   Chapter 16, Problem 16.10P 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

If the water table in Problem 16.9 drops down to 0.25 m below the foundation level, what would be the change in the factor of safety for the same gross allowable load?16.9 A square footing is shown in Figure 16.18. Determine the gross allowable load, Qall, that the footing can carry. Use Terzaghi’s equation for general shear failure (Fs = 4). Given: γ = 17 kN/m3, γsat = 19.2 kN/m3, c′ = 32 kN/m3, ϕ ′ = 26 ° , Df = 1 m, h = 0.5 m, and B = 1.5 m. Figure 16.18

To determine

Find the change in factor of safety for the estimated gross allowable load.

Explanation

Given information:

The factor of safety Fs is 4.0.

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

The saturated unit weight of the soil γsat is 19.2kN/m3.

The value of cohesion c is 32kN/m2.

The soil friction angle ϕ is 26°.

The depth of foundation Df is 1.0 m.

The depth of soil above water table h is 0.5 m.

The drop down depth of water table D is 0.25 m.

The width of footing B is 1.5 m.

Calculation:

For depth of water table at 0.5 m.

Determine the magnitude of q using the relation.

q=γ(Dfh)+γh=γ(Dfh)+(γsatγw)h (1)

Here, γ is the effective unit weight of soil and γw is the unit weight of water.

Take unit weight of water as 9.81kN/m3.

Substitute 17kN/m3 for γ, 1.0 m for Df, 0.5 m for h, 19.2kN/m3 for γsat, and 9.81kN/m3 for γw in Equation (1).

q=17(1.00.5)+(19.29.81)0.5=13.19kN/m2

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

qu=1.3cNc+qNq+0.4γBNγ

Here, Nc is the contribution of cohesion, Nq is the contribution of surcharge, and Nγ is the contribution of unit weight of soil.

Refer Table 16.1, “Terzaghi’s bearing-capacity factors–Nc, Nq, and Nγ” in the textbook.

Take the Nc as 27.09, Nq as 14.21, and Nγ as 9.84 for the ϕ value of 26°.

Substitute 32kN/m2 for c, 27.09 for Nc, 13.19kN/m2 for q, 14.21 for Nq, (19.29.81)kN/m3 for γ, 1.5 m for B, and 9.84 for Nγ.

qu=1.3×32×27.09+13.19×14.21+0.4×(19.29.81)×1.5×9.84=1,369.81kN/m2

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

qall=quB2Fs

Substitute 1,369.81kN/m2 for qu, 1.5 m for B, and 4.0 for Fs.

qall=1,369.81×1.524.0=771kN

For drop down depth of ground water table at 0

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