13NSCP 2015Pn = FerAg(505.3-1)The flexural buckling stress, Fer, is determined as follows:KL1. whenr1.71 (or F, 2 0.44F,)(or F. 2 0.44F,)FyFr = 0.Fy(505.3-2)KLE> 4.71Fy( or F. < 0.44F,)2.whenrFer = 0.877F.(505.3-3)whereF.= elastic critical buckling stress determined accordingto Eq. 505.3-4, Section 505.4, or the provisions ofSection 503.2, as applicable, MPa.F =KL(505.3-4)A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compressionmember. It is 7 m long, fixed at the top and bottom, and has additional support inthe weak direction 3 m from the top. Properties of the section are: A = 14,500mm2 , I = 258.6 x 10° mm“, Iy = 84.375 x 10° mmª1.Calculate the critical slenderness ratio of the member.А. 26.208C. 36.706B. 45.882D. 27.5292.Calculate the nominal axial load capacity of the column.B. 4 851 kNA. 3 104 kNC. 4 213 kND.5 344 kN3.Calculate the service axial dead load if the service axial live load is twice asthat of the dead load. Use LRFD.A. 1 354 kNC. 1 093 kNB. 992 kND.634 kN

Given data: W 12*79 of Grade60 ,Fy=415 MPa Length of member=7m (fixed at top and bottom) effective…

1. Calculate the critical slenderness ratio of the member. A. 26.208 B. 45.882 C. 36.706 D.27.529 2. Calculate the nominal axial load capacity of the column. A. 3 104 kN B. 4 851 KN C. 4 213 KN D.5 344 KN 3. Calculate the service axial dead load if the service axial live load is twice as that of the dead load. Use LRFD. A. 1 354 KN B. 992 KN C. 1 093 KN D.634 KN

1. Calculate the critical slenderness ratio of the member. A. 26.208 B. 45.882 C. 36.706 D.27.529 2. Calculate the nominal axial load capacity of the column. A. 3 104 kN B. 4 851 KN C. 4 213 KN D.5 344 KN 3. Calculate the service axial dead load if the service axial live load is twice as that of the dead load. Use LRFD. A. 1 354 KN B. 992 KN C. 1 093 KN D.634 KN

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 10.7.9P

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Concept explainers

Connections

In steel structures, connections are the members that join two or more steel members. The connections are the most important members of any steel structure, and hence, they are designed more carefully. The majority of the construction cost of steel structures is incurred due to the connections. The design of connections is done as per the recommendations and specifications of the American Institute of Steel Construction (AISC).

Question

13
NSCP 2015
Pn = FerAg
(505.3-1)
The flexural buckling stress, Fer, is determined as follows:
KL
1. when
r
1.71 (or F, 2 0.44F,)
(or F. 2 0.44F,)
Fy
Fr = 0.
Fy
(505.3-2)
KL
E
> 4.71
Fy
( or F. < 0.44F,)
2.
when
r
Fer = 0.877F.
(505.3-3)
where
F.= elastic critical buckling stress determined according
to Eq. 505.3-4, Section 505.4, or the provisions of
Section 503.2, as applicable, MPa.
F =
KL
(505.3-4)
A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression
member. It is 7 m long, fixed at the top and bottom, and has additional support in
the weak direction 3 m from the top. Properties of the section are: A = 14,500
mm2 , I = 258.6 x 10° mm“, Iy = 84.375 x 10° mmª
1.
Calculate the critical slenderness ratio of the member.
А. 26.208
C. 36.706
B. 45.882
D. 27.529
2.
Calculate the nominal axial load capacity of the column.
B. 4 851 kN
A. 3 104 kN
C. 4 213 kN
D.5 344 kN
3.
Calculate the service axial dead load if the service axial live load is twice as
that of the dead load. Use LRFD.
A. 1 354 kN
C. 1 093 kN
B. 992 kN
D.634 kN

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