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Storage tanks


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7. DESIGN OF COMPONENTSShell designBottom Plate designWind girder design


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Throughout the chemical, petrochemical and refineries gases, liquids and solids are stored, accumulated or processed in vessels of various shapes and sizes.

Such a large number of storage vessels or tanks are used by these industries that the design, fabrication and erection of these vessels have become a specialty of a number of companies.

Only very few companies in process industries now design storage vessels having large volumetric capacity.

However, the design of this equipment involves basic principles which are fundamental to the design of other types of equipment.

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Storage tanks are designed for internal pressure approximating atmospheric pressure, or internal pressure not exceeding the weight of the roof plates.

Higher internal pressure upto 2.5 psi is permitted when additional requirement are met.

Maximum operating temperature of 90º C is allowed however for higher temperature upto 260 º C allowable stress modification to be done.

Tank designed for one product can store other product of differing relative density(always of lesser density)

Height-to-diameter ratio is often a function of the processing requirements, available land area and height limitations.

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Following are the codes for the design of welded steel storage tanks,

API 650

BS 2654 For non-refrigerated, above ground storage

IS 803

API 12F For tanks for storage of Production liquids (upto 120m3)

API 653 For tank inspection, repair, alteration and reconstruction

API 620 For low-pressure storage tanks

IS 10987 For under-ground/above-ground storage of petroleum products

For design of reinforced tanks

BS 4994 For tanks in reinforced plastics

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The above ground storage tanks of large capacity are categorised based on the type of roof as follows,

Sim ply SupportedInternal ra fter typeInternal truss type

R after w ith C entra l co lum n

C one R oof D om e R oof

F ixed roof

S ing le D eckD ouble D eck

O pen top

S ingle D eckD ouble D eck

C losed top

F loating roof

S torage Tank

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Selection of specific type of tank and type of roof depends upon the intended service conditions, such as the product being stored, its vapor pressure and corrosive nature and anticipated weather and loading conditions.

Cone roof tanks are recommended for products with lower vapor pressure or with less emission control requirement.

Even for products with higher vapor pressure, cone roof tanks can be used if the product capacity is less and necessary venting and blanketing arrangements are provided.

Cone roof tanks are cheaper and easier to construct. Maintenance is very simple.

Floating roof tanks are recommended for storing products of higher volatility. The steel deck provide good insulation over the entire surface of the liquid.

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Following are the common plate material used for construction of tanks,

A 36 upto 40 mm

A 283 Gr C upto 25 mm

A 285 Gr C upto 25 mm

A 131 Gr A upto 12.5 mm

A 131 Gr B upto 25 mm

A 516 Gr 55,60,65,70 upto 40 mm

A 537 Cl1, Cl2 upto 45 mm

The minimum tensile strength of materials used in construction of tanks are between 55000 psi to 85000 psi.

Carbon content between 0.15% to 0.25%

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Low carbon steels are soft and ductile, easily sheared, rolled and formed into various shapes easily. Easy for welding and gives joints of uniform strength relatively free from localised stresses.

Low alloy, high strength steels are also used but it is more difficult to fabricate, because they have low ductility.

Plates shall be semi killed as minimum and fully killed and made to fine grain practice or normalised as required.

For material with minimum tensile strength upto 80 ksi, the manual metal arc-welding electrodes shall conform to E60 and E70 classification series.

For material with minimum tensile strength of 80 ksi to 85 ksi, electrodes shall conform to E80 series.

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SHELL DESIGN:As per Clause 3.6.3Shell thickness is calculated for two conditions, design conditions and hydrotest conditionDesign Condition:td = 2.6 D(H-1)G + CA


Hydrostatic Condition:tt = 2.6 D(H-1)G


where, td is the design thickness required in inches tt is the hydrostatic thickness required in inchesD is the diameter of the tank in feetH is the height of the tank in feetG is the specific gravity of the productCA is the corrosion allowance in inchesSd is the allowable stress for design condition in psi

St is the allowable stress for test condition in psi

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Method explained above is one-foot method of calculating the shell thickness. This method is very conservative method and gives a higher thickness.For tanks above 60 m dia, this method is not preferred.

Variable design point method is used for tanks above 60 m dia and if L/H ratio is less than or equal to 2, where L = (6Dt)0.5 where ‘t ‘ is the thickness of bottom shell course.This method normally provides a reduction in shell-course thickness and hence total material weight. Variable design point method is explained separately.

BOTTOM PLATE DESIGN:As per clause 3.4.1 of API 650All bottom plates shall have a minimum nominal thickness of 6 mm, exclusive of corrosion allowance.

Annular bottom plate design:For calculating the thickness of annular bottom plate the hydrostatic test stress in first shell course shall be calculated as given below

St = 2.6 D(H-1)Gt

where ‘t’ is thickness of first shell course.

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Thickness of annular plate shall be obtained from the table given below.

Radial width of annular plate at any point around the circumference of the tank shall be either Aw1 or Aw2, whichever is greater

Aw1= X + t + Y + L

where, X = 24 “ or as per Appendix E.4.2 X=0.0274 WL/GH whichever is greater

t = Provided thickness of the lowest course

Y = Projection of annular plate outside the shell

L = Annular-sketch plate lap

WL = Weight of tank contents

Aw2= 390 tb/ (HG)0.5

where, tb = Thickness of the annular plate

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WIND GIRDER DESIGN:As per Clause 3.9.7

Tanks of larger diameter may not have the necessary inherent rigidity to withstand wind pressure without deforming and excessively straining the shell. To avoid this suitable stiffening or wind girders are provided.

The maximum height of unstiffened shell H1 shall be calculated as follows:

H1 = 600,000 t ((t/D)3)1/2 (100/V)2 (As per Clause 3.9.7)

where, t = as ordered thickness of the top shell course(in.)

D = nominal tank diameter(ft)

V = wind velocity (mph)

After the maximum height of the unstiffened shell, H1 has been determined, the height of the transformed shell shall be calculated as follows:

Change the actual width each shell course into a transposed width of each shell course having the top shell thickness:

Wtr = W((tuniform/tact)5)0.5

where Wtr = Transposed width of each shell course, (in.)

W = Actual width of each shell course (in.)

tuniform = ordered thickness of top shell course (in.)

tact = ordered thickness of shell course for which transposed width is calculated (in.)

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Add the transposed widths of the courses. The sum of the transposed widths of the courses will give the height of the transformed shell

If Wtr is greater than H1 an intermediate wind girder is required.

For equal stability above and below the intermediate wind girder, the girder should be located at the mid height of the transformed shell.

If half the height of the transformed shell exceeds the than H1 a second intermediate wind girder shall be used to reduce the height of unstiffened shell to a height less than the maximum.

Overturning stability considering wind load shall be checked as follows:

Overturning moment from wind pressure shall not exceed two-thirds of the dead load resisting moment.

M less than or equal to 2/3(WD/2) for unanchored tanks

where, M = overturning moment from wind pressure

W = shell weight available to resist uplift, less any CA, plus dead weight supported by the shell minus simultaneous uplift from operating

conditions such as internal pressure

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The design procedure considers two response modes of the tanks and its contents:

a. The relatively high-frequency amplified response to lateral ground motion of the tank shell and roof, together with the portion of the liquid contents that moves in unison with the shell.

b. The relatively low-frequency amplified response of the portion of the liquid contents that moves in fundamental sloshing mode.

The design requires the determination of the hydrodynamic mass associated with each mode and the lateral force and overturning moment applied to the shell as a result of the response of the masses to lateral ground motion.

The overturning moment due to seismic forces applied to the bottom of the shell shall be determined as follows:

M = ZI (C1WSXS + C1WrXt+ C1W1X1 + C2W2X2)

where, Z = seismic zone factor

I = Importance factor as per Appendix E


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C1 C2 = lateral earth quake force coefficients

WS = Total weight of the tank shell(lb)

XS = Height from the bottom of the tank shell to the shell’s CG(ft)

Wr = Total weight of the tank roof(lb)

Ht = Total height of tank shell(ft)

W1 = Weight of the effective mass of the tank contents that move in unison with the tank shell(lb)

X1 = Height from the bottom of the tank shell to the centroid of lateral seismic force applied to W1 (ft)

W2 = Weight of the effective mass of the tank contents that move in unison in first sloshing mode(lb)

X2 = Height from the bottom of the tank shell to the centroid of lateral seismic force applied to W2 (ft)

Resistance to the over turning moment at the bottom of the shell may be provided by the weight of the tank shell and by anchorage of the tank shell or for unanchored tanks, the weight of a portion of the tank contents adjacent to

the shell.

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WL = 7.9tb(FbyGH)1/2

where, WL = maximum weight of the tank contents that may be used to resist the shell overturning moment, in lb/ft of shell circumference.

tb = thickness of the bottom plate under the shell(in.)

Fby = minimum specified yield strength of the bottom plate under the shell (lb/in.2)

G = design specific gravity of the liquid to be stored

Now, calculate Wt, weight of tank shell & portion of fixed roof supported by the shell, in lb/ft of shell circumference.

When M/[D2(Wt + WL )] is greater than 1.57 the tank is structurally unstable.

When the tank is unstable any one of the following measures shall be carried out:

a. Increase the thickness of the bottom plate tb under the shell.

b. Increase the shell thickness, t.

c. Change the proportions of the tank to increase the diameter and reduce the


d. Anchor the tank.

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Tank anchorage shall be provided if there exists a tendency for the shell and

the bottom plate, close to the shell, to lift off its foundations due to the following reasons,

Uplift on an empty tank due to internal design pressure counteracted by the effective weight of roof and shell.

Uplift due to internal design pressure in combination with wind loading counteracted by effective weight of roof and shell, plus the effective

weight of product considered .

The anchorage shall not be attached to the bottom plate only but principally to the shell.

The design shall accommodate movements of the tank due to thermal changes and hydrostatic pressure and reduce any induced stresses in the shell to a minimum.

If an anchored tank is not properly designed, its shell can be susceptible to tearing.Care should be taken to ensure that the strength of the anchorage attachments is greater than the specified minimum yield strength of the anchors so that the anchors yield before the attachment fail.


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The spacing between anchors shall not exceed 10ft. On tanks less than 50ft in diameter, the spacing between anchors shall not exceed 6ft.

Minimum diameter of anchor bolts shall be 1in. excluding corrosion all.

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Venting is required for all tanks.

The venting system provided shall cater for the following:

a) normal vacuum relief

b) normal pressure relief

c) emergency pressure relief

Normal venting is accomplished by a pressure relief valve, a vacuum relief valve, a pressure vacuum(PV) valve or an open vent with or without a flame-arresting device.

Emergency venting is by means of the following:

Larger or additional valves or open vents.

A gauge hatch that permits the cover to lift under abnormal internal pressure.

A manhole cover that lifts when exposed to abnormal internal pressure.

By means of frangible joint.


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Vent sizing is as per API 2000

Inbreathing (Vacuum Relief)

Required venting capacity for liquid movement out of tank : iQ1

5.6 ft3/hr for each 1 bbl/hr of maximum emptying rate

Required venting capacity for thermal inbreathing : iQ2

1 ft3/hr for each 1 bbl of tank capacity or 2ft3/hr for each 1 ft2 of total shell and roof area.

Required total venting capacity for inbreathing : iQt = iQ1 + iQ2

Outbreathing (Pressure Relief)

Required venting capacity for liquid movement into tank : oQ1

6 ft3/hr for each 1 bbl/hr of maximum filling rate

Required venting capacity for thermal outbreathing : oQ2

Should be 60% of the inbreathing requirement .

Required total venting capacity for outbreathing : oQt = oQ1 + oQ2

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Now select any standard venting device like open vent, open vent with flame arrester or Pressure/Vacuum valve

Once the type and size of venting device is selected, the flow capacity of the device iQc, oQc will be known.

Required number of device Ni shall be as follows:

For inbreathing Ni = iQt/iQc

For outbreathing No = oQt/oQc

Sample design of Roof Vent:


Tank capacity : 11,200 m3

Tank diameter : 35,000 mm (114.8 ft)

Tank height : 15,400 mm (50.5 ft)

Max. filling rate : 2,600 m3/hr (16,354 bbl/hr)

Max. emptying rate : 2,150 m3/hr (13,524 bbl/hr)

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For Inbreathing(Vacuum relief)

Required venting capacity for liquid movement out of tank : iQ1

iQ1 should be 5.6 ft3/hr for each 1 bbl/hr of maximum emptying rate.

iQ1 = 5.6 x 13,524 = 75,734 ft3/hr

= 2,145 m3/hr

Required venting capacity for thermal inbreathing : iQ2

iQ2 should be 2 ft3/hr for each 1 ft2 of total shell and roof area.(A=55,450 ft2 )

iQ2 = 2A = 2 x 55,450 = 110,900 ft3/hr

= 3,140 m3/hr

Required total venting capacity for inbreathing : iQt

iQt = iQ1 + iQ2

= 2,145 + 3,140

= 5,285 m3/hr

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For Outbreathing(Pressure relief)

Required venting capacity for liquid movement out of tank : oQ1

oQ1 should be 6.0 ft3/hr for each 1 bbl/hr of maximum filling rate.

oQ1 = 6.0 x 16,354 = 98,124 ft3/hr

= 2, 779 m3/hr

Required venting capacity for thermal outbreathing : oQ2

oQ2 should be 60% of the inbreathing requirement

oQ2 = 0.6 x 110,900 = 66,540 ft3/hr

= 1,884 m3/hr

Required total venting capacity for outbreathing : oQt

oQt = oQ1 + oQ2

= 2,779 + 1,884

= 4,663 m3/hr

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Venting Device capacity

Select a 12” Pressure/vacuum valve having the following capacity

iQc = 3,366 m3/hr

oQc = 5,600 m3/hr

Required set of PV valve:

For inbreathing : Ni

Ni = iQt/iQc = 5,285/3,366 = 1.57

For outbreathing : No

No = oQt/oQc = 4,663/5,600 = 0.84

Then, 2 sets of the above 12” venting device shall be provided for this tank.

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Roof plates shall be attached to the top angle of the tank with a continuous fillet weld on the top side only.

Frangible joint design:

In the event of excessive internal pressure build up failure occurs first in the roof to shell joint protecting the bottom to shell joint.

In most cases cone roofs are designed as frangible joints only.

Following are the design conditions for a frangible joint:

a. The continuous fillet weld between the roof plates and the top angle does not exceed 5 mm

b. The roof slope at the top-angle attachment does not exceed 1:6

c. The roof to compression-ring details are limited to those shown in figure.

d. Cross-sectional area of the roof-to-shell junction,A should be less than value calculated by the following Aa=W/201,000 tan θ

where W= Total weight of the shell & roof framing (but not the roof plate) supported by shell & roof


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Openings in tank shells larger than a NPS 2 nozzle shall be reinforced.

The minimum cross sectional area of the required reinforcement shall not be less than the product of the vertical diameter of the hole cut in the shell and the required plate thickness.

Reinforcement may be provided by one or any combination of the following:

a. The reinforcing plate

b. The portion of the neck

c. Excess shell-plate thickness.

d. The material in the nozzle neck. The area in the neck available for reinforcement shall be reduced by the ratio of allowable stress in the neck to shell.

The effective area of reinforcement provided by the neck is as follows,

a. The portion extending outward from the outside surface of the tank shell plate to a distance equal to four times the neck-wall thickness


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b. The portion lying within the shell-plate thickness.

c. The portion extending inward from the inside surface of the tank shell plate to a distance equal to four times the neck.

Sample case:

Manhole opening size : 511 mm

Manhole neck thick : 10 mm

Shell thickness required:5.958 mm, say 6 mm

Shell thick provided : 10 mm

Minimum c/s area of reinforcement required = 511 x 6

= 3066 mm2

Reinforcement provided:

A. By excess plate thickness

Reinforcement provided = 511 x (10-6) = 2044 mm2

B. By manhole neck

Neck thickness t = 10 mm, 4t = 40

Shell thickness = 10 mm

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Reinforcement provided by manhole neck = (40x10)+(10x10)

= 500 mm2

Reinforcement provided by A & B = 2044 + 500 = 2544 mm2

Balance required reinforcement = 3066 - 2544 = 522 mm2

Now the reinforcement plate OD can be selected such a way the total reinforcement area provided is higher than reinforcement required.

OD of RF plate as per code = 1055 mm

Selected OD of pad (to avoid fouling with weld seam) = 800 mm

Reinforcement provided by RF pad = (800 - 511) x 6 = 1734 mm2

which is greater than 522 mm2

Therefore provided reinforcement is OK

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All roofs and supporting structures shall be designed to support dead load plus a uniform live load of not less than 25 lb/ft2 of projected area.

Rafters shall be spaced so that in the outer ring, their centers are not more than 2л feet apart measured along the circumference of the tank.

Step 1:

calculate the total load/unit area W acting on the roof.

a. Live load & Vacuum load

25 lb /ft2 as per API 650 + any vacuum load if any

b. Dead load

Weight of roof plate and roof structures

Step 2:

Now minimum number of rafters required shall be calculated.

Say, a tank of 15 meter dia.

Then minimum number of rafters required shall be (л x 15)/1.915 = 24.6 nos


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Now say 30 rafters are provided.

Provide 15 primary rafter and 15 secondary rafter.

Step 3:

Calculate the total load acting on the primary rafter

Area of roof x W gives total load say P

Now P/15 gives the load per rafter

Since the roof is of cone type, loading is zero at the center and maximum at the periphery of the tank roof and is uniformly increasing nature from center to periphery.

This condition can be considered as hinged end condition. Ra = P/15

Step 4:


P/15 P/15


Ha Hb

Ra D/2=r Rb

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Sum of the moments about end C equal to zero

Ha x ht. + load x r x 2/3 = Ra x r Get value of Ha

Maximum bending moment M = 1/3 x r x load - 1/3 Ha x ht

Minimum section modulus required = M/All.Stress

Now select a structural member with higher sectional modulus than required.

Step 5:

Now check for Induced compressive axial stress and bending stress

Induced compressive axial stress =Ha/Ar

where Ar = cross sectional area of the member selected

Induced bending stress = M/Z

where Z = section modulus of the member selected.

If induced stress is less than allowable stress, then member size selected is OK

For allowable stress values refer table 5.1 and 6.1 of IS-800

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Typical Roof Structure Pattern:

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Providing adequate foundations is an important part of ensuring an economical and safe liquid storage tank installation.

Uneven foundation settlement on floating roof tank is a special problem as compared to fixed roof tank foundations.

The seals of floating roof tanks will compensate for reasonable variation in the tank diameter such as out-of-roundness of the shell. Extreme conditions will impair roof seal efficiency or cause jamming of the roof, which can be corrected by releveling the tank. Proper foundation design will avoid this problem.

At any tank site, the subsurface conditions must be known to estimate the soil bearing capacity and settlement that will be experienced.

The subgrade must capable of supporting the load of the tank and its contents.

The total settlement must not strain connecting piping or produce gauging inaccuracies, and the settlement should not continue to a point at which the tank bottom is below the surrounding ground surface.

The tank grade or surface on which a tank bottom will rest should be constructed at least 0.3 m above the surrounding ground surface.


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Clean washed sand 75 to100 mm deep is recommended as a final layer because it can be readily shaped to the bottom contour of the tank to provide maximum contact area and will protect the tank bottom from coming into contact large particles and debris.

The finished grade shall be crowned from its outer periphery to its center at a slope of one inch in ten feet.

The crown will partly compensate for slight settlement, which is likely to be greater at the center.

It will also facilitate cleaning and the removal of water and sludge.

Typical foundation types are earth foundation without a concrete ring wall and earth foundation with a concrete ring wall.

Foundation without a ring wall shall be adopted for small size tanks and on surface where adequate bearing capacity is available.

Tanks with heavy or tall shells and/or self-supported roofs impose a substantial load on the foundation under the shell.

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Advantages of Concrete ring wall

Provides better distribution of the concentrated load of the shell to produce a more uniform soil loading under the tank.

Provides a level, solid starting plane for construction of the shell.

Provides a better means of leveling the tank grade, and it is capable of preserving its contour during construction.

Disadvantages of Concrete ring wall

It doesn’t conform to differential settlements which may lead to high bending stress in the bottom plates adjacent to the ringwall.

Ringwall shall not be less than 300 mm thick.

The centerline diameter of the ringwall should equal the nominal diameter of the tank.

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Pan type floating roof is the first type used in the industry. As the name indicates, this roof looked very much like a shallow pan. The single deck sloped to the centre for drainage.

The pan roof could sink under heavy loads of water or snow or from leaks in the deck or drain. Since the single-deck was in direct contact with the stored liquid, the more volatile liquids would sometimes boil from the sun’s heat.

Pontoon type floating roof has a single deck with an annular pontoon divided by bulkheads into liquid-tight pontoon compartments.

The pontoon area was in excess of 50% of the total roof area. The top deck of the pontoon shaded the bottom deck which is in contact with the liquid.

The single deck area was designed to balloon upward to contain vapors produced by boiling. This reduced considerably the heat input and further boiling.

Double deck floating roof has two deck, one top and one bottom deck.

These two decks are separated by rim plates and bulk heads to form liquid-tight pontoon compartments.


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The top deck provides an insulating air space over the entire area and boiling

losses are held to a minimum.

The deck slopes to one or more drainage points and open emergency overflow drains protect the roof from excessive water loads.

Internal floating roofs is a fixed roof tank with a floating roof inside.

The fixed roof provides a shade from the sun, protection from the wind and also keeps the rain and snow off the floating roof.

Pontoon design:

Floating roofs shall have sufficient buoyancy to remain afloat on liquid with a specific gravity of 0.7 and with primary drains inoperative for the following conditions:

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a. 250 mm(10 in.) of rainfall in a 24-hour period with the roofs intact, except for double-deck roofs provided with emergency drains to keep water to a lesser volume that the roofs will safely support. Such emergency drains shall not allow the product to flow onto the roof.

b. Single-deck and any two adjacent pontoon compartments punctured in single-deck pontoon roofs and may any two adjacent compartments punctured in double-deck roofs, both roof types with no water or live load.


Following are the accessories of floating roofs:

Roof drain

Emergency drain

Bleeder vent

Rim vent

Foam seal

Supporting legs

Anti-rotation devices

Automatic tank gauging

Rolling ladder

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Roof drain:

Roof drains are for removing water from floating roofs in open top tanks.

These drains are made out of pipes with swing joint assembly.

These pipe drains are also called as flexible pipe drains as these pipes extends and shrinks with the varying level of the roof which depends on the product height.

Emergency drain:

Water automatically drains into the tank when it reaches a certain level on the roof. Rainwater cannot collect on the roof to endanger the safety of the floating roof .

Bleeder vent:

Vents the air from under a floating roof when the tank is being filled initially.

After the liquid rises enough to float the roof off its supports the vent automatically closes. When the tank is being emptied the vent is automatically opened just before the roof lands on its support.

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Rim vents:

Rim vents are provided to release any excess pressure in the rim space after the roof is floating.

Foam seals:

One of the important component of a floating roof is the primary seal between the floating roof and the tank shell.

A good seal closes the space effectively, yet permits normal roof movement while protecting against evaporation loses.

Supporting legs:

Floating roof shall be provided with supporting legs.

Legs fabricated from pipe shall be notched or perforated at the bottom to provide drainage.

The length of legs shall be adjustable from the top side of the roof.

The operating and cleaning position levels of the supporting legs shall be specified of fixing the adjustable positions.

The legs and attachments shall be designed to support the roof and a uniform live load of at least 1.2 kPa(25 lb/ft2)

Steel pads shall be used to distribute the leg loads on the bottom of the tank.

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Anti-rotation device:

Required to prevent floating roofs from rotating and damaging rolling ladder, pipe drains and seal.

A guided pole is used as anti-rotation device. The pole is fixed at the top and bottom and passes through a well. The guide pole can additionally used as gauging or sampling device.

Rolling ladder:

Rolling ladder provides safe and easy access from top of the tank to the floating roof.

On floating roof a runway is provided, over this runway the ladder provided with spark proof wheels will travel.

These ladders are provided with self-leveling treads.

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Design by this method gives shell thickness at design points that results in the calculated stresses being relatively close to the actual circumferential shell stresses.

To calculate the bottom-course thickness, preliminary values tpd and tpt for the design and hydrostatic test conditions shall first be calculated from the 1-foot method formula.

The bottom shell course thickness t1d and t1t for the design and hydrostatic test condition shall be calculated using the following formulae:

t1d = (1.06 - (0.463D/H)(HG/Sd)0.5 (2.6HDG/ Sd) + CA

t1t = (1.06 - (0.463D/H)(H/St)0.5 (2.6HD/ St)

To calculate the second-course thickness for both the design condition and the hydrostatic test condition, the value of the following ratio shall be calculated for the bottom course:

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h1 = height of the bottom shell course

r = nominal tank radius

t1 = actual thickness of the bottom shell course, less any thickness added for CA used to calculate t2

If the value of the ratio is less than or equal to 1.375,

t2 = t1

If the value of the ratio is greater than or equal to 2.625,

t2 = t2a

If the value of the ratio is greater than 1.375 but less than 2.625,

t2 = t2a + (t1 - t2 a)[2.1-(h1/1.25(rt1)0.5]


t2 = minimum design thickness of the second shell course excluding any CA

t2a = thickness of the second shell course as calculated for an upper shell course as calculated as described below

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The preceding formula for t2 is based on the same allowable stress being used for the design of the bottom and second courses.

For tanks where the value of the ratio is greater than or equal to 2.625, the allowable stress for the second course may be lower than the allowable stress for the bottom course.

To calculate the upper-course thickness for both the design condition and hydrostatic test condition, a preliminary value tu for the upper course thickness shall be calculated by the 1-foot method and then the distance x of the variable design point from the bottom of the course shall be calculated using the lowest value obtained from the following:

x1 = 0.61(rtu)0.5 + 3.84 CH

x2 = 12CH

x3 = 1.22(rtu)0.5


tu = thickness of the upper course at the girth joint

C = [K0.5(K-1)]/(1+K1.5)

K = tL/tu

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The minimum thickness tx for the upper shell courses shall be calculated for both the design condition (tdx) using the minimum value of x obtained as explained above

tdx = (2.6D(H-x/12)G)/ Sd) + CA

ttx = (2.6D(H-x/12)G)/ St)

The steps described above shall be repeated using the calculated value of tx as tu until there is little difference between the calculated values of tx in succession.

Repeating the steps twice is normally sufficient.

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Design DataType of Tank : Double deck floating roof

Diameter of Tank 'D' :92 m Height of Tank 'H'

:20m Product Stored :Crude OilDesign specific gravity 'G': 0.9

Corrosion Allowance 'C.A.' :0.03937 inches 1 mm

Course width : 2.5 m

Capacity of Tank : 132952.2 cu.m. 836176.3 barrels

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The bottom shell course thickness:

t1d = [ 1.06 - (0.463 D/H) sqrt(HG/Sd) ] (2.6HDG/Sd

= 1.59256 inches

= 40.45102 mm


D = Tank Diameter = 301.8336

H = Tank Height = 65.616

G = Product Specific Gravity = 0.9

Sd = Allowable stress for design condition

of bottom course= 28000 adding 0.03937 inches as corrosion allowance

t1d = 1.63193 inches

41.45102 mm

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The second shell course thickness:

h1/(rt1) 0.5 = 1.8327322


h1 = bottom course height = 98.424 inches

r = nominal tank radius = 1811.002 inches

t1 = bottom course thickness = 1.63193 inches

From API 650 Cl.

h1/ (rt1)0.5 applicable formula

1.375-2.625 t2 = t2a + (t1 - t2a)(2.1 - h1/1.25sqrt(rt1))

> 2.625 t2 = t2a

< 1.375 t2 = t1

Since h1/(rt1) 0.5 = 1.8327322

t2 = t2a + (t1 - t2a)(2.1 - h1/1.25sqrt(rt1))where,

t2 = minimum design thickness for second shell

t2a = thickness for second shell course as calculated for an upper course

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First trial:

Course # 2

H = 57.414 ft.

tu = (2.6D (H-1) G)/Sd

= 1.423024 inches

tl = 1.59256 inches

K = tl/tu = 1.119137

C= k^0.5 (k-1) / (1 + K^1.5)

= 0.05771

a = (rtu)^0.5

50.76514 inches

x1 = 0.61a + 3.84CH

= 43.69006 inches

x2 = 12CH

= 39.7604 inches

x3 = 1.22a

= 61.93347 inches

x = min ( x1, x2, x3)

= 39.7604 inches

x/12 = 3.313367 inches

tdx = (2.6D (H-x/12)G) / Sd

= 1.36467 inches

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Second trial:

H = 57.414

tu = 1.36467

tl = 1.59256

K = tl/tu


C= k^0.5 (k-1) / (1 + K^1.5)


a = (rtu)^0.5


x1 = 0.61a + 3.84CH

= 47.91824

x2 = 12CH

= 54.97837

x3 = 1.22a

= 60.65032

x = 47.91824

x/12 = 3.993187

tdx = (2.6D (H-x/12)G) / Sd


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Third trial:

H = 57.414

tu = 1.347522

tl = 1.59256

K = tl/tu


C= k^0.5 (k-1) / (1 + K^1.5)


a = (rtu)^0.5


x1 = 0.61a + 3.84CH

= 49.2095

x2 = 12CH

= 59.61084

x3 = 1.22a

= 60.26806

x = 49.2095

x/12 = 4.100791

tdx = (2.6D (H-x/12)G) / Sd

= 1.344808

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t2a = 1.344808

t2 = t2a + (t1 - t2a)(2.1 - h1/1.25sqrt(rt1))

1.501837 inches

38.14665 mm

Adding 0.0625 inches corrosion allowance

t2 = 1.541207 inches

39.14665 mm

The third shell course thickness:

First trial:

Course # 3

H = 49.212

tu = (2.6D(H-1)G)/Sd

1.216132 tl = 1.501837

As explained earlier repeat the steps and calculate the third shell course


Similarly, shell thickness of other courses are calculated

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This method of calculating the thickness of the shell is based on the assumption that the tank is filled with water and the tension in each ring is calculated at a point 12 in. above the center line of the lower horizontal joint of the horizontal row of welded plates being considered.

The hydrostatic pressure varies from a minimum at the top of the upper most course to a maximum at the bottom of the lowest course.

In determining the plate thickness for a particular course, a design based upon the pressure at the bottom of the course results in over-design for the rest of the plate. A design based upon the pressure at the top of the course would result in under-design.

However, some consideration should be given to the additional restraint offered by the plates adjoining a particular course.

In the lowest course, the plates of the vessel bottom offer considerable restraint to the bottom shell course.This additional restraint of the bottom edge is effective for an appreciable distance or height from the bottom of the lowest course.

In an intermediate course with a course of heavier plates below, the top of the heavier will be understressed.

Therefore, a design based upon the pressure at a height of 1 ft from the bottom of the course may be considered conservative.