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CONTENT PAGE
1 DESIGN DATA 1
3 WALL DESIGN 2
4 STIFFENER PROPERTIES 20
5 NOZZLES & OPENING 21
7 WEIGHT SUMMARY 38
8 WIND LOADING 39
9 TRANSPORTATION LOAD 40
10 LOAD AT BASE 41
11 LEG DESIGN 42
12 LEG BASEPLATE DESIGN 43
13 LIFTING LUG DESIGN CALCULATION 44
APPENDIX
ROARK'S FORMULA 47
PRESSURE VESSEL HANDBOOK 48
SHACKLE 49
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( Cd' x qz x Az ) x 103
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DESIGN DATA
DESCRIPTION :
TAG NO. : T-6500MANUFAC'S. SERIAL NO : PV-725
DIMENSION ( mm ) : 2520 mm (W) x 2520 mm (L) x 2020 mm (H)
DESIGN CODE : ASME SECT. VIII. DIV. I, (2004 EDITION) + ROARK'S FORMULA
CODE STAMPED : NO
THIRD PARTY : NO
PROPERTIES UNIT DATA
CAPACITY
CONTENT - SEAWATER / OIL
FLUID SPECIFIC GRAVITY - 1.00
DESIGN PRESSURE bar g FULL WATER + 0.05 / - 0.02 BARG
DESIGN TEMPERATURE, 60
HYDROSTATIC TEST PRESSURE bar g FULL OF WATER + 300mm STAND PIPE
IMPACT TEST - NO
RADIOGRAPHY - 10%
CORROSION ALLOWANCE mm in 3.0 0.12
mm
3
1.25 x 10
10
oC
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ROARK'S FORMULA
SIDE WALL DESIGN
ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
Tank Height, H = 39.37 in 1000 mm
Tank Width, W = 39.37 in 1000 mm
Tank Length, L = 59.06 in 1500 mm
Design Pressure = FULL WATER + 0.01 bar g
Design Temp. = 65
Material = SA 516M GR 485
As per Table 26 Case No.1a Chapter 10 of Roark's
Rectangular plate, all edges simply supported, with uniform loads over entire plate.
For Section , A (Worst Case)g = 9.81
1000
a = 19.69 in 500.0 mm
b = 19.69 in 500.0 mm
a/b = 1.0000
= 0.2994 Loading q= + Pa
= 0.0462 = 9810 + 750 = 0.4320 = 1.4225 + 0.1088 psi = 2.84E+07 psi = 1.5312 psi
t = 0.1969 in 5.0 mm
c.a = 0.1181 in 3 mmt (corr) = 0.3150 in 8.0 mm
At Center,
Maximum Deflection, =
= -1.25
= 1.25 mm < t/2 then O.K
= 4584 psi < 25081 psi. then OK
Material SA 516M GR 485
38002 psi
0.121
At center of long side,
Maximum reaction force per unit length normal to the plate surface,
R =
= 13.02 lb/in
= 1471.24 N/mm
o C
m/s2
water
= kg/m3
water
gH water
gh
N/m2
-(qb4)/Et3
Maximum Bending stress, = (qb2)/ t2
allowable
Yield Stress, y =
Stress Ratio, /y =
qb
S
a
S
S
Sb
B
1000
1500
500
500 x 3
500
A
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ROARK'S FORMULA
SIDE WALL DESIGN
ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
Tank Height, H = 39.37 in 1000 mm
Tank Width, W = 39.37 in 1000 mm
Tank Length, L = 59.06 in 1500 mm
Design Pressure = FULL WATER + 0.01 bar g
Design Temp. = 65
Material = SA 516M GR 485
As per Table 26 Case No.1a Chapter 10 of Roark's
Rectangular plate, all edges simply supported, with uniform loads over entire plate.
For Section , B (Worst Case)g = 9.81
1000
a = 24.80 in 630.0 mm
b = 26.50 in 673.0 mm
a/b = 0.9361
= 0.2749 Loading q= + Pa
= 0.0413 = 9810 + 750 = 0.4247 = 1.4225 + 0.1088 psi = 2.84E+07 psi = 1.5312 psi
t = 0.2756 in 7.0 mm
c.a = 0.1181 in 3 mmt (corr) = 0.3937 in 10.0 mm
At Center,
Maximum Deflection, =
= -1.33
= 1.33 mm < t/2 then O.K
= 3890 psi < 25081 psi. then OK
Material SA 36M
38001.5 psi
0.102
At center of long side,
Maximum reaction force per unit length normal to the plate surface,
R =
= 17.23 lb/in
= 1946.61 N/mm
o C
m/s2
water
= kg/m3
water
gH water
gh
N/m2
-(qb4)/Et3
Maximum Bending stress, = (qb2)/ t2
allowable
Yield Stress, y =
Stress Ratio, /y =
qb
S
a
S
S
Sb
B
674
2020
2520
673
630 X 4
A
673
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STIFFENER CALCULATION
For Horizontal
Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,
that is, at the middle of the beam.
Stiffener No. 1 (typical)
L = 500 mm = 19.69 in
30.14 lb/in = 500 mm = 19.7 in
Load q = 1.5312 psi
unit load W = q x psi
= 30.14 lb/in
FB 50 x 9
X
I = 0.7333
19.69 in
Bending Moment
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x = L/2 = 9.84 in
Maximum moment, =
= 1460 lb-in
M/I =
=
= 0.088
Use FB 50 x 9I/y = 1.296 > then O.K
Therefore, = 1127 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x =L/2= 9.84 in
384EI
= 0.003 < L/360 = 0.0547 in. then O.K
Therefore the size used is adequate.
in4
Mmax WL2/8
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
(5WL4)
WbWa
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STIFFENER CALCULATION
For Vertical
Stiffener No. 1 (typical)
L = 500 mm = 19.69 in
30.14 lb/in = 500 mm = 19.7 in
Load q = 1.5312 psi
unit load W = q x psi
= 30.14 lb/in
FB 50 x 9
X
I = 0.7333
19.69 in
Bending Moment
As per Table 8.1 Case 2d of Roark's (Uniformly increasing load)At x = 0.548L = 10.79 in
Maximum moment, =
= 251 lb-in
M/I =
=
= 0.015
Use FB 50 x 9
I/y = 1.296 > then O.K
Therefore, = 194 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2d of Roark's (Uniformly increasing load)
At x = 0.525L = 10.33 in
=
EI
= 0.0003 < L/360 = 0.0547 in. then O.K
Therefore the size used is adequate.
in4
Mmax 0.0215WL2
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
0.001309WL4
WbWa
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STIFFENER CALCULATION
For Horizontal
Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,
that is, at the middle of the beam.
Stiffener No. 1 (typical)
L = 630 mm = 24.80 in
40.57 lb/in = 673 mm = 26.5 in
Load q = 1.5312 psi
unit load W = q x psi
= 40.57 lb/in
FB 50 x 9
X
I = 0.7333
24.80 in
Bending Moment
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x = L/2 = 12.40 in
Maximum moment, =
= 3120 lb-in
M/I =
=
= 0.189
Use FB 50 x 9I/y = 1.296 > then O.K
Therefore, = 2408 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x =L/2= 12.40 in
384EI
= 0.010 < L/360 = 0.0689 in. then O.K
Therefore the size used is adequate.
in4
Mmax WL2/8
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
(5WL4)
WbWa
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STIFFENER CALCULATION
For Vertical
Stiffener No. 1 (typical)
L = 673 mm = 26.50 in
37.98 lb/in = 630 mm = 24.8 in
Load q = 1.5312 psi
unit load W = q x psi
= 37.98 lb/in
FB 50 x 9
X
I = 0.7333
26.50 in
Bending Moment
As per Table 8.1 Case 2d of Roark's (Uniformly increasing load)At x = 0.548L = 14.52 in
Maximum moment, =
= 573 lb-in
M/I =
=
= 0.035
Use FB 50 x 9
I/y = 1.296 > then O.K
Therefore, = 442 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2d of Roark's (Uniformly increasing load)
At x = 0.525L = 13.91 in
=
EI
= 0.0012 < L/360 = 0.0736 in. then O.K
Therefore the size used is adequate.
in4
Mmax 0.0215WL2
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
0.001309WL4
WbWa
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STIFFENER PROPERTIES
Size FB 50 x 9
Material, SS 316L
Yield Stress, 25000 psi
Allowable Stress, 16500 psi
Where :
d1 = 6 mm
d2 = 50 mm
b1 = 110 mm *
b2 = 9 mm
C
PART area (a) y a x y h
mm mm
1 657.89 3 1973.66 11.37 129.34 85094.47 1973.66
2 450 31 13950 16.63 276.46 124405.83 93750
TOTAL 1107.89 15923.66 209500.3 95723.66
Therefore,
14.37 mm
I = 305223.97 = 0.7333
Z (I/C) = 21235.94 = 1.2959
"*"b1 =
b1 = 110 mm take min. L = 1000 mm, so R:
R = 500 mm
y
allowable
h2 a x h2 bd3/12
mm2 mm3 mm2 mm4 mm4
C =
mm4 in4
mm3 in3
h1
b2
b1
h2
y2 C
d1
d2
y1 1
2
Plate
Stiffener
tr1 .56Rts
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ROARK'S FORMULA
BOTTOM PLATE DESIGN
ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
Tank Height, H 39.4 in 1000 mm
Tank Width, W 39.37 in 1000 mm
Tank Length, L 59.06 in 1500 mm
Design Pressure = FULL WATER + 0.01 BARG
Design Temp. = 65
Material = SA 516M GR 485
As per Table 26 Case No.1a Chapter 10 of Roark's
Assume rectangular plate, all edges simply supported, with uniform loads over entire plate
For Section , Each Section (Largest area)g = 9.81
1000
a = 19.69 in 500 mm
b = 19.69 in 500 mm
a/b = 1.0000
= 0.2874 Loading q= + 0.01 BARG
= 0.0444 = 9810 + 750 = 0.4200 = 1.4225 + 0.1088 psi = 2.80E+07 psi = 1.5312 psi
t = 0.1969 in 5.0 mm
c.a = 0.1181 in 3 `t (corr) = 0.3150 in 8.0 mm
At Center,
Maximum Deflection, =
= -1.21
= 1.21 mm < t/2 then O.K
= 4401 psi < 25081 psi. then OK
Material SA 516M GR 485
38001.5 psi
0.116
At center of long side,
Maximum reaction force per unit length normal to the plate surface,
R =
= 12.66 lb/in
o C
m/s2
water
= kg/m3
water
gh
N/m2
-(qb4)/Et3
Maximum Bending stress, = (qb2)/ t2
allowable
Yield Stress, y =
Stress Ratio, /y =
qb
S
a
S
S
Sb
A1000
1500
500 x 2
500 x 3
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STIFFENER CALCULATION (at Bottom Plate)
For Short Beam
Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,
that is, at the middle of the beam.
L = 500 mm = 19.69 in
30.14 lb/in = 500 mm = 19.7 in
Load q = 1.5312 psi
unit load W = q x psi
= 30.14 lb/in
FB 50 x 9
X
I = 0.733
19.69 in
Maximum bending moment,
At x = L/2 = 9.84 in
=
= 1460 lb-in
M/I =
=
= 0.088
Use FB 50 x 9
I/y = 1.296 > then O.K
Therefore, = 1127 psi < 16500 psi. then OK
Maximum Deflection at Center of Beam
At x = L/2 = 9.84 in
=
384EI
= 0.0028 < L/360 = 0.0547 in. then O.K
Therefore the size used is adequate.
in4
Mmax WL2/8
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
(5WL4)
WbWa
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ROARK'S FORMULA
ROOF CALCULATION
ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
Assume rectangular plate, all edges simply supported, with uniform loads over entire plate
Live load, LL = 0.2846 psi
Roof weight = 206.353 lb
Structure weight = 470 lb
Concentrated weight = 661 lb
Total dead load,TDL = 0.2910 psi
Total conc. load, CL = 0.2845 psi
Tank Width, W 39.37 in 1000 mm
Tank Length, L 59.06 in 1500 mm
g = 9.81
1000
a = 19.69 in 500.0 mm
b = 19.69 in 500.0 mm
a/b = 1.0000
= 0.29
= 0.0444 Loading q = LL + CL + TDL = 0.4200 = 0.860 psi
= 2.80E+07 psit = 0.12 in 3.0 mm
c.a = 0.12 in 3 mm
t (corr) = 0.24 in 6.0 mm
At Center,
Maximum Deflection, = All. Deflection =1500/300= 5.00 (max) mm
= -3.16 mm
= 3.16 mm < All. Deflection. O.K = 5.00
= 6866 psi < 25081 psi then OK
Material SA 516M GR 485
38001.5 psi
0.181
At center of long side,
Maximum reaction force per unit length normal to the plate surface,
R =
= 7.11 lb/in
m/s2
water
= kg/m3
-(qb4)/Et3
Maximum Bending stress, = (qb2)/ t2
allowable
Yield Stress, y =
Stress Ratio, /y =
qb
S
a
S
S
Sb
A
1000
1280
500 x 2
500 x 3
1500
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STIFFENER CALCULATION(at Roof Plate)
Short Beam
Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,
that is, at the middle of the beam.
L = 500 mm = 19.69 in
16.93 lb/in = 500 mm = 19.7 in
Load q = 0.860 psi
unit load W = q x psi
= 16.93 lb/in
FB 50 x 9
X
I = 0.7333
19.69 in
Maximum moment,
At x = L/2 = 9.84 in
=
= 2 lb-in
M/I =
== 9.487E-05
Use FB 50 x 9
I/y = 1.296 > then O.K
Therefore, = 1.208 psi < 16500 psi. then O.K
Maximum deflection at center of beam
At x = L/2 = 9.84 in
=
384EI
= 0.002 < L/360 = 0.0547 in. then O.K
Therefore the size used is adequate.
in4
Mmax WL2/8
/y
(I/y)required
M/
in3
in3 (I/y)required
allowable
max
(5WL4)
WbWa
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ROARK'S FORMULA
BASE WEIR PLATE DESIGN
ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
Weir Plate Height, H = 35.43 in 900 mm
Weir Plate Width, W = 98.43 in 2500 mm
Design Pressure = FULL WATER + 0.01 bar g
Design Temp. = 65
Material = SA 516M GR 485
As per Table 26 Case No.1a Chapter 10 of Roark's
Rectangular plate, all edges simply supported, with uniform loads over entire plate.
For Section , A (Worst Case)g = 9.81
1000
a = 24.61 in 625.0 mm
b = 17.72 in 450.0 mm
a/b = 1.3889
= 0.4487 Loading q= + Pa
= 0.0761 = 8829 + 750 = 0.4767 = 1.2802 + 0.1088 psi = 2.84E+07 psi = 1.3890 psi
t = 0.1969 in 5.0 mm
c.a = 0.1181 in 3 mmt (corr) = 0.3150 in 8.0 mm
At Center,
Maximum Deflection, =
= -1.22
= 1.22 mm < t/2 then O.K
= 5048 psi < 25081 psi. then OK
Material SA 36M
38001.5 psi
0.133
At center of long side,
Maximum reaction force per unit length normal to the plate surface,
R =
= 11.73 lb/in
= 1325.45 N/mm
o C
m/s2
water
= kg/m3
water
gH water
gh
N/m2
-(qb4)/Et3
Maximum Bending stress, = (qb2)/ t2
allowable
Yield Stress, y =
Stress Ratio, /y =
qb
S
a
S
S
Sb
900
2500
450
625 X 4
A
450
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STIFFENER CALCULATION
For Horizontal
Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,
that is, at the middle of the beam.
Stiffener No. 1 (typical)
L = 625 mm = 24.61 in
27.13 lb/in = 450 mm = 17.7 in
Load q = 1.5312 psi
unit load W = q x psi
= 27.13 lb/in
FB 50 x 9
X
I = 0.7333
24.61 in
Bending Moment
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x = L/2 = 12.30 in
Maximum moment, =
= 2053 lb-in
M/I =
=
= 0.124
Use FB 50 x 9I/y = 1.296 > then O.K
Therefore, = 1584 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x =L/2= 12.30 in
384EI
= 0.006 < L/360 = 0.0684 in. then O.K
Therefore the size used is adequate.
in4
Mmax WL2/8
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
(5WL4)
WbWa
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ROARK'S FORMULA
ADJUSTABLE WEIR PLATE DESIGN
ITEM NAME : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
Weir Plate Height, H = 23.62 in 600 mm
Weir Plate Width, W = 98.43 in 2500 mm
Design Pressure = FULL WATER + 0.01 bar g
Design Temp. = 65
Material = SA 516M GR 485
As per Table 26 Case No.1a Chapter 10 of Roark's
Rectangular plate, all edges simply supported, with uniform loads over entire plate.
For Section , A (Worst Case)g = 9.81
1000
a = 16.46 in 418.0 mm
b = 12.60 in 320.0 mm
a/b = 1.3063
= 0.4170 Loading q= + Pa
= 0.0698 = 5886 + 750 = 0.4672 = 0.8535 + 0.1088 psi = 2.84E+07 psi = 0.9622 psi
t = 0.1185 in 3.0 mm
c.a = 0.1181 in 3 mmt (corr) = 0.2366 in 6.0 mm
At Center,
Maximum Deflection, =
= -0.91
= 0.91 mm < t/2 then O.K
= 4535 psi < 25081 psi. then OK
Material SA 36M
38001.5 psi
0.119
At center of long side,
Maximum reaction force per unit length normal to the plate surface,
R =
= 5.66 lb/in
= 639.95 N/mm
o C
m/s2
water
= kg/m3
water
gH water
gh
N/m2
-(qb4)/Et3
Maximum Bending stress, = (qb2)/ t2
allowable
Yield Stress, y =
Stress Ratio, /y =
qb
S
a
S
S
Sb
600
2500
315
416
A
285
416
416416
418418
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STIFFENER CALCULATION
For Horizontal
Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero,
that is, at the middle of the beam.
Stiffener No. 1 (typical)
L = 418 mm = 16.46 in
19.29 lb/in = 320 mm = 12.6 in
Load q = 1.5312 psi
unit load W = q x psi
= 19.29 lb/in
FB 50 x 9
X
I = 0.7333
16.46 in
Bending Moment
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x = L/2 = 8.23 in
Maximum moment, =
= 653 lb-in
M/I =
=
= 0.040
Use FB 50 x 9I/y = 1.296 > then O.K
Therefore, = 504 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2e of Roark's (Uniform load on entire span)
At x =L/2= 8.23 in
384EI
= 0.001 < L/360 = 0.0457 in. then O.K
Therefore the size used is adequate.
in4
Mmax WL2/8
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
(5WL4)
WbWa
8/6/2019 Rect Tanks Sample Calculation
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STIFFENER CALCULATION
For Vertical
Stiffener No. 1 (typical)
L = 320 mm = 12.60 in
25.20 lb/in = 418 mm = 16.5 in
Load q = 1.5312 psi
unit load W = q x psi
= 25.20 lb/in
FB 50 x 9
X
I = 0.7333
12.60 in
Bending Moment
As per Table 8.1 Case 2d of Roark's (Uniformly increasing load)At x = 0.548L = 6.90 in
Maximum moment, =
= 86 lb-in
M/I =
=
= 0.005
Use FB 50 x 9
I/y = 1.296 > then O.K
Therefore, = 66 psi < 16500 psi. then O.K
Deflection
As per Table 8.1 Case 2d of Roark's (Uniformly increasing load)
At x = 0.525L = 6.61 in
=
EI
= 0.0000 < L/360 = 0.0350 in. then O.K
Therefore the size used is adequate.
in4
Mmax 0.0215WL2
/y
(I/y)required M/
in3
in3 (I/y)required
allowable
max
0.001309WL4
WbWa
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WIND LOADING - BS 6399 - PART 2 -1997
Terrain Category = 1
Region = D
Basic Wind Speed Vb = 50.00 m/s
Shielding Factor Ms = 1
Topographic Factor Sa = 1
Direction Factor Sd = 1
Probability Factor Sp = 1
Seasonal Factor Ss = 1
Terrain and Building Factor Sb = 1
Design Wind Speed Vz = 50.00 m/s ( Vb x Sa x Sd x Sp x Ss )
Effective (Design) Wind speed Ve = 50.00 m/s ( Vz x Sb )
Dynamic Pressure qz = 1.5325
Drag Coefficient Cd = 1
H = 1000.000 mm
D = 1000.000 mm
Az = 1000000.000
1000
H / D = 1.00
1000.000 Kar = 1
Cd' = 1 ( Cd x Kar )
Wind Force Fw = 1532.5 N
Height to COA h = 500.000 mm ( H / 2 )
Overturning Moment Mw = 766250 Nmm ( Fw x h )
Moment at the joint of the leg to the tank
550 mm = 998040 Nmm
kPa ( 0.613 x Ve2 x 10-3 )
mm2
( Cd' x qz x Az ) / 103
hT
= Mw1
Mw - hT( Fw - 0.5*qz*D*h
T)
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WEIGHT SUMMARY
ITEM : T-1060,T-1070,T-1080,T-1090 RECT. TANKS
JOB NO. JN05-320
QTY or
ITEM DESCRIPTION UNIT WT. WEIGHT
SIDE PLATE 2.520 m x 2.000 m x 10 thk 4 1562.5 kg
BASE PLATE 2.520 m x 2.520 m x 12 thk 1 590.6 kg
ROOF PLATE 2.520 m x 2.520 m x 8 thk 1 393.7 kg
PARTITION / WEIR PLATE - - - kg
STIFFENER
SIDE WALL FB 50 x 9 x 44.4 m 1 154.9 kg
ROOF PLATE FB 50 x 9 x ### m 1 52.7 kg
BOTTOM PLATE FB 50 x 9 x ### m 1 52.7 kg
WEIR PLATE 2.500 m x 1.400 m x 8 thk 1 217.0 kg
ANGLE 75 x 75 x 9t x 3.0 m 1 29.9 kg
NOZZLE / OPENINGS 500.0 kg
AND OTHERS
TOTAL WEIGHT 3554 kg
Liquid Weight 12701 kgWater Weight 12701 kg
EMPTY WEIGHT 3554 kg
OPERATING WEIGHT 16255 kg
FULL WATER WEIGHT 16255 kg
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TRANSPORTATION LOADS
TRANSPORTATION ACCELERATIONS
WEIGHTS
ERECTED .. We = 3554 kg -> 34865 N
OPERATING . Wo = 16255 kg -> 159460 N
FLOODED . Wf = 16255 kg -> 159460 N
VERTICAL = 13.73 m/s ( 1.4 x g )
LONGITUDINAL = 4.91 m/s ( 0.5 x g )
TRANSVERSE = 4.91 m/s ( 0.5 x g )
TRANSPORTATION FORCES
VERTICAL = 48811.4 N
HORIZONTAL = 17432.6 N
TRANSVERSE = 17432.6 N
AtV
AtH
AtT
FtV
( We x AtV
)
FtH
( We x AtH
)
FtT
( We x AtT
)
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LOADS AT BASE
WEIGHTS
Erected We = 3554 kg ------> 34865 N
Operating Wo = 16255 kg ------> 159460 N
Flooded Wf = 16255 kg ------> 159460 N
EXTERNAL LOADS
Wind Force Fw = 1533 N
Earthquake Force Feq = 0 N
Blasting Force Fb = 0 N
Client Specified Dynamic Force = 51831 N
( during tow-out and installation )
Wind Moment Mw = 766250 Nmm
Earthquake Moment Meq = 0 Nmm
Blasting Moment Mb = 0 Nmm
Client Specified Moment Mc = 68676 Nmm 1325 mm
( during tow-out and installation ) (from base)
Maximun Shear Force F = 51831 N >>> P = F/n = 12957.7 N
Maximun O/T Moment M = 766250 Nmm n = 4
HOLD DOWN BOLTS
Bolt Material.. = SA 193M GR B7
Bolt Yield Stress Sy = 207 MPa
Bolt UTS... Su = 507 MPa
Allowable Tensile Ft = 124.2 MPa
Allowable Shear Fs = 69 MPa
Bolt Size = M20
Bolt Number N = 4
Tensile Area. = 245
Shear Area = 225
Bolt PCD PCD = 2258 mm
AXIAL STRESS IN BOLT SHEAR STRESS IN BOLT
Load / Bolt, P = 4Mw - We Shear / Bolt, S = F
PCD.N N N x As
Load / Bolt = -8377 N fs = 57.59 MPa OK
** Since the value is -ve, therefore no axial stress Fs = 69 MPa
since fs < Fs the shear stress is OK
FD [( 0.5 x We )
2 + ( 1.4 x We )2 ]0.5
( FD
x COGerected
) COGerected
=
where, n = no of leg.
AT mm2
AS mm2
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LEG DESIGN
LEG DATA
Material.....= SA 36M
Yield Stress, Sy...= 248.2Allowable Axial Stress, fall....= 148.9
Allowable Bending Stress, fball.......= 165.5
LEG GEOMETRY :- ANGLE 90 x 90 x 8t
A = 1390
Ixx = 1040000
d = 50 mm
e = 25 mm
L = 450 mm
r = 11 mm
AXIAL STRESS
Axial Stress, fa = F / A = 28.68
BENDING STRESS
Bending Stress, fb = P x L x e = 16.58
Ixx
COMBINED STRESS
Combined Stress, f = (fa/fall + fb/fball) = 0.29
Since Combined Stress is < 1.00 The Leg Design is OK!
N/mm
2
N/mm2 ( 0.6 x Sy )
N/mm2 ( 2/3 x Sy )
mm2
mm4
N/mm2
N/mm2
e
d
X X
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LEG BASEPLATE DESIGN
Refer Dennis R Moss Procedure 3-10
tb = 3 x Q x F
4 x A x Fb
Q = Maximum Load / Support = 16255 N
F = Baseplate Width = 150 mm
A = Baseplate Length = 150 mm
Fb = Allowable Bending Stress = 163.68 MPa ( 0.66 Fy )
tb = 8.6 mm
Use Tb = 16 mm OK
BASE PLATE WELD CHECKING
Maximum stress due to Q & F = max(Q, F)/Aw = 10.80
< 86.9 OK
Weld leg size, g = 8.0 mm
Length of weld, l = 2*( 2*F + 2*A ) = 1200 mm
Area of weld, Aw = 0.5*g*l = 4800
Joint efficiency for fillet weld, E = 0.6 -
Welding stress for steel, fw = 144.8
Allowable stress for weld, fw = E*fw = 86.9Maximum vertical force, Q = 16254.9 N
Maximum horizontal force, F = 51831.0 N
N/mm2
N/mm2
mm2
N/mm2
N/mm
2
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LIFTING LUG DESIGN CALCULATION
tL
rL d
Weight of tank, We = 3,554 kg
= 34,865 N
Number of lifting lug, N = 4
1.1 LIFTING LUG
Distance, hc = 85 mm
Distance k = 106 mm
Distance J = 48 mmDistance M = 50 mm
Lug radius, rL = 50 mm
Diameter of hole, d = 40 mm
Lug thickness, tL = 15 mm
Plate thickness, tM = 9 mm
Length a = 100 mm
Length b = 60 mm
Pad length, Lp = 150 mm
Pad width, Wp = 100 mm
Pad thickness, tp = 6 mm
Angle, U (max) = 15
Shackle S.W.L : 4.75 tons
Type of shackle : Chain Shackles
Pin size, Dp = 22.225 mm
2.0 LIFTING LUG MATERIAL & MECHANICAL PROPERTIES
Material used =SA 240M GR 316 L
Specified yield stress, Sy = 248.22 N/mm
Impact load factor, p = 3.00
3.0 ALLOWABLE STRESSESAllowable tensile stress, St.all ( = 0.6 Sy ) = 148.93 N/mm
Allowable bearing stress, Sbr.all ( = 0.9 Sy ) = 223.40 N/mm
Allowable bending stress, Sbn.all ( = 0.66 Sy ) = 163.83 N/mm
Allowable shear stre = 99.29 N/mm( Cd' x qz x Az ) x 103
tp
Wpb
a
Lp
5
5
10
tM
hc
k
Fy
Pa
U
J
5
M
A A
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4.0 LIFTING LUG DESIGN - VERTICAL LIFTING
4.1 DESIGN LOAD
Design load , Wt ( = p.We ) = 104596 N
Design load per lug, W ( = Wt / N ) = 26149 N
Vertical component force, Fy = 26149 N
4.2 STRESS CHECK AT PIN HOLE
(a) Tensile Stress
Vertical component force, Fy = 26149 N
Cross sectional area of lug eye, Ae ( = 2*[ rL - d/2 ] x tL ) = 900 mm
Tensile stress, St ( = Fy / Ae ) = 29.05 N/mm
Since St < St.all, therefore the lifting lug size is satisfactory.
(b) Bearing Stress
Vertical component force, Fy = 26149 N
Cross sectional area of lug eye, Ae ( = Dp x tL ) = 333 mm
Bearing stress, Sbr ( = Fy / Ae ) = 78.44 N/mm
Since Sbr < Sbr.all,therefore the lifting lug size is satisfactory.
(c)Shear Stress
Vertical component force, Fy = 26149 N
Cross sectional area of lug eye, Ae ( = 2.(rL-d/2).tL ) = 900mm
Shear stress, Ss ( = Fy / Ae ) = 29.05 N/mm
Since Ss < Ss.all,therefore the lifting lug size is satisfactory.
5.0 STRESS CHECK AT SECTION A-A
(a) Bending Stress
Bending stress due to Pa ( = Fy x tan U ) = 7007 N
Bending moment, Mb ( = Pa x J ) = 336316 Nmm
= 3750
Bending stress, Sb ( = Mb/Z ) = 89.68 N/mmSince Sb < Sb.all, therefore the lifting lug size is satisfactory.
(b) Tensile Stress due to Fy
Cross section area, Ae (=2rL x tL) = 1500 mm
Tensile Stress, St (=Fy/Ae) = 17.43 N/mm
Since St < St.all, therefore the lifting lug size is satisfactory.
Combine Stress Ratio, CS (= St/St.all + Sb/Sbn.all) = 0.66
Since CS
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Allowable welding stress, Sa ( = E.Sa ) = 75.00 N/mm
Since Ssx < Sa, therefore the selected weld size i satisfactory .
7.0 DESIGN OF WELD SIZE AT PAD TO TANK JOINT
7.1 GENERAL
Weld leg , w = 6 mm
Weld throat thickness, tr = 4.2 mm
Fillet weld joint efficiency, E = 0.6
Allowable welding stress for steel grade 43 ( E-43 ) = 125 N/mm
7.2 CRITICAL WELD CROSS-SECTIONAL PROPERTIES
Area of weld, Aw ( = 2 tr ( Wp + Lp ) ) = 2121 mm
7.3 STRESS DUE TO FORCE Fy
Component force, Fy = 26149 N
Shear stress, Ssx ( = Fy / Aw ) = 12.33 N/mm
Allowable welding stress, Sa ( = E.Sa ) = 75.00 N/mm
Since Ssx < Sa, therefore the selected weld size i satisfactory .
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INTERPOLATION TABLE
RANGE
a/b 1.20 1.3333 1.40
0.3762 unknown 0.4530
0.0616 unknown 0.0770
0.4550 unknown 0.4780
= 0.4274
= 0.0719
= 0.4703
Recommended