Lesson - 1
Stairs
Introduction
From a structural viewpoint, the
staircase merely comprises slab/beam
elements, whose basic principles of
design have already been discussed
earlier in Concrete Structures 1.
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Introduction
Definition:
Types of Stairs
(a) straight stairs.
((b)) q
quarter-turn
stairs.
(c) dog-legged/half-
turn stairs.
(d) open-well stairs.
(e) spiral stairs.
(f) helicoidal stairs.
stairs
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Slab Type Stairs
Stair Classification
Depending on the predominant direction in which the
slab component of the stair undergoes flexure:
•stair
stair slab spanning transversely (stair widthwise);
•stair slab spanning longitudinally (along the incline).
Stair Slab Spanning Transversely :
•slab cantilevered from a spandrel beam or wall
•slab doubly cantilevered from a central spine beam
•slab supported between two stringer beams or walls
St i Slab
Stair Sl b S
Spanning
i llongitudinally
it di ll :
•single landing.
•double landing.
•folded.
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Stairs in Real Life
Straight stairs
Half-turn stairs
Stairs in Real Life
Quarter-turn stairs
Spiral stairs
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Stairs in Real Life
Open-well stairs
Helicoidal stairs
Stairs in Real Life
Transversely-spanning stairs Beam-supported stairs
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Stairs in Real Life
Floating stairs Multiple-landing
long stairs
Longitudinal Direction
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Longitudinal Direction
Longitudinal Direction
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Transverse Direction
Loads
Stair slabs are usually designed to resist gravity loads,
comprising dead loads and live loads.
Dead Loads:
•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)
•Live Loads:
•In general a uniformly distributed load of 5.0 kN/m2
•In residence live load of 3.0 kN/m2
•On tread slab a concentrated live load of 1.3 kN
applied at free end.
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Design Example 1
A straight staircase is made of structurally independent
tread slabs, cantilevered from a reinforced concrete
wall Given that the riser is 150 mm
wall. mm, tread is 300 mmmm,
and width of flight is 1.5 m, design a typical tread slab.
Apply the live loads specified for stairs liable to be
overcrowded. Use fc′ = 20 MPa and fy = 240 MPa
steel. Assume mild exposure conditions.
Design Example 1
Solution:
Given: R = 150 mm, T = 300 mm, W = 1.5 m
add 10 mm overlap , use B = T + 10 = 310 mm
mm.
Assume a slab thickness t = l/10 = 150/10 = 15 cm.
Dead Loads:
self weight of tread slab ≈ 25 kN/m3 × (0.15 × 0.31) m2
= 1.162 kN/m
finishes ≈ 0.6 kN/m × 0.31 m = 0.186 kN/m
2
wu,DL = 1.2 × (1.162 + 0.186) = 1.62 kN/m
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Design Example 1
Live Loads:
•Alternative I: wu,LL = (5.0 kN/m2 × 0.3 m) × 1.6
= 2.40
2 40 kN/m
•Alternative II: wu,LL = 1.3 kN × 1.6
= 2.08 kN (at free end)
Design Moment:
•At fixed end, Mu,DL = 1.62 × 1.52/2 = 1.823 kNm
u,LL = 2.4 × 1.5 /2 = 2.7 kNm
•Mu, 2
•Mu,LL = 2.08 × 1.5 = 3.12 kNm
Mu = 1.82 + 3.12 = 4.94 kNm
Design Example 1
Design of Main Bars:
Assuming a clear cover of 20 mm (mild exposure) and
a bar
b di diameter
t off 10 mm,
effective depth d = 150 – 20 – 10/2 = 125 mm.
Rn = Mu/(ϕbd2) = 4.94 × 106/(0.9 × 310 × 1252)
= 1.09 MPa
ρ = 0.85fc′/fy × (1 – (1 – 2Rn/0.85fc′)0.5)
= 0.85 × 20/240 × ((1 – ((1 – 2 × 1.09/0.85 × 20))0.5)
= 0.0047
As,req. = ρbd = 0.0047 × 310 × 125 = 182.125 mm2
Provide 3-D10 (As,prov. = 78.5 × 3 = 235.5 mm2)
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Design Example 1
Anchorage length required:
ld = 12/25 × fy/(fc′)0.5 × db
= 12/25 × 240/(20)0.5 × 10
= 257.59 mm (SNI 12.2)
Distributing transverse bars:
Ast,min = 0.002bt = 0.002 × 1000 × 150 = 300 mm2/m
Required Spacing for D8 bars = 50.3 × 1000/300
= 167.66
167 66 mm
Provide distributing transverse bars: D8-150 mm
Design Example 1
Check for shear:
Design (factored) shear force at support
Vu = (1.62 + 2.4) × 1.5 = 6.03 kN
ϕVc = ϕ0.17λ(fc′)0.5bd
= 0.75 × 0.17 × 1 × 200.5 × 310 × 125
= 22.09 kN
ϕVc > Vu ⇒ no shear reinforcement needed.
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Design Example 1
Design Example 2
Design the staircase slab shown. The stairs are simply
supported on beams provided at the first riser and at
the edge of the upper landing
landing. Assume a finish load of
0.8 kN/m and a live load of 5.0 kN/m2. Use fc′ = 20
2
MPa concrete and fy = 400 MPa steel. Assume mild
exposure conditions.
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Design Example 2
Solution:
Given: R = 150 mm, T = 300 mm
(R2 + T2)0.5
0 5 = 335
335.4
4 mm
Effective span l = 4.5 m.
Assume a waist slab thickness
≈ l/20 = 4500/20 = 225 mm, say 230 mm
Assuming 20 mm clear cover and 12mm main bars.
effective depth d = 230 – 20 – 12/2 = 204 mm
Design Example 2
Loads on going on projected plan area:
(1) selfweight of waist slab: 24×(0.230×335.4/300) m = 6.17 kN/m2
(2) selfweight of steps: 24×(0.5×0.15)
24 (0.5 0.15) = 1.80 kN/m2
(3) finishes (given) = 0.80 kN/m2
(4) live load (given) = 5.00 kN/m2
——————–
13.77 kN/m2
⇒ Factored load = 1.2(6.17+1.80+0.80)+1.6(5.00) = 18.52 kN/m2
Loads on landing:
self weight of slab : 24 × 0.23
(1) self-weight 0 23 = 5 52 kN/m2
5.52
(2) finishes = 0.80 kN/m2
(3) live loads = 5.00 kN/m2
⇒ Factored load = 1.2(5.52+0.80) + 1.6(5.00) = 15.58 kN/m2
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Design Example 2
Design Moment, considering 1-m wide strip of waist
slab:
R1 = (18
(18.52×3.45×(4.5
52×3 45×(4 5 – 1.725)/4.5)
1 725)/4 5)
+ (15.58×1.05×0.525/4.5)
R1 = 41.31 kN/m (width)
Max. factored moment occurs at the section of zero
shear, located at
x = 41.31/18.52 = 2.23 m from left support
Mu = (41.31×2.23) – (18.52× 2.232/2) = 46.07 kNm/m
Design Example 2
Main reinforcement:
Rn = Mu/(ϕbd2)
46 07 × 106/(0.9×
= 46.07 /(0 9× 1000 × 2042)
= 1.23 Mpa
ρ = 0.85fc′/fy × (1 – (1 – 2Rn/0.85fc′)0.5)
= 0.85 × 20/400 × (1 – (1 – 2 × 1.23/0.85 × 20)0.5)
= 0.0032
eq = ρbd = 0.0032 × 1000 × 204 = 652.8 mm
As, 2
s,req.
Provide D12-170 mm (664.7 mm2)
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Design Example 2
Distributing transverse bars:
Ast,min = 0.002bt = 0.002 × 1000 × 230 = 460 mm2
Required Spacing for D8 bars = 50.3 × 1000/460
= 109 mm
Provide distributing transverse bars: D8-100 mm
Design Example 2
Check for shear:
Check at d = 204 mm from the support
Vu = 41.31 – (18.52 + 0.204) = 37.53 kN/m
ϕVc = ϕ0.17λ(fc′)0.5bd
= 0.75 × 0.17 × 1 × 200.5 × 1000 × 204
= 478.96 kN
ϕVc > Vu ⇒ no shear reinforcement needed.
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Design Example 2
Design Example 2
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