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!' Structural <br /> Engineering & Design Inc. <br /> i <br /> 1815 Wright Ave La.Verne, CA 917 0 Tel Telm 909.596,1351 Fax 90U.526,7186 <br /> By: NIHAL Project: NORTHWEST PUMP Proiect#: <br /> Slab on Grade Configuration:TYPE A <br /> P slab <br /> a : ;----a-----; Concrete <br /> fc= 2,500 psi <br /> slab 1 t D IIb a :.;; . <br /> tslab=t= 5.0 in <br /> Cross tefF= 5.0 in <br /> `I ----c ---- <br /> Aisle = w <br /> x <br /> -N <br /> � ,• � Soil <br /> y I , a fsoil= 750 psf <br /> - � Down Aisle Movt= 85,862 in-lb <br /> SLAB ELEVATION Frame depth= 44.0 in <br /> Baseplate Plan View Sds= 1.101 <br /> Base Plate 0.2*Sds= 0.220 <br /> Effec,Baseplate width=B= 7.00 in width=a= 3,00 in <br /> Effec.Baseplate Depth=D= 5.00 in depth=b= 3.00 in R=B/D=-1.400 <br /> midway dist face of column to edge of plate=c= 5.00 in F'CA0.5= 50.00 psi <br /> Column Loads midway dist face of column to edge of plate=e= 4.00 in <br /> DEAD LOAD=D= 75 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 <br /> unfactored ASD load = 1.42016*75 lb+.1.42016*0.7*2000 lb+ 1* 1951 lb <br /> PRODUCT LOAD=P= 2,000 lb per column = 4,046 lb <br /> unfactoredASD load Load Case 2) (0.9-0.2Sds)D+(0.9-0.2Sds)*B*Papp+rho*E RMI SEC 2.2 EQTN 7 <br /> Papp= 1,340 lb per column = 0.67984*75 lb+0.67984*0.7* 1340 lb+ 1* 1951 lb <br /> P-seismic=E= (Movt/Frame depth) = 2,640 lb <br /> = 1,951 lb per column Load Case 3) 1.2*D+ 1.4*P RMI SEC 2.2 EQTN 1,2 <br /> unfactoredLlmlt State load = 1.2*75 lb+ 1.4*2000 lb <br /> B Uid00° = 2,890 lb <br /> rho Ix Load Case 4) 1.2*D+ 1.0*13+ 1.0E ACI318-14 Sec 5.3.1 <br /> Sds= 1.1008 = 4,041 lb Egtn 5.3,1e <br /> 1.2 + 0.2*Sds= 1.4202 Effective Column Load=Pu= 4,046 lb per column <br /> 0. 9-0.20Sds= 0.6798 <br /> Puncture <br /> Apunct= [(c+t)+(e+t)]*2*t <br /> = 190.0 inA2 <br /> Fpunctl= [(4/3 +8/(3*0)]*X*(F'cA0.5) fv/Fv Pu/(Apunct*Fpunct) <br /> 97.1 psi = 0.267 e 1 OK <br /> Fpunctl= 2.66*X*(F'cA0.5) <br /> = 79.8 psi <br /> Fpunct eff= 79.8 psi <br /> Slab Bending <br /> Pse=DL+PL+E= 4,046 lb <br /> Asoil= (Pse*144)/(fsoil) L= (Asoil)A0.5 y= (c*e)A0.5 + 2*t <br /> = 777 inA2 = 27.87 in = 14.5 in <br /> x= (L-y)/2 M= w*xA2/2 S-slab= 1*teff,,2/6 <br /> = 6.7 in = (fsoil*xA2)/(144*2) = 4.17 InA3 <br /> Fb= 5*(phi)*(fc)A0.5 = 116.9 in-lb fb/Fb= M/(S-slab*Fb) <br /> = 150.psi = 0.187 < 1,OK <br /> i <br /> TYPE A Page 3 of Zy 7/2/2024 <br />