Determine the collapse load for the simplysupported slab.3 m3 m

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Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter15: Retaining Walls, Braced Cuts, And Sheet Pile Walls

Section: Chapter Questions

Problem 15.12P

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For the following cases, determine the allowable gross vertical load-bearing capacity of the foundation. Use Terzaghi’s equation and assume general shear failure in soil. Use FS = 4. Parameters for Problem 6.1
A flexible circular footing of radius R carries a uniform pressure q. Find the depth (in terms of R) at which the vertical stress below the center is 20% of q.
A square column foundation has to carry a gross allowable load of 1805 kN (FS = 3). Given: Df = 1.5 m, =15.9 kN/m3, = 34, and c = 0. Use Terzaghis equation to determine the size of the foundation (B). Assume general shear failure.
In Problem 7.1, if there was no bedrock present for at least 12.0 ft below the foundation, what would the ultimate bearing capacity be? 7.1 A 7.5 ft wide rough continuous foundation is placed in the ground at 3.0 ft depth. There is bedrock present at 3 ft depth below the bottom of the foundation. The soil properties are c′ = 210 lb/ft2, ϕ′ = 25°, and γ = 115.0 lb/ft3. Determine the ultimate bearing capacity of the foundation.
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 Terzaghis equation for general shear failure (Fs = 4). Given: = 17 kN/m3, sat = 19.2 kN/m3, c = 32 kN/m3, =26, Df = 1 m, h = 0.5 m, and B = 1.5 m. Figure 16.18
A column foundation (Figure 16.23) is 3 m 2 m in plan. Given: Df = 1.5 m, =25, c = 70 kN/m2. Using Eq. (16.9) and FS = 3, determine the net allowable load [see Eq. (16.16)] the foundation could carry. FIG. 16.23
Redo Problem 7.6 using Vesic’s (1975) solution [Eq. (7.12)]. 7.6 A 2.0 m wide continuous foundation is placed at 1.5 m depth in a saturated clay where cu = 40 kN/m2 and γ = 18.5 kN/m3. At 2.0 m below the ground level, this clay layer is underlain by a stiffer clay where cu = 60 kN/m2 and γ = 19.0 kN/m3. What would be the maximum wall load allowed with FS = 3? Use Eq. (7.11).
A column foundation (Figure P6.9) is 3 m × 2 m in plan. Given: Df = 1.5 m, ф′ = 25°, c′ = 70 kN/m2. Using Eq. (6.28) and FS = 3, determine the net allowable load [see Eq. (6.24)] the foundation could carry. Figure P6.9
For the mat in Problem 16.15, what will be the depth, Df, of the mat for FS = 3 against bearing capacity failure? 16.15 Consider a mat foundation with dimensions of 18 m 12 m. The combined dead and live load on the mat is 44.5 MN. The mat is to be placed on a clay with cu = 40.7 kN/m2 and = 17.6 kN/m3. Find the depth, Df, of the mat for a fully compensated foundation.
Redo Problem 16.4 using the modified general ultimate bearing capacity Eq. (16.31). 16.1 A continuous footing is shown in Figure 16.17. Using Terzaghis bearing capacity factors, determine the gross allowable load per unit area (qall) that the footing can carry. Assume general shear failure. Given: = 19 kN/m3, c = 31kN/m2, =28, Df = 1.5 m, B = 2 m, and factor of safety = 3.5. Figure 16.17 16.4 Redo Problem 16.1 with the following: = 16.5 kN/m3, cu = 41 kN/m3, =0, Df = 1.5 m, and factor of safety = 5.
Refer to Problem 16.1. If a square footing with dimension 2 m 2 m is used instead of the wall footing, what would be the allowable bearing capacity? 16.1 A continuous footing is shown in Figure 16.17. Using Terzaghis bearing capacity factors, determine the gross allowable load per unit area (qall) that the footing can carry. Assume general shear failure. Given: = 19 kN/m3, c = 31kN/m2, =28, Df = 1.5 m, B = 2 m, and factor of safety = 3.5. Figure 16.17
The applied load on a shallow square foundation makes an angle of 20° with the vertical. Given: B = 5 ft, Df = 3 ft, γ = 115 lb/ft3, ф′ = 25°, and c′ = 600 lb/ft2. Use FS = 3 and determine the gross inclined allowable load. Use Eq. (6.28).

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Determine the collapse load for the simply supported slab. 3 m 3 m