Circular Steel Tank design calculation

April 2, 2018 | Author: Gautam Paul | Category: Soft Matter, Pressure, Fluid Mechanics, Materials Science, Nature


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Sheet1 Design of Molasses Tank:Given Data:Ht of Tank,h 9.85 m Dia of Tank,D 28 m Density of liquid,ρ 16.2 KN/m³ Grade of Plate 340 N/mm² Permissible stress 204 N/mm² Plate width at bot. Along height 2m Center of plate width 1m Total vol. Of liquid 98255.6 kN gravitational force,g 9.8 m/s² Solution:Hydrostatic Pressure on wall:Liquid pressure @ base 159.57 KN/m² Liquid pressure @ center of 2m wide plate 143.37 KN/m² U.D.L on half wall, 143.37 KN/m Total load on half circular wall,F 4014.36 KN Force on either end, 2007.18 KN Resistance by wall, Thk.xht.x(0.6fy) Therefore Thickness req 9.839 mm Provide plate thickness of wall,t 12 mm Thickness of roof plate 5 mm Hydrodynamic Pressure on wall:a) Spring mass model of tank Total liquid mass(m) mass of tank wall(mw) mass of roof(mt) h/D mi/m mc/m hi/h hi*/h hc/h hc*/h mi mc hi hi* hc hc* mt 9825.56 81619.71 83.20 24.64 0.35 0.4 0.55 0.375 1.125 0.51 1 3930.2 5404.1 3.69375 11.08125 5.0235 9.85 24.64 T KG T T T T m m T Linearised Pressure distribution on wall in circumferential direction by “Priestley et al” Page 1 71 For circular tanks. Seismic coefficient Response reduction Factor. qi =[(Ah)i X mi X g]/[πD/2] qc =[(Ah)c X mc X g]/[πD/2] Impulsive Time Period:All in M.007 For circular tanks.61 sec 2.05 Page 2 1 2 3 4 5 6 7 .5 0. max hydrodynamic force per unit circumferential length at Φ = 0.5 1.55 ≈ 1.Z Soil strata assumed to be medium Sa/g for med soil and T ≥ 0.70 0.1 2.230 0.92 sec Sa/g for med soil and T ≥ 0.021 142808. for convective mode is given by.R Importance factor.36/T 0. qi =[(Ah)i X mi X g]/[πD/2] 58946.K.I Zone factor. for impulsive mode is given by.55 ≈ 1. max hydrodynamic force per unit circumferential length at Φ = 0. No. max hydrodynamic force per unit circumferential length at Φ = 0.S unit √(t/D) √E Ti Hor.24 SL.Sheet1 For circular tanks.067 1 2 3 4 5 Convective Time Period:- 5. for impulsive and convective mode is given by. 0.36/T 0. qc =[(Ah)c X mc X g]/[πD/2] 121486. 59 N/m² bc=[(qc/h²)*(6hc-2h) 13073. (h/D)> (1/(Ahi) h/D 0.63 N/m² Anchorage Requirement Circular ground supported tanks shall be anchored to their foundation when. equivalent linear pressure distribution for convective mode is given by.85613177 8 9 10 11 12 13 14 Page 3 . ac=[(qc/h²)*(4h-6hc) 11593.11 N/m² For circular tanks.Sheet1 For circular tanks.351785714 (1/(Ahi) Anchorage not required 14. ai=[(qi/h²)*(4h-6hi) 10472. equivalent linear pressure distribution for impulsive mode is given by.8 N/m² bi=[(qi/h²)*(6hi-2h) 1496. Sheet1 Page 4 . 85 78.85 13.57 KN/m² Hydrostatic Pressure at 8.16 KN Page 5 .t 159.17 KN/m² Hydrostatic Pressure at 0.F @ 6.47 KN/m² 1.85m 3107.77 KN/m² Total load on half circular wall.Sheet1 Pressure Diagram:Hydrostatic Pressure at base Hydrodynamic impulsive pressure at base Hydrodynamic impulsive pressure at top Hydrodynamic convective pressure at base Hydrodynamic convective pressure at top Therefore total pressure intensity at the wall base Under earthquake condition.64 KN/m² 212.37 KN/m² Hydrostatic Pressure at 6.50 KN/m² 11.85 110.07 KN/m² 181.85 m Hydrostatic Pressure at base 159.57 KN/m² Hydrostatic Pressure at 2.59 KN/m² 13.23 KN/m² 14 mm THICKNESS OF WALL CALCULATION ALONG HEIGHT @ 2M INTERVAL Height of liquid in Tank 9. Hydrostatic pressure will govern the design Provide plate thickness of wall.97 KN/m² Hydrostatic Pressure at 4.85 143. 33% increase in permissible stresses is allowed Therefore.57 KN/m² 10.85 46. 37 28021.3786 HENCE OK Page 6 .89 14010.F @ 4.94 TOTAL WEIGHT OF WALL PLATE KN KN KN mm ≈ Mm ≈ Mm ≈ Mm ≈ 12 10 8 8 Deflection Perimeter.Δ 87.6 0.76 Total load on half circular wall.17 Thickness required @ 0.96 Total load on half circular wall.Sheet1 Total load on half circular wall.85m 7.85m 3.F @ 2.85m 0.85m 2199.95 10.e Change in length.57 KN/m² 200000000 KN/m² 0.39 Thickness required @ 2.62 Thickness required @ 4.965 m 159.E Strain.069 m 68.95 mm mm mm mm 156378600 156.L Stress Youngs mod.85m 385.56 Thickness required @ 6.85m 5.78 mm Now increased length become Therefore increased diameter Therefore increased radius Deflection Circumferential or tangential stress= (internal pressure) x D / (2 x t) 88033.00078189 internal pressure =rhoxgxh 156378.85m 1292.F @ 0. Sheet1 Circumferential stress 186165 Page 7 . Sheet1 WEIGHT(KG) 19335 Page 8 . Sheet1 16573 13811 11049 11049 71817 KG N/m2 N/m2 N/mm2 Page 9 .
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