5.5-9
The beam shown in Figure P5.5-9 is laterally supported at a, b, and c only. The given
loads are factored. Compute Cbfor the unbraced segment b-c.
30k
Loads are factored
ie/ft
Lateral bracing
at a, b, and conly
20/
30/
j.-lif-+
~
FIGURE
P5.5-9
5.5-9
CE 405: Design of Steel
Structures
Ch. 6: Welded Connections
V. Kodur
Professor
Dept of Civil & Env. Engineering
Michigan State University
CE 405 - Ch. 6
Slide # 1
Welded Connections
CLO - 8. Design welded tension connections
Types of Connections
Introduc
Michigan State University
Department of Civil and Environmental Engineering
CE 405: Design of Steel Structures
HW #1: Loads and Analysis
14 January 2015
(Due: Friday, January 26, 2015 at 4:30pm)
A two-dimensional (2D) building frame is shown in the follow
CE 405: Design of Steel
Structures
Ch. 4: Tension Member Design
V. Kodur
Professor
Dept of Civil & Env. Engineering
Michigan State University
CE 405 - Ch. 4
Slide # 1
Table of Contents
Typical Tension Members
Failure Modes
Introductory Concepts
Design Str
.
3.2-1
The tension member shown in Figure P3.2-1 is a PL ~ x 8 of A36 steel. The member is connected to a gusset plate with 1Ys-inch-diameterbolts. It is subjected to
the dead and live loads shown. Does this member have enough strength? Assume that
Ae= A
C~ 405: Design of Steel Structures'
HW #3: Load Combinations and Beams
16 September 2005
(Due: Friday, September 23,2005 at 4:30pm)
2.3-'
A column in the upper story of a building is subjected to a compressive load from the
following sources: dead load =3
Michigan State University
Department of Civil and Environmental Engineering
CE 405: Design of Steel Structures
HW #7: Tension Members
10 March 2014
(Due: March 17, 2014 at 4:30pm)
Problem 1
The tension member shown in Figure P3.2-1 is a PL 1/2 x 8 of A36
Michigan State University
Department of Civil and Environmental Engineering
CE 405: Design of Steel Structures
HW #8: Tension Members Design
18 March 2014
(Due: Wednesday, March 26, 2014 at 4:30pm)
Problem 1
A WT8 x 13 of A992 steel is used as a tension m
CE 405
Solutions to HW 5 (Columns)
Problem 1
Determine the axial compressive design strength by each of a. the following methods:
a. Use AISC Equation E3-2 or E3-3.
b. Use Table 4-1 from the Design Tables of the AISC Manual.
a)
KL 0.8(13 12 )
=
= 50.32 <
Michigan State University
Department of Civil and Environmental Engineering
CE 405: Design of Steel Structures
HW #5: Design of Columns
February 12, 2014
(Due: Wednesday, February 19, 2014 at 4:30 pm)
Problem 1
Determine the axial compressive design stren
CE 405: Design of Steel
Structures
Ch. 7: Structural Fire Protection
V. Kodur
Professor
Dept of Civil & Env. Engineering
Michigan State University
CE 405 - Ch. 7
Slide # 1
Outline
Structural Fire Safety
Background
Terminology
Fire Resistance
Evaluation
St
CE 405: Design of Steel
Structures
Ch. 3: Design of Compression Members
V. Kodur
Professor
Dept of Civil and Env. Engineering
Michigan State University
CE 405 - Ch.3
Slide # 1
Design of Compression Members
Short and long columns
Buckling load and buckling
CE 405: Design of Steel
Structures
Ch. 1:
Introduction to Design of Steel Structures
V. Kodur
Professor
Dept of Civil and Env. Engineering
Michigan State University
CE 405 - Ch. 1
Slide # 1
Outline
Design Process
Structural Members
Steel structures: Examp
CE 405: Design of Steel
Structures
Ch. 5: Bolted Connections
V. Kodur
Professor
Dept of Civil & Env. Engineering
Michigan State University
CE 405 - Ch. 5
Slide # 1
Bolted Connections
CLO - Design bolted tension connections
Types of Connections
Introductor
CE 405: Design of Steel
Structures
Ch. 2 : Design of Beams for Flexure
V. Kodur
Professor
Dept of Civil and Env. Engineering
Michigan State University
CE 405 - Ch. 2
Slide # 1
Design of Beams for Flexure
Introduction
Moment Curvature Response
Sectional Pr
HW 8 Solutions
Determine the block shear strength
Rn = (0.6 Fu Anv + UbsFu Ant) (0.6 FyAgv + UbsFu Ant)
Rn = 0.75 x (0.6 x 65 x 2.25 + 1.0 x 65 x 0.7125) = 100.55 kips
The upper limit on Rn is
(0.6 FyAgv + UbsFu Ant) = 0.75 x (0.6 x 50 x 3.375 + 1.0 x