How to Calculate Flat Plate Thickness of Flat Bottom Storage Tank

7/21/2016

CR4 - Thread: How to Calculate Flat Plate Thickness of Flat Bottom Storage Tank?

Design of Bottom Plate : Bottom Plate is assumed to be simply supported rectangular plate under uniform pressure ; The bottom plate has been provided with stiffeners at a span of 666 mm X 375 mm i.e. b1 = 0.666m & a1 = 0.375m.( a & b values will be as per your requirment) Therefore, b1 / a1 = 0.666/ 0.375 = 1.7 Now, as per Table 6, page 118 of TPS(theory of Plates) Mx = β' q1 a12 and (2.0) My = β'1 q1 a12 (3.0) where, Mx = Bending moment about 'X-axis' My = Bending moment about 'Y-axis' D.1.1 q1 = (Hydrostatic Pressure + Unit weight of Plate ) in kg/cm2 = Pressure due to water column of 2000 mm + Weight of plate / Plate area = [0.20 + 3000X2000x8x0.00000785/(300x200)] (considering 8mm thk. Plate)] = 0.20 + 0.00628 = 0.20628kg/cm2 β' = 0.0555(interpolated from Table 6, page 118 of TPS) β'1 = 0.0493 (interpolated from Table 6, page 118 of TPS) Therefore, putting the values in eqns. (2.0) & (3.0), we get Mx = 0.0555 x 0.20628 x (37.5)2 = 16.09 kg-cm and, My = 0.0493 x 0.20628 x (37.5)2 = 14.3 kg-cm Resultant Bending Moment, M1 = √( Mx 2 + My 2 ) = √( 16.09

2

+ 14.32 ) = 21.52 kg-cm

Again, as per Cl. BSB-8, page BSB-6&7 of 'Elementary Strength of Material' by Birinder Singh ; Section Modulus, Z1 = b4 x d2 = M (4.0) 6 Sd where b4 = 1 cm (Considering unit width of plate) http://cr4.globalspec.com/thread/69485/How-to-Calculate-Flat-Plate-Thickness-of-Flat-Bottom-Storage-Tank

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7/21/2016

CR4 - Thread: How to Calculate Flat Plate Thickness of Flat Bottom Storage Tank?

d = 't1' in cm. (i.e. thickness of plate to be obtained) Sd = Maximum Allowable Design Stress in kg/cm2 = 153 MPa = 153 X 10.19716/cm2 M = M1 = 21.52 kg-cm D.1.2 Therefore, putting the values in eqn. (4.0), we get 1 x t 12 = 21.52 6 153 X 10.19716 or, t1 = √[21.52x6/(1560.16)] = 0.287 cm = 2.87 mm (without corrosion allowance) Thus, thickness of bottom plate with corrosion allowance =t1+CA =2.87+2.0= 4.87mm Hence, Thickness of Bottom Plate provided as 8.0 mm. D.2 Stiffener Sizing Calculation for Bottom Plate : Total load on bottom plate, W1 = Total weight of water filled tank = 12376.8 kg = 0.206 kg/cm2 D.2.1 Considering, Section of stiffener as ISMC 75 with unit weight of 6.8 kg/m [as per table II, page 6 of SP-6(Part-I)]; Now, uniform load on stiffener section, w1 = W 1xa1 + unit wt. of section = 0.206x0.375x100 +0.068 =7.793 kg/cm Maximum length of un-stiffened section, l1 = b1 = 0.666m =66.6 cm Therefore, Considering UDL on simply supported Beam; Bending Moment on stiffener section, M1 = wl2/8 (as per Cl. SFBM-12, page SFBM-23 of 'Elementary Strength of Material' by Birinder Singh) = w1xl12/8 =7.793X66.62/8=4320.78 kg-cm Now, Allowable Bending Stress, σbc = 152 MPa (Refer Sl. No. C.1) = 152 x 10.19716 kg/cm2 Required section modulus, Z1 = M1/ σbc = 4320.78 = 2.787 cm3 152x10.19716 D.2.2 But, Section modulus of ISMC 75 = 20.27cm3 [as per table II, page 7 http://cr4.globalspec.com/thread/69485/How-to-Calculate-Flat-Plate-Thickness-of-Flat-Bottom-Storage-Tank

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7/21/2016

CR4 - Thread: How to Calculate Flat Plate Thickness of Flat Bottom Storage Tank?

of SP-6(Part-I)] > 2.787 cm3 and thus satisfactory. Hence, Actual Stiffener Size selected as ISMC 75. D.3 Deflection calculation of bottom plate : W1max = α q a4 = α q a4 (as per Eqn. No. 141, Page No. 117 of TPS) (5) D EI Where, W1max = Maximum deflection of plate in cm q = Uniformly distributed load, kg/cm2 /cm width of plate = q1 = 0.20628 kg/cm a = Smaller Side of the unstiffened area in cm = a1 = 0.375 m = 37.5 cm D = Flexural rigidity of the plate = EI [as per Eqn. 4, page 5 of TPS] α = Numerical factor depending upon ratio b/a E = Young's modulus in kg/cm2 = 2x106 kg/cm2 I = Moment of inertia in cm4 = (1/12)xbd3 [as per Cl. MOI-5, page MOI-4 of 'Elementary Strength of Material' by Birinder Singh] = (1/12) x 1 x 0.83 cm4 [Considering plate width,b = 1 cm & plate depth,d = 0.8 cm] Now, a = a1 = 0.375 m = 37.5 cm; b = b1 = 0.666m = 66.6 cm For b/a = 66.6/37.5 = 2, α = 0.00486 [as per Table 8, page 120 of TPS] D.3.1 Therefore, putting the values in eqn. (10.0), we get W1max = 0.00486 x 0.20628x (37.5)4 = 0.00061cm = 0.0061mm 2 x 106 x (1/12) x 37.5 x 0.83 D.3.2 Now, As per Cl. 3.13.1.2 of IS-800; Limiting vertical deflection, δ1 = a1/325 = 37.5/325 = 0.1154 cm. = 1.154 mm Since, W1max is less than δ1 , the 8.0mm thk. bottom plate is safe under deflection. reply me if u have any doubts

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