1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. ) -R9 X T6 W16 X 67 L4 X 3 x L3 X 3 x }
1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. ) -R9 X T6 W16 X 67 L4 X 3 x L3 X 3 x }
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter9: Moments And Products Of Inertia Of Areas
Section: Chapter Questions
Problem 9.13P: Figure (a) shows the cross section of a column that uses a structural shape known as W867...
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The calcul of this weight
![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%2Ff6c1ee62-b6fb-40d2-9659-237bb20bb00f%2Frpc9k8p_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
Transcribed Image Text:1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross
section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. )
91 x 6a-
W16 X 67
wL4X 3 x
L3 X 3 x
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