Steel Stair
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Transcript of Steel Stair
PROJECT : PAGE :
CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
Steel Stair Design Based on AISC 360-10
INPUT DATA & DESIGN SUMMARYCOLUMN SECTION (Tube or Pipe) = > HSS4X4X1/4
Tube A t h3.37 1.52 0.23 4.00
FLOOR BEAM - 1 = > W16X26
A d7.68 15.70 301.00 38.40 44.20
STRINGER - 1 (Channel or Tube) = > MC8X8.5
Channel A d2.50 8.00 23.30 5.82 6.95
STRINGER - 2 (Channel or Tube) = > C12X20.7
Channel A d6.08 12.00 129.00 21.50 25.60
LANDING BEAM - 1 (Channel or Tube) = > C8X18.75
Channel A d5.51 8.00 43.90 11.00 13.90
LANDING BEAM - 2 (Channel or Tube) = > C8X18.75
Channel A d5.51 8.00 43.90 11.00 13.90
DIMENSIONS H = 16 ft, story Ht
12 ft
6 ft
10 ft
NUMBER OF STORIES n = 2
GRAVITY LOAD DL = 50 psfLL = 100 psf
THE STAIR DESIGN IS ADEQUATE.
ANALYSIS
STRINGER - 1
33.69 deg, from horizontal
400 plf , projected
2.40 kips
7.20 ft-kips
36.00 ksi
12.49 ft-kips
> M [Satisfactory]E = 29000 ksi
0.46 in
< 0.72 in [Satisfactory]
LANDING BEAM - 1
450 plfP = 2.40 kips, from STRINGER - 1
4.65 kips
17.63 ft-kips
36.00 ksi
24.97 ft-kips
> M [Satisfactory]
0.14 in
rmin
Ix Sx Zx
Ix Sx Zx
Ix Sx Zx
Ix Sx Zx
Ix Sx Zx
L1 =
L2 =
L3 =
q =
w = 0.25 (DL / Cos q + LL) L3 =
R = 0.5 w L1 =
M = w L12 / 8 =
Fy =
Mn / Wb = Fy Zx / 1.67 =
DLL = 5 (wLL Cos q) (L1 / Cos q)4 / (384 E I) =
(L1 / Cos q) / 240 =
w = 0.5 (DL + LL) L2 =
R = 0.5 w L3 + P =
M = w L32 / 8 + P L3 / 2 =
Fy =
Mn / Wb = Fy Zx / 1.67 =
DLL = 5 wLL L34 / (384 E I) + PLL L3
3 / (24 E I) =
< 0.50 in [Satisfactory]L3 / 240 =
(cont'd)STRINGER - 2
w = 400 plf , projected, from STRINGER - 1P = 4.65 kips, from LANDING BEAM - 1
4.75 kips
4.70 kips
11.87 ft, from left
28.21 ft-kips
36.00 ksi
45.99 ft-kips
> M [Satisfactory]
0.24
< 0.90 in [Satisfactory]
LANDING BEAM - 2
w = 450 plf, from LANDING BEAM - 1
2.25 kips
5.63 ft-kips
36.00 ksi
24.97 ft-kips
> M [Satisfactory]
0.05 in
< 0.50 in [Satisfactory]
FLOOR BEAM - 1
L = 12 ftw = 600 plf, floor gravity loadP = P1 + P2 = 4.80 kips, total point loads, from STRINGER - 1
6.00 kips
25.21 ft-kips
50.00 ksi
110.28 ft-kips
> M [Satisfactory]
0.04 in
< 0.60 in [Satisfactory]
COLUMN
P = 6.95 kipsR = n P = 13.90 kipsKL = H = 16 ftK = 1.0
46 ksi126
18 ksi
16 ksi
31.80 kips
> R [Satisfactory]
17.98 / 1.67 = 10.77 ft-kips, (AISC 360-10 F)
0.18 W = 0.18 x ( 50 psf x 45.0 0.41 kips, ASD
F H / 4 = 1.62 ft-kips
0.57 < 1.0 [Satisfactory]
0.26 in
< 0.80 in [Satisfactory]
RL = [w L1 (0.5 L1 + L2) + P L2] / (L1 + L2) =
RR = [w L1 (0.5 L1) + P L1] / (L1 + L2) =
X = RL / w =
Mmax = RL X - (0.5 w X2 ) =
Fy =
Mn / Wb = Fy Zx / 1.67 =
DLL = 5 wLL (L1 + L2)4 / (384 E I) + PLL (L1 +L2)3 / (48 E I) =
(L1 + L2) / 240 =
R = 0.5 w L3 =
M = w L32 / 8 =
Fy =
Mn / Wb = Fy Zx / 1.67 =
DLL = 5 wLL L34 / (384 E I) =
L3 / 240 =
R = 0.5 w L + 0.5 P =
M = w L2 / 8 + P L3 / 4 =
Fy =
Mn / Wb = Fy Zx / 1.67 =
DLL = 5 wLL L4 / (384 E I) + PLL L3 / (48 E I) =
L / 240 =
Fy =K l / r =
Fe =
Fcr =
Pn / Wc = Fcr Ag / 1.67 =
Mc = Mn / Wb =
F = ft2 ) =(If no landing seismic load, F shall be zero.)
Mr =
DF = F H3 / (48 E I) =
H / 240 =
8, 0.2
9
, 0.22
r r r
c c c
r r r
c c c
P M PforP M P
P M PforP M P
PROJECT : PAGE :
CLIENT : DESIGN BY :
JOB NO. : DATE : REVIEW BY :
Steel Stair Design Based on AISC Manual 9th
INPUT DATA & DESIGN SUMMARYCOLUMN SECTION (Tube or Pipe) = > HSS4X4X1/4
Tube A t h3.37 1.52 0.23 4.00
FLOOR BEAM - 1 = > W16X26
A d7.68 15.70 301.00 38.40
STRINGER - 1 (Channel or Tube) = > MC8X8.5
Channel A d2.50 8.00 23.30 5.82
STRINGER - 2 (Channel or Tube) = > C12X20.7
Channel A d6.08 12.00 129.00 21.50
LANDING BEAM - 1 (Channel or Tube) = > C8X18.75
Channel A d5.51 8.00 43.90 11.00
LANDING BEAM - 2 (Channel or Tube) = > C8X18.75
Channel A d5.51 8.00 43.90 11.00
DIMENSIONS H = 16 ft, story Ht
12 ft
6 ft
10 ft
NUMBER OF STORIES n = 2
GRAVITY LOAD DL = 50 psfLL = 100 psf
THE STAIR DESIGN IS ADEQUATE.
ANALYSIS
STRINGER - 1
33.69 deg, from horizontal
400 plf , projected
2.40 kips
7.20 ft-kips
14.85 ksi
21.60 ksi > [Satisfactory]E = 29000 ksi
0.46 in
< 0.72 in [Satisfactory]
LANDING BEAM - 1
450 plfP = 2.40 kips, from STRINGER - 1
4.65 kips
17.63 ft-kips
19.23 ksi
21.60 ksi > [Satisfactory]
0.14 in
< 0.50 in [Satisfactory]
rmin
Ix Sx
Ix Sx
Ix Sx
Ix Sx
Ix Sx
L1 =
L2 =
L3 =
q =
w = 0.25 (DL / Cos q + LL) L3 =
R = 0.5 w L1 =
M = w L12 / 8 =
fb = M / Sx =
Fb = 0.6 Fy = fb
DLL = 5 (wLL Cos q) (L1 / Cos q)4 / (384 E I) =
(L1 / Cos q) / 240 =
w = 0.5 (DL + LL) L2 =
R = 0.5 w L3 + P =
M = w L32 / 8 + P L3 / 2 =
fb = M / Sx =
Fb = 0.6 Fy = fb
DLL = 5 wLL L34 / (384 E I) + PLL L3
3 / (24 E I) =
L3 / 240 =
(cont'd)STRINGER - 2
w = 400 plf , projected, from STRINGER - 1P = 4.65 kips, from LANDING BEAM - 1
4.75 kips
4.70 kips
11.87 ft, from left
28.21 ft-kips
15.75 ksi
21.60 ksi > [Satisfactory]
0.24
< 0.90 in [Satisfactory]
LANDING BEAM - 2
w = 450 plf, from LANDING BEAM - 1
2.25 kips
5.63 ft-kips
6.14 ksi
21.60 ksi > [Satisfactory]
0.05 in
< 0.50 in [Satisfactory]
FLOOR BEAM - 1
L = 12 ftw = 600 plf, floor gravity loadP = P1 + P2 = 4.80 kips, total point loads, from STRINGER - 1
6.00 kips
25.21 ft-kips
7.88 ksi
33.00 ksi > [Satisfactory]
0.04 in
< 0.60 in [Satisfactory]
COLUMN
P = 6.95 kipsR = n P = 13.90 kipsKL = H = 16 ftK = 1.0
46 ksi
112126
1.13
N/A
9.38
31.60 kips > R [Satisfactory]
RL = [w L1 (0.5 L1 + L2) + P L2] / (L1 + L2) =
RR = [w L1 (0.5 L1) + P L1] / (L1 + L2) =
X = RL / w =
Mmax = RL X - (0.5 w X2 ) =
fb = Mmax / Sx =
Fb = 0.6 Fy = fb
DLL = 5 wLL (L1 + L2)4 / (384 E I) + PLL (L1 +L2)3 / (48 E I) =
(L1 + L2) / 240 =
R = 0.5 w L3 =
M = w L32 / 8 =
fb = M / Sx =
Fb = 0.6 Fy = fb
DLL = 5 wLL L34 / (384 E I) =
L3 / 240 =
R = 0.5 w L + 0.5 P =
M = w L2 / 8 + P L3 / 4 =
fb = M / Sx =
Fb = 0.66 Fy = fb
DLL = 5 wLL L4 / (384 E I) + PLL L3 / (48 E I) =
L / 240 =
Fy =
Cc = (2p2Es/Fy)0.5 =K l / r =
F = (K l / r) / Cc =
Fa = {(1-F2/2)Fy / (5/3+3F/8-F3/8) = kis, for Cc > (Kl/r)
12p2Es/[23(Kl/r)2] = kis, for Cc < (Kl/r)
Rallow = A Fa =