Footings
Footings
Footings
Reinforced concrete foundations, or footings, transmit loads from a structure to the supporting soil.
Footings are designed based on the nature of the loading, the properties of the footing and the properties of
the soil. Design of a footing typically consists of the following steps: Determine the requirements for the
footing, including the loading and the nature of the supported Structure; select options for the footing and
determine the necessary soils parameters. This step is often completed by consulting with a Geotechnical
Engineer; and the geometry of the foundation is selected so that any minimum requirements based on soils
parameters are met.
Footings are structural elements that transmit column or wall loads to the underlying soil below the
structure. Footings are designed to transmit these loads to the soil without exceeding its safe bearing capacity,
to prevent excessive settlement of the structure to a tolerable limit, to minimize differential settlement, and to
prevent sliding and overturning. The settlement depends upon the intensity of the load, type of soil, and
foundation level. Where possibility of differential settlement occurs, the different footings should be designed
in such a way to settle independently of each other. Foundation design involves a soil study to establish the
most appropriate type of foundation and a structural design to determine footing dimensions and required
amount of reinforcement. Because compressive strength of the soil is generally much weaker than that of the
concrete, the contact area between the soil and the footing is much larger than that of the columns and walls.
Gross Pressure is the total pressure at the base of footing due to the applied column loads and footing self-
weight + overburden soil pressure (above the footing).
The Gross Soil Pressure combines the upward pressure and the added pressure due to the applied
load, equal to the applied load divided by the footing area. For the Net Soil Pressure, the overburden and self-
weight is canceled out by an equal and opposite upward pressure, leaving a Net Soil Pressure equal to the
applied load divided by the footing area.
When a Net soil pressure is specified the load is compared against a modified soil capacity as shown below:
Soil Capacity = Bearing Allowable (user-entered) + Footing Self Weight + Overburden (user-entered)
If the sum of Footing Self Weight and Overburden do not equal the soil weight which was present above
the footing/soil interface strata prior to excavation, then the modified soil capacity provided by the program is
incorrect. In this circumstance the Net Allowable Soil Bearing value should be specified as a Gross value in the
program, and the magnitude of Bearing Allowable should be increased by that pre-excavation soil weight.
PART 1 DESIGN AND ANLYSIS OF GROSS AND NET PRESSURE
In the design of concrete footing for shear and bending: USE THE
FACTORED NET PRESSURE ONLY.
PART 2 ANALYZING WIDE-BEAM AND PUNCHING SHEAR STRENGTHS OF THE FOOTING AND DETERMINE THE
REQUIRED NUMBER OF BARS IN EACH DIRECTION. CHECKING WIDE-BEAM SHEAR AND PUNCHING SHEAR
PART 3 CHECKING WIDE-BEAM SHEAR, PUNCHING SHEAR AND DETERMING THE NUMBER OF BARS IN EACH
SIDE.
Given:
/Νc = 4 ksi
/y = 60 ksi
Dead Load = D = 25 k/ft
Live Load = L = 12.5 k/ft
Wind O.T. = W = 4 k/ft
(axial load due to overturning under wind loading)
Seismic O.T. = E = 5 k/ft
(axial load due to overturning under earthquake loading)
Allowable soil bearing pressures:
D = 3 ksf = "a"
D + L = 4 ksf = "b"
D + L + (W or E) = 5 ksf = "c"
Bearing
Bearing resistance of footing
φbearing = 0.65
A/A221=
r c 1 2 1 B = φ(0.85 f ' A ) A / A
Br = 0.65(0.85)(4)(16)2 (2)
Br = 1131 k > 400 k OK
One-way action
Given:
n c u φV = φV > V OK
ƒ’c = 4 ksi
bw = 7.33 (12) = 88 in. and d = 15.5 in.
ƒy = 60 ksi
V2f'bdccw=
Dead Load = D = 180 k
V 0.75( 2 4000 )( 88 )( 15.5 ) / 1000 c φ =
Live Load = L = 100 k
= 129.4 k
Wind O.T. = W = 120 k
Vu =7.33 [(7.33/2) – (8+15.5)/12](7.5)
(axial load due to overturning under wind loading)
Allowable soil bearing pressures: in2
Due to D = 4 ksf = “a” Check for As,min= 0.0018 bh
Due to D + L = 6 ksf = “b” As,min= 0.0018(5)(12)(27) = 2.92 in2
Due to D + L + W = 8.4 ksf = “c” < 4.72 in2 OK
Design a rectangular footing with
an aspect ratio ≤ 0.6 Use 8 #7 bars distributed uniformly across
the entire 5ft width of footing
Sizing the footing. Ignoring the self-weight of the Note: εt = 0.041 > 0.005 for tension
footing; controlled sections and φ = 0.9.
D/a = 180/4 = 45 sq.ft.
(D+L)/b = 280/6 = 46.7 sq.ft. Calculate moment in the short direction,
(D + L + W)/c = 400/8.4 = 47.6 sq.ft. at the column face.
Controls Compute flexural tension reinforcement
Use 5 ft x 10 ft Mu = (10.2)(1.83)2 (10)/2 = 171.4 ft-k
A = 50 sq.ft. is OK φKn = Mu (12,000)/(bd2)
φKn = 171.4 (12,000)/[(10)(12)(22.5)2]
Required Strength U = 1.4D = 33.9 psi
= 1.4(180) For φKn = 33.9 psi, select ρ = 0.07%
= 252 k or (252/50) = 5.1 ksf As = ρbd = 0.0007 (10)(12)(22.5)
U = 1.2D + 1.6L = 1.89 in2
= 1.2(180) + 1.6(100)
= 376 k or (376/50) = 7.6 ksf Check for minimum reinforcement As,min= 0.0018
U = 1.2D + 1.6W + 1.0L bh
= 1.2(180) + 1.6(120) + 1.0(100) As,min= 0.0018(10)(12)(27) = 5.83 in2
= 508 k or 10.2 ksf (Controls) > 1.89 in2
U = 0.9D + 1.6W Use As = 5.83 in2
= 0.9(180) + 1.6(120) (Reinf. In central 5-ft band) / (total reinf.)
= 354 k or 7.1 ksf = 2/(β+1)
β = 10/5 = 2; and 2/(β+1) = 2/3
One-way action in the long direction is Reinf. In central 5-ft band = 5.83(2/3)
not a problem because the footing edge is = 3.89 in2
located within the potential critical
section for one-way shear. Use 7 #7 bars distributed uniformly across
the entire 5ft band.
Bearing Reinforcement outside the central band
Bearing resistance of footing = 5.83 – 7(0.6) = 1.63 in2
φbearing = 0.65 Use 6 #5 bars (3 each side) distributed
A/A221= uniformly outside the central band.
r c 1 2 1 B = φ(0.85 f ' A ) A / A
Br = 0.65(0.85)(4)(16)2 (2) FOOTINGS EXAMPLE 4 -
Br = 1131 k > 508 k OK Design of a pile cap.
Determine
Compute flexural tension reinforcement the size
Mu = (10.2)(4.33)2 (5)/2 = 479 ft-k and
φKn = Mu (12,000)/(bd2) reinforcing for
φKn = 479 (12,000)/[(5)(12)(23.5)2] a square
= 173.5 psi pile cap
For φKn = 173.5 psi, select ρ = 0.335%
As = ρbd = 0.00335 (5)(12)(23.5) = 4.72
that supports a 16 in. square column and is placed
on 4 piles. FOOTINGS EXAMPLE 5 - Design of a continuous
Given: footing with an overturning moment
ƒ’c = 5 ksi Determine the size and reinforcing bars for a
ƒy = 60 ksi continuous footing
Dead Load = D = 250 k under a 12-in. bearing wall, founded on soil, and
Live Load = L = 150 k subject to
16 x 16 in. reinforced concrete column loading that includes an overturning moment.
12 x 12 in. reinforced concrete piles Given:
(4 piles each @ 5 ft on centers) f’c = 4 ksi
fy = 60 ksi
Factored Loads: Depth from top of grade to bottom of footing = 3 ft
Column: Density of soil above footing = 100 pcf
U = 1.4D Density of footing concrete = 150 pcf
= 1.4(250) Vertical Dead Load = 15 k/ft (including wall weight)
= 350 k Horizontal wind shear = V = 2.3 k/ft (applied at 1 ft
U = 1.2D + 1.6L above
= 1.2(250) + 1.6(150) grade)
= 540 k = Vu (Controls) Allowable soil bearing pressure based on
Piles: unfactored loads
Pu = 540/4 = 135 k = Vu ignoring the = 4 ksf
self-weight of pile cap
Note:
The effective depth is conservative for the
two-way action but is O.K. considering
the overlapping of the critical sections
around the column and the piles
One-way action will not be a problem
because the piles are located within
potential critical sections for one-way
shear.