   Chapter 16, Problem 16.1P 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

A continuous footing is shown in Figure 16.17. Using Terzaghi’s bearing capacity factors, determine the gross allowable load per unit area (qall) that the footing can carry. Assume general shear failure. Given: γ = 19 kN/m3, c′ = 31kN/m2, ϕ ′ = 28 ° , Df = 1.5 m, B = 2 m, and factor of safety = 3.5. Figure 16.17

To determine

Find the gross allowable load per unit area (qall) that the footing can carry.

The gross allowable load per unit area (qall) that the footing can carry is 499.4kN/m2_.

Explanation

Given information:

The unit weight of the soil γ is 19kN/m3.

The value of cohesion c is 31kN/m2.

The soil friction angle ϕ is 28°.

The depth of foundation Df is 1.5 m.

The width of footing B is 2 m.

The factor of safety Fs is 3.5.

Calculation:

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

qu=cNc+qNq+12γBNγ=cNc+γDfNq+12γBNγ (1)

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 Nc as 31.61, Nq as 17.81, and Nγ as 13.70 for the ϕ value of 28°.

Substitute 31kN/m2 for c, 31.61 for Nc, 19kN/m3 for γ, 1.5 m for Df, 17.81 for Nq, 2 m for B, and 13.70 for Nγ in Equation (1).

qu=(31×31.61)+(19×1.5×17.81)+(12×19×2×13.70)=1,747.795kN/m2

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

qall=quFs

Substitute 1,747.795kN/m2 for qu and 3.5 for Fs.

qall=1,747.7953.5=499.4kN/m2

Therefore, the gross allowable load per unit area (qall) that the footing can carry is 499.4kN/m2_.

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