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STRENGTH CALCULATION ANALIZING FOR FRP PANEL TANK
DIMENSION : LxWxH = 5 M X 3 M X 1 M
Example : taken the largest volume for example calculation
Assume : - the laminate construction will be considered as alternating CSM and WR
- panel shall be considered as flat panel
Bottom thickness
The minimum thickness of bottom will be calculated by the following formula. Refer to Pressure vessel
hand book by Eugene F Megyesy
tb
Where :
tb
lb
S
H
G
= lb/[ 1.254 {S/0.036 G H}0.5]
=
=
=
=
=
minimum thickness of bottom, inch
maximum distance between support, inch
Stress value of plate, psi taken 24000 Psi (ASTM D 3299)
height of tank, inch
specific gravity of liquid, taken 1
Hence
tb
= 39.37/[1.254 { 24000/(0.036 x 1 x 78,73)}0.5] = 0.34 inch = 8.63 mm
Thickness of Bottom is 8-9 mm
The minimum thickness of shell will be calculated by the following formula
With condition, when a ≈ b
BARATA PRATAMA,CV - Engineering Document
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At the Shell
Where :
Ts1
= 0.7 b {0.036 G H/ S}0.5
b
G
H
S
=
=
=
=
Ts1
= 0.7 x 39.37 { (0.036 x 1 x 78,73)/24000}0.5 = 0.275 inch = 6.98 mm
vertical pitch, inch
specific gravity of liquid, taken 1
height of tank, inch
stress value of plate material, Psi
Hence
Thickness of shell is 6-7 mm
At the Cover Panel
The thickness of panel cover will be calculated by the following formula
t
= {(α P b4)/ E
lam}
0.25
-------- using deflection criteria.
Where :
t
= thickness of panel , mm
α
= constant appropriate to combination of edge condition and type of
loading
P
= uniform internal pressure N/mm2
B
= panel dimension, shortest span, in this case will be taken as 1000
mm
E lam
= young’s modulus (N/mm2) of laminate under consideration
Thickness of laminate will be calculated by the following formula
Thickness of laminate (mm) = material mass (kg/m 2) x thickness constant
For CSM 450 gr/m2
Resin to glass ratio = 2 1/3 :1 , ρ resin = 1.2
Thickness of glass
= 0.45 x 0.391 = 0.176 mm
Thickness of resin
= 2.33 x 0.45 x 0.833 = 0.873 mm
Thickness of laminate = ( 0.176 + 0.873 ) = 1.049 mm ≈ 1
For CSM 800 gr/m2
Resin to glass ratio = 1 :1.1 , ρ resin = 1.2
Thickness of glass
= 0.8 x 0.391 = 0.313 mm
Thickness of resin
= 1.1 x 0.8 x 0.833 = 0.733 mm
Thickness of laminate = ( 0.313 + 0.733 ) = 1.04 mm ≈ 1
BARATA PRATAMA,CV - Engineering Document
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Data
:
Dimension of panel : 1000 mm x 1000 mm
Liquid
: Water
See fig below .
Panel Cover (Top View)
1000 (a)
1000 (b)
Calculate length/breadth ratio (1000/1000) = 1.0 from The table BS 2994 ; 1987 we get value of α =
0.014
Where as , E
lam
=x
lam/t
E lam
X lam
t
= young’ modulus (N/mm2) of laminate under consideration
= unit modulus N/mm
= thickness of CSM or WR which be expected, mm
E
= (16000 N/mm)/1 mm = 16000 N/mm 2
P
Tc
lam
= design external pressure, static head of liquid up to zero (0) at the top.
That is why there are for some area at the cover for man inspection may be considered as live
load in this design.
Design load approximately= 750 N per 1 (one) section area ( 1000 x 1000)mm = 0.00075
N/mm2
= {(0.014 x 0.00075 x 10004)/16000}0.25 = 5.06 mm
Thickness of cover is 5-6 mm
BARATA PRATAMA,CV - Engineering Document
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Deflection of base frame (Base Steel Construction)
Frame will be subjected to uniform load on it. Force/Load per column on frame
(F) = 1 x 1 x 2 x 1000 x 10 = 20000 N.
System will be suppose as a beam supported at the ends and loaded uniformly
Where :
δ = deflection of beams, mm
W = load on beams, N
E = Modulus Elasticity, 200000 N/mm2
I = Moment of inertia of cross section of beams, mm4
L = Length of beam, mm
Allowable deflection of span is ∂ = 1/500 x 1000 = 2 mm
Assume ---- We use UNP 100
http://www.clag.org.uk/beam.html
Moment of inertia I
I
= bh3 ∕ 12
= 50 x 1003/12 = 4166667
δ
= 3.5FL3 ∕ 384EI
= 3.5 x 20000 (1000)3/384 x 200000 x 4166667 = 0.21 mm
Condition δ ≤ ∂
0.21 mm ≤ 2 mm -- OK
We use UNP 100
Load on Tie Rod will be calculated by the following formula is
P = ab 0.036 Gh
Where : a = horizontal pitch, inch
b = vertical pitch, inch
h = height of tank, inch
G = specific gravity of liquid, taken 1
P = 39.37 x 39.37 x 0.036 x 1 x 78,73 = 4393 lb
BARATA PRATAMA,CV - Engineering Document
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Required cross sectional area of tie rod. A = P/S
Where S = stress value of rod material, psi
A = 4393 /20000 = 0.21965 inch2
Π R2 = 0.21965
R = 0.468
D = 2 x 0.468 = 0.937 inch
We use 11/4 inch for diameter tie rod (Pipe)
===============================================================
SUMMARY CALCULATION
FRP PANEL TANK
Cover thickness
Shell thickness
Bottom thickness
= 5-6 mm
= 7-8 mm
= 8-9 mm
STEEL STRUCTURE
Steel base
Tie rod
Connecting plate
= UNP 100 x 50 x 6
= Galvanized pipe 11/4 inch
= MS Plate with Galvanized 200 x 200 x t 5 mm
CONCRETE STRUCTURE
(Not included, supply by Civil Constructor)
===============================================================
Jakarta, 16 April 2012
BARATA PRATAMA,CV - Engineering Document