randy kissell, p.e. tgb partnership - the...

Designing Aluminum Structures

1

1

Aluminum Structural Member Design

Randy Kissell, P.E.

TGB Partnership

Designing Aluminum Structural Members 2

Outline

1 Introduction2 Aluminum alloys and tempers3 Aluminum material properties4 Aluminum structural design overview5 Axial tension6 Axial compression7 Flexure


2

1. Introduction

Why learn aluminum structural design?To design aluminum structures

To better understand steel structural design

Designing Aluminum Structures 3


Examples of Aluminum Structures

Curtain walls and storefronts

Roofing and canopies

Space frames

Tanks and vessels (corrosive & cryogenic)

Portable structures (scaffolding, ladders)

Highway products (signs, light poles, bridge rail)


3


Science Land Egg (164’ wide)

courtesy of Temcor


Science Land Egg

courtesy of Temcor


4


The Aluminum Association

Founded in 1933, 50+ members are the major US producers

AA writes most standards on aluminum; has worldwide influence

Contact: www.aluminum.org

1525 Wilson Blvd, Suite 600

Arlington, VA 22209

703-358-2960; pubs 301-645-0756

2010 Aluminum Design Manual



5


Aluminum Design Manual (ADM)

Now issued every 5 years

Latest edition is ADM 2010

Previous editions in 2005, 2000

1st edition (1994) was compilation of several AA pubs previously issued separately; most importantly, the Specification for Aluminum Structures(SAS)


Aluminum Design Manual Contents

The Aluminum Design Manual (ADM) provides aluminum structural design tools

Part I – Specification for Aluminum Structures

Part II – Commentary

Part III – Design Guide

Part IV – Material Properties

Part V – Section Properties

Part VI – Design Aids

Part VII – Illustrative Examples


6


Specification for Aluminum Structures (SAS)

The Specification for Aluminum Structuresis Part I of the Aluminum Design Manual

SAS is also called “the Aluminum Specification”

Adopted in BOCA, UBC, SBC, and IBC

It’s the source of all aluminum structural design requirements in the US


Specification for Aluminum Structures (SAS)

2010 edition is a major rewriteIt’s a unified Specification, with both:

Allowable Strength Design For buildings and bridges

Load and Resistance Factor Design For buildings only Load factors from ASCE 7 (= 1.2D + 1.6L …)


7


2010 Specification for Aluminum Structures (SAS)

A. General ProvisionsB. Design RequirementsC. Design for StabilityD. Design of Members for TensionE. Design of Members for Compression F. Design of Members for FlexureG. Design of Members for ShearH. Design of Members for Combined Forces and TorsionJ. Design of Connections


2010 Specification for Aluminum Structures (SAS)

L. Design for ServiceabilityM. Fabrication and ErectionAppendices1. Testing3. Design for Fatigue4. Design for Fire Conditions5. Evaluation of Existing Structures6. Design of Braces for Columns and BeamsNew in 2010


8


2. Aluminum Alloys and Tempers

Wrought alloy designation system

Aluminum temper designation system

Material specifications


Aluminum Isn’t Just One Thing

Like other metals, aluminum comes in many alloys

Alloy = material with metallic properties, composed of 2 or more elements, of which at least one is a metal

Different aluminum alloys can have very different properties

Alloying elements are usually < 5%


9


Alloys

Alloys of iron are called “steel”Carbon < 0.5% or it’s cast iron

C, Cr, Ni, Mb, Cu, Ti used

Alloys of aluminum are called “aluminum alloys”

Si, Fe, Cu, Mn, Mg, Cr, Ni, Zn, Ti used


Designating Aluminum vs. Steel

Steel is identified by ASTM no. and gradeExample: ASTM A 709 Grade 50Steel ASTM specs are for one alloy and a few products (for example, shapes and plate)

Aluminum is identified by AA no., temper, and product

Example: 6061-T6 sheetAluminum ASTM specs are for many alloys and one product (e.g., ASTM B 209)


10


Wrought Alloy Designation System

Number Main Alloy Strength Corrosion

1xxx > 99% Al Fair Excellent

2xxx Cu High Fair

3xxx Mn Fair Good

4xxx Si Good Good

5xxx Mg Good Good

6xxx Mg Si Good Good

7xxx Zn High Fair

8xxx others


Wrought Alloy Key

1st digit denotes main alloying element

3rd and 4th digits are sequentially assigned

2nd digit denotes a variation

Example:2319 is AlCu alloy (2xxx), variation on 2219

2319 composition is identical to 2219 except slightly more Ti (grain refiner to improve weld strength); both have 6.3% Cu


11


1xxx Alloys (pure Al)

Common uses:Electrical conductors

Corrosive environments

Examples1060 (99.60% aluminum)

1100 (99.00% aluminum)

Pro: corrosion resistant, good conductors

Con: Not very strong


2xxx Alloys (Al-Cu)

Common UsesAircraft parts, skins

Fasteners

Example2024

Pro: Strong

Con: Not very corrosion resistant; hard to weld


12


3xxx Alloys (Al-Mn)

Common UsesRoofing and siding

Gutters and downspouts

Examples3003, 3004, 3105

Pro: Formable, good corrosion resistance

Con: Not that strong


4xxx Alloys (Al-Si)

Common UsesWelding and brazing filler metal

Example4043

Pro: Flows well

Con: Lower ductility


13


5xxx Alloys (Al-Mg)

Common UsesMarine applications

Welded plate structures

Examples5052, 5083

Pro: Strong, even when welded

Con: Hard to extrude; those with 3%+ Mg can have corrosion resistance issues


6xxx Alloys (Al-Mg2Si)

Common UsesStructural shapesPipe

Examples6061, 6063

Pro: Good combination of strength and corrosion resistance, very extrudableCon: Lose considerable strength when welded


14


7xxx Alloys (Al-Zn)

Common UsesAircraft parts

Two classesWith copper (example: 7075)

Without copper (example: 7005)

Pro: Very strong (7178-T6 Ftu = 84 ksi)

Con: Not very corrosion resistant;

hard to weld


How Alloys are Strengthened

Alloying elements (Mg is good example)

Treatment:Strain hardening (cold working)

Heat treatment

Heat treatable: 2xxx, 6xxx, 7xxx

Non-heat treatable: 1xxx, 3xxx, 5xxx


15


Annealed Condition

Before tempering, alloys start in the annealed condition (-O suffix)

Annealed condition is weakest but most ductile

Tempering increases strength, but decreases ductility

Most alloys are annealed by heating to 650oF (melting point is about 1100oF)


Strain Hardening

Mechanical deformation at ambient temps

For sheet and plate, deformation is by rolling to reduce the thickness

Some non-heat treatable alloys undergo a stabilization heat treatment

Purpose: to prevent age softening

Only used for some Al-Mg (5xxx) alloys


16


Strain Hardened Tempers

Temper Ftu (ksi) Description

5052-O 25 Annealed

5052-H32 31 ¼ hard

5052-H34 34 ½ hard

5052-H36 37 ¾ hard

5052-H38 39 Full hard


Effect of Strain Hardening 5052


17


Heat Treating

1) Annealed material is

solution heat treated6061-O is heated to 990oF, then quenched

Resulting temper is 6061-T4

2) Solution heat treated material is precipitation heat treated (artificially aged)

6061-T4 is heated to 350oF and held for 8 hrs

Resulting temper is 6061-T6


Heat Treatment Tempers

T1 through T4: naturally aged (6005-T1)

T5 through T9: artificially aged (6063-T6)

Artificial aging makes the stress-strain curve flatter after yield, which affects the inelastic buckling strength


18


Tempers Summarized

-H is for strain hardened tempers1xxx, 3xxx, 5xxx alloys

Higher 2nd digit: stronger, less ductile

-T is for heat treated tempers2xxx, 6xxx, 7xxx alloys

T4 = solution heat treated

T5 and greater = precipitation heat treated


ASTM Wrought Aluminum Specifications

B 209 Sheet and PlateB 210 Drawn Seamless TubesB 211 Bar, Rod, and WireB 221 Extruded Bars, Rods, Wire, Profiles and TubesB 241 Seamless Pipe and Seamless Extruded Tube B 247 Die Forgings, Hand Forgings, Rolled Ring Forgings


19


ASTM Wrought Aluminum Specifications

B 308 Standard Structural Profiles B 316 Rivet and Cold Heading Wire and Rod B 429 Extruded Structural Pipe and Tube B 928 High Magnesium Aluminum Alloy Sheet and Plate for Marine and Similar Service (has corrosion resistance reqs)There are others not included in SAS, including some used structurally


3. Aluminum Material Properties

StrengthsModulus of Elasticity, Poisson’s RatioDuctilityEffect of Welding on Properties


20


Aluminum’s Stress-Strain Curve

After yield, plastic behavior Linear region at low strainsslope is E

Stress

0 0.02 0.04 0.06 0.08 0.10

Strain

Fy

0.002


Yield Strength

Aluminum doesn’t exhibit a definite yield point like mild carbon steel – stress-strain curve is more like high strength steel’s

0.2% offset is used to define yield

Gage length for yield is 2 in. (50 mm)

Artificially aged tempers have different shape stress-strain curve vs. non-artificially aged tempers


21


Types of Strengths

Type of Stress Yield Ultimate

Tension Fty Ftu

Compression Fcy

Shear Fsu0.6Fty


Some Aluminum Alloy Strengths

Alloy-temper, product Fty

(ksi)

Ftu

(ksi)

5052-H32 sheet & plate 23 31

5083-H116 plate

< 1.5” thick

31 44

6061-T6 extrusions 35 38

6063-T5 extrusions < 0.5” thick

16 22


22


Modulus of Elasticity, Poisson’s Ratio

Modulus of Elasticity (Young’s Modulus) EMeasures stiffness and buckling strengthCompressive E = 1.02(Tensile E)Varies by alloy; Ec = 10,100 to 10,900 ksi for SAS alloysCompares to 29,000 ksi for steel

Poisson’s ratio νAverage value = 0.33

Shear Modulus G = 3800 ksi = E/[2(1+ ν)]


Ductility

Ductility : the ability of a material to withstand plastic strain before rupture

Fracture Toughness: Aluminum doesn’t have a transition temperature like steel

Elongation (e)

Notch-Yield Ratio =

(Ftu of standard notched specimens)/Fty

If notch-yield ratio > 1, that’s good


23


Effect of Welding

Other than alloying, strength comes from strain hardening or artificial aging

Heating (like welding) erases these effects

So welding reduces strengths:For -H tempers, to annealed (-O)

For -T tempers, to ≈ solution heat treated(-T4)

Reduction is least for 5xxx alloys

Some 2xxx, 7xxx aren’t weldable


Welded Strengths

Welded strengths are in SAS Table A.3.5Notation: add w to subscript

Ftuw = welded Ftu

AWS D1.2 Table 3.2 gives same Ftuw as SAS Table A.3.5

To qualify groove weld procedures, Ftuw must be achieved


24


Effect of Welding on Strength


4. Aluminum Structural Design Overview

Limit states

Strength limit state design:Allowable Strength Design (ASD)

Load and Resistance Factor Design (LRFD)

Analysis


25


Limit States

A structural engineer considers limit statesStatic strength available strength > required strength

Serviceability (deflection, vibration, etc.)

Fatigue


What a Structural Engineer Does

Analysis: determine forces, moments in the structure (required strength)

Use the same methods for all materials

But beware: since aluminum is more flexible than steel, 2nd order effects may be more significant

Design: proportion the aluminum structure to safely resist the loads (provide available strength)


26


ASD vs. LRFD

Allowable Strength Design (ASD):(strength)/(safety factor) > load effect

allowable strength > load effect

Load & Resistance Factor Design (LRFD):(strength)(resistance factor) > (load factor)(load effect)

design strength > (load factor)(load effect)

The difference is load factors


Safety/Resistance Factors for Aluminum Building Structures

Limit State Safety

Factor Ω

Resistance

Factor

Yield Ω = 1.65 = 0.90

Rupture Ω = 1.95 = 0.75

Fastener

rupture

1.2Ω = 2.34 = 0.65


27


Aluminum Safety Factors

Stress type yielding buckling or rupture

Axial tension 1.65 1.95

Bending tension 1.65 1.95

Axial compression 1.65 1.65

Bending compression 1.65 1.65

Shear 1.65 1.65


SAS Section C.2: Analysis Must Account for:

Axial, flexural, and shear deformations

Second-order effects (P-∆ and P-δ)

Geometric imperfections (use construction and fabrication tolerances)

Effect of inelasticity on flexural stiffness (use τb I in place of I )

Uncertainty in stiffness and strength (use 0.8E in place of E, i.e. 8000 ksi)


28

Loads for Analysis

For LRFD, use load factors prescribed for LRFD and use the resulting forces and moments produced by the analysis to determine the required strengths Pr

For ASD, use 1.6 times the ASD loads, then divide the resulting forces and moments produced by the analysis by 1.6 to determine the required strengths Pr


Loads for Analysis

Design method

LRFD ASD

LoadCombination

1.2D + 1.6L 1.6(D + L)

AnalysisResults

PLRFD PASD

Required Strength

PLRFD PASD /1.6

AvailableStrength

Rn

e.g. 28.5 ksiRn /Ωe.g. 19.5 ksi



29

Second-Order AnalysisMany structural analysis applications include 2nd order effects

Most applications address P-Δ (effect of loads acting on the displaced location of joints)

Some don’t include P-δ (effect of loads acting on the deflected shape of member between joints)

To check, compare results to benchmark examples given in 2005 AISC Spec commentary Section 7.3


P-Δ and P-δ Effects

P-Δ P-δ


P

P

P

P

δ

Δ


30

Geometric Imperfections


Tolerance on out-of-plumbnessspecified by the designer

This is the structureanalyzed

Effect of Inelasticity on Flexural Stiffness

Factor τb on flexural stiffness for inelasticity is a function of how highly stressed the structure is

τb ranges from:1.0 for Pr < 0.5Py /α = 0.31Py for ASD

0 for Pr = Py /α = 0.62Py for ASD

In between, τb = (4α Pr /Py)(1 – α Pr /Py)

α = 1.0 for LRFD, 1.6 for ASD



31


5. Axial Tension

SAS Chapter D covers axial tensionTensile limit state is reached at:

Rupture on the net section (Ω = 1.95)Yield on the gross section (Ω = 1.65)

Same criteria as in AISC for steelIt’s assumed that the net section exists only over a short portion of the member length, so yielding there won’t cause much elongation


Net and Gross Sections

Yield on the gross section

Rupture on thenet section


32


Allowable Tension Stress Example

6061-T6 Extrusions:

Fty = 35 ksi, Ftu = 38 ksi

Allowable stress on the gross section:

F /Ω = Fty /Ω = 35/1.65 = 21.2 ksi

Allowable stress on the net section:

F /Ω = Ftu /(Ω kt) = 38/[(1.95)(1.0)] = 19.5 ksi

Net section always governs


Allowable Tensile Stress Example

6063-T5 Extrusions:


Allowable stress on the gross section:

F /Ω = Fty / Ω = 16/1.65 = 9.7 ksi

Allowable stress on the net section:

F /Ω = Ftu /(Ω kt) = 22/[1.95)(1.0)] = 11.3 ksi

Gross or net section could govern


33


Tension Coefficient kt

kt is a notch sensitivity factor

For alloys in SAS, kt > 1 only for :2014-T6, 6005-T5, and 6105-T5, kt = 1.25

6066-T6 and 6070-T6, kt = 1.1


LRFD Tension Example

6061-T6 Extrusions:


LRFD design stress on the gross section:

F = y Fty = 0.90(35) = 31.5 ksi

LRFD design stress on the net section:

F = u Ftu /kt = 0.75(38)/(1.0) = 28.5 ksi

So just like ASD, net section governs.


34


Shear Lag in a Channel

flanges are not connected and not fully stressed at member end


Effective Area in Tension (Ae)

SAS Section D.3.2

If all parts of x-section aren’t connected to joint, full net area isn’t effective in tension

Example: Channel bolted through its web only (not flanges)

SAS addresses angles, channels, tees, zees, and I beams


35


Effective Net Area Ae

Effective net area =

where An = net areax = eccentricity in x directiony = eccentricity in y directionL = length of connection in load directionAe > An of connected elements


Effective Net Area Example

For a tee bolted through its flange only:

Other examples are in ADM Part II D.3.2

When only a single row of fasteners is used, L = 0 and Ae = An of connected elements only

neutral axis of tee

y


36


6. Axial Compression

Column = axial compression member SAS Chapter E addresses columnsColumn strength is the least of:

Member buckling strengthLocal buckling strengthInteraction between member buckling and local buckling strengths

Member Strength

Member’s compressive limit states areYielding (squashing)Inelastic bucklingElastic buckling



37


Yielding, Inelastic Buckling, and Elastic Buckling


Elastic Buckling

Elastic buckling stress = Fe = 0.852E / 2

E is the only material property that elastic buckling strength depends on = kL/r = largest slenderness ratio for buckling about any axisAll other things equal, Fe for aluminum is 1/3 Fe for steel since Ea = Es /3


38


Inelastic Buckling

Inelastic buckling strength = 0.85[Bc - Dc(kL/r)]Bc (y intercept) and Dc (slope) are buckling constants that depend on Fcy and ECalculate them by SAS equations in:

Table B.4.1 for O, H, T1 thru T4 tempersTable B.4.2 for T5 thru T9 tempers

Bc and Dc are tabulated in ADM Part VI Table 1-1 (unwelded) and 1-2 (welded)


Member Buckling

0.85 factor accounts for member out-of-straightness

k = 1 for all members (see Section C.3)

Allowable member buckling strengths really haven’t changed from 2005 SAS:


39


Yielding

Yield strength is simply Fcy

Yielding depends only on material strength


Buckling Strength vs. Slenderness

Inelastic Buckling Fc = 0.85(Bc – DckL/r)

Elastic Buckling Fc = 0.85π2E/(kL/r)2

Yielding Fc = Fcy

Stress

Slenderness ratio kL/r


40


Slenderness Limits S1, S2

S1 is the slenderness for which yield strength = inelastic buckling strengthS2 is the slenderness for which inelastic buckling strength= elastic buckling strengthSlenderness ratios (kL/r) are not limited by S1 and S2; S1 and S2 are just the limits of applicability of compressive strength equations


6061-T6 Column Strength

ADM Part VI, Table 2-19 gives allowable stresses based on SAS rules

Inelastic buckling allowable is always

< Fcy /Ω = 21.2 ksi, so S1 = 0

For kL/r < 66, Fc /Ω = 20.3 – 0.127(kL/r)

For kL/r > 66, Fc /Ω = 51,350/(kL/r)2

S2 = Cc = 66


41


Slenderness Limits Demonstrated

For kL/r = S2 = 66:Inelastic buckling allowable stress is

20.3 – 0.127(66) = 11.9 ksi

Elastic buckling allowable stress is

51,350/(66)2 = 11.8 ksi ≈ 11.9 ksi

Difference is only due to round off in allowable stress expressions


Column Example

What’s the allowable member buckling compressive stress for a column given:

6061-T6

Pinned-end support conditions

Length = 95”

Shape is AA Std I 6 x 4.03

rx = 2.53”, ry = 0.95”

No bracing


42


Column Example Answer

Column will buckle about minor axis since slenderness ratio kL/r is larger there:

kL/r = (1.0)(95”)/0.95” = 100

Since kL/r = 100 > S2 = 66 (buckling is elastic), so

Fc /Ω = 51,350/(100)2 = 5.1 ksi

We need to check local buckling, too


Flexural & Torsional Column Buckling

Flexural Buckling(lateral movement)

Torsional Buckling(twisting)

Deflected shape Undeflected shape


43


Types of Column Member Buckling

Flexural (lateral movement)

Torsional (twisting about longitudinal axis)

Torsional-flexural (combined effect)


Local Buckling

Local buckling is buckling of an element of a shape (i.e., a flange or web)Buckle length ≈ width of elementIf local buckling strength of all elements > yield strength, shape is compact, and local buckling isn’t a concernSince aluminum shapes vary widely (extrusions, cold-formed shapes), we can’t assume aluminum shapes are compact


44


Local Buckling

Web Buckling Flange Buckling

buckled shape


Elements of Shapes are Called:

Element

Flange or web

Component

Plate


45


Dividing a Shape Into Elements


Elements of Shapes

Cross sections can be subdivided into two types of elements:

Flat elements (slenderness = b/t )

Curved elements (slenderness = Rb /t )

Longitudinal edges of elements can be:Free

Connected to another element

Stiffened with a small element


46


Element Support Conditions

supported edge stiffened edge free edge


Elements in Uniform CompressionAddressed by the SAS

B.5.4.1 Flat element supported on one edge

(flange of an I beam or channel)

B.5.4.2 Flat element supported on both edges

(web of I beam or channel)

B.5.4.3 Flat element supported on one edge, other edge with stiffener

B.5.4.4 Flat element supported on both edges, with an intermediate stiffener

B.5.4.5 Curved element supported on both edges


47


Local Buckling Strengths

Yielding Fc = Fcy

Inelastic buckling Fc = Bp – Dp (kb/t)

Elastic buckling Fc = 2E /(kb/t)2

Postbuckling Fc = k2 (BpE)1/2/ (kb/t)

k = edge support factork = 5.0 for elements supported on 1 edge

k = 1.6 for elements supported on both edges

k2 = postbuckling factor ≈ 2 (Table B.4.3)


Postbuckling Strength

Only elements of shapes have postbuckling strength – members do not Postbuckling strength is not recognized by SAS for all types of elementsPostbuckling strength is only recognized for elements so slender that they buckle elastically


48


Element Strength vs. Slenderness


6061-T6 Column Flange Strength

Yielding Fc /Ω = Fcy /Ω

S1 = 6.7 Fc /Ω = 35/1.65 = 21.2 ksi

Inelastic buckling Fc /Ω = [Bp – Dp (5.0b/t)]/ Ω

S2 = 12 Fc /Ω = 27.3 – 0.910b/t

Elastic buckling Fc /Ω = [2E /(5.0b/t)2]/ Ω

S2 = 10.5 Fc /Ω = 2417 /(b/t)2

Postbuckling Fc /Ω = [k2 (BpE)1/2/(5.0b/t)]/ Ω

Fc /Ω = 186 /(b/t)


49


Column Local Buckling - Flange

Shape is AA Std I 6 x 4.03:flange slenderness = b/t

b/t = (4” – 0.19”)/2/0.29” = 6.6

S1 = 6.7 > 6.6 so

Fc /Ω = 21.2 ksi 6”

0.29”

0.19”

4”

R


Column Local Buckling - Web

Shape is AA Std I 6 x 4.03:web slenderness = b/t

b/t = [6” – 2(0.29”)]/0.19” = 28.5

S1 = 7.6 < 28.5 < 33 = S2, so

Fc /Ω = 27.3 – 0.291b/t =

Fc /Ω = 27.3 – 0.291(28.5) = 19.06”

0.29”

0.19”

4”

R


50


Weighted Average Allowable Compressive Stress of I 6 x 4.03Fcf /Ω = 21.2 ksiFcw /Ω = 19.0 ksiAf = 2(4”)(0.29”)

= 2.32 in2

Aw = (6” – 2(0.29”))(0.19”)= 1.03 in2

Fca /Ω = 21.2(2.32) + 19.0(1.03)(2.32 + 1.03)

= 20.5 ksi

6”

0.29”

0.19”

4”

R


The local buckling mode is shown to the right. Note, that there is no translation at the folds, only rotation. The load factor is 0.10, so elastic critical local buckling (Pcrl) occurs at 0.10Py in this member.


51


Local Buckling/Member Buckling Interaction

If the elastic local buckling stress < member buckling stress, member buckling stress must be reduced (SAS Section E.5)

Reduced member buckling stress is Frc:

Frc = (Fc)1/3(Fe)2/3

where Fc = elastic member buckling stress

Fe = elastic local buckling stress

This only governs if elements are very slender and postbuckling strength is used

Local/Member Buckling Interaction Example

Flange elastic buckling stress Fef

Web elastic buckling stress Few



52

Local/Member Buckling Interaction Example

Member elastic buckling stress Fc

Since Fe = 47.9 ksi > Fc = 8.5 ksi, the member buckling strength need not be reduced for interaction between local and member buckling


Column Design Summary

Column strength is the least of:Member buckling strength

Local buckling strength

Interaction between member and local buckling strengths



53


7. Flexure

Beam = flexural member

Beam strength is limited by:Rupture (Ω = 1.95)

Yielding (Ω = 1.65)

Buckling (local or member) (Ω = 1.65)

Buckling can be:Flexural stress buckling

Shear stress buckling

M M


Yielding and Rupture in Beams

neutral axis

yielding onsetcross section full yielding

Fy Fy

Fy Fy

MpMy

rupture

Fu

Fu

Mu


54


Shape Factors

Sec.

No.

Element Mp /My Mu /My

F.8.1.1 Uniform stress 1.0 1.0

F.6.1 Curved element 1.17 1.24

F.8.1.2 Flexural stress 1.30 1.42


Yielding and Rupture Example

6061-T6 extruded round tubeFor rupture, Fb /Ω = 1.24 Ftu /(ktΩ) for 6061-T6, Fb /Ω = 1.24(38)/[(1.0)(1.95)]

Fb /Ω = 24 ksiFor yielding, Fb /Ω = 1.17 Fty /Ωfor 6061-T6, Fb /Ω = 1.17(35)/(1.65)

Fb /Ω = 25 ksiSo Fb /Ω = 24 ksi (the lesser of the two)


55


Major Axis Bending

Top flange(in compression)buckles laterally

Bottom flange(in tension) staysIn place

Undeflected shape

Deflected shape at ultimate load

Load

Lateral-torsional buckling


Weak Axis Bending

Lateral buckling usually does not govern

Load

Undeflected shape

Deflected shape at ultimate load


56


Lateral-Torsional Buckling (LTB)Section Shape Slenderness

F.2 Open shapes

F.3 Closed shapes

F.4 Rectangular bars

F.5 Single Ls


Slenderness Ratio for Beams

Slenderness ratio depends on unbraced beam length Lb

Lb = length between bracing points or between a brace point and the free end of a cantilever beam. Braces:

restrain the compression flange against lateral movement, or

restrain the cross section against twisting

Appendix 6 addresses brace design


57


6061-T6 I-Beam Strength

Inelastic buckling Fc /Ω = [Bc – Dc (1.2Lb /ry)]/Ω

S2 = 79 Fc /Ω = 23.9 – 0.124 (Lb /ry)Elastic buckling

Fc /Ω = [2E /(1.2Lb /ry)2]/ ΩFc /Ω = 87,000 /(Lb /ry)2


LTB Example

What’s the allowable LTB stress for a beam given:

6061-T6

Length = 86”

Shape is AA Standard I 12 x 14.3

ry = 1.71”

No bracing


58


LTB Example Answer

Slenderness ratio is

Lb /ry = 86”/1.71” = 50.3

Since Lb /ry = 50.3 < 79 = S2,

Fc /Ω = 23.9 – 0.124 (Lb /ry)

Fc /Ω = 23.9 – 0.124 (50.3) = 17.7 ksi

Remember to check local buckling, too


Moment Variation Over Beam Length (Cb)

If moment varies along the beam, account for this using Cb. For doubly sym sections:

Cb = 12.5Mmax

(2.5Mmax + 3MA + 4MB + 3MC)

where MA = moment at ¼ point

MB = moment at midpoint

MC = moment at ¾ point


59


Moment Gradient Factor Cb

Cb = 1.0 (min)

Cb = 2.3

Cb = 1.14

M M

M

M

w


Moment Gradient Factor Cb

Replace unbraced length Lb with Lb /Cb for tubes and Lb /√Cb for single web beams

So Lb can be reduced to as little as

0.43Lb = Lb /2.3


60


Effective ry (= rye)

F.2.1 allows using ry for rye

It’s easy to determine, but conservative

When Lb / ry > 50, it’s worth determining rye

using F.2.2. It’s more work, but more accurate


Transverse Load LocationLoad acts toward shear center

(smallest bending strength)

Load acts away from shear center(greatest bending strength)

Load applied at neutral axis


61


rye for Shapes Symmetric About the Bending Axis

Load applied toward shear center

Load applied at shear center, or no load

Load applied away from shear center


I 12 x 14.3 Beam, 200” long

SAS Section

Apply

Load

rye

(in.)

Lb /rye Fb /Ω

(ksi)

F.2.1 – 1.71 117 6.4

F.2.2.1 above s.c. 1.80 111 7.0

F.2.2.3 at s.c. 2.08 96 9.4

F.2.2.1 at s.c. 2.10 95 9.6

F.2.2.1 below s.c. 2.45 82 13.1


62

Local Buckling of Beam Elements

Beam elements in uniform compression (flanges) are treated like column elements in uniform compression (see B.5.4)



How Can Beam Elements Be In Uniform Compression?

To be precise, they can’t

But the SAS idealizes elements parallel to the neutral axis as being in uniform compression

Example: the flange of an I-beam or channel - there is some variation in the compressive stress across the flange thickness, but it’s neglected


63


Beam Flange Stress

neutral axis

bending stresscross section

Compression

Tension


Beam Elements in SAS –Elements in Flexure (Webs)

B.5.5.1 Flat element - both edges supported

(web of I beam or channel)

B.5.5.2 Flat element - compression edge free, tension edge supported

B.5.5.3 Flat element with a longitudinal stiffener –both edges supported (see B.5.5.3 for stiffener requirements


64


Compression Bending Strength

Compression bending strength is the lesser of:

Member buckling strength

Local buckling strength


Weighted Average Bending Strength (SAS F.8.3)

ccf

ctfctw

ccw

If

Iw

If

tension side

compression side


65


Thank You

Please contact me with [email protected] Farmview Rd, Hillsborough, NC 27278www.tgbpartnership.com

For a more extensive aluminum seminar:go to www.asce.org/distancelearning; click on “View a complete list of ASCE courses”, scroll down alphabetically to “Aluminum Structural Design ” (or call 800-548-2723)

randy kissell, p.e. tgb partnership - the...

Documents