Stair Design. Stair Types Straight stairs.

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Stair Design

Transcript of Stair Design. Stair Types Straight stairs.

Stair Design

Stair TypesStraight stairs

Stair Type based on the structural loading type

Simply supported stair (transversely supported)

Longitudinally supported stair

Cantilever stair

Simply supported stair Design a straight flight stair in a residential building supported on reinforced concrete wall 1.5 m )(center to center)L.L = 3kN/m2Covering Material = 0.5kN/mThe riser are 16cm and goings are 30 cm

Loads and Analysis

cmh

cml

h

ave 5.165.7

8cos

5.78

5.720

150

20

22 1630

16

min

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

D.L (total) = 1.74 kN/m D.L (total) = 1.74 kN/m L.L =3*(0.3) =0.9 kN/mL.L =3*(0.3) =0.9 kN/m

0.30.16

0.34

Design for moment

stepeach for 121

9.01.89165*300*0018.0

0005.0139003 25 9.0 85.0

110211

420

2585.0

9.0

30030

1399.136.025.16

.1

22min

min2

6

lim

use

cmmmAs

assume

mmcmb

mmcmd

mkNM

w

u

uC VkNV 261000/3001396

2575.0

Design for shear

Longitudinal – Supported stairs

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

Longitudinal – Supported stairs (Example)

1 3 1

0,255,25

1,05

Loads and Analysis

cml

h 2220

525

2085.0min

D.L (D.L (slabslab) ) = 0.22*25 =5kN/m2 = 0.22*25 =5kN/m2 D.L (step) = 0.5*0.16*25=2 kN/m2D.L (step) = 0.5*0.16*25=2 kN/m2D.L (covering material) = 2 kN/m2D.L (covering material) = 2 kN/m2

D.L (flight) = 9 kN/m2 D.L (flight) = 9 kN/m2 D.L (landing) = 7 kN/m2D.L (landing) = 7 kN/m2L.L =3 kN/m2L.L =3 kN/m2

WWu u (flight) = 1.2(9)+1.6(3)=15.6kN/m(flight) = 1.2(9)+1.6(3)=15.6kN/mWWuu (landing) = (landing) = 1.2(7)+1.6(3)=13.2kN/m1.2(7)+1.6(3)=13.2kN/m

Design for moment

128

33.732.733194*1050*0036.0

0036.01940501 25 9.0 85.0

2.5210211

420

2585.0

9.0

1050105

1944.196.0222

.2.52

22min

2

6

lim

use

cmmmAs

assume

mmcmb

mmcmd

mkNM

w

u

uC VkNV 3.1271000/10501946

2575.0

Design for shear

Reinforcement Notes

Cantilever Stair

Cantilever stairs (Example)

Design a cantilever straight flight stair in a residential buildingL.L = 3kN/m2Covering Material = 0.5kN/mThe riser are 16cm and goings are 30 cm

cml

h 1510

150

10min

D.L(D.L(O.WO.W) =0.15*0.075*25 + 0.15*0.075*25=0.6 kN/m) =0.15*0.075*25 + 0.15*0.075*25=0.6 kN/mD.L (covering material) = 0.5 kN/mD.L (covering material) = 0.5 kN/m

D.L (total) = 1.1 kN/m D.L (total) = 1.1 kN/m L.L =3*(0.3) =0.9 kN/mL.L =3*(0.3) =0.9 kN/m

Wu = 1.2(1.1)+1.6(0.9) = 2.76 kN/mWu = 1.2(1.1)+1.6(0.9) = 2.76 kN/mVu = 2.76(1.5) = 4.14 kNVu = 2.76(1.5) = 4.14 kNMu = WMu = WuuLL22/2 = 3.1 kN.m/2 = 3.1 kN.m 15

15

7.5

Design for moment

102

72.061.71124*75*0077.0

0077.012457 25 9.0 85.0

1.310211

420

2585.0

9.0

755.7

1944.126.0215

.1.3

22min

2

6

lim

use

cmmmAs

assume

mmcmb

mmcmd

mkNM

w

u

uC VkNV 8.51000/751246

2575.0

Design for shear

Longitudinal – Supported stairs

Open-well stairs

General Example