Designing Aluminum Structures
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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
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1. Introduction
Why learn aluminum structural design?To design aluminum structures
To better understand steel structural design
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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)
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Science Land Egg (164’ wide)
courtesy of Temcor
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Science Land Egg
courtesy of Temcor
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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
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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)
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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
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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
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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 …)
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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
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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
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2. Aluminum Alloys and Tempers
Wrought alloy designation system
Aluminum temper designation system
Material specifications
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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%
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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
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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)
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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
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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
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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
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2xxx Alloys (Al-Cu)
Common UsesAircraft parts, skins
Fasteners
Example2024
Pro: Strong
Con: Not very corrosion resistant; hard to weld
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3xxx Alloys (Al-Mn)
Common UsesRoofing and siding
Gutters and downspouts
Examples3003, 3004, 3105
Pro: Formable, good corrosion resistance
Con: Not that strong
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4xxx Alloys (Al-Si)
Common UsesWelding and brazing filler metal
Example4043
Pro: Flows well
Con: Lower ductility
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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
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6xxx Alloys (Al-Mg2Si)
Common UsesStructural shapesPipe
Examples6061, 6063
Pro: Good combination of strength and corrosion resistance, very extrudableCon: Lose considerable strength when welded
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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
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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
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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)
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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
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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
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Effect of Strain Hardening 5052
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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
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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
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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
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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
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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
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3. Aluminum Material Properties
StrengthsModulus of Elasticity, Poisson’s RatioDuctilityEffect of Welding on Properties
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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
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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
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Types of Strengths
Type of Stress Yield Ultimate
Tension Fty Ftu
Compression Fcy
Shear Fsu0.6Fty
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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
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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)]
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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
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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
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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
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Effect of Welding on Strength
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4. Aluminum Structural Design Overview
Limit states
Strength limit state design:Allowable Strength Design (ASD)
Load and Resistance Factor Design (LRFD)
Analysis
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Limit States
A structural engineer considers limit statesStatic strength available strength > required strength
Serviceability (deflection, vibration, etc.)
Fatigue
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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)
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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
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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
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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
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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)
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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
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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
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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
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P-Δ and P-δ Effects
P-Δ P-δ
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P
P
P
P
δ
Δ
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Geometric Imperfections
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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
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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
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Net and Gross Sections
Yield on the gross section
Rupture on thenet section
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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
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Allowable Tensile Stress Example
6063-T5 Extrusions:
Fty = 16 ksi, Ftu = 22 ksi
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
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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
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LRFD Tension Example
6061-T6 Extrusions:
Fty = 35 ksi, Ftu = 38 ksi
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.
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Shear Lag in a Channel
flanges are not connected and not fully stressed at member end
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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
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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
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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
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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
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Yielding, Inelastic Buckling, and Elastic Buckling
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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
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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)
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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:
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Yielding
Yield strength is simply Fcy
Yielding depends only on material strength
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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
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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
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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
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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
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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
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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
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Flexural & Torsional Column Buckling
Flexural Buckling(lateral movement)
Torsional Buckling(twisting)
Deflected shape Undeflected shape
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Types of Column Member Buckling
Flexural (lateral movement)
Torsional (twisting about longitudinal axis)
Torsional-flexural (combined effect)
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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
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Local Buckling
Web Buckling Flange Buckling
buckled shape
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Elements of Shapes are Called:
Element
Flange or web
Component
Plate
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Dividing a Shape Into Elements
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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
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Element Support Conditions
supported edge stiffened edge free edge
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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
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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)
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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
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Element Strength vs. Slenderness
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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)
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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
Designing Aluminum Structures 98
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
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Designing Aluminum Structures 99
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
Designing Aluminum Structures 100
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.
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Designing Aluminum Structures 101
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
Designing Aluminum Structures 102
Designing Aluminum Structures
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
Designing Aluminum Structures 103
Column Design Summary
Column strength is the least of:Member buckling strength
Local buckling strength
Interaction between member and local buckling strengths
Designing Aluminum Structures 104
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Designing Aluminum Structures 105
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
Designing Aluminum Structures 106
Yielding and Rupture in Beams
neutral axis
yielding onsetcross section full yielding
Fy Fy
Fy Fy
MpMy
rupture
Fu
Fu
Mu
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Designing Aluminum Structures 107
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
Designing Aluminum Structures 108
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)
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Designing Aluminum Structures 109
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
Designing Aluminum Structures 110
Weak Axis Bending
Lateral buckling usually does not govern
Load
Undeflected shape
Deflected shape at ultimate load
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Designing Aluminum Structures 111
Lateral-Torsional Buckling (LTB)Section Shape Slenderness
F.2 Open shapes
F.3 Closed shapes
F.4 Rectangular bars
F.5 Single Ls
Designing Aluminum Structures 112
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
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Designing Aluminum Structures 113
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
Designing Aluminum Structures 114
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
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Designing Aluminum Structures 115
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
Designing Aluminum Structures 116
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
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Designing Aluminum Structures 117
Moment Gradient Factor Cb
Cb = 1.0 (min)
Cb = 2.3
Cb = 1.14
M M
M
M
w
Designing Aluminum Structures 118
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
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Designing Aluminum Structures 119
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
Designing Aluminum Structures 120
Transverse Load LocationLoad acts toward shear center
(smallest bending strength)
Load acts away from shear center(greatest bending strength)
Load applied at neutral axis
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Designing Aluminum Structures 121
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
Designing Aluminum Structures 122
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
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Local Buckling of Beam Elements
Beam elements in uniform compression (flanges) are treated like column elements in uniform compression (see B.5.4)
Designing Aluminum Structures 123
Designing Aluminum Structures 124
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
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Designing Aluminum Structures 125
Beam Flange Stress
neutral axis
bending stresscross section
Compression
Tension
Designing Aluminum Structures 126
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
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Designing Aluminum Structures 127
Compression Bending Strength
Compression bending strength is the lesser of:
Member buckling strength
Local buckling strength
Designing Aluminum Structures 128
Weighted Average Bending Strength (SAS F.8.3)
ccf
ctfctw
ccw
If
Iw
If
tension side
compression side
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Designing Aluminum Structures 129
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)
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