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~ ~ ~~ ~ r. ASOCIACION COSTARRICENSE DE~ I ~" S ~ I I .;0(;;-" INGENIERIA ESTRUCTURAL YSISMICAC!t:t ' - , ~~
Cu rso:
~ C i C~ I J J ) c r u ~ o DE N G f N ~V eMlES DE com PJeA
"Diseno de Estructuras de AceroLaminado en Frio"
Instructor:Dr. Roger A. LaBoube, Ph.D., P.E ..Universidad de Ciencia y Tecnologfa de
Missouri, USA
Auditorio Lanamme-UCRMartes 4 de setiem bre, 201 28:00 am-6:00 pm
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Produced for
AMERICAN IRON AND STEEL INSTITUT
DESIGN OF COLD-FORMEDSTEEL STRUCTURES
1 - 1
by
Roger LaBoube
Curators Teaching Professor EmeritusWei-Wen Yu Center for Cold-Formed Steel Structures
(Missouri University of Science and Technology)
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Wei-Wen Yu
Center for Cold Formed Steel Structure
1 - 2
[email protected], 573-341-4481
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STEEL DESIGN SPECIFICATIONS
Type of steel SpecificationCold-Formed, Carbon Steel AISI
1 - 3
-
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AISI Applicable in North America
AISI S100-07 with Supplements 1 and 2
Canada
Mexico
1 - 4
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Design Methods
United States:Allowable Strength Design and Load and
Resistance Factor Design
Canada:Limit States Design
1 - 5
ex co: owa e trengt es gn an oa anResistance Factor Design
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ORGANIZATION OF THE 2007 SPECIFICATION
A. General Provisions
B. Elements
C. Members
D. Structural Assemblies and Systems
1 - 6
.
F. Tests for Special Cases
G. Fatigue
Appendices for each country
Appendix 1 for Direct Strength Method
Appendix 2 for Second-Order Analysis
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KEY CHARACTERISTICS OF COLD-FORMED
STEEL STRUCTURES
Shapes are cold-formed from flat sheets or plates Original mechanical properties of steel are changed dueto the cold formin rocess
1 - 7
Standardized and customized shapes are available Thin material (in most cases less than 0.10inch)
Predominant cross-section failure mode is influenced bylocal buckling, followed by postbuckling strength increase
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SOME TYPICAL COLD-FORMED SHAPE
CROSS SECTIONSStuds or Joists
1 - 8
Other Shapes
In the United States structural HSS sections are typically designedby hot-rolled specification
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SOME TYPICAL COLD-FORMED PANEL
CROSS SECTIONSRoof Decks
Long-Span Decks
1 - 9
Floor Decks
Curtain Wall Panels
Ribbed Panels
Corrugated Sheets
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STEEL MATERIALS AND PRODUCTS
(Section A1.1)
The Specification Covers
Steel: Carbon and low alloy
1 - 10
Steel Products: Sheet, strip, plate, bar
Members: Cold formed to shape
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Limits of Applicability
(Section A1.2)
Nominal strength and stiffness shall be determined by
Chapters A through G and Appendices A and B.
As an alternate:
1 - 11
and Chapter F
Determine design strength or stiffness by
rational analysis. Direct Strength Method (Appendix 1)
Second Order Analysis (Appendix 2)
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TYPICAL APPROVED STEELS
(Section A2.1)
16 Approved Steels in Section A2.1.
ASTM A653Steel Sheet, Zinc Coated (Galvanized) orZinc-Iron Alloy-Coated (Galvannealed) by the Hot-DipProcess
1 - 12
Fy = 33 to 50 ksi Fu = 45 to 70 ksiFy = 80 ksi Fu = 82 ksi
ASTM A792Steel Sheet, 55% Aluminum-Zinc Alloy-Coated by the Hot-Dip ProcessFy = 33 to 50 ksi Fu = 45 to 70 ksiF
y
= 80 ksi Fu
= 82 ksi
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TYPICAL APPROVED STEELS
(Section A2.1)
ASTM A1011 Steel Sheet, Hot-Rolled, Carbon,
Structural, High Strength Low-Alloy with improvedFormability
1 - 13
ASTM A1003Steel Sheet, Carbon, Metallic- andNonmetallic-Coated for Cold Formed Framing
Members
(Replaced A653 for framing members)
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TYPICAL APPROVED STEELS
(Section A2.1)
ASTM A500Standard Specification for Cold-FormedWelded and Seamless Carbon Steel Structural Tubing in
Rounds and ShapesFy = 33 to 50 ksi Fu = 45 to 63 ksi
1 - 14
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OTHER STRUCTURAL QUALITY STEELS
(Section A2.2)
Published Material Specification with Specified
Mechanical and Chemical Properties
1 - 15
Minimum Ductility Requirements of Section A2.3
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REQUIRED DUCTILITY
(Section A2.3.1)
Fu/Fy1.08
1 - 16
Elongation 10% (two-inch gage length)
7% (eight-inch gage length)
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LOW DUCTILITY STEELS(Section A2.3.2)
ASTM A653, A792, A875 and A1008, Grade 80 material
May be used for deck and panel profiles
1 - 17
Limits on Fy and Fu
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GOVERNING MATERIAL THICKNESS
Design Thickness Bare steel flat productthickness, t (exclusive ofcoating)
1 - 18
Delivered Minimum Thickness Minimum thickness is0.95t, when measured onactual product (Section A2.4)
Design Formulas Account for corner andtolerance effects
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PROPERTIES OF STEELS
Key Measures of Strength and Deformability
Yield Stress Fy
1 - 19
u
Ratio of Tensile to Yield Fu/Fy
Elongation at Fracture u
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EFFECTS OF COLD-FORMING ON
MATERIAL PROPERTIES
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DESIGN YIELD STRESS(Section A7)
Use Virgin Steel Yield Stress, Fy
orAverage Yield Stress for Cross Section, Fya
1 - 21
e erm ne verage e ress rom er o eFollowing:
Full cross section tension test
Stub column test
Computation of average yield stress:
Fya = CFyc + (1 - C)Fyf Fuv
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MAJOR COLD-FORMED STEEL DESIGN CONCEPT
Effective Design Width
1 - 22
Partially Stiffened Element Unstiffened Elements
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STIFFENED AND UNSTIFFENED ELEMENTS:
EXAMPLES
Stiffened
1 - 23
Unstiffened
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ASDSTRENGTH REQUIREMENTS
(Section A4.1.1)
RRn/
1 - 24
= equ re s reng
Rn = Nominal strength
= Safety Factor
Rn/ = Design strength (allowable strength)
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LRFDSTRENGTH REQUIREMENTS
(Section A5.1.1)
Ru Rn
1 - 25
u= equ re s reng
Rn = Nominal strength
= Resistance factor
Rn = Design strength (factored resistance)
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Wei-Wen Yu
Center for Cold Formed Steel Structure
1 - 26
[email protected], 573-341-4481
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QUESTIONS?
1 - 27
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DESIGN OF COLD-FORMED STEEL STRUCTURES
USING THE 2007 NORTH AMERICAN SPECIFICATION
COMPRESSION ELEMENTS LOCAL BUCKLING
2 - 1
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THIN PLATE ELEMENTS IN COMPRESSION(Section B)
LOCAL BUCKLING CONSIDERATIONS
Elastic Critical Buckling
Post-Buckling
TYPES OF COMPRESSION ELEMENTS
2 - 2
Stiffened Partially Stiffened
Unstiffened
STRESS CONDITION
Uniform
Gradient
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STIFFENED PLATE ELEMENTS
(Section B)
2 - 3
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UNSTIFFENED PLATE ELEMENTS
(Section B)
2 - 4
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ELASTIC CRITICAL BUCKLING OF STIFFENED PLATE
2 - 5
cr
2
2 2f = kE
12(1- )(w/t)
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FREQUENTLY USED k VALUES
2 - 6
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POST-BUCKLING OF STIFFENED PLATE ELEMENT
2 - 7
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POST-BUCKLING STRESS DISTRIBUTION
2 - 8
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EFFECTIVE WIDTH CONCEPT
2 - 9
o
w
fdx = bf max
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CLASSIC EFFECTIVE WIDTH EXPRESSION
1
2
cr y
2
2 2f = F =4 E
12(1- )(b/t)
b = 1.9t E/Fy
2 - 10
3
4f
kE)
w
t(0.208-1
f
kE0.95t=b
maxmax
also, b = 1.9t E/fmax
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BASIC EFFECTIVE WIDTH EXPRESSION
(Section B2.1)
Eq.B2.1-1
Eq.B2.1-2
b = w when 0.673
0.673>w when=b
2 - 11
Eq.B2.1-3
Eq.B2.1-4
Eq.B2.1-5
Used in all cases of effective width considerations
f/F= cr
/
.
-1=
))(w/t-12(1
Ek=F 222
cr
BASIC EFFECTIVE WIDTH RELATIONSHIP
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BASIC EFFECTIVE WIDTH RELATIONSHIP
(Section B2.1)
2 - 12
Reduction factor, , vs. slenderness factor,
EFFECT OF LOCAL BUCKLING ON
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EFFECT OF LOCAL BUCKLING ONCOLUMN SECTION
ult
Ineffecitve
2 - 13
f = Fy
Aeff
Effective Section
reas
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IMPORTANT DEFINITIONS
2 - 14
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MAXIMUM FLANGE FLAT-WIDTH-TO-THICKNESS RATIOS
(Section B1.1)
(a) Maximum Flat-Width-to-Thickness Ratios
(1) Stiffened Compression Elements (Edge Stiffeners)
Simple lip (w/t 60) Section B1.1(a)(1)
2 - 15
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MAXIMUM FLANGE FLAT-WIDTH-TO-THICKNESS RATIOS
(Section B1.1)
Any other kind of stiffeners
i) when Is < Ia (w/t 60) Section B1.1(a)(1)i)
ii) when Is Ia (w/t 90) Section B1.1(a)(1)ii)
2 - 16
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(2) Stiffened Compression Elements (w/t 500)
Section B1.1(a)(2)
MAXIMUM FLANGE FLAT-WIDTH-TO-THICKNESS RATIOS
(Section B1.1)
2 - 17
MAXIMUM FLANGE FLAT WIDTH TO THICKNESS RATIOS
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MAXIMUM FLANGE FLAT-WIDTH-TO-THICKNESS RATIOS
(Section B1.1)
(3) Unstiffened Compression Elements (w/t 60)
Section B1.1(a)(3)
2 - 18
MAXIMUM WEB DEPTH TO THICKNESS RATIOS
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MAXIMUM WEB DEPTH-TO-THICKNESS RATIOS
(Section B1.2)
Section Depths
a) Unreinforced webs(h/t)max 200 Section B1.2(a)
b) Reinforced webs, satisfying Section C3.7.1
2 - 19
1) with bearing stiffeners(h/t)max 260 Section B1.2(b)(1)
2) with bearing and intermediate stiffeners
(h/t)max 300 Section B1.2(b)(2)
UNIFORMLY COMPRESSED STIFFENED ELEMENTS
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UNIFORMLY COMPRESSED STIFFENED ELEMENTS
(Section B2.1)
2 - 20
Figure B2.1-1
Use basic effective width expression with k = 4.0
EXAMPLE 2 1 BEAM SECTION
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Given: (s.c.e.)
Fy = 33 ksi; t = 0.105 in.; R = 2t
Determine:
1) Effective width of
EXAMPLE 2.1 BEAM SECTION
2 - 21
for strength determinationw = 8.0 6(0.105) = 7.37 in.
W = w/t = 7.37/0.105 = 70.2 < 500 OK
[B1.1(a)(2)]
Eq.B2.1-5ksi21.6)2.70()-12(1
E4=F 22
2
cr =
EXAMPLE 2 1 (C ti d)
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EXAMPLE 2.1 (Continued)
Eq.B2.1-4
Since > 0.673 Eq.B2.1-2
351.2=6.21
33=f/F= cr
/0.22-1=wb
=
2 - 22
Eq.B2.1-3
.in90.4=)37.7(650.6=b
0.665235.1/235.1
0.22-1= =
EXAMPLE 2 1 (Continued)
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EXAMPLE 2.1 (Continued)
2) Effective width for serviceability
Assume f = 0.6Fy = 19.8 ksi
Eq.B2.1-4570.9=
6.21
8.19=f/F= cr
2 - 23
Since > 0.673 Eq.B2.1-2
Eq.B2.1-3
in.5.93=)0.805(7.37=b
/0.22-1=;wb
=
805.0957.0/957.0
0.22-1= =
EXAMPLE 2 2 COLUMN SECTION
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Given: (s.c.e.)
Fy = 50 ksi; t = 0.075 in.; R = 2t
Determine:
Effective widths for
EXAMPLE 2.2 COLUMN SECTION
2 - 24
strength determination1) For w1 = 9.0 - 6 (0.075) = 8.55 in.
w1/t = 8.55/0.075 = 114 < 500 OK B1.1(a)(2)
Eq.B2.1-5ksi21.8)114()-12(1
E4=F 22
2
cr =
EXAMPLE 2 2 (Continued)
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Eq.B2.1-4
Since > 0.673 Eq.B2.1-2
EXAMPLE 2.2 (Continued)
47.2=21.8
50=f/F= cr
/0.22-1=;wb
= 369.047.2/0.22-1= =
2 - 25
Eq.B2.1-3
2) For w2 = 5.0 - 6(0.075) = 4.55 in.
w2/t = 4.55/0.075 = 60.7 < 500 OK
.in15.3=)55.8(90.36=w=b 11
.
EXAMPLE 2 2 (Continued)
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EXAMPLE 2.2 (Continued)
Eq.B2.1-5
Eq.B2.1-4
ksi9.28)7.60()-12(1
E4=F 22
2
cr =
f/F= cr 31.1=9.28
50=
2 - 26
Since > 0.673 Eq.B2.1-2
Eq.B2.1-3
.in89.2=)55.4(350.6=w=b 22
/0.22-1=;wb
= 0.63531.1/
31.1
0.22-1= =
UNIFORMLY COMPRESSED UNSTIFFENED ELEMENTS
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UNIFORMLY COMPRESSED UNSTIFFENED ELEMENTS
(Section B3.1)
2 - 27
UNIFORMLY COMPRESSED UNSTIFFENED ELEMENTS
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U O CO SS U S S
(Section B3.1)
2 - 28
Figure B3.1-1
Use basic effective width expression with k = 0.43
EXAMPLE 2 3 BEAM SECTION
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Given: (u.c.e.)
Fy = 50 ksi; t = 0.105 in.; R = 2t
Determine:
Effective width of compression
EXAMPLE 2.3 BEAM SECTION
2 - 29
flange for strength determinationw = 3.5 - 3t = 3.185 in.
w/t = 3.185/0.105 = 30.3 < 60 OK
Eq.B2.1-5
ksi5.12)3.30()-12(1
E43.0=F 22
2
cr =
EXAMPLE 2 3 (Continued)
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EXAMPLE 2.3 (Continued)
Eq.B2.1-4
Since > 0.673
0.2=5.12
50=f/F= cr
0.22
0.22
2 - 30
Eq.B2.1-3
Compression flange is only 44.5 % effective.
in..421=)185.3(450.4=w=b
-
..
0.2
-
UNIFORMLY COMPRESSED STIFFENED ELEMENTS
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UNIFORMLY COMPRESSED STIFFENED ELEMENTS
WITH CIRCULAR OR NON-CIRCULAR HOLES(Section B2.2)
2 - 31
UNIFORMLY COMPRESSED STIFFENED ELEMENTS
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WITH CIRCULAR OR NON-CIRCULAR HOLES
(Section B2.2)
NON-CIRCULAR HOLES:
i) Based on Section B2.1(a) at a stress FnAssume the web to consist of
unstiffened elements c , one on
2 - 32
each side of the perforation (k = 0.43)
Consider local buckling in edgestiffeners and flange elements, as
discussed before
ii) Or by conducting stub-column tests
based on AISI S902
UNSTIFFENED ELEMENTS AND EDGE STIFFENERSWITH STRESS GRADIENT (S i B3 2)
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WITH STRESS GRADIENT (Section B3.2)
i) If the stress decreasestoward the freeedge [Fig. B3.2-1(a)]
Eq.B3.2-234.0
578.0k+
=
2 - 33
= Abs[f2/ f1] Eq.B3.2-1
Figure B3.2-1(a)
UNSTIFFENED ELEMENTS AND EDGE STIFFENERSWITH STRESS GRADIENT (S ti B3 2)
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WITH STRESS GRADIENT (Section B3.2)
ii) If the stress increasestoward the freeedge [Fig. B3.2-1(b)]
k = 0.57 - 0.21+ 0.072 Eq.B3.2-3
2 - 34
Figure B3.2-1(b)
SECTION PROPERTIES
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use
SIMPLIFIED MIDLINE LINEAR METHOD
WITH
2 - 35
DUE CONSIDERATION TO LOCAL BUCKLINGIN ACCORDANCE WITH SECTION B
EXAMPLE 2.4 BEAM SECTION
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Given: (s.c.e.)
Fy = 50 ksi
2 - 36
Determine:
Effective moment of inertia for strength determination, Ixe.
Assume webs are fully effective
EXAMPLE 2.4 (Continued)
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Properties of 90corners
r = R + t/2 = 0.094 + 0.060/2 = 0.124 in.
u = 1.57r = 1.57(0.124) = 0.195 in.
2 - 37
c1
= 0.637r = 0.637(0.124) = 0.0790 in.
Dashed line is centerline
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EXAMPLE 2.4 (Continued)
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ELEM. L y Ly Ly2 I1
1 1.192 3.548 4.229 15.005 0.035
2 0.780 3.925 3.062 12.016
3 5.384 3.970 21.375 84.857
2 - 39
4 7.384 2.000 14.768 29.536 8.3885 2.573 0.030 0.077 0.002
6 0.390 0.075 0.029 0.002
17.703 43.540 141.418 8.423
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UNIFORMLY COMPRESSED ELEMENTSWITH SIMPLE LIP EDGE STIFFENERS
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WITH SIMPLE LIP EDGE STIFFENERS(Section B4)
1) Simple lip type
2 - 41
For Both Beam & Column Type Sections
ELEMENTS WITH SIMPLE LIP EDGE STIFFENERS(Section B4 Figure B4-1)
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(Section B4 Figure B4 1)
2 - 42
Use basic effective width expression with k varying between 0.43 & 4.0
COMPRESSION ELEMENTS WITH
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SIMPLE LIP EDGE STIFFENERS(Section B4)
For w/t 0.328S Eq.B4-7
Ia = 0 (no edge stiffener required)
-
( )f/E28.1=S
2 - 43
.
b1 = b2 = w/2 Eq.B4-2
ds = ds Eq.B4-3
COMPRESSION ELEMENTS WITH
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SIMPLE LIP EDGE STIFFENERS(Section B4)
For w/t > 0.328S
b1 = b/2(RI) Eq.B4-4
b2 = b b1 Eq.B4-5
2 - 44
ds = ds(RI) Eq.B4-6Eq.B4-7
Ia = 399t4[W/S 0.328]3 t4[115W/S + 5] Eq.B4-8
RI = Is/Ia 1 Eq.B4-9
W = w/t
( )f/E28.1=S
PLATE BUCKLING COEFFICIENTS k
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(Section B4 Table B4.1)
Simple Lip Edge Stiffener (140 40)
D/w 0.25 0.25 < D/w 0.8
n D5 n
2 - 45
where
n = 1/3 Eq.B4-11
S4
t/w
582.0
.. I= .
w
. I
=
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STIFFENED ELEMENTS UNDER STRESS GRADIENT(Section B2.3)
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Dimensions
2 - 47
Figure B2.3-2
STIFFENED WEBS UNDER STRESS GRADIENT(Section B2.3(a)(i))
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(a) Strength Determination
(i) Where f1 is in compression and f2 is in tension
= lf2/f1I (absolute value)Eq.B2.3-1
k = 4 + 2(1 + )3 + 2(1 + ) Eq.B2.3-2
2 - 48
b1 = be/(3 + ) Eq.B2.3-3b2 = be/2 when > 0.236 Eq.B2.3-4
b2 = be b1 when 0.236 Eq.B2.3-5
For ho/bo> 4
b1 = be/(3 + ) Eq.B2.3-6
b2 = be/(1 + ) b1 Eq.B2.3-7
EXAMPLE 2.6 BEAM SECTION
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Given:
Fy = 50 ksi; t = 0.090 in.; R = 2t
Determine:
Effective moment of inertia
2 - 49
or s reng e erm na on
Basic properties
r = 0.18 + 0.090/2 = 0.225 in.
u = 1.57r = 0.353 in.
c1 = 0.637r = 0.143 in.
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EXAMPLE 2.6 (Continued)
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RI = Is/Ia = 0.00706/0.0110 = 0.642
D/w = 1.25/3.96 = 0.316
( )3/1=n3/1Since
/0.
22
-1=;wb
= 7
48
0.=0
6.1/
0
6
.1
0.
22
-1=
EXAMPLE 2.6 (Continued)
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b1 = b/2(RI) = 2.96/2(0.642) = 0.950 in.
b2 = b b1 = 2.96 0.950 = 2.10 in.
Edge stiffener (u.c.e.) (B3.2)
.in96.2=)96.3(7480.=w=b
2 - 52
. . .
f1 = 50(7.23/7.5) = 48.2 ksi; = Abs[f2/f1]
f2 = 50(7.5 1.25)/7.5 = 41.7 ksi; = 41.7/48.2
= 0.865; k = 0.578/(0.865 + 0.34) = 0.480
ksi108=)9.10()-12(1
E480.0=F 22
2
cr
EXAMPLE 2.6 (Continued)
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Since < 0.673and lip is fully effective.in980.0=d=d
'
s
f/F= cr 668.0=108
2.48=
2 - 53
.in290.6=80(0.642)0.9=dR=d1.0
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24.0=)12(1+)12(1+4=k
1f/f=ksi48.2=f=fAssume
OK200 0.673
ksi7.24=)161()-12(1
E24=F 22
2
cr
40.1=48.2/24.7=f/F= cr
602.040.1/40.1
0.22-1/
0.22-1=;wbe =
=
=
EXAMPLE 2.6 (Continued)
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be = 0.602(14.46) = 8.71 in.
ho/bo = 15/4.5 = 3.33
For ho/bo 4
b1 = be/(3 + ) = 8.71/(3 + 1) = 2.18 in. (Eq. B2.3-3)
2 - 55
Since > 0.236
b2 = be/2 = 8.71/2 = 4.36 in. (Eq. B2.3-4)
(b1 + b2) = 2.18 + 4.36 = 6.54 in.
The compressed portion of web = [7.5 3(0.090)] = 7.23 in.
Since (b1 + b2) < 7.23 in., the web is subject to local buckling.
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EXAMPLE 2.6 (Continued)
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From a computer program - CFS
ycg = 8.29 in.
Ixe = 66.6 in.4
Sxe = 8.04 in.3
2 - 57
CONCLUSIONS
% Effective
Edge stiffener 64.2%
Compression flange 74.7%
Web 90.5%
ELEMENT WITH ONE INTERMEDIATE STIFFENER(Section B5.1)
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2 - 58
UNIFORMLY COMPRESSED STIFFENED ELEMENTS WITHSINGLE OR MULTIPLE INTERMEDIATE STIFFENERS
(S i B5 1)
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(Section B5.1)
2 - 59
C-SECTION WEBS WITH HOLES UNDER STRESS GRADIENT(Section B2.4)
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Provisions
(a) Strength Determination
2 - 60
When do/h < 0.38, the effective widths, b
1and b
2shall be
determined by Section B2.3(a) assuming no hole exists in the
web.
When do
/h 0.38, the effective width shall be determined by
Section B3.1(a) assuming the compression portion of the web
consists of an unstiffened element adjacent to the hole with
f = f1
as shown in Figure B2.3-1.
C-SECTION WEBS WITH HOLES UNDER STRESS GRADIENT(Section B2.4)
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(b) Serviceability Determination
The effective widths shall be determined by Section B2.3(b)assuming no hole exists in the web.
d = De th of web hole
2 - 61
b = Length of web holeb1, b2 = Effective widths defined in Figure B2.3-1
h = Depth of flat portion of the web measured along the
plane of the web
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QUESTIONS?
2 - 63
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FLEXURAL MEMBERS(Section C3)
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DESIGN CONSIDERATIONS
STRENGTH SERVICEABILITY1.) Flexure 1.) Deflection
4 - 2
. ear
3.) Web Crippling4.) Combined 1.) & 2.)
5.) Combined 1.) & 3.)
FLEXURAL MEMBERS(Section C3.1.1)
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Laterally Supported Members
Nominal Section Strength (Section C3.1.1)
4 - 3
a) Procedure I- Based on Initiation of Yielding
Mn = SeFy = My Eq.C3.1.1-1
b) Procedure II- Based on Inelastic Reserve Capacity(applicable for thicker HSS sections)
FLEXURAL MEMBERS(Section C3.1.1)
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b) Procedure II- Based on Inelastic Reserve Capacity
Conditions :
1. Laterally supported
2. Cold work of formin does not l
4 - 4
3. Compressive web portion to thickness ratio 1,4. Shear yielding governs the web (change)
5. 30
Nominal moment
Mn 1.25My ESe [Cyey]
FLEXURAL MEMBERS(Section C3.1.2)
Laterally Unsupported Members
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Laterally Unsupported Members
Section C3.1.2.1 - Lateral-Torsional Buckling forOpen Cross Section Members
4 - 5
. . . - -Closed Box Members
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OPEN CROSS SECTION FLEXURAL MEMBERS(Section C3.1.2.1)
Lateral Buckling Strength
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Lateral Buckling Strength
Elastic Lateral-Torsional Buckling
1
4 - 7
or
2
where
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OPEN CROSS SECTION FLEXURAL MEMBERS(Section C3.1.2.1)
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4 - 11
Maximum Unsupported Length, Lu
OPEN CROSS SECTION FLEXURAL MEMBERS(Commentary Section C3.1.2.1)
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pp gfor Singly-, Doubly-, and Point Symmetric Sections
4 - 12
or ng y-, an ou y- ymme r c ec ons
For Point-Symmetric Sections
OPEN CROSS SECTION FLEXURAL MEMBERS(Section C3.1.2.1)
Bending About Centroidal Axis
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Bending About Centroidal Axis
Perpendicular to Symmetry Axis forSingly-Symmetric Sections Only
4 - 13
where
and
EXAMPLE 4.1 BEAM SECTION
Given:
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Information of Ex. 2.6 of Lecture 2
Determine:1.) Nominal moment strength
4 - 14
initiation of yielding(Procedure I)
Sxe = 66.5/8.30 = 8.01 in.3
Mn = Sxe Fy
Mn = 8.01 (50) = 401 in.-k = 33.4 ft-k
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EXAMPLE 4.1 (Continued)
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4 - 16
0.56Fy = 28 ksi ; 2.78Fy = 139 ksi
EXAMPLE 4.1 (Continued)
Since 2.78Fy > Fe > 0.56Fy
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y e y
4 - 17
Now, use f = fc
= 46.0 ksi and calculate the effective sectionmodulus. From a computer program -- Sc = 8.33 in.
3
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DISTORTIONAL BUCKLING STRENGTHSection C3.1.4
For
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For
Mn= My (Eq. C3.1.4-1)
For d > 0.673
4 - 20
(Eq. C3.1.4-2)
My = SfyFySfy = Elastic section modulus of full unreduced section
Mcrd = SfFd
(Eq. C3.1.4-3)
DISTORTIONAL BUCKLING STRENGTHSection C3.1.4
General distortional buckling stress equation:
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General distortional buckling stress equation:
4 - 21
q. . . -
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CYLINDRICAL TUBULAR MEMBERS IN BENDING(Section C3.1.3)
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4 - 23
0.0714 0.318 0.441
SHEAR STRENGTH(Section C3.2)
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Section C3.2.1 Shear Strength of Webs Without Holes
Section C3.2.2 Shear Strength of C-Section Webs With Holes
4 - 24
SHEAR STRENGTH(Section C3.2.1)
SHEAR OF SOLID WEBS
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The Shear Strength Depends on:
The web slenderness ratio, h/t
4 - 25
The material properties
Use of transverse stiffeners
SHEAR STRENGTH(Section C3.2.1)
1) Unreinforced Webs
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1) Elastic shear buckling2) Inelastic shear buckling
Elastic shear buckling
4 - 26
For simply supported
edges along member
kv = 5.34
SHEAR STRENGTH(Section C3.2.1)
Inelastic Shear Buckling
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4 - 27
Specification uses 0.6
SHEAR STRENGTH(Section C3.2.1)
Nominal Shear Strength, VnV = A F Eq C3 2 1 1
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Vn = AwFv Eq.C3.2.1-1
(a) For
Eq.C3.2.1-2
4 - 28
Eq.C3.2.1-3
(c) For
Eq.C3.2.1-4
SHEAR STRENGTH(Section C3.2.1)
2) Reinforced Webs (Transverse Stiffeners)
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when a/h 1.0
4 - 29
when a/h > 1.0
SHEAR STRENGTH(Section C3.2.1
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4 - 30Shear Strength with Transverse Stiffeners
EXAMPLE 4.2 SHEAR OF SOLID WEBS
Given: Information of Ex. 2.6 of Lecture 2Determine:
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Determine:
1) The nominal shear strength of the unreinforced web
h = 14.46 in.; t = 0.090 in.; kv = 5.34
2
4 - 31
w . . . .
Since h/t > 84.7
Vn = AwFv = 1.30(5.49) = 7.14 kips
EXAMPLE 4.2 (Continued)
2) The nominal shear strength of the reinforced web satisfyingthe requirements of Section C3.6.1.
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a = 16 in.; a/h = 1.11
4 - 32
Since h/t > 108
Vn = AwFv = 1.30(8.86) = 11.5 kips61%increase in Vnwith stiffeners
SHEAR STRENGTH OF C-SECTION WEBS WITH HOLES(Section C3.2.2)
The nominal shear strength, Vn, determined from Section C3.2.1
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shall be multiplied by qs:
When c/t 54qs = 1.0 Eq.C3.2.2-1
4 - 33
When 5 c/t < 54
qs = c/(54t) Eq.C3.2.2-2
where
c = h/2 - do/2.83 for circular holes Eq.C3.2.2-3c = h/2 - do/2 for non-circular holes Eq.C3.2.2-4
SHEAR STRENGTH OF C-SECTION WEBS WITH HOLES(Section C3.2.2)
Provision Limits
(1) d /h 0 7
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(1) do/h < 0.7
(2) h/t 200
(3) Holes centered at mid-depth of web
4 - 34
ear stance etween o es n. mm
(5) Non-circular holes, corner radii 2t(6) Non-circular holes, do 2.5 in. (64 mm) and b 4.5 in. (114 mm)
(7) Circular hole diameters 6 in. (152 mm)
(8) do > 9/16 in. (14 mm)
FLEXURAL MEMBERS(Section C3.3)
COMBINED BENDING AND SHEAR
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4 - 35
High bending and shear simultaneously
STRENGTH FOR COMBINED BENDING AND SHEAR(Section C3.3.1 - ASD Method)
1) Unreinforced Webs
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2) Reinforced Webs
4 - 36
STRENGTH FOR COMBINED BENDING AND SHEAR(Section C3.3.2 - LRFD Method)
1) Unreinforced Webs
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2) Reinforced Webs
4 - 37
FLEXURAL MEMBERS(Section C3.3)
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4 - 38
WEB CRIPPLING(Section C3.4)
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4 - 39
WEB CRIPPLING(Section C3.4)
LOAD CASES:
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1. EOF -- End One Flange
2. IOF -- Interior One Flange
4 - 40
3. ETF -- End Two Flange
4. ITF -- Interior Two Flange
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WEB CRIPPLING(Section C3.4)
Basic Web Crippling Equation
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Eq.C3.4.1-1
4 - 42
Web crippling coefficients C, CR, CN, and Ch are given in the
appropriate tables for fastenedorunfastenedto the support.
R/t = inside bend radius ratioN/t = bearing length ratio
h/t = web slenderness ratio
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WEB CRIPPLING(Section C3.4)
SECTION TYPES (Fastened or unfastened to support)
Single Hat Sections(Table C3.4.1-4)
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g ( )
Limitations:h/t 200; N/t 200; N/h 2.0 and = 90;
limiting R/t values and resistance factors, & , given in table
- -
4 - 44
Limitations:h/t 200; N/t 210; N/h 3.0; 45 90
limiting R/t values and resistance factors, & , given in table
WEB CRIPPLING(Section C3.4)
Alternatively, for end-one-flange loading condition on a C- or Z-section,
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Pnc = Pn Eq. C3.4.1-2
is a function of the overhang length, purlin depth, and purlin thickness.
4 - 45
WEB CRIPPLING OF C-SECTION WEBS WITH HOLES(Section C3.4.2)
When a web hole is within the bearing length, a bearing stiffenershall be used.
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For C-sections with holes, multiply Eq.C3.4.1-1 by the reductionfactor, Rc.
4 - 46
. . . - . . -when any portion of a web hole is NOTwithin the bearing length:
Rc = 1.01 0.325do/h + 0.083x/h 1.0 Eq.C3.4.2-1N 1 in. (25 mm)
For IOF reaction (Eq.C3.4.1-1 with Table C3.4.1-2)
when any portion of a web hole is NOTwithin the bearing length:Rc = 0.90 0.047do/h + 0.053x/h 1.0 Eq.C3.4.2-2N 3 in. (76 mm)
WEB CRIPPLING OF C-SECTION WEBS WITH HOLES
(Section C3.4.2)
Provision Limits
(1) do/h < 0.7
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(2) h/t 200
(3) Holes centered at mid-depth of web
4 - 47
.
(5) Distance between end of member and edge of hole d(6) Non-circular holes, corner radii 2t
(7) Non-circular holes, do 2.5 in. (64 mm) and b 4.5 in. (114 mm)
(8) Circular hole diameters 6 in. (152 mm)
(9) do > 9/16 in. (14 mm)
COMBINED BENDING AND WEB CRIPPLING(Section C3.5.1 - ASD Method)
a) Shapes having single unreinforced webs
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Eq.C3.5.1-1
4 - 48
b) Shapes such as I-sections (high degree of restraint)
Eq.C3.5.1-2
COMBINED BENDING AND WEB CRIPPLING(Section C3.5.1 - ASD Method)
c) At Support Point of Two Nested Z-Sections
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4 - 49
Eq.C3.5.1-3
COMBINED BENDING AND WEB CRIPPLING(Section C3.5.2 LRFD Method)
a) Shapes having single unreinforced webs
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Eq.C3.5.2-1
4 - 50
b) Shapes such as I-sections (high degree of restraint)
Eq.C3.5.2-2
COMBINED BENDING AND WEB CRIPPLING(Section C3.5.2 - LRFD Method)
c) At Support Point of Two Nested Z-Sections
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4 - 51
Eq.C3.5.2-3
Provision limitsh/t 150N/t 140
R/t 5.5Fy 70 ksi
COMBINED BENDING AND TORSIONAL LOADING(Section C3.6)
For laterally unrestrained flexural members
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For laterally unrestrained flexural members
subject to both bending and torsional loading,Mn computed in accordance with Section
4 - 52
C3.1.1(a) shall be reduced by a reduction
factor.
R =
STIFFENERS
(Section C3.7)
Bearing Stiffeners (Section C3.7.1)
Attached to beam webs at points of concentrated load or
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Attached to beam webs at points of concentrated load orreaction, shall be designed as compression members.
Th n min l n n r l r r i n r n h P h ll
4 - 53
be the smaller of (a) or (b) as follows:
(a) Pn = AcFwy Eq.C3.7.1-1
(b) Pn = AeFn Eq.C4.1-1
with Ae replaced by Ab
c = 2.00 & c = 0.85
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DESIGN OF COLD-FORMED STEEL STRUCTURESUSING THE 2007 NORTH AMERICAN SPECIFICATION
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MEMBERS IN COMPRESSION
5 - 1
CONCENTRICALLY LOADED COMPRESSIONMEMBERS
Types of Compression Members
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Types of Compression Members
a) Doubly-symmetric
5 - 2
b) Singly-symmetric
c) Point-symmetric
d) Non-symmetric
ACTUAL FAILED STUD SPECIMEN
600S162-43
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DESIGN CONSIDERATIONS
a) Member behavior
i) Yielding (short & compact)
ii) O ll b kli
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ii) Overall buckling
~ Flexural (bending about one of the principal axes)
5 - 5
~
~ Torsional-flexural (simultaneous bending & twisting)
b) Element behavior
Local buckling of individual elements
NOMINAL AXIAL STRENGTH, Pn
a) For locally stable compression members
Pn = AgFnNo local buckling will occur before the nominal compressive
stress reaches the column buckling stress or the yield stress
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stress reaches the column buckling stress or the yield stress.
Hence, the gross area of the section is used.
5 - 6
Pn = AeFn Eq.C4.1-1
Local buckling will occur and the effective cross sectional areais used and is calculated at the nominal compressive bucklingstress.
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ELASTIC BUCKLING STRESS, Fe
Flexural Buckling (possible failure mode)The elastic flexural buckling stress is computed
by using the following expression:P
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y g g p
E2
5 - 8
(Eq.C4.1.1-1)( )KL r
e = 2/
P
KL=Le
ELASTIC BUCKLING STRESS, Fe
Torsional Buckling (possible failure mode)The elastic torsional buckling stress is computed
using the following expression:
P
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g g p
ECw
1 2
5 - 9
(Eq.C3.1.2.1-9)( )Ar K L
e
o t t t 2 2
P
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EFFECTIVE LENGTH FACTOR, K(Table C-C4.1-1 of Commentary)
(a) (b) (c) (d) (e) (f)
Buckled shape of column is
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Buckled shape of column is
shown by dashed line.
5 - 11
Theoretical K value
Recommended K valuewhen ideal conditions areapproximated
End condition code
Rotation fixed, Translation fixed
Rotation free, Translation fixed
Rotation fixed, Translation free
Rotation free, Translation free
0.5 0.7 1.0 1.0 2.0 2.0
0.65 0.80 1.2 1.0 2.1 2.0
LATERALLY UNBRACED STRUCTURES(Section C4.1 of Commentary)
When no lateral bracing against sidesway is present, such asin portal frames, the structure depends on its own bendingstiffness for lateral stability.
5
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4
5
5 - 12
Fig. C-C4.1-6 Fig. C-C4.1-5
PKL
L
P(I/L)beam
(I/L)column
K
1 2 3 4
0
1
2
3HingedBase
FixedBase
NONSYMMETRIC SECTIONS(Section C4.1.4)
For open shapes that have no symmetry, either about an axisor about a point, Fe shall be determined by a rational analysis
or from testing in accordance with Chapter F of the Standard.
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Analytically, tedious cubic equations have to be solved todetermine the torsional-flexural bucklin stress, as well, the
5 - 13
torsional warping constant, Cw, becomes quite complex to
solve. See Section 3 of part V (Supplementary Information) ofthe AISI Cold-Formed Steel Design Manual.
EXAMPLE 5.1 - CONCENTRICALLY LOADEDCOMPRESSION MEMBER
Given: The doubly-symmetric I-sectionmade up of 2 - 5.5CU1.25x045
channel sections. L = 4.5 ft, Fy = 33 ksi
R=0.1875
1.25 1.25
x
y
x
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Basic gross section properties (computer)
0.045 (in.)
5 - 14
g = 0.696 n. , w = 0.826 n. , = 0.000 0 n.
ry = 0.411 in., rx = 1.98 in., ro = 2.03 in.
Determine: The nominal compressive
strength, Pn
wPn
Pn
KL=Le
EXAMPLE 5.1 - (Continued)
1) Determine elastic buckling stress, Fe
a) Flexural buckling (Ky = 1.0)
K L/r = 4.5(12)/0.411 = 131 < 200 OK (Commentary)
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Ky
L/ry
4.5(12)/0.411 131 < 200 OK (Commentary)
Fe = 2E/(KyL/ry)
2 = 2 29500/(131)2 = 17.0 ksi Eq.C4.1.1-1
5 - 15
b) Torsional buckling (Kt = 1.0)
Fe = t = 1/(Aro2)[GJ + 2ECw/(KtLt)2] Eq.C3.1.2.1-9
Fe = 1/0.696/(2.03)2{11300(0.000470) +
2(29500)0.826/(54)2} = 30.6 ksi
Fe = 17.0 ksi, and flexural buckling controls.
EXAMPLE 5.1 - (Continued)
2) Determine nominal buckling stress, Fn
c = [Fy/Fe]1/2 = [33/17.0] 1/2 = 1.39 Eq.C4.1-4
Since c < 1.5
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Fn = (0.658 )Fy = [(0.658)(1.39x1.39)]33 = 14.7 ksi Eq.C4.1-2
c2
5 - 16
3) Determine effective area, Ae, at f = 14.7 ksi
Flange(u.c.e) --- w = 1.25 - (0.1875 + 0.045) = 1.018 in.W = w/t = 1.018/0.045 = 22.6 < 60 OK B.1.1(a)(3)
; Eq.B2.1-4&5crF/f= 22
2
cr W)1(12
E
kF -
=
;ksi4.22=)6.22(92.10
E43.0=F 2
2
cr809.0=4.22/7.14=
EXAMPLE 5.1 - (Continued)
Since > 0.673, b = w Eq.B2.1-2
= (1 - 0.22/ )/ = (1 - 0.22/0.809)/0.809 = 0.900 Eq.B2.1-3
b = 0.900(1.018) = 0.916 in.
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( )
Web(s.c.e) --- w = 5.5 - 2(0.1875 + 0.045) = 5.035 in.
5 - 17
W = w/t = 5.035/0.045 = 112 < 500 OK B.1.1(a)(2)
Eq.B2.1-4&5;F/f= cr 22
2
cr W)1(12Ek=F
;ksi50.8=)112(92.10
E4=F 2
2
cr
32.1=50.8/7.14=
EXAMPLE 5.1 - (Continued)
Since > 0.673, b = w Eq.B2.1-2
= (1 - 0.22/ )/ = (1 - 0.22/1.32)/1.32 = 0.631 Eq.B2.1-3
b = 0.631(5.035) = 3.18 in.
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( )
r = R + t/2 = 0.1875 + 0.045/2 = 0.210 in.
5 - 18
u = 1.57r = 1.57(0.210) = 0.330 in.
Ae = 0.045[4(0.916 + 0.330) + 2(3.18)] = 0.510 in.2
4) Determine nominal compressive strength, Pn
Pn = AeFn = 0.510(14.7) = 7.50 kips Eq.C4.1-1
EXAMPLE 5.2 - CONCENTRICALLY LOADEDCOMPRESSION MEMBER
Given: The point-symmetric Z-section4ZU1.25x060L = 3.0 ft (36 in.), Fy = 50 ksi.
0.060R=0.1875
1.25
x
y
x
x2
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Basic gross section properties (computer)
(in.)
5 - 19
so, see a e - o anua
Ag = 0.372 in.2
, Cw = 0.201 in.6
, J = 0.000446 in.4
rmin = 0.300 in., ro = 1.56 in.
Determine: The nominal compressivestrength, Pn
Pn
Pn
KL=Le
EXAMPLE 5.2 - (Continued)
1) Determine elastic buckling stress, Fe
a) Flexural buckling (K = 1.0)
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KL/rmin = 36/0.300 = 120 < 200 OK (Commentary)Fe =
2E/(KL/rmin)2 = 229500/(120)2 = 20.2 ksi Eq.C4.1.1-1
5 - 20
b) Torsional buckling (Kt = 1.0)
Fe = t = 1/(Aro2)[GJ + 2ECw/(KtLt)
2] Eq.C3.1.2.1-9
Fe = 1/0.372/(1.56)2{11300(0.000446) +
2(29500)(0.201)/(36)2} = 55.4 ksi
Fe = 20.2 ksi and flexural buckling controls
EXAMPLE 5.2 - (Continued)
2) Determine nominal buckling stress, Fn
Eq.C4.1-4
Since c > 1.5
2 2
57.12.20/50F/F eyc ===
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Fn = (0.877/c )Fy= [0.877/(1.57) ]50 = 17.8 ksi Eq.C4.1-3
3 Determine effective re A t f = 17.8 ksi
5 - 21
Flange (u.c.e) --- w = 1.25 - (0.1875 + 0.060) = 1.00 in.W = w/t = 1.00/0.060 = 16.7 < 60 OK B.1.1(a)(3)
Eq.B2.1-5ksi1.41
)7.16(92.10
E43.0
W)1(12
EkF
2
2
22
2
cr ===
-
658.01.41/8.17F/f cr ===
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EXAMPLE 5.2 - (Continued)
= (1 - 0.22/ )/= (1 - 0.22/0.755)/0.755 = 0.939 Eq. B2.1-3
b = 0.939(3.505) = 3.29 in.
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r = R + t/2 = 0.1875 + 0.060/2 = 0.218 in.
5 - 23
u = . r = . . = . n.
Ae = 0.060[3.29 + 2(0.341 + 1.00)] = 0.358 in.2
4) Determine nominal compressive strength, Pn
Pn = AeFn = 0.358(17.8) = 6.38 kips Eq.C4.1-1
EXAMPLE 5.3 - CONCENTRICALLY LOADEDCOMPRESSION MEMBER
Given: The singly-symmetric C-section3CS3x060L = 3.5 ft (42 in.), Fy = 50 ksi.
y
x
0.060
R=0.1875
x
0.75
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Basic gross section properties (computer)= 2 = 6 = 4
(in.)
5 - 24
ry = 1.12 in., rx = 1.27 in., ro = 3.33 in., xo = 2.87 in.
Determine: The nominal compressivestrength, P
n
Pn
Pn
KL=Le
.
EXAMPLE 5.3 - (Continued)
1) Determine elastic buckling stress, Fe
a) Flexural buckling (Ky = 1.0)
KyL/ry = 42/1.12 = 37.5 < 200 OK (Commentary)
F2
E/(K L/ )2
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Fe = E/(KyL/ry)= 2 29500/(37.5)2 = 207 ksi Eq.C4.1.1-1
5 - 25
b) Torsional-flexural buckling (Kt = Kx = 1.0)
t = Eq.C3.1.2.1-9
t = 1/0.593/(3.33)2{11300(0.000712) +
2(29500)(2.09)/(42)2}
t = 53.6 ksi
( )
12
2
2Ar
GJEC
K Lo
w
t t
+
EXAMPLE 5.3 - (Continued)
Fe = 1/(2){(ex + t) - [(ex + t)2 - 4ext]1/2} Eq.C4.1.2-1
KxL/rx = 42/1.27 = 33.1 < 200 OK (Commentary)
ex = 2E/(KxL/rx)
2 = 2 29500/(33.1)2 = 266 ksi Eq.C3.1.2.1-11
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= 1 - (xo/ro)2 = 1 - (2.87/3.33)2 = 0.257 Eq.C4.1.2-3
5 - 26
t = 53.6 ksi from before
Fe = 1/2/0.257{(266 + 53.6) - [(266 + 53.6)2 -
4(0.257)(266)(53.6)]1/2} Eq.C4.1.2-1
Fe = 46.3 ksi, and torsional-flexural buckling controls.
EXAMPLE 5.3 - (Continued)
2) Determine nominal buckling stress, Fn
c = [Fy/Fe]1/2 = [50/46.3]1/2 = 1.04 Eq.C4.1-4
Since c 1.5
(
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Fn = = [(0.658)(1.04x1.04)]50 = 31.8 ksi Eq.C4.1-2(0 658
2
. c Fy
5 - 27
3) Determine effective area, Ae , at f = 31.8 ksi
Flange (s.c.e) --- w = 3.00 - 2(0.1875 + 0.060) = 2.505 in.W = w/t = 2.505/0.060 = 41.8 < 60 OK B.1.1(a)(3)
Eq.B4-712.80.328S;0.398.31/E28.1S ===
EXAMPLE 5.3 - (Continued)
Since W > 0.328S, B4(a)
43
3s in.000636.0
12
)503.0(060.012/tdI ===
3841
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43
41a in.00213.0328.0
8.41)060.0(399I =
=
5 - 28
q. -
Eq.B4-9
( )444
2a .in00166.0539
8.41
115060.05S
t/w
115tI =
+=
+=
( ) 42a1aa .in00166.0I,IMinI ==
;383.0I/IR asI == 299.0505.2/75.0w/D ==
EXAMPLE 5.3 - (Continued)
Table B4.1
( ) 3/1=n3/1
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=505.2
5 - 29
Since > 0.673, b = w Eq.B2.1-2
= (1 - 0.22/ )/ = (1 - 0.22/0.855)/0.855 = 0.869 Eq.B2.1-3b = 0.869(2.505) = 2.18 in.
43.5ksi)8.41()-12(1
E85.2=
F 22
2
cr
=
550.8=
5.43
8.13=
EXAMPLE 5.3 - (Continued)
Edge stiffener (u.c.e.)
Eq.B2.1-5
ksi8.31f;38.8=60/0.0503.0=d/t =
ksi631)388()12(1
E
430F 22
2
cr
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ksi631)38.8()-12(143.0=F 22cr =
5 - 30
Eq.B2.1-4
Since 0.673, edge stiffener is fully effective Eq.B2.1-1
ds' = d = 0.503 in.
ds = RI ds' = 0.383(0.503) = 0.193 in.
4420.=
163
8.31=
Web (s.c.e) --- w = 3.0 - 2(0.1875 + 0.060) = 2.505 in.W = w/t = 2.505/0.060 = 41.8 < 500 OK B.1.1(a)(2)
Eq.B2.1-5
EXAMPLE 5.3 - (Continued)
ksi0.61
)841(9210
E4
W)1(12
EkF
2
2
22
2
cr ===
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)8.41(92.10W)1(12 -
5 - 31
Eq.B2.1-4
Since > 0.673, b = w Eq.B2.1-2
= (1 - 0.22/ )/= (1 - 0.22/0.722)/0.722 = 0.963 Eq.B2.1-3
b = 0.963(2.505) = 2.41 in.
722.061/8.31F/f cr ===
EXAMPLE 5.3 - (Continued)
r = R + t/2 = 0.1875 + 0.060/2 = 0.218 in.
u = 1.57r = 1.57(0.218) = 0.341 in.
Ae = 0.060[2.41 + 4(0.341) + 2(0.193 + 2.18)] = 0.511 in.2
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4) Determine nominal compressive strength, Pn
5 - 32
Pn = AeFn = 0.511(31.8) = 16.3 kips Eq.C4.1-1
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Flexural buckling stress:
Fe is determined according to Section C4.1.1 and the nominal
buckling strength Pn is then calculated from Section C4 1
CLOSED CYLINDRICAL TUBULAR MEMBERS INCOMPRESSION (Section C4.1.5)
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buckling strength, Pn, is then calculated from Section C4.1.
5 - 34
Ae = Ao + R( A - Ao) Eq.C4.1.5-1
Ao = Eq.C4.1.5-2
R = Fy/(2Fe) 1.0 Eq.C4.1.5-3
( ) ( ) yy FE
441.0t
DforAA667.0
tE/DF
037.0
+
QUESTIONS?
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5 - 35
DESIGN OF COLD-FORMED STEEL STRUCTURESUSING THE 2007 NORTH AMERICAN SPECIFICATION
COMBINED BENDING AND COMPRESSION
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COMBINED BENDING AND COMPRESSION
6 - 1
b x b y tM M T+ 10
COMBINED TENSILE AXIAL LOAD AND BENDING(Section C5.1)
ASD (C5.1.1)i) yielding of Eq. C5.1.1-1
tension flange
ii) failure of E C5 1 1 2
b x
nxt
b y
nyt
t
n
M
M
M
M
T
T+ + 10.
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b x y tM M T+ 10.ii) failure of Eq. C5.1.1-2
6 - 2
nx ny ncompress on ange
LRFD (C5.1.2)
i) yielding of Eq. C5.1.2-1tension flange
ii) failure of Eq. C5.1.2-2compression flange
M
M
M
M
T
Tux
b nxt
uy
b nyt
u
t n + + 10.
MM
MM
TT
ux
b nx
uy
b ny
u
t n + 10.
COMBINED COMPRESSIVE AXIAL LOAD AND BENDING(Section C5.2)
C5.2.1 ASDThe required strengths, P, Mx, and My shall satisfy thefollowing interaction equations:
i) Stability C5 2 1 1 c b mx x b my yP C M C M 10
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i) Stability C5.2.1-1 c b mx x b my yP C M C M+ + 10.
6 - 3
ii) Strength C5.2.1-2
When cP/Pn 0.15, the following equation may be used in
lieu of the above equations:
C5.2.1-3
n nx x ny y
c
no
b x
nx
b y
ny
PP
MM
M
M+ + 10.
c
n
b x
nx
b y
ny
P
P
M
M
M
M+ + 10.
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SECOND ORDER EFFECTS
ASD
> 0 Eq. C5.2.1-4
> 0 Eq C5 2 1-5 cP= 1
xc
Ex
P
P= 1
P
Mx
Mmax =Mx
x
Pu
Mux
Mmax =Mux
x
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> 0 Eq. C5.2.1-5y = 1
6 - 5
LRFD
> 0 Eq. C5.2.2-4
> 0 Eq. C5.2.2-5
Eq.C5.2.1-6, Eq.C5.2.1-7
Ey
Ex
u
x PP1= -
Ey
u
y P
P
1=-
PEI
K L
EI
K LEx
x
x x
Ey
y
y y
= = 2
2
2
2( );
( )P
P
Mx
Pu
Mux
ASD LRFD
If end moments are as shown on the previous slide, no additional
modification is necessary.
For unequal end moments M1 and M2 and compression membersin frames, the following modifications shall apply:
(a) For compression members in frames subject to jointt l ti ( id )
EFFECT OF MOMENTS (Cmx, Cmy)
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( ) p j jtranslation (sidesway)
6 - 6
Cm = 0.85
(b) For compression members in frames braced againstjoint translation and no transverse loading betweensupports
Cm = 0.6 + 0.4(M1/M2) (single curvature)Eq.C5.2.2-8
Cm = 0.6 - 0.4(M1/M2) (double curvature) Eq.C5.2.2-8
EFFECT OF MOMENTS (Cmx, Cmy)
(c) For compression members in frames braced against jointtranslation with transverse loading between supports
Cm may be determined by rational analysis, or in lieu of(1) for members with restrained ends C = 0 85
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(1) for members with restrained ends, Cm = 0.85,
6 - 7
(2) for members with unrestrained ends, Cm = 1.0.
Given:The closed box section8 x 8 x 0.105 as shownL = 12.0 ft (144 in.), Fy = 50 ksi.
Applied loads:i) Axial load PD = 4.00 kips, PL = 16.0 kips
EXAMPLE 6.1 - COMBINED AXIAL LOADAND BENDING
y
x
(in.)
t=0.105
R=0.1875
w
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i) Axial load PD 4.00 kips, PL 16.0 kips
8.0
6 - 8
n momen s D = . - ps, L = . - ps
Basic gross section properties (computer)
Ag = 3.27 in.2, Ix = Iy = 33.8 in.
4, rx = ry = 3.21 in.,
Sf = 8.44 in.3, r = R + t/2 = 0.24 in.,
u = 1.57r = 0.377 in., c = 0.637r = 0.153 in.Determine: The adequacy of the tubular
member using LRFDP
12 ft
M
M
EXAMPLE 6.1 - (Continued)
1) Check Interaction Equation C5.2.2-1
(Muy = 0)
Compute nominal axial strength, Pn,
0.1M
MC+
M
MC+
P
P
ynyb
uymy
xnxb
uxmx
nc
u
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6 - 9
Flexural buckling (K = 1.0)
Fn = = 43.2 ksi Eq.C4.1-2
Ae = 2.01 in.2
Pn = AeFn = 2.01(43.2) = 86.8 kips Eq.C4.1-1
( )0 6582
. c Fy
Compute nominal flexural strength, Mnx,based on lateral-torsional buckling C3.1.2.1Mnx = ScFc Eq.C3.1.2.1-1
~ Compression flange (1)
EXAMPLE 6.1 - (Continued)
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6 - 10
, = . .
~ Webs (3) (w/t = 70.6) C
ycg
22
4
3
4
3
1
5 T
f1
f2
ycg = 4.46 in. (Effective Section)Mnx = 313 in.-kips = 26.1 ft-kips Eq.C3.1.2.1-1
Compute x term (Cmx = 1.0)
x = 1 - Pu/PEx = 0.936 Eq.C5.2.2-4
EXAMPLE 6.1 - (Continued)
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Compute required strengths, Pu and MuxPu = 1.20PD + 1.60PL = 30.4 kips
Mux = 1.20MD + 1.60ML = 11.4 ft-kips
Check interaction equation C5.2.2-1Eq.C5.2.2-1
OK
0.1
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Com ute nominal axial stren th P
6 - 12
based on f = Fn = Fy C5.2
w = 7.415 in. (s.c.e.), w/t = 70.6 < 500 OK B1.1(a)(2) = 1.53 Eq.B2.1-4
Since > 0.673, b = w Eq.B2.1-2
= (1 - 0.22/ )/ = 0.560 Eq.B2.1-3
b = 0.560(7.415) = 4.15 in.
Ae = 4(0.105)[0.377 + 4.15] = 1.90 in.2
Pno = AeFn = 1.90(50) = 95.0 kips
Compute nominal flexural strength, Mnx
[From Part 1) above]C3.1.2.1
EXAMPLE 6.1 - (Continued)
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= - = - -
6 - 13
Check interaction equation C5.2.2-2
OK
0.1
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DESIGN OF COLD-FORMED STEEL STRUCTURESUSING THE 2007 NORTH AMERICAN SPECIFICATION
MEMBER BRACING
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STRAPPING
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TYPES OF BRACING
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STEEL BUILDING SYSTEMS
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TYPES OF BRACING
Member (beam or column)
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System (roof or wall)
MEMBER BRACING(Section D3)
Design Requirements:
Prevent lateral bending
Prevent twisting
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Prevent local crippling at attachments
MEMBER BRACING(Section D3.1)
SYMMETRIC SHAPES
Design considerations:
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Strength
Stiffness
MEMBER BRACING(Section D3.3)
BRACING OF AXIALLY LOADED COMPRESSION MEMBERS
Design considerations:
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Strength,
Stiffness,
BRACING OF C & Z SECTION BEAMS(Section D3.2)
Applications:
Top Flange Restrained by Deck or Sheathing
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MEMBER BRACING
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C-SECTION BRACING FORCES
F
P
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For a uniform loadper unit length, P:
PL = 1.5KaP
MEMBER BRACING
d
Pm=F
=
P
F
P
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P
F
MEMBER BRACING
x
xy
I2
IP=F PK=
P
F
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F
MEMBER BRACING(Section D3.2.1)
Neither Flange Restrained and the load acts through
the plane of the web:
Design brace force P :
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Design brace force, PL:
Uniform load, PL = 1.5 K'W Concentrated load, PL = 1.0 K'P + 1.4K'P(1- l/a)
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MEMBER BRACING(Section D3.2.2)
Neither Flange Restrained and the load does not act
through the plane of the web:
Design brace force, PL:
Uniform load
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Uniform load,
Concentrated load,
DIAPHRAGM BRACING
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DIAPHRAGM BRACING(Section D5)
Strength Consideration- LRFD
P d Sn
Where,
P = Factored loads on the diaphragm=
by calculation or test
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by calculation or test
Strength Consideration- ASD
P Sn/d
Where,P = Service loads on the diaphragm
D6 Metal Roof and Wall Systems
The provisions of Sections D6.1 through D6.3 shall apply tometal roof and wall systems that include cold-formed steel
purlins, girts, through-fastened roof systems and standing
seam roof panels.
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SYSTEM BRACING AND ANCHORAGE(Section D6.3)
Design brace force for typical roof systems is afunction of
Gravity load
C- or Z- purlins
Top flange attached to sheathing
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Top flange attached to sheathing(through fastened or standing seam panels)
Simple or continuous spans
For bracing arrangements not covered, test per Section F1
SYSTEM BRACING
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SYSTEM BRACING AND ANCHORAGE(Section D6.3.1)
=
=
p
i
j
N
1i total
j,ieffiLK
KPP
New Anchorage Device Equation:
Each anchorage device must resist PL
++= sin4Ccos
d
t)b25.0m(3C
dI
LI
1000
2CW1CP
2
xypi i
when every purlin is anchored
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ddI1000 x
( ) sysN1j
j,iefftotal KKKa
i+=
=
2
2
psysd
ELtN
1000
5CK =
1
p
p
aeff ELA6C
d
K
1K
j,i
ji,
+=
Effective lateral stiffness of all elements resisting Pi
Lateral stiffness of roof system, neglecting anchorage
device (purlin to rafter & panel to purlin connections)Effective lateral stiffness of anchorage device
when every purlin is anchored
by a rigid anchor
SYSTEM BRACING AND ANCHORAGE(Section D6.3.1)
References:
Sears and Murray, Proceedings of the Annual Stability
Conference, Structural Stability Research Council, April 2007
AISI design guideears ., ee , , an urray, . . , es gn u e or
C ld F d St l R f F i S t AISI D111
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Cold-Formed Steel Roof Framing Systems, AISI D111,
American Iron and Steel Institute, Washington, D.C.
MEMBER BRACING(Section D6.1.1)
Member flexural strength is a functionof:
C- or Z- purlins Roof and wall systems
Wind uplift or suction load
Simple or continuous span
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Simple or continuous span M
n= RS
eF
y For structural systems not
covered, test per Section F1
MEMBER BRACING(Section D6.1.1)
R = 0.60 for continuous span C-sections.R = 0.70 for continuous span Z-sections.
If adjacent spans vary by more than 20%R shall be taken from Table D6.1.1-1
TABLE D6.1.1-1Si l S C Z S ti R V l
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Simple Span C- or Z-Section R Values
Depth Range, in. (mm) Profile R
d 6.5 (165) C or Z 0.70
6.5 (165) < d 8.5 (216) C or Z 0.65
8.5 (216) < d 11.5 (292) Z 0.50
8.5 (216) < d 11.5 (292) C 0.40
MEMBER BRACING(Section D6.1.2)
Member Flexural Strength
C- or Z- Purlins
Gravity Load and Uplift Load
Top Flange Supporting Standing Seam Roof Panel
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Mn = RSeFy
MEMBER BRACING(Section D6.1.3)
Member Axial Strength
C- and Z- Sections Concentric Axial Load n Fl n A h D k r h hin
Attachment with Through Fasteners
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Attachment with Through Fasteners
Pn = C1C2C3 AE/29500 Eq.D6.1.3-1
C1C
2C
3reflect influence of fastener location, material thickness
and cross-section geometry.
QUESTIONS?
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DESIGN OF COLD-FORMED STEEL STRUCTURESUSING THE 2007 NORTH AMERICAN SPECIFICATION
CONNECTIONS
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TYPES OF CONNECTORS
Welds Bolts
Screws Other
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CONNECTIONS AND JOINTS(Section E)
Thickness Limits:Welds: Sheet thickness 3/16 in., Section E2
Bolts:
Sheet thickness 3/16 in., Section E3
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Screws: No explicit thickness limitations
WELDED CONNECTIONS(Section E2)
Arc Welds
Groove Welds
Arc Spot Welds
Fillet Welds
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Fillet Welds
Flare Groove Welds
Resistance Welds
(a) Tension or compression normal to weldPn = L te Fy Eq. E2.1-1
where,
Fy = Yield stress of lowest strength of base steel
DESIGN STRENGTH OF GROOVE WELDS IN BUTT JOINTS(Section E2.1)
e
L = Length of weld
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= 0.90
= 1.70
(b) Shear on effective weld area
Pn = L te (0.6Fxx) or Pn =
= 0.80, = 1.90 = 0.90, = 1.90
where,
DESIGN STRENGTH OF GROOVE WELDS IN BUTT JOINTS(Section E2.1)
( )L t Fe y/ 3
Fy = Yield stress of lowest strength of base steel
F = Tensile strength of electrode
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Fxx = Tensile strength of electrode
te = effective throat dimensionL = Length of weld
DESIGN STRENGTH OF ARC SPOT WELDS(Section E2.2)
Limitations
Maximum thickness of single sheet or combination of sheetsis 0.15 in.
Weld washers required for sheets less than 0.028 in. Minimum effective diameter, de = 3/8 in.
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DESIGN STRENGTH OF ARC SPOT WELDS(Section E2.2)
Definitions
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DESIGN SHEAR STRENGTH OF ARC SPOT WELDS(Section E2.2.1.1)
Minimum Edge Distance
emin = P/(Fut) for ASD
e i = P /(F t) for LRFD
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emin = Pu/(Fut) for LRFD
and vary with the Fu/Fy of the sheet
DESIGN SHEAR STRENGTH OF ARC SPOT WELDS(Section E2.2.1.2)
Shear Strength - Sheets Welded to Thicker Member:
Shear strength of weld
Tearing of connected partxx
2
e
nF75.0
4
dP
=
uan Ftd)C(=P
where the coefficient, C, varies from 1.40 to 2.20 depending on
the ratio (d /t)
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the ratio (da/t)
DESIGN SHEAR STRENGTH OF ARC SPOT WELDS(Section E2.2.1.3)
Shear Strength - Sheet-to-Sheet Connections:
Pn = 1.65 tdaFu
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= 2.20
= 0.70
DESIGN STRENGTH OF ARC SPOT WELDS(Section E2.2.2)
Tension
Tensile strength of weld:
Tearing of connected part:
xx
2e
n F4
dP
=
ua2
yun Ftd)F/F(8.0P =
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DESIGN STRENGTH OF ARC SEAM WELDS(Section E2.3)
Limit States
(a) Shear strength of weld: Pn = (de2/4 + Lde) 0.75Fxx
(b) Strength of connected part: Pn = 2.5tFu(0.25L + 0.96da)
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DESIGN STRENGTH OF FILLET WELDS(Section E2.4)
(a) Shear strength of weld
(b) T i f t d t
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(b) Tearing of connected part
1. Longitudinal Loading
2. Transverse Loading
DESIGN STRENGTH OF FLARE-BEVEL GROOVE WELDS(Section E2.5)
Transverse loading
Pn = 0.833tLFu
= 0.60
= .
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DESIGN STRENGTH OF FLARE GROOVE WELDS(Section E2.5)
Longitudinal loading
Tensile strength of weld (t > 0.10):
Pn = 0.75twLFxx = 0.60
= 2.55
Tearing of connected part:
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Pn = 0.75tLFu , for t tw < 2tPn = 1.50tLFu , for tw 2t = 0.55 = 2.80
RESISTANCE WELDS(Section E2.6)
Nominal Shear Strength of a Spot Weld
(a) For 0.01 in. t < 0.12 in.Pn = 144 t
1.47
b For 0.14 in. t 0.18 in.
Pn = 43.4 t + 1.93
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= 0.65 = 2.35
BOLTED CONNECTIONS(Section E3)
A307
A325
A354
A449
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A490
TYPES OF FAILURE OF BOLTED CONNECTIONS
(a) Longitudinal shear failure of sheet
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(b) Bearing failure of sheet
TYPES OF FAILURE OF BOLTED CONNECTIONS
(c) Tensile failure of sheet
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(d) Shear failure of bolt
SPACING AND EDGE DISTANCE REQUIREMENTS(Section E3.1)
Minimum center-to-center distance: 3d
Minimum distance from bolt center to edge or end: 1.5d
Stren th of connection:
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Pn = t e Fu
When Fu/Fy 1.08, = 2.0; = 0.70
When Fu/F
y< 1.08, = 2.22; = 0.60
DESIGN BEARING STRENGTH OF CONNECTED PART(Section E3.3)
When deformation around the bolt holes is not a designconsideration:
Pn = mfCdtFu Eq.E3.3.1-1
mf = modification factor for type of bearing connection
= 0.60, = 2.50; Table E3.3.1-1
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Thickness, t d/t C
0.024 t < 0.1875d/t < 10 3.0
10 d/t 22 4 0.1(d/t)
d/t > 22 1.8
BEARING MODIFICATION FACTOR, mf(Table E3.3.1-2)
Type of Bearing Connection mf
Single shear and outside sheets ofdouble shear connection with washers
under both bolt head and nut
1.00
Single shear and outside sheets ofdouble shear connection without 0.75
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washers under both bolt head and nut,or only one washer
Inside sheet of double shearconnection with or without washers 1.33
BOLTED CONNECTIONS IN BEARINGWithout Washer With Washer
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DESIGN BEARING STRENGTH OF CONNECTED PART(Section E3.3.2)
When deformation around the bolt holesisa designconsideration
Pn = (4.64t +1.53)dtFu Eq.E3.3.2-1 = un t convers on actor
= 1 for imperial units
= 0.0394 for metric units
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= 0.65 = 2.22
DESIGN TENSILE AND SHEAR STRENGTH OF BOLTS(Section E3.4)
Pn = Ab(Fnt or Fnv given in Table E3.4-1)
Type of Bolts Tensile Strength Shear Strength
Fnt
Fnv
, . . . . . . .( in. d < in.)
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A307 Bolts, Gr. A 0.75 2.25 45.0 0.65 2.40 27.0(d in.)
A325-N Bolts 0.75 2.0 90.0 0.65 2.40 54.0
A325-X Bolts 0.75 2.0 90.0 0.65 2.40 72.0
BOLTS SUBJECTED TO COMBINED SHEAR AND TENSION(Section E3.4)
Pn = AbFnt
Where the nominal tensile strength of a bolt subjected to a
combination of tension and shear is:nt = . v Table E3.4-2 (ASD)
fv = Computed shear stress in bolt
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Ab = Cross sectional area of bolt and are given in the specification
EXAMPLE 8.2 BOLTED CONNECTION
Given:
ASTM A307 bolts w/o washersd = 1/2 in.ASTM A653 Gr. 33; Fy = 33 ksi; Fu = 45 ksi
Determine:
The design capacity of the angle and bolt connecting element.
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EXAMPLE 8.2 (Continued)
1) Angle - Yielding of gross section
Tn = AgFy; Ag = (6 0.068)0.068 = 0.40 in.2 Eq.C2-1= (0.40)33 = 13.2 kips
Ta = Tn/ = 13.2/1.67 = 7.90 kips
2) Angle Fracture at connection
Pn = AeFu Eq.E3.2-8
Pa = Pn/ ; = 2.22
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Ae = UAn
U = 1.0 - 1.2 /L < 0.9, but shall not be less than 0.4 Eq.E3.2-9
U = 1.0 1.2(0.783/3) = 0.687 > 0.4 OK
An = Ag nAh = 0.40 1(0.5625)0.068 = 0.362 in.2
_
x
EXAMPLE 8.2 (Continued)
Ae = 0.687(0.362) = 0.249 in.2
Ta = 0.249(45)/2.22 = 5.04 kips3) Angle Block shear rupture E5.3
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Ant = [1 (0.5625)]0.068 = 0.0489 in.2
Anv = [4.5 - 1(0.5625)]0.068 = 0.249 in.2
Agt = 1(0.068) = 0.068 in.2
Agv = 4.5(0.068) = 0.306 in.2
EXAMPLE 8.2 (Continued)
FuAnt = 0.0489(45) = 2.20 kips
Rn = 0.6FyAgv + FuAnt = 0.6(33)(0.306)+2.20=8.26 kips -governs
Rn = 0.6FuAnv + FuAnt = 0.6(45)(0.249)+2.20=8.92 kips
n = . . = .
Angle controlled by block shear.
4) B lt ShE3.4
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4) Bolts ShearPn = AbF; Ab = d
2/4 = (1/2)2/4 = 0.196 in.2 Eq.E3.4-1
F = Fnv = 27 ksi Table E3.4-1
Pn
= 0.196(27) = 5.29 kips/bolt
Pn/ = 5.29/2.4 = 2.20 kips/bolt
EXAMPLE 8.2 (Continued)
Connection strength = 2(2.20) = 4.40 kips
5) Bolts Bearing E3.3
Pn = mf C Fudt
d/t = (1/2)/0.068 = 7.4, C = 3.0 Table E3.3.1-1
mf = 0.75 Table E3.3.1-2
Pn = 0.75 (3)(45)(1/2)(0.068) = 3.44 kips/bolt
Pn/ = 3.44/2.50 = 1.38 kips/bolt
Connection strength = 2(1.38) = 2.75 kips
6) Sh d di t i l t t i l E3 1
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6) Shear end distance in plate material E3.1Pn = teFu = 0.068(1.5)45 = 4.59 kips/bolt Eq.E3.1-1
Pn/ = 4.59/2.0 = 2.30 kips/bolt
Connection strength = 2(2.30) = 4.60 kips
Connection strength controlled by bearing = 2.75 kips
SCREW CONNECTIONS
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SCREW CONNECTIONS
Limitations
0.08 in. screw diameter 0.25 in. Thread forming or thread cutting With or without a self-drilling point
Diaphragm applications, Section D5
Spacing Minimum center-to-center distance: 3d
Minimum distance from screw center to edge or end: 1 5d
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Minimum distance from screw center to edge or end: 1.5d
ENDLAP SPLICE AND INTERMEDIATE FASTENING
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LIGHT WEIGHT STEEL FRAMING
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SELF-DRILLING SCREWS
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SCREW SIZES AND THICKNESS LIMITS
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LIMIT STATES OF SCREW CONNECTIONS
1. Longitudinal shear of sheet
2. Bearing failure of the sheet
3. Tilting of screw and subsequent tearing of sheet
4. Shear failure of screw
5. Tension pull-out of screw
6. Tension pull-over of sheet
f f
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7. Tension failure of screw
CONNECTION SHEAR LIMITED BY END DISTANCE(Section E4.3.2)
Pns = t e Fu Eq.E4.3.2-1
= 3.0
= 0 50
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= 0.50
CONNECTION SHEAR LIMITED BY TILTING AND BEARING(Section E4.3.1)
Tilting:
Pns = 4.2(t23d)1/2Fu2
Bearing:
P = 2.7 F
If t2/t1 2.5, bearingfailure alone governs