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Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design (1) ISOLATED FOOTING: Given Data: Service D.L = 250 k Service L.L = 200 k Allowable bearing capacity (qa) = 4 k/ft2 Depth of footing below ground level (z) = 4' Column size = 18? x 18? fc ' = 3 ksi fy = 40 ksi Figure 1.1: Isolated footing. Solution: - Step No 1: Sizes. Assume h =18?, d = h – 4 = 14? bo = 4(c + d) = 4 x (32) = 128 Since the space between the bottom of footing and the ground level is occupied partly by concrete and partly by soil, therefore, the pressure (W) of this material at 4 ft depth is: W = ?avgz =125 x 4 = 500 psf Prof Dr. Qaisar Ali Page 1 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design qe (Bearing pressure effective to carry column’s service load): qe = qa –W = 4 – 0.500 = 3.5 ksf Areq = Service load on column /qe Areq = (250 + 200) /3.5 =128.57 ft2 B x B = Areq B x B =128.57 B = 11.34' ˜ 11.5' Areq = 11.5' x 11.5' Step No 2: Loads. qu (Bearing pressure for strength design of footing): qu = Factored load on column/ Areq qu = (1.2 x 250 +1.6 x 200) / (11.5' x 11.5') = 4.68 ksf Step No 3: Analysis. (1) Punching Shear: Vup = qu [B2 – {(c + d)/12}2 ] = 4.68 x [11.52 – {(18 + 14)/12} 2 ] = 585.65 k Figure 1.2: Critical Perimeter for punching shear. Prof Dr. Qaisar Ali Page 2 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design (2) Beam Shear: Vub = qu {(B – c)/2 – d}B Vub = 4.68 x [{11.5 – (18/12)}/2 – (14/12)] x 11.5 = 206.31 k Figure 1.3: Section at distance “d” from column face for Beam shear. (3) Moment: Mu = qu B[{(B – c)/2} 2 ]/2 = 4.68 x 11.5 x [{11.5 - (18/12)}/2]2 /2 = 672.75 ft-k = 8073 in-k Figure 1.4: Section for Moment Calculation. Prof Dr. Qaisar Ali Page 3 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design Step No 4: Design. (1) Design for Punching Shear: Vup = 585.65 k F Vc = F4v (fc')bo d bo = 4(c + d) = 128 F Vc = 0.75 x 4 x v (3000) x 128 x 14/1000 = 294.45 k < 585.65 k Since FVc < Vup , therefore increase the depth of footing. Let, h = 24?, d = h – 4 = 20? Vup = 572 k F Vc = 499.52 k < 572 k (Again increase the depth) Let, h = 27?, d = 23? Vup = 564.29 k F Vc = 619.80 k > 564.29 k, O.K. Therefore the depth d = 23? is adequate for punching shear. (2) Design for Beam Shear: Vub = 165.95 k F Vc = F2v (fc')B*d F Vc = 0.75 x 2 x v (3000) x (11.5 x 12) x 23/1000 = 260.77 k > 165.95 k Therefore d = 23? is also adequate for beam shear. (3) Design for Flexure: Mu = 672.75 ft-k = 8073 in-k (i) Let a = 0.2d = 0.2 x 23 = 4.6? As = Mu / { F fy (d – a/2)} As = 8073/ {0.9 x 40 x (23 – 4.6/2)} = 10.83 in2 (ii) a = As fy / (0.85fc'B) a = 10.60 x 40/ (0.85 x 3 x 11.5 x 12) = 1.23? As = 8073/ {0.9 x 40 x (23 – 1.23/2)} = 10.02 in2 (iii) a = 10.02 x 40/ (0.85 x 3 x 11.5 x 12) = 1.138? Prof Dr. Qaisar Ali Page 4 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design As = 8073/ {0.9 x 40 x (23 – 1.138/2)} = 9.99 in2 (iv) a = 9.99 x 40/ (0.85 x 3 x 11.5 x 12) = 1.135? As = 8073/ {0.9 x 40 x (23 – 1.135/2)} = 9.99 in2 , O.K. • Check the minimum reinforcement: Asmin = {3v (fc ')/fy }B*d; but not less than (200/fy )B*d {3v (fc')/fy }B*d = {3v (3000)/40000} x (11.5 x 12) x 23 =13.03 in2 (200/fy )B*d = (200/40000) x (11.5 x 12) x 23 = 15.87 in2 . So Asmin = 15.87 in2 . As Asmin > As, thus Asmin governs. As= 15.87 in2 Using 1? F (#8) {#25, 25 mm}, with bar area Ab = 0.79 in2 No of bars = BAb /As = (11.5 x 12) x 0.79/15.87 = 6.86 ˜ 6? c/c Finally use #8 @ 6? c/c {#25 @ 150 mm c/c} both ways. Prof Dr. Qaisar Ali Page 5 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design GL 18" RCC Square column #8 @ 6" c/c (both ways) 4'-0" 3" (1:3:6) PCC Pad 27" 3" clear cover 11'-6" 12'-0" # 8 @ 6" c/c (bothways) 11'-6" 18" RCC Square column 11'-6" Figure 1.5: Drafting for Isolated footing. Prof Dr. Qaisar Ali Page 6 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design (2) COMBINED FOOTING (Ex 16.3, Nelson pg 562): Given Data: External column C1 (24? x 18?): D.L = 170 k L.L = 130 k Internal column C2 (24? x 24?): D.L = 250 k L.L = 200 k Footing depth = 6' below G.L fc ' = 3000 psi fy = 60000 psi Surcharge = 100 lb/ft2 qa = 6 k/ft 2 9" 18' 18" x 24" column 24" x 24" column h Length of footing 18" x 24" column 24" x 24" column B Figure 1.6: Combined footing. Prof Dr. Qaisar Ali Page 7 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design Solution: - Step No 1: Sizes. Service load for column C1: P1 = D + L = 170 + 130 = 300 k Service load for column C2: P2 = D + L = 250 + 200 = 450 k Net bearing capacity, qe = qa – {(weight of fill and concrete) + (surcharge)} An average unit weight of 125 pcf can be assumed for fill and concrete. qe = 6 – (0.125 x 6 + 0.1) = 5.15 k/ft2 Area of footing = (Sum of column loads)/qe = (300 + 450)/5.15 = 145.63 ft2 Location of resultant: P1 P2 R χ 18' Figure 1.7: Calculation for resultant load. R = P1 + P2 = 750 k Take moment about exterior column centerline. 750 x ? = 450 x 18 ? = 450 x 18/750 = 10.8 ft Therefore, length of footing = 2(10.8 + 0.75) = 23.1 ft = 23'-3? Width of footing (B) = 145.63/23.25 = 6.3' ˜ 6'-6? Assume depth of footing, h = 3'-5? = 41?, d = h – 3.5 = 37.5? Prof Dr. Qaisar Ali Page 8 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design Step No 2: Loads. Pu, ext = 1.2 x 170 + 1.6 x 130 = 412 k Pu, int = 1.2 x 250 + 1.6 x 200 = 620 k Total factored load (P) = 1032 k The net upward pressure (qu ) caused by the factored column load is: qu = P/ (Footing area) = 1032/ (23.25 x 6.5) = 6.83 k/ft 2 Net upward linear pressure/ foot in the longitudinal direction= 6.83 x 6.5= 44.4 k/ft. Step # 3: Analysis. (i) Punching Shear: Figure 1.8: Critical Perimeter for punching shear. Critical perimeter for column C1: bo = 2 x 36.75 + 61.5 = 135? Vu, ext = Pu, ext – qu{(36.75 x 61.5)/144} = 304.80 k Critical perimeter for column C2: bo = 4 x 61.3 = 246? Vu, int = Pu, int – qu{(61.5)/12}2 = 440.60 k (ii) Moment: The maximum negative moment between the columns occurs at a section of zero shears. Let “?” be the distance from the outer edge of the exterior column to this section, Vu = 44400? – 412000 = 0 ? = 9.28 ft (See Shear force diagram) Prof Dr. Qaisar Ali Page 9 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design 412000lb 620000 lb 44400 lb/ft 1'-6" 16'-3" 2'-0" 3'-6" 376000 lb 9.28' 135000 lb 345000 lb 3250000 lb-in 3100000 lb-in 19230000 lb-in Figure 1.9: Shear force & bending moment Diagram. The moment at 9.28 ft is: M (9.28) = {44400 x (9.282 /2) – 412000 x (9.28 – 0.75)} x 12 = –19230000 in- lb The moment at the right edge of the interior column is: Mint = 44400 x (3.52 /2) x 12 = 3263400 in- lb Step # 4: Design. (i) Design for punching shear: For column C1: Vu, ext = 304.80 k F Vc = 0.75 x 4 x v (3000) x 135 x 37.5/1000 = 831.85 k > Vu, ext , O.K. For column C2: Vu, int = 440.60 k F Vc = 0.75 x 4 x v (3000) x 246 x 37.5/1000 = 1515.82 k > Vu, int, O.K. Prof Dr. Qaisar Ali Page 10 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design (ii) Design for Moment: • Check the minimum reinforcement ratio: Asmin = {3v (fc ')/fy }B*d = (200/fy )B*d = (3 x v (3000) x 78 x 37.5/60000) = (200/40000) x 78 x 37.5 = 8.01 in2 < 9.75 in2 , Asmin = 9.75 in2 For M (9.28) = 19230 in-k Try 11, #9, As = 11.0 in2 a = As fy / (0.85fc 'B) = 11 x 60/ (0.85 x 3 x 78) = 3.31 in Md = F As fy (d – a/2) = 0.9 x 11 x 60 x (37.5 – 3.31/2) = 21309 in-k > M u, O.K. For Mint = 3263.4 in-k Try 16, #7, As = 9.62 in2 a = As fy / (0.85fc 'B) = 9.62 x 60/ (0.85 x 3 x 78) = 2.90 in Md = F As fy (d – a/2) = 0.9 x 9.62 x 60 x (37.5 – 2.90/2) = 18727 in-k > M u, O.K. Design of transverse beam under interior column: Width of transverse beam = 24 + 2(d/2) = 24 + d = 24 + 37.5 = 61.5? Net upward load per linear foot for transverse beam= 620000/6.5 = 95400 lb/ft Mu = 95400 x [{(6.5)-(24/12)}/2]2 x 12/2 = 2900000 in- lb = 2900 in-k Since the transverse bars are placed on top of the longitudinal bars, the actual value of “d t1 ” furnished is: dt1 = 37.5 – 1 = 36.5? Asmin = 0.0033 x 61.5 x 36.5 = 7.48 in2 Try 13, # 7, As = 7.80 in2 a = As fy / (0.85fc 'b) = 7.80 x 60/ (0.85 x 3 x 61.5) = 2.98 in Md = F As fy (dt1 – a/2) = 0.9 x 7.80 x 60 x (36.5 – 2.98/2) = 14746 in-k > M u, O.K. Prof Dr. Qaisar Ali Page 11 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design Design of transverse beam under exterior column: Width of transverse beam (b) = 18 + d/2 = 18 + 37.5/2 = 36.75? Net upward load per linear foot for transverse beam=412000/6.5=63384.61 lb/ft Mu = 63384.61 x [{(6.5)-(24/12)}/2]2 x 12/2 = 1925305 in- lb = 1925.305 in-k dt2 = 36.5? Asmin = 0.0033bdt2 Asmin = 0.0033 x 36.75 x 36.5 = 4.43 in2 Try 8, # 7, As = 4.80 in2 a = As fy / (0.85fc 'b) = 4.80 x 60/ (0.85 x 3 x 36.75) =1.44 in Md = F As fy (dt2 – a/2) = 0.9 x 4.80 x 60 x (36.5 – 1.44/2) = 9274 in-k > M u, O.K. Prof Dr. Qaisar Ali Page 12 of 13 Department of Civil Engineering N-W.F.P UET, Peshawar Reinforced Concrete Design 9" 18' 4'-6" 18" x 24" column 24" x 24" column 6'-0" 3" clear Supporting bars 3'-5" Supporting bars 3" clear 3' 2'-6" 2'-6" 23'-3" 8 # 7 (6'-0") 13 # 7 (6'-0") 18" x 24" column 11 # 9 {19'-6" (top)} 6'-6" 24" x 24" column A Figure 1.9: Drafting for Combined footing. Prof Dr. Qaisar Ali Page 13 of 13

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