DESIGN CALCULATION FOR HCl STORAGE TANK (Ø2500 X 4300mm(T/T) L) BASED ON BS-4994-1987
Rev
Date
1
15-06-2011
2
28-07-2011
Revision Description a. Design Data – Operating Pressure and Temperature b. Tank Capacity 26m3 to 25m3 c. Nozzle Loading onto N1 & N9 d. Live load on Platform 1000Pa to 2500 Pa e. Inclusion of Handrail Design Cage(Truss) web members rearrangement
Prepared by: ERM
ERM
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FRP TANK DESIGN BASED ON BS 4994 (1987)
A) Design Data Shape of tank: Cylindrical, horizontal, dish end. (0.30 Semi-Ellipse) Di = 2500mm Ø. L: 4300mm (T/T) Fluid handle: HCl, 9.9% concentration Specific gravity: 1.05 Operating temperature: Ambient. Design temperature: 200C(min) to 600C(max) Operating pressure: 0.015 barg (0.0015MPa) Design pressure: 0.34barg (0.034MPa) Design vacuum: -0.05barg (-0.005MPa) B) Material Data I.
Material Properties: Type of resin: Isopthalac resin. Ultimate tensile unit Strength, UTUS = 200 N/mm (CSM) = 250 N/mm (WR) = 500 N/mm (Unidirectional Filament) Unit modulus = 14,000 N/mm (CSM) = 16,000 N/mm (WR) = 28,000 N/mm (Unidirectional Filament) Maximum allowable strain, = 0.2%. (Clause 9.2.4) Note: Material strengths were based from minimum values provided in the BS 4994: 1987 Table 5.
C) Design Factor K= 3 x k1 x k2 x k3 x k4 x k5
(EQ 1)
where: 3:
represents a constant which allows for the reduction of material strength caused by long term loading.
k1: Handwork : 1.5 k2: Without thermoplastic lining: 1.6
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k3: Heat distortion temperature of resin: 1.0 k4: Cyclic loading: 1.1 k5: Without post cure: 1.5 K=3 x 1.5 x 1.6 x 1.0 x 1.1 x 1.5 = 11.88
D) Design Unit Loading 1.
Design unit loading, UZ a) Determine the Load limited allowable unit loading, uL. u uL = K (EQ 2) where u is UTUS from Table 5. 200 uL,CSM = 11.88 = 16.84 N/mm per kg/m2 glass.
250 = 21.04 N/mm per kg/m2 glass. 11.88 500 = 11.88 = 42.09N/mm per kg/m2 glass.
uL,WR =
uL,FW
b) Allowable strain, = 0.2% c) Strain limited allowable unit loading, uS. uS = Xz (EQ. 3) where Xz is the unit modulus from Table 5. uS,CSM = 14,000 x 0.2/100 = 28.0 N/mm per kg/m2 glass. uS,WR = 16,000 x 0.2/100 = 32.0 N/mm per kg/m2 glass. uS,FW = 28,000 x 0.2/100 = 56.0 N/mm per kg/m2 glass.
d) Design unit loading Uz. 3 of 21
Since UL < US, then the strain for each layer concerned shall be determined. (clause 9.2.6 b)
=
uL XZ
(EQ. 4)
16 .84 CSM = 14 ,000 x 100
= 0.12% WR =
21 .04 x 100 16 ,000
= 0.13%
42 .09 FW = 28 ,000 x 100
= 0.15% Therefore, the allowable strain, d = 0.12%. (Least of the two values) The Design unit loading for each layer, uZ, shall be determined from the formula below: uZ = Xz d
(EQ. 5)
uZ,CSM = 14,000 x 0.12/100 = 16.8 N/mm per kg/m2 glass. uZ,WR = 16,000 x 0.12/100 = 19.2 N/mm per kg/m2 glass. uZ,FW = 28,000 x 0.12/100 = 33.6 N/mm per kg/m2 glass. degrees) to the tank axis, the following formulas should be used: In circumferential direction: uZ = Xø d Fø
(EQ. 5a)
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In longitudinal direction: uZ = XX d FX
(EQ. 5b)
From Table 7, filament at winding angle 65°. Fø = 0.5,
FX = 0.5
From Figure 3, we can obtain the Unit modulus for winding angle 680, Xø = 18,000N/mm and XX = 4,400N/mm Therefore, (i) Circumferential uZ = Xø d Fø
(EQ 5a)
= 16000 x 0.12/100 x 0.5 = 9.60 N/mm per kg/m2 glass. (ii) Longitudinal uZ = XX d FX = 4400 x 0.12/100 x 0.5 (EQ 5b) = 2.52 N/mm per kg/m2 glass. Summary of Design unit loading: For CSM, uZ,CSM = 16.8 N/mm per kg/m2 glass. For WR, uZ,WR = 19.2 N/mm per kg/m2 glass. For Filament Winding, Circumferential uZ = 9.60 N/mm per kg/m2 glass. Longitudinal uZ= 2.64 N/mm per kg/m2 glass.
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E) Design For Construction. 1) Tank Shell Design. i) Circumferential unit load, QØ = pDi/2 (EQ 7) SG of liquid = 1.05 P = 0.034N/mm2 Di = 2500mm 0.034(2500) QØ = = 42.50N/mm 2 Use 8 layers of CSM as chemical barrier, 2 layer of WR and 4 layers of filament roving as reinforced layer. Check: Circumferential ULAMØ = (8 x 0.45 x 16.8) + (2 x 0.80 x 19.2) + (4 x 1.1 x 9.60) = 133.44N/mm > QØ ------ OK! ii) Longitudinal unit load, Qx =
pDi 4M + (EQ11) 4 Di 2
Tank Empty Weight = 1,530 kg (FRP Tank) Liquid Weight (full) = 1.05 x 1000kg/m3 x 25m3 = 26,250 kg Total Operating Weight = 27,780 kg W
= 27,780kg x 9.81m/s2 = 272,522N
Uniform load, w = 272,522N/5800mm = 46.99N/mm
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1300mm
1600mm
1600mm
1300mm
39.71 kN-m
39.71 kN-m 19.38 kN-m 14.50 kN-m
14.50 kN-m
9.93 kN-m
9.93 kN-m
61.09 kN 50.29 kN
24.89 kN
24.88 kN
50.29 kN 61.09 kN
Pressure due to liquid, p = 0.034N/mm2: Q=
4M pDi 4 Di 2
4(39.71x106 Nmm) 0.034(2500) 4 (2500) 2 Qx = 29.34N/mm ; 13.16 N/mm Q=
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Pressure due to vacuum, p = -0.005N/mm2: 4M pDi 4 Di 2 4(39.71x106 Nmm) 0.005(2500) Q= 4 (2500) 2 Qx = 4.96N/mm ; -11.21 N/mm
Q=
Using 8 layers of CSM, 2 layer of WR and 4 layers of filament roving, ULAMX = (8 x0.45 x 16.8) + (2 x 0.80 x 19.2) + (4 x 1.1 x 2.64) = 102.82N/mm > Qx -------- OK! iii) Compressive unit load. a) Due to Shear on saddle support: Qc =
111,380 N V = L 120 / 360 x x 2500
= 42.54 N/mm b) Due to weight onto nozzle N1 and N9: pDi Qc = ; 4 where p is computed from 20kg over effective area supporting nozzle (consider smaller area, so pressure is maximum) since N1 is smaller with Diameter = 80mm, effective diameter is 92mm; 92 mm 2 x = 6,648mm2 4 20 kgx9.81m / s 2 Therefore p = = 0.030 N/mm2 2 6,648 mm pDi 0.030x 2500 Qc = = =18.75 N/mm 4 4
Effective area =
Since compressive unit load due to saddle support (Qc= 42.54 N/mm) is greater than Qc due to nozzles N1 and N9 and also greater than Qx due to vacuum, therefore consider compressive load due to Shear on saddle supports.
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Permissible compressive unit load 0.6tX LAM Qp = (EQ 13) FDo 0.6(14 )(14 ,000 x0.45 x8 16 ,000 x0.80 x 2 4,400 x1.1x 4) Qp = 4(2500 2 x14 ) = 79.22N/mm > Qc ------- OK! iv)
Check minimum permissible thickness, tm, to prevent buckling due to external pressure or vacuum: L = 0.4 x 750 + 4300 = 4,900mm Do = 2500+2x14 = 2,528mm pvacuum = 0.005 N/mm2 F=4 X 14,000x0.45x8 16,000x0.80 x 2 4,400x1.10 x 4 ELAM = LAM (EQ14) t 14 = 6,811.43
L 4,900 1.94 ; Do 2528
E 1.35 LAM pF
0.17
E L Since < 1.35 LAM Do pF
6,811.43 1.35x 0.005x 4
0.17
11.77 ;
0.17
, then
0.4 pF L tm = Do x E LAM Do
0.40
0.4 x0.005x4 4,900 2,528 x 2,528 6,811.43
(EQ 16) 0.40
= 13.98mm < 14mm, OK! Therefore, the proposed construction of 8 layers CSM, 2 layers WR and 4 layers filament roving for shell is SAFE!
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2) Dish End Design. i) Against internal pressure: t hi Assume = 0.3 0.005 and Di Di hi Conservatively, use =0.25 Ks values from Table 11. Di Use Ks = 1.30 for semi ellipsoidal dome. QØ = 0.5pDi Ks = 0.5(0.034)(2500)(1.30)
(EQ 43)
QØ = 55.25 N/mm But
QØ U1, M1, N1 + U2 + M2 + N2 + … Using 11 layers of CSM + 3 layers of WR: Qact = 16.8x0.45x11 + 19.2x0.80x3 = 129.24 N/mm > QØ = 55.26 N/mm ---------OK!
ii) Check minimum permissible thickness against buckling, tm:
t m 1.7 Ro
pF E LAM
(EQ 19)
But Ro 0.5Do K e (EQ 46) and Ke = 1.50 (from Figure 15, BS4994:1987) Ro 0.5 x 2,528 x1.50 1,896
ELAM =
X LAM 14,000x0.45x11 16,000x0.80 x3 t 14 = 8,975
Therefore,
t m 1.7 Ro
pF 0.034x4 1.7 x1,896 =12.55 mm < 14mm, OK! E LAM 8,975
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iii) Check against vacuum, p =-0.005 N/mm2 QØ = 0.66pDi Ks (EQ 43) = 0.66(0.005)(2500)(1.30) = 10.725 N/mm < Qact, OK!
3) Mild Steel Saddle Support.
764
2800 A. Check Legs for compression: Total weight of tank, W = 272,522 N Factor of safety = 2 Design weight of tank = 545,044 N Load on each support set = 545,044 / 3 = 181,681 N Using 4 numbers of C-Channel 150 x 75 x 6.5 Maximum compressive load that center Leg will carry: = 66,617 N Allowable compression stress = 0.6Fy = 0.6 x 248 Fc = 148.8N/mm2 Actual compression stress =
P
A Where A = 2280mm (C-Channel cross section) 2
P
66,617 A 2280mm 2 = 29.22N/mm2 < Fc ------- OK!
c =
=
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B. Check Base against Compression: Total Compressive force to carry = 181,681 N Allowable compressive stress, Fc = 0.60Fy = 0.60x248MPa = 148.80 N/mm2 c =
Actual Compressive stress,
P A
=
181,681N = 6.49N/mm2
4) Design of M/S Cage
2800
1300
1600
1600
1300
5800
A. Platform Beam support FRP Grating, accessories loading Live Load
= 150 Pa = 2,500 Pa ========= Total = 2,650 Pa = 0.00265 N/mm2 Safety Factor = 2 Maximum distance between beams = 1,600 mm Beam length, l = 2,800 mm Design uniform load, w = 0.00265N/mm2 x 1,600mm x 2 w = 8.48 N/mm
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wl 2 8.48 x 2,800 2 = =8,310,400 N-mm 8 8 M 8,310 ,400 N mm SxReqd = = =50,772mm3 = 50.77cm3 0.66 Fy 0.66 x 248 MPa
Max bending Moment, M =
Use C-channel 125x65x6.0mm Thk (Sx = 89.40cm3>SxReqd), is OK! B. Design of Truss Members a. Truss self weight Assuming members using L100x100x6mmT, Mass per meter of member = 12.2kg/m x 9.81m/s2 = 119.68 N/m b. Handrails, and other accessories = 300Pa = 0.0003 N/mm2 Uniform load onto Top Chord of truss, w = 0.0003N/mm2 x 2800mm/2 = 0.42 N/mm c. Loading from Platform beam support: Reactions from beam support P1 = 0.00265N/mm2 x 2(SF) x 650mm x 2800mm / 2 = 4,823 N P2 = 0.00265N/mm2 x 2(SF) x 1450mm x 2,800mm / 2 = 10,759 N P3 = 0.00265N/mm2 x 2(SF) x 1600mm x 2,800mm / 2 = 11,872 N d. Tank assembly load Saddle support weight = 2,000 N per support Tank Empty weight = 1,530 kg Uniform load from empty tank = 1,530x9.81/5,800mm = 2.59 N/mm Support Reactions for saddle (see Shear Diagram on p. 14)
R1 = 2,770 N + 3,370 N = 6,140 N R2 = 1,370 N + 1,370 N = 2,740 N
Total load transferred to joints of truss: 13 of 21
T1 = R1 + 2,000N = 8,140 N T2 = R2 + 2,000N = 4,740 N
1300mm
1600mm
1600mm
1300mm
3.37 kN 2.77 kN
1.37 kN
1.37 kN
2.77 kN 3.37 kN
At Tank Operation:
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SUPPORT REACTIONS:
JOINT 2 3 4
FORCE-Y (kN) 12.01 26.01 12.01
MEMBER END FORCES Unit: Force: kN MEMBER
JT
AXIAL
SHEAR-Y
SHEAR-Z
1
1 2
2.58 -2.58
-.02 .13
.02 -.13
2
2 3
.00 .00
.09 .04
-.09 -.04
3
3 4
.00 .00
.04 .09
-.04 -.09
4
4 5
2.58 -2.58
.13 -.02
-.13 .02
5
6 7
.00 .00
.16 .33
-.16 -.33
6
7 8
-6.15 6.15
.32 .29
-.31 -.29
7
8 9
-6.15 6.15
.29 .32
-.29 -.32
8
9 10
-.01 .01
.33 .16
-.33 -.16
9
1 6
5.39 -5.06
.00 .00
.00 .00
10
2 7
11.69 -11.36
.01 -.01
-.01 .01
11
3 8
13.04 -12.70
.00 .00
.00 .00
12
4 9
11.69 -11.36
-.01 .01
.00 -.01
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13
5 10
5.39 -5.05
.00 .00
.00 .00
14
1 7
-5.95 6.28
.06 .05
-.06 -.05
15
3 7
7.34 -7.01
.07 .07
-.07 -.06
16
3 9
7.34 -7.01
.07 .07
-.07 -.06
17
5 9
-5.94 6.28
.06 .05
-.05 -.06
Maximum Axial Load = 12,700 N Maximum Shear = 330 N Check Member against Axial Load(Tension): Cross-sectional area = 1,560mm2 P 12 ,700 N T = = = 8.14 N/mm2 A 1,560 mm 2 Allowable Tensile stress, fT = 0.60Fy = 0.60 x 248 MPa = 148.80 N/mm2> T, OK! Check Member against Shear: P 330 N V = = = 0.21 N/mm2 A 1,560 mm 2 Allowable ==Shear stress, fV = 0.40Fy = 0.40 x 248 MPa = 99.20 N/mm2 > V, OK!
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During Lifting of Tank:
SUPPORT REACTIONS:
JOINT 7 9
FORCE-Y (kN) 14.00 14.00
MEMBER END FORCES Unit: Force: kN MEMBER
JT
AXIAL
SHEAR-Y
SHEAR-Z
1
1 2
.38 -.38
.06 .05
-.06 -.05
2
2 3
.37 -.37
.05 .08
-.05 -.08
3
3 4
.37 -.37
.08 .05
-.08 -.05
4
4 5
.38 -.38
.05 .06
-.05 -.06
5
6 7
.02 -.02
.20 .30
-.20 -.30
6
7 8
.00 .00
.32 .29
-.32 -.29
7
8 9
.00 .00
.29 .32
-.29 -.32
8
9 10
.02 -.02
.30 .20
-.30 -.20
9
1 6
.62 -.28
-.01 .01
.01 -.01
10
2 7
-8.28 8.62
.00 .00
.00 .00
11
3 8
1.15 -.82
.00 .00
.00 .00
12
4 9
-8.28 8.62
.00 .00
.00 .00
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13
5 10
.62 -.28
.01 -.01
-.01 .01
14
1 7
-.80 1.14
.05 .06
-.05 -.06
15
3 7
-3.58 3.92
.07 .07
-.07 -.07
16
3 9
-3.58 3.92
.07 .07
-.07 -.07
17
5 9
-.80 1.14
.05 .06
-.04 -.07
Maximum Axial Load = 862 N Maximum Shear = 320 N Check Member against Axial Load(Tension): Cross-sectional area = 1,560mm2 P 862 N T = = = 0.55 N/mm2 2 A 1,560 mm Allowable Tensile stress, fT = 0.60Fy = 0.60 x 248 MPa = 148.80 N/mm2> T, OK! Check Member against Shear: P 3260 N V = = = 0.20 N/mm2 A 1,560 mm 2 Allowable =Shear stress, fV = 0.40Fy = 0.40 x 248 MPa = 99.20 N/mm2 > V, OK! C. Lifting Lugs Check Plate Check 4 nos of Lifting Lugs to be weld connected to the Truss (Cage) Total Weight of Empty tank c/w M/SCage, ladder, hand rails etc: Empty Tank M/S Cage Total Lift Weight
= 1,530 kg = 2,300 kg ======== = 3,830 kg
Factor of Safety
=4
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Design Loading per Lifting Lugs, P = 3,830kg x 4 x 9.81m/s2 / 4 nos = 37,572.30 N
Using 12mm thick M/S Plate(A36 steel, Fy = 248MPa) a. Shear Stress Analysis Allowable Shear of plate Fv = 0.40Fy = 0.40(248) = 99.20 N/mm2 P 37 ,572 .30 = A (75 20 )12 = 56.93 N/mm2 < Fv --- OK!
Actual Shear stress on plate =
b. Tensile Stress Analysis Allowable Tensile of plate, Ft = 0.60Fy = 0.60(248) = 148.80 N/mm2 Actual Tensile stress on plate =
37 ,572 .30 P = = 16.92 N/mm2 (225 40 )12 A
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Welding Check a. Total welding length of Lifting lug to base plate, lLL = 2 x 225mm = 450mm Effective length = 0.70 x 450 = 315mm b.
Total welding length of base plate to Cage, lBP = 200 + 2x 100 + 100 = 500mm Effective length = 0.70 x 500 = 350mm
Using E60 electrode(Fu = 414MPa), Fw = 0.38Fu = 157.32 N/mm2 Since Fw of weld is higher than allowable shear capacity of plate, use plate’s value to be conservative. Therefore, Fw = Fv = 99.20 N/mm2 Throat of weld = 0.70 x 6mm Required length of weld =
37 ,572 .30 P = f w xthroat 99 .20 x(0.70 x6)
= 90.18 mm < effective length both on base plate and Lifting lugs, therefore, SAFE!
Since plate and weld check shows it can carry the weight, therefore material provided are SAFE!
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D. Handrail Design Load applied to handrail = 0.75 kN/m 1000
1000
1000
550
1100
550
Check for Horizontal Rail: M=
wL2 0.75 N / mmx 1000 mm 2 = = 93,750 N-mm 8 8
SxReqd =
M 93,750 Nmm = = 572.80 mm3 = 0.57cm3 fb 0.66 x 248MPa
Check for Vertical Rail: M = P x L = (0.75N/mm x 1000mm) x 1100mm = 825,000 N-mm SxReqd =
M 825,000 Nmm = = 5040 mm3 = 5.04cm3 fb 0.66 x 248MPa
Use of 40mm x 4mm Thk CHS (Sx = 5.92cm3) is SAFE!
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