Determine the design flexural strength in ft kips of a W12 x 35 of A242 (Fy = 50ksi) steel subject toa. Continuous lateral support. Blank 1b. An unbraced length of 10 ft with Cb = 1.0. Blank 2C. An unbraced length of 20 ft with Cb = 1.0. Blank 3W12x35...10.30in2d=12.500in.tw =0.300in.bf =6.580in.0.520in.T=10-1/9in.k=0.8200in.k1 =0.7500In.gage3 1/2In.itd'Af =1.740n.3.88205.00n.4Sx-45.80n.3IX=5.250n.ly=Sy =24.507.47In.43ry =1.540n.Zx =Zy =J=Cw =51.20n.3n.311.500.74in.4879n.6a =55.42n.Wno =19.60m2Sw=18.80n.49.75in43Qw =|25.40in.3

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Answered: tinuous lateral support. Blank 1…

tinuous lateral support. Blank 1 unbraced length of 10 ft with Cb = 1.0. Blank 2 unbraced length of 20 ft with Cb = 1.0. Blank 3 W12x35 A= 10.30 n2 12.500 in. tw 0.300 in. bf 6.580 in. 0.520

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter10: Plate Girders

Section: Chapter Questions

Problem 10.7.8P

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Question

Determine the design flexural strength in ft kips of a W12 x 35 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
W12x35
...
10.30
in2
d=
12.500
in.
tw =
0.300
in.
bf =
6.580
in.
0.520
in.
T=
10-1/9
in.
k=
0.8200
in.
k1 =
0.7500
In.
gage
3 1/2
In.
it
d'Af =
1.740
n.
3.88
205.00
n.4
Sx-
45.80
n.3
IX=
5.250
n.
ly=
Sy =
24.50
7.47
In.43
ry =
1.540
n.
Zx =
Zy =
J=
Cw =
51.20
n.3
n.3
11.50
0.74
in.4
879
n.6
a =
55.42
n.
Wno =
19.60
m2
Sw=
18.80
n.4
9.75
in43
Qw =|
25.40
in.3

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