DESIGN OF HELICAL STAIR CASE Width of stair case(b) = 1.2 mt Riser (r) = 0.17 mt Tread (T) = 0.3 mt Radius to the inside stairs (Ri)= 1.3 mt Radius to the outside stairs(Ro)= 2.5 mt Angle of inclination to the horizontal plan 25 degrees Minimum thickness of the slab(h) 0.22 mt Angle (B) turned through by stair 360 degrees Imposed load 3 kn/sqmt The radius of the centerline of the load (R1)=2(Ro^3-Ri^3)/(3*(Ro Therefore R1= 1.963 mt Then the radius of the center line of the stair (R2)=Ri+(1/2*b)= Therefore R2= 1.9 mt R1/R2= 1.963 / 1.9 .= 1.0332 mt b/h= 1.2 / 0.22 .= 5.4545 Assuming the mean thickness in plan of the stair (including tread The self weight of the stair= 0.35 * 25 .= 8.75 Thus the total ultimate load( 3 * 1.6 ) + 8.75 * Refer chart 178 in Reinf.Concrete Designer's Hand book BY Reynold k1= -0.494 k2= + 1.9608 k3= -1.6 Redundant moment acting tangentially at midspan (Mo)= k1*n*R2^2*b Horizontal redundant force at midspan (H) =k2*n*R2*b = Vertical moment at supports (Mvs)= k3*n*R2^2 *b = Effr. Depth of slab is given by B.M.=0.138fck*b*d*d Therfore d=SQRT(BM/(0.138fck*b)= 152.47 mm < 220 Adopt eff.depth a 200 mm and the over all depth a 220 mm Area of tension steel is given by BM=0.87fy*Ast(d-0.42Xm) where Ast=BM/(0.87fy*(d-0.42*0.48d)= 2049.8 sqmm Provide 16 at 100 mm c/c Ast= 2412 sqmm > 2050 sqmm Hence safe

Helical Stair Case

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DESIGN OF HELICAL STAIR CASE

Width of stair case(b) = 1.2 mtRiser (r) = 0.17 mtTread (T) = 0.3 mtRadius to the inside stairs (Ri)= 1.3 mtRadius to the outside stairs(Ro)= 2.5 mtAngle of inclination to the horizontal plan (a)= 25 degreesMinimum thickness of the slab(h) = 0.22 mtAngle (B) turned through by stair = 360 degreesImposed load = 3 kn/sqmt

The radius of the centerline of the load (R1)=2(Ro^3-Ri^3)/(3*(Ro^2-Ri^2))Therefore R1= 1.9632 mt

Then the radius of the center line of the stair (R2)=Ri+(1/2*b)=Therefore R2= 1.9 mt

R1/R2= 1.9632 / 1.9 .= 1.0332 mt

b/h= 1.2 / 0.22 .= 5.4545

Assuming the mean thickness in plan of the stair (including treads and finishes)= 0.35 mtThe self weight of the stair= 0.35 * 25 .= 8.75 kn/sqmtThus the total ultimate load(n)= ( 3 * 1.6 ) +( 8.75 * 1.4 )= 17.05 kn/sqmt

Refer chart 178 in Reinf.Concrete Designer's Hand book BY Reynolds & Steedmank1= -0.494 k2= + 1.9608 k3= -1.6Redundant moment acting tangentially at midspan (Mo)= k1*n*R2^2*b= -36.49 kn-mtHorizontal redundant force at midspan (H) =k2*n*R2*b = 76.224 kn Vertical moment at supports (Mvs)= k3*n*R2^2 *b = -118.2 kn-mt

Effr. Depth of slab is given byB.M.=0.138fck*b*d*d Therfore d=SQRT(BM/(0.138fck*b)= 152.47 mm < 220 mm Hence safeAdopt eff.depth as 200 mm and the over all depth as 220 mm

Area of tension steel is given by BM=0.87fy*Ast(d-0.42Xm) where Xm=0.48dAst=BM/(0.87fy*(d-0.42*0.48d)= 2049.8 sqmmProvide y 16 at 100 mm c/cAst= 2412 sqmm > 2050 sqmm Hence safe

Check for shearNominal shear stress Tv= Vu/bd = 76.224 *1000/( 1200 * 200 )=

.= 0.3176 N/sqmm% of tension steel= 100Ast/bd = 100 * 2411.5 / ( 1200 * 200 )=

.= 1.0048 %Shear sterength of M20 concrete for 1.0048 % steel Tc= 0.6097 N/sqmmShear strength for slab T 'c=kTc where k= 1.2 for 220 mm thick slabTherefore T 'c= 1.2 * 0.6097 .= 0.7316 N/sqmm > Tv Hence safe

Tempreture ReinforcementProvide 1- 12 mm bar as tempreture in each riser and in waist slab provid 0.12 % steelTherefore 0.12 / 100 * 1200 * 200 .= 288 sqmm / mtProvide 12mm bars @ 150 mm c./c