Design of Anchor Bolt

7/29/2019 Design of Anchor Bolt

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FOR

DRAIN VESSELS

A

Rev.

7/3/2006

Date

Signature

Date 15/2/2006 15/2/2006

SKPCS

15/2/2006

KJG

Client

Approval

Approved by

SKP

ByIssue Description

AIssued for

Comments

Doc. No. JO-MP/AE-734-CC-

Checked

By

Approved

DRAFT

By

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RB

CONTRACT NO. JO/HC32/MP04

DESIGN CALCULATION

V - 206 / V - 207

FormatNo.-

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10

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SUG

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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

SUG

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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

tNo.-

01/RevNo-01/Date.-

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Soil Report :

Approved SKPCS RB

Checked byIssued for Approval

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Date

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
mailto:+@sqrt(I9^2+I10^2)mailto:+@sqrt(I9^2+I10^2)


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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
mailto:+@sqrt(I9^2+I10^2)mailto:+@sqrt(I9^2+I10^2)


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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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Design of Anchor Bolt