Example 5.3: Design Example of a Rectangular Foundation forColumnThis section describes the design of a rectangular foundation to support a column havingdimensions of 0.4 m x 0.4 m in cross section. Other details are as follows:Dead load D= 290 kNLive load L= 110 kNDepth from the ground surface to the top of the foundation = 1.2 mAllowable gross soil-bearing capacity = 120 kN/m²Maximum width of foundation = B = 1.5 mf, = 413.7 MN/m²f=20.68 MN/m²Unit weight of soil = y = 17.27 kN/m³Unit weight of concrete = y₁ = 22.97 kN/m³General ConsiderationsFor this design, let us assume a foundation thickness of 0.45 m (Figure A.4a). The weightof foundation/m² = 0.45 y = (0.45) (22.97) = 10.34 kN/m², and the weight of soil abovethe foundation/m² = (1.2)y = (1.2) (17.27) = 20.72 kN/m² Hence, the net allowable soil-bearing capacity Iqata] =120-10.34-20.72 = 88.94 kN/m²The required area of the foundation = (D+L)/4(all) = (290 + 110)/88.94 =4.5 m². Hence, the length of the foundation is 4.5 m²/B = 4.5/1.5 = 3 m.The factored column load = 1.2D + 1.6L = 1.2(290) + 1.6(110) = 524 kN.The factored soil-bearing capacity, 4, = factored load/foundation area = 524/4.5 =116.44 kN/m².

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Answered: Example 5.3: Design Example of a…

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter17: Settlement Of Shallow Foundations

Section: Chapter Questions

Problem 17.6P

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Consider a continuous foundation of width B = 1.4 m on a sand deposit with c = 0, = 38, and = 17.5 kN/m3. The foundation is subjected to an eccentrically inclined load (see Figure 6.33). Given: load eccentricity e = 0.15 m, Df = 1 m, and load inclination = 18. Estimate the failure load Qu(ei) per unit length of the foundation a. for a partially compensated type of loading [Eq. (6.89)] b. for a reinforced type of loading [Eq. (6.90)]
Solve Problem 7.8 using Eq. (7.29). Ignore the post-construction settlement. 7.8 Solve Problem 7.4 with Eq. (7.20). Ignore the correction factor for creep. For the unit weight of soil, use γ = 115 lb/ft3. 7.4 Figure 7.3 shows a foundation of 10 ft × 6.25 ft resting on a sand deposit. The net load per unit area at the level of the foundation, qo, is 3000 lb/ft2. For the sand, μs = 0.3, Es = 3200 lb/in.2, Df = 2.5 ft, and H = 32 ft. Assume that the foundation is rigid and determine the elastic settlement the foundation would undergo. Use Eqs. (7.4) and (7.12).
The subgrade reaction of a sandy soil obtained from the plate load test (plate dimensions 1 m × 0.7 m) is 18 MN/m3. What will be the value of k on the same soil for a foundation measuring 5 m × 3.5 m?
A 2.0 m wide continuous foundation carries a wall load of 350 kN/m in a clayey soil where = 19.0 kN/m3, c = 5.0 kN/m2, and = 23. The foundation depth is 1.5 m. Determine the factor of safety of this foundation using Eq. (6.28).
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
Refer to Problem 8.8. If the full-sized foundation had dimensions of 70 ft × 30 ft, what will be the value of the coefficient of subgrade reaction? 8.8 From the plate load test (plate dimensions 1 ft × 1 ft) in the field, the coefficient of subgrade reaction of a sandy soil is determined to be 60 lb/in3. What will be the value of the coefficient of subgrade reaction on the same soil for a foundation with dimensions of 20 ft × 20 ft?
Repeat Problem 20.2 based on limit state design, using the factors given in Table 20.4. 20.2 A continuous foundation is required in a soil where c = 10 kN/m2, =26 and = 19.0 kN/m3. The depth of the foundation will be 1.0 m. The dead load and the live load are 600 kN/m and 400 kN/m, respectively. Determine the required width for the foundation based on allowable stress design with FS = 3, using Eq. (16.3) and Table 16.1.
Refer to Figure 5.2. Given: B = L = 1.75 m, Df = 1 m, H = 1.75 m, = 17 kN/m3, c = 0, and = 30. Using Eq. (5.6) and FS = 4, determine the gross allowable load the foundation can carry.
Repeat Problem 5.1 with the following data: B = 1.5 m, L = 1.5 m, Df = 1 m, H = 0.6 m, = 35, c = 0, and = 15 kN/m3. Use FS = 3. Refer to Figure 5.2 and consider a rectangular foundation. Given: B = 1.5 m, L = 2.5 m, Df = 1.2 m, H = 0.9 m, = 40, c = 0, and = 17 kN/m3. Using a factor of safety of 3, determine the gross allowable load the foundation can carry. Use Eq. (5.3).
It is proposed to place a 9 ft 9 ft foundation at 6 ft depth in a sandy soil, where the average N60 is 25 and the unit weight is 115.0 lb/ft3. Using Meyerhofs expressions presented in Section 9.6, estimate the allowable net pressure that would give 1.2 in. of settlement.
In Problem 7.3, if no bedrock was present for at least 4 m below the foundation, determine the ultimate bearing capacity. 7.3 A 1.5 m × 2.0 m rectangular foundation is placed at 1.0 m depth in sand where ϕ′ = 40° and y = 18.5 kN/m3. Bedrock is present at 1.0 m below the foundation. Using Eq. (7.3), determine the ultimate bearing capacity of the foundation.
Refer to the rectangular combined footing in Figure 10.1, with Q1 = 100 kip and Q2 = 150 kip. The distance between the two column loads L3 = 13.5 ft. The proximity of the property line at the left edge requires that L2 = 3.0 ft. The net allowable soil pressure is 2500 lb/ft2. Determine the breadth and length of a rectangular combined footing.

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