A strip wall footing 1.5 m wid
A strip wall footing 1.5 m wide is located 1.1 m belowthe ground surface.Supporting soil has a unit weight of 19.60 kN/m3 . The results oflaboratorytests on the soil samples indicated that the supporting soil hasthe followingproperties: c’ = 57.5 kN/m2and I’ = 25°Groundwater surface was notencountered during the subsurface exploration. Determine theallowablebearing capacity of the soil using a factor of safety of 2.5.
Answer:
Given, the width of foundation is B = 1.5 m.
The depth of foundation is, Df = 1.1 m
The properties of soil provided are c’ = 57.5 kn/m2 ,internal friction angle I’ = 25o
The unit weight of the soil is 19.6 kn/m3
Now, the ultimate bearing capacity as proposed by Terzaghi ismodified with some shape factor that depend on the shape of thefoundation, it is given by,
where, sc and 
arethe shape factors for cohesion and base respectively ,Nc, Nq and N
are thebearing capacity factors
The bearing capacity factors are obtained from the table belowwith the corresponding friction angle.
*here in the table thefriction angle is denoted as 
Thus the values are Nc = 20.72, Nq =10.66, N =10.88
The shape factors are equal to 1 for strip foundation
Thus the ultimate bearing capacity, = qult =57.5×20.72 + 19.6×1.1×10.66 + 0.5×19.6×1.5×10.88 = 1581.17kn/m2
Thus the allowable bearing capacity = qult / F.O.S =1581.17/ 2.5 = 632.46 kn/m2 ,since the factor of safetyis 2.5.
Thus the allowable bearing capacity is 632.46kn/m2
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