3. Compute the resisting moment with respect to the x-x axis for the A36 steel member shown. The cover plates are attached with continuous fillet welds, and the compression flange has full lateral support. (Use Fb= 24 ksi.) For the W18X 50: (section depth) d = 17.99 in. Ix = 800 in.4 10" X " plate W18 X 50 10" X " plate
3. Compute the resisting moment with respect to the x-x axis for the A36 steel member shown. The cover plates are attached with continuous fillet welds, and the compression flange has full lateral support. (Use Fb= 24 ksi.) For the W18X 50: (section depth) d = 17.99 in. Ix = 800 in.4 10" X " plate W18 X 50 10" X " plate
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter3: Tension Members
Section: Chapter Questions
Problem 3.4.1P
Related questions
Question
![3. Compute the resisting moment with respect to the x-x axis for the A36 steel member shown. The
cover plates are attached with continuous fillet welds, and the compression flange has full lateral
support. (Use F1= 24 ksi.)
For the W18X 50: (section depth) d = 17.99 in.
Ix
= 800 in 4
10" x " plate
W18 X 50
10" X" plate](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20cba98e-6744-4650-af87-e93c56f86f12%2F84935fd2-3b26-4958-94d5-7dca8dab4676%2F5rc25e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. Compute the resisting moment with respect to the x-x axis for the A36 steel member shown. The
cover plates are attached with continuous fillet welds, and the compression flange has full lateral
support. (Use F1= 24 ksi.)
For the W18X 50: (section depth) d = 17.99 in.
Ix
= 800 in 4
10" x " plate
W18 X 50
10" X" plate
![Appendix
Expressions for axially loaded compression member:
KI
For-< C. =
2T²E
[1 – (Kl/r)²/(2C? )]Fy
F, =
5 3(Kl/r) _ (Kl/r)³
8Cc
Fy
|21²E
1212E
KI
For-> Cc
Fa
Fy
23(KI/r)2
Pa =
FaA
Expressions for beam:
MR = F,S, in which S =-
Expressions for tension member:
Pt = A,F , in which F = 0.6Fy
P; = A,Ft , in which F = 0.5Fu
P = A.Ft , in which Ae = U An, F = 0.5Fu
P = A,F, + A,F; , in which Fy = 0.30Fµ , F; = 0.50Fu
P = nr,, in which r, = ApFv, ry = d t F,
TABLE 2-1 Values for Reduction Coefficient, U
U = 0.90
W, M, S shapes or their tees.
Connection is to the flanges.
Minimum of three bolts per line in the direction of
Case I
stress.
(min.)
(min.)
U = 0,85
All shapes and built-up cross sections not meeting
the requirements of case I. Minimum of three bolts
per line in the direction of stress.
Case II
U = 0.75
All members whose connections have only two bolts
per line in the direction of stress.
Case III
Units:
ksi : kips per square inch
kips: kilo pounds
lb/ft: pounds per foot
kips/ft: kilo pounds per foot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20cba98e-6744-4650-af87-e93c56f86f12%2F84935fd2-3b26-4958-94d5-7dca8dab4676%2F2q6daw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Appendix
Expressions for axially loaded compression member:
KI
For-< C. =
2T²E
[1 – (Kl/r)²/(2C? )]Fy
F, =
5 3(Kl/r) _ (Kl/r)³
8Cc
Fy
|21²E
1212E
KI
For-> Cc
Fa
Fy
23(KI/r)2
Pa =
FaA
Expressions for beam:
MR = F,S, in which S =-
Expressions for tension member:
Pt = A,F , in which F = 0.6Fy
P; = A,Ft , in which F = 0.5Fu
P = A.Ft , in which Ae = U An, F = 0.5Fu
P = A,F, + A,F; , in which Fy = 0.30Fµ , F; = 0.50Fu
P = nr,, in which r, = ApFv, ry = d t F,
TABLE 2-1 Values for Reduction Coefficient, U
U = 0.90
W, M, S shapes or their tees.
Connection is to the flanges.
Minimum of three bolts per line in the direction of
Case I
stress.
(min.)
(min.)
U = 0,85
All shapes and built-up cross sections not meeting
the requirements of case I. Minimum of three bolts
per line in the direction of stress.
Case II
U = 0.75
All members whose connections have only two bolts
per line in the direction of stress.
Case III
Units:
ksi : kips per square inch
kips: kilo pounds
lb/ft: pounds per foot
kips/ft: kilo pounds per foot
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