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 Fca Fci+YDƒNqFqsFqaFqi +0.5yB NyFysFyaFyi - YD, and q is the net average applied pressure on soil. = Figure, Tables, and Equations: 9.5.14c (1+0.1951+0.4%)+ (10.11) B qu(net) quq = 5.14c, 1+0.195 -q= (1+0.1952)(1 +0.4%) (10.12) all(net) = qu(net) FS = 1.713c, (1 + 0.1952) (1+0.42) (10.13) - For FS=3 General bearing capacity equation: qu = c'NcFcs FedFci+q'NqFqs Fqa Fai + 0.5yBNy FysFydFyi Shape factors by De Beer (1970) B N Fcs = 1 + 2)²) Depth factors by Hansen (1970) Inclination factors by Meyerhof (1963) and Hanna and Meyerhof (1981) Fei Fai = (1- = Fed 1+0.4(- 4 (1) B° Fqs B =1+) tan o' 900)2 Fqd = 1 + 2 tan q' (1 − sin ')² Dr - B Fyi β = (1 - Fyd = 1 B Fy, = 1-04(-) Bearing capacity factors Nc, Nq, Ny TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued) $' N. N₁ Ny $' No N₁ Ny 22 16.88 7.82 7.13 37 55.63 42.92 66.19 TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) 23 18.05 8.66 8.20 38 61.35 48.93 78.03 24 19.32 9.60 9.44 39 67.87 55.96 92.25 $' Ne Na Ny $' Ne N₁ Ny 25 20.72 10.66 10.88 40 75.31 64.20 109.41 0 5.14 1.00 0.00 11 8.80 2.71 1.44 26 22.25 11.85 12.54 41 83.86 73.90 130.22 1 5.38 1.09 0.07 12 9.28 2.97 1.69 27 23.94 13.20 14.47 42 93.71 85.38 155.55 2 5.63 1.20 0.15 13 9.81 3.26 1.97 28 25.80 14.72 16.72 43 105.11 99.02 186.54 3 5.90 1.31 0.24 14 10.37 3.59 2.29 29 27.86 16.44 19.34. 44 118.37 115.31 224.64 4 6.19 1.43 0.34 15 10.98 3.94 2.65 30 30.14 18.40 22.40 45 133.88 134.88 5 6.49 1.57 0.45 16 11.63 271.76 4.34 3.06 31 32.67 20.63 25.99 46 152.10 158.51 330.35 6 6.81 1.72 0.57 17 12.34 4.77 3.53 32 35.49 23.18 30.22 47 173.64 187.21 7 7.16 1.88 0.71 18 13.10 403.67 5.26 4.07 33 38.64 26.09 35.19 48 199.26 222.31 496.01 8 7.53 2.06 0.86 19 13.93 5.80 4.68 34 42.16 29.44 41.06 9 7.92 2.25 1.03 20 14.83 6.40 5.39 35 46.12 33.30 48.03. 55 49 229.93 265.51 613.16 50 266.89 319.07 762.89 10 8.35 2.47 1.22 21 15.82 7.07 6.20 36 50.59 37.75 56.31 (continued)

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 Fca Fci+YDƒNqFqsFqaFqi +0.5yB NyFysFyaFyi - YD, and q is the net average applied pressure on soil. = Figure, Tables, and Equations: 9.5.14c (1+0.1951+0.4%)+ (10.11) B qu(net) quq = 5.14c, 1+0.195 -q= (1+0.1952)(1 +0.4%) (10.12) all(net) = qu(net) FS = 1.713c, (1 + 0.1952) (1+0.42) (10.13) - For FS=3 General bearing capacity equation: qu = c'NcFcs FedFci+q'NqFqs Fqa Fai + 0.5yBNy FysFydFyi Shape factors by De Beer (1970) B N Fcs = 1 + 2)²) Depth factors by Hansen (1970) Inclination factors by Meyerhof (1963) and Hanna and Meyerhof (1981) Fei Fai = (1- = Fed 1+0.4(- 4 (1) B° Fqs B =1+) tan o' 900)2 Fqd = 1 + 2 tan q' (1 − sin ')² Dr - B Fyi β = (1 - Fyd = 1 B Fy, = 1-04(-) Bearing capacity factors Nc, Nq, Ny TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued) $' N. N₁ Ny $' No N₁ Ny 22 16.88 7.82 7.13 37 55.63 42.92 66.19 TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) 23 18.05 8.66 8.20 38 61.35 48.93 78.03 24 19.32 9.60 9.44 39 67.87 55.96 92.25 $' Ne Na Ny $' Ne N₁ Ny 25 20.72 10.66 10.88 40 75.31 64.20 109.41 0 5.14 1.00 0.00 11 8.80 2.71 1.44 26 22.25 11.85 12.54 41 83.86 73.90 130.22 1 5.38 1.09 0.07 12 9.28 2.97 1.69 27 23.94 13.20 14.47 42 93.71 85.38 155.55 2 5.63 1.20 0.15 13 9.81 3.26 1.97 28 25.80 14.72 16.72 43 105.11 99.02 186.54 3 5.90 1.31 0.24 14 10.37 3.59 2.29 29 27.86 16.44 19.34. 44 118.37 115.31 224.64 4 6.19 1.43 0.34 15 10.98 3.94 2.65 30 30.14 18.40 22.40 45 133.88 134.88 5 6.49 1.57 0.45 16 11.63 271.76 4.34 3.06 31 32.67 20.63 25.99 46 152.10 158.51 330.35 6 6.81 1.72 0.57 17 12.34 4.77 3.53 32 35.49 23.18 30.22 47 173.64 187.21 7 7.16 1.88 0.71 18 13.10 403.67 5.26 4.07 33 38.64 26.09 35.19 48 199.26 222.31 496.01 8 7.53 2.06 0.86 19 13.93 5.80 4.68 34 42.16 29.44 41.06 9 7.92 2.25 1.03 20 14.83 6.40 5.39 35 46.12 33.30 48.03. 55 49 229.93 265.51 613.16 50 266.89 319.07 762.89 10 8.35 2.47 1.22 21 15.82 7.07 6.20 36 50.59 37.75 56.31 (continued)

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter10: Mat Foundations

Section: Chapter Questions

Problem 10.8P

See similar textbooks