A strip footing is to be placed 2m below the surface or soil having a cohesion of 40 KPa, unit weight of 18.2 KN/m', and angle of friction of 10° Ne = 8.02 N = 1.94 Nc = 9.61 N = 2.69 N, = 0.56 N,' = 0.24 a) Assuming local shear failure, compute the ultimate bearing capacity of the footing if the width is 1.25 m. Ans. 287.21 KPa b) Considering a rectangular footing of 1.25m x 6m and a load factor of 2.5, determine the allowable bearing capacity under general shear failure. B. Qut = cN: (1+ 03) +yD,Ng + 0.5YBN,(1 – 0.2 %3D Ans. 204.97 KPa c) Find the allowable load that the rectangular footing could carry Ans. 1537.28 KN

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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A strip footing is to be placed 2m below the surface or soil having a
cohesion of 40 KPa, unit weight of 18.2 KN/m', and angle of friction of 10
Nc = 9.61
N = 2.69
Ne = 8.02
N = 1.94
N, = 0.56
N,' = 0.24
a) Assuming local shear failure, compute the ultimate bearing capacity of
the footing if the width is 1.25 m.
Ans. 287.21 KPa
b) Considering a rectangular footing of 1.25m x 6m and a load factor of
2.5, determine the allowable bearing capacity under general shear
failure.
cN. (1 + 0.3) + yD,N, + 0.5yBN, (1 – 0.25
Quit
Ans. 204.97 KPa
c) Find the allowable load that the rectangular footing could carry
Ans. 1537.28 KN
Transcribed Image Text:A strip footing is to be placed 2m below the surface or soil having a cohesion of 40 KPa, unit weight of 18.2 KN/m', and angle of friction of 10 Nc = 9.61 N = 2.69 Ne = 8.02 N = 1.94 N, = 0.56 N,' = 0.24 a) Assuming local shear failure, compute the ultimate bearing capacity of the footing if the width is 1.25 m. Ans. 287.21 KPa b) Considering a rectangular footing of 1.25m x 6m and a load factor of 2.5, determine the allowable bearing capacity under general shear failure. cN. (1 + 0.3) + yD,N, + 0.5yBN, (1 – 0.25 Quit Ans. 204.97 KPa c) Find the allowable load that the rectangular footing could carry Ans. 1537.28 KN
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