Shallow Foundation,,,,.pdf

3.0 Design of Shallow Foundations

Definition of the problem

And compare it with supporting material capacity to carry such a pressure

Definition of the problem

Footing should not collapse due to shear failure or excessively settle owing to material compressibility

3.0 Design Essentials ●

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Recall that the performance of any foundation is judged by its ability to resist both the bearing capacity failure and excessive settlement. it therefore means ................. Bearing Capacity Analysis

Settlement Analysis

Design of Shallow Foundations to meet both bearing capacity and settlement requirements

3.1 Spread footing: Geotechnical design ●

Design for Concentric Downward Loads

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Design for Eccentric or Moment Loads

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Design for Shear Loads

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Design for Wind or Seismic Loads

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Lightly Loaded Footings

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Footing on or Near Slopes

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Footings on Frozen Soils

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Footings on Soils Prone to Scour

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Footings on Rock

3.2 Design for concentric loads ●

Allowable Bearing Pressure Method ◦ (qa)bc to satisfy bearing capacity requirements ◦ (qa)s to satisfy settlement requirements ◦ (qa)design is the lower of (qa)bc and (qa)s

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Design Chart Method

3.3 Foundation depth ●

The foundation depth is among the first things to be decided on and this is based on geotechnical investigation recommendation, which itself is based on the nature of the subsurface conditions

3.4.1 Minimum depth: Square and rectangular footings

3.4.1 Minimum depth: Continuous footings

3.4.3 Other considerations for minimum depth

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Depth to “competent” soil

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Depth of frost penetration

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Presence of expansive soils

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Potential for scour

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Possibility of a shallow landslide

3.4.4 Other considerations for maximum depth

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Potential undermining existing foundations, utility lines etc.

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Desire to avoid ground water table

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Desire to avoid the need for excavation shoring

3.4

Allowable bearing pressure method

 Allowable bearing pressure, qa is the largest bearing pressure that satisfies both bearing capacity and settlement criteria.

 Select depth of embedment, D  Determine the allowable bearing pressure based on bearing capacity analysis using the smallest applied load

3.5

Allowable bearing pressure method

 Determine the allowable bearing pressure based on settlement analysis using the largest applied load

 Select qa as the lower value of the two allowable bearing pressures.

Worked out example: Design of shallow foundation for concentric loading

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The size of an isolated footing is to be limited to 1.5 m square. Calculate the depth at which the footing should be placed to take a load of 200 tonnes with a factor of safety 3. The soil is having angle of internal friction  = 30o ,  = 2.1 g/cc, weight of footing 5% of the external load. Nq = 22, N = 20.

Solution to example 1 The soil must be granular soil since it has no cohesion Allowable bearing capacity for square footing: qallow = 1/F { 1.3 cNc + D (Nq – 1) + 0.4  B N } + D Now total load = 200 + 5/100 x 200 = 210 tonnes qallow = 210/(1.5 x 1.5). Put values in the above eq.

VENJOH 2015-04-19 03:47:12

-------------------------------------------why isn't the load converted into newtons fromtonnes?

210/1.5 x 1.5 = 1/3 { 2.1 x D (22-1) + 0.4 x 2.1 x 1.5 x 20} + 2.1 x D = 14.7D + 8.4 + 2.1D 16.8D = 93.4 – 8.4 = 85   The

D = 85/16.8 = 5.6 meters footing should be placed at least 5.6 m below the ground level

Worked out example 2: Design of shallow foundation for concentric loading ●

The results of a plate load test in a sandy soil are given below. The size of the plate is 0.305x0.305m. Determine the size of square footing of a column foundation that should carry a load of 2500 kN with a maximum settlement of 25 mm.

Load/unit area (kN/m2) Settlement (mm)

200

400

600

700

5

12.5

28

60

Solution to example 2

Given: Size of the plate 0.305 x 0.305 m. Load v/s settlement , draw a curve or plot the data. Load on footing------2500 kN Load Max. settlement = 25mm 2 200

Settlement (mm)

10 30 50 70 80

400

600

800 kN/m

Solution to example 2

The relevant equation is: (Size in m)

Settlement of footing is determined through trial and error method – iteration procedure

Solution to example 2 What this means is that you assume the size of footing and find the corresponding amount of settlement where the trial agrees with the conditions given (25mm settlement) will be the size of footing.Thus, Table of results Qo , (kN)

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Assumed BF (cm)

Sp corresponding to Qo (mm) 4

SF from above Eq. (mm)

2500

400

qo Qo /BF 2 , (kN/m2) 156.25

2500

300

277.8

8

26.35

2500

320

244.10

6.8

22.70

2500

310

260.10

7.2

23.86

Recommended footing size is B 3.10m

13.8

Worked out example 3: Design of shallow foundation for concentric loading Plate load test performed on a uniform deposit of sand and the following observations were recorded. Load t/m2

5

Settlement (mm) 4.5

10

20

30

40

50

60

8.5

16

31.3

50

74

104

If the size of the plate was 30 x30cm, plot the load-settlement curve and determine the load on a footing 1.5m x 1.5m. Would the footing safely carry such a load if the settlement is not to exceed 50 mm?

Solution to example 3 ●

Given:

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Size of plate 30 x 30 cm

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Footing size 1.5 x 1.5 m

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Settlement of footing 50 mm

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Plot Load vs settlement

Solution to example 3

Load t/m2 10

Settlement (mm)

10

30 50 70 90

20

30

40

50

60

Solution to example 3

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Now from load – settlement curve the load corresponding to this settlement = 20 t/m2.

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 Safe load on this footing for 50mm settlement = 1.5 x 1.5 x 20 = 45 tons. Therefore the footing would be safe.

Worked out example 4: Design of shallow foundation for concentric loading

A load test was made with a 35 cm square plate at a depth of one meter below the ground level in soil with  = 0. The water table was located at a depth of 5m below the ground level. Failure occurred at a load of 5200 kg. What would be the ultimate bearing capacity per unit area for a 1.6m wide continuous footing with its base loaded at the same depth in the same soil. Unit weight of soil was 1.9 g/cm3 above water table. Note that for  = 0, Nc = 5.7, Nq = 1 , N = 0. Assume general shear failure.

Solution to example 4 For square footing, qf =1.3 c Nc + D Nq + 0.4  B N Taking the case of load test, B = 0.35cm , D = 1. Since  = 0 , Nq = 1 and N = 0 5200/(.35 2x1000) = 1.3 x c x 5.7 + 100 x 1.9 x1.0. 7.42 c = 42.5 – 1.9 = 40.6 c = 40.6/7.42 = 5.47 tonnes/m2 .

Solution to example 4 Now for continuous strip footing of width 1.0m qf = c Nc + D Nq + 0.5  B N = 5.47 x 5.7 + 1.9 x 1 x 1 + 0 = 31.2 + 1.9 = 33.1 ton/m2 .

3.6 Design for eccentric loads ●

When footings have overturning moments as well as axial loads, the resultant soil pressure may not coincide with the centroid of the footing as illustrated in the figure below.

3.6 Design for eccentric loads ●

Assuming the footing is sufficiently rigid, application of statics gives the resultant soil pressure equal to P, with the point of application through the centroid of the pressure diagram, and at an eccentricity to satisfy moment equilibrium.

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This same situation can be produced by having a column off-centre or at-centre columns to very tall buildings where wind action may induce appreciable bending moments at the base of the columns and foundations of retaining structures among others.

3.6 Design for eccentric loads ●

The linear non-uniform soil pressure diagram (see figure in the following slide) is obtained from superposition of compression and moment stresses.

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A uniform resultant soil pressure can be deliberately ensured by placing a column away from the centre. This solution is obviously valid only for moments which always act in the direction shown for that footing configuration. It is not a valid solution for wind moments, since reversals can occur.

3.6 Design for eccentric loads ●

It should be evident that a column can transmit a moment to the footing only if it is rigidly attached.

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There is a very serious question if a spread footing (unless very large in plan) can sustain an applied column moment without undergoing rotation.

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From structural analysis, if the footing rotates an amount , this results in moments in the opposite direction to that being applied by the column to develop.

3.6 Design for eccentric loads ●

Thus, any footing rotation reduces the moment applied to the footing with a corresponding change to the far-end moment on the column.

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Obviously, a sufficiently large rotation can reduce the footing moment to zero (but not less than zero).

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It is therefore important that the differential soil pressure across the footing is limited in order to avoid large rotations of the footings.

3.6 Design for eccentric loads ●

In the analysis of rigid footings, the soil pressure can be computed from principles of mechanics of materials for combined bending and axial loads.

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For moment about an axis perpendicular to the footing length L this gives,

3.6 Design for eccentric loads ●

Note that strictly the weight of the half of the footing on appropriate side of the moment axis should be used to decrease the applied moment.

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A slight reduction in maximum soil pressure will be obtained as well as a small increase in the minimum soil pressure if this is done.

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When the eccentricity is sufficiently large, the computed minimum soil pressure becomes negative, indicating a tensile stress state between the soil and the footing.

3.6 Design for eccentric loads ●

The soil tension on the footing is zero for cohesionless materials and is nearly zero for cohesive soil.

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It is standard practice to neglect any soil tensions; so if the soil pressure turns out to be negative we say this is a tension stress and neglect that portion of the footing from any load-carrying capacity for this load condition.

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To determine the maximum eccentricity to avoid development of tension stress (for the footing to be fully effective) the soil pressure equation is equated to zero, thus;

3.6 Design for eccentric loads ●

This middle third defines what is sometimes called the kern limit or dimensions illustrated below.

3.6 Design for eccentric loads ●

When the resultant soil pressure is exactly at kern limit, the toe pressure is a maximum; the heel pressure is zero and the average base pressure is qmax/2.

3.6 Design for eccentric loads ●

This differential pressure distribution will most certainly cause a base rotation and it is recommended to always increase the base length somewhat so that the average and maximum pressures have a ratio larger than 0.5, that is, heel pressure larger than zero and closer to the maximum toe value.

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Also note that the toe pressure qmax  qa as furnished by the geotechnical consultant.

3.6 Design for eccentric loads Eccentricity out of middle third of footing ●

There are occasions when it is not possible to keep the resultant soil pressure inside the middle one-third of the base.

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This situation occurs when one or more of the design load combinations substantially exceeds the overturning capacity.

3.6 Design for eccentric loads Eccentricity out of middle third of footing ●

For eccentricity e > L/6 with respect to one axis, an equation for the maximum soil pressure and the effective footing length L’ can be obtained from the figure below where it is obvious the base area is not fully effective by the amount L-L’.

3.6 Design for eccentric loads Eccentricity out of middle third of footing ●

The area of the pressure triangle (see figure) must equal to the vertical load P and acts at L’/3 from the toe through the triangle centroid.

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This point is at a distance e = M/P from the footing centre so that

3.6 Design for eccentric loads Eccentricity out of middle third of footing ●

Substituting for L’ into the expression for P and solving for q we obtain,

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With P, qa’ and eccentricity e fixed, we solve for B and L by trial and error to satisfy the equality

3.6 Design for eccentric loads Eccentricity out of middle third of footing ●

The soil pressure for footings with eccentricity about both axes (see figure below) can be computed when no footing separation occurs as:

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Or

3.6 Design for eccentric loads Eccentricity about both axes

3.6 Design for eccentric loads Eccentricity out of middle third of footing ●

Noting that the use of a negative sign gives the minimum pressure at one corner and a positive sign gives the maximum pressure qmax qa’, we can solve for optimum base dimensions as follows (and optimum being defined as a fully effective base area and one corner pressure = 0). (a) Set the relevant equation to zero and obtain

3.6 Design for eccentric loads Eccentricity out of middle third of footing (b) Substitute equation of the previous step in the main equation and using (+) obtain

(c) Obtain maximum qmax, by taking derivative of equation in the equation of step (b) with respect L and set it to zero and solve for L to obtain,

3.6 Design for eccentric loads Eccentricity out of middle third of footing

this gives B = 12ey when L is substituted back into equation obtained in step (a). With these values for B and L used in the flexural equation where B, L, ex and ey are as defined on the figure of eccentricity about both axes we will obtain optimum base dimensions where qqa’ and with the base area fully effective. This might not be the most economical base as illustrated in the following example.

Worked example 1: Design for eccentric loads A 24 m long retaining wall has a 6m wide foundation at a depth of 1.5m in a silty sand having c = 1.5 t/m2 ,  = 25o and  = 2 t/m3 . The wall carries a horizontal load of 25 ton/m run at a point 2m above the base and a centrally applied load of 100 t/m run. Determine the safety factor against general shear failure of the wall, (take Nc = 20.7, Nq = 10.7, N = 6.6)

Solution to example 1 Determine Z, e, B’, and L’ from: e/Z = tan = H/V = 25/100 = 0.25 , Z =2 m. e/Z = 0.25 , e = 0.25 x 2 = 0.5 ,  = 14o B = B- 2e = 5 , L = 24m A = 5 x24 =120 m2

H = 25 x 24  Ic = Iq = 0.616 I = Iq2 = 0.379

V =100 x 24

qf = c Nc Ic+ D Nq Iq+ 0.5  B N I = 52 Actual Bearing pressure for B = 100/5 = 20 Factor of safety = 52/20 = 2.6