Example 5.2: Design Example of a Square Foundation for a Colum Figure A.3a shows a square column foundation with the followin conditions: Live load=L=675 kN Dead load D= 1125 kN Allowable gross soil-bearing capacity 145 kN/m² Column size-0.5 m x 0.5 m -20.68 MN/m² 1,413.7 MN/m² Let it be required to design the column foundation. 1.25m-D L-675 AN D-1125 AN (a) Column section- 05 m x 0.5m

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).

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