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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.
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.
AMERICAN WOOD COUNCIL
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 1AD
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)
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.
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.
Reference and Adjusted Design Values for 4x16 Select Structural DF-L (size adjusted 4x12 values)
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 Factor
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
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
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.
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.
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.
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.
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.
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.
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.
Reference and Adjusted Design Values for 4x16 Select Structural DF-L (size adjusted 4x12 values)
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:
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
BSTRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 15
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.
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.
Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.
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.
Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.
Reference and Adjusted Design Values for 24F-1.8E structural glued laminated softwood timber
CD 1.15 CM 1.0 Ct 1.0 Temp up to 150 degrees F only occasionally
Cfu 1.0 Cc 1.0 Cb 1.0 CI 1.0 Cvr 1.0 (NDS Table 5.3.1)
E'x Ex CM Ct E'xmin Exmin CM Ct Note: Ex notation has changed to Ex app for 2018 NDS
E'x 1800000 psi E'xmin 950000 psi
Member length and properties
l 32 ft b 5 in d 30.25 in Initial iteration
Note: Beam length designated as lower case l instead of upper case L used in the Specification nomenclature
Ag b d Sxxb d2
6 Ixx
b d3
12 lsupport 6 in
Ag 151.3 in2 Sxx 762.6 in3 Ixx 11534 in4
Beam Stability Factor
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
b2 RB 13.375 Slenderness ratio for bending (3.3-5)
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 17G
LUED
LAMIN
ATED TIM
BER
BEAM
DESIG
N (ASD
) 1
.3
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.
Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.
Reference and Adjusted Design Values for 24F-1.8E structural glued laminated softwood timber
CD 1.15 CM 1.0 Ct 1.0 Temp up to 150 degrees F only occasionally
Cfu 1.0 Cc 1.0 Cb 1.0 CI 1.0 Cvr 1.0 (NDS Table 5.3.1)
E'x Ex CM Ct E'xmin Exmin CM Ct Note: Ex notation has changed to Ex app for 2018 NDS
E'x 1800000 psi E'xmin 950000 psi
Member length and properties
l 32 ft b 5 in d 30.25 in Initial iteration
Note: Beam length designated as lower case l instead of upper case L used in the Specification nomenclature
Ag b d Sxxb d2
6 Ixx
b d3
12 lsupport 6 in
Ag 151.3 in2 Sxx 762.6 in3 Ixx 11534 in4
Beam Stability Factor
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
b2 RB 13.375 Slenderness ratio for bending (3.3-5)
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:
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 21G
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:
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
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 25CO
MPR
ESSION
MEM
BER
S AND
COLU
MN
STABILITY CALCU
LATION
(ASD)
1.4
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.
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 27CO
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
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 29CO
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MEM
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ANALYSIS (ASD
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.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
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
Member length and properties
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.
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 31CO
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ANALYSIS (LR
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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.
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 33CO
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MEM
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ANALYSIS (LR
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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.
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 35CO
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BEN
DIN
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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
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)
lu 12inft l lu 144 in Laterally unsupported length
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 37CO
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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
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
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
Member length and properties
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
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 41CO
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BEN
DIN
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MPR
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(ASD)
1.7
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.
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
E1.7 COMBINED BENDING AND AXIAL COMPRESSION (ASD)42
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)
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 43CO
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BEN
DIN
G AN
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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)
E1.7 COMBINED BENDING AND AXIAL COMPRESSION (ASD)44
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
Determine Column Stability Factor Cp
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
Load Case 3: DL only
CD 0.90
Compression
Fc* Fc CD CM Ct CF Ci P3 PDead fc3P3Ag
Fc* 1395 psi P3 560 lbf fc3 68 psi
Determine Column Stability Factor Cp
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)
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 45CO
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BI-AXIAL B
END
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AXIAL COM
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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.
Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.
1.8 COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD)46
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.
Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.
Reference and Adjusted Design Values for No. 2 Southern Pine 2x4
Fc 1450 psi CM 1.0 Ct 1.0 CF 1.0 (NDS Table 4.3.1)
Cfu 1.10 Ci 1.0 Cr 1.0 Flat use factors are for weak 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
Member length and properties
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).
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 47CO
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BI-AXIAL B
END
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AXIAL COM
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N (ASD
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.8
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.
Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.
Reference and Adjusted Design Values for No. 2 Southern Pine 2x4
Fc 1450 psi CM 1.0 Ct 1.0 CF 1.0 (NDS Table 4.3.1)
Cfu 1.10 Ci 1.0 Cr 1.0 Flat use factors are for weak 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
Member length and properties
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
Determine Column Stability Factor Cp
FcE minFcE1
FcE2
FcE 728 psi Critical bucking design value for compression member
1.8 COMBINED BI-AXIAL BENDING AND AXIAL COMPRESSION (ASD)48
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
b2 RB 9.6 Slenderness ratio for bending (3.3-5) RB < 50 (NDS 3.3.3.7)
FbE1.20 E'min
RB2
FbE 6577 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 0.982 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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 55 LOAD
BEAR
ING
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:
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.
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
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
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS58
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
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS60
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 61 LOAD
BEAR
ING
WALL W
OO
D STU
D R
ESISTING
WIN
D AN
D G
RAVITY LOAD
S1
.9
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
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS62
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)
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:
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.
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS64
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.
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
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.
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
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 (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
Critical bucking design value for compression member
Compare Actual Compression Stress to Adjusted Compression Design Value
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.
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
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;
Compare Actual Compression Stress to Adjusted Compression Design Value
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.
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
Ke 1.0 Buckling length coefficient Ke for strong axis bending(pinned/pinned) column (Appendix G Table G1)
Compare Actual Compression Stress to Adjusted Compression Design Value
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 69 LOAD
BEAR
ING
WALL W
OO
D STU
D R
ESISTING
WIN
D AN
D G
RAVITY LOAD
S1
.9
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
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 (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
Critical bucking design value for compression member
Determine Column Stability Factor Cp for Load Combination 4a
Compare Actual Compression Stress to Adjusted Compression Design Value
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
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
Ke 1.0 Buckling length coefficient Ke for strong axis bending(pinned/pinned) column (Appendix G Table G1)
Stud dimensions, laterally unsupported lengths (NDS Figure 3F)and effective column lengths (3.7.1) for buckling in eachdirection. Subscript 1 is strong (but laterally unsupported) axis;
d1 d l1 L le Ke l1 12inft
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
Critical bucking design value for compression member
Determine Column Stability Factor Cp for Load Combination 4b
Compare Actual Compression Stress to Adjusted Compression Design Value
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
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
Determine Column Stability Factor Cp for Load Combination 5a
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
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS72
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 73 LOAD
BEAR
ING
WALL W
OO
D STU
D R
ESISTING
WIN
D AN
D G
RAVITY LOAD
S1
.9
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
CD6 1.6 Wind load duration factor CD controls for Load Combination 6a -NDS Appendix B Section B.2
Fc6a1* Fc CD6 CM Ct CF Ci Fc6a1* is reference compression design value adjusted with 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
Determine Adjusted Compression Design Value for Load Combination 6a1
Fc'6a1 Fc6a1* CP6 Fc'6a1 405 psi Adjusted compression design value for Load Combination 6a
Compare Actual Compression Stress with Adjusted Compression Design Value
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
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS74
Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures
Moment
wwindAWW 0.7516( )12
ft pmwfrsAWW Load Combination 6a1 includes 75% of the MWFRS WindLoad
wwindAWW 17.31 plf
MmwfrsAWWwwindAWW L2
812
inft Bending moment from out-of-plane MWFRS wind loads
MmwfrsAWW 9375 inꞏlbf
Determine Reference and Adjusted Bending Design Values for Load Combination 6a1
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'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
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
Check Combined Uniaxial Bending and Axial Compression
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)
Determing compression load in framing for load combination 6a2
P6a216( )12
ft Pdist10
P6a2 716 lbf
Determine Reference and Adjusted Compression Design Values for Load Combination 6a2
CD6 1.6 Wind load duration factor CD controls for Load Combination 6a -NDS Appendix B Section B.2
Fc6a2* Fc CD6 CM Ct CF Ci Fc6a2* is reference compression design value adjusted with alladjustment factors except the column stability factor CP,. Thefollowing calculations determine the column stabilty factor CP:Fc6a2* 2160 psi
Determine Column Stability Factor Cp for Load Combination 6a2
Determine Adjusted Compression Design Value for Load Combination 6a2
Fc'6a2 Fc6a2* CP6 Fc'6a2 405 psi Adjusted compression design value for Load Combination 6a
Compare Actual Compression Stress with Adjusted Compression Design Value
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
Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures
Moment
wwindBWW 0.7516( )12
ft pmwfrsBWW Load Combination 6a2 includes 75% of the MWFRS WindLoad
wwindBWW 14.74 plf
MmwfrsBWWwwindBWW L2
812
inft Bending moment from out-of-plane MWFRS wind loads
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 75 LOAD
BEAR
ING
WALL W
OO
D STU
D R
ESISTING
WIN
D AN
D G
RAVITY LOAD
S1
.9
Load Case 6a2: D + 0.75L + 0.75 W (wind perpendicular to ridge)
Determing compression load in framing for load combination 6a2
P6a216( )12
ft Pdist10
P6a2 716 lbf
Determine Reference and Adjusted Compression Design Values for Load Combination 6a2
CD6 1.6 Wind load duration factor CD controls for Load Combination 6a -NDS Appendix B Section B.2
Fc6a2* Fc CD6 CM Ct CF Ci Fc6a2* is reference compression design value adjusted with alladjustment factors except the column stability factor CP,. Thefollowing calculations determine the column stabilty factor CP:Fc6a2* 2160 psi
Determine Column Stability Factor Cp for Load Combination 6a2
Determine Adjusted Compression Design Value for Load Combination 6a2
Fc'6a2 Fc6a2* CP6 Fc'6a2 405 psi Adjusted compression design value for Load Combination 6a
Compare Actual Compression Stress with Adjusted Compression Design Value
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
Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures
Moment
wwindBWW 0.7516( )12
ft pmwfrsBWW Load Combination 6a2 includes 75% of the MWFRS WindLoad
wwindBWW 14.74 plf
MmwfrsBWWwwindBWW L2
812
inft Bending moment from out-of-plane MWFRS wind loads
1.9 LOADBEARING WALL WOOD STUD RESISTING WIND AND GRAVITY LOADS76
Determine Reference and Adjusted Bending Design Values for Load Combination 6a2
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'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
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
Check Combined Uniaxial Bending and Axial Compression
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)
Determing axial load in framing for load combination 7b
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
Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures
Moment
wwindBWW16( )12
ft pmwfrsBWW Load Combination 7b includes the MWFRS Wind Load
wwindBWW 19.65 plf
MmwfrsBWWwwindBWW L2
812
inft Bending moment from out-of-plane MWFRS wind loads
MmwfrsBWW 10642 inꞏlbf
Determine Reference and Adjusted Bending Design Values for Load Combination 7b
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'b7b Fb CD7 CM CL Ct CF Ci Cr F'b7b is adjusted bending design value for Load Combination 7b
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 77 LOAD
BEAR
ING
WALL W
OO
D STU
D R
ESISTING
WIN
D AN
D G
RAVITY LOAD
S1
.9
Load Case 7b: 0.6 D + 0.6 W (wind perpendicular to ridge)
Determing axial load in framing for load combination 7b
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
Determine Bending Stress from Out-of-Plane MWFRS Wind Pressures
Moment
wwindBWW16( )12
ft pmwfrsBWW Load Combination 7b includes the MWFRS Wind Load
wwindBWW 19.65 plf
MmwfrsBWWwwindBWW L2
812
inft Bending moment from out-of-plane MWFRS wind loads
MmwfrsBWW 10642 inꞏlbf
Determine Reference and Adjusted Bending Design Values for Load Combination 7b
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'b7b Fb CD7 CM CL Ct CF Ci Cr F'b7b is adjusted bending design value for Load Combination 7b
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 79 LOAD
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.
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.
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.
D 0.131 Fastener diameter (in.)
DH 0.281 Fastener head diameter (in.)
TL 1.5 Deformed Shank Length (in.)
tns 0.625 Net Side Member thickness (in.)
G 0.5 Specific gravity, main and side members (NDS Table 12.3.3A and 12.3.3B)
Checking Fastener Withdrawal
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.
D 0.131 Fastener diameter (in.)
DH 0.281 Fastener head diameter (in.)
TL 1.5 Deformed Shank Length (in.)
tns 0.4375 Net Side Member thickness (in.)
G 0.5 Specific gravity, main and side members (NDS Table 12.3.3A)
Checking Fastener Withdrawal
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)
2.2 SINGLE COMMON NAIL LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION 90
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.
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.
D 0.5 Fastener diameter (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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 93SIN
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
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.
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
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.
Define parameters:
Fem 5550 Main member Dowel Bearing Strength (NDS Table 12.3.3) (psi)
Fes 5550 Side member Dowel Bearing Strength (NDS Table 12.3.3) (psi)
2.4 SINGLE WOOD SCREW LATERAL DESIGN VALUE - DOUBLE WOOD TO WOOD CONNECTION96
Zdist
ZIm
ZIs
ZIIIs
ZIV
Zdist
1055
921
336
234
Creating an array with all Yield Mode Solutions
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 97SIN
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:
Fem 4800 Main member Dowel Bearing Strength (NDS Table 12.3.3) (psi)
Fes 4800 Side member Dowel Bearing Strength (NDS Table 12.3.3) (psi)
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)
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 101SIN
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.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 103SIN
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.
2.5 SINGLE BOLT LATERAL DESIGN VALUE - SINGLE SHEAR WOOD-TO-WOOD CONNECTION104
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.
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
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 107B
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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
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)
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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.
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
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)
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 111SEG
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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)
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
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
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)
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
Capacity 7455 Wall capacity (lbs)
Wall capacity exceeds demand, so design is OK.
Hold down capacity is based on induced unit shear (hold downs at each end of each panel)
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:
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)
Neglect deflection check.
Use ASCE 7-10 Minimum Design Loads for Buildings and Other Structures Simplified Approach todetermine seismic loads.
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 115SEG
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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.
Vs 4733 Applied shear load on each shear wall due to seismic force (lbs)
h 9 Wall height (ft)
(Note: Seismic force calculated using WFCM Table 2.6 and WFCM Commentary.)
Assume 7/16 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. 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.
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
Capacity 5320 Wall capacity (lbs)
Wall capacity exceeds demand, so design is OK.
Hold down capacity is based on induced unit shear of effective length of Full Height Segments.
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:
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, neglect deflection.
Use Minimum Design Loads for Building and Other Structures (ASCE 7-10) to determine loads.
3.3 Perforated Shear Wall Design - Wind
Check maximum segment length based on Aspect Ratio Limits
Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4)
Minimum segment length = Wall Height/Aspect Ratio
Lmin LminLmin93.5
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.
Design with Interior Gypsum
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)
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
STRUCTURAL WOOD DESIGN SOLVED EXAMPLE PROBLEMS 117PER
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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
Lmin LminLmin93.5
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.
Design with Interior Gypsum
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)
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
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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:
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)
Neglect deflection.
Use Minimum Design Loads for Building and Other Structures (ASCE 7-10) to determine loads.
Check maximum segment length based on Aspect Ratio Limits
Maximum aspect ratio for Wood Structural Panel Shear Walls = 3.5:1 (SDPWS 4.3.4)
Minimum segment length =Wall Height/Aspect Ratio
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.
Vs 4733 Applied shear load on each shear wall due to seismic force (lbs)
h 9 Wall height (ft)
(Note: Seismic force calculated using WFCM Table 2.6 and WFCM Commentary.)
Assume 7/16 in. thick Wood Structural Panel (WSP) Sheathing, 8d nails @ 4 in. o.c. 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.
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
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