Storage Tank Calculation

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    Design Code API 650 Eleventh Editions June 2007

    References

    Appendix BAppendix EFormulae for Stress And StrainTank Data Sheet

    Rules followed for designing All components are designed with API 650 June 2007

    edition Shell are designed as per one foot method Wind force and moment are calculated as per API 650 Seismic force and moment are calculated as per API 650

    CALCULATION

    1. DESIGN DATA TO BE COLLECTED AS PER DATA SHEET

    2. MATERIAL OF CONSTRUCTION

    Materials are selected for the following

    Shell/Bottom/Roof PlateRoof Structure/Curb Angle

    NozzlesFlangesBolting For NozzlesBolting for Structures

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    Wind Girder

    Plate specification and Minimum Yield strength, Minimum Tensile Strengthand Product Design Stress are selected from Table 5.6.2 for the shell plates.

    3. SHELL DESIGN CALCULATION

    Allowable Stress for Shell in Design Condition

    Sd = 2/3*Ys*Fy where Ys = Min. yield stressFy = Yield stress reduction factor Sd = 2/5*UTS UTS= Ultimate tensilesterength

    Sd = Table 5.2

    Three values of Sd are got and the least value is taken as Sd

    Allowable Stress For Shell in Hydrostatic Test Condition

    St = 3/4*Ys*Fy

    St = 3/7*UTS

    St = Table 5.2

    Three values of St are got and the least value is taken as St

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    4. Shell Thickness Calculation

    One foot Method

    The 1-foot method calculates the thicknesses required at design points 0.3 m(1 ft) above the bottom of each shell course. This method shall not be usedfor tanks larger than 60 m (200 ft) in diameter.

    4.9 x D x (H - 0.3) x Gtd = ---------------------------------- + C.A.

    Sd

    4.9 x D x (H - 0.3)tt = ------------------------------

    St

    Where,

    td = Design Shell Thickness in mmtt = Hydrostatic Test Shell Thickness in mmH = Design Liquid Height as per Cl. 5.6.3.2

    Minimum required shell thickness is selected from the table 5.6.1.1

    Thickness of each shell course is calculated using above formulae. Thethickness calculated must be greater than minimum thickness taken from thetable 5.6.3.2

    Thickness of each shell course is tabulated.

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    5. ANNULAR PLATE

    Radial Width of annular plateAnnular bottom plates shall have a radial width that providesat least 600 mm

    Therefore min required radial width = Projection outside the shell+Bottom plate

    Thickness+ Lap of bottom plate(Refer 5.5.2)

    Check for min. width of Annular PlateIn SI units:

    Wheretb = thickness of the annular plate (see 5.5.3), in mm,

    H = maximum design liquid level (see 5.6.3.2), in m,G = design specific gravity of the liquid to be stored.

    Thickness of annular plate

    Minimum thickness of the bottom plate is 6mm as per 5.4.1 (Excluding thecorrosion allowance)

    Hydrostatic test stress in the lowest course Sth is calculated. As per Sthvalue thickness of the annular plate is checked with Table 5.1

    6. Bottom PlateAs per API 650 5.4.1, the minimum required thickness shall be 6 mm + C.A.

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    All rectangular and sketch plates shall have a minimum nominal width of 1800 mm (72 in.)

    Tank bottoms requiring sloping shall have a minimum slope of 1:120upwards toward center of the tank.

    7. Design of Roof Plate Thickness:As per API 650 5.10.2.2, the minimum nominal thickness shall be 5 mm +C.A.

    Roof Slope / Angle 1 : 100

    8. Sizing of Top Curb AngleAs per API 650 5.1.5.9.e, min. curb angle of size is L 76 X 76 X 6.4 THK.

    9. Top Wind girder

    The Top Wind Girder is provided at: 1100 mm From Top Curb Angle as per 5.9.4

    As per API 650, 3.9.6 the required min. section modulus is,

    H2 = Height of Tank ShellVz=Design wind velocity

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    C = A(1)*Y(1) + A(2)*Y(2) + A(3)*Y(3)------------------------------------------------

    A(1)+A(2)+A(3)I= BD^3 +AY^2

    -------12

    Z =I/Ymax

    Zprovided > Zreqd.

    10. Intermediate Wind Girder

    Transformed width of each shell stake is given by

    Where,

    Wtr = transposed width of each shell course, mm (in.),W = actual width of each shell course, mm (in.),t uniform = as ordered thickness, unless otherwise specified, of the topshell course, mm (in.),

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    t actual = as ordered thickness, unless otherwise specified, of the shellcourse for which the transposed width is being calculated, mm (in.).

    The values are tabulated as follows

    The Cumm Wtr gives the Transformed Width Htr

    As per API 650 5.9.7.1, the maximum height of the unstiffened shell H1 is

    If Htr > H1 then intermediate wind girder must be provided

    The section modulus of the intermediate wind girder shall be based on the properties of the attached members and may include a portion of the tank shell for a distance above and below the attachment to the shell, in mm (in.)

    In SI units

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    11. WEIGHT CALCULATION

    Calculate Weight for each and every part of the tank

    12. WIND ANALYSIS

    Basic Wind Pressure on cylinder (Pcy) = P*Hc

    Basic Wind Pressure on roof (Pr) =P*Hc

    Hc = (v/160) ^2

    (The value of P is given below)

    (Refer 5.2.1 The design wind pressure shall be 0.86 kPa ( V /190)^2,([18lbf/ft2][ V /120]^2) on vertical projected areas of cylindrical surfaces and1.44 kPa (V/190)^2, ([30lbf/ft2][ V /120]^2) uplift on horizontal projectedareas of conical or doubly curved surfaces, where V is the 3-sec gustwindspeed.)

    Wind load on cylinder section (Fcyl) =Dw *H*Pcy

    Wind load on the roof section (Fr) =Ac*pr

    Where Dw = effective tank diameter H = height of the tank Ac =Effective cone area

    Resultant wind load F =Fcyl+Fc

    Resultant Moment M = F*Z + Fr*(H+Hd/3)

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    Hd=Dome height

    Effective wind load on tank Fe =1.1*F

    Effective wind moment on the base Me=1.1*M

    13. Check For Over Turning Due To Wind

    Overturning wind moment = Me

    Wt of shell + wt of roof +wt of plate =W1

    Force due to internal pressure =W2

    Wt available to resist uplift (W) = W1-W2

    Two third of the dead load resisting moment= 2/3 * (W*D/2)

    14. CALCULATION FOR SESMIC MOMENT FOR OPERATING CONDITION

    Find form factor (K) from Fig E-4

    Find effective mass of the contents from Fig E-2 (W1/Wt and W2/Wt)

    Find effective centroid of seismic force from Fig E-3 (X1/H and X2/H)

    Over turning at the base of the tank

    15. CHECK FOR OVERTURNING DUE TO SESMIC MOMENT

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    Max weight of the tank content resisting overturning moment isWl = 99*tb*(Fby*G*H)

    How ever overturning moment Wl should not exceed 196GHD

    Wheretb= Thickness of the bottom plate

    Fby= Yield Strength of the bottom plate