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Specifications EOP2o2/-00226 /6/6 5 Shl 72" RECEIVED SEP222021 CITY OF TIGARD BUILDING DIVISION Structural ngineering Design Project-Na.44.v : EDNETICS Proita-Nu r : 21-0917-12 �Ep P R OFFS �� C, INFF SRO Pam: 09/20/21 w�� � 62618PE Sfre Address. 1(01(05" SW 72NP op OREGON Cu(y/Sfv,'?i : PORTLANP, OR 97224 1)47EOFREG1\ J /NGQIAO Z� Scoff of-Work. : STORAGE RACK EXPIRES: 06-30-2022 09/22/2021 Mingqiao Zhu, PE/P.Eng 1428 N Shevlin Court Sewickley, PA 15143 Structural Engineering & Design I nc. 1815 Wright Ave La Verne. CA 91750 Tel:909.596.1351 Fax: 909.596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 TABLE OF CONTENTS Title Page 1 Table of Contents 2 Design Data and Definition of Components 3 Critical Configuration 4 Seismic Loads 5 to 6 Column 7 Beam and Connector 8 to 9 Bracing 10 Anchors 11 Base Plate 12 Slab on Grade 13 type a select-ed.xIs rage of t 9/18/202 I Structural Engineering & Design Inc. 1815 Wright Ave La Verne. CA 91750 Tel: 909.596.1351 Fax: 909.596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Design Data 1)The analyses herein conforms to the requirements of the: 2018 IBC Section 2209 2019 CBC Section 2209 ANSI MH 16.1-2012 Specifications for the Design of Industrial Steel Storage Racks"2012 RMI Rack Design Manual" ASCE 7-16,section 15.5.3 2)Transverse braced frame steel conforms to ASTM A570,Gr.55,with minimum strength, Fy=55 ksi Longitudinal frame beam and connector steel conforms to ASTM A570,Gr.55,with minimum yield, Fy=55 ksi All other steel conforms to ASTM A36, Gr.36 with minimum yield,Fy= 36 ksi 3)Anchor bolts shall be provided by installer per ICC reference on plans and calculations herein. 4)All welds shall conform to AWS procedures,utilizing E70xx electrodes or similar.All such welds shall be performed in shop,with no field welding allowed other than those supervised by a licensed deputy inspector. 5)The existing slab on grade is 8"thick with minimum 3500 psi compressive strength.Allowable Soil bearing capacity is 750 psf. The design of the existing slab is by others. 6) Load combinations for rack components correspond to 2012 RMI Section 2.1 for ASD level load criteria Definition of Components A Caiinn /� Beam a - � /` I I_� i_ ,to, promummumnommmommmems a{ ace Beam to Cbletim j://1 cbnt7ectcs Diagonal l Brace Fxa,� • HeBgttt Beam atue • king Base Piate and Anthers / Panel BeaneLength II I < gam' 'fig L.Frame4,f Depth Ftcsnt View Down Atte fLongit�inali Frame Sectarn A.Crass A`tsie f Transverse 'l Frame type a select-ed.xis Page 3 of 4 Q2 9/18/202 I Structural Engineering & Design Inc. 1815 Wright Ave La Verne, CA 91750 Tel 909.596.1351 Fax: 909.596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Configuration&Summary:Type A Selective Rack N. — 4 **RACK COLUMN REACTIONS ASO LOADS AXIAL DL= 50/b 36" 48" AXIAL LL= 4,000/b SEISMIC AXIAL Ps=+/- 1,536 lb BASE MOMENT= 0 in-lb 96" 96" N 60" 36" 4' 144" .I' -i` 42" -71' Seismic Criteria #Bm Lvls Frame Depth Frame Height #Diagonals Beam Length Frame Type Ss=0.852, Fa=1.159 2 42 in 96.0 in 2 144 in Single Row d Component Description STRESS Column Fy=55 ksi Hannibal IF1514-3x1-5/8x14ga P=4050 Ib, M=12572 in-lb 0.96-OK _ Column&Backer None None None N/A Beam Fy=55 ksi HMH 50140/5"Face x 0.071"thk Lu=144 in Capacity:4613 lb/pr 0.87-OK Beam Connector Fy=55 ksi Lvi 1:4 pin OK I Mconn=8370 in-lb Mcap=22664 in-lb 0.37-OK Brace-Horizontal Fy=55 ksi Hannibal 1-1/2x1-1/2x16ga 0.12-OK Brace-Diagonal Fy=55 ksi _ Hannibal 1-1/2x1-1/2x16ga 0.21-OK Base Plate Fy=36 ksi 8x5x3/8 Fixity=0 in-lb 0.49-OK Anchor 2 per Base 0.5"x 3.25"Embed HILTI KWIKBOLT TZ ESR 1917 Inspection Reqd(Net Seismic Uplift=324 lb) 0.125-OK Slab&Soil 8"thk x 3500 psi slab on grade.750 psf Soil Bearing Pressure 0.16-OK Level Load** Story Force Story Force Column Column Conn. Beam Per Level Beam Spcg Brace Transv Longit. Axial Moment Moment Connector 1 4,000 lb 60.0 in 36.0 in 346 lb 173 lb 4,050 lb 12,572 "# 8,370 "# 4 pin OK 2 4,000 lb 36.0 in 48.0 in 553 lb 276 lb 2,025 lb 2,487 "# 3,970 "# 4 pin OK (**Load defined as product weight per pair of beams Total: 899 lb 449 lb Notes type a select-ed.xls Page Lk- of `3 9/18/202 Structural Engineering & Design Inc. 1815 Wright Ave La Verne.CA 91750 Tel: 909.596.1351 Fax:909.596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Seismic Forces Configuration:Type A Selective Rack lateral analysis is performed with regard to the requirements of the 2012 RMI ANSI MN 16.1-2012 Sec 2.6&ASCE 7-16 sec 15.5.3 Ss= 0.852 Transverse(Cross Aisle)Seismic Load t 51= 0.390 V= Cs*Ip*Ws=Cs*Ip*(0.67*P*Prf+D) vt Fa= 1.159 Cs1= Sds/R Fv= 1.900 = 0.1646 Cs-max*Ip= 0.1646 Sds=2/3*Ss*Fa= 0,658 Cs2= 0.044*Sds Vmin= 0.015 Sd1=2/3*S1*Fv= 0.494 = 0.0290 Eff Base Shear=Cs= 0.1646 Transverse Elevation Ca=0.4*2/3*Ss*Fa= 0.2633 Cs3= 0.5*S1/R Ws= (0.67*PLRF1*PL)+DL(RMI 2.6.2) (Transverse,Braced Frame Dirt)R=4.0 = 0.0488 = 5,460 lb Ip 1.0 Cs-max= 0.1646 Vtransv=Vt= 0.1646* (100 lb+5360 lb) PRFI i Base Shear Coeff=Cs= 0.1646 Etransverse= 899 lb Pallet Height=hp= 48.0 in Limit States Level Transverse seismic shear per upright DL per Beam Lvl= 50 lb Level PRODUCT LOAD P P*0.67*PRFI DL hi wl*hi Fi Fi*(hi+hp/2) 1 4,000 lb 2,680 lb 50 lb 60 in 163,800 345.8 lb 29,047-# 2 4,000 lb 2,680 lb 50 lb 96 in 262,080 553.2 lb 66,384-# I sum: P=8000 lb 5,360 lb 100 lb W=5460 lb 425,880 899 lb 1=95,431 Longitudinal(Downaisle)Seismic Load Similarly for longitudinal seismic loads,using R=6.0 Ws= (0.67*PLRF2*P)+DL PRF2= 1.0 t', 1 `'``'" ""` r:.:1 Cs1=Sd1/(T*R)= 0.0823 = 5,460 lb (Longitudinal,Unbraced Dir.)R= 6.0 Cs2= 0.0290 Cs=Cs-max Ip* 0.0823 T= 1.00 sec (� 1 ���':'�N`"� "`= "" Cs3= 0.0325 Vlong= 0.0823*(100 lb+5360 lb) ""` r`';.[`.;::',:l 1,::,:�1 Cs-max= 0.0823 Elongltudinal= 449 lb Limit States Level Longlt.seismk shear per uptight Level PRODUC LOAD P P*0.67*PRF2 DL hi wi*hi Fi _Emit View 1 4,000 lb 2,680 lb 50 lb 60 in 163,800 172.7 lb 2 4,000 lb 2,680 lb 50 lb 96 in 262,080 276.3 lb I • sum: 5,360 lb 100 lb W=5460 lb 425,880 449 lb type a select-ed.xls Page ✓Of t75 9/15/202 1 Structural Engineering & Design Inc. 1815 Wright Ave La Verne.CA 91750 Tel: 909.596.1351 Fax:909.596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Downaisle Seismic Loads Configuration:Type A Selective Rack Determine the story moments by applying portal analysis.The base plate is assumed to provide no fixity. Seismic Story Forces Typical frame tnade Vlong= 449 lb Tributary area oftwo columns Vcol=Vlong/2= 225 lb of rack frame Niii, _____ _ ``F1= 173 lb - � �\' `::: ;':� mil F.` rap—re Typical Frame made F2= 276 lb �� \ l /oftwocolumns F3= 0 lb 0 0' ' �.�M ,,.`1:,:s' is i�'�+Tr t , MN TQtisw 1 96" Front View SLde_Yisys Seismic Story Moments Conceptual System COL Mbase-max= 0 in-lb <===Default capacity hl-eff= hl-beam clip height/2 Mbase-v= (Vcol*hlefF)/2 = 56 in Vcol tommommonli = 6,286 in-lb <===Moment going to base IF Mbase-eff= Minimum of Mbase-max and Mbase-v h2 = 0 in-lb PINNED BASE ASSUMED M 1-1= [Vcol*hleff]-Mbase-eff M 2-2= [Vcol-(F1)/2]*h2 ,�Ell °° = (225 lb* 56 in)-0 in-lb = [225 lb- 138.2 Ib]*36 in/2 t = 12,572 in-lb = 2,487 in-lb hi ! t 41, Mseis= (Mupper+Mlower)/2 Beam to Column Mseis(1-1)= (12572 in-lb+2487 in-lb)/2 Msels(2-2)= (2487 in-lb+0 in-lb)/2 Elevation = 7,529 in-lb = 1,243 in-lb rho= 1.0000 L_ Summary of Forces LEVEL hi Axial Load Column Moment** Mseismic** Mend-fixity Mconn** Beam Connector 1 60 in 4,050 lb 12,572 in-lb 7,529 in-lb 4,428 in-lb 8,370 in-lb 4 pin OK I 2 36 in 2,025 lb 2,487 in-lb 1,243 in-lb 4,428 in-lb 3,970 in-lb 4 pin OK i Mconn= (Mseismic+Mend-fixity)*0.70*rho Mconn-allow(4 Pin)= 22,664 in-lb **all moments based on limit states level loading type a select-ed.xls Page t of k' 9/1 8/202 I Structural Engineering & Design Inc. is 1815 Wright Ave La Verne. CA 91750 Tel: 909.596.1351 Fax:909.596.7186 By: Bs/Mqz Project: Ednetics Project*21-0917-12 Column(Longitudinal Loads) Configuration:Type A Selective Rack Section Properties Section: Hannibal IF1514-3x1-5/8x14ga h 3.000 in Aeff= 0.437 in^2 Iy= 0.169 in^4 Kx= 1.7 x _ Ix= 0.689 in^4 Sy= 0.205 In^3 Lx= 57.5 in ,, ¢I 5x= 0.459 in^3 ry= 0.621 In Ky= 1.0 . fir' i I 1.625 in rx= 1.255 in Fy= 55 ksi Ly= 36.0 in y r nj i1 10.075 n Qf= 1.67 Cmx= 0.85 Cb= 1.0 5 E= 29,500 ksi -{25 0.75 in Loads Considers loads at level 1 COLUMN DL= 50 lb Critical load cases are:RMI Sec 2.1 COLUMN PL= 4,000 lb Load Case 5::(1+0.105*Sds)D+0.75*(1.4+0.14Sds)*B*P+0.75*(0.7*rho*E)<=i.0,ASD Method Mcol= 12,572 in-lb axial load coeff: 0.78338505*P seismic moment weir: 0.5625*Mcol Sds= 0.6583 Load Case 6::(1+0.14*Sds)D+(0.85+0.14Sds)*8*P+(0.7*rho*E)<=1.0,ASD Method 1+0.105*Sds= 1.0691 axial load coeff.• 0.65951 seismic moment coeff.• 0.7*Mcol 1.4+0.14Sds= 1.4922 By analysis,Load case 6 governs utilizing loads as such 1+0.14Sds= 1.0922 0.85+0.14*Sds= 0.9422 Axial Load=Pax= 1.092162*50 lb+0.942162*0.7*4000 lb Moment=Mx= 0.7*rho*Mcol B= 0.7000 = 2,693 lb = 0.7* 12572 in-lb rho 1.0000 = 8,800 in-lb Axial Analysis KxLx/rx= 1.7*57.5"/1.255" KyLy/ry= 1*3670.621" Fe > Fy/2 = 77.9 = 58.0 Fn= Fy(1-Fy/4Fe) = 55 ksi*[1-55 ksi/(4*48 ksi)] Fe= n^2E/(KL/r)max^2 Fy/2= 27.5 ksi = 39.2 ksi = 48.0ksi Pa= Pn/Qc Pn= Aeff*Fn Qc= 1.92 = 17145 lb/1.92 = 17,145 lb = 8,930 lb P/Pa= 0.30 > 0.15 Bending Analysis Check: Pax/Pa+(Cmx*Mx)/(Max*px)5 1.0 P/Pao+ Mx/Max<_ 1.0 Pno= Ae*Fy Pao= Pno/Qc Myield=My= Sx*Fy = 0.437 in^2*55000 psi = 240301b/1.92 = 0.459 inA3*55000 psi = 24,030 lb = 12,515 lb = 25,245 in-lb Max= My/S?f Pcr= n^2EI/(KL)max^2 = 25245 in-lb/1.67 = nA2*29500 ksi/(1.7*57.5 in)^2 = 15,117 in-lb = 20,995 lb px= {1/[1-(Qc*P/Pcr)]}^-1 = {1/[1-(1.92*2693 lb/20995 Ib)]}^-1 = 0.75 Combined Stresses (2693 lb/8930 Ib)+(0.85*8800 In-Ib)/(15117 in-lb*0.75) = 0.96 < 1.0,OK (EQ C5-1) (2693 Ib/12515 Ib) + (8800 in-lb/15117 in-Ib)= 0.80 < 1.0,OK (EQ C5-2) **For comparison,total column stress computed for load case 5 is: 88.0% g loads 3186.996275 lb Axial and M= 6600 in-lb type a select-ed.xis Page L of 1 5 9/18/202 I Structural Engineering & Design Inc. 1815 Wright Ave La Verne. CA 91750 Tel: 909.596.1351 Fax: 909.596.7186 By: Bs/Mqz Project: Ednetics Project* 21-0917-12 BEAM Configuration:Type A Selective Rack DETERMINE ALLOWABLE MOMENT CAPACITY 2.75 in A)Check compression flange for local buckling(B2.1) 1.75 In 3 w= c-2*t-2*r _ = 1.75 in-2*0.071 in-2*0.071 in r f = 1.466 in 1.625 in w/t= 20.65 II l=lambda= [1.052/(k)^0.5]*(w/t)*(Fy/E)^0.5 Eq. B2.1-4 s.000in ti,! [1.052/(4)^0.5]* 20.65*(55/29500)1'0.5 = 0.469 <0.673, Flange is fully effective Eq. B2.1-1 0.071 in B)check web for local buckling per section b2.3 f1(comp)= Fy*(y3/y2)= 50.41 ksi f2(tension)= Fy*(y1/y2)= 102.17 ksi Y= f2/f1 Eq. B2.3-5 Beam= HMH 50140/5"Face x 0.071"thk = -2.027 Ix= 3.241 in^4 k= 4+ 2*(1-Y)^3+2*(1-Y) Eq. B2.3-4 Sx= 1.241 inA3 = 65.53 Ycg= 3.300 in flat depth=w= y1+y3 t= 0.071 in = 4.716 in w/t= 66.42253521 OK Bend Radius=r= 0.071 in 1=lambda= [1.052/(k)^0.5]*(w/t)* (f1/E)^0.5 Fy=Fyv= 55.00 ksi = [1.052/(65.53)^0.5] *4.716*(50.41/29500)^0.5 Fu=Fuv= 65.00 ksi = 0.357 <0.673 E= 29500 ksi be=w= 4.716 in b2= be/2 Eq B2.3-2 top flange=b= 1.750 in b1= be(3-Y) = 2.36 in bottom flange= 2.750 in = 0.938 Web depth= 5.P"Fy^ b1+b2= 3.298 in > 1.558 in,Web is fully effective Determine effect of cold working on steel yield point(Fya)per section A7.2 - t,(comp) Fya= C*Fyc+(1-C)*Fy (EQ A7.2-1) Lcorner=lc= (p/2)*(r+t/2) y2 0.167 in C= 2*Lc/(Lf+2*Lc) Lflange-top=Lf= 1.466 in = 0.186 In y3 m= 0.192*(Fu/Fy)-0.068 (EQ A7.2-4) depth = 0.1590 Bc= 3.69*(Fu/Fy)-0.819*(Fu/Fy)^2- 1.79 (EQ A7.2-3) = 1.427 since fu/Fv= 1.18 < 1.2 Ycg y1 and r/t= 1 < 7 0K 12(tanston) then Fyc= 8c* Fy/(R/t)^m (EQ A7.2-2) = 78.485 ksi Thus, Fya-top= 59.36 ksi (tension stress at top) Fya-bottom= Fya*Ycg/(depth-Ycg) yl= Ycg-t-r= 3.158 in = 115.22 ksi (tension stress at bottom) y2= depth-Ycg= 1.700 In Check allowable tension stress for bottom flange y3= y2-t-r= 1.558 in Lflange-bot=Lfb= Lbottom-2*r*-2*t = 2.466 in Cbottom=Cb= 2*Lc/(Lfb+2*Lc) = 0.119 Fy-bottom=Fyb= Cb*Fyc+(1-Cb)*Fyf = 57.80 ksi Fya= (Fya-top)*(Fyb/Fya-bottom) = 29.78 ksi if F= 0.95 Then F*Mn=F*Fya*Sx= 35.11 in-k O ' I • • Structural Engineering & Design Inc. 1815 Wright Ave La Verne. CA 91750 Tel:909.596.1351 Fax: 909,596,7186 , By: Bs/Mqz Project: Ednetics Project#: 21-0917-12 BEAM Contiguration: Type A Selective Rack RMI Section 5.2, PT II Section Beam= HMH 50140/5"Face x 0.071"thk Ix=Ib= 3.241 in^4 2.75 in Sx= 1.241 in^3 t= 0.071 in E= 29500 ksi \-1.75 in ,[ Fy=Fyv= 55 ksi F= 300.0 I` Fu=Fuv= 65 ksi L= 144 in r s Fya= 59.4 ksi Beam Level= 1 ______________ 1.625 in P=Product Load= 4,000 lb/pair D=Dead Load= 50 lb/pair 5.000 0.071 in 1.Check Bending Stress Allowable Loads Mcenter=F*Mn= W*L*W*Rm/8 W=LRFD Load Factor= 1.2*D+ 1.4*P+1.4*(0.125)*P RMI2.4 Item 8 FOR DL=2%of PL, W= 1.599 Rm= 1 -[(2*F*L)/(6*E*Ib+3*F*L)) t Ililllllllltlillliillpllillllllllllillll III 11 1 -(2*300*144 in)/[(6*29500 ksi*3.241 in^3)+(3*300*144 In)] = 0.877 • if F= 0.95 Then F*Mn=F*Fya*Sx= 69.98 in-k _ Thus,allowable load -'I ~ 11 per beam pair=W= F*Mn*8*(#of beams)/(L*Rm*W) I A,--- B1e am = 69.98 in-k*8*2/(144in*0.877* 1.599) Length = 5,545 lb/pair allowable load based on bending stress Mend= W*L*(1-Rm)/8 = (5545 lb/2)* 144 in*(1-0.877)/8 = 6,138 in-lb @ 5545 lb max allowable load = 4,428 in-lb ©4000 lb Imposed product load 2.Check Deflection Stress Allowable Loads Dmax= Dss*Rd Rd= 1 -(4*F*L)/(5*F*L+ 10*E*Ib) Allowable Deflection= L/180 = 1 -(4*300*144 in)/[(5*300*144 in)+(10*29500 ksi*3.241 in^4)] = 0.800 in = 0.853 in Deflection at Imposed Load= 0.694 in if Dmax= L/180 Based on L,/180 Deflection Criteria and Dss= 5*W*L^3/(384*E*Ib) L/180= 5*W*L^3*Rd/(384*E*Ib*#of beams) solving for W yields, W= 384*E*I*2/(180*5*L^2*Rd) = 384*3.241 in^4*2/[180*5*(144 in)^2*0.853) = 4,613 lb/pair allowable load based on deflection limits Thus,based on the least capacity of item 1 and 2 above: Allowable load= 4,613 lb/pair Imposed Product Load= 4,000 lb/pair Beam Stress= 0.87 Beam at Level 1 Structural Engineering & Design I nc. • 1815 Wright Ave I a Verne CA 91750 Tel.909 596 1351 Fax.909.596 7188 By: Bs/Mqz Project: Ednetics Project#: 21-0917-12 4 Pin Beam to Column Connection Type A Selective Rack I he beam end moments shown herein show the result ot the maximum induced fixed end monents torm seismic+static loads and the code mandated minimum value ot 1.5%(DL+PL) P11 P Mconn max= (Mseismic+Mend-fixity)*0.70*Rho v rho=.1O0Qp`. = 8,370 in-lb Load at level 1 Mir P 6" r 0 12" IM 1/2" Connector Type= 4 Pin Shear Capacity of Pin Pin Diam= 0.44 in Fy= 55,000 psi Ashear= (0.438 in)^2*Pi/4 = 0.1507 inA2 Pshear= 0.4*Fy*Ashear = 0.4*55000 psi*0.1507in^2 = 3,315 lb Bearing Capacity of Pin tcol= 0.075 in Fu= 65,000 psi Omega= 2.22 a= 2.22 Pbearing= alpha* Fu*diam*tcol/Omega = 2.22*65000 psi*0.438 In*0,075 in/2.22 = 2,135 lb <3315 lb Moment Capacity of Bracket Edge Distance=E= 1,00 in Pin Spacing= 2.0 in Fy= 55,000 psi C= P1+P2+P3+P4 tclip= 0.18 in Sclip= 0.127 in^3 = P1+P1*(4.576.5")+P1*(2.5"/6.5")+P1*(0.5"/6.5") = 2.154*P1 Mcap= Sclip*Fbending C*d= Mcap=2.154 d= E/2 = 0.127 in^3*0.66*Fy = 0.50 in = 4,610 in-lb Pclip= Mcap/(2.154*d) = 4610.1 in-lb/(2.154*0.5 in) Thus, P1= 2,135 lb = 4,281 lb Mconn-allow= [P1*6.5"+P1*(4.5"/6.5")*4.5"+P1*(2.5"/6.5")2,5"+P1*(0.576.5")*0.51 = 2135 LB*[6.5"+(4.576.5")*4.5"+(2.5"/6.5")*2.5"+(0.5"/6.5")*0.5"] = 22,664 in-lb > Mconn max, OK type a select-ed.xls Page f of l 6 9/18/202 I Structural Engineering & Design Inc. 1815 Wright Ave La Verne, CA 91750 Tel: 909.596.1351 Fax:909,596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 • Transverse Brace Configuration:Type A Selective Rack Section Properties Diagonal Member= Hannibal 1-1/2x1-1/2x16ga Horizontal Member= Hannibal 1-1/2x1-1/2x16ga Area= 0.273 in^2 hs-1.500 in Area= 0.273 In^2 r min= 0.496 In r min= 0.496 in 1.500 Fy= 55,000 psi 0_�"'-"� ''f Fy= 55,000 psi j--••—'--i- K= 1.0 1 11.5001.500 in K= 1.0 Qc= 1.92 ` r — 1 l=1.500 __ _J 14 -+ -0.250 in '-' 1 ¢-0.250 Frame Dimensions Bottom Panel Height=H= 48.0 in Clear Depth=D-B*2= 38.8 in Frame Depth=D= 42.0 in X Brace= NO Column Width=B= 1.6 in rho= 1.00 Diagonal Member -► 0 'Load Case 6::(1+ T85+0.19sds)*B*P+[0.7*rho*EJ<=1.0,ASD Method . D —►+ Vtransverse= 899 lb vb ��1 Vb=Vtransv*0.7*rho= 899 lb* 0.7* 1 (kl/r)= (k*Ldiag)/r min = 629 lb = (1 x 57.1 in/0.496 in) Ldiag= [(D-B*2)^2+(H-6")^2]^1/2 = 115.1 in Ldlag = 57.1 in Fe= pl^2*E/(kl/r)^2 H Pmax= V*(Ldiag/D)* 0.75 = 21,977 psi 41 tmax = 6421b axial load on diagonal brace member Since Fe<Fy/2, 3° tYPIIIIIIITIMM Pn= AREA*Fn Fn= Fe B . = 0.273 in^2*21977 psi = 21,977 psi tyoicai Panel_ = 6,000 lb Configuration Pallow= Pn/Q Check End Weld = 6000 lb/1.92 Lweld= 3.0 in = 3,125 lb Fu= 65 ksi tmin= 0.060 in Pn/Pallow= 0.21 <= 1.0 OK Weld Capacity= 0.75*tmin*L*Fu/2.5 = 3,510 lb OK Horizontal brace Vb=Vtransv*0.7*rho= 629 lb (kl/r)= (k* Lhoriz)/r min Fe= pi^2*E/(kl/r)^2 Fy/2= 27,500 psi = (1 x 42 in)/0.496 in = 40,584 psi = 84.7 in Since Fe>Fy/2, Fn=Fy*(1-fy/4fe) Pn= AREA*Fn Pallow= Pn/Qc = 36,366 psi = 0.273inA2*36366 psi = 9928 lb/1.92 = 9,928 lb = 5,171 lb Pn/Pallow= 0.12 <= 1.0 OK type a select-ed.xls Page f 6 of 6 9/18/202 I Structural Engineering & Design Inc. 1815 Wright Ave La Verne.CA 91750 Tel:909,596.1351 Fax: 909.596.7180 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Single Row Frame Overturning Configuration:Type A Selective Rack Loads Critical Load case(s): 1) RMI Sec 2.2, item 7: (0.9-0.2Sds)D+(0.9-0.20Sds)*B*Papp-E*rho hp Sds= 0.6583 v Vtrans=V=E=Qe= 899 lb (0.9-0.2Sds)= 0.7683 DEAD LOAD PER UPRIGHT=D= 100 lb (0.9-0.2Sds)= 0.7683 PRODUCT LOAD PER UPRIGHT=P= 8,000 lb B= h- tio>v mA_ H h Papp=P*0.67= 5,360 lb rho= 1.0000 Wst LC1=Wst1=(0.76834*D+0.76834*Papp*1)= 4,195 lb Frame Depth=Df= 42.0 in T I Product Load Top Level, Ptop= 4,000 lb Htop-Iv1=H= 96.0 in DL/Lvl= 50 lb #Levels= 2 Ia of-►I Seismic Ovt based on E,2:(Fi*hi)= 64,519 in-lb #Anchors/Base= 2 height/depth ratio= 2.3 in hp= 48.0 in SIDE ELEVATION A)Fully Loaded Rack h=H+hp/2= 120.0 in Load case 1: Movt= 2:(Fi*hi)*E*rho Mst= Wst1 *Df/2 T= (Movt-Mst)/Df = 64,519 in-lb = 4195 lb*42 in/2 = (64519 in-lb-88095 in-lb)/42 In = 88,095 in-lb = -561 lb No Uplift Net Seismic Uplift= -561 lb Ai Top Level Loaded Only Load case 1: 0 V1=Vtop= Cs*Ip*Ptop >=350 lb for H/D>6.0 Movt= [V1*h+V2* H/2]*rho = 0.1646*4000 lb = 79,798 in-lb = 658 lb T= (Movt-Mst)/Df V1eff= 658 lb Critical Level= 2 = (79798 in-lb-66154 in-lb)/42 In V2=VOA= Cs*Ip*D Cs*Ip= 0.1646 = 325 lb Net Uplift per Column = 16 lb Mst= (0.76834*D+0.76834*Ptop*1)*42 in/2 = 66,154 in-lb Net Seismic Uplift= 325 lb Anchor Check(2)0.5"x 3.25" Embed HILTI KWIKBOLT TZ anchor(s)per base plate. Special inspection is required per ESR 1917. Pullout Capacity=Tcap= 1,961 lb L.A.City Jurisdiction? NO Tcap*Phi= 1,961 lb Shear Capacity=Vcap= 2,517 lb Phi= 1 Vcap*Phi= 2,517 lb Fully Loaded: (224 lb/2517 Ib)^1 = 0.09 <= 1.2 OK Top Level Loaded: (162 Ib/1961 Ib)^1 +(164 lb/2517 Ib)^1 = 0.15 <= 1.2 OK type a select-ed.xls Page r( of 13 9/18/202 1 Structural Engineering & Design Inc. 1815 Wright Ave La Verne. CA 91750 Tel:909.596.1351 Fax:909.596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Base Plate Configuration:Type A Selective Rack Section 4--- a -' P Baseplate= 8x5x3/8 Eff Width=W= 8.00 in a= 3.00 in Mb Eff Depth=D= 5.00 in Anchor c.c. =2*a=d= 6.00 in Column Width=b= 3.00 in N=#Anchor/Base= 2 ( b 14. L Column Depth=dc= 1.63 in Fy= 36,000 psi �- w --; • L= 2.50 in • • Plate Thickness=t= 0.375 in Downalsle Elevation Down Aisle Loads Load Case 5. :(1+0.105*Sds)D+0.75*[(1.4+0.14Sds)*5*P+0.75*10.7*rho*E/<=1.0,ASD Method • COLUMN DL= 50 lb Axial=P= 1.0691215*50 lb+0.75* (1.492162*0.7*4000 lb) COLUMN PL= 4,000 lb = 3,187 lb Base Moment= 0 in-lb Mb= Base Moment*0.75*0.7*rho 1+0.105*Sds= 1.0691 = 0 in-lb*0.75*0.7*rho 1.4+0.14Sds= 1.4922 = 0 in-lb Effi 6= 6:100 j ` Axial Load P= 3,187 lb Mbase=Mb= 0 in-lb Effe Axial stress=fa = P/A=P/(D*W) M1= wLA2/2=fa*LA2/2 = 80 psi = 249 in-lb Moment Stress=fb = M/S=6*Mb/[(D*B^2] Moment Stress=fb2= 2*fb*L/W = 0.0 psi = 0.0 psi Moment Stress=fbl = fb-fb2 M2= fbl*L^2)/2 = 0.0 psi = 0 in-lb M3 = (1/2)*fb2*L*(2/3)*L=(1/3)*fb2*L^2 Mtotal= M1+M2+M3 = 0 in-lb = 249 in-lb/in S-plate= (1)(t^2)/6 Fb= 0.75*Fy = 0.023 in^3/in = 27,000 psi fb/Fb= Mtotal/[(S-plate)(Fb)] F'p= 0.7*Fc = 0.39 OK = 2,450 psi OK Tanchor= (Mb-(PLapp*0.75*0.46)(a))/[(d)*N/2] Tallow= 1,961 lb OK = -2,098 lb No Tension Cross Aisle Loads Critical load ease RMI See 2.1,Item 4:(1+O.11Sds)OL+(1+O.l4SOS)PL'075+EL"075<=1.0,ASP Method Check uplift load on Baseplate Check uplift forces on baseplate with 2 or more anchors per RMI 7,2.2. Pstatic= 3,187 lb 'When the base plate configuration consists of two anchor bolts located on either side .f the column and a net uplift force exists,the minimum base plate thickness Movt*0.75*0.7*rho= 33,872 in-lb Pseismic= Movt/Frame Depth -hall be determined based one design bending moment in the plate equal Frame Depth= 42.0 in = 806 lb to the uplift force on one anchor times 1/2 the distance from P=Pstatic+Pseismic= 3,993 lb I he centerline of the anchor to the nearest edge of the rack column" b=Column Depth= 1.63 in T c * L=Base Plate Depth-Col Depth= 2.50 in Ta Muhi a Mu mil I,. fa= P/A= P/(D*W) M= wLA2/2=fa*L^2/2 l b � ` = 100 psi = 312 in-lb/in Elevation Uplift per Column= 324 lb Sbase/in= (1)(t^2)/6 Fbase= 0.75*Fy Qty Anchor per BP= 2 = 0.023 in^3/in = 27,000 psi Net Tension per anchor=Ta= 162 lb c= 2.50 in fb/Fb= M/[(S-plate)(Fb)] Mu=Moment on Baseplate due to uplift= Ta*c/2 0.49 OK = 203 in-lb Splate= 0.117 in^3 fb Fb *0.75= 0.048 OK type a aelect-ed.xis Page I).- of (, ' 9/18/202 I Structural Engineering & Design Inc. 1815 Wright Ave La Verne. CA 91750 Tel:909.596.1351 Fax: 909,596.7186 By: Bs/Mqz Project: Ednetics Project#:21-0917-12 Slab on Grade Configuration:Type A Selective Rack P • slab a � Concrete r D ': b e , a f c= 3,500 psi r a ! tslab=t= 8.0 in slab :Cross teff= 8.0 in c -- Aisle hit. 'J 6 4- x -► [� C ' SOi �- y a fsoil= 750 psf .- L ► Down Aisle Movt= 79,008 in-lb SLAB ELEVATION Frame depth= 42.0 In Baseplate Plan View Sds= 0.658 Base Plate 0.2*Sds= 0.132 Effec.Baseplate 8.00 in width=a= 3.00 in D Effec.Baseplate Depth=D= 5.00 in depth=b= 1.63 in l3=B/D= 1.600 midway dist face of column to edge of plate=c= 5.50 in F'c^0.5= 59.20 psi Column Loads midway dist face of column to edge of plate=e= 3.31 in DEAD LOAD=D= 50 lb per column Load Case 1) (1.2+0.2Sds)D+(1.2+0.2Sds)*B*P+ rho*E RMI SEC 2.2 EQTN 5� unfactored ASV load = 1.33166*50 lb+ 1.33166*0.7*4000 lb+ 1 * 1881 lb PRODUCT LOAD=P= 4,000 lb per column = 5,676 lb unfactoredASDload Load Case 2) (0.9-0.2Sds)D+(0.9-0.2Sds)*B*Papp+rho*E RMI SEC 2,2 EQTN 7 Papp= 2,680 lb per column = 0.76834*50 lb+0.76834*0.7*2680 lb+1* 1881 lb P-seismic=E= (Movt/Frame depth) = 3,361 lb = 1,881 lb per column Load Case 3) 1.2*D+ 1.4*P RMI SEC 2,2 EQTN 1,2 unfactored Limit State load = 1.2*50 lb+ 1.4*4000 lb B €; = 5,660lb rho= Load Case 4) 1.2*D+ 1.0*P+ 1.0E ACI 318-14 Sec 5.3.1 Sds= 0.6583 = 5,941 lb Eqtn 5.3.1e 1.2+0.2*Sds= 1.3317 Effective Column Load=Pu= 5,941 lb per column 0.9-0.20Sds= 0.7683 Puncture Apunct= [(c+t)+(e+t)]*2*t = 397.0 inA2 Fpunctl= [(4/3 +8/(3*13)]*X*(F'c^0.5) fv/Fv= Pu/(Apunct*Fpunct) = 106.6 psi = 0.158 < 1 OK Fpunct2= 2.66*X*(F'c^0.5) = 94.5 psi Fpunct eff= 94.5 psi Slab Bending Pse=DL+PL+E= 5,941 lb Asoll= (Pse*144)/(fsoll) L= (Asoil)^0.5 y= (c*e)^0.5+2*t = 1,141 in^2 = 33.78 in = 20.3 in x= (L-y)/2 M= w*x^2/2 S-slab= 1*teff^2/6 = 6.8 In = (fsoil*x^2)/(144*2) = 10.67 in^3 Fb= 5*(phi)*(fc)^0.5 = 118.8 in-lb fb/Fb= M/(S-slab*Fb) = 177.48 psi = 0.063 < 1,OK type a select-ed.xls Page ( of t l! 9/18/202 I • BTF LLCI rustalictru:m4, 23150 S Viola Welch Rd Beavercreek OR 97004 971-371-6183 This will be storage for class I—IV products and no products will be stored above 12' high. This will not be high pile storage Thank you, Brian Ferrick 971-371-6183