In this project we are aimed at making an automatic stair climbing wheelchair with bed. A concept for a stair climbing wheelchair capable of moving in structured and unstructured environment…Full description
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The project deals with the designing and manufacturing of a tri wheeler stair climber, which can climb stairs with less effort as compared to manual work. The technical issues in designing of this vehicle are the stability and speed of the vehicle wh
Stair Case pressurisationFull description
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ANALYSIS & DESIGN OF STAIR CASE
Stair Climbing Training
DESIGN OF CONCRETE STAIRWAY Design Criteria : Materials strenght: fc' fy unit weight of concrete b
= = = =
20.76 mPa 276 mP mPa 3 24 kN/m 0.85
Service load: Live load……………………… Miscellaneous live load………… Floor finish and toppings………. Miscellaneous dead load….....
A. estimate the thickness thickness of the slab minimum thickness for simply supported slab to control deflection (NSCP 5th edition, Table 409-1)
landing slab
200
riser 0.25
tread
1600.00
Lc =
2,000.00
mm
Typical Stair
hmin = say:
Lc/20 = 2000/20 = hmin =
150
1. Service loads a. service dead load
100.00 mm mm
mm
m m 0 5 1
250 200
2
2
r + t
WDL = Wstep + Wslab + Wfloor finish +Wmiscellaneous dead load whare: W stem = 1/2 ( r ) wc =1/2 (200)(24) = 2.4 kPa W slab = h/t W slab =
r +t
3.0735 WDL =
( Wc) = 0.125 (24) kPA
(0.2) + (0.25) 0.25
(2.4)+ (2.1954)+ (1.1)+ (0.5) =
7.0735 kPa
b. service live load Wll = Wll stair +W miscellaneous live load Wll = 1.9 + 0.50 Wll = 2.4 kPa
2. Factored loads Wu = 1.4(DL) + 1.7(LL) Wu = 13.983 kPa note: Anlyze 1m strip of stair slab 3. Compute for the required effective depth of the stair slab. As required by the minimum thickness requirements of the ACI code. note: a) use a 12mm f main bars. b) use a 20mm minimum concrete cover. 1 m strip db/2 d
h
cc
d = h - (cc + db/2) d = 125 - (20 + 12/2) d= 74 mm
MOMENT STRENGHT : 2 2 Mu = 1/8 Wu Lc = 1/8 (13.983) (2250) Mu = 6.99145 kN-m
4.Compute for
min and
max
rmin = rmin =
fc' / 4(fy) = 0.004127
rmax =
(0.75) x {(0.85) ( fc' ) ( b ) ( 600)
1.4/fy = 0.005072
use!
fy ( fy + 600)
rmax = 5. Compute for
0.027917
w 2
Mu = f bd fc'w ( 1-0.59w) Mu = 6991450 N-mm f= 0.9 b= 1000 mm d= 74 mm fc' = 20.76 mPa 2 w - 0.59 w = 0.068334 w = 1.623579 w = 0.071336 use!
0.59 -1 3.14 12 10 20 100
6. Compare for
act
=
act =
w fc' fy 0.005366 >
min
<
max
ok!
7. Reinforcements: A. Main bars As = bd
r =
0.005366
b= d=
1000 74
using 12mmf : mm mm
2
(1000/s)(p*12 /4) = 335.0006
S= 284.691 > 3h As = 397.0621 mm therefore use 3h spacing = 450 use : 150mm therefore, use 12 mm main steel bars spaced @ 150 mm O.C 2
mm
B. Temperature bars Ast = 0.0018bh Ast = 0.0018(1000)(100) 2 mm Ast = 270
mm mm
using 10mmf : 2 (1000/s)(p*10 /4) = 180
S = 290.7407 < 5h ( ok ) therefore use actual spacing = 290.7407 mm use : 200mm therefore, use 10 mm temperature bars spaced @ 200 mm O.C