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Name of Project: hhghh General Notes:
Job No.: hhghh gkg
Client: yyy uyuu
Designed by: uuu hgh
Date: tyyuy
Sheet: 2 Of 2
Jerv Works Const.
STAIR DESIGN
Stair no. 2
specification
fy 40000 psi
fc' 3000 psi
unfactored service load
service Live Load 100 psf
service Dead Live 20 psf (excluding stair self-weight)
stair
ht. of rise, ft. 10 ft.
trial rise, 6.5" - 7.5" 6.75 in. actual rise 6.67 in.
tread, min 9 in. 11 in. (OK - rise and tread proportion)
slab h in. 8 in. >= 7.04 in. OK
conc. cover, suggested 3/4" 0.75 in.
main bar temp. bar
#5 #3
to reduced deflection
suggested p = .30 of .75Pb
reinforcement ratio
.75Pb > p > 200/fy
0.02784043 0.009 >= 0.005
run = 15.58'
top bar
# 5 - 8.5" c.c
top bar
# 5 - 8.5" c.c.
11 in. rise = 10'
bottom bar
# 5 - 7" c.c. 6.67 in. Results:
1. Height of rise = 10 ft.
slab h= 8 in. 2. Length of run = 15.58 ft.
3. Stair rise = 6.67 in.
temperature bar 4. Number of rise = 18
# 3- 7"c.c 5. Stair tread = 11 in.
6. number of tread = 17
7. Slab thickness = 8 in.
NO need for shear reinforcement
calculation:
no. of rise 18 no.
rise, in. 6.67 in.
t, tread = t + r = 17.7 10.83 in. >>== 9
OK
check, t*r = 70 to 75 70 < 73.37 < 75
OK
(OK - rise and tread proportion)
x= 15.58 ft.
y = rise - 1 stair rise 9.44416667 ft.
hence, stair span, ft. 14.6633333 ft.
considering 1-foot width
trial slab h 8 in.
weight of slab w = S (√r2 + t2)(12.5/t)*1 192.8673 plf
total WDL 212.8673 plf
wu = 1.4DL+1.7LL 468.01422 plf
calculation of d
B1 0.85
but not less than 0.65
hence B1 = 0.85
Pb = .85fc'B1/fy[87000/(87000+fy)] 0.03712057
Pmax = .75Pb 0.02784043
Pmin = 200/fy 0.005
adapted P 0.009
m = fy/.85fc' 15.6862745
Rn = Pfy(1-Pm/2) 334.588235 psi
s1
(-)M support (+)M middle (-)M support
ACI moment coefficient 1/16 1/10 1/16
Mu ft-lb = c*Wu*L^2 6289 32/97 10062.9303 6289.33142
max Mu 10062.9303 ft-lb
max Mn=Mu/Ǿ 11181.0336 ft-lb
required d = sqrt (Mn/Rn*b) 5.78076938 in.
required h = conc. Cover+bar/2+d 6.84 in.
simple support
min h factor = 20/[(1-(60000-fy)]/[60000+fy) 25
min h as per ACI Code 7.04 in.
min. h required 7.04 in.
actual h 8 in.
actual d = h-conc. Cover-bar/2 6.9375 in.
re-calculate based on actual d
s1
support middle support
Mn = Mu/.90 6988.14602 11181.0336 6988.14602
required Rn=Mn/bd^2 145.196444 232.31431 145.196444
required p=1/m(1-sqrt(1-2m*Rn/fy)) 0.00373959 0.00609967 0.00373959
hence p (considering min. 200/fy) 0.005 0.00609967 0.005
required As in^2 =pbd 0.41625 0.50779746 0.41625
min. As=2.1(sqrtfc'*b*d)/fy 0.02623687 0.02623687 0.02623687
adapted As 0.41625 0.50779746 0.41625
required main bar spacing in. 8.5 7 8.5
min. bar spacing=3*t or 18" 18 18 18
adapted spacing of bar 8.5 7 8.5
# 5- 8.5"c.c # 5- 7"c.c # 5- 8.5"c.c
shrinkage and temp. bar
shrinkage and temp. bar in^2 =.002*12*h 0.192 in2
spacing of temp. bars in. = h*12/total area of temp. ba 7 in. c. c.
min. no. of bar = 1 per tread 18 in. c. c.
hence spacing = 7 in. c. c.
# 3- 7"c.c
check for shear
max Vu=1.15*Wu*L/2 3946.02289 lb
concrete shear strength without shear reinforcement
= Ǿ*2*(sqrtfc')*b*d 7751.6435 lb >>>=== 3946.02289
NO need for shear reinforcement
main bar area in^2 diameter in. temp. bar area in^2 diameter in.
#3 0.11 0.375 #3 0.11 0.375
#4 0.2 0.5 #4 0.2 0.5
#5 0.31 0.625 0.11 0.375
#6 0.44 0.75
#7 0.6 0.875
#8 0.79 1
#9 1 1.128
# 10 1.27 1.27
# 11 1.56 1.41
0.31 0.625
