Determine the design flexural strength in ft kips of a W10X26 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0. Blank 2 C. An unbraced length of 20 ft with Cb 1.0. Blank 3 W10x26 A = 7.61 in. ^2 d = 10.300 in. tw = 0.260 in. bf = 5.770 in. tf = 0.440 in. T = 8-1/4 in. k = 0.7400 in. k1 = 0.6875 in. gage = 2-3/4 in. rt = 1.540 in. d/Af = 4.07 Ix = Sx = 144.00 in.14 27.90 in ^3 x = 4.350 in. ly = Sy = ry = Zx = 14.10 in ^4 4.89 in.^3 1.360 in. 31.30 in.^3 Zy = J = Cw = 7.50 in.43 0.40 in.^4 345 in 46 a = 47.14 in. Wno = 14.30 in ^2 Sw = Qf = 9.05 in.14 5.99 in ^3 Qw = 15.50 in 43

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Determine the design flexural strength in ft kips of a W10X26 of A242 (Fy = 50ksi) steel subject to
a. Continuous lateral support. Blank 1
b. An unbraced length of 10 ft with Cb = 1.0. Blank 2
C. An unbraced length of 20 ft with Cb 1.0. Blank 3
W10x26
A =
7.61
in. ^2
d =
10.300
in.
tw =
0.260
in.
bf =
5.770
in.
tf =
0.440
in.
T =
k =
8-1/4
in.
0.7400
in.
k1 =
0.6875
in.
gage =
2-3/4
in.
rt =
1.540
in.
d/Af =
4.07
Ix =
Sx =
144.00
in. 14
27.90
in ^3
rx =
4.350
in.
ly =
Sy =
ry =
Zx =
14.10
in.^4
4.89
in.^3
1.360
in.
31.30
in.^3
Zy =
J =
Cw =
7.50
in 43
0.40
in.^4
345
in ^6
a =
47.14
in
Wno =
14.30
in.^2
Sw =
Qf =
9.05
in.14
5.99
in ^3
Qw =
15.50
in 43
Transcribed Image Text:Determine the design flexural strength in ft kips of a W10X26 of A242 (Fy = 50ksi) steel subject to a. Continuous lateral support. Blank 1 b. An unbraced length of 10 ft with Cb = 1.0. Blank 2 C. An unbraced length of 20 ft with Cb 1.0. Blank 3 W10x26 A = 7.61 in. ^2 d = 10.300 in. tw = 0.260 in. bf = 5.770 in. tf = 0.440 in. T = k = 8-1/4 in. 0.7400 in. k1 = 0.6875 in. gage = 2-3/4 in. rt = 1.540 in. d/Af = 4.07 Ix = Sx = 144.00 in. 14 27.90 in ^3 rx = 4.350 in. ly = Sy = ry = Zx = 14.10 in.^4 4.89 in.^3 1.360 in. 31.30 in.^3 Zy = J = Cw = 7.50 in 43 0.40 in.^4 345 in ^6 a = 47.14 in Wno = 14.30 in.^2 Sw = Qf = 9.05 in.14 5.99 in ^3 Qw = 15.50 in 43
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