part3_bearingcapacity.pdf

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Part 3

Ultimate Bearing Capacity of

Shallow Foundations

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Shallow foundations are used for relatively low loads and where surficial soils are competent. A

foundation at a depth up to 4 times the foundation dimension may be called as “Shallow”.

Shallow Foundations

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Terzaghi (1943) derived bearing capacity equations based on Prandtl (1920) failure

mechanism and the limit equilibrium method for a footing at a depth Df below the

ground level of a homogeneous soil.

Terzaghi Solution To UBC

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Terzaghi Factors –General Shear Failure

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Allowable BC & safety factors

The net ultimate bearing capacity is defined as the ultimate load per unit

area at the foundation level that can be supported by the soil in excess of

the pressure caused by the surrounding soil and weight of concrete used

in the foundation. If the difference between the unit weight of concrete

used in the foundation and the unit weight of the surrounding soil is

assumed to be negligible. then

qu is the ultimate bearing capacity

qall is the allowable ultimate bearing capacity

Qall (g) is the allowable gross load = qall x Area of foundation

qu(net) is the net ultimate bearing capacity

q effective stress at foundation level ( for dry soil q = g h)

Qall (net) is the allowable net load = qall(net) x Area of foundation

,𝑞𝑎𝑙𝑙 𝑛𝑒𝑡 =
𝑞𝑢(𝑛𝑒𝑡)

𝐹𝑆

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Modifications for Ground Water

When ground water is present, The ultimate bearing

capacity equations must be modified. Three cases may

be encountered:

Dw = Depth of ground water

below surface

Df = Depth of foundation

below surface

Case 1: Dw< Df Use g = g  = effective unit weight of soil = gsat - gw Case 2: Df < Dw< (Df +B) Use g = gave= average effective unit weight of soil 1 ( ) ( ( )) ave w f w f D D B D D B g g g        Case 3: Dw ≥ (Df +B) Ground water table is deep and has no effect Use g 7 Example 3-1 A square foundation is 2m x 2m as shown. Determine the allowable gross load on the foundation. 8 Example 3-2 Consider example 3-1, assume the water table rises such that its depth below ground surface is 2.5 m. Determine the gross allowable load on the foundation. 9 Example 3-3 Consider example 3-1, assume the water table rises such that its depth below ground surface is 1 m. Determine the gross allowable load on the foundation. 10 General Bearing Capacity equation After the development of Terzaghi’s bearing capacity equation, several investigators worked in this area and refined the solution, the most popular of these solutions Meyerhof 1963, Hansen 1970, Vesic 1973). Different solutions show that the bearing capacity factors Nc and Nq do not change much. However, for a given value of f, the values of Ng obtained by different investigators vary widely. 1 2u c cs cd ci q qs qd qi s q i q c N F F F qN F F F BN F F F g g g g g   11 General Bearing Capacity Factors Used by Das text book 12 Meyerhof (1963) Bearing Capacity Factors f Nc Nq Ng f Nc Nq Ng 0 5.14 1 0 23 18.05 8.66 4.82 1 5.38 1.09 0.00 24 19.32 9.60 5.72 2 5.63 1.20 0.01 25 20.72 10.66 6.77 3 5.90 1.31 0.02 26 22.25 11.85 8.00 4 6.19 1.43 0.04 27 23.94 13.20 9.46 5 6.49 1.57 0.07 28 25.80 14.72 11.19 6 6.81 1.72 0.11 29 27.86 16.44 13.24 7 7.16 1.88 0.15 30 30.14 18.40 15.67 8 7.53 2.06 0.21 31 32.67 20.63 18.56 9 7.92 2.25 0.28 32 35.49 23.18 22.02 10 8.34 2.47 0.37 33 38.64 26.09 26.17 11 8.80 2.71 0.47 34 42.16 29.44 31.15 12 9.28 2.97 0.60 35 46.12 33.30 37.15 13 9.81 3.26 0.74 36 50.59 37.75 44.43 14 10.37 3.59 0.92 37 55.63 42.92 53.27 15 10.98 3.94 1.13 38 61.35 48.93 64.07 16 11.63 4.34 1.37 39 67.87 55.96 77.33 17 12.34 4.77 1.66 40 75.31 64.20 93.69 18 13.10 5.26 2.00 41 83.86 73.90 113.99 19 13.93 5.80 2.40 42 93.71 85.37 139.32 20 14.83 6.40 2.87 43 105.11 99.01 171.14 21 15.81 7.07 3.42 44 118.37 115.31 211.41 22 16.88 7.82 4.07 45 133.87 134.87 262.74 13 Example 3-4 Consider example 3-1, determine the gross allowable load on the foundation using (Das text book) equation 14 Example 3-5 Consider example 3-1, determine the gross allowable load on the foundation using (Meyerhof) equation 15 H Example 3-6 A square foundation is 2m x 2m as shown is subjected to vertical and horizontal load of 800 kN and 141 kN respectively. Determine the factor of safety against bearing capacity failure using (Das textbook equation)) 16 Example 3-7 A square foundation is 2m x 2m as shown is subjected to vertical and horizontal load of 800 kN and 141 kN respectively. Determine the factor of safety against bearing capacity failure using Meyerhof equation H 17 Along foundation width Along foundation length Eccentrically loaded foundations If e >B/6 tension developes. Since the tensile strength of

soil is zero, part of the foundation will not transmit loads

to the soil. You should avoid this situation by designing

foundation such that e