CE 3600

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CE 3600

Chapter 11

- Foundations -

When you finish this section, you should be able to:

1. Discuss Difference between Shallow and Deep Foundations2. Define the Bearing Capacity of a Soil

3. Determine the Bearing Capacity of a Soil under Dry, Partially

Submerged and Submerged Conditions.

The purpose of a foundation is to transmit the weight of a structure to the underlying

soil. A properly designed foundation must:

1. Minimize Settlement

2. Prevent Shear Failure

We've looked at and analyzed long-term settlement during Consolidation. We will

now look at shear failures and how to prevent them using deep foundations.

Bearing Capacity

Bearing capacity is the maximum load per unit area that a shallow footing can

support.

Consider ASTM D 1194 - Plate Load Test

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Figure 1 - Plate Bearing Test

A 1 ft2Plate is placed on a soil surface. A load P is placed on the plate and the settlement is measured as afunction of P . Two different types of shapes can be developed depending on the density of the soil. From this, we

can get a peak and residual bearing capacity.

Figure 2 - Plate Bearing Test Results for Loose and Dense Soils

Notice the shape of the curves; they look a lot like the curves seen in a direct shear test. This is because the direct

shear test reflects the conditions at failure.

General Shear - Dense Soils

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The figure below shows the case known as a General Shear Failure, which occurs in dense soils. The ultimate load

per unit length, Q" ult , is equal to the bearing capacity of the soil, qult , times the width of the footing, B. In general

shear, the soil can fail to the left or right if the footing can rotate, or both sides will fail if the footing is constrained

to prevent rotation. In practice, this can be determined by looking at the bulge or bulges that occur adjacent to the

footing.

There are three zones that fail. Zone 1 acts as a wedge, or an arrow point, that pushes into the soil. The wedge itself

remains intact and does not shear internally because it is fully supported. The angle is interesting, appearing

from model studies and field evaluation to be a value somewhere between for a rough base (friction between the

concrete and soil underneath) and for a smooth base.

Zone 1 pushes into Zone 2, however, and causes it to rotate. Zone 2 rotates around the corner or the edge of the

footing. It has the shape of a logarithmic spiral, which has the interesting property that a radial line from the center

of rotation makes an angle from the normal to the surface. Thus the force acting on the radial line properlyaccounts for the normal and tangential shear stresses. This section rotates into Zone 3, which is a passive failure

wedge as we saw in retaining wall analysis.

Figure 3- General Shear Failure

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Figure 4 - Logarithmic Spiral of Zone 2

As you remember from the direct shear test, the amount of deformation required to reach failure was the same no

matter what the normal stress was. Therefore, since all the zones are moving at the same time, when the deformationto failure is reached, all three Zones shear at once, and that represents the greatest stress that can be placed on the

footing. The peak value of bearing capacity is reached, and it starts to decrease again down to the residual value.

( Note a practical aspect here. In a testing frame, when the peak value is reached, the machine under strain control

slowly continues to deform until the load decreases down to residual strength. In normal situations, once the load is

applied (like the weight of a building), it will maintain that load (it is under load control, not strain control) until thegeometry of the system changes enough that the loads are relieved, i.e., a catastrophic failure.)

Local Shear - Less Dense Soil

If the soil is loose enough to not exhibit a peak strength, then it will fail at the residual value. In that case, the threeZones do not fail all at once, but rather progressively, starting along Zone 1, into Zone 2 and finally into Zone 3.

Because the failure is progressive, it is known as Local Shear. It requires more deformation than general shear,

because each location must undergo residual shear conditions.

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Figure 5 - Local (or Progressive) Shear

A fully formed failure surface develops, however, and continues to resist shearing as it is pushed further into the

soil.

Punching Shear

If the soil is so loose that it cannot sustain the shearing forces developed on the failure surface, the soil underneathwith collapse, causing the shear zone to progress downward, more or less vertically, until other constraints prohibit

further movement.

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Figure 6 - Punching Shear Failure

Terzaghi¶s Bearing Capacity Equation

One of Terzaghi¶s greatest strengths was knowing when he could borrow from other people and other work. When

Terzaghi was considering bearing capacity, he recognized that three different mechanisms were significant inresisting failure in the general shear case. These were:

1. The friction caused by the footing and the soil beneath the footing to develop normal and shear

stresses on the failure line (by Terzaghi),

2. The additional friction developed by surcharge soil and stresses acting on the plane at the bottom

of the footing (by Reissner), and

3. The resistance developed by cohesion along the base of the failure plane (by Prandtl).

He utilized solutions by Reissner and by Prandtl in a form of superposition with a third solution of his own. The

combination of terms gives a reasonable solution which is conservative for the assumptions used.

Strip Footing

For a long continuous strip footing, Terzaghi proposed

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.

This can be written more explicitly as:

Figure 7 - Definition Figure for Bearing Capacity Equation

The Bearing Capacity Factors, N

K , N

q and N

c are factors of J alone. Values are usually given in Tables and Figures,

such as Fig. 8, below. The angle of internal friction is determined from the soil below the bottom of the

foundation.

Notice that the unit weights show up in the different terms for different reasons. In all cases, the unit weights are the

effective unit weights. When the water is deep (below the failure zone), use the moist or wet unit weight. When the

water is at the base of the foundation, should be the buoyant or submerged unit

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weight . When the water is at the ground surface, then should also be basedon buoyant unit weight. The reduction in bearing capacity when water is near the footing is substantial, and the

worst scenario should be the controlling one in design.

Figure 8 - Bearing Capacity Factors

The bearing capacity, qult , is a stress acting under the footing. When the bearing capacity is multiplied by the width

of the footing, the ultimate load per unit length is determined. Normally, this is the value that the engineer andarchitect are most interested in. On the other hand, the ultimate load is a failure condition, the load at which the soil

will collapse. Generally, this is not considered a good situation, so for design, a factor of safety is introduced.

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and

Typical values for Factor of Safety in Bearing Capacity are FS = 2.5 for sand and FS = 3.0 for clay.

Bearing Capacities for Other Shapes

While many wall foundations can be described very well using strip footings, other footing shapes are common

under column loads or other special situations. McCarthy provides an excellent procedure for dealing with a range

of special situations using the Extended Bearing Capacity Equation, most texts use a simpler procedure to deal with

circular and square footings.

Square Footings

Circular Footings

Local Failures

Local failures occur when the soil density is low. Traditionally, this was handled by reducing the angle of internal

friction and the cohesion. When <= 28o, then the strength parameters are reduced by

and

Examples

Example 1

Find the allowable loading for the strip footing below.

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Solution:

Find the Bearing Capacity Factors from tables or figures.

NK = 56.3

Nq = 37.7

Nc = 50.6

Example 2

Example 1 with water at the base of the footing.

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Solution:

Example 3

Example 1 with water at the surface.

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The bearing capacity is almost half when the soil is saturated compared to dry.

Example 4

A square footing needs to support 400,000 pounds with a FS of 3.0. It will be buried 8 ft and water is deep. What

should be the length of each side?

Solution:

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NK = 22.4

Nq = 18.4

Nc = 30.1

B LHS RHS

4 75,000 40,615

5 48,000 41,645

5.5 39,670 42,160

5.3 42,720 41,950

5.35 41,925 42,000

Footing is 5.35 ft. square or 5'-4" square.

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