design of staircase

Name-Subhankar Chandra Kirtania

Class –B.C.E.[IV] ; Section-A1 ; Roll No-000610401009

1

Design Of Staircase

All Dimensions are in meter

Floor to Floor height of the building is 3.0 meter. And clear dimension for the stair-hall is

4.70mX3.50m. Now if we take Riser(R)=150 mm no of riser needed =

=20.We go for dog-legged

stair case with width of each flight taken as 1650mm. Now no of riser needed per flight

= 2.

/ 3 .Now taking tread width =300mm total space needed for tread

=(9X300)mm=2700mm.Now we Take mid-landing width 1650mm .

So effective length(le) =4950mm. Assuming overall depth(D)=

Providing overall depth of the waist slab,D = 130 mm.(Here waist slab of the stair is embedded

125mm inside 250mm brick wall)

Load calculation

On Landing portion-

Dead Load

i. s/w of waist slab = .13X25 = 3.25 KN/m2

ii. s/w of floor finish celling plaster etc.(assumed) = 1 KN/m2

_________________________________________________________________________

Total DL =4.25KN/m2



2

Now for 1.65 m wide waist slab Dead Load intensity on landing portion=(1.65X4.25)

KN/m=7.0125KN/m

Live Load

As per IS:456-200,cl-33.2,pg-63 live load intensity on the landing portion may be taken as

*( ) + ⁄ = 7.5 KN/m

Total loading intensity on landing portion ( )=(7.0125+7.5)KN/m=14.5125KN/m

On Inclined portion-

Dead Load

i. For s/w of waist-slab, floor-finish, ceiling-plaster

.

/ (

√

) (

√

)

ii. s/w of step = 12R=12X0.15

_________________________________________________________________

Total DL

Now for 1.65 m wide waist slab Dead Load intensity on Inclined portion

( 1.65)KN/m=10.8108KN/m

Total loading intensity on inclined portion ( )=( )

Let us consider maxm intensity of loading acting over the whole span –

Ma=Mb=.

/

W2=18.3108 KN/m W1=14.5125 KN/m



3

Ultimate mt. =Mu=

.

/

For M25 concrete & Fe-500 steel

= 0.46(from IS-456:2000,pg-70 for fy=500) and =25;

So Q=3.340152

√

( ) } As per IS:456-200,cl-33.2,pg-63 effective breadth could be

taken as (b+75)mm.

Assuming 10 bars & 20mm clear cover D=(98.66+5+20)=123.66mm 125mm Dassumed (Hence

OK)

Exact calculation

For case-

=29.632KN-m

For case-

, ( )-=6.3193KN-m

(a=1.825m,b=3.125m,s=2.7m,l=4.95m,w=3.7983KN)

, ( )-=4.218KN-m

Total--- MA=35.952 KN-m

MB=33.850 KN-m

MA>MB

Hence Ultimate moment(Mu)=1.5X35.952 KN-m=53.928 KN-m

W1=14.5125KN/m

CASE Ι

W=( W2 – W1)=3.7983 KN/m

CASE ΙΙ

(a) (b)

(s)

(l)



4

4

5

Dreqd=(96.74+20+5)=121.75mm Dprovided=130 mm

dprovided=(130-25)mm=105mm dreqd

________________________________________________________________________________________________________________

RA = 42.82 KN RB = 40.13 KN

Span moment will be maxm where shear force will be zero-

The distance from ‘A’ 2 .

/3

Maxm Span moment,

0( ) 2 .

/3 2

( )

31

= 18.5389 KN-m

MA=35.952

KN-m

MB=33.850 KN-m

case-ΙΙΙ RAΙΙΙ .

𝐴 𝐵𝑙

/ 𝐾𝑁 RBΙΙΙ .

𝐴 𝐵𝑙

/ 𝐾𝑁

case-ΙΙ RAΙΙ .

𝑤 𝑠 𝑏

𝑙/ 𝐾𝑁 RB

ΙΙ .𝑤 𝑠 𝑎

𝑙/ 𝐾𝑁

case-Ι RAΙ .

W l

/ 𝐾𝑁 RB

Ι .W l

/ 𝐾𝑁

W2=18.3108 KN/m

W1=14.5125KN/m



5

REINFORCEMENT DETAILS

(with the help of SP:16 1980:-table-3,page-49,for y=500 & page-230)

effective depth

For supports , dx=(D-20-4)mm=106 mm(for 8 distributer bars at bottom)

For span bottom, dy=(D-20)mm= 110 mm(for 12 main bars at bottom)

Here

=.

/

=257.4 mm2

Spacing reqd for 8 distributers =(

)

Spacing provd = 300 mm C/C

Condn

Moment

(KN-m)

Ultimate Moment

(KN-m)

4

( ) 5

Pt

=(%steel )

( )

(

)

Spacing(mm)

At support top ‘A’

35.952 53.93 2.7825 0.7544

1319.45 10@150mmC/C+

8@150mmC/C

(552.92+863.94)

=1416.86

At support top ‘B’

33.850 50.78 2.6246 0.7029

1238.12 10

@100mmC/C 1295.91

At mid span

bottom

18.5389 27.81 1.3324 0.328424

596.1 10

@200mmC/C 647.95

design of staircase

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