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Specifications 2010 ORSC: Wind Load Development, ASCE 7-05, Chp.6, MWFRS J2K Engineering, Inc. Method 1 -Simplified Procedure ( ASCE 7-05 Section 6.4) I. Review Building conditions for compliance with Section 6.4.1.1 provisions: 1.The structure is a "simple diaphragm" building. 2.The structure is a "low rise" building (mean roof height is less than 60 feet). 3.The structure is "enclosed". 4.The structure is a "regular" shaped building. 5.The structure is not a"flexible" building. 6.The structure does not have special response or site characteristics. 7.The structure has an approximately symetrical cross-section in each direction. 8. The structure has either a flat, gable, or hip roof profile with slope less than or equal to 45 degrees. 9.The structure is exempted from torsional load cases. II. Basic Wind speed,3-sec. gust ('10 OSSC Fig. 1609) V,:= 95mph III. Building Category (ASCE Table 1-1) Category II Building ("All" buildings, except ..) IV. Importance Factor(ASCE Sect. 6.5.5&Table 6-1) I:= 1.0 V. Exposure&Roughness Category(ASCE Sect. 6.5.6) Exposure"B" (urban &suburban &wooded areas) VI. Building height&exposure adjustment factor(ASCE Fig. 6-2) := 1.0 VII. Topographic factor (ASCE Sect. 6.5.7) KZt:= 1.0 mean roof height, h := 25ft+ Oin h =25 ft Roof Pitch, in/ft, P := 6 ft in (27 degree) g ) VIII. Simplified design wind pressures (ASCE Fig 6-2) (Horizontal loads are checked with roof zones (B & D)= 0 psf) qA:= 14.4psf qc := 11.5psf "net" design pressures: PA:= )�•IKt•l•gA PA= 14.4psf Pc 2'IKt-I•gc Pc= 11.5 psf IX. Building design pressures : The greater wind load is governed by the ORSC,therefore, ... USE: Pa:= 14.9psf Pc:= 14.9psf PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 3 of 59 t � - Wind Load Development (cont'd) J2K Engineering, Inc. Develop the horizontal wind loads for the building: A) Building END walls (Front/Rear Loading Direction) "applied"from REAR face of building: SMn :_ (Pa)•(9ft•70ft) SMn =9387 lb SLwr= Pa•9ft•67ft SLwr= 8985 lb TOTAL Wind Loading to END walls: Sena := SMn+ SLwr Sena = 18372 lb •0(- Building *Building "SIDE"walls (Left/Right Loading Direction): Wside Pa(9ft•52ft+ 0.5.34ft•6ft+ 9ft•16ft) TOTAL Wind Loading to "SIDE"walls: Wside= 10639 lb PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: . of ASCE 7-05 SEISMIC Load Development J2K Engineering', Inc. " Loads are developed in compliance with ASCE 7-05 Chapters 11 & 12 I. Seismic Ground Motion Values (Sec. 11.4 ASCE 7-05) 0.2 Sec Spectral Response Accel, Ss-0.95 (Fig. 1613.5(1) '10 OSSC) 1.0 Sec Spectral Response Accel, S1 := 0.45 (Fig. 1613.5(2) '10 OSSC) Site Soil Class, SSC:_ "D" (Sec. 11.4.2 ASCE 7-05) Site Coefficient, Fa, Fa= 1.12 (Table 11.4-1 ASCE 7-05) Site Coefficient, Fv, Fv := 1.55 (Table 11.4-2 ASCE 7-05) Short Period Maximum Acceleration, Sms=Fa-Ss Sms= 1.06 1 Sec Maximum Considered Accel, Smi := Fv•SI Smi =0.70 Short Period Design Acceleration, Sds=2•Sins± 3 Sds=0.71 1 Sec period Design Acceleration, Sdl := 2•Smi _ 3 Shc =0.47 Seismic Importance Factor, 1E= 1.0 (Sec. 11.5 ASCE 7-05) Seismic Design Category, SC:_ "D" (Table 11.6-1 ASCE 7-05) II. Seismic Design Requirements (Chet. 12 ASCE 7-05) Basic Seismic Force Resisting System SFRNI:= "Bearing Wall System" Bracing System BS := "Light Framed Wood Shear Walls" Building Type, BT:= "Regular" Response modifcation Factor, R=6.5 (Table 12.-1 ASCE 7-05) Overstrength Factor, S 3 Deflection Amplification Factor, cd:= 4 Redundancy Factor, p = 1.3 ("regular" building) (Sec. 12.3.4.2 ASCE 7-05) Ill. Seismic Load effects & Combinations (Sec. 12.4 ASCE 7-05) Horizontal Seismic load effect, Ey,= 0.7p•VF. (Sec. 12.4.2.1 ASCE 7-05) ("ASD" combination#5) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 5 of ASCE 7-05 SEISMIC Load Development (cont'd) J2K Engineering, Inc. w IV. Equivalent Lateral Force Procedure (Sec. 12.8 SCE 7-05) Approximate fundamental period, C, := 0.02 x:= 0.75 (Sec. 12.8.2 & 12.8.2.1 ASCE 7-05) hn := 16ft+ 6.67ft (11nx Ta= Cr — Ta= 0.208 ft� Seismic Response Coefficient, C – Sas s= R_ 1E CS= 0.109 << CONTROLS Sd] Csmax'= Csmax=0.34 Ta R_ IE) Seismic Base Shear, VE=Cs•WB V. Building Weight wroof= 15•psf•(34•ft•71•ft+ 14ft•60ft+ 28ft.8ft) wMnwalls= 10•psf•2•(70•ft+ 52ft)•9ft+ 9•ft•8•psf•(52•ft+ 2.9ft) wMnFlr= 12psf•(21ft•15ft+ 36ft•12ft+ 17ft•41ft+ 16ft•30ft) + 25psf•23ft•23ft WLwalI lOpsf•(2.17ft+ 70ft)•0.50.8ft+ 8psf•(14ft+ 5ft+ 8ft)•0.50.8ft WB =wroof+ wMnwalls+ wMnFlr+ wLwall WB= 120507 lb Seismic "DESIGN" Shear, VE= Cs•WB Eh =0.7p•VE Eh= 11967 lb <<< "CONTROLS" Lateral Design for: LEFT/RIGHT loading direction ... "WIND" loading controls FRONT/REAR load application direction. VI. Establish load distribution Wroof Aroof 2 15pSf Aoof=3478 ftE pr:= 0.7 P'Cs'(woof+ 0.50•wMnwalls) Aroof oof EUpr= 1.9psf AMnFIr=2453 ft2 ELS,,.:= 0.7•p•Cs•(wMnFlr + 0.50•wMnwalls+ WLwal]) - AMnFlr ELwr=2.2psf PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: G of Shearwall Evaluation J2K Engineering, Inc. As "previously established"the SEISMIC destributed load is: VP = 1.9.psf Wall Location: "REAR" Master BR Total Wall Length, Lr:= 32in Min.Wall Length, 11 1= 32in Height of shear wall, H1 9.0.ft LOADING: VwVP•(17ft•0 50.20ft+ 0.50.17ft•0.50.21ft) V�,=493 lb = Design unit shear to wall, Vw vlwr= 185 plf USE: Shear Wall=#0 Allowable shear-260;.plf(SEISMIC) 7/16"APA-rated struc'l wood panels ... ONE SIDE aw/ 8d nails"@°6" o.c. edges (12"field) uplift tension at wall Bld'g &wall weight, wt1 := 9ft•lOpsf-+'0.5.12ft.l5psf wt1 = 180 plf Add'I wt from Point Load, P:= 1.5ft(wti) uplift tension at end of wall, Ti := •[H1=11-vlwr-0.6•(0.5•wtl.1 -1- 11 P)1 T1 = 1356 lb PROVIDE: SIMPSONMSTA49 Strap Tie at EA. End of Shear Walls (2)2x6 post REQ'D at Wall Ends. (Detail: °1 ) Plate anchorage: Vw 'plate 15ft vplate= 33 plf Shear is LESS than 150 plf ... so fastening direct is OKAY. Curt_ es.) '+'"1p244.G 'r?.us5 %A" • PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: ''7.. of 51) Shearwall Evaluation J2K Engineering, Inc.' w As "previously established"the Upper level SEISMIC destributed load is: VP := 1.9•psf Wall Location: "REAR" Kitchen wall line Total Wall Length, LT:= 5.5ft Min. Wall Length, I := 5.5ft Height of shear wall, H := 9.0.ft LOADING: Vw:= VP.(05.17ft.0.50.21ft+ 0:50.23.5ft•37ft) Vw= 996 lb Design unit shear to wall, Vw A'tw,:= viw,.= 181 plf USE: Shear Wall#0 Lr Allowable shear=260 plf(SEISMIC) 7/16"APA-rated struc'I wood panels ONE SIDE w/8d nails @ 6"o.c. edges (12"field) uplift tension at wall Bldg &wall weight, wtl .= 9ft•10psf+_0.5.21;Oft 15psf wtl =247.5 plf Add'I wt from Point Load, P:= 1.5ft•(wtl) uplift tension at end of wall, T1 :_ •[H1 1i vlWr—0.6. 0:5•wtt 112+ 11 4)1 1 T1 =998 lb PROVIDE: SIMPSON MSTA49'Strap Tie atEA. End of Shear Walls (2)2x6 post"REQ'D at Wall Ends. (Detail: 1 ) Plate anchorage: VW "plate"- 26ft "plate= 38p1f Shear is LESS than 150 plf... so fastening direct is OKAY. (Dtl (,G') PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 8 of Shearw'all Evaluation J2K Engineering, Inc. - As "previously established"the Upper level SEISMIC destributed load is: VP := 1.9•psf Wall Location: Left/Right "INTERIOR"wall cine Total Wall Length, Lr:=_2.32in+2.36in Min. Wall Length, 11 = 32in Height of shear wall, H1 := 9.0•ft LOADING: VIA,:=-- VP•(0.5•17ft•37ft+ 0.50.27ft•34ft) V ,,= 1470 lb i Design unit shear to wall, V Vlwr;_ vicar= 130 plf USE:`Shear Wall#,0 Allowable:shear=260 plf; (SEISMIC) 7/16"APA-rated struc'l wood panels: ... ONE;SIDE - w/8d-nails @ z6":o.c.;edges (12"field) uplift tension at wall Bld'g &wall weight, wt : 9ft lopsf+ 6 Oft l5psf wt1 = 180 plf Add'I wt from Point Load, P:=:0.0ft•(wtl) uplift tension at end of wall, T1 := •[PI>-11•vi ,r- 0.6.0:5.wt1112-+ 11.P)] 1 T1 = 1023 lbPROVIDE: SIMPSON MSTA49 Strap Tie at EA. End of Shear Walls (2)2x6 post REQ`D at Wall Ends. (Detail: I ;@ rear Garage 9walls) PROVIDE: SIMPSON.LTTI9 Tension`T_ie at EA.`"End of Shear Walls (2) 2x6 post REQ'D at Wall Ends. (Detail: 2 @ rear Front Mstr walls;) Plate anchorage: Vw Vplate= 19 ft Vplate=77 plf Shear is LESS than 150 plf ... so fastening direct is OKAY. (Dtl #6 ) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: of 5 _ r ' Shearwall Evaluation J2K Engineering, Inc. As "previously established"the Upper level SEISMIC destributed load is: VP := 1.9•psf Wall Location: Left Front Mastr BR& REAR Garage wall line Total Wall Length, LT:= 7.5ft+"85ft Min. Wall Length, 11 := 7.5ft Height of shear wall, 'Ph := 9.0.ft LOADING: V�,�,:= VP•(0.5.15ft•70ft) Vi,+,= 997 lb Design unit shear to wall, Vw v�Wr viW.= 62 plf USE: Shear Wall#0 Allowable shear=260 plf(SEISMIC) 7/16"APA-rated struc'l wood panels ... ONE SIDE w/8d nails @ 6" o.c.edges (12"field) uplift tension at wall BId'g &wall weight, wtl s= 91t.l Opsf+ 7.0ft•15psf wt1 = 195 plf Add'I wt from Point Load, P:=,0.Oft. wtl) uplift tension at end of wall, T1 :_ 1 —0.6�O.S�wt� i2+ 1i'=P)1 wr T1 = 122 lb NOMINAL".Uplift, .... NO Holdown required Plate anchorage: Ver, plate 13ft+'14ft ''plate = 37 pif Shear is LESS than 150 plf...so fastening direct is OKAY. (Dtls #8 & 6 ) .t-'�'Q•d•G'T1Zu ss "C." !� PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 10 of J K Engineering, Iric. Shearwall Evaluation 2 As "previously established"the Upper level SEISMIC destributed load is: VP := 1.9•psf Wall Location: Front Office&Dining Room wall line Total Wall Length, LT:= 2.32in+ 5 5ft Min.Wall Length, 11 = 32m Height of shear wall, H1 := 9.0.ft LOADING: Vw:=VP•(16ft•34ft) Vw= 1034 lb Design unit shear to wall, V - w vlwr:= v =95 plf USE: Shear Wall#0 lwr Allowable shear=260 plf(SEISMIC) 7116">APA-rated struc'I wood,panels'... ONE SIDE wl`.8d nails @ .6":o.c.-:edges (12"field) uplift tension at wall Bld'g&wall weight, wtl := 9ft•l Opsf+5.5ft 15psf wtl = 172.5 plf Add'I wt from Point Load, P:= 1.5ft.(wti) uplift tension at end of wall, Ti := 1 [Hiliviwr- •0.6•(0.5 wt l 1 2 11 P)] ll . Tl =565 lb PROVIDE: 'SIMPSON,L"TT19 Holdown at EA.'End of`ShearWalls (2)2x6 post REQ'D at Wall Ends. (Detail: 2 ) ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTGwt:= [T1 - (i + 1.0ft)•(8in 24in)•140pcf] FTGwt=-119 lb NO additional footing volume required Plate anchorage: Vw 'plate= v 32 if 32ft plate= P Shear is LESS than 150 plf...so fastening direct is OKAY. Cres ECoG at 6) PROJECT: Ruecker Residence JOB No.: 11 - 048 PAGE: I I of a 9 * Shearwall Evaluation J2K Engineering, Inc.' As "previously established"the Upper level SEISMIC destributed load is: VP := 1.9•psf Wall Location: FrontGarage walls Total Wall Length, Lr:= 230in Min. Wall Length, 11 := 30in Height of shear wall, ;Hl := 8;5•ft LOADING: Vv,:= VP•(0.5032ft•26ft) Vw= 790 lb Design unit shear to wall, v lwr Vw v 158 if USE:iShear Wall#0 Lr Iwr-— h Allowable:shear=260 plf(SEISMIC) 7/16"APA-rated struc'I wood panels ... ONE SIDE w/8d nails @ 6"o.c.:edges (12"field) uplift tension at wall Bldg &wall weight, wt1 := 9ft•1Opsf+ 5.5ft•l5psf wt1 = 172.5plf Add'l wt from Point Load, P:= 1:5ft•(wt1) uplift tension at end of wall, Ti := 1i [HI 11 vicar-0.6 C0.5 wt1 112+ 11•13)] T1 = 1059 lb PROVIDE: SIMPSON LTT19 Holdown atEA. End of Shear Walls (2)..2x6 post REQ'D:at Wall Ends. (Detail: =4 ) ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTGwt := [T] - 01 + 0.0ft)•(8in•32in)•140pcfl FTGwt=437 lb Additional concrete weight provided, wgg:= 32•in tftg:= 15•in (depth req'd for Holdown) (below the "typical"footing) WTadd 140pcf 411 + 0.Oft)•(wftg•tftg- 8in•32in) WTadd = 544 lb PROVIDE: ;32"W x 15"'D Holdow Footing at Holdowns REQ'D Length: Shear wall length°+ 0.0 FT Ea. end Plate anchorage: PROVIDE: 1/2" Dia. A.B.:@ 48" o.c. Vw vplate'— 'Sft 'plate = 158plf (1)A.B.'(min ) [Dtl #4] PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 12. of Shearwall Evaluation J2K Engineering, Inc. As "previously established" the WIND destributed load is: VP := 14.9•psf Wall Location: LEFT Master BR& Bath wall line -2 Total Wall Length, LT 12ft+ 17ft " Min.Wall Length, 11 :=-12ft Height of shear wall, Hi := 9.O.ft -- LOADING: LOADING: Vw:= VP,(0.5.9.0ft•0.5017ft) Vw= 570 lb Design unit shear to wall, 'Vw vlwr:_ vicar=20 plf USE: Shear_Wall#0 �- Allowable shear=364 plf(WIND) 7116"APA-rated struc'I wood panels ... ONE SIDE .w/Sdmails;@ 6" o.c.--edges '(12"field) uplift tension at wall Bldg &wall weight, wt1.:= 9ft-.l Opsf+4'Oft•15psf wt1 = 150 plf Add'I wt from Point Load, P:= 0.0ft•(wti) uplift tension at end of wall, T1 : 11•[H1-11•viW.-0:6•(0.5•-wt1112•+ I1�Pjj Ti =-363 lb NO holdown required Plate anchorage: Vw uprate vplate=20plf Shear is LESS than 150 plf... so fastening direct is OKAY. ✓, (Dtl #8 ) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 13 of 55 Shearwall Evaluation J2K Engineering, Inc.4 • As "previously established"the WIND destributed load is: VP := 14.9-psf Wall Location: LEFT Office &Living Room wall line Total Wall Length, Lr:= 16ft+ 8ft Min. Wall Length, 11 := 8ft Height of shear wall, Hl 9.0ft LOADING: Vw:= VP•(0.5.9.0ft•0 50.45ft) VU,= 1509 lb Design unit shear to wall, Vw vlW,.:_ viW,.= 63 plf USE: Shear Wall.#0 -Allowable shear=364 plf(WIND) 7/16"APA-rated struc'I wood panels ... ONE SIDE w/8d nails @ 6" o.c.<edges (12"field) uplift tension at wall Bldg &wall weight, wti 9ft•1'Opsf+-0.50.11.Oft•15psf wt1 = 172.5 plf Add'I wt from Point Load, P:= 0.Oft•(wtt) uplift tension at end of wall, Ti 1 [H1 i Vlwr—0.6•C0.5•wti•1 2+ 1141 T1 = 152 lb "NOMINAL" uplift, ... NO holdown required --- Plate anchorage: Vw 'plate 1-11• plate= 63 plf Shear is LESS than 150 plf... so fastening direct is OKAY. (Dtl #8 ) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: j 4 of 5 a Shearwall Evaluation J2K Engineering, Inc. As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: LEFT Garage& Utility Room wall line Total Wall Length, LT:= 14ft+ 8ft+ 7 5ft Min.Wall Length, 11 := 7 5ft Height of shear wall, ]31 := 9.0.ft LOADING: Vw:= VP•(0.5.9.0ft•0.50.45ft) Vim,= 1509 lb Design unit shear to wall, Vw vicar:_ — vlwr=51 plf USE: Shear Wall#:0 Allowable shear=364 plf(WIND) 7116"APA-rated struc'I wood panels'... ONE SIDE wl8d nails @ 6"ox. edges (12"field) uplift tension at wall Bld'g &wall weight, wt1 = 9ft•10psf+ 6.0ft•l5psf wti = 180 plf Add'I wt from Point Load, P= 0.Oft•(wti) uplift tension at end of wall, T1 = [H1+1 vicar-0.6.0:5 wti•1 2+ 11-P)j T1 = 55 lb "NOMINAL" uplift, ...NO holdown required "v Plate anchorage: Vw uplate Vplate= 51 plf Shear is LESS than 150 plf... so fastening direct is OKAY. (Dtl #8 &6 ) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 1S of r ' Shearwall Evaluation J2K Engineering, Inca As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: RIGHT Utility Room wall line V r „s Total Wall Length, LT:= 10ft Min. Wall Length, 11 :_ '10.Oft Height of shear wall, H1 = 9.0 ft LOADING: Vv,,:= VP.(0.5.9.Oft.0.50.22ft) Vw= 738 lb Design unit shear to wall, Vw vlWr vlWr= 74 plf USE: Shear Wall#0 Allowable shear== 364 plf(WIND) 7/16"APA-rated struc'I wood panels ... ONE SIDE w/8d nails @ 6":o:c.:edges (12 field) uplift tension at wall Bldg &wall weight, wt1';= 9ft•lOpsf+ 6.Oft•15psf wt1 = 180 plf Add'l wt from Point Load, P:= 0.Oft.(wtl) uplift tension at end of wall, T1 1 [H1.11•viwr'— 0.6. 0.5•wt1 1 2+ 11=P)] T1 = 124 lb "NOMINAL" uplift__ NO holdown required Plate anchorage: Vw Vplate: Ll• voate = 74 Of Shear is LESS than 150 plf...so fastening direct is OKAY. (Dtl #6c ) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: `Co of 5'9 Shearwall Evaluation J2K Engineering, Inc. As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: RIGHT Garage wall line Total Wall Length, LT:= 2.13.5ft - Mina Wall Length, 11 := 13.5ft Height of shear wall, H1 := 9.0 ft LOADING: Vv,,:= VP-(0.5.9 0ft•0.50 2.2ft) Vw= 738 lb Design unit shear to wall, Vw vlwr vlwr=27 plf USE: Shear Wall#`0 Allowable shear=364 plf(WIND) 7/16"APA-rated struc'l wood panels....ONE:SIDE ,w/8d nails© 6" o.c. edges (12"field) uplift tension at wall Bldg &wall weight, wt1 := 9ft•1Opsf+ 0.50.22.Oft•15psf wt1 =255 plf Add'I wt from Point Load, P:= 0.Oft•(wti) uplift tension at end of wall, T1 ii [Hl 11 vlwr 0.6 0.5 wtl• 2+ 11•P)] T1 =—787 lb NO holdown required Plate anchorage: Vw Vplate 'plate=27 plf Shear is LESS than 150 plf... so fastening direct is OKAY. (Dtl #8 ) PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: t'•7 of 59 Shearwall Evaluation J2K Engineering, Inc: As "previously established"the LOWER SEISMIC destributed load is: VP := 2.2•psf Wall Location: Lower"REAR" BR#3 & Bath wall line Lateral load from UPPER walls: Vu1,:= 9961b Total Wall Length, LT:= l 8ft Min. Wall Length, 11 18ft Height of shear wall, !Hl := 8.O.ft LOADING: V,,,:= VP.0.50.12ft.26ft+ V VW= 1339 lb Design unit shear to wall, Vw vlwi:_ vlwr= 74 plf USE: Shear Wall#0 Allowable shear=.260 plf(SEISMIC) 7/16" APA-rated struc'l wood panels ... ONE SIDE wl 8d nails @ 6" o.c.edges (12"field) uplift tension at wall Bldg &wall weight, wt1,:= :17ft•1Opsf+'0.5.12ft•12psf wt1 =242 plf Add'I wt from Point Load, P:= Oft.(wt ) uplift tension at end of wall, T1 := 11 •[H1•11•v1 0.6•(0.5•wt1•i 2+ 11=P)1 wr T1 =—712 lb NO Holdown required Plate anchorage: V w Vplate:_ — 11' 'plate= 74 plf Shear is LESS than 150 plf... so fastening direct is OKAY. PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 68 of a , Shearwall Evaluation J2K Engineering, Inc. As "previously established"the LOWER SEISMIC destributed load is: VP := 2.2•psf Wall Location: Lower"REAR" BR# 1 & Bonus Room Lateral load from UPPER walls: V„p:= 4931b Total Wall Length, LT:= 4ft+4.5ft+ 2.3ft Min. Wall Length, 11 := 3ft Height of shear wall, 111 := 8.0•ft LOADING: Vw:= VP•0.50.17ft•32ft'+ Yup VW,= 1091 lb Design unit shear to wall, Vw vicar:_ Lr vh„,.= 75 plf USE: Shear Wall#'0 Allowable shear=260 plf(SEISMIC) 7/1!6"APA-rated;struc'l wood panels ... ONE SIDE w/8d nails @ 6"o.c. edges (12"field) uplift tension at wall Bld'g &wall weight, wt1 17ft•l0psf+'!0.5.12ft 12psf+ 0.5.171. 12psf wt1 =344 plf Add'I wt from Point Load, P:= 1.50ft•(wt') uplift tension at end of wall, T1 := 11 [Hl 11 v1w,-0.6(0 5 call 1 2-+ it P)] T1 =—17 lb NO Holdown required Plate anchorage: Vw Vplate vptate= 75 plf Shear is LESS than 150 plf ... so fastening direct is OKAY. PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: \9 of Ee A t Shearwall Evaluation J2K Engineering, Inc. As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: LEFT Bonus Room wall line Lateral load from UPPER walls: V„p:= 5701b Total Wall Length, LT:= 17ft '' Min.Wall Length, 11 := 17ft Height of shear wall, H1 := 8.Oft r LOADING: Vw:= VP•(9.5ft•0.50.19.25ft)+ V„p = 1932 lb Design unit shear to wall, Vw vlw:_ vlwr= 114 plf USE:'Shear Wall-#::0 Lir Allowable shear=364 plf(WIND) 7/16”APA-rated struc'I wood panels ... ONE SIDE w/8d nails @ 6”o.c. edges (12"field) uplift tension at wall Bldg &wall weight, wt1 c= 17ft'1 Opsf+4.Oft•15psf+4ft•12psf wt1 =278 plf Add'I wt from Point Load, P:='0.0ft.(wtl) uplift tension at end of wall, T1 :_ •[H1•11•vIWr—0.6.0.5-wt1•i 2+ 11-P)1 11 Ti =—508 lb NO holdown required Plate anchorage: Vw Vplate L7 vplate= 114 plf Shear is LESS than 150 plf ...so fastening direct is OKAY. PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 20 of S Shearwall Evaluation J2K Engineering, Ir c. As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: INTERIOR Bonus & BR wall line Lateral load from UPPER walls: V„,:= 0.50.150911 Total Wall Length, LT:= 13ft Min. Wall Length, 11 := 13ft Height of shear wall, H1 := 8.0.ft LOADING: Vw:= VP•(9.5ft•0.50.32ft)+ V_up Vq,=3019 lb Design unit shear to wall, Vw vicar:_ — vicar= 232 plf USE: Shear Wall#.0 Allowable:shear=364,plf(WIND) 7/16"APA-rated struc'I wood panels ... ONE SIDE w/8d nails.@ 6"o.c.edges (12"field) uplift tension at wall Bld'g &wall weight, wt1 8ft•8psf+4f-12psf wti = 112 plf Add'I wt from Point Load, P:= O.Oft•(wtl) uplift tension at end of wall, T1 := 11 [Hl 11'vlwr112+ 11•P)] T1 = 1421 lb PROVIDE: SIMPSON HTT4-Tension Tie at E.A. End of Shear Walls (2)2x6 post REQ'D at Wall:Ends. (Detail: #3 ) 46c ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTG,„:= [Ti –(11 + 1.Oft)•(lOin•12m)•140pcfl FTGw,=–212 lb NO additional weight required Plate anchorage: v0.5_2x_PL PROVIDE: 1/2" Dia. A.B. @ :24"<o.c. spcg:= v spcg=30.5 in lwr PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: Z 1 of 5 9 • Shearwall Evaluation J2K Engineering, Inc.' As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: RIGHT BR#1 wall line / Lateral load from UPPER walls: vim:= 0.01b Total Wall Length, 1'T:= 6ft / Min. Wall Length, 11 := 6ft Height of shear wall, H1 := 8.0.ft LOADING: Vw:= VP•(9.5ft•0.5024ft) +VUp VW= 1699 lb Design unit shear to wall, Vw Vicar vIwr=283 plf USE:'.Shear Wall#:0 Lir Allowable shear=364 plf(WIND) 7/16"APA-rated struc'l wood panels ... ONE SIDE w/8d nails @ ;6"o.c. edges (12"field) uplift tension at wall Bldg &wall weight, wti_:= 17ft•l Opsf+ 4ft•12psf+4ft•15psf wt1 =278 plf Add'I wt from Point Load, P:='0.Oft•(wt ) uplift tension at end of wall, T1 :_ �1-[H1.11•viwr'—0.6 (0.5 wti•'112+ li•P)] T1 = 1764 lb PROVIDE: SIMPSON HTT4 Tension Tie at EA.End of Shear Walls (2);2x6 post iREQ'D,at Wall Ends. (Detail: #`3 ) ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTGw,:= [T1 —(il + 1.0ft)•(8in.18in)•140pcfl FTGw,= 784 lb Additional concrete weight provided, wftg:= 16-in tftg:= 18•in (depth req'd for Holdown) (below the "typical"footing) WTadd = 140pcf•(11 + 1.Oft)•(wftg tftg— 8in•18in) WTadd = 980 lb PROVIDE: 16"W:x'18":D Holdow Footing at Holdowns REQ'D'Length: Shear wall length +'1.0'FT Ea. end Plate anchorage: Vo.s_2x_PL PROVIDE: 1/2" Dia. AB._@ 24": o.c. spcg spcg=25 in lwr PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 2.z. of 5. Shearwall Evaluation J2K Engineering, Inc. As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: LEFT Closet BR#3 wall line Lateral load from UPPER walls: V„1,:= 0.25.15091b Total Wall Length, Lr:— 7ft — Min.Wall Length, 11:= 7ft Height of shear wall, H1 := 8.O ft LOADING: Vw:= VP-(95ft•0.50.26ft) + Vup Vw=2217 lb Design unit shear to wall, Vw vlwr;_ ulwr=317p1f USE:Shear Wall:#;0 Allowable shear=364 plf(WIND) 7116"APA-rated struc'l wood panels ... ONE SIDE w/8d nails @ ;6" o,c. edges ;(12"'field) uplift tension at wall Bldg &wall weight, wt1 := 17ft•'lOpsf+w4ft•12psf+-4ft•l5psf wt1 =278 plf Add'I wt from Point Load, P: 0.Oft•(wti) uplift tension at end of wall, Ti 11 [HI 11 vlwr 0.6(0.5 wt1 'i 2+ 11 P) T1 = 1950 lb PROVIDE: SIMPSON HTT4TensionTie at EA.End of Shear Walls (2):2x6 post REQ'D at Wall Ends. (Detail: It 3 ) ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTG,,:= [Ti —(11 + 1.0ft)•(8in•18in)•140pcff FTGwt= 830 lb Additional concrete weight provided, wftg:= 16•in tftg:= 18-in (depth req'd for Holdown) (below the"typical"footing) WTadd:= 140pcf•(11 + 1.Oft)•(wftg•tftg— 8in•18in) WTadd = 1120 lb PROVIDE: 16"W x 18".D'Holdow footing at Holdowns REQ'D Length: Shear wall length+1.0 FT Ea. end Plate anchorage: VO 5_2x_PL PROVIDE: 1/2"Dia.A.B. @ 18" o.c. spcg vlwr spcg=22.4 in PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 2 3 of 5 ShearwalI Evaluation J2K Engineering, Inc.' As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: RIGHT BR#3 wall line a Lateral load from UPPER walls: Vup:= 7381b Total Wall Length, LT:= 4ft+ 9ft ` Min. Wall Length, 11 := 4ft ' Height of shear wall, H1 := 8.0-ft LOADING: Vw:= VP•(9.5ft•0.50•25ft) + V, Vw=2507 lb Design unit shear to wall, Vw vicar:_ vicar= 193 plf USE: Shear Wall#0 Allowable shear=364 plf(WIND) 7/16"APA-rated struc'I wood panels ... ONE SIDE w/8d nails @ ;6" o.c. edges (12 field) uplift tension at wall Bldg &wall weight, wti = 17ft•lopsf 4 4ft-12psf + b t.15psf wti =308 pif Add'I wt from Point Load, P:= 0.0ft•(wti) uplift tension at end of wall, T1 := I1•[HI•li•viWr—0.6•(0.5•wti•112+ T1 = 1173 lb PROVIDE: SIMPSON HTT4 Tension Tie at EA. End of Shear Walls (2).2x6;post'REQ'D,at';Wall Ends. (Detail: #-:3 ) ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTG,„:= [Ti —(li + 1.0ft)•(8in-18in)•140pcfl FTGW,=473 lb Additional concrete weight provided, wfig:= 24•in tftg:= 12•in (depth req'd for Holdown) (below the "typical"footing) WTadd 140pcf•(11 + 1.0ft)-(wftg-tfig— 8in•18in) WTadd= 700 lb PROVIDE: 24"W x 12"D Holdow Footing at Holdowns REQD Length: Shear wall length + 1.0 FT Ea. end Plate anchorage: V0s_2x_PL PROVIDE: 112" Dia. A.B.@ 24" o.c. spcg spcg= 36.7 in v iW, PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 2.y of S Shearwall Evaluation J2K Engineering, Inc. • As "previously established"the WIND destributed load is: VP := 14.9•psf Wall Location: Lower RIGHT Garage wall line e'v- Lateral load from UPPER walls: V„p:= 0.35.7381b Total Wall Length, LT:= 1lft ,// Min. Wall Length, 11 1aft Height of shear wall, H1 := 8 0•t LOADING: Vw:= VP•(9.5ft.0.50.9.5ft)+ V„p Vw= 931 lb Design unit shear to wall, Vw vicar:_ viw,.= 85 plf USE: Shear Wall:.#;0 Allowable shear-364:plf(WIND) 7/16"APA-ratedstruc7 wood panels ... ONE SIDE w/8d nails @ 6"o.c. edges (12"field) uplift tension at wall Bldg &wall weight, wt1 = 17ft•'10psf+4ft•12psf+ 6ft•l5psf wt1 =308 pif Add'I wt from Point Load, P= 0.Oft•(wti) uplift tension at end of wall, T1 : 11 [H1 11 vicar-0.6•(0.5•wtl•i12-+ 11-0 T1 =—340 lb NO holdown required ESTABLISH volume of concrete for holdown resistance, Concrete weight required, FTGWI:= [T] —(11 + 1.0ft)•(8in•18in)•140pcfl FTG,,=—2020 lb NO.additional weight required i Plate anchorage: V0.5_2x_PL PROVIDE: 1/2" Dia.A.B. @ 72 o.c. spcg spcg= 83.7 in iwr PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 2 5 of VI Maximum Spans Garage Floor System J2K Engineering, Inc. Allowable Load for 4 x 4 Column column unbraced length, I, := 8-ft+ 0-in End Reaction, R4 := R3 R4 =2968 lb Column Properties: A:= 12.25•inEmin := 470000•psi R4 check end bearing, fg := — A fg =242 psi Allowable bearing stress, Fg := 1.15.1300•psi Fg = 1495 psi F'g := 0.75Fg F'g = 1121 psi Reaction end bearing is OKAY without steel bearing surface check allowable axial load for member, le := 1.0.1u DOL := 1.00 CF:= 1.15 Fc:= 1300-psi F',:= DOL•CF•Fc d := 3.5•in 0.822•Emin F FcE := cE , := 0.8 £:= F' e c column stability factor, - -0.5 1 + C 1 + C�2 C CP :_ - -— 2•c � 2.c c_ CP=0.31 allowable axial load, Pa := CP•F',•A Pa = 5762 lb OKAY ... USE: 4 x 4 Hem-Fir No.2 for Columns for lengths up to 10 feet 0 inches AND reactions LESS THAN 5700 LBS. Project : Ruecker Residence JOB No.: 11 - 048 PAGE: 2 G'of 59 Y I a 1 M Maximum Spans Garage Floor System J2K Engineering, Inc. I) Exterior Beam with Concentrated Load at MID-SPAN @ Front Garage SPAN: Lg := L7 Lg =4 ft Loading(s): w9 5•L•DL w9=220 plf Pbm9:= 2000-lb Ag := 0.5•L9 Ag= 2 ft Design Requirements: Design Properties for No. 2 GRADE HEM-FIR P.T: M9max=2440 lb•ft V9max= 1238 lb LDF9 := 1.00 Eg := 0.9.0.95.1300000•psi L9 limit deflection to: As 3—60 Fgb:= 1.2.0.8.850•psi Fgv := 0.8.0.97.150psi Minimum Section Properties for Member: >>>THIS LOADING CONTROLS Mbr SIZE <<« 4 M9max 1.5•V9max 5•w9eq'L9 S9min •--- LD F9•F A9min := LDF F 19min E -�- 9 9b 9 9v 384s 4 s SELECTED SECTION: 4 x 10 (No. 2 Grade) S9min=35.88 in3 A9min= 15.96 in2 19min=40 in4 Sact=49.91 Aact= 32.38 'act= 230.8 Maximum Span for No. 2 Hem-Fir PT 4x10 r Front Garage floor Beam, L8 =4 ft Check Footing Requirements, Allowable Soil Pressure Qs= 1500 psf 1) Front Garage Footings -with 4x Beams Maximum Reaction R3 := 1.10•(0.5•LDL)•L9+ 2000•Ib Aft94:= Q3 Aftg4= 1.98ft2 <---- Minimum Ftg Size s R3 =2968 lb Provide: 20"x 20" x 10" Conc. Ftg's @ Exterior Posts. Project : Ruecker Residence JOB No.: 11 -048 PAGE: z7 of Maximum Spans Garage Floor System J2K Engineering, Inc. G) Uniformily Loaded Exterior Support Beam @ Front Garage SPAN: L7 := 4-ft L7 =4 ft Loading(s): w7 := 0.5L•(DL) w7 =220 plf Design Requirements: Design Properties for No. 2 GRADE HEM-FIR P.T: M7max =4401b•ft V7max=238 lb LDF7 := 1.00 E7 := 0.9.0.95.1300000-psi limit deflection to: L7 F7b:= 1.2.0.8.850-psi F7v := 0.8.0.97.150psi A7 — 360 Minimum Section Properties for Member: 4 M7max 1.5•V7max 5•w7•1_7 S7minA7min 17min SELECTED SECTION: 4 x 10 (No. 2 Grade) LDF7•F7b LDF7•F7v 384•E7 07 Sact=49.91 Aact=32.38 fact= 230.8 S7min = 6.47 in3 A7min=3.07 int 17min = 9 in4 H) Exterior Beam with Concentrated Load at SUPPORT-END of SPAN @ Front Garage SPAN: L8 := L7 L8 =4 ft Loading(s): wg := 0.5•L•DL w8 =220 plf Pbm8 := 2000-lb Ag := 12-in Ag= 1 ft Design Requirements: Design Properties for No. 2 GRADE HEM-FIR P.T: M8max =2008 lb•ft Vgmax = 1738 lb LDF8 := 1.00 Eg := 0.9.0.95.1300000•psi Lg Fgb:= 1.2.0.8.850•psi F8v := 0.8.0.97.150psi limit deflection to: As := 360 Minimum Section Properties for Member: 4 M8max 1.5•V8max 5•w8eq L8 SELECTED SECTION: 4 x 10 (No. 2 Grade) ---- S8min LDFB-Fab A8min LDF8.F8v 18min 384•Eg•A8 Sact=49.91 Aact= 32.38 lact= 230.8 S8min =29.53 in3 A8min =22.4 in2 18min =24 in4 64 Project : Ruecker Residence JOB No.: 11 -048 PAGE: 2 of 59 Maximum Spans Garage Floor System J2K Engineering, Inc. F) Interior Beam with Concentrated Load at MID-SPAN SPAN: L6 := L4 L6 = 6 ft Loading(s): w6 := L•DL w6 =440 plf Pbm=2000 lb A6 := 0.5•L6 A6= 3 ft Design Requirements: Design Properties for No.2 Grade DFL Lumber: Msmax=49801b•ft V6max= 1917 lb LDF6 := 1.00 E6 := 1300000•psi L6 F6b:= 875*Psi F6v:= 170-psi limit deflection to: A6 360 Minimum Section Properties for Member: 4 M6max 1.5•V6max 5•w6eq'L6 S6min•= LDF F 'Amin LDF F 16min 384E 4 s' 2b 6• 6v 6' 6 SELECTED SECTION: 4 x 12 (No. 2 Grade) S = 66.4 in3 A6min= 16.91 int I6min= 72 in4 Sact=73.83 Aact= 39.38 lact=415.3 6min Maximum Span for No.2-'6X10 Garage Floor Beam, L5= 6 ft Check Footing Requirements, Allowable Soil Pressure Qs:= 1500•psf 1) Exterior Garage Footings -with 6x Beams R1 Maximum Reaction := V5max2 Aftg2 QAftg2= 1.94 ft2 <----Minimum Ftg Size s R� =2916.7 lb 2) Interior Garage Footings -with 6x Beams R2 Maximum Reaction R2 := 1.10(L•DL)•L4+ 2000•Ib Aftg3:= Aftg3= 3.27ft2 <----Minimum Ftg Size Qs R2 =4904 Ib PROVIDE: 24" x 24"x 12" Conc. Ftg's @ Interior Posts & 18" x 18" x 10" Conc. Ftg's @ Exterior Posts. Project : Ruecker Residence JOB No.: 11 -048 PAGE: 29 of 9 r 1 Maximum Spans Garage Floor System J2K Engineering, Inc. D) Uniformily Loaded Interior Support Beam SPAN: L4 = 6.0-ft L4 = 6 ft Loading(s): w4 := L•(LL+ DL) w4= 840 plf Design Requirements: Design Properties for No.2 Grade DFL Lumber: M4max =37801b-ft V4max=25201b LDF4 := 1.00 E4 := 1300000-psi limit deflection to: L4 F4b:= 875'Psi F4,,:= 170•psi 44 360 Minimum Section Properties for Member: 4 M4max 1.5•V4max 5•w4•I•4 S4min A4min 14min SELECTED SECTION: 4 x 12 (No. 2 Grade) LDF4•F4b LDF4•Fav 384•E4.44 Sact= 73.83 in Aact= 39.38 lact=415.3 S4min=51.841n3 A4min =22.24 int 14min= 94 in4 E) Interior Beam with Concentrated Load at SUPPORT-END of SPAN Note: Members over 5-ft contain SPAN: L5 = L4 L5= 6 ft (2)- Concentrated Loads. Loading(s): w5 = L-DL w5=440 plf Pbm=20001b A5= 12-in A5 1 := 2ft Design Requirements: Design Properties for No.2 Grade DFL Lumber: M5max1 =4427lb•ft V5max2=2917 lb LDF5 := 1.00 E5 := 1300000-psi L5 F5b:= 875•psi F5S, := 170-psi limit deflection to: 45 := 360 Minimum Section Properties for Member: >>> THIS LOADING CONTROLS Mbr SIZE <<< 4 M5max2 1.5•V5max2 5'w5eg1'L5 SSmin LDFS•F5b A5min LDFS•FSV 15min 384•E5 45 SELECTED SECTION: 4 x 12 (No. 2 Grade) S 54.58 in3 A 25.74 in I 4 Sact= 73.83 Aact= 39.38 lact=415.3 5min= 5min= 5min = 110 in Project : Ruecker Residence JOB No.: 11 -048 PAGE: 30 of �� Maximum Spans Garage Floor System J2K Engineering, Inc. B) Floor Joist with Concentrated Load at MID-SPAN SPAN: L2 := L L2= 8 ft Loading(s): w2 := 12•in•DL w2 = 55p1f [Dist= 1000 lb A:= 0.5-L2 A=4 ft Design Requirements: Design Properties for Member to be sized: M2max=24401b-ft V2max= 720 Ib LDF2 := 1.00.1.15 E2 := 1600000•psi Minimum Section Properties for Member: F2b:= 900.psi F2v:= 180-psi L2 limit deflection to: A2 := — 360 >>>THIS LOADING CONTROLS Mbr SIZE <<< 4 M2max 1.5•V2max 5'W2eq'L2 S2min• LDF2•F2b A2min:= LDF2•F2v 12min := 384 E2•A2 SELECTED SECTION: 2 x 12 (No. 2 Grade) �- Sact= 31.64 Aact= 16.88 lact= 178.0 S2min =28.293 A2min= 5.22 int 12min=55 in ✓ C) Floor Joist with Concentrated Load at SUPPORT-END of SPAN SPAN: L3 := L L3 = 8 ft Loading(s): w3 := 12•in•DL w3 =55pif Pjst= 1000 lb A3 := 12•in A3= 1 ft Design Requirements: Design Properties for Member to be sized: M3max= 1082lb•ft V3max= 1045 lb LDF3:= 1.00.1.15 E3 := 1600000•psi Minimum Section Properties for Member: F3b:= 900.psi F3v:= 180.psi L2 limit deflection to: A3 3— 60 4 M3max 1.5 V3max 5'w3eq'L3 SELECTED SECTION: 2 x 12 (No. 2 Grade) S3min LDF3.F3b A3min LDF F 13min 384.E 4 3' 3b 3' 3v 3' 3 Sact= 31.64 Aact= 16.88 lact= 178.0 S3min= 12.55 in3 A3min=7.57 in2 13min= 55 in4 Project : Ruecker Residence JOB No.: 11 -048 PAGE: 3 1 of 59 Maximum Spans Garage Floor System J2K Engineering, Inc. Design values per 2005 Edition "National Design Specification for Wood Construction" Garage Floor Design Loadings: Uniform Live Load LL:= 50 psf Concentrated load of 2000 LBS to be distributed over 6 SQ IN area. Concentrated load for joists at 12" o.c.: Pist:= 1000•lb Concentrated load for beam: Pte:-- 2000•lb uniform Dead Load (w/3" of concrete+ 3/4"sht'g): DL= 55•psf "Floor Joist"Analysis A)Typical Floor Joist w/Uniform loadings SPAN: L= 8.0•ft Loading(s): w:= 12•in•(LL+ DL) w= 105 ptf Design Requirements: Design Properties for No. 2 Grade, DFL 2x12: Mmax= 8401b•ft Vmax=420 lb LDF:= 1.00.1.15 E := 1600000•psi Fb := 900•psi F,:= 180-psi Minimum Section Properties for Member: limit deflection to: A :_ 360 Mmax 1.5•Vmax 5-w-L4 Smin Amin Inn SELECTED SECTION: 2 x 12 (No. 2 Grade) LDF•Fb LDF•FV, 384•E•A Sact= 31.64 Aact= 16.88 lact= 178.0 Smin= 9.74 in3 Amin=3.04 in Imin=23 in4 Project : Ruecker Residence JOB No.: 11 - 048 PAGE: 3Z. of Jzrr tngineering, inc. Cantilever Garage/DW Retaining Wall 4-FOOT TALL (nom.) (Horizontal Backfill) The subject wall is designed to support the loadings from the gravel backfill approach slab and the floor slab framing associated with a residential garage. In compliance with the Oregon Residential Specialty Code and applicable sections o the Oregon Structural Speciality Code, the retaining wall has been designed for a 40 PSF uniform LIVE load and a 2000-LB concentrated load. 120 = Ds, density of soil (PCF) 1500 =Qs, allowable soil bearing pressure (PSF)w/o allowable increases 40 =Wsurcharge, uniform floor live load (PSF) 0.25 =Ka, active soil coefficient(30 PSF equiv. fluid pressure)(OSSC Table 1610.1) 0.35 = soil sliding resistance factor 8 =Tw, wall width (inches) 30 = R, heel base projection (inches) 10 =T, toe base width (inches) 48 =B, total base width (inches) 4.00 = H, total overall wall height(feet) 0 = Hs, stem exposure above finished grade (inches) 10 =Tb, base thickness (inches) 0 =C, backfill cover on toe of base(inches) 4.00 = Ha, height of soil at heel (feet) stability check PIECE FORCE ARM MOMENT (LBS) (FT) (FT-LBS) Bldg Load 320 3.33 1067 Backfill 950 1.25 1187 Stem 317 2.83 897 Base 500 2.00 1000 Ph 240 1.33 320 Psur 390 2.36 920 Total Fv= 2087 LBS Total M = 5392 FT-LBS Resultant Location = 31.01 inches from right end of base Resultant eccentricity from center of base, e = 7.01 inches Maximum eccentricity for Resultant= 8.00 inches Overturning Safety Factor : S. F. = 2.14 > 1.50 OKAY. SOIL PRESSURE CHECK : Max. soil pressure, qmax= 979 PSF Min. soil pressure, qmin = 65 PSF SLIDING RESISTANCE CHECK : Include balance of surcharge load in sliding resistance: 214 LBS Safety factor, S. F. = 1.5 > 1.5 Footing Geometry is OKAY Job No.: 11 - 048 PROJECT: Ruecker Residence Page: 33 of 9' JZK tngineerltng, Inc. Cantilever Garage/DW Retaining Wall 4-FOOT TALL (nom.) (Cont.) Evaluate Horizontal force applied to back of wall from 2000-lb point load positioned 12"from rear of the wall. See attached Fig. 11, from "Naval facilities Engineering Command" manual#7.02, Foundations and Earth Structures (Sept. '98)for the development of the horizontal load. From Pressure Diagram in Fig. 11, x=m H 4.00 FT = H = Ha 1 FT =x 0.25 = m =x/H 2000 LBS = Qp, Point load From Table in Fig. 11, PH(H/Qp)= 0.78 390 LBS/FT = PH =0.78Qp/H > Psur= 390 Point of application for force PH, 2.36 FT=0.59H = R Horizontal force from Uniform Pavement load: = 390 LBS The "Surcharge"laod on the wall is controlled by the Point Load Job No.: 11 -048 PROJECT: Ruecker Residence Page:34 of 5 Sec. 2.7 Designing for Soil Reinforcement 187 �` �` _ 117 = 0.1 0 1\\� \- 2 N 0._ ''.\ m = 0.5 N'` 0._ \♦,A...--177 = 0.6 N\ 1 \, \ \ / m = 0.7---7") l � \ m = O.Z� N 0.4 N 0.4 � II m = 0.3 m= 0.4 l 0 / O N / 0.6 0.6 / j /II; m R // m PH�PIA-Cp.) R 0.1 .601-1 / f0.3 .601 / 0.2 .78 .59H 0.8 // / 0.5 .56H 0'8 0.4 .78 .59H / 1f 0.7 .481 0.6 .45 .48H / / •1 1.0 1.0 0 0.2 0.4 0.6 0.8 1.0 0 0.5 1.0 1.5 (H2\ Value of a-y /H \ Value of cry Q P SQL/ Point load Q . X= n7H,, QP ) T Line load Q For m s 0.4: Z= WIro H X= niH� H' __ 0.20n '4IPy H r > \\\ \ QL/ (0.16 + n2)22 H Qy For m 0.4: R Py = 0.55 Q i H22' _ 028,7'2 ' -- Z= n H Q (0.16 + n )� Py Form > 0.4: \\\\\\\\\\� T P/ �H Form > 0.4: H _ (-Q H\ 1.28m2n /H \ 1.77m2n2 R Qy L/ (m- + n2)2 �x `QP\ (mz + n'-)3 / 0.64 - Q vy= Uycos'-(1.10) Resultant Py= QL _ - y Q ( 2 + 1) Er 3 P Pressure from line load Q vy (Boussinesq equation modified by experiment) X=n7H Section A-A Pressures from point load QP (Boussinesq equation modified by experiment) Figure 2.46 Lateral earth pressure due to a surface load. Left side is for line load; right side is for point load.(After NAVFAC[90]) G 3cP 59 ti t • J2K Engineering, Inc. CANTILEVER Garage/DW RETAINING WALL 4 FOOT TALL Design Information: concrete parameters: concrete strength, f,:= 3500.psi concrete density, D, := 150•pcf reinforcing steel strength, FY:= 60000.psi Wall Geometry: Maximum height of wall, H,„:= 4.ft+_0.in Stem exposure at outside of wall, HS:=-0•in Wall width, Tw:= 8•in Rear base projection, R:= 30•in Toe base width T:= 10.in Total base width, B =48in Base thickness Tb:= 10.in Point Load, Ped:=2000lb (distributed over 4-ft) Location of Pt. Load, Dp := 12in Horizontal Force, Phor 3901b Location of Horiz. Force, Dhor 2.36ft Soil Properties: Bearing pressure, Qs:= 1500.psf soil density, Ds:= 120.psf soil sliding resistance, FS:= 0.35 Active soil coefficient, 30•psf (Gravel backfill) ka Ds PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 3 of .tcl Walt Analysjs and Design (4-foot cant. wall) J2K Engineering, Inc. k A) Reinforcement for Stem : Loads are developed per 1 foot of wall length. Design load applied to wall after backfilling, "design" height of wall, Hd := HW— Hs—Tb Hd= 3.17ft 2 Fbf:= 0.5•ka•Ds•Hd Fbf= 1501b try#'5 bars: db := 0.625•in d:= TW—2•in— 0.5•db d= 5.69 in Ab:= 0.31.in2 n := 12± 18 (bars adjusted for spacing) design shear, Vstem= Fbf+ Phor Vstem = 540 lb 1.7Vstem = 11 psi <--109psi, OKAY 0.85T,•12in • design bending moment, Mc:= Hd3Fbf + Phor•(Dhor —Tb) Mc=754ft•lb nAID.Fy 1.7•Mc 2 a:= a=0.35 in Ac:= Ac= 0.05in 0.854c•12.in 0.9•Fy•(d—0.50•a) check minimum steel requirements (whichever is LESS), As:= n•Ab As=0.21 in2 200.psi•12•in•d -OR- 4 PROVIDE : #5 bars at 18" o.c. Armin Ascaic 3'Ac vertically in stem. FY A 0.23 in2 Ascaic=0.07 in2 smin= development length for#`5 bar(ACI 318 Chp 12) Imin =24 inches determine horizontal reinforcing; PROVIDE: #'5 bar @'18"o.c. Ahor = 0.0018•Tw.12•in Ahor =0.17 in2 horizontally in stem. (Ahor= 0.21 sq in) check steel/concrete ratios; Maximum steel allowed for wall with 2" cover to a#5 bar, 2 3000•psi 87000•psi Amax0.75. 0.85 • •d 12•in Fy Fy+ 87000.psi Asmax= 1.09 in2 steel area is OKAY PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: S-7 of =cll • Wall Analysis acid Design (4-foot cant. wall) J2K Engineering, Inc. B) Reinforcement for"Toe" Side of Base • maximum soil pressure: qs:= 979•psf for concrete design purposes only, consider the soil bearing resistance to be distributed over the Front 1/3 area only. Deduct the weight of the toe concrete from the total load. sum of vertical forces for TOE design, Fv:= 4304'lb Fvt:= Fv-T•Tb•1•ft•Dc Fvt=4200 lb try: #5 bars: dbt:= 0.625•in dt:= Tb - 3 in -0.5 dbt dt=6.69 in Abt:= 0.31.in2 nt := 12 18 (bars adjusted for spacing) design shear, Vtoe T•Fvt•3- B Vtoe= 2625 lb 1.7•Vtoe =44 psi <--109psi, OKAY 0.85.Tb•12in design bending moment, Mtoe 0.5•T2•Fvt•B Mtoe= 1094ft•lb -. nt'Abt•Fy 1.7•Mtoe at:= at=0.35 in Atoe Atoe=0.06 in 0.85 fo 12•in 0.9•Fy•(dt-0.5.at) check minimum steel requirements (whichever is LESS), Ast nt Abt Ast=0.21 int 200•psi•12•in•dt 4 EXTEND: the stem vertical bars into A .- F -OR- Asmin2 3•Atoe TOE of base: #5 bars at 18" o.c. Y Asmin= 0.27 in Asmin2=0.08 in C) Reinforcement for"Heel" Side of Base For concrete purposes only, develop the bending moment on the HEEL area of the base by applying the weight of the soil above the heel area to the concrete base section. soil wt at back of stem, cistern (Hw-Tb- Hs)•Ds cistern=380 plf for#5 bars: dbh:= 0.625•in dh := Tb-2•in - 0.5-dbh dh =7.69 in Abh:= 0.31.int nh := 12± 24 (bars adjusted for spacing) design shear, Vheel Cistern•R+ PLd Vheei = 1350 lb 1.7 Vheel -22 psi <--109psi, OKAY 5 0.85.Tb•12in D bending moment, Mheel (0.5.qstem.R2) + PLd' S Mheei= 1587ft•Ib ah := nh Abn• FY ah =0.26 in Aheel 1.7•Mheei Aheel=0.08 in 0.85 fo 12•in 0.9•Fy•(dh -0.50•ah) check minimum steel requirements (whichever is LESS), 200•psi•12•in•dh -OR- 4 Ash := nh-Abh Ash =0.16in2 ,vvw A,c F NA 3'Ahee� Y PROVIDE: #5 bars at 24" o.c. A 0.31 int A 0.11 int transverse in TOP of BASE. smin2= smin= PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE: 3& of -1 .21K Engin ring , Inc- a " - H" (MIN,) CONCRETE EXTENT OF LEVEL SLAB WBASE BACKFILL AT WALL 8'I DRIVEWAY / I SLOPE .: , _ - 1:"- -----1111 2" Ra --�I:'. COMPACTED ;11; CRUSHED AGGREGATE 1 BACKFILL #5 BARS (HORIZ) ® I8" 04. l<4 5 BARS (VERT) ® IS" oh. (W/90° BEND+EXTENSION INTO FTG) 50I' I O_0.0"0. 10"--/ Ni- 2"-O'' t 2"-O" WASHED ROCK °o°o°o° et (I8"xI8") WRAPPED ►4., p. �o°o°o°o' W/FILTER FABRIC 000000°0' iy 4" DIAM(MIN) /Id —'�f'�''t'1'�. I 5"MIN PERF.FDN DRAIN 10" — •l 4 6"(MAX) CLEAR \\/\ V� ce 4 CAST BASE AGAINST UNDISTURBED SOIL (TRANSVERSE) TR BOARS 4" Oh. m #5 BARS® 24" Oh. (LONGITUDINAL) RETAINING WALL W/GARAGE OR DRIVEWAY (AS REQ'D) 0 NTS * HORIZONTAL BACKFILL ONLY * CONCRETE STRENGTH (MIN.), FC = 3500 PSI 028 DAYS * REINFORCING STEEL GRADE 60 (24" LAP SPLICE LENGTH MIN) * APPLY WATERPROOF MEMBRANE ON BACKFILL SIDE WHEN ADJACENT TO HABITABLE SPACES * REFER TO "GUIDELINES FOR RETAINING WALL INSTALLATIONS" DETAIL FOR ADDITIONAL CRITERIA xpitProject: Job No.: 11–O4t5 Description: Page: 39 Of: 5 9 • J2K Engineering, Inc. • Cantilever Retaining Wall: 6-FOOT TALL (Horizontal Backfill) 120 = Ds, density of soil (PCF) 1500 = Qs, allowable soil bearing pressure (PSF) 0 = b, slope of backfill (degrees) 0.33 = Ka, active soil coefficient(35 PSF equiv. fluid pressure) 0.35 = soil sliding resistance factor 8 =Tw, wall width (inches) 40 = R, heel base projection (inches) 12 = T, toe base width (inches) 60 = B, total base width (inches) 7.00 = H, total overall wall heigth (feet) 6 = Hs, stem exposure above finished grade (inches) 12 =Tb, base thickness (inches) 0 = C, backfill cover on toe of base (inches) 6.50 = Ha, heigth of soil at heel (feet) stability check PIECE FORCE ARM MOMENT (LBS) (FT) (FT-LBS) Bldg Load 150 3.67 550 Backfill 2200 1.67 3667 Stem 600 3.67 2200 Base 750 2.50 1875 Ph 837 2.17 1813 Total Fv= 3700 LBS Total M = 10104 FT-LBS Resultant Location = 32.77 inches from right end of base Resultant eccentricity from center of base, e = 2.77 inches Maximum eccentricity for Resultant= 10.00 inches Overturning Safety Factor : S. F. = 5.67 > 1.50 OKAY. SOIL PRESSURE CHECK : Max. soil pressure, qmax= 945 PSF • Min. soil pressure, qmin = 535 PSF SLIDING RESISTANCE CHECK : Safety factor, S. F, = 1.5 > 1.5 Footing Geometry is OKAY CLIENT: Job No. PROJECT: Page: - j 6 FOOT TALL CANTILEVER RETAINING WALL 4 Design Information: concrete parameters: concrete strength, fc := 3000•psi concrete density, Dc := 150•pcf reinforcing steel strength, Fy := 60000-psi Wall Geometry: Maximum height of wall, HW := 7.ft+0-in Stem exposure at outside of wall, Hs := 6•in Wall width, Tw := 8-in Rear base projection, R := 40•in Toe base width T := 12•in Total base width, B = 60 in Base thickness Tb := 12•in Soil Properties: Bearing pressure, Qs := 1500•psf soil density, Ds := 120•psf soil sliding resistance, Fs := 0.35 Active soil coefficient, 40•psf ka Ds CLIENT: JOB No.: PROJECT: PAGE: '?-j-1 oF 5 Wall Analysis and Design (6-foot cant.wall) J1K tngineering, inc. A) Reinforcement for Stem : Loads are developed per 1 foot of wall length. Design load applied to wall after backfilling, "design" height of wall, Hd := Hw- Hs-Tb Hd = 5.5ft Fbf := 0.5•ka•Ds•Hd2 Fbf = 6051b try#4 bars, db := 0.5•in Ab := 0.20•in2 d := Tw-2-in-0.5-db d = 5.75in n := 12_ 18 (bars adjusted for spacing) n shear, VF V 1.6Vstem design stem := bf stem = 6051b = 11.86 psi <--109psi, OKAY 0.85Tw•12in design bending moment, Mc := Hd3Fbf Mc = 1109ft•Ib n-Ab•Fy 1.6-Mc 2 a := a = 0.26 in Ac := Ac = 0.07 in 0.85•fc•12.in 0.9•Fy•(d-0.50•a) check minimum steel requirements (whichever is LESS), As := n•Ab As = 0.13 in2 200•psi•12•in•d -OR- 4 Asmin = Ascalc -•Ac REQUIRES : #4 bars at 18" o.c. FY 3 vertically in stem. A Ascalc = 0.1 in2 smin = 0.23in2 development length for#4 bar(ACI 318 Chp 12) Imin=22 inches determine horizontal reinforcing; A := 0.00187w-12-in A 0.17in2 REQUIRES: #5 bar @ 18" o.c. horhor = horizontally in stem. (A.5= 0.31 sq in) check steel/concrete ratios; Maximum steel allowed for 8"wall with 2" cover to a#4 bar, As(max) := 0.75. 0.852.3000•psi 87000-psi •d•12•in As(max) = 1.11 in2 Fy (Fy+87000-psi) steel area is OKAY check steel required for two pour conditions, A 1.6Vstem A 0.03in2 PROVIDE: #3 bar @ 18" o.c. V 0.85•Fy•1.0.0.6 vertically in stem and footing for two pour conditions. 12" embement required (Av=0.11 sq in) CLIENT: JOB No.: PROJECT: PAGE: 4Z F Wall Analysis and Design (6-foot cant. wall) J2K Engineering,,Inc. B) Reinforcement for"Toe" Side of Base for concrete design purposes only, consider the soil bearing resistance to be distributed over the "toe" area only. Deduct the weight of the toe concrete from the total load. sum of vertical forces for TOE design, Fv := 3700lb Fvt := Fv-T.Tb•1•ft•De Fvt = 3550 lb try: #4 bars, dbt := 0.50•in dt := Tb-3•in -0.5•dbt dt = 8.75in Abt := 0.20.in2 nt := 12_ 18 (bars adjusted for spacing) 1.6•Vtoe design shear, Vtoe p <--109psi,_B Vtoe = 2130 lb = 27.84 psi psi, OKAY 0.85•Tb•12in design bending moment, Mtoe := 0.5•T•Fvt Mtoe = 1775ft•Ib nt.Abt.Fy 1.6•Mtoe 2 at := at= 0.26 in Atoe Atoe = 0.073 in 0.85•fc•12.in 0.9•Fy•(dt-0.5.at) check minimum steel requirements (whichever is LESS), Ast := nt'Abt Ast = 0.13in2 200 psi•12•in•dt 4 USE: #4 bars at 18"o.c. Asmin -OR- Asmin2 := 3'Atoe TRANSVERSE in bottom of Base FY (extend vertical bars into toe) Asmin = 0.35in2 Asmin2 = 0.1 in2 C) Reinforcement for"Heel" Side of Base For concrete purposes only,develop the bending moment on the HEEL area of the base by applying the weight of the soil above the heel area to the concrete base section. soil wt at back of stem, gstem = (Hw-Tb-Hs)•Ds stem = 660 plf try#4 bars: dbh := 0.5•in dh := Tb-2.in-0.5•dbh dh = 9.75in Abh := 0.20•in2 nh := 12± 12 (bars adjusted for spacing) 1.6•Vheel design shear, Vheel cistern•R Vheel = 22001b = 28.76 psi < 109psi, OKAY 0.85•Tb•12in bending moment, Mheel 0•5.gstem'R2 Mheel = 3667ft•Ib nh' y 1.6.Mheel 2 ah •= Abh'F ah = 0.39in Aheel Aheel = 0.14 in 0.85•fc•12•in 0.9•Fy•(dh-0.50•ah) check minimum steel requirements (whichever is LESS), 2 Ash nh'Abh Ash = 0.2 in 200•psi•12•in•dh -OR- 4 Asmin F Asmin2 3'Aheel PROVIDE: #4 bars at 12" o.c. y transverse in TOP of BASE A 0.39in2 Asmin2 = 0.18in2 smin = CLIENT: JOB No.: PROJECT: PAGE: 43F • J2K Engineering; Inc. • Cantilever Retaining Wall 8-Feet Tall (nom.) (Horizontal Backfill) 120 = Ds, density of soil (PCF) 1500 = Qs, allowable soil bearing pressure (PSF) 0 = b, slope of backfill (degrees) 40 = Equiv. Fluid Pressure (PCF) 0.33 = Ka, active soil coefficient 0.35 = soil sliding resistance factor 8 =Tw, wall width (inches) 56 = R, heel base projection (inches) 12 =T, toe base width (inches) 76 = B, total base width (inches) 9.00 = H, total overall wall heigth (feet) 6 = Hs, stem exposure above finished grade (inches) 12 =Tb, base thickness (inches) 0 = C, backfill cover on toe of base (inches) 8.50 = Ha, heigth of soil at heel (feet) stability check PIECE FORCE ARM MOMENT (LBS) (FT) (FT-LBS) Bldg Load 150 5.00 750 Backfill 4340 2.33 10127 Stem 800 5.00 4000 Base 950 3.17 3008 Ph 1445 2.83 4094 Total Fv = 6240 LBS Total M = 21979 FT-LBS Resultant Location = 42.27 inches from right end of base Resultant eccentricity from center of base, e = 4.27 inches Maximum eccentricity for Resultant= 12.67 inches Overturning Safety Factor : S. F. = 5.24 > 1.50 OKAY. SOIL PRESSURE CHECK : Max. soil pressure, qmax = 1317 PSF Min. soil pressure, qmin = 653 PSF SLIDING RESISTANCE CHECK : Safety factor, S. F. = 1.5 > 1.5 Footing Geometry is OKAY CLIENT: Job No.: PROJECT: PAGE: . .;, c,„ 5 CANTILEVER RETAINING WALL 8 FOOT TALL Design Information: concrete parameters: concrete strength, fc := 3000•psi concrete density, Dc := 150•pcf reinforcing steel strength, Fy := 60000•psi Wall Geometry: Maximum height of wall, HW := 9•ft+0•in Stem exposure at outside of wall, Hs := Gin Wall width, TW := 8•in Rear base projection, R := 56•in Toe base width T := 12•in Total base width, B = 76 in Base thickness Tb := 12•in Soil Properties: Bearing pressure, Qs := 1500•psf soil density, Ds := 120.psf soil sliding resistance, Fs := 0.35 Active soil coefficient, 40•psf ka :— Ds s CLIENT: JOB No.: PROJECT: PAGE: .4S o'f' 5 9 Wall Analysis and Design (8-foot cant. wall) JZK tnglneering, Inc. Y ( Y A) Reinforcement for Stern : Loads are developed per 1 foot of wall length. Design load applied to wall after backfilling, "design" height of wall, Hd := Hw- Hs-Tb Hd = 7.5ft Fbf := 0.5•ka•Ds•Hd2 Fbf = 1125Ib try#4 bars, db := 0.50•in Ab := 0.20•in2 d := TW-2•in-0.5•db d = 5.75in n := 12+ 12 (bars adjusted for spacing) 1.6Vstem design shear, Vstem Fbf Vstem = 1125lb = 22.06psi <--109psi, OKAY 0.85Tw•12in design bending moment, Mc := Hd3 bf Mc = 2813ft•Ib n.Ab.Fy 1.6•Mc a := a = 0.39in Ac := Ac = 0.18in2 0.85•fc•12•in 0.9•Fy•(d-0.50•a) check minimum steel requirements (whichever is LESS), As := n•Ab As = 0.2 int 200•psi•12•in•d - OR - Ascale - c 4 A Asmin REQUIRES : #4 bars at 12" o.c. = FY 3 vertically in stem. Asmin0.23in2 Ascalc = 0.2in2 = development length for#4 bar(ACI 318 Chp 12) Imin =22 inches determine horizontal reinforcing; 2 REQUIRES: #5 bar @ 18" o.c. Ahor 0.0020 8 in•12 in Ahor = 0.19in horizontally in stem. (A.5= 0.31 sq in) check steel/concrete ratios; • Maximum steel allowed for 8"wall with 2" cover to a#4 bar, As(max) := 0.75• 0.852.3000•psi 87000.psi •d•12 in As(max) = 1.11 int Fy (Fy+87000•psi) steel area is OKAY check steel required for two pour conditions, 1.6Vstem2 PROVIDE: #3 bar @ 18" o.c. Ay := Av = 0.06in vertically in stem and footing 0.85•F •1.0.0.6 for two pour conditions. 18" embement required (Av= 0.11 sq in) CLIENT: JOB No.: PROJECT: PAGE:.- 5 Wall Analysis and Design (8-foot cant.wall) J2K Engineering, Inc. B) Reinforcement for"Toe" Side of Base for concrete design purposes only, consider the soil bearing resistance to be distributed over the "toe" area only. Deduct the weight of the toe concrete from the total load. sum of vertical forces for TOE design, Fv := 6250.1b Fvt := Fv-T Tb•1•ft•Dc Fvt = 6100 lb try:#4 bars, dbt := 0.625•in dt := Tb-3•in-0.5•dbt dt= 8.69 in Abt := 0.31•int nt := 12- 18 (bars adjusted for spacing) 1.6-Vtoe <-109psi, OKAY design shear, Vtoe T•Fvt•3_B Vtoe = 2889 lb = 37.77 psi p 0.85•Tb-12in design bending moment, Mtoe = 0.5•T•Fvt Mtoe = 3050ft•Ib nt•Abt.Fy 1.6•Mtoe A 0.13in2 at := at= 0.41 in Atoe := toe = 0.85•fc•12.in 0.9.Fy•(dt-0.5•at) Ast := nt•Abt Ast = 0.21 int check minimum steel requirements (whichever is LESS), 200.psi•12•in•dt 4 USE: #5 bars at 18" o.c. Asmin -OR- Asmin2 3•Atoe TRANSVERSE in bottom of Base FY (extend vertical bars into toe) Asmin = 0.35in2 Asmin2 = 0.17 in 2 C) Reinforcement for"Heel"Side of Base For concrete purposes only,develop the bending moment on the HEEL area of the base by applying the weight of the soil above the heel area to the concrete base section. soil wt at back of stem, gstem (Hw-Tb-Hs)•Ds cistern = 900plf try#5 bars: dbh:= 0.625•in dh := Tb-2•in-0.5•dbh dh = 9.69in Abh= 0.31-in2 nh := 12_9 (bars adjusted for spacing) R Vheel = 42001b 1.6•Vheel p < 109psi, OKAY design shear, Vheel gstem = 54.9 psi 0.85•Tb•12in bending moment, Mheel 0•5•gstem•R2 Mheel = 9800ft•Ib nh•Abh•Fy 1.6.Mheel Aheel = 0.38in2 ah := ah = 0.81 in Aheel 0.9 F • dh-0.50 ah) 0.85•fc•12•in Y•� check minimum steel requirements (whichever is LESS), Ash nh•Abh Ash = 0.41 int 200•psi•12•in•dh -OR- 4 Asmin Asmin2 3'Aheel PROVIDE: #5 bars at 9" o.c. FY transverse in TOP of BASE AsminAsmin2 = 0.5in2 = 0.39in2 CLIENT: JOB No.: PROJECT: PAGE:-'ç7 cam' J2K Engineerin , Inc'. Cantilever Retaining Wall 10-FOOT TALL (Horizontal Backfill) 120 = Ds, density of soil (PCF) 1500 = Qs, allowable soil bearing pressure (PSF) 0 =Ab, slope of backfill (degrees) 40 =Equiv. Fluid Pressure 0.33 = Ka, active soil coefficient 0.35 = Fs, soil sliding resistance factor 8 =Tw, wall width (inches) 72 = R, heel base projection (inches) 16 =T, toe base width (inches) 96 = B, total base width (inches) 11.00 = Hw, total overall wall height (feet) 6 = Hs, stem exposure above finished grade (inches) 12 =Tb, base thickness (inches) 0 =C, backfill cover on toe of base (inches) horizontal backfill 10.50 = Hd, height of soil at heel (feet) stability check PIECE FORCE ARM MOMENT (LBS) (FT) (FT-LBS) Bldg Load 150 6.33 950 Backfill 6840 3.00 20520 Stem 1000 6.33 6333 Base 1200 4.00 4800 Ph 2205 3.50 7718 Total Fv= 9190 LBS Total M = 40321 FT-LBS Resultant Location = 52.65 inches from right end of base Resultant eccentricity from center of base, e = 4.65 inches Maximum eccentricity for Resultant= 16.00 inches Overturning Safety Factor: S. F. = 5.22 > 1.50 OKAY. SOIL PRESSURE CHECK : Max. soil pressure, qmax = 1483 PSF Min. soil pressure, qmin = 815 PSF SLIDING RESISTANCE CHECK : Safety factor, S. F. = 1.5 = 1.5 Footing Geometry is OKAY PROJECT: JOB No.: PAGE: 41. CANTILEVER RETAINING WALL 10 FOOT TALL Design Information: • concrete parameters: concrete strength, fc := 3000•psi concrete density, Dc := 150•pcf reinforcing steel strength, Fy := 60000•psi Wall Geometry: Maximum height of wall, Hw := 11•ft+0•in Stem exposure at outside of wall, Hs := 6•in Wall width, Tw := 8•in Rear base projection, R := 72•in Toe base width T := 16•in Total base width, B = 96 in Base thickness Tb := 12.in Soil Properties: Bearing pressure, Qs := 1500•psf soil density, Ds := 120•psf soil sliding resistance, FS := 0.35 Active soil coefficient, 40•psf ka :- Ds CLIENT: JOB No.: PROJECT: PAGE: •5 Wall Analysis and Design (10-foot cant. wall) JLK Cnglneering, inc. A) Reinforcement for Stem : Loads are developed per 1 foot of wall length. Design load applied to wall after backfilling, "design" height of wall, Hd := Hw— Hs—Tb Hd = 9.5ft Fbf := 0.54ka•Ds•Hd2 Fbf = 18051b try#5 bars, db := 0.625•in Ab := 0.31•in2 d := Tw-2•in -0.5•db d = 5.69in n := 12- 16 (bars adjusted for spacing) 1.6Vstem design shear, Vstem Fbf Vstem = 1805lb = 35.39psi <--109psi, OKAY 0.85Tw•12in Hd•Fbf5716ft•Ib design bending moment, Mc := 3 Mc = n•A F 1.6•Mc a := b y a = 0.46 in Ac := Ac = 0.37 in2 0.85•fc•12•in 0.9•Fy•(d-0.50•a) check minimum steel requirements (whichever is LESS), As := n•Ab As = 0.23in2 200•psi•12•in•d -OR- 4 Asmin Ascalc —3.Ac REQUIRES : #5 bars at 16" o.c. FY vertically in stem. Asmin0.23in2 Ascalc = 0.5in2 = development length for#5 bar Imin =26 inches determine horizontal reinforcing; Ahor0.0018 T 12 in Ahor = 0.17in2 REQUIRES: # 5 bar @ 18" o.c. w horizontally in stem. (A.5= 0.31 sq in) check steel/concrete ratios; Maximum steel allowed for 8"wall with 2" cover to a#4 bar, As(max) := 0.75• 0.852.3000•psi 87000•psi •d•12•in As(max) = 1.09in2 Fy (Fy+87000.psi) steel area is OKAY check steel required for two pour conditions, 1.6Vstem2 PROVIDE: #3 bar @ 14" o.c. Av := Av = 0.09in vertically in stem and footing 0.85•Fy 1.0.0.6 for two pour conditions. 18" embement required (Av=0.11 sq in) CLIENT: JOB No.: PROJECT: PAGE: , ,5 : , Wall Analysis and Design (10-foot cant.wall) J2K Engineering, Inc. t B) Reinforcement for"Toe" Side of Base a for concrete design purposes only, consider the soil bearing resistance to be distributed over the "toe" area only. Deduct the weight of the toe concrete from the total load. sum of vertical forces for TOE design, Fv := 9200.1b Fvt := Fv-T•Tb•1•ft•Dc Fvt = 9000 lb try:#5 bars, dbt := 0.625•in dt := Tb-3-in-0.5•dbt dt = 8.69in Abt := 0.31.in2 nt := 12- 11 (bars adjusted for spacing) 1.6.Vtoe design shear, Vtoe T'Fvt•3_B Vtoe = 4500 lb = 58.82 psi <__109 psi, OKAY 0.85•Tb.12in design bending moment, Mtoe 0.5•T•Fvt Mtoe = 6000ft•Ib at :- nt Abt 1.6•M toe at= 0.66 in Atoe = toe Atoe = 0.26 in2 0.85•fc•12•in 0.9•Fy•(dt-0.5.at) Ast := nt•Abt Ast = 0.34 in2 check minimum steel requirements (whichever is LESS), 200•psi•12•in•dt 4USE:#5 bars at 11" o.c. Asmin -OR- Asmin2 3'Atoe TRANSVERSE in bottom of Base FY (extend vertical bars into toe) Asmin = 0.35in2 Asmin2 = 0.34in 2 C) Reinforcement for"Heel" Side of Base For concrete purposes only, develop the bending moment on the HEEL area of the base by applying the weight of the soil above the heel area to the concrete base section. soil wt at back of stem, cistern (Hw-Tb-Hs)•Ds stem = 1140 plf I try#5 bars: dbh := 0.625•in dh := Tb-2•in-0.5•dbh dh = 9.69in Abh := 0.31•in2 nh := 12_9 (bars adjusted for spacing) 1.6•Vheel design shear, Vheel cistern-IR Vheel = 6840 lb = 89.41 psi < 109 psi, OKAY 0.85.Tb 12in bending moment, Mheel 0•5'gstem•R2 Mheel = 20520ft•Ib nh.Abh.Fy 1.6•Mheel 2 ah := ah = 0.81 in Aheel Aheel = 0.79 in 0.85•fc•12•in 0.9.Fy•(dh-0.50•ah) check minimum steel requirements (whichever is LESS), Ash nh•Abh Ash = 0.41 in2 200•psi•12.in-dh -OR- 4 Asmin F Asmin2 3'Aheel PROVIDE: #5 bars at 9" o.c. y transverse in TOP of BASE Asmin - A 0.39 in2 Asmin2 = 1.05in2 CLIENT: JOB No.: PROJECT: PAGE: i Sap-55 " . • U I<_ ft ng iirr iilncll , I raG- a, "H" (MIN.) 5/4"COMPACTED CRUSHED --1. / AGGREGATE PLACED IN EXTENT OF LEVEL8" 1/2' 0 A5O1 A.B. 6" MAX.LIFTS BACKFILL AT WALL_ / / 0 6'-0" O.G. (MAX) MIN.7" EMBED. N.N.o.) SLOPE(AS REQ'D) e,,, (MIN) 2" REINf. u=n-u=u ‘‘rell.- - COVER( .N.0.) 11=11=11=,[ �I11 "A"BARS(HORIZ.) =Illi U=. ;II; iiiiiMei gia ;T/131 B'q0° BBARS(VERT) (W/ END 4 EXTENSION INTO FTG,) IP . iwra, i RECT'D Z. 1. I 'F"—7 1°P .-;1/434 " 0011����/�A 4 DIAM(I"IIW S"(MIN.) PERF.FDN DRAIN .6n il 6"(MAX,) - CLEAR a CAST BASE AGAINST 'G"BARS UNDISTURBED 501E (TRANSVERSE) in "D" BARB (LONGITUDINAL) RETAINING WALL SCHEDULE H "A" "$" iv 'V E F 6 BARS BARS BARS BARS 64" *5 BARS *4 BARS *4 BARS *5 BARS 40" 12" 12" 6 I S"o.c. 0 IS"o.c. 0 12 o.c. 0 12"o.c. 8'-6" *5 BARS *5 BARS *5 BARS 445 BARS 56 a 12" 12" ® I S"o.c. 0 1E004. 0 cro.c. 0 I2"o.c. 101-6" *5 BARS *5 BARS *5 BARS 45 BARS 1211 16" 12" 0 I8"oh. to Ilb.c. 0 q"o.c. 0 12"o.c. RETAINING WALL (NON-GARAGE AREAS) 0 NTS * HORIZONTAL BACKFILL ONLY * CONCRETE STRENGTH(MIN),FG=3000 P51®25 DAYS * REINFORCING STEEL GRADE 60 (36'LAP SPLICE LENGTH MIN) 1 APPLY MIATERPR00F NER3RANE ON BACKFILL SIDE WHEN ADJACENT TO HABITABLE SPACES * REFER TO 'GUIDELINES FOR RETAINING N A.L INSTALLATIONS' DETAIL FOR ADDITIONAL CRITERIA * PROVIDE*5 BARS® 12'o.c.VERTICALLY IN STEM 4 FTG.®2 POUR CONDITIONS.EXTEND MIN.16" INTO STEM,q0°W/6"EXT.IN BTTM.OF FTG. Project:It Job No.: Description: Page: 52- Of: S •Deck Ledger Design &Analysis (w/0.250" Dia. Wood Screw) J2K Engineering Inc. A) Deck Geometry& Loadings deck width: Wd 6.011 cantilever width: Lea„, := 0.0ft deck length: Ld := 35.0ft length of ledger: Liedgr:= 26.0ft Total VERTICAL load to ledger: VTI := 0.50•Wd•Liedgc5Opsf VTI = 3900 lb w`{•Lci TOTAL horizontal LIVE load from occupant"impact” loading: Vocc :_ .225.1b.0.25 V 788 lb (15 sq ft per 225-lb occupant) 15.ft "" _— B) Deck Ledger Anchorage Develop Wood Screw Size, Spacing and Capacity for 1 1/2" Ledger Board Fastened to 1 1/2" Rim Joist or blkg. 1) Review Penetration Factors and WOOD Screw Diameter Criteria (N.D.S.Table 11L & Sec. 11.1.4.6) Penetration, Pact := L5'in Pact Maximum Screw Diameter, DmaT:= 6 Dmax=0.25 in No. 14 (1/4") is Maximum Screw Diameter (for 1 1/2" Rim Thickness) Check capacity of selected screw diameter: D:= 0.242•in Pact 2) Depth Adjustment Factor, Cd .... p= 10D , for Full Table Design Values. Cd:= Cd= 0.62 10•D 3) Screw Capacity: Per NDS Table 11L for"Hem-Fir" species lumber, Z:— 141 Ib Nominal shear value for fastener: "Load Duration Factor": CD:= 1.60 Allowable screw shear capacity: Z':= Z•CD•Cd Z'= 140 lb 4) Load Development to ledger: 4a) Vertical Loading to ledger, w _ VTI, I." i Lledgt ��I = 150plf Voce 0.5•t,Wd—Lcant 5b) Lateral Loading to ledger, w := L . v Leant W2= i 5 plf ledgr d— 5) Screw Spacing & Quantity Development : a= Angle of combined withdrawal and shear (also angle of combined shears from perp to grain) Allowable Screw Capacity (combined vertical &withdrawal), resultant force to ledger: w3 := wit+ w2 w3= 151 pif a := atan(w, _ w1) a = 5.8 deg Z'c:= W' PaC1 Z'_ Va "'Pact•eos(a�2 Z' + •in•sin(aY2) Z' = 141 lb C) Ledger Screw Spacing requirements 2•Z' 2.Z.Cd FASTEN: 2x8 P.T. Hem-Fir" Ledger to Rim Joists or Studs svt" w3 S\TL=22.3 in s`'' WI SV= 14 in using (2)#14 (0.242" Dia.) GALVN Wood Screws @ 12"o.c. (Position Screws 2.0" and 5.0"from Top of Ledger) Project: Ruecker Residence Deck JOB NO. : 11 -048 PAGE: .S of I9 J2K Engineering , Inc . MCI C e-t.ScZ..s1G.E.l,t_.C»`C eta?.l 1..D1 N.t-C 4 35-ET *4- 0,.7,,s 9E- 2Z aO v 5 ¶a to AP- Gel,Rr . "f3 LOG) n2.tt3 t' y * F�-' (zeci5 Les 1 S i.4ss �. = C 1-Z-71l=.›.;-"71 1 sre,ap = 2-c 9 'u3 9 j.ait Project: Job No.: I - O Lie Description: Page: '`+ of: 5 J2K Engineering , Inc . c> -"- t„6.L..V "MOD= # ' -V_`._ 1.--C #4*- 13.,i y .r # 1:::(- e5.-= M, Cc.. ',...\_(;)- —)*{-a.. --_, -kf” (.--...,:> (17.',- — -4- ,S4.. ---"--* **4t2_, pz,r--- L_ Pia, = 2.,Z)VD 1..Z7.7,, (aZ-51)2.::) o ✓ t ; I-to _ '47..e t L c>7:-r: —P1-57-AL .5---1 Z., Le z - ,L__ �` t cG d I d .' "TaCC)= CS 'ca..?C:= .0‘..7 ( .SLyell-- + -,-..;-,),,) iX--L,.0...S co /.+2, -- .,)1 �rz. - Z- -ao5 ""'" .*EC-it. G SO''Pcs,--= ( 0,z•C187--:-+ I 3 ---1:1 --X-Co- (.;:, •--a----4-?..--Z---Tc:22) 30 L C)-p --?— se. =. r- UG ed,./ (Z) 13- 'FT` - P--t:M.:-AG. z= 5 '-'Ft5 ',44(C) •c'Z's( \S'i:::+- 1 Z. --.1-.)) 1, '-' (C).-S ( .-S- ---1-"'\) O90 L e Project: Job No.: 10-048 .61it. Description: Page: 53 of: r J2K Engineering , Inc . 11 < IJ= O4C L.35 4. '„= �) �0 5 .D1=c<tG, v j ltl Amb - z._t �-r - ''� U �8h xla x 1��f 1=1 , 1�= o =F• (o-s (9, P••-:-1-)) Co.s C,IZT)) = I A FTZ- _� lk r : 1811 . .xi c3 ,x 1 Ott41t, Project: jest: Job No.: 00 "0-4-Sg Description: Page:Scc of: 5 Allowable Load for 6 x 6 Deck Post J2K Engineering, Inc. Design and Section Properties Hem-Fir species No. 1 Grade Lumber (N.D.S. Sections 3.6 &3.7) S66 27.73in3 A66:= 30.25•in2 Fb := 975psi Em := 470000•psi column unbraced length, I„ := 10.0•ft- 0.0ft Number of posts: n :- 2 Post Loadings, 35ft 7.75ft i 1 0 PH:= •225-Ib-0.25• - 15ft2 01) PH= 509 lb Pv:= 8251b+ 42651b (roof DL+ deck TL) Pv= 5090 lb Pv check end bearing, fg := fg = 168 psi A66 Allowable bearing stress, Reaction end bearing is OKAY Fg := 850 psi F'g .= 0.75Fg F'g =638 psi without steel bearing surface check allowable axial load for member, Ie := 1.2.1„ DOL := 1.6 d := 5.50 in Fc:= 850 psi Cf:= 1.0 Cfb:_ 1.0 _le =26 < 50 ... OKAY F,:= DOL-Cf•Fc d 0.822•E, FCE:- FCE �Ie2 FcE= 564 psi = 0.8 ^C:= Fc ("Designer" defined variable) d1 - -0.5 i 1 + C `- C columnstability factor, CP := 1 + C - - CP=0.371 F',:= Fe-CP 2•c _` 2•c i c_ F',= 504 psi allowable axial load, Pa := Cp•F',•A66 Pa= 5654 lb Check Bending, PH•(Iu -7.5ft) fb • DOL S66 fb= 344 psi 2 Check Combined Load (f \g/ fg \ <1.0, OKAY + fb ± Fb•DOL•Cfb• 1 -— = 0.43 �Fc) _ \ FcE/_ OKAY... USE: 6 x 6 Post for beam support. - LIMIT UNBRACED LENGTH to: 2' -0" (MAX.) PROJECT: Ruecker Residence Deck Job No.: 11 -048 PAGE:S7 of 5R' ' r Allowable Load for 2 x 6 Cross Brace J2K Engineering Inc. r V MEMBER CAPACITY CHECK for: "Hem-Fir" species No..2 grade P.T. lumber • 2 x 6 Section Properties: Alt,:= 8.25•in2 Lumber Strength Properties and Adjustment Factors: DOL:= 1.60 Cry := 1.1 CFt:= 1.3 Fe:= 1300-psi Ft := 525-psi Emin :=E470000.psi C1:= 0.8 Ci . 0.95 Cm,:= 0.80 CME:= 0.90 F',:= DOL'CF,•Cmc.Ci'F, F'L,= 1464 psi F't:= DOL•Cpt•C'.i•Ft F't = 874 psi E'min = CiF'CME'Emin E'min =401850psi Brace Geometry: h:= 9ft+ Oin lh:= 4.0ft+ 6in No. of Members: n:= 1 Total brace length, l„:_ Vh2+ 11,2 1 = 10.1 ñ TL�C:;;.1-A.t, CZ) check allowable axial load for member, Brace End Reaction, Rend �_ 1.414.0.25.2251b 35ft•[0.50 Oft+ 9.5f4 le:= 1.10.5•In d := m 1.5in 15ft` Rend= 1438 lb 0.822•E'min le FeE :_ d =44 2 FCE d! Ciu Fc Ci := 0.8 , 0.5 1 + Ci„ (1 + Cie_ CtnI column stability factor, Cr:_ - - 2•c1 2.c1 ) ct J Cr=0.126 allowable axial load, Pa:= CT,•F'e•A,fi•n pa= 1522 lb >--- Reid = 1438 lb allowable tensile load, Pat:= Ft•A2t,•n Pat=4331 lb (1) 2 x 6 brace is "OKAY"for cross brace member // where bracing is placed on ONE side of the post with screwed connection where bracing crosses. Also, provide (1) 2 x 6 horizontally ONE side of post at bottom of cross bracing system. Capacity of SDS 1/4"x 3 1/2"screws= 245 LBS PROVIDE: (3) SDS1/4"x 4"wood screw at ends .47 of ALL bracing members PROJECT: Ruecker Residence Deck JOB No.: 11 -048 PAGE:58 of S s ^ Footings for Cross - Braced Deck Posts J2K Engineering, Inc. V Design values per the 2005 Ed. of National Design Specification (NDS) "Design Values for Wood Construction". Maximum Loading to Posts RB 50901b Fpost 5091b Min Dia of footing for Bearing: Point of application for lateral force, h := 1.5•ft RB 0.5 Dimension of concrete footing, b:= 30•in bmin 1500•psfbmin=22 in Depth of embedment, dembed := 1.5•ft 3 Footing section properties:, Sftg:= b Aftg:= b2 6 RB Fpost-dembed < 1500 PSF ... OKAY Maximum soil pressure; qs:_ — + qS= 1108 psf Aftg Sftg Sliding resistance for footing; fs:_ 150—psf •b•dembed2 fs=844 lb > Fpost... OKAY PROVIDE: "30";x 30"-x 18 deep footing @ posts w/Cross Bracing Loading to OUTER Deck Posts Rp := 3090lb Fp 2551b Min Dia of footing for Bearing: Point of application for lateral force, ho:='1.O.ft ( Ro X0.5 Dimension of concrete footing, bo := 24.in bOmin 1500•psfi bomin= 17 to Depth of embedment, doembed 1.0.ft 3 Footing section properties:, Softg := —6° Aoftg:= bo2 RO FO'dpembed < 1500 PSF ... OKAY Maximum soil pressure; q 964 ps;_ + qos= psf Aoftg Softg Sliding resistance for footing; fos:= 150 psf•bo•doembed2 fps= 300 lb > Fpost ... OKAY PROVIDE: 24" x 24"x 12" deep footing @ OUTER posts w/Cross Bracing PROJECT: Ruecker Residence JOB No.: 11 -048 PAGE:5- of