Design of Stair Case

Staircase is an important component of a building providing access to different floors and roof of the building. It consists of a flight of steps (stairs) and one or more intermediate landing slabs between the floor levels. Different types of staircases can be made by arranging stairs and landing slabs. Staircase, thus, is a structure enclosing a stair. The design of staircase, therefore, is the application of the designs of the different elements of the staircase.

The following are some of the general guidelines to be considered while planning a staircase:

Vertical distance between two floors= 3.2 m

Assuming two flights

Height Of each flights= 3.2/2 = 1.6m

Clear Width Of each Step=1500mm

Rise Of each Step =150mm

Thread Of each Step=230mm

No of rise/flight=12

Going of each flight=12×230 = 2760mm

Width of Support=230mm

rade of Concrete=M₂₀

Grade of Steel =Fe₄₁₅

SPAN OF STAIRCASE:

Clear Span=6m

Effective Span=6000+230 =6230mm

DESIGN OF FLIGHT:

Bearing of the flight = 150mm

Effective horizontal Span=6.075m

THICKNESS OF SLAB:

Considering 50mm Per m of Span

1.6×50=300mm

2.Dead load=230×25×1×1 = 5750N/m²

3.Finish=50×24=1200 N/m² = 6950N/m²

Corresponding load/eq.moment of span

(R²+T²)/T)×L

Load/M =8297.41N/M

Waist slab&ceiling finish =8.300N/M

Dead load of slab =1/2×230×150×19=0.32KN/M

Live Load =4KN/M

Total Load =12.62KN/M

Factored Load =12.62×1.5=18.93KN/M

Factored load for 1.5M =18.93×1.6=30.288KN/M

MOMENT AND REACTION:

RA=50.19KN & RB=50.30KN

Moment @3.0M

50.3×3.0-(8.5×0.5×0.32)+(12.52×1×1.10)+(30.26×0.720)

M =95.68KNM

Mu=M_u limit

95.68⁶ =0.138×20×1000×d²

d =186.18mm<200mm

TO GET AST:

Ast = 50×(1-1√1-(4.6/20)×(95.68×10⁶/1000×200²))/(415/20))

=1.12%

Ast =1.12/100×1000×200 = 2240mm

No .of 16mm dia bars = 2240/(3.14(16)²/4) = 11 bars

DISTRIBUTION BARS

(0.12/100)×1000×200 =240mm²

Spacing of 6mm dia bars =(3.14(6)²/4)240×1000=120mmc/c

Provide 6 mm dia @ 120mmc/c as distribution steel

DESIGN OF LANDING:

Reaction from flight = R_b×1.5-(8,5×0.5)×0.5+(12.52×1) ×0.5

=68KNM

M_{u =} M_u limit

68×10⁶ = 0.138×20×1000×d²

d = 157 < 200mm

Ast = 50×(1-1√1-(4.6/20)×(68×10⁶/1000×200²))/(415/20)) =0.8%

Ast =0.8/100×1000×200=1600mm²

Spacing of 12 mm dia bars =(3.14(12)²/4)1600×1000= 70mm

Provide 12 mm dia @ 70mmc/c on landing


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