Helical Stair Case

DESIGN OF HELICAL STAIR CASE

Width of stair case(b) = Riser (r) = Tread (T) = Radius to the inside stairs (Ri)= Radius to the outside stairs(Ro)= Angle of inclination to the horizontal plan (a)= Minimum thickness of the slab(h) = Angle (B) turned through by stair = Imposed load =

1 .2 0 .1 7 0.3 1.3 2.5 25 0 .2 2 360 3

mt mt mt mt mt mt mt degrees mt degrees kn/sqmt

The radius of the centerline of o f the load (R1)=2(Ro^3-Ri^3)/(3*(Ro^2-Ri^2)) 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= b/h=

1.963

/

1 .9 . =

1.2

/

0 .2 2 . =

1.0332 mt 5.4545

Assu Assum ming ing the the mean ean thi thicckne kness in plan plan of the the sta stair ir (inc (inclu ludi ding ng trea treads ds and and fin finis ishe hes) s)= = 0.35 0.35 mt The self weight of the stair= 0 .3 5 * 25 .= .= 8.75 kn kn/sqmt Thus the total ultimate load(n)= ( 3 * 1.6 ) +( 8.75 * 1.4 )= 17.1 kn/sq Refer chart 178 in Reinf.Concrete Designer's Hand book BY Reynolds & Steedman k1= -0.494 k2= + 1.9608 k3= -1.6 Redu Redund ndan antt mome moment nt acti acting ng tang tangen enti tial ally ly at mids midspa pan n (Mo) (Mo)= = k1*n k1*n*R *R2^ 2^2* 2*b= b= -36. -36.49 49 kn-m kn-mtt Horizontal Horizontal redundant redundant force at midspan midspan (H) =k2*n*R2*b =k2*n*R2*b = 76.22 kn Vertical moment at supports (Mvs)= k3*n*R2^2 *b = -118.2 kn-mt 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 mm He Hence safe Adopt 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=0.48d Ast=BM/(0.87fy*(d-0.42*0.48d)= 2049.8 sqmm Provide y 16 at 100 mm mm c/c Ast= 2412 sqmm > 2050 sqmm Hence safe

Check for shear Nominal 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 % Shea hear ste sterength ngth of M20 concr ncrete for 1.00 1.0048 48 % steel Tc= Tc= 0.61 N/s N/sqmm Shea hear str strength for slab T 'c=kTc where k= 1.2 1.2 for 220 220 mm thi thick slab Therefore T 'c= 1 .2 * 0.6097 .= 0.7316 N/sqmm > Tv Hence safe Tempreture Reinforcement Prov Provid idee 112 mm bar bar as tem tempret pretur uree in each riser iser and and in wais waistt slab slab pro provid vid 0.12 0.12 % stee steell Therefore 0.12 / 100 * 1200 * 200 .= .= 288 sq sqmm / mt Provide 12mm bars @ 150 mm c./c

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