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REVIEWED BLD2023-0433+Structural_Calculations+4.6.2023_3.02.08_PM+3468266RECEIVED Apr 21 2023 c' of e1-11s oev-vMEry -vi- Tang Residence Structural Calculations REVIEWED BY CITY OF EDMONDS Prepared By: 840 Daley Street Edmonds, WA 98020 Snohomish County Sidesway Project No. 23010.01 SIDESWAY ENGINEERING BLD2023-0433 ow/o5/zv Tang Residence April 5, 2023 Project No. 23010.01 TABLE OF CONTENTS DESCRIPTION PAGE NO. Project Summary Gravity System Lateral System Wall Overlay 1.01 Design Loads 1.02 Roof & Floor Framing Design 1.02 - 1.06 Upper Floor Girder Truss Loads 1.04 Foundation Design 1.07 - 1.08 Seismic & Wind Loads 2.01 - 2.04 Shear Wall Design 2.05 - 2.16 Diaphragm & Chord Analysis 2.17 20305 87th Avenue W. SIDE5WAY Edmonds, WA 98026 -4160 (425)67673-4160 ENGINEERING Tang Residence April 5, 2023 Project No. 23010.01 Project Description Sidesway Engineering was retained by the homeowner to perform analysis and design as necessary to obtain a building permit for the proposed single family residence and ADU to be located at 840 Daley Street in Edmonds. The structure will be conventionally wood framed with pre -manufactured roof and floor trusses spanning to stud bearing walls and girder trusses with sawn lumber joists framing the crawl space. The structure will bear atop a continuous exterior footing at frost depth with interior spread footings in the crawl space and a slab on grade at the garage and ADU. The lateral force resisting system consists of sheathed roof and floor diaphragms spanning to various wood stud shear walls at the perimeter and interior. Scope of Work Provide gravity and lateral calculations for the proposed structure as required to obtain a building permit. Provide structural framing plans, notes, and details as required for permit. Design Criteria 2018 International Building Code (IBC) ASCE 7-16 Minimum Design Loads for Buildings and Other Structures Applicable Material Reference Standards (ACI, AISC, NDS) This is a Risk Category II structure designed for the following loads: Dead Loads: 15psf (roof), 15psf (floor), 9psf (walls) Snow Load: 25 psf Live Load: 40psf (floor), 60psf (deck) Wind Load: 100mph, Exposure'B', KZT = 1.0 (refer to wind loads) Seismic Load: R = 6.5 (wood s.w.), Site Class D, SDC D (refer to seismic loads) Project Summary The proposed single family residence with ADU as designed in the following calculations conforms to the 2018 IBC. Refer to the calculations and the construction drawings for structural framing requirements. 20305 871h Avenue W. SIDE WAY Edmonds, WA 98026 (425)673-4160 ENGINEERING Po© L.2Pc- L et ooz s . 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"C, --- LOoi- F-(Z-Aim 41) f C� _ 4 � `-� r t-(DfL-- = 7 � SC- �t x b � r✓-�t Z � �� � --��c �- .� ii � I' �ir-� Cr � � 'o b 5 Q ' �2� �-= E Z � 5 Itc�2 `7 liSG IX�Z o�42 w �r> sAL�� c�r�rr=o.�Z L�= sz`� 5 0. ti3 Ft,L, buKAPoc-z.T cA5�, r•JL-S`r L iiC--�/r zJs� f Sf2S�1Y6;O-� Description � Project No. ��S Z LO S t.•.t:r J !r 1 Dat? ! 7T ?3oioo( I D E WAY Project N ' r G Checked Sheet No. ENGINEERING Date I t0-,7 (6,4 t-J'T-rr"u 5 rBot►T Voo�7 VC7 3zXq D&LIE5 s1- /r4r c�Sc-c> x 6 ;Zv3� t.J — 0, `:rf rh -74 t2 N \ r�0 a P= (moo -vL-+ t cook ;7c 4 u34-vt_ f2t-4t Q V,h& 1�� 6x6 F z � /f :7 03 Y t � Gt Description ) By Project No. �— Date I D E WAY Project ('^� - Checked Sheet No. 5 Date ' 0 j ENGINEERING ( `i3(,.P` �-15Go5l. t(i GT2. 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S 3 ti5r 1Oz; l� RS c�3S `L=�Zt' yz vc 4- 667- 0 5 "In �sz ��63`c 3otiy `c `L=Z066 Y066P' c( ZZ3Z5` 3-Z J l 3G I lq6 s z163s� IZ = IQQQ n f Z c`3zs °Lf Ir3c�5` ELF 6-7 �3' S ° f 631 "f 3�S sL� � �J�7s f �(°r"f16z6 � ��a-1r✓ S-0C�C L,: ,i (p y = S °`f6 3r�` f3�S5t J► f I �3° f-ig341L a FC-X 1,4s Ps = °�z°+ IrS��4-4- 5`-+ g�Cc' � C-TC r5�kn) d �'�cc�� a Act 3z Ar�3 �i�d ��S 6 01=163cj °1k3 N � , L,c 5G- 4(ro Prim q CAS J 1J � zxio t� z ��Fb 'c `jZ a = -� ��7- r 5 us6 k x,z Li r SrQrc� C /, � Z_ 2466R 01 BSc i IJ S q d� d 7 U5'X%�* -Z Description By � Project No. Date 2 300,U( IDE WAY Project f � Checked Sheet No. 5 5 ENGINEERING Date I ,Q� �—�C— � � �" �.K3 r ,. �,-sG— 2 � �• � �3 rc�P c-•�� �3� � K G-n `"��Y c[ = 2 . t; r / c w `f df�GK r(,' S _ ' = 3�f3� t-SG-- o �x�2 "�� �� laii L O S e c& S 1 l( �� i 'G-rR-' � tZ- —� Q � � c.�. r-.c. � f h r � =''m7G ,� d -s_ 2 • � • l�av � r~- % 2oa� �� �%` �'ZGOc.e� � t�►J ((, � (S -E2 �� � �—� ��� �• (q � � �-trc.� -f l (`� � t S � � v) c-< � :f � `;o f G � . I . �, r"�"z � �= 7 ZZ PL,� FL - OF re au ' � � D.�3S�G` / �• in2 • (J� D �y• 2.� • t` mm D.�; W �� " Q , �. Q b urH2 Gt� �$ - / ` S36r V z wL�u 1t( �tS-f-2S> ..t- `�•�q-�- !t � �(SfKn) 'F Iso-(.S=(Z6Z�'t�f=���tLr Description By Project No. ... Date � —F30 (0. n- 1, 7 ? I D E WAY Project �.�! �� Checked Sheet No. 5 5ENGINEERING (� Date )�OF E Address: 840 Daley St Edmonds, Washington 98020 ASCE 7 Hazards Report Standard: ASCE/SEl7-16 Risk Category: 11 Soil Class: D - Default (see Section 11.4-3) Latitude: 47.813552 Longitude: -122.36814 Elevation: 173.09 ft (NAVD 88) F ,. NIA PGA ,, : 0.658 S"c 1.546 F pj., 1.2 Sup; : NIA Ir : 1 SD 1.03 C, _ 1.358 Ground motion hazard analysis may be required. See ASCEISEI 7-16 Section 11.4.8. Data Accessed: Tue Feb 28 2023 Date Source: USGS_Seismic Design Mats mot= = 3 o36 F-r7' Z�3�t�F't2 �IS�SF F/-Fc�PSr wAu1s� ��, Uescnption Seismic Soil Data and Seismic Weight "' )MD ' ". Date 02/28/22 23010.01 I jj C �A►a'�J Project Tang Residence checked sheet No. SjD L YY j - ---- — - --- ---- ----- ---- - ------ --- --- --- - - -- ENGINEERING MSFRS Seismic Loads Seismic Parameters: Latitude/Longitude = 47.813552, 122.36814 Risk Category = II (ASCE 7-16, Table 1.5-1) Importance Factor, Ie = 1.00 (ASCE 7-16, Table 1.5-2) Soils Site Class = D (Per geotech, else per 11.4.3) SDs = 1.03 (USGS) SI = 0.454 F„ = 1.85 SDI = 0.5599 Seismic Design Category - D (ASCE 7-16, Table 11.6-1, -2) Building Properties: Response Modification Coefficient, R = 6.5 (ASCE 7-16, Table 12.2-1) Overstrength Factor, no 3.0 (ASCE 7-16, Table 12.2-1) Deflection Amplification Factor, Cd = 4.0 (ASCE 7-16, Table 12.2-1) Fundamental Period, Ta = Cthn" Ct = 0.02 (ASCE 7-16, = 0.177 x= 0.75 Table 12.8-2) k = 1 (ASCE 7-16, Section 12.8.3) 1.5*TS = 0.815437 (If 1.5Ts < Ta, see 11.4.8) Seismic Response Coefficient, Cs: V = 0.044SDSI (Minimum) = 0.045 W V = (SDSIW)/R = 0.158 W GOVERNS V = (SD1I)/RT, (Maximum) - 0.487 W Vertical Distribution of Seismic Forces: Diaphragm DL Area wDL Story w;h; wXhX Force Fx Sum Level (psf) (ft`) (kips) Ht. (ft) (k-ft) E w;h;" (kips) FX Roof Framing 20 3036 60.7 18.3 1111 0.61 8.85 8.85 2nd Framing 25 2840 71.0 10.2 724 0.39 5.77 14.61 Base Shear (ULT) _ Base Shear (ASD) _ Diaphragm Design Forces: F = 131.7 1835 1.00 14.61 20.87 kips 14.61 kips Diaphragm w; F w; F; E F; i F; * wpx Fpx (Min) FpX (Max) FpX Level (kips) (kips) (kips) (kips) E w; 0.2SDSIwpx 0.4SoSIwpx Govern Roof Framing 60.7 60.7 8.8 8.8 8.8 8.76 17.51 8.85 2nd Framing 71.0 131.7 5.8 14.6 7.9 10.24 20.48 10.24 r Seismic Vert Distribution JMD Date 2/28/2023 23010.01 ID E WAY Project Tang Residence Checked Sheet No. Date ENGINEERING ZI �7, 2 S© Fr ss D j Soo yt. �{SLL--l4, ZG. or- wr� k 2-r = I. o �san,e A.01 a �..;e 0'to A56' •d/2'-120 ca.nn,;ng ^t n9':.,'76 hrc. a; x.xyge: abate :- �.:ne SI-rE Description Kzt Determination BY JMD Project No. ---------- - --------------- - _ __ —_ ___---- Date 02/28/23 02/28/23 2 0 eD( ID E WAY Project Tang Residence Checked sheet No. ENGINEERING Date Z �+j 9sF 5� �0r2-rtK /59C.-rm Etc-vartzo�)s por A A 4 =10,1 r-TZ g : 101 1001 Acu-AS ;::l7 I i— A::: 13P ( w 1131p(Z 5IDE WAY ENGINEERING WPM i EL&Ua-,ram p ,8o# 11 I�r- - 10 l i r'r z. 4- $fit" I t OFT 71 ,, Ir r�JG j tJS �- zZOr z LAF FVGC. © NoR r/Sn4r�r ooF NormaT.r CcP 5617Ar w�•� E� _ �= oo Zb�S a T iption Wind Loads 7MD p" ---------------------------------- Deis---02/28/23 23010.01 Tang Residence 2.oy L 4f SLq4 EG2 G-n���s d-�-�- .: w�� l.o��s Noy srto�..�►� Lt 60-A L Ls,JEOA W(T2 = Z746 M 30 3 6 F-( L,�4 E & 3o36�z 3e5r, F CZ �Z!3Z �... - 1G .r .. r3 n 'AJOiuS i Lzac- p� . S32F� = z KZs L�aC—509 rz� `i 39q -3v36 z1 t Description Roof Shear Wall Lengths By JMD Project No. Dare 02/28/23 23010.01 I D E C S 5WAY Project----y � 6- ��s --- — —` Checked Sheet No. -__ --- ENGINEERING Dare -2,� 9. 6 PL C Sw 6 14 DC-C� (AAT c rod F r o.o S z S. Z C-LLD -A 21gO � r�fz- /3 r. (Z) Z' Z cJ ll c c 5 u e 6 DC--Q -A-< 1-fa s L-5(6 2 — 3 L W A-l-L � -T"-,,j S FL-r - & r-t /-- & -T 2,Oq) Sc,✓ G oG- a "'. Tc- Cl." 4 ` 3z .1 r � = 11I Z FLU LA4 -r &- �—c�.zs utiSr 37- e4 �� 7 Fold- C"i L= I I06; Z 3 a 4 EQ LAL o, 7- 1, z 1 1r Z t / 3 ID�JA�t 3r� 5W6 o zz`WA-u,o ee--r f�lzfiLr=-Z2`•�3.1 o.�e-(rs 6-���s� 8,r�,I�t: "�s� 3 6 I P4-F 3:73Pi-r L�6 r l Py�3,Q2 r r � (�s �.!i•34� - 3ex� �. - sc r�STwjL 3kZ� ON G C-mJ 1) z� r-f 0e Description By, - Project No. Date �3 O �0. o1 5 I D E WAY 3 z 23 Project Chec ed Sheet No. ENGINEERING Date 2,D(,7, Gjr-( (--/ ra- tA)J &-(- EDO� CAN T-j�1-1-C-0 V= ;.�7--';' -&=-Iq ( V= 17-3, z F-L, r 5� ( , 4DC—,Q C.AA—C & ------------ L Z ",,1,4 U- 4 T., 1-t , - 0.25' 61.)Ak- (v V = Ll,>qq ;v- Z&% 2 Z,'3 c/) JZ 7ti �XtJ T�-lJ6— � -V=- 13 ( , 7- Pl- F -S eJ 6 A De--C� --A--r6 /14 / "r.- D. -'t 6 65 1/ -+ q, g. o , 2 t-- 1, 6 , '-I- < K-- ,.-I -f- 3 ---Z— r 0 0 1. ,6,'20 r It 3 36 Description By Project No. Date I D E WAY Project Checked Sheet No. Date ENGINEERING Project Information Code: 20181BC Date: 3/1/2023 Project: Tang Residence Wall Line: Roof Line A 2.8' Walls Shear Wall Calculation Variables VI 1006 lbfl Opening 1 Li 2.80 ft ha 1.10 L2 2.80 ft hp 5.10 hwn, 8.20 ft hb 2.00 L-11 12.02 ft Lol 6.42 1. Hold-down forces: H = Vh*„ i,/L*,*li 686 Ibf 2. Unit shear above + below opening First opening: val = vb1 = H/(h,+hb) = 221 plf 3. Total boundary force above +_below openings First opening: 01 = vat x (Lol) = 1421 Ibf 4. Corner forces F1 = 01(Ll)/(Ll+L2) = 711 Ibf F2 = Ol(L2)/(Ll+L2) = 711 Ibf 5. Tributary length of openings Tl = (Ll*Lo1)/(Ll+L2) = 3.21 ft T2 = (L2*Lol)/(Ll+L2) = 3.21 ft P2=hdL2= 1.82 1 N/A 6. Unit shear beside opening vl = (V/L)(Ll+Tl)/Ll = 180 plf v2 = (V/L)(T2+L2)/L2 = 180 pif Check vl*Ll+v2*L2=V? 1006 lbf OK 7. Resistance to corner forces R1=vl*L1= 5031bf R2=v2*L2= 503 lbf 8. Difference corner force + resistance RI-F1= -208 Ibf R242 = -208lbf 9. Unit shear in comer zones vcl = (R1-F3)/L1 = -74 pif vc2 = (R2-F2)/L2 = -74 pif Check Summary of Shear Values for One Opening line 1: vcl(h,+hb)+vl(h,)=H? -230 916 686 Ibf Line 2: va1(h,+hb)-vcl(h,+hb)-vl(ho)=0? 686 -230 916 0 Line 3: va1(h,+he)-vc2(h,+hj,1(h.)=0? 686 -230 916 0 line 4: vc2(h,+hb)+v2(ho)=H? -230 916 686 Ibf Req. Sheathing Capacity 221 pif Req. Strap Force 711 Ibf Req. HD Force (H) 686 Ibf Req. Shear Wall Anchorage Force (v, .,) _._ 84 pif Design Summary* 4-Term Deflectionj 0.056 in. I / 3-Term Deflection 0. 101 in. rm ift % �. 4- Term 5to 11Arift % : 0.002% .i,6)� N j ,.. (t C/to �t��� W/0 !+ 7l e2 � 686—(9.q6-04- *The Design Summary assumes that the shear wall is designed as blocked. f✓ v —��/ OJ o 1-1D 2�Q Project Information Code: 20181BC Date: 3/1/2023 Designer: JMD Client: Project: Tang Residence Wall Line: Roof Line A 23' Wails arrrs�rsntt�strm V (Ib! d tr y a L-00 Shear Wall Calrulatinn Variahla V1 826 lbfJ Opening 1 1.11 2.30 It I h, 1.10 ft L2 2.30 ft h, 5.10 ft hw.n 8.20 ft hb 2.00 ft L..0 11.02 ft Loi 6.42 ft 1. Hold-down forces: H = Vh,,yi/L,,,i, 615 Ibf 2. Unit shear above + below opening First opening: vat = vb1= H/(h,+hb) = 198 plf 3. Total boundary force above + below openings First opening: 01= val x (Lol) = 1273 Ibf 4. Comer forces F1= 01(Ll)/(Li+L2) = 636lbf F2 = 01(L2)/(L1+L2) = 636 Ibf S. Tributary length of openings T1= (l1*Lol)/(LS+L2) = 3.21 It T2 = (L2*Loi)/(L3+L2) = 3.21 ft V Ub)b) P2=h,/L2= 2.22 1 0.902 6. Unit shear beside opening vl = (V/L)(Ll+Tl)/L1= 190 plf v2 = (V/L)(T2+L2)/L2 = 180 pif Check vl*Li+v2*L2=V? 826lbf OK 7. Resistance to comer forces R1=v1*L1= 413lbf R2=v2*L2= 413lbf B. Difference corner force + resistance 11143 = -223 Ibf R242 = -223 Ibf 9. Unit shear in comer zones vcl = (R7-Fl)/L1= -97 plf vc2 = (R2-F2)/L2 = -97 plf Check Summary of Shear Values for One Opening Line 1: vcl(h,+hb)+vl(h,)=H? -301 916 615lbf Line 2: va1(h,+hb)-vc1(h,+hb)-v1(h,)=0? 6 615 -301 916 0 Line 3: va1(h,+h,l-vc2(h.+he)-v1(he)=0? 615 -301 916 0 Line 4: vc2(h,+hb)+v2(h,)=H? -301 916 615 ibf Design Summary* Req. Sheathing Capacity 199 plf M 4-Term Deflection I 0.080 in. 3-Term Deflection 0.130 in. Req. Strap Force 636 Ibf 4-Term story Drift % 0.003 % 3-Term Story Drift % 0.005 % Req. F Force (H) 75 Ibf Req. Shear Wall Anchorage Force (v,,,,) 75 pif / �t / C / �y` C/�J l 7 ( 0/1 **Req. Sheathing Capacity has been adjusted per the "The that the Is designed as biocked2 Aspect Ratio Adjustment Factor Design Summary assumes shear wail d Project Information Code: 20181BC Date: 3/2/2023 Designer: 1MD Client: Project: Tang Residence Wall Une: Roof Line B FT Walls 1. Hold-down forces: H = Vha/Lw,i, 8581bf 2. Unit shear above + below opening First opening: vat = vbl = H/(h,+hb) = 168 pIf 3. Total boundary force above + below openings First opening: 01= val x (Lol) = 393 Ibf 4. Corner forces F1= Ol(L1)/(Ll+L2) = 221 Ibf F2 = 01(L2)/(L1+L2) = 171 Ibf S. Tributary length of T1= (Ll*Lo1)/(Li+L2) = 1.32 ft T2 = (L2*Lol)/(Ll+L2) = 1.02 ft 6. Unit shear beside opening v1= (V/L)(Ll+Tl)/L1= 142 pIf v2 = (V/L)(T2+L2)/L2 = 142 pIf Check vl*LS+v2*L2=V? 9761bf OK 7. Resistance to corner forces R1=v1*L1= 5501bf R2 = v2*L2 = 426 Ibf 8. Difference corner force + resistance 11143 = 3291bf 11242 = 2541bf 9. Unit shear in comer zones vcl = (Rl-Fl)/L1= 85 pIf vc2 - (R2-F2)/L2 = 85 pif Check Summary of Shear Values for One Opening Line 1: vcl(h,+hb)+vl(h,)=H? 433 426 858 Ibf Line 2: va1(h,+hb)-vc1(h,+hb)-v1(ho)=0? 858 433 426 0 Line 3: va1(h,+h&vc2(h,+hb)-v1(h,)=0? s�� 858 433 426 0 Line 4: vc2(h,+hb)+v2(h,)=H? 433 426 858 Ibf Design Summary* Req. Sheathing Capacity t 168 If 4-Term Deflection 0.033 in. 3-Term Deflection 0.065 in. Req. Strap Force 221 Ibf 4-Term Story Drift % 0.001 % 3-Terri Story Drift % Req. HD Force (H) 858 Ibf < S� 1p'" iVi.L _ •. fA,Pc n �� d Req. Shear Wall Anchorage Force (v,*„ j 106 pif \fir k .The Design Summary assumes that the shear wall Is des*�as blocked. Z c� 14 T,J u'uve"f F.C"ck Ra,f NGt-�1W�gc, L; :. Nb O-Z-60 3f 3 Q W Zn� D LOADS No7 �fftx�J LTG-�i1L IOA D Lza�—� -70 -f- 5� - !bo8r--r -7-T3.7.� ?�40 iL 0 A 9 r 19 ,v zr,�0 lyry f s� .236 =18�� Zgko ? L1 N C-10 2L, ;L7oerz =3R e+ i Rx.� f Z lei 7$ FzZ = 7- g o z Description Upper Floor Shear Wall Lengths JMD Project No. _--------_ _..�_____.—�-------_--------- ------_...._--_---- Dare 0_,: 23 23010.01 I D E WAY Project Tang Residence ENGINEERING 3 so Pl..F are swf 5.7 ra M- '� ` ,uA t4 5 L� f4)6j " Pa 'fq C, � CZ� 3 f `lW4Q, S u_SG— Fo Fell- �S(4 C-C--r- Sw6 GI06--C�V'Ar� �( tt z c-t4c—�o otz �S�r�r�� U!6 1�$V�C-Ce%i°L"t�r G-l(ll�8d -Z �� 1r 3 5't y,/A-GU-f2AtJ5rz-7Z owy r--'40 Aav 3 -o` h}��'�' =Z��,� P�-F • 3 �►a.z _ cv,�/G (�S.Z-f-R.fo.z)�i,S=Z��I -�77q =also, z, s' Jbtc. 7-09,7-PLf=—cc- ✓G-t�--'�G- ,, Pi—(-- 41, � t • !to,z ' _ 62. Y6- (lS • 16.3 • `� 7� S A � y' 75 r - o- S 3OZY' uSC P Dli 1( &A <--d-0 O (- ' - 'f LAV 3g150 3L 6` wAU- '�- 'i g'wdc-- Esc, Fa9-661- T-"iQ5a.id 5W-3 I-J/ KOLkZ E-tl �,.�� v(Z-- Szr-(DfO, �tgv �! Sa N��� NETS- Lfn � -f ZBY,�P�F �.ybrI o,zG.y6 o, y� (t S •2t 10.2� z - 3c� S z �`:: a S� S� >•-r j f �ZB�.�•V.i IC�.Z _�st6 z Zf6o�. Sc Dl Description By3 Project No. ... S1-f t-� fZ- W d •- C � Date Z30 CO- 01 I D E WAY Project Checked Sheet No. 5 ENGINEERING Date Z.12 y�C- X te a- C(-s j��T �- i K� i�.z' — d.y.� -�IS•6 --� q'►�3>� 6,2 ,�Zq�� L= �►E �/= 3 �? -rr7 ,&V = z3 , 1 1 'l - 16 �. 6 A?v 4- o_ 16�y 'y.sv (z o. s C- LS (-Q3 C,q �ASv 4, 13 Z, ` r o6c- NC- `f =I Iqz -� �z =Z1 84( , ue, I-tpLc55 (-rfll0 L-xoeo \/= 7k/ - 5t A t i Uri FTb,5 iF�tv- ABvC- Ov C-'R,:—, , �Jaft-ri'C C2tSl d6✓ Z(- 2 ',� t-��' wd� ra E-^� �P = von �-} Z32 � PLi= .� �, z t• � o.z' ,�. o. YG - �5 - � � -t- � • �o z' � :�6 t Ag ue 4 Z37.2 i &r-. 7_ �s ` to Z � _ 0 A-:-- Ls�E 3 V= 3ZZ� = �f S 'l/ _ -7(m ��� , : s�6 .t p�c� Ou &RTu nl G--� � = ��, � PL% - 6 - l �-2 � li• Y6 •� �S •) � f-F�-- Lt LS r A-t P 0 � � ft Jet u 9'-0 F�E��= -��6( : 16r lJo Hp C�d� Description r Bya Project No. INt� .... Date I DE WAY Project Checked Sheet No. ���►� �-c-sue- �c- ENGINEERING Date 2.13 Project Information Code: 20181BC Date: 3/2/2023 Client: Project: Tang Wall Une: Upper Floor Line A T-11" Walls V (lby) —� Shear Wall Calculation Variables 1. Hold-down forces: H = Vh,/1,i 1133 Ibf 2. Unit shear above + below opening First opening: vat = vb1= H/(ha+hb) = 203 pif 3. Total boundary force above+ below openings First opening: 01= val x (1.01) = 1997 Ibf 4. Corner forces F1= 01(Ll)/(Ll+L2) = 998 lbf F2 = 01(L2)/(L1+L2) = 998 Ibf S. Tributary length of openings T1= (Ll*Lol)/(L3+L2) = 4.92 ft T2 = (L2*Lol)/(Ll+L2) = 4.92 ft Check Summary of Shear Values for One Line 1: vcl(ha+hb)+vl(ho)=H? Line 2: va1(h,+hb)-VC1(ha+hb)-v1(ho)=0? .Line 3: val(h,+he)-vc2(ha+he)-vl(h,)=07 Line 4: vc2(ha+hb)+v2(ho)=H7 Req. Sheathing Capacity Req. Strap Force Req. HD Force (H) Req. Shear Wall Anchorage Force (vm„) 6. Unit shear beside opening v1= (V/L)(Ll+Tl)/LS = 241 pif v2 = (V/L)(T2+L2)/L2 = 241 pif Checkvl*Ll+v2*L2=V? 18931bf ON 7. Resistance to corner forces R1=v1*L1= 947 lbf R2 = v2*L2 = 947 Ibf 8. Difference corner force + resistance R1-F3= -52 lbf R2-F2 = -52 Ibf 9. Unit shear in comer zones vc1 = (Rl-Fl)/L1= -13 pif vc2 = (R2-F2)/L2 = -13 plf .74 1207 11331bf 1133 -74 1207 0 1133 -74 1207 0 C7 -74 1207 11331bf Design Summary* 241 pif 4-Tenn Deflection 0.101 in. 3-Term Deflection 0.168 in. 998 Ibf 4Term�/Syytory Drift/% II0.003 % \/� t 3L-Term Story Drift % [0..005 % nn V17p1f ggt3 . 6''{.6. SSG L7l)� wl(1i) C-LL �7Y 13 plf It 33 — o.Kb (15 -k -6116 ' -The Design Summary assumes that the shear wall is designed as blocked. Project Information Code: 20181BC Date: 3/2/2023 Designer: tMD Client: Uooer Floor Line A 2'-7" & 3'-5" Walls V LI L2 h.aii L—, 10.58 ft 12.33 ft V (Ibj Shear Wall Calculation Variables upening i h, 2.58 ft ho 5.00 ft hb 3.00 ft W1 6.33 ft 1. Hold-down forces: H = Vh,,,,i,/L„„ , 1242 Ibf 2. Unit shear above + below opening First opening: vat = vbl = HAIn.+hb) = 223 pif 3. Total boundary force above + below openings First opening: 01 = val x (Lol) = 1409lbf 4. Corner forces F1 = 01(Ll)/(Ll+L2) = 606lbf F2 = Ol(L2)/(Ll+L2) = 803lbf 5. Tributary length of openings T1 = (Ll*Lo1)/(L1+L2) = 2.72 ft T2 = (L2*Lol)/(Ll+L2) = 3.61 ft P1=hdL1= 1.94 N/A P2=hdL2= IAA 1 N/A 6. Unit shear beside opening vI = (V/L)(Ll+Tl)/Ll = 241 plf v2 = (V/L)(T2+L2)/L2 = 241 pif Check vl*Ll+v2*L2=V? 1448lbf ON 7. Resistance to corner forces R1=vi*L1= 623lbf R2 = v2*1.2 = 825 Ibf 8. Difference corner force + resistance RI-Fl = 17 Ibf R2-F2 = 22 Ibf 9. Unit shear in corner zones vcl = (RI-F1)/L1 = 6 plf vc2 = (R2-F2)/L2 = 6 pif Check Summary of Shear Values for One Opening Line 1: vcl(h,+hb)+vi(hp)=H? 36 1207 1242lbf Line 2: val(h,+hb)-vci(h,+hb)-vi(ho)=O? 1242 36 1207 0 1242 36 1207 0 Line 3: va1(h.+h„j-v12(h•+he)-v1(hej=0? w Line 4: vc2(h,+hb)+v2(hj=H? "'`"""'"'' 36 1207 1242 Ibf Req. Sheathing Capacity 241 plf Req. Strap Force 803 Ibf Req. HD Force (H) 1242 Ibf Req. Shear Wall Anchorage Force (v, .,) 117 pif Design Summary* 4-Term Deflection 0.115 in. 3-Term Deflection 0.182 in. 4-Term Story Drift %I 0.004 % 3-Term Story Drift % 0.006 % 61- l 6 = ) �d c✓� 6-34 I .The Design Summary assumes that the shear wall is designed as blocked. Gc 56 xr � e4r C—W L57KO� Z.is Project Information Code: 20181BC Date: 3/2/2023 Designer: 1MD Client: Project: Tang Wall Line: Upper Floor Line B 3'-6" & 4'-6" Walls Shear Wall Calculation Variables 1. Hold-down forces: H = Vh_,JLw,ii 22251bf 2. Unit shear above + below opening First opening: vat = vbl = H/(h,+hb) = 366 pif 3. Total boundary force above + below openings First opening: 01- val x (Lol) = 9151bf 4. Comer forces F1 = 01(Ll)/(Ll+L2) = 400 Ibf F2 = Ol(L2)/(Ll+L2) = 5151bf S. Tributary length of openings T1= (Ll*Lol)/(L3+L2) = 1.09 It T2 = (L2*Lol)/(Ll+L2) = 1.41 ft 6. Unit shear beside opening v1 = (V/L)(Ll+Tl)/L1= 290 pif v2 = (V/L)(T2+L2)/L2 = 290 pif Checkvl*L3+v2*L2=V? 23181bf OK 7. Resistance to comer forces R3=v1*L1= 10141bf R2=v2*1.2= 13041bf 8. Difference corner force + resistance 11143 = 614lbf R242 = 789 Ibf 9. Unit shear in comer tones vcl = (RI-Fl)/Ll = 175 pif vc2 = (R2-F2)/L2 = 175 pif Check Summary of Shear Values for One Opening Line 1: vc1(h,+hb)+v1(hJ=H? 1066 1159 2225 Ibf Line 2: va1(h,+hb)-vc1(h,+hb)-v1(ho)=0? 2225 1066 1159 0 Line 3c va1(h,+h,)•ve2(h,+hJ.- 1(ho)=0? 2225 1066 W 77 1159 0 Line 4: vc2(h,+hb)+v2(K,)=H? 1066 1159 22251bf Design Summary* Req. Sheathing Capacity 366 If 4-Term Deflection 0.119 in. 3-Term Deflection 0.181 in. Ibf Req. Strap Force��2225 Req. HD Force (H)1bfReq. 4-Term Story Drift % 0. % p\ ' % 3-Term Story Drift % 0.006 % Shear Wall Anchorage Force (v,„„jpIf o.Lc 6 Ct5 `(f 'f • to.z)'+�?f = (8 58�� `� S� *The Design Summary assumes that the shear wall Is designed as blocked. Z us 6-kA�y (/ -1!X t H 6-1, I-cD D a 6 _ar7 o36 2 - 6 T I Z !/SKG.4T�'t�rJ� ���Z r� /0 ZoFT� 2 SKI # �= z5��� zd�o 3S �S !r r (( 2 # Description ra By Project No. Date 2 I D E WAY Project ` � Checked Sheet No. F Date -Zv 1 ENGINEERING