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SHALLOW FOUNDATIONS

Those that transmit structural loads to the near-surface soils.

Include spread footing foundation and mat foundation.

1. Square spread footings

Those footings that have a plan dimension of B x B (square) with a depth D from

the ground to the bottom of the footing and a thickness T.

2. Rectangular spread footing

Footings with a plan dimension of B x L, where L is the longest dimension

Useful when obstructions prevent the construction of a square footing.

3. Circular spread footing

Footings that are round in a plan view

Frequently used as foundations for light standards, flagpoles and power

transmission lines.

4. Continuous spread footing(strip footing or wall footings)

Used to support bearing walls.

5. Rings spread footing

Continuous footing that have wrapped into a circle.

Commonly used to support the walls of above-ground circular storage tanks.

6. Mat foundation (Floating or raft foundation)

A very large spread footing that usually encompasses the entire footprint of the

structure.

Usually made of reinforce concrete Used when dealing with the following conditions:

o The structural loads are so high or the soil condition is so poor that

spread footings would be largeo The soil is very erratic and prone to excessive differential settlements

o The structural loads are erratic, and thus increase the likelihood of

excessive differential settlements.o The lateral loads are uniformly distributed through the structure and thus

may cause differential horizontal movements in spread footings or pilecaps.

o The uplift loads are larger than spread footings can accommodate.

o The bottom of the structure is located below the groundwater table, so

waterproofing is an important concern.

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Bearing Pressure (Gross)

Is the contact force per unit area along the bottom of a shallow foundation.

q = duAWfP

+

where:o q = is bearing pressure

o P = vertical column load

o Wf = weight of foundation, including the weight of soil above the

foundation, if anyo A = base area of foundation (B2 for square foundation, or BL for

rectangular foundation)

o ud= pore water pressure

For continuous footings, the bearing pressure is:

q =du

B

b

Wf

b

P

+

where:

b = unit length (usually 1 meter strip or 1 foot strip)

B = width of the continuous footing

Example:

The 1.52 square footing shown in figure supports a column load of 445 kN. Compute forthe bearing pressure.

445 kN

0.91 mts. 0.76 mts.

= 18.83 kN/m3

0.45 mts.

1.52 mts.

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Solution:

For conservative assumption, the concrete is assumed to extend from the ground

surface to a depth D. (conservative because soil has a lower unit weight)

concrete = 23.5 kN/m3

Wf = 1.52(1.52)(0.76+0.45)(23.5) = 65.7 kNA = 1.52(1.52) = 2.3104 m2

ud = w zw = 9.81 kN/m3(0.76+0.45-0.91) = 2.943 kN/m2

q = 943.23104.2

7.65445

+kN

q = 218.101 kN/m2

Example:

A 0.70 mts. wide continuous footing supports a wall load of 110 kN/m. The bottom of thisfooting is at a depth of 0.50 mts below the adjacent ground surface and the soil has aunit weight of 17.5 kN/m3. The groundwater table is at a depth of 10 mts below theground surface. Compute for the bearing pressure.

Solution:

concrete = 23.5 kN/m3

(unit weight of concrete)P/b = 110 kN/mAssume a 1 meter stripWf/b = 0.70 m(0.50 m)(23.5) = 8.225 kN/m

Since water table is way below the base of the foundationud = 0

q = 070.0

/225.8/110

+

m

mkNmkN

q = 168.9 (kPa) or kN/m2

NET BEARING PRESSURE

The difference between the gross bearing pressure and the initial vertical

effective stress at depth D.

Measure of the increase in vertical effective stress at the depth D.

q = q z or q = qsoil(D)

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Example:

The mat foundation in the figure is to be 50 m wide, 70 mts. long, and 1.8 mts. thick. Thesum of the column and wall loads is 805,000 kN. Compute the average bearing

pressure, then compare it with the initial effective stress in the soil immediately below themat.

5.0 mts. 8.7 m=19.0kN/m3

Mat or floating foundation

1.8 mts.

concrete = 23.5 kN/m3

Wf = 50 m(70 m)(1.8 m)(23.5) = 148,050 kNA = 50(70) = 3,500 m2

ud = w zw = 9.81 kN/m3(8.7-5.0) = 36.297 kN/m2

q = 297.36500,3

050,148000,805

+kN

q = 236.003 kN/m2

The vertical effective stress at a depth D before construction was:

= (H) uwhere:

= unit weight of soilH = total depth to the bottom of foundationu = pore water pressure

= 19.0 kN/m3(8.7 m) 36.297 kN/m2

= 129.003 kPa

The increase in the vertical effective stress below the mat as a result of excavating andconstructing the building is:

q = q zq = 236.003 129.003q = 107 kPa

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Problem: (To be submitted soon)

1. A 400 kN vertical downward column load acts at the centroid of a 1.5 m.

square footing. The bottom of this footing is 0.40 mts. below the groundsurface and the top is flush with the ground surface. The groundwater table isat a depth of 3 meters below the ground surface. Compute the bearingpressure.

2. The mat foundation in the fig. is 45 meters wide and 90 meters long. It has aweight of 140,000kN. The sum of the applied structural loads is 1,300,000kN. Compute the average bearing pressure with the groundwater table atposition A. Repeat the computation with the groundwater table at position B.

Explain why these two values ofqare different.

6.5 mts

B

15 mts 21 mts

=17.5 kN/m3

A

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BEARING FAILURES OF SHALLOW FOUNDATIONS

1. General Shear Failure Most common and occurs on soils that are relatively incompressible and

reasonably strong, in rocks and in saturated, normally consolidated claysthat are loaded rapidly where the undrained condition prevails.

Failure surface is well defined and failure occurs quite suddenly.

Bulges may appear on both sides, but ultimate failure occur only on one

of the side, accompanied by rotation of the foundation.

General Shear Failure

2. Punching Shear Failure

Opposite extreme of the general shear failure, which occurs in very loose

sands, in thin crust of strong soil underlain by a very weak soil, or in weakclays loaded under slow, drained conditions.

High compressibility of the above mentioned soils causes large settlements

and poorly defined vertical shear surfaces. Little or no bulging occurs at the ground surface and failure develops

gradually.

Punching shear failure

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3. Local Shear Failure

An intermediate case between the general shear and the punching shear

failure.

Shear surfaces are well-defined under the foundation and becomes vague

near the ground surface.

Small bulge may occur, but considerable settlement is necessary, before a

clear shear surface forms near the ground.

Unlike the general shear, sudden failure does not occur.

Foundation just continues to sink even deeper into the ground.

Local shear failure

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