American Wood Council
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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
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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
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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
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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
Outline• Document organization• Overview of LRFD Concept• Chapter‐by‐chapter description• Changes from previous editions• Examples
Overview of LRFDOverview of LRFD• Design process• Design concepts• Comparison with ASD
Design Process
Demand Demand ≤≤ CapacityCapacity
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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
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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
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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
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Find range of reliability index φ such that for fixed factored loads, a target β is achieved
Reliability Index Calibration
Example:
Bending strength resistance of 2x8
Reliability Index Calibration
es s a ce o 8lumber subjected to snow load
Example:
Bending strength resistance of 2x8
Reliability Index Calibration
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
Reliability Index Calibration
%(lower tail)
φ = 0.85givesβ = 2.6 to 2.8
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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
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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:
NDS 2005 LRFD Specification
RN = RASDASD
RN = φ λ KF RASD
LRFD
RASD reference strengths
Format Conversion Factor KF:
NDS 2005 LRFD Specification
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
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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
Application - LRFD vs. ASD
Beam Example - UDL Simply Supported
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
Application - LRFD vs. ASD
Beam Example - UDL Simply Supported
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
≤
demand capacitydemand capacity
LRFD ASD
DISPLACEMENT
Serviceability Limit State
Application - LRFD vs. ASD
Beam Example - UDL Simply Supported
LRFD ASDDISPLACEMENT
L 5 wL L4
360 384 E I≥ L 5 wL L4
360 384 E I≥
demandcapacity demandcapacity
LRFD ASD
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Outline• Document organization• Overview of LRFD Concept• Chapter‐by‐chapter description• Changes from previous editions• Examples
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')
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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
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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
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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
Chapter 3 – Behavioral Equations
• 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
Chapter 3 – Behavioral Equations
• Columns– CP column stability
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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′
demand capacitydemand capacity
(ΣαQ) ≤ A Fc KF λ φc CP CM Ct (ΣQ) ≤ A Fc CD CP CM Ct
Chapter 3 – Column Equations
Column Example
PDead Load = 5500 lbs
Live Load = 31500 lbs
A, S, I
Normal Time Duration
L = 16 ft (each direction)
Ends pinned
L
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Chapter 3 – Column Equations
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
Chapter 3 – Column Equations
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
Chapter 3 – Column Equations
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
Chapter 3 – Column Equations
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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Chapter 3 – Column Equations
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
Chapter 3 – Column Equations
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
Chapter 3 – Column Equations
Column ExampleLRFD ASDCAPACITY
R ti 818FcE1223F Eαc Ratios1550818
FF
*c
cE =
= 0.46 = 0.53
26781223
FF
*c
cE =
Chapter 3 – Column Equations
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
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Chapter 3 – Column Equations
LRFD ASDCOMPRESSION
Safety Limit State
Column Example – Axial Load only
LRFD ASDCOMPRESSIONP P’≤ P P’≤
demand capacitydemand capacity
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
Chapter 3 – Behavioral Equations
• wood and tension perpendicular to grain – Not recommended per NDS 3.8.2
initiators:• notches• moment connections• hanging loads
Chapter 3 – Behavioral Equations
• Combined bi‐axial bending and axial compression
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Chapter 3 – Behavioral Equations
• 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
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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
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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
Chapter 4 – Finger‐Jointed Lumber
• 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
Chapter 4 – Finger‐Jointed Lumber
• NON‐HRA– Adhesive not rated for heat resistance
– Designated on grade stamp
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Chapter 4 – Finger‐Jointed Lumber
• HRA marks absent?– Treat same as NON‐HRA
??
Chapter 4 – Finger‐Jointed Lumber
• Other Stamp Designations– Exterior Use allowed– Structural applications are not limited
– Must meet HRA criteria in rated assemblies
Chapter 4 – Finger‐Jointed Lumber
• Other Stamp Designations– STUD USE ONLY or – VERT USE ONLY
– Limited to use where bending or tension stresses are of short duration
Chapter 4 – Finger‐Jointed Lumber
• Older Stamps– Old grade marks
• Obliterated
– New finger‐jointed grade stamps apply
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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
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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
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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
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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
Chapter 8 – Structural Composite Lumber (SCL)
• Adjustment factors– CV – volume
• Not cumulative with lateral stability factor, CCL
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Chapter 8 – Structural Composite Lumber (SCL)
• Adjustment factors–Cr = 1.04
• Cr is different than lumber (Cr lumber = 1.15) • Applied to FbApplied to Fb
Chapter 8 – Structural Composite Lumber (SCL)
• 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
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Chapter 9 – Wood Structural Panels (WSP)
• Adjustment factors– CG ‐ grade & construction
Chapter 9 – Wood Structural Panels (WSP)
• Adjustment factors– Cs ‐ panel size– CM ‐ wet service– Ct ‐ temperaturet p
Chapter 9 – Wood Structural Panels (WSP)
• 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
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Chapter 10 – Mechanical Connections
• “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
Chapter 10 – Mechanical 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.
Chapter 10 – Mechanical Connections
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
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Local Stresses in Fastener Groups
• Closely spaced fasteners– brittle failure– lower capacity
wood failure mechanisms need to be considered in design
Local Stresses in Fastener Groups
• Properly spaced fasteners– increased ductility– higher capacity
spread out the fasteners!
Local Stresses in Fastener Groups
•• Appendix EAppendix E NDS Expressions– Net tension:
nettNT AFZ '' =
– Row tear-out:
∑=
=
=row
i
i
n
iRTRT
viRT
ZZ
tsFnZ
1
''
min''
Local Stresses in Fastener Groups
•• 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
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Local Stresses in Fastener Groups
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.
Local Stresses in Fastener Groups
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″
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Local Stresses in Fastener Groups
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″
Local Stresses in Fastener Groups
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″
Local Stresses in Fastener Groups
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″
Local Stresses in Fastener Groups
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″
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Local Stresses in Fastener Groups
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″
Chapter 10 – Mechanical Connections
• 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
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Group Action Factor, Cg
• 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
Group Action Factor, Cg
• 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
Group Action Factor, Cg
• 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
Group Action Factor, Cg
• Equation method Example
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Group Action Factor, Cg
• Tabulated values
Am = gross x-sectional area of main member, in2
As = sum of gross x-sectional areas of all side members, in2
Group Action Factor, Cg
• 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
Chapter 10 – Mechanical Connections
• Wet Service Factor, CM
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Wet Service Factor, CM
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
Wet Service Factor, CM
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
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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
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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
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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
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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
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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
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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
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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
Nailed Tension Tie – LRFD & ASD
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
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Nailed Tension Tie – LRFD & ASD
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
Nailed Tension Tie – LRFD & ASD
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
Nailed Tension Tie – LRFD & ASD
CAPACITYSafety Limit State
LRFD ASDw Z’≤ W Z’≤w Z≤ W Z≤
demand capacitydemand capacity
3,840 lbs 4,000 lbs≤ 2,400 lbs 2,976 lbs≤
0.96 0.81Demand / Capacity Ratio
16 Nails 13 Nails
Nailed Tension Tie – LRFD & ASD
• 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)
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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
Chapter 13 – Timber Rivets
• Many applications
Photos courtesy of Gary Williams, Timber Systems Limited.
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Chapter 13 – Timber Rivets
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
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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
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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.
Timber Rivets ‐ Design3. Check wood capacity
• Qw equation (13.2‐3) ‐ perpendicular to grain
Qw = qw p0.8 CΔ
qw = Table 13.2.2A
CΔ = Table 13.2.2B
Capacities in table are for one ¼″ plate with associated rivets.
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Timber Rivets ‐ Design3. Check wood capacity
• Perpendicular to grain
qw = Table 13.2.2Aqw
Capacities in table are for one ¼″ plate with associated rivets.
Timber Rivets ‐ Design3. Check wood capacity
• 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′
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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)
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)• Ct Temperature
(Table 10.3.4)
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)• Ct Temperature
(Table 10.3.4)• Cst Metal Side Plate
Factor (Table 13.2.3)
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) 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
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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
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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
Timber Rivets – Example 1
Timber Rivets – Example 1
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Timber Rivets – Example 1
Timber Rivets – Example 1
Timber Rivets – Example 1
Solution
Timber Rivets – Example 1
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Timber Rivets – Example 1
Timber Rivets – Example 1
Timber Rivets – Example 1
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.
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Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
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Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
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Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
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Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
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Timber Rivets – Example 2
Timber Rivets – Example 2
Timber Rivets – Example 2
DES110:DES110:Connection DesignConnection Design
••www.awc.orgwww.awc.org••More comprehensiveMore comprehensive••CEU’s availableCEU’s available
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• 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
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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
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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
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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
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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
Fire Design Example (ASD)
• 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:
Fire Design Example (ASD)
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
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Calculate the design resisting moment:M’ = F’b Ss = (2371)(162) / 12= 32,009 ft-lb
Structural Capacity Check: M’ > Mmax
Fire Design Example (ASD)
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
Fire Design Example (ASD)
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
Fire Design Example (ASD)
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
Fire Design Example (ASD)
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
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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
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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
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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!!
= + +
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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
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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