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COMPLIANCE INFO_2022
EnvironmentalHealth
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2200 - Hazardous Waste Program
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PR0513687
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COMPLIANCE INFO_2022
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Entry Properties
Last modified
11/28/2022 2:53:56 PM
Creation date
4/25/2022 9:36:55 AM
Metadata
Fields
Template:
EHD - Public
ProgramCode
2200 - Hazardous Waste Program
File Section
COMPLIANCE INFO
FileName_PostFix
2022
RECORD_ID
PR0513687
PE
2228
FACILITY_ID
FA0009181
FACILITY_NAME
LODI HONDA
STREET_NUMBER
1700
Direction
S
STREET_NAME
CHEROKEE
STREET_TYPE
LN
City
LODI
Zip
95240
CURRENT_STATUS
01
SITE_LOCATION
1700 S CHEROKEE LN
P_LOCATION
02
P_DISTRICT
004
QC Status
Approved
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SJGOV\gmartinez
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EHD - Public
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Lodi Honda <br /> See Attachment "8" 1700 S. Cherokee Lane, Lodi, CA 95240 <br /> Tank Information <br /> _ (Normal Operation• Short <br /> -- Wt tanktull= 4714 _ lbs <br /> 2.03 ft Wt tank empty= 1086 _ lbs <br /> FOT wi Wt Liquid= 3628 _ lbs <br /> 5.180 ft �� Htank= 5.2900 _ ft <br /> 5.290 ft Governing Wtenk= 4.000 _ ft <br /> 2.650 ft <br /> 2.03 ft > Hllquid 5.1800 ft <br /> w�aa"a _ <br /> hOTM= 2.6500 _ ft <br /> Armt= 2.030 ft <br /> F __ Armr= 2.030 _ft <br /> P -- Tank Length (long)= 4.000 _ft <br /> 4.00 ft Tank Width (short)= 4.000 _ft <br /> FAnchor Pull Out= 0 lbs <br /> 1. Calculate Overturning it Resisting RA rients 3/8"Trubolt Wedge Anchor(2-518"min embedment) <br /> MOT = FOT(hOTM)= 2.650 xFOT <br /> NR=Wtenk empty x Armt+Wtuqu;d x Armr + Faachcr x ArmA= 1,086 x 2.030+3,628 x 2.030+ 0 x 4.000 <br /> M = <br /> 2. Set „T = M 9,569 ft-lbs <br /> 2.650 For = 9„569 <br /> FOT= 9,569 / 2.650 = 3,611 Ibs <br /> 3. Calculate Seismic Design Force <br /> Fp= ((0.4(ap)(SDS)(IP)(Wp))/(RP/IP)) x (1+2(z/h)) Seismic Design Force (ASCE 7-16, 13.3-1) <br /> where: <br /> ap = 1.0 We = 4714 <br /> SDs = 0.535 Re = 1.50 <br /> Ip = 1.5 1+2(Z/h) = 1.0 when z is at or below the <br /> FP (min) = 0.30 x SDsx W base result is 1.0 <br /> (tank full)X Ile Minimum Seismic Design Force (ASCE ;-16 13.3-3) <br /> FP (min)= 0.30 x 0.535 x 4714 x 1.50 = 1135 Ibs <br /> FP (max) = 1.6 X SDS X W(tank full)X Jr, Maximum Seismic Design Force (ASCE 7-16 13.3-2) <br /> Fp (max)= 1.6 x 0.535 x 4,714 x 1.50 = 6053 Ibs <br /> FP = ((0.4x1.0x0.535x1.5x4,714) / (1.5 / 1.5)) x (1.0) = 1513 Ibs <br /> Governing Seismic Design Force (FP) =C� Ibs <br /> 4. Calculate Safety Factor <br /> SF = Fot / FP= 3,611 / 1,513 = 2.39 >1.5 OK - Tank is Sf <br /> 5. Calculate Resistance to Sliding <br /> V < W tan 300 = 4,714 x tan 30° 2722 > Design Force of 1513 <br /> Tank does not sling <br /> QROFESS�Ip <br /> FVLANC y �j <br /> Fth <br /> eismic/Anchoring Findings: The tank as designed and operated, and r, <br /> e containment area in which it is located proved to bo seismically sounn <br /> EXP.&-�= <br /> CIVIL <br /> FpF ChE�FO <br />
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