mathcad - spreader beam design calculations as per dnv 5th ver

Spreader Beam AnalysisDesign Calculation

Doc No:EC000120-8

MAGNUM SUBSEA SYSTEMS

SPREADER BEAM ANALYSIS-DESIGN

CALCULATION

Rev Date Description Prepared

By

Checked By Approved

By

Status

A 20/05/2013 Spreader Beam –

Calculation

Analysis

Susee Tay Zar Ravi

Customer:

DOF

Contract No.

Document Title:

Spreader Beam-Design Calculation

Document No.:

DOF10011-25

Rev:

00

1


Doc No:EC000120-8

1.0 SCOPE:

This document calculated the design of Spreader Bar,Padeye and the Slings

2.0 REFERENCES

2.1 REFERENCE DOCUMENTS

DNV 2.7-1: DNV STANDARD FOR CERTIFICATION No.2.71.

API 17D Annex-K:PADEYE designed based on API 17D Annex-K

3.0 ASSUMPTIONS

1.DAF is Considered as 3 for designing the Spreader bar design

2.Design Factor is Considered as 5 for Sling design

Spreader Beam is designed based on the self weight of Jumper spool

Load and the Connector Weight.Factor of safety is considered as 3.

Weights of the Jumper spool and

Connector:

Overall weight of the Jumper includes Connector :4315kg

1st Connector Weight C1 1761.3kg:=

2nd Connector Weight

C2 1761.3kg:=

Jumper pipe Weight includes fluid weight J1 933.33kg:=

Design Factor DAF 3:=

1st Connector weight with FOS A1 C1 DAF⋅ 5283.9 kg=:=

2nd Connector Weight with FOS B1 C2 DAF⋅ 5283.9 kg=:=

Jumper with Fluid weight includes FOS act at the

COG: JFOS J1 DAF⋅ 2800 kg=:=

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Calculating the load acting at the top and bottom side of the Spreader Beam

due to the jumper load and Connector Weight

Moment about A,

RB 6713.45kg:=

RA 6654.55kg:=

The reaction force RA and RB are the force which are lifting the Jumper Spool

load and this would be acting downwards of the spreader beam.

Self weight of the spreader Beam SW=6630kg.

Dynamic Amplification Factor=3

Self weight has been shared on the spreader beam with below specified

loacations with DAF as 3.

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Considering the Self Weight of the Spreader Beam to find out the Reaction force at X

and Y: Rx and Ry

Moment about Rx;

15.378 RY =6713.45 (16.339)+6630 (13.39)+6630(8.65)+6630(3.91)-6654.55(0.961)

Ry 17905.09kg:=

Rx 15352.91kg:=

The reaction force Rx and Ry are the weight which should be lifted by the top side

padeye.

Shear Force

at Ra=-6654.55kg Shear Force at Ra

Shear Force at Rxat Rx=8698.36kg

at Sw1=2068.36kg Shear Force at Sw1

at Sw2=-4561.64kg Shear Force at Sw2

at Sw3=-11191.64kg Shear Force at Sw3

at Ry=6713.45kg Shear Force at Ry

at Rb=0 Shear Force at Rb

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Bending Moment

Bending Moment at Raat Ra=0

Bending Moment at Rxat Rx= -6395.02kg m

at Sw1= 19253.75kg m Bending Moment at Self Weight1



at Ry= -25265.85kg m Bending Moment at Ry

at Rb= 0kg m Bending Moment at Rb

Maximum Bending Moment occur at Sw2=29053.47kgm

Designing the Spreader beam dia and thickness based on the maximum bending

moment

Outer Diameter of Pipe OD 16in:=

Thickness of the pipe t 19.11mm:= t 0.752 in⋅=

ID OD 2 t⋅−( ):=

Inner Diameter of the pipe ID 0.368 m=

YOD

2:= Y 0.203 m=

Moment of Inertia Iπ

64OD

4ID

4−( ):=

I 4.37 104−

× m4

=

Maximum Bending Moment Mmax Mmax 29053.47kg m⋅:=

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Maximum Stress acting on the Spreader

Beamσ

Mmax Y⋅ g⋅

I:=

σ 132.482 MPa⋅=

Yield Stress σy 344MPa:=

Allowable Stress:σall σall2

3σy⋅:=

σall 229.333 MPa⋅=

FOSσall

σ1.731=:= Factor Of Safety

Designing Padeye at the top of the Spreader Beam

Reaction force acting at two ends of the top side of the Spreader Beam

Rx 15352.91kg 15352.91 kg=:=

Ry 17905.91kg 17905.91 kg=:=

MGW Rx Ry+ 33258.82 kg=:=

MGW1 33.258tonne:=

MGW1 MGW1 g⋅ 326.15 kN⋅=:=

Padeye -In Plane Loads (Per DNV 2.7-3 Section 3.5.4)

α 30°:= Sling angle from vertical

No significant uncertainity in CoGPL 0.5:=

Resultant Sling force (RSF) on each

Padeye for single point lift.RSF1.2 PL⋅ MGW1⋅

cos α( )225.963 kN⋅=:=

RSF 23.042 tonnef⋅=

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Padeye Design

σypadeye1 50ksi 3.447 108

× Pa=:= padeye material yield

Allowable stress (Per DNV 2.7-3,section

3.4.3)σa 0.85 σypadeye1⋅ 2.93 10

8× Pa=:=

E 200 109Pa⋅:= Elastic Modulas for steel

Shackle Selection

Selecting G-2130 Shackle from crossby catalogue 25 ton load limit.

A 2.88in:= Shackle jaw width

B 2.04in:= Shackle pin dia

F 4.19in:= Shackle Flange Width

Padeye geometry

t1

0.75 A⋅ 2.16 in⋅=:= Minimum Padeye thickness(API 17D-K 2.3.3)

t2

0.9 A⋅ 2.592 in⋅=:= Maximum Padeye thickness (API 17D-K2.3.3)

t 60mm:= Selected Padeye Thickness

tp 40mm:= Main Plate thickness

tc 10mm:= Cheek Plate thickness

Dh 1.06 B⋅ 54.925 mm⋅=:=Maximum Padeye Hole Dia (Per API 17D-K.2.3.2)

Dh 55mm 2.165 in⋅=:=

Dpin 2.04in:=

DCheek 100mm:=

Rcheek

DCheek

21.969 in⋅=:=

a 5mm:= Weld throat thickness

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R1

1.75 Dh⋅ 3.784 in⋅=:= Padeye Min Radius (API 17D-K.2.3.4)

R2

2 Dh⋅ 4.325 in⋅=:= Padeye Maximum Radius (API 17D-K.2.3.4)

Rpl 4.3in:= Selected Padeye Radius

Rpad

Rpl tp⋅ 2 Rcheek⋅ tc⋅+

t3.523 in⋅=:=

ht

21.181 in⋅=:= Weld height(API 17D-K.2.3.5)

Clearance (API 17D K.2.3.5)C 1in:=

HF

2h+

C+:=Distance from base to center of hole (API 17D-K.2.3.5)

H 4 in⋅=

β 30°:= Assuming Padeye with 60deg tapered sides

L 2Rpad

cos β( )H h−( ) tan β( )⋅+

12 in⋅=:= Length of Padeye (API 17D-K.2.3.6)

Bearing Pressure

σbearing 0.18

RSF1

Dpin

1

Dh

−

⋅ E⋅

t⋅:=

σbearing 1.651 108

× Pa=

Factor of Safety of Bearing Pressure should

be more than 1 SFbearing

σa

σbearing

1.775=:=

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Tear Out Stress (DNV 2.7-3,AppendixA)

σto22 RSF⋅

2 Rpad Dh−( )⋅ t⋅109.224 MPa⋅=:=

SFto2

σa

σto2

2.683=:= Factor of Safety of Tear out stress should be

more than 1

Cheek Plate Welds:

σch

RSF tc⋅

t DCheek⋅ a⋅75.321 MPa⋅=:=

SFCheek

σa

σch

3.89=:=

Combined Streass as per DNV 2.7-3 A.6:

Resultant Sling Force(RSF) Fsling RSF 225.963 kN⋅=:=

Sling Angle from Vertical θ 30°:=

Padeye Length L 297.42 mm⋅=

Padeye Thicknesst 60 mm⋅=

Padeye Hole Dia Dh 55 mm⋅=

Padeye Material yield Strength σy 355MPa:=

Allowable σe 0.85 σy⋅:=

σe 301.75 MPa⋅=

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Vertical Inplane Load FVsling Fsling cos θ( )⋅:=

FVsling 195.69 kN⋅=

Horizontal Inplane Load FHsling Fsling sin θ( )⋅:=

FHsling 112.982 kN⋅=

Fop Fsling .05⋅:=Design out of Plane Load

Fop 11.298 kN⋅=

Tensile Stress:

Tensile Stress due to in-plane vertical load σt1

FVsling

L t⋅ Dh t⋅( )−:=

σt1 13.454 MPa⋅=

Utilisation Uσt1

σt1

σe

:= Uσt1 0.045=

Shear Stress:

Equivalent Shear Force: Fs FHsling2

Fop2

+

:=

Fs 113.545 kN⋅=

τs

Fs

L t⋅ Dh t⋅( )− :=

Equivalent Shear Stress

τs 7.806 MPa⋅=

UtilisationUτs

τs

σe

:= Uτs 0.026=

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Bending Stress due to out of plane horizontal force:

Bending Moment Arm Lba

H Dh+

2:=

Lba 81.806 mm⋅=

Effective Moment of Inertia IopL t

3⋅

12:=

Iop 5.354 106

× mm4

⋅=

Mop Fop Lba⋅:=Out Plane Bending Moment

Mop 0.924 kN m⋅⋅=

Out Plane Bending Stress σbop Mop

t

2

Iop

⋅:=

σbop 5.179 MPa⋅=

Bending Stress due to In-Plane Horizontal Force:

In Plane Bending Moment: Mip FHsling Lba⋅:=

Mip 9.243 kN m⋅⋅=

Effective Moment of Inertia Iip tL

3

12⋅:=

Iip 1.315 104−

× m4

=

In Plane Bending Stress: σbip Mip

L

2

Iip

⋅:=

σbip 10.449 MPa⋅=

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Total Bending Stress: σbt σbip σbop+:=

σbt 15.628 MPa⋅=

Combined Stress

σcs σt1 σbt+( )2

3 τs2

⋅+:=Combined Stress

σcs 32.071 MPa⋅=

Utilisation Uσcs

σcs

σe

:=

Uσcs 0.106=

Weld Shear Stress due to Horizontal Component of the Force (API 17D,K.3.3.3.3)

α1 60°:=

Fh RSF sin α1( )⋅:=

Horizontal Component ForceFh 195.69 kN⋅=

x sin 45°( ) h⋅:=

x 0.835 in⋅= Mean Weld bead size

Aw 2 x L t+( )⋅[ ] 0.015 m2

=:= Total average throat area

τh

Fh

Aw

1.29 107

× Pa=:= Stress due to horizontal component of force

(K.3.3.3.3)

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τallowable1

1.44σy⋅ 2.465 10

8× Pa=:= Allowable shear stress ,equation K.21,API

17D Section K 3.3.3.3

Safety Factor should be more than 1.44(API 17D

SEC K.3.3.3.3)SFτh

τallowable

τh

19.103=:=

Weld Shear Stress due to Vertical Load Test

LOADtest 2.5 MGW⋅ 8.315 104

× kg=:=

Fh.Load.. LOADtest g⋅ 815.394 kN⋅=:=

τh.Load.test

Fh.Load..

Aw

5.377 107

× Pa=:=

SFτ.h.load.test

τallowable

τh.Load.test

4.585=:= Factor of Safety should be more than

1.44.(API 17D Sec K.3.3.3.3)

Tensile Stress due to Vertical C omponent of Force at throat of the weld (API

K.3.3.3.4)

FV RSF cos α( )⋅:=

Vertical Component of ForceFV 1.957 10

5× N=

Tensile Stress due to vertical com ponent on

throat of weldσV

FV

Aw

1.29 107

× Pa=:=

Factor of Safety should be more than

1.67.(API 17D Sec K.3.3.3.4) SFσ.H

σa

σV

22.707=:=

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Design of Sling based BASED on RSF:

RSF 23.042 tonnef⋅= Resultant Sling Force

MBL 5 RSF⋅ 115.209 tonnef⋅=:= Maximum Breaking Load

Nominal Dia 40mm,6*36 CLASS IWRC,1960 Grade, MAXIMUM BREAKING LOAD 115

TONNE.

Designing a Padeye at the bottom of the Spreader Beam

Reaction force acting at two ends of the connector

Ra 6654.55kg 6654.55 kg=:=

Rb 6713.45kg 6713.45 kg=:=

MGW Ra Rb+ 13368 kg=:=

MGWunitless 13.368:=

MGW 13.368tonne:=

MGW1 MGW g⋅ 131.095 kN⋅=:=

Padeye -In Plane Loads (Per DNV 2.7-3 Section 3.5.4)

α 0°:= Sling angle from vertical

No significant uncertainity in CoGPL 0.5:=

RSF1.2 PL⋅ MGW1⋅

cos α( )78.657 kN⋅=:= Resultant Sling force (RSF)

on each Padeye for single

point lift.


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Padeye Design

σy 50ksi 3.447 108

× Pa=:= padeye material yield

Allowable stress (Per DNV 2.7-3,section 3.4.3)σa 0.85 σy⋅ 2.93 10

8× Pa=:=

E 200 109Pa⋅:= Elastic Modulas for steel

Shackle Selection

Selecting G-2130 Shackle from crossby catalogue 8.5 ton load limit.

A 1.69in:= Shackle jaw width

B 1.15in:= Shackle pin dia

F 2.38in:= Shackle Flange Width

Padeye geometry

t1

0.75 A⋅ 1.268 in⋅=:= Minimum Padeye thickness(API 17D-K 2.3.3)

t2

0.9 A⋅ 1.521 in⋅=:= Maximum Padeye thickness (API 17D-K2.3.3)

t 38mm:= Selected Padeye Thickness

Dh 1.06 B⋅ 30.963 mm⋅=:=Maximum Padeye Hole Dia (Per API 17D-K.2.3.2)

Dh 31mm 1.22 in⋅=:=

R1

1.75 Dh⋅ 2.133 in⋅=:= Padeye Min Radius (API 17D-K.2.3.4)

R2

2 Dh⋅ 2.438 in⋅=:= Padeye Maximum Radius (API 17D-K.2.3.4)

R 2.4in:= Selected Padeye Radius

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ht

20.748 in⋅=:= Weld height(API 17D-K.2.3.5)

Clearance (API 17D K.2.3.5)C 1in:=

Distance from base to center of hole (API 17D-K.2.3.5)H

F

2h+

C+:=

H 3 in⋅=

β 30°:= Assuming Padeye with 60deg tapered sides

L 2R

cos β( )H h−( ) tan β( )⋅+

8 in⋅=:= Length of Padeye (API 17D-K.2.3.6)

Bearing Pressure (Apendix A,DNV 2.7-3)

σbearing 0.045RSF E⋅

Dh t⋅⋅ 164.546 MPa⋅=:=

Factor of Safety of Bearing Pressure should be

more than 1 SFbearing

σa

σbearing

1.781=:=

Tear Out Stress (DNV 2.7-3,AppendixA)

σto22 RSF⋅

2 R Dh−( )⋅[ ] t⋅69.004 MPa⋅=:=

SFto2

σa

σto2

4.247=:= Factor of Safety of Tear out stress should be

more than 1

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Combined Streass as per DNV 2.7-3 A.6:

Resultant Sling Force(RSF) Fsling1 RSF 78.657 kN⋅=:=

Sling Angle from Vertical θ 30°:=

Padeye Length L 205.012 mm⋅=

Padeye Thicknesst 38 mm⋅=

Padeye Hole Dia Dh 31 mm⋅=

Padeye Material yield Strength σy1 355MPa:=

Allowable σe1 0.85 σy⋅:=

σe 301.75 MPa⋅=

Vertical Inplane Load FVsling1 Fsling1 cos θ( )⋅:=

FVsling1 68.119 kN⋅=

Horizontal Inplane Load FHsling1 Fsling1 sin θ( )⋅:=

FHsling1 39.329 kN⋅=

Fop1 Fsling1 .05⋅:=Design out of Plane Load

Fop1 3.933 kN⋅=

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Doc No:EC000120-8

Tensile Stress:

Tensile Stress due to in-plane vertical load σt1

FVsling1

L t⋅ Dh t⋅( )−:=

σt1 10.302 MPa⋅=

Utilisation Uσt1

σt1

σe

:=

Uσt1 0.034=

Shear Stress:

Equivalent Shear Force: Fs1 FHsling12

Fop12

+

:=

Fs1 39.525 kN⋅=

τs1

Fs1

L t⋅ Dh t⋅( )− :=

Equivalent Shear Stress

τs1 5.977 MPa⋅=

Uτs1

τs1

σe

:=Utilisation Uτs1 0.02=

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Bending Stress due to out of plane horizontal force:

Bending Moment Arm Lba1

H Dh+

2:= Lba1 52.813 mm⋅=

Effective Moment of Inertia Iop1L t

3⋅

12:= Iop1 9.375 10

5× mm

4⋅=

Mop1 Fop Lba1⋅:=Out Plane Bending Moment Mop1 0.597 kN m⋅⋅=

Out Plane Bending Stress σbop1 Mop1

t

2

Iop1

⋅:= σbop1 12.094 MPa⋅=

Bending Stress due to In-Plane Horizontal Force:

In Plane Bending Moment: Mip1 FHsling1 Lba1⋅:=

Mip1 2.077 kN m⋅⋅=

Effective Moment of Inertia Iip1 tL

3

12⋅:=

Iip1 2.729 105−

× m4

=

In Plane Bending Stress: σbip1 Mip1

L

2

Iip1

⋅:=σbip1 7.803 MPa⋅=

Total Bending Stress: σbt1 σbip1 σbop1+:=

σbt1 19.896 MPa⋅=

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Combined Stress

σcs1 σt1 σbt1+( )2

3 τs2

⋅+:=Combined Stress

σcs1 33.087 MPa⋅=

Utilisation Uσcs1

σcs

σe

:=

Uσcs1 0.106=

Weld Shear Stress due to Vertical Load Test

LOADtest1 2.5 MGW⋅ 3.342 104

× kg=:=

Fh.Load.1. LOADtest1 g⋅ 327.738 kN⋅=:=

τh.Load.test1

Fh.Load.1.

Aw

2.161 107

× Pa=:=

SFτ.h.load.test1

τallowable

τh.Load.test1

11.407=:= Factor of Safety should be more than

1.44.(API 17D Sec K.3.3.3.3)

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Tensile Stress due to Vertical C omponent of Force at throat of the weld (API

K.3.3.3.4)

FV1 RSF cos α( )⋅:=

Vertical Component of ForceFV1 7.866 10

4× N=

Tensile Stress due to vertical com ponent on throat of

weldσV1

FV1

Aw

5.187 106

× Pa=:=

Factor of Safety should be more than

1.67.(API 17D Sec K.3.3.3.4) SFσ.H1

σa

σV1

56.492=:=

Design of Sling based on RSF:

RSF 78.657 kN⋅= Resultant Sling Force


MBL1 5 RSF⋅ 40.104 tonnef⋅=:= Maximum Breaking Load

Nominal Dia 24mm,6*36 CLASS IWRC,1960 Grade MAXIMUM BREAKING LOAD 41

TONNE.

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Calculating stress acting on each Stud of the Clamp mounted on the Spreader

Beam

Resultant Sling Force P 225.963kN:=

θ 60°:=

Vertical Load Pv P sin θ( )⋅:= Pv 195.69 kN⋅=

Horizontal Load PH P cos θ( )⋅:= PH 112.982 kN⋅=

Number of Studs N 5:=

Total No. of Studs 6.Consider N =5;assume i f 1stud fails.

Stud Dia D 0.875in:=

Area of Stud Asπ

4D

2⋅:=

As 3.879 104−

× m2

=

Tensile Stess due to Vertical Load acting on each Stud:

Vertical Load acting on each stud Pt

Pv

N39.138 kN⋅=:=

Tensile Stress σt

Pt

As

:=σt 100.885 MPa⋅=

Tensile Strength of Stud σts 125ksi:=

σys 105ksi:=Yield strength of bolt

σall 0.60 σts⋅:=Allowable Tensile Strength σall 517.107 MPa⋅=

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Safety Factor SF1

σall

σt

:=

SF1 5.126=

Utility Ratio UR1

σt

σall

:=

UR1 0.195=

Shear Stres due to Horizontal Force acting on each Stud

Horizontal Load acting on each bolt Ps

PH

N22.596 kN⋅=:=

Shear strength acting on each stud τs

Ps

As

58.246 MPa⋅=:=

τall1

1.44σys⋅:=

Allowable Shear Strength

τall 502.743 MPa⋅=

SF2

τall

τs

8.631=:=Factor of Safety

Utility Ratio UR2

τs

τall

0.116=:=

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Preload Calculation for the Studs using in the Clamp:

D =Stud DiameterD 0.875in:= D 0.022 m=

P =Thread Pitch P 2.822mm:= P 2.822 103−

× m=

Asπ

4D 0.9743 P⋅( )−[ ]

2:= As Per API 6A Annex D D.3 Equations

As =Stress area As 2.979 104−

× m2

=

As =Stress area of one stud

Combined Stress

Tensile Stress acting on one stud σt 100.885 MPa⋅=

Shear Stress acting on one stud τs 58.246 MPa⋅=

σtotal σt2

3 τs2

⋅+:=

σtotal 142.672 MPa⋅=

F σtotal As⋅:= As Per API 6A Annex D D.3 Equations

F 4.25 104

× N= F =Force acting on one stud.

E 0.8028in:= E =Pitch Diameter of the thread

E 0.02 m=

f 0.13:= f =friction Coefficient

S1

cos 30°( ):= S 1.1547=

H 1.5 D⋅ 3.175mm+:= H 36.512 mm⋅= H =Hex size (Nut)

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K 3.175mm:= K =Nut internal Chamfer

T1F E⋅ P π f⋅ E⋅ S⋅+( )⋅

2 π E⋅ P f⋅ S⋅−( )F f⋅

H D+ K+

4

⋅+:= As Per API 6A Annex D D.3

Equations

T1 170.217 J=Torque required for one stud and nut

Calculating Hoop Stress of the Pipe

Force acting on the Clamp P1 225.963kN:=

Thickness of the Pipe tp 19.11mm:=

Force acted along the length

of the pipeLp 23.6in:=

Hoop Stress acted on the pipe length which is mounted by the clamp σθ

σθP1

tp Lp⋅:=

σθ 19.726 MPa⋅=

σy 50ksi:=Yield Strength of the pipe: σy

σc 0.60 σy⋅:=Compression Stress:σc σc 206.843 MPa⋅=

FSσc

σθ

10.486=:=Factor of safety FS

Utility Ratio UR URσθ

σc

0.095=:=

END

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mathcad - spreader beam design calculations as per dnv 5th ver

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