# CALCULATING WIND LOADS ON LOW-RISE STRUCTURES PER 2015 WFCM ENGINEERING PROVISIONS (STD342-1)

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5 WOOD FRAME CONSTRUCTION MANUAL 11 Figure 1.1 Basic Wind Speeds for One- and Two-Family Dwellings Based on 700 yr Return Period 3-second Gust Basic Wind Speeds 1 GENERAL INFORMATION (from ASCE 7-10 Figure 6.5-1A with permission from ASCE)

7 WOOD FRAME CONSTRUCTION MANUAL 35 Figure.a Typical Lateral Framing Connections Cross Sec on End View Ra er Ceiling Joist ENGINEERED DESIGN Stud Floor Joist Stud Floor Joist

8 36 ENGINEERED DESIGN Figure.b Shear Connection Locations

9 WOOD FRAME CONSTRUCTION MANUAL 37 Figure.c Typical Wind Uplift Connections Cross Sec on End View Ra er Ceiling Joist ENGINEERED DESIGN Stud Floor Joist Stud Floor Joist

10 38 ENGINEERED DESIGN Figure.d Overturning Detail

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13 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 3 Table C1.1 ASD Velocity Pressure, q (psf), for Exposures B, C, and D and 33' MRH Exposure Category ASD Velocity Pressure, q (psf) 700-yr. Wind Speed 3-second gust (mph) Exposure B Exposure C Exposure D and q = 0.6 q h q h = K z K zt K d V K z (33 ft) = 0.7 K zt = 1.0 ASCE 7-10 Table (MWFRS), Table (C&C) at mean roof height (MRH) of 33 ft No topographic effects K d = 0.85 ASCE 7-10 Table GENERAL INFORMATION Design wind pressures in ASCE 7-10 are based on an ultimate 700-year return period. Since the WFCM uses allowable stress design, forces calculated from design wind pressures are multiplied by 0.60 in accordance with load combination factors per ASCE For example, the ASD velocity pressure, q, at 150 mph for Exposure B is calculated as follows: q = 0.6 (0.0056)(0.7)(1.0)(0.85)(150) (lbs/ft ) = 1.15 (lbs/ft ) In order to use the 015 WFCM with basic wind speeds from the 015 International Residential Code (IRC), see the wind speed conversion Table C1. based on the following calculations: Equating wind pressures calculated using ASCE 7-10 wind speeds with those from the 015 IRC. Velocity pressure for the ASCE 7-05 basic wind speed of 90 mph (Exposure B) is calculated as follows: q = (0.7)(0.85)90 = 1.7 psf ASD velocity pressure using the ASCE 7-10 wind speed of 116 mph (Exposure B) is calculated as follows: q = (0.60)[0.0056(0.7)(0.85)116 ] = 1.7 psf On the basis of equating wind pressures, the 90 mph ASCE 7-05 basic wind speed is equivalent to the 116 mph ASCE 7-10 basic wind speed. Table C1. Wind Speed Conversion Table ASCE 7-05 Basic Wind Speeds (mph) Equivalent ASCE 7-10 Basic Wind Speeds (mph) Wind speed contour maps in the 015 IRC show the 90 mph contour as covering approximately the same geographical area as that for the 115 mph wind speed contour in ASCE The velocity pressure for the 115 mph (Exposure B) ASCE 7-10 wind speed (1.4 psf) however, is slightly less than the velocity pressure corresponding to the 90 mph 015 IRC (Exposure B) wind speed (1.7 psf). Note that the worst case of internal pressurization is used in design. Internal pressure and internal suction for MWFRS are outlined in WFCM Tables C1.3A and C1.3B, respectively. Pressure coefficients and loads for wind parallel and perpendicular to ridge are tabulated. Parallel to ridge coefficients are used to calculate wind loads acting perpendicular to end walls. Perpendicular-toridge coefficients are used to calculate wind loads acting perpendicular to side walls. Pressures resulting in shear, uplift, and overturning forces are applied to the building as follows:

15 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 5 opening protection or special glazing requirements may be required by the local authority to ensure that the building envelope is maintained. C1.1.6 Ancillary Structures Design of ancillary structures is outside the scope of this document. 1 Table C1.3A ASCE 7-10 Exposure B Main Wind-Force Resisting System (MWFRS) Loads, p (psf), for an Enclosed Building, 33' Mean Roof Height with Internal Pressure, 150 mph (700-yr. 3-second gust), Exposure B q = 1.15 psf (See Table C1.1) GENERAL INFORMATION with internal pressure Interior Zones Exterior Zones Roof Angle E E 3E 4E 5E 6E GC pf GC pi p (psf) GC pf GC pi p (psf) (6:1) GC pf GC pi p (psf) GC pf GC pi p (psf) GC pf GC pi p (psf)

16 6 GENERAL INFORMATION Table C1.3B ASCE 7-10 Exposure B Main Wind-Force Resisting System (MWFRS) Loads, p (psf), for an Enclosed Building, 33' Mean Roof Height with Internal Suction, 150 mph (700-yr. 3-second gust), Exposure B q = 1.15 psf (See Table C1.1) with internal suction Interior Zones Exterior Zones Roof Angle E E 3E 4E 5E 6E GC pf GC pi p (psf) (6:1) GC pf GC pi p (psf) GC pf GC pi p (psf) GC pf GC pi p (psf) GC pf GC pi p (psf)

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23 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 15 Footnote 1: Tabulated loads are based on end zone pressures. Where the requirements of footnote 1 are met, tabulated loads may be decreased as follows: Given mph, Exposure B, 33' MRH, 0 degree roof slope, 1' roof span, ' overhangs, 0 psf roof/ceiling dead load, 1" o.c. End Zone uplift force is: 118 plf Interior Zone uplift force based on calculation similar to exterior zone is: 88 plf Reduction factor: = (88 plf) / (118 plf) = 0.75 ENGINEERED DESIGN Dead Load Wind Uplift Force Uplift Force Reaction (F W ) Leeward Wall

26 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 17 Summing moments about the windward top of wall: Required capacity of ridge connections spaced at 16" o.c. ΣM W = 0 = - [(-475.) + (108)][9] - [-77.][5.5] F Rx + 18F Ry = F Rx + 18F Ry Summing moments about the leeward top of wall: ΣM L = 0 = [(-70.0) + (108)][9] + [-157.5][5.5] F Rx + 18F Ry = F Rx + 18F Ry Setting 18F Ry (leeward) = 18F Ry (windward) ΣM W = 0 = F Rx = F Rx 1F Rx = = 336 plf (16 in./1in./ft) = 447 lbs = T(1.33) (WFCM Table.B Footnote 3) Footnote 1: For framing not located within W/5 or 6' of corners, wind pressures are reduced. Horizontal force at ridge: p WR = 1.15[ ] = psf p LR = 1.15[ ] = psf Using the reduced horizontal wind force and solving for F Rx : ENGINEERED DESIGN Solving for F Rx : F Rx = -336 plf Maximum horizontal tension force (T) per the figure of Table.B: T = 336 plf (WFCM Table.B) F Rx = 18 plf Dividing the reduced load by the tabulated load: = 18 / 336 = 0.65 WFCM Table.B specifies a slightly conservative multiplier of 0.70 to cover all cases.

27 WOOD FRAME CONSTRUCTION MANUAL 65 Table.C Rake Overhang Outlooker Uplift Connection Loads 700-yr. Wind Speed 3-second gust (mph) Outlooker Spacing (in.) Uplift Connection Loads (lbs) 1,, Tabulated outlooker uplift connection loads assume a building located in Exposure B with a mean roof height of 33 feet. For buildings located in other exposures, or with mean roof heights less than 33 feet, the tabulated values shall be multiplied by the appropriate adjustment factor in Section Tabulated outlooker uplift connection loads are based on foot overhangs. For overhangs less than feet, tabulated values shall be permitted to be multiplied by [(' + OH) / 4'] (OH measured in feet). 3 For overhangs located in Zone per the figures of Table.4, tabulated uplift loads shall be permitted to be multiplied by Outlooker overhang length shall be limited to 0 inches. See footnote to calculate reduced uplift connection load. ENGINEERED DESIGN 4 o.c. Nail Spacing L Connec on per Sec on...1 & Perimeter Zone panel field nailing 4 Perimeter Zone panel edge nailing Lesser of L/ or Required Blocking Minimum x4 Outlooker Gable Endwall

28 18 ENGINEERED DESIGN Table.C Rake Overhang Outlooker Uplift Connection Loads Description: Uplift loads at the connection of the rake overhang outlooker to endwall or rake truss. Procedure: Calculate wind pressures based on C&C pressure coefficients assuming Zone 3 and Zone 3 Overhang wind loads per the Figure of Table.4. Sum moments about the first interior truss to calculate the uplift connection load. Background: Outlooker connection loads are based on Table.4 suction pressures. Example: Given mph, Exposure B, 4" o.c. outlooker spacing, 4" truss spacing, ' overhang. For Zone 3 Roof Overhangs the suction pressure is: From WFCM Table.4: 78.3 psf For Zone 3 Roof the suction pressure is: From WFCM Table.4: 63.0 psf Summing moments about the first interior truss, the uplift force at the connector: Thus, where Zone occurs, tabulated loads may be multiplied by the following factor: U zone /U zone 3 U zone /U zone 3 Required Blocking = (419 lbs) /(697 lbs) = 0.60 (a conservative multiplier of 0.65 is specified in Table. Footnote 3) Wind Uplift Forces Uplift Connector Gable Endwall First Interior Truss or Rafter Outlooker Footnote 4: For the footnoted cases in Table.C, the outlooker overhang length, L overhang, is limited to 0 inches based on the bending capacity of the outlooker. w = 13.3 psf ΣM = 0 = (78.3/1)(36 in.)(4 in.) + (63.03/1)(4 in.) (1 in.) - (U)(4 in in.) U = [(78.6/1)(36 in.)(4 in.) + (63.03/1)(4 in.)(1 in.)]/(4 in. 3.5 in.) U = 349 lbs/ft Required capacity of connections spaced at 4" o.c. U = 349 lbs/ft (4 in./1 in./ft) U = 697 lbs (WFCM Table.C) Footnote 3: Tabulated loads may be reduced when they occur in Zone per the Figure of Table.4. The required capacity for a connector in Zone with Zone overhang wind loads for the overhanging section is: U zone = 419 lbs L overhang 1/ 3.5 M < F b S x L The bending moment, M, due to the uplift forces on the portion of the cantilevered outlooker is: M = (wl )/ = (13.3/1)(L )/ = 5.51 (L ) in.-lbs/ft

29 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 19 where: M Req. < F b ' S x w = 13.3 psf (WFCM Table.4 Zone mph) L = L overhang + 1/ in. (sheathing thickness) in. (width of stud) L overhang = L - 4 in. F b ' S x for a No., x4 Southern Pine Outlooker is: F b ' S x = F b C D C F C r S x = (1,100 psi)(1.6)(1.0)(1.15)(3.065) = 6,199 in.-lbs (Note: In the NDS Supplement, bending design values for Southern Pine are already size-adjusted, so the size factor, C F, is equal to 1.0.) For outlookers spaced at 4" o.c. (5.51)(L ) in.-lbs/ft (ft) 6,199 in.-lbs L 3.7 in. Therefore L overhang in. = 19.7 in. ~0 in. WFCM Table.C limits outlooker overhang lengths to 0" for 4" outlooker spacing and the 195 mph, 180 mph, and 175 mph (700 yr, 3-second gust) Exposure B wind load cases. NOTE: WFCM Table 3.4C provides additional outlooker overhang limits for Exposure C wind load cases. ENGINEERED DESIGN

30 WOOD FRAME CONSTRUCTION MANUAL 67 Table.4 Roof and Wall Sheathing Suction Loads (For Sheathing and Sheathing Attachment) 700 yr. Wind Speed 3 second gust (mph) Sheathing Location Zone Zone Zone Zone 3 Overhang Zone Zone The dimension, a, is measured as 10% of the minimum building dimension, but not less than 3 feet. Tabulated framing loads assume a building located in Exposure B with a mean roof height of 33 feet. For buildings located in other exposures, the tabulated values shall be multiplied by the appropriate adjustment factor in Section Dual Slope Roof Suction Pressure (psf) ENGINEERED DESIGN

31 ENGINEERED DESIGN Table.4 Roof and Wall Sheathing Suction Loads Description: Procedure: Suction pressure on roof sheathing. Calculate wind pressures based on C&C pressure coefficients. Background: Roof sheathing suction loads are tabulated for dual-slope roofs based on ASCE 7-10 Figures 30.4-A, B, and C. The roof slope creating the greatest suction loads for slopes between 7 o and 45 o are tabulated. Wall sheathing suction loads are tabulated based on ASCE 7-10 Figure Example: 150 mph, Exposure B, 33' MRH Zone 3 Overhang: GC p = ([log(A/100)] / [log(10/100)]) = -3.7 For all Zones GC pi = +/ Calculate roof pressures: (Negative pressures denote suction) Zone 1: p = 1.15 ( ) = -5.0 psf (WFCM Table.4) where: p = q(gc p - GC pi ) Zone : p = 1.15( ) = psf (WFCM p = pressure on the roof/wall Table.4) q = 1.15 psf (See Table C1.1) Zone Overhang: p = 1.15(-.) = psf GC p = external pressure coefficient for individual roof/wall areas GC pi = +/ internal pressure coefficient for enclosed buildings ASCE 7-10 states that for cladding and fasteners, the effective wind area shall not be greater than the tributary area for an individual fastener. Maximum loads occur for effective wind areas 10 ft or less. Sheathing suction loads are therefore based on an effective wind area of 10 ft for all roof and wall zones. From ASCE 7-10 Figures 30.4-A, B, and C: Zone 1: Zone : GC p = ([log(A/100)] / [log(10/100)]) = -1.0 GC p = ([log(A/100)] / [log(10/100)]) = -1.8 Zone Overhang: Zone 3: GC p = -. GC p = ([log(A/100)] / [log(10/100)]) = -.8 Zone 3: p = 1.15( ) = psf Zone 3 Overhang: (WFCM Table.4) p = 1.15(-3.7) = psf (WFCM Table.4) From ASCE 7-10 Figure : Zone 5: Zone 4: GC p = [(log(A/100)) / (log(10/100))] = -1.4 GC p = [(log(A/100)) / (log(10/100))] = -1.1 For all Zones GC pi = +/ Calculate wall pressures: (Negative pressures denote suction) Zone 4: Zone 5: p = 1.15( ) = -7.1 psf p = 1.15( ) = psf (WFCM Table.4) (WFCM Table.4)

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33 68 ENGINEERED DESIGN Table.5A Lateral Diaphragm Loads from Wind - Perpendicular to Ridge (For Calculating In-Plane Shear in Roof and Floor Diaphragm) Roof / Ceiling Diaphragm Shear Wall Floor Diaphragm w roof w floor w floor w floor B L 700 yr. Wind Speed second gust (mph) Roof Pitch Roof Span (ft) Unit Lateral Loads for Roof Diaphragm, w roof, (plf) 1,3,4,5 0:1 1: :1 3: : : : : : : : : : Unit Lateral Loads for Floor Diaphragm, w floor, (plf) 1,,3, See footnotes 1-5.

34 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 3 Table.5A Lateral Diaphragm Loads from Wind - Perpendicular to Ridge (Roof and Floor Shear) Description: For calculating lateral loads in roof and floor diaphragms from wind forces acting perpendicular to the roof ridge. Procedure: Compute wall and roof wind pressures. Collect forces into diaphragms. Background: Lateral loads are based on MWFRS and are a function of roof slope. For loads into floor diaphragms, the MWFRS coefficients are based on a worst case 0 degree roof slope. Minimum 10 psf (ASD) wall pressure and minimum 5 psf (ASD) roof pressure specified by ASCE 7-10 Section are checked. Example: Given mph, Exposure B, 33' MRH, 6:1 roof slope, 4' roof span, 10' wall height. Since these pressures act in the same direction, they will sum to 19.9 psf at the interior zone and 6.7 psf at the end zone. Calculate the average pressure on the wall assuming a building length (parallel to ridge), L, equal to the roof span (4 ft): where: p = [19.9(L-X) + 6.7(X)] / L = [19.9 psf (18 ft) psf (6 ft)] / 4 = 1.59 psf L = Building Length (parallel to ridge) X = End Zone Length Calculate the roof pressures for a 6:1 roof pitch: ENGINEERED DESIGN Floor Diaphragm Roof Diaphragm B L w roof w floor w floor w floor Interior Zone End Zone Windward Leeward Windward Leeward GC pf GC pi p (psf) Since these pressures act in opposite directions, they will sum to 7.4 psf at the interior zone and 8.4 psf at the end zone. Calculate the average pressure on the roof: Calculate wind forces in roof and floor diaphragms: p = q(gc pf - GC pi ) where: p = pressure on the roof/walls where: p = [7.4(L-X) + 8.4(X)] / L = [7.4(18 ft) + 8.4(6 ft)] / 4 = 7.65 psf W = Building Length (parallel to ridge) X = End Zone Length q = 1.15 psf (See Table C1.1) Calculate wall pressures: Interior Zone End Zone Windward Leeward Windward Leeward GC pf GC pi p (psf) Calculate the lateral load on the roof diaphragm: The roof diaphragm will take load from half the wall below and load directly applied to the vertical projection of the roof diaphragm. w roof = 1.59(10/ ft) [(6/1)(4/) ft] w roof = 1.59(5 ft) (6 ft) = 154 plf (WFCM Table.5A)

35 4 ENGINEERED DESIGN Calculate average pressure on the wall given the maximum MWFRS coefficients occur at a 0 degree roof slope per ASCE Interior Zone End Zone Windward Leeward Windward Leeward GC pf GC pi p (psf) p = [0.3(18 ft) (6 ft) +]/4 =.8 psf Summing the wall pressure for a 10' high wall and adding an extra 1' to account for the depth of the floor joists, calculate the lateral load on the floor diaphragm: w floor =.8(11) = 51 plf (WFCM Table.5A) Since these pressures act in the same direction, they will sum to 0.3 psf at the interior zone and 30.3 psf at the end zone.

36 WOOD FRAME CONSTRUCTION MANUAL 69 Table.5B Lateral Diaphragm Loads from Wind - Parallel to Ridge (For Calculating In-Plane Shear in Roof and Floor Diaphragm) w roof Roof / Ceiling Diaphragm Shear Wall w floor w floor w floor B L Floor Diaphragm ENGINEERED DESIGN 700 yr. Wind Speed 3 second gust (mph) Roof Pitch 0:1 1:1 :1 3:1 4:1 5:1 6:1 7:1 8:1 9:1 10:1 11:1 1:1 See footnotes 1-5. Roof Span (ft) Unit Lateral Loads for Roof Diaphragm, w roofll, (plf) 1,3,4, Unit Lateral Loads for Floor Diaphragm, w floorii, (plf) 1,,3,

38 70 ENGINEERED DESIGN Footnotes to Tables.5A and.5b 1 The total shear load equals the tabulated unit lateral load multiplied by the building length perpendicular to the wind direction. Tabulated unit lateral loads are based on 10 foot wall heights and a 1 foot floor depth. For other wall heights, H, tabulated values for floor diaphragms shall be permitted to be used when multiplied by (H+1)/11. 3 Tabulated unit lateral loads assume a building located in Exposure B with a mean roof height of 33 feet. For buildings located in other exposures, the tabulated values shall be multiplied by the appropriate adjustment factor in Section Hip roof systems shall be designed using Table.5A for both orthogonal directions. 5 Shear capacity requirements for roof diaphragms, shear walls, and floor diaphragms shall be calculated as follows: Calculating Total Shear Capacity Requirements V roof, V floor Calculating Diaphragm Unit Shear Capacity Requirements v roof, v floor Calculating Total Shear Wall Shear Capacity Requirements V wall Wind Perpendicular to Ridge ( w from Table.5A) V w ( L) roof roof (lbs) Vfloor () i wfloor() i ( L) Wind Parallel to Ridge ( w from Table.5B) V ( ) roof wroof B V ( ) floor ( i) wfloor ( i) B Wind Perpendicular to Ridge Vroof vroof ( B) V floor( i) v floor( i) ( B) (plf) v v Wind Parallel to Ridge roof floor ( i) Vroof ( L) V floor ( i ( L) ) (plf) Wind Perpendicular to Ridge Wind Parallel to Ridge Shear Wall Bracing Roof & Ceiling V V V wall Vroof wall roof V V Shear Wall Bracing Roof/Ceiling & 1 Floor V wall V V roof floor (1) wall V V roof floor(1) Shear Wall Bracing Roof/Ceiling & Floors V wall V V V roof floor (1) floor () wall V V V roof floor(1) floor()

39 WOOD FRAME CONSTRUCTION MANUAL 71 Table.5C Lateral Diaphragm Loads from Wind - Parallel to Ridge (For Attic Floor or Ceiling Diaphragm When Bracing Gable Endwall) Tribut ary Area of Ceiling Diaphragm w 700 yr. Wind Speed 3 second gust (mph) Roof Pitch 0:1 1:1 :1 3:1 4:1 5:1 6:1 7:1 8:1 9:1 a c floor / ceiling 10:1 11:1 1:1 A c Floor / Ceiling Diaphragm Roof Span (ft) Unit Lateral Loads for Attic Floor/Ceiling Diaphragm, w attic floor/ceiling (plf) 1,,3,4,5,6,7, W IND ENGINEERED DESIGN See footnotes 1-8.

42 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 7 Interior Zone End Zone Windward Leeward Windward Leeward GC pf GC pi ±0.18 ±0.18 ±0.18 ±0.18 positive internal pressure (psf) negative internal pressure (psf) Reduction Factor = 1.3 / [1.3 - (-.3)] = 0.84 Footnote 7: The lateral load on the roof diaphragm from Table.5B may be reduced as follows using this example: w roof (with attic floor/ceiling diaphragm) = w roof (without attic floor/ceiling diaphragm) w attic floor/ceiling = = 3 plf. ENGINEERED DESIGN

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44 WOOD FRAME CONSTRUCTION MANUAL 81 Table.9A Exterior Wall Stud Bending Stresses from Wind Loads (Cont.) 700 yr. Wind Speed 3 second gust (mph) Stud Size Wall Height 8 ft 10 ft 1 ft 14 ft 16 ft 18 ft 0 ft 1 3 Stud Spacing x4 x6 x8 x4 x6 x8 x4 x6 x8 x4 x6 x8 x4 x6 x8 Induced f b (psi) 1,,3 1 in in in in in in in in in in in in in in in in in in in in in Tabulated bending stresses assume a building located in Exposure B with a mean roof height of 33 feet. For buildings located in other exposures, the tabulated values shall be multiplied by the appropriate adjustment factor in Section Tabulated bending stresses shall be permitted to be multiplied by 0.9 for framing not located within 3 feet of corners for buildings less than 30 feet in width (W), or within W/10 of corners for buildings greater than 30 feet in width. The tabulated bending stress (f b ) shall be less than or equal to the allowable bending design value (F b ') ENGINEERED DESIGN

45 COMMENTARY TO THE WOOD FRAME CONSTRUCTION MANUAL 33 Table.9A Exterior Wall Stud Bending Stresses from Wind Loads Description: Procedure: Bending stress in wall studs due to wind load. Compute wind pressures using C&C coefficients and calculate stud requirements. Background: As in Table.4, peak suction forces are very high. Defining the effective wind area and the tributary area of the wall stud is key to computing the design suction. Stud span equals the wall height minus the thickness of the top and bottom plates. For a nominal 8' wall, the height is: 97 1/8" - 4.5" = 9 3/8". Two cases have been checked in these tables. For C&C wind pressures, the bending stresses are computed independent of axial stresses. In addition, the case in which bending stresses from MWFRS pressures act in combination with axial stresses from wind and gravity loads must be analyzed. For buildings limited to the conditions in this Manual, the C&C loads control the stud design. Example: Given mph, Exposure B, 33' MRH, 10' wall height, x4 studs, 16" o.c. stud spacing. Calculations from Table.10 for exterior wall induced moments from wind loads showed the applied bending moment for this case to be ft-lbs. Substituting this bending moment into a bending stress calculation: f b = M (tabulated) /S = (1)/3.065 = 1,80 psi f b (Tabulated) = 1,80 psi (WFCM Table.9A) Footnote : See Commentary Table.1 for calculation of footnotes. ENGINEERED DESIGN

46 WOOD FRAME CONSTRUCTION MANUAL 85 Table.10 Exterior Wall Induced Moments from Wind Loads 700 yr. Wind Speed 3 second gust (mph) Wall Height 8 ft 10 ft 1 ft 14 ft 16 ft 18 ft 0 ft 1 Stud Spacing Induced Moment (ft lbs) 1, 1 in in in in in in in in in in in in in in in in in in in in in Tabulated induced moments assume a building located in Exposure B with a mean roof height of 33 feet. For buildings located in other exposures, the tabulated values shall be multiplied by the appropriate adjustment factor in Section Tabulated induced moments shall be permitted to be multiplied by 0.9 for framing not located within 3 feet of corners for buildings less than 30 feet in width (W), or within W/10 of corners for buildings greater than 30 feet in width. ENGINEERED DESIGN

47 36 ENGINEERED DESIGN Table.10 Exterior Wall Induced Moments From Wind Loads Description: Applied moment on wall due to wind loads. Procedure: Calculate the applied moment based on C&C wind pressures. Background: Applied suction force is dependent on tributary areas. Example: Given mph, Exposure B, 33' MRH, 10' wall height, 16" o.c. stud spacing. Substituting this uniform load, w, into a bending calculation for a simply supported member: M = wl 8 ( 39 48) ( 375 ) = 1 8 = 466 ft lbs p max = 9.61 psf (See Commentary to Table.1) w = p max (16 in./1in./ft) = plf M (Tabulated) = 466 ft-lbs (A value of 465ft-lbs is shown in WFCM Table.10 difference due to rounding in calculation of p max )

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49 96 ENGINEERED DESIGN Table.14A Rafter Spans for 0 psf Live Load L/D LL = 180 L/D LL = 40 L/D LL = 360 Attached Ceiling x4 x6 x8 x10 x1 DL = 10 psf DL = 0 psf No Attached Flexible Finish Brittle Finish Maximum Span 1,,3 Ceiling (including gypsum board) (including plaster and stucco) Rafter f b f b E 4 E 4 E 4 Spacing (psi) (psi) (psi) (psi) (psi) (ft-in.) (ft-in.) (ft-in.) (ft-in.) (ft-in.) ,46 84,37 16, ,885 38, , , , , , , ,619 1,011, ,000 1, ,107 94,809 1,414, * 1,00 1,600 99,516 1,39,355 1,859, * 1,400 1,867 1,171,34 1,561,765,34, * 1 in. 1,600,133 1,431,084 1,908,111,86, * 6-0* 1,800,400 1,707,630,76,840 3,415, * 6-0*,000,667,000,000,666,667 4,000, * 6-0*,00,933,307,379 3,076,506 4,614, * 6-0*,400 3,00,69,068 3,505,44 5,58, * 6-0* 6-0*,600 3,467,964,456 3,95,608 5,98, * 6-0* 6-0*,800 3,733 3,313,005 4,417,340 6,66, * 6-0* 6-0* 3,000 4,000 3,674,35 4,898,979 7,348, * 6-0* 6-0* ,77 73, , ,919 06, , , , , , , ,37 876, ,000 1,333 61,37 816,497 1,4, ,00 1, ,984 1,073,313 1,609, ,400 1,867 1,014,396 1,35,58,08, * 16 in. 1,600,133 1,39,355 1,65,473,478, * 1,800,400 1,478,851 1,971,801,957, *,000,667 1,73,051,309,401 3,464, * 6-0*,00,933 1,998,49,664,33 3,996, * 6-0*,400 3,00,76,840 3,035,787 4,553, * 6-0*,600 3,467,567,94 3,43,059 5,134, * 6-0*,800 3,733,869,146 3,85,58 5,738, * 6-0* 3,000 4,000 3,181,981 4,4,641 6,363, * 6-0* ,000 66, , ,41 188,56 8, , , , , , , , ,000 1, , ,356 1,118, ,00 1, , ,796 1,469, ,400 1,867 96,013 1,34,684 1,85, in. 1,600,133 1,131,371 1,508,494,6, * 1,800,400 1,350,000 1,800,000,700, *,000,667 1,581,139,108,185 3,16, *,00,933 1,84,144,43,19 3,648, *,400 3,00,078,461,771,81 4,156, * 6-0*,600 3,467,343,608 3,14,811 4,687, * 6-0*,800 3,733,619,160 3,49,14 5,38, * 6-0* 3,000 4,000,904,738 3,87,983 5,809, * 6-0* ,71 59,68 89, , ,655 5, , , , , , ,08 715, ,000 1, , ,667 1,000, ,00 1, ,67 876,356 1,314, ,400 1,867 88,51 1,104,335 1,656, in. 1,600,133 1,011,99 1,349,38,03, ,800,400 1,07,477 1,609,969,414, ,000,667 1,414,14 1,885,618,88, *,00,933 1,631,564,175,418 3,63, *,400 3,00 1,859,03,478,709 3,718, *,600 3,467,096,187,794,916 4,19, *,800 3,733,34,648 3,13,531 4,685, * 3,000 4,000,598,076 3,464,10 5,196, * 6-0* * Spans (horizontal projection) are limited to 6 feet in length. Check sources for availability of lumber in lengths greater than 0 feet. See footnotes 1-4.

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