Design of Slab:

For 6m x 4m (Case 6)tmin = {2(4) + 2(6)}/180 = .1111muse t = 130 mm

Loads:Dead load:Wudl = (weight of slab + weight of ceiling)1.4Wudl = c x t + 0.2 = 24(.13)(1.4) + 0.2(1.4)Wudl = 4.396 kPaLive load:Wull = (live load for residential buildings = 2)1.7Wull = 3.4 kPaTotal factored load:wut = 7.796 kPaconsidering 1m stripwut = 7.796 kN/m

m= 4/6 = .667 saym = .65from table 8.3, 8.4 and 8.5ca,neg = 0.077cb,neg = 0.014ca,dl = 0.032cb,dl = 0.006ca,ll = 0.053cb,ll = 0.010

Negative moments @ continuous edge:-Ma = 0.077 (7.796) (4)2 = 9.605 kN.m-Mb = 0.014 (7.796) (6)2 = 3.929 kN.mPositive moments:Along short direction:+Ma,dl = 0.032 (4.396) (4)2 = 2.251 kN.m+Ma,ll + = 0.053(3.4) (4)2 = 2.883 kN.m+Ma = 5.134 kN.mAlong long direction:+Mb,dl = .006 (4.396) (6)2 = 0.95 kN.m+Mb,ll = 0.010(3.4) (6)2 = 1.224 kN.m+Mb = 2.174 kN.m

f'c fymin = 1.4/414 = 0.0038As = x b x dmm2+Mb = 2.174 kN.md = 92 mmRu = 0.285 MPa = 0.0007use min = 0.00338As = 311.1111 mm2Use 12 mm barsN = 2.75S = 363.53 mmSay s = 350 mm o.c.

Along long direction:-Mb = 3.929 kN.mRu = 0.516 MPa = .00126use min = 0.00338As = 311.11 mm2N = 2.751s = 363.527 mm say s = 350 mm o.c.nodeFx (kN)Column 941 DL35 1052.124-0.113-3.493-0.011 10.635-0.316 2 LL35 193.162-0.023 0.075-0.001-0.193-0.066

COLUMN NO. 94 DESIGN PER ACI 318-08 - AXIAL + BENDING

Fy - 413.7 MPa, Fc - 27.6 MPA, SQRE SIZE - 400.0 X 400.0 MMS, TIED ONLY MINIMUM STEEL IS REQUIRED. AREA OF STEEL REQUIRED = 1600.0 SQ. MM BAR CONFIGURATION REINF PCT. LOAD LOCATION PHI --------------------------------------------------------------------------------------------- 8 - 16 MM 1.005 1 END 0.650 (PROVIDE EQUAL NUMBER OF BARS ON EACH FACE)

TIE BAR NUMBER 12 SPACING 256.00 MM

From the shown results the most stressed support is the node 49 with axial load Pdl = 904.488 kN and Pll = 145.483 kN and a moment along x-axis Mdl = 8.453 kN.m and Mll = 1.377 kN.m.The bending along z and y axes are negligible.Data:depth of 1.5 m from ground levelf'c = 27.5 MPafy = 276 MPac = 24 kN/m3s = 15.2 kN/m3

Finding the depth of footing:qu = Pu / Aftg ;Pu = 1.4(904.488) + 1.7(145.483)Pu = 1513.6043 kNqu = 1513.6043/2.52qu = 242.18 kPaRequired steel Area:d = 325 mmMu = 242.18 x 2.5 x 105 x 105/2Mu = 333.754 kN.mRu = Mu / x Ru x b x d2Ru = 1.404 MPa = [0.85(27.5)/276] [1-1-2(1.404)/0.85(27.5)] = 0.005251min = 0.00509use = 0.005251As = x b x dAs = 4266.47 mm2N=4As/d2 ; using 20 mm diam barsN = 13.58N = 14

Checking if dowel are needed:Actual bearing strength = Pu = 1513.6043 kNPermissible bearing stress:A2/A1 = 6.25 use 2 x 0.85 x fc A1 x (A2/A1 < 2) = 5236000 N= 5236 kN > Pu (No need for dowels)Minimum Area of dowelsArea = 0.005(400)2 = 800 mm2N = 4(800)/(16)2N = 4 (atleast 4 bars from column to be extended into footing)Data:Live load = 2 kPa + 1.47 kPa = 3.47 kPac = 24 kN/m3fc =27.5 MPafy = 413.7 MPaL = 1.4 mRise = 200 mmRun = 250 mmSolution:Minimum stair thickness to satisfy deflection requirements is given by:hmin = l/10 = 1500/10 = 150 mm < 200 mm (OK)let slab waist t = 75 mmsolving for loads:a) Dead loadsWeight of step = x 0.25 x 0.2 x 24 = 0.6 kN/mWeight of slab = 0.075 x 0.3202 x 24 = .58 kN/mb) Live loads3.47(0.25) = 0.8675 kN/m

Factored loadWu = 1.4(0.6 + 0.58) + 1.7(0.8675) = 3.127 kN/mShear forceVu = 3.127(1.4)/2 = 2.189 kNd = 200 20 6 = 174 mmVc = .85(1/6) 27.5 x 250 x 174Vc = 32.32 kN > Vu (OK)

Steel reinforcement:Mu = wL2/8 = 3.127x1.42 / 8 = .766 kN.mRu = Mu / x b x d2Ru = 0.112 MPa = 0.00027use min = 0.00338As = 0.00338 x 250 x 174As = 147.21 mm2

N = 1.3Use N = 2 of 12mm bar for each stepFor shrinkage reinforcement (use 8mm diam. temperature bars):As = 0.0018(1000)(75)As = 135 mm2 N = 2.69Spacing = 372.34 mmSay spacing = 350 mm o.c.