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shear strength shear strength (su) of 120 kPa. Given the depth of bottom of the mat D₁ = 3.0 m, (a) determine the factor of safety for bearing capacity (Use Fs = quq where the generalized bearing capacity equation qu = c'NcFesFca Fci+q'NqFas Fad Fai + 0.5yBNy FysFydFyi). (b) determine the factor of safety against bearing capacity by adopting the concept of compensated foundation (Use Fs = qnet(u)/(q-Di)). (c) if a fully compensated foundation, i.e. no net increase of stress on the soil at the bottom of the mat foundation, is desired, what would be the depth of the foundation?

shear strength shear strength (su) of 120 kPa. Given the depth of bottom of the mat D₁ = 3.0 m, (a) determine the factor of safety for bearing capacity (Use Fs = quq where the generalized bearing capacity equation qu = c'NcFesFca Fci+q'NqFas Fad Fai + 0.5yBNy FysFydFyi). (b) determine the factor of safety against bearing capacity by adopting the concept of compensated foundation (Use Fs = qnet(u)/(q-Di)). (c) if a fully compensated foundation, i.e. no net increase of stress on the soil at the bottom of the mat foundation, is desired, what would be the depth of the foundation?

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A mat foundation with a flat slab has dimensions of 15 m × 25 m. The total load of the building is 240 × 103 kN. The mat foundation is situated in a thick clay layer with a saturated unit weight of 18.5 kN/m³ and an undrained shear strength shear strength (su) of 120 kPa. Given the depth of bottom of the mat Dƒ = 3.0 m, (a) determine the factor of safety for bearing capacity (Use Fs = qulaq where the generalized bearing capacity equation qu = c'NcFcsFca Fci + q'NqFqsFqaFqi + 0.5yBNyFysFyaFyi). (b) determine the factor of safety against bearing capacity by adopting the concept of compensated foundation (Use Fs = qnet(u)(q-D)). (c) if a fully compensated foundation, i.e., no net increase of stress on the soil at the bottom of the mat foundation, is desired, what would be the depth of the foundation?
Transcribed Image Text:A mat foundation with a flat slab has dimensions of 15 m × 25 m. The total load of the building is 240 × 103 kN. The mat foundation is situated in a thick clay layer with a saturated unit weight of 18.5 kN/m³ and an undrained shear strength shear strength (su) of 120 kPa. Given the depth of bottom of the mat Dƒ = 3.0 m, (a) determine the factor of safety for bearing capacity (Use Fs = qulaq where the generalized bearing capacity equation qu = c'NcFcsFca Fci + q'NqFqsFqaFqi + 0.5yBNyFysFyaFyi). (b) determine the factor of safety against bearing capacity by adopting the concept of compensated foundation (Use Fs = qnet(u)(q-D)). (c) if a fully compensated foundation, i.e., no net increase of stress on the soil at the bottom of the mat foundation, is desired, what would be the depth of the foundation?
9net(u) = quq = 5.14c1+ (1+0.4²) General bearing capacity equation qu= c'NcFesFcaFci + q'NaFas FaaFai +0.5yBNyFysFya Fyi Shape factors by De Depth factors by Hansen (1970) Beer (1970) B No Fcs = 1+ N Fas = 1 + Fys 0 1 2 3 4 5 6 7 8 9 10 B = 10.4( Ne 5.14 5.38 5.63 5.90 6.19 6.49 6.81 7.16 7.53 7.92 8.35 tan o' 0.195B L TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) N₁ Na 1.00 2.71 1.09 2.97 1.20 3.26 1.31 3.59 1.43 3.94 1.57 4.34 1.72 1.88 2.06 2.25 2.47 Ne 16.88 18.05 19.32 20.72 22.25 23.94 25.80 27.86 30.14 32.67 35.49 38.64 42.16 46.12 50.59 Fad = 1 + 2 tano (1 — sin y)²: qd B Fyd = 1 Fcd = 1 + 0.4( Ny 0.00 0.07 0.15 0.24 0.34 0.45 0.57 0.71 0.86 1.03 1.22 7.13 8.20 9.44 10.88 12.54 14.47 16.72 19.34 22.40 25.99 30.22 35.19 41.06 48.03 56.31 = 11 12 13 14 15 16 17 18 19 20 21 TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued) φ' Na $' Na 22 7.82 42.92 23 8.66 48.93 24 9.60 55.96 25 10.66 64.20 26 11.85 73.90 27 13.20 85.38 28 14.72 99.02 29 16.44 115.31 30 18.40 134.88 31 20.63 158.51 32 23.18 187.21 33 26.09 222.31 34 29.44 265.51 35 33.30 319.07 36 37.75 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Ne 8.80 9.28 9.81 10.37 10.98 11.63 12.34 13.10 13.93 14.83 15.82 For saturated clay: p For a method: fav = acu For method: f = (₁ +2c₂) Ne 55.63 61.35 67.87 75.31 83.86 93.71 105.11 118.37 133.88 152.10 173.64 199.26 229.93 266.89 Apqp = ApCu Nc 4.77 5.26 5.80 6.40 7.07 Ny 1.44 1.69 1.97 2.29 2.65 3.06 3.53 4.07 4.68 5.39 6.20 (continued) Ny 66.19 78.03 92.25 109.41 130.22 155.55 186.54 224.64 271.76 330.35 403.67 496.01 613.16 762.89 Inclination factors by Meyerhof (1963) and Hanna and Meyerhof (1981) Fci = Fai = (1-2 Fyi = (1 - B² Bº Ne 9 for p = 0 TABLE 12.10 Variation of A with Pile Embedment Length, L Embedment length, L (m) 0 5 10 15 20 25 30 35 40 50 60 70 80 90 Coyle and Castello (1981): Embedment ratio, L/D 0 10 20 30 40 50 60 70 Qp = q'N₁ Ap 10 T T T T T e $' Bearing capacity factor, N 20 32° 36° = 30° 34° λ 0.5 0.336 0.245 0.200 0.173 0.150 0.136 0.132 0.127 0.118 0.113 0.110 0.110 0.110 38° TABLE 12.11 Variation of a (Interpo- lated Values Based on Terzaghi et al., 1996) 40° Cu Pa ≤0.1 0.2 0.3 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.4 2.8 40 60 80 100 200 Note: Pa ≈100 kN/m² Qs = atmospheric pressure Embedment ratio, L/D = : (Ko'tan 8')pL 0.15 0.2 0 5 10 15 20 25 30 a 35 36 1.00 0.92 0.82 0.74 0.62 0.54 0.48 0.42 0.40 0.38 0.36 0.35 0.34 0.34 Earth pressure coefficient, K 1.0 $' 30° 31° 32° 33° 8=0.80' 2 34° 35⁰ 36° 5
Transcribed Image Text:9net(u) = quq = 5.14c1+ (1+0.4²) General bearing capacity equation qu= c'NcFesFcaFci + q'NaFas FaaFai +0.5yBNyFysFya Fyi Shape factors by De Depth factors by Hansen (1970) Beer (1970) B No Fcs = 1+ N Fas = 1 + Fys 0 1 2 3 4 5 6 7 8 9 10 B = 10.4( Ne 5.14 5.38 5.63 5.90 6.19 6.49 6.81 7.16 7.53 7.92 8.35 tan o' 0.195B L TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) N₁ Na 1.00 2.71 1.09 2.97 1.20 3.26 1.31 3.59 1.43 3.94 1.57 4.34 1.72 1.88 2.06 2.25 2.47 Ne 16.88 18.05 19.32 20.72 22.25 23.94 25.80 27.86 30.14 32.67 35.49 38.64 42.16 46.12 50.59 Fad = 1 + 2 tano (1 — sin y)²: qd B Fyd = 1 Fcd = 1 + 0.4( Ny 0.00 0.07 0.15 0.24 0.34 0.45 0.57 0.71 0.86 1.03 1.22 7.13 8.20 9.44 10.88 12.54 14.47 16.72 19.34 22.40 25.99 30.22 35.19 41.06 48.03 56.31 = 11 12 13 14 15 16 17 18 19 20 21 TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued) φ' Na $' Na 22 7.82 42.92 23 8.66 48.93 24 9.60 55.96 25 10.66 64.20 26 11.85 73.90 27 13.20 85.38 28 14.72 99.02 29 16.44 115.31 30 18.40 134.88 31 20.63 158.51 32 23.18 187.21 33 26.09 222.31 34 29.44 265.51 35 33.30 319.07 36 37.75 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Ne 8.80 9.28 9.81 10.37 10.98 11.63 12.34 13.10 13.93 14.83 15.82 For saturated clay: p For a method: fav = acu For method: f = (₁ +2c₂) Ne 55.63 61.35 67.87 75.31 83.86 93.71 105.11 118.37 133.88 152.10 173.64 199.26 229.93 266.89 Apqp = ApCu Nc 4.77 5.26 5.80 6.40 7.07 Ny 1.44 1.69 1.97 2.29 2.65 3.06 3.53 4.07 4.68 5.39 6.20 (continued) Ny 66.19 78.03 92.25 109.41 130.22 155.55 186.54 224.64 271.76 330.35 403.67 496.01 613.16 762.89 Inclination factors by Meyerhof (1963) and Hanna and Meyerhof (1981) Fci = Fai = (1-2 Fyi = (1 - B² Bº Ne 9 for p = 0 TABLE 12.10 Variation of A with Pile Embedment Length, L Embedment length, L (m) 0 5 10 15 20 25 30 35 40 50 60 70 80 90 Coyle and Castello (1981): Embedment ratio, L/D 0 10 20 30 40 50 60 70 Qp = q'N₁ Ap 10 T T T T T e $' Bearing capacity factor, N 20 32° 36° = 30° 34° λ 0.5 0.336 0.245 0.200 0.173 0.150 0.136 0.132 0.127 0.118 0.113 0.110 0.110 0.110 38° TABLE 12.11 Variation of a (Interpo- lated Values Based on Terzaghi et al., 1996) 40° Cu Pa ≤0.1 0.2 0.3 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.4 2.8 40 60 80 100 200 Note: Pa ≈100 kN/m² Qs = atmospheric pressure Embedment ratio, L/D = : (Ko'tan 8')pL 0.15 0.2 0 5 10 15 20 25 30 a 35 36 1.00 0.92 0.82 0.74 0.62 0.54 0.48 0.42 0.40 0.38 0.36 0.35 0.34 0.34 Earth pressure coefficient, K 1.0 $' 30° 31° 32° 33° 8=0.80' 2 34° 35⁰ 36° 5
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