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Report 'S(,F2O v - 0O2(0 Z. —?SO O 5 Nt.s Tech Lin • RECEIVED SNOV 102014 tructural CITY OF TIGARD RUING DIVISION Concepts n ineerin 1615 Yeager Ave La Verne, CA. 91750 Tel: 909.596.1351 Fax: 909.596.7186 e-mail: mail @sceinc.net O PRp, �Nc° GIN :; 0 725/ PE Project Name : LENNOX /EGON f R. 11.2C)°. Project Number : 0-103014-3LV SP EXPIRES: 12/31/2015 Date : 10/30/14 OCT 3 0 2014 Street Address : 7500 SW TECH CENTER DRIVE SUITE 110 City/State : TIGARD OR 97224 Scope of Work : SELECTIVE RACK Structural Concepts Engineering 1200 Nl. Jefferson Ste, Ste F Anaheim, CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV 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 B SELECT-LENNOX Page .2 of / 3 1 0/30/2014 Structural Concepts Engineering 1200 N.Jefferson Ste_Ste F Anaheim. CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Design Data 1)The analyses conforms to the requirements of the 2013 CBC and the 2008 RMI/ANSI MH 16.1 Rack Design Manual Steel Storage Racks(RMI)and the ASCE 7-05,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 E70>o(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 slab on grade is 5"thick with minimum 2500 psi compressive strength.Soil bearing capacity is 750 psf. Definition of Components A Column Beam • I I I _ Horizontal Brace Beam to Column Connector Diagonal Brace Frame Height Beam Spacing Base Plate and I Anchors I Panel Beam Height Length H Frame i Depth Front View: Down Aisle Section A: Cross Aisle • (Longitudinal) Frame (Transverse ) Frame TYPE B SELECT-LENNOX Page 3 of /3 10/30/2014 ' Str tural kroncepts :- Engineering `- 1200 N. Jefferson Ste. Ste F Anaheim, CA 52807 Tel: 714.637330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Configuration&Summary:TYPE B SELECTIVE RACK ', _ _ ,-, - _ T 1- **RACK COLUMN REACTIONS ASD LOADS 48" 42" AXIAL DL= 105/b AXIAL LL= 6,000/b „ f SEISMIC AXIAL Ps=t/- 2,416 lb BASE MOMENT= 0 in-lb 144" 144" 42., 42" .4 96" ,�. 42" f Seismic Criteria #Bm Lvls Frame Depth Frame Height # Diagonals Beam Length Frame Type Ss=0.972,Fa=1.111 3 42 in 144.0 in 3 96 in Single Row Component Description STRESS Column Fy=55 ksi ANDRSN I300/3x3x14ga P=6105 Ib,M=12533 in-lb 0.72-OK Column&Backer None None None N/A Beam Fy=55 ksi AND 41140/4.125"Face x 0.075"thk Lu=96 in Capacity: 5998 lb/pr 0.67-OK Beam Connector Fy=55 ksi Lvl 1: 3 pin OK I Mconn=9831 in-lb Mcap=12691 in-lb 0.77-OK Brace-Horizontal Fy=55 ksi Andrsn 1-1/2x1-1/4x14ga 0.25-OK Brace-Diagonal Fy=55 ksi Andrsn 1-1/2x1-1/4x14ga 0.34-OK Base Plate Fy=36 ksi 8x5x3/8 I Fixity=0 in-lb 0.92-OK Anchor 2 per Base 0.5"x 3.25"Embed HILTI KWIKBOLT TZ ESR 1917 Inspection Reqd(Net Seismic Uplift=857 Ib) 0.4-OK Slab&Soil 5"thk x 2500 psi slab on grade. 750 psf Soil Bearing Pressure 0.85-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 48.0 in 42.0 in 166 lb 93 lb 6,105 lb 12,533 "# 9,831 "# 3 pin OK 2 4,000 lb 48.0 in 42.0 in 332 lb 186 lb 4,070 lb 5,570 "# 6,385 "# 3 pin OK 3 4,000 lb 48.0 in 42.0 in 498 lb 279 lb 2,035 lb 3,342 "# 4,296 "# 3 pin OK Total: 995 lb 557 lb Notes TYPE E.SELECT-LENNOX Page V of /3 10/30/2014 ' Structural Concepts Engineering 1200 N. Jefferson Ste.Ste F Anaheim. CA 92807 Tel: 714.632J330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-1030143LV Seismic Forces Configuration:TYPE B SELECTIVE RACK Lateral analysis is performed with regard to the 2013 CBC Sec.2208.1,2008 RMI/ANSI MH 16.1 Sec 2.6&ASCE 7-05 sec 15.5.3 Ss= 0.972 Transverse(Cross Aisle)Seismic Load r14 51= 0.421 V= Cs*Ip*Ws=Cs*Ip*(0.67*LL*PLrf+DL) --- t Fa= 1.111 Cs1= [Sds/R]*0.67 0e7 rc.e.,,-„e, rSe u.Sec utsnx States Redtm n s>ASP Loading 1r= Fv= 1.579 = 0.1206 Cs-max*Ip= 0.1206 FM Sds=2/3*Ss*Fa= 0.720 Cs2= 0.14*Sds*0.67 V,,,,= 0.015 ,, Sd1=2/3*S1*Fv= 0.443 PAM = 0.0695 Eff Base Shear=Cs= 0.1206 Transver tiort Ca=0.4*2/3*Ss*Fa= 0.2880 Cs3= [0.5*S1/R] *0.67 Ws= (0.67*PLRFI*PL)+DL (Transverse,Braced Frame Dir.)R= 4,0 = 0.0353 = 8,250 lb Ip= 1.0 .......... ._............... Cs-max= 0.1206 Vtransv=Vt= 0.1206 * (210 lb+8040 lb) PLRF1= 10 Base Shear coeff=Cs= 0.1206 EL= 995 lb Pallet Height=hp= 48.0 in ASD Format Loads ASD Level Transverse seismic shear per upright DL per Beam Lvl= 70 lb Level PRODUC LOAD/LVL,PL PL*0.67*PLrf DL hi wi*hi Fi Fi*(hi+hp/2) 1 4,000 lb 2,680 lb 70 lb 48 in 132,000 165.8 lb 11,938-# 2 4,000 lb 2,680 lb 70 lb 96 in 264,000 331.7 lb 39,804-# 3 4,000 lb 2,680 lb 70 lb 144 in 396,000 497.5 lb 83,580-# I sum: 12,000 lb 8,040 lb 210 lb W=8250 lb 792,000 995 lb 2=135,322 Longitudinal(Downaisle)Seismic Load !Similarly for longitudinal seismic loads,using R=6.0 Ws= (0.67*PLRF2*PL) +DL PLRF2= 1.0 1,,.,,...I 0 Q Cs1=Sd1/(T*R)= 0.0495 = 8,250 lb (Longitudinal,Unbraced Dir.)R= 6.0 Cs2= 0.0675 Cs=Cs-max*Ip= 0.0675 T= 1.00 sec !,„.. Cs3= 0.0235 Vlong= 0.0675* (210 lb+ 8040 lb) . L >: k: Cs-max= 0.0675 EL= 557 lb ASD Level Longit,seismic shear per upright Level PRODUC LOAD/LVL,PL PL*0.67*PLrf DL hi wi*hi Fi Front View 1 4,000 lb 2,680 lb 70 lb 48 in 132,000 92.8 lb 2 4,000 lb 2,680 lb 70 lb 96 in 264,000 185.7 lb 3 4,000 lb 2,680 lb 70 lb 144 in 396,000 278.5 lb I sum: 8,040 lb 210 lb W=8250 lb 792,000 557 lb TYPE 5 SELECT-LENNOX Page 17 of i_j 1 0/30/20 I4 • Structural Concepts i Eng ineering 1200 N.Jefferson Ste.Ste F Anaheim. CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Downaisle Seismic Loads Configuration:TYPE B SELECTIVE RACK Determine the story moments by applying portal analysis.The base plate is assumed to provide no fixity. Seismic Story Forces Typical frame made Vlong= 557 lb Tributary area oftwocolumns Vcol=Vlong/2= 279 lb of rack frame F1= 93 1b i I_ [. h- T I I l "I - Typical Frame ma,ie F2= 186 lb 0f two columns -1, F3= 279 lb -►1•-` '-4 E H H f _ Top View 96" Front View Side View Seismic Story Moments Conceptual System Mbase-max= 0 in-lb <===Default capacity hl-eff= h1-beam clip height/2 Mbase-v= (Vcol*hieff)/2 = 45 in Vcol it = 6,266 in-lb <__=Moment going to base � , Mbase-eff= Minimum of Mbase-max and Mbase-v h2 = 0 in-lb PINNED BASE ASSUMED M Mr 1-1= [Vcol*hleff]-Mbase-eff M 2-2= [Vcol-(F1)/2] * h2 f_ � = (279 lb*45 in)-0 in-lb = [279 lb-92.9 Ib]*48 in/2 �''g' = 12,533 in-lb = 5,570 in-lb h1 hieff IF I' Mseis= (Mupper+Mlower)/2 Beam to Column Elevation Mseis(1-1)= (12533 in-lb+5570 in-lb)/2 Mseis(2-2)= (5570 in-lb+3342 in-lb)/2 = 9,051 in-lb = 4,456 in-lb Summary of Forces LEVEL hi Axial Load Column Moment Mseismic Mend-fixity Mconn** Beam Connector 1 48 in 6,105 lb 12,533 in-lb 9,051 in-lb 4,057 in-lb 9,831 in-lb 3 pin OK 2 48 in 4,070 lb 5,570 in-lb 4,456 in-lb 4,057 in-lb 6,385 in-lb 3 pin OK 3 48 in 2,035 lb 3,342 in-lb 1,671 in-lb 4,057 in-lb 4,296 in-lb 3 pin OK I Mconn= (Mseismic+Mend-fixity)*0.75 Mconn-allow(3 Pin)= 12,691 in-lb TYPE 0 SELECT-LENNOX Page of of (3 1 0/30/201 4 Structural Concepts Engineering 1200 N. Jefferson Ste.Ste F_Anaheim. CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Column(Longitudinal Loads) Configuration:TYPE B SELECTIVE RACK Conforms to the requirements of Chapter C5 of the AISI Cold Formed Steel Design Manual for combined bending and axial loads, Section Properties Section: ANDRSN I300/3x3x14ga k 3.000 in el Aeff= 0.643 inA2 Iy= 0.749 inA4 Kx= 1.7 X Ix= 1.130 inA4 Sy= 0.493 inA3 Lx= 45.9 in Sx= 0.753 inA3 ry= 1.080in Ky= 1.0 y_._. __. , � 3.000 in rx= 1.326 in Fy= 55 ksi Ly= 42.0 in 10.075 in f flf= 1.67 Cmx= 0.85 Cb= 1.0 4_ • 4 E= 29,500 ksi 14_0.75 in Loads Considers loads at level 1 Critical load case RMI Sec 2.1,item 4:(1+0.115ds)DL +(1+0.14505)PL*0.75+EL*0.75<=1.0,ASD Method COLUMN DL= 105 lb COLUMN PL= 6,000 lb Axial Load=P= (1.079189*105Ib)+(1.100786*60001b*0.75) Moment=Mx= Mcol*0.75 Mcol= 12,532 in-lb = 5,067 lb = 12532 in-lb * 0.75 0.11Sds= 0.079189 = 9,399 in-lb 0.14Sds= 0.100786 Axial Analysis KxLx/rx= 1.7*45.94"/1.326" KyLy/ry= 1*42"/1.08" Fe > Fy/2 = 58.9 = 38.9 Fn= Fy(1-Fy/4Fe) = 55 ksi*[1-55 ksi/(4*83.9 ksi)] Fe= nA2E/(KL/r)maxA2 Fy/2= 27.5 ksi = 46.0 ksi = 83.9ksi Pa= Pn/4c Pn= Aeff*Fn f2c= 1.92 = 29571 Ib/1.92 = 29,571 lb = 15,402 lb P/Pa= 0.33 > 0.15 Bending Analysis Check: P/Pa +(Cmx*Mx)/(Max*px) <_ 1.0 P/Pao+Mx/Max<_ 1.0 Pno= Ae*Fy Pao= Pno/Qc Myield=My= Sx*Fy = 0.643 inA2*55000 psi = 353651b/1.92 = 0.753 inA3* 55000 psi = 35,365 lb = 18,419 lb = 41,415 in-lb Max= My/S f Pcr= nA2EI/(KL)maxA2 = 41415 in-Ib/1.67 = nA2*29500 ksi/(1.7*45.94 in)A2 = 24,799 in-lb = 53,941 lb px= {1/[1-(2c*P/Pcr)]}A-1 = {1/[1-(1.92*5067lb/53941lb)]}A-1 = 0.82 Combined Stresses (5067 lb/15402 Ib) +(0.85*9399 in-lb)/(24799 in-Ib*0.82) = 0.72 < 1.0,OK (EQ C5-1) (5067 lb/18419 Ib)+(9399 in-lb/24799 in-Ib)= 0.65 < 1.0,OK (EQ C5-2) TYPE 5 SELECT-LENNOX Page 7 of /mil 10/30/2014 StrAtural Vonce is ,,,z4 pry engineering 1200 N.Jefferson Ste,Ste F Anaheim,CA 92807 Tel:714.632.7330 Fax:714.632.7763 By: BOB Project: LENNOX Project#:0-103o14-3i-v BEAM Configuration:TYPE B SELECTIVE RACK DETERMINE ALLOWABLE MOMENT CAPACITY 2.75 in A)Check compression flange for local buckling(82.1) -1.75 in w= c-2*t-2*r = 1.75 in-2*0.075 in-2*0.075 in r 1______________� f = 1.450 in 1.625 in WA= 19.33 1=lambda= [1.052/(k)^0.5] *(w/t)*(Fy/E)^0.5 Eq. B2.1-4 4.125 in= [1.052/(4)^0.5]* 19.33* (55/29500)^0.5 0.075 in = 0.439 <0.673,Flange is fully effective Eq.B2.1-1 B)check web for local buckling per section b2.3 f1(comp)= Fy*(y3/y2)= 49.12 ksi f2(tension)= Fy*(y1/y2)= 100.88 ksi Y= f2/f1 Eq.B2.3-5 Beam= AND 41140/4.125"Face x 0.075"thk = -2.054 Ix= 2.196 in^4 k= 4+2*(1-Y)^3 +2*(1-Y) Eq. B2.3-4 Sx= 1.021 inA3 = 67.08 Ycg= 2.723 in flat depth=w= y1+y3 t= 0.075 in = 3.825 in Bend Radius=r= 0.075 in w/t= 51 OK Fy=Fyv= 55.00 ksi l=lambda= [1.052/(k)^0.5]*(w/t)*(fl/E)^0.5 Fu=Fuv= 65.00 ksi = [1.052/(67.08)^0.5] *3.825*(49.12/29500)^0.5 E= 29500 ksi = 0.267 < 0.673 top flange=b= 1.750 in be=w= 3.825 in b2= be/2 Eq B2.3-2 bottom flange= 2.750 in b1= be(3-Y) = 1.91 in Web depth= 4.125 in = 0.757 Fy bl+b2= 2.667 in > 1.2525 in,Web is fully effective fl(comp) Determine effect of cold working on steel yield point(Fva)per section A7.2 Fya= C*Fyc +(1-C)*Fy (EQ A7.2-1) Lcorner=Lc= (p/2)*(r+t/2) 0.177 in C= 2*Lc/(Lf+2*Lc) y2 Lflange-top=Lf= 1.450 in = 0.196 in 13 m= 0.192*(Fu/Fy)-0.068 (EQ A7.2-4) depth = 0.1590 rf Bc= 3.69*(Fu/Fy)-0.819*(Fu/Fy)A2- 1.79 (EQ A7.2-3) = 1.427 since fu/Fv= 1.18 < 1.2 Ycg y1 and r/t= 1 < 7 OK f2(ten on) then Fyc= Bc*Fy/(R/t)^m (EQ A7.2-2) = 78.485 ksi Thus, Fya-top= 59,61 ksi (tension stress at top) yl= Ycg-t-r= 2.573 in • Fya-bottom= Fya*Ycg/(depth-Ycg) y2= depth-Ycg= 1.403 in = 115.71 ksi (tension stress at bottom) y3= y2-t-r= 1.253 in Check allowable tension stress for bottom flange Lflange-bot=Lfb= Lbottom-2*r*-2*t = 2.450 in Cbottom=Cb= 2*Lc/(Lfb+2*Lc) = 0.126 Fy-bottom=Fyb= Cb*Fyc+(1-Cb)*Fyf = 57.97 ksi Fya= (Fya-top)*(Fyb/Fya-bottom) = 29.86 ksi if F= 0.95 Then F*Mn=F*Fya*Sx= 28.96 in-k Strurai once is - - engineering 1200 N.Jefferson Ste,Ste F Anaheim,CA 92807 Tel:714.632.7330 Fax:714.632.7763 By: BOB Project: LENNOX Project#:0-103014-3w BEAM Configuration:TYPE B SELECTIVE RACK RMI Section 5.2, PT II Section Beam= AND 41140/4.125"Face x 0.075"thk Ix=Ib= 2.196 inA4 2.75 in Sx= 1.021 inA3 t= 0.075 in E= 29500 ksi 1.75 in Fy=Fyv= 55 ksi F= 0.0 / �_,� Fu=Fuv= 65 ksi L= 96 in Fya= 59.6 ksi 1.625 in 4.125 in 41.-1 Bending I 0.075 in I Mcenter=F*Mn= W*L*W*Rm/8 W=LRFD Load Factor= 1.2*DL+ 1.4*PL+1.4*(0.125)*PL RMI 2.Z item 5 FOR DL=2%of PL, W= 1.599 Rm= 1-[(2*F*L)/(6*E*Ib+3*F*L)] 1 -(2*0*96 in)/[(6*29500 ksi*2.196 in^3)+(3*0*96 in)] = 1 if F= 0.95 Then F*Mn=F*Fya*Sx= 57.82 in-k Thus,allowable load per beam pair=W= F*Mn*8*(#of beams)/(L*Rm*W) = 57.82 in-k*8*2/(96in* 1* 1.599) = 6,027 lb/pair Mend= W*L*(1-Rm)/8 = (6027 lb/2)*96 in* (1-1)/8 = in-lb Deflection Dmax= Dss*Rd Rd= 1-(4*F*L)/(5*F*L+ 10*E*Ib) = 1-(4*0*96 in)/[(5*0*96 in)+(10*29500 ksi*2.196 in^4)] = 1.000 in • if Dmax= L/180 solving for W yields, and Dss= 5*W*L^3/(384*E*Ib) W= 384*E*I*2/(180*5*L^2*Rd) = 384*2.196 in^4*2/[180*5*(96 in)^2*1) = 5,998 lb/pair L/180= 5*W*LA3*Rd/(384*E*Ib*#of beams) Allowable load= 5,998 lb/pair -+n1 Ir.+I me.I nninoorinn 1200 N.Jefferson Ste, Ste F Anaheim, CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-1 030 I 4- 3 Pin Beam to Column Connection TYPE B SELECTIVE RACK I he beam end moments shown herein show the result of the maximum induced fixed end monents torm seismic+static loads and the code mandated minimum value of 1.5u/o(DL+PL) Mconn max= (Mseismic+Mend-fxity)*0.75 0 el = 9,831 in-lb Load at level 1 0 E2 0 P3 \ 1/2" 411-U c X12" Connector Type= 3 Pin Shear Capacity of Pin Pin Diam= 0.44 in Fy= 55,000 psi Ashear= (0.438 in)A2*Pi/4 = 0.1507 inA2 Pshear= 0.4*Fy*Ashear = 0.4* 55000 psi *0.1507inA2 = 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 tclip= 0.18 in Sclip= 0.127 inA3 = P1+P1*(2.5"/4.5")+P1*(0.5"/4.5") = 1.667*P1 Mcap= Sclip*Fbending C*d= Mcap= 1.667 d= E/2 = 0.127 inA3*0.66*Fy = 0.50 in = 4,610 in-lb Pclip= Mcap/(1.667*d) = 4610.1 in-lb/(1.667*0.5 in) Thus,P1= 2,135 lb = 5,531 lb Mconn-allow= [P1*4.5"+P1*(2.574.5")*2.5"+P1*(0.5"/4.5")*0.51 = 2135 LB*[4.5"+(2.5"/4.5")*2.5"+(0.5"/4.5")*0.51 = 12,691 in-lb > Mconn max, OK TYPED SELECT-LENNOX Page ( of /3 10/30/201 4 Structural Concepts Engineering 1200 N. Jefferson Ste.Ste F Anaheim. CA 92807 Tel• 714.632.7330 Fax: 714.632.7763 • By: BOB Project: LENNOX Project#: 0-103014-3LV Transverse Brace Configuration:TYPE B SELECTIVE RACK Section Properties Diagonal Member= Andrsn 1-1/2x1-1/4x14ga Horizontal Member= Andrsn 1-1/2x1-1/4x14ga Area= 0.278 inA2 1.0-1 .500 Area= 0.278 in^2 °--1 .500 r min= 0.405 in r min= 0.405 in I Fy= 55,000 psi Fy= 55,000 psi K= 1.0 11 .250 K= 1.0 I i t .250 c2c= 1.92 0.25 -al 10-0.25 Frame Dimensions Bottom Panel Height=H= 42.0 in Clear Depth=D-B*2= 36.0 in Frame Depth=D= 42.0 in X Brace= NO Column Width=B= 3.0 in Diagonal Member Critical load case RMI Sec 2.1,item 4:(1+0.11Sds)DL+(1+0.14SD5)PL*0.75+EL*0.75<-1.0,ASD Method, DL=PL=0 for upright brace members D Vb=Vtransv= 995 lb vb Mak= (kl/r)= (k*Ldiag)/r min Ldiag= [(D-B*2)^2+ (H-6")^21^1/2 = (1 x 50.9 in/0.405 in) = 50.9 in = 125.7 in Ldiag Pmax= V*(Ldiag/D)*0.75 Fe= pi^2*E/(kl/r)^2 = 904 lb = 18,427 psi SINCE Fe<Fy/2, Fn= Fe 3^Y Pn= AREA*Fn = 18,427 psi B = 0.278 in^2* 18427 psi Typical Panel = 5,123 lb Confiauration Check End Weld Lweld= 2.5 in Pallow= Pn/52 Fu= 65 ksi = 5123 lb/1.92 tmin= 0.075 in = 2,668 lb Weld Capacity= 0.75*tmin*L*Fu/2.5 = 3,656 lb OK Pn/Pallow= 0.34 <= 1.0 OK Horizontal brace Pmax=V= 746 lb (kl/r)= (k*Lhoriz)/r min Fe= pi^2*E/(kl/r)^2 Fy/2= 27,500 psi = (1 x 42 in)/0.405 in = 27,075 psi = 103.7 in Since Fe<Fy/2,Fn=Fe Pn= AREA*Fn Pallow= Pn/c2c = 27,075 psi = 0.278in^2*27075 psi = 7527 lb/1.92 = 7,527 lb = 3,920 lb Pn/Pallow= 0.25 <= 1.0 OK TYPE 5 SELECT-LENNOX Pace /D of 73 10/30/2014 ' Structural Concepts pp' Engineering 1200 N. Jefferson Ste.Ste F Anaheim. CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Single Row Frame Overturning Configuration:TYPE B SELECTIVE RACK Loads Load case per RMI Sec 2.1, item 3: (0.6-0.11Sds)DL+(0.6-0.14Sds)PLapp*0.75-EL*0.75 hp p' Vtrans=V= 995 lb • (0.6-0.11Sds)= 0.5208 vpl I DEAD LOAD PER UPRIGHT=DL= 210 lb (0.6-0.14Sds)= 0.4992 PL PER UPRIGHT=PL= 12,000 lb PLapp=PL*0.67= 8,040 lb Frame Depth=D= 42.0 in H h Nst=(0.520811*DL+0.499214*PLapp*0.75)= 3,119 lb Htop-Iv1=H= 144.0 in PL©TOP= 4,000 lb #Levels= 3 DL/Lvl= 70 lb #Anchors/Base= 2 T h= 168.0 in :(Fi*hi)= 135,322 in-lb hp= 48.0 in 1-41-D Total Dead Load per Bay=DL= 210 lb Load Case 1: Fully Loaded rack SIDE ELEVATION Vtrans= 995 lb Movt= :(F*hi)*0.75 Mst= Wst*D/2 T= (Movt-Mst)/D = 101,492 in-lb = 3119 lb*42 in/2 = (101492 in-lb-65499 in-lb)/42 in = 65,499 in-lb = 857 lb Net Uplift per column Load Case 2:Top Level Loaded Only ()tical Level= 3 h= 168.0 in V1=Vtop= Cs*Ip*Ws Pinned Base Movt= [V1*h+V2*H/2]*0.75 = 0.1206*(4000 lb) = [482 lb*168 in+25 lb*144 in/2]*0.75 = 482 lb = 62,082 in-lb V2=VDL= Cs*Ip*DL = 25 lb T= (Movt-Mst)/D Mst= (0.520811*DL+0.499214*PL*0.75)*D/2 = (62082 in-lb-33747 in-lb)/42 in = (70 lb*3*0.520811 +4000 lb*0.499214*0.75)*42 in/2 = 675 lb = 33,747 in-lb Net Uplift per column Anchor Check(2)0.5"x 3.25"Embed HILTI KWIKBOLT T2 anchor(s)per base plate. Special inspection is required per ESR 1917. Pullout Capacity=Tcap= 1,250 lb L.A.City Jurisdiction? NO Shear Capacity=Vcap= 1,840 lb Phi= 1 Tcap*Phi= 1,250 lb Vcap*Phi= 1,840 lb Fully Loaded: (429 Ib/1250 Ib)^1 +(249 lb/1840 Ib)^1 = 0.48 <= 1.2 OK Top Level Loaded: (338 lb/1250 lb)^1 +(121 lb/1840 Ib)^1 = 0.34 <= 1.2 OK TYPE 5 SELECT-LENNOX Page // of /3 C/30/20 14 Structural Concepts Fp- Engineering 1200 N. Jefferson Ste.Ste F Anaheim. CA 92807 Tel: 714.632.7330 Fax: 714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Base Plate Configuration:TYPE B SELECTIVE RACK P Section a -1 Baseplate= 8x5x3/8 O, Eff Width=W= 8.00 in a= 3.00 in 111/11, Mb Eff Depth=D= 5.00 in Anchor c.c. =2*a=d = 6.00 in .% fmi Column Width=b= 3.00 in N=#Anchor/Base= 2 I , I b L Column Depth=dc= 3.00 in Fy= 36,000 psi ��r- W L= 2.50 in Plate Thickness=t= 0.375 in Downaisle Elevation Down Aisle Loads Critical load case RMI Sec 2.1,item 4:(1+0.115ds)DL +(1+0.14505)PL*0.75+EL*0.75 c=.1.0,A5D Method COLUMN DL= 105 lb COLUMN PL= 6,000 lb Axial=P= (1.079189*105 Ib)+(1.100786*6000 lb*0.75) Mb= Base Moment*0.75 Base Moment= 0 in-lb = 5,067 lb = 0 in-lb* 0.75 0.11Sds= 0.079189 = 0 in-lb 0.14Sds= 0.100786 Axial Load P = 5,067 lb Mbase=Mb = 0 in-lb Effec. Effec. I Axial stress=fa = P/A= P/(D*W) M1= wL^2/2=fa*L^2/2 = 127 psi = 396 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*LA2)/2 P' = 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 = 396 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*F'c 0.63 OK = 1,750 psi OK Tanchor= (Mb-(PLapp*0.75*0.46)(a))/[(d)*N/2] Tallow= 1,250 lb OK = -1,560 lb No Tension Cross Aisle Loads C^t.« RMS..24iwn*rl."sse, c+a.o.:+sos," al"s<=:.o„AS°M.nae Check uplift load on Baseplate Check uplift forces on baseplate with 2 or more anchors per RMI 7.2.2. Pstatic= 5,067 lb When the base plate configurason 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= 101,492 in-lb Pseismic= Movt/Frame Depth all be determined based on a design bending moment in the plate equal Frame Depth= 42.0 in = 2,416 lb o the uplift force on one anchor times 1/2 the distance from P=Pstatic+Pseismic= 7,483 lb he centerline of the anchor to the nearest edge of the rack column" b=Column Depth= 3.00 in T c • L=Base Plate Depth-Col Depth= 2.50 in Ta Mu Ta fa = P/A= P/(D*W) M= wLA2/2=fa*L^2/2 b ;, = 187 psi = 585 in-lb/in Elevation Uplift per Column= 857 lb Sbase/in= (1)(t^2)/6 Fbase= 0.75*Fy Qty Anchor per BP= 2 = 0.023 inA3/in = 27,000 psi Net Tension per anchor=Ta= 429 lb c= 2.50 in fb/Fb= M/[(S-plate)(Fb)] Mu=Moment on Baseplate due to uplift= Ta*c/2 = 0.92 OK = 536 in-lb Splate= 0.117 inA3 [fb/Fb]*0.75= 0.127 OK TYPE 13 SELECT-LENNOX Page 71-- of /j i 0/30/20,4 Structural Concepts E. _ Engineering 1200 N. Jefferson Ste.Ste F Anaheim. CA 92807 Tel: 714.632,7330 Fax:714.632.7763 By: BOB Project: LENNOX Project#: 0-103014-3LV Slab on Grade Configuration:TYPE B SELECTIVE RACK P slab a I 1 Concrete D a b e f0 2,500 psi slab t I ;. tslab=t= 5.0 in Cross tell= 5.0 in IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII61UUIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII •----� ----' Aisle phi=Sb= 0.6 x -1.-I I~ c - I Soil r- fsoil= 750 psf II.. L - Down Aisle Movt= 101,491 in-lb SLAB ELEVATION Frame depth= 42.0 in Baseolate Plan view Sds= 0.720 Base Plate 0.2*Sds= 0.144 Effec.Baseplate width=B= 8.00 in width=a= 3.00 in midway dist face of column to edge of plate=c= 530 in Effec.Baseplate Depth=D= 5.00 in depth=b= 3.00 in midway dist face of column to edge of plate=e=4.00 in Column Loads DEAD LOAD=DL= 105 lb per column Load Case 1) (1.2+0.2Sds*DL)+(0.85+0.2Sds*PL)+1.5*EL RMI SEC 2.2 EQTN 5 unfactored ASD load = 1.34398*105 lb+0.99398*6000 lb+ 1.5*2416 lb PRODUCT LOAD=PL= 6,000 lb per column = 9,729 lb unfactoredASO load Load Case 2) (0.9-0.2Sds)DL+(0.9-0.2Sds)*PLapp+ 1.5EL RMI SEC 2.2 EQTN 6 P-seismic=EL= (Movt/Frame depth) = (0.75602*105 Ib)+(0.75602*4020 Ib)+(1.5*2416 lb) = 2,416 lb per column = 6,743 lb unfactored ASD load Load Case 3) 1.2*DL+ 1.4*PL RMI SEC 2.2 EQTN 2 = 1.2*105 lb+ 1.4*6000 lb = 8,526 lb Effective Column Load=Pu= 9,729 lb per column Puncture Apunct= [(c+t)+(e+t)]*2*t Fpunct= 2.66*phi*sgrt(fc) = 195.0 in^2 = 79.8 psi fv/Fv= Pu/(Apunct*Fpunct) = 0.625 < 1 OK Slab Bending Pse=DL+PL+E= 9,729 lb Asoil= (Pse*144)/(fsoil) L= (Asoil)^0.5 y= (c*e)^0.5 +2*t r = 1,868 inA2 = 43.22 in = 14.7 in . x= (L-y)/2 M= w*x^2/2 S-slab= 1*teff^2/6 = 14.3 in = (fsoil*x^2)/(144*2) = 4.17 in^3 Fb= 5*(phi)*(fc)^0.5 = 529.9 in-lb fb/Fb= M/(S-slab*Fb) = 150. psi = 0.848 < 1,OK TYPE 5 SELECT-LENNOX Page /. of 1) 10/30/2014 Walter L. Cook PE Fire Protection Engineering Since 1991 January 16, 2015 s Josh Welch Engineered Products 11314 SE 31st Ave Milwaukie OR 97222 RE: Determination of Commodity Class of Materials Stored Evaluation of the Requirement for Fire Protection Lennox Parts Distribution,Tigard OR Dear Josh, In response to requirements of the City of Tigard I offer the following evaluation and determination submitted under my Oregon engineers'seal. We conclude that the existing sprinkler density is adequate for rack storage of Class III commodities up to 12'in height to the top of materials. The matter we investigated is the fire protection requirements and specifically the requirement for sprinklers in the storage area where racking has been installed,we did not evaluate other Fire & Life Safety features of the building. In January I conducted a field survey and subsequent evaluation of the commodity storage. I conclude the following; • Based on observation and review of materials stored onsite it appears that these are primarily Class II & Ill commodities, • The commodities proposed for rack storage includes copper coils,fiberglass air filters and HVAC controls in metal enclosures. All commodities are stored in boxes on racks which shall be configured to allow storage up to 12' in height to the top of materials, • The existing sprinkler system is OH II with a density of 0.21 GPM/1,500 Ft 2. This level of protection would allow storage,where the height of the Class II &Ill commodity does not exceed 12', The first step in determining if sprinkler coverage is required is the occupancy classification. Paragraph 311.2 of the OSSC defines storage of the commodities contemplated for this site to be Group 5-1,as such automatic sprinkler protection is required if the fire area of the building exceeds 12,000 Ft 2,which is true for this building. • Fire sprinklers were installed in 1999;the density is OH II or 0.21 gpm/1,500 sf 2,according to the riser tag. The tag also indicates storage is limited to 12'. WER-! PHONE (503)333-3470 13467 SW Summerwood Dr E-MAIL WALT.L.000KOGMAIL.COM Tigard,OR 97223 Page 2 January 16,2015 Josh Welch Engineered Products The warehouse space includes storage in the following manner; • Storage at floor level of commodities staged for shipment in the front half of the warehouse,existing sprinkler coverage is adequate, • Storage of Class II-Ill commodities on racks along the southern and western wall. The total rack storage area is less than 500 sf 2. Racks shall be configured to allow storage up to 12' in height,top of materials. • Storage of Class Ill commodities on shelves in the central western area, less than 12' in height is allowed under current sprinkler protection. Chapter 16 of NFPA 13 specifies the required sprinkler density for rack storage of these commodities. First for storage up to 12' in height 16.2.1.2.1 refers to Chapter 13 for sprinkler density and storage conditions. Based on Table 13.2.1 any of the commodities in this warehouse can be stored on racks where the top of materials does not exceed 12', in accordance with the conditions specified in the table,see attached Table 13.2.1 from NFPA 13. Based on the findings presented herein,the existing sprinkler system is sufficient for storage of Class II&Ill commodities not to exceed 12',top of materials. Once you have reviewed my findings, please contact me if you need to explore further options. Sincerely Walter L. Cook PE