stair design. stair types straight stairs
TRANSCRIPT
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
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
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