design of hcl frp storage tank

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


(ii) Longitudinal uZ= XXdFX

= 4400 x 0.12/100 x 0.5 (EQ 5b)


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!

design of hcl frp storage tank

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