Question

A strip wall footing 1.5 m wide is located 1.1 m below the ground surface.
Supporting soil has a unit weight of 19.60 kN/m3 . The results of laboratory
tests on the soil samples indicated that the supporting soil has the following
properties: c’ = 57.5 kN/m2
and I’ = 25°Groundwater surface was not
encountered during the subsurface exploration. Determine the allowable
bearing capacity of the soil using a factor of safety of 2.5.

Answer 1

Given, the width of foundation is B = 1.5 m.

The depth of foundation is, D_f = 1.1 m

The properties of soil provided are c' = 57.5 kn/m² , internal friction angle I' = 25^o

The unit weight of the soil is 19.6 kn/m³

Now, the ultimate bearing capacity as proposed by Terzaghi is modified with some shape factor that depend on the shape of the foundation, it is given by,

  

where, s_c and are the shape factors for cohesion and base respectively , N_c, N_q and N are the bearing capacity factors

The bearing capacity factors are obtained from the table below with the corresponding friction angle.

*here in the table the friction angle is denoted as

Thus the values are N_c = 20.72, N_q = 10.66, N = 10.88

The shape factors are equal to 1 for strip foundation

Thus the ultimate bearing capacity, = q_ult = 57.5x20.72 + 19.6x1.1x10.66 + 0.5x19.6x1.5x10.88 = 1581.17 kn/m²

Thus the allowable bearing capacity = q_ult / F.O.S = 1581.17/ 2.5 = 632.46 kn/m² ,since the factor of safety is 2.5.

Thus the allowable bearing capacity is 632.46 kn/m²