Eng. Khalid Stair-Design

8/2/2019 Eng. Khalid Stair-Design

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Stair Design


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Stair Types

Straight stairs


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Stair Type based on the structural

loading type

Simply supported stair

(transversely supported)Longitudinally supported stair

Cantilever stair


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Simply supported stair

Design a straight flight stair in a residential building supported onreinforced concrete wall 1.5 m )(center to center)

L.L = 3kN/m2

Covering Material = 0.5kN/m

The riser are 16cm and goings are 30 cm


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Loads and Analysis

cmh

cml

h

ave5.16

5.78

cos

5.78

5.720150

20

22 1630

16

min

D.L(O.W) =0.34*0.075*25 + 0.5*0.16*0.3*25=1.24 kN/mD.L (covering material) = 0.5 kN/m

D.L (total) = 1.74 kN/m

L.L =3*(0.3) =0.9 kN/m

0.30.16

0.34


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Design for moment

stepeachfor121

9.01.89165*300*0018.0

0005.0139003259.085.0

1102

11420

2585.0

9.0

30030

1399.136.025.16

.1

22

min

min2

6

lim

use

cmmmAs

assume

mmcmb

mmcmd

mkNM

w

u

uCVkNV 261000/300139

6

2575.0

Design for shear


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LongitudinalSupported stairs


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Design a U- stair in a residential building L.L = 3 kN/m2Covering Material = 2 kN/m2

The riser are 16 cm and goings are 30 cm

LongitudinalSupported stairs (Example)


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Loads and Analysis

cml

h 2220

525

2085.0min

D.L (slab) = 0.22*25 =5kN/m2

D.L (step) = 0.5*0.16*25=2 kN/m2D.L (covering material) = 2 kN/m2

D.L (flight) = 9 kN/m2

D.L (landing) = 7 kN/m2

L.L =3 kN/m2

Wu(flight) = 1.2(9)+1.6(3)=15.6kN/m

Wu(landing) = 1.2(7)+1.6(3)=13.2kN/m


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Design for moment

128

33.732.733194*1050*0036.0

0036.01940501259.085.0

2.52102

11420

2585.0

9.0

1050105

1944.196.0222

.2.52

22

min

2

6

lim

use

cmmmAs

assume

mmcmb

mmcmd

mkNM

w

u

uCVkNV 3.1271000/1050194

6

2575.0

Design for shear


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Reinforcement Notes


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Cantilever Stair


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Cantilever stairs (Example)

Design a cantilever straight flight stair in a residential buildingL.L = 3kN/m2

Covering Material = 0.5kN/m

The riser are 16cm and goings are 30 cm


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cm

l

h 1510

150

10min

D.L(O.W) =0.15*0.075*25 + 0.15*0.075*25=0.6 kN/m

D.L (covering material) = 0.5 kN/m

D.L (total) = 1.1 kN/m

L.L =3*(0.3) =0.9 kN/m

Wu = 1.2(1.1)+1.6(0.9) = 2.76 kN/m

Vu = 2.76(1.5) = 4.14 kN

Mu = WuL2/2 = 3.1 kN.m 15

15

7.5


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Design for moment

102

72.061.71124*75*0077.0

0077.012457259.085.0

1.3102

11420

2585.0

9.0

755.7

1944.126.0215

.1.3

22

min

2

6

lim

use

cmmmAs

assume

mmcmb

mmcmd

mkNM

w

u

uCVkNV 8.51000/75124

6

2575.0

Design for shear


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LongitudinalSupported stairswell stairs-Open


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General Example

Eng. Khalid Stair-Design

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