,min= 0.0033*8.25*12*22.5=7.35 in2

Design of Short-Span Steel Under Interior Column

Assuming steel spread ver width = column width + (2) (( d)/2)

= 20in. + (2) (( 25.5 in )/2) = 42.5 in.

Design of Short-Span Steel Under Interior Column

8 k/f ((3.29 ft )/2) = 314.9 ft-k

Soil stress used in determining short span steel

Design of Short-Span Steel Under Interior Column

As = ( M )/(φƒy(d-( a)/2))= ( 314.9 * 12 )/(0.9*60* (22.5 - ( 2)/2))=3.25 in2

### FEM Modeling of Combined Footing

Combined footing may be easily designed by FEM modeling in ETABS, SAFE or any other software.

- Calculate applied pressure (p)
- Calculate settlement (s)
- Calculate subgrade modulus (p/s)
- Determine thickness of footing (beam shear may govern)
- Model the footing, analyse and design

### Alternative concept of combined footing

1. Find cg of two column loads

2. Determine length of footing so that cg of footing and load are at approximately same location

3. Determine width of footing for required area

1. If pf is the factored contact pressure of footing

2. Upward line load below T-beam = pf x footing width

3. Upward pressure below flange of T-beam = pf

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