A. Cylindrical Tank, Fixed-Roof with Rafter & Column (cont.)

Transcription

1 According to API 650 Code, Edition Sept Page : 23 of Seismic Design. [APPENDIX E, API 650] 9.1. Overturning Moment due to Seismic forces applied to bottom of tank shell, M = Z I (C1 Ws Xs + C1 Wr Ht + C1 W1 X1 + C2 W2 X2) Ft-lb Kg-M where, Z = Seismic force factor, depends on seismic zone [Table E-2] 0.15 Seismic zone of site 2A I = Importance factor 1 C1 = Lateral earthquake force coefficient [Para. E.3.3.3] 0.6 C2 = Lateral earthquake force coefficient [Para. E.3.3.3] = 0.75 S / T when T <= = S / T 2 when T > S = Site coefficient [Table E-3] 1.5 T = Natural period of the 1st sloshing mode = k (D 0.5 ) Sec. k = Factor depends on ratio D/H [Figure E-4] D = Nominal tank dia Ft M H = Max. design liquid level Ft M Ratio D/H Since T > 4.5 seconds, Use C Ws = Total weight of tank shell lb Ton Xs = Height from bottom of tank shell to shell's center of gravity Ft M Wr = Total weight of tank roof (fixed or floating) + portion of the lb Ton snow load, if any, specified by the purchaser Ht = Total height of tank shell Ft M W1 & W2 = Weight of effective mass of tank contents that move in unison with tank shell [Para. E & Figure E-2] D/H Find ratio W 1 / W T by knowing D/H [Figure E-2] Find ratio W 2 / W T by knowing D/H [Figure E-2] 0.52 W T = Total weight of tank contents = Tank volume x specific gr lb Ton (specific gravity of the product to be specified by the purchaser) Tank volume 5.99E+12 Ft M 3 W 1 = Ratio ( W 1 / W T ). W T lb Kg W 2 = Ratio ( W 2 / W T ). W T lb Kg

2 According to API 650 Code, Edition Sept Page : 24 of Seismic Design. (cont.) [APPENDIX E, API 650] X 1 = Height from bottom of tank shell to the centroid of lateral seismic force applied to W 1 [E & Figure E-3] X 2 = Height from bottom of tank shell to the centroid of lateral seismic force applied to W 2 [E & Figure E-3] Ft Ft D/H Find ratio X 1 / H by knowing D/H [Figure E-3] Find ratio X 2 / H by knowing D/H [Figure E-3] 0.56 X 1 = Ratio ( X 1 / H ). H Ft M X 2 = Ratio ( X 2 / H ). H Ft M 9.2. Resistance to Overturning Moment at bottom of tank shell, W L [Para. E-4] Resistance to the overturning moment at the bottom of the shell may be provided by the weight of the tank shell and by the anchorage of the tank shell, or, for unanchored tanks, the weight of a portion of the tank contents adjacent to the shell. For unanchored tanks, the portion of the contents that may be used to resist overturning depends on the width of the bottom plate under the shell that lifts off the foundation and may be determined as follows : w L = 7.9 t b ( F by G H ) 0.5 lb/ft However, w L shall not exceed 1.25 G H D lb/ft Where, w L = max. weight of the tank contents that may be used to resist the shell overturning moment, w L1 = 7.9 t b ( F by G H ) lb/ft w L2 = 1.25 G H D lb/ft Use w L = Min ( w L1, w L2 ) lb/ft t b = Thickness of the bottom plate under the shell INCH MM F by = Min. specified yield strength of the bottom plate under the shell PSI G = Design specific gravity of the liquid to be stored 1

3 According to API 650 Code, Edition Sept Page : 25 of Seismic Design. (cont.) [APPENDIX E, API 650] 9.3. Shell Compression. [Para. E-5] 9.3.a. Unanchored Tanks [Para. E.5.1] Value of M / [ D 2 (w t + w L )] where, b = max. longitudinal compressive force at the bottom of the shell lb/ft (lb/ft of shell circumference) wt = weight of tank shell and the portion of the fixed roof lb/ft supported by the shell (lb/ft of shell circumference) = wt / 3.14 D I. When M / [ D 2 (w t + w L )] <= b = w t M / D lb/ft II. When < M / [ D 2 (w t + w L )] <= b may be computed from the value of (b + w L )/ (w t + w L ) obtained from Figure E-5. Value obtained from Figure E-5 2 b = ( w t + w L ). Value from Figure E-5 - w L lb/ft III. When 1.5 < M / [ D 2 (w t + w L )] <= b = ( w + w ) M 1 2 D ( w t + wl ) t L 0. 5 w L lb/ft VI. For M / [ D 2 (w t + w L )] > or when b / 12 t > F a Value b / 12 t PSI < Fa Value Fa PSI OK In case b/12 t > Fa, the tank is structurally unstable. It is necessary to take one of the following measures : a. Increase the thickness of the bottom plate under the shell, tb, to increase wl without exceeding the limitations of E.4.1 and E.4.2. b. Increase the shell thickness, t (see Item 9.5 for shell courses). c. Change the proportions of the tank to increase the dia. and reduce the height. d. Anchor the tank in accordance with E.6.

4 According to API 650 Code, Edition Sept Page : 26 of Seismic Design. (cont.) [APPENDIX E, API 650] 9.3.b. Anchored Tanks [Para. E.5.2] For anchored tanks, the max. longitudinal compressive force at the bottom of shell, b = w t M / D lb/ft 9.4. Max. Allowable Shell Compression [Para. E.5.3] The max. longitudinal compressive stress in the shell, b / 12 t, shall not exceed the max. allowable stress, Fa determined by the following formulas for Fa, which take into account the effect of internal pressure due to the liquid contents. G H D2 / t > When G H D 2 / t 2 >= 10 6, Fa = 10 6 t / D PSI - When G H D 2 / t 2 < 10 6, Fa = 10 6 t / 2.5 D (G H ) PSI However, Fa shall not be greater than 0.5 F ty 0.5 F ty PSI OK where, t = thickness of the bottom shell course, excl. any corr. allow INCH F a = Max. allowable longit. compressive stress in the shell PSI F ty = Min. specified yield strength of the bottom shell course PSI 9.5. Upper Shell Course [Para. E.5.4] If the thickness of the lower shell course calculated to resist the seismic overturning moment is greater than the thickness required for hydrostatic pressure, both excluding any corrosion allowance, then the calculated thickness of each upper shell course for hydrostatic pressure shall be increased in the same proportion, unless a special analysis is made to determine the seismic overturning moment and corresponding stresses at the bottom of each upper shell course Anchorage of Tanks [Para. E.6] When anchorage is provided, it shall be designed to provide the following min. anchorage resistance in lb/ft of shell circumference : Min. anchorage resistance = M / D 2 - w t lb/ft

5 According to API 650 Code, Edition Sept Page : 27 of Seismic Design. (cont.) [APPENDIX E, API 650]

6 According to API 650 Code, Edition Sept Page : 28 of Overturning Moment due to Wind, Mw. Mw1 = P (Ar Xr + As Xs) lb-ft Kg-M Where, P = Wind pressure PSI 50 Kg/M 2 Ar = Projected area over roof = 0.5 Do (Ht - H) INCH M 2 Xr = Height from bottom of tank shell to center of gravity Ft M of roof = H (Ht - H) As = Projected area of shell = Do H INCH M 2 Xs = Height from bottom of tank shell to shell's center of gravity Ft M = (H1+H2)/2 + (H2+H3)/2 + x where x (see next page) M Do = D + 2 t M ( Up- lift for wind ) = 4 Mw / 3.14 D lb/ft Kg/M Total load, W (operating: full liq. p = 1+ wind) Ton ( Dead Load ) = W / 3.14 D lb/ft Kg/M where W = weight of tank exc. weight of floating roof & bottom Ton Ratio of Up-lift load / Dead load < 1 Since Up-lift (128 Kg/M) < Dead load (2838 Kg/M) Not necessary of Anchor Bolts. 11. Anchor Bolt Strength Wind Load, Pw = 0.7 x P x Do x H lb Kg Where, P = Wind Pressure 50 Kg/M 2 Do = Tank diameter M H = Tank Height M Wt = Total Tank Weight (excluding corrosion allowance) Kg Wt' = Tank Weight except Bottom Plate (excl. corr. allow.) Kg Design pressure atmospheric Overturning Moment, Mw2 = Pw. Xs lb-ft Kg-M Overturning moment, Mw = Max. (Mw1, Mw2) lb-ft Kg-M Anti-Overturning Moment, Rw = Wt'. D/ lb-ft Kg-M Ratio of Mw / Rw < 1 Since Mw < Rw Anchor Bolts are not required.

7 According to API 650 Code, Edition Sept Page : 29 of 34 To find tank shell's CG : Take Moment about CG : W1 [(H1+H2)/2 + (H2+H3)/2 + x] + W2 [(H2+H3)/2 + x] + W3 [x] = W4 [(H3+H4)/2 - x] + W5 [(H3+H4)/2 - x + (H4+H5)/2] + W6 [(H3+H4)/2 - x + (H4+H5)/2 + (H5+H6)/2] + W7 [(H3+H4)/2 - x + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2] + W8 [(H3+H4)/2 - x + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2 + (H7+H8)/2] W1 [(H1+H2)/2 + (H2+H3)/2] + W1 x + W2 [(H2+H3)/2] + W2 x + W3 x = W4 [(H3+H4)/2] - W4 x + W5 [(H3+H4)/2 + (H4+H5)/2] - W5 x + W6 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2] - W6 x + W7 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2] - W7 x + W8 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2 + (H7+H8)/2] - W8 x x (W1 + W2 + W3) + W1 [(H1+H2)/2 + (H2+H3)/2] = W4 [(H3+H4)/2] + W5 [(H3+H4)/2 + (H4+H5)/2] + W6 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2] Figure (13) + W7 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2] + W8 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2 + (H7+H8)/2] - x (W4 + W5 + W6 + W7 + W8) x (W1 + W2 + W3) + x (W4 + W5 + W6 + W7 + W8) + W1 [(H1+H2)/2 + (H2+H3)/2] + W2 [(H2+H3)/2] = W4 [(H3+H4)/2] + W5 [(H3+H4)/2 + (H4+H5)/2] + W6 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2] + W7 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2] + W8 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2 + (H7+H8)/2] x =[W4 [(H3+H4)/2] Ft M + W5 [(H3+H4)/2 + (H4+H5)/2] + W6 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2] + W7 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2] + W8 [(H3+H4)/2 + (H4+H5)/2 + (H5+H6)/2 + (H6+H7)/2 + (H7+H8)/2] - W1 [(H1+H2)/2 + (H2+H3)/2] - W2 [(H2+H3)/2]] / (W1+W2+W3+W4+W5+W6+W7+W8)

8 According to API 650 Code, Edition Sept Page : 30 of Sliding. Sliding force by wind, Ps = 0.7 x P x D x H lb Kg Frictional Resistance Force by wind, Rs = 0.3 Wt lb Kg Ratio Ps / Rs < 1 Since Ps < Rs Anchor bolts are not required Anchor Bolts Provided. Number of anchor bolts, n 12 Size of anchor bolts (UNC) 0.75 INCH MM Anchor bolt sectional area, A 2.36 CM 2 Allowable load (for 3/4" bolt) : - Tensile, fat lb 1950 Kg - Shearing, fas lb 1350 Kg Anchor Bolt Load. Tensile load per each bolt, ft = (4 Mw / n D) - Wt' / n Since ft < fat Shearing load per each bolt, fs = (Ps - Rs)/n Since fs < fas Kg/Bolt The anchor bolt is sufficient Kg/Bolt The anchor bolt is sufficient 12. Fixed-Roof Inspection Hatches. [API-650, Appendix H.65.3] Inspection hatches shall be located on tank fixed roof to permit visual inspection of the seal region. - Max. spacing of inspection hatches per code 75 Ft M - Min. spacing of inspection hatches per code 4 Ft M Circle dia. of inspection hatches = D - 6, Ft Ft M Calculated No. of inspection hatches = 3.14(D-6)/ Assume actual No. of inspection hatches 6 Actual spacing of inspection hatches Ft M < 75 Ft OK > 4 Ft OK

9 According to API 650 Code, Edition Sept Page : 31 of Total Weight of Tank. a. Weight of Shell 1st Course, W1 = 3.14 D t H p Ton 2nd Course, W Ton 3rd Course, W Ton 4th Course, W Ton 5th Course, W Ton 6th Course, W Ton 7th Course, W7 0 Ton 8th Course, W8 0 Ton Sub-Total (a) lb Ton b. Weight of Bottom Plate, R1 = D/2 - Annular width, Rw M L1 = 2 R1 Sin (O/2) M h1 = L1 / 2 R1 Tan (O/2) M Total Bottom plate Area, A = 0.5 L1. h1. n M 2 Total bottom Plate Weight, W = A. ta. p Sub-Total (b) lb Ton Figure (14) c. Weight of Annular Bottom Plate lb Ton d. Weight of Roof Plate = Roof area * t p lb Ton Roof conical area = 3.14 * (D/2) * sq. root ((D/2) 2 + h 2 ) M 2 'where, h = Ht - H = Cone height = Rafter Slope * Do/ M Rafter slope = 1/16 [see Figure (2)] 01:30

10 According to API 650 Code, Edition Sept Page : 32 of 34 e. Weight of Rafters Ton Total length of rafters (H beam 250x125x6/9) M Weight of one meter of beam Kg f. Weight of Column, Pipe 20" NPS, Sch Ton g. Weight of Top Wind Girder Ton Circumferential length of 2 angles 6x4x3/8" (160x100x10mm) M Weight of one meter of angle Kg Total area of top wind girder, A M 2 Weight of top wind girder, W = A * P * t Ton h. Weight of Intermediate Wind Girder 5 Ton Wind Girder ID M Wind Girder OD M i. Weight of Manholes, Cleanout Doors & Nozzles 6 Ton j. Weight of Ladders & Platforms 5 Ton k. Weight of brackets (cooling system supports) Ton No. of brackets per level = 3.14 D / Support Total brackets for two levels of cooling system + level for foam 84 Support Bracket dimensions 400x850x100x8 600x850x8 Weight of one bracket Kg l. Weight of Inspection Hatches. Ton No. of inspection hatches, Nih 6 Weight of nozzle Weight of flange (or loose cover) m. Weight of Manways (for fixed-roof tanks). Ton At least one manhole, min 24" ID shall be provided for access to the tank interior. [API-650, Appendix H.6.5.1] No. of manholes located at fixed roof, 24" NPS 2 Weight of nozzle Weight of flange Total Tank Weight Ton