Design and Structural Components of Staircases

Design Of Staircases
Part -1 ~Pritesh Parmar
priteshparmar5.6.1998@gmail.com

Introduction :
 The staircase is an important component of a building, and often the only means of
access between the various floors in the building.
 It consists of a flight of steps, usually with one or more intermediate landings
(horizontal slab platforms) provided between the floor levels.

Component of staircase :
 Flight : The inclined slab of staircase is called flight.
 Landing : It is level platform at the top or bottom of a flight between floor.
 Riser : Vertical projection of the step (i.e., the vertical distance between two
neighbouring steps) is called Riser.
 Tread : The horizontal top portion of a step (where the foot rests) is termed Tread.
 Nosing : It is the projecting part of the tread beyond the face of riser.

Component of staircase :
 Waist slab : The slab below steps in the stair case is called waist slab.
 Soffit : Underside of a stair is known as soffit.
 Going : The horizontal projection (plan) of an inclined flight of steps, between the
first and last risers, is termed going.
 The steps in the flight can be designed in a number of ways: with waist slab, with
tread-riser arrangement (without waist slab) or with isolated tread slab.

Typical flight in a staircase :
Plan
Elevation
Waist Slab Type

Typical flight in a staircase :
Tread Riser Type
Isolated Tread
. Slab Type

Type of staircase :
 A wide variety of staircases are met with in practice.
 Some of the more common geometrical configurations are :
• straight stairs (with or without intermediate landing)
• quarter-turn stairs
• dog-legged stairs
• open well stairs
• spiral stairs
• helicoidal stairs

Structural Classification :
 Structurally, staircases may be classified largely into two categories, depending on
the predominant direction in which the slab component of the stair undergoes
flexure:
 1. stair slab spanning transversely (stair widthwise);
 2. stair slab spanning longitudinally (along the incline).

Stair slab spanning transversely :
 This category generally includes:
1. slab cantilevered from a spandrel beam or wall ;
2. slab doubly cantilevered from a central spine beam ;
3. slab supported between two stringer beams or walls.

 When the slab is supported at the two sides by means of ‘stringer beams’ or
masonry walls , it may be designed as simply supported, but reinforcement at the
top should be provided near the supports to resist the ‘negative’ moments that
may arise on account of possible partial fixity.
 In the case of the cantilevered slabs , it is economical to provide isolated treads
(without risers). However, the tread-riser type of arrangement and the waist slab
type are also sometimes employed in practice, as cantilevers.
 The spandrel beam is subjected to torsion (‘equilibrium torsion’), in addition to
flexure and shear.
 The slab supports gravity loads by bending essentially in a transverse vertical plane,
with the span along the width of the stair.

 When the slab is doubly cantilevered from a central (spine) beam, it is essential to
ensure, by proper detailing, that the slab does not separate from the beam when
loaded on one side only.
 This can be done by anchoring the slab reinforcement into the beam, so that the
same reinforcement acts as a stirrup in the beam.
 Alternative arrangements are possible; however, it should be ensured that the
beam stirrups are ‘closed’, to provide desired torsional resistance.

 It may be noted that, although the stair slab spans Transversel , the supporting
spandrel/spine/stringer beams span Longitudinally along the incline of the stair,
framing into supporting columns.

Stair slab spanning longitudinally :
 In this case, the supports to the stair slab are provided parallel to the riser at two or
more locations, causing the slab to bend longitudinally between the supports.

Stair slab spanning longitudinally :

Effective span of stair:
 IS 456 :2000 Cl. 33.1 , Pg. No : 63

Effective span of stair:
 Another case frequently encountered in residential and office buildings is that of
the landings supported on three sides. This case has not been explicitly covered by
the IS Code. The ACI Code and BS Code also do not have any special provision as
yet for this condition. However, recent studies (based on experiments as well as
finite element analysis) reveal that the flight essentially spans between the landing-
going junctions, with hogging moments developing at these junctions.

Load on stair slab:
 Stair slabs are usually designed to resist gravity loads, comprising Dead and Live
load.
 In the case of cantilevered tread slabs, the effects of seismic loads should also be
investigated. The vertical vibrations induced by earthquakes may induce flexural
stresses of considerable magnitude. It is desirable to provide bottom steel in the
cantilever slabs (near the support locations) to counter the possibility of reversal of
stresses.

Dead Load:
 The components of the dead load to be considered comprise:
 • self-weight of stair slab (tread/tread-riser slab/waist slab);
 • self-weight of step (in case of ‘waist slab’ type stairs);
 • self-weight of tread finish (usually 0.5 – 1.0 kN/m2).
 The unit weight of reinforced concrete for the slab and step may be taken as
25kN/m3 as specified in the Code (Cl. 19.2.1).

Dead Load:
 Live loads are generally assumed to act as uniformly distributed loads on the
horizontal projection of the flight, i.e., on the ‘going’. The Loading Code [IS 875 :
1987 (Part II)] recommends,

Distribution of Load:
 IS 456 : 2000 Cl. 33.2 , Pg. No 63

Waist Slab Spanning Longitudinally :
 Slab thickness t may be taken as approximately l/20 for simply supported end
conditions and l/25 for continuous end conditions.
 The normal load component wn causes flexure in vertical planes containing the
span direction (parallel to the longitudinal axis of the slab), and the tangential load
component wt causes axial compression (of low order) in the slab.
 The main bars are placed longitudinally, and designed for the bending moments
induced in the vertical planes along the slab span.
 The distributor bars are provided in the transverse directions.
 These moments may be conveniently computed by considering the entire vertical
load w acting on the projected horizontal span (going), rather than considering the
normal load component wn acting on the inclined span s

 Example :
Design the staircase slab, shown in fig. The stairs are simply supported on beams
provided at the first riser and at the edge of the upper landing. Assume a finish load of
0.8 kN/m2 and a live load of 5.0 kN/m2. Use M 20 concrete and Fe 415 steel.

 Example :
• Riser = 150 mm , Tread = 300 mm
• Effective span =(Left support width)/2+Going+Landing width-(Right support .
width)/2
=150+3000+1500-150
= 4500 mm / 4.5 m
• Assume waist slab thickness = l/20 = 4500/20 = 225 mm ≈ 250 mm .
• Initially considering main bar of dia. 12 mm,
Effective cover = 20+12/2 = 26 mm.
• Effective depth , d = D - effective cover = 250 – 26 = 224 mm .

 Example :
• Load on going on projected plan area :
1) Self weight of waist slab = 25 KN/m3 X 0.250 m X 335.4/300 = 6.99 KN/m2
Where,
25 KN/m3(Density of concrete)
0.250 m (Waist slab thickness)
335.4 mm(inclined flight distance)=(Riser2 + Tread2 )^0.5 )=(1502 + 3002 )^0.5
. = 335.4 mm

 Example :
2) Self weight of step = 25 KN/m3 X (0+0.15)/2 m = 1.88 KN/m2
Where,
(0+0.15)/2 m (Avg. thickness of step above waist slab)

 Example :
3) Floor finish load = 0.8 KN/m2 …………………(given)
4) Live load = 5 KN/m2 …………………(given)
……………………………………………………………………………..
Total load on going = 14.67 KN/m2

 Example :
• Total load on going = 14.67 KN/m2
• Factored load = 1.5 ( F.o.s ) X 14.67
= 22 KN/m2
Considering width of stair = 1 m
• Factored load on going = 22 KN/m

 Example :
• Load on landing:
1) Self weight of slab = 25 KN/m3 X 0.250 m = 6.25 KN/m2
Where,
0.250 m (Landing slab thickness)

 Example :
• Load on landing :
……………………………………………………………………………..

 Example :
• Total load on landing = 12.05 KN/m2
= 18.075 KN/m2
• Factored load on going = 18.075 KN/m

 Example :
Loading on staircase can be represent as :
18.075 KN/m22 KN/m
54.61

 Example :
Reaction on left support R1 =
∑M @ right support =0,
R1 X 4.5 = 18.075 X 1.052 / 2 + 22 X 3.45 X (3.45/2 + 1.05)
R1 =49.02 KN
• Maximum moment will occurs at point of zero shear,
Let, at distance x from left support, Shear force = 0 ,
49.02 – 22 * x = 0 => x = 2.228 m from left support.

 Example :
Now Bending moment @ x = 2.228 m from left support,
Mu = 49.02 * 2.228 -22 * 2.2282/2
= 54.61 KNm .
Main reinforcement design :
Mu = 54.61 KNm
For Fe415
for balanced case, Mu,lim =0.138*fck*b*d2 => dreq. = 140.66 mm < dprovided = 224
mm
Ok.
Note :Based on trial and error depth of stair can be reduced considering flexure and deflection
criteria.

 Example :
Percentage of steel in tension, Pt = 50*(1-(1-4.6*Mu/ fck*b*d2 )^0.5)/(fck/fy)
fck
= 20 Mpa
fy
= 415 Mpa
b = 1000 mm
d = 224 mm
Mu = 54.61 KNm Ans. Pt = 0.33 %

 Example :
Area of steel in tension , Ast req. = Pt X b X d /100
= 0.33*1000*224/100
= 739.2 mm2
Ast min. = 0.12 X b X d /100
= 0.12*1000*250/100
= 300 mm2 < Ast req.
Spacing =ast /Ast *1000 ast = d2/4
= *122/4
=113.097 mm2

 Example :
Spacing =ast /Ast *1000
=113.097*1000/ 739.2
=152.99 mm
 Provide 12 mm dia. Bar @ 150 mm c/c.

 Example :
Distribution reinforcement design :
Ast min. = 0.12 X b X d /100
= 0.12*1000*250/100
= 300 mm2
Considering 8 mm dia. Bar,
= *82/4
=50.26 mm2

 Example :
=50.26*1000/ 300
=167.53 mm

 Example :
 Check for shear :
• As reinforcement of stair are confined by compression , shear should be checked at
‘d’ distance from support, ( IS 456 :2000 Cl. 22.6.2.1 )
Shear force at distance ‘d’ from left support :
Vu = 49.02 – (21.17 × 0.224) = 44.47 kN
Shear stress , Tv = Vu /b*d
= 0.197 N/mm2

 Example :
Shear strength of concrete , Tc‘= Tc * k
k = 1.1 ( IS 456:2000 Cl. 40.2.1.1 )
Tc = 0.4 N/mm2 (For M20 grade and Pt = 0.33)( IS 456:2000 , Table -19)
Tc‘= 0.4 * 1.1 = 0.44N/mm2 >>> Tv Safe.
 Nominal reinforcment are provided in order to prevent cracks, shown in detailing.

 Example :
Deflection check :
Basic l/d ratio = 20 (IS 456 :2000 , Cl. 23.2.1)
Modification factor (IS 456:2000 fig 4 , Pg. 38)
OR MF = 1/(1+0.625 * log10 Pt)
MF = 1/(1+0.625 * log10 0.33) (you can check by putting Pt = Pt,provided , Pt = Pt,req .
is critical )
MF = 1.43
Span / effective depth = 20*1.43 = 28.6
Actual span / effective depth = 4500/224 = 20.089 << 28.6 Deflection is in
control.

 Example :
Check for development length :
M1/V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C))
As reinforcements are confined by compressive reaction M1/V increased by 30 %.
M1=moment capacity for section
V = Shear force at support
Pt provided = Ast provided*100/b*d
Ast provided =ast / Spacing *1000
=113.097*1000/ 150
=753.98 mm
Pt provided = 0.34 %

 Example :
Pt = 50*(1-(1-4.6*Mu/ fck*b*d2 )^0.5)/(fck/fy)  Mu =57.22 KNm
Lo = 8*dia of bar ( Anchorage value for 90degree bent )
Ld = 47 dia. For fe415 and M20 (IS 456 : 2000, Cl. 26.2.2.1)
M1/V + Lo = 57.22*1000/49.02+8*12=1263.27 >> 47*12=564 mm OK

 Example :
Detailing :
150
160
150
250
250

 Example :
Design a (‘waist slab’ type) dog-legged staircase for an office building, given the following
data:
• height between floor = 3.2 m;
• riser = 160 mm, tread = 270 mm;
• width of flight = landing width = 1.25 m
• live load = 5.0 kN/m2
• finishes load = 0.6 kN/m2
Assume the stairs to be supported on 230 mm
thick masonry walls at the outer edges
of the landing, parallel to the risers.
Use M 20 concrete and Fe 415 steel.

 Example :
• Effective span = C/C distance between supports
=230+2*1250+270*9
= 5160 mm / 5.16 m
• For economy in design landing slab is taken as 200mm as landing are subjected to
low shear force and bending moment compared to going.

 Example :
1) Self weight of waist slab = 25 KN/m3 X 0.280 m X 314/270 = 8.14 KN/m2
Where,
314 mm(inclined flight distance)=(Riser2 + Tread2 )^0.5 )=(1602 + 2702 )^0.5
. = 314 mm

 Example :
2) Self weight of step = 25 KN/m3 X (0+0.16)/2 m = 2.0 KN/m2
Where,
(0+0.16)/2 m (Avg. thickness of step above waist slab)

 Example :
……………………………………………………………………………..

 Example :
= 23.61 KN/m2

 Example :
Where,

 Example :
……………………………………………………………………………..

 Example :
= 15.90 KN/m2

 Example :
23.61 c
71.40

 Example :
As loading is symmetric,half-half load transferred to supports,
R1 = (2*15.90*1.365+23.61*2.43)/2
R1 =50.38 KN
• Maximum moment will occurs at point of zero shear(i.e at mid span),

 Example :
Now Bending moment @ mid span ,
Mu = 50.39*2.58 -15.90*1.365*(2.58-1.365/2)-23.61*(2.58-1.365)^2/2
= 71.40 KNm .
Mu = 71.40 KNm
For Fe415
mm
Ok.
criteria.

 Example :
fck
= 20 Mpa
fy
= 415 Mpa
b = 1000 mm
d = 254 mm
Mu = 71.40 KNm Ans. Pt = 0.33 %

 Example :
= 0.33*1000*254/100
= 838.2 mm2
Ast min. = 0.12 X b X d /100
= 0.12*1000*280/100
= *122/4
=113.097 mm2

 Example :
=113.097*1000/ 838.2
=134.92 mm

 Example :
Ast min. = 0.12 X b X d /100
= 0.12*1000*280/100
= 336 mm2
= *82/4
=50.26 mm2

 Example :
=50.26*1000/ 336
=149.588 mm

 Example :
Vu = 49.33 – (15.90 × 0.254) = 45.29 kN
= 0.178 N/mm2

 Example :
k = 1.0 ( IS 456:2000 Cl. 40.2.1.1 )
Tc‘= 0.4 * 1.0 = 0.4 N/mm2 >>> Tv Safe.
 Nominal reinforcment are provided in order to prevent cracks.

 Example :
Deflection check :
OR MF = 1/(1+0.625 * log10 Pt)
is critical )
MF = 1.43
control.

 Example :
M1/V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C))
=113.097*1000/ 130
=869.97 mm

 Example :
Pt = 50*(1-(1-4.6*Mu/ fck*b*d2 )^0.5)/(fck/fy)  Mu =73.57 KNm
M1/V + Lo = 73.57*1000/50.38+8*12=1541.66 >> 47*12=564 mm OK

 Example :
Detailing :
12mm @ 130 c/c
12mm @ 130 c/c
8mm @ 140 c/c
280

 Example :
Detailing :
• Some nominal main reinforcement ( 10 mm dia. @ 220mm c/c) is provided in the
landing slab near the support at the top to resist possible ‘ negative ‘ moments on
account of partial fixity.
• 8 mm dia. @ 250mm c/c is also provided as distribution reinforcement.
• From crossing of bar , the bars must be extent up to Ld(Development length, IS 456
: 2000, Cl. 26.2.1)

 Example :
Design a (‘waist slab’ type) dog-legged staircase
for an office building, given the following data:
• height between floor = 3.2 m;
• riser = 160 mm, tread = 270 mm;
• width of flight = landing width = 1.25 m
• live load = 5.0 kN/m2
• finishes load = 0.6 kN/m2
Assume the landing to be supported
On two edges perpendicular to the risers.
Use M 20 concrete and Fe 415 steel.

 Example :
• Effective span = C/C distance between landing
=2*625+2430
= 3680 mm / 3.68 m

 Example :
1) Self weight of waist slab = 25 KN/m3 X 0.185 m X 314/270 = 5.38 KN/m2
Where,
314 mm(inclined flight distance)=(Riser2 + Tread2 )^0.5 )=(1602 + 2702 )^0.5
. = 314 mm

 Example :
……………………………………………………………………………..

 Example :
= 19.47 KN/m2

 Example :
Where,

 Example :
……………………………………………………………………………..

 Example :
= 15.35 KN/m2
 Only 50% load is acts longitudionally, (as landing slab is one way) i.e 15.35/2 = 7.68
KN/m

 Example :
c

 Example :
Design of waist slab :
R1 = (7.68*0.625+19.47*2.43)/2
R1 =28.46 KN

 Example :
Mu = 28.46*3.68/2 -7.68*0625*(1.84-0.625/2)-19.47*1.215^2/2
= 30.69 KNm .
Mu = 30.69 KNm
For Fe415
mm
Ok.
criteria.

 Example :
fck
= 20 Mpa
fy
= 415 Mpa
b = 1000 mm
d = 159 mm
Mu = 30.69 KNm Ans. Pt = 0.37 %

 Example :
= 0.37*1000*159/100
= 588.3 mm2
Ast min. = 0.12 X b X d /100
= 0.12*1000*185/100
= *122/4
=113.097 mm2

 Example :
=113.097*1000/ 588.3
=192.25 mm

 Example :
Ast min. = 0.12 X b X d /100
= 0.12*1000*185/100
= 222 mm2
= *82/4
=50.26 mm2

 Example :
=50.26*1000/ 222
=226.39 mm

 Example :
Vu = 28.46 – (7.68 × 0.159) = 27.23 kN
= 0.172 N/mm2

 Example :
k = 1.23 ( IS 456:2000 Cl. 40.2.1.1 )
Tc‘= 0.416 * 1.23 = 0.511 N/mm2 >>> Tv Safe.

 Example :
Deflection check :
OR MF = 1/(1+0.625 * log10 Pt)
is critical )
MF = 1.36
control.

 Example :
M1/V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C))
=113.097*1000/ 190
=595.54 mm

 Example :
Pt = 50*(1-(1-4.6*Mu/ fck*b*d2 )^0.5)/(fck/fy)  Mu =31.47 KN/m
M1/V + Lo = 31.47*1000/28.46+8*12=1201.76 >> 47*12=564 mm OK

 Example :
Design of landing slab :
Load :
1) Direct loaing on landing : 15.35*1.25(width of landing) =19.19KN/m
2) Load from going : (19.47*2.43)*0.5 (Half load transferd to lower and half load to
upper landing)
……………………………………………………………………………………………...
Total load = 42.85KN/m

 Example :
Design of landing slab :
R1 = 42.85*2.60/2
R1 =55.70 KN

 Example :
Mu = 42.85*2.60^2/8
= 36.20 KNm .
Mu = 36.20 KNm
For Fe415 , b=1.25m
mm
Ok.
Note :Based on trial and error depth of landing can be reduced considering flexure and
deflection criteria.

 Example :
fck
= 20 Mpa
fy
= 415 Mpa
b = 1250 mm
d = 159 mm
Mu = 36.20 KNm Ans. Pt = 0.342 %

 Example :
= 0.342*1000*159/100
= 544 mm2
Ast min. = 0.12 X b X d /100
= 0.12*1000*185/100
= *122/4
=113.097 mm2

 Example :
=113.097*1000/ 544
=207.89 mm

 Example :
Ast min. = 0.12 X b X d /100
= 0.12*1000*185/100
= 222 mm2
= *82/4
=50.265 mm2

 Example :
=50.265*1000/ 222
=226.39 mm

 Example :
k = 1.23 ( IS 456:2000 Cl. 40.2.1.1 )
Tc‘= 0.4032 * 1.23 = 0.495 N/mm2 >>> Tv Safe.

 Example :
Deflection check :
OR MF = 1/(1+0.625 * log10 Pt)
is critical )
MF = 1.414
control.

 Example :
M1/V+Lo <= Ld (IS 456 :2000 , Cl. 26.2.3.3 (C))
=113.097*1000/ 200
=565.485 mm

 Example :
Pt = 50*(1-(1-4.6*Mu/ fck*b*d2 )^0.5)/(fck/fy)  Mu =45 KNm
M1/V + Lo = 45*1000/55.70+8*12=903.899 >> 47*12=564 mm OK

 Example :
Detailing :
8mm @ 250 c/c220 c/c
300
8mm @ 220 c/c
8mm @ 220 c/c

Thank you
Presentation is limited to design of waist slab type stair case , different type
of staircase design will cover in next presentation.

Design and Structural Components of Staircases

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