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Specifications T0cO1 �� ,\ �ow/a t Lr2 9 1i \. 1r I'.'II I' 2 i I'3) 114 1 I ; oikilEWINIENERIE ,. STRUCTURAL CALCULATIONS RECEIVEI) NEW ADDITION 11958 SW 125th Ct., Tigard, OR 97223 MAY 1 0 2018 CHRIS & JESSICA ANDERSON CITY OF TIGARD BUILDING DIVISION May 9, 2018 Project No. 180145 78 Pages ,s> „,,, 5�(0)—v.PROic�L` ,:6-. off Q, •• 909P' ,`"%,, , .� o.jArir 734 rk., IRE c411P" EFFTRES:i'ap. zo/.' ***LIMITATIONS*** ENGINEER WAS RETAINED IN A LIMITED CAPACITY FOR THIS PROJECT DESIGN IS BASED UPON INFORMATION PROVIDED BY THE CLIENT, WHO IS SOLELY RESPONSIBLE FOR THE ACCURACY OF THE SAME. NO RESPONSIBILITY AND/OR LIABILITY IS ASSUMED BY, OR IS TO BE ASSIGNED TO THE ENGINEER FOR ITEMS BEYOND THAT SHOWN ON THESE SHEETS. 11611725 710,9eea nt, #17 'Pa ftcauuen, V,4 9g6E4 503, 737,4344 _ca{ tttezu.e.pact l).si a:Map.-'t:naaua.rr3 i :l=xlt t:ttl+ ;:. rth rwk,.ts o‘;,.:11.is;c a rear sgx�`u .un,arrear}.phis:"ttttii�tste ... usGs Design Maps Summary Report User—Specified Input Report Title Anderson Residence 5..,t torr S . 1s2t':::w:. Building Code Reference Document ASCD 7-:0;tariddi !+1LS...S ra.:41::::et4,0,..s frL,t 24/:&: Site Coordinates 45.433734N, 122.605S9£W Site Soil Classification Site Class D 'Sti(1 Sail• Risk Category I/Ii/III • J-4III Sb Oto *Portland Beaverton i Lake Oswego 44, lu Alei tin • Sherwood • • Oregian City USGS—Provided Output Ss = 0.963t Sns = 1,074y S„.5 zit 0.716q S, = 0.423 9 S = 0.667 q 5 = 0.445 y For information on hove the SS and S1 values above have heen calculated from probabilistic Frisk-targeted)and deterministic or::lun,i motions in the tfre,:tic"n of ma'irrclrr hori!:antai resruunee, please_return to the application and seleet the`2009 P,EF-IRP _ode reference document Mt'F,Rcspon>a-'puUtruan pont i Response spectrum si t Period.3-tut'• Peri d.'T Isee. For Fn K. T arid Cvalues, pleantr s rear Rtkt:a. I Ii et.in 3ttsn e>a pttucet St tt'e U.S_Cr4.1k gtc4 a.trey, tl o'sea t'u narznnky,topres.re{x{ni led.,e k">CP:.:ta.raCy of tt.+ .'.=ttl•erCYs-rh tefitiS eel,i stb tatuie setatxa-£Mace I. {4e11G8_ 1:311 PM '=VA' /'v u,!'tt,2q_Le( Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC i Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 1 of 78 SEISMIC FORCES (ASCE7) SEISMIC FORCES(ASCE 7-10) Tedds calculation version 3.0.10 Site parameters Site class D Mapped acceleration parameters (Section 11.4.1) at short period Ss=0.963 at 1 sec period Si =0.423 Site coefficientat short period (Table 11.4-1) Fa= 1.115 at 1 sec period (Table 11.4-2) = 1.577 Spectral response acceleration parameters at short period(Eq. 11.4-1) Sims=Fa x Ss= 1.074 at 1 sec period (Eq. 11.4-2) SM1 = F„x Si =0.667 Design spectral acceleration parameters(Sect 11.4.4) at short period (Eq. 11.4-3) Sas= 2/3 x SMs=0.716 at 1 sec period (Eq. 11.4-4) Sol =2/3 x SM1 =0.445 Seismic design category Risk category(Table 1.5-1) II Seismic design category based on short period response acceleration (Table 11.6-1) D Seismic design category based on 1 sec period response acceleration (Table 11.6-2) D Seismic design category D Approximate fundamental period Height above base to highest level of building hn= 17 ft From Table 12.8-2: Structure type All other systems Building period parameter Ct Ct=0.02 Building period parameter x x=0.75 Approximate fundamental period (Eq 12.8-7) Ta=Ct x (hn)0 x 1sec/(1ft)x=0.167 sec Building fundamental period (Sect 12.8.2) T=Ta=0.167 sec Long-period transition period Tt_= 16 sec Seismic response coefficient Seismic force-resisting system(Table 12.2-1) A. Bearing_Wall_Systems 15. Light-frame(wood)walls sheathed with wood structural panels Response modification factor(Table 12.2-1) R=6_5 f' V_ %�t «L�rt:tp 1'�'f_" I Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 2 of 78 Seismic importance factor(Table 1.5-2) le= 1.000 Seismic response coefficient(Sect 12.8.1.1) Calculated (Eq 12.8-2) C5_calc=SDS/(R/le)=0.1101 Maximum (Eq 12.8-3) Cs_max=SD1 /(T x (R/le)) =0.4086 Minimum (Eq 12.8-5) Cs_nsn= max(0.044 x SDS x le,0.01) =0.0315 Seismic response coefficient Cs=0.1101 Seismic base shear(Sect 12.8.1) Effective seismic weight of the structure W= 12.8 kips Seismic response coefficient Cs=0.1101 Seismic base shear(Eq 12.8-1) V= Cs x W= 1_4 kips Vertical distribution of seismic forces(Sect 12.8.3) Vertical distribution factor(Eq 12.8-12) Cvx=wx x hxk/E(w; x h;k) Lateral force induced at level i(Eq 12.8-11) Fx=Cvx x V Vertical force distribution table Portion of Distribution effective Height from exponent Vertical Lateral force Level base to Level i seismic weight related to distribution induced at hx assigned to (ft), Level i(kips), building period, factor, Cvx Level i (kips), Fx wx 1 8.0 8.3 1.00 0.465 0.7 2 17.0 4.5 1.00 0.535 0.8 • _ ___ Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 3 of 78 Design wind force: vuit= 135 mph Vasd=Vult*il(0.6)=104.571 mph WIND LOADING (ASCE7-10) WIND LOADING(ASCE7-10) in accordance with ASCE7-10 incorporating Errata No. 1 and Errata No.2 Using the directional design method Tedds calculation version 2.0.20 _V_ N 12ft-- -NI. 14 46.1 ft 101 Plan Elevation Building data Type of roof Gable Length of building b= 12.00 ft Width of building d =46.08 ft Height to eaves H = 17.00 ft Pitch of roof ao= 18.9 deg Mean height h =20.94 ft General wind load requirements Basic wind speed V= 105.0 mph Risk category II /'�'f' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)7374344 By R.P. Date 5/7/2018 Page 4 of 78 Velocity pressure exponent coeff(Table 26.6-1) Kd=0.85 Exposure category(c1.26.7.3) B Enclosure classification (c1.26.10) Enclosed buildings Internal pressure coef+ve(Table 26.11-1) GCp;_p=0.18 Internal pressure coef-ve(Table 26.11-1) GCp;_fl=-0.18 Gust effect factor Gf=0.85 Topography Topography factor not significant Kzt= 1.0 Velocity pressure equation q =0.00256 x KZ x KZt x Kd x V2 x 1psf/mph2 Velocity pressures table z(ft) K (Table 27.3-1) qz(psi) 15.00 0.57 13.67 15.00 0.57 13.67 17.00 0.59 14.15 20.94 0.63 15.06 24.89 0.66 15.81 Peak velocity pressure for internal pressure Peak velocity pressure-internal (as roof press.) qi= 15.06 psf Pressures and forces Net pressure p=q x Gf x Cpe-qi x GCp; Net force Fw= p x Aref Roof load case 1 -Wind 0, GCp; 0.18, -cpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref FW (ft) (psi) (psf) (ft2) (kips) A(-ve) 20.94 -0.44 15.06 -8.39 292.26 -2.45 B (-ve) 20.94 -0.58 15.06 -10.11 292.26 -2.95 Total vertical net force Fw,v= -5.11 kips Total horizontal net force Fw,h=0.16 kips Walls load case 1 -Wind 0,GCp;0.18, -cpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psi) (psi) (ft2) (kips) Ai 15.00 0.80 13.67 6.59 180.00 1.19 A2 17.00 0.80 14.15 6.91 24.00 0.17 B 20.94 -0.21 15.06 -5.37 204.00 -1.10 C 20.94 -0.70 15.06 -11.67 965.19 -11.26 D 20.94 -0.70 15.06 -11.67 965.19 -11.26 Overall loading Projected vertical plan area of wall Avert w o= b x H =204.00 ft2 - ? Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC I Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 1 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 5 of 78 Projected vertical area of roof Avert_r_o= b x d/2 x tan(ao) =94.67 ft2 Minimum overall horizontal loading Fw,total_min=Pmin_w x Avert_w_o+pmin_r x Avert_r_o=4.02 kips Leeward net force F=FW,we=-1.1 kips Windward net force Fw= Fw,wA_l + Fw,wA_2= 114 kips Overall horizontal loading Fw,tota =max(Fw- Fi+ Fw,h, Fw,total_min) =4.0 kips Roof load case 2-Wind 0,GCpi-0.18, -Ocpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psi) (psi) (ft2) (kips) A(+ve) 20.94 0.00 15.06 2.66 292.26 0.78 B(+ve) 20.94 -0.58 15.06 -4.69 292.26 -1.37 Total vertical net force Fw,v=-0.56 kips Total horizontal net force Fw,h=0.70 kips Walls load case 2-Wind 0, GCp;-0.18,-Ocpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psi) (psi) (ft2) (kips) At 15.00 0.80 13.67 12.01 180.00 2.16 A2 17.00 0.80 14.15 12.33 24.00 0.30 B 20.94 -0.21 15.06 0.05 204.00 0.01 C 20.94 -0.70 15.06 -6.25 965.19 -6.03 D 20.94 -0.70 15.06 -6.25 965.19 -6.03 Overall loading Projected vertical plan area of wall Avert_w_o=b x H =204.00 ft2 Projected vertical area of roof Avert_r_o=b x d/2 x tan(ao)=94.67 ft2 Minimum overall horizontal loading Fw,total_min=pmin_w x Avert_w_o+pmin_r x Avert_r_o=4.02 kips Leeward net force Fi=Fw,wB=000 kips Windward net force Fw= Fw,wA_l + Fw,wA_2=2.5 kips Overall horizontal loading Fw,total=max(Fw-Fi+ Fw,h, Fw,total_min) =4_0 kips Roof load case 3 -Wind 90, GCp;0.18, -cpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psi) (psi) (ft2) (kips) A(-ve) 20.94 -1.13 15.06 -17.17 510.10 -8.76 B(-ve) 20.94 -0.70 15.06 -11.67 74.42 -0.87 Total vertical net force Fw,v=-9.11 kips Total horizontal net force Fw,h=0.00 kips Walls load case 3-Wind 90,GCp;0.18,-cpe �, Vr y f r r ,'r_ Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P, Date 5/7/2018 Page 6 of 78 Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psf) (psf) (ft2) (kips) Ai 15.00 0.80 13.67 6.59 691.25 4.55 A2 15.00 0.80 13.67 6.59 0.00 0.00 A3 24.89 0.80 15.81 8.04 273.97 2.20 B 20.94 -0.50 15.06 -9.11 965.19 -8.79 C 20.94 -0.70 15.06 -11.67 204.00 -2.38 D 20.94 -0.70 15.06 -11.67 204.00 -2.38 Overall loading Projected vertical plan area of wall Avert_w_90=d x H +d2 x tan(ao)/4=965.19 ft2 Projected vertical area of roof Avert_Loo 90=0.00 ft2 Minimum overall horizontal loading Fw,totai min=Pmin_w x Avert_w_90+pmin_r x Avert_r_90=15.44 kips Leeward net force FI=Fw,wB= -8.8 kips Windward net force Fw= Fw,WA_1 +Fw,wA 2+ Fw,wA_3=668 kips Overall horizontal loading Fw,total=max(Fw- FI + Fw,h, Fw,total_min) = 15.5 kips Roof load case 4-Wind 90, GCp;-0.18,+cpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psf) (psf) (ft2) (kips) A(+ve) 20.94 -0.18 15.06 0.41 510.10 0.21 B (+ve) 20.94 -0.18 15.06 0.41 74.42 0.03 Total vertical net force Fw,v=0.22 kips Total horizontal net force Fw,h=0.00 kips Walls load case 4-Wind 90, GCpi-0.18, +cpe Ref. Ext pressure Peak velocity Net pressure Area Net force Zone height coefficient cpe pressure qp p Aref Fw (ft) (psf) (psf) (ft2) (kips) Ai 15.00 0.80 13.67 12.01 691.25 8.30 A2 15.00 0.80 13.67 12.01 0.00 0.00 A3 24.89 0.80 15.81 13.46 273.97 3.69 B 20.94 -0.50 15.06 -3.69 965.19 -3.56 C 20.94 -0.70 15.06 -6.25 204.00 -1.27 D 20.94 -0.70 15.06 -6.25 204.00 -1.27 Overall loading - -- -- Projected vertical plan area of wall Avert_w_90=d x H +d2 x tan(a0)/4 =965.19 ft2 Projected vertical area of roof Avert r 90=0.00 ft2 Minimum overall horizontal loading Fw,total_min=pmin_w x Avert_w_90+pmin_r x Avert_r_90= 15.44 kips Leeward net force Ft=Fw,WB= -3.6 kips Windward net force Fw= Fw,wA_1 + Fw,WA_2+ Fw,wA_3= 12.0 kips Project New Addition Job# 180145 Address 11958 SW 125th Ct., Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr,#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 7 of 78 Overall horizontal loading max(Fw-Fi+Fw,h, Fw,totai_m;n)=115.5 kips n• F. { a } B C r q r Y_ 46.1 46.1 ft ►i 4. 12 ft ► 4 -12 ft--► Side face Windward face Leeward face „ ( ,2 /'f Project New Addition Job# 180145 f Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC 1 Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 8 of 78 yrl------12A--- -' S i r � E ft* Mat A-. N— — 46-ft r 1 Ems` I 12 ft 'f -------_ ._-__-_.--- -------46.1 ft Side face Leeward face 'r.}iutti,j%'/'�' Project New Addition Job# 180145 Address 11958 SW 125th CL,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 9 of 78 Resolution of reaction R4 into RI & R2: R4 reactions: R4_wind= .756 kips R4_seismic= .246 kips Distance to centroid of structure: d =6 ft Length of structure: L=28.75 ft Moment about centroid: Mcentroid_wind= R4_wind*d=4.536 kip_ft Mcentroid_seismic= R4_seismic*d=1.476 kip_ft Reactions at Ri &R2: Rwind= Mcentroid_wind/L=0.158 kips Rseismic=Mcentroid_seismic/L=0.051 kips WOOD SHEAR WALL R1 DESIGN (NDS) WOOD SHEAR WALL DESIGN (NDS) In accordance with NDS2015 allowable stress design and the segmented shear wall method Tedds calculation version 1.2.02 Panel details Structural wood panel sheathing on one side Panel height h=8 ft Panel length b= 12 ft Total area of wall A=96 ft2 D+S W+Eq it i 11 i �i ill i 1 II TI-II i� I I I it ( I II ! 1 II II I I II I! Ii! II Ii II I II I II III III I I ( i 'I I' 'I i t I I I 'I 11 itif I II II II I Ii 56 il 1 II �� ;I II II II If ISI II I Ii I �I I II I �. , II i1 II II I� Ij I i II ! . IIS Ch1 it ,I I1 I I 11 I, n h ♦_ li 11 L II 11 IL 11IC 2 a i 2- V' V• O O A ---- 12' cRSV�S 14*,,,ftiny i"e Project New Addition Job# 180145 it I P7177,477M:1:4.4.1471 I ; Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 10 of 78 Panel construction Nominal stud size 2"x 6" Dressed stud size 1.5"x 5.5" Area of studs As=8.25 in2 Stud spacing s= 16 in Nominal end post size 2 x 2"x 6" Dressed end post size 2 x 1.5"x 5.5" Area of end posts Ake= 16.5 in2 Hole diameter Dia =0.625 in Net area of end posts Aen= 14.625 in2 Nominal collector size 2 x 2"x 6" Dressed collector size 2 x 1.5"x 5.5" Service condition Dry Temperature 100 degF or less Anchor stiffness ka=30000 lb/in From NDS Supplement Table 4A-Reference design values for visually graded dimension lumber(2"-4"thick) Species and grade Douglas Fir-Larch, stud grade, 2"&wider Specific gravity G =0.50 Tension parallel to grain Ft=450 lb/in2 Compression parallel to grain Fc=850 lb/in2 Modulus of elasticity E = 1400000 lb/in2 Minimum modulus of elasticity Emtn=510000 lb/in2 Sheathing details Sheathing material 15/32"wood panel oriented strandboard sheathing Fastener type 8d common nails at 6"centers From SDPWS Table 4.3A Nominal Unit Shear Capacities for Wood-Frame Shear Walls-Wood-based Panels Nom.seismic shear capacity vs=520 lb/ft Nom.wind shear capacity vW=730 lb/ft Apparent shear stiffness Ga= 13 kips/in Loading details Dead load D=247.5 lb/ft Snow load S =412.5 lb/ft Self weight of panel SWt= 12 lb/ft2 In plane wind load W=1465 lbs Service wind load factor fwsery= 1.00 In plane seismic load Eq=297 lbs SDs=0/16 From ASCE 7-10-c1.2.4.1 Basic combinations Load combination no.1 D +0.6W Load combination no.2 D +0.7E Load combination no.3 D +0.75Lf+0.45W+0.75(Lr or S or R) Load combination no.4 D +0.75Lf+0.525E+ 0.75S Load combination no.5 0.6D+0.6W Load combination no.6 0.6D+ 0.7E Adjustment factors Load duration factor Co= 1.60 Size factor tension CFt= 1.00 Size factor compression CFc= 1.00 Wet service factor tension CMt= 1.00 Temp.factor tension Ctt= 1.00 Wet serv.factor compression Civic= 1.00 Temp.factor compression Ctc= 1.00 Wet serv.factor mod.elasticity CME= 1.00 Temp.factor mod.of elasticity CtE= 1.00 Incising factor G= 1.00 Buckling stiffness factor CT= 1.00 Column stability factor CP=0.70 , Project New Addition Job# 180145 r Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 11 of 78 From SDPWS Table 4.3.4 Maximum Shear Wall Aspect Ratios Max.shear wall aspect ratio 3.5 Shear wall length b=12 ft Shear wall aspect ratio h/b=0.667 Segmented shear wall capacity Max.wind shear force Vw_max=0.879 kips Wind shear capacity Vw=4.38 kips Vwmax/Vw=0.201 PASS-Shear capacity for wind load exceeds maximum shear force Max.seismic shear force Vs_max=0.208 kips Seismic shear capacity Vs=3.12 kips Vs max I Vs=0.067 PASS-Shear capacity for seismic load exceeds maximum shear force Chord capacity for chords 1 and 2 Shear wall aspect ratio h/b=0.667 Shear force for tension V=0.879 kips Maximum tensile force T=0.449 kips Max.applied tensile stress ft=31 lb/in2 Design tensile stress Ft=720 lb/in2 ft/Ft'=0.043 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force,compression V=0.879 kips Max. compressive force C=0.815 kips Max. applied comp.stress fo=49 lb/in2 Design compressive stress Fc'=945 lb/in2 fo/Fc'=0.052 PASS-Design compressive stress exceeds maximum applied compressive stress Hold down force Chord 1 T, =0.449 kips Chord 2 T2=0.449 kips Wind load deflection Shear wall deflection 6sww=0.096 in Deflection limit Aw_allow=0.24 in 6sww/Aw allow=0.398 PASS-Shear wall deflection is less than deflection limit Seismic load deflection Shear wall deflection fisws=0.071 in Deflection limit ds_anow=2.4 in fisws/As avow=0.029 PASS-Shear wall deflection is less than deflection limit Use Simpson MSTC28 holdown device '( Project New Addition Job# 180145 '17', Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 12 of 78 WOOD SHEAR WALL R2 DESIGN (NDS) WOOD SHEAR WALL DESIGN (NDS) In accordance with NDS2015 allowable stress design and the segmented shear wall method Tedds calculation version 1.2.02 Panel details Structural wood panel sheathing on one side Panel height h =8 ft Panel length b= 12 ft Total area of wall A=96 ft2 D+S ♦VVVi ii***+++++i+iii ►++i W+E a si ;�Hi rI 11,1 „, , 1 �� 1T r 111 11 r 1, , i 1'i , 1 I ,i, , , , H ,I �, , 11 I1 li , ,, 1, I , , 1 1,1 E. 1, !j i 1 ) I 11 ! 1 , I 1 i , � �1I 11 ii ,1 1 11 ,I I,I II I II � �I i, , , 1 ii 1 it II 1, ,, ,i 1 ,, „ „ I I , ii I, I I! I , I _,', tali r! li I it 1 j :Ch2 • _Q a Y • Y T T i O O s 12' s Panel construction Nominal stud size 2”x 6" Dressed stud size 1.5"x 5.5" Area of studs A5=8.25 in2 Stud spacing s= 16 in Nominal end post size 2 x 2"x 6" Dressed end post size 2 x 1.5"x 5.5" Area of end posts Ae= 16.5 in2 Hole diameter Dia=0.625 in Net area of end posts Aen= 14.625 in2 Nominal collector size 2 x 2"x 6" Dressed collector size 2 x 1.5"x 5.5" Service condition Dry Temperature 100 degF or less Anchor stiffness ka=30000 lb/in From NDS Supplement Table 4A-Reference design values for visually graded dimension lumber(2"-4"thick) Species and grade Douglas Fir-Larch, stud grade, 2"&wider Specific gravity G =0.50 Tension parallel to grain Ft=450 lb/in2 Compression parallel to grain Fc=850 lb/in2 1 „ -.:6)-_,-,---TS / ,,,Bing ILLY Project New Addition Job# 180145 ,t7274 Address 11958 SW 125th Ct., Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 - Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 13 of 78 Modulus of elasticity E= 1400000 lb/in2 Minimum modulus of elasticity Emin=510000 lb/in2 Sheathing details Sheathing material 15/32"wood panel oriented strandboard sheathing Fastener type 8d common nails at 6"centers From SDPWS Table 4.3A Nominal Unit Shear Capacities for Wood-Frame Shear Walls -Wood-based Panels Nom.seismic shear capacity vs=520 lb/ft Nom.wind shear capacity vw=730 lb/ft Apparent shear stiffness Ga=13 kips/in Loading details Dead load D=215.62 lb/ft Snow load S=359.375 lb/ft Self weight of panel Swt= 12 lb/ft2 In plane wind load W=1465 lbs Service wind load factor fwsery= 1.00 In plane seismic load Eq=297 lbs SDs=0.716 From ASCE 7-10 -c1.2.4.1 Basic combinations Load combination no.1 D+0.6W Load combination no.2 D+0.7E Load combination no.3 D+0.75Lf+0.45W+0.75(L1 or S or R) Load combination no.4 D+0.75Lt+0.525E+0.75S Load combination no.5 0.6D+ 0.6W Load combination no.6 0.6D+0.7E Adjustment factors Load duration factor CD=1.60 Size factor tension CF,= 1.00 Size factor compression CFC= 1.00 Wet service factor tension CM,= 1.00 Temp.factor tension Ctt= 1.00 Wet serv.factor compression Civic= 1.00 Temp.factor compression Ctc= 1M0 Wet serv.factor mod.elasticity CME= 1.00 Temp.factor mod.of elasticity C,E= 1.00 Incising factor CI=1.00 Buckling stiffness factor CT= 1.00 Column stability factor CP=0.70 From SDPWS Table 4.3.4 Maximum Shear Wail Aspect Ratios Max.shear wall aspect ratio 3.5 Shear wall length b= 12 ft Shear wall aspect ratio h/b=0.667 Segmented shear wall capacity Max.wind shear force Vw_max=0.879 kips Wind shear capacity Vw=4.38 kips Vw max/Vw=0.201 PASS-Shear capacity for wind load exceeds maximum shear force Max.seismic shear force Vs_max=0.208 kips Seismic shear capacity Vs=3.12 kips Vs_max/Vs=0.067 PASS-Shear capacity for seismic load exceeds maximum shear force Chord capacity for chords 1 and 2 Shear wall aspect ratio h/b=0.667 Project New Addition Job# 180145 • Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 14 of 78 Shear force for tension V=0.879 kips Maximum tensile force T=0.461 kips Max.applied tensile stress ft=32 lb/in2 Design tensile stress Ft'=720 lb/in2 ft/Ft'=0.044 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force,compression V=0.879 kips Max. compressive force C=0.794 kips Max. applied comp.stress fc=48 lb/in2 Design compressive stress Fe'=945 lb/in2 fo/Fc'=0.051 PASS-Design compressive stress exceeds maximum applied compressive stress Hold down force Chord 1 Ti =0.461 kips Chord 2 T2=0.461 kips Wind load deflection Shear wall deflection 6sww=0.096 in Deflection limit ow aVfow=0.24 in bsww/Aw allow=0.399 PASS-Shear wall deflection is less than deflection limit Seismic load deflection Shear wall deflection bsws=0.072 in Deflection limit As_allow=2.4 in bsws/As_allow=0.03 PASS-Shear wall deflection is less than deflection limit Use Simpson MSTC28 holdown device ) Project New Addition Job# 180145 7 e:Wi. Address 11958 SW 125th Ct., Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 15 of 78 I WOOD SHEAR WALL R3 DESIGN (NDS) WOOD SHEAR WALL DESIGN (NDS) In accordance with NDS2015 allowable stress design and the segmented shear wall method Tedds calculation version 1.2.02 Panel details Structural wood panel sheathing on one side Panel height h=8 ft Panel length b=28.75 ft " D+S V*VVVi4"'VVVV*V*V***ii++i�V***V*VV*Viii+Vi+4+Vi+i44ili"i *V W+E { in1u1r1 'W f4 jr II �I 11 I''_I _ — 11I1 1 , 1I11 4'11111. II I 1 I 1 11 I I 1 11 1111 itII h 11 m 1 k� II II i li 11I 11 Ii 1 ' m 11 1 11 WTI i I 11 11 1 _ __ 1� I I (�I I 1 1 1 11 I• 11 1 it 11 11 r ==r1; i 11 (I 1111'1 11,1 H I ;I 11 I� 1 '1 III 1 x;11 1 11 Ch1 : r8: I Chs `-- �H6 _L_ 11 il_ I ..il. _. t14 I 1L11 Ch10 1 I_x 12 ia O p p 't -3'----P `,T,A--2'-_04...__4'1.004"----♦ i-.--- ---6'4.996"-- ----_it.. ..._.-9'-- -----._..... -M2'5.004"P of Panel opening details Width of opening wo, =0_5 ft Height of opening ho, =7 ft Height to underside of lintel 101 =7 ft Position of opening Poi =3 ft Width of opening wog=2 ft Height of opening hoz=6_5 ft Height to underside of lintel 102=7 ft Position of opening P02=4.333 ft Width of opening Wo3=0.5 ft Height of opening h03=7 ft Height to underside of lintel 103=7 ft Position of opening P03= 10.417 ft Width of opening W04=9 ft Height of opening h04=4.5 ft Height to underside of lintel 104=7 ft Position of opening PO4= 17.333 ft Total area of wall A= 169.5 ft2 Panel construction Nominal stud size 2"x 6" Dressed stud size 1.5"x 5.5" Area of studs A5=8.25 in2 Stud spacing s= 16 in Nominal end post size 2 x 2"x 6" Dressed end post size 2 x 1.5"x 5.5" Area of end posts Ae= 16.5 in2 Hole diameter Dia=0.625 in Net area of end posts Aen= 14.625 in2 Nominal collector size 2 x 2"x 6" Dressed collector size 2 x 1.5"x 5.5" Service condition Dry Temperature 100 degF or less Anchor stiffness ka=30000 lb/in � 1 cleG'v 1 , t tl2y /'/'C' Project New Addition Job# 180145 F Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 i By R.P. Date 5/7/2018 Page 16 - of 78 From NDS Supplement Table 4A-Reference design values for visually graded dimension lumber(2"-4"thick) Species and grade Douglas Fir-Larch, stud grade,2"&wider Specific gravity G=0.50 Tension parallel to grain Ft=450 lb/in2 Compression parallel to grain Fc=850 lb/in2 Modulus of elasticity E= 1400000 lb/in2 Minimum modulus of elasticity Emin=510000 lb/in2 Sheathing details Sheathing material 15/32"wood panel oriented strandboard sheathing Fastener type 8d common nails at 6"centers From SDPWS Table 4.3A Nominal Unit Shear Capacities for Wood-Frame Shear Walls -Wood-based Panels Nom.seismic shear capacity vs=520 lb/ft Nom.wind shear capacity vw=730 lb/ft Apparent shear stiffness Ga= 13 kips/in Loading details Dead load D=30 lb/ft Snow load S=50 lb/ft Self weight of panel SWt= 12 lb/ft2 In plane wind load W=756 lbs Service wind load factor fWsery= 1.00 In plane seismic load Eq=246 lbs SDs=0.716 From ASCE 7-10-c1.2.4.1 Basic combinations Load combination no.1 D+0.6W Load combination no.2 D+0.7E Load combination no.3 D+0.75Lf+0.45W+0.75(Lr or S or R) Load combination no.4 D+0.75Lf+0.525E +0.75S Load combination no.5 0.6D+0.6W Load combination no.6 0.6D+0.7E Adjustment factors Load duration factor CD= 1.60 Size factor tension CR= 1.00 Size factor compression CFC= 1.00 Wet service factor tension CMt= 1.00 Temp.factor tension Ctt= 1.00 Wet serv.factor compression Cmc= 1.00 Temp.factor compression Cte= 1.00 Wet serv.factor mod.elasticity CME= 1.00 Temp.factor mod.of elasticity CtE= 1.00 Incising factor C;=1.00 Buckling stiffness factor Cr= 1.00 Column stability factor CP=0.70 From SDPWS Table 4.3.4 Maximum Shear Wall Aspect Ratios Max.shear wall aspect ratio 3.5 Segment 1 wall length bi =3 ft Shear wall aspect ratio h/131 =2.667 Segment 2 wall length b2=0.833 ft Shear wall aspect ratio h/b2=9.604 Segment 3 wall length b3=4.084 ft Shear wall aspect ratio h/b3= 1.959 Segment 4 wall length b4=6.416 ft Shear wall aspect ratio h/b4= 1.247 Segment 5 wall length b5=2.417 ft Shear wall aspect ratio h/b5=3.31 Segmented shear wall capacity-Equal deflection method Wind loading: Y i Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 17 of 78 Segment 4 stiffness Ica=6.577 kips/in Shear capacity,widest seg. vswwa=365 plf Deflection at capacity load Scap=0.356 in Seg. 1 stiffness k1 =2.165 kips/in Seg. 1 shear at Scap Vdswwl =256.97 plf Seg 1 shear capacity Vsww1 =334.58 plf Vdswwl /Vswwl =0.768 PASS-Segment shear capacity exceeds segment unit shear at(Cap Seg. 3 stiffness k3=3.457 kips/in Seg. 3 shear at Scap Vdsww3=301.44 plf Seg 3 shear capacity vsww3=365 plf Vdsww3/Vsww3=0.826 PASS-Segment shear capacity exceeds segment unit shear at Scap Seg. 5 stiffness k5=1.538 kips/in Seg. 5 shear at Scap Vdsww5=226.59 plf Seg 5 shear capacity vsww5=305.24 plf vdsww5/Vsww5=0.742 PASS-Segment shear capacity exceeds segment unit shear at&ap Max.wind shear force Vw_max=0.454 kips Wind shear capacity Vw=4.892 kips Vw_max/Vw=0.093 PASS-Shear capacity for wind load exceeds maximum shear force Seismic loading: Segment 4 stiffness ka=6.577 kips/in Shear capacity,widest seg. vswsa=260 plf Deflection at capacity load Scap=0.254 in Seg. 1 stiffness k, =2.165 kips/in Seg. 1 shear at Scap vdswsl = 183.05 plf Seg 1 shear capacity vswsi =238.33 plf Vdswsi /Vsws1 =0.768 PASS-Segment shear capacity exceeds segment unit shear at Scap Seg. 3 stiffness k3=3.457 kips/in Seg. 3 shear at Scap vdsws3=214.72 plf Seg 3 shear capacity Vsws3=260 plf Vdsws3/Vsws3=0.826 PASS-Segment shear capacity exceeds segment unit shear at Scap• Seg. 5 stiffness k5=1.538 kips/in Seg. 5 shear at Scap Vdswa5= 161.41 plf Seg 5 shear capacity Vsws3=217.43 plf vdsws5/Vsws3=0.742 PASS-Segment shear capacity exceeds segment unit shear at Scap Max.seismic shear force Vs_max=0.172 kips Seismic shear capacity Vs=3.484 kips Vs_max/Vs=0.049 PASS-Shear capacity for seismic load exceeds maximum shear force Chord capacity for chords 1 and 2 Shear wall aspect ratio h/bi =2.667 Shear force for tension V=0.454 kips Maximum tensile force T=0.140 kips Max.applied tensile stress ft= 10 lb/in2 Design tensile stress Ft' =720 lb/in2 ft/Ft=0.013 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=0.454 kips Max. compressive force C=0.275 kips Max. applied comp.stress fc=17 lb/in2 Design compressive stress Fc'=945 lb/in2 ff/Fc'=0.018 PASS-Design compressive stress exceeds maximum applied compressive stress Chord capacity for chords 3 and 4 Shear wall aspect ratio h/b2=9.604 Segment not considered,shear wall aspect ratio exceeds maximum allowable. t' _t'x I Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC i Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P, Date 5/7/2018 Page 18 of 78 Chord capacity for chords 5 and 6 Shear wall aspect ratio h/b3= 1.959 Shear force for tension V=0.454 kips Maximum tensile force T=0.173 kips Max.applied tensile stress ft= 12 lb/in2 Design tensile stress Ft =720 lb/in2 ft/Ft=0.016 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=0.454 kips Max. compressive force C=0.308 kips Max. applied comp.stress fc= 19 lb/in2 Design compressive stress Fc'=945 lb/in2 fc/Fc'=0.020 PASS-Design compressive stress exceeds maximum applied compressive stress Chord capacity for chords 7 and 8 Shear wall aspect ratio h/b4= 1.247 Shear force for tension V=0.454 kips Maximum tensile force T=0.220 kips Max.applied tensile stress ft= 15 lb/in2 Design tensile stress Ft =720 lb/in2 ft/Ft=0.021 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=0.454 kips Max. compressive force C =0.355 kips Max. applied comp.stress fa=22 lb/in2 Design compressive stress Fc' =945 lb/in2 fc/Fa'=0.023 PASS-Design compressive stress exceeds maximum applied compressive stress Chord capacity for chords 9 and 10 Shear wall aspect ratio h/b5=3.31 Shear force for tension V=0.454 kips Maximum tensile force T=0.118 kips Max.applied tensile stress ft=8 lb/in2 Design tensile stress Ft'=720 lb/in2 ft/Ft=0.011 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=0.454 kips Max. compressive force C=0.252 kips Max. applied comp.stress fc= 15 lb/in2 Design compressive stress Fa'=945 lb/in2 fc/Fc'=0.016 PASS-Design compressive stress exceeds maximum applied compressive stress 1 Collector capacity Maximum shear force Vmax=0.454 kips Max. force in collector Pcaii=0.129 kips Max.applied tensile stress ft=8 lb/in2 Design tensile stress Ft =720 lb/in2 ft/Ft=0.011 PASS-Design tensile stress exceeds maximum applied tensile stress Max.applied comp.stress fa=8 lb/in2 Design compressive stress Fd= 1360 lb/in2 fa/Fc'=0.006 PASS-Design compressive stress exceeds maximum applied compressive stress Hold down force Chord 1 Tt =0.14 kips Chord2 T2=0.14 kips Chord 5 T5=0.173 kips Chord 6 Te=0.173 kips f ' t ' / si�Erta /'(Y' Project New Addition Job# 180145 Address 11958 SW 125th Ct., Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 t By R.P. Date 5/7/2018 Page 19 of 78 Chord 7 T7=0.22 kips Chord 8 Ta=0.22 kips Chord 9 T9=0.118 kips Chord 10 Tio=0.118 kips Wind load deflection Shear wall deflection 8sww=0.053 in Deflection limit Aw avow=0.24 in Ssww/Aw allow=0.221 PASS-Shear wall deflection is less than deflection limit Seismic load deflection Shear wall deflection 8sws=0.065 in Deflection limit As allow=2.4 in Ssws/As allow=0.027 PASS-Shear wall deflection is less than deflection limit Use Simpson MSTC28 holdown device Resolution of reaction R8 into R5 & R6: Re reactions: Rewind= 1.62 kips Re_seismic= .699 kips Distance to centroid of structure: d =6 ft Length of structure: L=46.0833 ft Moment about centroid: Mcentroid_wind=R8_wind*d=9.720 kip_ft Mcentroid_seismic= Re_seismic*d =4.194 kip_ft Reactions at R5&Re: Rwind= Mcentroid_wind/L=0.211 kips Rseismic=Mcentroid_seismic/L=0.091 kips Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 20 of 78 WOOD SHEAR WALL R5 DESIGN S GN (NDS) WOOD SHEAR WALL DESIGN (NDS) In accordance with NDS2015 allowable stress design and the segmented shear wall method Tedds calculation version 1.2.02 Panel details Structural wood panel sheathing on one side Panel height h =8.25 ft Panel length b=8 ft Total area of wall A=66 ft2 D+S W+Eq 4- --_- ► �s1 T -. i— — 11 1- IL,� il I II ,, ii 1 IIi I 1 i II I ba j ! i i �( It it iII it c I� I�1 i I 11 !II I .chi i-------, ---__1 ----- 11 I sCh2 — — = a CO a Y ..Y O Q 0) N N - 8' —-- —l^i Panel construction Nominal stud size 2"x 6" Dressed stud size 1.5"x 5.5" Area of studs AS=8.25 in2 Stud spacing s= 16 in Nominal end post size 2 x 2"x 6" Dressed end post size 2 x 1.5"x 5.5" Area of end posts Ae= 16.5 in2 Hole diameter Dia=0.625 in Net area of end posts Aen=14.625 in2 Nominal collector size 2 x 2"x 6" Dressed collector size 2 x 1.5"x 5.5" Service condition Dry Temperature 100 degF or less Anchor stiffness ka=30000 lb/in From NDS Supplement Table 4A-Reference design values for visually graded dimension lumber(2" -4"thick) Species and grade Douglas Fir-Larch, stud grade,2"&wider Specific gravity G =0.50 Tension parallel to grain Ft=450 lb/in2 Compression parallel to grain Fc=850 lb/in2 -�`- /7 }Ming Pi r' Project New Addition Job# 180145 j ,# Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 21 of 78 Modulus of elasticity E= 1400000 lb/in2 Minimum modulus of elasticity Emm= 510000 lb/in2 Sheathing details Sheathing material 15/32"wood panel oriented strandboard sheathing Fastener type 8d common nails at 4"centers From SDPWS Table 4.3A Nominal Unit Shear Capacities for Wood-Frame Shear Walls-Wood-based Panels Nom.seismic shear capacity vs=760 lb/ft Nom.wind shear capacity vw= 1065 lb/ft Apparent shear stiffness Ga=19 kips/in Loading details Dead load D= 130 lb/ft Snow load S =216.68 lb/ft Self weight of panel Swt= 12 lb/ft2 In plane wind load W=4836 lbs Service wind load factor fwsery= 1.00 In plane seismic load Eq=790 lbs SDs=0.716 From ASCE 7-10 -c1.2.4.1 Basic combinations Load combination no.1 D +0.6W Load combination no.2 D +0.7E Load combination no.3 D+0.75Lr+0.45W+0.75(Lr or S or R) Load combination no.4 D+0.75Lt+0.525E+0.75S Load combination no.5 0.6D+0.6W Load combination no.6 0.6D+0.7E Adjustment factors Load duration factor CD= 1.60 Size factor tension CFt= 1.00 Size factor compression CFc= 1.00 Wet service factor tension CMt= 1.00 Temp.factor tension Ott= 1.00 Wet serv.factor compression CMc= 1.00 Temp.factor compression Ctc= 1.00 Wet serv.factor mod.elasticity CME= 1.00 Temp.factor mod.of elasticity CtE= 1.00 Incising factor Ci=1.00 Buckling stiffness factor CT= 1.00 Column stability factor CP=0.67 From SDPWS Table 4.3.4 Maximum Shear Wall Aspect Ratios Max.shear wall aspect ratio 3.5 Shear wall length b=8 ft Shear wall aspect ratio h/b= 1.031 Segmented shear wall capacity Max.wind shear force Vw_max= 2.902 kips Wind shear capacity Vw=4.26 kips Vw max/Vw=0.681 PASS-Shear capacity for wind load exceeds maximum shear force Max.seismic shear force Vs_max=0.553 kips Seismic shear capacity Vs=3.04 kips Vs max/Vs=0.182 PASS-Shear capacity for seismic load exceeds maximum shear force Chord capacity for chords 1 and 2 Shear wall aspect ratio h/b= 1.031 . � '�v 1 na ttiii.=y 1_I I Project New Addition Job# 180145 it Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 < By R.P. Date 5/7/2018 Page 22 of 78 1 Shear force for tension V=2.902 kips Maximum tensile force T=2.901 kips Max.applied tensile stress ft= 198 lb/in2 Design tensile stress Ft' =720 lb/in2 ft/Ft'=0.275 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=2.902 kips Max. compressive force C =3.145 kips Max. applied comp.stress fc= 191 lb/in2 Design compressive stress Fe'=916 lb/in2 fo/ =0.208 PASS-Design compressive stress exceeds maximum applied compressive stress Hold down force Chord 1 Ti =2.901 kips Chord 2 T2=2.901 kips Wind load deflection Shear wail deflection 8sww=0.445 in Deflection limit Aw anew=0.495 in 8sww/Aw allow=0.9 PASS-Shear wall deflection is less than deflection limit Seismic load deflection Shear wall deflection Ssws=0.284 in Deflection limit As avow=2.475 in Ssws/As allow=0.115 PASS-Shear wall deflection is less than deflection limit Use Simpson MSTC48B3 holdown device /_ Project Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 I Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 23 of 78 WOOD SHEAR WALL R6 DESIGN (NDS) WOOD SHEAR WALL DESIGN (NDS) In accordance with NDS2015 allowable stress design and the segmented shear wall method Tedds calculation version 1.2.02 Panel details Structural wood panel sheathing on one side Panel height h=8 ft Panel length b= 12 ft D+S w+Eq Si r-- i 1Ir,- - 1 2 li tt III IG 1 =� j i II il) ! it I III Ij I1 (I!' it'lI 11 I ! I i) 1 I I 1I B 111 II )j II?j 1 IIIi 'i III II 11 I III IIIi-I .7- rl- r,-- ---(�j !; !I 11 I I �, ! I I 11 II ,'I SII I �i iI i� 11 �� II ! I ail Ill; ih I ,I 1,_ ;I II,� a 11l _Ti Ch1 II_Il. _.._11 Jh2 * li —_.11.1.r h3 IL J._Cho a a to 2- Coco coc CV N N N 'I 1- 4 4' 11.4 4' ._ . 4' Panel opening details Width of opening woi =4 ft Height of opening hot =4 ft Height to underside of lintel lot =7 ft Position of opening Poi =4 ft Total area of wall A=80 ft2 Panel construction Nominal stud size 2"x 6" Dressed stud size 1.5"x 5.5" Area of studs As=8.25 in2 Stud spacing s= 16 in Nominal end post size 2 x 2"x 6" Dressed end post size 2 x 1.5"x 5.5" Area of end posts As=16.5 in2 Hole diameter Dia=0.625 in Net area of end posts Aen= 14.625 in2 Nominal collector size 2 x 2"x 6" Dressed collector size 2 x 1.5"x 5.5" Service condition Dry Temperature 100 degF or less Anchor stiffness ka=30000 lb/in I; di)&V..S ,Oslictrtin9 Ye Project New Addition Job# 180145 142,E 1 ` Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 24 of 78 From NDS Supplement Table 4A-Reference design values for visually graded dimension lumber(2"-4"thick) Species and grade Douglas Fir-Larch, stud grade,2"&wider Specific gravity G =0.50 Tension parallel to grain Ft=450 lb/in2 Compression parallel to grain Fc=850 lb/in2 Modulus of elasticity E=1400000 lb/in2 Minimum modulus of elasticity Emin=510000 lb/in2 Sheathing details Sheathing material 15/32"wood panel oriented strandboard sheathing Fastener type 8d common nails at 2"centers From SDPWS Table 4.3A Nominal Unit Shear Capacities for Wood-Frame Shear Walls-Wood-based Panels Nom.seismic shear capacity vs= 1280 lb/ft Nom.wind shear capacity vw=1790 Ib/ft Apparent shear stiffness Ga=39 kips/in Loading details Dead load D=240 lb/ft Snow load S =359.38 lb/ft Self weight of panel SWt= 12 lb/ft2 In plane wind load W=4836 lbs Service wind load factor fwsery= 1.00 In plane seismic load Eq=790 lbs Sos=0.716 From ASCE 7-10-c1.2.4.1 Basic combinations Load combination no.1 D+0.6W Load combination no.2 D+0.7E Load combination no.3 D+0.75Lt+0.45W+ 0.75(Lr or S or R) Load combination no.4 D+0.75Lt+0.525E +0.75S Load combination no.5 0.6D+0.6W Load combination no.6 0.6D+0.7E Adjustment factors Load duration factor Co= 1.60 Size factor tension CFt= 1.00 Size factor compression CFC= 1.Q0 Wet service factor tension CMt=1.00 Temp.factor tension Ctt= 1.00 Wet serv.factor compression CMC= 1.00 Temp.factor compression Ctc= 1.00 Wet serv.factor mod.elasticity CME= 1.00 Temp.factor mod.of elasticity CtE= 1.00 Incising factor C; =1.00 Buckling stiffness factor CT= 1.00 Column stability factor CP=0.70 From SDPWS Table 4.3.4 Maximum Shear Wall Aspect Ratios Max.shear wall aspect ratio 3.5 Segment 1 wall length b, =4 ft Shear wall aspect ratio h/b, =2 Segment 2 wall length b2=4 ft Shear wall aspect ratio h/b2=2 Segmented shear wall capacity-Equal deflection method Wind loading: Segment 2 stiffness k2=5.11 kips/in Shear capacity,widest seg. Vsww2=895 plf Deflection at capacity load 6cap=0.701 in k' ` .+ Project New Addition Job# 180145 Address 11958 SW 125th Ct., Tigard,OR 97223 .RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 25 of 78 Seg. 1 stiffness ki =5.11 kips/in Seg. 1 shear at bcap vdswwi =895 plf Seg 1 shear capacity vswwi = 895 plf vdswwi /vswwi = 1.000 PASS-Segment shear capacity exceeds segment unit shear at kap Max.wind shear force Vw_max=2.902 kips Wind shear capacity Vw=7.16 kips Vw max/Vw=0.405 PASS-Shear capacity for wind load exceeds maximum shear force Seismic loading: Segment 2 stiffness k2=5.11 kips/in Shear capacity,widest seg. vsws2=640 plf Deflection at capacity load 6cap=0.501 in Seg. 1 stiffness ki =5.11 kips/in Seg. 1 shear at 6Cap Vdswsl =640 plf Seg 1 shear capacity vswsi =640 plf vdswsi /vswsi =1.000 PASS-Segment shear capacity exceeds segment unit shear at Soap Max.seismic shear force Vs_max=0.553 kips Seismic shear capacity Vs= 5.12 kips Vs_max/Vs=0.108 PASS-Shear capacity for seismic load exceeds maximum shear force Chord capacity for chords 1 and 2 Shear wall aspect ratio h/bi =2 Shear force for tension V=2.902 kips Maximum tensile force T=2.767 kips Max.applied tensile stress ft= 189 lb/in2 Design tensile stress Ft=720 lb/in2 ft/Ft=0.263 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=2.902 kips Max. compressive force C =3.126 kips Max. applied comp.stress fc=189 lb/in2 Design compressive stress Fc'=945 lb/in2 fc/Fc'=0.200 PASS-Design compressive stress exceeds maximum applied compressive stress Chord capacity for chords 3 and 4 Shear wall aspect ratio h/b2=2 Shear force for tension V=2.902 kips Maximum tensile force T=2.767 kips Max.applied tensile stress ft= 189 lb/in2 Design tensile stress Ft =720 lb/in2 ft/Ft =0.263 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force,compression V=2.902 kips Max. compressive force C =3.126 kips Max. applied comp.stress fc=189 lb/in2 Design compressive stress Fc'=945 lb/in2 fc/Fc'=0.200 PASS-Design compressive stress exceeds maximum applied compressive stress Collector capacity Maximum shear force Vmax=2.902 kips Max.force in collector Pccii=0.484 kips Max.applied tensile stress ft=29 lb/in2 Design tensile stress Ft'=720 lb/in2 ft/Ft' =0.041 PASS-Design tensile stress exceeds maximum applied tensile stress Max.applied comp.stress fc=29 lb/in2 Design compressive stress Fc'= 1360 lb/in2 fc/Fc'=0.022 PASS-Design compressive stress exceeds maximum applied compressive stress ti -AL 1'_; i t i` , ,'r Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 26 of 78 Hold down force Chord 1 Ti =2.767 kips Chord 2 T2=2.767 kips Chord 3 T3=2.767 kips Chord 4 T4=2.767 kips Wind load deflection Shear wall deflection bsww=0.464 in Deflection limit 4w_allow=0.48 in bsww/4w auow=0.967 PASS-Shear wall deflection is less than deflection limit Seismic load deflection Shear wall deflection bsws=0.282 in Deflection limit As allow=2.4 in bsws/4s allow=0.117 PASS-Shear wall deflection is less than deflection limit Use.Simpson HDU2-SDS2.5 holdown device c4,4 ftt :,9lltiProject New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 27 of 78 WOOD SHEAR WALL R7 DESIGN (NDS) WOOD SHEAR WALL DESIGN (NDS1 In accordance with NDS2015 allowable stress design and the segmented shear wall method Tedds calculation version 1.2.02 Panel details Structural wood panel sheathing on one side Panel height h=8 ft Panel length b=28.75 ft D+S i' 'trii"Vi**ii'*1+Y1iiVVIviIIVviVV*iVvi•V*VVVVOIYVvvVV VYiV*ii n+E . —R ; _ _ r_ _,7 -r = 1 T1 ' i2 Is3 r _ i 1,-,,.,L � -t . L -- R r ` , '2 i il i I I. I li I I I 11 iI� 1 11 III III I1 I II H. 11 I1 Id li I 1 i t I III 11 III 1I ,;11 I � Ii I, it i 11 fl' �o I 1, it .il, , .I, - 11 I I1 I 1 111111 1II !i III ! f t lilt I1 11 I I I,II 1 I I I1li 11 ! iiI II, to 1 SCh Cr2_� �Ch1 -- Ctr4} Nh . Il Ch6 Ch7 G h8J 1 1 T 1 0 I 1 1 I w 1 1 l a 1 Y l a I _ 1 ! y ,I M M V M N N d d O d O O _.1. 1'6" V-...2 6"--N-__. ._-____- --9'-._...__.-__._._. __.*A--2'6' .1 --_._4'9.504" ___---M-----5'3.996"-------s4--3'1.5"--N Panel opening details Width of opening woi =2.5 ft Height of opening hot =6.55 ft Height to underside of lintel lot =7 ft Position of opening Pot = 1.55 ft Width of opening W02=2®5 ft Height of opening h02=665 ft Height to underside of lintel 102=7 ft Position of opening Po2= 13 ft Width of opening w03=5.333 ft Height of opening ho3=7 ft Height to underside of lintel los=7 ft Position of opening P03=20.292 ft Total area of wall A= 160.169 ft2 Panel construction Nominal stud size 2"x 6" Dressed stud size 1.5"x 5.5" Area of studs As=8.25 in2 Stud spacing s= 16 in Nominal end post size 2 x 2"x 6" Dressed end post size 2 x 1.5"x 5.5" Area of end posts Ae=16.5 in2 Hole diameter Dia=0.625 in Net area of end posts Aen= 14.625 in2 Nominal collector size 2 x 2"x 6" Dressed collector size 2 x 1.5"x 5.5" Service condition Dry Temperature 100 degF or less Anchor stiffness ka=30000 lb/in From NDS Supplement Table 4A-Reference design values for visually graded dimension lumber(2" -4"thick) Species and grade Douglas Fir-Larch, stud grade, 2" &wider Specific gravity G=0.50 Tension parallel to grain Ft=450 lb/in2 Compression parallel to grain F0=850 lb/in2 '�1t C P flrq /' / ' !' Project New Addition Job# 180145 , 14 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC ! Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 4 1 Vancouver,WA 98684 (503)737-4344 ! By R.P. Date 5/7/2018 Page 28 of 78 Modulus of elasticity E= 1400000 lb/in2 Minimum modulus of elasticity Emin=510000 lb/in2 Sheathing details Sheathing material 15/32"wood panel oriented strandboard sheathing Fastener type 8d common nails at 6"centers From SDPWS Table 4.3A Nominal Unit Shear Capacities for Wood-Frame Shear Walls -Wood-based Panels Nom.seismic shear capacity vs=520 lb/ft Nom.wind shear capacity vw=730 lb/ft Apparent shear stiffness Ga= 13 kips/in Loading details Dead load D= 120 lb/ft Snow load S=50 lb/ft Self weight of panel SH,t= 12 lb/ft2 In plane wind load W= 1620 lbs Service wind load factor fwsery= 1.00 In plane seismic load Eq=699 lbs SDS=0.716 From ASCE 7-10-c1.2.4.1 Basic combinations Load combination no.1 D +0.6W Load combination no.2 D+0.7E Load combination no.3 D +0.75Lf+0.45W+0.75(Lr or S or R) Load combination no.4 D+0.75Lf+0.525E +0.75S Load combination no.5 0.6D+0.6W Load combination no.6 0.6D+0.7E Adjustment factors Load duration factor CD= 1.60 Size factor tension OFt= 1.00 Size factor compression CFC= 1.00 Wet service factor tension CMt=1.00 Temp.factor tension Ott=1.00 Wet serv.factor compression Cmc= 1.00 Temp.factor compression Ctc= 1.00 Wet serv.factor mod.elasticity CME= 1.00 Temp.factor mod.of elasticity CtE= 1.00 Incising factor Ci=1.00 Buckling stiffness factor CT= 1.00 Column stability factor CP=0.70 From SDPWS Table 4.3.4 Maximum Shear Wall Aspect Ratios Max.shear wall aspect ratio 3.5 Segment 1 wall length bi =1_5 ft Shear wall aspect ratio h/bi =5.333 Segment 2 wall length b2=9 ft Shear wall aspect ratio h/b2=0.889 Segment 3 wall length b3=4.792 ft Shear wall aspect ratio h/b3= 1.669 Segment 4 wall length b4=3.125 ft Shear wall aspect ratio h/b4=2.56 Segmented shear wall capacity-Equal deflection method Wind loading: Segment 2 stiffness k2=10.32 kipslin Shear capacity,widest seg. vsww2=365 plf Deflection at capacity load 6Cap=0.318 in Seg. 3 stiffness k3=4.366 kips/in Seg. 3 shear at 6cap vdsww3=290.03 plf Seg 3 shear capacity vsww3=365 plf Vdsww3/Vsww3=0.795 Project New Addition Job# 180145 a; Address 11958 SW 125th Ct., Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 29 of 78 PASS-Segment shear capacity exceeds segment unit shear at 8cap I Seg.4 stiffness Ica=2.307 kips/in Seg.4 shear at Scap Vdsww4=234.96 plf Seg 4 shear capacity vsww4=339.45 plf vdsww4/Vsww4=0.692 PASS-Segment shear capacity exceeds segment unit shear at*cap Max.wind shear force Vw_max=0.972 kips Wind shear capacity Vw=5.409 kips Vw_max/Vw=0.18 PASS-Shear capacity for wind load exceeds maximum shear force Seismic loading: Segment 2 stiffness k2=10.32 kips/in Shear capacity,widest seg. vsws2=260 plf Deflection at capacity load Scap=0.227 in Seg. 3 stiffness k3=4.366 kips/in Seg. 3 shear at 6cap Vdsws3=206.59 plf Seg 3 shear capacity vsws3=260 plf Vdsws3/Vsws3=0.795 PASS-Segment shear capacity exceeds segment unit shear at*cap Seg.4 stiffness k4=2.307 kips/in Seg.4 shear at kap vdsws4= 167.37 plf Seg 4 shear capacity Vsws4=241.8 plf Vdsws4/Vsws4=0.692 PASS-Segment shear capacity exceeds segment unit shear at*cap Max.seismic shear force Vs max=0.489 kips Seismic shear capacity Vs=3.853 kips Vs max/Vs=0.127 PASS-Shear capacity for seismic load exceeds maximum shear force Chord capacity for chords 1 and 2 Shear wall aspect ratio h/bi =5.333 Segment not considered,shear wall aspect ratio exceeds maximum allowable. Chord capacity for chords 3 and 4 Shear wall aspect ratio h/b2=0.889 Shear force for tension V=0.972 kips Maximum tensile force T=0.438 kips Max.applied tensile stress ft=30 lb/in2 Design tensile stress Ft' =720 lb/in2 ft/Ft'=0.042 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force,compression V=0.972 kips Max. compressive force C =0.669 kips Max. applied comp.stress fc=41 lb/in2 Design compressive stress Fc' =945 lb/in2 fc/Fc'=0.043 PASS-Design compressive stress exceeds maximum applied compressive stress Chord capacity for chords 5 and 6 Shear wall aspect ratio h/b3= 1.669 Shear force for tension V=0.972 kips Maximum tensile force T=0.331 kips Max.applied tensile stress ft=23 lb/in2 Design tensile stress Fi =720 lb/in2 ft/Ft=0.031 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force,compression V=0.972 kips Max. compressive force C=0.561 kips Max. applied comp.stress fc=34 lb/in2 Design compressive stress Fc'=945 lb/in2 fc/Fc'=0.036 PASS-Design compressive stress exceeds maximum applied compressive stress Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 30 of 78 Chord capacity for chords 7 and 8 Shear wall aspect ratio h/b4=2.56 Shear force for tension V=0.972 kips Maximum tensile force T=0.251 kips Max.applied tensile stress ft= 17 lb/in2 Design tensile stress Ft' =720 lb/in2 ft/Ft'=0.024 PASS-Design tensile stress exceeds maximum applied tensile stress Shear force, compression V=0.972 kips Max. compressive force C=0.482 kips Max. applied comp.stress fc=29 lb/in2 Design compressive stress Fc'=945 lb/in2 fo/Fc'=0.031 PASS-Design compressive stress exceeds maximum applied compressive stress Collector capacity Maximum shear force Vmax=0.972 kips Max. force in collector Pcon=0.154 kips Max.applied tensile stress ft=9 lb/in2 Design tensile stress Ft =720 lb/in2 — ft/Ft'=0.013 PASS-Design tensile stress exceeds maximum applied tensile stress Max.applied comp.stress fo=9 lb/in2 Design compressive stress Fc'= 1360 Ib/int fo/Fc'=0.007 PASS-Design compressive stress exceeds maximum applied compressive stress Hold down force Chord 3 T3=0.438 kips Chord4 T4=0.438 kips Chord 5 T5=0.331 kips Chord 6 Ts=0.331 kips Chord 7 T7=0.251 kips Chord 8 Ts=0.251 kips Wind load deflection Shear wall deflection Ssww=0.093 in Deflection limit Aw allow=0.24 in Ssww/Aw allow=0.387 PASS-Shear wall deflection is less than deflection limit Seismic load deflection Shear wall deflection Ssws=0.157 in Deflection limit Os allow=2.4 in Ssws/As allow=0.065 PASS-Shear wall deflection is less than deflection limit Use Simpson DTT2Z-SDS2.5 holdown device Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)7374344 By R.P. Date 5/7/2018 Page 31 of 78 FLOOR JOISTS ANALYSIS & DESIGN (NDS) STRUCTURAL WOOD MEMBER ANALYSIS&DESIGN (NDSI In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 toad Envelope•Combination 1 0.077— OD L 17 A 1 3 �p_ft €3,1--dF.4o,ent Enveiape 1.377— . 1.4 ft l 12 A 1 B kips Shear Force Envelope 0.5 0.459— 0- -0 459 ft I 12 A 1 B Applied loading Beam loads Dead self weight of beam x 1 Dead full UDL 20 lb/ft Live full UDL 53 lb/ft Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Span 1 Dead x 1.00 Live x 1.00 Support B Dead x 1.00 Js',S ' 'i stt/tin� 1710 Project New Addition Job# 180145 to i Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 32 of. 78 Live x 1.00 Analysis results Design moment M= 1377.4 lb ft Design shear F =459.1 lb Total load on member Wtot=918.3 lb Reaction at support A RA_max=459 lb RA_min=459 lb Unfactored dead load reaction at support A RA_Dead= 140 lb Unfactored live load reaction at support A RA_Live=319 lb Reaction at support B RB_max=459 lb RB min=459 lb Unfactored dead load reaction at support B Rs_Dead=140 lb Unfactored live load reaction at support B RB_Live=319 lb ILJ -V, .4 1.5" .. _.._.-- --4 -► Sawn lumber section details Nominal breadth bnom=2 in Dressed breadth b= 1.5 in Nominal depth dnom= 10 in Dressed depth d=9.25 in Number of sections N = 1 Breadth of member bb= 1.5 in Lumber grading No.2 Douglas Fir-Larch Member details Service condition 121.1 Length of span Ls1 = 12 ft Length of bearing Lb=4 in Load duration Ten years The beam is one of three or more repetitive members Bearing perpendicular to grain -c1.3.10.2 Adjusted compression Fc_perp=625 lb/in2 Applied compression fo_perp=77 lb/in2 PASS-Design compressive stress exceeds applied compressive stress at bearing Strength in bending-c1.3.3.1 Design bending stress Fb'= 1139 lb/in2 Actual bending stress fb=773 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain -c1.3.4.1 Design shear stress FV= 180 lb/in2 Actual shear stress fv=50 lb/in2 — PASS-Design shear stress exceeds actual shear stress Deflection-c1.3.5.1 } Allowable deflection 6adm=0.432 in Total deflection &b sl =0.226 in PASS- Total deflection is less than design deflection Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson I 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 33 of 78 Use 2x10 DF/L#2 at 16" o.c. FLOOR BEAM #1 ANALYSIS & DESIGN (AISC360) STEEL BEAM ANALYSIS& DESIGN (AISC360-10) In accordance with AISC360 14th Edition published 2010 using the ASD method Tedds calculation version 3.0.12 load Envelope-Combination 1 aacai 0.0 ft I 17.5 A 1 0 klp ft Gann,a.r.s r-toEn.v`t:s 14.138— 13:1 it i 173 I. A 1 a kips Shear Force Envelope 3.232— 31 3232— -3.2 ft I 173 i... A 1 8 Support conditions Support A Vertically restrained Rotationally free Support B Vertically restrained Rotationally free Applied loading Beam loads Dead self weight of beam x 1 Y ' Cry Project New Addition Job# 180145 pt Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 34 of 78 Dead full UDL 0.105 kips/ft Live full UDL 0.239 kips/ft Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Roof live x 1.00 Snow x 1.00 Dead x 1.00 Live x 1.00 Roof live x 1.00 Snow x 1.00 Support B Dead x 1.00 Live x 1.00 Roof live x 1.00 Snow x 1.00 Analysis results Maximum moment Mmax= 14.1 kips_ft Mmin=0 kips_ft Maximum shear Vmax=3.2 kips Vmin=-3.2 kips Deflection bmax=0.5 in 8min=0 in Maximum reaction at support A RA_max=3.2 kips RA min=32 kips Unfactored dead load reaction at support A RA_Dead= 1_1 kips Unfactored live load reaction at support A RA_Live=2.1 kips Maximum reaction at support B RB max=3.2 kips Re_min=3.2 kips Unfactored dead load reaction at support B RB_Dead= 1.11 kips Unfactored live load reaction at support B RB_Live=2.1 kips Section details Section type W 6x25(AISC 14th Edn (v14.1)) ASTM steel designation A36 Steel yield stress Fy=36 ksi Steel tensile stress FU=58 ksi Modulus of elasticity E =29000 ksi r «<`:=:y L Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684. (503)737-4344 By R.P. Date 5/7/2018 Page 35 of 78 o r l 1 -►i 4-0.32" co 1 r 1 } fl -----------0.08".-------Safety factors Safety factor for tensile yielding city= 1.67 Safety factor for tensile rupture Qtr=2.00 Safety factor for compression S2c= 1.67 Safety factor for flexure Ste= 1.67 Safety factor for shear S2v= 1.50 Lateral bracing Span 1 has continuous lateral bracing Classification of sections for local buckling -Section B4.1 Classification of flanges in flexure-Table B4.1 b(case 10) Width to thickness ratio bf/(2 x tf) =6.68 Limiting ratio for compact section kpff=0.38 x J[E/Fy]= 10.79 Limiting ratio for non-compact section krff= 1.0 x J[E/Fy] =28.38 Compact Classification of web in flexure-Table B4.1 b(case 15) Width to thickness ratio (d-2 x k)/tw= 15.53 Limiting ratio for compact section Xpwf=3.76 x 4[E/Fyj= 106/2 Limiting ratio for non-compact section krwf=5.70 x '/[E/Fy]= 161.78 Compact Section is compact in flexure Design of members for shear-Chapter G Required shear strength Vr=max(abs(Vmax), abs(Vmin)) =3.232 kips Web area Aw=d x tw=2.042 in2 Web plate buckling coefficient kv=5 Web shear coefficient-eq G2-2 Cv= 1.000 Nominal shear strength -eq G2-1 Vn=0.6 x Fy x Aw x Cv=44.099 kips Allowable shear strength Vc=Vn/S2v=29.399 kips PASS-Allowable shear strength exceeds required shear strength ck' ivs 44l'Einy /'!'C Project New Addition Job# 180145 ° Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC ; Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 36 of 78 Design of members for flexure in the major axis-Chapter F Required flexural strength Mr= max(abs(Msi_max), abs(Msi_min)) = 14.138 kips_ft Yielding -Section F2.1 Nominal flexural strength for yielding-eq F2-1 Mnyld=Mp= Fy x Zx=56.7 kips_ft Nominal flexural strength Mn= Mnyld=56.700 kips_ft Allowable flexural strength Mc= Mn/C2b=33.952 kips_ft PASS-Allowable flexural strength exceeds required flexural strength Design of members for vertical deflection Consider deflection due to dead, live, roof live and snow loads 1 Limiting deflection 6iiim= Ls1 /360=0.583 in Maximum deflection span 1 6=max(abs(8max), abs(6min)) =0.503 in PASS-Maximum deflection does not exceed deflection limit Use W6x25 i "_-.k =:.'C Project New Addition Job# 180145 , Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 37 of 78 FLOOR BEAM #2 ANALYSIS & DESIGN (NDS) STRUCTURAL GLUED LAMINATED TIMBER(GLULAM)MEMBER ANALYSIS &DESIGN(NDS) " In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Load Envelope-Combination 1 4.511- ft I 17.833 I A 1 8 kip_ft 8erdrg Moment Envelope 46568- _._..____--- .. 42'i.6 ft I —�. 17.833 I: A 1 B kips Shur Fake&metope R1 8.070-. r__ -..-_._..._.,._ -___ .4 �_ 0.0-A - - 1 A -8533 i 41.5 ft I 17.833 I A 1 B Applied loading Beam loads Dead self weight of beam x 1 i Dead full UDL 306 lb/ft Snow full UDL 350 lb/ft Dead point load 1070 lb at 118.00 in Live point load 2094 lb at 118.00 in 1 Wind point load 1347 lb at 118.00 in Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Snow x 1.00 r, f dt'-__-VS .0 ,'«4't«,y /'/'i' 1 Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611.NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 38 of 78 Wind x 1.00 Span 1 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support B Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Analysis results Design moment M =46568.0 lb_ft Design shear F=8533.2 lb Total load on member Wtot= 16602.8 lb Reaction at support A RA_max=8070 lb RA_min=8070 lb Unfactored dead load reaction at support A RA_Dead=3405 lb Unfactored live load reaction at support A RA_Uve=939 lb Unfactored snow load reaction at support A RA_snow=3121 lb Unfactored wind load reaction at support A RA_wind=604 lb Reaction at support B RB_max=8533 lb Rs min=8533 lb Unfactored dead load reaction at support B RB_Dead=3515 lb Unfactored live load reaction at support B RB Live= 1155 lb Unfactored snow load reaction at support B RB_snow=3121 lb Unfactored wind load reaction at support B RB_wnd=743 lb co 45 51. ► 4., 4- Glulam section details Net finished section breadth b=555 in Net finished section depth d= 16.5 in Number of sections N =1 Alignment of laminations Horizontal Stress class 24F-V4 DF/DF Member details Service condition Length of span Ls, = 17.833 ft Length of bearing Lb=4 in Load duration Two months Bearing perpendicular to grain -c1.3.10.2 Adjusted compression Fc_perp'=650 lb/in2 Applied compression fc_perp=388 lb/in2 PASS-Design compressive stress exceeds applied compressive stress at bearing Project New Addition Job# 180145 ( Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684503 ( )737-4344 i By R.P. Date 5/7/2018 Page 39 of 78 Strength in bending-c1.3.3.1 Design bending stress Fb=2698 lb/in2 Actual bending stress fb=2239 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain-c1.3.4.1 Design shear stress Fv' =305 lb/in2 Actual shear stress fv= 141 lb/in2 PASS-Design shear stress exceeds actual shear stress Deflection -c1.3.5.1 Allowable deflection 8adm=0.749 in Total deflection 8b sl =0.661 in PASS- Total deflection is less than design deflection Use 5'/z"x 16 1/2"24F-V4 GLB /Yr Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 ! By R.P. Date 5/7/2018 Page 40 of` 78 MAIN FLOOR CANTILEVER JOISTS ANALYSIS & DESIGN (NDS) STRUCTURAL WOOD MEMBER ANALYSIS& DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Load Envekipe-Combination2 0075- 0.0 ft I 4 12 1. A 1 B 2 C k;p ft Bending Moment Envebpe -0.150- i>.2 " I 0.085-+ 0.1 ft 4 2 A 1 8 2 C LiNs Shear Force Envelope 0.150- .0.1 0.2 0.07 -0.188- -0.2 ft I 4 I 2 A 1 8 2 C Applied loading Beam loads Dead self weight of beam x 1 Dead full UDL 20 lb/ft Live full UDL 53 Ib/ft Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Span 1 Dead x 1.00 Live x 1.00 Support B Dead x 1.00 Live x 1.00 `t R AQ %7!i�tul'& CP6 [ Project New Addition Job# 180145 .i `s Address 11958 SW 125th Ct.,Tigard,OR 97223 RN$Consulting LLC Client Chris&Jessica Anderson .11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 41 of 78 Span 2 Dead x 1.00 Live x 1.00 Support C Dead x 1.00 Live x 1.00 Analysis results Design moment M = 150.3 lb_ft Design shear F= 187..9 lb Total load on member Wtot=450.9 lb Reaction at support A RA_max= 113 lb RA_min= 113 lb Unfactored dead load reaction at support A RA_Dead=33 lb Unfactored live load reaction at support A RA_LiVe=80 lb Reaction at support B Rc_max=338 lb RB min=338 lb Unfactored dead load reaction at support B RB_Dead=99 lb Unfactored live load reaction at support B RB_Live=239 lb Reaction at support C Rc_max=0 lb Rc_min=0 lb Unfactored dead load reaction at support C Rc_Dead=0 lb Unfactored live load reaction at support C Rc_uve=0 lb --✓1.5"y.. y..__._,q..._--_. Sawn lumber section details Nominal breadth bnom=2 in Dressed breadth b= 1.5 in Nominal depth dnom=6 in Dressed depth d =555 in Number of sections N =1 Breadth of member bb= 1.5 in Lumber grading No.2 Douglas Fir-Larch Member details Service condition Length of span 1 Lxi =4 ft Length of span 2 L52=2 ft Length of bearing Lb=4 in • Load duration Ten years The beam is one of three or more repetitive members Bearing perpendicular to grain-c1.3.10.2 Adjusted compression Fc_perp=684 lb/in2 Applied compression fc_perp=56lb/in2 PASS-Design compressive stress exceeds applied compressive stress at bearing Strength in bending-c1.3.3.1 Design bending stress Fb' = 1346 lb/in2 Actual bending stress fb=239 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain-c1.3.4.1 Design shear stress Fv.=180 lb/in2 Actual shear stress f5=34 lb/in2 Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/712018 Page 42 of 78 PASS-Design shear stress exceeds actual shear stress Deflection-c1.3.5.1 Allowable deflection Sadm=0.144 in Total deflection 6b s2=0.008 in • PASS- Total deflection is less than design deflection Use 2x6 DF/L#2 at 16"o.c. • MAIN FLOOR BEAM ANALYSIS & DESIGN (NDS) STRUCTURAL WOOD MEMBER ANALYSIS&DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 • toad Entrbpe-Combination 1 0225-4 G- rt I 8 I A B kip h SerMine Moment Envelope 0.0- 1.797) 1.$ B kips Shear Force Envelope 9 0.899- °- ._ fI -0.899- €t I -0.9 s A 1 B Applied loading Beam loads Dead self weight of beam x 1 Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 43 of 78 Dead full UDL 60 lb/ft Live full UDL 160 lb/ft Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Span 1 Dead x 1.00 Live x 1.00 Support B Dead x 1.00 Live x 1.00 Analysis results Design moment M=1797.4 lb ft Design shear F=898.7 lb Total load on member Wtot= 1797.4 lb Reaction at support A RA_max=899 lb R,n_mm=899 lb Unfactored dead load reaction at support A RA_Dead=259 lb Unfactored live load reaction at support A RA_Live=640 lb Reaction at support B RB_max=899 lb RB_min=899 lb Unfactored dead load reaction at support B RB_Dead=259 lb Unfactored live load reaction at support B RB_uve=640 lb N----3.5"--r 4" p Sawn lumber section details Nominal breadth bnom=4 in Dressed breadth b=3.5 in Nominal depth dnom=6 in Dressed depth d =5_5 in Number of sections N =1 Breadth of member bb=3.5 in Lumber grading Select Structural Douglas Fir-Larch Member details Service condition Length of span Lsi =8 ft Length of bearing Lb=4 in Load duration Ten years Bearing perpendicular to grain-ci.3.10.2 Adjusted compression Fc_perp=625 Ib/int Applied compression fc_perp=64 lb/in2 PASS-Design compressive stress exceeds applied compressive stress at bearing Strength in bending -c1.3.3.1 Design bending stress Fb'=1950 Ib/int Actual bending stress fb= 1222 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain-c1.3.4.1 Design shear stress Fv'= 180 lb/in2 Actual shear stress fv=70 lb/in2 PP j Project New Addition Job# 180145 { Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 44 of 78 PASS-Design shear stress exceeds actual shear stress Deflection-c1.3.5.1 Allowable deflection badm=0.288 in Total deflection bb s1 =0.225 in PASS- Total deflection is less than design deflection Use 4x6 DF/L#2 at 48" o.c. MAIN FLOOR RIM BEAM ( R5 ANALYSIS & DESIGN (NDS) STRUCTURAL WOOD MEMBER ANALYSIS & DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 ea 34 Envelope-Ec.,:_-_.per a•+ 1 '441-- . 0,0—A a to is m -2301— • ft I 1.633 1 633 I 1.633 I 1.633 I 3.633 I 1.633 I 1.633 I 1.633 I 1.633 I 1.633 I A 1 B 2 C 3 Co 4 E 5 F 6 G 7 it 8 1 9 1 10 K. kip_ft Bending Moment Enve#ope -0.6581 '0.7 -0S 0-4 -0.1 -0.1 -0.1 -0.2 0.0— .:� .... . . �� 0.1 0.0 0.0 0.1 0. 0.1 • 0.D.2 0.757- 0.8 ft 1 1:633 I 1.633 I 1.633 I 3.633 ( 1.633 I 1.633 I 1.633 I 1.633 I 1.633 I 1.633 I A 1 B 2 C 3 D4 65 F 66 7 H8 1 9 1 10 K kips Shear Force Envelope 2.213— 2.2 2.0 2.1 0.2 03 0.3 0.6 0.5 0.2 0.5 -0.3 -03 -03 -02 t1.A -0A -1.259-- -0.6 �-.`� -0.8 -13 ft I 1.633 I 1.633 I 1.633 I 1.633 11.633 I 1.633 I 1.633 I 1.633 I 1.633 I 1.633 A 192 C 3 0 4 E 5 F 6 6 N 8 1 9 1 10 K C i Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 ' By R.F. Date 5/7/2018 Page 45 of 78 Applied loading Beam loads Dead self weight of beam x 1 Dead full UDL 184 lb/ft Live full UDL 40 lb/ft Snow full UDL 122 lb/ft Wind point load 2901 lb at 105.00 in Wind point load-2901 lb at 192.00 in Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 1 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support B Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 2 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support C Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 3 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support D Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x1.00 Span 4 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support E Dead x 1.00 digit /J7Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)7374344 By R.P. Date 5/7/2018 Page 46 of 78 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 5 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support F Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 6 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support G Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 7 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support H Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 8 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support 1 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Span 9 Dead x 1.00 Live x 1.00 J Snow x 1.00 Wind x 1.00 Support J Dead x 1.00 Live x 1.00 1 Snow x 1.00 i rdt,=y%'//' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737.4344 By R.P. Date 5/7/2018 Page 47 of 78 Wind x 1.00 Span 10 Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Support K Dead x 1.00 Live x 1.00 Snow x 1.00 Wind x 1.00 Analysis results Design moment M=756.5!bit Design shear F=2213.5 lb Total load on member Wtot=5715.7 lb Reaction at support A RD_max=224 lb RA_min=224 lb Unfactored dead load reaction at support A RA_Dead= 121 lb Unfactored live load reaction at support A RA_Live=26 lb Unfactored snow load reaction at support A RA_snow=78 lb Unfactored wind load reaction at support A RA_Wind=-1 lb Reaction at support B RD_max=655 lb Re_min=655 lb Unfactored dead load reaction at support B Rs_Dead=348 lb Unfactored live load reaction at support B Re Jive=74 lb Unfactored snow load reaction at support B Rs_snow=226 lb Unfactored wind load reaction at support B Re_wind=7 lb Reaction at support C Rc_max=523 lb Rc_min=523 lb Unfactored dead load reaction at support C RC_Dead=296 lb Unfactored live load reaction at support C RD_Live=63 lb , Unfactored snow load reaction at support C Rc_snow= 192 lb Unfactored wind load reaction at support C RC_Wind=-28 lb Reaction at support D RD_max=683 lb RD_min=683 lb Unfactored dead load reaction at support D RD_Dead=310 lb Unfactored live load reaction at support D RD_Live=66 lb Unfactored snow load reaction at support D RD_snow=201 lb Unfactored wind load reaction at support D RD wind= 106 lb Reaction at support E RE_max= 176 lb RE_min= 176 lb , Unfactored dead load reaction at support E RE_Dead=306 lb Unfactored live load reaction at support E RE_Live=65 lb Unfactored snow load reaction at support E RE_snow= 198 lb Unfactored wind load reaction at support E RE_wind=-394 lb Reaction at support F Rr_max=2811 lb RF min=2811 lb Unfactored dead load reaction at support F RF_Dead=308 lb Unfactored live load reaction at support F RF_uve=65 lb Unfactored snow load reaction at support F RF_snow= 199 lb Unfactored wind load reaction at support F RF_wind=2238 lb Reaction at support G RD_max= 1782 lb RG_min= 1782 lb Unfactored dead load reaction at support G RG_Dead=306 lb ,R,.-..'11::S ! Unfactored live load reaction at support G RG_Live=65 lb Unfactored snow load reaction at support G RG_Snow= 198 lb Unfactored wind load reaction at support G RG_wind= 1212 lb Reaction at support H RH_max=228 lb RH min=228 lb Unfactored dead load reaction at support H RH_Dead=310 lb Unfactored live load reaction at support H RH_uVe=66 lb Unfactored snow load reaction at support H RH snow=201 lb Unfactored wind load reaction at support H RH_Wind=-349 lb Reaction at support I Ri_max=869 lb Ri min=869 lb Unfactored dead load reaction at support I Ri_Dead=296 lb Unfactored live load reaction at support I RI_Live=63 lb Unfactored snow load reaction at support I RI_Snow= 192 lb Unfactored wind load reaction at support I Ri_wnd=318 lb Reaction at support J RJ_max=-299 lb R. min=-299 lb Unfactored dead load reaction at support J RJ_Dead=348 lb Unfactored live load reaction at support J RJ_Live=74 lb Unfactored snow load reaction at support J RJ snow=226 lb Unfactored wind load reaction at support J RJ_wind=-947 lb Reaction at support K RK_max=-1936 lb RK min= -1936 lb Unfactored dead load reaction at support K RK Dead= 121 lb Unfactored live load reaction at support K RK Live=26 lb Unfactored snow load reaction at support K RK snow=78 lb Unfactored wind load reaction at support K RK wind=-2161 lb 4_-3„ r 4 4" lo, Sawn lumber section details Nominal breadth bnom=2 in Dressed breadth b= 1.5 in Nominal depth dnom=6 in Dressed depth d =5.5 in Number of sections N =2 Breadth of member bb=3 in Lumber grading No.2 Douglas Fir-Larch Member details Service condition Length of span 1 Lsi =1.633 ft Length of span 2 Ls2= 1.633 ft Length of span 3 Ls3=1.633 ft Length of span 4 Ls4= 1.633 ft Length of span 5 Lss= 1.633 ft Length of span 6 Ls6= 1.633 ft Length of span 7 Ls7= 1.633 ft Length of span 8 Lss= 1.633 ft ii-i-V-1 44; ,,tu,iy e_.'t Project New Addition Job# 180145 St 3 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC . Client Chris&Jessica Anderson .11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 49 of 78 Length of span 9 Lys= 1.633 ft Length of span 10 Lsio= 1.633 ft Length of bearing Lb=4 in Load duration Ten minutes Bearing perpendicular to grain-c1.3.10.2 Adjusted compression Fc_perp=684 lb/in2 Applied compression fc_perp=234 lb/in2 PASS-Design compressive stress exceeds applied compressive stress at bearing Strength in bending -c1.3.3.1 Design bending stress Fb'=1872 lb/in2 Actual bending stress fb=600 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain-c1.3.4.1 Design shear stress FJ =288 lb/in2 Actual shear stress f„=201 Ib/int PASS-Design shear stress exceeds actual shear stress Deflection -c1.3.5.1 Allowable deflection 8adm=0.059 in Total deflection 8b_s6=.0.903 in PASS- Total deflection is less than design deflection Use dbl.2x6 DF/L#2 1 1 JV?JVS .p,e14,t/Ei,19_ILA Project New Addition Job# 180145 , Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 50 of 78 CANTILEVER ROOF RAFTER ANALYSIS & DESIGN (NDS) STRUCTURAL WOOD MEMBER ANALYSIS&DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Load Envelope-Combination 1 0.05'1--1 0.0— ft l s I 4 A 1 8 2 C kipit Bending Moment Envelope -0.453— -0.5 0.255— ---_._.-. ft I 8 q (. A 1 8 2 C kips Shear Force Envelope 0.227— 0,2 0.2 0.0— -0.283— ft 1 8 -0.3 4 A 1 8 2 • Applied loading Beam loads Dead self weight of beam x 1 Dead full UDL 20 lb/ft Snow full UDL 33 lb/ft Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Snow x 1,00 Span 1 Dead x 1.00 Live x 1.00 Snow x 1.00 Lye Project New Addition Job# 180145 Address 11958 SW 125th Ct., Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 51 of 78 Support B Dead x 1.00 Live x 1.00 Snow x 1.00 Span 2 Dead x 1.00 Live x 1.00 Snow x 1.00 Support C Dead x 1.00 Live x 1.00 Snow x 1.00 Analysis results Design moment M=453.1 'bit Design shear F =283.2 lb Total load on member Wtoc=679.7 lb Reaction at support A RA_max= 170 lb RA_min= 170 lb Unfactored dead load reaction at support A RA_Dead=70 lb Unfactored snow load reaction at support A RA_snow= 100 lb Reaction at support B RB_max=510 lb RB_min=510 lb Unfactored dead load reaction at support B RB_Dead=210 lb Unfactored snow load reaction at support B RB_snow=300 lb Reaction at support C Rc_max=0 lb Rc_min=0 lb Unfactored dead load reaction at support C RC_Dead=0 lb Unfactored snow load reaction at support C Rc_snow=0 lb r_ --4. 4-1.5' Sawn lumber section details Nominal breadth bnom= 2 in Dressed breadth b= 1.5 in Nominal depth dnom= 10 in Dressed depth d =9.25 in Number of sections N = 1 Breadth of member bb= 1.5 in Lumber grading No.2 Douglas Fir-Larch Member details Service condition Length of span 1 L51 =8 ft Length of span 2 Ls2=4 ft Length of bearing Lb=4 in Load duration Two months The beam is one of three or more repetitive members Bearing perpendicular to grain -c1.3.10.2 Adjusted compression Fc_perp'=684 lb/in2 Applied compression fc_perp=85 lb/in2 r Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 52 of 78 PASS-Design compressive stress exceeds applied compressive stress at bearing Strength in bending -c1.3.3.1 Design bending stress Fb'= 1309 lb/in2 Actual bending stress fb=254 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain -c1.3.4.1 Design shear stress F5'=207 Ib/int Actual shear stress fv=31 lb/in2 PASS-Design shear stress exceeds actual shear stress Deflection-c1.3.5.1 Allowable deflection dadm=0.288 in Total deflection 6b s2=0.020 in PASS- Total deflection is less than design deflection Use 2x10 DF/L#2 at 16" o.c. HIP BEAM ANALYSIS & DESIGN (NDS) STRUCTURAL WOOD MEMBER ANALYSIS&DESIGN(NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Load Envelope-combination 1 3.34 -. U.D.- it I 11 335' A 5 1 2 { xSp ft Bending Moment Envelope -0 7 0.0-, , 3826— -��`— _L _ A 11.33 5.657 j 1 R 2 ii'':::k 1 i,yt ,1 �'%'�' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 53 of 78 Shear force Envelope kips 1.5 1A62- 43 -1.153- •1.2 ft i 11.33 �_...� 5.667 A g 2 t Applied loading Beam loads Dead self weight of beam x 1 Dead full VDL 140 lb/ft to 0 lb/ft Snow full VDL 200 lb/ft to 0 lb/ft Load combinations Load combination 1 Support A Dead x 1.00 Live x 1.00 Snow x 1.00 Span 1 Dead x 1.00 Live x 1.00 Snow x 1.00 Support B Dead x 1.00 Live x 1.00 Snow x 1.00 Span 2 Dead x 1.00 Live x 1.00 Snow x 1.00 Support C Dead x 1.00 Live x 1.00 Snow x 1.00 Analysis results Design moment M=3425.6 Ib_ft Design shear F= 1462.0 lb Total load on member Wtot= 2959.2 lb Reaction at support A RA_max= 1462 lb RA_min= 1462 lb Unfactored dead load reaction at support A RA_Dead=612 lb Unfactored snow load reaction at support A RA_snow=850 lb Reaction at support B RB_max= 1497 lb Rs_min= 1497 lb Unfactored dead load reaction at support B RB Dead=647 lb Unfactored snow load reaction at support B RB_Snow=850 lb Reaction at support C Rc_max=0 lb Rc_min=0 lb Unfactored dead load reaction at support C RC_Dead=0 lb Unfactored snow load reaction at support C Rc_snow=0 lb 'c1V 0 ,itiny22C Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 54 of 78 I Vg Sawn lumber section details Nominal breadth bnom=2 in Dressed breadth b= 1.5 in Nominal depth dnom= 12 in Dressed depth d= 11.25 in Number of sections N = 1 Breadth of member bb= 1.5 in Lumber grading Select Structural Douglas Fir-Larch Member details Service condition Length of span 1 Lsi = 11.33 ft Length of span 2 Ls2=5.667 ft Length of bearing Lb=4 in Load duration Two months Bearing perpendicular to grain-c1.3.10.2 Adjusted compression Fc_perp=684 Ib/int Applied compression fc_perp=250 Ib/int PASS-Design compressive stress exceeds applied compressive stress at bearing Strength in bending -c1.3.3.1 Design bending stress Fb= 1725 lb/in2 Actual bending stress fb= 1299 lb/in2 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain -c1.3.4.1 Design shear stress Fv'=207 lb/in2 Actual shear stress f„= 130 lb/in2 PASS-Design shear stress exceeds actual shear stress Deflection -c1.3.5.1 Allowable deflection Scam=0.408 in Total deflection 8b s2=-0.296 in PASS- Total deflection is less than design deflection Use 2x12 DF/L Sel. Str. �t�J Mutt 2 P/''/' Project New Addition Job# 180145 t : 4 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 55 of 78 RcIV ' t g tie I Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 56 of 78 A-FRAME RAFTER MEMBER DESIGN (NDS) STRUCTURAL WOOD MEMBER DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Analysis results Design moment in major axis Mx=4254 Ib_ft Design shear F =842 lb N W 1 it 3"--0- Sawn "-►Sawn lumber section details Nominal breadth of sections bnom=2 in Dressed breadth of sections b= 1.5 in Nominal depth of sections dnom= 10 in Dressed depth of sections d=9.25 in Number of sections in member N =2 Overall breadth of member bb= N x b= 3 in 1 Species,grade and size classification Douglas Fir-Larch, No.1 grade, 2"&wider Bending parallel to grain Fb= 1000 lb/in2 Tension parallel to grain Ft=675 lb/in2 Compression parallel to grain F0= 1500 lb/in2 Compression perpendicular to grain Fc_perp=625 lb/in2 Shear parallel to grain Fy= 180 lb/in2 Modulus of elasticity E = 1700000 lb/in2 Modulus of elasticity, stability calculations Emin=620000 lb/in2 Mean shear modulus Gdef= E/16= 106250 lb/in2 Member details Service condition Load duration Two months Section properties Cross sectional area of member A= N x b x d=27.75 in2 Section modulus Sx= N x b x d2/6=42.78 in3 Sy=d x (N x b)2/6= 13.87 in3 Second moment of area lx= N x b x d3/12= 197.86 in4 ly=d x (N x b)3/ 12=20.81 in4 Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)7374344 i By R.P. Date 5/7/2018 Page 57 of 78 Adjustment factors Load duration factor-Table 2.3.2 CD= 1.15 Temperature factor-Table 2.3.3 Ct=1.00 Size factor for bending-Table 4A CFb= 1.10 Size factor for tension-Table 4A CF,= 1.10 Size factor for compression-Table 4A CFc= 1.00 Flat use factor-Table 4A Cru= 1.20 Incising factor for modulus of elasticity-Table 4.3.8 CiE= 1.00 Incising factor for bending, shear,tension &compression-Table 4.3.8 C.= 1.00 Incising factor for perpendicular compression-Table 4.3.8 Cic_perp= 1.00 Repetitive member factor-c1.4.3.9 Cr= 1.00 Bearing area factor-c1.3.10.4 Cb= 1.00 Depth-to-breadth ratio dnom/(N x bnom) = 2.50 -Beam is fully restrained Beam stability factor-c1.3.3.3 CL= 1.00 Strength in bending -c1.3.3.1 Design bending stress Fb'= Fb x CD X Ct X CL x CFb x Ci x Cr= 1265 lb/in2 Actual bending stress fb=MX/SX= 1193 lb/in2 fb/Fb'=0.943 PASS-Design bending stress exceeds actual bending stress Strength in shear parallel to grain-c1.3.4.1 Design shear stress FJ=F„x CD X Ct x C,=207 lb/in2 Actual shear stress-eq.3.4-2 ff=3 x F/(2 x A)=46 lb/in2 f„/Fv'=0.220 PASS-Design shear stress exceeds actual shear stress Use dbl.2x10 DF/L#1 Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 58 of 78 A-FRAME TENSION MEMBER DESIGN (NDS) STRUCTURAL WOOD MEMBER DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7 04 Analysis results Design axial tension P= 1191 lb A- T- ► 1.5" I- Sawn lumber section details Nominal breadth of sections bnom=2 in Dressed breadth of sections b= 1.5 in Nominal depth of sections dnom=6 in Dressed depth of sections d=5.5 in Number of sections in member N = 1 Overall breadth of member bb= N x b= 1.5 in Species, grade and size classification Douglas Fir-Larch, No.2 grade,2"&wider Bending parallel to grain Fb=900 lb/in2 , Tension parallel to grain Fc=575 lb/in2 Compression parallel to grain Fc= 1350 lb/in2 Compression perpendicular to grain Fc_perp=625 lb/in2 Shear parallel to grain Fv= 180 lb/in2 Modulus of elasticity E= 1600000 lb/in2 Modulus of elasticity, stability calculations Emin=580000 lb/in2 Mean shear modulus Gdet= E/16= 100000 lb/in2 Member details Service condition 211 Load duration Two months The beam is one of three or more repetitive members Section properties Cross sectional area of member A= N x b x d=8.25 in2 Section modulus SX=N x b x dz/6=7.56 in3 S,=dx (Nxb)z/6=2.06in3 Second moment of area lX= N x b x d3/12=20.80 in4 ly=d x (N x b)3/12=1.55 in4 ,t 0444,2 /' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting.LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 59 of 78. Adjustment factors Load duration factor-Table 2.3.2 CD= 1.15 Temperature factor-Table 2.3.3 Ct= 1.00 Size factor for bending-Table 4A CFb= 1.30 Size factor for tension-Table 4A Cn= 1.30 Size factor for compression-Table 4A CFC= 1.10 Flat use factor-Table 4A Cfu=1.15 Incising factor for modulus of elasticity-Table 4.3.8 CIE= 1.00 Incising factor for bending, shear,tension&compression-Table 4.3.8 Ci=1.00 Incising factor for perpendicular compression-Table 4.3.8 Cic_perp=1.00 Repetitive member factor-c1.4.3.9 Cr= 1.15 Bearing area factor-c1.3.10.4 Cb= 1.00 Depth-to-breadth ratio dnom/(N x bnom) =3.00 -Beam is fully restrained Beam stability factor-c1.3.3.3 CL= 1.00 Tension parallel to grain-c1.3.8.1 Design tensile stress Ft= Ft x CD x Ct x Crt x Ci=860 Ib/int Applied tensile stress ft= P/A= 144 Ib/int ft/Ft'=0.168 PASS-Design tensile stress exceeds applied tensile stress Use 2x6 DF/L#2 w/(5)Simpson 1/4" x 3" SDS screws ea.end(mina 1'e Project New Addition Job# 180145 It =7","7,F;14-14-L,7127; Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 60 of 78 BEAM #1 SUPPORT COLUMN MEMBER DESIGN (NDS) STRUCTURAL WOOD MEMBER DESIGNSTRUCTURAL WOOD MEMBER DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Analysis results Design axial compression P=3164 lb a in ei • r — 3.5".------* Sawn lumber section details Nominal breadth of sections bnom=4 in Dressed breadth of sections b=3.5 in Nominal depth of sections dnom=4 in Dressed depth of sections d=3.5 in Number of sections in member N= 1 Overall breadth of member bb= N x b=3.5 in Species,grade and size classification Douglas Fir-Larch, No.2 grade, 2"&wider Bending parallel to grain Fb=900 lb/in2 Tension parallel to grain Ft=575 lb/in2 Compression parallel to grain Fe= 1350 lb/in2 Compression perpendicular to grain Fc_perp=625 lb/in2 Shear parallel to grain F"= 180 lb/in2 Modulus of elasticity E= 1600000 lb/in2 Modulus of elasticity, stability calculations Emin=580000 lb/in2 Mean shear modulus Gdef= E/16= 100000 lb/in2 Member details Service condition Load duration Ten years Unbraced length in x-axis LX=8 ft Effective length factor in x-axis Kx= 1 Effective length in x-axis Lex= Lx x Kx=8 ft Unbraced length in y-axis Ly= 1 ft Effective length factor in y-axis KY= '1 Effective length in y-axis Ley= Ly x Ky=1 ft Section properties Cross sectional area of member A= N x b x d= 12.25 in2 }, ---Y♦ � ..:.,„z.1,,.:,_./ `fi' Project New Addition Job# 180145 •n7_177"..44711 42L-74 �i j Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 •Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 61 of 78 Section modulus Sx=N x b x d2/6=7.15 in3 Sy=d x (N x b)2/6=7.15 in3 Second moment of area lx= N x b x d3/ 12= 12.51 in4 ly=d x (N x b)3/12= 12.51 in4 Adjustment factors Load duration factor-Table 2.3.2 CD= 1.00 Temperature factor-Table 2.3.3 Ct= 1.00 Size factor for bending-Table 4A CFb= 1.50 Size factor for tension-Table 4A CFt= 1.50 Size factor for compression-Table 4A CFc= 1.10 Flat use factor-Table 4A Cru= 1.00 Incising factor for modulus of elasticity-Table 4.3.8 CiE= 1.00 Incising factor for bending,shear,tension &compression-Table 4.3.8 Ci= 1.00 • Incising factor for perpendicular compression-Table 4.3.8 Cic_perp= 1.00 Repetitive member factor-c1.4.3.9 Cr= 1.00 Bearing area factor-c1.3.10.4 Cb= 1.00 Adjusted modulus of elasticity for column stability Emin= Emin x CME x Ct x CIE=580000 lb/in2 Reference compression design value Fc*= Fc x CD x Cmc x Ct x CFc x Ci = 1485 lb/in2 Critical buckling design value for compression FcE=0.822 x Emin'/(Lex/d)2=634 lb/in2 c=0.80 Column stability factor-eq.3.7-1 Cp=(1 +(FcE/F0*))/(2 x c)- -\l[((1 +(FcE/Fe'))/(2 x c))2-(FcE/Fc*)/c}=;0:38 Depth-to-breadth ratio dnom/(N x bnom) = 1.00 - Beam is fully restrained Beam stability factor-c1.3.3.3 CL= 1.00 Strength in compression parallel to grain -c1.3.6.3 Design compressive stress Fc' = Fc x CD x Ct x CFc x Ci x CP=564 lb/in2 Applied compressive stress fc= P/A=258 lb/in2 fc/Fc'=0.458 PASS-Design compressive stress exceeds applied compressive stress Use 4x4 DF/L#2 1 L 1LS t4uutin9 1'/'6; Project New Addition Job# 180145 v.47 .2 t 4 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R,P. Date 5/7/2018 Page 62 of 78 BEAM #1 SUPPORT FOOTING ANALYSIS & DESIGN (ACI318) FOUNDATION ANALYSIS& DESIGN (ACI318) In accordance with ACI318-14 Tedds calculation version 3.2.07 FOOTING ANALYSIS Length of foundation Lx=2 ft Width of foundation Ly=2 ft Foundation area A= Lx x Ly=4 ft2 Depth of foundation h = 10 in Depth of soil over foundation h$on=0 in Density of concrete ycono= 150.0 lb/ft3 iI ) 1 lI f l ( l ( ll ( 1j I (i j ( 0.916 ksf L_L-L..-I 11.1.1.-1..1.�L..I L-1 I.L_1 1. 1-1 L_.1_-�. �-J.1._L �>11 J 0.916 ksf - 1 f - ! • —� y J 1 x r--- 0 916 0.916 ksf I I1.1.1 1 1 L-WI...I_I 1 -L..I 1 [1_L I_L_.[I i 1_L_1..f11 0.916 ksf Column no.1 details Length of column lxi =4.00 in Width of column ly1 =4.00 in position in x-axis xi = 12.00 in position in y-axis y1 = 12.00 in Soil properties Gross allowable bearing pressure gallow Gross= 1.5 ksf Density of soil ysou= 120.0 lb/ft3 ', tl ck?'/V T LEu2 Ift' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)7374344 By R.P. Date 5/7/2018 Page 63 of 78 Angle of internal friction (1)19=30.0 deg Design base friction angle Ebb=30.0 deg Coefficient of base friction tan(6bb) =0.577 Foundation loads Self weight Fswt= h x yconc= 125 psf Column no.1 loads Dead load in z FDZ1 = 1_1 kips Live load in z FLz, =2.1 kips Footing analysis for soil and stability Load combinations per ASCE 7-10 1.0D(0.262) 1.0D + 1.0L(0.611) Combination 2 results: 1.0D+ 1.0L Forces on foundation Force in z-axis Fdz=7D x A x Fswt+yD x FDz1 +YL X FLz, =3.7 kips Moments on foundation Moment in x-axis, about x is 0 Max=yo x A x Fswt x Lx/2+yo x (FDzi x x1) +yL x (FLzrx xi) =3.7 kip_ft Moment in y-axis, about y is 0 Mdy=yD x A x Fswt x Ly/2+yD x (FDz1 x yi) +yL x (FLz1 x yi)=3.7 kip_ft Uplift verification Vertical force Fdz=3.664 kips PASS-Foundation is not subject to uplift Bearing resistance Eccentricity of base reaction Eccentricity of base reaction in x-axis edx= Mdx/Fdz-Lx/2=0 in Eccentricity of base reaction in y-axis edy= Mdy I Fdz-Ly/2=0 in Pad base pressures q, = Fdz x (1 -6 x edx/Lx-6 x edy/Ly)/(Lx x Ly) =0.916ksf q2= Fdz x (1 -6xedx/Lx+6xedy/Ly)/(Lx xLy) =0.916ksf q3= Fdz X (1 +6xedx/Lx-6xedy/Ly)/(Lx xLy) =0.916ksf q4= Fdzx (1 +6 x edx/Lx+6 x edy/Ly)/(Lx x Ly) =0.916 ksf Minimum base pressure gmin=min(gi,g2,g3,q4)=0.916 ksf Maximum base pressure qmax=max(ql,q2,q3,q4)=0.916 ksf Allowable bearing capacity Allowable bearing capacity gauow=gauow_Gross=1.55 ksf qmax/gauow=0.611 PASS-Allowable bearing capacity exceeds design base pressure '1" Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 64 of 78 FOOTING DESIGN (AC1318) In accordance with ACI318-14 Material details Compressive strength of concrete co=3000 psi Yield strength of reinforcement fy=60000 psi Cover to reinforcement cnom=3 in Concrete type Normal weight Concrete modification factor k= 1.00 Column type Concrete Analysis and design of concrete footing Load combinations per ASCE 7-10 1.4D (0.028) 1.2D + 1.6L+0.5Lr(0.087) Combination 2 results: 1.2D+ 1.6L+0.5Lr Forces on foundation Ultimate force in z-axis Fuz=yo x A x F5wt+yo x FDz1 +YL x FLz1 =5_2 kips Moments on foundation Ultimate moment in x-axis, about x is 0 MUX=yo x A x Fs,t x LX/2 +yo x (FDz1 x xi) +yi_x (FLz1 x xi) =552 kip_ft Ultimate moment in y-axis, about y is 0 Muy=yD x A x Fs,1 x Ly/2 +yo x (Foz1 x yi)+yL x (FLz1 x yi)=552 kip_ft Eccentricity of base reaction Eccentricity of base reaction in x-axis eux= Mux/Fuz- LX/2=0 in Eccentricity of base reaction in y-axis euy= Muy I Fuz-Ly/2 =0 in Pad base pressures qui = Fuzx (1 6 x eux/LX 6xeuy/Ly)/(LX x Ly)= 1.309 ksf qu2= Fuzx (1 -6xeuX/LX+6xeuy/Ly)/(Lx x Ly) = 1.309ksf qua= Fuzx (1 +6xeux/Lx-6x euy/Ly)/(Lx x Ly) = 1.309ksf qua= Fuzx (1 +6 x eux/Lx+6 x euy/Ly)/(Lx x Ly) = 1.309 ksf Minimum ultimate base pressure qumin= min(qul,qu2,qu3,qua)=1.309 ksf Maximum ultimate base pressure qumax=max(qul,qu2,qu3,qua) =1.309 ksf Shear diagram,x axis(kips) 2.3 0 ti -2.3 I V % r ;t' ''/1 Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 ,Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 65 of 78 Moment diagram,x axis(kip_ft) 0 0 1.2 Moment design,x direction, positive moment Ultimate bending moment Mu.x.max=0.805 kip_ft Tension reinforcement provided 2 No.5 bottom bars (17.3 in c/c) Area of tension reinforcement provided Asx.bot.prov=0.62 in2 Minimum area of reinforcement(8.6.1.1) As.min=0.0018 x Ly x h=0.432 in2 PASS-Area of reinforcement provided exceeds minimum Maximum spacing of reinforcement(8.7.2.2) smax=min(2 x h, 18 in) =18 in PASS-Maximum permissible reinforcement spacing exceeds actual spacing Depth to tension reinforcement d =h-Cnom-41x.bot/2 =6.688 in Depth of compression block a=Asx.bot.prov x fy/(0.85 x fo x Ly)=0.608 in Neutral axis factor 131 =0.85 Depth to neutral axis c=a/(31 =0.715 in Strain in tensile reinforcement(8.3.3.1) Et=0.003 x d/c-0.003=0.02506 PASS- Tensile strain exceeds minimum required,0.004 Nominal moment capacity Mn=Asx.bot.prov x fy x (d-a/2)= 19.789 kip_ft Flexural strength reduction factor Qtr= min(max(0.65+ (Et-0.002) x (250/3), 0.65), 0.9)=0.900 Design moment capacity OMn=tim x Mn=17.81 kip_ft Mu.x.max/4Mn=0.045 PASS-Design moment capacity exceeds ultimate moment load One-way shear design, x direction Ultimate shear force Vu.x=0.76 kips • Depth to reinforcement dv=h-Cnom-cx.bot/2 =6.688 in Shear strength reduction factor =0.75 Nominal shear capacity(Eq. 22.5.5.1) Vn=2 x x'I(fo x 1 psi) x Ly x dv= 17.582 kips Design shear capacity 4Vn=4v x Vn=13.186 kips Vu.x/4Vn=0.058 PASS-Design shear capacity exceeds ultimate shear load Two-way shear design at column 1 Depth to reinforcement dv2=6.375 in Shear perimeter length (22.6.4) lxp= 10.375 in Shear perimeter width(22.6.4) ly,= 10.375 in Shear perimeter(22.6.4) bo=2 x IND+2 x lyp=41.500 in Shear area Ap=lxp x lyp=107.641 in2 Surcharge loaded area Asur=Ap-Ix1 x ly1 =91.641 in2 Ultimate bearing pressure at center of shear area qup.avg= 1.309 ksf Ultimate shear load Fur,=yD X FDz1 +yL x FLz1 +YD x Ap x Fswt-qup.avg x Ap=3.768 kips Ultimate shear stress from vertical load vug=max(Fup/(bo x dv2),0 psi) = 14.244 psi f i_;,, / ,'c Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 66 of 78 Column geometry factor(Table 22.6.5.2) R =ly1 /lxi = 1.00 Column location factor(22.6.5.3) as=40 Concrete shear strength(22.6.5.2) Vcpa= (2 +4/13) x 2,,x Ain x 1 psi) =328.634 psi , vcpb=(as x dv2/bo+2) x X x Ain x 1 psi) =446.097 psi Vcpc=4 x X.x 4(fc x 1 psi) =219.089 psi Vcp=min(vopa,vcpb,vcpc)=219.089 psi Shear strength reduction factor (1)v=0.75 Nominal shear stress capacity(Eq.22.6.1.2) tin=Vcp=219.089 psi Design shear stress capacity(8.5.1.1(d)) Ovn=(i)v x vn= 164.317 psi Vug/On=0.087 PASS-Design shear stress capacity exceeds ultimate shear stress load Shear diagram,y axis(kips) 2.3 0 0 -2.3 Moment diagram,y axis(kip_ft) ° a 1.2 Moment design,y direction, positive moment Ultimate bending moment Mu.y.max=0.805 kip_ft Tension reinforcement provided 2 No.5 bottom bars (17.3 in c/c) Area of tension reinforcement provided Asy.bot.prov=0.62 in2 Minimum area of reinforcement(8.6.1.1) As min=0.0018 x Lx x h=0.432 in2 PASS-Area of reinforcement provided exceeds minimum Maximum spacing of reinforcement(8.7.2.2) smax=min(2 x h, 18 in) =18 in PASS-Maximum permissible reinforcement spacing exceeds actual spacing Depth to tension reinforcement d =h-cnom-tjtx.bot-4y.bot/2=6.063 in Depth of compression block a =Asy.bot.prov x fy/(0.85 x fc x Lx) =0.608 in Neutral axis factor i31 =0.85 Depth to neutral axis c=a/(31 =0.715 in Strain in tensile reinforcement(8.3.3.1)• Et=0.003 x d/c 0.003=0.02243 PASS- Tensile strain exceeds minimum required, 0.004 Nominal moment capacity Mn=Asy.bot.prov x fy x (d-a/2)= 17.852 kip_ft Flexural strength reduction factor (1)f=min(max(0.65 + (£t-0.002) x (250/3), 0.65), 0.9)=0.900 Design moment capacity 4Mn=(Of x Mn= 16.066 kip_ft t Mu.y.max/4Mn=0.050 r-IV� -.1-'A,0-1,29 P/'C Project New Addition Job# 180145 rra, ; j os.zro; s Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,'WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 67 of 78 PASS-Design moment capacity exceeds ultimate moment load . One-way shear design,y direction Ultimate shear force Vuy=0.76 kips Depth to reinforcement dv=h-Cnom-(1)x bpi-()y.bot/2 =6.063 in Shear strength reduction factor (1)v=0.75 Nominal shear capacity(Eq. 22.5.5.1) Vn=2 x 7 x 'Aro x 1 psi) x Lx x dv= 15.939 kips Design shear capacity =delv x Vn= 11.954 kips Vu.y/(IAA=0.064 PASS-Design shear capacity exceeds ultimate shear load Two-way shear design at column 1 Depth to reinforcement dv2=6.375 in Shear perimeter length (22.6.4) lxp= 10.375 in Shear perimeter width (22.6.4) lyp= 10.375 in Shear perimeter(22.6.4) bo=2 x lxp+2 x lyp=41.500 in Shear area Ap=lxp x lyp= 107.641 in2 Surcharge loaded area Asur=Ap-Ix1 x ly1 =91.641 in2 Ultimate bearing pressure at center of shear area qup.avg= 1.309 ksf Ultimate shear load Fup=yD x FDz1 +yL x FLz1 +yo x Ap x Fswt-qup.avg x Ap=3.768 kips Ultimate shear stress from vertical load vug=max(Fup/(b0 x dv2),0 psi)= 14.244 psi Column geometry factor(Table 22.6.5.2) (3=ly1 /lxi =1.00 Column location factor(22.6.5.3) as=40 Concrete shear strength (22.6.5.2) Vcpa=(2+4/13) x 2.x -(fc x 1 psi) =328.634 psi Vcpb= (as x dv2/bo+2) x A,x Arc x 1 psi)=446.097 psi v0p0=4xXx4(f0x 1 psi) =219.089 psi Vcp= min(vcpa,Vcpb,Vcpc)=219.089 psi Shear strength reduction factor (w=0.75 Nominal shear stress capacity(Eq.22.6.1.2) vn=Vcp=219.089 psi Design shear stress capacity(8.5.1.1(d)) = v x vn= 164.317 psi Vug/(bVn=0.087 PASS-Design shear stress capacity exceeds ultimate shear stress load !Y.! Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr,#17 Vancouver,WA 98684 (503)737-4344 ? By R.P. Date 5/7/2018 Page 68 of 78 I; A ! .. «- - -- - - -._ —_. ;. 2 No.5 bottom ba (17.3 in c/c) I I V / I 111 i T I f ` } } 2 No.5 bottom bars(17.3 in c/c) h'- 1 i Wolin- p f'l' Project New Addition •7,r. Job# 180145 / v Address 11958 SW 125th Ct., Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 1 By R.P. Date 5/7/2018 Page 69 of 78 BEAM #2 SUPPORT COLUMN MEMBER DESIGN (NDS) STRUCTURAL WOOD MEMBER DESIGN (NDS) In accordance with the ANSI/AF&PA NDS-2015 using the ASD method Tedds calculation version 1.7.04 Analysis results Design axial compression P=8069 lb A- - • 3.5" h Sawn lumber section details Nominal breadth of sections bnom=4 in Dressed breadth of sections b=3_5 in Nominal depth of sections dnom=6 in Dressed depth of sections d=5.5 in Number of sections in member N = 1 Overall breadth of member bb= N x b=3.5 in Species,grade and size classification Douglas Fir-Larch, No.2 grade,2"&wider Bending parallel to grain Fb=900 lb/int Tension parallel to grain Fc=575 lb/in2 • Compression parallel to grain Fo= 1350 lb/in2 Compression perpendicular to grain Fc_perp=625 lb/in2 Shear parallel to grain Fv= 180 lb/in2 Modulus of elasticity E= 1600000 lb/in2 Modulus of elasticity, stability calculations Emin=580000 lb/in2 Mean shear modulus Gdef=E/16= 100000 lb/in2 Member details Service condition Dry Load duration Ten minutes Unbraced length in x-axis Lx=8 ft Effective length factor in x-axis Kx= 1 Effective length in x-axis Lex= Lx x Kx=8 ft Unbraced length in y-axis Ly= 1 ft Effective length factor in y-axis Ky= 1 Effective length in y-axis Ley= Ly x Ky= 1 ft Section properties Cross sectional area of member A= N x b x d= 19.25 in2 ,4ifiug./1A. Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 70 of 78 Section modulus Sx=N x b x d2/6= 17.65 in3 Sy=dx(Nxb)2/6= 11W23in3 Second moment of area lx= N x b x d3/12 =48.53 in4 ly=d x (N x b)3/12= 19.65 in4 Adjustment factors Load duration factor-Table 2.3.2 Co= 1.60 Temperature factor-Table 2.3.3 Ct= 1.00 Size factor for bending-Table 4A CFb= 1.30 Size factor for tension-Table 4A CFt= 1.30 Size factor for compression-Table 4A CFc= 1.10 Flat use factor-Table 4A Cfu= 1.05 Incising factor for modulus of elasticity-Table 4.3.8 C;E= 1.00 Incising factor for bending, shear,tension&compression-Table 4.3.8 G=1.00 Incising factor for perpendicular compression-Table 4.3.8 Cic_perp=1.00 Repetitive member factor-c1.4.3.9 Cr= 1.00 Bearing area factor-c1.3.10.4 Cb= 1.00 Adjusted modulus of elasticity for column stability Emin'=Emin x CME x Ct x CIE=580000 lb/in2 Reference compression design value Fc* = Fc x Co x CMC x Ct x CFc x C;=2376 lb/in2 Critical buckling design value for compression FcE=0.822 x Emin/(Lex/d)2=1565 lb/in2 c=0.80 Column stability factor-eq.3.7-1 CP=(1 + (FCE/Fc*))/(2 x c)-A/[((1 + (FCE/Fc*))/(2 x c))2- (FcE/Fc*)/Oj=0.54 Depth-to-breadth ratio dnom/(N x bnom) = 1.50 -Beam is fully restrained Beam stability factor-c1.3.3.3 CL= 1.00 Strength in compression parallel to grain -cI.3.6.3 Design compressive stress Fo'= Fo x Cox Ct x CFc x C; x CP= 1272 lb/in2 Applied compressive stress fc=P/A=419 lb/in2 fc/Fc'=0.330 PASS-Design compressive stress exceeds applied compressive stress Use 4x6 DF/L#2 IP / f+ „eti,y t't'r' Project New Addition Job# 180145 • Address 11958 SW 125th Ct., Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr,#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 71 of 78 BEAM #2 SUPPORT FOOTING ANALYSIS & DESIGN (ACI318) FOUNDATION ANALYSIS& DESIGN (ACI318) In accordance with ACI318-14 Tedds calculation version 3.2.07 FOOTING ANALYSIS Length of foundation Lx=225 ft Width of foundation LY=2.5 ft Foundation area A= Lx x LY=6.25 ft2 Depth of foundation h = 10 in Depth of soil over foundation hsoll=0 in Density of concrete yconc= 150.0 lb/ft3 1.2ksf '----- --.____----------__....... .._.1. _----..-,.., 1.2ksf I Y I x 12 ksf J T Ti 1 T IT f T 1.2 ksf � � Column no.1 details Length of column lx1 =4.00 in Width of column lyl =6.00 in position in x-axis x1= 15.00 in position in y-axis yi = 15.00 in Soil properties Gross allowable bearing pressure gallow_Gross= 1.5 ksf Density of soil Tsai=120.0 ib/ft3 Angle of internal friction db=30.0 deg AL-1LS' C', «f'tbi,i'i'i' . Project New Addition Job# 180145 I Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)7374344 1 By R.P. Date 5/7/2018 Page 72 of 78 Design base friction angle 5bb=30.0 deg Coefficient of base friction tan(8bb)=0.577 Self weight Fswt=h x yconc= 125 psf Column no.1 loads Dead load in z FDz1 =334 kips Live load in z FLz1 =009 kips Snow load in z Fszi =3_1 kips Wind load in z Fw71 =00.6 kips Footing analysis for soil and stability Load combinations per ASCE 7-10 1.0D(0.447) 1.0D+ 1.0L(0.547) 1.0D+ 1.0S (0.779) 1.0D+0.75L+0.75Lr(0.522) 1.0D'+0.75L+0.75S (0.771) 1.0D +0.6W(0.485) i 1.0D+0.75L+0.75Lr+0.45W(0.551) 1.00 +0.75L+0.75S+0.45W(0.800) 1.00 +0.75L+0.75R +0.45W(0.551) 0.6D+0.6W(0.307) Combination 12 results: 1.0D+ 0.75L+0.75S +0.45W Forces on foundation Force in z-axis Fdz=yD x A x Fswt+yD x FDz1 +YL X FLz1 +ys x Fszi +yw x Fwzi =7_5 kips Moments on foundation Moment in x-axis, about x is 0 Mdx=yD x A x Fsw(x LX/2 +yD x (FDz1 x xi)+yL x (FLz1 x x1)+ys x (Fszi x xi) +yw x (Fwzi xxi) =994kip_ft Moment in y-axis, about y is 0 Mdy=yD X A x Fswc x Ly/2+yo x (FDz1 x yi) +yi_x (FLz1 x yi) +ys x (Fszi x yi)+yw x (Fwzi x yi) =9_4 kip_ft Uplift verification Vertical force Fdz=7.503 kips PASS-Foundation is not subject to uplift Bearing resistance Eccentricity of base reaction Eccentricity of base reaction in x-axis edx=Max/Fdz- L/2=0 in Eccentricity of base reaction in y-axis edy= Mdy/Fdz- Ly/2 =0 in Pad base pressures q1 = Fdz X (1 -6xedx/LX-6xedv/Ly)/(Lx XLy)= 1.2ksf q2= Fdz X (1 -6xedx/Lx+6xed /L L = v v)/(L X xy) 1_2 ksf q3= Fdz x (1 +6xedx/Lx-6xedy/Ly)/(Lx xLy) = 1_2ksf q4= Fdz x (1 +6xedx/Lx+6xedyl Ly)/(Lx x Ly) = 11.22ksf Minimum base pressure groin=min(g1,g2,q3,q4) =1.22 ksf =l` Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard, OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 73 of 78. Maximum base pressure qmax= max(g1,g2,q3,q4) = 1.22 ksf Allowable bearing capacity Allowable bearing capacity gaoow=ganow_cross=1.55 ksf qmax/gauow=0.800 PASS-Allowable bearing capacity exceeds design base pressure FOOTING DESIGN (AC1318) In accordance with AC1318-14 • Material details Compressive strength of concrete fo=3000 psi Yield strength of reinforcement fy=60000 psi Cover to reinforcement cnom=3 in Concrete type Normal weight Concrete modification factor = 1.00 Column type Concrete Analysis and design of concrete footing Load combinations per ASCE 7-10 1.4D (0.079) 1.2D+ 1.6L+0.5Lr(0.092) 1.2D+ 1.6L+0.5S(0.118) 1.2D+ 1.0L+ 1.6S (0.166) 1.2D + 1 6Lr+0.5W(0.073) 1.2D + 1.6S+0.5W(0.155) 1.2D + 1.6R +0.5W(0.073) 1.2D+ 1.0L+0.5Lr+ 1.0W(0.093) 1.2D+ 1.0L+0.5S+ 1.0W(0.119) 1.2D+ 1.0L+0.5R+ 1.0W(0.093) 0.9D+ 1.0W(0.066) Combination 6 results: 1.2D+1.0L+ 1.6S Forces on foundation Ultimate force in z-axis Fuz=yo X A x Fswt+yD x FDz1 +YL X FLz1 +ys x Fszi = 11.0 kips 11 Moments on foundation Ultimate moment in x-axis, about x is 0 Mux=yD x A x Fswt x Lx/2+yo x (FDzt x xi) +yL x (FLz1 x xi) +ys x (Fszi x xi) = 13.7kip_ft Ultimate moment in y-axis, about y is 0 Muy=yD x A x Fswt x Ly/2 +yo x (FDz1 x yi) +yL x (FL21 x yi) +ys x (Fszi x y1)= 13.7 kip_ft Eccentricity of base reaction Eccentricity of base reaction in x-axis eux= Mux/Fuz- Lx/2=0 in Eccentricity of base reaction in y-axis euy=Muy I Fuz-Ly 1 2=0 in Pad base pressures qui = Fuz x (1 -6xeux/Lx-6xeuy/Ly)/(Lx xLy) = 1.753 ksf qu2= Fuz x (1 -6xeux/Lx+6xeuy/Ly)/(Lx xLy)= 1.753ksf 144E.11z9 I'!'/' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 74 of 78 qua= Fuz x (1 +6 x eux/Lx-6xeuy/Ly)/(Lx x Ly) = 1.753 ksf qua=Fuzx (1 +6xeux/Lx+6xeuy/Ly)/(Lx xLy)=1.753ksf Minimum ultimate base pressure qumin= min(qul,qu2,qu3,qua) = 1.753 ksf Maximum ultimate base pressure qumax= max(qul,qu2,qu3,qua) =1.753 ksf Shear diagram,x axis(kips) 5 0 -5 Moment diagram,x axis(kip_ft) ° o 3.1 Moment design,x direction, positive moment Ultimate bending moment Muxmax=2.352 kip_ft Tension reinforcement provided 3 No.5 bottom bars(11.6 in c/c) Area of tension reinforcement provided Asx.bot.prov=0.93 in2 Minimum area of reinforcement(8.6.1.1) As.min=0.0018 x Ly x h=0.54 in2 PASS-Area of reinforcement provided exceeds minimum Maximum spacing of reinforcement(8.7.2.2) Smax=min(2 x h, 18 in) =18 in PASS-Maximum permissible reinforcement spacing exceeds actual spacing Depth to tension reinforcement d=h-cnom-(1)X.bot/2=6.688 in Depth of compression block a=Asx.bot.prov x fy/(0.85 x fc x Ly) =0.729 in Neutral axis factor 81 =0.85 Depth to neutral axis c=a/131 =0,858 in Strain in tensile reinforcement(8.3.3.1) st=0.003 x d/c-0.003=0.02038 PASS- Tensile strain exceeds minimum required,0.004 Nominal moment capacity Mn=Asx.bot.prov x fy x (d-a/2) =29.401 kip_ft Flexural strength reduction factor (i)f=min(max(0.65+(et-0.002) x (250/3),0.65), 0.9)=0.900 Design moment capacity 4/Mn=4/f x Mn=26.461 kip_ft Mu.x.max/4/Mn=0.089 PASS-Design moment capacity exceeds ultimate moment load One-way shear design,x direction Ultimate shear force Vu.x=2.317 kips Depth to reinforcement dv=h-Cnom-4x.bot/2=6.688 in Shear strength reduction factor (1)v=0.75 Nominal shear capacity(Eq. 22.5.5.1) Vn=2 x x 'I(fc x 1 psi) x Ly x dv=21.977 kips Design shear capacity (I)Vn=(1)v x Vn= 16.483 kips c 1 rr iliiny L/'2 Project New Addition Job# 180145 v. Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 75 of 78 Vux/4Vn=0.141 PASS-Design shear capacity exceeds ultimate shear load Two-way shear design at column 1 Depth to reinforcement dv2=6.688 in Shear perimeter length (22.6.4) Ixp= 10.688 in Shear perimeter width (22.6.4) lyp= 12.687 in Shear perimeter(22.6.4) bo=2 x lxp+2 x lyp=46.750 in Shear area Ap=Ixp x lyp=135.598 in2 Surcharge loaded area Asur=Ap-Ix, x ly, = 111.598 in2 Ultimate bearing pressure at center of shear area qup.av9= 1.753 ksf Ultimate shear load Fup=yo x Foz1 +71 x FLz1 +ys x Fszi +yo x Ap x Fswt-qup.av9 x Ap=8.509 kips Ultimate shear stress from vertical load vu9=max(Fup/(b0 x dv2),0 psi) =27.217 psi Column geometry factor(Table 22.6.5.2) (3=ly1 /Ix1 = 1.50 Column location factor(22.6.5.3) as=40 Concrete shear strength (22.6.5.2) vcpa= (2+4/(3) x x 1I(fc x 1 psi) =255.604 psi vcpa= (as x dv2/bo+2) x x 4(f0 x 1 psi) =422.947 psi vcpc=4 x .x'J(f0 x 1 psi) =219.089 psi v00= min(vcpa,vcpb,vcpc)=219.089 psi Shear strength reduction factor (1)v=0.75 Nominal shear stress capacity(Eq. 22.6.1.2) vn=vcp=219.089 psi Design shear stress capacity(8.5.1.1(d)) =(l)v x vn= 164.317 psi vu9/4vn=0.166 PASS-Design shear stress capacity exceeds ultimate shear stress load Shear diagram, y axis (kips) 5 0 0 -5 Moment diagram,y axis(kip_ft) 0 0 3.1 Moment design,y direction, positive moment Ultimate bending moment Mu.y.max=2.006 kip_ft Tension reinforcement provided 3 No.5 bottom bars (11.6 in c/c) Area of tension reinforcement provided Asy.bot.prov=0.93 int Minimum area of reinforcement(8.6.1.1) As min=0.0018 x Lx x h =0.54 in2 PASS-Area of reinforcement provided exceeds minimum cJ Jvj ��,,,/E,.,,z 1�'(' Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 76 of 78 Maximum spacing of reinforcement(8.7.2.2) smax=min(2 x h, 18 in) = 18 in PASS-Maximum permissible reinforcement spacing exceeds actual spacing Depth to tension reinforcement d=h-Cnom-ttx.bot-ty.bot/2=6.063 in Depth of compression block a=Asy.bot.prov x fy/(0.85 x fc x Lx) =0.729 in Neutral axis factor 131 =0.85 Depth to neutral axis c=a/131 =0.858 in Strain in tensile reinforcement(8.3.3.1) Et=0.003 x d/c-0.003=0.01819 PASS- Tensile strain exceeds minimum required, 0.004 Nominal moment capacity Mn=Asy.bot.prov x fy x (d-a/2)=26.495 kip_ft Flexural strength reduction factor = min(max(0.65+ (Et-0.002) x (250/3), 0.65), 0.9)=0.900 Design moment capacity 4Mn=4t x Mn=23.845 kip_ft Mu.y.max/4/Mn=0.084 PASS-Design moment capacity exceeds ultimate moment load One-way shear design, y direction Ultimate shear force Vu.y= 1.985 kips Depth to reinforcement dv= h-Cnom-ttx.bot-4 y.bot/2=6.063 in Shear strength reduction factor (I)v=0.75 Nominal shear capacity(Eq. 22.5.5.1) Vn=2 x x i(fc x 1 psi) x Lx x dv= 19.923 kips Design shear capacity 0/0=4v x Vn= 14.943 kips Vu.y/4Vn=0.133 PASS-Design shear capacity exceeds ultimate shear load Two-way shear design at column 1 Depth to reinforcement dv2=6.375 in Shear perimeter length (22.6.4) lig)= 10.375 in Shear perimeter width (22.6.4) lyp= 12.375 in • Shear perimeter(22.6.4) bo=2 x lxp+2 x lyp=45.500 in Shear area Ap=lxp x I yp=128.391 in2 Surcharge loaded area Asu,=Ap-1.1 x ly1 =104.391 in2 Ultimate bearing pressure at center of shear area qup.avg= 1.753 ksf Ultimate shear load Fup=yo x Foz1 +yL X FLz1 +ys x Fszi +yo x Ap x Fswt-qup.avg x Ap=8.589 kips Ultimate shear stress from vertical load vug=max(Fup/(bo x dv2),0 psi)=29.612 psi Column geometry factor(Table 22.6.5.2) 13 =ly, /lx1 = 1.50 Column location factor(22.6.5.3) as=40 Concrete shear strength (22.6.5.2) vcpa= (2 +4/13)x ti x 4(fo x 1 psi) =255.604 psi vcpb= (as x dv2/bo+2) x x Al(fc x 1 psi) =416.510 psi vcpc=4 x x J(fc x 1 psi) =219.089 psi vcp= min(vcpa,vcpb,vcpc) =219.089 psi Shear strength reduction factor (1)v=0.73 Nominal shear stress capacity(Eq. 22.6.1.2) vn=vcp=219.089 psi Design shear stress capacity(8.5.1.1(d)) 4vn=¢v X vn= 164.317 psi vug/4vn=0.180 Project New Addition Job# 180145 Address 11958 SW 125th Ct., Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 5/7/2018 Page 77 of 78 PASS-Design shear stress capacity exceeds ultimate shear stress load 3 No.5 bottom bars(11.6 in c/c) A , 3 No.5 bottom bars(11.6 in c/c) 1 CfeeN `af; yrti,ay-Lee Project New Addition Job# 180145 Address 11958 SW 125th Ct.,Tigard,OR 97223 RNS Consulting LLC Client Chris&Jessica Anderson 11611 NE Angelo Dr.#17 Vancouver,WA 98684 (503)737-4344 By R.P. Date 51712018 Page 78 of 78 r -- I- 1T ms o E i:-Fa GARAGEi 10-8 1/2" r— —i- I ----J � fu 1° I Ik1 9 Q POWDER t UNDRY a Cr M .� ._'\ ., ROOM 7-Lt% Si) Darn Orn fi1, DINING ENTRY V A f V I I } 'I•o Y AREA 1 BEDROOM 4 1„ f rorn�0 wm° - H �6 4. J'1 HALL ,,m Flx @y ,8'-3•x i0'-,D• nFM1N F �,o 'i - iA,02,TaWOR 7-11' Wul p w 30 .oa irt la f � -•� CLOSET. 1 y 4. ma .7_aNig R... tr-°•' B_„WR `,.v " . 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