isolated footing foundation design .pptx

isolated footing foundation design of a water sump

1. Foundation for water sump 1 2 m 2 m 0.5 m Gk =148.675 KN Qk = 180.256 KN Isolated rectangular footing- all 4 columns Eurocode 03 2 m unit weight of soil = 20 KN/m3 Unit weight of conc. = 25 KN/m3 C' = 7KN/m3 Friction angle = 25 deg Undrained cohesion =128.03 KN/m3 Bearing capacity of soil = 250 KN/m2 soil type - Brown medium stiff clay Column size = 0.3 m x 0.4 m Brown medium stiff clay layer 1.8 m Basement rock layer

2. Weight of soil above foundation (W S ) = (2 x 2-0.3x0.4) x 2 x 20 kN = 160 kN Self-weight of foundation (W F ) = (2 x2x0.5 + 0.4 x0.3x2) x 25 kN = 50 kN Total weight of foundation (W T ) = W S + W F = 160 + 50 = 210 kN Eccentricity (e) = 0 So, B = B’ = L = 2 m GEO LIMIT STATE bearing capacity check

3. DA 1/combination 1- compliance check(A1+M1+R1) partial factors γ G,unf = 1.35, γ Q,unf = 1.5, γ Ø ’ = 1.0, γ c ’ = 1.0, γ RV = 1.0 Design material properties : 1.35 Gk + 1.5 Qk Total designed vertical force = foundation load + total vertical force = (1.35 x 210) + (1.35 x 148.675 + 1.5 x 180.256) = 754.6 KN Total bearing capacity ( R ) = (250 KN/m2) x (2 x2) Rd = 1000 /1 = 1000 KN Over design factor = 1000/754.6 = 1.32 > 1 ok. C u.d = c’/ γ cu = 7/1.0 = 7 kPa Ø’ d = tan -1 (tan Ø’/ γ Ø ’ ) = tan -1 (tan 25/1.0) = 25

4. DA 1/combination 2- compliance check(A2+M2+R1) Over design factor = 1.68 >1 ok Settlement check Immediate settlement Q app = ( Gk+Qk+W )/ A‘ (without factors) From the graphs, M1=0.35 , M0 =0.94 = (148.675+180.256+210)/4 Eu =500 Cu = 134.73 KN = 500 x 128.03 =64040 KN/m2 Si =M1 x M0 x qapp x B/ Eu = 0.35x 0.94 x 134.73 x 2/64040 Depth factor =0.73 = 1.38 mm Rigidity factor = 0.8 Actual Immediate settlement = si x D.F. x R.F. = 1.38 x 0.8 x 0.73 = 0.81mm

5. Consolidation settlement P= 328.9338 KN Considering a piece of the foundation divided in to 4, P new = p/4= 82.23 KN L’ new= B’ new= 1m effective OBP at the base (q) = r x D = 20 x 2.5 = 50 KN/m2 The OBP @ depth of compressible soil layer(H) = 50 x 20% = 10 KN/m2 50/1.8 = 10/(1.8-H) H = 1.44 m Considering Z1 =0.5 m ,Z2 = 1.22     m= B' new/Z n= L'new / Z

6. dz z m n Iz ∆σ σ i σ f range mv sc                       0.5 0.5 2 2 0.06 32.33603 60 92.33603 0-80 0.11 1.778481                 80-160 0.18 2.910242                       0.72 1.22 0.819672 0.819672 0.022 11.85654 74.4 86.25654 0-80 0.11 0.939038                 80-160 0.18 1.536608                       TOTAL CONSOLIDATION SETTLEMENT 7.16437 Take R.F. =0.8 D.F. = 0.8 Lateral Strain = 0.7 Therefore actual consolidation settlement = 7.164 x 0.8 x0.8 x0.7 = 3.21 mm Therefore total settlement of the foundation = immediate settlement + consolidation settlement = 0.81 + 3.21 = 4.01 mm