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 = 35Â°. The water table is at a great depth. The unit weight of soil 7= 114 lb/ft3. Determine: (1) the net Full Article…

# Bearing capacity problem example 5

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

# Bearing capacity problem example 4

If the water table inÂ Ex. 1Â occupies any of the positions (a) 1.25 m below ground level orÂ (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 inÂ Ex. Full Article…

# Bearing capacity problem example 3

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

# Bearing capacity problem example 2

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

# Bearing capacity problem example 1

A strip footing of width 3 m is founded at a depth of 2 m below the ground surface in a (c – 0) soil having a cohesion c = 30 kN/m2 and angle of shearing resistance 0 = 35Â°. The water table is at a depth of 5 m below ground level. The moist Full Article…

# Lateral Earth Pressure retaining wall problem example 2

For the wall given in Example 1, determine the total passive pressure P e under seismicÂ conditions. What is the additional pressure due to the earthquake? Solution From Eq. (11.91),

# Lateral Earth Pressure retaining wall problem example 1

A gravity retaining wall is required to be designed for seismic conditions for the active state. The following data are given: Solution From Eq. (11.79) For all practical purposes, the point of application of Pae may be taken as equal to H/2 aboveÂ the base of the wall or 4 m above the base in this Full Article…

# Passive Earth Pressure problem example 2

For the data given in Example 11.15, determine the reduction in passive earth pressure for aÂ curved surface of failure if 8 = 30Â°. 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…