Mathcad - Spreader Beam Design Calculations as Per DNV 5th Ver

Spreader Beam Analysis Design Calculation Doc No:EC000120-8

MAGNUM SUBSEA SYSTEMS SPREADER BEAM ANALYSIS-DESIGN CALCULATION

Rev

Date

Description

A

20/05/2013

Spreader Beam – Calculation Analysis

Prepared By Susee

Customer:

Checked By Tay Zar

Approved By Ravi

Status

Contract No.

DOF Document Title:

Spreader Beam-Design Calculation Document No.:

DOF10011-25

Rev: 00

1


1.0 SCOPE: This docum ent 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 Jum per includes Connector :4315kg 1st Connector Weight

C1 := 1761.3kg

2nd Connector Weight

C2 := 1761.3kg

Jum per pipe Weight includes fluid weight Design Factor

J1 := 933.33kg

DAF := 3

1st Connector weight with FOS

A1 := C1 ⋅ DAF = 5283.9 kg

2nd Connector Weight with FOS

B1 := C2 ⋅ DAF = 5283.9 kg

Jum per with Fluid weight includes FOS act at the COG:

JFOS := J1 ⋅ DAF = 2800 kg

2


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 R A and R B 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. Dynam ic Am plification Factor=3 Self weight has been shared on the spreader beam with below specified loacations with DAF as 3.

3


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 R Y =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 at Rx=8698.36kg

Shear Force at Ra Shear Force at Rx

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

4


Bending Moment at Ra=0 at Rx= -6395.02kg m

Bending Mom ent at Ra Bending Mom ent at Rx

at Sw1= 19253.75kg m

Bending Mom ent at Self Weight1

at Sw2= 29053.47kg m


at Sw3= 7438.29kg m


at Ry= -25265.85kg m

Bending Mom ent at Ry

at Rb= 0kg m

Bending Mom ent at Rb

Maximum Bending Moment occur at Sw2=29053.47kgm Designing the Spreader beam dia and thickness based on the m aximum bending m oment Outer Diam eter 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 Y :=

Moment of Inertia

Maximum Bending Moment Mm ax

I :=

OD 2 π 64

Y = 0.203 m

(OD4 − ID4)

−4 4

I = 4.37 × 10

m

Mmax := 29053.47kg⋅ m

5


Maximum Stress acting on the Spreader Beam

σ :=

Mmax ⋅ Y ⋅ g I

σ = 132.482 ⋅ MPa

Yield Stress

Allowable Stress:σall

σy := 344MPa σall :=

2 3

⋅ σy

σall = 229.333 ⋅ MPa FOS :=

σall σ

= 1.731

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) Sling angle from vertical

α := 30°

No significant uncertainity in CoG

PL := 0.5 RSF :=

1.2 ⋅ PL⋅ MGW1 cos( α)

= 225.963 ⋅ kN

Resultant Sling force (RSF) on each Padeye for single point lift.

RSF = 23.042⋅ tonnef

6


Padeye Design 8

σypadeye1 := 50ksi = 3.447 × 10 Pa 8

σa := 0.85⋅ σypadeye1 = 2.93 × 10 Pa 9

padeye m aterial yield Allowable stress (Per DNV 2.7-3,section 3.4.3) Elastic Modulas for steel

E := 200⋅ 10 Pa Shackle Selection

Selecting G-2130 Shackle from crossby catalogue 25 ton load lim it. A := 2.88in

Shackle jaw width

B := 2.04in

Shackle pin dia

F := 4.19in

Shackle Flange Width

Padeye geometry t := 0.75⋅ A = 2.16⋅ in

Minim um Padeye thickness(API 17D-K 2.3.3)

t := 0.9 ⋅ A = 2.592⋅ in

Maximum Padeye thickness (API 17D-K2.3.3)

t := 60mm

Selected Padeye Thickness

1 2

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)

D h := 55mm = 2.165⋅ in D pin := 2.04in D Cheek := 100mm Rcheek :=

D Cheek

a := 5mm

2

= 1.969⋅ in Weld throat thickness

7


R := 1.75⋅ Dh = 3.784⋅ in

Padeye Min Radius (API 17D-K.2.3.4)

R := 2 ⋅ Dh = 4.325⋅ in

Padeye Maximum Radius (API 17D-K.2.3.4)

Rpl := 4.3in

Selected Padeye Radius

1

2

Rpad :=

h :=

t 2

Rpl⋅ tp + 2 ⋅ R cheek⋅ tc t

= 3.523⋅ in

Weld height(API 17D-K.2.3.5)

= 1.181⋅ in

Clearance (API 17D K.2.3.5)

C := 1in H :=

 F + h + C   2 

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

H = 4 ⋅ in Assuming Padeye with 60deg tapered sides

β := 30°

 Rpad  + ( H − h) ⋅ tan( β) = 12⋅ in  cos(β) 

L := 2 

Length of Padeye (API 17D-K.2.3.6)

Bearing Pressure RSF⋅  σbearing := 0.18⋅



1 D pin

−

1

⋅E 

Dh 

t 8

σbearing = 1.651 × 10 Pa σa SFbearing := = 1.775 σbearing

Factor of Safety of Bearing Pressure should be more than 1

8


Tear Out Stress (DNV 2.7-3,AppendixA)

2 ⋅ RSF σto2 := = 109.224 ⋅ MPa 2 ⋅ Rpad − D h  ⋅ t

(

)

σa SFto2 := = 2.683 σto2

Factor of Safety of Tear out stress should be m ore than 1

Cheek Plate Welds: RSF⋅ tc σch := = 75.321⋅ MPa t⋅ D Cheek⋅ a σa SFCheek := = 3.89 σch 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 Thickness

Padeye Hole Dia

t = 60⋅ mm D h = 55⋅ mm

Padeye Material yield Strength

σy := 355MPa

Allowable

σe := 0.85⋅ σy σe = 301.75 ⋅ MPa

9


Vertical Inplane Load

FVsling := Fsling⋅ cos( θ) FVsling = 195.69 ⋅ kN

Horizontal Inplane Load

FHsling := Fsling⋅ sin( θ) FHsling = 112.982 ⋅ kN

Design out of Plane Load

Fop := Fsling⋅ .05 Fop = 11.298 ⋅ kN

Tensile Stress: Tensile Stress due to in-plane vertic al load

FVsling σt1 := L⋅ t − D h⋅ t

(

)

σt1 = 13.454 ⋅ MPa Utilisation

σt1 Uσt1 := σe

Uσt1 = 0.045

Shear Stress: Equivalent Shear Force:

Fs :=

2 2 F  Hsling + Fop 

Fs = 113.545 ⋅ kN

Equivalent Shear Stress

Fs τs := L⋅ t − D h⋅ t 



(

)

τs = 7.806⋅ MPa Utilisation

Uτs :=

τs σe

Uτs = 0.026

10


Bending Stress due to out of plane horizontal force: Bending Mom ent Arm

Lba :=

H + Dh 2

Lba = 81.806 ⋅ mm 3

Effective Moment of Inertia

L⋅ t Iop := 12

6

Iop = 5.354 × 10 ⋅ mm Out Plane Bending Moment

4

Mop := Fop⋅ Lba Mop = 0.924⋅ kN⋅ m t

Out Plane Bending Stress

2

σbop := Mop⋅ Iop σbop = 5.179⋅ MPa

Bending Stress due to In-Plane Horizontal Force: In Plane Bending Mom ent:

Mip := FHsling⋅ Lba Mip = 9.243⋅ kN⋅ m 3


L Iip := t⋅ 12 −4 4

Iip = 1.315 × 10

m

L

In Plane Bending Stress:

2

σbip := M ip⋅ Iip σbip = 10.449⋅ MPa

11


Total Bending Stress:

σbt := σbip + σbop σbt = 15.628 ⋅ MPa

Combined Stress Combined Stress

σcs :=

(σt1 + σbt)2 + 3⋅τs2

σcs = 32.071 ⋅ MPa Utilisation

σcs Uσcs := 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 Com ponent Force

Fh = 195.69 ⋅ kN x := sin( 45°) ⋅ h

Mean Weld bead size

x = 0.835⋅ in

Aw := 2 [ x⋅ ( L + t) ] = 0.015 m

Fh 7 τh := = 1.29 × 10 Pa Aw

2

Total average throat area

Stress due to horizontal component of force (K.3.3.3.3)

12


1 8 τallowable := ⋅ σy = 2.465 × 10 Pa 1.44

SFτh :=

τallowable τh

Allowable shear stress ,equation K.21,API 17D Section K 3.3.3.3 Safety Factor should be m ore than 1.44(API 17D SEC K.3.3.3.3)

= 19.103

Weld Shear Stress due to Vertical Load Test

4

LOADtest := 2.5 ⋅ MGW = 8.315 × 10 kg Fh.Load.. := LOAD test⋅ g = 815.394 ⋅ kN τh.Load.test :=

Fh.Load.. Aw

7

= 5.377 × 10 Pa

τallowable SFτ.h.load.test := = 4.585 τh.Load.test

Factor of Safety should be m ore than 1.44.(API 17D Sec K.3.3.3.3)

Tensile Stress due to Verti cal C omponent of Force at throat of the weld (API K.3.3.3.4) FV := RSF⋅ cos( α) 5

FV = 1.957 × 10 N FV 7 σV := = 1.29 × 10 Pa Aw σa SFσ.H := = 22.707 σV

Vertical Com ponent of Force

Tensile Stress due to vertical com ponent on throat of weld


13


Design of Sling based BASED on RSF: Resultant Sling Force

RSF = 23.042⋅ tonnef 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 CoG PL := 0.5 1.2 ⋅ PL⋅ MGW1 Resultant Sling force (RSF) RSF := = 78.657 ⋅ kN cos( α) on each Padeye for single point lift. RSF = 8.021⋅ tonnef

14


Padeye Design 8

padeye m aterial yield

σy := 50ksi = 3.447 × 10 Pa 8

σa := 0.85⋅ σy = 2.93 × 10 Pa 9

E := 200⋅ 10 Pa

Allowable stress (Per DNV 2.7-3,section 3.4.3)

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 t := 0.75⋅ A = 1.268⋅ in

Minim um Padeye thickness(API 17D-K 2.3.3)

t := 0.9 ⋅ A = 1.521⋅ in

Maximum Padeye thickness (API 17D-K2.3.3)

t := 38mm

Selected Padeye Thickness

1 2

Dh := 1.06⋅ B = 30.963⋅ mm

Maximum Padeye Hole Dia (Per API 17D-K.2.3.2)

D h := 31mm = 1.22⋅ in R := 1.75⋅ Dh = 2.133⋅ in

Padeye Min Radius (API 17D-K.2.3.4)

R := 2 ⋅ Dh = 2.438⋅ in

Padeye Maximum Radius (API 17D-K.2.3.4)

R := 2.4in

Selected Padeye Radius

1

2

15


h :=

t 2

Weld height(API 17D-K.2.3.5)

= 0.748⋅ in

Clearance (API 17D K.2.3.5)

C := 1in

H :=

 F + h + C 2   

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

H = 3 ⋅ in Assuming Padeye with 60deg tapered sides

β := 30° L := 2 

R

 cos(β)

+ ( H − h) ⋅ tan( β) = 8 ⋅ in

Length of Padeye (API 17D-K.2.3.6)



Bearing Pressure (Apendix A,DN V 2. 7-3)

σbearing := 0.045⋅

RSF⋅ E Dh⋅ t

= 164.546 ⋅ MPa

σa SFbearing := = 1.781 σbearing

Factor of Safety of Bearing Pressure should be m ore than 1

Tear Out Stress (DNV 2.7-3,AppendixA) 2 ⋅ RSF σto2 := = 69.004⋅ MPa [ 2 ⋅ ( R − Dh) ] ⋅ t σa SFto2 := = 4.247 σto2

Factor of Safety of Tear out stress should be m ore than 1

16


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 Thickness

Padeye Hole Dia

t = 38⋅ mm D h = 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

Design out of Plane Load

Fop1 := Fsling1 ⋅ .05 Fop1 = 3.933⋅ kN

17


Tensile Stress: FVsling1 σt1 := L⋅ t − D h⋅ t

Tensile Stress due to in-plane vertic al load

(

)

σt1 = 10.302 ⋅ MPa σt1 Uσt1 := σe

Utilisation

Uσt1 = 0.034

Shear Stress: Equivalent Shear Force:

Fs1 :=

2 2 F  Hsling1 + Fop1 

Fs1 = 39.525 ⋅ kN

Equivalent Shear Stress

Fs1 τs1 := L⋅ t − D h⋅ t 



(

)

τs1 = 5.977⋅ MPa

Utilisation

Uτs1 :=

τs1 σe

Uτs1 = 0.02

18


Bending Stress due to out of plane horizontal force:

Bending Mom ent Arm

Lba1 :=

H + Dh

Lba1 = 52.813 ⋅ mm

2

3


Out Plane Bending Moment

L⋅ t Iop1 := 12

5

Iop1 = 9.375 × 10 ⋅ mm

4

Mop1 := Fop⋅ Lba1 Mop1 = 0.597⋅ kN⋅ m t

Out Plane Bending Stress

2

σbop1 := M op1⋅ Iop1

σbop1 = 12.094 ⋅ MPa

Bending Stress due to In-Plane Horizontal Force: In Plane Bending Mom ent:

Mip1 := FHsling1⋅ Lba1 Mip1 = 2.077⋅ kN⋅ m 3


L Iip1 := t⋅ 12

−5 4

Iip1 = 2.729 × 10

m

L 2

In Plane Bending Stress:

σbip1 := M ip1⋅ Iip1

Total Bending Stress:

σbt1 := σbip1 + σbop1

σbip1 = 7.803⋅ MPa

σbt1 = 19.896⋅ MPa

19


Combined Stress Combined Stress

σcs1 :=

( σt1 + σbt1) 2 + 3⋅τs2

σcs1 = 33.087 ⋅ MPa σcs Uσcs1 := σe

Utilisation

Uσcs1 = 0.106

Weld Shear Stress due to Vertical Load Test 4

LOADtest1 := 2.5 ⋅ MGW = 3.342 × 10 kg Fh.Load.1. := LOAD test1 ⋅ g = 327.738 ⋅ kN τh.Load.test1 :=

Fh.Load.1. Aw

7

= 2.161 × 10 Pa

τallowable SFτ.h.load.test1 := = 11.407 τh.Load.test1


20


Tensile Stress due to Verti cal C omponent of Force at throat of the weld (API K.3.3.3.4) FV1 := RSF⋅ cos( α) 4

FV1 = 7.866 × 10 N FV1 6 σV1 := = 5.187 × 10 Pa Aw σa SFσ.H1 := = 56.492 σV1

Vertical Com ponent of Force

Tensile Stress due to vertical com ponent on t hroat of weld


Design of Sling based on RSF: RSF = 78.657⋅ kN

Resultant Sling Force

RSF = 8.021⋅ tonnef MBL1 := 5 ⋅ RSF = 40.104⋅ tonnef

Maximum Breaking Load

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

21


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 if 1stud f ails. Stud Dia

D := 0.875in

Area of Stud

π 2 As := ⋅ D 4 −4 2

As = 3.879 × 10

m

Tensile Stess due to Vertical Load acting on each Stud:

Vertical Load acting on each stud

Tensile Stress

Pt σt := As

Tensile Strength of Stud Yield strength of bolt Allowable Tensile Strength

Pv Pt := = 39.138⋅ kN N

σt = 100.885 ⋅ MPa

σts := 125ksi σys := 105ksi σall := 0.60⋅ σts σall = 517.107 ⋅ MPa

22


Safety Factor

SF1 :=

 σall   σ   t 

SF1 = 5.126 Utility Ratio

σt UR1 := σall

UR1 = 0.195

Shear Stres due to Horizontal Force acting on each Stud

Horizontal Load acting on each bolt

Shear strength acting on each stud

Allowable Shear Strength

PH Ps := = 22.596 ⋅ kN N

Ps τs := = 58.246⋅ MPa As

1 τall := ⋅σ 1.44 ys τall = 502.743 ⋅ MPa

Factor of Safety

τall SF2 := = 8.631 τs

Utility Ratio

τs UR2 := = 0.116 τall

23


Preload Calculation for the Studs using in the Clamp: D =Stud Diameter D := 0.875in

D = 0.022 m

P =Thread Pitch P := 2.822mm

P = 2.822 × 10

π 2 As := [ D − ( 0.9743 ⋅ P) ] 4

As Per API 6A Annex D D.3 Equations

−3

m

−4 2

A s =Stress area

As = 2.979 × 10

m

A s =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

2

σtotal :=

2

σt + 3 ⋅ τs

σtotal = 142.672 ⋅ MPa As Per API 6A Annex D D.3 Equations

F := σtotal⋅ As 4

F = 4.25 × 10 N

F =Force acting on one stud.

E := 0.8028in E = 0.02 m

E =Pitch Diameter of the thread

f := 0.13

f =friction Coefficient

S :=

1 cos( 30°)

S = 1.1547

H := 1.5 ⋅ D + 3.175mm

H = 36.512⋅ mm

H =Hex size (Nut)

24


K =Nut internal Chamfer

K := 3.175mm

T1 :=

+ F⋅ f ⋅ 

F⋅ E⋅ ( P + π⋅ f ⋅ E⋅ S)



2 ( π⋅ E − P⋅ f ⋅ S)

T1 = 170.217 J

H + D + K 4

 

As Per API 6A Annex D D.3 Equations

Torque required for one st ud and nut

Calculating Hoop Stress of the Pipe Force acting on the Clam p

P1 := 225.963kN

Thickness of the Pipe

tp := 19.11mm

Force acted along the length of the pipe

Lp := 23.6in

Hoop Stress acted on the pipe length which is mounted by the clamp σθ P1 σθ := t p ⋅ Lp

Yield Strength of the pipe: σy

σy := 50ksi

Compression Stress:σc

σc := 0.60⋅ σy

Factor of safety FS

FS :=

Utility Ratio UR

UR :=

σc σθ σθ σc

σθ = 19.726⋅ MPa

σc = 206.843 ⋅ MPa

= 10.486

= 0.095

END

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Mathcad - Spreader Beam Design Calculations as Per DNV 5th Ver

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