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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3 Given: T- shape foundation as shown in figure is loaded with a uniform load of 120 kN/m². Required: the increment in vertical stress at point (P) at a depth of 5m. 0.6 0.6 0.6 0.6 3m m n I, 1.5 0.3 0.0629 1.2 0.1431 9m 3m 0.1069
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).
Soil mechanics
A rigid foundation is subjected to a vertical column load, P = 700 kN, as shown in Figure below. Estimate the elastic settlement due to the net applied pressure, Δσ, on the foundation. Given: B = 2m; L = 2m; Df = 1.5m; H = 3m; Es = 13,500 kN/m2; and µs = 0.35. (Useful formula and table can be found from Appendix)
The plan of a mat foundation is shown in Figure 1. Calculate the soil pressure at points A, B, C, D, E and F. (Note: All column sections are planned to be 450mmX450mm). Determine factor of safety of the Mat.
Refer to Figure 7.1. A flexible foundation measuring 1.5 m x 3 m is supported by a saturated clay. Given: Df = 1.2 m, H = 3 m, Es (clay) = 600 kN/m2, and qo = 150 kN/m2. Determine the average elastic settlement of the foundation.
In the following example, a rectangular foundation must be designed whose B/L ratio must be equal to 0.5.The ultimate allowable load Qadm that must be transmitted to the ground is 520 KN, using a safety factor of 2.0. Designing with Meyerhoff theory, assume that the soil undergoes a general shear failure process. For this problem, propose values of B and L that meet the B/L ratio until Qadm is found to be equal to 520 kN.
Question 1) For a shallow foundation measuring (1.7 m x 2.2 m) as shown below: , A. Estimate the elastic settlement proposed by Mayerhof. Then, B. Estimate the elastic settlement proposed by Bowles, if the water table rises 1.5 m. Then, Use yw=10 kN/m³ qnet= 1.2 MN/m2 G.S 1.5 m Sand Yd=16 kN/m³ Ysat= 17 kN/m3 %3D 2.5 m N60=52 V W.T. Silty Sand Ya=18 kN/m³ Ysat = 18.5 kN/m? N60=52 3.5 m Sand Ya=19 kN/m3 Ysat = 22 kN/m³ e, = 0.4, Ae=0.04 , o'= 194 kN/m2 5 m Cc= 0.3, Cs= 0.2 , Ca= 0.05 N60=60 CS Scanned with CamScanner
A rigid shallow foundation 1 m 1 m in plan is shown in Figure below. Calculate the elastic settlement at the center of the foundation. Ag = 200 kN/m2 1 m 1m x1 m E, (kN/m²)| + 8000 Hy = 0.3 2 + 6000 - 3 + 10,000 6/3/2021 ECG 4313 FOUNDATION DESIGN
绘图编辑 3.12 For a strip foundation, the width is 1.2m, buried depth is 1.2m, the standard value of the load applied on the foundation F=155KN/m, the soil layer is distributed as follows: from the ground to the depth of 1.2m is the land fill, =18KN/m³; the depth in the range of 1.2~1.8m is silty clay, fak=155kPa, E,=8.1Mpa, y=19KN/m³; below the depth of 1.8m is the mucky clay, fak=102kPa, E,=2.7Mpa. Try to check the base pressure and the bearing capacity of the weak sub-layer. Ground F=155KN/m OLand fill 18KN/m³ 1.2m O Silty clay La-155kPa;E-8.1Mpa;y-19KN/m³ -0, 1.0 O Mucky clay -0, H1.0 Sak-102kPa;E,-2.7Mpa 1.2m
Raft foundations

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