NDS Structural Wood Design Examples 2015/2018 Edition

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EXAMPLES Structural Wood Design Examples 2015/2018 EDITION Updates and Errata While every precaution has been taken to ensure the accuracy of this document, errors may have occurred during development. Updates or Errata are posted to the American Wood Council website at www.awc.org. Technical inquiries may be addressed to [email protected].

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2015/2018 Structural Wood Design Examples

First Web Version: August 2019

ISBN 978-1-940383-51-4

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Copyright Permission American Wood Council 222 Catoctin Circle, SE, Suite 201 Leesburg, VA 20175 [email protected] Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS iii

FOREWORD

This document is intended to aid instruction in structural simple to complex and cover many design scenarios. In design of wood structures using both Allowable Stress the solutions where a particular provision of the NDS Design (ASD) and Load and Resistance Factor Design or SDPWS is cited, reference is made to the document (LRFD). It contains design examples and complete and corresponding provision number, e.g. NDS 4.3.1. solutions calculated using ASD and LRFD. Solutions have It is intended that this document be used in conjunction been developed based on the 2015 and 2018 National with competent engineering design, accurate fabrication, Design Specification®(NDS®) for Wood Construction, and adequate supervision of construction. Neither the and the 2015 Special Design Provisions for Wind and American Wood Council, the International Code Council, Seismic (SDPWS), as appropriate. References are also nor their members assume any responsibility for errors or made to the 2015 and 2018 Wood Frame Construction omissions in this document, nor for engineering designs, Manual (WFCM) for One- and Two- Family Dwellings. plans, or construction prepared from it. Those using this Copies of these standards produced by the American document assume all liability arising from its use. The Wood Council can be obtained at www.awc.org/codes- design of engineered structures is within the scope of standards/publications. Procedures will be applicable for expertise of licensed engineers, architects, or other licensed both 2015 and 2018 versions of the NDS and WFCM, professionals for applications to a particular structure. unless otherwise noted. Example problems range from American Wood Council

Solutions have been developed using the MathCAD® 15 software by PTC (https://www.ptc.com/). Some formatting is the result of the program layout, for example the use of “:=” denotes an assigned value, while an “=” denotes a calculated value. Examples may contain notes or comments for instances where program constraints have led to the use of non-standardized terms or subscripts.

TABLE OF CONTENTS

Problem Page Problem Page Foreword...... iii Connection Problem Set...... 80 2.1a Withdrawal - Plain Shank Nail 80 Characteristic Problem Set...... 1 2.1b Fastener Uplift - Roof Sheathing Ring 1.1 Adjustment Factors (ASD and LRFD) 1 Shank Nail - 5/8" WSP (2018 NDS Only) 82 1.2a Simply Supported Beam Capacity Check 2.1c Fastener Uplift - Roof Sheathing Ring (ASD) 7 Shank Nail - 7/16" WSP (2018 NDS Only) 84 1.2b Simply Supported Beam Capacity Check 2.2 Common Nail Lateral - Single Shear (LRFD) 11 Wood-to-wood 86 1.3 Glued Laminated Timber Beam Design 2.3 Withdrawal - Lag Screw 91 (ASD) 16 2.4 Wood Screw Lateral - Double Shear 1.4 Compression Members - 4x4 and 6x6 Wood-to-wood 93 (ASD) 23 2.5 Bolt Lateral - Single Shear Wood-to-Wood 97 1.5a Compression Member - 2x6 Stud (ASD) 28 2.6 Bolted Wood-to-Wood Tension Splice 1.5b Compression Member - 2x6 Stud (LRFD) 31 Connection 105 1.6 Bending and Axial Tension (ASD) 34 1.7 Bending and Axial Compression (ASD) 40 Shear Wall Problem Set...... 110 1.8 Bi-Axial Bending and Axial Compression 3.1 Segmented Shear Wall - Wind 110 (ASD) 45 3.2 Segmented Shear Wall - Seismic 114 1.9 Loadbearing Wall Wood Stud Resisting 3.3 Perforated Shear Wall - Wind 116 Wind and Gravity Loads 55 3.4 Perforated Shear Wall - Seismic 119

1.1 Adjustment Factors (ASD and LRFD) 1.1

E1.1 - Adjustment Factors (ASD and LRFD) ADJUSTMENT FACTORS (ASD AND LRFD)

A No. 1 Douglas Fir-Larch (DF-L) nominal 2x6 is used for a floor joist @ 16" o.c. (supporting only dead and live loads) for an exterior deck. The in-service moisture content is greater than 19%, and the member is not subject to elevated temperatures. The ends are held in place with full-depth blocking. Determine the adjusted bending design value (F'b) , adjusted shear design value (F'v), adjusted tension design vaue (F't) and modulii of elasticity (E' and Emin') for the member using both Allowable Stress Design (ASD) and Load and Resistance Factor Design (LRFD). Assume lumber is incised.

E1.1 - Adjustment Factors (ASD and LRFD)

Reference and Adjusted Design Values for No. 1 DF-L 2x6 (NDS Supplement Table 4A)

Fb  1000 psi E 1700000 psi Emin  660000psi (NDS Supplement Table 4A)

Ft  800 psi Fv  180psi

Determine Adjusted Bending Design Value (F'b) using ASD factors

Load Duration Factor

CD  1.0 (NDS Table 2.3.2) Wet Service Factor

CMb  0.85 (NDS Supplement Table 4A Adjustment factors) Temperature Factor

Ct  1.0 (NDS Table 2.3.3) Beam Stability Factor

CL  1.0 (NDS 3.3.3.2 and 4.4.1.2)

Size Factor

CF  1.3 (NDS Supplement Table 4A) Flat Use Factor

Cfu  1.0 (NDS Supplement Table 4A) Incising Factor

Ci  0.8 (NDS Table 4.3.8)

Repetitive Member Factor

Cr  1.15 (NDS 4.3.9)

F'bASD  FbCDCMbCtCLCFCfuCiCr

F'bASD  1017 psi Adjusted ASD Bending Design Value (F'b)

Determine Adjusted Bending Design Value (F'b) using LRFD factors

Format Conversion Factor ADJUSTMENT FACTORS (ASD AND LRFD)

KF  2.54 (NDS Table 4.3.1) Resistance Factor  0.85 (NDS Table 4.3.1) ϕ Time Effect Factor (NDS Appendix N.3.3 - use combination 1.2D+1.6L+0.5(L or S or R), L  0.8 r λ from occupancy)

F'  F C C C C C C C K   bLRFD b Mb t L F fu i r F ϕ λ

F'bLRFD  1756 psi Adjusted LRFD Bending Design Value (F'b)

Determine Adjusted Shear Parallel to Grain Design Value (F'v) using ASD factors Load Duration Factor

CD  1.0 (NDS Table 2.3.2) Wet Service Factor

CMV  0.97 (NDS Supplement Table 4A)

Temperature Factor

Ct  1.0 (NDS Table 2.3.3)

Incising Factor

Ci  0.8 (NDS Table 4.3.8)

F'vASD  FvCDCMVCtCi Adjusted ASD Shear Parallel to Grain Design Value (F'v) F'vASD  140 psi

Determine Adjusted Shear Parallel to Grain Design Value (F'v) using LRFD factors

Format Conversion Factor

KFv  2.88 (NDS Table 4.3.1) Resistance Factor  0.75 (NDS Table 4.3.1) ϕv Time Effect Factor  0.8 (NDS Appendix N.3.3 - use combination 1.2D+1.6L+0.5(L or S or R), λ r L from occupancy)

F'vLRFD  FvCMVCtCiKFv v ϕ λ Adjusted LRFD Shear Parallel to Grain Design Value (F'v)

F'vLRFD  241 psi

Determine Adjusted Tension Parallel to Grain Design Value (F't) using ASD factors Load Duration Factor

CD  1.0 (NDS Table 2.3.2) Wet Service Factor

CMt  1.0 (NDS Supplement Table 4A)

Temperature Factor

Ct  1.0 (NDS Table 2.3.3)

Size Factor

CF  1.3 (NDS Supplement Table 4A) Incising Factor

Ci  0.8 (NDS Table 4.3.8)

F'tASD  FtCDCMtCtCFCi Adjusted ASD Tension Parallel to Grain Design Value (F't) F'tASD  832 psi

Determine Adjusted Tension Parallel to Grain Design Value (F't) using LRFD factors

Format Conversion Factor

KFt  2.70 (NDS Table 4.3.1) Resistance Factor  0.8 (NDS Table 4.3.1) ϕt Time Effect Factor  0.8 (NDS Appendix N.3.3 - use combination 1.2D+1.6L+0.5(L or S or R), L λ r from occupancy)

F'tLRFD  FtCMtCtCFCiKFt t ϕ λ Adjusted LRFD Tension Parallel to Grain Design Value (F't)

F'tLRFD  1438 psi

Determine Adjusted Modulus of Elasticity (E') (same for ASD and LRFD)

Wet Service Factor

CME  0.9 (NDS Supplement Table 4A) Temperature Factor

CtE  1.0 (NDS Table 2.3.3)

Incising Factor ADJUSTMENT FACTORS (ASD AND LRFD)

CiE  0.95 (NDS Table 4.3.8)

E' E CMECtECiE

E' 1453500 psi Adjusted ASD/LRFD Modulus of Elasticity (E')

Determine Beam and Column Stability Adjusted Modulus of Elasticity (Emin') with ASD Factors

Wet Service Factor

CME  0.9 (NDS Supplement Table 4A) Temperature Factor

CtE  1.0 (NDS Table 2.3.3)

Incising Factor

CiE  0.95 (NDS Table 4.3.8) Buckling Stiffness Factor

CT  1.0 (NDS 4.4.2)

E'minASD  EC MECtECiECT

E'minASD  1453500 psi Adjusted ASD Beam and Column Stability Modulus of Elasticity (Emin')

Determine Beam and Column Stability Adjusted Modulus of Elasticity (Emin') with LRFD Factors

Wet Service Factor

CME  0.9 (NDS Supplement Table 4A) Temperature Factor

CtE  1.0 (NDS Table 2.3.3)

Incising Factor

CiE  0.95 (NDS Table 4.3.8) Buckling Stiffness Factor

CT  1.0 (NDS 4.4.2)

Format Conversion Factor

KFE  1.76 (NDS Table 4.3.1) Resistance Factor  0.85 (NDS Table 4.3.1) ϕE

E'  EC C C C K  minLRFD ME tE iE T FE ϕE

E'minLRFD  2174436 psi Adjusted LRFD Beam and Column Stability Modulus of Elasticity (Emin')

1.2a Simply Supported Beam Capacity Check (ASD)

E1.2a - Simply Supported Beam Capacity Check (ASD) 1.2A

bending. Assume normal load duration. The beam is supported on a 2x4 top plate. Lateral support is provided SIMPLY SUPPORTED BEAM CAPACITY CHECK (ASD) only at the ends of the member and the ends are considered pinned.

Check beam's capacity to resist shear stress from maximum (moment controlled) load; determine deflection from maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations.

E1.2a - Simply Supported Beam Capacity Check (ASD)

A Select Structural Douglas Fir-Larch (DF-L) nominal 4X16 beam on a 20 ft span supports a hoist located at the center of the span. Determine the maximum allowable load on the hoist (including its weight) based on bending. Assume normal load duration. The beam is supported on a 2x4 top plate. Lateral support is provided only at the ends of the member and the ends are considered pinned.

Check beam's capacity to resist shear stress from maximum (moment controlled) load; determine deflection from maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations.

Reference and Adjusted Design Values for 4x16 Select Structural DF-L (size adjusted 4x12 values)

Fb  1500 psi E 1900000 psi Emin  690000 psi (Table 4A) Fc  625 psi Fv  180 psi ⊥ CD  1.0 CM  1.0 Ct  1.0 Cfu  1.0 Cr  1.0 Ci  1.0 CT  1.0 CF  1.0 (Table 4A 14" and wider)

E' E CMCtCi E'min  EminCMCtCiCT E' 1900000 psi E'min  690000 psi

Member dimensions and properties

l 20 ft b 3.5 in d 15.25 in wbearing  3.5 in (width of bearing) 3 2 bd bd I  A  bd S  12 g 6 2 3 4 Ag  53.38 in S 135.66 in I 1034 in

Beam Stability Factor

F'b*  FbCDCMCtCFCiCr F'b* is adjusted bending design value with all adjustment factors except the beam stability factor CL and flat use factor F'b*  1500 psi Cfu applied.

in l  12 l l  240 in Laterally unsupported length u ft u lu  15.7 lu/d >7 (Table 3.3.3) d

le  1.37 lu  3d le  375 in (Table 3.3.3)

led R  R  21.6 Slenderness ratio for bending (3.3-5) 1.2A B 2 B b

1.20 E'min F  F  1776 psi Critical bucking design value for bending (3.7.1) SIMPLY SUPPORTED BEAM CAPACITY CHECK (ASD) bE 2 bE RB

2  FbE    FbE   FbE  1    1       F'b*    F'b*   F'b*  C      L 1.9  1.9  0.95

CL  0.876

F'b  F'b*CfuCL

F'b  1313 psi Adjusted bending design value with all adjustment factors

Determine Maximum Moment Allowed on Beam

Maximum total moment is the adjusted bending design value F'b times the section modulus S S M  F'  M  14849 ftꞏlbf max b in max 12 ft

Determine Maximum Hoist Load P Maximum hoist load P is determined from subtracting moment due to beam weight from the maximum total moment allowed on the beam and solving for hoist load P. Load P creates a moment on beam length L of PL/4. Assume density of beam material is 37.5 lbs/ft3 (110% of tabulated of the specific gravity G for Southern Pine).

lbf b d  37.5 w    w  13.9 plf Note: ρ 3 beamweight ρ in in beamweight ft 12 12 wbeamweight is self weight of beam ft ft 2  wbeamweight ()l Mbeamweight is moment due to self M  M  695 ftꞏlbf beamweight 8 beamweight weight Mallow is maximum allowable Mallow  Mmax  Mbeamweight Mallow  14154 ftꞏlbf moment due to applied hoist load

Mallow P4 l

Result: The total allowable concentrated moment-limited midspan load (hoist plus payload) is P 2831 lbf

Check Beam's Capacity to Resist Shear from Maximum (bending controlled) Load E1.2b - Simply Supported Beam Capacity Check (LRFD) P V  V 1415 lbf 2 A Select Structural Douglas Fir-Larch (DF-L) nominal 4X16 beam on a 20 ft span supports a hoist located at the 3V center of the span. Determine the maximum allowable load on the hoist (including its weight) based on f  f  40 psi bending. Assume load combination 1.2D+1.6L applies (λ=0.8). The beam is supported on a 2x4 top plate. v 2b d v Lateral support is provided only at the ends of the member and the ends are considered pinned. F'v  FvCDCMCtCi F'v  180 psi fv

Note: NDS Section 4.3.12 allows Fc to be increased by Cb as specified in Section 3.10.4. That increase was ⊥ not used in this example.

Check Deflection Total deflection is the combination of deflection from beam weight and deflection from the applied crane load. Deflection from beam weight is considered long term deflection. Deflection from crane load may be considered short-term. 4 5 wbeamweight ft  in     l 12   0.025 in Δbeam_weight 384 EI 12 in  ft  Δbeam_weight 3 P  in   l 12   0.415 in Δcrane_load 48 EI  ft  Δcrane_load    0.44 in Δtotal Δbeam_weight Δcrane_load Δtotal

Calculate Span/Deflection Ratio in 12 l ft  545 L/Δtotal ratio Δtotal

1.2b Simply Supported Beam Capacity Check (LRFD)

E1.2b - Simply Supported Beam Capacity Check (LRFD)

A Select Structural Douglas Fir-Larch (DF-L) nominal 4X16 beam on a 20 ft span supports a hoist located at the center of the span. Determine the maximum allowable load on the hoist (including its weight) based on bending. Assume load combination 1.2D+1.6L applies (λ=0.8). The beam is supported on a 2x4 top plate. 1.2B Lateral support is provided only at the ends of the member and the ends are considered pinned.

Check beam's capacity to resist shear stress from maximum (moment controlled) load; determine deflection SIMPLY SUPPORTED BEAM CAPACITY CHECK (LRFD) from maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations.

E1.2b - Simply Supported Beam Capacity Check (LRFD)

A Select Structural Douglas Fir-Larch (DF-L) nominal 4X16 beam on a 20 ft span supports a hoist located at the center of the span. Determine the maximum allowable load on the hoist (including its weight) based on bending. Assume load combination 1.2D+1.6L applies (λ=0.8). The beam is supported on a 2x4 top plate. Lateral support is provided only at the ends of the member and the ends are considered pinned.

Check beam's capacity to resist shear stress from maximum (moment controlled) load; determine deflection from maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations.

Reference and Adjusted Design Values for 4x16 Select Structural DF-L (size adjusted 4x12 values)

Fb  1500 psi E 1900000 psi Emin  690000 psi (Table 4A) Fc  625 psi Fv  180 psi ⊥  0.8 C  1.0 C  1.0 λ M t Cfu  1.0 Cr  1.0 Ci  1.0 CT  1.0 CF  1.0 (Table 4A 14" and wider)

K  2.54  0.85 Fb ϕb Subscript "b" denotes bending, "v" denotes K  2.88  0.75 shear, "c " denotes compression Fv ϕv ⊥ perpendicular to grain

KFc  1.67 c  0.90 ⊥ ϕ ⊥ K  1.76  0.85 FEmin ϕEmin E' E C C C E'  E C C C C K  M t i min min M t i T FEmin ϕEmin E' 1900000 psi E'min  1032240 psi

Member length and properties

Beam Stability Factor F'b* is adjusted bending design value with all adjustment F'  F C C C C C K   b* b M t F i r Fb ϕb λ factors except the beam stability factor CL and flat use factor Cfu applied. The following calculations determine the beam F'b*  2591 psi stabilty factor CL: Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 13

in l  12 l l  240 in Laterally unsupported length u ft u lu  15.7 lu/d >7 (Table 3.3.3) d 1.2B

le  1.37 lu  3d le  375 in (Table 3.3.3) SIMPLY SUPPORTED BEAM CAPACITY CHECK (LRFD)

led R  R  21.6 Slenderness ratio for bending (3.3-5) B 2 B b

1.20 E'min F  F  2657 psi Critical bucking design value for bending (3.7.1) bE 2 bE RB

2  FbE    FbE   FbE  1    1       F'b*    F'b*   F'b*  C      L 1.9  1.9  0.95

CL  0.827

F'b  F'b*CfuCL

F'b  2143 psi F'b is adjusted bending design value with all adjustment factors.

Determine Maximum Moment Allowed on Beam

Maximum total moment is the adjusted bending design value F'b times the section modulus S S M  F'  M  24230 ftꞏlbf max b in max 12 ft

Note: lbf  37.5 b d wbeamweight is self weight of beam ρ 3 w    ft beamweight ρ in in 12 12 ft ft Mbeamweight is moment due to self weight M is maximum allowable w  13.9 plf allow beamweight moment due to applied hoist load 2 1.2wbeamweight()l Mbeamweight  Mbeamweight  834 ftꞏlbf 1.2 factor for load combination 1.2D+1.6L 8 applied here

Mallow  Mmax  Mbeamweight

Mallow  23396 ftꞏlbf

Mallow P4 1.6 factor for load combination 1.2D+1.6L l 1.6 applied here Result: The total allowable concentrated moment-limited midspan load (hoist plus payload) is P 2925 lbf

Check Beam's Capacity to Resist Shear from Maximum (bending controlled) Load 1.6P V  1.6 factor applied here for live load V 2340 lbf 2 3V f  f  66 psi v 2b d v F'  F C C C K   F'  311 psi f

F'c  Fc CMCtCiKFc  c F'c  939 psi fc < F'c OK ⊥ ⊥ ⊥ ϕ ⊥ ⊥ ⊥ ⊥

Note: NDS Section 4.3.12 allows Fc to be increased by Cb as specified in Section 3.10.4. That increase was not used in ⊥ this example.

3 SIMPLY SUPPORTED BEAM CAPACITY CHECK (LRFD) P  in   l 12   0.429 in Δcrane_load 48 EI  ft  Δcrane_load    0.454 in Δtotal Δbeam_weight Δcrane_load Δtotal

Calculate Span/Deflection Ratio in 12 l ft  529 L/Δtotal ratio Δtotal

1.3 Glued Laminated Timber Beam Design (ASD)

E1.3 - Glued Laminated Timber Beam Design (ASD) E1.3 - Glued Laminated Timber Beam Design (ASD)

Design a simple roof supporting beam spanning 32 ft, with 5000 lb loads (1000 lb dead load (DL) + 4000 lb Design a simple roof supporting beam spanning 32 ft, with 5000 lb loads (1000 lb dead load (DL) + 4000 lb snow load (SL)) applied by purlins at 8 ft on center (at 1/4 points plus the ends). Member has lateral supports snow load (SL)) applied by purlins at 8 ft on center (at 1/4 points plus the ends). Member has lateral supports at the ends and compression edge supports at the purlin locations. Beam supports are 6 inches long. at the ends and compression edge supports at the purlin locations. Beam supports are 6 inches long. Assume dry service conditions. Temperature is less than 100 degrees (F) but occasionally may reach 150 Assume dry service conditions. Temperature is less than 100 degrees (F) but occasionally may reach 150 degrees (F). Use 24F-1.8E structural glued laminated (glulam) Southern Pine timber. degrees (F). Use 24F-1.8E structural glued laminated (glulam) Southern Pine timber.

Reference and Adjusted Design Values for 24F-1.8E structural glued laminated softwood timber

Fbx+  2400 psi Fbx-  1450 psi (NDS Supplement Table 5A) Fc x  650 psi Fvx  265 psi ⊥ Ft  1100 psi Fc  1600 psi Ex  1800000 psi Exmin  950000 psi CD  1.15 CM  1.0 Ct  1.0 Temp up to 150 degrees F only occasionally Cfu  1.0 Cc  1.0 Cb  1.0 CI  1.0 Cvr  1.0 (NDS Table 5.3.1)

E'x  ExCM Ct E'xmin  ExminCMCt Note: Ex notation has changed to Ex app for 2018 NDS

E'x  1800000 psi E'xmin  950000 psi

Member length and properties l  32 ft b  5 in d  30.25 in Initial iteration

Note: Beam length designated as lower case l instead of upper case L used in the Specification nomenclature 2 3 bd bd A  bd S  I  l  6 in g xx 6 xx 12 support 2 3 4 Ag 151.3 in Sxx 762.6 in Ixx 11534 in

Beam Stability Factor

Fbx+*  Fbx+CDCMCtCcCI F'b* is adjusted bending design value with all adjustment factors except the beam stability factor CL flat use factor Cfu Fbx+*  2760 psi and volume factor CV applied.

in l  12 8 ft l 96 in Laterally unsupported length u ft u

le  1.54 lu le 148 in (Table 3.3.3)

led R  R  13.375 Slenderness ratio for bending (3.3-5) B 2 B b

E1.3 - Glued Laminated Timber Beam Design (ASD)

Design a simple roof supporting beam spanning 32 ft, with 5000 lb loads (1000 lb dead load (DL) + 4000 lb snow load (SL)) applied by purlins at 8 ft on center (at 1/4 points plus the ends). Member has lateral supports at the ends and compression edge supports at the purlin locations. Beam supports are 6 inches long. Assume dry service conditions. Temperature is less than 100 degrees (F) but occasionally may reach 150 degrees (F). Use 24F-1.8E structural glued laminated (glulam) Southern Pine timber.

Notes: 1.3 Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations. GLUED LAMINATED TIMBER BEAM DESIGN (ASD) Reference and Adjusted Design Values for 24F-1.8E structural glued laminated softwood timber

E'x  ExCM Ct E'xmin  ExminCMCt Note: Ex notation has changed to Ex app for 2018 NDS

E'x  1800000 psi E'xmin  950000 psi

Member length and properties l  32 ft b  5 in d  30.25 in Initial iteration

Beam Stability Factor

in l  12 8 ft l 96 in Laterally unsupported length u ft u

le  1.54 lu le 148 in (Table 3.3.3) led R  R  13.375 Slenderness ratio for bending (3.3-5) B 2 B b

1.20 E'xmin F  F  6373 psi Critical bucking design value for bending (3.3.3.8) bE 2 bE RB

2 Note: CL must be ≤ 1.0 per NDS 5.3-1  FbE    FbE   FbE  1    1       Fbx+*    Fbx+*   Fbx+*  C      L 1.9  1.9  0.95

CL  0.965

Volume Factor

x 20 Southern Pine

 1    1 1 1 C  min  V  x x x   21 ft   12 in   5.125 in          l   d   b  

CV  0.936 CL and Cv shall not apply simulatenously (5.3.5). CV is less than CL. CV controls Adjusted Bending Design Value

  CL   F'  F min C  b  bx+*   fu   CV 

F'b  2584 psi F'b is adjusted bending design value with all adjustment factors.

Assume Beam Weight and Determine Section Modulus Required to Resist Bending

Wood density can be estimated based on NDS Supplement 3.1.3. Maximum total moment is the adjusted bending design value F'b times the section modulus, S lbf w  40 Estimated self weight of beam beamweight ft

P  5000 lbf Load from purlin  2   l wbeamweightl  in M  P  12 M 1021440 inꞏlbf est  2 8  ft est

Mest S  3 reqd S 395 in S is < 762.6 in3 so try a smaller F'b reqd reqd section for efficiency

Try a 5 X 22 member

b2  5 in d2  22 in Beam dimension for trial section. Subscript "2" used to denote second iteration 2 b d 3 2 2  A  b  d S  b2 d2 g2 2 2 xx2 6 I  xx2 12

2 3 4 1.3 Ag2 110 in Sxx2 403.3 in Ixx2 4437 in GLUED LAMINATED TIMBER BEAM DESIGN (ASD)

led2 R  R  11.406 Slenderness ratio for bending (3.3-5) B2 2 B2 b2 1.20 E'xmin F  F  8763 psi Critical bucking design value for bending (3.3.3.8) bE2 2 bE2 RB2

2  FbE2    FbE2   FbE2  1  1         Note: C must be ≤ 1.0 per NDS 5.3-1  Fbx+*    Fbx+*   Fbx+*  L C      L2 1.9  1.9  0.95

CL2  0.978

 1    1 1 1    CV2 min x x x  21 ft   12 in   5.125 in           l   d2   b2  

CV2  0.951 CL and Cv shall not apply simulatenously (5.3.5). CV is less than CL. CV controls Adjusted Bending Design Value

  CL2   F'  F min C  b2  bx+*   fu   CV2  

F'b2  2625 psi F'b2 is adjusted bending design value with all adjustment factors for the trial section considered. lbf w  30 beamweight2 ft Self weight for 5 X 22 beam estimated per NDS Supplement 3.1.3  2   l wbeamweight2l  in M  P  12 2  2 8  ft

M2 1006080 inꞏlbf Rtotal  12988 lbf

F'  F C C M2 c x c x M t S  ⊥ ⊥ reqd2 F' b2 F'c x  650 psi 3 ⊥ S 383 in 383 in3 < 403 in3 reqd2 Rtotal Okay fc  ⊥ b2lsupport Shear Parallel to Grain f  433 psi c Actual compression stress perpendicular to grain is less than The two 5000 pound purlin loads at the ends of the beam are within "d" of the supports and can be ignored for shear ⊥ the adjusted compression perpendicular to grain design value. (3.4.3.1(a)). Shear determined from remaining purlin loads OK 3P At Purlins Vpurlins  2 Purlins are supported by saddle style hangers that transfer commpressive loads to the top of the beam. Determine the area of the hangers required to support each purlin without creating actual compression stresses V  7500 lbf purlins greater than the adjusted compression perpendicular to grain design value  l  1  P Vbeamweight2  wbeamweight2  d2  lsupport Ahanger  2  2  F'c x ⊥ Vbeamweight2  417.5 lbf 2 Ahanger 7.69 in

Vtotal  Vpurlins  Vbeamweight2 Assuming that purlins frame in from both sides of the beam, the width of the hanger can be calculated as follows:

V  7918 lbf total Ahanger whanger  0.50 b2 Adjusted (F v ') and Actual (f v ) Shear Parallel to Grain whanger 3.08 in Cvr  1.0 (NDS 5.3.10)

F'v  FvxCDCMCtCvr

F'v  305 psi 3 Vtotal fv   2 b2d2 fv < F'v Actual shear stress parallel to grain less than adjusted OK fv  108 psi Compression Perpendicular to Grain At Bearing Ends The bearing ends of the beam transmit all the purlin loads so the two 5000 pound purlin loads at the ends of the beam are included in the bearing load calculations 1 R  5 P purlins 2

Rpurlins  12500 lbf 1   lsupport  R  w l2   beamweight2 2 beamweight2   2  Rbeamweight2  487.5 lbf

Rtotal  Rpurlins  Rbeamweight2

M2 1006080 inꞏlbf Rtotal  12988 lbf

F'  F C C M2 c x c x M t S  ⊥ ⊥ reqd2 F' b2 F'c x  650 psi 3 ⊥ S 383 in 383 in3 < 403 in3 reqd2 Rtotal Okay fc  ⊥ b2lsupport

Shear Parallel to Grain 1.3 f  433 psi c Actual compression stress perpendicular to grain is less than The two 5000 pound purlin loads at the ends of the beam are within "d" of the supports and can be ignored for shear ⊥ the adjusted compression perpendicular to grain design value. (3.4.3.1(a)). Shear determined from remaining purlin loads OK GLUED LAMINATED TIMBER BEAM DESIGN (ASD) 3P At Purlins Vpurlins  2 Purlins are supported by saddle style hangers that transfer commpressive loads to the top of the beam. Determine the area of the hangers required to support each purlin without creating actual compression stresses V  7500 lbf purlins greater than the adjusted compression perpendicular to grain design value  l  1  P Vbeamweight2  wbeamweight2  d2  lsupport Ahanger  2  2  F'c x ⊥ Vbeamweight2  417.5 lbf 2 Ahanger 7.69 in

Vtotal  Vpurlins  Vbeamweight2 Assuming that purlins frame in from both sides of the beam, the width of the hanger can be calculated as follows:

V  7918 lbf total Ahanger whanger  0.50 b2 Adjusted (F v ') and Actual (f v ) Shear Parallel to Grain whanger 3.08 in Cvr  1.0 (NDS 5.3.10)

F'v  FvxCDCMCtCvr

Rpurlins  12500 lbf

1   lsupport  R  w l2   beamweight2 2 beamweight2   2  Rbeamweight2  487.5 lbf

Rtotal  Rpurlins  Rbeamweight2

A 3-1/8 wide purlin hanger is adequate. Note: The compression perpendicular to grain design value F' can be c⊥ increased by the bearing area factor Cb (5.3.12). For 3 inch bearing the factor is:

lb 0.375 in lb  3 in Cb  Cb  1.125 lb 1 whanger 2.735 in Cb

Using the bearing factor Cb confirms that a 3 inch wide hanger across the beam would be adequate.

At this stage of the calculations, the span of the beam can be reviewed. The 32 foot span was based on the center to center distance between supports. The length of the span used in design is the face to face distance plus 1/2 of the required bearing length at the ends (3.2.1).

In the example, the face to face distance is 32 ft minus 6 inches or 31.5 feet. At the end of the beam the required bearing distance is 12,998 lbs/(5 inches * 650 psi) or 4 inches. At the interior face, half the purlin load is assumed to be transferred to the beam end. Required length in bearing is (2500 lbs + 7500 lbs + 488 lbs)/5 inches * 650 psi) or 3.25 inches. The two required bearing lengths and the face to face distance produces a span of 31.5 ft + 1/2 (4.00/12) + 1/2 (3.25/12) or 31.8 feet which 99.4% of the center to center span. The shorter span reduces moments and bending stresses by 1.25%. The reduction is considered insufficient to allow the use of the next smaller beam.

Deflection The specification does not include specific deflection limits for roofs. In some applications, deflections may be critical and the designer may wish to limit deflections. For this example, a deflection limit of L/240 has been selected.

Dead load deflection is usually calculated to determine the desired camber of the beam. The recommended camber is usually 150% of the dead load deflection. Deflection for the 5000 lb concentrated loads and the beam weight is: 4  wbeamweight2   in  3 5 l 12   in  in  ft  19 P12 l   12  ft ft       Δpurlin Δbeamweight 384 E'xIxx2 384E'xIxx2 1.754 in 0.089 in  0.375in Δpurlin Δbeamweight Δcamber    Δtotal Δpurlin Δbeamweight Δcamber Note: for this example 3/8" camber will be specified 1.468 in Δtotal

Length/Deflection Ratio

in l 12 ft  262 L/Δ >240. The length deflection ratio satisfies specified criteria Δtotal

1.4 Compression Members and Column Stability Calculation (ASD)

E1.4 - Compression Members and Column Stability Calculation (ASD)

Compare the axial compression capacity of a nominal 4x4 and nominal 6x6 post being used for an interior column (only carrying gravity loads - dead load (DL) + floor live load (LL)). Both members are No. 2 Southern Pine and have a length of 10 feet. Both ends are assumed to be pinned (Ke=1.0 - NDS 3.7.1.2). Assume all members are loaded concentrically. 1.4 COMPRESSION MEMBERS AND COLUMN CALCULATION STABILITY (ASD)

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E1.4 - Compression Members and Column Stability Calculation (ASD)

Reference and Adjusted Design Values - 4x4 Post

Fc  1450psi (NDS Supplement Table 4B) E 1400000psi

Emin  510000psi

CF  1.0 Size factor (NDS Supplement Table 4B)

CM  1.0 Moisture factor (NDS Supplement Table 4B) Ct  1.0 Temperature factor (NDS Table 2.3.3)

C  1.0 i Incising factor (NDS Table 4.3.8)

CT  1.0 Buckling Stiffness factor (NDS 4.4.2)

CD  1.0 Load Duration factor (NDS Table 2.3.2)

6 E' E CMCtCi E' 1.4 10 psi 5 E'min EminCMCiCtCT E'min 5.1 10 psi

d 3.5in Actual member dimensions = 3.5" x 3.5"

2 Area 12.25in

Ke  1.0

Length 120in

le  Ke Length

le 120 in le  34.3 Needs to be less than 50 (NDS 3.7.1.3) d

E1.4 - Compression Members and Column Stability Calculation (ASD) Column Stability Factor Calculation Compare the axial compression capacity of a nominal 4x4 and nominal 6x6 post being used for an interior column (only carrying gravity loads - dead load (DL) + floor live load (LL)). Both ()0.822 E'min members are No. 2 Southern Pine and have a length of 10 feet. Both ends are assumed to be F  F 357 psi (NDS 3.7.1) cE 2 cE pinned (Ke=1.0 - NDS 3.7.1.2). Assume all members are loaded concentrically.  l   e   d  Reference and Adjusted Design Values - 4x4 Post

Fc  1450psi Fc*  FcCDCMCtCFCi Fc* 1450 psi (NDS Supplement Table 4B) E 1400000psi 1.4

csawn  0.8 (NDS 3.7.1) Emin  510000psi COMPRESSION MEMBERS AND COLUMN CALCULATION STABILITY (ASD) C  1.0 Size factor (NDS Supplement Table 4B) F 2  FcE   FcE   FcE  1   1    Moisture factor (NDS Supplement Table 4B)       CM 1.0  Fc*   Fc*   Fc*  C    C  1.0 Temperature factor (NDS Table 2.3.3) P   t 2csawn  2csawn  csawn C  1.0 i Incising factor (NDS Table 4.3.8)

 CT  1.0 Buckling Stiffness factor (NDS 4.4.2) CP 0.232

C  1.0 Load Duration factor (NDS Table 2.3.2) D Axial Buckling Capacity F'  F C C C C C C 6 c c D M t F i P E' E CMCtCi E' 1.4 10 psi 5 F'c 336 psi E'min EminCMCiCtCT E'min 5.1 10 psi

P  F'c Area d 3.5in Actual member dimensions = 3.5" x 3.5" P 4120 lbf 2 Area 12.25in Bearing Capacity Ke  1.0 fc  F'c fc  336 psi Actual compression stress parallel to grain assuming column loaded to 100% calculated Length 120in 0.75 Fc*  1088 psi buckling capacity l  K  Length e e fc is less than Fc* (NDS 3.10.1) so bearing parallel to grain is OK. fc is less than 0.75 Fc*, so no rigid bearing insert is required per NDS 3.10.1.3 le 120 in le  34.3 Needs to be less than 50 (NDS 3.7.1.3) d

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Reference and Adjusted Design Values - 6x6 Post Note: "2" subscript indicates 6x6; differentiates between 6x6 and Fc2  525psi 4x4 properties and calculations (NDS Supplement Table 4D) E2  1200000psi

Emin2  440000psi

CF2  1.0 Size factor (NDS Supplement Table 4D)

CM2  1.0 Moisture factor (NDS Supplement Table 4D)

Ct2  1.0 Temperature factor (NDS Table 2.3.3)

Ci2  1.0 Incising factor (NDS Table 4.3.8)

CT2  1.0 Buckling Stiffness factor (NDS 4.4.2)

CD2  1.0 Load Duration factor (NDS Table 2.3.2) 6 E'2  E2CM2Ct2Ci2 E'2  1.2 10 psi 5 E'min2  Emin2CM2Ci2Ct2CT2 E'min2  4.4 10 psi

d2  5.5in Actual member dimensions = 5.5" x 5.5"

2 Area2  30.25in

Ke2  1.0

Length2  120in

le2  Ke2 Length2

le2 120 in le2  21.8 d2 Needs to be less than 50 (NDS 3.7.1.3)

Column Stability Factor Calculation

0.822 E'min2 F  F 760 psi (NDS 3.7.1) cE2 2 cE2  le2    d2 

Fc2*  Fc2CD2CM2Ct2CF2Ci2 Fc2* 525 psi

(NDScsawn2 3.7.1) 0.8 1.4

2 COMPRESSION MEMBERS AND COLUMN CALCULATION STABILITY (ASD)  FcE2  FcE2  FcE2 1   1      Fc2*   Fc2*   Fc2*  CP2      2csawn2  2csawn2  csawn2

CP2  0.801

Axial Buckling Capacity

F'c2  Fc2CD2CM2Ct2CF2Ci2CP2

F'c2 421 psi

P2  F'c2 Area2

P2 12725 lbf

Bearing Capacity

fc2  F'c2 fc2  421 psi Actual compression stress parallel to grain assuming column loaded to 100% calculated buckling capacity 0.75 Fc2*  394 psi

fc is less than Fc* (NDS 3.10.1) so bearing parallel to grain is OK. fc is greater than 0.75 Fc*, so rigid bearing insert such as 20 gage metal plate is required per NDS 3.10.1.3. 4x4 Post Capacity = 4120 lbs (no bearing plate required)

6x6 Post Capacity = 12725 lbs (bearing plate required) Note: for eccentrically loaded columns, see NDS Chapter 15

1.5a Compression MemberAnalysis (ASD)

E1.5a - Compression Member Analysis (ASD)

A No 2 Spruce Pine Fir (SPF) nominal 2X6 interior bearing stud, 91.5 inches long, sheathed on both sides with gypsum board, carries dead load and snow load from the roof. Determine CP and the allowable compression parallel to grain design value (Fc') for the stud. Assume studs are placed 16" on center and top and bottom plates are the same grade and species. Determine axial loads based on buckling and bearing limit states.

E1.5a - Compression Member Analysis (ASD)

Reference and Adjusted Design Values for No. 2 SPF 2x6

Fc  1150 psi Emin 510000 psi Fc  425 psi (NDS Table 4A) ⊥

CD  1.15 CM  1.0 Ct  1.0 CF  1.1 Ci  1.0 CT  1.0 (NDS Table 4.3.1) 1.5A E'min  Emin CMCtCiCT E'min  510000 psi F'c  Fc CMCtCi F'c  425 psi ⊥ ⊥ ⊥ COMPRESSION MEMBER ANALYSIS (ASD) Member length and properties l 91.5 in b 1.5 in d 5.5 in

Column Stability Factor Fc*  FcCDCMCtCFCi Fc* is adjusted bending design value with all adjustment factors except the column stability factor CP , Fc*  1455 psi Effective lengths of compression member in planes of lateral le2  0 le1  l support. Strong axis buckling controls. See NDS A.11.3 regarding lateral support of the weak axis due to gypsum le2 le1 sheathing.  0  16.636 b d le<50 OK (NDS 3.7.1.4)

0.822 E'min F  F  1515 psi Critical bucking design value for compression cE 2 cE  l  members (3.7.1.5)  e1  d  c 0.8 Sawn lumber (3.7.1.5)

2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      Column Stability Factor (3.7-1) P 2c  2c  c

CP  0.705

F'c  Fc*CP

F'c  1025 psi F'c is adjusted compression parallel to grain design value with all adjustment factors.

Determine Axial Loads Based on Buckling and Bearing

PBuckling  bd F'c PBuckling  8458 lbf PBearing  bd F'c PBearing  3506 lbf Perpendicular to grain bearing ⊥ PBearing2  bd Fc* PBearing2  12002 lbf Parallel to grain bearing

Cb  1.25 PBearingIncreased  bd F'c Cb PBearingIncreased  4383 lbf ⊥ Controlling Value

Note: With a 3:1 snow to dead load ratio, this translates to 3,287 lbs snow load and 1,096 lbs dead load.

1.5b Compression MemberAnalysis (LRFD)

E1.5b - Compression Member Analysis (LRFD)

A No 2 Spruce Pine Fir (SPF) nominal 2X6 interior bearing stud, 91.5 inches long, sheathed on both sides with gypsum board, carries dead load and snow load from the roof (assume load combination 1.2D + 1.6S, λ=0.8). Determine CP and the allowable compression parallel to grain design value (Fc') for the stud. Assume studs are placed 16" on center and top and bottom plates are the same grade and species. Determine axial loads based on buckling and bearing limit states. 1.5B COMPRESSION MEMBER ANALYSIS (LRD)

E1.5b - Compression Member Analysis (LRFD)

Reference and Adjusted Design Values for No. 2 SPF 2x6

Fc  1150 psi Emin  510000 psi Fc  425 psi (NDS Table 4A) ⊥

 0.8 C  1.0 C  1.0 C  1.1 C  1.0 C  1.0 (NDS Table 4.3.1 and Appendix N) λ M t F i T K  2.40  0.9 K  1.76  0.85 Fc ϕc FEmin ϕEmin

Kc  1.67 c  0.9 ⊥ ϕ ⊥

E'  E C C C C K  E'  762960 psi min min M t i T FEmin ϕEmin min

F'c  Fc CMCtCiKc  c F'c  638.775 psi ⊥ ⊥ ⊥ ϕ ⊥ ⊥

Member length and properties l 91.5 in b 1.5 in d 5.5 in

Column Stability Factor F  F C C C C K   F * is adjusted bending design value with all adjustment c* c M t F i Fc ϕc λ c factors except the column stability factor CP , Fc*  2186 psi

le2  0 le1  l Effective lengths of compression member in planes of lateral support. Strong axis buckling controls le2 le1  0  16.636 <50 OK (NDS 3.7.1.4) b d 0.822 E'min F  F  2266 psi Critical buckling design value for compression cE 2 cE  l  members (NDS 3.7.1.5)  e1  d 

c 0.8 Sawn lumber (NDS 3.7.1.5)

2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      Column Stability Factor (NDS 3.7-1) P 2c  2c  c

CP  0.703 Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 33

F'c  Fc*CP

F'c  1537 psi F'c is adjusted compression design value with all adjustment factors.

Determine Axial Loads Based on Buckling and Bearing

PBuckling  bd F'c PBuckling  12683 lbf PBearing  bd F'c PBearing  5270 lbf Perpendicular to grain bearing ⊥ PBearing2  bd Fc* PBearing2  18034 lbf Parallel to grain bearing

Note: Bearing area factor (Cb) can be used to increase the bearing controlled load on interior studs. The bearing factor for the 1-1/2 bearing length measured parallel to grain is 1.25 (NDS Equation 3.10-2 and Table 3.10.4) C  1.25 b 1.5B PBearingIncreased  bd F'c Cb PBearingIncreased  6587 lbf ⊥ Controlling Value COMPRESSION MEMBER ANALYSIS (LRD)

Note: With a 3:1 snow to dead load ratio, this translates to 3,294 lbs snow load and 1098 lbs dead load.

1.6 Combined Bending and Axial Tension (ASD)

E1.6 - Combined Bending and Axial Tension Loading of a Truss Chord Member (ASD)

A No. 2 Hem-Fir nominal 2x8 is considered for use as the bottom chord of a 24-ft roof truss (12 ft between panel points). The chord will be subject to a uniform dead load of 8 psf as well as tension forces (assuming pinned connections) of 880 lb from roof wind loads (WL), 880 lb from roof live (RLL) and 1420 lb from dead loads (DL). Trusses are to be spaced 4 ft on center. Framing will have a 19% (max) moisture content. Check the adequacy of the bottom chord member for bending and tension for the appropriate load cases.

Note: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations.

E1.6 - Combined Bending and Axial Tension Loading of a Truss Chord Member (ASD)

Note: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations.

Reference and Adjusted Design Values for No. 2 Hem-Fir 2x8

Fb  850 psi E 1300000 psi Emin  470000 psi (NDS Supplement Table 4A)

Ft  525 psi CM  1.0 Ct  1.0 CF  1.2 (NDS Table 4.3.1) Cfu  1.0 Ci  1.0 Cr  1.0

CT  1.0 1.6 E' E CMCtCi E'min  EminCMCtCiCT E' 1300000 psi E'min  470000 psi COMBINED BENDING AND AXIAL TENSION (ASD)

Member length and properties l  12 ft b  1.5 in d  7.25 in 2 bd A  bd S  g 6 2 3 Ag 10.875 in S 13.141 in

Applied Loads lbf w  8 w  4 ft T  880 lbf T  880 lbf T  1420 lbf D 2 trib wind Live Dead ft

Load Case 1: DL + RLL + WL

CD  1.6 NDS Appendix B Section B.2 (non-mandatory)

Tension

Ft' FtCDCMCtCFCi Adjusted tension parallel to grain design value for short duration loads (NDS 2.3.1 and 4.3.1) Ft' 1008 psi

T1  Twind  TLive  TDead Subscripts refer to Load Case

T1 3180 lbf T1 ft1  Tensile stress in bottom chord Ag

ft1 292 psi Ft' 1008 psi Actual tension stress is less than adjusted tension parallel to design value. OK (NDS 3.8.1)

Bending F'b* is adjusted bending design value with all adjustment F'  F C C C C C C b* b D M t F i r factors except the beam stability factor CL and flat use factor Cfu applied. The following calculations determine the beam F'b*  1632 psi stabilty factor CL:

Determine Beam Stability Factor CL (NDS 3.3.3) in l  12  l l 144 in Laterally unsupported length u ft u lu  19.9 lu/d >7 (NDS Table 3.3.3) d

le  1.63 lu 3d le 256.5 in (NDS Table 3.3.3)

led R  R  28.75 RB < 50 OK (NDS 3.3.3.7) B 2 B b

1.20 E'min F  F  682 psi (NDS 3.3.3.6) bE 2 bE RB

2  FbE    FbE   FbE  1    1       F'b*    F'b*   F'b*  C      (NDS Equation 3.3-6) L 1.9  1.9  0.95

CL  0.404 Resulting beam stability factor CL.

F'b  F'b*CL Cfu F'b is the fully adjusted bending design value with all adjustment factors including the beam stability factor CL F'  660 psi b and flat use factor applied

F'b** F'b Since Cv does not apply to solid sawn lumber, F'b** is equal to F'b F'b** 660 psi 2 in w w l 12 D trib ft Bending resulting from dead load M  max 8

Mmax 6912 inꞏlbf

Mmax f  f  526 psi b S b

fb 526 psi F'b 660 psi Ok. Actual bending stress fb does not exceed adjusted bending design value F'b

Combined Bending and Axial Tension 1.6 ft1 fb   0.61 <1.0. Ok (NDS Equation 3.9-1)

Ft' F'b* COMBINED BENDING AND AXIAL TENSION (ASD) fb  ft1  0.354 <1.0 ok (NDS Equation 3.9-2) F'b**

Load Case 2: DL+RLL Load Case 3: DL only C  1.25 C  0.9 D Appendix B Section B.2 (non-mandatory). Roof Live Load is D a construction load. Tension Tension Ft' FtCDCFCMCiCt Ft' FtCDCMCtCFCi Adjusted tension parallel to grain design value for short duration loads (NDS 2.3.1 and 4.3.1) Ft' 567 psi Ft' 787.5 psi T3  TDead T2  TLive  TDead T3 1420 lbf T 2300 lbf 2 T3 ft3  T2 Ag f  t2 A g ft3 131 psi Ft' 567 psi ft2 211 psi Ft' 787 psi Bending F'b*  FbCDCMCtCFCiCr F'b*  FbCDCMCtCFCiCr F'b*  1275 psi F'  918 psi 2 b*  FbE    FbE   FbE  1  1   2        FbE    FbE   FbE   F'b*    F'b*   F'b*  1  1   C          L    F'b*    F'b*   F'b*  1.9  1.9  0.95 C      L 1.9  1.9  0.95 CL  0.509 CL  0.674 F'b  F'b*CL Cfu F'b  F'b*CL Cfu F'b 649 psi F'b 619 psi F'b** F'b F'b** F' F'b** 649 psi b F'b** 619 psi fb 526 psi F'b 649 psi fb 526 psi F'b 619 psi Combined Bending and Axial Tension ft2 fb Combined Bending and Axial Tension   0.68 <1.0. ok Ft' F'b* ft3 fb   0.8 <1.0. ok fb  ft2  0.48 <1.0 ok Ft' F'b* F'b** fb  ft3  0.64 <1.0 ok F'b** Results: No 2 Hem-Fir 2 x 8 satisfies NDS Criteria for combined bending and axial tension.

Load Case 3: DL only

CD  0.9

Tension

Ft' FtCDCFCMCiCt Ft' 567 psi

T3  TDead T3 1420 lbf T3 ft3  Ag

ft3 131 psi Ft' 567 psi

Bending

F'b*  FbCDCMCtCFCiCr

F'  918 psi

b* 1.6 2  FbE    FbE   FbE  1    1       F'b*    F'b*   F'b*  COMBINED BENDING AND AXIAL TENSION (ASD) C      L 1.9  1.9  0.95

CL  0.674

F'b  F'b*CL Cfu

F'b 619 psi

F'b** F'b

F'b** 619 psi

fb 526 psi F'b 619 psi

Combined Bending and Axial Tension

ft3 fb   0.8 <1.0. ok Ft' F'b* fb  ft3  0.64 <1.0 ok F'b** Results: No 2 Hem-Fir 2 x 8 satisfies NDS Criteria for combined bending and axial tension.

1.7 Combined Bending and Axial Compression (ASD)

E1.7 - Combined Bending and Axial Compression (ASD) E1.7 - Combined Bending and Axial Compression (ASD) No. 1 Southern Pine nominal 2x6 beam-columns are being designed to carry an axial compressive load of 840 No.lb (snow) 1 Southern and 560 Pine lb nominal (dead) plus2x6 beam-columnsa 25 psf wind load are on being their designed narrow face. to carry Columns an axial are compressive 9 ft long and load spaced of 840 4 lbfeet (snow) o.c. andTheir 560 ends lb (dead)are held plus in position a 25 psf and wind lateral load supporton their isna providedrrow face. along Columns the entire are 9narrow ft long face.and spaced 4 feet o.c. Their ends are held in position and lateral support is provided along the entire narrow face. Check the adequacy of the beam-column for bending and compression for the appropriate load cases. Check the adequacy of the beam-column for bending and compression for the appropriate load cases. Notes: Notes:Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements Loadon load cases combinations. used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements onThe load column combinations. being considered is not subjected to especially severe service conditions or extraordinary hazard. The column being considered is not subjected to especially severe service conditions or extraordinary hazard. Reference and Adjusted Design Values for No. 1 Southern Pine 2x6

Fb  1350 psi E 1600000 psi Emin  580000 psi (NDS Supplement Table 4B)

Fc  1550 psi CM  1.0 Ct  1.0 CF  1.0 (NDS Table 4.3.1)

Member length and properties l 9 ft b 1.5 in d 5.5 in 2 2 bd db A  bd S  S  g x 6 y 6 2 3 3 Ag  8.25 in Sx  7.562 in Sy  2.062 in

Applied Loads lbf w  25 w  4 ft P  840 lbf P  560 lbf strong 2 trib snow Dead ft lbf w w  100 strong trib ft

wweak  0 Load applied to weak axis of beam column. In this example, no load is applied.

E1.7 - Combined Bending and Axial Compression (ASD) E1.7 - Combined Bending and Axial Compression (ASD)

No. 1 Southern Pine nominal 2x6 beam-columns are being designed to carry an axial compressive load of 840 No. 1 Southern Pine nominal 2x6 beam-columns are being designed to carry an axial compressive load of 840 lb (snow) and 560 lb (dead) plus a 25 psf wind load on their narrow face. Columns are 9 ft long and spaced 4 lb (snow) and 560 lb (dead) plus a 25 psf wind load on their narrow face. Columns are 9 ft long and spaced 4 feet o.c. Their ends are held in position and lateral support is provided along the entire narrow face. feet o.c. Their ends are held in position and lateral support is provided along the entire narrow face.

Check the adequacy of the beam-column for bending and compression for the appropriate load cases. Check the adequacy of the beam-column for bending and compression for the appropriate load cases.

Notes: Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations. on load combinations. The column being considered is not subjected to especially severe service conditions or extraordinary hazard. The column being considered is not subjected to especially severe service conditions or extraordinary hazard.

Reference and Adjusted Design Values for No. 1 Southern Pine 2x6

Fb  1350 psi E 1600000 psi Emin  580000 psi (NDS Supplement Table 4B)

Fc  1550 psi CM  1.0 Ct  1.0 CF  1.0 (NDS Table 4.3.1)

Cfu  1.0 Ci  1.0 Cr  1.0 Flat use factors are for weak axis bending. However, since lateral CT  1.0 c 0.8 support is provided along the entire narrow face, Cfu = 1.0 E'min  EminCMCtCiCT factor "c" in column stability factor CP 1.7 equation for sawn lumber. (3.7.1) E'min  580000 psi COMBINED BENDING AND AXIAL COMPRESSION (ASD) Member length and properties l 9 ft b 1.5 in d 5.5 in 2 2 bd db A  bd S  S  g x 6 y 6 2 3 3 Ag  8.25 in Sx  7.562 in Sy  2.062 in

Applied Loads lbf w  25 w  4 ft P  840 lbf P  560 lbf strong 2 trib snow Dead ft lbf w w  100 strong trib ft

wweak  0 Load applied to weak axis of beam column. In this example, no load is applied.

Load Case 1: DL + SL + WL

CD  1.6 Appendix B Section B.2 (non-mandatory)

Compression

Fc*  FcCDCMCtCFCi Fc* is reference compression parallel to grain design value adjusted with all adjustment factors except the column stability factor C . The following calculations determine the Fc*  2480 psi P, column stabilty factor CP: P  P  P 1 snow Dead Subscripts refer to Load Case

P1  1400 lbf P1 fc1  Ag

fc1  170 psi Actual compression stress in beam column

Determine Effective Lengths and Critical Buckling Design Values

Ke  1.0 Buckling length coefficient Ke for rotation free/translation fixed (pinned/pinned) column (NDS Appendix G Table G1)  in  d1  d l1  l le1  Kel112  Beam dimensions, laterally unsupported lengths (Figure  ft  3F) and effective column lengths (NDS 3.7.1) for buckling 7  in  in each direction. Subscript 1 is strong (but unsupported) d  b l  ft l  K l 12  2 2 12 e2 e 2  ft  axis; Subscript 2 is the weak but laterally supported axis.

le1  108 in le2  0 Effective lengths in each axis. Le2 can be assumed to equal 0 per NDS A.11.3.

 le1 l  max  l  108 in Controlling effective length e   e  le2

0.822 E'min Critical bucking design value for compression member (NDS 3.7.1) F  F  1236 psi cE 2 cE  le1    d1 

led R  R  16 Slenderness ratio for bending (NDS 3.3-5) B 2 B b

1.20 E'min F  F  2636 psi Critical bucking design value for bending (NDS 3.7.1) bE 2 bE RB

Determine Column Stability Factor Cp

2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      P 2c  2c  c

CP  0.433 Column Stability Factor (3.7-1)

Fc' Fc*CP Fc' 1073 psi Adjusted compression parallel to grain design value

fc1  170 psi Fc' 1073 psi Actual compression stress fc does not exceed adjusted compression design value F'c

Bending

CL  1.0 Depth to breadth (d/b) ratio (6/2 = 3.0). 2

F'b1  FbCDCMCLCtCFCiCr F'b is adjusted bending design value with all adjustment factors. F'b is calculated for strong axis bending only since F'b1  2160 psi the weak axis is laterally supported.

2 in wstrongwtrib l 12 ft Actual bending stress in strong direction resulting 1.7 M  max1 8 from wind load applied to narrow face of beam-column

Mmax1  12150 inꞏlbf COMBINED BENDING AND AXIAL COMPRESSION (ASD)

Mmax1 fb1  fb1  1607 psi Sx

Actual bending stress fb1 does not exceed adjusted fb1  1607 psi F'b1  2160 psi bending design value F'b

Combined Bending and Axial Compression

2  fc1 fb1     0.89 < 1.0 ok (NDS 3.9-3) Checking NDS 3.9-4 is not necessary  Fc'    fc1  since the weak axis is fully braced. F'b11      FcE 

fc1  170 psi FcE  1236 psi Actual compression stress is less than critical bucking design values for strong axis buckling. OK (NDS 3.9.2)

fb1  1607 psi FbE  2636 psi Actual bending stress does not exceed adjusted parallel to grain design value. OK (NDS 3.9.2)

Load Case 2: DL+SL Load duration is not included in E' calculation. It is important to evaluate CD  1.15 multiple combinations such that all load duration effects are considered.

Compression P2 Fc*  FcCDCMCtCFCi P2  Psnow  PDead fc2  Ag

Fc*  1782 psi P2  1400 lbf fc2  170 psi

Determine Column Stability Factor Cp 2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      P 2c  2c  c

CP  0.555 Fc' Fc*CP Fc' 990 psi

fc2  170 psi Fc' 990 psi Actual compression parallel to grain stress fc does not exceed adjusted compression parallel to grain design value F'c. OK Load Case 3: DL only

CD  0.90

Compression P3 Fc*  FcCDCMCtCFCi P3  PDead fc3  Ag

Fc*  1395 psi P3  560 lbf fc3  68 psi

Determine Column Stability Factor Cp

2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      P 2c  2c  c

CP  0.648

Fc' Fc*CP

Fc' 904 psi

fc3  68 psi Fc' 904 psi Actual compression stress does not exceed adjusted compression parallel to grain design values for strong axis buckling. OK (NDS 3.9.2) Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 45

1.8 Combined Bo-Axial Bending and Axial Compression (ASD)

E1.8 - Combined Bi-axial Bending and Axial Compression (ASD)

A No. 2 Southern Pine nominal 2x4 oriented flatwise is being considered for use as a member within the top

chord of a parallel chord gable end truss. The member is 3 ft long (between panel points) and will be subjected BENDING AND AXIAL COMPRESSION (ASD) COMBINED BI-AXIAL to axial compression forces of 300 lb dead load (DL) and 600 lb snow load (SL), concentrated loads of 50 lb (DL) and 100 lb (SL) at the midpoint of the member on its wide face and 120 lb wind load (WL) at the midpoint of the member on its narrow face. Lateral support is provided only at the ends of the member and the ends are considered pinned.

Check the adequacy of the beam-column for bi-axial bending and axial compression for the appropriate load cases.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations. 1.8

Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL 46 1.8 COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD)

E1.8 - Combined Bi-axial Bending and Axial Compression (ASD)

A No. 2 Southern Pine nominal 2x4 oriented flatwise is being considered for use as a member within the top chord of a parallel chord gable end truss. The member is 3 ft long (between panel points) and will be subjected to axial compression forces of 300 lb dead load (DL) and 600 lb snow load (SL), concentrated loads of 50 lb (DL) and 100 lb (SL) at the midpoint of the member on its wide face and 120 lb wind load (WL) at the midpoint of the member on its narrow face. Lateral support is provided only at the ends of the member and the ends are considered pinned.

Check the adequacy of the beam-column for bi-axial bending and axial compression for the appropriate load cases.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements on load combinations. Reference and Adjusted Design Values for No. 2 Southern Pine 2x4

Fb  1100 psi E 1400000 psi Emin  510000 psi (NDS Supplement Table 4B)

Fc  1450 psi CM  1.0 Ct  1.0 CF  1.0 (NDS Table 4.3.1)

Cfu  1.10 Ci  1.0 Cr  1.0 Flat use factors are for weak axis bending (NDS 3.9.2) c 0.8 Factor "c" in column stability factor CP equation for sawn lumber. (NDS 3.7.1) E'min  EminCMCtCi

E'min  510000 psi

Member length and properties l 3 ft b 1.5 in d 3.5 in 2 2 bd db A  bd S  S  g x 6 y 6 2 3 3 Ag  5.25 in Sx  3.062 in Sy  1.312 in

Applied Loads Axial Weak (y) Axis Strong (x) Axis Applied loads. Subscripts depict load DL  300 lbf SL  100 lbf WL  120 lbf Axial Weak Strong type (DL-dead load, SL-snow load and SLAxial  600 lbf DLWeak  50 lbf WL-wind load) and application in relation to the member (applied to strong or weak axis).

E1.8 - Combined Bi-axial Bending and Axial Compression (ASD) Load Case 1: DL + SL + WL Subscripts refer to Load Case A No. 2 Southern Pine nominal 2x4 oriented flatwise is being considered for use as a member within the top C  1.6 NDS Appendix B Section B.2 (non-mandatory) chord of a parallel chord gable end truss. The member is 3 ft long (between panel points) and will be subjected D to axial compression forces of 300 lb dead load (DL) and 600 lb snow load (SL), concentrated loads of 50 lb Compression BENDING AND AXIAL COMPRESSION (ASD) COMBINED BI-AXIAL (DL) and 100 lb (SL) at the midpoint of the member on its wide face and 120 lb wind load (WL) at the midpoint of the member on its narrow face. Lateral support is provided only at the ends of the member and the ends are Fc*  FcCDCMCtCFCi Fc* is reference compression parallel to grain design value considered pinned. adjusted with all adjustment factors except the column stability factor C F  2320 psi P. Check the adequacy of the beam-column for bi-axial bending and axial compression for the appropriate load c* cases. P1  DLAxial  SLAxial

Notes: P  900 lbf Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirements 1 on load combinations. P1 f  f  171 psi c1 c1 Actual compression stress in beam-column Reference and Adjusted Design Values for No. 2 Southern Pine 2x4 Ag

Fb  1100 psi E 1400000 psi Emin  510000 psi (NDS Supplement Table 4B) Determine Effective Lengths and Critical Buckling Design Values K  1.0 Buckling length coefficient K for rotation free/translation fixed Fc  1450 psi CM  1.0 Ct  1.0 CF  1.0 (NDS Table 4.3.1) e e (pinned/pinned) column (NDS Appendix G Table G1) Cfu  1.10 Ci  1.0 Cr  1.0 Flat use factors are for weak axis  in  bending (NDS 3.9.2) d  d l  l l  K l 12  Beam dimensions, laterally unsupported lengths (Figure 1 1 e1 e 1  ft  c 0.8 3F) and effective column lengths (NDS 3.7.1) for buckling Factor "c" in column stability factor CP in in each direction. Subscript 1 is strong (but unsupported) d  b l  l l  K l 12  equation for sawn lumber. (NDS 3.7.1) 2 2 e2 e 2   axis; Subscript 2 is the weak but laterally supported axis. E'min  EminCMCtCi  ft  le1  36 in le2  36 in Effective lengths in each axis E'min  510000 psi l l e1 e2 le/d is less than 50 for each axis (NDS 3.7.1.4)  10  24 1.8 Member length and properties d1 d2 l 3 ft b 1.5 in d 3.5 in 0.822 E'min Critical bucking compression design values for compression F  F  3963 psi 2 2 cE1 2 cE1 member in planes of lateral support (NDS 3.7.1) bd db l A  bd S  S   e1 g x 6 y 6    d1  2 3 3 A  5.25 in S  3.062 in S  1.312 in g x y 0.822 E'min F  F  728 psi cE2 2 cE2  le2 Applied Loads    d2  Axial Weak (y) Axis Strong (x) Axis Applied loads. Subscripts depict load DL  300 lbf SL  100 lbf WL  120 lbf Determine Column Stability Factor Cp Axial Weak Strong type (DL-dead load, SL-snow load and SL  600 lbf DL  50 lbf WL-wind load) and application in relation Axial Weak  FcE1 to the member (applied to strong or weak F  min  F  728 psi Critical bucking design value for compression member cE   cE axis).  FcE2 2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      P 2c  2c  c CP  0.29 Column Stability Factor (3.7-1) Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL 48 1.8 COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD)

' Determine Adjusted Compression Parallel to Grain Design Value Fc

Fc' Fc*CP Fc' 673 psi Adjusted compression parallel to grain design value

fc1  171 psi Fc' 673 psi Actual compression stress fc does not exceed adjusted compression design value F'c (NDS 3.6.3). OK

Narrow Face (Strong Axis) Bending - (load parallel to wide face)

F'b*  FbCDCMCtCFCiCr F'b* is adjusted bending design value with all adjustment factors except the beam stability factor CL and flat use factor F'b*  1760 psi Cfu applied.

Determine Strong Axis Beam Stability Factor CL and Adjusted Bending Design Values in l  12 l l  36 in Laterally unsupported length u ft u lu  10.3 lu/d >7 (Table 3.3.3) d

le  1.37 lu  3d le  59.8 in (Table 3.3.3)

led R  R  9.6 Slenderness ratio for bending (3.3-5) RB < 50 (NDS 3.3.3.7) B 2 B b

1.20 E'min F  F  6577 psi Critical bucking design value for bending (3.7.1) bE 2 bE RB

2  FbE    FbE   FbE  1    1       F'b*    F'b*   F'b*  C      L 1.9  1.9  0.95

CL  0.982 Resulting beam stability factor CL. As an alternative, 4.4.1.2 (b) allows CL = 1.0 for d/b = (4/2) when the ends are held in position by full depth solid blocking, bridging, hangers, nailing, bolting or other suitable means.

F'b1  FbCDCMCLCtCFCiCr F'b1 is adjusted edgewise bending design value with all adjustment factors.

F'b1  1729 psi

Determine Bending Load and Resulting Bending Stress - Strong Axis Bending

in WL l12 Strong ft Actual bending stress in strong direction resulting M  1Strong 4 from wind load applied to narrow face of beam-column BENDING AND AXIAL COMPRESSION (ASD) COMBINED BI-AXIAL

M1Strong  1080 inꞏlbf

M1Strong fb1  Sx

fb1  353 psi F'b1  1729 psi Ok. Actual bending stress fb does not exceed adjusted edgewise compressive design value F'b1

Wide Face (Weak Axis) Adjusted Bending Design Value - (load parallel to narrow face)

CL  1.0 Since d

Flat use factor Cfu applies for weak axis bending. Cfu  1.1

F'b2  FbCDCMCLCtCfuCFCiCr F'b2 is adjusted flatwise bending design value with all adjustment factors.

F'b2  1936 psi

Determine Bending Load and Resulting Bending Stress - Weak Axis Bending 1.8

in DL  SL l12 Weak Weak ft Bending stress in strong direction resulting from dead M  1Weak 4 and snow load applied to narrow face of beam-column

M1Weak  1350 inꞏlbf

M1Weak fb2  Sy

fb2  1029 psi F'b2  1936 psi Ok. Actual bending stress fb does not exceed adjusted bending design value F'b (NDS 3.3.1)

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Combined Bi-axial Bending and Axial Compression

2  fc1 fb1 fb2      0.98 < 1.0 ok (3.9-3)  Fc'    fc1   2 F'  1    fc1   fb1   b1    F'  1     FcE1 b2         FcE2  FbE   2 fc1  fb1      0.24 < 1.0 ok (3.9-4) FcE2  FbE 

fc1  171 psi FcE1  3963 psi Actual compression stress is less than critical bucking design values for both weak and strong axis buckling. Ok fc1  171 psi FcE2  728 psi

fb1  353 psi FbE  6577 psi Actual bending stress is less than critical buckling design value. Ok

Load Case 2: DL+SL

CD  1.15 P2  DLAxial  SLAxial P2  900 lbf COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD) COMBINED BI-AXIAL P1 fc2  Ag Compression

Fc*  FcCDCMCtCFCi

Fc*  1667 psi fc2  171 psi

CP  0.387 Fc' Fc*CP Fc' 646 psi

fc2  171 psi Fc' 646 psi Actual compression stress fc does not exceed adjusted compression design value F'c OK 1.8 Weak Axis Bending

F'b2  FbCDCMCLCtCfuCFCiCr

F'b2  1392 psi in DL  SL l12 Weak Weak ft Bending stress in strong direction resulting from snow M  2Weak 4 load applied to narrow face of beam-column

M2Weak  1350 inꞏlbf

M2Weak fb2  Sy

fb2  1029 psi F'b2  1392 psi Ok. Actual bending stress fb does not exceed adjusted bending design value F'b Strong Axis Bending

Wind load is zero. No strong axis bending (fb1 in following equations set to zero)

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Combined Bending and Compression Load Case 3: DL only 2 C  0.90  fc2 fb10 fb2 D      1.04 Fc' f 2 ~1.0 OK (NDS 3.9-3)     c2     F'  1    fc2   fb1 0   Engineering judgement Compression b1    F'  1   FcE1 b2       required for rounding. P    FcE2 FbE 3       Fc*  FcCDCMCtCFCi P3  DLAxial fc3  Ag Checking NDS Equation 3.9-4 not necessary since f =0 b1 Fc*  1305 psi P3  300 lbf fc3  57 psi

f  171 psi F  3963 psi Actual compression stress is less than critical bucking design c2 cE1 Determine Column Stability Factor Cp values for both weak and strong axis buckling. OK   fc2 171 psi FcE2 728 psi 2  FcE    FcE   FcE  1    1      F F F fb2  1029 psi FbE  6577 psi Actual bending stress is less than critical buckling design value.  c*    c*   c*  OK C      P 2c  2c  c

CP  0.473

Fc' Fc*CP

Fc' 617 psi

fc3  57 psi Fc' 617 psi Actual compression parallel to grain stress fc does not exceed adjusted compression parallel to grain design value F'c OK Weak Axis Bending

F'b2  FbCDCMCLCtCfuCFCiCr

F'b2  1089 psi in DL l12 Weak ft Bending stress in strong direction resulting from dead M  3Weak 4 load applied to narrow face of beam-column

M3Weak  450 inꞏlbf

M3Weak fb3  Sy

fb3  343 psi F'b2  1089 psi Actual bending stress fb does not exceed adjusted bending design value F'b OK Strong Axis Bending

Wind load is zero. No strong axis bending (fb1 = 0)

Load Case 3: DL only

CD  0.90

Compression BENDING AND AXIAL COMPRESSION (ASD) COMBINED BI-AXIAL P3 Fc*  FcCDCMCtCFCi P3  DLAxial fc3  Ag

Fc*  1305 psi P3  300 lbf fc3  57 psi

Determine Column Stability Factor Cp

2  FcE    FcE   FcE  1    1       Fc*    Fc*   Fc*  C      P 2c  2c  c

CP  0.473

Fc' Fc*CP

Fc' 617 psi

fc3  57 psi Fc' 617 psi Actual compression parallel to grain stress fc does not exceed adjusted compression parallel to grain design value F' OK c 1.8 Weak Axis Bending

F'b2  FbCDCMCLCtCfuCFCiCr

F'b2  1089 psi in DL l12 Weak ft Bending stress in strong direction resulting from dead M  3Weak 4 load applied to narrow face of beam-column

M3Weak  450 inꞏlbf

M3Weak fb3  Sy

fb3  343 psi F'b2  1089 psi Actual bending stress fb does not exceed adjusted bending design value F'b OK Strong Axis Bending

Wind load is zero. No strong axis bending (fb1 = 0)

Copyright © American Wood Council. Downloaded/printed pursuant to License Agreement. No reproduction or transfer authorized. AMERICAN WOOD COUNCIL 54 1.8 COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD)

Combined Bending and Compression

2  fc3 fb10 fb3      0.35 < 1.0 OK (NDS 3.9-3) Fc' f 2     c3     F'  1    fc3   fb1 0   b1    F'  1     FcE1 b2         FcE2  FbE  

Checking NDS Equation 3.9-4 not necessary since fb1=0

fc3  57 psi FcE1  3963 psi Actual compression stress is less than critical bucking design values for both weak and strong axis buckling. OK fc3  57 psi FcE2  728 psi

fb3  343 psi FbE  6577 psi Actual bending stress is less than critical buckling design value. OK

1.9 Loadbearing Wall Wood Stud Resisting Wind and Gravity Loads

E1.9 - Loadbearing Wall Wood Stud Resisting Wind and Gravity Loads Background: The 2 story home considered in the Design of Wood Frame Buildings for High Wind, Snow, and Seismic Loads (WFCM Workbook) has a Foyer with a vaulted ceiling. The south bearing wall of the Foyer must support gravity loads from the roof and attic above, must resist reactions from uplift wind forces on the roof and must resist LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD out-of-plane wind pressures. The home is located in an area with a basic wind speed of 160 mph - Exposure B.

The Foyer originally had a 4 foot wide plant shelf at the second floor level. The plant shelf was platform framed as part of the floor diaphragm and limited stud length to one story. The resulting configuration was within the limitations of the prescriptive provisions of the Wood Frame Construction Manual (WFCM) and the wall framing could be determined from Chapter 3 of the WFCM.

Removing the plant shelf requires the wall to be balloon framed and will increase the stud lengths to 19 feet where they are no longer within the limitations of the prescriptive provisions of the WFCM.

Goal: Determine requirements for studs that are balloon framed from the first floor to the roof. Approach: Analyze wall framing as part of the Main Wind Force Resisting System (MWFRS) exposed to in-plane and out-of-plane load combinations specified by ASCE 7 Minimum Design Loads for Buildings and Other Structures. Analyze wall framing as Components and Cladding (C&C) exposed to out of plane C&C wind pressures only. Design wall framing per the National Design Specification (NDS) for Wood Construction.

In this example, the following loads are assumed:

Roof Loads Attic/Ceiling

Dead Load 10 psf Dead Load 15 psf Live Load 20 psf Attic Live Load 30 psf Ground Snow Load 30 psf Rain & Earthquake effects not considered in the analyses

Additionally, the following design assumptions apply: 1.9 Stud spacing = 16" o.c. Exterior Sheathing = Wood Structural Panels Interior Sheathing = 1/2" Gypsum Wallboard

The analysis involves an iterative approach. Initial values are selected for the member properties (depth, number of members and their species group and grade); initial analyses are completed and stresses and deflections determined and compared to allowable values. The member properties are then varied and analyses repeated until stress and deflection criteria are satisfied.

Note: As stated in the Foreword, examples are applicable to both the 2015 and 2018 versions of the NDS and WFCM unless otherwise noted. This example also references ASCE 7 and is applicable to both the 2010 and 2016 versions. Where section, figure, and table designations vary between the two versions of ASCE 7, they are noted. Reference to the International Building Code (IBC) is also applicable to the 2015 and 2018 versions.

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Reference and Adjusted Design Values (values were revised during iterations - final values are for No. 2 SP 2x8)

Fb  925 psi E 1400000 psi Emin 510000 psi (NDS Supplement Table 4B)

Fc  1350 psi Ft  550psi

CM  1.0 Ct  1.0 (NDS Table 4.3.1) C used to represent Wall Stud C  1.0 C  1.0 C  1.25 r fu i r Repetitive Member Factors (SDPWS 3.1.1.1) CF  1.0 CT  1.0 c 0.8 factor "c" in column stability factor C E'min Emin C C C C E'min 510000 psi P M t i T equation for sawn lumber. (NDS 3.7.1)

CF Factor is included in Table 4B values Member Properties n1 b 1.5 in d 7.25 in n is the number of full length studs 2 3 nb d nb d A  nb d S  I  g x 6 x 12 2 3 4 Ag  10.9 in Sx  13.1 in Ix  47.6 in

Home Dimensions

L 19 ft length of balloon-framed studs W 32 ft building width Wovhg  2 ft width of roof overhang

Determine Distributed Loads Supported by the South Wall Dead and Live Loads lbf 1 lbf 1 w  15  16ft w  10  36ft DLAttic 2 2 DLRoof 2 2 ft ft wDLAttic  120 plf wDLRoof  180 plf lbf 1 lbf 1 w  30  16ft w  20  36ft LLAttic 2 2 LLRoof 2 2 ft ft wLLAttic  240 plf wLLRoof  360 plf

wtotalDead  wDLAttic  wDLRoof

wtotalDead  300 plf

Rain Load Earthquake Load Rain and earthquake loads are included in ASCE 7 load R 0 plf E  0 plf l combinations). The subscript for the earthquake load is used to differentiate the earthquake load from the modulus of elasticity

Snow Load lbf p  30 C  1.0 C  1.0 I  1.0 Subscript "s" added to Ct to distinguish the g 2 e ts s ft temperature factor for snow load calculations from Ct for design value calculations LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD pf  0.7 CeCtsIspg Flat roof snow load pf  21 psf Slope factor per ASCE 7-10 Figure 7-2 (ASCE 7-16 Figure C  1.0 s 7.4-1) p  C p p  21 psf sBal s f sBal Balanced snow load (applied to the entire roof) psUnBal  Ispg psUnBal  30 psf Unbalanced snow load (applied to leeward side of roof with no snow on windward side roof) 1 p  36ft  378 plf Load on south wall from a balanced snow condition sBal 2 3 psUnBal 18ft  405 plf Load on south wall from an unbalanced snow 4 condition  1  psBal 36ft  2  Controlling snow condition w  max  snow 3  p  18ft   sUnBal 4  wsnow  405 plf Unbalanced snow load controls

Calculate MWFRS Wind Loads MWFRS Wind Pressures are calculated using the Envelope Procedure contained in Chapter 28 of ASCE 7. The wind pressure equation is:

p = q [(GC )-(GC )] (ASCE 7-10 Eq. 28.4-1 or ASCE 7-16 Eq. 28.3-1)

h pf pi 1.9

Where: qh is the velocity pressure GCpf is the external pressure coefficient for the surface being analyzed and GCpi is the internal pressure coefficient

Determine Velocity Pressure qh Note: The 160 Exp B velocity pressures qh in the V 160 WFCM is 24.06 psf and is based on a 33 ft MRH where the velocity pressure exposure coefficient K for Kz  0.70 z Exp B is 0.72. Kd  0.85 ASCE 7 Table 26.6-1 Kzt  1.0 ASCE 7 Section 26.8.2 Kz for Exp B evaluated at 25 ft MRH in this example is 2 lbf 0.70 per ASCE 7-10 Table 28.3.1(ASCE 7-16 Table q  (0.60 ) 0.00256K K K V  26.10-1) and produces a slightly lower velocity h z zt d 2 ft pressure of 23.4 psf. qh  23.4 psf The 0.60 factor in the velocity pressure equation incorporates ASCE 7 load factors for allowable stress design (ASD) load combinations

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Determine MWFRS Roof Pressure Coefficients (GCpf)

ASCE 7-10 Figure 28.4-1 (ASCE 7-16 Figure 28.3-1) shows the external pressure coefficient for interior and end zones for two load cases. Load Case A is for wind perpendicular to the ridge; Load Case B is for wind parallel to the ridge. By observation, GCpi = -0.18 results in the highest positive reaction at roof supports for Case A and GCpi = +0.18 results in the highest positive reaction at roof supports for Case B. Zone 2 (windward) Zone 3 (leeward) Roof Overhang Internal

Load Case A GCpfAWW  0.21 GCpfALW  0.43 GCpOH  0.70 GCpi  0.18

(ASCE 7-10 Section (ASCE 7-10 Table 26.11-1 / 28.4.3 / ASCE 7-16 ASCE 7-16 Table 26.13-1) Section 28.3.3)

Determine MWFRS Wind Pressures on Roof for Load Cases A and B Load Case A - wind perpendicular to ridge windward roof overhang

WROA  qhGCpfAWW  GCpOH

WROA  11.5psf leeward roof windward roof LRA  qhGCpfALW  GCpi WRA  qhGCpfAWW  GCpi LRA  5.8psf WRA  9.1 psf

Load Case B - wind parallel to ridge

Zone 2 (windward) Zone 3 (leeward) Roof Overhang Internal LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD

Load Case B GCpfBWW  0.69 GCpfBLW  0.37 GCpOH  0.70 GCpi  0.18 windward roof overhang

WROB  qhGCpfBWW

WROB  16.1psf windward roof leeward roof WRB  qhGCpfBWW  GCpi LRB  qhGCpfBLW  GCpi WRB  20.4psf LRB  12.9psf 1.9

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Determine Wind Load Reactions

Reactions at the top of the bearing wall are determined by summing overturning moments about the top of leeward wall for both load cases and determining the controlling reaction to use in the design. Horizontal projections are used in the analysis.

Note: The component of the overturning moment that results from wind pressures on the leeward roof overhang was not considered because: (1) it has a short (1 ft) moment arm and (2) the uplift pressures on the overhang occur downwind of the leeward wall and reduce the net overturning moment reaction slightly. This approach provides slightly conservative results.

Load Case A 1   Wovhg   RwindwardA  WovhgW  WROA   W   2   W 3W W 1W     WR   LR     2 4 A 2 4 A  

RwindwardA  62 plf

Load Case B 1   Wovhg   RwindwardB  WovhgW  WROB   W   2   W 3W W 1W     WR   LR     2 4 B 2 4 B  

RwindwardB  329plf

Note: The uplift reaction on the windward wall can be determined from WFCM Table 2.2A by interpolating the uplift connection loads between the 24 and 36 foot roof spans for the 0 psf roof/ceiling dead load and multiplying the uplift by 0.75 to account for the wall framing not being located in an exterior zone (footnote 1). That approach produces an uplift reaction of 391 plf which is higher than the results of these calculations. The higher reactions result primarily because uplift values in Table 2.2A are based on the worst case (20o) roof slope. The velocity pressure being calculated at 33 ft instead of 25 ft also contributes to slightly higher values.

Determine MWFRS Wind Pressures on Walls for Load Cases A and B

External and internal pressure coefficients (GCpf and GCpi) are from ASCE 7-10 Figure 28.4-1 and Table 26.11-1

(ASCE 7-16 Figure 28.3-1 and Table 26.13-1). LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD

Load Case A - wind perpendicular to ridge

Zone 1 (windward) Zone 4 (leeward) Internal

Load Case A GCpfwallAWW  0.56 GCpfwallALW  0.37 GCpi  0.18 windward wall leeward wall pmwfrsAWW  qhGCpfwallAWW  GCpi pmwfrsALW  qhGCpfwallALW  GCpi pmwfrsAWW  17.3 psf pmwfrsALW  4.4psf

Load Case B - wind parallel to ridge Zone 1 (windward) Zone 4 (leeward) Internal

GC  0.45 GC  0.45 GC  0.18 Load Case B pfwallBWW pfwallBLW pi

windward wall leeward wall

pmwfrsBWW  qhGCpfwallBWW  GCpi pmwfrsBLW  qhGCpfwallBLW  GCpi

pmwfrsBWW  14.7psf pmwfrsBLW  14.7psf 1.9

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Determine the Distributed Loads Supported by the Bearing Wall for ASCE 7 ASD Load Combinations ASCE 7 (section 2.4.1) includes the following ASD load combinations. Respective NDS load duration factors are shown in [brackets] next to load combination.

1. D [CD = 0.9] 2. D + L [CD = 1.0] 3. D + (Lr or S or R) 3a. D + (Lr) [CD = 1.25] 3b. D + (S) [CD = 1.15]

4. D + 0.75L + 0.75(Lr or S or R) 4a. D + 0.75L + 0.75(Lr) [CD = 1.25] 4b. D + 0.75L + 0.75(S) [CD = 1.15] 5. D + (0.6W or 0.7E) 5a. D + (0.6W or 0.7E) (wind parallel to ridge) [CD = 1.6] 5b. D + (0.6W or 0.7E) (wind perpendicular to ridge) [CD = 1.6] 6. D + 0.75L + 0.75(0.6W or 0.7E) + 0.75(Lr or S or R) 6a1. D + 0.75L + 0.75(0.6W) + 0.75(Lr or S) (wind parallel to ridge) [CD = 1.6] 6a2. D + 0.75L + 0.75(0.6W) + 0.75(Lr or S) (wind perpendicular to ridge) [CD = 1.6] 6b. D + 0.75L + 0.75(0.7E) + 0.75S [CD = 1.6] 7. 0.6D + 0.6W 7a. 0.6D+ 0.6W (wind parallel to ridge) [CD = 1.6] 7b. 0.6D+ 0.6W (wind perpendicular to ridge) [CD = 1.6] 8. 0.6D + 0.7E [CD = 1.6]

where D = dead load L = live load Lr = roof live load W = wind load (note the 0.6 load factor has already been included in the velocity pressure qh) S = snow load R = rain load E = earthquake load

Load combination are included in the following array (note that rain and earthquake load are neglected (E1= R = 0)

 w   totalDead  w  w  totalDead LLAttic  LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD    wtotalDead  wLLRoof   w  w   totalDead snow   wtotalDead  0.75 wLLAttic  0.75 wLLRoof    1  wtotalDead  0.75 wLLAttic  0.75 wsnow    1 300 wtotalDead  RwindwardA   2 540  wtotalDead  RwindwardB    3 660 w 4 705 Pdist    LLRoof  w  0.75 w  0.75 R  0.75 max   totalDead LLAttic windwardA   5 750  wsnow    6 784  w   LLRoof  Pdist  7 362 plf w  0.75 w  0.75 R  0.75 max  totalDead LLAttic windwardB     wsnow  8 -29   w 9 831   LLRoof    wtotalDead  0.75 wLLAttic  0.75 max  10 537  wsnow      11 784    0.6 wtotalDead RwindwardA  12 242    0.6 wtotalDead  RwindwardB  13 -149   14 180  0.6wtotalDead 

Load Combinations 1, 2, 3, 4a, 4b, 6b, and 8 model gravity only loads (dead load, live load and/or snow load). Load

Combinations 5, 6a1, 6a2, 7a, and 7b include MWFRS wind loads. Load Combinations are keyed to the array as 1.9 follows:

Load Combo Pdist 11 22 3a 3 3b 4 4a 5 4b 6 5a 7 5b 8 6a1 9 6a2 10 6b 11 7a 12 7b 13 8 14

All combinations that include wind will use a 1.6 load duration factor. Since load combinations 1-4, 6b (which is identical to 4), and 8 each have different load duration factors, those combinations will be analyzed individually.

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Analyze Framing for Load Combinations 1-4 and 6b (compute actual and allowable stresses and deflections. Iterate material properties to develop design) The bearing walls must resist distributed loads from the attic floor and roof and out-of-plane MWFRS wind loads proportional to the width of their tributary areas. The analyses are conducted for 16 inch stud spacing. The number of studs on either side of the framed openings in the south wall shall be determined from the WFCM Table 3.23C. Reductions allowed by WFCM Section 3.4.1.4.2 and Table 3.23D are acceptable.

By inspection, Load Combination 1 controls over Load Combination 8, so Load Combination 8 will not be analyzed. Load Combination 1: D

Determing compression force in framing for load combination 1

()16 P1  ftPdist  P1  400 lbf Compression force in the framing on each side of the wall 12  1 openings for Load Combination 1.

Calculate Reference & Adjusted Compression Design Values for Load Combination 1

CD1  0.9 Dead load duration factor CD for Load Combination 1. NDS Appendix B Section B.2 Fc1*  FcCD1CMCtCFCi Fc1* is reference compression design value for Load Combination 1 adjusted with all adjustment factors except the Fc1*  1215 psi column stability factor CP

Ke  1.0 Buckling length coefficient Ke for strong axis bending (pinned/pinned) column (Appendix G Table G1)

 in  d1  d l1  L le  Kel112  Stud dimensions, laterally unsupported lengths (NDS Figure 3F)  ft  and effective column lengths (NDS 3.7.1) for buckling in each direction. Subscript 1 is strong (but laterally unsupported) axis;

le  228 in Effective length in the strong axis. Assume gypsum wallboard is adequately connected to the studs and provides lateral support (NDS A.11.3)

0.822 E'min F  F  424 psi cE 2 cE  le     d1  Critical bucking design value for compression member

Determine Column Stability Factor Cp for Load Combination 1

2  FcE    FcE   FcE  1    1       Fc1*    Fc1*   Fc1*  C      P1 2c  2c  c

CP1  0.319 Column Stability Factor for Load Combination 1 (NDS 3.7-1)

Compare Actual Compression Stress to Adjusted Compression Design Value LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD

Fc'1  Fc1*CP1 Fc'1  388 psi Adjusted compression design value P1 fc1  fc1  37 psi Actual compression stress Ag Actual compression stress fc1 is less than the adjusted fc1  0.09 compression design value F'c1. Ratio of actual stress to Fc'1 adjusted compression design value < 1. OK

Load Combination 2: D + L

Determing compression force in framing for load combination 2

()16 P2  ftPdist  P2  720 lbf Compression force in the framing on each side of the wall 12  2 openings for Load Combination 2.

Calculate Reference & Adjusted Compression Design Values for Load Combination 2

CD2  1.0 Live load duration factor CD for Load Combination 2. NDS Appendix B Section B.2 Fc2*  FcCD2CMCtCFCi Fc2* is reference compression design value for Load Combination 2 adjusted with all adjustment factors except the Fc2*  1350 psi column stability factor CP 1.9

Ke  1.0 Buckling length coefficient Ke for strong axis bending (pinned/pinned) column (Appendix G Table G1)

 in  d1  d l1  L le  Kel112  Stud dimensions, laterally unsupported lengths (NDS Figure 3F)  ft  and effective column lengths (3.7.1) for buckling in each direction. Subscript 1 is strong (but laterally unsupported) axis;

Effective length in the strong axis. Assume gypsum wallboard is adequately connected to the studs and provides lateral support (NDS A.11.3) le  228 in

0.822 E'min F  F  424 psi cE 2 cE  le     d1  Critical bucking design value for compression member

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Determine Column Stability Factor Cp for Load Combination 2

2  FcE    FcE   FcE  1    1       Fc2*    Fc2*   Fc2*  C      P2 2c  2c  c

CP2  0.29 Column Stability Factor for Load Combination 2 (NDS 3.7-1)

Compare Actual Compression Stress to Adjusted Compression Design Value

Fc'2  Fc2*CP2 Fc'2  392 psi Adjusted compression design value P2 fc2  fc2  66 psi Actual compression stress Ag Actual compression stress fc2 is less than the adjusted fc2  0.17 compression design value F'c2. Ratio of actual stress to Fc'2 adjusted compression design value < 1. OK

Load Combination 3a: D + L r

Determing compression force in framing for load combination 3a

()16 P3a  ftPdist  P3a  880 lbf Compression force in the framing on each side of the wall 12  3 openings for Load Combination 3.

Calculate Reference & Adjusted Compression Design Values for Load Combination 3a

CD3a  1.25 Roof live load duration factor CD for Load Combination 3a. NDS Appendix B Section B.2 Fc3a*  FcCD3aCMCtCFCi Fc3a* is reference compression design value for Load Combination 3a adjusted with all adjustment factors except the Fc3a*  1688 psi column stability factor CP

Ke  1.0 Buckling length coefficient Ke for strong axis bending (pinned/pinned) column (Appendix G Table G1)

le  228 in Effective length in the strong axis. Assume gypsum wallboard is adequately connected to the studs and provides lateral support (NDS A.11.3) 0.822 E'min F  F  424 psi cE 2 cE LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD  le     d1  Critical bucking design value for compression member

Determine Column Stability Factor Cp for Load Combination 3a

2  FcE    FcE   FcE  1    1       Fc3a*    Fc3a*   Fc3a*  C      P3a 2c  2c  c

CP3a  0.237 Column Stability Factor for Load Combination 3a (NDS 3.7-1)

Fc'3a  Fc3a*CP3a Fc'3a  399 psi

Compare Actual Compression Stress to Adjusted Compression Design Value

P3a fc3a  fc3a  81 psi Actual compression stress Ag

Actual compression stress fc3a is less than the adjusted fc3a  0.2 compression design value F'c3a. Ratio of actual stress to Fc'3a adjusted compression design value < 1. OK 1.9

Load Combination 3b: D + S

Determing compression force in framing for load combination 3b

()16 P3b  ftPdist  P3b  940 lbf Compression force in the framing on each side of the wall 12  4 openings for Load Combination 3b.

Calculate Reference & Adjusted Compression Design Values for Load Combination 3b

CD3b  1.15 Snow load duration factor CD for Load Combination 3b. NDS Appendix B Section B.2 Fc3b*  FcCD3bCMCtCFCi Fc3b* is reference compression design value for Load Combination 3b adjusted with all adjustment factors except the Fc3b*  1552 psi column stability factor CP

Ke  1.0 Buckling length coefficient Ke for strong axis bending (pinned/pinned) column (Appendix G Table G1)

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Effective length in the strong axis. Assume gypsum wallboard le  228 in is adequately connected to the studs and provides lateral support (NDS A.11.3)

0.822 E'min F  F  424 psi cE 2 cE  le     d1  Critical bucking design value for compression member

Determine Column Stability Factor Cp for Load Combination 3b

2  FcE    FcE   FcE  1    1       Fc3b*   Fc3b*  Fc3b* C      P3b 2c  2c  c

CP3b  0.255 Column Stability Factor for Load Combination 3b (NDS 3.7-1)

Compare Actual Compression Stress to Adjusted Compression Design Value

Fc'3b  Fc3b*CP3b Fc'3b  397 psi Adjusted compression design value P3b fc3b  fc3b  86 psi Actual compression stress Ag Actual compression stress fc3b is less than the adjusted fc3b  0.22 compression design value F'c3b. Ratio of actual stress to Fc'3b adjusted compression design value < 1. OK

Load Combination 4a: D + 0.75 L + 0.75 L r

Determing compression force in framing for load combination 4a Compression force in the framing on each side of the wall openings for Load Combination 4a ()16 P4a  ftPdist  P4a  1000 lbf 12  5 Adjusted compression design value

Calculate Reference & Adjusted Compression Design Values for Load Combination 4a

CD4a  1.25 roof live load duration factor CD for Load Combination 4a. NDS Appendix B Section B.2

Fc4a*  FcCD4aCMCtCFCi Fc4a* is reference compression design value for Load Combination 4a adjusted with all adjustment factors except the Fc4a*  1688 psi column stability factor CP LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD Ke  1.0 Buckling length coefficient Ke for strong axis bending (pinned/pinned) column (Appendix G Table G1)

Effective length in the strong axis. Assume gypsum wallboard le  228 in is adequately connected to the studs and provides lateral support (NDS A.11.3)

0.822 E'min F  F  424 psi cE 2 cE  le     d1  Critical bucking design value for compression member

Determine Column Stability Factor Cp for Load Combination 4a

2  FcE    FcE   FcE  1    1      F F F  c4a*    c4a*   c4a*  1.9 C      P4a 2c  2c  c

CP4a  0.237 Column Stability Factor for Load Combination 4a (NDS 3.7-1)

Compare Actual Compression Stress to Adjusted Compression Design Value

Fc'4a  Fc4a*CP4a Fc'4a  399 psi Adjusted compression design value

P4a fc4a  fc4a  92 psi Actual compression stress Ag

Actual compression stress fc4a is less than the adjusted fc4a  0.23 compression design value F'c4a. Ratio of actual stress to Fc'4a adjusted compression design value < 1. OK

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Load Combination 4b: D + 0.75 L + 0.75 S

Determing compression force in framing for load combination 4b

()16 P4b  ftPdist  P4b  1045 lbf Compression force in the framing on each side of the wall 12  6 openings for Load Combination 4b.

Calculate Reference & Adjusted Compression Design Values for Load Combination 4b

CD4b  1.15 Snow load duration factor CD for Load Combination 4b. NDS Appendix B Section B.2 Fc4b*  FcCD4bCMCtCFCi Fc4b* is reference compression design value for Load Combination 4b adjusted with all adjustment factors except the Fc4b*  1552 psi column stability factor CP

Ke  1.0 Buckling length coefficient Ke for strong axis bending (pinned/pinned) column (Appendix G Table G1)

Stud dimensions, laterally unsupported lengths (NDS Figure 3F) in   and effective column lengths (3.7.1) for buckling in each d1  d l1  L le  Kel112   ft  direction. Subscript 1 is strong (but laterally unsupported) axis;

Effective length in the strong axis. Assume gypsum wallboard l  228 in e is adequately connected to the studs and provides lateral support (NDS A.11.3)

0.822 E'min F  F  424 psi cE 2 cE  le     d1  Critical bucking design value for compression member

Determine Column Stability Factor Cp for Load Combination 4b

2  FcE    FcE   FcE  1    1       Fc4b*   Fc4b*  Fc4b* C      P4b 2c  2c  c

CP4b  0.255 Column Stability Factor for Load Combination 4b (NDS 3.7-1)

Compare Actual Compression Stress to Adjusted Compression Design Value

Fc'4b  Fc4b*CP4b Fc'4b  397 psi Adjusted compression design value P4b fc4b  fc4b  96 psi Actual compression stress Ag Actual compression stress fc4b is less than the adjusted fc4b  0.24 compression design value F'c4b. Ratio of actual stress to Fc'4b adjusted compression design value < 1. OK

Load Combination Applied Stress/ Allowable Stress 1 0.09 2 0.17 3a 0.20

3b 0.22 Table comparing ratio of applied stress to allowable LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD 4a 0.23 stresses for gravity load controlled combinations 4b 0.24

Load Case 5a: D + 0.6 W (wind parallel to ridge) Determing compression load in framing for load combination 5a ()16 P5a  ftPdist  P5a  483 lbf 12  7

Determine Reference and Adjusted Compression Design Values for Load Combination 5a

CD5  1.6 Wind load duration factor CD controls for Load Combination 5a - NDS Appendix B Section B.2

Fc5a*  FcCD5CMCtCFCi Fc5a* is reference compression design value adjusted with all adjustment factors except the column stability factor CP,. The Fc5a*  2160 psi following calculations determine the column stabilty factor CP:

Determine Column Stability Factor Cp for Load Combination 5a

2  FcE    FcE   FcE  1    1       Fc5a*    Fc5a*   Fc5a*  C      P5 2c  2c  c

CP5  0.188 Column Stability Factor for Load Combination 5a (NDS 3.7-1) 1.9

Determine Adjusted Compression Design Value for Load Combination 5a

Fc'5a  Fc5a*CP5 Fc'5a  405 psi Adjusted compression design value for Load Combination 5a

Compare Actual Compression Stress with Adjusted Compression Design Value P5a f  c5a f  44 psi Actual compression stress f does not exceed adjusted Ag c5a c5a compression design value F'c5a. Ratio of actual Fc'  405 psi 5a compression stress to adjusted compression design value < 1. OK fc5a  0.11 Fc'5a

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Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures Moment ()16 w  ftp Load Combination 5a includes the MWFRS Wind Load windAWW 12 mwfrsAWW

wwindAWW  23.08 plf 2 wwindAWWL in M  12 Bending moment from out-of-plane MWFRS wind loads mwfrsAWW 8 ft

MmwfrsAWW  12500 inꞏlbf

Determine Reference and Adjusted Bending Design Values for Load Combination 5a

CL  1.0 Depth to breadth (d/b) ratio 2

F'b5a  FbCD5CMCLCtCFCiCr F'b5a is adjusted bending design value for Load Combination 5a F'b5a  1850 psi

Compare Actual Bending Stress with Adjusted Bending Design Values for Load Combination 5a

MmwfrsAWW fb5a  fb5a  951 psi Bending stress resulting from out-of-plane MWFRS wind Sx loads F'b5a  1850 psi

fb5a Ok. Actual bending stress f is less than adjusted bending  0.51 b5a F'b5a design value F'b5a. Ratio of actual bending stress to adjusted bending design value < 1

Check Combined Uniaxial Bending and Axial Compression

2  fc5a  fb5a     0.59 < 1.0 ok (NDS 3.9-3)  Fc'5a   fc5a  F'b5a1      FcE 

Actual compression stress fc5a does not exceed adjusted compression design value in plane of lateral support for fc5a  44 psi FcE  424 psi edgewise bending FcE. OK

Load Case 5b: D + 0.6 W (wind perpendicular to ridge) Determing axial load in framing for load combination 5b ()16 P  ftP  P  39lbf Note: by inspection, Load Combination 7b controls and is analyzed 5b dist 5b LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD 12  8 below.

Load Case 6a1: D + 0.75L + 0.75 W (wind parallel to ridge) Determing compression load in framing for load combination 6a1

()16 P6a1  ftPdist  P6a1  1107 lbf 12  9

Determine Reference and Adjusted Compression Design Values for Load Combination 6a1

CD6  1.6 Wind load duration factor CD controls for Load Combination 6a - NDS Appendix B Section B.2

Fc6a1*  FcCD6CMCtCFCi Fc6a1* is reference compression design value adjusted with all adjustment factors except the column stability factor CP,. The Fc6a1*  2160 psi following calculations determine the column stabilty factor CP:

Determine Column Stability Factor Cp for Load Combination 6a1

2  FcE    FcE   FcE  1    1       Fc6a1*    Fc6a1*   Fc6a1*  C      P6 2c  2c  c

CP6  0.188 Column Stability Factor for Load Combination 6a1 (NDS 3.7-1) 1.9

Determine Adjusted Compression Design Value for Load Combination 6a1

Fc'6a1  Fc6a1*CP6 Fc'6a1  405 psi Adjusted compression design value for Load Combination 6a

Compare Actual Compression Stress with Adjusted Compression Design Value P6a1 f  c6a1 f  102 psi Actual compression stress f does not exceed adjusted Ag c6a1 c6a1 compression design value F'c6a1. Ratio of actual compression Fc'  405 psi 6a1 stress to adjusted compression design value < 1. OK fc6a1  0.25 Fc'6a1

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Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures Load Case 6a2: D + 0.75L + 0.75 W (wind perpendicular to ridge) Moment Determing compression load in framing for load combination 6a2 ()16 w  0.75 ftp Load Combination 6a1 includes 75% of the MWFRS Wind ()16 windAWW mwfrsAWW P6a2  ftPdist  P6a2  716 lbf 12 Load 12  10 wwindAWW  17.31 plf Determine Reference and Adjusted Compression Design Values for Load Combination 6a2 2 w L windAWW in C  1.6 Wind load duration factor CD controls for Load Combination 6a - M  12 Bending moment from out-of-plane MWFRS wind loads D6 mwfrsAWW 8 ft NDS Appendix B Section B.2 M  9375 inꞏlbf mwfrsAWW Fc6a2*  FcCD6CMCtCFCi Fc6a2* is reference compression design value adjusted with all adjustment factors except the column stability factor CP,. The Determine Reference and Adjusted Bending Design Values for Load Combination 6a1 Fc6a2*  2160 psi following calculations determine the column stabilty factor CP:

 Depth to breadth (d/b) ratio 2

MmwfrsAWW Fc'  F C Fc'  405 psi Adjusted compression design value for Load Combination 6a fb6a1  fb6a1  713 psi Bending stress resulting from out-of-plane MWFRS wind 6a2 c6a2* P6 6a2 Sx loads F'b6a1  1850 psi Compare Actual Compression Stress with Adjusted Compression Design Value fb6a1 Ok. Actual bending stress f is less than adjusted bending P  b6a1 6a2 0.39 f  F' design value F' . Ratio of actual bending stress to adjusted c6a2 f  66 psi Actual compression stress f does not exceed adjusted b6a1 b6a1 Ag c6a2 c6a2 bending design value < 1 compression design value F'c6a2. Ratio of actual Fc'  405 psi 6a2 compression stress to adjusted compression design value < 1. OK fc6a2 Check Combined Uniaxial Bending and Axial Compression  0.16 Fc'6a2 2  fc6a1  fb6a1 Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures     0.57 < 1.0 ok (NDS 3.9-3)  Fc'6a1    fc6a1  Moment F'b6a11      FcE  ()16 wwindBWW  0.75 ftpmwfrsBWW Load Combination 6a2 includes 75% of the MWFRS Wind 12 Load    Actual compression stress fc6a1 does not exceed adjusted wwindBWW 14.74 plf compression design value in plane of lateral support for fc6a1  102 psi FcE  424 psi 2 edgewise bending F . OK wwindBWWL in cE M  12 Bending moment from out-of-plane MWFRS wind loads mwfrsBWW 8 ft MmwfrsBWW  7982inꞏlbf

Load Case 6a2: D + 0.75L + 0.75 W (wind perpendicular to ridge) Determing compression load in framing for load combination 6a2 ()16 P6a2  ftPdist  P6a2  716 lbf LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD 12  10

Determine Reference and Adjusted Compression Design Values for Load Combination 6a2

CD6  1.6 Wind load duration factor CD controls for Load Combination 6a - NDS Appendix B Section B.2

Fc6a2*  FcCD6CMCtCFCi Fc6a2* is reference compression design value adjusted with all adjustment factors except the column stability factor CP,. The Fc6a2*  2160 psi following calculations determine the column stabilty factor CP:

Determine Column Stability Factor Cp for Load Combination 6a2

2  FcE    FcE   FcE  1    1       Fc6a2*    Fc6a2*   Fc6a2*  C      P6 2c  2c  c

CP6  0.188 Column Stability Factor for Load Combination 6a2 (NDS 3.7-1)

Determine Adjusted Compression Design Value for Load Combination 6a2

Fc'6a2  Fc6a2*CP6 Fc'6a2  405 psi Adjusted compression design value for Load Combination 6a

Compare Actual Compression Stress with Adjusted Compression Design Value 1.9 P6a2 f  c6a2 f  66 psi Actual compression stress f does not exceed adjusted Ag c6a2 c6a2 compression design value F'c6a2. Ratio of actual Fc'  405 psi 6a2 compression stress to adjusted compression design value < 1. OK fc6a2  0.16 Fc'6a2

Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures Moment ()16 wwindBWW  0.75 ftpmwfrsBWW Load Combination 6a2 includes 75% of the MWFRS Wind 12 Load wwindBWW  14.74plf 2 wwindBWWL in M  12 Bending moment from out-of-plane MWFRS wind loads mwfrsBWW 8 ft

MmwfrsBWW  7982inꞏlbf

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Determine Reference and Adjusted Bending Design Values for Load Combination 6a2 Load Case 7b: 0.6 D + 0.6 W (wind perpendicular to ridge) Determing axial load in framing for load combination 7b

CL  1.0 Depth to breadth (d/b) ratio 2

MmwfrsBWW fb6a2  fb6a2  607 psi Bending stress resulting from out-of-plane MWFRS wind Compare Actual Tension Stress with Adjusted Tension Design Value Sx loads F'  1850 psi b6a2 P7b f  f  18psi Actual tension stress ft7b does not exceed adjusted t7b A t7b fb6a2 Ok. Actual bending stress f is less than adjusted bending g tension design value F' . Ratio of actual tension stress  0.33 b6a2 t7b to adjusted tension design value < 1. OK F'b6a2 design value F'b6a2. Ratio of actual bending stress to adjusted F't7b  880 psi bending design value < 1 ft7b  0.02 F't7b

Check Combined Uniaxial Bending and Axial Compression

Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures 2  fc6a2  fb6a2     0.42 < 1.0 ok (NDS 3.9-3) Moment Fc' f  6a2    c6a2  ()16 F'  1  b6a2    wwindBWW  ftpmwfrsBWW Load Combination 7b includes the MWFRS Wind Load   FcE  12 wwindBWW  19.65plf Actual compression stress fc6a2 does not exceed adjusted 2 w L f  66 psi F  424 psi compression design value in plane of lateral support for windBWW in c6a2 cE MmwfrsBWW  12 Bending moment from out-of-plane MWFRS wind loads edgewise bending FcE. OK 8 ft MmwfrsBWW  10642inꞏlbf

Load Case 7a: 0.6 D + 0.6 W (wind parallel to ridge) Determine Reference and Adjusted Bending Design Values for Load Combination 7b Determing compression load in framing for load combination 7a CL  1.0 Depth to breadth (d/b) ratio 2

F'b7b  FbCD7CMCLCtCFCiCr F'b7b is adjusted bending design value for Load Combination 7b F'b7b  1850 psi

Load Case 7b: 0.6 D + 0.6 W (wind perpendicular to ridge) Determing axial load in framing for load combination 7b ()16       P7b ft Pdist  P7b 199 lbf LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD 12  13

Determine Adjusted Tension Design Value for Load Combination 7b

CD7  1.6 Wind load duration factor CD controls for Load Combination 7b - NDS Appendix B Section B.2

F't7b  FtCD7CMCtCFCi F't7b is adjusted tension design value

F't7b  880 psi

Compare Actual Tension Stress with Adjusted Tension Design Value

P7b ft7b  ft7b  18psi Actual tension stress ft7b does not exceed adjusted Ag tension design value F't7b. Ratio of actual tension stress to adjusted tension design value < 1. OK F't7b  880 psi ft7b  0.02 F't7b

Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures Moment ()16 wwindBWW  ftpmwfrsBWW Load Combination 7b includes the MWFRS Wind Load

12 1.9 wwindBWW  19.65plf 2 wwindBWWL in M  12 Bending moment from out-of-plane MWFRS wind loads mwfrsBWW 8 ft

MmwfrsBWW  10642inꞏlbf

Determine Reference and Adjusted Bending Design Values for Load Combination 7b

CL  1.0 Depth to breadth (d/b) ratio 2

F'b7b  FbCD7CMCLCtCFCiCr F'b7b is adjusted bending design value for Load Combination 7b F'b7b  1850 psi

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Compare Actual Bending Stress with Adjusted Bending Design Values for Load Combination 7b

MmwfrsBWW fb7b  fb7b  810 psi Bending stress resulting from out-of-plane MWFRS wind Sx loads F'b7b  1850 psi

fb7b Ok. Actual bending stress f is less than adjusted bending  0.44 b7b F'b7b design value F'b7b. Ratio of actual bending stress to adjusted bending design value < 1

Check Combined Uniaxial Bending and Axial Tension

ft7b fb7b   0.417 < 1.0 ok (NDS 3.9-1) F't7b F'b7b

fb7b  ft7b  0.448 < 1.0 ok (NDS 3.9-2) F'b7b

Load Combination Applied Stress/ Allowable Stress Table comparing ratio of applied stress to allowable 5a 0.59 stresses for combined bending and axial load 6a1 0.57 combinations. By inspection and comparison to 6a2 0.42 gravity load combinations analyzed earlier, Load 7b 0.45 Combination 5a controls so far with combined dead plus MWFRS loads parallel to ridge. Check Adequacy of Framing to Resist Components and Cladding (C&C) loads Calculate C&C Pressures on Wall

CDCC  1.6 CD for C&C loading

Determine External C&C Pressure Coefficient 2 Note: "EWA" is used for Effective Wind Area since "A" is a L 1 EWA   EWA 120 previously defined variable. Per ASCE 7 Chapter 26, EWA need 3 2 ft not be less than (L)2/3   EWA   log  The south wall is in Zone 4. ASCE 7-10 Figure 30.4-1 / ASCE   500   GC (EWA )  0.8  0.3  7-16 Figure 30.3-1. The equation for GC for Zone 4 for 10 ft2 < p  10  p  log   2   500   EWA < 500 ft

GCp(EWA )  0.909 External pressure coefficient for full height studs in Foyer wall

pCC(EWA )q hGCp(EWA )  GCpi  Equation for C&C pressures for framing in the Foyer wall qh  23.4 psf By observation negative external pressure coefficients (GCp) are GC (EWA )  GC   1.089 greater than positive external pressure coefficients. So negative  p pi  external pressures and positive internal pressures (windward) create the greatest C&C pressures

pCC(EWA )  25.48psf C & C pressures for full height framing in the south wall

Apply C&C Pressures to Wall Framing and Check Bending and Deflection LOADBEARING WALL LOADS WOOD WIND AND GRAVITY RESISTING STUD Bending ()16 w  ftp ()EWA w  34 plf CC 12 CC CC

2 in w L 12 CC ft M  CC 8 MCC  18399 inꞏlbf

Determine Reference and Adjusted Bending Design Values for C&C loading

F'bCC  FbCDCCCMCLCt()1.0 CFCiCr F'bCC  1850 psi

Compare Actual Bending Stress with Adjusted Bending Design Values for C&C loading MCC Ok. Actual bending stresses that result from C&C pressures f  f  1400 psi bCC bCC f do not exceed adjusted bending design value F' Sx bCC bCC

F'bCC  1850 psi fbCC  0.76 The fb/F'b ratio is greater for C&C loading than for MWFRS F'bCC loading. C&C controls for strength calculations.

Determine Deflection for C&C loading 1.9 4 5 0.70 wCC ft in    L 12  IBC Table 1604 footnote (f) allows the wind load used in ΔCC   384 CrEIx 12 in  ft  deflection calculations to be 0.42 times the C&C load. A factor of 0.70 is applied since a 0.60 factor has already been CC  0.84 in Δ incorporated into the velocity pressure qh. in L 12 ft OK - Span to deflection ratio is greater than L / 180  273 ΔCC

Results - Framing the south wall of the Foyer using No.2 SP 2x8 studs on 16 inch centers is adequate to resist ASCE 7 loads.

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2.1a Withdrawal Design Value - Plain Shank Nail

E2.1a - Withdrawal Design Value - Plain Shank Nail

Using 2015/2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawal design value in pounds (capacity) of an 8d common smooth-shank nail in the connection below. Assume all adjustment factors are unity.:

Main member: Spruce-Pine-Fir Nominal 4x (Actual dimension 3.5 in.) (G = 0.42)

Side member: 12 gage (0.105 in. thick) ASTM A653 Grade 33 steel side plate

Fastener Dimensions: 8d common nail (NDS Table L4) Length = 2.5 in. Diameter = 0.131 in.

Using 2015/2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference WITHDRAWAL DESIGN VALUE - PLAIN SHANK NAIL withdrawal design value in pounds (capacity) of an 8d common smooth-shank nail in the connection below. Assume all adjustment factors are unity.:

Main member: Spruce-Pine-Fir Nominal 4x (Actual dimension 3.5 in.) (G = 0.42)

Side member: 12 gage (0.105 in. thick) ASTM A653 Grade 33 steel side plate

Fastener Dimensions: 8d common nail (NDS Table L4) Length = 2.5 in. Diameter = 0.131 in. D 0.131 Fastener diameter (in.)

G 0.42 Specific gravity (NDS Table 12.3.3A)

L 2.5 Nail Length (in.)

Ls  0.105 Side Member thickness (in.)

pt  LL s Nail penetration into main member (in.)

pt  2.395

5 2 W 1380 G D NDS Equation 12.2-3

W 20.7 Reference withdrawal design value. Compare to NDS Table 12.2C, W = 21 lbs/in

Wp t  49.5 Reference withdrawal design value based on nail penetration into main member (lbs)

See NDS Table 11.3.1 for application of additional adjustment factors for connections based on end use conditions.

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2.1b Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail (2018 NDS Only)

E2.1b - Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail in 5/8" WSP (2018 NDS Only) Using 2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawal design value in pounds (capacity) and head pull-through design value in pounds (capacity) of a 0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a Douglas Fir-Larch nominal 2x6 with a 5/8 in. thick Douglas-Fir Wood Structural Panel (plywood or oriented strand board) side member. Assume all adjustment factors are unity.

Main member: Douglas Fir-Larch (DFL) 2x6 (G = 0.5) Side member: 5/8 in. thick Wood Structural Panel (WSP) (G = 0.5)

Fastener Dimensions: Dash No. 05 (NDS Table L6) Length = 3 in. Diameter = 0.131 in. Head diameter = 0.281 in. TL = 1.5 in.

E2.1b - Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail in 5/8" WSP (2018 NDS Only) 2.1B Using 2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawal design value in pounds (capacity) and head pull-through design value in pounds (capacity) of a 0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a Douglas FASTENER UPLIFT CAPACITY - ROOF SHEATHING NDS ONLY) RING SHANK NAIL (2018 Fir-Larch nominal 2x6 with a 5/8 in. thick Douglas-Fir Wood Structural Panel (plywood or oriented strand board) side member. Assume all adjustment factors are unity.

Main member: Douglas Fir-Larch (DFL) 2x6 (G = 0.5) Side member: 5/8 in. thick Wood Structural Panel (WSP) (G = 0.5)

Fastener Dimensions: Dash No. 05 (NDS Table L6) Length = 3 in. Diameter = 0.131 in. Head diameter = 0.281 in. TL = 1.5 in. D 0.131 Fastener diameter (in.)

DH  0.281 Fastener head diameter (in.) TL 1.5 Deformed Shank Length (in.)

tns  0.625 Net Side Member thickness (in.)

G 0.5 Specific gravity, main and side members (NDS Table 12.3.3A and 12.3.3B)

Checking Fastener Withdrawal 2 W 1800 G D NDS Equation 12.2-5 W 59 Reference withdrawal design value. Compare to NDS Table 12.2E, W = 59 lbs/in

W TL  88 Reference withdrawal design value based on deformed shank fastener penetration (TL) in main member (lbs) Checking Fastener Head Pull-Through

tns  0.625

2.5DH  0.703 2.5DH greater than tns, so NDS Equation 12.2-6a applies 2 W  690 D G t NDS Equation 12.2-6a H π H ns

WH  95 Head pull-through design value (lbs). Compare to NDS Table 12.2F, W H = 95 lbs Fastener head pull-through design value of 95 lbs is greater than withdrawal design value of 88 lbs; withdrawal controls design capacity. See NDS Table 11.3.1 for application of additional adjustment factors for connections based on end use conditions.

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2.1c Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail (2018 NDS Only

E2.1c - Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail in 7/16" WSP (2018 NDS Only) Using 2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawal design value in pounds (capacity) and head pull-through design value in pounds (capacity) of a 0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a Douglas Fir-Larch nominal 2x6 with a 7/16 in. thick Douglas-Fir Wood Structural Panel (plywood or oriented strand board) side member. Assume all adjustment factors are unity.

Main member: Douglas Fir-Larch (DFL) 2x6 (G = 0.5) Side member: 7/16 in. thick Wood Structural Panel (WSP) (G = 0.5)

Fastener Dimensions: Dash No. 05 (NDS Table L6) Length = 3 in. Diameter = 0.131 in. Head diameter = 0.281 in. TL = 1.5 in.

E2.1c - Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail in 7/16" WSP (2018 NDS Only) Using 2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawal design value in pounds (capacity) and head pull-through design value in pounds (capacity) of a 2.1C 0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a Douglas Fir-Larch nominal 2x6 with a 7/16 in. thick Douglas-Fir Wood Structural Panel (plywood or oriented strand board) side member. Assume all adjustment factors are unity.

FASTENER UPLIFT CAPACITY - ROOF SHEATHING NDS ONLY) RING SHANK NAIL (2018 Main member: Douglas Fir-Larch (DFL) 2x6 (G = 0.5) Side member: 7/16 in. thick Wood Structural Panel (WSP) (G = 0.5)

Fastener Dimensions: Dash No. 05 (NDS Table L6) Length = 3 in. Diameter = 0.131 in. Head diameter = 0.281 in. TL = 1.5 in. D 0.131 Fastener diameter (in.)

DH  0.281 Fastener head diameter (in.)

TL 1.5 Deformed Shank Length (in.)

tns  0.4375 Net Side Member thickness (in.)

G 0.5 Specific gravity, main and side members (NDS Table 12.3.3A) Checking Fastener Withdrawal

2 W 1800 G D NDS Equation 12.2-5 W 59 Reference withdrawal design value. Compare to NDS Table 12.2E, W = 59 lbs/in Reference withdrawal design value based on deformed shank fastener W TL  88 penetration (TL) in main member (lbs) Checking Fastener Head Pull-Through

tns  0.438

2.5DH  0.703 2.5DH greater than tns, so NDS Equation 12.2-6a applies 2 W  690 D G t NDS Equation 12.2-6a H π H ns

WH  67 Head pull-through design value (lbs). Compare to NDS Table 12.2F, W H = 67 lbs Fastener head pull-through design value of 67 lbs is less than withdrawal design value of 88 lbs; fastener head pull-through controls design capacity. See NDS Table 11.3.1 for application of additional adjustment factors for connections based on end use conditions.

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2.2 Single Common Nail Lateral Design Value - Single Shear Wood-to-wood Connection

E2.2 - Single Common Nail Lateral Design Value - Single Shear Wood-to-wood Connection Using the 2015/2018 NDS yield limit equations in section 12.3, determine the Allowable Stress Design (ASD) reference lateral design value of a single shear connection with the following configuration. Assume all adjustment factors are unity.

Main member Nominal 3x Douglas Fir-Larch (Actual thickness = 2.5 in.) (G = 0.5) (NDS Table 12.3.3A)

Side member Nominal 1x Douglas Fir-Larch (Actual thickness = 0.75 in.) (G = 0.5) (NDS Table 12.3.3A)

Fastener Dimensions: 10d Common Nail (NDS Table L4) D = 0.148 in. Length = 3 in.

Main member

Nominal 3x Douglas Fir-Larch (Actual thickness = 2.5 in.) (G = 0.5) (NDS Table 12.3.3A) 2.2

Side member Nominal 1x Douglas Fir-Larch (Actual thickness = 0.75 in.) (G = 0.5) (NDS Table 12.3.3A) SINGLE COMMON NAIL LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION

Fastener Dimensions: 10d Common Nail (NDS Table L4) D = 0.148 in. Length = 3 in.

Define parameters:

Fem  4650 Main member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fes  4650 Side member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fem R  e R  1 Fes e

Fyb  90000 Fastener dowel bending yield strength (psi) (NDS Table I1)

D 0.148 Nail Diameter (in.)

Tip 2 D Length of tapered fastener tip (in.) (NDS 12.3.5.3b)

Ls  0.75 Side member Dowel Bearing Length (in.) (NDS 12.3.5) Tip L  3L  Main member Dowel Bearing Length (in.) (NDS 12.3.5.3) m s 2

Lm  2.1 2015 NDS 12.1.6.5 (2018 NDS 12.1.6.4) Requires minimum main member penetration equal to 6D, Lm > 0.89 in.

Rd  2.2 Reduction Term (NDS Table 12.3.1B)

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Calculate k 1 , k 2 , and k 3 (NDS Table 12.3.1A)

Lm Rt  Rt  2.803 Ls

 2  2 2 3   Re  2R e 1R t  Rt   Rt Re   Re1R t  k1  1R e

k1  0.935

2 2F yb1 2R e D k  1  21 R  2 e 2 3F emLm

k2  1.047

2 21 Re  2F yb2R e D k3  1     Re 2   3F emLs

k3  1.347

Yield Mode Calculations (NDS Table 12.3.1A)

Mode Im

DL mFem ZIm  Rd

ZIm  658 Yield Mode Im Solution (lbs)

Mode Is

DL sFes ZIs  Rd

ZIs  235 Yield Mode Is Solution (lbs)

Mode II

k1DLsFes ZII  Rd

ZII  219 Yield Mode II Solution (lbs) 2.2 Mode IIIm

k2DLmFem ZIIIm  SINGLE COMMON NAIL LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION 1 2R e Rd

ZIIIm  230 Yield Mode IIIm Solution (lbs)

Mode IIIs

k3DLsFem ZIIIs  2R e Rd

ZIIIs  105 Yield Mode IIIs Solution (lbs)

Mode IV

2  D  2F emFyb    ZIV    Rd  31 Re

Z  118 IV Yield Mode IV Solution (lbs)

 ZIm     658  Z  Is   235   Z     II   219  Zdist  Zdist  Creating an array with all Yield Mode Solutions  ZIIIm   230      105  ZIIIs       118   ZIV 

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E2.3 - Withdrawal Design Value - Lag Screw Z min Zdist

Z 105 Minimum value of all Yield Modes provides Z-reference lateral Using 2015/2018 NDS provisions (NDS 12.2), calculate the Allowable Stress Design (ASD) withdrawal design value of a lag screw in the connection below. Assume all adjustment factors design value (lbs). Mode IIIs controls. Compare to NDS Table 12N are unity. value = 105 lbs. See NDS Table 11.3.1 for application of

additional adjustment factors for connections based on end use Main member: conditions. Southern Pine Nominal 6x (Actual thickness = 5.5 in.) (G = 0.55) (NDS Table 12.3.3A)

Side member: Southern Pine Nominal 2x (Actual thickness = 1.5 in.) (G = 0.55) (NDS Table 12.3.3A)

Fastener Dimensions: 1/2 in. diameter lag screw (NDS Table L2) Length = 4 in. Tip Length = 0.3125 in.

2.3 Withdrawal Design Value - Lag Screw

E2.3 - Withdrawal Design Value - Lag Screw

Using 2015/2018 NDS provisions (NDS 12.2), calculate the Allowable Stress Design (ASD) withdrawal design value of a lag screw in the connection below. Assume all adjustment factors are unity.

Main member: Southern Pine Nominal 6x (Actual thickness = 5.5 in.) (G = 0.55) (NDS Table 12.3.3A)

Side member:

Southern Pine Nominal 2x (Actual thickness = 1.5 in.) (G = 0.55) (NDS Table 12.3.3A) 2.3

Fastener Dimensions:

1/2 in. diameter lag screw (NDS Table L2) WITHDRAWAL DESIGN VALUE - LAG SCREW Length = 4 in. Tip Length = 0.3125 in.

E2.3 - Withdrawal Design Value - Lag Screw E2.4 - Single Wood Screw Lateral Design Value - Double Using 2015/2018 NDS provisions (NDS 12.2), calculate the Allowable Stress Design (ASD) Shear Wood-to-wood Connection withdrawal design value of a lag screw in the connection below. Assume all adjustment factors Using 2015/2018 NDS yield limit equations in section 12.3, determine the allowable stress are unity. design (ASD) reference lateral design value of a double shear connection with the following

configuration. Assume all adjustment factors are unity. Main member: Southern Pine Nominal 6x (Actual thickness = 5.5 in.) (G = 0.55) (NDS Table 12.3.3A) Main member

Actual 3 in. Structural Composite Lumber Member (G = 0.5) (NDS 12.3.3.3) Side member: Southern Pine Nominal 2x (Actual thickness = 1.5 in.) (G = 0.55) (NDS Table 12.3.3A) Side members

Nominal 2x Douglas Fir-Larch (DFL) (Actual thickness = 1.5 in.) (G = 0.5) (NDS 12.3.3A) Fastener Dimensions: 1/2 in. diameter lag screw (NDS Table L2) Fastener Dimensions: Length = 4 in. Number 10 Wood Screw (NDS Table L3) Tip Length = 0.3125 in. D = 0.19 in. Dr= 0.152 in. D 0.5 Fastener diameter (in.) Length = 6 in. tip 0.3125 Fastener tapered tip length (in.) G 0.55 Specific gravity (NDS Table 12.3.3A)

L4 Lag screw length (in.)

Ls  1.5 Side Member thickness (in.)

pt  LL s  tip Lag screw penetration into main member (in.) Note: Per Table L2, the unthreaded body length, S = 1.5". p  2.188 t Therefore, the threaded portion begins at the wood-to-wood 3 interface. Note pt also matches Table L2 dimension T-E = 2- /16". Longer lag screws have different thread length dimensions requiring evaluation to determine actual thread penetration.

3 3 2 4 W 1800 G D NDS Equation 12.2-1

W 436.6 Compare to NDS Reference Withdrawal Design Value Table 12.2A, W = 437 lbs/in.

Wp t  955 Withdrawal design value based on main member penetration (lbs)

See NDS Table 11.3.1 for application of additional adjustment factors for connections based on end use conditions.

2.4 Single Wood Screw Lateral Design Value - Double Shear Wood-to-wood Connection

E2.4 - Single Wood Screw Lateral Design Value - Double Shear Wood-to-wood Connection Using 2015/2018 NDS yield limit equations in section 12.3, determine the allowable stress design (ASD) reference lateral design value of a double shear connection with the following configuration. Assume all adjustment factors are unity.

Main member Actual 3 in. Structural Composite Lumber Member (G = 0.5) (NDS 12.3.3.3)

Side members Nominal 2x Douglas Fir-Larch (DFL) (Actual thickness = 1.5 in.) (G = 0.5) (NDS 12.3.3A)

Fastener Dimensions: Number 10 Wood Screw (NDS Table L3) D = 0.19 in. 2.4 Dr= 0.152 in. Length = 6 in. SINGLE WOOD SCREW LATERAL DESIGN VALUE - DOUBLE SHEAR WOOD-TO-WOOD CONNECTION

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Main member Actual 3 in. Structural Composite Lumber Member (G = 0.5) (NDS 12.3.3.3)

Side members Nominal 2x Douglas Fir-Larch (DFL) (Actual thickness = 1.5 in.) (G = 0.5) (NDS 12.3.3A)

Fastener Dimensions: Number 10 Wood Screw (NDS Table L3) D = 0.19 in. Dr= 0.152 in. Length = 6 in.

Define parameters:

Fem  5550 Main member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fes  5550 Side member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fem Re  Re  1 Fes

tm  3.0 Main Member thickness (in.)

ts  1.5 Side Member thickness (in.) Fyb  80000 Fastener dowel bending yield strength (psi) (NDS Table I1)

D 0.19 Screw Diameter (in.)

Dr  0.152 Screw Root Diameter (in.) Lscrew  6 Screw Length (in.)

Tip 2 D Length of tapered fastener tip (in.) (NDS 12.3.5.3b) L  t L  3 m m m Main member Dowel Bearing Length (in.) (NDS 12.3.5.3) Tip Ls  Lscrew  Lm  ts  Side member (holding the screw tip) Dowel Bearing 2 Length (in.) (NDS 12.3.5)

Ls  1.31 NDS 12.1.5.6 requires minimum 6D penetration, Ls>1.14 in. OK

Rd  10D 0.5 Rd  2.4 Reduction Term (NDS Table 12.3.1B)

Calculate k 3 (NDS Table 12.3.1A) (k 1 and k 2 not used)

2 21 Re  2F yb2R e Dr k3  1     Re 2   3F emLs

k3  1.095

Yield Mode Calculations (NDS Table 12.3.1A)

Mode Im 2.4 DrLmFem ZIm  Rd SINGLE WOOD SCREW LATERAL DESIGN VALUE - DOUBLE SHEAR WOOD-TO-WOOD CONNECTION

ZIm  1055 Yield Mode Im Solution (lbs)

Mode Is

2DrLsFes ZIs  Rd

ZIs  921 Yield Mode Is Solution (lbs)

Mode IIIs

2k3DrLsFem ZIIIs  2R e Rd

ZIIIs  336 Yield Mode IIIs Solution (lbs)

Mode IV  2   2Dr  2F emFyb ZIV     Rd  31 Re

Z  234 IV Yield Mode IV Solution (lbs)

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 Z   Im   1055  Z    Is   921  Zdist    Zdist  Creating an array with all Yield Mode Solutions ZIIIs  336         234   ZIV 

Z min Zdist

Z 234 Minimum value of all Yield Modes provides Z-reference lateral design value (lbs). Mode IV controls. There are no tabulated values in the NDS to compare. Note that a single shear wood screw design value in NDS Table 12L is 117 lbs, which is half the value calculated here, since it is also Mode IV controlled. See NDS Table 11.3.1 for application of additional adjustment factors for connections based on end use conditions.

2.5 Single Bolt Lateral Design Value - Single Shear Wood-to-Wood Connection

E2.5 - Single Bolt Lateral Design Value - Single Shear Wood-to-Wood Connection Using the 2015/2018 NDS Yield Limit Equations (NDS 12.3), determine the Allowable Stress Design (ASD) reference lateral design value of a single shear connection with the following configuration:

Main member Nominal 4x Hem-Fir (Actual thickness = 3.5")

Side member Nominal 4x Hem-Fir (Actual thickness = 3.5")

Both members loaded parallel to grain G = 0.43 for Hem-Fir (NDS Table 12.3.3A)

Fastener Dimensions: 1/2 in. diameter bolt 8 in. Bolt with 1.5 in. thread length (NDS Table L1) 2.5

SINGLE BOLT LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION

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Main member Nominal 4x Hem-Fir (Actual thickness = 3.5")

Side member Nominal 4x Hem-Fir (Actual thickness = 3.5")

Both members loaded parallel to grain G = 0.43 for Hem-Fir (NDS Table 12.3.3A)

Fastener Dimensions: 1/2 in. diameter bolt 8 in. Bolt with 1.5 in. thread length (NDS Table L1)

Define parameters:

Fem  4800 Main member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fes  4800 Side member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fem R  e R  1 Fes e

Fyb  45000 Fastener dowel bending yield strength (psi) (NDS Table I1)

D 0.5 Bolt Diameter (in.) Per NDS 12.3.7.2, check that threads are less than 1/4 the bearing length in the member holding the threads. In this case, 3.5 in./4 > 0.5 in. Therefore, OK to use D instead of Dr in calculations.

Ls  3.5 Side member Dowel Bearing Length (in.) (NDS 12.3.5)

Lm  3.5 Main member Dowel Bearing Length (in.) (NDS 12.3.5)

Rd1  4.0

R  3.6 d2 Reduction Terms (NDS Table 12.3.1B)

Rd3  3.2

Calculate k 1 , k 2 , and k 3 (NDS Table 12.3.1A)

Lm Rt  Rt  1 Ls

 2  2 2 3   Re 2R e  1R t  Rt  Rt  Re  Re 1R t  k1  1R e

k1  0.414

2 2F yb1 2R e D k  1  21 R  2 e 2 3F emLm 2.5

k2  1.093 SINGLE BOLT LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION 2 21 Re  2F yb2R e D k3  1     Re 2   3F emLs

k3  1.093

Yield Mode Calculations (NDS Table 12.3.1A)

Mode Im

DL mFem ZIm  Rd1

ZIm  2100 Yield Mode Im Solution (lbs)

Mode Is

DL sFes ZIs  Rd1

ZIs  2100 Yield Mode Is Solution (lbs)

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Mode II

k1DLsFes ZII  Rd2

ZII  966 Yield Mode II Solution (lbs)

Mode IIIm

k2DLmFem ZIIIm  1 2R e Rd3

ZIIIm  957 Yield Mode IIIm Solution (lbs)

Mode IIIs

k3DLsFem ZIIIs  2R e Rd3

ZIIIs  957 Yield Mode IIIs Solution (lbs)

Mode IV

2  D  2F emFyb    ZIV    Rd3  31 Re

Z  663 IV Yield Mode IV Solution (lbs)

 ZIm     ZIs   Z   II  Zdist  Creating an array with all Yield Mode Solutions  ZIIIm     ZIIIs     ZIV 

Minimum value of all Yield Modes provides Z-reference lateral Z min Z Z 663 dist design value (lbs). Mode IV controls.

Repeat same problem, but solve using Technical Report 12 - General Dowel Equations for Calculating Lateral Connection Values (TR-12) Equations for comparison

qs  Fes D Side member dowel bearing resistance, lbs/in.

qm  Fem D Main member dowel bearing resistance, lbs/in.

3 FybD Side and Main member dowel resistance (equal due to equivalent dowel M  6 diameter in both members), in.-lbs

gap 0 Gap between member shear planes, in. 2.5 SINGLE BOLT LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION

The limiting wood stresses used in the yield model are based on the load at which the load-deformation curve from a fastener embedment test intersects a line represented by the initial tangent modolus offset 5% of the fastener diameter. The reduction term, Rd, reduces the values calculated using the yield limit equations to approximate estimates of the nominal proportional limit design values in previous NDS editions.

Yield Mode Calculations (TR-12 Table 1-1)

Mode Im

PIm  qm Lm

TR-12 Yield Mode Im Solution (lbs) PIm  8400

Mode Is

PIs  qs Ls

TR-12 Yield Mode Is Solution (lbs) PIs  8400

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Mode II 1 1 AII   4q s 4q m Ls Lm B   gap  II 2 2 2 2 qsLs qmLm C   II 4 4

2 BII  BII 4A II CII PII  2A II

PII  3479 TR-12 Yield Mode II Solution (lbs)

Mode IIIm

1 1 AIIIm   2q s 4q m Lm B  gap  IIIm 2 2 qmLm C  M  IIIm 4

2 BIIIm  BIIIm 4A IIIm CIIIm PIIIm  2A IIIm

PIIIm  3062 TR-12 Yield Mode IIIm Solution (lbs)

Mode IIIs 1 1 AIIIs   4q s 2q m Ls B  gap  IIIs 2 2 qsLs C  M  IIIs 4 2 BIIIs  BIIIs 4A IIIs CIIIs PIIIs  2A IIIs

PIIIs  3062 TR-12 Yield Mode IIIs Solution (lbs)

Mode IV 1 1 AIV   2q s 2q m

BIV  gap

CIV  M  M 2 BIV  BIV 4A IV CIV PIV  2A IV

PIV  2121 TR-12 Yield Mode IV Solution (lbs)

P   Im   R 

d1 2.5    PIs   R  d1 SINGLE BOLT LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION    2100  P  II   2100  Converting from TR-12 "P" values to NDS "Z"  R    values and creating an array. Shows TR-12  d2   966  Zdist2  Zdist2  results equal NDS results for each Yield  PIIIm   957  Mode. All values in units of lbs.     957  Rd3    663  P     IIIs   Rd3     PIV     Rd3 

Z  min Z 2 dist2 Z value from TR-12 equations is equivalent to Z value from NDS equations and comparable to NDS Table 12A value Zparallel = 660 Z2  663 lbs. See NDS Table 11.3.1 for application of additional adjustment factors for connections based on end use conditions.

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 2100   2100  For comparison, Z shows NDS  2100   2100  dist     equation results, Zdist2 shows TR-12  966   966  Z  Z  equation results, for Modes Im, Is, II, dist  957  dist2  957      IIIm, IIIs, and IV, respectively. All  957   957  values in units of lbs.  663   663 

2.6 Bolted Wood-to-Wood Tension Splice Connection Capacity

E2.6 - Bolted Wood-to-Wood Tension Splice Connection Capacity

Determine the ASD Capacity of a tension splice shown in the Figure below.

Assume 1" diameter x 5" long bolts Main and side members are nominal 2x12 No. 2 Southern Pine (G = 0.55) Dead and construction live loads apply Normal moisture and temperature conditions Members loaded parallel to grain

3-5/8" P P 4" 3-5/8"

7" 4" 4" 4" 4" 4" 4" 7" 38" 2.6 P P BOLTED WOOD-TO-WOOD TENSION SPLICE CONNECTION CAPACITY

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E2.6 - Bolted Wood-to-Wood Tension Splice Geometry Factor (C Δ ) Connection Capacity Geometry factor provides reduction of reference lateral design values for less than full end distance, edge distance, or spacing (NDS 12.5.1). Smallest calculated geometry factor will Determine the ASD Capacity of a tension splice shown in the Figure below. control.

Assume 1" diameter x 5" long bolts End distance Main and side members are nominal 2x12 No. 2 Southern Pine (G = 0.55) NDS 12.5.1.2(a), Table 12.5.1A Dead and construction live loads apply Normal moisture and temperature conditions Minimumend  7 Minimum end distance required for CΔ = 1.0 (in.) Members loaded parallel to grain Minimum end distance = 7D for softwoods loaded parallel to grain 3-5/8" Provided  4 Provided end distance (in.) P P end 4" Providedend 3-5/8" C end  Δ Minimumend

7" 4" 4" 4" 4" 4" 4" 7" C end  0.57 38" Δ Geometry factor based on end distance

Spacing between bolts P P NDS 12.5.1.2(c), Table 12.5.1B

Minimumbolt  4 Minimum spacing between bolts required for CΔ = 1.0 (in.) Minimumum spacing = 4D for parallel to grain load Define parameters: Provided  4 bolt Provided spacing between bolts (in.) D 1.0 Bolt Diameter (in.)

s4 Fastener row spacing (in.) C bolt  1.0 Provided spacing greater than or equal to minimum spacing for CΔ = 1.0 Δ Ls  1.5 Side member Dowel Bearing Length (NDS 12.3.5) Edge distance Lm  1.5 Main member Dowel Bearing Length (NDS 12.3.5.3) NDS 12.5.1.3, Table 12.5.1C Minimum  1.5 Minimum edge distance required for C = 1.0 (in.) C  1.25 Construction load duration (NDS Table 2.3.2) edge Δ D Minimumum edge distance = 1.5D for parallel to grain load and L/D≤6

CM  1.0 Ct  1.0 Providededge  3.625 Provided edge distance between bolts (in.)

Group Action Factor (C g )  Provided edge distance greater than or equal to minimum spacing for C = 1.0 2 C edge 1.0 Δ As  2 16.875 Area of side members (In ) Δ 2 Am  16.875 Area of main member (In )

NDS Table 11.3.6A Footnote 1 states when As/Am > 1.0, use Am/As and Am instead of As. Assuming 3 fasteners per row, interpolate results from Table 11.3.6A

Cg  0.97

Geometry Factor (C Δ )

Geometry factor provides reduction of reference lateral design values for less than full end distance, edge distance, or spacing (NDS 12.5.1). Smallest calculated geometry factor will control.

End distance NDS 12.5.1.2(a), Table 12.5.1A

Minimumend  7 Minimum end distance required for CΔ = 1.0 (in.) Minimum end distance = 7D for softwoods loaded parallel to grain

Providedend  4 Provided end distance (in.)

Providedend C end  Δ Minimumend

C end  0.57 Δ Geometry factor based on end distance

Spacing between bolts 2.6 NDS 12.5.1.2(c), Table 12.5.1B

Minimum  4 Minimum spacing between bolts required for CΔ = 1.0 (in.)

bolt BOLTED WOOD-TO-WOOD TENSION SPLICE CONNECTION CAPACITY Minimumum spacing = 4D for parallel to grain load Provided  4 bolt Provided spacing between bolts (in.)

C bolt  1.0 Provided spacing greater than or equal to minimum spacing for CΔ = 1.0 Δ

Edge distance NDS 12.5.1.3, Table 12.5.1C

Minimumedge  1.5 Minimum edge distance required for CΔ = 1.0 (in.) Minimumum edge distance = 1.5D for parallel to grain load and L/D≤6

Providededge  3.625 Provided edge distance between bolts (in.)

C edge  1.0 Provided edge distance greater than or equal to minimum spacing for CΔ = 1.0 Δ

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Spacing between rows NDS 12.5.1.3, Table 12.5.1D

Minimumrow  1.5 Minimum spacing between rows required for CΔ = 1.0 (in.) Minimumum spacing = 1.5D for parallel to grain load Provided  4 row Provided spacing between bolts (in.)

C row  1.0 Provided spacing greater than or equal to minimum spacing for CΔ = 1.0 Δ

C  min C endC boltC edgeC row Δ Δ Δ Δ Δ C  0.57 Δ Reference Lateral Design Value

Z 2310 Reference Lateral Design Value (NDS Table 12F)

Adjusted Multiple Bolt Capacity

n6 n = number of bolts on each side

Zadj  nZ CDCgC CMCt Δ Adjusted ASD lateral design value (lbs) Zadj  9603

Local Stresses in Fastener Groups NDS Appendix E calls for net section tension, row tear-out, and group tear-out capacity checks. The only applicable adjustment factor for these checks will be the load duration factor, CD.

Net Section Tension Check

Anet  (11.25 1.0625 2) 1.5 Net area = (member depth - bolt hole width) x member width Note: hole size includes 1/16" oversizing per 2015/2018 NDS 12.1.3.2 A  13.7 net Net area (in.2)

Ft  450 Nominal tension parallel to grain design value (psi) Ftadj  FtCDCMCt

Ftadj  562.5 Adjusted tension parallel to grain design value (psi)

ZNTadj  FtadjAnet

ZNTadj  7699 Adjusted Net Section Tension Capacity (lbs)

Row Tear-Out Check t 1.5 Member thickness (in.)

Fv  175 Nominal shear parallel to grain design value (psi)

Fvadj  FvCDCMCt Adjusted shear parallel to grain design value (psi) Fvadj  219

scritical  4 scritical is the minimum of the end distance and the in-row bolt spacing

ni  3 Number of fasteners in a single row

ZRTiadj  niFvadjtscritical Adjusted row tear-out capacity for single row (lbs) ZRTiadj  3938

ZRTadj  2n iFvadjtscritical Adjusted row tear-out capacity for two rows (lbs) 2.6 ZRTadj  7875

Group Tear-Out Check BOLTED WOOD-TO-WOOD TENSION SPLICE CONNECTION CAPACITY

ZRT1adj  ZRTiadj ZRT1' is row tear-out capacity for 1st row of bolts,ZRT2' is row tear-out capacity for 2nd row of bolts, both values in lbs ZRT2adj  ZRTiadj

Agroupnet  (4 1.0625 ) 1.5 Agroupnet is equal to distance between bolt rows times member thickness (in.2)

Ftadj  562.5 Adjusted tension parallel to grain design value (psi)

ZRT1adj ZRT2adj Z    F A GTadj 2 2 tadj groupnet

ZGTadj  6416 Adjusted group tear-out capacity (lbs) Connection capacity is minimum of calculated Capacity min Z Z Z Z adj NTadj RTadj GTadj capacities (lbs). Note group tear-out controls. However, increasing the spacing between bolt Capacity 6416 rows, up to a maximum of 5" for dimension lumber, can increase group tear-out capacity. Bolt sizes could be reduced to more closely match net section capacities.

3.1 Segmented Shear Wall Design - Wind

E3.1 - Segmented Shear Wall Design - Wind Design with Interior Gypsum Using the 2015 Special Design Provisions for Wind and Seismic (SDPWS), design the first floor wall shown in the diagram below as a segmented shear wall for a two-story house using Allowable Check maximum segment length based on Aspect Ratio Limits Stress Design (ASD) provisions Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4)

Design wind speed = 160 mph (700-yr wind speed, 3-second gust) Minimum segment length =Wall Height/Aspect Ratio Exposure Category B 9 Building dimensions: L  L  2.6 Minimum full height wall segment length (ft) min 3.5 min L = 40 ft All full height segments satisfy aspect ratio requirements. Maximum aspect ratio for WSP shear W = 32 ft walls to avoid capacity adjustments = 2:1, so 3 foot wide segments will require capacity Roof pitch = 7:12 adjustments. Top plate to ridge height = 9.3 ft V  5520 Applied shear load on each shear wall due to wind force (lbs) Wall height = 9 ft w Door height = 7 ft 6 in. h9 Wall height (ft) Window height = 4.5 ft (Note: Shear load calculated using Table 2.5B of 2015 Wood Frame Construction Manual) Stud spacing = 16 in. o.c. Studs are Southern Pine (G=0.55) Assume 15/32 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind) Check design with and without interior gypsum and neglect deflection.

Use ASCE 7-10 Minimum Design Loads for Buildings and Other Structures to determine wind 1065 ASD Shearwall Capacity, WSP (lbs/ft) v  v  532.5 loads. wASDWSP 2 wASDWSP (SDPWS 4.3.3)

Maximum aspect ratio for gypsum wallboard = 2:1 (SDPWS 4.3.4) and segments exceeding 1.5:1 shall be blocked. Since no segment in the wall meets the 1.5:1 aspect ratio, assume all gypsum panels will be blocked.

Assume 1/2 in. thick Gypsum Wallboard (GWB) Sheathing, 5d cooler nail @ 7 in. o.c. edge spacing, 16 in. o.c. studs, blocked

SDPWS Table 4.3C nominal capacity = 250 lb/ft 250 ASD Shearwall Capacity, Gypsum (lbs/ft) v  v  125 wASDGWB 2 wASDGWB (SDPWS 4.3.3)

Capacity of WSP and GWB can be added (SDPWS 4.3.3.3.2)

vWCASD  vwASDWSP  vwASDGWB

vWCASD  657.5 ASD Shearwall Capacity, WSP and GWB combined (lbs/ft) Need minimum 5520 lbs capacity to resist applied wind shear. Provide full height segments on either side of door for 10 ft of full height sheathing.

Capacity 2 5vWCASD

Capacity 6575 Wall capacity (lbs) Wall capacity exceeds demand, so design is OK.

Design with Interior Gypsum SEGMENTED SHEAR WALL DESIGN - WIND Check maximum segment length based on Aspect Ratio Limits Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4) Minimum segment length =Wall Height/Aspect Ratio 9 L  L  2.6 Minimum full height wall segment length (ft) min 3.5 min All full height segments satisfy aspect ratio requirements. Maximum aspect ratio for WSP shear walls to avoid capacity adjustments = 2:1, so 3 foot wide segments will require capacity adjustments.

Vw  5520 Applied shear load on each shear wall due to wind force (lbs) h9 Wall height (ft) (Note: Shear load calculated using Table 2.5B of 2015 Wood Frame Construction Manual)

Assume 15/32 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind)

1065 ASD Shearwall Capacity, WSP (lbs/ft) v  v  532.5 wASDWSP 2 wASDWSP (SDPWS 4.3.3)

Assume 1/2 in. thick Gypsum Wallboard (GWB) Sheathing, 5d cooler nail @ 7 in. o.c. edge spacing, 16 in. o.c. studs, blocked

SDPWS Table 4.3C nominal capacity = 250 lb/ft 250 ASD Shearwall Capacity, Gypsum (lbs/ft) v  v  125 wASDGWB 2 wASDGWB (SDPWS 4.3.3)

Capacity of WSP and GWB can be added (SDPWS 4.3.3.3.2)

vWCASD  vwASDWSP  vwASDGWB

Capacity 2 5vWCASD

Capacity 6575 Wall capacity (lbs) Wall capacity exceeds demand, so design is OK.

Hold down capacity is based on induced unit shear (hold downs at each end of each panel)

L  10 eff Effective length of Full Height Segments (ft)  Vw  T   h T 4968 Required Hold down capacity (lbs)  Leff  Hold down would need to be combined with 2nd floor hold down requirements. Dead load offset has been neglected in this example. Design without Interior Gypsum

All full height segments satisfy aspect ratio requirements. Maximum aspect ratio for WSP shear walls to avoid capacity adjustments = 2:1, so 3 foot wide segments will require capacity adjustments.

Vw  5520 Applied shear load on each shear wall due to wind force (lbs) h9 Wall height (ft) (Note: Shear load calculated using Table 2.5B of 2015 Wood Frame Construction Manual)

Assume 15/32 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind) 1065 vwASDWSP  vwASDWSP  532.5 ASD Shearwall Capacity, WSP 2 (lbs/ft) (SDPWS 4.3.3) Provide two 5 ft wide by 9 ft height panels and two 3 ft wide by 9 ft height panels. Three ft wide panels require adjustment per SDPWS 4.3.3.4.1. Capacity of each 3 ft wide segment is multiplied by 2bs/h adjustment factor.

bs  3 Width of panel requiring capacity adjustment (ft)  2bs  Capacity 2 5v   (2 )3( )v wASDWSP  h  wASDWSP

Capacity 7455 Wall capacity (lbs) Wall capacity exceeds demand, so design is OK.

Hold down capacity is based on induced unit shear (hold downs at each end of each panel)

 2bs  Leff  25()  (2 )3()Leff  14 Provided effective Full Height Sheathing Length (ft)  h  SEGMENTED SHEAR WALL DESIGN - WIND

 Vw  T   h T 3549 Required Hold down capacity (lbs) (SDPWS 4.3.6.1.2)  Leff 

Hold down would need to be combined with 2nd floor hold down requirements. Dead load offset has been neglected in this example. (SDPWS 4.3.6.4.2)

3.2 Segmented Shear Wall Design - Seismic

E3.2 - Segmented Shear Wall Design - Seismic Using the 2015 Special Design Provisions for Wind and Seismic (SDPWS), design the first floor wall shown in the diagram below as a segmented shear wall for a two-story house using Allowable Stress Design (ASD) provisions

Seismic Design Category D1 (Based on 2015 IRC) Building dimensions:

L = 40 ft W = 32 ft Roof pitch = 7:12 Top plate to ridge height = 9.3 ft Wall height = 9 ft Door height = 7 ft 6 in. Window height = 4.5 ft Stud spacing = 16 in. o.c. Studs are Southern Pine (G=0.55)

Neglect deflection check.

Use ASCE 7-10 Minimum Design Loads for Buildings and Other Structures Simplified Approach to determine seismic loads.

Check maximum segment length based on Aspect Ratio Limits Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4) 3.2 Minimum segment length =Wall Height/Aspect Ratio 9 L  L  2.6 Minimum full height wall segment length (ft) SEGMENTED SHEAR WALL DESIGN - SEISMIC min 3.5 min

Vs  4733 Applied shear load on each shear wall due to seismic force (lbs) h9 Wall height (ft)

(Note: Seismic force calculated using WFCM Table 2.6 and WFCM Commentary.) Assume 7/16 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. Studs @ 16 in. o.c. triggers Footnote 2 which allows for use of 15/32 in. panel shear values. SDPWS Table 4.3A nominal capacity = 760 lbs/ft (Seismic). Unlike wind design, gypsum capacity is not included for seismic shear wall design. 760 v  v  380 ASD Shearwall Capacity (lbs/ft) (SDPWS 4.3.3) sASDWSP 2 sASDWSP

Provide two 5 ft wide by 9 ft height panels and two 3 ft wide by 9 ft height panels. Three ft wide panels require adjustment per SDPWS 4.3.3.4.1. Capacity of each 3 ft wide segment is multiplied by 2bs/h adjustment factor.

bs  3 Width of panel requiring capacity adjustment (ft)  2bs  Capacity 2 5v   ()3()v2  sASDWSP  h  sASDWSP

Capacity 5320 Wall capacity (lbs) Wall capacity exceeds demand, so design is OK.

Hold down capacity is based on induced unit shear of effective length of Full Height Segments.

 2bs  L  25()  ()3()2 L  14 Provided effective Full Height Sheathing Length (ft) eff  h  eff

Vs T   h T 3043 Required Hold down capacity (lbs) (SDPWS 4.3.6.1.2) Leff Hold down would need to be combined with 2nd floor hold down requirements. Dead load offset has been neglected in this example. (SDPWS 4.3.6.4.2)

3.3 Perforated Shear Wall Design - Wind

E3.3 - Perforated Shear Wall Design - Wind Check maximum segment length based on Aspect Ratio Limits Using the 2015 Special Design Provisions for Wind and Seismic (SDPWS), design the first floor Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4) wall shown in the diagram below as as perforated shear wall (PSW) for a two-story house using Minimum segment length = Wall Height/Aspect Ratio Allowable Stress Design (ASD) provisions 9 L  Minimum full height wall segment length (ft) Design Wind Speed = 160 mph (3 sec. gust, 700 year return) min 3.5 Exposure B All full height segments satisfy aspect ratio requirements. Maximum aspect ratio for WSP shear Building dimensions: walls to avoid capacity adjustments = 2:1, so 3 foot wide segments will require capacity adjustments. L = 40 ft W = 32 ft Design with Interior Gypsum Roof pitch = 7:12 Top plate to ridge height = 9.3 ft Vw  5520 Applied shear load on each shear wall due to wind force (lbs) Wall height = 9 ft h9 Wall height (ft) Door height = 7 ft 6 in. Window height = 4.5 ft (Note: Shear load calculated using Table 2.5B of 2015 Wood Frame Construction Manual) Stud spacing = 16 in. o.c. Studs are Southern Pine (G=0.55) Assume 15/32 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind) Check design with and without interior gypsum, neglect deflection. 1065 v  v  532.5 ASD Shear wall Capacity (lbs/ft) (SDPWS 4.3.3) Use Minimum Design Loads for Building and Other Structures (ASCE 7-10) to determine loads. wASDWSP 2 wASDWSP

Assume 1/2 in. thick Gypsum Wallboard (GWB) Sheathing, 5d cooler nail @ 7 in. o.c. edge spacing, 16 in. o.c. studs, blocked

SDPWS Table 4.3C nominal capacity = 250 lb/ft

250 vwASDGWB  vwASDGWB  125 ASD Shear wall Capacity, Gypsum (lbs/ft) 2 (SDPWS 4.3.3)

Capacity of WSP and GWB can be added (SDPWS 4.3.3.3.2)

vWCASD  vwASDWSP  vwASDGWB

vWCASD  657.5 ASD Shear wall Capacity, WSP and GWB combined (lbs/ft)

Calculate PSW Shear Capacity Adjustment Factor (Co)

 23   L  25() 4 3 i  9   Effective length of Full Height Segments (ft) using Li  18 adjustment from SDPWS 4.3.4.3

Ltot  40 Wall length (ft)

Check maximum segment length based on Aspect Ratio Limits Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4) Minimum segment length = Wall Height/Aspect Ratio 9

L  Minimum full height wall segment length (ft) 3.3 min 3.5 All full height segments satisfy aspect ratio requirements. Maximum aspect ratio for WSP shear walls to avoid capacity adjustments = 2:1, so 3 foot wide segments will require capacity PERFORATED SHEAR WALL DESIGN - WIND adjustments.

Design with Interior Gypsum

Vw  5520 Applied shear load on each shear wall due to wind force (lbs) h9 Wall height (ft) (Note: Shear load calculated using Table 2.5B of 2015 Wood Frame Construction Manual)

Assume 15/32 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind)

1065 v  v  532.5 ASD Shear wall Capacity (lbs/ft) (SDPWS 4.3.3) wASDWSP 2 wASDWSP

Assume 1/2 in. thick Gypsum Wallboard (GWB) Sheathing, 5d cooler nail @ 7 in. o.c. edge spacing, 16 in. o.c. studs, blocked

SDPWS Table 4.3C nominal capacity = 250 lb/ft

250 vwASDGWB  vwASDGWB  125 ASD Shear wall Capacity, Gypsum (lbs/ft) 2 (SDPWS 4.3.3)

Capacity of WSP and GWB can be added (SDPWS 4.3.3.3.2)

vWCASD  vwASDWSP  vwASDGWB

vWCASD  657.5 ASD Shear wall Capacity, WSP and GWB combined (lbs/ft)

Calculate PSW Shear Capacity Adjustment Factor (Co)

 23   L  25() 4 3 i  9   Effective length of Full Height Segments (ft) using Li  18 adjustment from SDPWS 4.3.4.3

Ltot  40 Wall length (ft)

2 Ao  4( 4.5 3 ) () 6 7.5 Area of openings (ft ) Ao  99 1 r  Ao r 0.621 (SDPWS Eqn. 4.3-6) 1  hL i

 r  Ltot Co    Co  0.78 (SDPWS Eqn. 4.3-5)  3 2r Li

vPSW  vWCASDCo vPSW  516 Perforated Shearwall (PSW) Capacity (lbs/ft)

Vw LPSW  LPSW  10.7 Required length of Full Height Sheathing (LPSW) (ft) vPSW

Provided effective length 18 ft of FHS is greater than required 10.7 ft

Vwh T 1T Required Hold-down capacity (lbs) (SDPWS Eqn. 4.3-8) CoLi Design without Interior Gypsum

Vw  5520 Wind reaction on shear wall (lbs) h9 Wall height (ft)

Assume 15/32 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind)

1065 v  v  532.5 ASD Shear wall Capacity (lbs/ft) (SDPWS 4.3.3) wASDWSP 2 wASDWSP

Co  0.78 Calculated PSW Shear Capacity Adjustment Factor (same as above)

vPSW  vwASDWSPCo Perforated Shearwall (PSW) Capacity (lbs/ft) vPSW  418 Vw LPSW  10.7 Required length of Full Height Sheathing (LPSW) (ft) vPSW Provided effective length 18 ft of FHS is greater than required 13.2 ft

Vwh T 3519 Required Hold-down capacity (lbs) (SDPWS Eqn. 4.3-8) CoLi Hold-down would need to be combined with 2nd floor hold-down requirements. Dead load offset has been neglected in this example (SDPWS 4.3.6.4.2)

3.4 Perforated Shear Wall Design - Seismic

E3.4 - Perforated Shear Wall Design - Seismic Using the 2015 Special Design Provisions for Wind and Seismic (SDPWS), design the first floor wall shown in the diagram below as as perforated shear wall for a two-story house using Allowable Stress Design (ASD) provisions.

Seismic Design Category D1 (Based on 2015 IRC)

Building dimensions: 3.4 L = 40 ft W = 32 ft

Roof pitch = 7:12 PERFORATED SHEAR WALL DESIGN - SEISMIC Top plate to ridge height = 9.3 ft Wall height = 9 ft Door height = 7 ft 6 in. Window height = 4.5 ft Stud spacing = 16 in. o.c. Studs are Southern Pine (G=0.55)

Neglect deflection.

Use Minimum Design Loads for Building and Other Structures (ASCE 7-10) to determine loads.

Check maximum segment length based on Aspect Ratio Limits Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4) Minimum segment length =Wall Height/Aspect Ratio 9 L  L  2.6 Minimum full height wall segment length (ft) min 3.5 min All full height segments satisfy aspect ratio requirements. Maximum aspect ratio for WSP shear walls to avoid capacity adjustments = 2:1, so 3 foot wide segments will require capacity adjustments.

Vs  4733 Applied shear load on each shear wall due to seismic force (lbs)

h9 Wall height (ft) (Note: Seismic force calculated using WFCM Table 2.6 and WFCM Commentary.)

Assume 7/16 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. edge spacing. Studs @ 16 in. o.c. triggers Footnote 2 allows for use of 15/32 in. panel shear values. SDPWS Table 4.3A nominal capacity = 760 lbs/ft (Seismic). Unlike wind design, gypsum capacity is not included for seismic shear wall design. 760 v  v  380 ASD Shearwall Capacity (lbs/ft) (SDPWS 4.3.3) sASDWSP 2 sASDWSP

Calculate PSW Shear Capacity Adjustment Factor (Co)

 23   L  25() 4 3 i  9   Effective length of Full Height Segments (ft) using adjustment from SDPWS 4.3.4.3 Li  18

Ltot  40 Total wall length (ft)

2 Ao  4( 4.5 3 ) () 6 7.5 Area of openings (ft )

Ao  99 1 r  Ao (SDPWS Eqn. 4.3-6) 1  hL i

 r  Ltot Co    (SDPWS Eqn. 4.3-5)  3 2r Li

Co  0.78

vPSW  vsASDWSPCo vPSW  298 Perforated Shearwall (PSW) Capacity (lbs/ft)

Vs FHSLPSW  LPSW  15.9 Required length of Full Height Sheathing (FHS) (ft) vPSW

Provided effective length 18 ft of FHS is greater than required 15.9 ft 3.4

Hold down capacity for Perforated Shearwalls specified in SDPWS Eqn. 4.3-8 PERFORATED SHEAR WALL DESIGN - SEISMIC

Vsh T 1T T 3017 Required Hold down capacity (lbs) CoLi

Hold down would need to be combined with 2nd floor hold down requirements. Dead load offset has been neglected in this example. (SDPWS 4.3.6.4.2)

NOTES

AMERICAN WOOD COUNCIL American Wood Council

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