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FOR
DRAIN VESSELS
A
Rev.
7/3/2006
Date
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Date 15/2/2006 15/2/2006
SKPCS
15/2/2006
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CONTRACT NO. JO/HC32/MP04
DESIGN CALCULATION
V - 206 / V - 207
FormatNo.-
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M/S JOINT OPERATIONS
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TABLE OF CONTENTS
JOB SPECIFICATIONS
APPLICABLE CODES & REFERENCES
DESIGN BASIS
PRIMARY LOAD CASES
LOAD COMBINATIONS
DESIGN CALCULATIONS FOR DRAIN VESSELS
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JOB SPECIFICATIONS :
Joint Operations Specifications - Exhibit - B
SECTION - 01069 Climatic and Environmental conditions
SECTION - 01091 Reference Standards
SECTION - 02200 Earthwork
SECTION - 03100 Concrete Reinforcement
SECTION - 03300 Cast - in - place Concrete
SECTION - 05500 Metal Fabrications
MATERIAL SPECIFICATIONS :
Grade of concrete K - 250
Grade of Reinforcement FY 425
Anchor Bolts ASTM A307 GR.B
APPLICABLE CODES :
BS 8110 (part - 1:1997) Structural use of Concrete
Code of Practice for design and construction
REFERENCES :
Mechnical data sheet (Doc. No. JO-MP/AE-734-M-307)
General Layout
JO - MP/AE - 734 - PP - 002 (Plot plan)
Topographical Survey Drawings
Net Safe Bearing Capacity of Soil considered as perGulf Inspection International Report .
Forma
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Reinforced Concrete Designer's Handbook
By 'Reynolds and James '
JO - MP/AE - 734 - M - 308 (General
Arrangement of Drain Vessels. Item
No. : SUG - V - 206 / V - 207)
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DESIGN BASIS :
1) Equipment Loads
The Loading given in Loading data sheet
2) Live Load
a ) Live Load intensity considered at top of vessel is 3.0 KN/m2
Wind and Seismic loads are not applicable in this design
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r mary oa ases :
Load Case 1 : Erection / Empty Load DL
Load Case 2 : Live Load on Vessel LL
Load Case 3 : Operating Load( wt.of fluid only) OP
Load Case 4 : Hydro Test Load( wt.of fluid only) HT
Load Combinations :
Load Combination shall be evaluated in accordance with BS : 8110 : Part 1 : 1997
1) Operating case
b) DL + OP + LL
1.4 + 1.6 + 1.6
2) Hydrotesting case
b) DL + LL + HT
1.0 + 1.0 + 1.0
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DESIGN OF ANCHOR BOLT
Fy = -3.35 t
Mx = 2.05 tm
Mz = 1.04 tm
Fz = 1.95 t
Fx = 0.99 tNb = 4 nos.
q = 0.3 m
Px = Mx/q
= 2.05/0.3
= 6.8333 t
F1 = Px/2
= 6.8333/2
= 3.41665 t
Pz = Mz/q
= 1.04/0.3
= 3.4667 t
F2 = Pz/2
= 3.4667/2
= 1.73335 t
F3 = Fy/4
= -3.35/4
= -0.8375 t
Ft = F1+F2+F3
= 3.41665+1.73335+(-0.8375)
= 4.3125 t 43125 N
Fs = (Fx+Fz)/4
= (0.99+1.95)/4
= 0.735 t 7350 N
Db = 24 mm
ar = 361.9112 mm2
ttper = 413.68 N/mm
tsper = 68.94 N/mm
tscal = Fs/ar
= 7350/361.911168
= 20.3089 N/mm2
ttcal = Ft/ar
= 43125/361.911168
= 119.1591 N/mm2
ttcal/ttper+tscal/tsper = 119.1591/413.68+20.3089/68.94
= 0.582635
0.5826 < 1
CONCLUSION: Provide 4 no.s-M24 4.6 grade bolts
DESIGN OF BASE PLATE
e/l > 0.33
P = 3.35
M = sqrt(Mx2+Mz
2)
=
= 2.298717
e = M/P
= 0.686184 m 686.1842 mm
L = 0.4 m 400 mm
B = 0.4 m 400 mm
L = 0.2096 m 209.6 mm
B = 0.2058 m 205.8 mm
e/L = 1.71546
0 of 0
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1.7155 > 0.333333
Base plate with large eccentricity
Cs = 203.2 mm
h = 88.400 mm
Sp = 300 mm
M25 = 25 N/mm2
fc = 18 N/mm2
ft = 242.25 N/mm2
m = 280/(3xfc)
= 5.185185
d' = 50 mmd = 350 mm
x = d/(1+ft/(mxfc))
= 97.34294 mm
z = (1-k/3)xd
= 316.4 mm
Moment @ B Txd = 1/2xxxfcxx/3+M-PxL/2
= 28428.89 PROBLEM
T = 10.97 t
THICKNESS OF BASE PLATE
scbc = 265 N/mm2
A = 0.16 m2
Z = 0.01 m3
P/A = 20.9375 t/m2
M/Z = 229.8717 t/m2
P/A+M/Z = 250.8092
P/A-M/Z = -208.934
Max.B.M.M1 = 0.92384 tm 9238397
t = sqrt(M1x6/(scbcperxB))
= 22.86762 mm
CHERK WITH ROARKS
b = 195 mm
a = 88.4 mm
a/b = 0.453333
q = 2.36 N/mm2
b1 = 0.631
t = sqrt(b1xqxb2/scbcper)= 14.618 mm
CONCLUSION: Length = 400 mm
Width = 400 mm
Depth = 24 mm
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q t/m2
P1
P2
h
h/2
h/3
Kayh Kaq
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DESIGN OF FOUNDATION FOR BLOWERS
Assumptions:
Length of footing Lf = 1500 mm 1.5 mWidth of footing Wf = 1500 mm 1.5 m
Depth of footing Df = 250 mm 0.25 m
Density of footing r = 2.5 t/m2
Self weight of footing Sw = LfxWfxDfxr
= 1.5x1.5x0.25x2.5 t/m
= 1.40625 t
Critical load combination = DL+WL (+Z) -0.265 t t
Max tension @ support = -0.265 t
Max compression @ support = 0.265 t
Mx = 0.995 kN-m 0.158 1.184
Mz = 0.593 kN-m 1.500 m
Total Load P = -0.265+0.265+0.9375x1.5
= 1.406 t
= 14.063 kN
Taking moment @ A MA = -0.265x0.158+0.265x(0.158+)+0.9375x1.5x1.5/2
= 1.055 t-m
ex = Mz/P
= 0.593/14.063
= 0.042167
ez = Mx/P
= 0.995/14.063
= 0.070753
C3 = 1.7 From Handbook of concrete Engg.by Fintel
qp = 0.625
0.265
0.93750
0.158
A B
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Corner pressure Cp = C3xqp
=
= 1.063 t/m2
10 t/m2
> 1.063 t/m2
OK
Factored U.D.L w = 1.594 t/m
BMmax = wxLf /8
= 1.59375x1.5^2/8
= 0.448242 t-m 4397256 N-mm
Clear Cover = 75 mm 0.075 m
Effective Depth d = Df-Clear Cover
= 250-75= 175 mm 0.175 m
fcu = 25 N/mm2
fy = 460 N/mm2
Assume 1m Width b = 1000 mm 1.000 m
K = M/(bxd2xfcu)
= 4397256/(1000x175 2x25)
= 0.00574 N/mm2
K' = 0.156 N/mm2
0.156 N/mm2 > 0.00574 N/mm2 Compressive reinforcement is not required
Z = d(0.5+sqrt(0.25-K/0.9))
= 175x(sqrt(0.25-0.00574/0.9))
= 173.877 mm
173.88 mm > 166.25 mmAs = M/(0.95xfyxZ)
= 4397256/(0.95x460x166.25))
= 60.52554 mm2
Minimum reinforcement Asmin = 0.13% of Ag
= 0.13%1000x250
= 325 mm2
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Provide 10 f bars A1bar = 78.54 mm
Spacing = A1bar/Astx1000
= 78.54/325x1000
= 241.6615 mm 0.242 m
Provide 10 f @ 200 mm /C (Both way)
SUMMARY
Size of footing = 1.5 m x 1.5 m x 0.25 m
Reinforcement Bottom = 10 f @ 200 mmC/C (Both way)
Reinforcement Top = 10 f @ 200 mmC/C (Both way)
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Rev Date