1. Calculation of soil bearing capacity:

Transcription

1 1. Calculation of soil bearing capacity: Calculations for allowable bearing capacity of soil according to Terzaghi equation q ult =cncs c + q N q γ BN γ S γ Consider Depth of Foundation = 2.0 m From soil investigation log sheet, SPT Value, N = 6 Type of soil : Cohesion-less φ considered = 25 Degree N c = 25.1, N q = 12.7 and N g = 9.70 ( Ref. to Table 4 2, Page 222, Foundation Analysis & Design By JOSEPH E. BOWLES) against f value 25 Degree. For square footing S c = 1.3 & S γ = 0.8 ( Ref. Table 4-1, Page 220, Foundation Analysis & Design -By JOSEPH E. BOWLES) For Cohesionless soil c = 0 q =Overburden pressure =γ sub *D f = 16.0 kn/sqm. γ sub = Submerged density of soil = 8.0 kn/cum. Allowable bearing capacity of soil q a at D f =2.0 meter ( Factor of Safety =3.0) B (m) Nc Nq Nγ Sc Sγ γ sub (kn/cum) c (kn/sqm) q (kn/sqm) q u (kpa) q a ( kpa ) FS= Loads From Superstructure onto Foundation Top ( at each Pedestal Top ): Ref. Drawing No. : SMN/BNGRS/EqDes-2 Load Details : Vertical Loads : Weight of Equipments = Weight of Structure = Wt. of Man + Erection Kit = Total Vertical Load, Fz = Horizontal Loads : Wind load on Conductor = Wind load on Bushing = Wind load on Base Frame = Wind load on Support = Total Horizontal Load, Fx = Moments at Plinth level, M: 9.40 kn 2.50 kn 5.00 kn kn ( Wind load is critical that's why seismic loading combination is not taken into consideration) 4.30 kn 2.00 kn 6.30 kn kn kn.m 1.2 Soil Data: Moment ( kn.m ) Horizontal Load ( kn ) Vertical M x M y F x F y Load ( kn ) *1.2= 2.4 2*1.2= So Net Allowable bearing capacity of soil is considered : kn/m2 Unit weight of soil : 18 kn/m Material Properties : Concrete... fc'= 20 N/mm 2 Reinforcing Steel..fy= 415 N/mm 2 Concrete Clear Cover.= 60 mm Unit Weight of Concrete.= 24 KN/m 3 Frustum angle : 15 Degree Water Table from EGL : 3 m Designed by : Md. Giasuddin Page 2 of 6 Date : 25 April '11

2 1.4 Foundation Geometry : Df = 2000 t' = FSYL. FGL SECTION ON A-A Y 625 A X 2000 A 2600 FOUNDATION PLAN Length of Pad; L = 2.60 m Breath of Pad; B = 2.00 m Pedestal Size; a x = 0.38 m Pedestal Size; a y = 0.75 m Height of Pedestal above EGL; t' = 0.45 m Pad Thickness; t = 0.25 m Foundation Depth; D f = 2.00 m Height of Pedestal T = 2.13 m 1.5 Action Forces on Foundation : Weight calculation : Weight of Pedestal : = 0.38*0.75*2.125*24*2 = kns Weight of Pad : = 2.6*2*0.25*24 = 31.2 kns Weight of Overburden Soil : =(2.6*2-2*0.38*0.75)*( )*18 = kns Total Weight = kns Total Submerged Weight = 97.2 kns Loads at Base Level : Designed by : Md. Giasuddin Page 3 of 6 Date : 25 April '11

3 F zb =2* F z +Total Weight = kn M b =2* M+2*F x *(t'+d f ) = kn.m Moment Moment kn.m Horizontal Load kn Vertical at Base F M x M y F x F z kn y M xb M yb : Stability Check : 2.1 : Stability Check for Soil Bearing Capacity : Q = F zb = kn F' zb = kn ; Considered Submerged Condition A = LxB = 5.20 m 2 e x = M xb /F' zb = m e y = M yb /F' zb = m L /6 = >e B /6 = >e Q 6e 6e x y q = 1 When ex/ y L/6 ma x A ± ± = < L B q (max) = kn/sqm q (min) = kn/sqm Compression Compression Net Vertical F zb kn 247 Gross allowable soil pressure =44.34+γ s D f = kn/m 2 ; So consideration is OK. Net upward thrust qu = *2= kn/m : Stability Check for Overturning : Maximum Overturning Moment = M yb = 0.00 kn.m Total Vertical Submerged Weight = F' zb = kn Min. Resisting Moment = F' zb *B/2 = kn.m So FOS against Overturning =174.19/24.8 = #DIV/0! ; Which is more than 1.5, so OK. 3. Structural Design : 3.1 : Design of Pedestal : Design Loads: Compression =8+0.6*0.6*3.125*24 = kns M x =M x +F y *3.125 = kn.m M y = kn.m h' Resultant Moment = M'/M'=M+β x y x M y b' ( Ref. BS 8110 : Part 1; Page 3/39 ) β = 1 So Resultant Moment = kn.m P Mc Max. or Min. stress on the section = σ max/ min = ± A I I = bh 3 /12 = mm 4 Maximum stress on the section = N/mm 2 Compression Maximum stress on the section = N/mm 2 Tension Both tensile and compressive stresses are within acceptable limit, so no rebar is required from structural point of view. But as per code Minimun Rebar = 0.004*Ag = 0.004*600*600 mm2. = 1440 mm2. ( Ref. BS 8110 : Part 1, Table 3.27; Page 3/47 ) So Use 12 nos. of dia 16mm Bar for Vertical Reinforcement. Use dia 8mm 250mm c/c for tie. 4.0 Design of the Pad : Check for Flexural Shear : Loading diagram results from Upward pressure: Designed by : Md. Giasuddin Page 4 of 6 Date : 25 April '11 2.2*37.44 = kn/m. Upward Pressure by Soil

4 Upward Pressure by Soil 2.2*37.44 = kn/m kn kn kn kn Shear Force Diagram kn kn 8.38 kn-m 8.38 kn-m kn-m kn-m kn-m Bending Moment Diagram Clear Cover = 60; so d = mm = 175 mm Maximum Shear ; V = Kns Use FS = So V u = Kns Where, b= 2000 mm So, v c = V c /bd = N/mm 2 AS per ACI Shear Stress applied to concrete should be less than 0.17 f c ' N/mm 2 In present case which is coming 0.76 Mpa. This is much greater than applied stress so consideration is quite Ok Check for Punching Shear : Vertical Load is very less; so check for Punching is not necessary Check for Bending Moment : Maximum Moment = kn.m (1.5*23.17) Let us check with minimum reinforcement. As per ACI code, Ratio of minimum reinforcement ( in SI unit) is given by =1.4/fy ρ min = ρfy M=φρf u ybd ;Where φ = 0.9 f' c u d= = ρfy φρfyb( ) f' c Which is less than d provide ; so OK M mm ; d Provided = 175 mm Reinforcement Calculation : Top Reinforcement : Along Long Direction : M des =1.5*8.38 = KN.m Assuming depth of stress block, a = 2.1 mm Area of steel, As = M* /(0.9*fy*(d-a/2)) = mm2 (Ref. -Design of concrete structure, By-Nilson & Winter,Page 83,10th Ed.) Check for a a = As*fy/(0.85*fc'*b) = 2.4 mm Consideration is OK, So As = mm 2 Designed by : Md. Giasuddin Page 5 of 6 Date : 25 April '11

5 As per Code Min Rebar Required = bt = 900 mm 2 So Nos. of Bars = 12 Nos. So Spacing = ( )/12 = 189 mm Say 170 mm Along Short Direction : As per Code Min Rebar Required = bt = 1170 mm 2 So Nos. of Bars = 15 Nos. So Spacing = ( )/28 = 349 mm Say 170 mm Bottom Reinforcement : Along Long Direction : M des =1.5*23.17 = KN.m Assuming depth of stress block, a = 6.0 mm Area of steel, As = M* /(0.9*fy*(d-a/2)) = mm2 (Ref. -Design of concrete structure, By-Nilson & Winter,Page 83,10th Ed.) Check for a a = As*fy/(0.85*fc'*b) = 6.6 mm Consideration is OK, So As = mm 2 As per Code Min Rebar Required = bt = 900 mm 2 So Nos. of Bars = 12 Nos. So Spacing = ( )/12 = 171 mm Say 170 mm. Along Short Direction : As per Code Min Rebar Required = bt = 1170 mm 2 So Nos. of Bars = 15 Nos. So Spacing = ( )/28 = 349 mm Say 170 mm. Designed by : Md. Giasuddin Page 6 of 6 Date : 25 April '11