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Flexure strength reduction factor centreline with horizontal " Φ " Spac
kN/m^2 Nos Asb Bottom full width 2 legged Closed hoops steel 1703 4082 4082 2154 143 5.122 13 150 Reinforcement Dia of top of bars Dia of bottom of bars Dia of closed hoop bars as distribution 20 20 14 mm Nos Total Ultimate Load "w" 7
Bottom steel "Asb"mm 3819 1608 Closed hoops per m 5. Top steel "Ast" mm^2 Vertical angle at the step Riser "R" 0.175 m 24 Provided Spac mm mm Date Plan angle of the stair "α " 132 degrees Weight of reinforced concrete "γrc" kN/m^3 Rev. 0 DESIGN OF HELICOIDAL STAIR Geometrical Properties Material Properties PROJECT TITLE: LOCATION: Check Tread "T" 0.35 m width of steps and Landing"B"or bw 1.95 m Strength of steel "fy" Strength of concrete "fc" 35 N/mm^2 m 0. 420 N/mm^2 Inner radius " Ri" 5.562 m Clear cover to concrete 25 mm kN/m^2 Modular Ratio "n = Es/Ec" Thickness of slab "h" 280 mm Shear strength reduction factor 0. Outer Radius "Ro" = Ri+B 7. 200000 N/mm^2 Loading Modulus of Elasticity of concrete "Ec" 27805. Eff. Depth "d" 248.5 Modulus of Elasticity of steel "Es" kN/m^2
Required Self Weight of slab 8. Floor Finish 1 weight of steps 2. Live load 4 kN/m^2 Total dead Load "Udl" N/mm^2 kN/m^2 Ast top full width 12.16 13 150 Total service Load "tsl" 15.2984 kN/m^2 Total Ultimate Load for full width"wu"
kN/m^2 kN/m
36 degrees
Date Rev. 0 DESIGN OF HELICOIDAL STAIR Geometrical Properties Material Properties PROJECT TITLE: LOCATION: Check Radius of centreline of loading =R 1 =2/3{(Ro^3 -Ri^3 )/(Ro^2 -Ri^2 )} = m Radius of centreline of steps = R 2 =(Ri+Ro)/2 = m k2 = F9-(F9-F16)/0.1(R1/R2-1) k1 = F19-(F19-F22)/0.1(R1/R2-1) k3 = F27-(F27-F32)/0.1(R2/R1-1) Bending moment at midspan "Mo" = k1wuR 22 = Horizontal thrust at midspan = "H" = k2wuR 2 = Vertical moment at support "Mvs" = k3wuR 22 Positive vertical moment "Mvp" at θ = α/3 = 44 Mvp =MoCos θ+HR 2 πθ/180 tanΦ sinθ-wuR 12 (1-cosθ) = Lateral moment "Ml" at θ = 3α/8 = 49. Ml = MoSinθSinΦ-HR 2 π/180tanΦCosθSinΦ-HR 2 CosΦSinθ+wuR 1 Sinθ-R 2 πθ/180) = Torsional Moment "T" at θ = 3α/8 = 49. T = (MoSinθ-HR 2 πθ/180CosθtanΦ+wuR 12 Sinθ-wuR 1 R 2 πθ/180)CosΦ+HR 2 SinθSinΦ= Axial Thrust "N" at θ = 3α/8 = (^) 49. N = -HSinθCosΦ-wuR 1 πθ/180SinΦ = Vertical Shearforce "Vv" at θ = α/2= 66 Vv = -HSinθSinΦ+wuR 1 πθ/180CosΦ = Lateral Shearforce "Vh" at θ = 0 Vh = Hcosθ = Vu = 1.2sqrt(Vv^2 +Vh^2 ) = Shear strength of concrete = ΦcVc = 0.75Sqrt(fc)B.d/6 = ΦcVc > Vu O.K Vertical negative bending of flights Mu = 1.2*Mvs = ω 0. ρ 0. Ast1 = 3083 mm 2 for full width
kNm -10.
kNm kn
Calculations -992. kNm kNm
-231.
-276. kNm kN kN kN kN kN kNm
Date Rev. 0 DESIGN OF HELICOIDAL STAIR Geometrical Properties Material Properties PROJECT TITLE: LOCATION: Check = 0. 3.68 Cross section O.K At/s = Tu/(фt1.7Aoh*fy) At/s = mm^2 /mm/2 legs = 0.29 mm^2 /mm/2 legs minimum torsional reinforcement = 1.7 mm^2 /mm/2 legs = 1.63 mm^2 /mm/2 legs Hence minimum torsion reinforcement = 1.7 mm^2 /mm/2 legs dia of stirrups used = 14 mm using 2-legged stirrups required spacing = s = mm c/c required Nos of closed stirrups /m = 5. Provided Nos of closed stirrups /m = 7
Date Rev. 0 DESIGN OF HELICOIDAL STAIR Geometrical Properties Material Properties PROJECT TITLE: LOCATION: Check Design of Longitudinal Torsion reinforcement Al = 606 mm Minimum Longitudinal torsion reinforcement = mm^2 Required torsion steel = 606 mm^2 Provided 2 bars at top edges of dia = 20 Provided 2 bars at bottom edges of dia = 20 Longitudinal torsion steel provided = 1256 mm^2 > required OK Tension design of flights Factored Tension = Nu = 1.2*N = Astn =Nu/(Φtfy)= mm 2 Astn to be provided on one face = mm^2 Total Top steel of flight Ast-top = Ast1+Astn = mm^2 Required dia 20 mm 12.2 Nos at top As required = mm^2 Provided dia 20 mm 13 Nos at top As provided= mm^2 Total Bottom steel of flight Asb =Asb1+Astn = mm^2 Required dia 20 mm 5.12 Nos As required = mm^2 Provided dia 20 mm 13 Nos As provided= mm^2
at bottom 4082 at bottom 1608.
278.2 kN 736
