The document discusses the design of storage tanks. It covers general considerations for tank design codes, types of tanks, selection of tanks, material specifications, and design of various tank components like shells, bottoms, roofs, foundations. It also discusses seismic analysis, anchorage requirements, venting, and floating roof tank accessories. Key aspects covered include allowable stresses and corrosion allowances for materials, thickness calculations using different methods, wind girder design, and anchorage design considering uplift forces.

1. STORAGE TANKS
1. INTRODUCTION
2. GENERAL
3. DESIGN CODES
4. TYPE OF TANKS
5. SELECTION OF TANKS
6. MATERIAL SPECIFICATIONS
7. DESIGN OF COMPONENTS
Shell design
Bottom Plate design
Wind girder design
8. SEISMIC ANALYSIS

2. 9. ANCHORAGE REQUIREMENT
10. VENTING OF TANKS
11. ROOF TO SHELL JOINT DETAIL
12. REINFORCEMENT REQUIREMENT
13. ROOF STRUCTURE DESIGN
14. FOUNDATION DESIGN CONSIDERATION
15. TYPE OF FLOATING ROOF AND ITS ACCESSORIES
16. CALCULATION OF THICKNESS BY VARIABLE POINT METHOD

3. INTRODUCTION
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.

4. GENERAL
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.

5. DESIGN CODES
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

6. TYPES OF TANKS
The above ground storage tanks of large capacity are categorised based
on the type of roof as follows,
Simply Supported
Internal rafter type
Internal truss type
Rafter with Central column
Cone Roof Dome Roof
Fixed roof
Single Deck
Double Deck
Open top
Single Deck
Double Deck
Closed top
Floating roof
Storage Tank

9. SELECTION OF TANK
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.

10. MATERIAL SPECIFICATIONS
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%

11. 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.

12. DESIGN OF TANK COMPONENTS
SHELL DESIGN:As per Clause 3.6.3
Shell thickness is calculated for two conditions, design conditions and
hydrotest condition
Design Condition:
td = 2.6 D(H-1)G + CA
Sd
Hydrostatic Condition:
tt = 2.6 D(H-1)G
St
where, td is the design thickness required in inches
tt is the hydrostatic thickness required in inches
D is the diameter of the tank in feet
H is the height of the tank in feet
G is the specific gravity of the product
CA is the corrosion allowance in inches
Sd is the allowable stress for design condition in psi
St is the allowable stress for test condition in psi

13. 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 650
All 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)G
t
where ‘t’ is thickness of first shell course.

14. 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

18. 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.)

19. 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

20. 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
SEISMIC ANALYSIS

21. 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.

22. 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
height.
d. Anchor the tank.

23. 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.
ANCHORAGE REQUIREMENT

24. 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.

25. 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.
VENTING OF TANKS

26. 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

27. 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:
Condition:
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)

28. 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

29. 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

30. 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.

31. 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
ROOF TO SHELL JOINT

32. 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
REINFORCEMENT REQUIREMENT

33. 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

34. 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

35. 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
ROOF STRUCTURE DESIGN

36. 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:
C
P/15 P/15
ht
Ha Hb
Ra D/2=r Rb

37. 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

38. Typical Roof Structure Pattern:

39. 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.
TANK FOUNDATION DESIGN
CONSIDERATIONS

40. 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.

41. 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.

42. 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.
TYPE OF FLOATING ROOFS AND ITS
ACCESSORIES

43. 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:

44. 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.
ACCESSORIES:
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

45. 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.

46. 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.

47. 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.

51. CALCULATION OF THICKNESS BY VARIABLE POINT
METHOD:
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:

52. h1/(rt1)0.5
where,
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]
where,
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

53. 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
where,
tu = thickness of the upper course at the girth joint
C = [K0.5(K-1)]/(1+K1.5)
K = tL/tu

54. 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.

55. Design Data
Type of Tank : Double deck floating roof
Diameter of Tank 'D' :92 m
Height of Tank 'H':20m
Product Stored :Crude Oil
Design 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

56. The bottom shell course thickness:
t1d = [ 1.06 - (0.463 D/H) sqrt(HG/Sd) ] (2.6HDG/Sd
= 1.59256 inches
= 40.45102 mm
Where,
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

57. The second shell course thickness:
h1/(rt1) 0.5 = 1.8327322
Where,
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.3.6.4.5
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

58. 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

59. Second trial:
H = 57.414
tu = 1.36467
tl = 1.59256
K = tl/tu
1.166992
C= k^0.5 (k-1) / (1 + K^1.5)
0.079798
a = (rtu)^0.5
49.71338
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
1.347522

60. Third trial:
H = 57.414
tu = 1.347522
tl = 1.59256
K = tl/tu
1.181843
C= k^0.5 (k-1) / (1 + K^1.5)
0.086522
a = (rtu)^0.5
49.40005
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

61. 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
thickness.
Similarly, shell thickness of other courses are calculated

62. ONE-FOOT METHOD DESCRIPTION:
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.