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EC LI PS E CAL‘55 sv k • ECLIPSE - ENGINEERING . C O M ENGINEERING RECEIVED SEP 2 0 2018 01FwC COPY OF Structural Calculations BUiLDINIG DGARD ®1VIJION SEP 17 2018 Pip PROOFS Steel Storage Shelving ��4ti� s' By Pipp Mobile Storage Systems, Inc. tea$-E PIPP PO #37704 SO #92045 jr19Z(§) v- -J,, P�P yOkass.+O - 920 xpiration Qac ' Stance Washington Square 9455 SW Washington Square Rd. — Space #A-15 Portland, Oregon 97223 Prepared For: Pipp Mobile Storage Systems, Inc. 2966 Wilson Drive NW Walker, MI 49544 Please note: The calculations contained within justify the seismic resistance of the storage shelving, the fixed and mobile base supports, and the connection to the existing partition walls for both lateral and overturning forces as required by the 2014 Oregon Structural Specialty Code. These storage shelves are not accessible to the general public. m_...,;.s> "k -yf.; SPOKANE tc�-r.�., 113 West Mend,SLOB,Missals,MT 59082 913WsmnsnAve,Stele 102 Whitefish 64759937 421 West 9lVeraneAve.,506421 Spokane,WA 99201 376 SW Bid Dnhe,5ufe8,Bend,OR 97702 111 SW ColunteaStreet Sole 1090 Pot0.b,OR 97201 Phan:(406)721-5733.Fax(406)5524788 Phone:(406)8823715•Fax(406)5524768 Phone:(509)922-7731.Fax(406)552-4768 Phan:(541)3898859-Fax:(406)552-4768 Phan:(503)595-1229•Fax(406)552-4768 5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong, PE Pipp Mobile STEEL STORAGE SHELF DESIGN 2014 OSSC & ASCE 7-10 - 13.3.1 & 15.5.3.4 Design Vertical Steel Posts at Each Corner - Shelving Dimensions: Are Shelving Units set as Single Depth(1)or Back to Back(2)? N„ := 1 12-SHELF UNITS Total Height of Shelving Unit- ht:= 10.00•ft plf:= lb-ft 1 Width of Shelving Unit- w:= 4.00•ft psf:= lb-ft—2 Depth of Shelving Unit- d := NU•(1.25•ft) = 1.25ft pcf:= lb-ft—3 Number of Shelves- N := 12 kips := 1000-lb ht ksi:= kips•in 2 Vertical Shelf Spacing- S:= = 10.91•in N - 1 Shelving Loads - Maximum Live Load on each shelf is 40 lbs: - Weight Load in Design Live Dead Load per shelf- psf- Load on Shelf- on Shelf- Wtl:= Nu•(40-10 =401b LLI:= W4 =8•psf LL:= LL—8•psf DL := 2.50 psf w•d Section Properties of Double Rivet 14 Gauge Steel 'L' Post : Modulus of Elasticity of Steel- E := 29000.ksi Steel Yield Stress- Fy:= 33•ksi Physical Dimensions of L Post: Density of Steel- psteel:= 490•pcf L Post Width-out-to-out- bi:= 1.500-in L Post Depth-out-to-out- di:= 1.500-in Radius at Corners- R0:= 0.188.in Post Thickness(14 Gauge)- t:= 0.0750•in L Post Width-End-to- IF- L Post Depth-End-to-IF- bi,:= bi —t= 1.425-in di,:= di —t= 1.425•in Radius of Gyration in x and y- rX:= 0.5390•in ry:= 0.5390-in Section Modulus in x and y SX:= 0.0396 in3 Sy:= 0.0396•in3 Moment of Inertia in x and y IX:= 0.0406 in4 Iy:= 0.0406-in4 Full&Reduced Cross Sectional Area's- APf:= 0.225•in2 Apr:= 0.138•in2 Length of Unbraced Post- L,:= S = 10.91.in Ly-•= S = 10.91-in Lt:= S= 10.91-in Effective Length Factor- KX:= 1.7Ky:= 1.7 Kt:= 1.7 Weight of Post- Vertical DL on Post- Vertical LL on Post- W := steel•A h =7.66lb P DL•w•d•N + W =45.161b P := LL •w tl N = 1201b 4-Na P 4•Np Total Vertical Load on Post- Pp := Pd + Pi = 165 lb PPE:= Pd + 0.67•Pi= 126 lb 1 ' , 5 EC LI PS E STANCE 911012018 ENGINEERING PORTLAND,OR Rolf Armstrong, PE Floor Load Calculations : Weight of Mobile Carriage: W0:= 0•Ib Total Load on Each Unit: W:= NU•4•Pp + W0=660.631b Area of Each Shelf Unit: AU := w•(d+ 6•in) =7ft2 Floor Load under Shelf: PSF := W =94•psf Au NOTE:SHELVING LIVE LOAD IS CONSISTENT WITH 100 psf REQ'D FOR RETAIL FLOOR LOADING Find the Seismic Load using Full Design Live Load - ASCE-7 Seismic Design Procedure: Building's Risk Category- BRC:= 2 Importance Factor- IE:= 1.0 Determine Ss and S1 from maps- Sa:= 0.976 S1:= 0.425 Determine the Site Class- SSC:= "D" E Determine Fa and F, - Fa = 1.110 Fv= 1.575 Determine Sips and SD1_ SDs:= 3-(Fa•Ss) 0. 722 Sot.:= 3 (FvSl) =0.446 J Seismic Design Category- SDC= "D" Structural System-Section ASCE-7 Sections 13.3.1815.5.3.4.: 4.Steel Storage Racks R := 4.0 520 := 2 Cd := 3.5 Rp := R aP := 2.5 IP := 1.0 Total Vertical DL WP Total Vertical LL Load on Shelf Wd DL•w•d + N0.4•N = 151b Load on Shelf W; LL•w•d =40 lb Seismic Analysis Procedure per ASCE-7 Sections 13.3.1815.5.3.4: Average Roof Height- hr:= 20.0•ft Height of Rack Attachment- z:= 0•ft Grow For Ground floor) 0.4 4.a •S 1 Seismic Base Shear Factor- Vt:= 1 + 2•— =0.18 Rp hr Ip Shear Factor Boundaries =0.217 Vtmax:= 1.6•SDs-lp = 1.155 mr3 p Seismic Coefficient- Vt:= min(max(Vtmin,Vt) Vtmax) =0.217 Overstrength Factor- 52:= 2.0 NOTE:By ASCE 7-10 Section 13.3.1,0 does not apply for vertically cantilevered architectural systems. 2 0. EC LI PE STANCE 9/10/2018 • ENGINEERING PORTLAND, OR Rolf Armstrong, PE Seismic Loads Continued : ASS LRFD For ASD,Shear maybe reduced- Vp := 0.7•Vt= 0.152 Vp4) := Vt= 0.217 Seismic DL Base Shear- Vtd := Vp•Wd•N = 27.39 lb Vtd4, := Vp)•Wd•N =39.12 lb DL Force per Shelf: Fd := Vp•Wd = 2.28 lb Fdq, := Vp�•Wd = 3.261b Seismic LL Base Shear- Vt1:= Vp•Wi•N =72.781b Vti, := Vp�•WI•N = 103.971b LL Force per Shelf: F1 := Vp•Wi = 6.061b Fid := Vpo.W1=8.661b 0.67*LL Force per Shelf: F1.67:= 0.67•Vp•Wi=4.06 lb F167,p := 0.67•Vp4,•W1= 5.8 lb Force Distribution per ASCE-7 Section 15.5.3.3: - Operating Weight is one of Two Loading Conditions-Condition#1:Each Shelf Loaded to 67% of Live Weight: Cumulative Heights of Shelves- • H1:= 0.0.S+ 1.0•S + 2.0.S+ 3.0•S + 4.0•S + 5.0•S + 6.0.5+ 7.0•S + 8.0-S+ 9.0•S H2 := 10.0•S + 11.0•S H := H1 + H2 = 60.00ft Total Moment at Shelf Base- Mt:= H•Wd + H•0.67•W1 = 2511.13ft•Ib Total Base Shear- V1 := Vtd+ 0.67-Vti=76.151b V14, := Vtdo + 0.67•Vtic,= 108.78 lb Vertical Distribution Factors for Each Shelf- Wd•0.0•S+ Wi•0.67.0.0•S Wd•1.0•S + Wi•0.67.1.0.S C1 := = 0.000 C2 :_ = 0.015 Mt Mt F1 := Ci-(V1) = 0.00 Flit, := Cl (Vl�) =0.00 F2 := C2•(V1) = 1.15 Ib F20 := C2•(V1,p) = 1.65 Ib Wd•2.0•S+ W1•0.67.2.0•S Wd•3.0.S + W1-0.67-3.0.S C3 :_ =0.030 C4:_ =0.045 Mt Mt F3 := C3•(V1) = 2.31 lb F30 := C3•(V1o) =3.30lb F4:= C4•(V1) = 3.461b F40 := C4•(V1,6) =4.941b Wd•4.0•S+ Wi•0.67.4.0.S Wd•5.0•S+ Wi•0.67.5.0•S C5 := =0.061 C6 :_ = 0.076 Mt Mt F5 := C5•(V1) =4.61 lb F5d, := C5•(V16) =6.59 lb F6 := C6•(V1) = 5.77 lb F6d) := C6.0/10) =8.241b Wd•6.0•S+ Wi•0.67•6.0•S Wd•7.0•S + WI.0.67.7.0.S C7 :_ = 0.091 C8 :_ =0.106 Mt Mt F7 := C7•(V1) = 6.921b F7o := C7•(V1o) =9.891b F8 := C8•(V1) =8.081b F8,6 := C8•(V1,p) = 11.541b 3 5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong,PE • Wd•8.O•S + WI.0.67.8.0•S Wd•9.0•S+ W1.0.67.9.0•S C9 :_ =0.121 C10:_ = 0.136 Mt Mt F9 := C9.(V1) =9.231b F9,,i, := C9•(V10) = 13.191b Flo:= C10•(0 = 10.381b F10�:= C10•(V10) = 14.831b Wd•10.0•S+ Wi•0.67.10.0•S Wd•11.0•S+ Wi•0.67.11.0•S Cu.— =0.152 C12:— =0.167 Mt Mt F11:= C11•(V1) = 11.54lb F14,, := C11•(VO = 16.481b F12:= C12•(V1) = 12.69lb F12� := C12•(V1(1)) = 18.13lb C1 + C2 + C3 + C4 + C5 + C6 + C7 + C8 + C9 + C10+ C11+ C12= 1 Coefficients Should total 1.0 Force Distribution Continued : Condition#2:Top Shelf Only Loaded to 100% of Live Weight Total Moment at Base of Shelf- Mta:= H-Wd + (N — 1)•S•Wi = 1303ft•lb Total Base Shear- V2 := Vtd + F1=33 lb V2,4) := Vtd, + Fid =48 lb Wd•0.0•S + 0 W 0.0•S Wd•1.0•S + 0•Wi•1.0-S Cla:_ = 0 C2a:— =0.011 Mta Mta Fla Cla•(V2) =0 Fiat := Cla•(V2(6) =0 F2a:= C2a4V2) =0.41b F20 := C2a•(V24)) =0.5 lb Cla+ C2a+ C3a+ C4a+ C5a+ Cga+ C7a+ Cga+ Cga+ Olga+ Clla+ C12a= 1 Coefficients Should total 1.0 Condition #1 Controls for Total Base Shear By Inspection,Force Distribution for intermediate shelves without LL are negligible. Moment calculation for each column is based on total seismic base shear. Column at center of rack is the worst case for this shelving rack system. Column Design in M 1 .S•max(V ,V ) =8.65ft•lb Bending Stress f • MS =2.62•ksi Short Direction : s N •4 2 1 2 on Column- bx• Sx N0.4 Allowable Ratio of Allowable I fbxMUST BE LESS Bending Stress- Fb := 0.6•Fy= 19.8•ksi Ultimate Stress- 0.13 THAN 1.0 Bending at the Base of Each Column is Adequate 4 '" - E( LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND, OR Rolf Armstrong, PE Deflection of Shelving Bays-worst case is at the bottom bay-the following is the list of shears used in deflection equations. Vp1 := V1 — F1=76 lb Vp2 := Vp1 — F2 =75 lb Vp3 := Vp2 — F3 =73 lb Villa V2 — Fla= 331bVp2a := Via — F2a=33 lb Vp3a V 2a — F3a=321b 1 max(Vp1,Vpia�•S3 max(Vp2,Vp2a)•SS3 O1 :_ =0.0017•in 02 :_ = 0.002.in Nu.4 12•E•l Nu.4 12•E•l Da := 0.05.ht=6•in Ot:= Ol + 02 + 03 + O4+ 05 + Os + 07 + O8 + 01()+ 011+ Al2=0.0134•in if(Ot<Da,"Deflection is Adequate" ,"No Good") = "Deflection is Adequate" Note:The deflection shall not exceed 5%Ht,so shelving deflection is adequate. Moment at Rivet Connection: Shear on Ms dr 2•Tr each rivet dr:= 0.25 in Vr:= =69.22 Ib Ar:= =0.0491•in2 1.5 in 4 Steel Stress Vr Ultimate Stress on Rivet Omega Factor on Rivet fv:= A = 1.41 ksi (SAE C1006 Steel) Fur 47.9ksi (ASD)- S2r:= 2.0 r Allowable Stress 0.563•Fur Ratio of Allowable/ fv on Rivet- Fvr - = 13.48•ksi Ultimate Stress- — =0.10 MUST BE LESS THAN 1.0 S2r Fvr RIVET CONNECTION IS ADEQUATE FOR MOMENT CONNECTION FROM BEAM TO POST Seismic Uplift on Shelves : Seismic Vertical E := 0.2.S (DL+ LL)•w•d =7.58 lb Vertical Dead D := (DL + LL).w•d = 52.50 lb Component: v — os Load of Shelf: Note:since the shelf LL is used to generate the seismic uplift force,it may also be used to calculate the net uplift load. For an empty shelf,only the DL would be used,but the ratio of seismic uplift will be the same. Net Uplift Load on Shelf: Fu := Ev— 0.6.D Fu =—23.92 lb Note: This uplift load is for the full shelf. Each shelf will be connected at each corner. Number of Shelf Uplift Force Fu Connections: Nc 4 per Corner: Fuc := N Fuc = 5.98 lb Ne NOTE:Since the uplift force is negative,a mechanical connection is not required. 5 ry EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND, OR Rolf Armstrong,PE STEEL STORAGE SHELF DESIGN - cont'd Find Overturning Forces : Total Height of Shelving Unit- ht= 10ft Width of Shelving Unit- w=4ft Depth of Shelving Unit- d= 1.25ft WORST CASE Number of Shelves- N = 12 Vertical Shelf Spacing- S= 10.9091•in Height to Top Shelf Height to Shelf (N + 1) CenterofG tsptCenterofG- h,:_ •S = 5.9091ft h := h = 10ft 2 From Vertical Distribution of Seismic Force previously calculated-Controlling Load Cases: ASD Ma:= F1.0.0•S+ F2.1.0•S+ F3.2.0•S+ F4.3.0•S + F5.4.0•S + F6.5.0•S+ F7•6.0•S Moments- Mb := F8.7.0•8+ F9.8.0•S+ F10•9.0•S + F11.10.0•S + F12•11.0•S LRFD Macb := Flo•0.0•S + F2d,•1.0•S + F3�•2.0-S + F44,•3.0•S + F50•4.0•S+ F6d,•5.0•S + F7m.6.0.S Moments- Mbo := F80•7.0•S+ F9,•8.0•S + F100•9.0•S+ F11o•10.0•S+ F12b•11.0•S For Screws-ASD For Anchors-LRFD Weight of Shelf and 67%of LL- W1:= N.(0.6-0.14•SDS) (Wd + 0.67•W1) = 250.57 lb W1b := N.(0.9 -0.2.SDS)•(Wd + 0.67•W1) = 379.48 lb Overturning Shelf Ma Mb530.71ft•Ib M M + M 758.16ft-lb and 67%of LL 1 = 1� a� b� _ Seismic Shelf and 67% ° 1 (M W ) M W of LLTension&Shear Tl 2• dl -21 J= 149.641b T1� := 1 • d� - 2 J= 208.39 lb V1 =76.15 lb V1 = 108.78 lb Weight of Shelf and 100%Top Shelf- W2 :_ (0.6 - 0.14•Sos).(Wd•N + W1) = 110.07 lb W24, := (0.9 - 0.2•SDS)•(Wd•N + W1) = 166.711b erturning Shelf and 100%Top Shelf- M2 := Vtd•hc+ F htop = 222.47ft•lb M20 = Vtdcb• c+ Fi�•htop =317.81ft.lb (M W 1 1M W Seismic Shelf and 100% T2 := 2 2 1 2�1 - = 61.47 lb T20 := • -�� =85.45 lb of LLTension&Shear- 2 \ d 2 ) 2 \ d 2 ) V2 =33.45 lb Val) =47.79 lb 1 max(M1,M2) Force on Column Screws&Anchors: TE:= 2• d = 212 lb r V1 V2 Tension Single - Tsmax max 4 , 4 ,0 lb,= 19.041b Tsmaxd, := max(T1d,,T2,,t),0•Ib) =208.391b Shear Single- max(T T 0•Ib� = 149.641b V maxiV-� , V = 27.191b Vsmax - 1> 2 smax0 -4 , 4 J Tension Double- Tdmax 2•Tsmax=38.071b Tdmax(1) 2•Tsmax0=4171b Shear Double- Vdmax:= 2•Vsmax = 299.28 lb Vdmao 2•Vsmao = 54.39 lb 6 / EC LI PS E STANCE 9/10/2018 - ENGINEERING PORTLAND,OR Rolf Armstrong, PE Find Allowable Axial Load for Column : Allowable Buckling Stresses- IT2•E Qex / 2:= o-ex = 241.77.ksi Kx'Lx 1 rx Distance from Shear Center t••dio2•bic2 to CL of Web via X-axis e0 4.1en= 1.9043 in x Distance From CL Web to x,:= 0.649•in — 0.5•t x0= 0.6115.in Centroid- Distance From Shear Center xo := xc+ en xo = 2.5158•in to Centroid- Polar Radius of Gyration- r0 :=Irx2 + ry2 + x02 r0 = 2.6287 in Torsion Constant- J := 3•(2.bi•t3 + di.t3) J=0.00063.in4 t.bi3.d12 /3•bi•t+ 2.di•t1 Warping Constant- Cw:= Cw= 0.0339•in6 12 6•bi.t+ di•t ) Shear Modulus- G := 11300•ksi _ 1 1T2•E•C 1 6t:= • G•J + `� at= 37.0826-ksi 2 Apr ro _ (Kt-Lt)— J rx \2 := 1 — � I 13=0.0841 ro ) Fet 210•[(6ex+ at) — (0-e)( + Qt)2 —4'1•6ex'6t Fet=32.4691.ksi Elastic Flexural Buckling Stress- Fe := if(Fet <6ex,Fet,6ex) Fe =32.4691.ksi Allowable Compressive Stress- Fn := if Fe > Fy,Fy• 1 — Fy ), F�1 Fn =24.6151 ksi 2 4•Fe Factor of Safety for Axial Comp.- 120 := 1.92 7 ECLPS E STANCE 9/10/2018 E N G I N E E R I N G PORTLAND,OR Rolf Armstrong, PE Find Effective Area - Determine the Effective Width of Flange- Flat width of Flange- wf:= b1 —0.51 wf= 1.4625•in Flange Plate Buckling Coefficient- kf:= 0.43 w F Flange Slenderness Factor- Xf:- 12 f n Xf=0.9114 � kt t E • 0.221 1 Pf 1 — I Pf= 0.8323 • Xf J Xf Effective Flange Width- be := if(Xf> 0.673,pf'wf,wf) be = 1.2173•in Determine Effective Width of Web: Flat width of Web- ww:= d1 —t ww= 1.425•in Web Plate Buckling Coefficient- kw:= 0.43 I Web Slenderness Factor- Xw:= 1'0 w2 ww J E Xw= 0.888 • 0.221 1 Pw:= 1 — Pw=0.8471 • Xw ) Xw Effective Web Width- he := if(Xw> 0.673, pw•ww,ww) he = 1.2071•in Effective Column Area- Ae := t•(he + be) Ae =0.1818•in2 Nominal Column Capacity- Pn := Ae•Fn Pn =4476 lb Pn Allowable Column Capacity- Pa:_ — Pa =2331 lb Slo Check CombinedTr 2 Stresses - Pax x Pc,=33786.94 lb { Kx Lx)2 Pcr Pcrx Pcr=33786.94 lb 19o'Pp 1 Magnification Factor- a:= 1 — =0.991 Cm:= 0.85 Pcr ) Combined 'pp + Cm•fbx -0.18 pp +E+TE Cn,fb MUST BE LESS 0.2 THAN 1.0 Stresses: Pa Fb•a I. a .1 5.1 Final Design: 14 GA.'L' POSTS ARE ADEQUATE FOR REQD COMBINED AXIAL AND BENDING LOADS NOTE: Pp is the total vertical load on post. PpE is the seismic vertical load on post,using 67%live load. 8 5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND, OR Rolf Armstrong, PE STEEL BASE CLIP ANGLE DESIGN -A1018 PLATE STEEL Tension(Uplift) Force Yield Stress of at Corner: T := 50 Ib Angle Steel: FyP := 36.ksi Thickness of Angle: to:= 0.075•in 14 ga Foot Plate Width of Angle Leg: ba := 1.25•in Length of Angle La := 1.375•in Section: Distance out to L:= 0.75•in Section Modulus ba•ta2 Tension Force: of Angle Leg: Se :_ 6 =0.0012 in3 Design Moment Bending Stress M on Angle: M := T�L=3.125 ft•lb on Angle: fb S = 32.ksi e Allowable Bending F — 0.90.E 32.4•ksi Ratio of Pb =0 988 MUST BE LESS THAN 1.00 Stress: b vp = Allowable Loads: F b Ultimate Tensile Gross Area of Strength of Clip: Fup:= 65•ksi the Clip: Agc := be•to = 0.0938.int Effective Net Area of the Clip: A„ := Agc —Eta•(0.375.in)] =0.0656.in2 Limiting Tensile Strength of Clip: Tcmao := min[(0.90•Fyp•Agc),(0.75.Fup-Aec)] = 3037.5 lb {f�Tcmax� >Tsmax4,,"Checks Okay" ,"No Good") = "Checks Okay" 14 GA. ANGLE CLIP WILL DEFORM PRIOR TO ANCHOR PULLING OUT OF CONCRETE, BUT NOT WILL NOT TEAR COMPLETELY THROUGH, THEREFORE CLIPS ARE ADEQUATE. BEARING STRENGTH OF SCREW CONNECTIONS - AISI E.4.3.1 Omega for Bearing(ASD)- S2 := 3.00 Stu := 2.35 Specified Tensile Stress of Clip&Post,Respectively- Fel:= 51ksi Fut:= 51ksi Diameter of Screw- dss:= 0.25in 14 GA Clip Thickness- ts1:= 0.075in 14 GA Post Thickness- ts2 := 0.075in Nominal Bearing Strength- Single Screw-ASD Double Screw-ASD 4.2.Fu2 dss.tsz 311 (AISI C-E4.3-3) Pns min 2.7 Fur dss ts1 = 22001b Pnd:= 2'Pns=44001b. 2.7 Fuz•dss•tsz )) Allowable Bearing Strength- Pas := Pns =733.3 lb Pad:= Pnd = 1466.5 lb Qs Qs 9 5 E(' LI PS E STANCE 9/10/2018 ENGLNEERING PORTLAND, OR Rolf Armstrong,PE SCREW CONNECTION CAPACITIES (1/4" SCREW IN 14 GA STEEL): Note:Values obtained from'Scafco'Labels using an 0=3.00 Single Screw-ASD Double Screw-ASD Allowable Tensions,Pullout- Tsst 227lb Tsdt 2•Tsst=454 lb Allowable Tensions,Pullover- Tssv 6561b Tsdv 2•Tssv= 1312 lb Allowable Shear- Vss:= 6001b Vsd := 2•Vss= 1200 lb The allowable shear values for(1) 1/4"dia.screw exceeds the allowable bearing strength of Ref Attached'Scafco'Table the connection. Therefore,bearing strength governs for screw connection capacity. for V&T Values BOLT CONNECTION CAPACITIES (3/8" DIA. x 3", MIN., HILTI KB-TZ WITH 2 5/16" EMBEDMENT): Single Anchor-LRFD Double Anchor-LRFD Ref Attached'HILTI'PROFIS Allowable Tension Force- Tas := 1107-lb Tad:= 2149-lb calcs fort/&T Values 5 Allowable Shear Force- Vas := 1466•Ib Vad:= 2055-lb (: 3 USE: HILTI KB-TZ ANCHOR (or equivalent)-318" x 3", min., long anchor with 2 5116" embedment installed per the requirements of Hilti to fasten fixed shelving units to existing concerete slab. Use 1J4" dia. screw to fasten base to 14 GA shelf member. DETERMINE ALLOWABLE TENSION/SHEAR FORCES FOR CONNECTION: Single Screw-ASD Double Screw-ASD Allowable Tension Force- Tasi Tssv= 6561b Tas2 Tsdv= 13121b Allowable Shear Force- Vas1:= min(Vss,Pas) = 600 lb Vas2:= min(Vsd,Pad) = 1200 lb ...ae /Usmax�l� Wall Supported S2•Vsmax, �., Combined Loading smax� + 0.06 <1.00 Shear Loading = 0.04 <1.00 (Single Anchor) Tas ) � Vas ) OKAY (Single Anchor)- Vas OKAY Combined Loading Stu •/Vsma" + 0.71.Tsmax 1 0.19 <1.00 Tension Pullout Tsmax _0 08 <1.00 (Single Screw)- 1.10 - ` Vast Tasl ) OKAY (Single Screw) Tsst OKAY Wall Supported iV Combined Loading �Tdmax4"I + /Vdmaxo I� 0.07 <1.00 Shear Loading dma"� =0.05 <1.00 (Double Anchor)- — OKAY (Double Anchor)- Vad OKAY Tad ) Vad ) Combined Loading IZu Vdmax• + 0.71.Tdmax 1=0.19 <1.00 Tension Pullout 0.08 <1.00 (Double Screw)- 1.10 Sts Vas2 Tas2 ) OKAY (Double Screw) Tsdt OKAY 10 •.5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong,PE Connection from Steel Shelf to Wall Seismic Analysis Procedure per ASCE-7 Section 13.3.1: Average Roof Height- hr= 20ft Height of Shelf Attachments- zb := z+ ht zb = 10 ft (At Top for fixed racks connected to walls) 0.4•ap•SDS ( zb Seismic Base Shear Factor- Vt:= • 1 + 2•— Vt=0.361 Rp hr) Ip Shear Factor Boundaries- Vtmin := 0.3•Sps•Ip =0.217 Vtmax := 1.6•Sips.Ip = 1.155 Seismic Coefficient- Vt:= min(max(Vtmin,Vt),Vtmax) =0.361 Number of Shelves- N = 12 Weight per Shelf- Wti =40 lb Total Weight on Rack- WT:= 4•(Pd + 0.67•Pi) WT= 502.23 lb Seismic Force at top and bottom- T 0.7 Vt WT ,,:= 2 T„=63.451b Connection at Top: Standard Stud Spacing- Sstud := 16-in Width of Rack- w=4 ft Number of Connection Points on each rack- Force on each connection point Nu:= max 2. floor W stud��1 =3 Fu:= —v = 21.15 lb S Capacity per inch of lb Required Fc embedment into wood Nailer- Ws= 135.—in Embedment Depth- ds:_ Ws For Steel Studs: Pullout Capacity for#10 Screw Ratio of Allowable Loads it Fc MUST BE in 20 go studs(per Scafco)- 120:= 84.1b for screws into walls- T20 =0.25 <1.0 Connection at Bottom: Ratio of Allowable Loads 11.TV MUST BE 0.09 for anchors into slab- <1.0 0 7•Vad MIN #10 SCREW ATTACHED TO EXISTING WALL STUD IS ADEQUATE TO RESIST SEISMIC FORCES ON SHELVING UNITS. EXPANSION BOLT IS ADEQUATE AT THE BASE. 11 5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND, OR Rolf Armstrong,PE • Pipp Mobile STEEL STORAGE SHELF DESIGN 2014 OSSC & ASCE 7-10 - 13.3.1 & 15.5.3.4 Design Vertical Steel Posts at Each Corner - Shelving Dimensions: Are Shelving Units set as Single Depth(1)or Back to Back(2)? Nu := 1 6-SHELF UNITS Total Height of Shelving Unit- ht:= 10.00•ft plf:= lb.ft 1 Width of Shelving Unit- w:= 3.00•ft psf:= Ib•ft 2 Depth of Shelving Unit- d:= Nu•(1.50•ft) = 1.5ft pcf:= Ib•ft 3 Number of Shelves- N := 6 kips := 1000•Ib Vertical Shelf Spacing- S:= ht = 24.in ksi := kips in 2 N - 1 Shelving Loads - Maximum Live Load on each shelf is 75 lbs: Weight Load in Design Live Dead Load per shelf- psf- Load on Shelf- on Shelf- W4:= Nu•(75.10 =751b LL1:= Wfj =16.6667 psf LL:= LLQ = 16.6667•psf DL := 2.50•psf w•d Section Properties of Double Rivet 14 Gauge Steel 'L' Post : Modulus of Elasticity of Steel- E:= 29000.ksi Steel Yield Stress- FY:= 33.ksi Physical Dimensions ofLPost: Density of Steel- psteel:= 490-pcf L Post Width-out-to-out- bi := 1.500•in L Post Depth-out-to-out- di:= 1.500•in Radius at Corners- Re:= 0.188•in Post Thickness(14 Gauge)- t:= 0.0750•in L Post Width-End-to-IF- L Post Depth-End-to-IF- bi, := bi—t= 1.425-in di,:= di—t= 1.425•in Radius of Gyration in x and y- rx:= 0.5390•in ry:= 0.5390 in S - Sx:= 0.0396•in3 Sy:= 0.0396•in3_ Moment of Inertiainxandy- Ix:= 0.0406•in4 Iy:= 0.0406•in4 Full&Reduced Cross Sectional Area's- APf:= 0.225•in2 Apr:= 0.138 int Length of Unbraced Post- Lx:= S=24.00-in Ly:= S=24.00•in Lt:= S = 24.00•in Effective Length Factor- Kx:= 1.7Ky:= 1.7 Kt:= 1.7 Weight of Post- Vertical DL on Post- Vertical LL on Post- DL•w•d•N LL•w-d•N Wp := psteel•Apf•ht=7.66 lb Pd : 4 N + WP = 24.53 lb Pi:= = 112.5 lb Nu 4-Nu Total Vertical Load on Post- Pp := Pd + P1 = 137 lb PpE:= Pd + 0.67•P1= 100lb 12 ECLI PS E STANCE 9/10/2018 • ENG I EER I N G PORTLAND,OR Rolf Armstrong,PE Floor Load Calculations : Weight of Mobile Carriage: W,:= 40•Ib Total Load on Each Unit: W:= N,•4•Pp + W,= 588.121b Area of Each Shelf Unit: Au := w•(d + 6•in) =6ft2 Floor Load under Shelf: PSF =98•psf Au NOTE:SHELVING LIVE LOAD IS CONSISTENT WITH 100 psf REQ'D FOR RETAIL FLOOR LOADING Find the Seismic Load using Full Design Live Load - ASCE-7 Seismic Design Procedure: Building's Risk Category- BRC:= 2 Importance Factor- IE:= 1.0 Determine Ss and Si from maps- SS= 0.976 S1 =0.425 Determine the Site Class- SSC := "D" Determine Fa and F, - Fa = 1.110 F,= 1.575 2 2 Determine SDS and SDI._ SDS:= 3,(Fa Ss) = 0.722 SD1 := 3 •(F„•Si) =0.446 Seismic Design Category- SDC= "D Structural System-Section ASCE-7 Sections 13.3.1&15.5.3.4.: 4.Steel Storage Racks R := 4.0 S2 := 2 Cd := 3.5 RP := R au := 2.5 Ip := 1.0 Total Vertical DL Wp Total Vertical LL Load on Shelf Wd DL•w•d + NU•4 N = 161b Load on Shelf W1 LL-w•d =751b Seismic Analysis Procedure per ASCE-7 Sections 13.3.1&15.5.3.4: Average Roof Height- hr= 20 ft Height of Rack Attachment- z=0 Grow For Ground floor) Seismic Base Shear Factor- Vt 0.4 ap•SDS •/1 + 2• '=0.18 Rp hr Ip = 1.155Shear Factor Boundaries =0.217 Vtmaxtmin . DSp 1.6 SDS Ip Seismic Coefficient Vt min(max(Vtmin>Vt). Vt, ax) =0.217 Overstrength Factor- := 2.0 NOTE:By ASCE 7-10 Section 13.3.1, does not apply for vertically cantilevered architectural systems. 13 n E( I P E STANCE 9/10/2018 ENGINEERING PORTLAND, OR Rolf Armstrong, PE Seismic Loads Continued : ASD LRFD For ASD,Shear may be reduced- Vp := 0.7.Vt=0.152 Vpd, := Vt=0.217 Seismic DL Base Shear- Vtd := Vp•Wd•N = 14.881b Vtdd, := Vpd,•Wd•N =21.251b DL Force per Shelf: Fd := Vp•Wd = 2.481b Fdd, := Vpd,•Wd =3.54lb Seismic LL Base Shear- Vt1:= Vp•W1•N =68.231b Vtl4 := Vp4•W1•N =97.471b LL Force per Shelf: F1 := Vp•W1 = 11.37 Ib F1d, := Vpd,•W1= 16.24 Ib 0.67*LL Force per Shelf: F1.67:= 0.67•Vp•W1 = 7.62 lb F167d,:= 0.67•Vp4•W1 = 10.88 lb Force Distribution per ASCE-7 Section 15.5.3.3: Operating Weight is one of Two Loading Conditions Condition#1:Each Shelf Loaded to 67% of Live Weight: • Cumulative Heights of Shelves- H1:= 0.0•S+ 1.0•S+ 2.0•S+ 3.0•S + 4.0•S+ 5.0-S H2 := 0 H := H1 + H2 = 30.00ft Total Moment at Shelf Base- Mt:= H•Wd + H•0.67•W1 = 1998.13ft•lb Total Base Shear- V1 :=Vtd + 0.67.V11=60.591b V10 := Vtd4 + 0.67•Vtl4=86.56 lb Vertical Distribution Factors for Each Shelf- Wd•O.O•S+ Wl•0.67 0.0•S Wd•1.0•S + WI.0.67.1.0•S C1:= =0.000 C2 :_ =0.067 Mt Mt F1:= C1 (V1) =0.00 F10-:= C1.(V14) = 0.00 F2 := C2•(V1) =4.041b F24, := C2 (Vic,) = 5.771b Wd•2.0•S + W1.0.67.2.0•S Wd•3.0•S+ W1•0.67.3.0•S C3 := =0.133 C4:= =0.200 Mt Mt / / F3 := C3•(Vl) =8.08 lb F34, := C3•(Vld,) = 11.54 lb F4:= C4•\Vl� = 12.12Ib F44, := C4•lVld,) = 17.311b Wd•4.0•S + W1 0.67 4.0 S Wd•5.0•S+ W1.0.67.5.0•S C5 :_ =0.267 C6 :_ =0.333 Mt Mt F5 := C5•(V1) = 16.16 Ib F54, := C5•(V�4,) = 23.081b F6 := C6.0/1) = 20.201b F64, := C6•�V1�) = 28.85 Ib C1 + C2 + C3 + C4+ C5+ C6 = 1 Coefficients Should total 1.0 14 00 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong, PE Force Distribution Continued : Condition#2:Top Shelf Only Loaded to 100%of Live Weight Total Moment at Base of Shelf- Mta:= H•Wd + (N — 1)•S-Wi = 1241ft•Ib Total Base Shear- V2 := Vtd + Fi= 26 lb V2, := Vtdo + F1 =371b Wd•0.0•S+ 0•Wi•0.0-S Wd•1.0•S + 0•W1.1.0•S Cla:_ = 0 C2a:_ = 0.026 Mta Mta Fla Cla•(V2) = 0 Fla•r1;•:= Cia•(V2�) =0 F2a C2a'(V2) = 0.71b F2a0:= C2a'(V20) = 1 lb Cla+ C2a+ C3a+ C4a+ C5a+ C6a= 1 Coefficients Should total 1.0 Condition #1 Controls for Total Base Shear By Inspection,Force Distribution for intermediate shelves without LL are negligible. Moment calculation for each column is based on total seismic base shear. Column at center of rack is the worst case for this shelving rack system. Column Design inM _ 1 .S•max(V V � = 15.15ft•Ib Bending Stress f Ms— =4.59•ksi Short Direction: s N •4 2 1' 2 on Column bx:= S u x Allowable Ratio of Allowable/ fbx MUST BE LESS Bending Stress Fb := 0.6 Fy= 19.8•ksi Ultimate Stress- 0'23 THAN 1.0 F6` Bending at the Base of Each Column is Adequate • 15 5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong, PE Deflection of Shelving Bays-worst case is at the bottom bay-the following is the list of shears used in deflection equations. Vol := V1 — F1=611b Vo2 := Vo1 — F2 =57lb Vo3 := Vo2 — F3 =481b Vola V2 — Fia=261b VA2a Vola — F2a=261b VO3a VO2a — F3a=241b 1 max(VA1,Vola)•S3 1 max(Vo2,Vo2a).S33 O1:_ =0.0148•in 02 :_ =0.014•in Nu•4 12•E•I Nu•4 12.E•I, Da := 0.05.ht=6•in Ot:= O1+ 02 + 03 + A4 + 05 + 06 =0.0543•in if(Ot<Da,"Deflection is Adequate" ,"No Good") _ "Deflection is Adequate" Note:The deflection shall not exceed 5%Ht,so shelving deflection is adequate. Moment at Rivet Connection: Shear on Ms r d 2•'rr each rivet- dr:= 0.25 in Vr:= = 121.18 lb Ar := =0.0491•in2 1.5 in 4 Steel Stress Vr Ultimate Stress on Rivet Omega Factor on Rivet- f := A = 2.47•ksi (SAE C1006 Steel) Fur 47.9ksi (ASD)- SZr:= 2.0 r Allowable StressF — 0.563•Fur = 1348 ksi Ratio of Allowable 1 fv =0 18 MUST BE LESS THAN 1.0 on Rivet yr'— . Ultimate Stress S2r Fvr RIVET CONNECTION IS ADEQUATE FOR MOMENT CONNECTION FROM BEAM TO POST Seismic Uplift on Shelves Seismic Vertical E := 0.2•S (DL+ LL)•w•d = 12.451b Vertical Dead D :_ (DL + LL)•w•d=86.251b Component: v' os Load of Shell: Note:since the shelf LL is used to generate the seismic uplift force,it may also be used to calculate the net uplift load. For an empty shelf,only the DL would be used,but the ratio of seismic uplift will be the same. Net Uplift Load on Shelf: Fu := Ev—0.6.D Fu =—39.30 lb Note: This uplift load is for the full shelf. Each shelf will be connected at each corner. Number of Shelf Uplift Force Fu Connections: Nc 4 per Corner: Fuc := N Fuc —- 9.82 Ib c NOTE:Since the uplift force is negative,a mechanical connection is not required. 16 E•-•61 I PS E STANCE 9/10/2018 • ENG I EER I N G PORTLAND,OR Rolf Armstrong, PE STEEL STORAGE SHELF DESIGN - cont'd Find Overturning Forces : Total Height of Shelving Unit- ht= 10ft Width of Shelving Unit- w=3 ft Depth of Shelving Unit- d= 1.5ft WORST CASE Number of Shelves- N =6 Vertical Shelf Spacing- S= 24•in Height to Top Shelf Height to Shelf (N + 1) Center of G- htop ht= 10 ft Center of G - h,:- 2 •S =7 ft From Vertical Distribution of Seismic Force previously calculated-Controlling Load Cases: ASD Ma := F1.0.0•S + F2.1.0•S+ F3.2.O-S + F4.3.0•S+ F5.4.0•S + F6.5.0•S Moments- Mb 0 LRFD Mao Flo•0.0-S + F20•1.0•S + F3,i,•2.0•S + F40•3.0•S + F5d,•4.0•S+ F60•5.0•S Moments- Mbo 0 For Screws-ASD For Anchors-LRFD Weight of Shelf and 67%of LL- W1 := N.(0.6 -0.14.SDs)•(Wd + 0.67•W1) = 199.381b W10 := N.(0.9 - 0.2•Sps)•(Wd + 0.67•W1) =301.961b Overturning Shelf M1Ma + Mb444.32ft-lb M M + M 634.75ft•lb and 67%of LL = 1� a� b� _ Seismic Shelf and 67% 1 M1 W11 1 M10 W1 1 of LL Tension&Shear- Ti 2' d -2 J=98.26 lb Tl� := 2 d - 2 )= 136.09 lb V1 =60.59 lb V10 =86.56 lb Weight of Shelf and 100%Top Shelf- • W2 :_ (0.6 -0.14.SDs)•(Wd•N + w1) =86.381b W20 := (0.9 - 0.2•SDs)•(Wd•N + W1) = 130.811b Overturning Shelf and 100%Top Shelf- M2 Vtd•hc+ Frhtop = 217.85ft•lb M2o := Vtdo•hc+ Flo•htop =311.22ft•lb Seismic Shelf and 100% T = 1•/M2 -W2 I1 •/M�0 W20)_ 2 :- -• - = 51.021b Ted, := - - - 71.041b of LL Tension&Shear- 2 d 2 J 2 d 2 ) V2 =26.25 lb V20 =37.50 lb maxM M2) Force on Column Screws&Anchors: TE:= 1 • dl = 148 lb Tension Single - Ts„ := max/V1 , V-2 ,0•Ib1= 15.151b Ism*, := max(T10,T2,,,0•Ib) = 136.09lb �4 4 ) Shear Single- Vsmax := max(T1,12,0-10 =98.26 lb Vsmax,, := max/Vl� , V-20 ,=21.641b 4 4 Tension Double- Tdmax 2•Tsmax =30.29 lb Tdmax(¢ 2•Tsmaxo = 272 lb Shear Double- Vdmax:= 2.Vsmax= 196.52 lb Vdmax 2•V43.28 lb � �= smax� = 17 4.. 5 EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong,PE ' Find Allowable Axial Load for Column : Allowable Buckling Stresses- _ 72•E hex• vex=49.95•ksi /Kx.Lx12 rx ) 2 2 Distance from Shear Center e t•dic .bic e,— 1.9043 in to CL of Web via X-axis °'_ 4•Ix Distance From CL Web to x,:= 0.649.in— 0.5.t x,= 0.6115.in Centroid- . Distance From Shear Center xp := x,+ e, x0 = 2.5158.in to Centroid- Polar Radius of Gyration- ro := Jrx2 + ry2 + x02 r0 = 2.6287 in Torsion Constant- J := 3 J2-13113 + di•t3) J=0.00063•in4 t•b13•d12 (3•b0+ 2 di•t) 6 Warping Constant CW:= C,„,,=0.0339•in 12 6.b0+ di•t Shear Modulus- G:= 11300.ksi 1 72•E•C 1 . vt:_ • G J + at= 13.611.ksi Apr•r02 _ (Kt•Lt)2 ix \2 13:= 1 - ° 13=0.0841 pro J 1 Fet:_ 2 �•L( ex+ 6t) —Jwex+ Cr)2 —4•110-ex.4 Fet= 10.8522•ksi Elastic Flexural Buckling Stress- Fe := if(Fet<vex,Fet,vex) Fe = 10.8522•ksi _ Allowable Compressive Stress- Fn := if Fe > Fy ,Fy• 1 — Fy 1,F� F, = 10.8522•ksi 2 4.Fe) Factor of Safety for Axial Comp.- 120 := 1.92 18 ECLIPSESTANCE 911012018 ENGINEERING• PORTLAND,OR Rolf Armstrong, PE Find Effective Area — Determine the Effective Width of Flange- Flat width of Flange- wf:= b1 — 0.51 wf= 1.4625-in Flange Plate Buckling Coefficient- kf:= 0.43 Flange Slenderness Factor- 1.052 wt Fn g �f:_ t E Xf=0.6052 0.221 1 Pf:= 1 — I pf= 1.0517 Xf ) Xf Effective Flange Width- be := if(Xf> 0.673,pf-Wf,wf) be = 1.4625.in Determine Effective Width of Web: Flat width of Web- ww:= d1 —t w = 1.425-in Web Plate Buckling Coefficient- kN := 0.43 Web Slenderness Factor- 1.052 Ww Fn �`w= t E Xw=0.5897 0.221 1 Pw 1 — pw= 1.0632 ) Xw Effective Web Width- he := if(X > 0.673,pw.ww,ww) he = 1.425-in Effective Column Area- Ae := t-(he + be) Ae =0.2166-in2 Nominal Column Capacity- Pn := Ae•Fn Pn =2350 lb Pn Allowable Column Capacity- Pa :_ Pa = 1224 lb 0 Check Combined Stresses - i2•E.Ix Pcrx Pcrx=6980.77 lb Kx-Lx)2 Per Pcrx Pcr=6980.77 lb Magnification Factor a:= 1 — )=0.962 Cm :— 0.85 Pcr Combined Pp Cm•fbx. PpE+ TE Cm'fbx MUST BE LESS Stresses: — + 0 32, • + 0:41 Pa Fb'a. Pa Fbcx THAN 1.0 Final Design: 14 GA. 'L' POSTS ARE ADEQUATE FOR REQD COMBINED AXIAL AND BENDING LOADS NOTE:Pp is the total vertical load on post. PpE is the seismic vertical load on post,using 67%live load. 19 5 EC LI PS E STANCE 9/10/2018 E N G I N E E R I N G PORTLAND,OR Rolf Armstrong,PE STEEL BASE CLIP ANGLE DESIGN -A1018 PLATE STEEL Tension(Uplift) Force Yield Stress of at Corner: T := 50.1b Angle Steel: Fyp := 36 ksi Thickness of Angle: to:= 0.075.in 14 ga Foot Plate Width of Angle Leg: ba := 1.25.in Length of Angle La:= 1.375•in Section: 2 Se Distance out to L:= 0.75.in Section Modulus ba•ta =0.0012 in3 Tension Force: of Angle Leg: 6 Design Moment Bending Stress M on Angle: M := T•L=3.125 ft•Ib on Angle: fb := S = 32.ksi e Allowable Bendingf Stress: Fb := 0.90•Fyp =32.4.ksi Alaowable Loads: F =0.988 MUST BE LESS THAN LOU b . Ultimate Tensile Gross Area of Fg ba'ta 0.0938 int Strength of Clip: up:= 65 ksi the Clip: Ac := = Effective Net 2 Area of the Clip: Aec := Agc —Lta•(0.375.in)] =0.0656.in Limiting Tensile Strength of Clip: Tcmax,0 := min[(0.90•Fyp•Agc),(0.75•Fup•AeC)] = 3037.5 lb if(Tcmax0, >Tsmax ,"Checks Okay "No Good") = "Checks Okay" 14 GA. ANGLE CLIP WILL DEFORM PRIOR TO ANCHOR PULLING OUT OF CONCRETE, BUT NOT WILL NOT TEAR COMPLETELY THROUGH, THEREFORE CLIPS ARE ADEQUATE. BEARING STRENGTH OF SCREW CONNECTIONS - AISI E.4.3.1 Omega for Bearing(ASD)- 12s:= 3.00 Stn := 2.35 Specified Tensile Stress of Clip&Post,Respectively- Ful:= 51ksi Fut:= 51ksi ' Diameter of Screw- dss:= 0.25in 14 GA Clip Thickness- ts1:= 0.075in 14 GA Post Thickness- ts2:= 0.075in Nominal Bearing Strength- Single Screw-ASD Double Screw-ASD ri 4.2•Fu2.i dss'ts2311 (AISI C-E4.3-3) Pns := min 2.7•Fui dss'tsi = 22001b Pnd:= 2.13n, =44001b 2.7•Fu2.dss.ts2 )) P Allowable Bearing Strength- Pas:= Pns =733.31b Pad:= rid = 1466.51b Sls us 20 5 EC. LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND, OR Rolf Armstrong,PE SCREW CONNECTION CAPACITIES (1/4"4) SCREW IN 14 GA STEEL): Note:Values obtained from'Scafco'tabels using an 02=3.00 Single Screw-ASD Double Screw-ASD Allowable Tensions,Pullout- Tsst 2271b Tsdt 2.Tsst =454 lb Allowable Tensions,Pullover- Tssv 6561b Tsdv 2•Tssv= 1312 lb Allowable Shear- Vss:= 600lb Vsd := 2•Vss= 12001b The allowable shear values for(1)114'dia.screw exceeds the allowable bearing strength of Ref Attached'Scafco'Table the connection. Therefore,bearing strength governs for screw connection capacity. for V&T Values • BOLT CONNECTION CAPACITIES (3/8" DIA. x 3", MIN., HILTI KB—TZ WITH 2 5/16" EMBEDMENT): Single Anchor-LRFD Double Anchor-LRFD Ref Attached'HILT!'PROFIS calcs for V&T Values Allowable Tension Force- Tas := 1107•Ib Tad:= 2149.Ib 5 Allowable Shear Force- Vas := 1466•Ib Vad:= 2055•Ib C 3 USE: HILTI KB-TZ ANCHOR (or equivalent)-318"x 3", min., long anchor with 2 5/16" embedment installed per the requirements of Hilti to fasten fixed shelving units to existing concerete slab. Use 114" dia.screw to fasten base to 14 GA shelf member. DETERMINE ALLOWABLE TENSION/SHEAR FORCES FOR CONNECTION: Single Screw-ASD Double Screw-ASD Allowable Tension Force- Tasi:= Tssv=656 lb Tas2 Tsdv= 1312 lb Allowable Shear Force Vast:= min(Vss Pas = 600lb Vas2:= Min(Vsd Pad = 1200 lb Combined Loading "Tsmax01 (Vsmax0) Wall Supported S2•Vsmaz� (Single Anchor)- + =0.03 <1.00 Shear Loading =0,03 <1.00 Tas ) Vas ) OKAY (Single Anchor)- Vas OKAY Qu "Vsmax Tsmax 1 Tension Pullout Tsmax Combined Loading + 0.71. _ <1.00 —1)&„,1 0.13 <1.00 • (Single Screw)- 1.10•SZs ' Vasz Tasi OKAY (Single Screw) Tsst • OKAY Wall Supported it V �' Combined Loading Tdmaxcp 1 Vdmax41 <1.00 Shear Loading dmaxcp a 04 <1.00 (Double Anchor) Tad ) + Vad ) 0.03 OKAY (Double Anchor)-t Vad ,a OKAY Combined Loading 12u 1Vdmax Tdmax Tension Pullout rd-max; + 0.71 = 0.13 <1.00 :07 <1.00 (Double Screw)- 1.10 11s ` Vas2 Tas2 ) OKAY (Double Screw)- Tsdt OKAY • 21 w .-- EC LI PS E STANCE 9/10/2018 ENGINEERING PORTLAND,OR Rolf Armstrong, PE STEEL ANTI-TIP CUP AND ANTI-TIP TRACK DESIGN Tension(Uplift) Force on each side- T := 2•Vdmax=393.05 lb Connection from Shelf to Carriage=1/4"diameter bolt through 14ga.steel: Capacity of 1/4"diam.screw in 14 ga.steel- Z,:= 715.1b if(T <2•Z,,"(2) 1/4" Bolts are Adequate","No Good") ="(2) 1/4" olts are Adequate" Use 3/16"Diameter anti-tip device for connection of carriage to track Yield Stress of Angle Steel- Thickness of Anti-tip Head- Width of Anti-tip Rod+Radius- Width of Anti-tip Head- Fy:= 36•ksi to:= 0.090.in br:= 0.25•in ba := 0.490-in - Area of Anti- 2 Area of Anti- 7T•br2 2 tip Weld Aw:= IT•br•(0.094.in)•cos(45•deg) =0.052•in tip Rod Air:= 4 =0.049•in Stress on Weld fw:= T =7.5291-ksi Stress on T Connection- A,,, rod_ fr:_ — =8.0071 ksi Ar Ratios of fw to Fy fw =0.2091 fr =0.2224 fw0.3585 The stress on the bolt head is less &fr to Fy: Fy Fy 0.3 (70 ksi) than the weld and material capacity. 0.85.13, — br Width of Anti-tip Flange- La:_ = 0.083-in Tension Force per Flange leg- Ti := 0.5•T 2 • b t2 Bending Moment on Leg- M1:= Ti La =0.682ft•lb Section Modulus of Leg- Si := a a =0.001•in3 2 6 Bending Stress on Leg- fb := M1 = 12.366•ksi Ratio of Allowable Loads- fb =0.40 M10T BE Si 0.85•Fy . Thickness of Aluminum Track Width of Anti-Tip track- L:= 5.1•in (average thickness)- tt:= 0.33.in L•tt23 Spacing of Bolts- Stb:= 22.5•in Section Modulus of Track- St:= — =0.093•in 6 Design Moment on Track- T•Stb BendingStress on Track- f M 11.942 ksi for continuous track section M ba — — 8 St Allowable Stress t• f of Aluminum- Fb := 21•ksi Ratio of Allowable Loads- [ ba =0.57 Fb. '''''m7.0044.6' Ratio of Allowable Loads (Single Anchor)- ,: for continuous track section t., ANTI-TIP CLIP STEEL CONNECTION AND TRACK ARE ADEQUATE 22 E+ LI PE STANCE 9/10/2018 • ENGINEERING PORTLAND,OR Rolf Armstrong,PE Connection from Steel Shelf to Wall Seismic Analysis Procedure per ASCE-7 Section 13.3.1: Average Roof Height- hr= 20ft Height of Shelf Attachments- zb := z+ ht zb = 10 ft (At Top for fixed racks connected to walls) Seismic Base Shear Factor- 0.4 ap•Sos zb Vt:= • 1 + 2.— Vt= 0.361 Rp hr J 1p Shear Factor BoundariesVtmin 0.3 Sps 1p =0.217 Vtmax 1.6•Sos•lp = 1.155 Seismic Coefficient- Vt:= min(max(Vtmin,Vt),Vtmax) =0.361 Number of Shelves- N =6 Weight per Shelf- Wti= 75 lb Total Weight on Rack- WT:= 4•(Pd + 0.67.Pi) WT= 399.63 lb 0.7•Vt.WT Seismic Force at top and bottom- Tv:= 2 Tv= 50.49 lb Connection at Top: Standard Stud Spacing –stud 16-in Width of Rack- w=3ft • Number of Connection Points on each rack- Force on each connection point- Nc:= maxr2, \'floor\ W 2 Fc:= Nc = 25.25lb Sdd)JJ Capacity per inch of lb Required Fc embedment into wood Nailer- Ws 135 in Embedment Depth- — =0.187 in :r a : ..,. ks_ For Steel Studs: Pullout Capacity for#10 Screw Ratio of Allowable Loads Fc }' MUST BE in 20 ga studs(per Scafco)- T20:= 84 Ib for screws into walls- e 30 <1.0 T , ¢, Connection at Bottom: Ratio of Allowable Loads ; � Tv MUST BE 0.07- for anchors into slab- <1.0 0.7•Vad MIN #10 SCREW ATTACHED TO EXISTING WALL STUD IS ADEQUATE TO RESIST SEISMIC FORCES ON SHELVING UNITS. EXPANSION BOLT IS ADEQUATE AT THE BASE. 23 USGS Design Maps Summary Report User-Specified Input Report Title 18-09-154 Mon September 10, 2018 21:44:50 UTC Building Code Reference Document 2012/2015 International Building Code (which utilizes USGS hazard data available in 2008) Site Coordinates 45.44968°N, 122.78663°W Site Soil Classification Site Class D - "Stiff Soil" Risk Category I/II/III RN:JR It .11111 sb Oro Are ortl tan u Beaverton f" {• ?r • -X- Tigard. _ Lake Oswego 47* u, 1atin 414, USGS-Provided Output Ss = 0.976g SMS = 1.083g SDs = 0.722g S1 = 0.425 g SM1 = 0.669 g SD1 = 0.446 g • For information on how the SS and Si values above have been calculated from probabilistic (risk-targeted) and deterministic ground motions in the direction of maximum horizontal response, please return to the application and select the"2009 NEHRP" building code reference document. NICE, Response Spectrum Design Response Spectrum 11ra 0;K+ 0.72 0231 0[+1 0.77 0.56 'it OM 0.us 1, 0.35 i 0,40 P .41 032 0.33 0.34 0.22 0.16 0.11 001 ODD OOO 0X% 0.31 1,10 OW 0243 10) 1.31 1,4+7 lit: 1247 200 olrl 0.3.7 0.10 Oda? 0231) 1)2* 1.37 1.10 1W 1147 24X) Pafad,T(sec) Perkxl,T(see) Although this information is a product of the U.S.Geological Survey,we provide no warranty,expressed or implied,as to the accuracy of the data contained therein.This tool is not a substitute for technical subject-matter knowledge. ,, screw and Weld Capacities SSMA Screw Capacities Table Notes 1. Capacities based on AISI 5100 Section E4. 6. Pull-out capacity is based on the lesser of pull-out capacity in 2. When connecting materials of different steel thicknesses or sheet closest to screw tip or tension strength of screw. tensile strengths, use the lowest values. Tabulated values 7. Pull-over capacity is based on the lesser of pull-over capacity for assume two sheets of equal thickness are connected. sheet closest to screw header or tension strength of screw. 3. Capacities are based on Allowable Strength Design (ASD) and 8. Values are for pure shear or tension loads. See AISI Section E4.5 include safety factor of 3.0. for combined shear and pull-over. 4. Where multiple fasteners are used, screws are assumed to 9. Screw Shear(Pss), tension (Pts), diameter, and head diameter have a center-to-center spacing of at least 3 times the nominal are from CFSEI Tech Note (F701-12). diameter(d). 10.Screw shear strength is the average value, and tension strength 5. Screws are assumed to have a center-of-screw to edge-of-steel is the lowest value listed in CFSEI Tech Note (F701-12). dimension of at least 1.5 times the nominal diameter(d) of the 11.Higher values for screw strength (Pss, Pts), may be obtained by screw, specifying screws from a specific manufacturer. Allowable Screw Connection Capacity(lbs) ' ` #6 Screw #8 Screw #10 Screw #12 Screw 3/4"Screw Thickness Design FY Fu (Pss=643 lbs,Pts=419 lbs) (Pss=1278 lbs,Pts=586 lbs) (Pss=1644 lbs,Pts=1158 lbs) (Pss=2330 lbs,Pts=2325 lbs) (Pss=3048 lbs,Pts=3201 lbs) (Mils) Thickness Yd Tensile (ksiiel) (ksi) 0.138"dia,0.272"Head 0.164"dia,0.272"Head 0.190"dia,0.340"Head 0.216"dia,0.340"Head 0.250"dia,0.409"Head Shear Pull-Out Pull-Over Shear Pull-Out Pull-Over Shear Pull-Out Pull-Over Shear Pull-Out Pull-Over Shear Pull-Out Pull-Over 18 0.0188 33 33 44 24 84 48 29 84 52 33 105 55 38 105 60 44 127 27 0.0283 33 33 82 37 127 89 43 127 96 50 159 102 57 159 110 66 191 30 0.0312 33 33 95 40 140 103 48 140 111 55 175 , 63 175 127 73 211 1 33 0.0346 33 45 151 61 140 164 72 195 177 84 265 188 95 265 203 110 318 43 0.0451 33 45 214 79 140 244 94 195 263 109 345 280 124 345 302 144 415 54 0.0566 33 45 214 100 140 344 118 195 370 137 386 394 156 433 424 180 521 68 0.0713 33 45 214 125 140 426 149 195 523 173 386 557 196 545 600 227 656 97 0.1017 33 45 214 140 140 426 195 195 548 246 386 777 280 775 1,016 324 936 118 0.1242 33 45 214 140 140 426 195 195 548 301 386 777 342 775 1,016 396 1,067 54 0.0566 50 65 214 140 140 426 171 195 534 198 386 569 225 625 613 261 752 68 0.0713 50 65 214 140 140 426 195 195 548 249 386 777 284 775 866 328 948 97 0.1017 50 65 214 140 140 426 195 195 548 356 386 777 405 775 1,016 468 1,067 118 0.1242 50 65 214 I 140 140 426 195 195 , 548 386 386 t 777 494 775 1,016 572 1,067 Weld Capacities Table Notes 1. Capacities based on the AISI S100 Specification Sections E2.4 for 6. Transverse capacity is loading in perpendicular direction of the fillet welds and E2.5 for flare groove welds. length of the weld. 2. When connecting materials of different steel thicknesses or 7. For flare groove welds, the effective throat of weld is tensile strengths, use the lowest values. conservatively assumed to be less than 2t. 3. Capacities are based on Allowable Strength Design (ASD). 8. For longitudinal fillet welds, a minimum value of EQ E2.4-1, 4. Weld capacities are based on E60 electrodes. For material E2.4-2, and E2.4-4 was used. thinner than 68 mil, 0.030"to 0.035" diameter wire electrodes 9. For transverse fillet welds, a minimum value of EQ E2.4-3 and may provide best results. E2.4-4 was used. 5. Longitudinal capacity is considered to be loading in the direction 10.For longitudinal flare groove welds, a minimum value of of the length of the weld. EQ E2.5-2 and E2.5-3 was used. Allowable Weld Capacity(lbs!in) Thickness Design FY Fu ; Fillet Welds Flare Groove Welds (Mils) Thickness Yield Tensile (ksi) (ksi)" Longitudinal Transverse Longitudinal Transverse 43 0.0451 33 45 499 864 544 663 54 0.0566 33 45 626 1084 682 832 68 0.0713 33 45 789 1365 859 1048 97 0.1017 33 45 1125 1269 -, 54 0.0566 50 65 905 1566 985 1202 68 0.0713 50 65 1140 1972 1241 1514 97 0.1017 50 65 1269 1269 'Weld capacity for materia/thickness greater than 0.10"requires engineering judgment to determine/eg of welds, W1 and W2. . . - ..,..�. ,. ,.,-... MII[ATII www.hntl.us Profis Anchor 2.7.6 www.hee.us Profis Anchor 2.7.6 Company: ECLIPSE ENGINEERING,INC. Page: 1 Company: ECLIPSE ENGINEERING,INC. Page: 2 Specifier: Project: Specifier: Project: Address: 376 SW Bluff Dr.,Suite 8 Sub-Project I Pos.No.: Address: 376 SW Bluff Dr.,Suite 8 Sub-Project I Pos.No.: Phone I Fax: 541-389-96591 Date: 5/14/2018 Phone I Fax: 541.389-96591 Date: 5/14/2018 E-Mai: E-Mal: specifiers comments: 2 Proof I Utilization(Governing Cases) Design values[Ib] Utilization 1 Input data ,,....,, Loading Proof Load Capacity fafg4,[Y4I Status Tension Pullout Strength 500 1,107 46/- OK Anchor type and diameter: Kwik Bolt TZ-CS 3/8(2) :n. Shear Steel Strength 300 1,466 -/21 OK Effective embedment depth: 1-1,,,,,,=2.000 in.,h,wm=2.313 in. Material: Carbon Steel Leading Ba 13,, { Utilization pyv[Yo] Status Evaluation Service Report: ESR-1917 SAFE---ET Combined tension and shear loads 0.452 0.205 5/3 34 OK Issued I Valid: 5/1/20171 5/1/2019 Proof: Design method ACI 318-14/Mech. Stand-off instatafion: -(Recommended plate thickness:not caicdated) 3 Warnings Profile: no prole •Please consider all details and hints/warnings given in the detated report! Base material: cracked concrete,2500,f'=2,500 psi;h=4.000 in. I Fastening meets the design criteria! Installation: hammer drilled hole,Installation condition:Dry Reinforcement: tension:condition B,shear:condition B;no supplemental s[itting reinforcement present 4 Remarks;Your Cooperation Duties edge reinforcement:none or<No.4 bar •Any and all information and data contained in the Software concem solely the use of Hiti products and are based on the principles,formulas and Seismic loads(cat.C,D,E,or F) Tension load:yes(17.2.3.4.3(d)) security regulations in accordance with Hiles technical directions and operating,mounting and assembly instructions,etc.,that must be strictly Shear load:yes(17.23.5.3(c)) complied with by the user.All figures contained therein are average figures,and therefore use-specific tests are to be conducted prior to using the relevant Hiti product The results of the calculations carried out by means of the Software are based essentially on the data you put in. Therefore,you bear the sole responsiblity for the absence of errors,the completeness and the relevance of the data to be put in by you. Geometry[in.]&Loading[Ib,in.lb] Moreover,you bear sde responsiblity for having the results of the calculation checked and bleared by an expert,particularly with regard to compliance with applicable norms and permits,prior to using them for your specific rashly.The Software serves only as an aid to interpret norms Z and permits without any guarantee as to the absence of errors,the correctness and the relevance of the results or suitablay for a specific application. •You must take all necessary and reasonable steps to prevent or limit damage caused by the Software.In particular,you must arrange for the regular backup of programs and data and,if applicable,carry out the updates of the Software offered by His on a regular basis.It you do not use the AutoUpdate function of the Software,you must ensure that you are using the current and thus up-to-date version of the Software in each case by carrying out manual updates via the Hiti Website. Hiti wit not be liable for consequences,such as the recovery of lost or damaged data or programs,arising from a culpable breach of duty by you. gt g U ./'r i 1L I C1 0 Pi Input data and resits must be checked for agreement with the existing conditions and for piauslblkyl Input data and results must be checked for agreement with the existing conditions and for piausiblity! PRORS Anchor(c)2003.2009 HIO AG,FL-9494 Schaan HIS is a registered Trademark of HIS AG,Steen, PRORS Anchor(c)2003.2009115 AG,FL-9494 Sidman HIS Is a registered Trademark of Hili AG,Schaan - T • www.hlltL us - Profis Anchor 2.7.6 www.hllti.us Profis Anchor 2.7.6 Company: ECLIPSE ENGINEERING,INC. Page: 1 Company: ECLIPSE ENGINEERING,INC. - Page: 2 Specifier: Project: Specifier: - Project: Address: 376 SW Bluff Dr.,Suite 8 Sub-Project I Pos.No. Address: 376 SW Bluff Dr.,Suite 8 Sub-Project I Pos.No.: Phone I Fax: 541-389-96591 Date: 5/14/2018 Phone I Fax: 541.389-96591 Date: 5/14/2018 E-Mal: E-Mai: Speclfler's comments: 2 Proof I Utilization(Governing Cases) Design values[lb] Utilization 1 Input data Loading Proof Load Capacity Ba I By CAl Status a Tension Concrete Breakout Strength 1,000 2,149 47/- OK Anchor type and diameter: Kwlk Bolt TZ-CS 3/8(2) ea.n, ,,,-„I....*„„ {it„ 4i.Ti Effective embedment depth: h,,,,,,=2.000 in.,11,0„,=2.313 in. Shear Concrete edge future in direction x+ 700 2,055 -/35 OK Material: Carbon Steel S4AFE.--ET' Loading Bx By Utilization p.,ENStatus Evaluation Service Report: ESR-1917 r Combined tension and shear loads 0.465 0.341 5/3 45 OK Issued I Valid: 5/1/20171 5/1/2019 Proof: Design method ACI 318-14/Mech. Stand-off installation: eh,=0.000 in.(no stand-off);t=0.074 in. 3 Warnings Anchor plate: Ix xi,x t=3.000 in.x 7.000 in.x 0.074 in.;(Recommended plate thickness:not calculated •Please consider al detals and hints/wamings given in the detaled report! Probe: no prole Fastening meets the design criteria! Base material: cracked concrete,2500,f.'=2,500 psi;h=4.000 in. Installation: hammer drilled hole,Installation condition:Dry 4 Remarks;Your Cooperation Duties Reinforcement: tension:condition B,shear:condition B;no supplemental splitting reinforcement present Any• and all information and data contained in the Software concern solelythe use of Hifi products and are based on the principles,formulas and edge reinforcement:none or<No.4 bar security regulations in accordance with Hiti's technical directions and operating,mounting and assembly instructions,etc.,that must be strictly Seismic loads(cat C,D,E,or F) Tension load:yes(17.2.3.4.3(d)) complied with by the user.P11 figures contained therein are average figures,and therefore use-specific tests are to be conducted prior to using the relevant Hifi product.The results of the calculations carried out by means of the Software are based essentially on the data you put in. Shear load:yes(17.2.3.5.3(c)) Therefore,you bear the sole responsibiity for the absence of errors,the completeness and the relevance of the data to be put in by you. Moreover,you bear sde responsiblity for having the results of the calculation checked and deared by an expert,particularly with regard to Geometry[In.]&Loading[lb,in.ib] compliance with applicable norms and permits,prior to using them for your specific faclity.The Software serves oily as an aid to interpret norms and permits without any guarantee as to the absence of errors,the correctness and the relevance of the results or suitablity for a specific Z application. •You must take all necessary and reasonable steps to prevent or limit damage caused by the Software.In particular,you must arrange for the regular backup of programs and data and,if applicable,carry out the updates of the Software offered by Hlti on a regular basis.If you do not use the AutoUpdate function of the Software,you must ensure that you are using the current and thus up-to-date version of the Software in each case by carrying out manual updates via the Hifi Website. Hiti wit not be liable for consequences,such as the recovery of lost or damaged data or programs,arising from a culpable breach of duty by you. t D B S —'S ,- ,r-- �.iD74 J ? "`j 4 ___I \II a e . f x \\IT '. 11 Input data and resits must be checked for agreement wah the existing condnons and for plausibly! Input data and res.4ts must be checked for agreement with the eesdng conditions and for pausibilty! PRORS Anchor(c)2003.2009 FRS AG,FL-9494 Schaaf, Hifi is a registered Trademark of Hlt!AG,Schoen PROFIS Anchor(c)2003-2009 Hid AG.FL-9454 Schwan His is a registered Trademark of Hlti AG.Schaaf, 1.m vemm Aaa4.4»b4db<Qbaea mmm4<o41m!o< 4<ae ve44 Abba aaa4 4<4< �I Iz 2966 WILSON DRIVE NW . GRANDRADS,MI49534 -LaUIIIII'! 11, FAX: 616.791.9916 - - - - - - WWW.PIPPMOBILE.COM MOBILE STORAGE SYSTEMS INC. E-MAIL: CUSTOMERSERV@PIPPMOBILE.COM INSTALLATION INSTRUCTIONS i� UNITS WILL BE 10'HIGH WITH 6 SHELVES 5 OPENINGS SHELVING UNIT PARTS: O O © © OD OE O OG 0 0 0., 41# toet_ r,l)r- O ,'UPRGIft TUPRGHT DOUBLE DOUBLE RIVET DOUBLE RIVET •BAR SHELF MATE IAL HAN--‘ r„,; 3*-- t. G INSTALLATDN RIVET UPSDEDOWN LOW PROFLE i RE ca• A tk 4 A 46, r r 4t itt, v f T AT al gy I I 1 !tit, f ? f it fig . Q EACH CORNERA /4 4 A . 7 1 1 1A 1 fA f x ©EACH LEVEL A .. _ O N CT I 111*.rj ©EACH LEVEL N 43 ' EACH LEVEL E apy 0c3 EACH LEVEL© ,� N ._.. ("1 G --4 / �''� 13 E4ELETS OPEN EACH LEVEL NiGI GIm ~ O O O ril ,.._-.( 4 - , ,...,, *1 ,, ' rc,.- N ,��fy:;,. In N fD 1 114 APPROX.21'OPEN EACH LEVEL co X ^ry? H- / I (.,_-g, iffliff Iik _ / P If a : corn v 3 , x \u 0 .. , comes ? T / Ln D- (1) v v v m 7- T 104,---- ,� ;� O 3 3 3 v 0 o Q \ q 36 OR 4R' 0 EACH LEVEL N t- n MI O ^0 I ( 1) f,�, N K I< in . 4 r B ♦ R