How to calculate staircase steel quantity?

Number of bar calculation

No. of bar = [ (width of flight – clear cover) / spacing ] + 1

= [ 1200 – 2 side cover 25 mm / 100 ] + 1

= [ 1200 – 50 / 100 ] + 1

= 12.5 say 13 nos.

Crank bar = 7 no.

Normal bar = 6 no.

Calculation of bar shape no. 1

a= L/7 + 0.42 D……….where D = Depth of slab

a= 4500/7 + 0.42 (150)

a= 706 mm

c=

by proportion theorem

AB/DE = AC/DF

Cⁱ/292 = 1800/150      ……..where Cⁱ say C Dash

Cⁱ = 3504 mm

BC/EF = AC/DF

BC/250 = 1800/150

BCⁱ (Base) = 3000 mm or 3 m

 d from ∆ JIK and ∆ DEF

JI/DE = JK/FE

JI/292 = 100/250

JI = 116.85 mm

GI = GJ + JI

GI = (150 – 25) + (116.85)

GI = 241.85 mm

Again,

From ∆ GIH and ∆ DEF

GH/FE = GI/FD

GH/250 = 241.85/150

GH = 403.08 mm

Therefor, d = 1000 – 403.08 = 596.9 say 597 mm

 b

 b = L – L/7 – (BCⁱ) – d

 b = 4500 – 642.857 – 3000 – 597

 b = 260 mm

cutting length = a + b + c + d + development length – bend

= 706 + 260 + 3504 + 597 + (2 x 390) – (4 x 1d) – (2 x 2d)

= 5847 – (4 x 1 x 10) – (2 x 2 x 10)

= 5767 mm

= 5.767 m

No. of bars = 7

Total length of bar = 5.767 x 7 = 40.4 m

Calculation of bar shape no. 2

 e= L/7 + b

= 4200/7 + 260

= 643 + 260

= 903 mm

f [ ‘f’ is equal to ‘c’ ]

f = c

f = 3504 mm

g [ ‘g’ is equal to ‘d’ ]

g = d

g = 597 mm

cutting length = c + f + g + development length – bend

= 903 + 3504 + 597 + (390) + (220) – (2 x 1 x d) – (1 x 2 x d)

= 5614 – (2 x 1 x 10) – (1 x 2 x 10)

= 5574 mm

= 5.574 m

No. of bars = 6 no.

Total length = 5.574 x 6 = 33.44 m


Calculation of bar shape no. 3

 In ∆ MNO & ∆ DEF

MN/FE = MO/DE or MN/292 = 76/250 or MN = 87.6 mm

Again,

In ∆ MNP & ∆ DEF

MP/DE = MN/DF or MP/250 = 87.6/150 or MP = 146 mm

Again,

In ∆PQR & ∆ DEF

PR/DE = QR/DF or PR/250 = 150/150 or PR = 250 mm

Extra bar = h + i

= 450 + (MP + PR + d)

= 450 + (146 + 250 + 597)

= 1443 mm

Cutting length = extra bar + development length – bend

= 1443 + 220 – (1 x 1 x d)

= 1443 + 220 – (1 x 1 x 10)

= 1653 mm

= 1.653 m

No. of bar =[ (width of flight – clear cover) / spacing ] + 1

= [(1200 – 50)] + 1

= 6.75 say 7 no.

For 7 no. of bar length = 1.653 x 7 = 11.571 m

Calculation of binder

Cutting length = width of flight – 2 cover

= 1200 – 2 x 25

= 1150 mm

= 1.15 m

No. of binder in waist slab =[ length of waist slab / spacing ] + 1

= [3504 / 150] + 1

= 24.36 say 24 no.

For 24 no. of bar length = 1.15 x 24 = 27.6 m

LANDING : 1

Bottom binder bar

Cutting length = length – cover

= 2500 – 2 x 25

= 2450 mm

= 2.45 m

No. of binder = e/spacing + 1

= 903/150 + 1….explaination for 903.[e = L/7 + b = 643 + 260 = 903]

= 7 nos.

Therefor,

Total length = 2.45 x 7 = 17.15 m

TOP binder bar

No. of binder = [(L/7)/ Spacing ] + 1

= (643/150) + 1

= 5 nos.

Total length = 2.45 x 5 = 12.25 m

LANDING : 2

Binder for extra bar

Landing cutting length = length – cover

= 2500 – 50

= 2450 mm

= 2.45 m

No. of bar = (847/150) + 1…….[PR + d = 250 + 597 = 847 mm]

= 6.64 say 7 nos.

Therefor.

Total length = 2.47 x 7 = 17.15 m

Under waist slab

No. of bar = (450/150) + 1

= 4 nos.

Therefor,

Total length = 1.15 x 4 = 4.6 m

Binder for TOP

Cutting length = 2.45 m

No. of bars = (d/150) + 1

= (597/150) + 1

= 5 no.

Total length = 5 x 2.45 = 12.25 m