design of dog legged stair case in rcc
In this we given both procedure and problem so scroll below to view both topics
àCaluclate
number of steps =(Height of 1st flight)/(rise )
rise may be assumed as 150 mm = 0.15 m
Let us assume a beam of bearing
width and provide at bottom and at mid
loading.
àThickness of stair slab
l/d=20 X M.F ; d= L/(20 X M.F)
Let us assume percentage of steel as Pt=0.4%
Let
us provide slab thickness =200 mm (overall depth )
àCaluclation of loadings
i) Self weight ii) Live
load iii) floor finish
iv) weight of steps =1/2x
clear cover + effective depth
Total loading
àUltimate
loading per length =Total Load x F.O.S(1.5)
Assume the stair as a simply supported beam with UDL
By using this calculate the
Bending moment (Mud)
àCheck for the depth of stair case
Mud=0.36
x fck x
0.48 (1-0.42 x 0.48)x bd2
d=? Find d value by the above formulae
If dprovided
> d
hence safe
àCaluclation of reinforcement
Mu=0.87
x fy x Ast x(1-(Ast X fy)/(fck X bd)) X d
find Ast value Ast=?
find the spacing(s)=(1000 Xπ/4 X∅)/Ast here, ∅-dia of bars
à Check
for deflection
à Design temperature reinforcement
Astmin=0.12/100 X bd
Find the spacing (S)=(1000 X π/4 X ∅)/Ast
provide the centre to centre reinforcement
Problem no:1
Q) Design a dog-legged stair case for a
residential building the stair roof is available 3m x 5m distance between floor
is 3.23m ,consider live load as 3KN/m2 and
floor finish is 1 KN/M2,use M20 mix and Fe415 grade steel?
SOL:
àNumber
of steps in 1st flight =1.5/0.15=3m
height distance =(11-1)x0.3=3m
tread=300mm
Let
us assume a beam of bearing width =300 mm
provide at bottom and at mid loading as shown in figure
àThickness of stair slab
Let us assume a beam of bearing width and provide at bottom and at mid loading.
L/d=20 for simply supported beam as per cl 23.2.1 pg:37
d=5000/(20 X1.3)
=192.3 mm
Let us assume percentage of tension
reinforcement =0.4%
Let us provide slab thickness =200
mm(overall depth) =0.2m
Let us choose 10 mm dia bars
Assuming clear cover =15mm (for stair
case 15 mm is ok)
Effective depth of slab =200-15-10/2 =180 mm
àCaluclation
of loading
self weight of slab =0.2 x 25 = (5 KN/M2)
Given Live or super load = 3 (KN/M2)
Floor complete finish = 1 (KN/M2)
Weight of steps = =1/2x0.15x18
= 1.35 KN/M2
Total loading =10.35 KN/M
RA+RB=48.56
RA=RB
[email protected]=41.84x2.5-15.5x1(2.5-0.5)-17.56x(2.5-1-1.5/2)
Mmax=53.8 KN-M(sagging)
Mud=53.8 KN-M
àCheck for the depth of stair case
=> This formulae will be available in IS 456-2000 code book
Mud=0.36
x fck x
0.48 (1-0.42 x 0.48)x bd2
53.8x106=0.36x20x0.48(1-0.42x0.48)x1000xd2
d=140 mm
dprovided(180mm)
> d
Hence safe
àCaluclation of reinforcement
=> This formulae will be available in IS 456-2000 code book
Mu=0.87 x fy x Ast x(1-(Ast X fy)/(fck X bd)) X d
53.8x106=0.87x415xAst
x(1-(Ast X 415)/(20 X 1000X180))x180
Ast=927.64 mm
Spacing(S)=(1000 X π/4 X ∅)/Ast=(1000 X π/4 X 102)/927.64=84.66
~ 80 mm c/c
àCheck for defflecction
Pt= (100 X Ast)/bd =(100 X 927.64 )/(1000 X 180)=
0.55%
àDesign of temperature reinforcement
Astmin=0.12/100xbxD=0.12/100x1000x200
=240 mm2
Spacing(S)=(1000 X π/4 X ∅)/Astmin
=(1000 X π/4 X 102)/240
=
327 mm
so
provide 10 mm dia bars
at 320 c/c
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