nds structural wood design examples 2015/2018 edition€¦ · 1.2a simply supported beam capacity...

EXAMPLESStructural Wood Design Examples2015/2018 EDITION

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EXAMPLESStructural Wood Design Examples2015/2018 EDITION

AMERICAN WOOD COUNCIL

STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMSii

2015/2018 Structural Wood Design Examples

First Web Version: August 2019

ISBN 978-1-940383-51-4

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STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS iii

FOREWORDThis document is intended to aid instruction in structural design of wood structures using both Allowable Stress Design (ASD) and Load and Resistance Factor Design (LRFD). It contains design examples and complete solutions calculated using ASD and LRFD. Solutions have been developed based on the 2015 and 2018 National Design Specification®(NDS®) for Wood Construction, and the 2015 Special Design Provisions for Wind and Seismic (SDPWS), as appropriate. References are also made to the 2015 and 2018 Wood Frame Construction Manual (WFCM) for One- and Two- Family Dwellings. Copies of these standards produced by the American Wood Council can be obtained at www.awc.org/codes-standards/publications. Procedures will be applicable for both 2015 and 2018 versions of the NDS and WFCM, unless otherwise noted. Example problems range from

simple to complex and cover many design scenarios. In the solutions where a particular provision of the NDS or SDPWS is cited, reference is made to the document and corresponding provision number, e.g. NDS 4.3.1. It is intended that this document be used in conjunction with competent engineering design, accurate fabrication, and adequate supervision of construction. Neither the American Wood Council, the International Code Council, nor their members assume any responsibility for errors or omissions in this document, nor for engineering designs, plans, or construction prepared from it. Those using this document assume all liability arising from its use. The design of engineered structures is within the scope of expertise of licensed engineers, architects, or other licensed professionals for applications to a particular structure.

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.


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STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMSiv



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS v

TABLE OF CONTENTSProblem Page Problem Page

Foreword ............................................................ iii

Characteristic Problem Set ............................ 11.1 Adjustment Factors (ASD and LRFD) 11.2a Simply Supported Beam Capacity Check

(ASD) 71.2b Simply Supported Beam Capacity Check

(LRFD) 111.3 Glued Laminated Timber Beam Design

(ASD) 161.4 Compression Members - 4x4 and 6x6

(ASD) 231.5a Compression Member - 2x6 Stud (ASD) 281.5b Compression Member - 2x6 Stud (LRFD) 311.6 Bending and Axial Tension (ASD) 341.7 Bending and Axial Compression (ASD) 401.8 Bi-Axial Bending and Axial Compression

(ASD) 451.9 Loadbearing Wall Wood Stud Resisting

Wind and Gravity Loads 55

Connection Problem Set ...............................802.1a Withdrawal - Plain Shank Nail 802.1b Fastener Uplift - Roof Sheathing Ring

Shank Nail - 5/8" WSP (2018 NDS Only) 822.1c Fastener Uplift - Roof Sheathing Ring

Shank Nail - 7/16" WSP (2018 NDS Only) 842.2 Common Nail Lateral - Single Shear

Wood-to-wood 862.3 Withdrawal - Lag Screw 912.4 Wood Screw Lateral - Double Shear

Wood-to-wood 932.5 Bolt Lateral - Single Shear Wood-to-Wood 972.6 Bolted Wood-to-Wood Tension Splice

Connection 105

Shear Wall Problem Set ...............................1103.1 Segmented Shear Wall - Wind 1103.2 Segmented Shear Wall - Seismic 1143.3 Perforated Shear Wall - Wind 1163.4 Perforated Shear Wall - Seismic 119



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMSvi


E1.1 - 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 onlydead and live loads) for an exterior deck. The in-service moisture content is greater than 19%, andthe member is not subject to elevated temperatures. The ends are held in place with full-depthblocking. 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 usingboth Allowable Stress Design (ASD) and Load and Resistance Factor Design (LRFD). Assumelumber is incised.


STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 1AD

JUSTM

ENT FACTO

RS (ASD

AND

LRFD

)1

.1

1.1 Adjustment Factors (ASD and LRFD)


E1.1 - 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 onlydead and live loads) for an exterior deck. The in-service moisture content is greater than 19%, andthe member is not subject to elevated temperatures. The ends are held in place with full-depthblocking. 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 usingboth Allowable Stress Design (ASD) and Load and Resistance Factor Design (LRFD). Assumelumber is incised.

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)


1.1 ADJUSTMENT FACTORS (ASD AND LRFD)2


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

Format Conversion Factor

KF 2.54 (NDS Table 4.3.1)

Resistance Factor

ϕ 0.85 (NDS Table 4.3.1)

Time Effect Factor

(NDS Appendix N.3.3 - use combination 1.2D+1.6L+0.5(Lr or S or R), Lfrom occupancy)

λ 0.8

F'bLRFD Fb CMb Ct CL CF Cfu Ci Cr KF 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



Wet Service Factor

CMV 0.97 (NDS Supplement Table 4A)

Temperature Factor


Incising Factor


F'vASD Fv CD CMV Ct CiAdjusted 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


KFv 2.88 (NDS Table 4.3.1)

Resistance Factor

ϕv 0.75 (NDS Table 4.3.1)

Time Effect Factor

λ 0.8 (NDS Appendix N.3.3 - use combination 1.2D+1.6L+0.5(Lr or S or R),L from occupancy)

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



JUSTM

ENT FACTO

RS (ASD

AND

LRFD

)1

.1


F'vLRFD 241 psi

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



Wet Service Factor

CMt 1.0 (NDS Supplement Table 4A)

Temperature Factor


Size Factor

CF 1.3 (NDS Supplement Table 4A)

Incising Factor


F'tASD Ft CD CMt Ct CF CiAdjusted 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


KFt 2.70 (NDS Table 4.3.1)

Resistance Factor

ϕt 0.8 (NDS Table 4.3.1)

Time Effect Factor

λ 0.8 (NDS Appendix N.3.3 - use combination 1.2D+1.6L+0.5(Lr or S or R), Lfrom occupancy)

F'tLRFD Ft CMt Ct CF Ci KFt ϕt 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

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


Temperature Factor


Incising Factor


Buckling Stiffness Factor

CT 1.0 (NDS 4.4.2)

E'minASD E CME 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


Temperature Factor


Incising Factor


Buckling Stiffness Factor

CT 1.0 (NDS 4.4.2)



JUSTM

ENT FACTO

RS (ASD

AND

LRFD

)1

.1



KFE 1.76 (NDS Table 4.3.1)

Resistance Factor

ϕE 0.85 (NDS Table 4.3.1)

E'minLRFD E CME CtE CiE CT KFE ϕE

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





STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 7SIM

PLY SUPPO

RTED

BEAM

CAPACITY CHECK

(ASD)

1.2

A

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 thecenter of the span. Determine the maximum allowable load on the hoist (including its weight) based onbending. Assume normal load duration. The beam is supported on a 2x4 top plate. Lateral support is providedonly 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 deflectionfrom 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 requirementson load combinations.

1.2a Simply Supported Beam Capacity Check (ASD)



1.2A SIMPLY SUPPORTED BEAM CAPACITY CHECK (ASD)8

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 thecenter of the span. Determine the maximum allowable load on the hoist (including its weight) based onbending. Assume normal load duration. The beam is supported on a 2x4 top plate. Lateral support is providedonly at the ends of the member and the ends are considered pinned.



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)

Ib d3

12Ag b d S

b d2

6

Ag 53.38 in2 S 135.66 in3 I 1034 in4


F'b* Fb CD CM Ct CF Ci Cr F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. F'b* 1500 psi

lu 12inft l lu 240 in Laterally unsupported length

lud

15.7 lu/d >7 (Table 3.3.3)

le 1.37 lu 3 d le 375 in (Table 3.3.3)



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 9SIM

PLY SUPPO

RTED

BEAM

CAPACITY CHECK

(ASD)

1.2

A

RBle d

b2 RB 21.6 Slenderness ratio for bending (3.3-5)

FbE1.20 E'min

RB2

FbE 1776 psi Critical bucking design value for bending (3.7.1)

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

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

Mmax F'bS

12inft

Mmax 14849 ftꞏlbf

Determine Maximum Hoist Load P

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

ρ 37.5lbf

ft3 wbeamweight ρ

b

12inft

d

12inft

wbeamweight 13.9 plf Note:wbeamweight is self weight of beam

Mbeamweight is moment due to selfweight Mallow is maximum allowablemoment due to applied hoist load

Mbeamweightwbeamweight l( )2

8 Mbeamweight 695 ftꞏlbf

Mallow Mmax Mbeamweight Mallow 14154 ftꞏlbf

P 4Mallow

l

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



1.2A SIMPLY SUPPORTED BEAM CAPACITY CHECK (ASD)10

Check Beam's Capacity to Resist Shear from Maximum (bending controlled) Load

VP2

V 1415 lbf

fv3 V2 b d

fv 40 psi

F'v Fv CD CM Ct Ci F'v 180 psi fv<F'v okay

Check Compression Perpendicular to Grain at Bearing Points

fc⊥V

b wbearing fc⊥ 116 psi

F'c⊥ Fc⊥ CM Ct Ci F'c⊥ 625 psi fc⊥< F'c⊥ okay

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 consideredshort-term.

Δbeam_weight5

384 E I

wbeamweight12

ftin l 12

inft

4 Δbeam_weight 0.025 in

Δcrane_loadP

48 E Il 12

inft

3 Δcrane_load 0.415 in

Δtotal Δbeam_weight Δcrane_load Δtotal 0.44 in

Calculate Span/Deflection Ratio

12 linft

Δtotal545 L/Δtotal ratio

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 thecenter of the span. Determine the maximum allowable load on the hoist (including its weight) based onbending. 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.





1.2

BSTRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 11

SIMPLY SU

PPOR

TED B

EAM CAPACITY CH

ECK (LR

FD)





1.2b Simply Supported Beam Capacity Check (LRFD)



1.2B SIMPLY SUPPORTED BEAM CAPACITY CHECK (LRFD)12





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

λ 0.8 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)

KFb 2.54 ϕb 0.85

Subscript "b" denotes bending, "v" denotesshear, "c⊥" denotes compressionperpendicular to grain

KFv 2.88 ϕv 0.75

KFc⊥ 1.67 ϕc⊥ 0.90

KFEmin 1.76 ϕEmin 0.85

E' E CM Ct Ci E'min Emin CM Ct Ci CT KFEmin ϕEmin

E' 1900000 psi E'min 1032240 psi

Member length and properties

l 20 ft b 3.5 in d 15.25 in wbearing 3.5 in (width of bearing)

Ib d3

12Ag b d S

b d2

6

Ag 53.38 in2 S 135.66 in3 I 1034 in4

Beam Stability FactorF'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. The following calculations determine the beamstabilty factor CL:

F'b* Fb CM Ct CF Ci Cr KFb ϕb λ

F'b* 2591 psi



1.2


SIMPLY SU

PPOR

TED B

EAM CAPACITY CH

ECK (LR

FD)


lud

15.7 lu/d >7 (Table 3.3.3)

le 1.37 lu 3 d le 375 in (Table 3.3.3)

RBle d


FbE1.20 E'min

RB2


CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.827

F'b F'b* Cfu CL

F'b 2143 psi F'b is adjusted bending design value with all adjustmentfactors.

Determine Maximum Moment Allowed on Beam

Maximum total moment is the adjusted bending design value F'b times the section modulus S

Mmax F'bS

12inft

Mmax 24230 ftꞏlbf



1.2B SIMPLY SUPPORTED BEAM CAPACITY CHECK (LRFD)14

Determine Maximum Hoist Load P

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

Note:wbeamweight is self weight of beam

Mbeamweight is moment due to selfweight Mallow is maximum allowablemoment due to applied hoist load

ρ 37.5lbf

ft3 wbeamweight ρ

b

12inft

d

12inft

wbeamweight 13.9 plf

Mbeamweight1.2wbeamweight l( )2

8 Mbeamweight 834 ftꞏlbf 1.2 factor for load combination 1.2D+1.6L

applied here

Mallow Mmax Mbeamweight

Mallow 23396 ftꞏlbf

P 4Mallowl 1.6

1.6 factor for load combination 1.2D+1.6Lapplied 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

V1.6P2

1.6 factor applied here for live load V 2340 lbf

fv3 V2 b d

fv 66 psi

F'v Fv CM Ct Ci KFv ϕv λ F'v 311 psi fv<F'v OK

Check Compression Perpendicular to Grain at Bearing Points

fc⊥V

b wbearing fc⊥ 191 psi

F'c⊥ Fc⊥ CM Ct Ci KFc⊥ ϕc⊥ F'c⊥ 939 psi fc⊥< F'c⊥ OK

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.



1.2


SIMPLY SU

PPOR

TED B

EAM CAPACITY CH

ECK (LR

FD)

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 consideredshort-term.

Δbeam_weight5

384 E I

wbeamweight12

ftin l 12

inft

4 Δbeam_weight 0.025 in

Δcrane_loadP

48 E Il 12

inft

3 Δcrane_load 0.429 in

Δtotal Δbeam_weight Δcrane_load Δtotal 0.454 in

Calculate Span/Deflection Ratio

12 linft

Δtotal529 L/Δtotal ratio



1.3 GLUED LAMINATED TIMBER BEAM DESIGN (ASD) 16

1.3 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 lbsnow load (SL)) applied by purlins at 8 ft on center (at 1/4 points plus the ends). Member has lateral supportsat 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 150degrees (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


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

Ag b d Sxxb d2

6 Ixx

b d3

12 lsupport 6 in

Ag 151.3 in2 Sxx 762.6 in3 Ixx 11534 in4


Fbx+* Fbx+ CD CM Ct Cc CI F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL flat use factor Cfuand volume factor CV applied. Fbx+* 2760 psi

lu 12inft 8 ft lu 96 in Laterally unsupported length

le 1.54 lu le 148 in (Table 3.3.3)

RBle d




STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 17G

LUED

LAMIN

ATED TIM

BER

BEAM

DESIG

N (ASD

) 1

.3




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


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

Ag b d Sxxb d2

6 Ixx

b d3

12 lsupport 6 in

Ag 151.3 in2 Sxx 762.6 in3 Ixx 11534 in4


Fbx+* Fbx+ CD CM Ct Cc CI F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL flat use factor Cfuand volume factor CV applied. Fbx+* 2760 psi

lu 12inft 8 ft lu 96 in Laterally unsupported length

le 1.54 lu le 148 in (Table 3.3.3)

RBle d





FbE1.20 E'xmin

RB2

FbE 6373 psi Critical bucking design value for bending (3.3.3.8)

Note: CL must be ≤ 1.0 per NDS 5.3-1

CL

1FbEFbx+*

1.9

1FbEFbx+*

1.9

2 FbEFbx+*

0.95

CL 0.965

Volume Factor

x 20 Southern Pine

CV min

1

21 ftl

1x 12 in

d

1x

5.125 in

b

1x

CV 0.936 CL and Cv shall not apply simulatenously (5.3.5). CV isless than CL. CV controls

Adjusted Bending Design Value

F'b Fbx+* minCL

CV

Cfu

F'b 2584 psi F'b is adjusted bending design value with all adjustmentfactors.

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 isthe adjusted bending design value F'b times the section modulus, S

wbeamweight 40lbfft

Estimated self weight of beam

P 5000 lbf Load from purlin

Mest Pl2

wbeamweight l2

8

12

inft Mest 1021440 inꞏlbf

SreqdMestF'b

Sreqd 395 in3 Sreqd is < 762.6 in3 so try a smallersection for efficiency




LUED

LAMIN

ATED TIM

BER

BEAM

DESIG

N (ASD

) 1

.3

Try a 5 X 22 member

b2 5 in d2 22 in Beam dimension for trial section. Subscript "2" used to denotesecond iteration

Ag2 b2 d2 Sxx2b2 d2

2

6 Ixx2

b2 d23

12

Ag2 110 in2 Sxx2 403.3 in3 Ixx2 4437 in4

RB2le d2

b22

RB2 11.406 Slenderness ratio for bending (3.3-5)

FbE21.20 E'xmin

RB22

FbE2 8763 psi Critical bucking design value for bending (3.3.3.8)

Note: CL must be ≤ 1.0 per NDS 5.3-1 CL2

1FbE2Fbx+*

1.9

1FbE2Fbx+*

1.9

2 FbE2Fbx+*

0.95

CL2 0.978

CV2 min

1

21 ftl

1x 12 in

d2

1x

5.125 in

b2

1x

CV2 0.951 CL and Cv shall not apply simulatenously (5.3.5). CV isless than CL. CV controls

Adjusted Bending Design Value

F'b2 Fbx+* minCL2

CV2

Cfu

F'b2 2625 psi F'b2 is adjusted bending design value with all adjustmentfactors for the trial section considered.

wbeamweight2 30lbfft

Self weight for 5 X 22 beam estimated per NDSSupplement 3.1.3

M2 Pl2

wbeamweight2 l2

8

12

inft




M2 1006080 inꞏlbf

Sreqd2M2F'b2

Sreqd2 383 in3 383 in3 < 403 in3

Okay

Shear Parallel to Grain

The two 5000 pound purlin loads at the ends of the beam are within "d" of the supports and can be ignored for shear(3.4.3.1(a)). Shear determined from remaining purlin loads

Vpurlins3 P2

Vpurlins 7500 lbf

Vbeamweight2 wbeamweight2l2

d212lsupport

Vbeamweight2 417.5 lbf

Vtotal Vpurlins Vbeamweight2

Vtotal 7918 lbf

Adjusted (F v ') and Actual ( f v ) Shear Parallel to Grain

Cvr 1.0 (NDS 5.3.10)

F'v Fvx CD CM Ct Cvr

F'v 305 psi

fv32

Vtotalb2 d2

fv < F'v Actual shear stress parallel to grain less than adjusted OKfv 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 beamare included in the bearing load calculations

Rpurlins125 P

Rpurlins 12500 lbf

Rbeamweight212wbeamweight2 l 2

lsupport2

Rbeamweight2 487.5 lbf

Rtotal Rpurlins Rbeamweight2

Rtotal 12988 lbf

F'c⊥x Fc⊥x CM Ct

F'c⊥x 650 psi

fc⊥Rtotal

b2 lsupport

fc⊥ 433 psi Actual compression stress perpendicular to grain is less thanthe adjusted compression perpendicular to grain design value.OK

At Purlins

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 stressesgreater than the adjusted compression perpendicular to grain design value

AhangerP

F'c⊥x

Ahanger 7.69 in2

Assuming that purlins frame in from both sides of the beam, the width of the hanger can be calculated as follows:

whangerAhanger0.50 b2

whanger 3.08 in




LUED

LAMIN

ATED TIM

BER

BEAM

DESIG

N (ASD

) 1

.3

M2 1006080 inꞏlbf

Sreqd2M2F'b2

Sreqd2 383 in3 383 in3 < 403 in3

Okay

Shear Parallel to Grain

The two 5000 pound purlin loads at the ends of the beam are within "d" of the supports and can be ignored for shear(3.4.3.1(a)). Shear determined from remaining purlin loads

Vpurlins3 P2

Vpurlins 7500 lbf

Vbeamweight2 wbeamweight2l2

d212lsupport

Vbeamweight2 417.5 lbf

Vtotal Vpurlins Vbeamweight2

Vtotal 7918 lbf

Adjusted (F v ') and Actual ( f v ) Shear Parallel to Grain

Cvr 1.0 (NDS 5.3.10)

F'v Fvx CD CM Ct Cvr

F'v 305 psi

fv32

Vtotalb2 d2

fv < F'v Actual shear stress parallel to grain less than adjusted OKfv 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 beamare included in the bearing load calculations

Rpurlins125 P

Rpurlins 12500 lbf

Rbeamweight212wbeamweight2 l 2

lsupport2

Rbeamweight2 487.5 lbf

Rtotal Rpurlins Rbeamweight2

Rtotal 12988 lbf

F'c⊥x Fc⊥x CM Ct

F'c⊥x 650 psi

fc⊥Rtotal

b2 lsupport

fc⊥ 433 psi Actual compression stress perpendicular to grain is less thanthe adjusted compression perpendicular to grain design value.OK

At Purlins

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 stressesgreater than the adjusted compression perpendicular to grain design value

AhangerP

F'c⊥x

Ahanger 7.69 in2

Assuming that purlins frame in from both sides of the beam, the width of the hanger can be calculated as follows:

whangerAhanger0.50 b2

whanger 3.08 in




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

lb 3 in Cblb 0.375 in

lb Cb 1.125

1Cb

whanger 2.735 in

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 thecenter to center distance between supports. The length of the span used in design is the face to face distanceplus 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 requiredbearing distance is 12,998 lbs/(5 inches * 650 psi) or 4 inches. At the interior face, half the purlin load is assumedto be transferred to the beam end. Required length in bearing is (2500 lbs + 7500 lbs + 488 lbs)/5 inches * 650psi) 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 reducesmoments and bending stresses by 1.25%. The reduction is considered insufficient to allow the use of the nextsmaller beam.

Deflection

The specification does not include specific deflection limits for roofs. In some applications, deflections may be criticaland 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 isusually 150% of the dead load deflection. Deflection for the 5000 lb concentrated loads and the beam weight is:

Δpurlin

19 P 12 linft

3

384 E'x Ixx2 Δbeamweight

5wbeamweight2

12inft

l 12inft

4

384E'x Ixx2

Δpurlin 1.754 in Δbeamweight 0.089 in Δcamber 0.375in

Δtotal Δpurlin Δbeamweight ΔcamberNote: for this example 3/8" camber will be specified

Δtotal 1.468 in

Length/Deflection Ratio

l 12inft

Δtotal262 L/Δ >240. The length deflection ratio satisfies specified criteria



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 23CO

MPR

ESSION

MEM

BER

S AND

COLU

MN

STABILITY CALCU

LATION

(ASD)

1.4

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 foran interior column (only carrying gravity loads - dead load (DL) + floor live load (LL)). Bothmembers are No. 2 Southern Pine and have a length of 10 feet. Both ends are assumed to bepinned (Ke=1.0 - NDS 3.7.1.2). Assume all members are loaded concentrically.



1.4 COMPRESSION MEMBERS AND COLUMN STABILITY CALCULATION (ASD)24



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)

Ci 1.0 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)

E' E CM Ct Ci E' 1.4 106 psi

E'min Emin CM Ci Ct CT E'min 5.1 105 psi

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

Area 12.25in2

Ke 1.0

Length 120in

le Ke Length

le 120 in

led

34.3 Needs to be less than 50 (NDS 3.7.1.3)




MPR

ESSION

MEM

BER

S AND

COLU

MN

STABILITY CALCU

LATION

(ASD)

1.4



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)

Ci 1.0 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)

E' E CM Ct Ci E' 1.4 106 psi

E'min Emin CM Ci Ct CT E'min 5.1 105 psi

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

Area 12.25in2

Ke 1.0

Length 120in

le Ke Length

le 120 in

led

34.3 Needs to be less than 50 (NDS 3.7.1.3)

Column Stability Factor Calculation

FcE0.822 E'min( )

led

2 FcE 357 psi (NDS 3.7.1)

Fc* Fc CD CM Ct CF Ci Fc* 1450 psi

csawn 0.8 (NDS 3.7.1)

CP

1FcEFc*

2csawn

1FcEFc*

2csawn

2 FcEFc*

csawn

CP 0.232

Axial Buckling Capacity

F'c Fc CD CM Ct CF Ci CP

F'c 336 psi

P F'c Area

P 4120 lbf

Bearing Capacity

fc F'c fc 336 psi Actual compression stress parallel to grainassuming column loaded to 100% calculatedbuckling capacity0.75 Fc* 1088 psi

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



1.4 COMPRESSION MEMBERS AND COLUMN STABILITY CALCULATION (ASD)26

Reference and Adjusted Design Values - 6x6 Post Note: "2" subscript indicates 6x6;differentiates between 6x6 and4x4 properties and calculationsFc2 525psi

(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)

E'2 E2 CM2 Ct2 Ci2 E'2 1.2 106 psi

E'min2 Emin2 CM2 Ci2 Ct2 CT2 E'min2 4.4 105 psi

d2 5.5inActual member dimensions = 5.5" x 5.5"

Area2 30.25in2

Ke2 1.0

Length2 120in

le2 Ke2 Length2

le2 120 in

le2d2

21.8Needs to be less than 50 (NDS 3.7.1.3)




MPR

ESSION

MEM

BER

S AND

COLU

MN

STABILITY CALCU

LATION

(ASD)

1.4

Column Stability Factor Calculation

FcE20.822 E'min2

le2d2

2 FcE2 760 psi (NDS 3.7.1)

Fc2* Fc2 CD2 CM2 Ct2 CF2 Ci2 Fc2* 525 psi

(NDS 3.7.1)csawn2 0.8

CP2

1FcE2Fc2*

2csawn2

1FcE2Fc2*

2csawn2

2 FcE2Fc2*

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 assumingcolumn 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 perNDS 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 MEMBER ANALYSIS (ASD)28

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 withgypsum board, carries dead load and snow load from the roof. Determine CP and the allowable compressionparallel to grain design value (Fc') for the stud. Assume studs are placed 16" on center and top and bottomplates are the same grade and species. Determine axial loads based on buckling and bearing limit states.




MPR

ESSION

MEM

BER

ANALYSIS (ASD

)1

.5A

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 withgypsum board, carries dead load and snow load from the roof. Determine CP and the allowable compressionparallel to grain design value (Fc') for the stud. Assume studs are placed 16" on center and top and bottomplates are the same grade and species. Determine axial loads based on buckling and bearing limit states.

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)

E'min Emin CM Ct Ci CT E'min 510000 psi

F'c⊥ Fc⊥ CM Ct Ci F'c⊥ 425 psi


l 91.5 in b 1.5 in d 5.5 in

Column Stability FactorFc* 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 lateralsupport. Strong axis buckling controls. See NDS A.11.3regarding lateral support of the weak axis due to gypsumsheathing.

le2 0 le1 l

le2b

0le1d

16.636 le<50 OK (NDS 3.7.1.4)

FcE0.822 E'min

le1d

2 FcE 1515 psi Critical bucking design value for compression

members (3.7.1.5)

c 0.8 Sawn lumber (3.7.1.5)

CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c Column Stability Factor (3.7-1)

CP 0.705

F'c Fc* CP

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



1.5A COMPRESSION MEMBER ANALYSIS (ASD)30

Determine Axial Loads Based on Buckling and Bearing

PBuckling b d F'c PBuckling 8458 lbf

PBearing b d F'c⊥ PBearing 3506 lbf Perpendicular to grain bearing

PBearing2 b d Fc* PBearing2 12002 lbf Parallel to grain bearing

Note: Bearing area factor (Cb) can be used to increase the bearing controlled load on interior studs. The bearingfactor for the 1-1/2 bearing length measured parallel to grain is 1.25 (Equation 3.10-2 and Table 3.10.4) Cb 1.25

PBearingIncreased b d 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.




MPR

ESSION

MEM

BER

ANALYSIS (LR

D)

1.5

B

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 withgypsum 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 areplaced 16" on center and top and bottom plates are the same grade and species. Determine axial loads basedon buckling and bearing limit states.



1.5B COMPRESSION MEMBER ANALYSIS (LRFD)32

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 withgypsum 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 areplaced 16" on center and top and bottom plates are the same grade and species. Determine axial loads basedon buckling and bearing limit states.

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

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

λ 0.8 CM 1.0 Ct 1.0 CF 1.1 Ci 1.0 CT 1.0 (NDS Table 4.3.1 and Appendix N)

KFc 2.40 ϕc 0.9 KFEmin 1.76 ϕEmin 0.85

Kc⊥ 1.67 ϕc⊥ 0.9

E'min Emin CM Ct Ci CT KFEmin ϕEmin E'min 762960 psi

F'c⊥ Fc⊥ CM Ct Ci Kc⊥ ϕc⊥ F'c⊥ 638.775 psi


l 91.5 in b 1.5 in d 5.5 in

Column Stability FactorFc* Fc CM Ct CF Ci KFc ϕc λ Fc* is adjusted bending design value with all adjustment

factors except the column stability factor CP ,Fc* 2186 psi

le2 0 le1 l Effective lengths of compression member in planes of lateralsupport. Strong axis buckling controls

le2b

0le1d

16.636 <50 OK (NDS 3.7.1.4)

FcE0.822 E'min

le1d

2 FcE 2266 psi Critical buckling design value for compression

members (NDS 3.7.1.5)

c 0.8 Sawn lumber (NDS 3.7.1.5)

CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c Column Stability Factor (NDS 3.7-1)

CP 0.703




MPR

ESSION

MEM

BER

ANALYSIS (LR

D)

1.5

B

F'c Fc* CP

F'c 1537 psi F'c is adjusted compression design value with all adjustmentfactors.

Determine Axial Loads Based on Buckling and Bearing

PBuckling b d F'c PBuckling 12683 lbf

PBearing b d F'c⊥ PBearing 5270 lbf Perpendicular to grain bearing

PBearing2 b d 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 bearingfactor for the 1-1/2 bearing length measured parallel to grain is 1.25 (NDS Equation 3.10-2 and Table 3.10.4) Cb 1.25

PBearingIncreased b d F'c⊥ Cb PBearingIncreased 6587 lbf Controlling Value

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)34

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 panelpoints). The chord will be subject to a uniform dead load of 8 psf as well as tension forces (assuming pinnedconnections) 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 adequacyof 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 forrequirements on load combinations.




MB

INED

BEN

DIN

G AN

D AXIAL TEN

SION

(ASD)

1.6

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 panelpoints). The chord will be subject to a uniform dead load of 8 psf as well as tension forces (assuming pinnedconnections) 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 adequacyof 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 forrequirements 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

E' E CM Ct Ci E'min Emin CM Ct Ci CT

E' 1300000 psi E'min 470000 psi


l 12 ft b 1.5 in d 7.25 in

Ag b d Sb d2

6

Ag 10.875 in2 S 13.141 in3

Applied Loads

wD 8lbf

ft2 wtrib 4 ft Twind 880 lbf TLive 880 lbf TDead 1420 lbf




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 shortduration loads (NDS 2.3.1 and 4.3.1)

Ft' 1008 psi

T1 Twind TLive TDead Subscripts refer to Load Case

T1 3180 lbf

ft1T1Ag

Tensile stress in bottom chord

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

BendingF'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. The following calculations determine the beamstabilty factor CL:

F'b* Fb CD CM Ct CF Ci Cr

F'b* 1632 psi

Determine Beam Stability Factor CL (NDS 3.3.3)


lud

19.9 lu/d >7 (NDS Table 3.3.3)

le 1.63 lu 3 d le 256.5 in (NDS Table 3.3.3)

RBle d

b2 RB 28.75 RB < 50 OK (NDS 3.3.3.7)

FbE1.20 E'min

RB2

FbE 682 psi (NDS 3.3.3.6)

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95 (NDS Equation 3.3-6)

CL 0.404 Resulting beam stability factor CL.




MB

INED

BEN

DIN

G AN

D AXIAL TEN

SION

(ASD)

1.6

F'b F'b* CL Cfu F'b is the fully adjusted bending design value with alladjustment factors including the beam stability factor CLand flat use factor applied F'b 660 psi

F'b** F'b Since Cv does not apply to solid sawn lumber, F'b** isequal to F'b

F'b** 660 psi

Bending resulting from dead loadMmax

wD wtrib l2 12inft

8

Mmax 6912 inꞏlbf

fbMmaxS

fb 526 psi

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

Combined Bending and Axial Tension ft1Ft'

fbF'b*

0.61 <1.0. Ok (NDS Equation 3.9-1)

fb ft1

F'b**0.354 <1.0 ok (NDS Equation 3.9-2)




Load Case 2: DL+RLL

CD 1.25 Appendix B Section B.2 (non-mandatory). Roof Live Load isa construction load.

Tension

Ft' Ft CD CM Ct CF Ci Adjusted tension parallel to grain design value for shortduration loads (NDS 2.3.1 and 4.3.1)

Ft' 787.5 psi

T2 TLive TDead

T2 2300 lbf

ft2T2Ag

ft2 211 psi Ft' 787 psi


F'b* 1275 psi

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.509

F'b F'b* CL Cfu

F'b 649 psi

F'b** F'b

F'b** 649 psi

fb 526 psi F'b 649 psi

Combined Bending and Axial Tension ft2Ft'

fbF'b*

0.68 <1.0. ok

fb ft2

F'b**0.48 <1.0 ok

Load Case 3: DL only

CD 0.9

Tension

Ft' Ft CD CF CM Ci Ct

Ft' 567 psi

T3 TDead

T3 1420 lbf

ft3T3Ag


Bending


F'b* 918 psi

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.674

F'b F'b* CL Cfu

F'b 619 psi

F'b** F'b

F'b** 619 psi


Combined Bending and Axial Tension

ft3Ft'

fbF'b*

0.8 <1.0. ok

fb ft3

F'b**0.64 <1.0 ok

Results: No 2 Hem-Fir 2 x 8 satisfies NDS Criteria for combined bending and axial tension.




MB

INED

BEN

DIN

G AN

D AXIAL TEN

SION

(ASD)

1.6


CD 0.9

Tension

Ft' Ft CD CF CM Ci Ct

Ft' 567 psi

T3 TDead

T3 1420 lbf

ft3T3Ag


Bending


F'b* 918 psi

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.674

F'b F'b* CL Cfu

F'b 619 psi

F'b** F'b

F'b** 619 psi


Combined Bending and Axial Tension

ft3Ft'

fbF'b*

0.8 <1.0. ok

fb ft3

F'b**0.64 <1.0 ok

Results: No 2 Hem-Fir 2 x 8 satisfies NDS Criteria for combined bending and axial tension.



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 840lb (snow) and 560 lb (dead) plus a 25 psf wind load on their narrow face. Columns are 9 ft long and spaced 4feet 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.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations. 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 axisbending. However, since lateralsupport is provided along theentire narrow face, Cfu = 1.0

CT 1.0 c 0.8

E'min Emin CM Ct Ci CT factor "c" in column stability factor CPequation for sawn lumber. (3.7.1)

E'min 580000 psi


l 9 ft b 1.5 in d 5.5 in

Ag b d Sxb d2

6 Sy

d b2

6

Ag 8.25 in2 Sx 7.562 in3 Sy 2.062 in3

Applied Loads

wstrong 25lbf

ft2 wtrib 4 ft Psnow 840 lbf PDead 560 lbf

wstrong wtrib 100lbfft

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)40





1.7 Combined Bending and Axial Compression (ASD)




MB

INED

BEN

DIN

G AN

D AXIAL CO

MPR

ESSION

(ASD)

1.7












Cfu 1.0 Ci 1.0 Cr 1.0 Flat use factors are for weak axisbending. However, since lateralsupport is provided along theentire narrow face, Cfu = 1.0

CT 1.0 c 0.8

E'min Emin CM Ct Ci CT factor "c" in column stability factor CPequation for sawn lumber. (3.7.1)

E'min 580000 psi


l 9 ft b 1.5 in d 5.5 in

Ag b d Sxb d2

6 Sy

d b2

6

Ag 8.25 in2 Sx 7.562 in3 Sy 2.062 in3

Applied Loads

wstrong 25lbf

ft2 wtrib 4 ft Psnow 840 lbf PDead 560 lbf

wstrong wtrib 100lbfft

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 valueadjusted with all adjustment factors except the columnstability factor CP,. The following calculations determine thecolumn stabilty factor CP:

Fc* 2480 psi

P1 Psnow PDead Subscripts refer to Load Case

P1 1400 lbf

fc1P1Ag

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)

d1 d l1 l le1 Ke l1 12inft

Beam dimensions, laterally unsupported lengths (Figure3F) and effective column lengths (NDS 3.7.1) for bucklingin each direction. Subscript 1 is strong (but unsupported)axis; Subscript 2 is the weak but laterally supported axis. d2 b l2

712

ft le2 Ke l2 12inft

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

le maxle1

le2

le 108 in Controlling effective length

Critical bucking design value for compression member (NDS 3.7.1)FcE

0.822 E'min

le1d1

2 FcE 1236 psi

RBle d

b2 RB 16 Slenderness ratio for bending (NDS 3.3-5)

FbE1.20 E'min

RB2

FbE 2636 psi Critical bucking design value for bending (NDS 3.7.1)




MB

INED

BEN

DIN

G AN

D AXIAL CO

MPR

ESSION

(ASD)

1.7

Determine Column Stability Factor Cp

CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

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 adjustedcompression design value F'c

Bending

CL 1.0 Depth to breadth (d/b) ratio (6/2 = 3.0). 2<d/b<4 End restraints for the beam-column satisfy 4.4.1.2 (b)

F'b1 Fb CD CM CL Ct CF Ci Cr F'b is adjusted bending design value with all adjustmentfactors. F'b is calculated for strong axis bending only sincethe weak axis is laterally supported.F'b1 2160 psi

Actual bending stress in strong direction resultingfrom wind load applied to narrow face of beam-columnMmax1

wstrong wtrib l2 12inft

8

Mmax1 12150 inꞏlbf

fb1Mmax1Sx

fb1 1607 psi

Actual bending stress fb1 does not exceed adjustedbending design value F'b

fb1 1607 psi F'b1 2160 psi

Combined Bending and Axial Compression

fc1Fc'

2 fb1

F'b1 1fc1FcE

0.89 < 1.0 ok (NDS 3.9-3) Checking NDS 3.9-4 is not necessarysince the weak axis is fully braced.

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

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




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

Compression

Fc* Fc CD CM Ct CF Ci P2 Psnow PDead fc2P2Ag

Fc* 1782 psi P2 1400 lbf fc2 170 psi


CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c

CP 0.555

Fc' Fc* CP

Fc' 990 psi

fc2 170 psi Fc' 990 psi Actual compression parallel to grain stress fc does notexceed adjusted compression parallel to grain design valueF'c. OK


CD 0.90

Compression

Fc* Fc CD CM Ct CF Ci P3 PDead fc3P3Ag

Fc* 1395 psi P3 560 lbf fc3 68 psi


CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c

CP 0.648

Fc' Fc* CP

Fc' 904 psi

fc3 68 psi Fc' 904 psi Actual compression stress does not exceed adjustedcompression parallel to grain design values for strong axisbuckling. OK (NDS 3.9.2)




MB

INED

BI-AXIAL B

END

ING

AND

AXIAL COM

PRESSIO

N (ASD

)1

.8

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 topchord of a parallel chord gable end truss. The member is 3 ft long (between panel points) and will be subjectedto 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 midpointof the member on its narrow face. Lateral support is provided only at the ends of the member and the ends areconsidered pinned.

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




1.8 COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD)46








Cfu 1.10 Ci 1.0 Cr 1.0 Flat use factors are for weak axisbending (NDS 3.9.2)

c 0.8 Factor "c" in column stability factor CPequation for sawn lumber. (NDS 3.7.1)E'min Emin CM Ct Ci

E'min 510000 psi


l 3 ft b 1.5 in d 3.5 in

Ag b d Sxb d2

6 Sy

d b2

6

Ag 5.25 in2 Sx 3.062 in3 Sy 1.312 in3

Applied Loads

Axial Weak (y) Axis Strong (x) Axis

Applied loads. Subscripts depict loadtype (DL-dead load, SL-snow load andWL-wind load) and application in relationto the member (applied to strong or weakaxis).

DLAxial 300 lbf SLWeak 100 lbf WLStrong 120 lbf

SLAxial 600 lbf DLWeak 50 lbf




MB

INED

BI-AXIAL B

END

ING

AND

AXIAL COM

PRESSIO

N (ASD

)1

.8








Cfu 1.10 Ci 1.0 Cr 1.0 Flat use factors are for weak axisbending (NDS 3.9.2)

c 0.8 Factor "c" in column stability factor CPequation for sawn lumber. (NDS 3.7.1)E'min Emin CM Ct Ci

E'min 510000 psi


l 3 ft b 1.5 in d 3.5 in

Ag b d Sxb d2

6 Sy

d b2

6

Ag 5.25 in2 Sx 3.062 in3 Sy 1.312 in3

Applied Loads

Axial Weak (y) Axis Strong (x) Axis

Applied loads. Subscripts depict loadtype (DL-dead load, SL-snow load andWL-wind load) and application in relationto the member (applied to strong or weakaxis).

DLAxial 300 lbf SLWeak 100 lbf WLStrong 120 lbf

SLAxial 600 lbf DLWeak 50 lbf

Load Case 1: DL + SL + WL Subscripts refer to Load Case

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

Compression

Fc* Fc CD CM Ct CF Ci Fc* is reference compression parallel to grain design valueadjusted with all adjustment factors except the columnstability factor CP.Fc* 2320 psi

P1 DLAxial SLAxial

P1 900 lbf

fc1P1Ag

fc1 171 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)

d1 d l1 l le1 Ke l1 12inft

Beam dimensions, laterally unsupported lengths (Figure3F) and effective column lengths (NDS 3.7.1) for bucklingin each direction. Subscript 1 is strong (but unsupported)axis; Subscript 2 is the weak but laterally supported axis. d2 b l2 l le2 Ke l2 12

inft

le1 36 in le2 36 in Effective lengths in each axis

le/d is less than 50 for each axis (NDS 3.7.1.4) le1d1

10le2d2

24

Critical bucking compression design values for compressionmember in planes of lateral support (NDS 3.7.1) FcE1

0.822 E'min

le1d1

2 FcE1 3963 psi

FcE20.822 E'min

le2d2

2 FcE2 728 psi


FcE minFcE1

FcE2

FcE 728 psi Critical bucking design value for compression member

CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

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.



Determine Adjusted Compression Parallel to Grain Design Value F'c

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 adjustedcompression 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 adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. F'b* 1760 psi

Determine Strong Axis Beam Stability Factor CL and Adjusted Bending Design Values


lud

10.3 lu/d >7 (Table 3.3.3)

le 1.37 lu 3 d le 59.8 in (Table 3.3.3)

RBle d

b2 RB 9.6 Slenderness ratio for bending (3.3-5) RB < 50 (NDS 3.3.3.7)

FbE1.20 E'min

RB2


CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

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 inposition 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 alladjustment factors.

F'b1 1729 psi




MB

INED

BI-AXIAL B

END

ING

AND

AXIAL COM

PRESSIO

N (ASD

)1

.8

Determine Bending Load and Resulting Bending Stress - Strong Axis Bending

Actual bending stress in strong direction resultingfrom wind load applied to narrow face of beam-columnM1Strong

WLStrong l 12inft

4

M1Strong 1080 inꞏlbf

fb1M1Strong

Sx

fb1 353 psi F'b1 1729 psi Ok. Actual bending stress fb does not exceed adjustededgewise compressive design value F'b1

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

CL 1.0 Since d<b (2 in < 4 in) CL=1.0

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 alladjustment factors.

F'b2 1936 psi

Determine Bending Load and Resulting Bending Stress - Weak Axis Bending

Bending stress in strong direction resulting from deadand snow load applied to narrow face of beam-columnM1Weak

DLWeak SLWeak l 12inft

4

M1Weak 1350 inꞏlbf

fb2M1Weak

Sy

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




Combined Bi-axial Bending and Axial Compression

fc1Fc'

2 fb1

F'b1 1fc1FcE1

fb2

F'b2 1fc1FcE2

fb1FbE

2

0.98 < 1.0 ok (3.9-3)

fc1FcE2

fb1FbE

2

0.24 < 1.0 ok (3.9-4)

fc1 171 psi FcE1 3963 psi Actual compression stress is less than critical bucking designvalues 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




MB

INED

BI-AXIAL B

END

ING

AND

AXIAL COM

PRESSIO

N (ASD

)1

.8

Load Case 2: DL+SL

CD 1.15

P2 DLAxial SLAxial P2 900 lbf

fc2P1Ag

Compression

Fc* Fc CD CM Ct CF Ci

Fc* 1667 psi fc2 171 psi


CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c

CP 0.387

Fc' Fc* CP

Fc' 646 psi

fc2 171 psi Fc' 646 psi Actual compression stress fc does not exceed adjustedcompression design value F'c OK

Weak Axis Bending

F'b2 Fb CD CM CL Ct Cfu CF Ci Cr

F'b2 1392 psi

Bending stress in strong direction resulting from snowload applied to narrow face of beam-columnM2Weak

DLWeak SLWeak l 12inft

4

M2Weak 1350 inꞏlbf

fb2M2Weak

Sy

fb2 1029 psi F'b2 1392 psi Ok. Actual bending stress fb does not exceed adjustedbending design value F'b

Strong Axis Bending

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




Combined Bending and Compression

fc2Fc'

2 fb1 0

F'b1 1fc2FcE1

fb2

F'b2 1fc2FcE2

fb1 0

FbE

2

1.04 ~1.0 OK (NDS 3.9-3)Engineering judgementrequired for rounding.

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

fc2 171 psi FcE1 3963 psi Actual compression stress is less than critical bucking designvalues for both weak and strong axis buckling. OK


fb2 1029 psi FbE 6577 psi Actual bending stress is less than critical buckling design value. OK


CD 0.90

Compression

Fc* Fc CD CM Ct CF Ci P3 DLAxial fc3P3Ag



CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c

CP 0.473

Fc' Fc* CP

Fc' 617 psi

fc3 57 psi Fc' 617 psi Actual compression parallel to grain stress fc does notexceed adjusted compression parallel to grain design valueF'c OK

Weak Axis Bending


F'b2 1089 psi

Bending stress in strong direction resulting from deadload applied to narrow face of beam-columnM3Weak

DLWeak l 12inft

4

M3Weak 450 inꞏlbf

fb3M3Weak

Sy

fb3 343 psi F'b2 1089 psi Actual bending stress fb does not exceed adjusted bendingdesign value F'b OK

Strong Axis Bending

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




MB

INED

BI-AXIAL B

END

ING

AND

AXIAL COM

PRESSIO

N (ASD

)1

.8


CD 0.90

Compression

Fc* Fc CD CM Ct CF Ci P3 DLAxial fc3P3Ag



CP

1FcEFc*

2 c

1FcEFc*

2 c

2 FcEFc*

c

CP 0.473

Fc' Fc* CP

Fc' 617 psi

fc3 57 psi Fc' 617 psi Actual compression parallel to grain stress fc does notexceed adjusted compression parallel to grain design valueF'c OK

Weak Axis Bending


F'b2 1089 psi

Bending stress in strong direction resulting from deadload applied to narrow face of beam-columnM3Weak

DLWeak l 12inft

4

M3Weak 450 inꞏlbf

fb3M3Weak

Sy

fb3 343 psi F'b2 1089 psi Actual bending stress fb does not exceed adjusted bendingdesign value F'b OK

Strong Axis Bending

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




Combined Bending and Compression

fc3Fc'

2 fb1 0

F'b1 1fc3FcE1

fb3

F'b2 1fc3FcE2

fb1 0

FbE

2

0.35 < 1.0 OK (NDS 3.9-3)

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 designvalues for both weak and strong axis buckling. OK


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



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 55 LOAD

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WALL W

OO

D STU

D R

ESISTING

WIN

D AN

D G

RAVITY LOAD

S1

.9

1.9 Loadbearing Wall Wood Stud Resisting Wind and Gravity Loads

E1.9 - Loadbearing Wall Wood Stud Resisting Wind and Gravity LoadsBackground: The 2 story home considered in the Design of Wood Frame Buildings for High Wind, Snow, and SeismicLoads (WFCM Workbook) has a Foyer with a vaulted ceiling. The south bearing wall of the Foyer must support gravityloads from the roof and attic above, must resist reactions from uplift wind forces on the roof and must resistout-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 partof the floor diaphragm and limited stud length to one story. The resulting configuration was within the limitations of theprescriptive provisions of the Wood Frame Construction Manual (WFCM) and the wall framing could be determinedfrom 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 wherethey 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-planeand 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. Designwall 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 psfLive Load 20 psf Attic Live Load 30 psfGround Snow Load 30 psfRain & Earthquake effects not considered in the analyses

Additionally, the following design assumptions apply:

Stud spacing = 16" o.c.Exterior Sheathing = Wood Structural PanelsInterior Sheathing = 1/2" Gypsum Wallboard

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

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



1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS56

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


Fc 1350 psi Ft 550psi

CM 1.0 Ct 1.0 (NDS Table 4.3.1)

Cr used to represent Wall StudRepetitive Member Factors (SDPWS3.1.1.1)

Cfu 1.0 Ci 1.0 Cr 1.25

CF 1.0 CT 1.0 c 0.8

factor "c" in column stability factor CPequation for sawn lumber. (NDS 3.7.1)

E'min Emin CM Ct Ci CT E'min 510000 psi

CF Factor is included in Table 4B values

Member Properties

n 1 b 1.5 in d 7.25 in n is the number of full length studs

Ag n b d Sxn b d2

6 Ix

n b d3

12

Ag 10.9 in2 Sx 13.1 in3 Ix 47.6 in4

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 So uth Wall

Dead and Live Loads

wDLAttic 15lbf

ft2

12 16 ft wDLRoof 10

lbf

ft2

12 36 ft

wDLAttic 120 plf wDLRoof 180 plf

wLLAttic 30lbf

ft2

12 16 ft wLLRoof 20

lbf

ft2

12 36 ft

wLLAttic 240 plf wLLRoof 360 plf

wtotalDead wDLAttic wDLRoof

wtotalDead 300 plf

Rain Load Earthquake LoadRain and earthquake loads are included in ASCE 7 loadcombinations). The subscript for the earthquake load isused to differentiate the earthquake load from themodulus of elasticity

R 0 plf El 0 plf




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.9

Snow Load

pg 30lbf

ft2 Ce 1.0 Cts 1.0 Is 1.0 Subscript "s" added to Ct to distinguish the

temperature factor for snow load calculations from Ct fordesign value calculations

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 Figure7.4-1)Cs 1.0

psBal Cs pf psBal 21 psf Balanced snow load (applied to the entire roof)

psUnBal Is pg psUnBal 30 psf Unbalanced snow load (applied to leeward side of roofwith no snow on windward side roof)

psBal12 36 ft 378 plf Load on south wall from a balanced snow condition

psUnBal34 18 ft 405 plf Load on south wall from an unbalanced snow

condition

Controlling snow conditionwsnow max

psBal12 36 ft

psUnBal34 18 ft

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. Thewind pressure equation is:

p = qh[(GCpf)-(GCpi)] (ASCE 7-10 Eq. 28.4-1 or ASCE 7-16 Eq. 28.3-1)

Where: qh is the velocity pressureGCpf is the external pressure coefficient for the surface being analyzed andGCpi is the internal pressure coefficient

Determine Velocity Pressure qh Note: The 160 Exp B velocity pressures qh in theWFCM is 24.06 psf and is based on a 33 ft MRHwhere the velocity pressure exposure coefficient Kz forExp B is 0.72.

Kz for Exp B evaluated at 25 ft MRH in this example is0.70 per ASCE 7-10 Table 28.3.1(ASCE 7-16 Table26.10-1) and produces a slightly lower velocitypressure of 23.4 psf.

The 0.60 factor in the velocity pressure equationincorporates ASCE 7 load factors for allowable stressdesign (ASD) load combinations

V 160

Kz 0.70

Kd 0.85 ASCE 7 Table 26.6-1

Kzt 1.0 ASCE 7 Section 26.8.2

qh 0.60( ) 0.00256 Kz Kzt Kd V2lbf

ft2

qh 23.4 psf




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 zonesfor 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 = +0.18results 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 Section28.4.3 / ASCE 7-16Section 28.3.3)

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

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 psfWRA 9.1 psf




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.9

Load Case B - wind parallel to ridge

Zone 2 (windward) Zone 3 (leeward) Roof Overhang Internal

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




Determine Wind Load Reactions

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

Note: The component of the overturning moment that results from wind pressures on the leeward roof overhang wasnot considered because: (1) it has a short (1 ft) moment arm and (2) the uplift pressures on the overhang occurdownwind of the leeward wall and reduce the net overturning moment reaction slightly. This approach providesslightly conservative results.

Load Case A

RwindwardA1W

Wovhg WWovhg

2

WROA

W2

3 W4

WRA W2

1 W4

LRA

RwindwardA 62 plf

Load Case B

RwindwardB1W

Wovhg WWovhg

2

WROB

W2

3 W4

WRB W2

1 W4

LRB

RwindwardB 329 plf

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




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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).

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

GCpfwallBWW 0.45 GCpfwallBLW 0.45 GCpi 0.18Load Case B

windward wall leeward wall

pmwfrsBWW qh GCpfwallBWW GCpi pmwfrsBLW qh GCpfwallBLW GCpi

pmwfrsBWW 14.7 psf pmwfrsBLW 14.7 psf




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 areshown 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 + 0.6W7a. 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]

whereD = dead loadL = live loadLr = roof live loadW = wind load (note the 0.6 load factorhas already been included in the velocitypressure qh)S = snow loadR = rain loadE = earthquake load

Load combination are included inthe following array (note that rainand earthquake load areneglected (E1= R = 0)




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Pdist

wtotalDead

wtotalDead wLLAttic

wtotalDead wLLRoof

wtotalDead wsnow

wtotalDead 0.75 wLLAttic 0.75 wLLRoof

wtotalDead 0.75 wLLAttic 0.75 wsnow

wtotalDead RwindwardA

wtotalDead RwindwardB

wtotalDead 0.75 wLLAttic 0.75 RwindwardA 0.75 maxwLLRoof

wsnow

wtotalDead 0.75 wLLAttic 0.75 RwindwardB 0.75 maxwLLRoof

wsnow

wtotalDead 0.75 wLLAttic 0.75 maxwLLRoof

wsnow

0.6 wtotalDead RwindwardA

0.6 wtotalDead RwindwardB

0.6wtotalDead

Pdist

11234567891011121314

300540660705750784362-29831537784242-149180

plf

Load Combinations 1, 2, 3, 4a, 4b, 6b, and 8 model gravity only loads (dead load, live load and/or snow load). LoadCombinations 5, 6a1, 6a2, 7a, and 7b include MWFRS wind loads. Load Combinations are keyed to the array asfollows:

Load Combo Pdist1 12 23a 33b 44a 54b 65a 75b 86a1 96a2 106b 117a 127b 138 14

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




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 loadsproportional to the width of their tributary areas. The analyses are conducted for 16 inch stud spacing. The numberof 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

P116( )12

ft Pdist1

P1 400 lbf Compression force in the framing on each side of the wallopenings 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. NDSAppendix B Section B.2

Fc1* Fc CD1 CM Ct CF CiFc1* is reference compression design value for LoadCombination 1 adjusted with all adjustment factors except thecolumn stability factor CP

Fc1* 1215 psi

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

d1 d l1 L le Ke l1 12inft

Stud dimensions, laterally unsupported lengths (NDS Figure 3F)and effective column lengths (NDS 3.7.1) for buckling in eachdirection. Subscript 1 is strong (but laterally unsupported) axis;

le 228 in Effective length in the strong axis. Assume gypsum wallboardis adequately connected to the studs and provides lateralsupport (NDS A.11.3)

FcE0.822 E'min

led1

2 FcE 424 psi

Critical bucking design value for compression member

Determine Column Stability Factor Cp for Load Combination 1

CP1

1FcEFc1*

2 c

1FcEFc1*

2 c

2 FcEFc1*

c




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

Compare Actual Compression Stress to Adjusted Compression Design Value

Fc'1 Fc1* CP1 Fc'1 388 psi Adjusted compression design value

fc1P1Ag

fc1 37 psi Actual compression stress

Actual compression stress fc1 is less than the adjustedcompression design value F'c1. Ratio of actual stress toadjusted compression design value < 1. OK

fc1Fc'1

0.09

Load Combination 2: D + L

Determing compression force in framing for load combination 2

P216( )12

ft Pdist2

P2 720 lbf Compression force in the framing on each side of the wallopenings 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. NDSAppendix B Section B.2

Fc2* Fc CD2 CM Ct CF CiFc2* is reference compression design value for LoadCombination 2 adjusted with all adjustment factors except thecolumn stability factor CP

Fc2* 1350 psi



Stud dimensions, laterally unsupported lengths (NDS Figure 3F)and effective column lengths (3.7.1) for buckling in eachdirection. Subscript 1 is strong (but laterally unsupported) axis;

Effective length in the strong axis. Assume gypsum wallboardis adequately connected to the studs and provides lateralsupport (NDS A.11.3)

le 228 in

FcE0.822 E'min

led1

2 FcE 424 psi





Determine Column Stability Factor Cp for Load Combination 2

CP2

1FcEFc2*

2 c

1FcEFc2*

2 c

2 FcEFc2*

c

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


Fc'2 Fc2* CP2 Fc'2 392 psi Adjusted compression design value

fc2P2Ag

fc2 66 psi Actual compression stress

Actual compression stress fc2 is less than the adjustedcompression design value F'c2. Ratio of actual stress toadjusted compression design value < 1. OK

fc2Fc'2

0.17

Load Combination 3a: D + L r

Determing compression force in framing for load combination 3a

P3a16( )12

ft Pdist3

P3a 880 lbf Compression force in the framing on each side of the wallopenings 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 CiFc3a* is reference compression design value for LoadCombination 3a adjusted with all adjustment factors except thecolumn stability factor CP

Fc3a* 1688 psi



Stud dimensions, laterally unsupported lengths (NDS Figure 3F)and effective column lengths (NDS 3.7.1) for buckling in eachdirection. Subscript 1 is strong (but laterally unsupported) axis;




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le 228 in Effective length in the strong axis. Assume gypsum wallboardis adequately connected to the studs and provides lateralsupport (NDS A.11.3)

FcE0.822 E'min

led1

2 FcE 424 psi


Determine Column Stability Factor Cp for Load Combination 3a

CP3a

1FcEFc3a*

2 c

1FcEFc3a*

2 c

2 FcEFc3a*

c

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

Fc'3a Fc3a* CP3a Fc'3a 399 psi


fc3aP3aAg

fc3a 81 psi Actual compression stress

Actual compression stress fc3a is less than the adjustedcompression design value F'c3a. Ratio of actual stress toadjusted compression design value < 1. OK

fc3aFc'3a

0.2

Load Combination 3b: D + S

Determing compression force in framing for load combination 3b

P3b16( )12

ft Pdist4

P3b 940 lbf Compression force in the framing on each side of the wallopenings 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 CiFc3b* is reference compression design value for LoadCombination 3b adjusted with all adjustment factors except thecolumn stability factor CP

Fc3b* 1552 psi








le 228 in

FcE0.822 E'min

led1

2 FcE 424 psi


Determine Column Stability Factor Cp for Load Combination 3b

CP3b

1FcEFc3b*

2 c

1FcEFc3b*

2 c

2 FcEFc3b*

c

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


Fc'3b Fc3b* CP3b Fc'3b 397 psi Adjusted compression design value

fc3bP3bAg

fc3b 86 psi Actual compression stress

Actual compression stress fc3b is less than the adjustedcompression design value F'c3b. Ratio of actual stress toadjusted compression design value < 1. OK

fc3bFc'3b

0.22

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 wallopenings for Load Combination 4a

P4a16( )12

ft Pdist5

P4a 1000 lbf 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




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Fc4a* Fc CD4a CM Ct CF Ci Fc4a* is reference compression design value for LoadCombination 4a adjusted with all adjustment factors except thecolumn stability factor CP

Fc4a* 1688 psi





le 228 in

FcE0.822 E'min

led1

2 FcE 424 psi



CP4a

1FcEFc4a*

2 c

1FcEFc4a*

2 c

2 FcEFc4a*

c

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


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

fc4aP4aAg

fc4a 92 psi Actual compression stress

Actual compression stress fc4a is less than the adjustedcompression design value F'c4a. Ratio of actual stress toadjusted compression design value < 1. OK

fc4aFc'4a

0.23




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

Determing compression force in framing for load combination 4b

P4b16( )12

ft Pdist6

P4b 1045 lbf Compression force in the framing on each side of the wallopenings 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 CiFc4b* is reference compression design value for LoadCombination 4b adjusted with all adjustment factors except thecolumn stability factor CP

Fc4b* 1552 psi





le 228 in

FcE0.822 E'min

led1

2 FcE 424 psi


Determine Column Stability Factor Cp for Load Combination 4b

CP4b

1FcEFc4b*

2 c

1FcEFc4b*

2 c

2 FcEFc4b*

c

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


Fc'4b Fc4b* CP4b Fc'4b 397 psi Adjusted compression design value

fc4bP4bAg

fc4b 96 psi Actual compression stress

Actual compression stress fc4b is less than the adjustedcompression design value F'c4b. Ratio of actual stress toadjusted compression design value < 1. OK

fc4bFc'4b

0.24




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Load Combination Applied Stress/ Allowable Stress1 0.092 0.173a 0.203b 0.224a 0.234b 0.24

Table comparing ratio of applied stress to allowablestresses for gravity load controlled combinations

Load Case 5a: D + 0.6 W (wind parallel to ridge)

Determing compression load in framing for load combination 5a

P5a16( )12

ft Pdist7

P5a 483 lbf

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 alladjustment factors except the column stability factor CP,. Thefollowing calculations determine the column stabilty factor CP:Fc5a* 2160 psi


CP5

1FcEFc5a*

2 c

1FcEFc5a*

2 c

2 FcEFc5a*

c

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

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

fc5aP5aAg

fc5a 44 psi Actual compression stress fc5a does not exceed adjustedcompression design value F'c5a. Ratio of actualcompression stress to adjusted compression design value< 1. OK

Fc'5a 405 psi

fc5aFc'5a

0.11




Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures

Moment

wwindAWW16( )12

ft pmwfrsAWW Load Combination 5a includes the MWFRS Wind Load

wwindAWW 23.08 plf

MmwfrsAWWwwindAWW L2

812

inft Bending moment from out-of-plane MWFRS wind loads

MmwfrsAWW 12500 inꞏlbf

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

CL 1.0 Depth to breadth (d/b) ratio 2<d/b<4 End restraints for thebeam-column satisfy NDS 4.4.1.2 (b) and sheathing/gypsumwall board nailing provides lateral support for the compressionedges NDS 4.4.1.2 (c)

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

fb5aMmwfrsAWW

Sx fb5a 951 psi Bending stress resulting from out-of-plane MWFRS wind

loadsF'b5a 1850 psi

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

fb5aF'b5a

0.51

Check Combined Uniaxial Bending and Axial Compression

fc5aFc'5a

2 fb5a

F'b5a 1fc5aFcE

0.59 < 1.0 ok (NDS 3.9-3)

Actual compression stress fc5a does not exceed adjustedcompression design value in plane of lateral support foredgewise bending FcE. OK

fc5a 44 psi FcE 424 psi




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Load Case 5b: D + 0.6 W (wind perpendicular to ridge)

Determing axial load in framing for load combination 5b

P5b16( )12

ft Pdist8

P5b 39 lbf Note: by inspection, Load Combination 7b controls and is analyzedbelow.

Load Case 6a1: D + 0.75L + 0.75 W (wind parallel to ridge)

Determing compression load in framing for load combination 6a1

P6a116( )12

ft Pdist9

P6a1 1107 lbf

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


Fc6a1* Fc CD6 CM Ct CF Ci Fc6a1* is reference compression design value adjusted with alladjustment factors except the column stability factor CP,. Thefollowing calculations determine the column stabilty factor CP:Fc6a1* 2160 psi

Determine Column Stability Factor Cp for Load Combination 6a1

CP6

1FcE

Fc6a1*

2 c

1FcE

Fc6a1*

2 c

2 FcEFc6a1*

c

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

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


fc6a1P6a1Ag

fc6a1 102 psi Actual compression stress fc6a1 does not exceed adjustedcompression design value F'c6a1. Ratio of actual compressionstress to adjusted compression design value < 1. OKFc'6a1 405 psi

fc6a1Fc'6a1

0.25





Moment

wwindAWW 0.7516( )12

ft pmwfrsAWW Load Combination 6a1 includes 75% of the MWFRS WindLoad

wwindAWW 17.31 plf

MmwfrsAWWwwindAWW L2

812


MmwfrsAWW 9375 inꞏlbf

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


F'b6a1 Fb CD6 CM CL Ct CF Ci Cr F'b6a1 is adjusted bending design value for Load Combination6a1 F'b6a1 1850 psi

Compare Actual Bending Stress with Adjusted Bending Design Values for Load Combination 6a1

fb6a1MmwfrsAWW

Sx fb6a1 713 psi Bending stress resulting from out-of-plane MWFRS wind

loadsF'b6a1 1850 psi

Ok. Actual bending stress fb6a1 is less than adjusted bendingdesign value F'b6a1. Ratio of actual bending stress to adjustedbending design value < 1

fb6a1F'b6a1

0.39


fc6a1Fc'6a1

2 fb6a1

F'b6a1 1fc6a1FcE

0.57 < 1.0 ok (NDS 3.9-3)

Actual compression stress fc6a1 does not exceed adjustedcompression design value in plane of lateral support foredgewise bending FcE. OK

fc6a1 102 psi FcE 424 psi

Load Case 6a2: D + 0.75L + 0.75 W (wind perpendicular to ridge)


P6a216( )12

ft Pdist10

P6a2 716 lbf





CP6

1FcE

Fc6a2*

2 c

1FcE

Fc6a2*

2 c

2 FcEFc6a2*

c





fc6a2P6a2Ag

fc6a2 66 psi Actual compression stress fc6a2 does not exceed adjustedcompression design value F'c6a2. Ratio of actualcompression stress to adjusted compression design value< 1. OK

Fc'6a2 405 psi

fc6a2Fc'6a2

0.16


Moment

wwindBWW 0.7516( )12

ft pmwfrsBWW Load Combination 6a2 includes 75% of the MWFRS WindLoad

wwindBWW 14.74 plf

MmwfrsBWWwwindBWW L2

812


MmwfrsBWW 7982 inꞏlbf




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Load Case 6a2: D + 0.75L + 0.75 W (wind perpendicular to ridge)


P6a216( )12

ft Pdist10

P6a2 716 lbf





CP6

1FcE

Fc6a2*

2 c

1FcE

Fc6a2*

2 c

2 FcEFc6a2*

c





fc6a2P6a2Ag

fc6a2 66 psi Actual compression stress fc6a2 does not exceed adjustedcompression design value F'c6a2. Ratio of actualcompression stress to adjusted compression design value< 1. OK

Fc'6a2 405 psi

fc6a2Fc'6a2

0.16


Moment

wwindBWW 0.7516( )12

ft pmwfrsBWW Load Combination 6a2 includes 75% of the MWFRS WindLoad

wwindBWW 14.74 plf


812






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


F'b6a2 Fb CD6 CM CL Ct CF Ci Cr F'b6a2 is adjusted bending design value for Load Combination6a2 F'b6a2 1850 psi

Compare Actual Bending Stress with Adjusted Bending Design Values for Load Combination 6a2

fb6a2MmwfrsBWW

Sx fb6a2 607 psi Bending stress resulting from out-of-plane MWFRS wind

loadsF'b6a2 1850 psi

Ok. Actual bending stress fb6a2 is less than adjusted bendingdesign value F'b6a2. Ratio of actual bending stress to adjustedbending design value < 1

fb6a2F'b6a2

0.33


fc6a2Fc'6a2

2 fb6a2

F'b6a2 1fc6a2FcE

0.42 < 1.0 ok (NDS 3.9-3)

Actual compression stress fc6a2 does not exceed adjustedcompression design value in plane of lateral support foredgewise bending FcE. OK

fc6a2 66 psi FcE 424 psi

Load Case 7a: 0.6 D + 0.6 W (wind parallel to ridge)

Determing compression load in framing for load combination 7a

P7a16( )12

ft Pdist12

P7a 323 lbf Note: by inspection, Load Case 5a controls and calculations will not berepeated for Load Case 7a.

Load Case 7b: 0.6 D + 0.6 W (wind perpendicular to ridge)


P7b16( )12

ft Pdist13

P7b 199 lbf

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

ft7bP7bAg

ft7b 18 psi Actual tension stress ft7b does not exceed adjustedtension design value F't7b. Ratio of actual tension stressto adjusted tension design value < 1. OK

F't7b 880 psift7bF't7b

0.02


Moment

wwindBWW16( )12

ft pmwfrsBWW Load Combination 7b includes the MWFRS Wind Load

wwindBWW 19.65 plf


812



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


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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Load Case 7b: 0.6 D + 0.6 W (wind perpendicular to ridge)


P7b16( )12

ft Pdist13

P7b 199 lbf

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

ft7bP7bAg

ft7b 18 psi Actual tension stress ft7b does not exceed adjustedtension design value F't7b. Ratio of actual tension stressto adjusted tension design value < 1. OK

F't7b 880 psift7bF't7b

0.02


Moment

wwindBWW16( )12

ft pmwfrsBWW Load Combination 7b includes the MWFRS Wind Load

wwindBWW 19.65 plf


812



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


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




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

fb7bMmwfrsBWW

Sx fb7b 810 psi Bending stress resulting from out-of-plane MWFRS wind

loadsF'b7b 1850 psi

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

fb7bF'b7b

0.44

Check Combined Uniaxial Bending and Axial Tension

ft7bF't7b

fb7bF'b7b

0.417 < 1.0 ok (NDS 3.9-1)

fb7b ft7b

F'b7b0.448 < 1.0 ok (NDS 3.9-2)

Load Combination Applied Stress/ Allowable Stress5a 0.596a1 0.576a2 0.427b 0.45

Table comparing ratio of applied stress to allowablestresses for combined bending and axial loadcombinations. By inspection and comparison togravity load combinations analyzed earlier, LoadCombination 5a controls so far with combineddead 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

Note: "EWA" is used for Effective Wind Area since "A" is apreviously defined variable. Per ASCE 7 Chapter 26, EWA neednot be less than (L)2/3

EWAL2

31

ft2 EWA 120

The south wall is in Zone 4. ASCE 7-10 Figure 30.4-1 / ASCE7-16 Figure 30.3-1. The equation for GCp for Zone 4 for 10 ft2 <

EWA < 500 ft2

GCp EWA( ) 0.8 0.3log

EWA500

log10500

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

pCC EWA( ) qh 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) aregreater than positive external pressure coefficients. So negativeexternal pressures and positive internal pressures (windward)create the greatest C&C pressures

GCp EWA( ) GCpi 1.089




BEAR

ING

WALL W

OO

D STU

D R

ESISTING

WIN

D AN

D G

RAVITY LOAD

S1

.9

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

Bending

wCC16( )

12ft pCC EWA( ) wCC 34 plf

MCC

wCC L2 12inft

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

Ok. Actual bending stresses that result from C&C pressuresfbCC do not exceed adjusted bending design value F'bCC

fbCCMCCSx

fbCC 1400 psi

F'bCC 1850 psifbCCF'bCC

0.76 The fb/F'b ratio is greater for C&C loading than for MWFRSloading. C&C controls for strength calculations.

Determine Deflection for C&C loading

ΔCC5

384 Cr E Ix

0.70 wCC

12

ftin L 12

inft

4 IBC Table 1604 footnote (f) allows the wind load used in

deflection calculations to be 0.42 times the C&C load. A factorof 0.70 is applied since a 0.60 factor has already beenincorporated into the velocity pressure qh.

ΔCC 0.84 in

OK - Span to deflection ratio is greater than L / 180L 12

inft

ΔCC273

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



2.1A WITHDRAWAL DESIGN VALUE - PLAIN SHANK NAIL80

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

Using 2015/2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) referencewithdrawal design value in pounds (capacity) of an 8d common smooth-shank nail in theconnection 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.

2.1a Withdrawal Design Value - Plain Shank Nail



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 81W

ITHD

RAW

AL DESIG

N VALU

E - PLAIN SH

ANK

NAIL

2.1

A E2.1a - Withdrawal Design Value - Plain Shank NailUsing 2015/2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) referencewithdrawal design value in pounds (capacity) of an 8d common smooth-shank nail in theconnection 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 L Ls Nail penetration into main member (in.)

pt 2.395

W 1380 G

52

D NDS Equation 12.2-3

W 20.7 Reference withdrawal design value. Compare to NDS Table12.2C, W = 21 lbs/in

W pt 49.5 Reference withdrawal design value based on nailpenetration into main member (lbs)

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



2.1B FASTENER UPLIFT CAPACITY - ROOF SHEATHING RING SHANK NAIL (2018 NDS ONLY)82

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 withdrawaldesign value in pounds (capacity) and head pull-through design value in pounds (capacity) of a0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a DouglasFir-Larch nominal 2x6 with a 5/8 in. thick Douglas-Fir Wood Structural Panel (plywood ororiented 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.

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



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 83FASTEN

ER U

PLIFT CAPACITY - RO

OF SH

EATHIN

G R

ING

SHAN

K N

AIL (2018 ND

S ON

LY)2

.1B

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 withdrawaldesign value in pounds (capacity) and head pull-through design value in pounds (capacity) of a0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a DouglasFir-Larch nominal 2x6 with a 5/8 in. thick Douglas-Fir Wood Structural Panel (plywood ororiented 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.


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

W 1800 G2 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 fastenerpenetration (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

WH 690 π DH G2 tns NDS Equation 12.2-6a

WH 95 Head pull-through design value (lbs). Compare to NDS Table 12.2F, WH = 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 adjustmentfactors for connections based on end use conditions.



2.1C FASTENER UPLIFT CAPACITY - ROOF SHEATHING RING SHANK NAIL (2018 NDS ONLY84

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 withdrawaldesign value in pounds (capacity) and head pull-through design value in pounds (capacity) of a0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a DouglasFir-Larch nominal 2x6 with a 7/16 in. thick Douglas-Fir Wood Structural Panel (plywood ororiented 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.

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



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 85FASTEN

ER U

PLIFT CAPACITY - RO

OF SH

EATHIN

G R

ING

SHAN

K N

AIL (2018 ND

S ON

LY)2

.1C

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 withdrawaldesign value in pounds (capacity) and head pull-through design value in pounds (capacity) of a0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a DouglasFir-Larch nominal 2x6 with a 7/16 in. thick Douglas-Fir Wood Structural Panel (plywood ororiented 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.


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

W 1800 G2 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 fastenerpenetration (TL) in main member (lbs)W TL 88

Checking Fastener Head Pull-Through

tns 0.438

2.5DH 0.703 2.5DH greater than tns, so NDS Equation 12.2-6a applies

WH 690 π DH G2 tns NDS Equation 12.2-6a

WH 67 Head pull-through design value (lbs). Compare to NDS Table 12.2F, WH = 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 ofadditional adjustment factors for connections based on end use conditions.



2.2 SINGLE COMMON NAIL LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION 86

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 AllowableStress Design (ASD) reference lateral design value of a single shear connection with thefollowing 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.

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



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 87SIN

GLE CO

MM

ON

NAIL LATER

AL DESIG

N VALU

E - SING

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.2

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 AllowableStress Design (ASD) reference lateral design value of a single shear connection with thefollowing 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.

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)

ReFemFes

Re 1

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)

Lm 3 LsTip2

Main member Dowel Bearing Length (in.) (NDS 12.3.5.3)

Lm 2.1

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

Rd 2.2 Reduction Term (NDS Table 12.3.1B)




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

RtLmLs

Rt 2.803

k1Re 2 Re

2 1 Rt Rt

2

Rt

2 Re3

Re 1 Rt

1 Re

k1 0.935

k2 1 2 1 Re 2 Fyb 1 2 Re D2

3 Fem Lm2

k2 1.047

k3 12 1 Re

Re

2 Fyb 2 Re D2

3 Fem Ls2

k3 1.347

Yield Mode Calculations (NDS Table 12.3.1A)

Mode Im

ZImD Lm Fem

Rd

ZIm 658 Yield Mode Im Solution (lbs)

Mode Is

ZIsD Ls Fes

Rd

ZIs 235 Yield Mode Is Solution (lbs)




GLE CO

MM

ON

NAIL LATER

AL DESIG

N VALU

E - SING

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.2

Mode II

ZIIk1 D Ls Fes

Rd

ZII 219 Yield Mode II Solution (lbs)

Mode IIIm

ZIIImk2 D Lm Fem

1 2 Re Rd

ZIIIm 230 Yield Mode IIIm Solution (lbs)

Mode IIIs

ZIIIsk3 D Ls Fem

2 Re Rd

ZIIIs 105 Yield Mode IIIs Solution (lbs)

Mode IV

ZIVD2

Rd

2 Fem Fyb

3 1 Re

ZIV 118 Yield Mode IV Solution (lbs)

Zdist

ZIm

ZIs

ZII

ZIIIm

ZIIIs

ZIV

Zdist

658

235

219

230

105

118

Creating an array with all Yield Mode Solutions




Z min Zdist

Z 105 Minimum value of all Yield Modes provides Z-reference lateraldesign value (lbs). Mode IIIs controls. Compare to NDS Table 12Nvalue = 105 lbs. See NDS Table 11.3.1 for application ofadditional adjustment factors for connections based on end useconditions.

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



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 91W

ITHD

RAW

AL DESIG

N VALU

E - LAG SCR

EW2

.3

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 factorsare 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) 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



2.3 WITHDRAWAL DESIGN VALUE - LAG SCREW92

E2.3 - Withdrawal Design Value - Lag ScrewUsing 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 factorsare 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) Fastener Dimensions: 1/2 in. diameter lag screw (NDS Table L2) Length = 4 in. Tip Length = 0.3125 in.


tip 0.3125 Fastener tapered tip length (in.)

G 0.55 Specific gravity (NDS Table 12.3.3A)

L 4 Lag screw length (in.)

Ls 1.5 Side Member thickness (in.)

pt L Ls tip Lag screw penetration into main member (in.)

Note: Per Table L2, the unthreaded body length, S = 1.5".Therefore, the threaded portion begins at the wood-to-woodinterface. Note pt also matches Table L2 dimension T-E = 2-3/16".Longer lag screws have different thread length dimensionsrequiring evaluation to determine actual thread penetration.

pt 2.188

W 1800 G

32

D

34

NDS Equation 12.2-1

W 436.6 Compare to NDS Reference Withdrawal Design Value Table12.2A, W = 437 lbs/in.

W pt 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 enduse conditions.

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 stressdesign (ASD) reference lateral design value of a double shear connection with the followingconfiguration. 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. Dr= 0.152 in. Length = 6 in.




GLE W

OO

D SCR

EW LATER

AL DESIG

N VALU

E - DO

UB

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.4





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



2.4 SINGLE WOOD SCREW LATERAL DESIGN VALUE - DOUBLE WOOD TO WOOD CONNECTION94





Define parameters:



ReFemFes

Re 1

tm 3.0 Main Member thickness (in.)

ts 1.5 Side Member thickness (in.)


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)

Lm tm Lm 3 Main member Dowel Bearing Length (in.) (NDS 12.3.5.3)

Ls Lscrew Lm tsTip2

Side member (holding the screw tip) Dowel BearingLength (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)




GLE W

OO

D SCR

EW LATER

AL DESIG

N VALU

E - DO

UB

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.4

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

k3 12 1 Re

Re

2 Fyb 2 Re Dr2

3 Fem Ls2

k3 1.095


Mode Im

ZImDr Lm Fem

Rd


Mode Is

ZIs2Dr Ls Fes

Rd


Mode IIIs

ZIIIs2k3 Dr Ls Fem

2 Re Rd


Mode IV

ZIV2Dr

2

Rd

2 Fem Fyb

3 1 Re




2.4 SINGLE WOOD SCREW LATERAL DESIGN VALUE - DOUBLE WOOD TO WOOD CONNECTION96

Zdist

ZIm

ZIs

ZIIIs

ZIV

Zdist

1055

921

336

234


Z min Zdist

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




GLE B

OLT LATER

AL DESIG

N VALU

E - SING

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.5

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 AllowableStress Design (ASD) reference lateral design value of a single shear connection withthe 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



2.5 SINGLE BOLT LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION98

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 AllowableStress Design (ASD) reference lateral design value of a single shear connection withthe 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)

Define parameters:



ReFemFes

Re 1


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 holdingthe 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

Rd2 3.6 Reduction Terms (NDS Table 12.3.1B)

Rd3 3.2




GLE B

OLT LATER

AL DESIG

N VALU

E - SING

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.5

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

RtLmLs

Rt 1

k1Re 2 Re

2 1 Rt Rt

2

Rt

2 Re3

Re 1 Rt

1 Re

k1 0.414

k2 1 2 1 Re 2 Fyb 1 2 Re D2

3 Fem Lm2

k2 1.093

k3 12 1 Re

Re

2 Fyb 2 Re D2

3 Fem Ls2

k3 1.093


Mode Im

ZImD Lm Fem

Rd1


Mode Is

ZIsD Ls Fes

Rd1





Mode II

ZIIk1 D Ls Fes

Rd2

ZII 966 Yield Mode II Solution (lbs)

Mode IIIm

ZIIImk2 D Lm Fem

1 2 Re Rd3

ZIIIm 957 Yield Mode IIIm Solution (lbs)

Mode IIIs

ZIIIsk3 D Ls Fem

2 Re Rd3


Mode IV

ZIVD2

Rd3

2 Fem Fyb

3 1 Re


Zdist

ZIm

ZIs

ZII

ZIIIm

ZIIIs

ZIV


Minimum value of all Yield Modes provides Z-reference lateraldesign value (lbs). Mode IV controls.Z min Zdist Z 663




GLE B

OLT LATER

AL DESIG

N VALU

E - SING

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.5

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.

Side and Main member dowel resistance (equal due to equivalent doweldiameter in both members), in.-lbsM

Fyb D3

6

gap 0 Gap between member shear planes, in.

The limiting wood stresses used in the yield model are based on the load at which theload-deformation curve from a fastener embedment test intersects a line represented by theinitial tangent modolus offset 5% of the fastener diameter. The reduction term, Rd, reduces thevalues calculated using the yield limit equations to approximate estimates of the nominalproportional 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




Mode II

AII1

4 qs

14 qm

BIILs2

gapLm2

CIIqs Ls

2

4

qm Lm2

4

PIIBII BII

2 4 AII CII

2 AII

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

Mode IIIm

AIIIm1

2 qs

14 qm

BIIIm gapLm2

CIIIm Mqm Lm

2

4

PIIImBIIIm BIIIm

2 4 AIIIm CIIIm

2 AIIIm

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

Mode IIIs

AIIIs1

4 qs

12 qm

BIIIs gapLs2

CIIIs Mqs Ls

2

4

PIIIsBIIIs BIIIs

2 4 AIIIs CIIIs

2 AIIIs




GLE B

OLT LATER

AL DESIG

N VALU

E - SING

LE SHEAR

WO

OD

-TO-W

OO

D CO

NN

ECTION

2.5

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

Mode IV

AIV1

2 qs

12 qm

BIV gap

CIV M M

PIVBIV BIV

2 4 AIV CIV

2 AIV

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

Converting from TR-12 "P" values to NDS "Z"values and creating an array. Shows TR-12results equal NDS results for each YieldMode. All values in units of lbs.

Zdist2

PImRd1

PIsRd1

PIIRd2

PIIImRd3

PIIIsRd3

PIVRd3

Zdist2

2100

2100

966

957

957

663

Z2 min Zdist2 Z value from TR-12 equations is equivalent to Z value from NDSequations and comparable to NDS Table 12A value Zparallel = 660lbs. See NDS Table 11.3.1 for application of additional adjustmentfactors for connections based on end use conditions.

Z2 663




For comparison, Zdist shows NDSequation results, Zdist2 shows TR-12equation results, for Modes Im, Is, II,IIIm, IIIs, and IV, respectively. Allvalues in units of lbs.

Zdist

2100

2100

966

957

957

663

Zdist2

2100

2100

966

957

957

663



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 105B

OLTED

WO

OD

-TO-W

OO

D TEN

SION

SPLICE CON

NECTIO

N CAPACITY

2.6

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

4"

7"

38"

P P

P P

3-5/8"

4" 4" 4" 4" 7" 4" 4"

3-5/8"

2.6 Bolted Wood-to-Wood Tension Splice Connection Capacity



2.6 BOLTED WOOD-TO-WOOD TENSION SPLICE CONNECTION CAPACITY106

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

4"

7"

38"

P P

P P

3-5/8"

4" 4" 4" 4" 7" 4" 4"

3-5/8"

Define parameters:

D 1.0 Bolt Diameter (in.)

s 4 Fastener row spacing (in.)

Ls 1.5 Side member Dowel Bearing Length (NDS 12.3.5)

Lm 1.5 Main member Dowel Bearing Length (NDS 12.3.5.3)

CD 1.25 Construction load duration (NDS Table 2.3.2)

CM 1.0 Ct 1.0

Group Action Factor (C g )

As 2 16.875 Area of side members (In2)

Am 16.875 Area of main member (In2)

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 enddistance, edge distance, or spacing (NDS 12.5.1). Smallest calculated geometry factor willcontrol.

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.)

CΔendProvidedendMinimumend

CΔend 0.57 Geometry factor based on end distance

Spacing between bolts

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

Providedbolt 4 Provided spacing between bolts (in.)

CΔbolt 1.0 Provided spacing greater than or equal to minimum spacing for CΔ = 1.0

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.)

CΔedge 1.0 Provided edge distance greater than or equal to minimum spacing for CΔ = 1.0




OLTED

WO

OD

-TO-W

OO

D TEN

SION

SPLICE CON

NECTIO

N CAPACITY

2.6

Geometry Factor (C Δ )

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

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.)

CΔendProvidedendMinimumend

CΔend 0.57 Geometry factor based on end distance

Spacing between bolts

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

Providedbolt 4 Provided spacing between bolts (in.)

CΔbolt 1.0 Provided spacing greater than or equal to minimum spacing for CΔ = 1.0

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.)

CΔedge 1.0 Provided edge distance greater than or equal to minimum spacing for CΔ = 1.0



2.6 BOLTED WOOD-TO-WOOD TENSION SPLICE CONNECTION CAPACITY108

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

Providedrow 4 Provided spacing between bolts (in.)

CΔrow 1.0 Provided spacing greater than or equal to minimum spacing for CΔ = 1.0

CΔ min CΔend CΔbolt CΔedge CΔrow

CΔ 0.57

Reference Lateral Design Value

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

Adjusted Multiple Bolt Capacity

n 6 n = number of bolts on each side

Zadj n Z CD Cg CΔ CM CtAdjusted 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 widthNote: hole size includes 1/16" oversizing per 2015/2018 NDS12.1.3.2

Anet 13.7 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)




OLTED

WO

OD

-TO-W

OO

D TEN

SION

SPLICE CON

NECTIO

N CAPACITY

2.6

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 thein-row bolt spacing

ni 3 Number of fasteners in a single row

ZRTiadj ni Fvadj t scriticalAdjusted row tear-out capacity for single row (lbs)

ZRTiadj 3938

ZRTadj 2 ni Fvadj t scriticalAdjusted row tear-out capacity for two rows (lbs)

ZRTadj 7875

Group Tear-Out Check

ZRT1adj ZRTiadjZRT1' is row tear-out capacity for 1st row of bolts,ZRT2' is rowtear-out capacity for 2nd row of bolts, both values in lbsZRT2adj 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)

ZGTadjZRT1adj

2

ZRT2adj2

Ftadj Agroupnet

ZGTadj 6416 Adjusted group tear-out capacity (lbs)

Connection capacity is minimum of calculatedcapacities (lbs). Note group tear-out controls.However, increasing the spacing between boltrows, up to a maximum of 5" for dimensionlumber, can increase group tear-out capacity.Bolt sizes could be reduced to more closelymatch net section capacities.

Capacity min Zadj ZNTadj ZRTadj ZGTadj

Capacity 6416



3.1 SEGMENTED SHEAR WALL DESIGN - WIND110

3.1 Segmented Shear Wall Design - Wind

E3.1 - Segmented Shear Wall Design - WindUsing the 2015 Special Design Provisions for Wind and Seismic (SDPWS), design the first floorwall shown in the diagram below as a segmented shear wall for a two-story house using AllowableStress Design (ASD) provisions

Design wind speed = 160 mph (700-yr wind speed, 3-second gust)Exposure Category BBuilding dimensions:

L = 40 ftW = 32 ftRoof pitch = 7:12Top plate to ridge height = 9.3 ftWall height = 9 ftDoor height = 7 ft 6 in.Window height = 4.5 ftStud spacing = 16 in. o.c.Studs are Southern Pine (G=0.55)

Check design with and without interior gypsum and neglect deflection.

Use ASCE 7-10 Minimum Design Loads for Buildings and Other Structures to determine windloads.

Design with Interior Gypsum

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

Lmin93.5

Lmin 2.6 Minimum full height wall segment length (ft)

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

Vw 5520 Applied shear load on each shear wall due to wind force (lbs)

h 9 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. edgespacing. SDPWS Table 4.3A nominal capacity = 1065 lbs/ft (Wind)

ASD Shearwall Capacity, WSP (lbs/ft)(SDPWS 4.3.3)vwASDWSP

10652

vwASDWSP 532.5

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

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

SDPWS Table 4.3C nominal capacity = 250 lb/ft

ASD Shearwall Capacity, Gypsum (lbs/ft)(SDPWS 4.3.3)vwASDGWB

2502

vwASDGWB 125

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 oneither 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.



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 111SEG

MEN

TED SH

EAR W

ALL DESIG

N - W

IND

3.1





Lmin93.5







ASD Shearwall Capacity, WSP (lbs/ft)(SDPWS 4.3.3)vwASDWSP

10652

vwASDWSP 532.5




ASD Shearwall Capacity, Gypsum (lbs/ft)(SDPWS 4.3.3)vwASDGWB

2502

vwASDGWB 125



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 oneither side of door for 10 ft of full height sheathing.

Capacity 2 5 vWCASD





3.1 SEGMENTED SHEAR WALL DESIGN - WIND112

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

Leff 10 Effective length of Full Height Segments (ft)

TVwLeff

h

T 4968 Required Hold down capacity (lbs)

Hold down would need to be combined with 2nd floor hold down requirements. Dead load offsethas been neglected in this example.

Design without Interior Gypsum




Lmin93.5







vwASDWSP10652

vwASDWSP 532.5 ASD Shearwall Capacity, WSP(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 widepanels require adjustment per SDPWS 4.3.3.4.1. Capacity of each 3 ft wide segment ismultiplied by 2bs/h adjustment factor.

bs 3 Width of panel requiring capacity adjustment (ft)

Capacity 2 5 vwASDWSP2bsh

2( ) 3( ) vwASDWSP



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




MEN

TED SH

EAR W

ALL DESIG

N - W

IND

3.1

Leff 2 5( )2bsh

2( ) 3( ) Leff 14 Provided effective Full Height Sheathing Length (ft)

TVwLeff

h

T 3549 Required Hold down capacity (lbs) (SDPWS 4.3.6.1.2)

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



3.2 SEGMENTED SHEAR WALL DESIGN - SEISMIC114

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

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


Neglect deflection check.

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

3.2 Segmented Shear Wall Design - Seismic




MEN

TED SH

EAR W

ALL DESIG

N - SEISM

IC3

.2




Lmin93.5



Vs 4733 Applied shear load on each shear wall due to seismic force (lbs)


(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. edgespacing. Studs @ 16 in. o.c. triggers Footnote 2 which allows for use of 15/32 in. panel shearvalues. SDPWS Table 4.3A nominal capacity = 760 lbs/ft (Seismic). Unlike wind design, gypsumcapacity is not included for seismic shear wall design.

vsASDWSP7602

vsASDWSP 380 ASD Shearwall Capacity (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 widepanels require adjustment per SDPWS 4.3.3.4.1. Capacity of each 3 ft wide segment ismultiplied by 2bs/h adjustment factor.

bs 3 Width of panel requiring capacity adjustment (ft)

Capacity 2 5 vsASDWSP2bsh

2( ) 3( ) vsASDWSP



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

Leff 2 5( )2bsh

2( ) 3( ) Leff 14 Provided effective Full Height Sheathing Length (ft)

TVsLeff

h T 3043 Required Hold down capacity (lbs) (SDPWS 4.3.6.1.2)

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



3.3 PERFORATED SHEAR WALL DESIGN - WIND116

E3.3 - Perforated Shear Wall Design - WindUsing the 2015 Special Design Provisions for Wind and Seismic (SDPWS), design the first floorwall shown in the diagram below as as perforated shear wall (PSW) for a two-story house usingAllowable Stress Design (ASD) provisions

Design Wind Speed = 160 mph (3 sec. gust, 700 year return)Exposure BBuilding dimensions:


Check design with and without interior gypsum, neglect deflection.

Use Minimum Design Loads for Building and Other Structures (ASCE 7-10) to determine loads.

3.3 Perforated Shear Wall Design - Wind



Minimum segment length = Wall Height/Aspect Ratio

Lmin LminLmin93.5

Minimum full height wall segment length (ft)







vwASDWSP10652

vwASDWSP 532.5 ASD Shear wall Capacity (lbs/ft) (SDPWS 4.3.3)




vwASDGWB2502

vwASDGWB 125 ASD Shear wall Capacity, Gypsum (lbs/ft)(SDPWS 4.3.3)



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

Calculate PSW Shear Capacity Adjustment Factor (Co)

Li 2 5( ) 42 39

3

Effective length of Full Height Segments (ft) usingadjustment from SDPWS 4.3.4.3Li 18

Ltot 40 Wall length (ft)



STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 117PER

FOR

ATED SH

EAR W

ALL DESIG

N - W

IND

3.3



Minimum segment length = Wall Height/Aspect Ratio

Lmin LminLmin93.5

Minimum full height wall segment length (ft)







vwASDWSP10652





vwASDGWB2502

vwASDGWB 125 ASD Shear wall Capacity, Gypsum (lbs/ft)(SDPWS 4.3.3)



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


Li 2 5( ) 42 39

3

Effective length of Full Height Segments (ft) usingadjustment from SDPWS 4.3.4.3Li 18

Ltot 40 Wall length (ft)



3.3 PERFORATED SHEAR WALL DESIGN - WIND118

Ao 4 4.5 3( ) 6 7.5( ) Area of openings (ft2)

Ao 99

r1

1Aoh Li

r 0.621 (SDPWS Eqn. 4.3-6)

Cor

3 2r

LtotLi

Co 0.78 (SDPWS Eqn. 4.3-5)

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

LPSWVw

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

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

T 1 TTVw h Co Li

Required Hold-down capacity (lbs) (SDPWS Eqn. 4.3-8)

Design without Interior Gypsum

Vw 5520 Wind reaction on shear wall (lbs)



vwASDWSP10652


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

vPSW vwASDWSP Co

Perforated Shearwall (PSW) Capacity (lbs/ft)vPSW 418

LPSW 10.7LPSWVw

vPSW Required length of Full Height Sheathing (LPSW) (ft)


T 3519TVw h Co Li

Required Hold-down capacity (lbs) (SDPWS Eqn. 4.3-8)

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




FOR

ATED SH

EAR W

ALL DESIG

N - SEISM

IC3

.4

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

Seismic Design Category D1 (Based on 2015 IRC)

Building dimensions:


Neglect deflection.

Use Minimum Design Loads for Building and Other Structures (ASCE 7-10) to determine loads.

3.4 Perforated Shear Wall Design - Seismic



3.4 PERFORATED SHEAR WALL DESIGN - SEISMIC120




Lmin93.5



Vs 4733 Applied shear load on each shear wall due to seismic force (lbs)


(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. edgespacing. 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 capacityis not included for seismic shear wall design.

vsASDWSP7602

vsASDWSP 380 ASD Shearwall Capacity (lbs/ft) (SDPWS 4.3.3)


Li 2 5( ) 42 39

3

Effective length of Full Height Segments (ft) usingadjustment from SDPWS 4.3.4.3

Li 18

Ltot 40 Total wall length (ft)

Ao 4 4.5 3( ) 6 7.5( ) Area of openings (ft2)

Ao 99

r1

1Aoh Li

(SDPWS Eqn. 4.3-6)

Cor

3 2r

LtotLi

(SDPWS Eqn. 4.3-5)

Co 0.78




FOR

ATED SH

EAR W

ALL DESIG

N - SEISM

IC3

.4

vPSW vsASDWSP Co vPSW 298 Perforated Shearwall (PSW) Capacity (lbs/ft)

FHS FHSLPSWVs

vPSW LPSW 15.9 Required length of Full Height Sheathing (FHS) (ft)


Hold down capacity for Perforated Shearwalls specified in SDPWS Eqn. 4.3-8

T 1 TTVs h Co Li

T 3017 Required Hold down capacity (lbs)

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



122

NOTES

American Wood Council

AWC Mission StatementTo increase the use of wood by assuring the broadregulatory acceptance of wood products, developingdesign tools and guidelines for wood construction,and influencing the development of public policiesaffecting the use and manufacture of wood products.

ISBN 978-1-940383-05-7

9 781940 383057

08-19

American Wood Council222 Catoctin Circle, SE, Suite 201Leesburg, VA [email protected]

International Code CouncilGovernment Affairs Office500 New Jersey Avenue, NW, 6th FloorWashington, DC 20001-2070Regional Offices: Birmingham, AL; Chicago, IL; Los Angeles, CA1-888-422-7233www.iccsafe.org

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