Style of Stair Scenario – Engineering Feed

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Style of Stair Scenario

Staircase is an important element of a developing supplying accessibility to distinctive flooring and roof of the developing. It is composed of a flight of methods (stairs) and one or far more intermediate landing slabs among the ground degrees. Different varieties of staircases can be built by arranging stairs and landing slabs. Staircase, consequently, is a construction enclosing a stair. The layout of staircase, as a result, is the application of the patterns of the distinct features of the staircase.

The following are some of the standard tips to be deemed although setting up a staircase:


Vertical length amongst two flooring= 3.2 m

Assuming two flights

Height Of every single flights= 3.2/2 = 1.6m

Apparent Width Of every Phase=1500mm

Rise Of each and every Step =150mm

Thread Of each individual Stage=230mm

No of increase/flight=12

Likely of just about every flight=12×230 = 2760mm

Width of Guidance=230mm

Quality of Concrete=M20

Grade of Steel =Fe415


SPAN OF STAIR Circumstance:

Distinct Span=6m

Productive Span=6000+230 =6230mm


Design and style OF FLIGHT:

Bearing of the flight = 150mm

Efficient horizontal Span=6.075m



Looking at 50mm For every m of Span


2.Useless load=230×25×1×1 = 5750N/m2

3.Finish=50×24=1200 N/m2 = 6950N/m2

Corresponding load/eq.second of span


Load/M =8297.41N/M

Waistline slab&ceiling complete =8.300N/M

Useless load of slab =1/2×230×150×19=.32KN/M

Dwell 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


Minute AND Reaction:

RA=50.19KN & RB=50.30KN

Second @3.0M


M =95.68KNM

Mu=Mu limit

95.686 =.138×20×1000×d2

d =186.18mm<200mm


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


Ast =1.12/100×1000×200 = 2240mm

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


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


Mu = Mu 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))


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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