Learning Objectives - · PDF fileLRFD vs. ASD • Three new notations -φ,λ, and...

American Wood Council

1

STD 104:STD 104:

Copyright © 2007-2011 American Wood Council. All rights reserved.

STD 104:STD 104:ASD and LRFD ASD and LRFD

with the 2005 NDSwith the 2005 NDS®®

“The Wood Products Council” is a Registered Provider with The American Institute of Architects Continuing Education Systems (AIA/CES). Credit(s) earned on completion of this program will be reported to AIA/CES for AIA members. Certificates of Completion for both AIA members and non-AIA members are available upon request.

This program is registered with AIA/CES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product.

Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation.

Copyright Materials

This presentation is protected by US and International Copyright laws. Reproduction,

distribution, display and use of the presentation without written permission of the speaker iswithout written permission of the speaker is

prohibited.

© The Wood Products Council 2011

Learning ObjectivesAt the end of this program, participants will be able to:

1) Learn how to design structural framing and connections using the latest design standard provisions2) Review the basis for the design equations used in ASD2) Review the basis for the design equations used in ASD and LRFD design including reliability and determining design values3) Discuss behavior equations for beams and columns including relevant adjustment factors such as load duration, and wet service4) Focus on the design of wood connections including the use of nails, staples, lag screws, bolts, and timber rivets


2

History

The first NDS (1944) was based on allowable stress design (ASD).

ASD through 2001 NDS1944

1962

1973

1977 1991

1968

1971

1982

1986

1991

1997

2001

Load Resistance Factor Design

ASD and LRFD in 2005 NDS


3

Outline• Document organization• Overview of LRFD Concept• Chapter‐by‐chapter description• Changes from previous editions• Examples

NDS® 2005 and Supplement

2001• 16 Chapters• 13 Appendices

2005• 16 Chapters• 14 Appendices

What’s changed?

NDS 2005 Chapters200520051 General Requirements for Building Design1 General Requirements for Building Design2 Design Values for Structural Members2 Design Values for Structural Members3 Design Provisions and Equations3 Design Provisions and Equations4 Sawn Lumber4 Sawn Lumber5 Structural Glued Laminated Timber5 Structural Glued Laminated Timber6 Round Timber Poles and Piles6 Round Timber Poles and Piles6 Round Timber Poles and Piles6 Round Timber Poles and Piles7 Prefabricated Wood I7 Prefabricated Wood I--JoistsJoists8 Structural Composite Lumber8 Structural Composite Lumber9 Wood Structural Panels9 Wood Structural Panels10 Mechanical Connections10 Mechanical Connections11 Dowel11 Dowel--Type FastenersType Fasteners12 Split Ring and Shear Plate Connectors12 Split Ring and Shear Plate Connectors13 Timber Rivets13 Timber Rivets14 Shear Walls and Diaphragms14 Shear Walls and Diaphragms15 Special Loading Conditions15 Special Loading Conditions16 Fire Design of Wood Members16 Fire Design of Wood Members

20052005A Construction and Design PracticesA Construction and Design PracticesB Load DurationB Load DurationC Temperature EffectsC Temperature EffectsD Lateral Stability of BeamsD Lateral Stability of BeamsE Local Stresses in Fastener GroupsE Local Stresses in Fastener GroupsF Design for Creep and Critical Deflection F Design for Creep and Critical Deflection

ApplicationsApplications

NDS 2005 Appendices

Important!

ppppG Effective Column LengthG Effective Column LengthH Lateral Stability of ColumnsH Lateral Stability of ColumnsI Yield Limit Equations for ConnectionsI Yield Limit Equations for ConnectionsJ Solution of Hankinson EquationJ Solution of Hankinson EquationK Typical Dimensions for Split Ring and Shear K Typical Dimensions for Split Ring and Shear

Plate ConnectorsPlate ConnectorsL Typical Dimensions for Standard Hex Bolts, Hex L Typical Dimensions for Standard Hex Bolts, Hex

Lag Screws, Wood Screws, Common, Box, Lag Screws, Wood Screws, Common, Box, and Sinker Nailsand Sinker Nails

M Manufacturing Tolerances for Rivets and Steel M Manufacturing Tolerances for Rivets and Steel Side Plates for Timber Rivet Connections Side Plates for Timber Rivet Connections

N Appendix for Load and Resistance Factor N Appendix for Load and Resistance Factor Design (LRFD)Design (LRFD) -- MandatoryMandatory


4

200520051 1 Sawn Lumber Grading AgenciesSawn Lumber Grading Agencies

NDS 2005 Supplement

2 2 Species CombinationsSpecies Combinations3 3 Section PropertiesSection Properties44 Design Values Design Values

-- Lumber and TimberLumber and Timber-- NonNon--North American Sawn LumberNorth American Sawn Lumber-- Structural Glued Laminated TimberStructural Glued Laminated Timber-- MSR and MELMSR and MEL


Overview of LRFDOverview of LRFD• Design process• Design concepts• Comparison with ASD

Design Process

Demand Demand ≤≤ CapacityCapacity


5

Design ProcessDesign Process

LoadSupport ConditionsGeometry

DemandDemandGeometryMaterialsMaterialsPerformanceFireEconomicsAesthetics….

CapacityCapacity

Design Concepts

Two Limit State concerns:

• safety against failure or collapse

• Serviceability (performance in service)

Serviceability

• Unfactored loads• Mean (avg) material

strength values

LRFD - Safety

• Factored loads• Material strength

values - modified


6

σx

σx = standard deviation

xσxx = mean

σx

SCL

ncy

Property Variability

COVx =x

Material Property Values

Visually Graded Visually Graded LumberLumber

MSR Lumber

Glulam

I-Joist

Load

Rel

ativ

e Fr

eque

n

Statistical ModelNormal Distribution Curve for Load or Resistance

Based on actual physical measurements - data sets

Statistical Model

Normal Distribution Curves for Load, S , and Resistance, R

failure

Statistical Model

Normal Distribution Curves for Safety Function, Z

fZ = fR - fS

22SRz σ+σ=σ

mZ = mR - mS

z

zmσ

=β


7

LRFD - Probability of Failure

Pf = one failure expected for x number of structures designed and built with a given β

β Pf

5.2 1 : 10,000,0004.7 1 : 1,000,0004.2 1 : 100,000and built with a given β ,3.7 1 : 10,0003.2 1 : 1,0002.7 1 : 1002.2 1 : 10

LRFD - Range on β

β Range for Wood Strength

Low Typical Highβ 2.4 2.6 2.9Pf 1 : 25 1 : 63 1 : 251

LRFD Safety Design Equation

Demand Demand ≤≤ CapacityCapacityΣΣ αα Q Q ≤≤ λλ φφ RRnn

ββ

n

i=1

Reliability Index Calibration

Example:

Bending strength of 2x8 lumber o 8 u besubjected to snow load


8

Find range of reliability index φ such that for fixed factored loads, a target β is achieved


Example:

Bending strength resistance of 2x8


es s a ce o 8lumber subjected to snow load

Example:

Bending strength resistance of 2x8


lumber subjected to snow load

Critical region for matching distribution type to test data

Find a range of reliability index, φ, such that for fixed factored loads, a target β is achieved.

15% data


%(lower tail)

φ = 0.85givesβ = 2.6 to 2.8


9

Wh t t th ASD?Wh t t th ASD?What stays the same as ASD?What stays the same as ASD?

Allowable Stress Design

• Same basic equation format

• Same adjustment factorsfactors

• Same behavioral equations

• Formatted for compatibility

What changes from ASD?What changes from ASD?

LRFD vs. ASD

• Three new notations - φ, λ, and KF• Design loads (factored) for safety are

bigger• Design loads (unfactored) for serviceability

are the same• Material resistance values are bigger• Load Duration factor changes to Time

Effect Factor


10

LRFD vs. ASD

Theoretical safety margin applied to material stresses

ASDapplied stress ≤ allowable stress

Tested material strength

Estimated loads

Design Load

Adjusted Resistance≤

Design values

LRFD vs. ASD

Member performance factor

LRFDfactored load ≤ factored resistanceLoad factors to account for variations in loads

Tested member resistance

Estimated loads

Factored Design Load

Factored Design Resistance

≤

Design values

2005 NDS LRFD Standard

Factored Load Combinations ASCE 7-02

F = flood H = hydrostatic

NDS 2005 LRFD Specification


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NDS 2005 LRFD Specificationλ tied to ASCE 7-02 Factored Load Equations:

Format Conversion Factor KF:


RN = RASDASD

RN = φ λ KF RASD

LRFD

RASD reference strengths

Format Conversion Factor KF:


RN = φ λ KF RASD

2005 NDS LRFD Specification• Why use LRFD for wood?

– ease of designing with multiple materials that use an LRFD basis (steel or concrete)

– more rational treatment of loads rather than penalizing materialrather than penalizing material strength for unknowns on loads

– realize efficiencies with:• multiple transient live loads• extreme event loads

– ASD load combinations have not been maintained in deference to LRFD load combinations


12

Q

Application - LRFD vs. ASD

Beam Example - UDL Simply Supported

A S IL

LRFD ASDDEMAND LOADS

Safety

Serviceability

wf = Σ αQ w = Σ Q

wL= Σ QL wL= Σ QL

A, S, I

LRFD ASDSHEAR

Safety Limit State 1



LRFD ASDSHEAR

wf L 2 λ φv KF F′v A2 3≤ w L 2 F′v CD A

2 3≤

Prime denotes inclusion of applicable C factors except CD

demand capacitydemand capacity

FLEXURE

Safety Limit State 2



LRFD ASD

wf L2 λ φb KF F′b S8

≤

U

Prime denotes inclusion of applicable C factors except CD

w L2 F′b CD S8

≤


LRFD ASD

DISPLACEMENT

Serviceability Limit State



LRFD ASDDISPLACEMENT

L 5 wL L4

360 384 E I≥ L 5 wL L4

360 384 E I≥

demandcapacity demandcapacity

LRFD ASD


13


Chapter 1 ‐ TerminologyBasic requirements for checking strength are revised to use terminology applicable to both ASD and LRFD

Example:“3.3.1 The actual bending stress or moment shall not exceed the adjusted allowable bending design value.” 1In equation format, this takes the standard form fb ≤ Fb'

– “allowable” (typically associated with ASD) replaced by adjusted• more generally applicable to either ASD or LRFD• better describe applying adjustment factors to reference design values

– Reference design values (Fb, Ft, Fv, Fc, Fc⊥, E, Emin) are multiplied by adjustment factors to determine adjusted design values (Fb', Ft', Fv', Fc', Fc⊥', E', Emin')

1

Chapter 1 – Design Loads

• references loads in accordance with minimum load standards such asstandards, such as ASCE 7 – 02

Chapter 2 – Adjustment Factors

• Applicable to ALL defined wood products• Adjusts from reference to site conditions

– CD, λ time‐dependentC i– CM wet service

– Ct temperature


14

Chapter 2 – Adjustment Factors

• Wet Service Factor, CM

Wet Service Conditions

20

25

30

C %

Temp 30 deg F

0

5

10

15

0 20 40 60 80 100

Relative Humidity %

Woo

d EM

C Temp 30 deg FTemp 70 deg FTemp 130 deg F

90

100

110

stur

e C

onte

nt

Impact Strength

Wet Service Conditions

40

50

60

70

80

12 14 16 18 20 22 24 26 28 30

Moisture Content of Wood (%)

%St

reng

th a

t 12%

Moi

p S gModulus of ElasticityModulus of RuptureCrushing Strength

Wet Service Factor, CM

• values found in the NDS Supplement for lumber


15

Chapter 3 – Behavioral Equations

• ASD vs LRFD – adjusted stresses from reference

ASD F′n = Fn CD x adjustment factors

LRFD F′n = Fn KF φn λ x adjustment factors


• Beams – CL beam stability

Chapter 3 – Behavioral Equations• Beams

– FbE equivalence

2b

'bE

2b

'min

bE REK

RE20.1F ==

– Emin adjusted for safety for both ASD and LRFD processes

2005 NDS

2001 NDS


• Columns– CP column stability


16

Chapter 3 – Behavioral Equations• Columns

– FcE equivalence

2e

'cE

2e

'min

cE

dl

EK

dl

E822.0F⎟⎠⎞

⎜⎝⎛

=

⎟⎠⎞

⎜⎝⎛

=

– Emin adjusted for safety for both ASD and LRFD processes

dd ⎟⎠

⎜⎝

⎟⎠

⎜⎝

2005 NDS

2001 NDS

Chapter 3 – Behavioral Equations• Emin

– FcE equivalence

66.1/))COV(645.11(E03.1E Emin −=

E = reference MOE1.03 = adjustment factor to convert E to a pure bending

basis (shear-free) (use 1.05 for glulam)1.66 = factor of safetyCOVE = coefficient of variation in MOE (NDS Appendix F)

Chapter 3 – Column Equations

LRFD ASD

Safety Limit State

Column Example – Axial Load only

LRFD ASD

P ≤ P′ P ≤ P′


(ΣαQ) ≤ A Fc KF λ φc CP CM Ct (ΣQ) ≤ A Fc CD CP CM Ct


Column Example

PDead Load = 5500 lbs

Live Load = 31500 lbs

A, S, I

Normal Time Duration

L = 16 ft (each direction)

Ends pinned

L


17


Column Example

LRFD ASDLOADS

Safety P Σ Q P = Σ QSafety P = Σ α Q= 1.2 D + 1.6 L= 1.2 (5500) + 1.6 (31500)= 57000 lbs

P = Σ Q= D + L= 5500 + 31500= 37000 lbs


Column Example Try 6-3/4″ x 9″ Glulam visually graded western species, 16F-1.3E

GEOMETRYSection X-X Y-Yd = 9 inb = 6.75 inA = 61 in2

Pinned endKed = 1.0Ld = 16 ftLed = Ked Ld

Y YPinned endKeb = 1.0Lb = 16 ftLeb = Keb Lb

X-X

Y-Y

Slenderness = max

= 28

⎟⎠⎞

⎜⎝⎛

dL,

bL edeb


Column Example

SERVICE CONDITIONS LRFD ASD

Try 6 3/4″ x 9″ Glulam visually graded western species, 16F-1.3E

Adjustment Factors

Time-dependent (normal)Wet-service (dry) CMTemperature (normal) Ct

λ = 0.81.01.0

CD = 1.01.01.0

MATERIALS


Column Example

LRFD ASD

Try 6 3/4” x 9” Glulam visually graded western species, 16F-1.3E

FcEEminc (Glulam)

φc (compression)φs (stability)KF compressionKF stability

1,550 psi1,500,000 psi780,000 psi0.9

0.900.852.16 / φc = 2.401.5 / φs = 1.76

1,550 psi1,500,000 psi780,000 psi0.9


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Column Example

LRFD ASDCAPACITY

Crushing F * F K λ φ C C F * F C C CCrushing Fc* = Fc KF λ φc CM Ct= (1,550)(2.40)(0.8)(0.9)(1.0 all)= 2,678 psi

P0 = A Fc*= (61)(2,678)= 163,382 lbs

Fc* = Fc CD CM Ct= (1,550)(1.0)(1.0 all)= 1,550 psi

P0 = A Fc*= (61)(1,550)= 94,550 lbs


Column Example

LRFD ASDCAPACITYBuckling E′min = Emin KF φs CM Ct E′min = Emin CM Ct

2

'min

cE s)Slendernes(0.822EF =

2(28))(780000)822.0(

=

= 818 psi

2

'min

cE s)Slendernes(0.822EF =

2(28))(1166880)822.0(

=

= 1,223 psi

min min F s M t min min M t

= 780,000 psi= (780,000)(1.76)(0.85)(1.0)= 1,166,880 psi


Column ExampleLRFD ASDCAPACITY

R ti 818FcE1223F Eαc Ratios1550818

FF

*c

cE =

= 0.46 = 0.53

26781223

FF

*c

cE =


Column Example

LRFD ASDCAPACITY

C11 *

cE*

cE*

cEFF2

FF

FF

⎥⎤

⎢⎡ ++

Cpc2c2c

C ccc FFFp −

⎥⎥⎦⎢

⎢⎣

−=

= 0.43 = 0.48

P′ = A Fc* Cp

= (61)(2,678)(0.43)= (61) (1,146)= 69,914 lbs

P′ = A Fc* Cp= (61)(1,550)(0.48)= (61)(744)= 45,384 lbs


19


LRFD ASDCOMPRESSION

Safety Limit State

Column Example – Axial Load only

LRFD ASDCOMPRESSIONP P’≤ P P’≤


57,000 lbs 69,914 lbs≤ 37,000 lbs 45,384 lbs≤

0.82 0.82Load / Capacity Ratio

Chapter 3 – Behavioral Equations• Tension members (tension parallel to grain)

ASD F′t = Ft CD x adjustment factors

LRFD F′t = Ft KF φt λ x adjustment factors


• wood and tension perpendicular to grain – Not recommended per NDS 3.8.2

initiators:• notches• moment connections• hanging loads


• Combined bi‐axial bending and axial compression


20


• Combined bending and axial ‐ compression

Chapter 3 – Behavioral Equations• Bearing perpendicular to grain

– F′c⊥= Fc⊥CM Ct Ci Cb (ASD)– F′c⊥= Fc⊥CM Ct Ci Cb Kf φc λ (LRFD)

– Cb bearing area factor

same as NDS 2001

Chapter 4 – Lumber• Design values

– Visually graded lumber– MSR / MEL– Timber– Decking

Chapter 4 – Lumber• Lumber adjustment factors– CF ‐ size factor– Cfu ‐ flat use fu

– Ci ‐ incising– CT ‐ buckling stiffness– Cr ‐ repetitive member


21

Chapter 4 – Lumber• Lumber adjustment factors– CF ‐ size factor– Cfu ‐ flat usefu

Chapter 4 – Lumber• Lumber adjustment factors– Ci ‐ incising– CT ‐ buckling stiffnessT

Chapter 4 – Lumber• Lumber adjustment factors– Cr – repetitive member

Chapter 4 – Lumber• Adjustment factors

– Cf form factor removed

Why?– derived from plastic deformation in

ll l i h Csmall clear specimens that may not be applicable to full‐size members

– applicability to standard wood products was limited (not allowed in poles & piles – it’s built into the reference design value)

Cf


22

Chapter 4 – Lumber• Example

– F′t = Ft CD CF (ASD)– F′t = Ft CF KF φt λ (LRFD)– Unincised, axially loaded

tension member in l i tnormal environment

Chapter 4 – Finger‐Jointed Lumber

• Widely accepted for use by IBC and IRC

• Interchangeable with lid l bsolid sawn lumber

with certain limitations:– HRA/NON‐HRA– Moisture– Load conditions


• HRA– Heat Resistant Adhesive– Designated on grade stamp

– Used where fire rated assemblies are requiredassemblies are required by code

• Exterior walls• Dwelling unit separations• Commercial tenant separations


• NON‐HRA– Adhesive not rated for heat resistance

– Designated on grade stamp


23


• HRA marks absent?– Treat same as NON‐HRA

??


• Other Stamp Designations– Exterior Use allowed– Structural applications are not limited

– Must meet HRA criteria in rated assemblies


• Other Stamp Designations– STUD USE ONLY or – VERT USE ONLY

– Limited to use where bending or tension stresses are of short duration


• Older Stamps– Old grade marks

• Obliterated

– New finger‐jointed grade stamps apply


24

Chapter 5 – Glued Laminated Timber

• Design values added to NDS Supplement• Reformatted glulam radial tension values• Shear values increased 10%

Chapter 5 – Glulam• Design values

– Frt radial tension

Chapter 5 – Glulam• Adjustment factors

– CV volume– Not cumulative with CL

Chapter 5 – Glulam• Adjustment factors

– Cc curvature– Applies to Fb– Curved portion of pbending member

– Not applied to straight portion of member


25

Chapter 5 – Glulam• Example

– F′c = Fc CD CP (ASD)– F′c = Fc CP KF φc λ (LRFD)– Axially loaded compression

member in normal environment

Chapter 6 – Poles & Piles• Poles ‐ post‐frame

• Piles ‐foundations

Chapter 6 – Poles & Piles• Design values

– No changes from 2001 NDS

Chapter 6 – Poles & Piles• Adjustment factors

– LRFD provisions


26

Chapter 6 – Poles & Piles• Adjustment factors

– Cu ‐ untreated– Ccs ‐ critical section– Csp ‐ single pilesp g p

Chapter 6 – Poles & Piles• Example

– F′c = Fc CD Csp (ASD)– F′c = Fc Csp KF φc λ (LRFD)– Single, axial load, treated, full

lateral support, normal i tenvironment

Chapter 7 – I‐joists

• Design values– M, V, EI, K – no changes

• Evaluation ReportsC i i d i– Contain proprietary design

Chapter 7 – I‐joists• Adjustment factors

– LRFD provisions


27

Chapter 7 – I‐Joists• Adjustment factors

– Cr = 1.0 • revised to agree with ASTM D5055‐02

• factor of 1.0 maintained for clarity transitioning from 2001 NDS

Chapter 7 – I‐joists• Example

– M′r = Mr CD (ASD)– M′r = Mr KF φb λ (LRFD)– Full lateral support, ppbending member, normal environment

Chapter 8 – Structural Composite Lumber (SCL)

• Design values in evaluation reports– Note less variability (low COV)– No changes from 2001 NDS

E l ti R t• Evaluation Reports– Contain proprietary design σ

x

x = mean

COVx =σxx

Material Property Values

Visually Graded Visually Graded LumberLumber

MSR Lumber

GlulamI-JoistSCL

Load

Rel

ativ

e Fr

eque

ncy


• Adjustment factors– CV – volume

• Not cumulative with lateral stability factor, CCL


28


• Adjustment factors–Cr = 1.04

• Cr is different than lumber (Cr lumber = 1.15) • Applied to FbApplied to Fb


• Example– F′b = Fb CD CV (ASD)– F′b = Fb CV KF φb λ (LRFD)– Full lateral support, bending

member, normal environment

Chapter 9 – Wood Structural Panels (WSP)• Design values – obtain from an approved source

– FbS– FtA– Fvtv– Fss– FcA– EI– EA– Gvtv– Fc⊥

Chapter 9 – Wood Structural Panels (WSP)

• Adjustment factors– CG ‐ grade & construction– Cs ‐ panel size– CM ‐ wet serviceM

– Ct ‐ temperature


29


• Adjustment factors– CG ‐ grade & construction


• Adjustment factors– Cs ‐ panel size– CM ‐ wet service– Ct ‐ temperaturet p


• Example– FbS′ = FbS CD (ASD)– FbS′ = FbS KF φb λ (LRFD)– Non‐structural I, >24″ width, loaded in bending, normal environment

Chapter 10 – Mechanical Connections

• Design issues• Reference design values

– Contained in subsequent chapters• Chapter 11 – dowel‐type connectors (nails, bolts, lag/wood p yp ( gscrews)

• Chapter 12 – split rings and shear plates• Chapter 13 – timber rivets

• Adjustment factors


30


• “full design value” terminology revised– Examples:

• Minimum spacing for full design value• Minimum end distance for full design value

– alternate placement descriptions ensure full design value is p p gdeveloped (local stresses)

– apply section 10.1.2 (Appendix E) to checkwood strength at connections


• “full design value” terminology revised– Examples: 10.2.2 Multiple Fastener Connections

When a connection contains two or more fasteners of the same type and similar size, each of which exhibits the same yield mode (see Appendix I), the total adjusted design value for the connection shall be the sum of the adjusted design values for each individual fastener. Local stresses in connections using multiple fasteners shall be checked in accordance with principles of engineering mechanics (see 10.1.2).

11.1.2.4 Edge distance, end distance, and fastener spacing required to develop full design values shall not be less than the requirements in be in accordance with Tables 11.5.1A-D.


10.1.2 Stresses in Members at ConnectionsStructural members shall be checked for load carrying capacity at connections in accordance with all applicable provisions of this standard including 3.1.2, 3.1.3, and 3.4.3.3.L l i i i l i lLocal stresses in connections using multiple fasteners shall be checked in accordance with principles of engineering mechanics.One method for determining these stresses is provided in Appendix E.

2005 NDS Appendix E

•• Appendix EAppendix E Local Stresses in Fastener Local Stresses in Fastener GroupsGroups

• new topic introduced in 2001 NDS– retroactive to all previous editions of the NDS


31

Local Stresses in Fastener Groups

• Closely spaced fasteners– brittle failure– lower capacity

wood failure mechanisms need to be considered in design


• Properly spaced fasteners– increased ductility– higher capacity

spread out the fasteners!


•• Appendix EAppendix E NDS Expressions– Net tension:

nettNT AFZ '' =

– Row tear-out:

∑=

=

=row

i

i

n

iRTRT

viRT

ZZ

tsFnZ

1

''

min''


•• Appendix EAppendix E NDS Expressions– Group tear-out

netgrouptbottomRTtopRT

GT AFZZZ −

−− ++= '''

'22

• applicable to ALL editions of the NDS• Appendix E available free from www.awc.org

netgrouptGT 22


32


Truss Bottom Chord and Splice• Design the bottom chord of a sawn lumber

commercial/industrial truss to support a tensile force (T) of 20,000 lbs (D+L)

Example

• Assume a dry moisture service condition, un-incised material and a load duration factor of 1.0.


Truss Bottom Chord and Splice• The chord includes connections with two rows of 7/8 inch bolts (in a 1/16 inch oversized hole) spaced per NDS Section 11.5 for full design values.

Example

Check the local stresses to verify your member size selection.

Local Stresses in Fastener GroupsSolution• Try 4x12 No.1 Hem Fir:

T′ = 24,600 lbs

Local Stresses in Fastener GroupsSolution• Single shear steel splice plate• Neglect eccentricity• Rows spaced at 1/3 depth – within NDS limits

3.75″

3.75″

3.75″


33


SolutionUsing Appendix E provisions:Net Section Tension

Net cross section area = (3.5)(11.25 - 2(0.875 + 0.0625)) = 32.8 in.2

ZNT′ = 625(32.8) = 20,500 lbs > 20,000 lbs OK

11.25”3.5″

7/8″ + 1/16″

7/8″ + 1/16″


SolutionUsing Appendix E provisions:• Row Tear-out Capacity

From the NDS Supplement

7D = 7(0.875″) = 6.125″

3.753.75″ > 1.5D = 1.31> 1.5D = 1.31″

From the NDS SupplementF′v = 150psiCritical spacing scritical is lessor of end distance (7D)or fastener spacing (4D) = 4(0.875″) = 3.5 inches

ZRT′ = nrow ni Fv′ t scritical = (2)(8)(150)(3.5)(3.5) = 29,400 lbs > 20,000 lbs OK

4D = 4(0.875″) = 3.5″


SolutionUsing Appendix E provisions• Group Tear-out Capacity

Assume: 11 25″

3.75″”

Assume:– uniform row spacing – edge distance = 3.75 inches

ZGT′ = ZRT′/2 + Ft′ Agroup-net

= (29,400) / 2 + 625(3.5) [11.25 – 2(3.75) - (0.875 + 0.0625)]= 20,850 lbs > 20,000 lbs OK

The design is still acceptable. Design is net section critical.

11.253.75″


Alternate: maximum spacing between rowsUsing Appendix E provisions:• Group Tear-out Capacity

Assume: 11.25″

1.5D = 1.31″

– uniform row spacing – edge distance = 1.31 inches

ZGT′ = ZRT′/2 + Ft′ Agroup-net= (29,400) / 2 + 625(3.5) [11.25 – 2(1.31) - (0.875 + 0.0625)]= 31,527 lbs > 20,000 lbs OK

The design is still acceptable. Capacity dramatically increases with increased row spacing.

11.25

1.5D = 1.31″


34


Alternate: minimum spacing between rows

Using Appendix E provisions• Group Tear-out Capacity

Assume:– uniform row spacing 11 25″

4.97″

1.5D = 1.31” – inter-row distance = 1.31 inches

ZGT′ = ZRT′/2 + Ft′ Agroup-net= (29,400) / 2 + 625(3.5) [1.31]= 17,566 lbs < 20,000 lbs NG

The design is unacceptable. Spacing between rows is too tight!

11.254.97″


• Adjustment factors– Cg– CΔ

C– Cd– Ceg– Cst– Cdi– Ctn

Group Action Factor, CgMultiple fastener connections• accounts for load distribution within the connection• Split rings, shear plates, dowels < 1″• tabulated values still in the NDS• can calculate Cg if outside tabulated range

Group Action Factor, Cg

Cg definitions• row of fasteners

– 2 or more split ring or shear plate connector units aligned in the direction of load

– 2 or more bolts of same diameter loaded in2 or more bolts of same diameter loaded in direction of load

– 2 or more lag screws of same type and size loaded in direction of load


35


• Equation method

( )( )[ ] ⎥⎦⎤

⎢⎣⎡

−+

⎥⎦

⎤⎢⎣

⎡+−++

−=

mR

mmmRnmmC EA

nnEA

n

g 11

111)1(

2

2

where:

( )( )[ ]⎦⎣ EA

ss

mm

mm

ssEA AE

AEAEAER or oflessor the=

12 −−= uum

⎥⎦

⎤⎢⎣

⎡++=

ssmm AEAEsu 112

1 γ


• Load / slip modulus, γ (lb/in.)

D = diameter of bolt or lag screw (in.) γ (lb/in)

Bolts, lag screws: wood-to-metal connections

(270,000)(D1.5)

Bolts, lag screws: wood-to-wood connections

(180,000)(D1.5)

2-1/2″ split ring & 2-5/8″ shear plate

400,000

4″ split ring & 4″ shear plate

500,000


• Equation method ExampleFind Cg for two rows of 1″ diameter bolts spaced 4″ apart in a wood-to-wood double shear splice connection using 2x12’s for main and side members

E 1400000 iE 1400000 iW d D t

γ 1.8 105×lbfin

=γ 180000lbf

in2.5D1.5⋅:=

D 1in:=

n 10:=s 4in:=

Am 16.875in2=

Am

As0.5=As 2 1.5⋅ in 11.25⋅ in:=Am 1.5in 11.25⋅ in:=

Es 1400000psi:=Em 1400000psi:=Wood Data

Fastener Data

Load / Slip


• Equation method Example


36


• Tabulated values

Am = gross x-sectional area of main member, in2

As = sum of gross x-sectional areas of all side members, in2


• Tabulated Values Example

• As/Am > 1.0, so use Am/As= 0.5 to enter column 1 of the table

• also, use Am for column 2 according to Note 1 (Am = 16.875 in2)

• read across to column for 10 fasteners in a row

• interpolate Cg = 0.665 ≈ 0.669 from calculation

Group Action Factor , Cg

Not applicable here - unit loads acting along the length of the member are not axial

Anchor Bolts andWashers as required

Bottom Plate


• Wet Service Factor, CM


37


Saturated

19% MC

• Dowel-type connectors• bolts• drift pins• drift bolts• lag screws

Dry

CM 1.0 0.7 0.4* Lateral load (*CM=0.7 for D<1/4″)1.0 0.7 1.0 Withdrawal load - lag & wood screws only1.0 0.25 0.25 Withdrawal load - nails & spikes

• lag screws• wood screws• nails

fabrication MCin-service MC

Saturated

19% MC

CCMM = 1.0 if:= 1.0 if:

1 fastener


Dry

CM 0.4 Lateral load (D>1/4″)

fabrication MCin-service MC

2+ fasteners

split splice plates

Fasteners on Common Splice Plate

Good detailing to allow for shrinkage

Chapter 11 – Dowel‐type Fasteners

ASD and LRFD accommodated through Table 10.3.1– Bolts– Lag screws– Wood screws– Wood screws– Nails & spikes


38

Fastener Values

Included in U.S. design literatureFastener Type Reference Bolts NDS or NER Lag Screws NDS or NER

dWood Screws NDS or NER Nails & Spikes NDS or NER Split Ring Connectors NDS Shear Plate Connectors NDS Drift Bolts & Drift Pins NDS Metal Plate Connectors NER Hangers & Framing Anchors NER Staples NER

National Evaluation Reports (NER) are developed for proprietary products

Fastener Bending Yield

Fastener Bending Yield Test

Center-Point Bending Test

Load

Dowel Bearing Strength


39

Yield Limit Equations

•4 Modes•Single & double shear

•Wood-to-wood•Wood-to-Steel•Wood-to-Concrete

Yield Limit Equations

Reduction Factors, Rd

I

II

III

IV

Nail Types and Designations

Nails and nomenclature• short• box nail• ring nail• ring nail• common nail• sinker• power‐driven• roofing

Nail Types and DesignationsNail types described in Appendix L


40

Nail Types and DesignationsNail capacity tables in 2005 NDS

Penetration Depth Factor, Cd• Removed in 2001 NDS • Penetration is built in to Yield Limit Equations:

– m main member dowel bearing length– s side member dowel bearing length – Rt = m / s

applies to lag screws wood screws and nailsapplies to lag screws, wood screws, and nails• Tables for lag screws, wood screws, and nails:

– Based on 8D, 10D, and 10D penetrations, respectively– Reduced penetration? Use Table Footnotes

Fastener Penetration

Fastener Type Minimum

Lag Screws 4D

Wood Screws 6DWood Screws 6D

Nails & Spikes 6D

NDS 11.3.4 “…The length of dowel bearing shall notinclude the length of the tapered tip of a fastener forfastener penetration lengths less than 10D.”

Penetration ‐ Nails

• P = 6D (NDS 11.1.5.5)6D


41

Penetration – Wood Screws• P = 6D (NDS 11.1.4.6) ‐ includes tip

6D

Penetration – Lag Screws

• P = 4D (NDS 11.1.3.6) ‐ does not include tip4D

Dowel DiameterReduced Body Diameter• Dr used to calculate capacity regardless of shear plane location (NDS 11.3.6.1)

ar p

lane

shea

Dr

Lateral Capacity – Lag ScrewsFull Body Diameter• Dr used to calculate capacity if threads are in or near the shear plane (NDS 11.3.6.2)

use Dr

shear plane


42

Lateral Capacity – Lag Screws

• Why no threads near the shear plane?Because the induced maximum moment can occur in the threads if the shear plane is not located sufficiently into the shank away from the threads

Lateral Capacity – Lag Screws

• Info on where to locate the shear plane

Lateral Capacity – Wood Screws

Rolled Thread• Dr used to calc capacity if shear plane is anywhere (NDS 11.3.6.1)

shea

r pla

ne

Lateral Capacity – Wood Screws

Cut Thread• Dr used to calc capacity if shear plane is in the threads or a little past (NDS 11.3.6.3) (can use D otherwise see TR12 for details)use Duse DD otherwise – see TR12 for details)

shea

r pla

ne

use Druse D

~ 3D to 4Dr


43

Toe‐nailing

Withdrawal– Ctn = 0.67

LateralLateral– Ctn = 0.83

Chapter 11 – Connections• Capacity definitions

– Z||– Zm⊥

– Zs⊥Z– Z⊥

Spacing, End, & Edge Distances

Nail Capacity Calculations

• Example: Shear Wall Chord Tie with Nails– ASD & LRFD


44

Nailed Tension Tie

Design connection ties between first and second floor shear wall chords. Floor framing consists of

How many nails for this connection?A653

g9.5” deep pre‐fabricated wood I‐joists. Walls are 2x6, dry Douglas Fir‐Larch studs spaced at 16″ o.c. The specified wind overturning force is 2.4 kips.

Nailed Tension Tie – LRFD & ASD

Try:• ASTM A653 Grade metal strap16 gage x 2.5” wide• 2 rows staggered 10d common nails


2005 NDS Table 11P

Z = 116 lbs

Nailed Tension Tie – LRFD & ASD2005 NDS Table 11P, Table 10.3.1, and Appendix N

LRFD ASDADJUSTMENT FACTORSTime dependentWet service CMTemperature Ct

λ= 1.0 Table N3 CD = 1.61.0 1.01 0 1 0Temperature Ct

Group Action CgGeometry CΔEnd grain CegDiaphragm CdiToe nail CtnPenetration 3” > 10D

Format conversion KFResistance φz

1.0 1.01.0 1.01.0 1.01.0 1.01.0 1.01.0 1.01.0 1.0

2.16 / φz = 3.32 Table N10.65 Table N2


45


LRFD ASD

DEMAND – Wind Overturning

w = αot Wot= (1.6)(2,400)

w = Wot= 2 400 lbs (1.6)(2,400)

= 3,840 lbs 2,400 lbs


LRFD ASD

CAPACITY

Z’ = Z φ λ KF (all C factors)

Safety Limit State

Z’ = Z C (all C factors)Z = Z φz λ KF (all C factors) = (116)(0.65)(1.0)(3.32)(1.0)= 250 lbs

Z = Z CD (all C factors) = (116)(1.6)(1.0)= 186 lbs

Nails Needed

n = w / Z’ = (3,840) / (250)= 15.4 → 16 nails

n = w / Z’ = (2,400) / (186)= 12.9 → 13 nails


CAPACITYSafety Limit State

LRFD ASDw Z’≤ W Z’≤w Z≤ W Z≤


3,840 lbs 4,000 lbs≤ 2,400 lbs 2,976 lbs≤

0.96 0.81Demand / Capacity Ratio

16 Nails 13 Nails


• Why the ASD / LRFD discrepancy?– 2005 NDS format conversion does not benefit LRFD in the wind‐only case

– Real benefits are realized with combinedReal benefits are realized with combined multiple transient live loads (e.g. wind + snow + live)


46

Connection Calculator

AWC.org

Chapter 12 – Split Rings and Shear Plates

• Capacity tables ‐ unchanged

Availability:www.clevelandsteel.com

Chapter 13 – Timber Rivets

• Capacity tables and details ‐ unchanged


• Many applications

Photos courtesy of Gary Williams, Timber Systems Limited.


47


Rivets• Steel AISI 1035• Rockwell Hardness C32-39• Fu = 145 ksi• Hot-dipped galvanized

Plates• Steel ASTM A36• Hot-dipped galvanized if in

wet service

Timber Rivet System

timber rivet cross section

Timber Rivet System

perforated steel plates

Timber Rivet System

One‐ or two‐sided connectionb/2

pp

b

Metalside plates


48

Timber Rivet System

loading to grainLoad parallel to grain

es

sq

epep

Load perpendicular to grain

es

sp

a q

Rivit rows

e s

sp

a p

P

Rivit rows

Q

es

ep

sq

eq

es

Timber Rivet System

angle to grain valuesLoad at angle to grain

ep

sp

s

e

a

q

q

Metal side plates

Rivit rows

N

q

Rows align parallel with the direction of loading on the plate

Timber Rivets ‐ Design• Four strength limit states:

– Rivet yielding• Pr – parallel to grain• Qr – perpendicular to grain

– Wood failure• Pw – parallel to grain• Qw – perpendicular to grain

– Plate yielding• Enhanced ductility

Lowest value governs designPhoto courtesy of Gary Williams, Timber Systems Limited.

Timber Rivets ‐ Design• Design process is iterative• Variables:

– Plate thickness– Rivet lengthRivet length– Rivet spacing parallel to wood grain– Number of rows of rivets– Number of rivets in each row


49

Timber Rivets ‐ Design1. Assume a trial design

• Must accommodate rivet array• Tabulated minimum edge and end distances

(Table 13.3.2 and Figure 13A)

Timber Rivets ‐ Design2. Check rivet yield

• Pr equation (13.2‐1) ‐ parallel to grain

Pr = 280 p0.32 nR nCp = rivet length – plate thickness – 1/8″p = rivet length plate thickness 1/8

• Qr equation (13.2‐2) ‐ perpendicular to grain

Qr = 160 p0.32 nR nC

Capacities above are based on a single rivet through a single plate.

Timber Rivets ‐ Design3. Check wood capacity

• Pw Tables (13.2.1A ‐ F) ‐ parallel to grain

Capacities in table are for one ¼″ plate with associated rivets.


• Qw equation (13.2‐3) ‐ perpendicular to grain

Qw = qw p0.8 CΔ

qw = Table 13.2.2A

CΔ = Table 13.2.2B



50


• Perpendicular to grain

qw = Table 13.2.2Aqw



• Geometry Factor

CΔ = Table 13.2.2BΔ

• Based on ratio:

qc

p

S)1n(e−

Timber Rivets ‐ Design4. Find governing strength limit state capacity

• Min (Pr, Qr, Pw, Qw)

Pr or Qr governing means rivet yield = ductilityr r g g y yPw or Qw governing means wood capacity = less

ductility

Timber Rivets ‐ Design5. Determine

adjusted capacity for site conditions P′ or Q′


51

Timber Rivets ‐ Design5. Determine adjusted

capacity for site conditions P′ or Q′• Table 10.3.1 – check

for applicability of:• CM Wet service

(Table 10.3.3)




(Table 10.3.3)• Ct Temperature

(Table 10.3.4)




(Table 10.3.3)• Ct Temperature

(Table 10.3.4)• Cst Metal Side Plate

Factor (Table 13.2.3)


capacity for site conditions P′ or Q′• Table 10.3.1 – check for

applicability of:• CM Wet service (Table

10 3 3) P′ = P n C C C10.3.3)• Ct Temperature (Table

10.3.4)• Cst Metal Side Plate

Factor (Table 13.2.3)

P′ = Pgoverning nP CM Ct Cst

Q′ = Qgoverning nP CM Ct


52

Timber Rivets ‐ Design6. Determine final capacity

• ASD(P′ or Q′) CD CD (Table 2.3.2)

Note: CD applies only if wood strength governs!D

• LRFDλ φz KF (P′ or Q′) λ φz KF (Appendix

N)

Timber Rivets ‐ Design7. Calculate demand / capacity ratios

demand 0.1capacitydemand

< OK!

Timber Rivets ‐ Design7. Design no good?

0.1capacitydemand

> NG!

try:– Add more rivets– Increase rivet spacing parallel to grain– Increase plate thickness– Increase rivet length

and repeat the trial starting at Step 1.

Timber Rivets – Design Examples

• Tension Splice

• Beam Hanger


53

Timber Rivets – Example 1• Tension Splice

– Determine the number of rivets required for this wet service tension splice. The glulam members are 3"x9" Southern Pine. The tensile force is due to dead plus snow load. pUse 1‐1/2" rivets and 3/16" steel side plates, and the 2005 NDSprovisions. Note that the designer must also check member capacity and rivet plate capacity.

Timber Rivets – Example 1




54




Solution



55




Timber Rivets – Example 2• Beam Hanger

– Design the beam hanger connection shown using timber rivets. The glulam beam and girder are untreated Douglas Fir glulam. The factored beam reaction i d d d l l d dis due to dead plus snow loads, and service conditions are wet. Use 2‐1/2" rivets and 1/4" plate, and 2005 NDSprovisions. Note that the designer must also confirm the bearing area for the roof beam support, and consider possible uplift conditions in the design of this connection.


56






57






58






59






60




DES110:DES110:Connection DesignConnection Design

••www.awc.orgwww.awc.org••More comprehensiveMore comprehensive••CEU’s availableCEU’s available


61

• enabling language for shear wall and diaphragm design

• design information and values in:

Chapter 14 – Shear Walls and Diaphragms

values in:ANSI / AF&PA SDPWS-2005 standard

ANSI / AF&PA SDPWS‐2005

• WIND & SEISMIC standard– Special design provisions for wind and

seismic loads– Values for a wide variety of panel products

200520051 Designer Flowchart1 Designer Flowchart2 General Design Requirements2 General Design Requirements3 Members and Connections3 Members and Connections4 Lateral Force Resisting Systems4 Lateral Force Resisting SystemsAppendix AAppendix AReferencesReferencesCommentaryCommentary

Chapter 15 – Special Loading

• Built‐up columns– Revised to correct limitation on short built‐up columns

15.3.2.2…. Each ratio shall be used to calculate a column stability factor, CP, per section 15.3.2.4 and the smaller CPshall be used in determining the allowable compression design value parallel to grain, Fc', for the column. Fc' for built-up columns need not be less than Fc' for the individual laminations designed as individual solid columns per section 3.7.

Chapter 16 – Fire Design of Wood Members

Applies to ASD only


62

Chapter 16 – Fire (ASD)• Fire resistance up to

two hours– Columns– Beams – Tension Members

C bi d L di– Combined Loading

• Additional special provisions for glulam

Chapter 16 – Fire (ASD)

TR10 - Calculating the Fire Resistance of Exposed Wood MembersE d f l d d•Expands uses for large, exposed wood

members

•Expands applicability of current methods to other EWP’s (SCL)

•Expands use of large, exposed wood members to 2 hour fire endurance applications

Chapter 16 – Fire

• Superior fire performance of heavy timbers– attributed to the charring effect of wood

B fit f h i• Benefits of charring– an insulating char layer is formed – protects the core of the section

Chapter 16 – Fire

• Experimental charring rates measured in various parts of the world appear to be consistent– North America ‐ Standard fire endurance test ASTM E‐119

many other countries comparable fire exposure in ISO 834– many other countries ‐ comparable fire exposure in ISO 834

• Effects of fire on adhesives– synthetic glues used in the manufacture of glulam do not adversely

affect performance


63

Mechanics‐Based Design Method

• expands the use of large exposed wood members:– loading conditions

– fire exposures

– mechanical propertiesmechanical properties

– stress interactions

– expanded range of wood products

Design Considerations

• predicts reduced cross‐sectional dimensions

• adjusts for charring at the corners

• accounts for the loss of strength and stiffness in the heated zone

Analog for Cross‐Sectional Dimensions

Estimating Cross‐sectional Dimensions due to Charring

• 4‐Sided Exposure (i.e. columns) b = B - 2βt d = D - 2βt

• 3‐Sided Exposure (i.e. beams) b = B - 2βt d = D - βt

• 2‐Sided Exposure (i.e. decking) b = B - βt d = D - βt

where:β is the char rate of the materialt is the fire exposure time


64

Model for Charring of Wood

• Nonlinear char model used ‐ nominal linear char rate input.

• To account for rounding at corners and reduction of strength and stiffness of the heated zone, the nominal char rate values, βn, are increased 20%.

βeff = 1.2 βn

t 0.187

where:βeff is the effective char rate (in/hr), adjusted for exposure time, t

βnis the nominal linear char rate (in/hr), based on 1‐hr exposuret is the exposure time (hrs)

Effective Char Rates and Char Layer Thickness

Required Fire Effective Char Effective Char LayerEndurance Rate, βeff Thickness, αchar

(hr) (in/hr) (in)

(for βn = 1.5 inches/hour)

1-Hour 1.80 1.81½-Hour 1.67 2.52-Hour 1.58 3.2

Design for Member Capacity

Dead Load + Live Load ≤ K * Allowable Design Capacity

where:K is a factor to adjust from allowable design capacity to

average ultimate capacity

Allowable Design Stress to Average Ultimate Strength Adjustment Factor

Member Capacity KBending Moment Capacity, in-lb. 2.85Tensile Capacity lb 2 85Tensile Capacity, lb. 2.85Compression Capacity, lb. 2.58Beam Buckling Capacity, lb. 2.03Column Buckling Capacity, lb. 2.03


65

Predicted Time vs. Fire Test Observed Time

120

140

160

min

utes

)

Mechanics-Based Model PredictionWood Beams Exposed on 3-Sides

0

20

40

60

80

100

120

Pre

dict

ed T

ime

to F

ailu

re (m

0 20 40 60 80 100 120 140 160

Observed Time to Failure (minutes)

• Douglas fir glulam beams– Span L = 18 feet– Spaced at s = 6 feet

Fire Design Example (ASD)

• Design Load– qlive = 100 psf– qdead = 15 psf

• Timber decking nailed to the compression edge of beams provides lateral bracing

Size the beam for required bending strength for 1 hour fire duration

For the structural design of the beam, calculate the induced moment:

• Beam load:wtotal = s (qdead + qlive) = (6’)(15+100) = 690 plf


• Induced demand moment:Mmax = wtotal L2 / 8 = (690)(18)2 / 8 = 27,945 ft-lb

Select a 6-3/4” x 12” 24F-V4 Douglas-fir glulam beamTabulated bending stress, Fb, equal to 2400 psi

Calculate the beam section modulus:


Ss = BD2/6 = (6.75)(12)2 / 6 = 162.0 in3

Calculate the adjusted allowable bending stress:Assuming: CD = 1.0, CM = 1.0, Ct = 1.0, CL = 1.0, CV = 0.99F’b = Fb CD CM Ct (lesser of CL or CV)

= 2400(1.0)(1.0)(1.0)(0.99) = 2371 psi


66

Calculate the design resisting moment:M’ = F’b Ss = (2371)(162) / 12= 32,009 ft-lb

Structural Capacity Check: M’ > Mmax


32,009 ft-lb > 27,945 ft-lb

For the fire design of the wood beam:• the loading is unchanged• therefore, the maximum moment is unchanged• the fire resistance must be calculated


From NDS Table 16.2.1, find charring depth αchar for 1 hour duration:

Required Fire Effective Char Effective Char LayerEndurance Rate, βeff Thickness, αchar

(hr) (in/hr) (in) 1-Hour 1.80 1.8

1½-Hour 1.67 2.5 2-Hour 1.58 3.2


Substitute in residual cross-section dimensions for 3-sided beam into the section modulus relation, i.e.:

• 3-Sided Exposure (i.e. beams) b = B - 2βt d = D - βtB 2 D= B - 2αchar = D - αchar

Calculate charred beam section modulus exposed on 3-sides:Sf = (B-2αchar)(D- αchar)2 / 6 = (6.75 - 2(1.8))(12-1.8)2 / 6

= 54.6 in3


Calculate the adjusted allowable bending stress (some adjustment factors don’t apply and may have been other than 1.0 before):F’b = Fb(lesser of CL or CV) = 2400 (0.99) = 2371 psi

C l l t t th i ti t i h d tiCalculate strength resisting moment using charred cross-section:M’ = K F’b Sf = (2.85)(2371)(54.6) / 12 = 30,758 ft-lb

Fire Capacity Check: M’ > Mmax

30,758 ft-lb > 27,945 ft-lb


67

2005 NDS Fire Design

• Full-scale test results indicate that the mechanics-based method conservatively estimates the fire endurance time of all exposed wood members.

• Given the theoretical derivation of the new mechanics based design• Given the theoretical derivation of the new mechanics-based design method, it is easily incorporated in current wood structural design provisions.

• Incorporation of new mechanics-based method in the NDS assists in the proper design of all exposed wood members for standard fire exposures.

20052005A Construction and Design PracticesA Construction and Design PracticesB Load DurationB Load DurationC Temperature EffectsC Temperature EffectsD Lateral Stability of BeamsD Lateral Stability of BeamsE Local Stresses in Fastener GroupsE Local Stresses in Fastener GroupsF Design for Creep and Critical Deflection F Design for Creep and Critical Deflection

ApplicationsApplications

Appendices Layout

Important!

ppppG Effective Column LengthG Effective Column LengthH Lateral Stability of ColumnsH Lateral Stability of ColumnsI Yield Limit Equations for ConnectionsI Yield Limit Equations for ConnectionsJ Solution of Hankinson EquationJ Solution of Hankinson EquationK Typical Dimensions for Split Ring and Shear K Typical Dimensions for Split Ring and Shear

Plate ConnectorsPlate ConnectorsL Typical Dimensions for Standard Hex Bolts, Hex L Typical Dimensions for Standard Hex Bolts, Hex

Lag Screws, Wood Screws, Common, Box, Lag Screws, Wood Screws, Common, Box, and Sinker Nailsand Sinker Nails

M Manufacturing Tolerances for Rivets and Steel M Manufacturing Tolerances for Rivets and Steel Side Plates for Timber Rivet Connections Side Plates for Timber Rivet Connections

N Appendix for Load and Resistance Factor N Appendix for Load and Resistance Factor Design (LRFD)Design (LRFD) -- MandatoryMandatory

Appendix N new!• Load and Resistance Factor Design

– source for new variablesASTM D5457 – Standard Specification for Computing the Reference Resistance of Wood‐Based Materials and Structural Connections for Load and Resistance Factor Design

– tabulates K conversion factors to convert from ASD reference values (see– tabulates KF conversion factors to convert from ASD reference values (see NDS Supplement) to LRFD reference values

– tabulates resistance factors φ

– tabulates time effect factors λ for load combinations listed in:ASCE 7‐02 – Minimum Design Loads for Buildings and Other Structures• NDS clarified for cases involving hydrostatic loads (H) and for cases where H is not in combination with L, use λ = 0.6

2005 NDS Supplement

• Updated to include latest reference values for:– visually graded lumber and timber– mechanically graded lumber

glued laminated timber– glued laminated timber


68

2005 NDS Supplement ‐ Emin

• Emin addition for reference MOE for beam and column stability:– visually graded lumber and timber

mechanically graded lumber– mechanically graded lumber– glued laminated timber

• Represents 5% lower exclusion shear‐free E value so that design value adjustments are not part of the basic design equation for column and beam stability

2005 NDS Supplement ‐ Lumber

Visually graded dimension lumber (Table 4A)• Four new species added:

– Alaska cedar (Alaska & Western states)– Alaska Hemlock (Alaska & Western states)( )– Alaska Yellow Cedar (Alaska‐grown only)– Baldcypress

2005 NDS Supplement ‐ Timber

Visually graded timber (Table 4D)• Two new species added:

– Alaska cedar (Alaska & Western states)– Baldcypressyp

2005 NDS Supplement – Foreign Species

non‐North American Species (Table 4F)• Several new species added:

– Montane pine (South Africa)– Norway Spruce (Romania and the Ukraine)y p ( )– Silver fir (Germany, NE France, and Switzerland)– Southern pine (Misiones Argentina)– Southern pine (Misiones Argentina free of heart center and

medium grain density


69

2005 NDS Supplement ‐MSR and MEL

Mechanically graded dimension lumber (Table 4C)• New design values added:

– Table 4C Footnote 2 – new G, Fv, Fc⊥ values for MSR and MELTable 4C new E values for MSR and MEL– Table 4C new Emin values for MSR and MEL

2005 NDS Supplement ‐ Glulam

Structural glued laminated timber (Table 5A)• New design values added:

– Table 5A new Emin values added– Table 5A 16F stress class – revised Ft, Fc, G– Table 5A values now match Table 5A‐Expanded values

• Species groups for split ring and shear plate connectors removed (NDS Table 12A values inappropriate) – use G of the wood located on the face receiving the connector with NDS Table 12A assignment of species group.

• Fv values increased for prismatic members (Footnote d revised) – use of test‐based shear values removing the 10% reduction used previously (AITC and APA).

• Fv values increased for non‐prismatic members unchanged (AITC and APA).

• Non‐prismatic Frt (radial tension) for D.fir‐L, and SP glulam increased slightly

2005 NDS Supplement ‐ Glulam

Structural glued laminated timber (Table 5B)• New combinations added for Southern Pine with more

information on slope of grain differences.• Fbx design values reformatted to include footnoted table

adjustments for special tension laminations.• Fvy columns consolidated and values updated with Table 5A

info.

Changes from previous editions

• A new format for the future – ASD and LRFD• NDS is one volume!!

= + +


70

2005 Wood Design Package

• ANSI/AF&PA NDS‐2005 National Design Specification (NDS) for Wood Construction with Commentary and Supplement

• ANSI/AF&PA SDPWS‐2005 – Special Design Provisions for Wind and Seismic ‐ withProvisions for Wind and Seismic with Commentary

• ASD/LRFD Manual for Engineered Wood Construction

• Structural Wood Design Solved Example Problems (Workbook)

Special Design Provisions for Wind and Seismic

• WIND & SEISMIC standard references NDS provisions– Shear wall and diaphragm design– Values for a wide variety of panelValues for a wide variety of panel products

– Commentary included

ASD/LRFD Manual

• Non‐mandatory information– All products covered by NDS– 40 connection details– Comprehensive fire chapter

• Manual Chapters correspond to NDS Chapters

Example Problems ‐Workbook

• 40 examples updated to the 2005 NDS

• ASD and LRFD


71

NDS 2005 Summary• format changes to accommodate addition of LRFD :

– Revised terminology– Expanded applicability of adjustment factor tables– Re‐format of radial tension design values– Revised format of beam and column stability provisions (addition of

Emin property)– Addition of NDS Appendix N – Load and Resistance Factor Design

• other changes introduced in the 2005 Edition:– Removal of form factor– Revision of repetitive member factor for I‐joists– Revision of full‐design value terminology– Clarification of built‐up column provisions

NDS 2005 Supplement Summary

• changes in design value tables :– Emin values added for all materials– Fv values for prismatic glulam increased– minor re‐formattingg– updated to include latest reference values for:

• visually graded lumber and timber• mechanically graded lumber • glued laminated timber

2005 Wood Design Package

This concludes The American Institute of Architects Continuing

Education Systems Course

Questions?

Wood Products Council 866.966.3448 [email protected]

Presented by: Scott Lockyear, PEPhone/Text: 615-439-0624

[email protected]/in/scottlockyear

Presented by: Scott Lockyear, PEPhone/Text: 615-439-0624

[email protected]/in/scottlockyear

Learning Objectives - · PDF fileLRFD vs. ASD • Three new notations -φ,λ, and...

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