Diseño de Estructuras de Acero Laminado en Frio (R LaBoube)

7/30/2019 Diseo de Estructuras de Acero Laminado en Frio (R LaBoube)

1/354

~ ~ ~~ ~ 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


2/354

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)


3/354

Wei-Wen Yu

Center for Cold Formed Steel Structure

1 - 2

[email protected], 573-341-4481


4/354

STEEL DESIGN SPECIFICATIONS

Type of steel SpecificationCold-Formed, Carbon Steel AISI

1 - 3

-


5/354

AISI Applicable in North America

AISI S100-07 with Supplements 1 and 2

Canada

Mexico

1 - 4


6/354

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


7/354

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


8/354

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


9/354

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


10/354

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)


13/354

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


14/354


(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)


15/354


(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)


27/354

Wei-Wen Yu

Center for Cold Formed Steel Structure

1 - 26

[email protected], 573-341-4481


28/354

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


32/354

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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(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)


(Section B1.1)

2 - 17

MAXIMUM FLANGE FLAT WIDTH TO THICKNESS RATIOS


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(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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(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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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 =



53/354

Eq.B2.1-4

Since > 0.673 Eq.B2.1-2


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

.



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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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(Section B3.1)

2 - 27



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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


57/354

Given: (u.c.e.)

Fy = 50 ksi; t = 0.105 in.; R = 2t

Determine:

Effective width of compression


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 =



58/354


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

-



59/354


WITH CIRCULAR OR NON-CIRCULAR HOLES(Section B2.2)

2 - 31



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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


63/354

use

SIMPLIFIED MIDLINE LINEAR METHOD

WITH

2 - 35

DUE CONSIDERATION TO LOCAL BUCKLINGIN ACCORDANCE WITH SECTION B



64/354

Given: (s.c.e.)

Fy = 50 ksi

2 - 36

Determine:

Effective moment of inertia for strength determination, Ixe.

Assume webs are fully effective



65/354

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


66/354



67/354

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)


70/354

(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

=


74/354

STIFFENED ELEMENTS UNDER STRESS GRADIENT(Section B2.3)


75/354

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



77/354

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.


78/354



79/354

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=



80/354

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



81/354

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


82/354

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 =

=

=



83/354

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.


84/354



85/354

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)


86/354

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)


88/354

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)


89/354

(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


90/354


91/354

QUESTIONS?

2 - 63


92/354

FLEXURAL MEMBERS(Section C3)


93/354

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)



95/354

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]


Laterally Unsupported Members


96/354

Laterally Unsupported Members

Section C3.1.2.1 - Lateral-Torsional Buckling forOpen Cross Section Members

4 - 5

. . . - -Closed Box Members


97/354

OPEN CROSS SECTION FLEXURAL MEMBERS(Section C3.1.2.1)

Lateral Buckling Strength


98/354

Lateral Buckling Strength

Elastic Lateral-Torsional Buckling

1

4 - 7

or

2

where


99/354


100/354


101/354



102/354

4 - 11

Maximum Unsupported Length, Lu

OPEN CROSS SECTION FLEXURAL MEMBERS(Commentary Section C3.1.2.1)


103/354

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


Bending About Centroidal Axis


104/354

Bending About Centroidal Axis

Perpendicular to Symmetry Axis forSingly-Symmetric Sections Only

4 - 13

where

and


Given:


105/354

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


106/354



107/354

4 - 16

0.56Fy = 28 ksi ; 2.78Fy = 139 ksi


Since 2.78Fy > Fe > 0.56Fy


108/354

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


109/354


110/354

DISTORTIONAL BUCKLING STRENGTHSection C3.1.4

For


111/354

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:


112/354

General distortional buckling stress equation:

4 - 21

q. . . -


113/354

CYLINDRICAL TUBULAR MEMBERS IN BENDING(Section C3.1.3)


114/354

4 - 23

0.0714 0.318 0.441

SHEAR STRENGTH(Section C3.2)


115/354

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


116/354

The Shear Strength Depends on:

The web slenderness ratio, h/t

4 - 25

The material properties

Use of transverse stiffeners


1) Unreinforced Webs


117/354

1) Elastic shear buckling2) Inelastic shear buckling

Elastic shear buckling

4 - 26

For simply supported

edges along member

kv = 5.34


Inelastic Shear Buckling


118/354

4 - 27

Specification uses 0.6


Nominal Shear Strength, VnV = A F Eq C3 2 1 1


119/354

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


2) Reinforced Webs (Transverse Stiffeners)


120/354

when a/h 1.0

4 - 29

when a/h > 1.0

SHEAR STRENGTH(Section C3.2.1


121/354

4 - 30Shear Strength with Transverse Stiffeners

EXAMPLE 4.2 SHEAR OF SOLID WEBS

Given: Information of Ex. 2.6 of Lecture 2Determine:


122/354

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


2) The nominal shear strength of the reinforced web satisfyingthe requirements of Section C3.6.1.


123/354

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


124/354

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


125/354

(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


126/354

4 - 35

High bending and shear simultaneously

STRENGTH FOR COMBINED BENDING AND SHEAR(Section C3.3.1 - ASD Method)



127/354

2) Reinforced Webs

4 - 36

STRENGTH FOR COMBINED BENDING AND SHEAR(Section C3.3.2 - LRFD Method)



128/354

2) Reinforced Webs

4 - 37

FLEXURAL MEMBERS(Section C3.3)


129/354

4 - 38

WEB CRIPPLING(Section C3.4)


130/354

4 - 39


LOAD CASES:


131/354

1. EOF -- End One Flange

2. IOF -- Interior One Flange

4 - 40

3. ETF -- End Two Flange

4. ITF -- Interior Two Flange


132/354


Basic Web Crippling Equation


133/354

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


134/354


SECTION TYPES (Fastened or unfastened to support)

Single Hat Sections(Table C3.4.1-4)


135/354

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


Alternatively, for end-one-flange loading condition on a C- or Z-section,


136/354

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.


137/354

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


138/354

(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


139/354

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


140/354

4 - 49

Eq.C3.5.1-3

COMBINED BENDING AND WEB CRIPPLING(Section C3.5.2 LRFD Method)

a) Shapes having single unreinforced webs


141/354

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


142/354

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


143/354

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


144/354

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


145/354


146/354

DESIGN OF COLD-FORMED STEEL STRUCTURESUSING THE 2007 NORTH AMERICAN SPECIFICATION


147/354

MEMBERS IN COMPRESSION

5 - 1

CONCENTRICALLY LOADED COMPRESSIONMEMBERS

Types of Compression Members


148/354

Types of Compression Members

a) Doubly-symmetric

5 - 2

b) Singly-symmetric

c) Point-symmetric

d) Non-symmetric

ACTUAL FAILED STUD SPECIMEN

600S162-43


149/354

5 - 3


150/354

DESIGN CONSIDERATIONS

a) Member behavior

i) Yielding (short & compact)

ii) O ll b kli


151/354

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


152/354

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.


153/354

ELASTIC BUCKLING STRESS, Fe

Flexural Buckling (possible failure mode)The elastic flexural buckling stress is computed

by using the following expression:P


154/354

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


155/354

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


156/354

EFFECTIVE LENGTH FACTOR, K(Table C-C4.1-1 of Commentary)

(a) (b) (c) (d) (e) (f)

Buckled shape of column is


157/354

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


158/354

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.


159/354

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


160/354

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)


161/354

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.


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


162/354

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=


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.


163/354

( )

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=


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.


164/354

( )

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


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


165/354


(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



a) Flexural buckling (K = 1.0)


166/354

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



Eq.C4.1-4

Since c > 1.5

2 2

57.12.20/50F/F eyc ===


167/354

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 ===


168/354


= (1 - 0.22/ )/= (1 - 0.22/0.755)/0.755 = 0.939 Eq. B2.1-3

b = 0.939(3.505) = 3.29 in.


169/354

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


Pn = AeFn = 0.358(17.8) = 6.38 kips Eq.C4.1-1


Given: The singly-symmetric C-section3CS3x060L = 3.5 ft (42 in.), Fy = 50 ksi.

y

x

0.060

R=0.1875

x

0.75


170/354

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

.



a) Flexural buckling (Ky = 1.0)

KyL/ry = 42/1.12 = 37.5 < 200 OK (Commentary)

F2

E/(K L/ )2


171/354

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

+


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


172/354

= 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.



c = [Fy/Fe]1/2 = [50/46.3]1/2 = 1.04 Eq.C4.1-4

Since c 1.5

(


173/354

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 ===


Since W > 0.328S, B4(a)

43

3s in.000636.0

12

)503.0(060.012/tdI ===

3841


174/354

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 ==


Table B4.1

( ) 3/1=n3/1


175/354

=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=


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


176/354

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


ksi0.61

)841(9210

E4

W)1(12

EkF

2

2

22

2

cr ===


177/354

)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 ===


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


178/354


5 - 32

Pn = AeFn = 0.511(31.8) = 16.3 kips Eq.C4.1-1


179/354

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)


180/354

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?


181/354

5 - 35


COMBINED BENDING AND COMPRESSION


182/354

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.


183/354

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


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


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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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)



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, = . .

~ 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



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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



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= - = - -

6 - 13

Check interaction equation C5.2.2-2

OK

0.1


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MEMBER BRACING


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STRAPPING


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7 - 2


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TYPES OF BRACING


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7 - 4

STEEL BUILDING SYSTEMS


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TYPES OF BRACING

Member (beam or column)


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7 - 6

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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7 - 12

For a uniform loadper unit length, P:

PL = 1.5KaP

MEMBER BRACING

d

Pm=F

=

P

F

P


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7 - 13

P

F

MEMBER BRACING

x

xy

I2

IP=F PK=

P

F


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7 - 14

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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7 - 15

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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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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7 - 18


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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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7 - 20

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 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


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 Flexural Strength

C- or Z- Purlins

Gravity Load and Uplift Load

Top Flange Supporting Standing Seam Roof Panel


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7 - 28

Mn = RSeFy


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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CONNECTIONS


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8 - 1

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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8 - 5

= 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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8 - 9

emin = Pu/(Fut) for LRFD

and vary with the Fu/Fy of the sheet


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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8 - 10

the ratio (da/t)


Shear Strength - Sheet-to-Sheet Connections:

Pn = 1.65 tdaFu


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8 - 11

= 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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8 - 15

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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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


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


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


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


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 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

Diseño de Estructuras de Acero Laminado en Frio (R LaBoube)

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