   Chapter 16, Problem 16.18P Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939

Solutions

Chapter
Section Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939
Textbook Problem

Refer to the footing in Problem 16.16. Determine the gross ultimate load the footing can carry using the Patra et al. (2015) reduction factor method for rectangular foundations given in Eqs. (16.47), (16.49), and (16.50).16.16 A square footing on sand is subjected to an eccentric load as shown in Figure 16.20. Using Meyerhof’s effective area concept, determine the gross allowable load that the footing could carry with Fs = 4. Given: γ = 16 kN/m3, c′ = 0, ϕ′ = 29°, Df = 1.3 m, B = 1.75 m, and x = 0.25 m. Use Eqs. (16.32) through (16.42) for shape, depth, and inclination factors. To determine

Find the gross ultimate load that the footing can carry using the reduction factor method.

Explanation

Given information:

The unit weight of the soil γ is 16kN/m3.

The value of cohesion c is 0.

The soil friction angle ϕ is 29°.

The location of depth of footing base Df is 1.3 m.

The width of the footing B is 1.75 m.

The value of eccentricity e is 0.25 m.

The factor of safety value Fs is 4.

Calculation:

Determine the reduction factor Rk using the relation.

Rk=a(eB)k (1)

Here, a and k are the functions of the embedment ratio DfB.

Determine the function a using the relation.

a=(BL)21.6(BL)+2.13

Substitute 1.75 m for B and 1.75 m for L.

a=(1.751.75)21.6(1.751.75)+2.13=1.53

Determine the function k using the relation.

k=0.3(BL)20.56(BL)+0.9

Substitute 1.75 m for B and 1.75 m for L.

k=0.3(1.751.75)20.56(1.751.75)+0.9=0.64

Substitute 1.53 for a, 0.25 m for e, 1.75 m for B, and 0.64 for k in Equation (1).

Rk=1.53(0.251.75)0.64=0.44

Determine the shape factor λcs using the relation.

λcs=1+(BL)(NqNc)

Here, Nq and Nc are the bearing-capacity factors.

Refer Table 16.2, “Bearing-capacity factors Nc, Nq, and Nγ” in the textbook.

For ϕ=29°;

The values of Nc is 27.86, Nq is 16.44, and Nγ is 19.34.

Substitute 1.75 m for B, 1.75 m for L, 16.44 for Nq, and 27.86 for Nc.

λcs=1+(1.751.75)(16.4427.86)=1.421

Determine the depth factor λcd using the relation.

λcd=1+0.4tan1(DfB)

Substitute 1.3 m for Df and 1.75 m for B.

λcd=1+0.4(1.31.75)=1.297

Determine the shape factor λqs using the relation.

λqs=1+(BL)tanϕ

Substitute 1.75 m for B, 1.75 m for L, and 29° for ϕ.

λqs=1+(1.751.75)tan29°=1.554

Determine the depth factor λqd using the relation.

λqd=1+2tanϕ(1sinϕ)2(DfB)

Substitute 29° for ϕ, 1

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