   Chapter 16, Problem 16.15P 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 Problem 16.13. Design the size of the footing using the modified general ultimate bearing capacity Eq. (16.31).16.13 A square footing (B × B) must carry a gross allowable load of 1160 kN. The base of the footing is to be located at a depth of 2 m below the ground surface. If the required factor of safety is 4.5, determine the size of the footing. Use Terzaghi’s bearing capacity factors and assume general shear failure of soil. Given: γ = 17 kN/m3, c′ = 48 kN/m2, ϕ′ = 31°.

To determine

Find the size of the footing using the modified general ultimate bearing capacity equation.

Explanation

Given information:

The gross allowable load (qall) is 1,160 kN.

The location of depth of footing base Df is 2 m.

The factor of safety Fs is 4.5.

The unit weight of the soil γ is 17kN/m3.

The value of cohesion c is 48kN/m2.

The soil friction angle ϕ is 31°.

Calculation:

Trial 1.

Consider width of footing B as 1.25 m.

Determine the shape factor λcs using the relation.

λcs=1+(BL)(NqNc)

Here, Nq and Nc are bearing-capacity factors.

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

For ϕ=31°;

The value of Nc is 32.67, Nq is 20.63, and Nγ is 25.99.

Substitute 1.25 m for B, 1.25 m for L, 20.63 for Nq, and 32.67 for Nc.

λcs=1+(1.251.25)(20.6332.67)=1.63

The ratio of (DfB) is greater than 1.0.

Determine the depth factor λcd using the relation.

λcd=1+0.4tan1(DfB)

Substitute 2 m for Df and 1.25 m for B.

λcd=1+0.4tan1(21.25)=1.404

Determine the shape factor λqs using the relation.

λqs=1+(BL)tanϕ

Substitute 1.25 m for B, 1.25 m for L, and 31° for ϕ.

λqs=1+(1.251.25)tan31°=1.6

Determine the depth factor λqd using the relation.

λqd=1+2tanϕ(1sinϕ)2tan1(DfB)

Substitute 31° for ϕ, 2 m for Df, and 1.25 m for B.

λqd=1+2tan31°(1sin31°)2tan1(21.25)=1.286

Determine the depth factor λγs using the relation.

λγs=10.4(BL)

Substitute 1.25 m for B and 1.25 m for L.

λγs=10.4(1.251.25)=0.6

Determine the ultimate bearing capacity of the soil (qu) using the relation.

qu=cNcλcsλcdλci+qNqλqsλqdλqi+12γBNγλγsλγdλγi=cNcλcsλcdλci+γDfNqλqsλqdλqi+12γBNγλγsλγdλγi (1)

Here, λγd is the depth factor, λci,λqi,andλγi are the inclination factors, and Nγ is the contribution of unit weight of soil.

For a vertical loading, all inclination factors are equal to one.

Substitute 48kN/m2 for c, 32.67 for Nc, 1.63 for λcs, 1.404 for λcd, 1.0 for λci, 17kN/m3 for γ, 2 m for Df, 20.63 for Nq, 1.6 for λqs, 1.286 for λqd, 1.0 for λqi, 1.25 m for B, 25.99 for Nγ, 0.6 for λγs, 1.0 for λγd, and 1.0 for λγi in Equation (1).

qu={(48×32.67×1.63×1.404×1)+(17×2×20.63×1.6×1.286×1.0)+(12×17×1.25×25.99×0.6×1×1)}=5,197.7kN/m2

Determine the gross allowable load (qall) that the footing can carry using the relation.

qall=quB2Fs

Substitute 5,197.7kN/m2 for qu, 1

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