Spread Footing Reference: CSA A23.3-14-Given:- Dimension Width of footing: b = 2400 mm Length of footing: L = 2500 mm Thickness of footing: t = 500 mm Embedment of footing: h = 1800 mm concrete cover and the radius of rebars: cc = 90 mm - Material Specified compressive strength of concrete: f'c = 25 MPa Minimum yield strength of rebars: fy = 400 MPa Soil unit weight of the embedment: g = 16 kN/m3 Allowable soil pressure: qa = 300 kPa Reinforcement provided for Mx per meter: Asx = 1500 mm2 Reinforcement provided for My per meter: Asy = 1500 mm2 - Column Information Width: bc = 450 mm Length: Lc = 450 mm Specified compressive strength of concrete: f'c2 = 25 MPa - Specified Loads Specified axial load (compression +): P0 = 450 kN Specified bending moment about x: Mx = 8 kN.m Specified bending moment about y: My = 15 kN.m - Factored Loads Factored axial load (compression +): Pf = 500 kN Factored bending moment about x: Mfx = 10 kN.m Factored bending moment about y: Mfy = 20 kN.m -Results:- Minimum reinforcement per meter: Asmin = 0.002 t .(1000) = 0.002 x (0.5 x 1,000) x 1,000 =1000 mm2, Cl. 7.8.1- Total Bearing Pressure for Specified Loading Eccentricity of loading: ey = Mx / [P0 + g.(h - t).b.L + 24.t.b.L] = 8.0 / [(450.0) + 16.0 x (1.8 - 0.5) x 2.4 x 2.5 + 24 x 0.5 x 2.4 x 2.5] =0.012 m Eccentricity of loading: ex = My / [P0 + g.(h - t).b.L + 24.t.b.L] = 15.0 / [(450.0) + 16.0 x (1.8 - 0.5)x 2.4 x 2.5 + 24 x 0.5 x 2.4 x 2.5] =0.023 m- ey ≤ b / 6 , OK ex ≤ L / 6 , OK Maximum pressure: Qmax = g.(h - t) + 24.t + P0 / (b.L) + 6. Mx / (L.b2) + 6. My / (b.L2) = 16.0 x (1.8 - 0.5) + 24 x 0.5 + (450.0) / (2.4 x 2.5) + 6 x 8.0 / (2.5 x 2.42) + 6 x 15.0 / (2.4 x 2.52) =117 kPa Minimum pressure: Qmin = g.(h - t) + 24.t + P0 / (b.L) - 6. Mx / (L.b2) - 6. My / (b.L2) = 16.0 x (1.8 - 0.5) + 24 x 0.5 + (450.0) / (2.4 x 2.5) - 6 x 8.0 / (2.5 x 2.42) - 6 x 15.0/ (2.4 x 2.52) =98 kPa Qmax < qa , OK- Net Bearing Pressure for Factored Loading Eccentricity of loading: ey = Mfx / Pf = 10.0 / 500.0 =0.02 m Eccentricity of loading: ex = Mfy / Pf = 20.0 / 500.0 =0.04 m- ey ≤ b / 6 , OK ex ≤ L / 6 , OK Maximum Net Factored Pressure: Qmax = Pf / (b. L) + 6. Mfx / (L. b2) + 6. Mfy / (b. L2) = 500.0 / (2.4 x 2.5) + 6 x 10.0 / (2.5 x 2.42) + 6 x 20.0 / (2.4 x 2.52) =95 kPa Minimum Net Factored Pressure: Qmin = Pf / (b. L) - 6. Mfx / (L. b2) - 6. Mfy / (b. L2) = 500.0 / (2.4 x 2.5) - 6 x 10.0 / (2.5 x 2.42) - 6 x 20.0/ (2.4 x 2.52) =71 kPa- Design net pressure in direction for Mfx: Qf1 = Pf / (b.L) + 6 Mfx / (L.b2) + 6 Mfy / (b.L2) = 500.0 / (2.4 x 2.5) + 6 x 10.0 / (2.5 x 2.42) + 6 x 20.0 / (2.4 x 2.52) =95 kPa Qf2 = Pf / (b.L) - 6 Mfx / (L.b2) + 6 Mfy / (b.L2) = 500.0 / (2.4 x 2.5) - 6 x 10.0 / (2.5 x 2.42) + 6 x 20.0 / (2.4 x 2.52) =87 kPa Cantilever length of footing: y = (b - bc) / 2 = (2.4 - 0.45)x 1,000 / 2 =975 mm Pressure at the edge of column: Qf3 = Qf2 + (Qf1 - Qf2).(b - y) / b = 87 + (95 - 87)(2.4 - 0.975) / 2.4 =92 kPa Equivalent uniform net pressure: Qf0 = (Qf3 + Qf1) / 2 = (92 + 95) / 2 =94 kPa- Check the one-way shear in direction for Mfx: d = t - cc = (0.5 - 0.09) x 1,000 =410 mm dv = Max. [0.9 d, 0.72 t] = Max. [0.9 x 0.41 x 1,000, 0.72 x 0.5 x 1,000] =369 mm Shear: Vf = (y - d) .(1m) . Qf0 = (0.975 - 0.41) x 1 x 93.8 =53.0 kN Shear coefficient: k = 230 / (1000 + dv) = 0.17 Shear resistance: Vc = k.fc.f'c0.5 x 1000 x dv = 0.17 x 0.65 x 250.5 x 1,000 x 0.369 =201.5 kN, Cl. 11.3.4 Vf ≤ Vc , OK- Design the flexural reinforcement in direction for Mfx: Shear: Vf = y. (1m). Qf0 = 0.975 x 1 x 93.8 =91.5 kN Factored moment: Mf = Qf0. y2 / 2 = 93.8 x 0.9752/2 =44.59 kN.m/m d . Vf / Mf < 1.0, Not deep beam a1 = 0.85 - 0.0015 f'c = 0.85 - 0.0015 x 25 =0.81 Cl. 10.1.7 b1 = 0.97 - 0.0025 f'c = 0.97 - 0.0025 x 25 =0.91 Cl. 10.1.7 Recommended reinforcement per meter: Asx = Mf.(106) / (0.9 fs.fy.d x 1000) = 355 mm2 a = fs. Asx. fy / (fc.a1. f'c .1m) = 0.85 x 1500.0 x 400 / (0.65 x 0.81 x 25 x 1,000) =39 mm c = a / b1 = 39 / 0.91 =43 mm c/d ≤ 700 / (700 + fy), OK, Cl. 10.5.2 Flexural resistance: Mr = fs. Asx. fy.(d - a/2) = 0.85 x 1500.0 x 400 x [(0.41 x 1,000) - 39 / 2] x 10-6 =199.3 kN.m/m Mr ≥ Mf, the reinforcement is OK.- Design net pressure in direction for Mfy: Qf1 = Pf / (b.L) + 6 Mfx / (L.b2) + 6 Mfy / (b.L2) = 500.0 / (2.4 x 2.5) + 6 x 10.0 / (2.5 x 2.42) + 6 x 20.0 / (2.4 x 2.52) =95 kPa Qf2 = Pf / (b.L) + 6 Mfx / (L.b2) - 6 Mfy / (b.L2) = 500.0 / (2.4 x 2.5) + 6 x 10.0 / (2.5 x 2.42) - 6 x 20.0 / (2.4 x 2.52) =80 kPa Cantilever length of footing: y = (L - Lc) / 2 = (2.5 - 0.45)x 1,000 / 2 =1025 mm Pressure at the edge of column: Qf3 = Qf2 + (Qf1 - Qf2).(L - y) / L = 80 + (95 - 80)(2.5 - 1.025) / 2.5 =89 kPa Equivalent uniform net pressure: Qf0 = (Qf3 + Qf1) / 2 = (89 + 95) / 2 =92 kPa- Check the one-way shear in direction for Mfy: Shear: Vf = (y - d). (1m). Qf0 = (1.025 - 0.41) x 1 x 92.2 =56.7 kN Shear coefficient: k = 230 / (1000 + dv) = 0.16 Shear resistance: Vc = k. fc .f'c0.5 (1m). dv = 0.16 x 0.65 x 250.5 x 1,000 x 0.369 =217.4 kN, Cl. 11.3.4 Vf ≤ Vc , OK- Design the flexural reinforcement in direction for Mfy: Shear: Vf = y. (1m). Qf0 = 1.025 x 1 x 92.2 =94.5 kN Factored moment: Mf = Qf0 . y2 / 2 = 92.2 x 1.0252/2 =48.44 kN.m/m d . Vf / Mf < 1.0, Not deep beam a1 = 0.85 - 0.0015 f'c = 0.85 - 0.0015 x 25 =0.81 b1 = 0.97 - 0.0025 f'c = 0.97 - 0.0025 x 25 =0.91 a = fs.Asy.fy / [fc.a1. f'c . (1m)] = 0.85 x 1500.0 x 400 / (0.65 x 0.81 x 25 x 1,000) =39 mm c = a / b1 = 39 / 0.91 =43 mm Recommended reinforcement per meter: Asx = Mf / (0.9 fs. fy.d) = 48.4 x106 / (0.9 x 0.85 x 400 x 0.41 x 1,000) =386 mm2 c/d ≤ 700 / (700 + fy), OK, Cl. 10.5.2 Flexural resistance: Mr = fs. Asy. fy.(d - a/2) = 0.85 x 1500.0 x 400 x [(0.41 x 1,000) - 39 / 2] x 10-6 =199.3 kN.m/m Mr ≥ Mf, the reinforcement is OK.- Two-way shear: The critical section is at d/2 from the edge of the column for the four sides. Effecitve length: Le = Min[ L, (Lc + d) ] = Min[ 2.5, 0.86] =860 mm Effective width: be = Min[ b, (bc + d) ] = Min[ 2.4, 0.86] =860 mm Shear perimeter: b0 = 2(Le + be) = 2 x (0.86 + 0.86) x 1000 =3440 mm Polar moment of inertia about axis Y: Jy = 2Le .d3/12 + 2d .Le3/12 + 2 be .d .Le2/4 = 2 x 0.86 x 0.413/12 + 2 x 0.41 x 0.863/12 + 2 x 0.86 x 0.41 x 0.862/4 = 0.18 m4 Polar moment of inertia about axis X: Jx = 2be .d3/12 + 2d .be3/12 + 2 Le .d .be2/4 = 2 x 0.86 x 0.413/12 + 2 x 0.41 x 0.863/12 + 2 x 0.86 x 0.41 x 0.862/4 = 0.18 m4 gvy = 1 - 1 / [1 + (2/3)(Le/be)0.5] = 1 - 1 / [1 + (2/3) x (0.86/0.86)0.5] =0.4 gvx = 1 - 1 / [1 + (2/3)(be/Le)0.5] = 1 - 1 / [1 + (2/3) x (0.86/0.86)0.5] =0.4 Shear stress for two-way shear: vf = Pf/(b0.d)+ gvy. Mfy .(Le/2)/Jy + gvx. Mfx .(be/2)/Jx = 500.0 x 10-3 /(3.44 x 0.41) + 0.4 x 20.0 x (0.86/2) x 10-6 / 0.18 + 0.4 x 10.0 x (0.86/2) x 10-6/ 0.18 = 0.35 kPa Shear resistance: vc = Min [ 0.38 x 1, (1 + 2bc/Lc)0.19, (4d/b0 + 0.19)].fc.f'c0.5 = Min[0.38 x 1, (1 + 2 x 0.45 / 0.45) x 0.19 , (4 x 0.41 / 3.4 + 0.19) ] x 0.65 x 250.5 = Min [1.2, 1.9, 2.2] =1.2 kPa, Cl. 13.3.4.1 vf ≤ vc , two-way shear is OK.- Dowel design: Because Mfx or Mfy > 0 then, the reinforcement of column must cross the interface.
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