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Tag Archives: Problem example

Bearing capacity problem example 7

A rectangular footing of size 10 x 20 ft is founded at a depth of 6 ft below the ground level in a cohesive soil (0 = 0) which fails by general shear. Given: ysal =114 lb/ft3, c = 945 lb/ft2. The water table is close to the ground surface. Determine q , q and  Full Article…

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Bearing capacity problem example 6

A rectangular footing of size 10 x 20 ft is founded at a depth of 6 ft below the ground surface in a homogeneous cohesionless soil having an angle of shearing resistance 0 = 35°. The water table is at a great depth. The unit weight of soil 7= 114 lb/ft3. Determine: (1) the net  Full Article…

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Bearing capacity problem example 5

A square footing fails by general shear in a cohesionless soil under an ultimate load of Qult – 1687.5 kips. The footing is placed at a depth of 6.5 ft below ground level. Given 0 = 35°, and 7=110 Ib/ft3, determine the size of the footing if the water table is at a great depth (Fig. Ex.  Full Article…

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Bearing capacity problem example 4

If the water table in Ex. 1  occupies any of the positions (a) 1.25 m below ground level or (b) 1.25 m below the base level of the foundation, what will be the net safe bearing pressure? Assume ysat = 18.5 kN/m3, /(above WT) = 17.5 kN/m3. All the other data remain the same as given in Ex.  Full Article…

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Bearing capacity problem example 3

If the water table in Ex. 1 rises to the ground level, determine the net safe bearing pressure of the footing. All the other data given in Ex. 12.1 remain the same. Assume the saturated unit weight of the soil ysat= 18.5 kN/m3. Solution When the WT is at ground level we have to use the submerged  Full Article…

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Bearing capacity problem example 2

If the soil in Ex. 1 fails by local shear failure, determine the net safe bearing pressure. All the other data given in  Ex. 1 remain the same. Solution For local shear failure:

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Passive Earth Pressure problem example 2

For the data given in Example 11.15, determine the reduction in passive earth pressure for a curved surface of failure if 8 = 30°. Solution For a plane surface of failure P from Eq. (11.76) is It is clear from the above calculations, that the soil resistance under a passive state gives highly erroneous values for  Full Article…

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