Lecture 19 – Page 1 of 10

Lecture 19 – Wall Footings All structural load-bearing walls must bear on footings. These footings in turn distribute the loads to the soil. If the loading on the soil exceeds the soil bearing capacity, undesirable settlement will occur, resulting in cracked walls, uneven floors, or potentially structural failure in extreme overloading.

Lecture 19 – Page 2 of 10

Wall Footing Requirements: 1) Must distribute the loads evenly to the soil. 2) Bottom of footing must be deeper than the frost penetration or else frost

heave will occur. 3) Must be well-drained to prevent wash-out. 4) Must bear on undisturbed (or engineered) soil.

Lecture 19 – Page 3 of 10

Typical Allowable Soil Bearing Values per IBC Table 1804.2

Soil Type: Allowable Soil Bearing (qa)*: Bedrock 12,000 PSF Sedimentary and foliated rock 4,000 PSF Sandy gravel and/or gravel 3,000 PSF Sand, silty sand, clayey sand, silty gravel, clayey gravel 2,000 PSF Clay, sandy clay, silty clay, clayey silt, silt, sandy silt 1,500 PSF Organics, peat, top soil NOT Recommended for use

as footing bearing * = An increase of one-third is permitted when using alternate load combinations in

IBC Section 1605.3.2 that include wind or earthquake loads. Example GIVEN: An 8” wide load-bearing poured-in-place concrete wall is to bear on a wall footing with the following:

• Service dead load on wall (incl. wall weight) = 8000 PLF • Service live load on wall = 6000 PLF • Allowable soil bearing = qa = 3000 PSF • Depth to frost = 3’-0” below finished grade • Soil unit weight = 110 PCF • Concrete f’c = 3000 PSI • Main steel tension reinforcing = #5 grade 60 bars • Longitudinal shrinkage/temperature bars = #4 grade 60 bars • Concrete cover = 3” per ACI

REQUIRED: Design the wall footing, including dimensions, and bar requirements.

Step 1 – Determine preliminary thickness of footing:

Preliminary footing thickness ≈ 1½(Wall width) ≈ 1½(8”) ≈ 12” TRY Preliminary footing thickness = 12”

Lecture 19 – Page 4 of 10

Step 2 – Determine effective soil bearing, qe:

qe = effective soil bearing = reduced soil bearing capacity resulting from weight of footing and weight of soil above footing (overburden) = (Allowable soil bearing) – (weight of footing) – (weight of soil) = qa – (ftg. wt.) – (soil wt.)

qe = qa – (ftg. wt.) – (soil wt.) = 3000 PSF – (1’-0” thick(150 lb/ft3)) + ((2’-0” thick(110 lb/ft3)) = 3000 PSF – 150 PSF – 220 PSF qe = 2630 PSF

Finished grade

qe

1’-0”

2’-0”

Dep

th to

fros

t = 3

’-0”

Soil overburden

Lecture 19 – Page 5 of 10

Step 3 – Determine required width of footing, wf:

Footing width = wf = e

total

qP

wf = PSFPLF

263014000

wf = 5.32 feet

USE footing width = 5’-6”

qe = 2630 PSF

Ptotal = SERVICE LOADS =PDL + PLL = 8000 PLF + 6000 PLF = 14,000 PLF

Footing width = wf

Lecture 19 – Page 6 of 10

Step 4 – Determine soil bearing pressure for strength design, qu:

qu = Factored soil bearing pressure acting on footing

= f

LLDL

wPP )(6.1)(2.1 +

= ft

PLFPLF50.5

)6000(6.1)8000(2.1 +

qu = 3490 PSF

Step 5 – Determine minimum depth of footing, dmin based on shear:

d = footing thickness – concrete cover – ½(Bar dia.)

= 12” – 3” – ½ ⎟⎠⎞

⎜⎝⎛ "85

= 8.69”

Vu = ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

22_

2dWidthWallw

q fu

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛

−2/"12

"69.8

2/"12"8

2'5.53490

ftftPSF

Vu = 7170 Lbs.

Footing thickness

d

#5 bar

Lecture 19 – Page 7 of 10

dmin = Minimum depth to reinforcement based on shear

= )"12('2)85.0( c

u

fV

= )"12(30002)85.0(

.7170PSI

Lbs

dmin = 6.4” < 8.69” OK

Step 6 – Determine factored soil pressure moment, Mu at face of wall: MA = Moment at point “A”

= qu(Footing overhang) ⎟⎠⎞

⎜⎝⎛

2_ overhangFooting

= 3490 PSF(2.42’) ⎟⎠⎞

⎜⎝⎛

2'42.2

MA = 10,219 FT-LB

Overhang

Footing overhang = 2.42’

qu = 3490 PSF

Footing width = 5.5 ft.

0.67’

Point “A”

Isolate footing overhang and treat as if it were a cantilevered beam

Point “A”

Lecture 19 – Page 8 of 10

Step 7 – Determine required tension steel, As in footing:

Recalling from Lecture 4 – Design Aid Table 1:

Concrete f’c = 3000 PSI, Grade 60 bars

2bdM u

φ = 2)"69.8)("12(9.0

)/"12)(10219( ftLbFt −

= 150.4 PSI

Use ρmin = 0.0033 → since 2bdM u

φ = 150.4 PSI < 190.3 PSI

As = ρbd = (0.0033)(12”)(8.69”) As = 0.34 in2 per 1’-0” length of footing

Lecture 19 – Page 9 of 10

Step 8 – Determine spacing of #5 tension bars:

Spacing = ⎟⎟⎠

⎞⎜⎜⎝

⎛

s

s

AbarperA __

"12

= ⎟⎟⎠

⎞⎜⎜⎝

⎛2

2

34.0_5_#_31.0"12

inbarperin

= 10.9” Use #5 bars spacing = 10” o.c.

Step 9 – Determine longitudinal shrinkage/temperature bars:

As temp = 0.0018bh = 0.0018(5.5’ x 12”/ft)(12”) = 1.43 in2

Number of bars = barperA

A

s

s

__

= barperin

in_4_#_20.0

43.12

2

= 7.15 bars Use 8 - #4 longitudinal temperature/shrinkage bars

h = 12”

Footing width “b” = 5.5 ft.

Lecture 19 – Page 10 of 10

Step 10 – Draw “Summary Sketch”:

NOTES: 1) Conc. f’c = 3000 PSI. 2) All bars grade 60 3) Wall reinf. & dowels not shown 4) Allow. Soil bearing = 3000 PSF

8” poured-in-place conc. wall centered over footing (dowel bars not shown) Finished grade – sloped

away from bldg.

5’-6”

12”

#5 bars @ 10” o.c.

8 - #4 longitudinal bars

3’-0

” min

imum

3” conc. cover all around

Formed “Keyway”

Footing to bear on undisturbed soil