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8/13/2019 Design of HCl FRP Storage TANK
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DESIGN CALCULATIONFOR
HCl STORAGE TANK(2500 X 4300mm(T/T) L) BASED ON BS-4994-1987
Rev Date Revision Description Prepared by:
1 15-06-2011
a. Design DataOperating Pressure and Temperature
ERM
b. Tank Capacity 26m3to 25m3
c. Nozzle Loading onto N1 & N9
d. Live load on Platform 1000Pa to 2500 Pa
e. Inclusion of Handrail Design
2 28-07-2011 Cage(Truss) web members rearrangement 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 .L: 4300mm (T/T)
Fluid handle: HCl, 9.9% concentration
Specific gravity: 1.05Operating temperature: Ambient.
Design temperature: 200C(min) to 60
0C(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, UZa) Determine the Load limited allowable unit loading, uL.
uL= K
u
(EQ 2)where u is UTUS from Table 5.
uL,CSM= 88.11
200
= 16.84 N/mm per kg/m2 glass.
uL,WR=88.11
250= 21.04 N/mm per kg/m2 glass.
uL,FW = 88.11
500
= 42.09N/mm per kg/m2 glass.
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.
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Since UL < US, then the strain for each layer concerned shall be
determined. (clause 9.2.6 b)
= Z
L
X
u
(EQ. 4)
CSM=000,14
84.16
x 100
= 0.12%
WR=000,16
04.21x 100
= 0.13%
FW=000,28
09.42
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= X dF (EQ. 5a)
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In longitudinal direction:
uZ= XXdFX (EQ. 5b)
From Table 7, filament at winding angle 65.
F = 0.5, FX= 0.5
From Figure 3, we can obtain the Unit modulus for winding
angle 680,
X = 18,000N/mm and XX= 4,400N/mm
Therefore,
(i) Circumferential uZ= X dF (EQ 5a)
= 16000 x 0.12/100 x 0.5
= 9.60 N/mm per kg/m2 glass.
(ii) Longitudinal uZ= XXdFX
= 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, Q= pDi/2 (EQ 7)
SG of liquid = 1.05
P = 0.034N/mm
2
Di = 2500mm
Q=2
)2500(034.0= 42.50N/mm
Use 8 layers of CSM as chemical barrier, 2 layer of WR and 4 layers of filamentroving as reinforced layer.
Check:
Circumferential ULAM= (8 x 0.45 x 16.8) + (2 x 0.80 x 19.2) + (4 x 1.1 x 9.60)= 133.44N/mm > Q ------ OK!
ii) Longitudinal unit load, Qx=4
pDi +2
4DiM
(EQ11)
Tank Empty Weight = 1,530 kg (FRP Tank)
Liquid Weight (full) = 1.05 x 1000kg/m3x 25m
3
= 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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39.71 kN-m
19.38 kN-m
39.71 kN-m
61.09 kN
50.29 kN
14.50 kN-m 14.50 kN-m
24.89 kN
24.88 kN
61.09 kN
50.29 kN
9.93 kN-m 9.93 kN-m
1300mm 1600mm 1600mm 1300mm
Pressure due to liquid, p = 0.034N/mm2:
Q =4
pDi
2
4
Di
M
Q =4
)2500(034.0
2
6
)2500(
)1071.39(4
Nmmx
Qx = 29.34N/mm ; 13.16 N/mm
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Pressure due to vacuum, p = -0.005N/mm2:
Q =4
pDi
2
4
Di
M
Q =4
)2500(005.0 2
6
)2500()1071.39(4
Nmmx
Qx = 4.96N/mm ; -11.21 N/mm
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=
L
V=
2500360/120
380,111
xx
N
= 42.54 N/mm
b) Due to weight onto nozzle N1 and N9:Qc=
4
pDi;
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;
Effective area =4
92 2 xmm= 6,648mm
2
Therefore p =2
2
648,6
/81.920
mm
smkgx= 0.030 N/mm
2
Qc=4
pDi=
4
2500030.0 x=18.75 N/mm
Since compressive unit load due to saddle support (Qc= 42.54 N/mm) isgreater 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
Qp =FDo
tXLAM6.0 (EQ 13)
Qp =)1422500(4
)41.1400,4280.0000,16845.0000,14)(14(6.0
x
xxxxxx
= 79.22N/mm > Qc ------- OK!
iv) Check minimum permissible thickness, tm, to prevent buckling due toexternal pressure or vacuum:
L = 0.4 x 750 + 4300 = 4,900mmDo = 2500+2x14 = 2,528mm
pvacuum = 0.005 N/mm2
F = 4
ELAM=
14
410.1400,4280.0000,16845.0000,14 xxxxxx
t
XLAM (EQ14)
= 6,811.43
94.12528
900,4
oD
L;
77.11
4005.0
43.811,635.135.1
17.017.0
x
x
pF
ELAM ;
SinceoD
L Q= 55.26 N/mm ---------OK!
ii) Check minimum permissible thickness against buckling, tm:
LAM
omE
pFRt 7.1 (EQ 19)
But eoo KDR 5.0 (EQ 46)
and Ke = 1.50 (from Figure 15, BS4994:1987)
896,150.1528,25.0 xxRo
ELAM=14
380.0000,161145.0000,14 xxxx
t
XLAM
= 8,975
Therefore,
975,8
4034.0896,17.17.1
xx
E
pFRt
LAM
om =12.55 mm < 14mm, OK!
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764
2800
iii) Check against vacuum, p =-0.005 N/mm2
Q= 0.66pDi Ks (EQ 43)= 0.66(0.005)(2500)(1.30)
= 10.725 N/mm < Qact, OK!
3) Mild Steel Saddle Support.
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 6.5Maximum 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 =A
P
Where A = 2280mm2(C-Channel cross section)
c =A
P=
22280
617,66
mm
= 29.22N/mm2< Fc ------- OK!
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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/mm
2
Actual Compressive stress, c =A
P=
mmmmx
N
102800
681,181= 6.49N/mm
2
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Max bending Moment, M =8
2wl
=8
800,248.8 2x=8,310,400 N-mm
SxReqd=Fy
M
66.0=
MPax
mmN
24866.0
400,310,8 =50,772mm
3= 50.77cm
3
Use C-channel 125x65x6.0mm Thk (Sx = 89.40cm3>SxReqd), is OK!
B. Design of Truss Membersa. 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/mm2Uniform 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 NP2 = 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 loadSaddle 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:
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T1 = R1 + 2,000N = 8,140 N
T2 = R2 + 2,000N = 4,740 N
3.37 kN
2.77 kN
1.37 kN
1.37 kN
3.37 kN
2.77 kN
1300mm 1600mm 1600mm 1300mm
At Tank Operation:
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SUPPORT REACTIONS:
JOINT FORCE-Y (kN)
2 12.01
3 26.014 12.01
MEMBER END FORCES
Unit: Force: kN
MEMBER JT AXIAL SHEAR-Y SHEAR-Z
1 1 2.58 -.02 .022 -2.58 .13 -.13
2 2 .00 .09 -.09
3 .00 .04 -.04
3 3 .00 .04 -.04
4 .00 .09 -.09
4 4 2.58 .13 -.13
5 -2.58 -.02 .02
5 6 .00 .16 -.16
7 .00 .33 -.33
6 7 -6.15 .32 -.31
8 6.15 .29 -.29
7 8 -6.15 .29 -.29
9 6.15 .32 -.32
8 9 -.01 .33 -.33
10 .01 .16 -.16
9 1 5.39 .00 .00
6 -5.06 .00 .00
10 2 11.69 .01 -.017 -11.36 -.01 .01
11 3 13.04 .00 .00
8 -12.70 .00 .00
12 4 11.69 -.01 .00
9 -11.36 .01 -.01
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13 5 5.39 .00 .00
10 -5.05 .00 .00
14 1 -5.95 .06 -.06
7 6.28 .05 -.05
15 3 7.34 .07 -.07
7 -7.01 .07 -.06
16 3 7.34 .07 -.07
9 -7.01 .07 -.06
17 5 -5.94 .06 -.05
9 6.28 .05 -.06
Maximum Axial Load = 12,700 NMaximum Shear = 330 N
Check Member against Axial Load(Tension):
Cross-sectional area = 1,560mm2
T=A
P=
2560,1
700,12
mm
N= 8.14 N/mm
2
Allowable Tensile stress, fT= 0.60Fy
= 0.60 x 248 MPa = 148.80 N/mm2>T, OK!
Check Member against Shear:
V=A
P=
2560,1
330
mm
N= 0.21 N/mm2
Allowable ==Shear stress, fV= 0.40Fy= 0.40 x 248 MPa = 99.20 N/mm
2> V, OK!
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During Lifting of Tank:
SUPPORT REACTIONS:
JOINT FORCE-Y (kN)
7 14.00
9 14.00
MEMBER END FORCES
Unit: Force: kN
MEMBER JT AXIAL SHEAR-Y SHEAR-Z
1 1 .38 .06 -.062 -.38 .05 -.05
2 2 .37 .05 -.05
3 -.37 .08 -.08
3 3 .37 .08 -.08
4 -.37 .05 -.05
4 4 .38 .05 -.05
5 -.38 .06 -.06
5 6 .02 .20 -.20
7 -.02 .30 -.30
6 7 .00 .32 -.32
8 .00 .29 -.29
7 8 .00 .29 -.29
9 .00 .32 -.32
8 9 .02 .30 -.30
10 -.02 .20 -.20
9 1 .62 -.01 .01
6 -.28 .01 -.01
10 2 -8.28 .00 .007 8.62 .00 .00
11 3 1.15 .00 .00
8 -.82 .00 .00
12 4 -8.28 .00 .00
9 8.62 .00 .00
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13 5 .62 .01 -.01
10 -.28 -.01 .01
14 1 -.80 .05 -.05
7 1.14 .06 -.06
15 3 -3.58 .07 -.07
7 3.92 .07 -.07
16 3 -3.58 .07 -.07
9 3.92 .07 -.07
17 5 -.80 .05 -.04
9 1.14 .06 -.07
Maximum Axial Load = 862 N
Maximum Shear = 320 N
Check Member against Axial Load(Tension):
Cross-sectional area = 1,560mm2
T=A
P=
2560,1
862
mm
N= 0.55 N/mm
2
Allowable Tensile stress, fT= 0.60Fy
= 0.60 x 248 MPa = 148.80 N/mm2>T, OK!
Check Member against Shear:
V=
A
P=
2
560,1
3260
mm
N= 0.20 N/mm2
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 = 1,530 kgM/S Cage = 2,300 kg
========
Total Lift Weight = 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 AnalysisAllowable Shear of plate Fv = 0.40Fy
= 0.40(248)= 99.20 N/mm
2
Actual Shear stress on plate =A
P=
12)2075(
30.572,37
= 56.93 N/mm2< Fv --- OK!
b. Tensile Stress AnalysisAllowable Tensile of plate, Ft = 0.60Fy
= 0.60(248)= 148.80 N/mm
2
Actual Tensile stress on plate =A
P=
12)40225(
30.572,37
= 16.92 N/mm
2
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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
plates value to be conservative.
Therefore, Fw = Fv = 99.20 N/mm2
Throat of weld = 0.70 x 6mm
Required length of weld =
xthroatf
P
w
=
)670.0(20.99
30.572,37
xx
= 90.18 mm < effective length both on baseplate and Lifting lugs, therefore, SAFE!
Since plate and weld check shows it can carry the weight, thereforematerial provided are SAFE!
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D. Handrail Design
Load applied to handrail = 0.75 kN/m
1000 1000 1000
1100
550
550
Check for Horizontal Rail:
M =8
2wL
=8
1000/75.0 2mmmmxN= 93,750 N-mm
SxReqd=fb
M=
MPax
Nmm
24866.0
750,93= 572.80 mm
3= 0.57cm
3
Check for Vertical Rail:
M = P x L = (0.75N/mm x 1000mm) x 1100mm
= 825,000 N-mm
SxReqd=fbM =
MPaxNmm
24866.0000,825 = 5040 mm3= 5.04cm3
Use of 40mm x 4mm Thk CHS (Sx = 5.92cm3) is SAFE!