2. For a 50-story (above the ground) building with 4 floors of basement (D = 12.0 m), the average floor weight (including dead load and live load) is 15 kN/m². Assume the height of each floor of the basement is 3 m (including the thickness of the slab). The mat foundation (with a dimension of 30 m x 20 m) is situated in a thick clay layer with a saturated unit weight of 18 kN/m³ and an undrained shear strength (Su) of 30 kPa (remains constant with depth). In addition, the clay has an effective friction angle o' = 28° while c' = 0. Answer the following questions: (A) Would the mat foundation of the building be fully compensated? (B) The factor of safety for bearing capacity in a short-term condition. (C) The factor of safety for bearing capacity in a long-term condition. (Use FS = qu(net)/q, where the generalized bearing capacity equation is given by qu(net) = c'NcFcs Fcd Fci + YDNqFqsFqa Fai + 0.5yB Ny Fys Fya Fyi - YDf and q is the net average applied pressure on soil). Figure, Tables, and Equations: qu5.14c 1+0.195- 52) (1+0.4%)+9 9.-5.14c14 5.14c. (1+0.195)(1 +0.4%) Qu(net) qu-q= 5.14c, 1+ 0.195 Pile: Meyerhof's method Q₁ = 4,9 = 4,q'N₁₁ ≤ A¸¶₁ q₁ = 0.5ptan o' TABLE 12.6 Interpolated Values of N Based on Meyerhof's Theory (10.11) B Soil friction angle, ' (deg) 20 21 N₁₁ (10.12) all(net) Qu(net) FS = 1.713c,(1+0.195)(1 +0.4%) (10.13) B 26 27 28 General bearing capacity equation qu= c'NcFcs Fca Fci+q'NqFqsFqaFqi + 0.5yBNy Fys Fya Fyi Shape factors by De Beer Depth factors by Hansen (1970) (1970) 29 Inclination factors by Meyerhof (1963) and Hanna and Meyerhof (1981) 33 2222222222223 12.4 13.8 15.5 17.9 24 21.4 25 26.0 29.5 34.0 39.7 46.5 30 56.7 68.2 81.0 96.0 Fcs = 1+ B N L-Ne 34 Во 115.0 Fcd = 1 +0.4() Fci = Fqi = (1 900)2 35 143.0 36 168.0 B = Fas 1+) tan op' Fqd = 1 + 2 tan op' (1 - sin p')² - Df 37 194.0 Fyi = (1 - B 38 231.0 B Fys = 1-0.4(7) Fyd = 1 39 276.0 40 346.0 41 420.0 42 525.0 Shallow foundation: TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) 43 650.0 TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued) 44 780.0 45 930.0 $' N₂ N Ny $' N No N₁ $' No 22 N₁ Ny $' No Na 16.88 7.82 7.13 37 55.63 42.92 66.19 Ny 23 18.05 8.66 8.20 38 61.35 48.93 78.03 0 5.14 1.00 0.00 11 8.80 2.71 1.44 24 19.32 9.60 9.44 39 67.87 55.96 92.25 1 5.38 1.09 0.07 12 9.28 2.97 1.69 The critical depth for skin friction in piles 25 20.72 10.66 10.88 40 75.31 64.20 109.41 2 5.63 1.20 0.15 13 9.81 3.26 1.97 26 22.25 11.85 12.54 41 83.86 73.90 130.22 3 5.90 1.31 0.24 14 10.37 3.59 2.29 27 23.94 13.20 14.47 42 93.71 85.38 155.55 4 6.19 1.43 0.34 15 10.98 3.94 2.65 28 25.80 14.72 16.72 43 105.11 99.02 186.54 L'≈ 15D 5 6.49 1.57 0.45 16 11.63 29 4.34 27.86 16.44 19.34 44 118.37 115.31 224.64 3.06 30 30.14 18.40 22.40 45 133.88 134.88 271.76 6 6.81 1.72 0.57 17 12.34 4.77 3.53 31 32.67 20.63 25.99 46 152.10 158.51 330.35 7 7.16 1.88 0.71 18 13.10 5.26 4.07 32 35.49 23.18 30.22 47 173.64 187.21 403.67 The unit frictional resistance or the unit skin friction 8 7.53 2.06 0.86 19 13.93 5.80 4.68 33 38.64 26.09 35.19 48 199.26 222.31 496.01 9 7.92 2.25 1.03 20 14.83 6.40 5.39 34 42.16 29.44 41.06 49 229.93 265.51 613.16 10 8.35 2.47 1.22 21 15.82 7.07 6.20 35 46.12 33.30 48.03 50 266.89 319.07 762.89 (continued) 36 50.59 37.75 56.31 f = Ko tan 8'

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)]

A 300 mm 450 mm plate was used in carrying out a plate loading test in a sand, during which the plate settled 5 mm under the applied pressure of 250 kN/m2. a. What is the coefficient of subgrade reaction for a 300 mm wide square plate? b. What would be the coefficient of subgrade reaction of a 2 m 3 m foundation?

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.