Design Of Dog Legged Staircase 143m5c

Design of Dog Legged Staircase Problem 1) Design of a dog legged staircase (Fig1) with the following properties. a) b) c) d) e) f)

fy = 415Mpa , fck= M20 Height between floors = 9’-0”(2.74m) Riser = 6.35”(160mm) Tread = 11”(280mm) Live Load = 3.0KN/m2 Finish Load = 1KN/m2

Solution: Assume

( dL )=20

d=

(for slab), hence

Thickness of waist slab (W) =

hence d ( new )=146−15−

m =124.25 mm 125 mm ( 2.485 20 )

( 122 )=146 mm 150 mm

125+15+

,

( 122 )129 mm 130 mm

1. Calculation of Dead Load(load per meter width) B2 =R2+T2, Given R=160mm and T=280mm B= (1602+2802)1/2=322.5mm Steps =

0.5 ( 0.160 x 0.280 ) =0.0224 m 2

Waist slab= ( 0.322 x 0.150 ¿=0.0483 m2 Total Area = 0.0707m2 DL of step section, 1m in width and 280mm in plan length = DL per m2 on plan =

1.77 x

0.0707 x 25 KN =1.77 KN / m m3

=6.321 KN /m 2 ( 1000 280 )

Finish load = 1KN/m2 Total DL = (6.321+1)KN/m2=7.321KN/m2 LL per m2 = 3.5KN/m2 Total Load =

1.5 ( ¿+ DL ) =1.5 ( 7.321+3.5 ) =16.23KN/m2

2. Calculation of Bending Moment Calculate BM (Mu) using = (wl2/8) or use STAAD pro, Mu=13.1Kn/m2 Mu=0.138fckbd2, Hence

d=√(

Mu ) 70 mm<130 ( provided ) Hence OK 0.138 fckbd 2 =

3. Area of Steel

( x )(1−0.416 ( dx ))

Mu= 0.36 fck d

Dividing by

We get

∴

( 0.36 ) x {0.416

6.6 Mu/ fckbd ¿

( dx )=1.2−√ {(1.2)

2

( dx ) }fckbd

)=

2

-

bd2

2.4

2

( dx )−( dx )

6.68 Mu/ fckbd ¿

2

2

) hence

6.68 Mu/ fckbd

2

=

6.68 x 13.1/20 x 1000 x 130 2 ) =0.51 ¿

( dx )=1.2−√ {(1.2)

2

-0.15=0.23569

x ( )} , Hence z = 130x0.901 =117.13mm, Hence A =M /0.87fy.z Also z=d {1-0.416 d st u Ast= (13.1x106/0.87x415x117.13)=310mm2 per meter width Total steel required = 310 x 1.0668m =330.71mm2 Area of steel provided: Use 10mm bars @150c/c, no of bars = (1066/150) =7nos % of steel provided = (7x78.5/1066x130) x100 =0.4% of steel Distribution steel = 0.0012xbt (for Fe415) =0.0012x1000x150=180mm2or 200mm2(required)