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APPROVED BLD STRUC CALCS BLD2019-095113228 NE 20th St, Suite 100, Bellevue, WA 98005 Phone (425) 614-0949 Fax (425) 614-0950 STRUCTURAL CALCULATIONS FOR: Edmonds Apartment Building 23830 Edmonds Way, Edmonds, WA 98026 Client: Hayes Wilson Lund Architects, LLC 385 101 st AVE SE Bellevue, WA 98004 By rill Armour. S.E.. SECB Mitchell Beck, P.E. Konrads Leitis, E.I.T. 7/25/19 PLAN REVIEW ACCEPTANCE FOR COMPLIANCE WITH THE APPLICABLE CONSTRUCTION CODES IDENTIFIED BELOW. ® BUILDING ® STRUCTURAL ® MECHANICAL ® PLUMBING ELECTRICAL ® ENERGY ® ACCESSIBILITY FIRE PLAN REVIEW ACCEPTANCE OF DOCUMENTS DOES NOT AUTHORIZE CONSTRUCTION TO PROCEED IN VIOLATION OF ANY FEDERAL, STATE, OR LOCAL REGULATIONS. BY: ��' DATE:07/29/2020 WEST COAS CODE CONSULTANTS, INC. AUE No. 18456 Date 7/11/2019 Design Criteria Code Summary Building Code: 2015 International Building Code Supplemental Code: 2010 ASCE 7 Risk Category: II Standard -Occupancy Building Seismic Loading: Ss: 1.3957 Sas: 0.9305 Ie: 1.00 Site Class: D S1: 0.5514 Sal: 0.5514 R: 6.50 Seismic Design Category: D Cs: 0.1431 Wind Loading: Vas: 110 mph Exposure Category: B Risk Category: II Kzt: 1.00 Design Loads Dead Loads: (Floor Structure: 9-1/2" 1-joists @ 16" O.C. 1.9 psf 3/4" Sheathing 2.8 psf 1/2" GWB 2.2 psf 3/4" gyperete 6.5 psf Carpete + Pad 2.0 psf M&E 1.0 psf Misc 1.6 psf Cement Board at Walkways 5.6 psf Total Interior Floor 18 psf Total Exterior Floor 24 psf Roof: Gangnail Trusses @ 24" O.C. 2.5 psf 5/8" Sheathing 1.8 psf 1/2" GWB 2.2 psf Roofing 3.0 psf Insullation 1.0 psf MEP 2.0 psf Misc. 2.5 psf Total 15.0 psf Live Load: Roof. 20.0 psf Snow: 25.0 psf Terrace: 100.0 psf Floor: 40.0 psf Walkway: 100.0 psf Storage: 125.0 psf Decks: 60.0 psf Soil Properties: Allowable Vertical Bearing = 1500.0 psf Deflection Criteria Stair: Floor Roof Walls Live Load: L/ 480 Live Load: L/ 480 Live Load: L/ 240 Flexible Finishes: L/ 180 Total Load: L/ 360 Total Load: L/ 240 Total Load: L/ 180 Brittle Finishes: L/ 360 Supporting Glass: L/ 240 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: DC 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 IBC Seismic Main Seismic Force Resisting Systems I IBC 2015 Risk Category: II Standard -Occupancy Building Fa = 1.00 Sms = 1.396 Ss= 1.396 hn = 15 ft Fv = 1.50 Sint = 0.827 S1= 0.551 Ct= 0.02 Ie = 1.00 SDC: D Sds= 0.930 Shc= 0.551 Site Class = D le: 1 98026 1 Use Zip Code Value Seismic Variables from USGS Seismic Variables from User Input Max Loc Max Ss = 1.3957 47.89; 122.32 Ss = - S1 = 0.5514 47.89,-122.32 S1 = - PGA = 0.5858 47.89,-122.32 Min Lee Ss = 1.2608 47.78,-122.35 S1 = 0.4920 47.78,-122.32 PGA = 0.5091 47.78,-122.35 See ASCE 12.8 Equiv. Lateral Force Procedure A15. Light -Frame (wood) Walls Sheathed with Wood Structural Panels Rated for Shear Resistance R= 6.5 no= 3.0 Cd= 4.0 Cs = 0.143 T = 0.152 Seconds Cs max = 0.557 Cs min = 0.010 S 1 < 0.6 Cs min = Not Applicable S 1 >= 0.6 Cs min = 0.010 V = 0.143 * W 13228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 Project: Edmonds Apartment Building Pg. No: MSFRS 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wind Loads - Main Wind Force Resisting System ASCE 7-10 Chapter 27 - Directional Procedure Environmental Criteria & Site Characteristics Risk Category: II Table 1.5-1 Basic Wind Speed: 110 Section 26.5.1 Exposure Category: B Section 26.7.3 Hurricane Prone: No Wind Load Parameters C Terrain Type: User Input Direction: Downwind of Crest Wind Directionality Factor, Kd: 0.85 Table 26.6-1 Gust Effect Factor, G: 0.85 Section 26.9 upwind of crest Pos. Lh: 0.00 ft Dist. Upwind of Crest to Half Ht of Hill or Escarp. H: 0.00 ft Ht. of Hill or Excarp. Relative to the upwind terrain x: 0.00 ft Dist. (Upwind or Downwind) from the crest to the building site z: 31.50 ft Ht above Local Ground Level h: 31.50 ft Mean Ht above Local Ground Level Kz: 0.71 K,,: Not Appl. Equation 26.8-1 K,: 1.00 *User Input based on local jurisdiction criteria au qz: 18.7 psf Eq. 27.3-1 qh: 18.7 psf Eq. 27.3-1 13228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 Project: Edmonds Apartment Building Pg. No: MWFRS 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wind Loads - Main Force Resisting System ASCE 7-10 Chapter 27, Directional Procedure Building Characteristics Building Type: Enclosed Buildings Length: 89.00 ft Longer Dimension Width: 67.00 ft Shorter Dimension Mean Height: 31.50 ft Roof Area: 5,963.00 sq ft GCpi: 0.18 Table 26.11-1 -0.18 Table 26.11-1 Wind Direction X Roof Angle: 20 LB: 1.33 (Length/Width) Cp Use w/ Windward Wall: 0.8 qz Leeward Wall: -0.44 qh Side Wall: -0.70 qh Figure 27.4-1 Windward Roof: C Use w/ Normal to Ridge for 0 >= 10 Degrees -0.34 0.12 qh Leeward Roof CP Use w/ Normal to Ridge for 0 >= 10 Degrees -0.6 qh Flat Roof or Distance From Windward Use w/ gh Parallel to Ridge Edge Cp Normal to ridge for 0 to h/2 -0.90 -0.18 0 < 10 Degrees or h/2 to h -0.90 -0.18 Parallel to ridge for h to 2h -0.50 -0.18 all >2h -0.30 -0.18 Wind Dir. Y Length Wind Dir. X © Width A❑ Mean Ht L I- Wind Direction Y Roof Angle: 20 LB: 0.75 (Width/Length) Cp Use w/ Windward Wall: 0.8 qz Leeward Wall: -0.50 qh Side Wall: -0.70 qh Figure 27.4-1 Windward Roof: Cp Use w/ Normal to Ridge for O >= 10 Degrees -0.38 0.04 qh Leeward Roof Cp Use w/ Normal to Ridge for O >= 10 Degrees -0.6 qh Flat Roof or Distance From Windward Use w/ gh Parallel to Ridge Edge Cp Normal to ridge for 0 to h/2 -0.90 -0.18 0 < 10 Degrees or h/2 to h -0.90 -0.18 Parallel to ridge for h to 2h -0.50 -0.18 all >2h -0.30 -0.18 13228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 Project: Edmonds Apartment Building Pg. No: MWFRS 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wind Loads - Main Force Resisting System ASCE 7-10 Chapter 27, Directional Procedure Wind Pressure Calculations Zg: 1200 a: 7.00 Wind Pressure on Walls Wind Direction X GCp: 0.68 GCPi: 0.18 or -0.18 Px Leeward: -7.0 psf Internal Px;: 3.4 psf -3.4 psf LRFD AM Ht (ft) Kz Px (psf) P (psf) P (psf) 0 0.57 10.29 17.29 10.37 5 0.57 10.29 17.29 10.37 15 0.57 10.29 17.29 10.37 25 0.67 11.91 18.90 11.34 35 0.73 13.11 20.10 12.06 #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! Wind Pressure on Walls Wind Direction Y GCP: 0.68 GCPi: 0.18 or -0.18 or Px Leeward: -7.9 psf Pxj: 3.4 psf -3.4 psf 1.RFD ASD Ht (ft) Kz Px (psf) P (psf) P (psf) 0 0.57 10.29 18.24 10.94 5 0.57 10.29 18.24 10.94 15 0.57 10.29 18.24 10.94 25 0.67 11.91 19.86 11.91 35 0.73 13.11 21.06 12.63 #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! #VALUE! 3228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 roject: Edmonds Apartment Building Pg. No: MWFRS 3 Lddress: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 'lient: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wind Pressure on Roofs (X-X) Windward Forces: (LRFD) Location: Cp 0 to h/2 -0.34 or 0.12 h/2 to h -0.34 or 0.12 h to 2h -0.34 or 0.12 >2h -0.34 or 0.12 Location: P (psf) 0 to h/2 -5.4 or 1.9 h/2 to h -5.4 or 1.9 h to 2h -5.4 or 1.9 >2h -5.4 or 1.9 Leeward Forces: (LRFD) Cp P (psf) -0.6 -9.5 Forces Parallel to Ridge: (LRFD) Location: Cp 0 to h/2 -0.9 or -0.18 h/2 to h -0.9 or -0.18 h to 2h -0.5 or -0.18 >2h -0.3 or -0.18 Location: P(psf) 0 to h/2 -14.3 or -2.9 h/2 to h -14.3 or -2.9 h to 2h -7.9 or -2.9 >2h -4.8 or -2.9 Windward Forces: (ASD) (0.6WL) Location: P (psf) h/2 to h -3.2 or 1.1 h to 2h -3.2 or 1.1 >2h -3.2 or 1.1 0 -3.2 or 1.1 Leeward Forces: (ASD) (0.6WL) P (psf) -5.7 Forces Parallel to Ridge: (ASD) (0.6WL) Location: P(psf) 0 to h/2 -8.6 or -1.7 h/2 to h -8.6 or -1.7 h to 2h -4.8 or -1.7 >2h -2.9 or -1.7 Wind Pressure on Roofs (Y-Y) Windward Forces: (LRFD) Location: Cp 0 to h/2 -0.38 or 0.04 h/2 to h -0.38 or 0.04 h to 2h -0.38 or 0.04 >2h -0.38 or 0.04 Location: P (psf) 0 to h/2 -6.0 or 0.6 h/2 to h -6.0 or 0.6 h to 2h -6.0 or 0.6 >2h -6.0 or 0.6 Leeward Forces: (LRFD) Cp P (psf) -0.6 -9.5 Forces Parallel to Ridge: (LRFD) Location: Cp 0 to h/2 -0.9 or -0.18 h/2 to h -0.9 or -0.18 h to 2h -0.5 or -0.18 >2h -0.3 or -0.18 Location: P(psf) 0 to h/2 -14.3 or -2.9 h/2 to h -14.3 or -2.9 h to 2h -7.9 or -2.9 >2h -4.8 or -2.9 Windward Forces: (ASD) (0.6WL) Location: P (psf) h/2 to h -3.6 or 0.4 h to 2h -3.6 or 0.4 >2h -3.6 or 0.4 0 -3.6 or 0.4 Leeward Forces: (ASD) (0.6WL) P (psf) -5.7 Forces Parallel to Ridge: (ASD) (0.6WL) Location: P(psf) 0 to h/2 -8.6 or -1.7 h/2 to h -8.6 or -1.7 h to 2h -4.8 or -1.7 >2h -2.9 or -1.7 13228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 Project: Edmonds Apartment Building Pg. No: MWFRS 4 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wind Loads - Components and Cladding, Walls ASCE 7-10 Chapter 30, Part 1 - Low Rise Buildings Design Criteria Risk Category: II Basic Wind Speed: 110 mph Exposure Category: B Hurricane Prone: No Building Type: Enclosed Buildings Length: 89.00 ft Width: 67.00 ft Height: 31.50 ft Component Exp. Category: B Wind Load Calculation - Walls Wind Directionality Factor, Kd: Gust Effect Factor, G: Zg: a: Kz: Kzt: qh: Longer Dimension Shorter Dimension 0.85 Table 26.6-1 0.85 Section 26.9 Roof Angle: > 10 Degrees 1200 Gcpi: 0.18 7.00 ft -0.18 0.71 Length B Width 1.00 Zone 5 Extent: 6.70 11 18.71 psf External Pressure Coefficients n Ht Area Location 0 10 20 50 120 200 500 4&5 1.00 1.00 0.95 0.88 0.81 0.77 0.70 GCpf. 4 -1.10 -1.10 -1.05 -0.98 -0.91 -0.87 -0.80 5 -1.40 -1.40 -1.29 -1.15 -1.02 -0.94 -0.80 * Roof is Over 10 Degrees, No Modification of GCp LOW Per ASCE 7-10 LRFD Load Cases (Buildine <= 60ft1 Area Location 0 10 20 50 120 200 500 4&5 22.1 22.1 21.1 19.8 18.5 17.8 16.5 p 4 -23.9 -23.9 -23.0 -21.7 -20.4 -19.6 -18.3 5 -29.6 -29.6 -27.5 -24.9 -22.4 -20.9 -18.3 0.6W Per ASCE 7-05 ASD Load Cases Area Location 0 10 20 50 120 200 500 4&5 13.2 13.2 12.7 11.9 11.1 10.7 9.9 p 4 -14.4 -14.4 -13.8 -13.0 -12.2 -11.8 -11.0 5 -17.7 -17.7 -16.5 -14.9 -13.5 -12.6 -11.0 Fig. 30.4-1 EQ 30.4-1 EQ 30.4-1 13228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 Project: Edmonds Apartment Building Pg. No: CCWa11 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 c*- A I iaK 0 Ut 13228 NE 211th St, Saite 100, Bellevue, WA 980051 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Address: 23830 Edmonds Way, Edmonds, WA Date: Client: Ha es Wilson Lund Architects, LLC Job No.: 19456 v FAI .(A a 3228 NE 20tb St, Suite 100, BeUe e, WA 980051(425) 614-0949 Snjea: Edmonds Apartment Building Pg. No: Lddass: 23830 Edmonds Way, Edmonds, WA Date: :Bent: Haynes Wilson Lund Architects, LLC Job No.: 19456 u V) \6 CL LJ 13228 NE 20th St, Suite 100, Bellevue, WA 98005 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: At. Address: 23830 Edmonds Way, Edmonds, WA Date: _ wy Client: Haynes Wilson Lund Architects, LLC Job No.: 18456 PANEL NAILING SILL PLATE TO CONC SILL PLATE TO WOOD SHEAR TRANSFER MARK LEVEL SHEATHING SIZE EDGE NAILING FIELD NAILING PLATE ANCHOR BOLTS PLATE NAILING CLIPS Q Upper 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Middle 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II CS 10d 6"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC Q Upper 15/32 STRUCT II C5 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Middle 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II C5 10d 4"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC Q3 Upper 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6' OC A35 @ 16" OC Middle 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6' OC A35 @ 16" OC Lower 15/32 STRUCT II C5 10d 6"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC a4 Upper 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16 OC Middle 15/32 STRUCT II C5 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II CS 10d 4"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC - 55 Upper 15/32 STRUCT II C5 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16 OC Middle 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II C5 10d 6"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC A66 Upper 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16 OC Middle 15/32 STRUCT II CS 10d 6"oc - 12" OC 2X {2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II CS 10d 6"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC Q Upper 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6 OC A35 @ 16 OC Middle 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II CS 10d 6"oc 12" OC 3X OR (2) 2X 5/8' DIA @ 48" OCAs Upper 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6' OC A35 @ 16 OC Middle 15/32 STRUCT II CS 10d 6"oc 12" OC 2X (2) ROWS 10d @ 6" OC A35 @ 16" OC Lower 15/32 STRUCT II CS 10d 6"oc 12" OC 3X OR (2) 2X 5/8" DIA @ 48" OC - Shear Wall Analysis Shearwall Delta 1 7.89 8.75 8.75 ft ft ft Vertical Loads: Roof DL: 2nd Floor DL: 3rd Floor DL: Wall DL: T1 T1 T2 T2 Ld 14.33 ft 15 psf 18 psf 18 psf 10 Dsf 3rd Floor Wall - Local Overturn MOT: 7.732 k-ft Wall Wt: 1.131 k Roof Weight: 3.671 k MR: 34.41 k-ft Uplift: 0.00 k @ T1 Roof Fr: 0.98 k 3rd Flr �'J 0.9 k Total F3: 1.88 k 2nd Flr *==[--> 0.9 k Total F2: 2.78 k Foundation MOT: 82.11 k-ft Wall Wt: 3.638 k Wall Trib: 25.39 ft 2nd Flr Wt: 1.236 k 2nd Flr Trib: 4.79 ft 3rd Flr Wt: 1.236 k 3rd Flr Trib: 4.79 ft Roof Weight: 3.671 k Roof Trib: 17.08 ft MR: 70.08 k-ft Uplift: 2.943 k Wall Trib: 7.89 ft Roof Trib.: 17.08 ft 2nd Floor Wall - Local Overtu MOT: 32.76 k-ft Wall Wt: 2.385 k Roof Weight: 3.671 k 3rd Flr Wt: 1.236 k MR: 43.39 k-ft Uplift: 0.49 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 17.08 ft 3rd Flr Trib: 4.79 ft 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: Shear Wall 1 ress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 nt: Hayes Wilson Lund Architects, LLC Job No.: 18456 7.89 8.75 8.75 ft ft ft Ah T1 T1 T2 T2 t 19.83 ft Roof Fr: 2.1 k 3rd Flr 1.9 k Total F3: 4 k 2nd Flr 1.9 k Total F2: 5.9 k Foundation MOT: 174.9 k-ft Wall Wt: 5.035 k 2nd Flr Wt: 0.714 k 3rd Flr Wt: 0.714 k Roof Weight: 0 k MR: 64.08 k-ft 7.25 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 2 ft 3rd Flr Trib: 2 ft Roof Trib: 0 ft MOT: 69.94 k-ft Wall Wt: 3.3 k Roof Weight: 0 k 3rd Flr Wt: 0.714 k MR: 32.72 k-ft Uplift: 2.67 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 0 ft 3rd Flr Trib: 2 ft 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: Shear Wall 2 kddress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 ,ient: Hayes Wilson Lund Architects, LLC Job No.: 18456 7.89 8.75 8.75 ft ft ft Ah T1 T1 T2 T2 t 14.33 ft Roof Fr: 1.2 k 3rd Flr 1.1 k Total F3: 2.3 k 2nd Flr 1.1 k Total F2: 3.4 k Foundation MOT: 100.5 k-ft Wall Wt: 3.638 k 2nd Flr Wt: 1.236 k 3rd Flr Wt: 1.236 k Roof Weight: 3.671 k MR: 70.08 k-ft Uplift: 4.29 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 4.79 ft 3rd Flr Trib: 4.79 ft Roof Trib: 17.08 ft MOT: 40.09 k-ft Wall Wt: 2.385 k Roof Weight: 3.671 k 3rd Flr Wt: 1.236 k MR: 43.39 k-ft Uplift: 1.03 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 17.08 ft 3rd Flr Trib: 4.79 ft 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: Shear Wall 3 ress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 nt: Hayes Wilson Lund Architects, LLC Job No.: 18456 (Worst Case) 7.89 8.75 8.75 II ft ft Ah T1 T1 T2 T2 t 16.67 ft Roof Fr: 2.539 k 3rd Flr 2.18 k Total F3: 4.719 k 2nd Flr 2.18 k Total F2: 6.899 k Foundation MOT: 207.4 k-ft Wall Wt: 4.233 k 2nd Flr Wt: 1.725 k 3rd Flr Wt: 1.725 k Roof Weight: 1.615 k MR: 77.5 k-ft 10.16 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 5.75 ft 3rd Flr Trib: 5.75 ft Roof Trib: 6.46 ft MOT: 83.54 k-ft Wall Wt: 2.774 k Roof Weight: 1.615 k 3rd Flr Wt: 1.725 k MR: 36.58 k-ft Uplift: 3.89 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 6.46 ft 3rd Flr Trib: 5.75 ft 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: Shear Wall 4 kddress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 ,ient: Hayes Wilson Lund Architects, LLC Job No.: 18456 (Worst Case) 7.89 8.75 8.75 II ft ft Ah T1 T1 T2 T2 t 19.33 ft Roof Fr: 2.944 k 3rd Flr 2.523 k Total F3: 5.467 k 2nd Flr 2.523 k Total F2: 7.991 k Foundation MOT: 240.3 k-ft Wall Wt: 4.908 k 2nd Flr Wt: 2.001 k 3rd Flr Wt: 2.001 k Roof Weight: 3.311 k MR: 118.1 k-ft 9.23 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 5.75 ft 3rd Flr Trib: 5.75 ft Roof Trib: 11.42 ft MOT: 96.83 k-ft Wall Wt: 3.217 k Roof Weight: 3.311 k 3rd Flr Wt: 2.001 k MR: 63.09 k-ft Uplift: 3.21 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 11.42 ft 3rd Flr Trib: 5.75 ft 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: Shear Wall 5 kddress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 ,ient: Hayes Wilson Lund Architects, LLC Job No.: 18456 7.89 8.75 8.75 ft ft ft T1 T1 T2 T2 t 12.75 ft Roof Fr: 0.7 k 3rd Flr 0.66 k Total F3: 1.36 k 2nd Flr 0.66 k Total F2: 2.02 k Foundation MOT: 59.25 k-ft Wall Wt: 3.237 k 2nd Flr Wt: 0.23 k 3rd Flr Wt: 0.23 k Roof Weight: 0.191 k MR: 24.78 k-ft 3.66 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 1 ft 3rd Flr Trib: 1 ft Roof Trib: 1 ft MOT: 23.55 k-ft Wall Wt: 2.122 k Roof Weight: 0.191 k 3rd Flr Wt: 0.23 k MR: 14.74 k-ft Uplift: 1.21 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 1 ft 3rd Flr Trib: 1 ft 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: Shear Wall 6 kddress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 ,ient: Hayes Wilson Lund Architects, LLC Job No.: 18456 7.89 8.75 8.75 ft ft ft Ah T1 T1 T2 T2 t 22 ft Roof Fr: 1.4 k 3rd Flr 1.3 k Total F3: 2.7 k 2nd Flr 1.3 k Total F2: 4 k Foundation MOT: 117.8 k-ft Wall Wt: 5.586 k 2nd Flr Wt: 1.584 k 3rd Flr Wt: 1.584 k Roof Weight: 2.257 k MR: 121.1 k-ft 2.16 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 4 ft 3rd Flr Trib: 4 ft Roof Trib: 6.84 ft MOT: 46.92 k-ft Wall Wt: 3.661 k Roof Weight: 2.257 k 3rd Flr Wt: 1.584 k MR: 65.1 k-ft Uplift: 0.38 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 6.84 ft 3rd Flr Trib: 4 ft 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: Shear Wall 7 iddress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 'lient: Hayes Wilson Lund Architects, LLC Job No.: 18456 7.89 8.75 8.75 ft ft ft Ah T1 T1 T2 T2 t 21.83 ft Roof Fr: 1.4 k 3rd Flr 1.3 k Total F3: 2.7 k 2nd Flr 1.3 k Total F2: 4 k Foundation MOT: 117.8 k-ft Wall Wt: 5.543 k 2nd Flr Wt: 0.393 k 3rd Flr Wt: 0.393 k Roof Weight: 0.327 k MR: 72.65 k-ft 3.58 k Floor Wall - Local Wall Trib: 25.39 ft 2nd Flr Trib: 1 ft 3rd Flr Trib: 1 ft Roof Trib: 1 ft MOT: 46.92 k-ft Wall Wt: 3.633 k Roof Weight: 0.491 k 3rd Flr Wt: 0.589 k MR: 45.01 k-ft Uplift: 0.96 k @ T2 Wall Trib: 16.64 ft Roof Trib.: 1.5 ft 3rd Flr Trib: 1.5 ft 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: Shear Wall 8 iddress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 'lient: Hayes Wilson Lund Architects, LLC Job No.: 18456 North Shearwall for Terrace Floor 11.67 ft T1 Vertical Loads: 2nd Floor DL: 18 psf Wall DL: 10 Dsf T1 14.92 ft Wall Global Overturning: MOT: 14 k-ft Wall Wt: 1.816 k 2nd Flr Wt: 0.269 k MR: 15.55 k-ft Uplift: 0.33 k Terrace F4===* Fr: 1.2 k Wall Trib: 12.17 ft 2nd Flr Trib: 1 ft Plan South Shearwall for Terrace Floor Terrace F Fr: 1.2 k 11.67 ft T1 T1 5.75 ft Vertical Loads: Wall Global Overturning: 2nd Floor DL: 18 psf MOT: 14 k-ft Wall DL: 10 psf Wall Wt: 0.901 k Wall Trib: 15.67 ft 2nd Flr Wt: 0.104 k 2nd Flr Trib: 1 ft MR: 2.888 k-ft Uplift: 2.25 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: Shear Wall 9 ress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 nt: Hayes Wilson Lund Architects, LLC Job No.: 18456 East Shearwall for Terrace Floor 11.67 ft T1 Vertical Loads: 2nd Floor DL: 18 psf Wall DL: 10 Dsf T1 20.75 ft Wall Global Overturning: MOT: 14 k-ft Wall Wt: 3.252 k 2nd Flr Wt: 3.33 k MR: 68.28 k-ft Uplift: 0.00 k Terrace F4===* Fr: 1.2 k Wall Trib: 15.67 ft 2nd Flr Trib: 8.915 ft Plan West Shearwall for Terrace Floor Terrace F*===* Fr: 1.2 k 11.67 ft T1 T1 9.5 ft Vertical Loads: Wall Global Overturning: 2nd Floor DL: 18 psf MOT: 14 k-ft Wall DL: 10 psf Wall Wt: 1.156 k Wall Trib: 12.17 ft 2nd Flr Wt: 1.524 k 2nd Flr Trib: 8.915 ft MR: 12.73 k-ft Uplift: 0.71 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: Shear Wall 10 ress: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 nt: Hayes Wilson Lund Architects, LLC Job No.: 18456 (Worst Case) Uplift: Loc. On Ftg Simpson Holdown to Use: Holdown Capacity: Check Shearwall Delta 1 2942.6 lb Midwall Simpson STHD14 3695 lb OK Shearwall Delta 2 7245.7 lb Corner Simpson HDU11 9535 lb OK Shearwall Delta 3 4291.4 lb Midwall Simpson HDU4 5465 lb OK Shearwall Delta 4 10160.2 lb Midwall Simspon HDU14 10770 lb OK Shearwall Delta 5/6 3663.9 lb Midwall Simpson STHD14 3695 lb OK Shearwall Delta 7 2159.0 lb Corner Simpson STHD14 3695 lb OK Shearwall Delta 8 3578.2 lb Midwall Simpson STHD14 3695 lb OK Terrace: Plan North 329.8 lb Corner Simpson STHD14 3695 lb OK Terrace: Plan South 2246.4 lb Corner Simpson STHD14 3695 lb OK Terrace: Plan East 0.0 lb Corner No Strap Needed N/A lb OK Terrace: Plan West 705.2 lb Corner Simpson STHD14 3695 lb OK *Note: All Strap Capacities based on 6" Thick Stemwall Tension Strap Design at 2nd Floor: (Worst Case) Uplift: Simpson Strap to Use: Strap Capacity: Check Shearwall Delta 1 493.9 lb Simpson MSTC28 1540 lb OK Shearwall Delta 2 2670.8 lb Simpson MSTC40 3080 lb OK Shearwall Delta 3 1032.7 lb Simpson MSTC28 1540 lb OK Shearwall Delta 4 3888.8 lb Simpson MSTC52 4620 lb OK Shearwall Delta 5/6 1213.7 lb Simpson MSTC28 1540 lb OK Shearwall Delta 7 376.2 lb Simpson MSTC28 1540 lb OK Shearwall Delta 8 960.3 lb Simpson MSTC28 1540 lb OK *Note: All Strap Capacities based on max 16" clear span Tension Strap Design at 3rd Floor (Worst Case) Uplift: Simpson Strap to Use: Strap Capacity: Check Shearwall Delta 1 0.0 lb No Strap Needed N/A lb OK Shearwall Delta 2 385.4 lb Simpson MSTC28 1540 lb OK Shearwall Delta 3 0.0 lb No Strap Needed N/A lb OK Shearwall Delta 4 339.4 lb Simpson MSTC28 1540 lb OK Shearwall Delta 5/6 77.9 lb Simpson MSTC28 1540 lb OK Shearwall Delta 7 0.0 lb No Strap Needed N/A lb OK Shearwall Delta 8 0.0 lb No Strap Needed N/A lb OK *Note: All Strap Capacities based on max 16" clear span 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Shear Wall 11 Address: 23820 Edmonds Way Edmonds, WA 98026 Date: 6/12/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 1 Loadine Summa OTM = 82.11 k-ft Pwall = 9.781 k Pleft = 0 k Pright = 0 k Footing Geometry a = 3.58 ft b = 14.33 ft c= 5 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 82.11 k-ft Resisting Moment = 139.1 k-ft Factor of Safety = 1.694 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 0.862 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall Width = 6 in Height = 12 in Length = 22.91 ft Weight = 1.718 k u Soil Above Footin fA OTM Pleft Pright Pwall Concrete Footing a b c Width = 18 in Height = 6 in Length = 22.91 ft Weight = 1.89 k Desim Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 5 As,min req'd = 0.95 inA2 # of Bars in Top Layer = 3 # of Bars in Bot. Layer = 2 Slab on Grade Weight Width = 18 in Thickness: 4 in Length = 22.91 ft Weight = 1.661 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: SW Footing 1 ress: 23820 Edmonds Way Date: 18456 nt: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 2 Loading Summary OTM = 174.9 k-ft Pwall = 6.463 k Pleft = 1.125 k Pright = 0.938 k Footin¢ Geometry a = 3.58 ft b = 19.83 ft c= 1 ft Depth = 1 ft Width = 3.5 ft Check Overturning and Bearing Pressure Overturning Moment = 174.9 k-ft Resisting Moment = 273.9 k-ft Factor of Safety = 1.566 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 0.975 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall Width = 6 in Height = 12 in Length = 24.41 ft Weight = 1.831 k Footing & Stem at Left End Stem Width = 6 in Stem Height = 12 in Length = 3 ft Weight = 0.225 k rN OTM Pleft Pright Pwall CConcrete Footing a b c Soil Above Footin Width = 36 in Height = 6 in Length = 24.41 ft Weight = 4.028 k Ftg Width = 24 in Thickness = 12 in Length = 3 ft Weight = 0.9 k Desien Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 5 As,min req'd = 1.663 inA2 # of Bars in Top Layer = 5 # of Bars in Bot. Layer = 3 Slab on Grade Weight Width = 18 in Thickness: 4 in Length = 24.41 ft Weight = 1.77 k Footing & Stem at Left End Stem Width = 6 in Ftg Width = 24 in Stem Height = 12 in Thickness = 12 in Length = 2.5 ft Length = 2.5 ft Weight = 0.188 k Weight = 0.75 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SW Footing 2 Address: 23820 Edmonds Way Date: 18456 Client: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 3 Loading Summary OTM = 100.5 k-ft Pwall = 9.781 k Pleft = 0.469 k Pright = 0.469 k Footin¢ Geometry a = 4.92 ft b = 14.33 ft c = 4.92 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 100.5 k-ft Resisting Moment = 170.8 k-ft Factor of Safety = 1.7 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 0.947 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall Width = 8 in Height = 12 in Length = 24.17 ft Weight = 2.417 k Footing & Stem at Left End Stem Width = 6 in Stem Height = 12 in Length = 1.25 ft Weight = 0.094 k Soil Above Footin rN OTM Pleft Pright Pwall Concrete Footing a b c Width = 16 in Height = 6 in Length = 24.17 ft Weight = 1.772 k Ftg Width = 24 in Thickness = 12 in Length = 1.25 ft Weight = 0.375 k Desien Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 0.95 inA2 # of Bars in Top Layer = 3 # of Bars in Bot. Layer = 2 Slab on Grade Weight Width = 18 in Thickness: 4 in Length = 24.17 ft Weight = 1.752 k Footing & Stem at Left End Stem Width = 6 in Ftg Width = 24 in Stem Height = 12 in Thickness = 12 in Length = 1.25 ft Length = 1.25 ft Weight = 0.094 k Weight = 0.375 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SW Footing 3 Address: 23820 Edmonds Way Date: 18456 Client: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 4: Part 1 Loading Summary OTM = 207.4 k-ft Pwall = 9.299 k Pleft = 0 k Pright = 0 k Footin¢ Geometry a = 4.92 ft b = 16.67 ft c = 3.25 ft Depth = 1 ft Width = 4.5 ft Check Overturning and Bearing Pressure Overturning Moment = 207.4 k-ft Resisting Moment = 321.6 k-ft Factor of Safety = 1.551 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 0.87 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall fA OTM Pleft Pright Pwall EConcrete Footing 104 a b c Soil Above Footin Desim Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 2.138 inA2 # of Bars in Top Layer = 6 # of Bars in Bot. Layer = 4 Slab on Grade Weight Width = Width = 48 in Width = 18 in Height = 6 in Thickness: 4 in Length = 24.84 ft Length = 24.84 ft Weight = 5.465 k Weight = 1.801 k 6 in Height = 12 in Length = 24.84 ft Weight = 1.863 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: SW Footing 4 ress: 23820 Edmonds Way Date: 18456 nt: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 4: Part 2 Loading Summary OTM = 240.3 k-ft Pwall = 12.22 k Pleft = 0 k Pright = 0 k Footin¢ Geometry a = 3.25 ft b = 19.33 ft c= 5 ft Depth = 1 ft Width = 4.5 ft Check Overturning and Bearing Pressure Overturning Moment = 240.3 k-ft Resisting Moment = 396.5 k-ft Factor of Safety = 1.65 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 0.784 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall fA OTM Pleft Pright Pwall EConcrete Footing 104 a b c Soil Above Footin Desim Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 2.138 inA2 # of Bars in Top Layer = 4 # of Bars in Bot. Layer = 4 Slab on Grade Weight Width = Width = 48 in Width = 18 in Height = 6 in Thickness: 4 in Length = 27.58 ft Length = 27.58 ft Weight = 6.068 k Weight = 2 k 6 in Height = 12 in Length = 27.58 ft Weight = 2.069 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: SW Footing 5 ress: 23820 Edmonds Way Date: 18456 nt: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 5/6: Loadine Summa OTM = 59.25 k-ft Pwall = 3.887 k Pleft = 0 k Pright = 0 k Footing Geometry a= 5 ft b = 12.75 ft c= 5 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 59.25 k-ft Resisting Moment = 137.2 k-ft Factor of Safety = 2.315 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 0.622 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall fA OTM Pleft Pright Pwall r- Concrete Footing a b c Soil Above Footin Desim Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 5 As,min req'd = 0.95 inA2 # of Bars in Top Layer = 2 # of Bars in Bot. Layer = 2 Slab on Grade Weight Width = Width = 18 in Width = 18 in Height = 6 in Thickness: 4 in Length = 22.75 ft Length = 22.75 ft Weight = 1.877 k Weight = 1.649 k 6 in Height = 12 in Length = 22.75 ft Weight = 1.706 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: SW Footing 6 ress: 23820 Edmonds Way Date: 18456 nt: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 7: Loading Summary OTM = 117.8 k-ft Pwall = 11.01 k Pleft = 0 k Pright = 0 k Footing Geometry a= 2 ft b = 22 ft c= 2 ft Depth = 1 ft Width = 2.75 ft Check Overturning and Bearing Pressure Overturning Moment = 117.8 k-ft Resisting Moment = 207 k-ft Factor of Safety = 1.757 > F Concrete Footing a b c Design Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 5 As,min req'd = 1.307 inA2 1.5 OK # of Bars in Top Layer = 3 # of Bars in Bot. Layer = 3 Pressure under left end = 0 ksf Pressure under right end = 0.689 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall at Shear Wall Width = 16 in Height = 12 in Length = 26 ft Weight = 5.2 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SW Footing 7 Address: 23820 Edmonds Way Date: 18456 Client: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Delta 8 Loading Summary OTM = 117.8 k-ft Pwall = 6.656 k Pleft = 0.75 k Pright = 0.75 k Footine Geometry a= 2 ft b = 21.83 ft c= 2 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 117.8 k-ft Resisting Moment = 186.2 k-ft Factor of Safety = 1.58 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 1.013 ksf Allowable Bearing Pressure = 1.5 ksf OK Slab Weight w/ Shearwall Width = 24 in Thick: 4 in Length = 25.83 ft Weight = 2.583 k Footing & Stem at Left End Stem Width = 6 in Stem Height = 12 in Length = 2 ft Weight = 0.15 k fA OTM Pleft Pright Pwall Concrete Footing a b c Stemwall at Shear Wall Width = 8 in Height = 12 in Length = 25.83 ft Weight = 2.583 k Ftg Width = 24 in Thickness = 12 in Length = 2 ft Weight = 0.6 k Desien Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 0.95 inA2 # of Bars in Top Layer = 2 # of Bars in Bot. Layer = 2 Footine & Stem at Left End Stem Width = 6 in Stem Height = 12 in Length = 2 ft Weight = 0.15 k Ftg Width = 24 in Thickness = 12 in Length = 2 ft Weight = 0.6 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SW Footing 8 Address: 23820 Edmonds Way Date: 18456 Client: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Terrace (North): Loading Summary OTM = 14 k-ft Pwall = 2.084 k Pleft = 0 k Pright = 0 k Footing Geometry a = 0.5 ft b = 14.92 ft c= 2 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 14 k-ft Resisting Moment = 56.9 k-ft Factor of Safety = 4.063 > rN OTM Pleft Pright Pwall IConcrete Footing 104 a b c Design Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 0.95 inA2 1.5 OK # of Bars in Top Layer = 2 # of Bars in Bot. Layer = 2 Pressure under left end = 0.049 ksf Pressure under right end = 0.326 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall on Shearwall Width = 6 in Height = 12 in Length = 17.42 ft Weight = 1.307 k 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: SW Footing ! address: 23820 Edmonds Way Date: 18456 'lient: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Terrace (South) Loadine Summa OTM = 14 k-ft Pwall = 1.005 k Pleft = 0.75 k Pright = 0.75 k Footing Geometry a= 2 ft b = 5.75 ft c= 1 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 14 k-ft Resisting Moment = 21.13 k-ft Factor of Safety = 1.508 > 1.5 OK Pressure under left end = 0 ksf Pressure under right end = 1.17 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall on Shearwall Width = 8 in Height = 12 in Length = 8.75 ft Weight = 0.875 k Footing & Stem at Left End Stem Width = 6 in Stem Height = 12 in Length = 2 ft Weight = 0.15 k fA OTM Pleft Pright Pwall EConcrete Footing 104 a b c Ftg Width = 24 in Thickness = 12 in Length = 2 ft Weight = 0.6 k Desien Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 0.95 inA2 # of Bars in Top Layer = 2 # of Bars in Bot. Layer = 2 Footin¢ & Stem at Left End Stem Width = 6 in Stem Height = 12 in Length = 2 ft Weight = 0.15 k Ftg Width = 24 in Thickness = 12 in Length = 2 ft Weight = 0.6 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SW Footing 10 Address: 23820 Edmonds Way Date: 18456 Client: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Terrace (East): Loading Summary OTM = 14 k-ft Pwall = 6.581 k Pleft = 0 k Pright = 0 k Footing Geometry a= 1 ft b = 20.75 ft c= 1 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 14 k-ft Resisting Moment = 97.04 k-ft Factor of Safety = 6.93 > rN OTM Pleft Pright Pwall EConcrete Footing 104 a b c Design Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 0.95 inA2 1.5 OK # of Bars in Top Layer = 2 # of Bars in Bot. Layer = 2 Pressure under left end = 0.106 ksf Pressure under right end = 0.269 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall on Shearwall Width = 6 in Height = 12 in Length = 22.75 ft Weight = 1.706 k 3228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 'roject: Edmonds Apartment Building Pg. No: SW Footing 1 address: 23820 Edmonds Way Date: 18456 'lient: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Shear Wall Footing Analysis Shear Wall: Shearwall Terrace (West) Loading Summary OTM = 14 k-ft Pwall = 2.681 k Pleft = 0 k Pright = 0 k Footing Geometry a= 1 ft b = 9.5 ft c= 1 ft Depth = 1 ft Width = 2 ft Check Overturning and Bearing Pressure Overturning Moment = 14 k-ft Resisting Moment = 24.8 k-ft Factor of Safety = 1.771 > rN OTM Pleft Pright Pwall EConcrete Footing 104 a b c Design Cont. Footing Reinforcement Cont. Reinf. Top+Bot. Bar Size 4 As,min req'd = 0.95 inA2 1.5 OK # of Bars in Top Layer = 2 # of Bars in Bot. Layer = 2 Pressure under left end = 0 ksf Pressure under right end = 0.574 ksf Allowable Bearing Pressure = 1.5 ksf OK Stemwall on Shearwall Width = 6 in Height = 12 in Length = 11.5 ft Weight = 0.863 k 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 r ect:Edmonds Apartment Building Pg. No: SW Footing 12 ress: 23820 Edmonds Way Date: 18456 nt: Hayes Wilson Lund Architects, LLC Job No.: 7/10/2019 Wind Loads - Components and Cladding, Roofs ASCE 7-10 Chapter 30, Part 1 - Low Rise Buildings o esign eria v� Risk Category: II Basic Wind Speed: 110 mph Exposure Category: B Hurricane Prone: No Building Type: Enclosed Buildings Length: 89.00 ft Width: 67.00 ft Height: 31.50 ft Component Exp. Category: B Longer Dimension Shorter Dimension © ' _O. Iaar Gable and Hip Roofs, Zone Extents Wind Load Calculation - Flat, Gable or Hip Roofs Wind Directionality Factor, Kd: 0.85 Table 26.6-1 $ Gust Effect Factor, G: 0.85 Section 26.9 Roof Angle: 70< e < = 270 h Zg: 1200 0.18 Gcpi: a: 7.00 ft -0.18 Roof Angle Kz: 0.71 Kzt: 1.00 Zone 5 Extent: 6.70 ft qh: 18.71 psf Parapet Height Above Deck: 0.00 ft Roof External Pressure Coefficients Area (ft^2) Location 0 10 20 50 100 200 500 1 -0.9 -0.9 -0.9 -0.8 -0.8 -0.8 -0.8 2 -1.7 -1.7 -1.6 -1.4 -1.2 -1.2 -1.2 GCpf 3 ' -2.6 -2.6 -2.4 -2.2 -2.0 -2.0 -2.0 1, 2 & 3 1 0.5 0.5 0.4 0.4 0.3 0.3 0.3 2 & 3 NA NA NA NA NA NA NA 1) Low or No Parapet or Angled Roof, No Impact Overhang External Pressure Coefficients Area (ft^2) Location 0 10 20 50 100 200 500 1 NA NA NA NA NA NA NA GCpf 2 -2.2 -2.2 -2.2 -2.2 -2.2 -2.2 -2.2 3 -3.7 -3.7 -3.4 -2.9 -2.5 -2.5 -2.5 3228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 roject: Edmonds Apartment Building Pg. No: CCRoof 1 Lddress: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Iient: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wind Loads - Components and Cladding, Roofs ASCE 7-10 Chapter 30, Part 1 - Buildings Less than 60ft Roof Wind Loads LOW Per ASCE 7-10 LRFD Load Cases Area (ft^2) Location 0 10 20 50 100 200 500 1 -20.20 -20.20 -19.27 -18.33 -18.33 -18.33 -18.33 2 -35.17 -35.17 -33.30 -29.55 -25.81 -25.81 -25.81 P (psf) 3 1 -52.00 -52.00 -48.26 -44.52 -40.78 -40.78 -40.78 1, 2 & 3 12.72 12.72 11.04 10.47 8.98 8.98 8.98 2 & 3 NA NA NA NA NA NA NA 0.6W Per ASCF, 7-10 ASD i,nad Cases Area (ft^2) Location 0 10 20 50 100 200 500 1 -12.12 -12.12 -11.56 -11.00 -11.00 -11.00 -11.00 2 -21.10 -21.10 -19.98 -17.73 -15.49 -15.49 -15.49 P (psf) 3 1 -31.20 -31.20 -28.96 -26.71 -24.47 -24.47 -24.47 1, 2 & 3 7.63 7.63 6.62 6.28 5.39 5.39 5.39 2 & 3 NA NA NA NA NA NA NA Overhang Wind Loads 1.OW Per ASCF, 7-10 LRFD Land Cases Area (ft^2) Location 0 10 20 50 100 200 500 1 N/A N/A N/A N/A N/A N/A N/A P (psf) 2 -44.5 -44.5 -44.5 -44.5 -44.5 -44.5 -44.5 3 -72.6 -72.6 -67.0 -57.6 -50.1 -50.1 -50.1 *Assumes a GCPi of 0.18 0.6W Per ASCE 7-10 ASD Load Cases Area (ft^2) Location 0 10 20 50 100 200 500 1 N/A N/A N/A N/A N/A N/A N/A P (psf) 2 -26.7 -26.7 -26.7 -26.7 -26.7 -26.7 -26.7 3 -43.5 -43.5 -40.2 -34.6 -30.1 -30.1 -30.1 *Assumes a GCPi of 0.18 JA8UE 13228 NE 20th St, Suite 100, Bellevue, WA 98005 I (425) 614-0949 Project: Edmonds Apartment Building Pg. No: CCRoof 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/11/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Roof Framing Lavout Plans I I I overhaming per MFR I I a°P \ overtraming per MFR f furies per MFR / � \ Q 24^ O.C. HDR I I I I I / I I. overtmming par MFR RB' I I I I I I I I I I I I I I RR2 I I I I I I I I I I I I I I I I I I Raoi Trusses per MFR I ¢ roof trusses Per MFR a'" O.C. HDR HPRI I I I I I I I I I I I I I I _HDR HDR — — HDR 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Layout-1 Address: 23830 Edmonds Way, Edmonds, WA Date: 6-13-19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 3rd Floor Framing Layout Plans T--7 II 3 2 w 3B1 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Layout-2 Address: 23830 Edmonds Way, Edmonds, WA Date: 6-13-19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 2nd Floor Framing Layout Plans 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Layout-3 Address: 23830 Edmonds Way, Edmonds, WA Date: 6-13-19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member Load Takeoff L Trib DL LL SL WLH WL (-) (ft) (ft) (psf) (psf) (psf) (psf) (psf) RRl 12.17 1.333 15 20 25 12.7 72.6 RR2 10.25 1.33 15 20 25 12.7 72.6 RBI 9.167 6.083 15 20 25 12.7 72.6 RB2 16.08 6.083 15 20 25 12.7 72.6 RB3 19.08 4.875 15 20 25 12.7 72.6 RB4 13.83 10.17 15 20 25 12.7 72.6 RB5 8.75 10.08 15 20 25 12.7 72.6 RB6 5 17 15 20 25 12.7 72.6 MI 4.833 1.333 24 60 25 22.1 29.6 M2 6 1.33 24 60 25 22.1 29.6 M3 6.583 1.33 24 60 25 22.1 29.6 M4 5.167 1.333 24 100 25 22.1 29.6 M5 5.5 1.33 24 100 25 22.1 29.6 M6 4.5 1.33 24 100 25 22.1 29.6 M7 5.25 1.33 24 100 25 22.1 29.6 3B1 11.75 2.417 24 60 25 22.1 29.6 3B2 4.417 1.33 24 60 25 22.1 29.6 3B3 9.5 12 24 60 25 22.1 72.6 3B4 9.25 2.875 24 60 25 22.1 29.6 3B5 19 2.75 24 100 25 22.1 29.6 3B6 13.67 5.083 24 100 25 22.1 29.6 3B7 9 5.25 24 100 25 22.1 29.6 3B8 4.75 1.33 24 100 25 22.1 29.6 3B9 5.25 1.33 24 100 25 22.1 29.6 3B10 6.417 3.25 24 60 25 22.1 29.6 3B11 6.5 1.333 24 60 25 22.1 29.6 3B12 10.83 9.5 18 40 0 0 0 3B13 18.12 11.5 18 40 0 0 0 3B14 10.25 6.5 24 100 25 22.1 29.6 1 2J1 4.833 1.333 24 60 25 22.1 29.6 2 2J2 6 1.33 24 60 25 22.1 29.6 3 2J3 6.583 1.33 24 60 25 22.1 29.6 4 2J4 5 1.33 24 100 25 22.1 29.6 5 2J5 5.5 1.33 24 100 25 22.1 29.6 6 2J6 5 1.33 24 100 25 22.1 29.6 7 2J7 5.25 1.33 24 100 25 22.1 29.6 8 2J8 3.75 1.33 24 100 25 22.1 29.6 9 2J9 3.5 1.33 24 100 25 22.1 29.6 10 2R1 8 1.33 15 20 25 12.7 72.6 11 2R2 19 1.33 15 20 25 12.7 72.6 12 2B1 11.75 2.417 24 60 25 22.1 29.6 13 2B2 4.417 1.33 24 60 25 22.1 29.6 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Takeoff-1 kddress: 23830 Edmonds Way, Edmonds, WA Date: 6-13-19 Ment: Hayes Wilson Lund Architects, LLC Job No.: 18456 L Trib DL LL SL WL(+) WL (-) (ft) (ft) (psf) (psf) (psf) (psf) (psf) 14 2B3 9.25 3 24 60 25 22.1 29.6 15 2B4 6.5 1.333 24 60 25 22.1 29.6 16 2B5 6.417 3.25 24 60 25 22.1 29.6 17 2B6 19 2 24 60 25 12.7 72.6 18 2B7 9.75 1.75 24 100 25 22.1 29.6 19 2B8 9.75 5.75 24 100 25 22.1 29.6 20 2B9 10 9.25 24 100 25 22.1 29.6 21 2B10 10 5.25 24 100 25 22.1 29.6 22 2B11 13.67 5.083 24 100 25 22.1 29.6 23 2B 12 9 5.25 24 100 25 22.1 29.6 24 2B13 0 0 0 0 0 0 0 Loading Through RISA 25 2B 14 4.75 6 24 100 25 22.1 29.6 26 2B 15 8.75 5.25 24 100 25 22.1 29.6 27 2B 16 8.75 5.75 24 100 25 22.1 72.6 28 2B17 8.75 6.75 15 20 25 12.7 72.6 29 2B 18 10.83 9.5 18 40 0 0 0 30 2B19 18.12 11.5 18 40 0 0 0 31 2B20 9.25 4.5 24 125 0 22.1 29.6 32 2B21 9.25 4.5 24 125 0 22.1 29.6 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Takeoff-2 Address: 23830 Edmonds Way, Edmonds, WA Date: 6-13-19 Zlient: Hayes Wilson Lund Architects, LLC Job No.: 18456 APA Floor Joists Takeoff Unit Floor Joists: Distributed Load: Max Span: 13.50 ft Try: 9 1/2 - PRI-20 Dead Load: 18.0 psf Spacing: 16 in O.C. Weight = 2.20 plf Live Load: 40.0 psf Left Brg: 1.75 in Mallow = 30.24 k-in Durration Factor: 1 Right Brg: 1.75 in Vallow = 1.12 k EI = 132 inA2 - lb x 10^6 Point Load: Mmax = 21.14 k-in OK K = 4.94 lbs x 10^6 Dead Load: 0 lb Vmax = 0.52 k OK Live Load: 0 Ib Location: 0.00 ft ALL = 0.326 in -> L/ 498 OK ATL = 0.472 in -> L/ 343 OK Storage Floor Joists: Distributed Load: Max Span: 9.00 ft Try: 9 1/2 - PRI-20 Dead Load: 18.0 psf Spacing: 16 in O.C. Weight = 2.20 plf Live Load: 125.0 psf Left Brg: 1.75 in Mallow = 30.24 k-in Durration Factor: 1 Right Brg: 1.75 in Vallow = 1.12 k EI = 132 in^2 - lb x 10^6 Point Load: Mmax = 23.17 k-in OK K = 4.94 lbs x 10^6 Dead Load: 0 lb Vmax = 0.86 k OK Live Load: 0 lb Location: 0.00 ft ALL = 0.219 in -> L/ 493 OK ATL = 0.251 in -> L/ 431 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Joists-1 Address: 23830 Edmonds Way, Edmonds, WA Date: 6-13-19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member RR1 Induced Loading P= 0k MX = -10.2 k-in My = 0 k-in _ VX = -0.28 k - Vy = 0 k 10 Span = 12.17 ft Try: 2x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 3.372 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.60 (NDS Seer. 2.3.2) CM (Fb) = 1.00 (NDS Seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 13.88 in E = 1700000 psi CF = 1.20 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 81.4) IX = 98.9 in Ea in = 620000 psi CL = 0.27 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 2.60 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 21.39 in Fc.pe1F = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cb, = 1.00 (NDS Sect. 4.3.7) Sy = 3.47 in Ft = 675 psi Member Analysis Unbraced Length,l = 12 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.5 ksi f,, X_X = 0.0 ksi Fc* = 321.2 psi RB = 24.50 Fb, X-X = 0.59 ksi OK Fv, X-X = 0.29 ksi OK FcE,X = 2046 psi FbE = 3 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 2046 psi Fb1 = 0.59 ksi Fb, y-y = 0.59 ksi OK FV, y-y = 0.29 ksi OK Cp = 0.97 Fb2 = 0.59 ksi C = 0.8 Check 0.81 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.81 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F c = 0.31 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 240 TL: L/ 180 Comb. ALL= 0.098in-> L/ 1495 OK USE WO No. 1 Douglas Fir -Larch Comb. ADL = 0.059 in -> L / 2491 Comb. ATL w/ K,r= 0.215 in -> L / 679 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member RR2 Induced Loading P= 0k MX = -7.24 k-in My = 0 k-in VX = -0.24 k Vy = 0 k Span = 10.25 ft Try: 2x8 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.643 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.60 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 7.25 in Fb = 1000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 10.88 in E = 1700000 psi CF = 1.20 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) IX = 47.6 in Emir = 620000 psi CL = 0.40 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 2.04 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 13.14 in Fc.peCp = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cb, = 1.00 (NDS Sect. 4.3.7) Sy = 2.72 in Ft = 675 psi Member Analysis Unbraced Length,l = 10 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.6 ksi f,, X_X = 0.0 ksi Fc* = 479.3 psi RB = 19.91 Fb, X-X = 0.88 ksi OK Fv, X-X = 0.29 ksi OK FcE,X = 1771 psi FbE = 5 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 1771 psi Fb1 = 0.88 ksi Fb, y-y = 0.88 ksi OK FV, y-y = 0.29 ksi OK Cp = 0.94 Fb2 = 0.88 ksi c = 0.8 Check 0.63 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.63 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F c = 0.45 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 240 TL: L/ 180 Comb.ALL= 0.102 in-> L/ 1206 OK USE 2x8 No. 1 Douglas Fir -Larch Comb. ADL = 0.061 in -> L/ 2010 Comb. ATL w/ K,r= 0.224 in -> L/ 548 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member RB1 Induced Loading P= Ok M= 37.38 k-in My = 0 k-in Vx = 1.359 k - Vy = 0 k Span = 9.167 ft Try: 4x8 No.1 Douglas Fir-Larch-Re/erence20l5NDS Shape Properties Self Weight = 6.168 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.15 (AW Sect. 2.3.2) CM (Fb) = 1.00 (NDs Sect. 8.1.4) d = 7.25 in Fb = 1000 psi Ct = 1.00 (NDSSeet. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) A = 25.38 in' E _ ##### psi CF = 1.30 (NDSSeet.. 4.3.6) CM M = 1.00 (NDSSect. 8.1.4) Ix = 111.1 in4 E.in _ ##### psi CL = 1.00 (NDSSeet. 3.3.3) Ci (Fb) = 1.00 (NDSSect. 8.1.4) 15, = 25.90 in4 F = 180 psi Cb = 1.19 (NDSSeet. 3.10.4) C;(E) = 1.00 (NDSSect. 8.1.4) Sx = 30.66 in FcPem = 625 psi C, = 1.15 (NDSSeet. 4.3.9) Ctu = 1.00 (NDSSect. 4.3.7) Sy = 14.80 in3 Ft = 675 psi Member Analysis Unbraced Length,1 = 1 ft Bearing Length, lb, = 2 in Bearing Length, lb, = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.2 ksi f, x_x = 0.1 ksi Fc* = 932.2 psi RB = 3.08 Fb, x-x = 1.72 ksi OK F,,, x_x = 0.21 ksi OK FcE,x = 104693 psi FbE = 215 ksi fb, y_y = 0.0 ksi f,,, y_y = 0.0 ksi FcE,y = 104693 psi Fbl = 1.72 ksi Fb, y_y = 1.72 ksi OK F, y_y = 0.21 ksi OK Cp = 1.00 Fb2 = 1.72 ksi c = 0.8 Check 0.71 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.71 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.93 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y= 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 240 TL: L/ 180 Comb. ALL= 0.128in-> L/ 860 0- USE 4x8 No. 1 Douglas Fir -Larch Comb. ODL= 0.077 in- L / 1434 Comb. OTL w/ K« = 0.281 in -> L / 391 OK r r AVE 13228 NE 20th St, Suite 100, Bellevue, WA 98005 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 3 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 P= Ok Mx = 98.7 k-in My = 0 k-in Vx = 2.69 k Vy = 0 k Span = 16.08 ft Try: I b = 3 1/8 in d = 12 in I Stress Class = 24F-1.8E Un-Balanced •nefere> 2015 NDS Lumber Species = Western Species Shape Properties I Self Weight = 9.115 plf b = 3.125 in Kcr = 1.50 Ex = ##### psi Adjustment Factors d = 12.00 In Fbx(+)= 2400 psi Ey = ##### psi CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A= 37.50 in' Fbx(_) 1450 psi Ex,,�i = ##### psi Ct= 1.00 (NDSSect. 2.3.3) CM(E)= 1.00 (NDS Sect. 8.1.4) Ix = 450 in Eby- 1450 psi Ey, "jn = ##### psi Cy = 1.00 (NDS Sect. 5.3.6) CM M = 1.00 (NDS Sect. 8.1.4) Iy = 31 in Fix = 265 psi Ft = 1100 psi CL = 1.00 (NDS Sect. 3.3.3) CM (F'c, pe,.) = 1.00 (NDS Sect. 8.1.4) Sx = 75 in F,y = 230 psi Cb = 1.19 (NDS Sect. 3.10.4) Cf, = 1.16 (NDS Sect. 5.3.7) Sy = 19.5 in Fc.perp = 650 psi Ce = 1.00 (NDS Sect. 5.3.8 C1= I.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,1 = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis IShear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.3 ksi f, x_x = 0.1 ksi Fc* = 647 psi RB = 4.43 Fb, x-x = 2.39 ksi OK F, x_x = 0.27 ksi OK FcE,x = 439476 psi FbE = 58 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 393215 psi Fbl = 2.39 ksi Fb, y_y = 1.68 ksi OK F, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.55 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.55 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.4 ksi F'c, perp., x-x = 0.77 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL = 0.195 in -> L / OK USE 3 2 5 �J 1 X 12 " G D GB Comb. ADL= 0.195 in-> Comb. 990 L/ 990 .1 ATL w/ K, = 0.811 in -> L / 238 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 4 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member RB3 P= 0k Mx = 129.8 k-in My = 0 k-in Vx = 2.268 k Vy = 0 k Span = 19.08 ft Try: I b = 5 1/8 in d = 10.5 in I Stress Class = 24F-1.8E Un-Balanced •aefere> 2015 NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = ##### psi Adjustment Factors d = 10.50 In Fbx(+)= 2400 psi Ey = ##### psi CD = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A= 53.81 in' Fbx(_) 1450 psi Ex,,�i = ##### psi Ct= 1.00 (NDSSect. 2.3.3) CM(E)= 1.00 (NDS Sect. 8.1.4) Ix = 494 in Eby- 1450 psi Ey, "jn = ##### psi Cy = 1.00 (NDSSect. 5.3.6) CM M = 1.00 (NDSSect. 8.1.4) Iy = 118 in Fix = 265 psi Ft = 1100 psi CL = 1.00 (NDS Sect. 3.3.3) CM (F'c, pe1p.) = 1.00 (NDS Sect. 8.1.4) Sx = 94 in F,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.petp = 650 psi Cc = 1.00 (NDS Sect. 5.3.8 C1= 1.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,1 = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis IShear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.4 ksi f, x_x = 0.0 ksi Fc* = 746.3 psi RB = 2.53 Fb, x-x = 2.76 ksi OK F, x_x = 0.30 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi FbI = 2.76 ksi Fb, y_y = 1.83 ksi OK F, y_y = 0.26 ksi OK Cp = 1.00 Fb2 = 1.83 ksi c = 0.9 Check 0.50 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.50 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.77 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.in-> L/ OK USE 5.125" x 10.5" GLB Comb. ADL = 0.245 245 in -> L / 934 934 Comb. ATL w/ Kc, = 0.776 in -> L / 295 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 5 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 P= 0k Mx = 142.3 k-in My = 0 k-in Vx = 3.428 k Vy = 0 k Span = 13.83 ft Try: I b = 5 1/8 in d = 10.5 in I Stress Class = 24F-1.8E Un-Balanced •nefere> 2015 NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = ##### psi Adjustment Factors d = 10.50 In Fbx(+)= 2400 psi Ey = ##### psi CD = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A= 53.81 in' Fbx(_) 1450 psi Ex,,�i = ##### psi Ct= 1.00 (NDSSect. 2.3.3) CM(E)= 1.00 (NDS Sect. 8.1.4) Ix = 494 in Eby- 1450 psi Ey, "jn = ##### psi Cy = 1.00 (NDSSect. 5.3.6) CM M = 1.00 (NDSSect. 8.1.4) Iy = 118 in Fix = 265 psi Ft = 1100 psi CL = 1.00 (NDS Sect. 3.3.3) CM (F'c, pe1p.) = 1.00 (NDS Sect. 8.1.4) Sx = 94 in F,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.petp = 650 psi Cc = 1.00 (NDS Sect. 5.3.8 C1= 1.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,1 = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis IShear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.5 ksi f, x_x = 0.1 ksi Fc* = 746.3 psi RB = 2.53 Fb, x-x = 2.76 ksi OK F, x_x = 0.30 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi FbI = 2.76 ksi Fb, y_y = 1.83 ksi OK F, y_y = 0.26 ksi OK Cp = 1.00 Fb2 = 1.83 ksi c = 0.9 Check 0.55 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.55 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.3 ksi F'c, perp., x-x = 0.77 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.in-> L/ OK USE 5.125" x 10.5" GLB Comb. ADL = 0.141 141 in -> L / 117176 Comb. ATL w/ Kc, = 0.447 in -> L / 371 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 6 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member RB5 Induced Loading P= Ok Mx = 56.45 k-in My = 0 k-in Vx = 2.151 k - Vy = 0 k Span = 8.75 ft Try: 4x10 No.1 Douglas Fir-Larch-Re/erence20l5NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.15 (AW Sect. 2.3.2) CM (Fb) = 1.00 (NDs Sect. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDSSeet. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) A = 32.38 in' E _ ##### psi CF = 1.20 (NDSSeet.. 4.3.6) CM M = 1.00 (NDSSect. 8.1.4) Ix = 230.8 in4 E.in _ ##### psi CL = 1.00 (NDSSeet. 3.3.3) Ci (Fb) = 1.00 (NDSSect. 8.1.4) 15, = 33.05 in4 F = 180 psi Cb = 1.19 (NDSSeet. 3.10.4) C;(E) = 1.00 (NDSSect. 8.1.4) Sx = 49.91 in Fc Petp = 625 psi C, = 1.15 (NDSSeet. 4.3.9) Ctu = 1.00 (NDSSect. 4.3.7) Sy = 18.89 in3 Ft = 675 psi Member Analysis Unbraced Length,1 = 1 ft Bearing Length, lb, = 2 in Bearing Length, lb, = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.1 ksi f, x_x = 0.1 ksi Fc* = 860.1 psi RB = 3.48 Fb, x-x = 1.58 ksi OK F, x_x = 0.21 ksi OK FcE x = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi f,,, y_y = 0.0 ksi FcE,y = 170421 psi Fbl = 1.58 ksi Fb, y_y = 1.58 ksi OK F, y_y = 0.21 ksi OK Cp = 1.00 Fb2 = 1.58 ksi c = 0.8 Check 0.71 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.71 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.86 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.3 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y= 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 240 TL: L/ 180 Comb. ALL= 0.085in-> L/ 1239 0- USE 4x10 No. 1 Douglas Fir -Larch Comb. ODL= 0.051 in- L / 2066 Comb. OTL w/ K« = 0.186 in -> L / 563 OK r r AVE 13228 NE 20th St, Suite 100, Bellevue, WA 98005 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 7 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 P= Ok M. = 37.38 k-in My = 0 k-in Vx = 1.359 k Vy = 0 k Lwww�wwwwwm Try: I b = 3 1/8 in d = 9 in I Stress Class = 24F-1.8E I Un-Balanced *Reference 2015 NDS Lumber Species = Western Species Shape Properties Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 Ex = ##### psi Adjustment Factors d = 9.00 in Fbx(+)= 2400 psi Ey = ##### psi Co = 1.15 (NDS Sect. 2.3.2) CM (F,) = 1.00 (NDS Sect. 8.1.4) A = 28.13 in' Fbx(_)= 1450 psi Ex. min = ##### psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 190 in4 Fby 1450 psi Ey, min = ##### psi CV = 1.00 (NDSSect.. 5.3.6) CM M = 1.00 (NDSSect. 8.1.4) ly = 23 in4 F_ = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, pelp.) = 1.00 (NDSSect. 8.1.4) Sx = 42 in Fvy = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cffi = 1.16 (NDSSect. 5.3.7) Sy = 14.6 in3 Fc.peip = 650 psi Cc = 1.00 (NDS Sect. 5.3.8 CI = 1.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,1 = 1.3 ft Bearing Length, lb,. = 2 in Bearing Length, lb, = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb x_x = 0.9 ksi fv, x_x = 0.0 ksi Fc* = 744.6 psi RB = 3.84 Fb, x-x = 2.75 ksi OK F x_x = 0.30 ksi OK FcE x = 377 psi FbE = 77 ksi fb, y_y = 0.0 ksi f,,, y_y = 0.0 ksi FcE,y = 377 psi FbI = 2.75 ksi Fb, y_y = 1.93 ksi OK Fv, y_y = 0.26 ksi OK Cp = 0.47 Fb2 = 1.93 ksi c = 0.9 Check 0.32 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.32 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.35 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.77 ksi OK fc, perp. y-y= 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.0in- L / 15OK USE 3.125" x 9" GLB Comb. ADL= 0.042 42 in -> L / 2593 93 Comb. ATL w/ Kc,= 0.134 in -> L / 819 OK r JA U EM 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: A 8 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3J1 Induced Loading P= 0k Mx = 3.9237 k-in _ My = 0 k-in Vx = 0.2706 k - - Vy = 0 k Span = 4.833 ft Try: 2x6 No. 2 Douglas Fir -Larch "Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors "Reference NDS Manual # of Members = 1 b = 1.5 In Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in' E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM M = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in Emi = 580000 psi CL = 0.99 (NDS Sect. 3.3.3) Ct (Fb) = 1.00 (NDS Sect. 8.1.4) ly = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 7.56 in Fe perp = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cf, = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.5 ksi f,,, x_x = 0.0 ksi Fe* = 805.2 psi RE = 6.25 Fb, x_x = 1.33 ksi OK 17, x_x = 0.18 ksi OK FeE,x = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi f,,, y_y = 0.0 ksi FeE,y = 56364 psi Fb1 = 1.33 ksi Fb, y_y = 1.33 ksi OK 17, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c= 0.8 Check 0.39 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.39 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.80 ksi fc / F'c = 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK fc, perp. y-y= 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.03 in -> L / 1965 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL= 0.012 in -> L/ 4913 Comb. ATL w/ Kef = 0.053 in -> L / 1092 OK AAU 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member M2 Induced Loading P= 0k Mx = 6.033 k-in MY = 0 k-in _ Vx = 0.335 k * - Vy = 0 k 10 Span = 6 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NOS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 In Kcr = 2.00 Cn = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (ADS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in' E = ##### psi CF = 1.30 (NDSSect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in En in = ##### psi CL = 0.99 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) IY = 1.55 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci(E) = 1.00 (NDS Sect. 8.1.4) Sx = 7.56 in Fc.per, = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) CM = 1.00 (NDSSect. 4.3.7) SY = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb,x = 2 in Bearing Length, lb Y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.8 ksi f, x_x = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, x-x = 1.33 ksi OK F, x_x = 0.18 ksi OK FcE,x = 56364 psi FbE = 49 ksi fb, Y_, = 0.0 ksi f, Y_, = 0.0 ksi FcE,y = 56364 psi Fbl = 1.33 ksi Fb, y_y = 1.33 ksi OK F, Y_Y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.60 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.60 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc / F'c = 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK fc, perp. y-y= 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.07in-> L/ 1030 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL= 0.028 in -> L/ 2574 Comb. OTL w/ Kc, = 0.126 in -> L / 572 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member M3 Induced Loading P= Ok Mx = 7.26 k-in - My = 0 k-in Vx = 0.37 k - Vy = 0 k Span = 6.58 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2ols NOs Shape Properties Self Weight = 2.01 plf Adjustment Factors *Reference NDSMan 1 # of Members = 1 I b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) A = 8.25 in' E = #### psi CF = 1.30 (NDssect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in E ni = #### psi CL = 0.99 (NDS Sect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 7.56 in Fc.perp = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cr = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb,. = 2 in Bearing Length, lb, = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.0 ksi f, x_x = 0.0 ksi Fc* = 805 psi RB = 6.25 Fb, x_x = 1.33 ksi OK F, x_x = 0.18 ksi Ob FcE,x = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi f, y_y = 0.0 ksi FcE,y = 56364 psi Fb1 = 1.33 ksi Fb, y_y = 1.33 ksi OK F, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.72 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.72 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.80 ksi fc / F'c = 0.00 OK f c, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.1 in -> L / 779 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL= 0.04 in -> L/ 1949 Comb. ATL w/ Kc, = 0.18 in -> L/ 433 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 3 ddress: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Haves Wilson Lund Architects. LLC Job No.: 18456 Member 3J4 Induced Loading P= Ok Mx = 6.619 k-in My = 0 k-in Vx = 0.427 k - Vy = 0 k 10 Span = 5.167 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference2015NDS Shape Properties Self Weight = 2.005 plf Adjustment *Reference NDS Manual Factors # of Members = 1 b = 1.5 In Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) A = 8.25 in` E _ ##### psi CF = 1.30 (NDSSect. 4.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Ix = 20.8 in Em;n = ##### psi CL = 0.99 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) C; (E) = 1.00 (NDS Sect. 8.1.4) Sx = 7.56 in Fe.pep = 625 psi Ct = 1.15 (NDS Sect. 4.3.9) Cfi, = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb,. = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.9 ksi f,, x_x = 0.1 ksi Fe* = 805.2 psi RB = 6.25 Fb, x-x = 1.33 ksi OK F,, x_x = 0.18 ksi OK F�E,x = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi f, y_y = 0.0 ksi F�E,y = 56364 psi Fbt = 1.33 ksi Fb, y_y = 1.33 ksi OK F, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.66 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.66 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.80 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y= 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.064 in -> L / 965 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL= 0.015 in -> L/ 4022 Comb. ATL w/ K,, = 0.095 in -> L/ 652 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 4 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member M5 Induced Loading P= 0k Mx = 7.483 k-in My = 0 k-in Vx = 0.454 k - - Vy = 0 k _ Span = 5.5 ft Try: 2x6 No. 2 Douglas Fir -Larch "Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors "Reference NDS Manual # of Members = 1 b = 1.5 In Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 In Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in' E = ##### psi CF = 1.30 (NDS Sect. 4.3.6) CM M = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in En; = ##### psi CL = 0.99 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 8.1.4) h, = 1.55 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDSSect. 8.1.4) Sx = 7.56 in Fe.Perp = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cf, = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb,. = 2 in Bearing Length, lb,,, = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x-x = 1.0 ksi f, x-x = 0.1 ksi Fe* = 805.2 psi RE = 6.25 Fb, x_x = 1.33 ksi OK F, x_x = 0.18 ksi OK FeE,x = 56364 psi FbE = 49 ksi fb,y-y = 0.0 ksi f,y-,, = 0.0 ksi FcE,y = 56364 psi Fb1 = 1.33 ksi Fb, y_y = 1.33 ksi OK F, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.74 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.74 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.80 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.082 in -> L / 802 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ODL = 0.02 in -> L / 3342 Comb. OTL w/ K,, = 0.122 in -> L / 542 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 5 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member M6 Induced Loading P= 0k Mx = 5.009 k-in MY = 0 k-in Vx = 0.371 k * - Vy = 0 k 10 Span = 4.5 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NOS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 In Kcr = 2.00 Cn = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (ADS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in' E = ##### psi CF = 1.30 (NDSSect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in En in = ##### psi CL = 0.99 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) IY = 1.55 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci(E) = 1.00 (NDS Sect. 8.1.4) Sx = 7.56 in Fc.per, = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) CM = 1.00 (NDSSect. 4.3.7) SY = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb,x = 2 in Bearing Length, lb Y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.7 ksi f, x_x = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, x-x = 1.33 ksi OK F, x_x = 0.18 ksi OK FcE,x = 56364 psi FbE = 49 ksi fb, Y_, = 0.0 ksi f, Y_, = 0.0 ksi FcE,y = 56364 psi Fbl = 1.33 ksi Fb, y_y = 1.33 ksi OK F, Y_Y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.50 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.50 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc / F'c = 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK fc, perp. y-y= 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.037in-> L/ 1464 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL= 0.009 in -> L/ 6101 Comb. OTL w/ Kc, = 0.055 in -> L / 989 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 6 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member M7 Induced Loading P= Ok Mx = 6.818 k-in My = 0 k-in Vx = 0.433 k - Vy = 0 k Span = 5.25 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NOs Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDSMan 1 # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NOS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) A = 8.25 in' E _ ##### psi CF = 1.30 (NDSSect. 4.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Ix = 20.8 in E ni = ##### psi CL = 0.99 (NDS Sect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) ly = 1.55 in F, = 180 psi Cb = 1.19 (NDSSect. 3.10.4) Ci (E) = 1.00 (NOssect. 8.1.4) Sx = 7.56 in Fc.peTp = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cf, = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length, 1 = 1 ft Bearing Length, lb,. = 2 in Bearing Length, lb,, = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.9 ksi f,,, x_x = 0.1 ksi Fc* = 805.2 psi RB = 6.25 Fb, x-x = 1.33 ksi OK F, x_x = 0.18 ksi OK FcE,x = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi f,,, y_y = 0.0 ksi FcE,y = 56364 psi Fbl = 1.33 ksi Fb, y_y = 1.33 ksi OK F, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.68 OK (NDSEg. 3.9-3) Bi-axial Analysis = 0.68 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.80 ksi fc/Fc= 0.000K f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.068in-> L/ 922 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL= 0.016 in -> L/ 3842 Comb. ATL w/ Kc, = 0.101 in -> L/ 623 OK AVE 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 7 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member3B2 Induced Loading P= 0k Mx = 3.269 k-in MY = 0 k-in _ Vx = 0.247 k A - VY = 0 k - Span = 4.417 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference zols NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 81.4) A = 32.38 in' E _ ##### psi CF = 1.20 (NDS Sect. 4.3.e) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 230.8 in E,nb, _ ##### psi CL = 1.00 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 8.1.4) ly = 33.05 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 49.91 in Fe.per = 625 psi C, = 1.15 (NDSSect. 4.3.9) Cf� = 1.00 (NDS Sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length,1 = 1 ft Bearing Length, lb,. = 2 in Bearing Length, lb y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.1 ksi f, x_x = 0.0 ksi F�* = 748.2 psi RE = 3.48 Fb, x-x = 1.38 ksi OK F, x_x = 0.18 ksi OK FcE,x = 170421 psi FbE = 169 ksi fb, Y_Y = 0.0 ksi f, Y_, = 0.0 ksi FcE.Y = 170421 psi Fb1 = 1.38 ksi Fb, y_y = 1.38 ksi OK F, Y_Y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.05 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.05 OK Bi-axial Analysis = 0.0 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.002 in _> L / 30442 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 7E-04 in -> L/ 76105 Comb. ATL w/ K,, = 0.003 in -> L / 16912 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 9 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B3 P= 0k Mx = 136.5 k-in my = 0 k-in Vx = 4.788 k Vy = 0 k Span = 9.5 ft Try: I b = 3 1/8 in d = 10.5 in Stress Class = 24F-1.8E Un-Balanced *Refer wn 2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 7.975 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 5.1.4) A = 32.81 in Fbx(_) 1450 psi Ex, ,,Ii� = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 301 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 27 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 57 in F", = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf� = 1.16 (NDSSect. 5.3.7) Sy = 17.1 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 2.4 ksi f,, x_x = 0.1 ksi Fc* = 647.4 psi RE = 4.15 Fb, x-x = 2.39 ksi OK F,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 66 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb1 = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.99 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.99 OK Bi-axial Analysis = 0.6 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.8 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL = 0.243 in -> L / 469 OK T T � � 3.125 �J I � � 1 t � 1 � �TLD X � O D Comb. ADL= 0.097 in -> L / 1172 • • Comb. ATL w/ Ket= 0.389 in -> L / 293 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 21 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B4 Induced Loading P= 0k JL Mx = 31 k-in My = 0 k-in - - Vx = 1.117 k - - Vy = 0 k Span = 9.25 ft Try: 4x10 No. 1 Douglas Fir -Larch 'Reference zols NDs Shape Properties Self Weight = 7.869 plf Adjustment Factors 'Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 NDS Seat. 2.3.3) CM (E) = 1.00 (NDS Seat. 8.1.4) A = 32.38 in` E _ ##### psi CF = 1.20 (NDsseet. 4.3.6) CM (V) = 1.00 (NDS Sect. 81.4) Ix = 230.8 in E,n;,, _ ##### psi CL = 1.00 NDS Sect. 3.3.3) Ci (Fb) = 1.00 NDS Sect. 81.4) Iy = 33.05 in F, = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDSSeet. 81.4) Sx = 49.91 in F,.pem = 625 psi Ct = 1.15 NOS Sect. 4.3.9) Cs, = 1.00 NOS Sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.6 ksi fv, x_x = 0.0 ksi Fe* = 748.2 psi RB = 3.48 Fb, x-x = 1.38 ksi OK Fv, x-x = 0.18 ksi OK FcE,x = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FeE,y = 170421 psi Fbi = 1.38 ksi Fb, y_y = 1.38 ksi OK F, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.45 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.45 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.000K fc, perp. x-x = 0.2 ksi F c, perp., x-x = 0.74 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.072in-> L/ 1533 USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.029 in -> L / 3833 Comb. ATL w/ Kc, = 0.13 in -> L / 852 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 11 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 P= Ok Mx = 184.7 k-in My = 0 k-in Vx = 3.24 k Vy = 0 k Try: I b = 5 1/8 in d = 10.5 in I Stress Class = 24F-1.8E I Un-Balanced *Reference 2015 NDS Lumber Species = Western Species Shape Properties Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = ##### psi Adjustment Factors d = 10.50 in Fbx(+)= 2400 psi Ey = ##### psi Co = 1.00 (LADS Sect. 2.3.2) CM (F,) = 1.00 (NDS Sect. 8.1.4) A = 53.81 in' Fbx(_)= 1450 psi Ex. min = ##### psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 494 in4 Fby 1450 psi Ey, min = ##### psi CV = 1.00 (NDSSect.. 5.3.6) CM M = 1.00 (NDSSect. 8.1.4) ly = 118 in4 F_ = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, pelp.) = 1.00 (NDSSect. 8.1.4) Sx = 94 in Fvy = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cffi = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in3 Fc.peip = 650 psi Cc = 1.00 (NDS Sect. 5.3.8 CI = 1.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,1 = 1.3 ft Bearing Length, lb,. = 2 in Bearing Length, lb, = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb x_x = 2.0 ksi fv, x_x = 0.1 ksi Fc* = 649.1 psi RB = 2.53 Fb, x-x = 2.40 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f,,, y_y = 0.0 ksi FcE,y = 301055 psi FbI = 2.40 ksi Fb, y_y = 1.59 ksi OK Fv, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.82 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.82 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.3 ksi F'c, perp., x-x = 0.77 ksi OK fc, perp. y-y= 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.in- L / OK USE 5.125" x 10.5" GLB Comb. ADL= 0.217 217 in -> L / 104048 Comb. ATL w/ Kc,= 1.232 in -> L / 185 OK 6 u 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 12 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B6 P= 0k Mx = 176.6 k-in my = 0 k-in Vx = 4.307 k Vy = 0 k Span = 13.67 ft Try: I b= 51/8 in d= 10.5 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 53.81 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 494 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 118 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 94 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.9 ksi f,,, x_x = 0.1 ksi Fc* = 649.1 psi RE = 2.53 Fb, x-x = 2.40 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb, = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.78 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.78 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.4 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.448 in -> L / 366 OK USE 5.125" x 10.5" GLB Comb. ADL = 0.108 in -> L / 1524 Comb. ATL w/ Ka= 0.61 in -> L / 269 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 13 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B7 P= 0k Mx = 79.1 k-in my = 0 k-in Vx = 2.93 k Vy = 0 k Span = 9 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 9.00 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 28.13 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 42 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.16 (NDSSect. 5.3.7) Sy = 14.6 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.9 ksi f,,, x_x = 0.1 ksi Fc* = 647.8 psi RE = 3.84 Fb, x-x = 2.39 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.78 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.78 OK Bi-axial Analysis = 0.4 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.5 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.227 in -> L / 476 OK USE 3.125" x 9" GLB Comb. ADL = 0.054 in -> L / 1984 Comb. ATL w/ Ker= 0.308 in -> L / 350 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 14 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B8 P= Ok Mx = 5.582 k-in my = 0 k-in Vx = 0.392 k Vy = 0 k Span = 4.75 ft Try: I b = 3 1/8 in d = 10.5 in Stress Class = 24F-1.8E Un-Balanced *Refer wn 2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 7.975 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 5.1.4) A = 32.81 in Fbx(_) 1450 psi Ex, ,,Ii� = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 301 in Fby 1450 psi Ey,,,,;,, = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 27 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 57 in F", = 230 psi Cb = 1.09 (NDSSect. 3.10.4) Cf� = 1.16 (NDSSect. 5.3.7) Sy = 17.1 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 4 in Bearing Length, lb,y = 4 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.1 ksi f,, x_x = 0.0 ksi Fc* = 647.4 psi RE = 4.15 Fb, x-x = 2.39 ksi OK F,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 66 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb1 = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.04 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.04 OK Bi-axial Analysis = 0.0 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.71 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.71 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL = 0.003 in -> L / 20304 OK T T � � 3.125 �J I � � 1 t � 1 � �TLD X � O D Comb. ADL= 7E-04 in -> L / 84599 • • Comb. ATL w/ Ket= 0.004 in -> L / 14929 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 15 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B9 P= 0k Mx = 6.818 k-in my = 0 k-in Vx = 0.433 k Vy = 0 k Span = 5.25 ft Try: I b = 3 1/8 in d = 10.5 in Stress Class = 24F-1.8E Un-Balanced *Refer wn 2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 7.975 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 5.1.4) A = 32.81 in' Fbx(_) 1450 psi Ex, ,,Ii� = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 301 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 27 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 57 in F", = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf� = 1.16 (NDSSect. 5.3.7) Sy = 17.1 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.1 ksi f,, x_x = 0.0 ksi Fc* = 647.4 psi RE = 4.15 Fb, x-x = 2.39 ksi OK F,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 66 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi FcE,y = 301055 psi Fb1 = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.05 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.05 OK Bi-axial Analysis = 0.0 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.004 in -> L / 15038 OK USE 3.125" x 10.5" GLB Comb. ADL = 0.001 in -> L / 62657 Comb. ATL w/ Ket= 0.006 in -> L / 11057 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 16 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B10 Induced Loading P= 0k Mx = 16.86 k-in My = 0 k-in Vx = 0.876 k - Vy = 0 k _ Span = 6.417 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference tors rvDs Shape Properties Self Weight = 7.869 plf Adjustment Factors "Reference NDS Manaar # of Members = 1 b = 3.5 in Kcr = 2.00 Co = 1.00 (rvDS sect. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8. 1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDssect. 2.3.3) CM (E) = 1.00 (NDssect. 8.1.4) A = 32.38 in' E = ##### psi CF = 1.20 OVDS&a.. 4.3.6) CM (M = 1.00 (NDSS-t. 8.1.4) Ix = 230.8 in Ern = ##### psi CL = 1.00 (NDssect. 3.3.3) Ci (Fb) = 1.00 (NDssect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 OVDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect 8.1.4) S= 49.91 in Faperp = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cf� = 1.00 (NDssect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.3 ksi fv, x_x = 0.0 ksi Fe* = 748.2 psi RB = 3.48 Fb, x-x = 1.38 ksi OK Fv, x_x = 0.18 ksi OK FcE,x = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FeE,y = 170421 psi Fbt = 1.38 ksi Fb, y_y = 1.38 ksi OK Fv, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.25 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.25 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb.ALL= 0.019 in-> L/ 4063 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.008 in -> L/ 10156 Comb. ATL w/ Kit= 0.034 in -> L / 2257 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 17 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B11 Induced Loading P= 0k Mx = 7.096 k-in My = 0 k-in Vx = 0.364 k A - Vy = 0 k No Span = 6.5 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference tors rvDs Shape Properties Self Weight = 7.869 plf Adjustment Factors "Reference NDS Manaar # of Members = 1 b = 3.5 in Kcr = 2.00 Co = 1.00 (rvDS sect. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8. 1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDssect. 8.1.4) A = 32.38 in' E _ ##### psi CF = 1.20 OVDS&a.. 4.3.6) CM (M = 1.00 (NDSS-t. 8.1.4) Ix = 230.8 in Ern = ##### psi CL = 1.00 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDssect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 OVDS sect. 3.10.4) Ci (E) = 1.00 (NDS seer. 8.1.4) Sx = 49.91 in Faperp = 625 psi Cr = 1.15 (NDSSect. 4.3.9) Cf� = 1.00 (NDssect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.1 ksi fv, x_x = 0.0 ksi Fe* = 748.2 psi RB = 3.48 Fb, x-x = 1.38 ksi OK Fv, x_x = 0.18 ksi OK FcE,x = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FeE,y = 170421 psi Fbt = 1.38 ksi Fb, y_y = 1.38 ksi OK Fv, y_y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.10 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.10 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.74 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.008 in-> L/ 9529 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.003 in -> L/ 23822 Comb. ATL w/ Kit = 0.015 in -> L/ 5294 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 18 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B12 Induced Loading P= 0k Mx = 97 k-in My = 0 k-in _ Vx = 2.98 k ! - Vy = 0 k Span = 10.83 ft Try: PSL 3 1/2 x 9 1/2 Parallam 2.0E (Beam) *Reference Weyerhauser Product Information Shape Properties Self Weight = 8.082 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 3.50 in Kcr = 1.5 Co = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 9.50 in Fb = 2900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 33.25 in' E = 2000000 psi CF = 1.00 *Reference Weyerhauser CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 250.07 in Enlin = 1016535 psi CL = 1.00 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.94 in Fv = 290 psi Cb = 1.38 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 52.65 in Fe.P-p = 750 psi Cr = 1.00 (NDS Sect. 4.3.9) Cr = 1.00 (NDS Sect. 4.3.9) Sy = 19.40 in Ft = 2025 psi Member Analysis Slenderness Check Unbraced Length, l x = 1.333 ft Kex = 1.00 Lx / d = 1.684 OK Unbraced Length, luy= 1 ft Key = 1.00 Ly / d = 3.429 OK Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.8 ksi f,, x_x = 0.1 ksi Fc* = 750 psi RE = 3.05 Fb, x-x = 3.99 ksi OK F3, x_x = 0.29 ksi OK FeE,x = 294726.0839 psi FbE = 131 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi FeF,y = 523696 psi Fb1 = 3.99 ksi Fb, y_y = 3.99 ksi OK FV, y_y = 0.29 ksi OK Cp = 1.00 Fb2 = 3.99 ksi c = 0.9 Check 0.46 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.46 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.4 ksi F'c, perp., x-x = 0.75 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.75 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb. ALL= 0.235in-> L/ 552 OK USE (1) 3 1/2 x 9 1/2 PSL Comb. ADL = 0.106 in -> L/ 1227 Comb. ATL w/ K,t= 0.394 in -> L/ 330 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 19 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B13 Induced Loading P= 0k Mx = 174 k-in My = 0 k-in = _ - Vx = 3.47 k ! - Vy = 0 k Span = 18.12 ft Try: PSL 7 x 9 1/2 Parallam 2.0E (Beam) *Reference Weyerhauser Product Information Shape Properties Self Weight = 16.16 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 7.00 in Kcr = 1.5 Cn = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 9.50 in Fb = 2900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 66.50 in' E = 2000000 psi CF = 1.00 *Reference Weyerhauser CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 500.14 in Enlin = 1016535 psi CL = 1.00 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 271.54 in Fv = 290 psi Cb = 1.38 (NDS Sect. 3.10.4) C, (E) = 1.00 (NDS Sect. 8.1.4) Sx = 105.29 in Fc.P-p = 750 psi Cr = 1.00 (NDS Sect. 4.3.9) Cr = 1.00 (NDS Sect. 4.3.9) Sy = 77.58 in Ft = 2025 psi Member Analysis Slenderness Check Unbraced Length, hex = 1.333 ft Kex = 1.00 Lx / d = 1.684 OK Unbraced Length, 1 y= 1 ft Key = 1.00 Ly / d = 1.714 OK Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.7 ksi f,, x_x = 0.1 ksi Fc* = 750 psi RE = 1.53 Fb, x-x = 3.99 ksi OK F3, x_x = 0.29 ksi OK FeE,x = 294726.0839 psi FbE = 524 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi F�F,y = 523696 psi Fbi = 3.99 ksi Fb, y_y = 3.99 ksi OK FV, y_y = 0.29 ksi OK Cp = 1.00 Fb2 = 3.99 ksi c = 0.9 Check 0.41 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.41 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.75 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.75 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb. ALL= 0.361 in-> L/ 602 OK USE (1) 7 x 9 1/2 PSL Comb. ADL = 0.162 in -> L/ 1342 Comb. ATL w/ K,t= 0.604 in -> L/ 360 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 20 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 3B14 P= 0k Mx = 127 k-in my = 0 k-in Vx = 4.131 k Vy = 0 k Span = 10.25 ft Try: I b= 51/8 in d= 10.5 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 53.81 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 494 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 118 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 94 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.3 ksi f,,, x_x = 0.1 ksi Fc* = 649.1 psi RE = 2.53 Fb, x-x = 2.40 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb, = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.56 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.56 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.4 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 480 TL: L/ 240 Comb. DLL= 0.181 in -> L / 678 OK USE 5.125" x 10.5" GLB Comb. ADL = 0.044 in -> L / 2825 Comb. ATL w/ Ka= 0.247 in -> L / 499 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 21 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J1 Induced Loading P= 0k MX = 3.924 k-in My = 0 k-in VX = 0.271 k A Vy = 0 k Span = 4.833 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.5 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbl = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.39 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.39 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.03 in-> L/ 1965 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.012 in -> L/ 4913 Comb. ATL w/ K,r= 0.053 in -> L/ 1092 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J2 Induced Loading P= 0k MX = 6.033 k-in My = 0 k-in _ VX = 0.335 k Vy = 0 k Span = 6 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfi, = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.8 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.60 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.60 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.07in-> L/ 1030 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.028 in -> L / 2574 Comb. ATL w/ K,r= 0.126 in -> L / 572 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J3 Induced Loading P= 0k MX = 7.263 k-in My = 0 k-in VX = 0.368 k Vy = 0 k 10 Span = 6.583 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 1.0 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi C = 0.8 Check 0.72 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.72 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.101 in-> L/ 779 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.041 in -> L / 1949 Comb. ATL w/ K,r= 0.182 in -> L / 433 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 3 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J4 Induced Loading P= 0k MX = 6.185 k-in My = 0 k-in _ VX = 0.412 k Vy = 0 k Span = 5 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfi, = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.8 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.61 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.61 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.056in-> L/ 1067 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.013 in -> L/ 4448 Comb. ATL w/ K,r= 0.083 in -> L/ 721 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 4 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J5 Induced Loading P= 0k MX = 7.483 k-in My = 0 k-in _ VX = 0.454 k - Vy = 0 k Span = 5.5 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 1.0 ksi f,, X_X = 0.1 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.74 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.74 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.2 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.082 in-> L/ 802 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.02 in -> L / 3342 Comb. ATL w/ K,r= 0.122 in -> L / 542 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 5 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J6 Induced Loading P= 0k MX = 6.185 k-in My = 0 k-in _ VX = 0.412 k - Vy = 0 k Span = 5 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.8 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.61 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.61 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.056in-> L/ 1067 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.013 in -> L/ 4448 Comb. ATL w/ K,r= 0.083 in -> L/ 721 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 6 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J7 Induced Loading P= 0k MX = 6.818 k-in My = 0 k-in _ VX = 0.433 k - Vy = 0 k Span = 5.25 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.9 ksi f,, X_X = 0.1 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.68 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.68 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.068in-> L/ 922 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.016 in -> L / 3842 Comb. ATL w/ K,r= 0.101 in -> L / 623 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 7 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J8 Induced Loading P= 0k MX = 3.479 k-in My = 0 k-in _ VX = 0.309 k - Vy = 0 k Span = 3.75 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.5 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.35 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.35 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.018in-> L/ 2530 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.004 in -> L / 10543 Comb. ATL w/ K,r= 0.026 in -> L / 1710 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 8 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2J9 Induced Loading P= 0k MX = 3.03 k-in My = 0 k-in _ VX = 0.289 k - Vy = 0 k Span = 3.5 ft Try: 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 20.8 in Emir = 580000 psi CL = 0.99 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.55 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc Pem = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.06 in Ft = 575 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.4 ksi f,, X_X = 0.0 ksi Fc* = 805.2 psi RB = 6.25 Fb, X-X = 1.33 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 56364 psi FbE = 49 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 56364 psi Fbi = 1.33 ksi Fb, y-y = 1.33 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.33 ksi c = 0.8 Check 0.30 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.30 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.80 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb.ALL= 0.013 in-> L/ 3112 OK USE 2x6 No. 2 Douglas Fir -Larch Comb. ADL = 0.003 in -> L/ 12967 Comb. ATL w/ K,r= 0.02 in -> L/ 2103 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 9 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2R1 Induced Loading P= 0k MX = 6.224 k-in My = 0 k-in _ VX = 0.259 k Vy = 0 k 10 Span = 8 ft Try: 2x8 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.643 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 Cn = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 7.25 in Fb = 900 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 10.88 in E = 1600000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 47.6 in Emir = 580000 psi Ci = 0.99 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 2.04 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 13.14 in Fc.pe1E = 625 psi C, = 1.15 (NDS sect. 4.3.9) Cfu = 1.00 (NDS sect. 4.3.7) Sy = 2.72 in Ft = 575 psi Member Analysis Unbraced Length, l = 1 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.5 ksi f,, X_X = 0.0 ksi Fc* = 850.9 psi RE = 7.18 Fb, X-X = 1.41 ksi OK Fv, X-X = 0.21 ksi OK FcE,X = 97938 psi FbE = 37 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 97938 psi Fb1 = 1.41 ksi Fb, y-y = 1.41 ksi OK FV, y-y = 0.21 ksi OK Cp = 1.00 Fb2 = 1.41 ksi c = 0.8 Check 0.34 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.34 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F c = 0.85 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.04in-> L/ 2388 OK USE 2x8 No. 2 Douglas Fir -Larch Comb. ADL = 0.024 in -> L / 3980 Comb. ATL w/ Kcr= 0.088 in -> L / 1085 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 10 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2R2 Induced Loading P= 0k MX = 35.11 k-in My = 0 k-in _ VX = 0.616 k - Vy = 0 k 10 Span = 19 ft Try: 2x12 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 4.102 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 Cn = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 11.25 in Fb = 900 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 16.88 in E = 1600000 psi CF = 1.00 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 178.0 in Emir = 580000 psi Ci = 0.98 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 3.16 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 31.64 in Fc.pe1F = 625 psi C, = 1.15 (NDS sect. 4.3.9) Cb, = 1.00 (NDS sect. 4.3.7) Sy = 4.22 in Ft = 575 psi Member Analysis Unbraced Length, l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 1.1 ksi f,, X_X = 0.0 ksi Fc* = 705.3 psi RB = 8.94 Fb, X-X = 1.17 ksi OK Fv, X-X = 0.21 ksi OK FcE,X = 235821 psi FbE = 24 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 235821 psi Fbl = 1.17 ksi Fb, y_y = 1.17 ksi OK FV, y_y = 0.21 ksi OK Cp = 1.00 Fb2 = 1.17 ksi c = 0.8 Check 0.95 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.95 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.70 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.2 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.342 in-> L/ 666 OK USE 2x12 No. 2 Douglas Fir -Larch Comb. ADL = 0.205 in -> L / 1110 Comb. ATL w/ Kcr= 0.753 in -> L / 303 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 11 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B1 Induced Loading P= 0k MX = 42.04 k-in My = 0 k-in _ VX = 1.193 k - Vy = 0 k Span = 11.75 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS seer. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 32.38 in E = 1700000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 230.8 in Emir = 620000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 49.91 in Fc.peCp = 625 psi Cr = 1.15 (NDS sect. 4.3.9) Cb, = 1.00 (NDS sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.8 ksi f,, X_X = 0.0 ksi Fc* = 748.2 psi RB = 3.48 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 170421 psi Fbl = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.61 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.61 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.2 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.158in-> L/ 890 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.063 in -> L/ 2224 Comb. ATL w/ Kcr= 0.285 in -> L/ 494 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 12 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2112 Induced Loading P= 0k MX = 3.269 k-in My = 0 k-in VX = 0.247 k A Vy = 0 k Span = 4.417 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS seer. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 32.38 in E = 1700000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 230.8 in Emir = 620000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 49.91 in Fc.peCp = 625 psi Cr = 1.15 (NDS sect. 4.3.9) Cb, = 1.00 (NDS sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.1 ksi f,, X_X = 0.0 ksi Fc* = 748.2 psi RB = 3.48 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 170421 psi Fbl = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.05 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.05 OK Bi-axial Analysis = 0.0 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.0 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.002 in-> L/ 30442 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 7E-04 in -> L / 76105 Comb. ATL w/ Kcr= 0.003 in -> L / 16912 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 13 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2113 Induced Loading P= 0k MX = 32.34 k-in My = 0 k-in _ VX = 1.166 k - Vy = 0 k Span = 9.25 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS seer. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 32.38 in E = 1700000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 230.8 in Emir = 620000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 49.91 in Fc.peCp = 625 psi Cr = 1.15 (NDS sect. 4.3.9) Cb, = 1.00 (NDS sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.6 ksi f,, X_X = 0.0 ksi Fc* = 748.2 psi RB = 3.48 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 170421 psi Fb1 = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.47 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.47 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.2 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.076in-> L/ 1469 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.03 in -> L / 3673 Comb. ATL w/ Kcr= 0.136 in -> L / 816 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 14 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2114 Induced Loading P= 0k MX = 7.096 k-in My = 0 k-in _ VX = 0.364 k Vy = 0 k Span = 6.5 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS seer. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 32.38 in' E = 1700000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 230.8 in Emir = 620000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 49.91 in Fc.perp = 625 psi C, = 1.15 (NDS sect. 4.3.9) Cfu = 1.00 (NDS sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.1 ksi f,, X_X = 0.0 ksi Fc* = 748.2 psi RB = 3.48 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 170421 psi Fb1 = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.10 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.10 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.008in-> L/ 9529 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.003 in -> L/ 23822 Comb. ATL w/ Kcr= 0.015 in -> L/ 5294 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 15 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2114 Induced Loading P= 0k MX = 7.096 k-in My = 0 k-in _ VX = 0.364 k Vy = 0 k Span = 6.5 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS seer. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 32.38 in' E = 1700000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 230.8 in Emir = 620000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 49.91 in Fc.perp = 625 psi C, = 1.15 (NDS sect. 4.3.9) Cfu = 1.00 (NDS sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.1 ksi f,, X_X = 0.0 ksi Fc* = 748.2 psi RB = 3.48 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 170421 psi Fb1 = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.10 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.10 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.1 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.008in-> L/ 9529 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.003 in -> L/ 23822 Comb. ATL w/ Kcr= 0.015 in -> L/ 5294 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 16 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B6 Induced Loading P= 0k Mx = 16 k-in My = 0 k-in Vx = 1.08 k ! - Vy = 0 k 10 Span = 5 ft Try: PSL 7 x 11 7/8 Parallam 2.0E (Beam) *Reference Weyerhauser Product Information Shape Properties Self Weight = 20.2 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 7.00 in Kcr = 1.5 Cn = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 11.88 in Fb = 2900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 83.13 in' E = 2000000 psi CF = 1.00 *Reference Weyerhauser CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 976.83 in Enlin = 1016535 psi CL = 1.00 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 339.43 in Fv = 290 psi Cb = 1.38 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 164.52 in Fc.P-p = 750 psi Cr = 1.00 (NDS Sect. 4.3.9) Cr = 1.00 (NDS Sect. 4.3.9) Sy = 96.98 in Ft = 2025 psi Member Analysis Slenderness Check Unbraced Length, hex = 1.333 ft Kex = 1.00 Lx / d = 1.347 OK Unbraced Length, luy= 1 ft Key = 1.00 Ly / d = 1.714 OK Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.1 ksi f5,, x_x = 0.0 ksi Fc* = 750 psi RE = 1.71 Fb, x-x = 3.99 ksi OK F,, x_x = 0.29 ksi OK FeE,x = 460509.506 psi FbE = 419 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi F�F,y = 818274 psi Fbt = 3.99 ksi Fb, y_y = 3.99 ksi OK FV, y_y = 0.29 ksi OK Cp = 1.00 Fb2 = 3.99 ksi c = 0.9 Check 0.02 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.02 OK Bi-axial Analysis = 0.0 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.75 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.75 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb. ALL = 9E-04 in -> L / 69463 OK r U S L) (1) 7 X 1 1 7/8 P S L Comb. ADL = 0.002 in -> L/ 26717 Comb. ATL w/ K,t= 0.004 in -> L/ 14176 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 17 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2117 Induced Loading P= 0k MX = 30.94 k-in My = 0 k-in _ VX = 1.058 k - Vy = 0 k Span = 9.75 ft Try: 4x10 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS Seer. 2.3.2) CM (Fb) = 1.00 (NDS Seer. 8.1.4) d = 9.25 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 32.38 in E = 1600000 psi CF = 1.20 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 230.8 in Emir = 580000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 5.1.4) SX = 49.91 in Fe.pe1F = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 18.89 in Ft = 575 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.6 ksi f,, X_X = 0.0 ksi Fc* = 748.3 psi RB = 3.48 Fb, X-X = 1.24 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 159426 psi FbE = 159 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 159426 psi Fb1 = 1.24 ksi Fb, y-y = 1.24 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.24 ksi C = 0.8 Check 0.50 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.50 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.2 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.096in-> L/ 1214 OK USE 4x10 No. 2 Douglas Fir -Larch Comb. ADL = 0.023 in -> L/ 5060 Comb. ATL w/ K,r= 0.143 in -> L/ 821 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: B 10 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B8 P= 0k Mx = 101.7 k-in my = 0 k-in VX = 3.476 k Vy = 0 k Span = 9.75 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 EX = 1800000 psi Adjustment Factors d = 9.00 in Fbx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 5.1.4) A = 28.13 in' Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 5.1.4) IX = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fe, perp.) = 1.00 (NDSSect. 8.1.4) SX = 42 in F"Y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.16 (NDSSect. 5.3.7) Sy = 14.6 in Fe.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 2.4 ksi f,,, x_x = 0.1 ksi Fc* = 647.8 psi RE = 3.84 Fb, X-X = 2.39 ksi OK F,,, X_X = 0.27 ksi OK FcE,X = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi C = 0.9 Check 1.01 NG (NDSEq. 3.9-3) Bi-axial Analysis = 1.01 OK Bi-axial Analysis = 0.5 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.6 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.342 in -> L / 342 OK �T/� Comb. ADL = 0.082 in -> L / 1425 l� Comb. ATL w/ Ka= 0.465 in -> L / 251 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 19 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B9 P= 0k Mx = 172.1 k-in my = 0 k-in VX = 5.735 k Vy = 0 k Span = 10 ft Try: I b= 51/8 in d= 10.5 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 EX = 1800000 psi Adjustment Factors d= 10.50 in Fbx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 53.81 in' Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) IX = 494 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 118 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) SX = 94 in F"Y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.8 ksi f,,, x_x = 0.1 ksi Fc* = 649.1 psi RE = 2.53 Fb, X-X = 2.40 ksi OK F,,, X_X = 0.27 ksi OK FcE,X = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi FcE,y = 301055 psi Fb, = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.76 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.76 OK Bi-axial Analysis = 0.4 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.6 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.234 in -> L / 513 OK USE 5.125" x 10.5" GLB Comb. ADL = 0.056 in -> L / 2138 Comb. ATL w/ Ka= 0.318 in -> L / 377 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 20 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B10 P= 0k Mx = 97.65 k-in my = 0 k-in VX = 3.255 k Vy = 0 k Span = 10 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 EX = 1800000 psi Adjustment Factors d = 9.00 in Fbx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) A = 28.13 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) IX = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDS Sect. 5.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDS Sect. 3.3.3) CM (Fc, perp.) = 1.00 (NDS Sect. 8.1.4) SX = 42 in F"Y = 230 psi Cb = 1.19 (NDS Sect. 3.10.4) Cf, = 1.16 (NDS Sect. 5.3.7) Sy = 14.6 in Fc.perp = 650 psi Cc = 1.00 (NDS Sect. 5.3.8 CI = 1.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 2.3 ksi f,,, x_x = 0.1 ksi Fc* = 647.8 psi RE = 3.84 Fb, X-X = 2.39 ksi OK F,,, X_X = 0.27 ksi OK FcE,X = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.97 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.97 OK Bi-axial Analysis = 0.4 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.5 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.346 in -> L / 347 OK r USE 3.125 �� x 9" GLB Comb. ADL = 0.083 in -> L / 1446 Comb. ATL w/ Ker= 0.47 in -> L/ 255 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 21 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B11 P= 0k Mx = 176.6 k-in my = 0 k-in Vx = 4.307 k Vy = 0 k Span = 13.67 ft Try: I b= 51/8 in d= 10.5 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 53.81 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 494 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 118 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 94 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.9 ksi f,,, x_x = 0.1 ksi Fc* = 649.1 psi RE = 2.53 Fb, x-x = 2.40 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb, = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.78 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.78 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.4 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.448 in -> L / 366 OK USE 5.125" x 10.5" GLB Comb. ADL = 0.108 in -> L / 1524 Comb. ATL w/ Ka= 0.61 in -> L / 269 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 22 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B12 P= 0k Mx = 79.1 k-in my = 0 k-in Vx = 2.93 k Vy = 0 k Span = 9 ft Try: I b = 5 1/8 in d = 9 in Stress Class = 24F-1.8E Un-Balanced *Refer wn 2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 11.21 plf b = 5.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 9.00 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 5.1.4) A = 46.13 in Fbx(_) 1450 psi Ex, ,,Ii� = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 311 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 101 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 69 in F"y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 39.4 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.1 ksi f,, x_x = 0.1 ksi Fc* = 649.2 psi RE = 2.34 Fb, x-x = 2.40 ksi OK F,, x_x = 0.27 ksi OK FcE,x = 247205 psi FbE = 208 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fbt = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.48 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.48 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.3 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.138 in -> L / 781 OK USE 5.125" x 9" GLB Comb. ADL = 0.033 in -> L / 3254 Comb. ATL w/ Ket= 0.188 in -> L / 574 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 23 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B13 P= 0k Mx = 46.6 k-in my = 0 k-in Vx = 1.971 k Vy = 0 k Span = 9.23 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 9.00 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 28.13 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 42 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.16 (NDSSect. 5.3.7) Sy = 14.6 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.1 ksi f,,, x_x = 0.1 ksi Fc* = 647.8 psi RE = 3.84 Fb, x-x = 2.39 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.46 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.46 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.3 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.189 in -> L / 586 OK USE 3.125" x 9" GLB Comb. ADL = 0.037 in -> L / 2994 Comb. ATL w/ Ker= 0.245 in -> L / 453 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 24 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B14 Induced Loading P= 0k MX = 25.18 k-in My = 0 k-in _ VX = 1.767 k Vy = 0 k Span = 4.75 ft Try: 4x10 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.869 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.00 (NDS seer. 2.3.2) CM (Fb) = 1.00 (NDS seer. 8.1.4) d = 9.25 in Fb = 1000 psi Ct = 1.00 (NDS sect. 2.3.3) CM (E) = 1.00 (NDS sect. 8.1.4) A = 32.38 in E = 1700000 psi CF = 1.20 (NDS sect. 4.3.6) CM (v) = 1.00 (NDS sect. 8.1.4) IX = 230.8 in Emir = 620000 psi CL = 1.00 (NDSSect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.05 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E)= 1.00 (NDS Sect. 8.1.4) SX = 49.91 in Fc.pe1F = 625 psi Cr = 1.15 (NDS sect. 4.3.9) Cb, = 1.00 (NDS sect. 4.3.7) Sy = 18.89 in Ft = 675 psi Member Analysis Unbraced Length, 1. = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.5 ksi f,, X_X = 0.1 ksi Fc* = 748.2 psi RB = 3.48 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.18 ksi OK FcE,X = 170421 psi FbE = 169 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 170421 psi Fb1 = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.18 ksi OK Cp = 1.00 Fb2 = 1.38 ksi c = 0.8 Check 0.37 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.37 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F c = 0.75 ksi fc/Fc= 0.00 OK f c, perp. x-x = 0.3 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb.ALL= 0.018in-> L/ 3255 OK USE 4x10 No. 1 Douglas Fir -Larch Comb. ADL = 0.004 in -> L / 13562 Comb. ATL w/ Kcr= 0.026 in -> L / 2199 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 25 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B15 P= 0k Mx = 74.76 k-in my = 0 k-in Vx = 2.848 k Vy = 0 k Span = 8.75 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 9.00 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 28.13 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 42 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.16 (NDSSect. 5.3.7) Sy = 14.6 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.8 ksi f,,, x_x = 0.1 ksi Fc* = 647.8 psi RE = 3.84 Fb, x-x = 2.39 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.74 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.74 OK Bi-axial Analysis = 0.4 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.5 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.203 in -> L / 518 OK USE 3.125" x 9" GLB Comb. ADL = 0.049 in -> L / 2159 Comb. ATL w/ Ka= 0.276 in -> L / 381 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 26 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B16 P= 0k Mx = 81.88 k-in my = 0 k-in Vx = 3.119 k Vy = 0 k Span = 8.75 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 9.00 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect 2.3.2) CM (Fb) = 1.00 (NDSSect. 8.1.4) A = 28.13 in Fbxt_j 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 8.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 42 in F,,y = 230 psi Cb = 1.19 (NDSSect. 3.10.4) Cf, = 1.16 (NDSSect. 5.3.7) Sy = 14.6 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.9 ksi f,,, x_x = 0.1 ksi Fc* = 647.8 psi RE = 3.84 Fb, x-x = 2.39 ksi OK F,,, x_x = 0.27 ksi OK FcE,x = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.39 ksi Fb, y_y = 1.68 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.68 ksi c = 0.9 Check 0.81 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.81 OK Bi-axial Analysis = 0.4 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.5 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.222 in -> L / 473 OK USE 3.125" x 9" GLB Comb. ADL = 0.053 in -> L / 1971 Comb. ATL w/ Ker= 0.302 in -> L / 348 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 27 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B17 P= 0k Mx = 37.79 k-in my = 0 k-in VX = 1.44 k Vy = 0 k Span = 8.75 ft Try: I b= 31/8 in d= 9 in Stress Class = 24F-1.8E Un-Balanced *2cfimwe2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 6.836 plf b = 3.125 in Kcr = 1.50 EX = 1800000 psi Adjustment Factors d = 9.00 in Fbx(+)= 2400 psi Ey = 1600000 psi CD = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) A = 28.13 in Fbx(_) 1450 psi Ex,,,,,,, = 950000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) IX = 190 in Fby 1450 psi Ey,,,,i„ = 850000 psi Cv = 1.00 (NDS Sect. 5.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Iy = 23 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDS Sect. 3.3.3) CM (Fc, perp.) = 1.00 (NDS Sect. 8.1.4) SX = 42 in F"Y = 230 psi Cb = 1.19 (NDS Sect. 3.10.4) Cf, = 1.16 (NDS Sect. 5.3.7) Sy = 14.6 in Fc.perp = 650 psi Cc = 1.00 (NDS Sect. 5.3.8 CI = 1.00 (NDS Sect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.9 ksi f,,, x_x = 0.1 ksi Fc* = 744.6 psi RE = 3.84 Fb, X-X = 2.75 ksi OK F,,, X_X = 0.30 ksi OK FcE,X = 247205 psi FbE = 77 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 221184 psi Fb, = 2.75 ksi Fb, y_y = 1.93 ksi OK FV, y_y = 0.26 ksi OK Cp = 1.00 Fb2 = 1.93 ksi c = 0.9 Check 0.33 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.33 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.74 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.77 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.77 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. DLL= 0.065 in -> L / 1612 OK USE 3.125" x 9" GLB Comb. ADL = 0.039 in -> L / 2687 Comb. ATL w/ Ka= 0.124 in -> L / 848 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 28 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B18 Induced Loading P= 0k Mx = 97 k-in My = 0 k-in _ Vx = 2.98 k ! - Vy = 0 k Span = 10.83 ft Try: PSL 3 1/2 x 9 1/2 Parallam 2.0E (Beam) *Reference Weyerhauser Product Information Shape Properties Self Weight = 8.082 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 3.50 in Kcr = 1.5 Co = 1.00 (NDS Sect. 2.3.2) CM (F,)= 1.00 (NDS Sect. 8.1.4) d = 9.50 in Fb = 2900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 33.25 in' E = 2000000 psi CF = 1.00 *Reference Weyerhauser CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 250.07 in Enlin = 1016535 psi CL = 1.00 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 33.94 in Fv = 290 psi Cb = 1.38 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 52.65 in Fe.P-p = 750 psi Cr = 1.00 (NDS Sect. 4.3.9) Cr = 1.00 (NDS Sect. 4.3.9) Sy = 19.40 in Ft = 2025 psi Member Analysis Slenderness Check Unbraced Length, l x = 1.333 ft Kex = 1.00 Lx / d = 1.684 OK Unbraced Length, luy= 1 ft Key = 1.00 Ly / d = 3.429 OK Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.8 ksi f,, x_x = 0.1 ksi Fc* = 750 psi RE = 3.05 Fb, x-x = 3.99 ksi OK F3, x_x = 0.29 ksi OK FeE,x = 294726.0839 psi FbE = 131 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi FeF,y = 523696 psi Fb1 = 3.99 ksi Fb, y_y = 3.99 ksi OK FV, y_y = 0.29 ksi OK Cp = 1.00 Fb2 = 3.99 ksi c = 0.9 Check 0.46 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.46 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.4 ksi F'c, perp., x-x = 0.75 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.75 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb. ALL= 0.235in-> L/ 552 OK USE (1) 3 1/2 x 9 1/2 PSL Comb. ADL = 0.106 in -> L/ 1227 Comb. ATL w/ K,t= 0.394 in -> L/ 330 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 29 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B19 Induced Loading P= 0k Mx = 174 k-in My = 0 k-in = _ - Vx = 3.47 k ! - Vy = 0 k Span = 18.12 ft Try: PSL 7 x 9 1/2 Parallam 2.0E (Beam) *Reference Weyerhauser Product Information Shape Properties Self Weight = 16.16 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 7.00 in Kcr = 1.5 Cn = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 9.50 in Fb = 2900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 66.50 in' E = 2000000 psi CF = 1.00 *Reference Weyerhauser CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 500.14 in Enlin = 1016535 psi CL = 1.00 (NDSSect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 271.54 in Fv = 290 psi Cb = 1.38 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) Sx = 105.29 in Fc.P-p = 750 psi Cr = 1.00 (NDS Sect. 4.3.9) Cr = 1.00 (NDS Sect. 4.3.9) Sy = 77.58 in Ft = 2025 psi Member Analysis Slenderness Check Unbraced Length, hex = 1.333 ft Kex = 1.00 Lx / d = 1.684 OK Unbraced Length, 1 y= 1 ft Key = 1.00 Ly / d = 1.714 OK Bearing Length, ib,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 1.7 ksi f,, x_x = 0.1 ksi Fc* = 750 psi RE = 1.53 Fb, x-x = 3.99 ksi OK F3, x_x = 0.29 ksi OK FeE,x = 294726.0839 psi FbE = 524 ksi fb, y_y = 0.0 ksi fl, y_y = 0.0 ksi F�F,y = 523696 psi Fb1 = 3.99 ksi Fb, y_y = 3.99 ksi OK FV, y_y = 0.29 ksi OK Cp = 1.00 Fb2 = 3.99 ksi c = 0.9 Check 0.41 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.41 OK Bi-axial Analysis = 0.2 OK fc = 0.00 ksi Bearing Analysis F'c = 0.75 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.75 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.75 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb. ALL= 0.361 in-> L/ 602 OK USE (1) 7 x 9 1/2 PSL Comb. ADL = 0.162 in -> L/ 1342 Comb. ATL w/ K,t= 0.604 in -> L/ 360 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 30 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B20 P= 0k Mx = 197.8 k-in my = 0 k-in Vx = 7.127 k Vy = 0 k Span = 9.25 ft Try: I b = 5 1/8 in d = 10.5 in Stress Class = 24F-1.8E Un-Balanced *Refer wn 2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 5.1.4) A = 53.81 in Fbxt_j 1450 psi Ex, ,,Ii� = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 494 in Fby 1450 psi Ey,,,,;,, = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 118 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 94 in F", = 230 psi Cb = 1.06 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 6 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 2.1 ksi f,, x_x = 0.1 ksi Fc* = 649.1 psi RE = 2.53 Fb, x-x = 2.40 ksi OK F,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb1 = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.88 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.88 OK Bi-axial Analysis = 0.5 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.69 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.69 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.208 in -> L / 533 OK USE 5.125" x 10.5" GLB Comb. ADL = 0.077 in -> L / 1442 Comb. ATL w/ Ket= 0.324 in -> L / 343 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 31 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member 2B21 P= 0k Mx = 197.8 k-in my = 0 k-in Vx = 7.127 k Vy = 0 k Span = 9.25 ft Try: I b = 5 1/8 in d = 10.5 in Stress Class = 24F-1.8E Un-Balanced *Refer wn 2015NDS Lumber Species = Western Species Shape Properties I Self Weight = 13.08 plf b = 5.125 in Kcr = 1.50 Ex = 1800000 psi Adjustment Factors d= 10.50 in T"bx(+)= 2400 psi Ey = 1600000 psi CD = 1.00 (NDSSect. 2.3.2) CM (Fb) = 1.00 (NDSSect. 5.1.4) A = 53.81 in Fbx(_) 1450 psi Ex, ,,Ii� = 950000 psi Ct = 1.00 (NDSSect. 2.3.3) CM (E) = 1.00 (NDSSect. 8.1.4) Ix = 494 in Fby 1450 psi Ey,,,,;,, = 850000 psi Cv = 1.00 (NDSSect. 5.3.6) CM (V) = 1.00 (NDSSect. 5.1.4) Iy = 118 in F, = 265 psi Ft = 1100 psi CL = 1.00 (NDSSect. 3.3.3) CM (Fc, perp.) = 1.00 (NDSSect. 8.1.4) Sx = 94 in F", = 230 psi Cb = 1.06 (NDSSect. 3.10.4) Cf, = 1.10 (NDSSect. 5.3.7) Sy = 46.0 in Fc.perp = 650 psi Cc = 1.00 (NDSSect. 5.3.8 CI = 1.00 (NDSSect. 5.3.9) Member Analysis Unbraced Length,l = 1.3 ft Bearing Length, lb,x = 6 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 2.1 ksi f,, x_x = 0.1 ksi Fc* = 649.1 psi RE = 2.53 Fb, x-x = 2.40 ksi OK F,, x_x = 0.27 ksi OK FcE,x = 336474 psi FbE = 178 ksi fb, y_y = 0.0 ksi f" y_y = 0.0 ksi FcE,y = 301055 psi Fb1 = 2.40 ksi Fb, y_y = 1.59 ksi OK FV, y_y = 0.23 ksi OK Cp = 1.00 Fb2 = 1.59 ksi c = 0.9 Check 0.88 OK (NDSEq. 3.9-3) Bi-axial Analysis = 0.88 OK Bi-axial Analysis = 0.5 OK fc = 0.00 ksi Bearing Analysis F'c = 0.65 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.69 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.69 ksi OK Deflection CheckILL: L/ 240 TL: L/ 180 Comb. ALL= 0.208 in -> L / 533 OK USE 5.125" x 10.5" GLB Comb. ADL = 0.077 in -> L / 1442 Comb. ATL w/ Ket= 0.324 in -> L / 343 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: C 32 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Concrete Wall Footing Design Typ 3 Story Bear Roof Loading 13rd Floor 12nd Floor DL = 15 psf DL = 24 psf DL = 24 psf LL = 20 psf LL = 100 psf LL = 100 psf SL = 25 psf SL = 25 psf SL = 25 psf wL 12.7 psf wL 12.7 psf wL 12.7 psf wL 72.6 psf wL 72.6 psf wL 72.6 psf 0.7E = 0 psf 0.7E = 0 psf 0.7E = 0 psf Trib = 15 ft Trib = 5 ft Trib = 5 ft Wall DL= 10 psf Wall DL= 10 psf Wall DL= 10 psf Wall H= 10 ft Wall H= 10 ft Wall H= 10 ft Load Combinations (ASD) 1) D = 2465 plf 2) D + L = 3765 plf 3) D + S = 3090 plf 4) D + 0.75L + 0.755 = 3909 plf 5) D + W = 2783 plf 6a) D + 0.75L+0.75W+0.755 = 4147 plf 6b) D + 0.75L + 0.75E + 0.755 = 3909 plf 7) 0.61) + W(-) = 462 plf 8) 0.61) + E = 462 plf eck Footing For Bearing on Soil Strata Governing Load = 4147 plf Footing Area = 3 ft2 3earing Pressure = 1382 psf OK eck Footing Depth Required for Shear Vu = 1728 lb d= 1.52 in hreg'd = 4.52 in OK termine Required Reinforcing Steel Area Cant. Length = 1.25 ft Mu = 1.08 k-ft Try: 3 #5 Bars AS = 0.93 in AS,,,;,, = 0.778 in OK a= 1.368 d = 9 in Mn = 3 8.67 k-ft (oMn = 34.8 OK Wall Materal Properties Concrete fc = 4000 psi Concrete Weight Y = 150 pcf Allow. Soiling Bearing = 1500 psf Soil Weight Y = 110 pcf Rebar, fy = 60 ksi Combinations Properties Stem Wall Thickness = 6 in Stem Wall Height = 2 ft Stem Wall Weight = 150 plf Strip Footing Width = 3 ft Strip Footing Thickness = 1 ft Strip Footing Weight = 450 plf 1) 1.41) = 3451 plf 2) 1.21) + 1.6L +0.55 = 5351 plf 3a) 1.21) + 1.6S + L = 5258 plf 3b) 1.21) + 1.6S + 0.84W = 4225 plf 4) 1.2D + 1.67W + L + 0.55 = 5038 plf 5) 1.21) + 1.4E + L + 0.25 = 4383 plf 6) 0.91) + 1.67W(-) = 816 plf 7) 0.91) + 1.4E = 0 plf of Reinforcing Bars `Pt: 1 T,: 1 `,: 0.8 A: 1 cb = 3 in ( cb + Kt,)/ db = 2.5 Bar Type: Straight # of Bar Dia. = 22.77 Id- —'d = 14.23 in OK (Soil Weight Soil Depth = 2 ft Soil Weight = 1100 plf 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Wall Footing 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Concrete Wall Footing Design Typ 3 Story Non Roof Loading 13rd Floor 12nd Floor DL = 15 psf DL = 24 psf DL = 24 psf LL = 20 psf LL = 100 psf LL = 100 psf SL = 25 psf SL = 25 psf SL = 25 psf wL 12.7 psf wL 12.7 psf wL 12.7 psf wL 72.6 psf wL 72.6 psf wL 72.6 psf 0.7E = 0 psf 0.7E = 0 psf 0.7E = 0 psf Trib = 2 ft Trib = 2 ft Trib = 2 ft Wall oL= 10 psf Wall DL= 10 psf Wall DL= 10 psf Wall H= 10 ft Wall H= 10 ft Wall H= 10 ft Load Combinations (ASD) 1) D = 2346 plf 2) D + L = 2786 plf 3) D + S = 2496 plf 4) D + 0.75L + 0.755 = 2789 plf 5) D + W = 2422 plf 6a) D + 0.75L+0.75W+0.755 = 2846 plf 6b) D + 0.75L + 0.75E + 0.755 = 2789 plf 7) 0.61) + W(-) = 112.8 plf 8) 0.61) + E = 112.8 plf eck Footing For Bearing on Soil Strata Governing Load = 2846 plf Footing Area = 2.5 112 3earing Pressure = 1138 psf OK eck Footing Depth Required for Shear Vu = 1138 lb d = 1.00 in hreg'd = 4.00 in OK termine Required Reinforcing Steel Area Cant. Length = 1 ft Mu = 0.569 k-ft Try: 3 #4 Bars AS = 0.6 in As, n ii = 0.648 in OK a = 0.882 d = 9 in Mn = 25.68 k-ft (oMn = 23.11 OK Wall ral Properties Concrete fc = 4000 psi Concrete Weight Y = 150 pcf Allow. Soiling Bearing = 1500 psf Soil Weight Y = 110 pcf Rebar, fy = 60 ksi Combinations Properties Stem Wall Thickness = 6 in Stem Wall Height = 3 ft Stem Wall Weight = 225 plf Strip Footing Width = 2.5 ft Strip Footing Thickness = 1 ft Strip Footing Weight = 375 plf 1) 1.41) = 3284 plf 2) 1.21) + 1.6L +0.55 = 3594 plf 3a) 1.21) + 1.6S + L = 3495 plf 3b) 1.21) + 1.6S + 0.84W = 3119 plf 4) 1.2D + 1.67W + L + 0.55 = 3432 plf 5) 1.21) + 1.4E + L + 0.25 = 3285 plf 6) 0.91) + 1.67W(-) = 219 plf 7) 0.91) + 1.4E = 0 plf of Reinforcing Bars `Pt: 1 T,: 1 `,: 0.8 A: 1 cb = 3 in ( cb + Kt,)/ db = 2.5 Bar Type: Straight # of Bar Dia. = 22.77 Id- -'d = 11.38 in KII (Soil Weight Soil Depth = 3 ft Soil Weight = 1320 plf 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Wall Footing 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Concrete Wall Footing Design Typ 3 Story Interior Bearing Wall Roof Loading 13rd Floor 12nd Floor Materal Properties Footing Properties DL = 15 psf DL = 24 psf DL = 24 psf LL = 20 psf LL = 100 psf LL = 100 psf Concrete fc = 4000 psi Stem Wall Thickness = 0 in SL = 25 psf SL = 25 psf SL = 25 psf Concrete Weight Y = 150 pcf Stem Wall Height = 0 ft wL (+) = 12.7 psf wL (+) = 12.7 psf wL (+) = 12.7 psf Stem Wall Weight = 0 plf wL (-) = 72.6 psf wL (-) = 72.6 psf wL (-) = 72.6 psf Allow. Soiling Bearing = 1500 psf 0.7E = 0 psf 0.7E = 0 psf 0.7E = 0 psf Soil Weight Y = 110 pcf Strip Footing Width = 3 ft Trib = 0 ft Trib = 12.5 ft Trib = 12.5 ft Strip Footing Thickness = 1 ft Wall DL= 10 psf Wall DL= 10 psf Wall DL= 10 psf Rebar, fy= 60 ksi Strip Footing Weight= 450 plf Wall H= 0 ft Wall = 10 ft Wall = 10 ft Load Combinations (ASD) Load Combinations (LRFD) Soil Weight 1) D = 1250 plf 1) 1.41) = 1750 plf Soil Depth = 0 ft 2) D + L = 3750 plf 2) 1.21) + 1.6L +0.5S = 5813 plf Soil Weight = 0 plf 3) D + S = 1875 plf 3a) 1.21) + 1.6S + L = 5000 plf 4) D + 0.75L + 0.755 = 3594 plf 3b) 1.21) + 1.6S + 0.84W = 2767 plf 5) D + W = 1568 plf 4) 1.213 + 1.67W + L + 0.55 = 4686 plf 6a) D + 0.75L+0.75W+0.755 = 3832 plf 5) 1.21) + 1.4E + L + 0.25 = 4125 plf 6b) D + 0.75L + 0.75E + 0.755 = 3594 plf 6) 0.91) + 1.67W(-) = 2311 plf 7) 0.61) + W(-) = 1335 plf 7) 0.91) + 1.4E = 0 plf 8) 0.61) + E = 1335 plf Check Footing For Bearing on Soil Strata Check Development Length of Reinforcing Bars Governing Load = 3832 plf `Pt: 1 Footing Area = 3 ft2 `,: 1 Bearing Pressure = 1277 psf OK `,: 0.8 A: 1 Check Footing Depth Required for Shear Vu = 1916 lb cb = 3 in d = 1.68 in ( cb + Kt,)/ db = 2.5 hmq'd = 4.68 in OK Bar Type: Straight # of Bar Dia. = 22.77 Determine Required Reinforcing Steel Area ld. —'d = 11.38 in OK Cant. Length = 1.5 ft Mu = 1.437 k-ft Try: 4 #4 Bars As = 0.8 in AS,,,,;,, = 0.778 in OK a= 1.176 d = 9 in Mn = 33.65 k-ft roMn = 30.28 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Wall Footing 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Header Check Induced Loading P = 5.75 k MX = 0 k-in My = 0 k-in - vX = 0 k vy = 0 k Span = 8 ft Try: 4x4 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.977 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 3.50 in Fb = 1000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 12.25 in E = 1700000 psi CF = 1.50 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 12.5 in Emir = 620000 psi CL = 1.00 (NDS Sect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Ty = 12.51 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDSSect. 8.1.4) Sx = 7.15 in Fc Pem = 625 psi C, = 1.00 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 7.15 in Ft = 675 psi Member Analysis Unbraced Length,1 = 8.75 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, x_x = 0.0 ksi f,, x_X = 0.0 ksi Fc* = 1078 psi RB = 5.48 Fb, x-x = 1.73 ksi OK Fv, X-X = 0.21 ksi OK FcE x = 566.3 psi FbE = 68 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 566 psi Fbi = 1.73 ksi Fb, y-y = 1.73 ksi OK FV, y-y = 0.21 ksi OK Cp = 0.45 Fb2 = 1.73 ksi c = 0.8 Check 0.93 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.00 OK Bi-axial Analysis = 0.0 OK fc = 0.47 ksi Bearing Analysis F'c = 0.49 ksi fc / F'c = 0.97 OK f c, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb.ALL= 0.01 in-> L/ 9600 OK USE 4x4 No. 1 Douglas Fir -Larch Comb. ADL = 0.01 in -> L / 9600 Comb. ATL w/ K,r= 0.01 in -> L / 9600 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Lumber 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member Description Induced Loading P= 17k Mx = 0 k-in My = 0 k-in - vX = 0 k - vy = 0 k Span = 10 ft Try: 6x6 No. 1 (C) Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.352 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 5.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.5 in Fb = 1200 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 30.25 in' E = 1600000 psi CF = 1.00 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 76.3 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 76.26 in Fv = 170 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 5.1.4) SX = 27.73 in Fc Pem = 625 psi C, = 1.00 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 27.73 in Ft = 825 psi Member Analysis Unbraced Length,l = 1 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.0 ksi f,, X_X = 0.0 ksi Fc* = 625 psi RB = 1.48 Fb, X-X = 1.20 ksi OK Fv, X-X = 0.17 ksi OK FcE,X = ##### psi FbE = 880 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = ##### psi Fbl = 1.20 ksi Fb, y-y = 1.20 ksi OK FV, y-y = 0.17 ksi OK Cp = 1.00 Fb2 = 1.20 ksi c = 0.8 Check 0.81 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.00 OK Bi-axial Analysis = 0.0 OK fc = 0.56 ksi Bearing Analysis F c = 0.62 ksi fc/Fc= 0.90 OK fc, perp. x-x = 0.0 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.01 in -> L/ 12000 OK USE 6x6 No. 1 (C) Douglas Fir -Larch Comb. ADL = 0.01 in -> L/ 12000 Comb. ATL w/ K,r= 0.03 in -> L/ 4000 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Timber 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Member Description Induced Loading P = 21.5 k Mx = 0 k-in My = 0 k-in - vX = 0 k - vy = 0 k Span = 20 ft Try: 6x6 No. 1 (C) Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 7.352 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 5.5 in Kcr = 2.00 CD = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.5 in Fb = 1200 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 30.25 in E = 1600000 psi CF = 1.00 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) IX = 76.3 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 76.26 in Fv = 170 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 5.1.4) SX = 27.73 in Fc Pem = 625 psi C, = 1.00 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 27.73 in Ft = 825 psi Member Analysis Unbraced Length,l = 2 ft Bearing Length, lb,X = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 0.0 ksi f,, X_X = 0.0 ksi Fc* = 718.8 psi RB = 2.09 Fb, X-X = 1.38 ksi OK Fv, X-X = 0.20 ksi OK FcE,X = 25038 psi FbE = 440 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 25038 psi Fbl = 1.38 ksi Fb, y-y = 1.38 ksi OK FV, y-y = 0.20 ksi OK Cp = 0.99 Fb2 = 1.38 ksi c = 0.8 Check 0.99 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.00 OK Bi-axial Analysis = 0.0 OK fc = 0.71 ksi Bearing Analysis F c = 0.71 ksi fc/Fc= 0.99 OK fc, perp. x-x = 0.0 ksi F c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.25 in -> L/ 960 OK USE 6x6 No. 1 (C) Douglas Fir -Larch Comb. ADL = 0.25 in -> L/ 960 Comb. ATL w/ Kcr = 0.75 in -> L/ 320 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Timber 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Worst Case Interior Wall Post Induced Loading P = 6.94 k MX = 3.47 k-in My = 0 k-in - VX = 0 k - Vy = 0 k Span = 8.75 ft Try: 4x6 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 4.679 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 3.5 in Kcr = 2.00 CD = 1.15 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 1000 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 19.25 in E = 1700000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (v) = 1.00 (NDS Sect. 8.1.4) Ix = 48.5 in Emir = 620000 psi CL = 1.00 (NDS Sect. 3.3.3) G, (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 19.65 in Fv = 180 psi Cb = 1.19 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 5.1.4) Sx = 17.65 in Fc Pem = 625 psi C, = 1.00 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 11.23 in Ft = 675 psi Member Analysis Unbraced Length, l = 8.75 ft Bearing Length, lb,x = 2 in Bearing Length, lb,y = 2 in Flexural Analysis I Shear Analysis Compression Analysis Combined Axial + Bending fb, x_X = 0.2 ksi f,, x_X = 0.0 ksi Fc* = 934.4 psi RB = 6.87 Fb, x-x = 1.50 ksi OK Fv, X-X = 0.21 ksi OK FcE,x = 1398 psi FbE = 43 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 1398 psi Fbi = 1.50 ksi Fb, y-y = 1.50 ksi OK FV, y-y = 0.21 ksi OK Cp = 0.81 Fb2 = 1.50 ksi c = 0.8 Check 0.36 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.13 OK Bi-axial Analysis = 0.0 OK fc = 0.36 ksi Bearing Analysis F'c = 0.76 ksi fc / F'c = 0.48 OK f c, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 480 TL: L/ 240 Comb.ALL= 0.01 in-> L/ 10500 OK USE 4x6 No. 1 Douglas Fir -Larch Comb. ADL = 0.01 in -> L / 10500 Comb. ATL w/ K,r= 0.01 in -> L / 10500 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Lumber 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 2x6 Wood Stud Walls (Grade to 2nd Out of Plane Load = 17.7 psf (Zone 5) Out of Plane Load = 0 psf (Parapet) Axial Load = 3697 plf Stud Spacing = 12 in O.C. Moment, Main Span = 6.192 k-in Moment, Cantilever = 0 k-in Axial Load = 8.318 k 0 8.75 ft ft R2 = 77.44 lb RI = 77.44 lb Try: I 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors # of Members = 2 *Reference NDS Manual b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 16.50 in' E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 41.6 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 81.4) IY = 3.09 in Fv = 180 psi Cb = 1.38 (NDS Sect. 3.10.4) C; (E) = 1.00 (NDS Sect. 8.1.4) SX = 15.13 in Fc.pep = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cf = 1.00 (NDS Sect. 4.3.7) SY = 4.13 in Ft = 575 psi Member Analysis Slenderness Check Deflection CheckILL: L/ 600 TL: L/ 600 Unbraced Length, 1 X= 8.75 ft KeX = 1.00 LX / d = 19.09 OK Main Span Unbraced Length, luY= 1 ft K,, = 1.00 Ly / d = 8 OK Wind Def. = 0.02 in -> L/ 4276 Bearing Length, lb,, = 2 in Cantilever Bearing Length, lb,Y = 2 in Wind Def. = N/A in -> 2L / N/A fb, X_X = 0.4 ksi Fb, X-X = 1.35 ksi OK fb, Y-Y = 0.0 ksi Fb, y-y = 1.35 ksi OK Bi-axial Analvsis = 0.3 OK Shear Analysis ICompression Analysis Combined Axial + Bending f, X_X = 0.0 ksi Fc* = 812.5 psi RE = 5.42 FV, X-X = 0.18 ksi FcE,X = 1308 psi FbE = 24 ksi FcE,Y = ##### psi Fbi = 1.35 ksi fv, Y-Y = 0.0 ksi Cp = 0.89 Fb2 = 1.35 ksi F, Y_Y = 0.18 ksi OK c = 0.9 Check 0.43 OK (NDS Eq. 3.9-3) fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.86 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.86 ksi OK Bi-axial Analvsis = 0.0 OK fc = 0.25 ksi F'c = 0.72 ksi fc / F'c = 0.35 OK USE (2) No. 2 2x6 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SL Wall 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 2x6 Wood Stud Walls (2nd Flr to 3rd Out of Plane Load = 17.7 psf (Zone 5) Out of Plane Load = 0 psf (Parapet) Axial Load = 2961 plf Stud Spacing = 16 in O.C. Moment, Main Span = 4.684 k-in Moment, Cantilever = 0 k-in Axial Load = 3.948 k 0 8.75 ft ft R2 = 103.3 lb RI = 103.3 lb Try: I 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in' E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) IY = 1.55 in Fv = 180 psi Cb = 1.38 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc.pep = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cffi = 1.00 (NDS Sect. 4.3.7) SY = 2.06 in Ft = 575 psi Member Analysis Slenderness Check Deflection CheckILL: L/ 600 TL: L/ 600 Unbraced Length, 1 X= 8.75 ft KeX = 1.00 LX / d = 19.09 OK Main Span Unbraced Length, luY= 1 ft K,, = 1.00 Ly / d = 8 OK Wind Def. = 0.07 in -> L / 1604 Bearing Length, lb,, = 2 in Cantilever Bearing Length, lb,Y = 2 in Wind Def. = N/A in -> 2L / N/A fb, X_X = 0.6 ksi Fb, X-X = 1.35 ksi OK fb, Y-Y = 0.0 ksi Fb, y-y = 1.35 ksi OK Bi-axial Analvsis = 0.5 OK Shear Analysis ICompression Analysis Combined Axial + Bending f, X_X = 0.0 ksi Fc* = 812.5 psi RE = 5.42 FV, X-X = 0.18 ksi FcE,X = 1308 psi FbE = 24 ksi FcE,Y = ##### psi Fbi = 1.35 ksi fv, Y-Y = 0.0 ksi Cp = 0.89 Fb2 = 1.35 ksi F, Y_Y = 0.18 ksi OK c = 0.9 Check 0.90 OK (NDS Eq. 3.9-3) fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.86 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.86 ksi OK Bi-axial Analvsis = 0.1 OK fc = 0.48 ksi F'c = 0.72 ksi fc / F'c = 0.66 OK USE (1) No. 2 2x6 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SL Wall 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 2x6 Wood Stud Walls (3rd Flr to Roof Out of Plane Load = 17.7 psf (Zone 5) Out of Plane Load = 0 psf (Parapet) Axial Load = 2224 plf Stud Spacing = 16 in O.C. Moment, Main Span = 3.653 k-in Moment, Cantilever = 0 k-in Axial Load = 2.965 k 0 7.83 ft ft R2 = 92.39 lb RI = 92.39 lb Try: I 2x6 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.005 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 5.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 8.25 in' E = 1600000 psi CF = 1.30 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 20.8 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) IY = 1.55 in Fv = 180 psi Cb = 1.38 (NDS Sect. 3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 7.56 in Fc.pep = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cffi = 1.00 (NDS Sect. 4.3.7) SY = 2.06 in Ft = 575 psi Member Analysis Slenderness Check Deflection CheckILL: L/ 600 TL: L/ 600 Unbraced Length, 1 X= 7.83 ft KeX = 1.00 LX / d = 17.08 OK Main Span Unbraced Length, luY= 1 ft K,, = 1.00 Ly / d = 8 OK Wind Def. = 0.04 in -> L / 2238 OK Bearing Length, lb,, = 2 in Cantilever Bearing Length, lb,Y = 2 in Wind Def. = N/A in -> 2L / N/A fb, X_X = 0.5 ksi Fb, X-X = 1.35 ksi OK fb, Y-Y = 0.0 ksi Fb, y-y = 1.35 ksi OK Bi-axial Analvsis = 0.4 OK Shear Analysis ICompression Analysis Combined Axial + Bending f, X_X = 0.0 ksi Fc* = 812.5 psi RE = 5.42 FV, X-X = 0.18 ksi FcE,X = 1634 psi FbE = 24 ksi FcE,Y = ##### psi Fbi = 1.35 ksi fv, Y-Y = 0.0 ksi Cp = 0.92 Fb2 = 1.35 ksi F, Y_Y = 0.18 ksi OK c = 0.9 Check 0.59 OK (NDS Eq. 3.9-3) fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.86 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.86 ksi OK Bi-axial Analvsis = 0.1 OK fc = 0.36 ksi F'c = 0.75 ksi fc / F'c = 0.48 OK USE (1) No. 2 2x6 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SL Wall 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 2x4 Wood Stud Walls (Grade to Out of Plane Load = 5 psf (interior) Out of Plane Load = 0 psf (Parapet) Axial Load = 3382 plf Stud Spacing = 6 in O.C. Moment, Main Span = 1.133 k-in Moment, Cantilever = 0 k-in Axial Load = 1.691 k 0 8.75 ft ft R2 = 10.94 lb RI = 10.94 lb Try: I 2x4 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 1.276 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 3.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 5.25 in' E = 1600000 psi CF = 1.50 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 5.4 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 81.4) IY = 0.98 in Fv = 180 psi Cb = 1.38 (NDS Sect. 3.10.4) C; (E) = 1.00 (NDS Sect. 8.1.4) SX = 3.06 in Fc.pQ1p = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cffi = 1.00 (NDS Sect. 4.3.7) SY = 1.31 in Ft = 575 psi Member Analysis Slenderness Check Deflection CheckILL: L/ 600 TL: L/ 600 Unbraced Length, 1 X= 8.75 ft KeX = 1.00 LX / d = 30 OK Main Span Unbraced Length, luY= 1 ft K,, = 1.00 Ly / d = 8 OK Wind Def. = 0.03 in -> L / 3901 OK Bearing Length, lb,, = 2 in Cantilever Bearing Length, lb,Y = 2 in Wind Def. = N/A in -> 2L / N/A fb, X_X = 0.4 ksi Fb, X-X = 1.55 ksi OK fb, Y-Y = 0.0 ksi Fb, y-y = 1.55 ksi OK Bi-axial Analvsis = 0.2 OK Shear Analysis ICompression Analysis Combined Axial + Bending f, X_X = 0.0 ksi Fc* = 937.5 psi RE = 4.32 FV, X-X = 0.18 ksi FcE,X = 529.7 psi FbE = 37 ksi FcE,Y = 40558 psi Fbi = 1.55 ksi fv, Y-Y = 0.0 ksi Cp = 0.51 Fb2 = 1.55 ksi F, Y_Y = 0.18 ksi OK c = 0.9 Check 0.69 OK (NDS Eq. 3.9-3) fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.86 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.86 ksi OK Bi-axial Analvsis = 0.0 OK fc = 0.32 ksi F'c = 0.48 ksi fc / F'c = 0.67 OK USE (1) No. 2 2x4 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SL Wall 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 2x4 Wood Stud Walls (L2 to L3 Out of Plane Load = 5 psf (interior) Out of Plane Load = 0 psf (Parapet) Axial Load = 1691 plf Stud Spacing = 12 in O.C. Moment, Main Span = 0.997 k-in Moment, Cantilever = 0 k-in Axial Load = 1.691 k 0 8.75 ft ft R2 = 21.88 lb RI = 21.88 lb Try: I 2x4 No. 2 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 1.276 plf Adjustment Factors # of Members = 1 *Reference NDS Manual b = 1.5 in Kcr = 2.00 CD = 1.00 (NDS Sect. 2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 3.50 in Fb = 900 psi Ct = 1.00 (NDS Sect. 2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 5.25 in' E = 1600000 psi CF = 1.50 (NDS Sect. 4.3.6) CM (V) = 1.00 (NDS Sect. 8.1.4) Ix = 5.4 in Emir = 580000 psi CL = 1.00 (NDS Sect. 3.3.3) C; (Fb) = 1.00 (NDS Sect. 81.4) IY = 0.98 in Fv = 180 psi Cb = 1.38 (NDS Sect. 3.10.4) C; (E) = 1.00 (NDS Sect. 8.1.4) SX = 3.06 in Fc.pep = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cffi = 1.00 (NDS Sect. 4.3.7) Sy = 1.31 in Ft = 575 psi Member Analysis Slenderness Check Deflection CheckILL: L/ 600 TL: L/ 600 Unbraced Length, 1 X= 8.75 ft KeX = 1.00 LX / d = 30 OK Main Span Unbraced Length, luY= 1 ft K,, = 1.00 Ly / d = 8 OK Wind Def. = 0.05 in -> L / 1950 Bearing Length, lb,, = 2 in Cantilever Bearing Length, lb,Y = 2 in Wind Def. = N/A in -> 2L / N/A fb, X_X = 0.3 ksi Fb, X-X = 1.55 ksi OK fb, Y-Y = 0.0 ksi Fb, y-y = 1.55 ksi OK Bi-axial Analvsis = 0.2 OK Shear Analysis ICompression Analysis Combined Axial + Bending f, X_X = 0.0 ksi Fc* = 937.5 psi RE = 4.32 FV, X-X = 0.18 ksi FcE,X = 529.7 psi FbE = 37 ksi FcE,Y = 40558 psi Fbi = 1.55 ksi fv, Y-Y = 0.0 ksi Cp = 0.51 Fb2 = 1.55 ksi F, Y_Y = 0.18 ksi OK c = 0.9 Check 0.66 OK (NDS Eq. 3.9-3) fc, perp. x-x = 0.0 ksi F'c, perp., x-x = 0.86 ksi OK fc, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.86 ksi OK Bi-axial Analvsis = 0.0 OK fc = 0.32 ksi F'c = 0.48 ksi fc / F'c = 0.67 OK USE (1) No. 2 2x4 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: SL Wall 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wood Stair Stringer Design up to 8'-0" Long Induced Loading P = 0.034 k Trib. = 16 inches MX = 13.44 k-in LL = 100 psf My = 0 k-in DL = 5 psf VX = 0.56 k wLL = 133.3 plf . Vy = 0 k w, DL + LL = 140 plf Span = 8 ft *Effective Member Try: I 2x6 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.188 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 Cn = 1.00 (NDS Sect. C2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 6.00 in Fb = 1000 psi Ct = 1.20 (NDS Sect. C2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 9.00 in E = 1700000 psi CF = 1.20 (NDS Sect. 4.3.6) CM (e) = 1.00 (NDS Sect. 8.1.4) IX = 27.0 in Emir = 620000 psi CL = 1.00 (NDS Sect. C3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.69 in Fv = 180 psi Cb = 1.19 (NDSSect. C3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 9.00 in Fe.pe1F = 625 psi C, = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.25 in Ft = 675 psi Member Analysis Unbraced Length, 1 = 8.0 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 1.5 ksi f,, X_X = 0.1 ksi Fc* = 750 psi RB = 16.00 Fb, X-X = 1.66 ksi OK FV, X-X = 0.22 ksi OK FcE,X = 1991 psi FbE = 8 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 1991 psi Fbi = 1.66 ksi Fb, y-y = 1.66 ksi OK FV, y-y = 0.22 ksi OK Cp = 0.91 Fb2 = 1.66 ksi c = 0.8 Check 0.90 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.90 OK Bi-axial Analysis = 0.3 OK fc = 0.00 ksi Bearing Analysis F'c = 0.68 ksi fc/F'c= 0.01 OK fc, perp. x-x = 0.2 ksi F'c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 360 TL: L/ 240 Comb. ALL= 0.268in-> L/ 360 OK Use Notched 2x12 DF#1 @ 16" o.c. Comb. ADL = 0.013 in -> L / 7172 Comb. ATL w/ K,= 0.294 in -> L / 326 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Stringer 1 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456 Wood Stair Stringer Design up to 13'-0" Long Induced Loading P = 0 k Trib. = 4 inches MX = 8.873 k-in LL = 100 psf My = 0 k-in DL = 5 psf VX = 0.228 k wLL = 33.33 plf . Vy = 0 k w, DL + LL = 35 plf Span = 13 ft *Effective Member Try: I 2x6 No. 1 Douglas Fir -Larch *Reference 2015 NDS Shape Properties Self Weight = 2.188 plf Adjustment Factors *Reference NDS Manual # of Members = 1 b = 1.5 in Kcr = 2.00 Cn = 1.00 (NDS Sect. C2.3.2) CM (Fb) = 1.00 (NDS Sect. 8.1.4) d = 6.00 in Fb = 1000 psi Ct = 1.20 (NDS Sect. C2.3.3) CM (E) = 1.00 (NDS Sect. 8.1.4) A = 9.00 in E = 1700000 psi CF = 1.20 (NDS Sect. 4.3.6) CM (e) = 1.00 (NDS Sect. 8.1.4) IX = 27.0 in Emir = 620000 psi CL = 1.00 (NDS Sect. C3.3.3) Ci (Fb) = 1.00 (NDS Sect. 8.1.4) Iy = 1.69 in Fv = 180 psi Cb = 1.19 (NDSSect. C3.10.4) Ci (E) = 1.00 (NDS Sect. 8.1.4) SX = 9.00 in Fe.pe1F = 625 psi Cr = 1.15 (NDS Sect. 4.3.9) Cfu = 1.00 (NDS Sect. 4.3.7) Sy = 2.25 in Ft = 675 psi Member Analysis Unbraced Length, l = 13.0 ft Bearing Length, lb,, = 2 in Bearing Length, lb,y = 2 in Flexural Analysis Shear Analysis Compression Analysis Combined Axial + Bending fb, X_X = 1.0 ksi f,, X_X = 0.0 ksi Fc* = 750 psi RB = 20.40 Fb, X-X = 1.66 ksi OK FV, X-X = 0.22 ksi OK FcE,X = 753.9 psi FbE = 5 ksi fb, y_y = 0.0 ksi fv, y_y = 0.0 ksi FcE,y = 754 psi Fbi = 1.66 ksi Fb, y-y = 1.66 ksi OK FV, y-y = 0.22 ksi OK Cp = 0.69 Fb2 = 1.66 ksi c = 0.8 Check 0.60 OK (NDS Eq. 3.9-3) Bi-axial Analysis = 0.60 OK Bi-axial Analysis = 0.1 OK fc = 0.00 ksi Bearing Analysis F'c = 0.52 ksi fc / F'c = 0.00 OK fc, perp. x-x = 0.1 ksi F'c, perp., x-x = 0.74 ksi OK f c, perp. y-y = 0.0 ksi F'c, perp., y-y = 0.74 ksi OK Deflection Check LL: L/ 240 TL: L/ 240 Comb. ALL= 0.47 in -> L / 334 OK Use Notched (1) 2x12 DF No. 1 @ 4" o.c. Comb. ADL = 0.023 in -> L / 6685 Comb. ATL w/ K,r= 0.513 in -> L / 304 OK 13228 NE 20th St, Suite 100, Bellevue, WA 98005 1 (425) 614-0949 Project: Edmonds Apartment Building Pg. No: Stringer 2 Address: 23830 Edmonds Way, Edmonds, WA Date: 7/25/19 Client: Hayes Wilson Lund Architects, LLC Job No.: 18456