kupdf.com_118784294-lifting-beam-deign-staad-82187560-design-calc-10t-spreader-beam.pdf

DESIGN STRENGTH CALCULATION

10 TON SPREA SPREADER DER B EAM EA M

11/02/2012

Page 1 of 20

TABLE TAB LE OF CONTENT CONTENT

1.

STAAD Pro : Spreader Beam Analysis for 4-points and 2-points lifting

Page

3

Analysis ( Introduction). 2.

STAAD Pro : Design basis & Loading Assumptions

3.

STAAD Pro : Loading assumptions assumptions & Steel Design analysis

4

4.

STAAD Pro : Material Specification & Conclusion

4

5.

STAAD Input File

5

6.

STAAD Output Result & Steel Design analysis result

7.

Additional Calc for 2 point Lifting analysis (Lug & Spreader Sizing)

11 – 15

8.

Additional Calc for 4 point Lifting analysis (Lug & Spreader Sizing)

16 – 20

11/02/2012

3–4

6 – 10

Page 2 of 20

STAAD Pro : Spreader Beam Analys is f or 4-point s and 2-point s lif tin g

INTRODUCTION

I)

Scope

This document sets out the structural design criteria for 10 ton spreader beam. The spreader beam has been checked for structural safety and integrity for all loads anticipated during, LIFTING condition. The integrity of the structure has been checked based on the applicable codes and standards approach.

II)

Analysis Methodolog y

The actions effects of the various loads have been determined by linear analysis taking into account the relevant material properties and geometric effects. The three dimensional nature of the structure have been taken into account in the interpretation of the results of the analysis. A linear static stress analysis was performed on the 3-D modeled frame structure using STAAD Pro software. For the analysis the modeled structure was loaded with its own self-weight together with all other loads (Non-modeled structure, Equipment & Piping) and applied at appropriate position as described in the load calculation sheets. The analysis and code check of the structure was primarily based on the guidelines from the Document provided, BS 5950 and API RP2A.

III)

Boundary Condition on Computer Model

For 4-points & 2-points Lifting, the spreader beam was considered to have pinned support location for appropriate computer run.

IV)

Design Basis and Loading Assumpti ons

1.

Loadings: 

2.

Selfweight of the Structure has been generated by the software itself with a 20% contingency factor.

Loadings Combination: 

The Skid has been analyzed for LIFTING condition (4-points & 2-Points). The different load combination taken for these analyses are as follows:

11/02/2012

Page 3 of 20

3.

LIFTING condition: 3.1

Dead Load + Dry weight of Spreader Beam + 20% contingency factor

3.2

2.5 x ( Dead weight of Spreader Beam )

For 4 point s Lifti ng Analysis. 3.3

1.5 x (Dead weight of Spreader Beam)

For 2 points Lifting Analysis.

V)

Steel Design Analys is 

The Steel Design Code use = BS5950 : 2000.



Spreader beam deflection were limited to less than L/300 as per DNV 2.7-1 Specification.



Loading assumption : Point load were assigned & divided equally based on the design load that located on the spreader beam

VI)

Material Specific ation A106 Gr B (min yield 240 Mpa) - Pipes sections

VII)

Resul t Print out and Conclu sion 

All structural members were found to be adequate and safe during lifting. The stresses and

deflections

were

all

below

allowable.

The

BS

5950

code-check

was

incorporated

in the STAAD design and included in the print-out for the Unity check of all members.

11/02/2012

Page 4 of 20

STAAD I NPUT STAAD SPACE DXF I MPORT OF 10- TON- SPREADER- BAR. DXF START J OB I NFORMATI ON ENGI NEER DATE 3- FEB- 12 J OB NAME 10- TON- SPREADER- DNV CALC J OB CLI ENT XXX ENGI NEER NAME NAS CHECKER NAME ALLEN CHECKER DATE 3- FEB- 12 J OB NO 10TON J OB REV 0 J OB REF SSL- STK- 01211301. DWG END J OB I NFORMATI ON I NPUT WI DTH 79 I NPUT WI DTH 72 I NPUT WI DTH 72 UNI T MMS NEWTON J OI NT COORDI NATES 16 0 3340 0; 17 6000 3340 0; MEMBER I NCI DENCES 1 16 17; ** ** ** ** ** * DEFI NI TI ON OF MEMBER PROPERTI ES ** ** ** ** ** ** ** ** ** ** DEFI NE MATERI AL START I SOTROPI C STEEL E 199947 POI SSON 0. 3 DENSI TY 7. 68191e- 005 ALPHA 6e- 006 DAMP 0. 03 END DEFI NE MATERI AL UNI T MMS KG MEMBER PROPERTY CANADI AN ** ** ** ** ** * DEFI NI TI ON OF PI PE 8I N SCH 160 ** ** ** ** ** ** ** ** ** * 1 TABLE ST PI PE OD 219. 1 I D 173. 08 UNI T MMS NEWTON CONSTANTS MATERI AL STEEL ALL * ** ** ** ** ** DEFI NI TI ON OF BOUNDARY CONDI TI ON & LOADI NG * ** ** ** ** * ** ** ** ** ** UNI T MMS KG SUPPORTS 16 17 PI NNED LOAD 1 LOADTYPE Dead TI TLE 4 PTS LI FTI NG ( SELFWEI GHT + 20% CONTI NGENCY) SELFWEI GHT Y - 1. 2 MEMBER LOAD 1 CON GY - 2500 125 109. 55 1 CON GY - 2500 1000 109. 55 1 CON GY - 2500 5000 109. 55 1 CON GY - 2500 5875 109. 55 LOAD 2 LOADTYPE Dead TI TLE 2 PTS LI FTI NG ( SELFWEI GHT + 20% CONTI NGENCY) SELFWEI GHT Y - 1. 2 MEMBER LOAD 1 CON GY - 5000 5875 109. 55 1 CON GY - 5000 125 109. 55 LOAD COMB 3 COMBI NATI ON LOAD CASE 1&2 ( 4 POI NTS LI FTI NG) 1 2. 5 2 2. 5 LOAD COMB 4 COMBI NATI ON LOAD CASE 1&2 ( 2 POI NTS LI FTI NG) 1 1. 5 2 1. 5 ** ** ** ** ** *DEFI NI TI ON OF DESI GN CODE ANALYSI S ** ** ** ** ** ** ** ** PERFORM ANALYSI S PARAMETER 1 CODE BS5950 ** ** ** ** ** ** ** PI PE MEMBER ** ** ** ** * 240 MPA ** ** ** ** ** ** ** ** ** PY 240 ALL ************************************************************** MAI N 1 ALL RATI O 1 ALL TRACK 1 ALL BEAM 1 ALL DFF 300 ALL CHECK CODE ALL FI NI SH *************************************************************

11/02/2012

Page 5 of 20

Job No

Sheet No

Client

0

Part

Software licensed to Job Title

Rev

1

10TON

10-TON-SPREADER-DNV CALC

Ref

SSL-STK-01211301.DWG

By

NAS

File

XXX

Date

Chd

3-FEB-12

10 Ton Spreader Bar.std

Date/Time

ALLEN

11-Feb-2012 19:27

Job Information

Engineer

Checked

Name:

NAS

ALLEN

Date:

3-FEB-12

3-FEB-12

Structure Type

Approved

SPACE FRAME

Number of Nodes

2

Highest Node

17

Number of Elements

1

Highest Beam

1

Number of Basic Load Cases

2

Number of Combination Load Cases

2

Included in this printout are data for:

All

The Whole Structure

Included in this printout are results for load cases: Type L/C

Name

Primary

1

4 PTS LIFTING (SELFWEIGHT + 20% C

Primary

2

2 PTS LIFTING (SELFWEIGHT + 20% C

Combination

3

COMBINATION LOAD CASE 1&2 (4 POI

Combination

4


Supports Node

X

Y

Z

rX

rY

-

-

rZ -

(kip/in)

(kip/in)

(kip/in)

(kip ft/deg) (kip ft/deg) (kip ft/deg)

16

Fixed

Fixed

Fixed

-

-

-

17

Fixed

Fixed

Fixed

-

-

-

Basic Load Cases Number

Name

1

4 PT S LIFT ING (SELFW EIGHT + 20% C

2

2 PT S LIFT ING (SELFW EIGHT + 20% C

Combination Load Cases Comb.

Combination L/C Name

Primary

3


1

4 PTS LIFTING (SELFW EIGHT + 20% C

2.50

2


2.50

1


1.50

2


1.50

4


Print Time/Date: 11/02/2012 19:29 11/02/2012

Primary L/C Name

STAAD.Pro V8i (SELECTseries 1) 20.07.06.35

Factor

Print Run 1 of 5 Page 6 of 20

Job No

Sheet No

Client

0

Part


Rev

2

10TON


Ref


By

NAS

File

XXX

Date

Chd

3-FEB-12


Date/Time

ALLEN

11-Feb-2012 19:27

Beam Loads : 1 4 PTS LIFTING (SELFWEIGHT + 20% CONTINGENCY) Beam

Type

Direction

Fa

Da

Fb

Db

(mm) 1

Ecc. (mm)

CON

kg

GY

-2.5E+3

125.000

-

-

109.550

CON

kg

GY

-2.5E+3

999.998

-

-

109.550

CON

kg

GY

-2.5E+3

5E+3

-

-

109.550

CON

kg

GY

-2.5E+3

5.87E+3

-

-

109.550

Selfweight : 1 4 PTS LIFTING (SELFWEIGHT + 20% CONTINGENCY) Di rec ti on Y

Fac to r -1.200

Beam Loads : 2 2 PTS LIFTING (SELFWEIGHT + 20% CONTINGENCY) Beam

Type

Direction

Fa

Da

Fb

Db

(mm) 1

Ecc. (mm)

CON

kg

GY

-5E+3

125.000

-

-

109.550

CON

kg

GY

-5E+3

5.87E+3

-

-

109.550

Selfweight : 2 2 PTS LIFTING (SELFWEIGHT + 20% CONTINGENCY) Di rec ti on Y

Fac to r -1.200

Print Time/Date: 11/02/2012 19:29 11/02/2012



Job No

Sheet No

Client

0

Part


Rev

3

10TON


Ref


By

NAS

File

XXX

Date

Chd

3-FEB-12


Date/Time

ALLEN

11-Feb-2012 19:27

-2e+003 kg -2e+003 kg

n17

-2e+003 kg -2e+003 kg

# 1

n16

Y X Z

Load 1

4 PTS LIFTING

Print Time/Date: 11/02/2012 19:29 11/02/2012



Job No

Sheet No

Client

0

Part


Rev

4

10TON


Ref


By

NAS

File

XXX

Date

Chd

3-FEB-12


Date/Time

ALLEN

11-Feb-2012 19:27

-5e+003 kg

n17

-5e+003 kg

# 1

n16

Y X Z

Load 2

2 PTS LIFTING

Print Time/Date: 11/02/2012 19:29 11/02/2012



Job No

Sheet No

Client

0

Part


Rev

5

10TON


Ref


By

NAS

File

XXX

Date

Chd

3-FEB-12


Date/Time

ALLEN

11-Feb-2012 19:27

Steel Design (Track 2) Beam 1 Check 1 ALL UNI TS ARE - KG MEMBER

MMS

( UNLESS OTHERWI SE NOTED)

RESULT/ CRI TI CAL COND/ RATI O/ LOADI NG/ FX MY MZ LOCATI ON ======================================================================= 1 ST

TABLE

PI P E

PASS BS- 4. 8. 3. 2 0. 077 3 0. 00 0. 00 11591765. 00 3000. 00 ======================================================================= MATERI AL DATA Gr ade of st eel = S 275 Modul us of el ast i ci t y = 200 kN/ mm2 Desi gn St r engt h ( py) = 2353 N/ mm2 SECTI ON PROPERTI ES ( uni t s - cm) Member Lengt h = 600. 00 Gr oss Ar ea = 141. 75 Net Ar ea = 141. 75

Moment of i ner t i a Pl ast i c modul us El ast i c modul us Ef f ect i ve modul us Shear Ar ea

: : : : :

DESI GN DATA ( uni t s - kN, m) Sect i on Cl ass

:

Moment Capaci t y Reduced Moment Capaci t y Shear Capaci t y

: : :

z - z ax i s 6906. 883 888. 797 630. 478 630. 478 85. 050

Ef f . Ar ea = 141. 75 y- y axi s 6906. 883 888. 797 630. 478 630. 478 85. 050

BS5950- 1/ 2000 SEMI - COMPACT z- z

ax i s 1483. 9 1483. 9 12010. 4

y- y axi s 1483. 9 1483. 9 12010. 4

BUCKLI NG CALCULATI ONS ( uni t s - kN, m) ( axi s nomencl at ure as per desi gn code) LTB check unnecessar y f or t hi s secti on CRI TI CAL LOADS FOR EACH CLAUSE CHECK ( uni t s- kN, m) : CLAUSE RATI O LOAD FX VY VZ MZ MY BS- 4. 2. 3- ( Y) 0. 022 3 264. 8 BS- 4. 8. 3. 2 0. 077 3 0. 0 264. 8 0. 0 113. 7 0. 0 BS- 4. 8. 3. 3. 1 0. 077 3 0. 0 113. 7 0. 0 BS- 4. 8. 3. 3. 3 0. 077 3 0. 0 113. 7 0. 0 Tor si on and def l ect i ons have not been consi dered i n t he desi gn.

Print Time/Date: 11/02/2012 19:29 11/02/2012



LIFTING SPREADER PIPE SIZING CALCULATIONS (2-POINTS LIFTING) ITEM :

10 TONNE SPREADER BEAM

Actual Load at spreader beam ( + Spreader beam weight )

=

5,640 kg

Component force acting on beam, F

=

Impact factor

=

2

Pipe size

:

8 in SCH 160

Outer diameter of pipe, D

=

219.1 mm

Thickness of pipe,

=

23.01 mm

Outer Radius of pipe, R

=

109.55 mm

Inner radius of pipe, r

=

86.54 mm

Section modulus of pipe, Zx-x

=

630,478 mm3

Second Moment of pipe, I

=

4,309,117 mm4

Cross section area of pipe, A

=

14175 mm2

Unbraced length of member, L

=

5750 mm

Modulus Of Elasticity , E

=

2.0E+05 N/mm²

Spreader Pipe Weight

=

Material used

=

Specified yield stress, Sy

=

82,990 N

1) PIPE SIZING

639.8 kg A 106.Gr.B 241.32 N/mm²

F

a) Bending Stress

L R2

R1

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam. (See Appendix)

Total bending moment, M ( = F*L /6)

=

Bending stress, Sb ( = M / Zx-x )

=

126.15 N/mm²

Max Bending stress = Fx * L / ( 60 * E * I )

=

305.19 N/mm²

Allowable bending stress, Sb.all ( = 0.66Sy )

=

159.27 N/mm²

3

Since Sb

<

Sb.all,therefore the pipe size is

79,531,790 Nmm

satisfactory.

b) Compressive Stress

Compressive force, Fc

=

82,990 N

Compressive stress, Sc = Fc / A

=

5.85 N/mm²

Allowable compressive stress, Sc.all ( = 0.6Sy )

=

144.79 N/mm²

c) Combined stresses,

U

=

Sc Sc.all

Since U

11/02/2012

<

+

Sb

=

0.83

Sb.all 1, therefore the pipe size is

satisfactory.

Page 11 of 20


10 TONNE SPREADER BEAM Fz

2) LUG SIZING

P

rL(s) tL(s)

d(s)

a Fx

Fyl

hL(s)

wL(s)

Lug radius, rL(s)

=

50 mm

Lug thickness, tL(s)

=

15 mm

Lug base width, wL(s)

=

200 mm

Diameter of hole, d(s)

=

30 mm

Distance from lug hole to base, hL(s)

=

150 mm

Collar plate thickness, tcp

=

0 mm

Collar ring diameter, Dcp

=

0 mm

Clearance btw shackle & lug size

Result

Lug thickness, tL(s) A =

42.9

mm

15

mm

OK

50

mm

OK

Lug radius, rL(s) C=

84

mm

Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTAB LE

Per PTS Secti on 6.3

Check a) Lug hole diameter, d shall be Max of

i) Dp + 3mm

=

28.40 mm

ii) Dp X 1.05

=

26.67 mm

=

31.40 mm

b) Lug hole diameter, d shall be less than < (Dp + 6mm) Dp =

30

result

a) = satisfactory b) = satisfactory

Hole,d Diameter of hole, d(s)

btw

28.40

30

OK

31.40

No of lug eye,

=

Maximum combined force acting on lug eye, Fc

=

2 41495 N

=

4230 kg

Material used

=

A 36


=

248.21 N/mm²

Allowable bending stress, fbx.all ( = 0.66Sy ) _In Plane

=

163.82 N/mm²

Allowable bending stress, fby.all ( = 0.75Sy ) _Out Of Plane

=

186.16 N/mm²

Allowable tensile stress, St.all ( = 0.6Sy )

=

148.93 N/mm²

Allowable bearing stress, Sbr.all ( = 0.9Sy )

=

223.39 N/mm²

Allowable shear stress, Ss.all ( = 0.4Sy )

=

99.28 N/mm²

LIFTING LUG MATERIAL & MECHANICAL PROPERTIES

11/02/2012

Page 12 of 20



SHACKLES

Shackle rating ( S.W.L )

:

Type of shackle

6.5 tons

BOLT Type Anchor shackle G2130

Pin size, Dp

=

MAXIMUM SLING TENSION ON PADEYE

FACTOR OF SAFETY

25.40 mm

Ts

=

27,663 N

F.O.S.

=

2.00

P

=

55,326 N

a

=

60.00 Deg.

b

=

0.00 Deg.

DESIGN LOAD:

SLING TENSION

P = FOS * Ts

LIFTING ANGLE ACTUAL OUT OF PLANE ANGLE VERTICAL FORCE ON PADEYE

Fz = P * sin a

Fz

=

47,914 N

OUT OF PLANE FORCE

Fyl = 5% of P

Fyl

=

2,766 N

HORIZONTAL FORCE ON PADEYE

Fx = P * cos a

Fx

=

27,663 N

exl

=

0.00

My

=

4,149,485

N-mm

Mx

=

414,948

N-mm

Horizontal dist.PIN CL to N.A. 1

STRESS CHECK AT BASE a) Moment Calc at distance , H

In Plane Moment

My = ( Fx*H ) - ( Fz*ex l)

Out of plane moment

Mx = ( FyI*hL )

b) Tensile Stress

Maximum tensile force, ft = Fz / [ tL(s) * wL(s) ]

=

16 N


=

148.93 N/mm²

<

Since ft

St.all, therefore the lug size is

satisfactory.

c) Bending stress (In Plane)

Maximum bending stress , fbx = ( 6*Mx ) / ( wL(s) * [(tL(s)+tcp)^ 2] )

=

55 N/mm²


=

163.82 N/mm²

<

Since fbx

fbx.all,therefore the lug size is

satisfactory.

d) Bending stress (Out of Plane)

Maximum bending stress , fby = ( 6*My ) / [ tL(s) +(2*tcp)] * [ wL(s)^2 ] )

=

41 N/mm²


=

186.16 N/mm²

Since fby

<

fby.all,therefore the lug size is

satisfactory.

e) Unity Check : Combined stresses,

U

=

Since U

11/02/2012

St/St.all + fby/fby.all + fb x/fbx.all

<

1, therefore the lug size is

=

0.67

satisfactory.

Page 13 of 20



f) SHEAR stress (In Plane)

Maximum SHEAR stress , fsx = Fx / [ wL(s) * tL(s) ]

=

9 N/mm²


=

99.28 N/mm²

Since fsx

<

Ss.all,therefore the lug size is

satisfactory.

g) Bending stress (Out of Plane)

Maximum SHEAR stress , fsy = Fyl / [ wL(s) * tL(s) ]

=

1 N/mm²


=

99.28 N/mm²

Since fsx 1.1

<


satisfactory.

CHECKING VON-MISES CRITERIA a) Sum of stress in X-PLANE

fx = St + fby =

81.01 N/mm²

b) Sum of stress in Y-PLANE

fy = St + fbx =

94.85 N/mm²

fxy = SQRT [ (fsx^2)+(fsy^2) ] =

9.27 N/mm²

Fcomb = SQRT [ (fx^2)+(fy^2)-(fx+fy+3fxy^2) ] =

122.99 N/mm²

c) Therefore, average Shear stress d) Maximum Combined stress

Allowable combined stress : Fcomb.all ( = 0.66Sy ) Since fsx 2

<

=


163.82 N/mm²

satisfactory.

ST RESS CHECK AT PIN H OL E a) Tensile Stress

Maximum tensile force, P

=

At = [ 2 * ( t L(s )* ( rL (s) - d (s)/2 ))] + Cross sectional area of lug eye, [ 2 * ( tc p* (( Dcp/2) - d(s)/2 ))] + [ 2 * ( t cp* (( Dcp /2) - d(s)/2 ))]

=

Tensile stress, St

=

39.52 N/mm²


=

148.926 N/mm²

Since St

<


b) Shear Stress

41495 N 1050 mm²

satisfactory.

`

Maximum shear force, P

=


=

Shear stress, Ss

=

39.52 N/mm²


=

99.28 N/mm²

Since Ss

<


41495 N 1050 mm²

satisfactory.

(c) Unity check, Combine Stresses

Ss U=

----------------

St +

----------------

Ss.all Therefore, the lifting lug size is

11/02/2012

=

0.66 is < than 1

St.all Satisfactory.

Page 14 of 20



3) WELD SIZE CALCULATIONS

Weld leg used,

=

8 mm

Weld throat thickness used, tr

=

6 mm

Filler metal material

:

E-43

Fillet weld joint efficiency, E

=

0.49

Welding stress for steel grade 43 ( E-43 ),

=

125 N/mm²

Allowable welding stress,Sw

=

61.25 N/mm²

Maximum tensile force,Ft

=

41495 N

Area of weld, Aw = 2*(tL+wL)*tr

=

2408 mm²

Tensile stress, St = [(Ft/Aw)]

=

17.23 N/mm²

a) Tensile Stress

Since St

<

Sw,therefore weld leg is

satisfactory.

(b) Shear stres Maximum shear force,Ft

=

Shear stress, Ss = (Ft/Aw)

=

17.23 N/mm²

Allowable welding stress for steel grade 43 ( E-43 ), Sw

=

61.25 N/mm²

Since Ss

<

Sw,therefore weld leg dimension is

41495 N

SATISFACTORY.

(c) Bending stress Maximum bending force,Fb

=

Bending stress, Sb = [(Fb/Aw)]

=

3.41 N/mm²


=

61.25 N/mm²

Since Sb

11/02/2012

<


46723 N

SATISFACTORY.

Page 15 of 20



Actual Load at spreader beam ( + Spreader beam weight )

=

2,945 kg

Component force acting on beam, F

=

Impact factor

=

2

Pipe size

:

8 in SCH 160

Outer diameter of pipe, D

=

219.1 mm

Thickness of pipe,

=

23.01 mm

Outer Radius of pipe, R

=

109.55 mm

Inner radius of pipe, r

=

86.54 mm

Section modulus of pipe, Zx-x

=

630,478 mm3

Second Moment of pipe, I

=

4,309,117 mm4

Cross section area of pipe, A

=

14175 mm2

Unbraced length of member, L

=

4000 mm

Modulus Of Elasticity , E

=

2.0E+05 N/mm²

Spreader Pipe Weight

=

Material used

=


=

57,782 N

1) PIPE SIZING

445.1 kg A 106.Gr.B 241.32 N/mm²

a) Bending Stress F

L R2

R1

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam. (See Appendix)

Total bending moment, M ( = F*L / 6 )

=

Bending stress, Sb ( = M / Zx-x )

=

61.10 N/mm²

Max Bending stress = Fx * L / ( 60 * E * I )

=

71.54 N/mm²

Allowable bending stress, Sb.all ( = 0.66Sy )

=

159.27 N/mm²

3

Since Sb

<

Sb.all,therefore the pipe size is

38,521,646 Nmm

satisfactory.

b) Compressive Stress

Compressive force, Fc

=

57,782 N

Compressive stress, Sc = Fc / A

=

4.08 N/mm²

Allowable compressive stress, Sc.all ( = 0.6Sy )

=

144.79 N/mm²

c) Combined stresses,

U

=

Sc Sc.all

Since U

11/02/2012

<

+

Sb

=

0.41

Sb.all 1, therefore the pipe size is

satisfactory.

Page 16 of 20


10 TONNE SPREADER BEAM Fz

2) LUG SIZING

P

rL(s) tL(s)

d(s)

a Fx

Fyl

hL(s)

wL(s)

Lug radius, rL(s)

=

50 mm

Lug thickness, tL(s)

=

15 mm

Lug base width, wL(s)

=

200 mm

Diameter of hole, d(s)

=

30 mm

Distance from lug hole to base, hL(s)

=

150 mm

Collar plate thickness, tcp

=

0 mm

Collar ring diameter, Dcp

=

0 mm

Clearance btw shackle & lug size

Result

Lug thickness, tL(s) A =

42.9

mm

15

mm

OK

50

mm

OK

Lug radius, rL(s) C=

84

mm

Since A & C clearance against Lug size , Therefore the Lug is is ACCEPTAB LE

Per PTS Secti on 6.3

Check a) Lug hole diameter, d shall be Max of

i) Dp + 3mm

=

28.40 mm

ii) Dp X 1.05

=

26.67 mm

=

31.40 mm

b) Lug hole diameter, d shall be less than < (Dp + 6mm) Dp =

30

result

a) = satisfactory b) = satisfactory

Hole,d Diameter of hole, d(s)

btw

28.40

30

OK

31.40

No of lug eye,

=

Maximum combined force acting on lug eye, Fc

=

2 28891 N

=

2945 kg

Material used

=

A 36


=

248.21 N/mm²


=

163.82 N/mm²


=

186.16 N/mm²


=

148.93 N/mm²

Allowable bearing stress, Sbr.all ( = 0.9Sy )

=

223.39 N/mm²


=

99.28 N/mm²

LIFTING LUG MATERIAL & MECHANICAL PROPERTIES

11/02/2012

Page 17 of 20



SHACKLES

Shackle rating ( S.W.L )

:

Type of shackle

6.5 tons

BOLT Type Anchor shackle G2130

Pin size, Dp

=

MAXIMUM SLING TENSION ON PADEYE

FACTOR OF SAFETY

25.40 mm

Ts

=

14,446 N

F.O.S.

=

2.00

P

=

28,891 N

a

=

90.00 Deg.

b

=

0.00 Deg.

DESIGN LOAD:

SLING TENSION

P = FOS * Ts

LIFTING ANGLE ACTUAL OUT OF PLANE ANGLE VERTICAL FORCE ON PADEYE

Fz = P * sin a

Fz

=

28,891 N

OUT OF PLANE FORCE

Fyl = 5% of P

Fyl

=

1,445 N

HORIZONTAL FORCE ON PADEYE

Fx = P * cos a

Fx

=

0 N

exl

=

My

=

0

N-mm

Mx

=

216,684

N-mm

Horizontal dist.PIN CL to N.A. 1

0.00

STRESS CHECK AT BASE a) Moment Calc at distance , H

In Plane Moment

My = ( Fx*H ) - ( Fz*ex l)

Out of plane moment

Mx = ( FyI*hL )

b) Tensile Stress

Maximum tensile force, ft = Fz / [ tL(s) * wL(s) ]

=

10 N


=

148.93 N/mm²

<

Since ft


satisfactory.

c) Bending stress (In Plane)

Maximum bending stress , fbx = ( 6*Mx ) / ( wL(s) * [(tL(s)+tcp)^ 2] )

=

29 N/mm²


=

163.82 N/mm²

<

Since fbx

fbx.all,therefore the lug size is

satisfactory.

d) Bending stress (Out of Plane)

Maximum bending stress , fby = ( 6*My ) / [ tL(s) +(2*tcp)] * [ wL(s)^2 ] )

=

0 N/mm²


=

186.16 N/mm²

Since fby

<

fby.all,therefore the lug size is

satisfactory.

e) Unity Check : Combined stresses,

U

=

Since U

11/02/2012

St/St.all + fby/fby.all + fb x/fbx.all

<

1, therefore the lug size is

=

0.24

satisfactory.

Page 18 of 20



f) SHEAR stress (In Plane)

Maximum SHEAR stress , fsx = Fx / [ wL(s) * tL(s) ]

=

0 N/mm²


=

99.28 N/mm²

Since fsx

<


satisfactory.

g) Bending stress (Out of Plane)

Maximum SHEAR stress , fsy = Fyl / [ wL(s) * tL(s) ]

=

0 N/mm²


=

99.28 N/mm²

Since fsx 1.1

<


satisfactory.

CHECKING VON-MISES CRITERIA a) Sum of stress in X-PLANE

fx = St + fby =

27.52 N/mm²

b) Sum of stress in Y-PLANE

fy = St + fbx =

56.41 N/mm²

fxy = SQRT [ (fsx^2)+(fsy^2) ] =

0.48 N/mm²

Fcomb = SQRT [ (fx^2)+(fy^2)-(fx+fy+3fxy^2) ] =

62.08 N/mm²

c) Therefore, average Shear stress d) Maximum Combined stress

Allowable combined stress : Fcomb.all ( = 0.66Sy ) Since fsx 2

<

=


163.82 N/mm²

satisfactory.

ST RESS CHECK AT PIN H OL E a) Tensile Stress

Maximum tensile force, P

=


=

Tensile stress, St

=

27.52 N/mm²


=

148.926 N/mm²

Since St

<


b) Shear Stress

28891 N 1050 mm²

satisfactory.

`

Maximum shear force, P

=


=

Shear stress, Ss

=

27.52 N/mm²


=

99.28 N/mm²

Since Ss

<


28891 N 1050 mm²

satisfactory.

(c) Unity check, Combine Stresses

Ss U=

----------------

St +

----------------

Ss.all Therefore, the lifting lug size is

11/02/2012

=

0.46 is < than 1

St.all Satisfactory.

Page 19 of 20



3) WELD SIZE CALCULATIONS

Weld leg used,

=

8 mm

Weld throat thickness used, tr

=

6 mm

Filler metal material

:

E-43

Fillet weld joint efficiency, E

=

0.49

Welding stress for steel grade 43 ( E-43 ),

=

125 N/mm²

Allowable welding stress,Sw

=

61.25 N/mm²

Maximum tensile force,Ft

=

28891 N

Area of weld, Aw = 2*(tL+wL)*tr

=

2408 mm²

Tensile stress, St = [(Ft/Aw)]

=

12.00 N/mm²

a) Tensile Stress

Since St

<

Sw,therefore weld leg is

satisfactory.

(b) Shear stres Maximum shear force,Ft

=

Shear stress, Ss = (Ft/Aw)

=

12.00 N/mm²


=

61.25 N/mm²

Since Ss

<


28891 N

SATISFACTORY.

(c) Bending stress Maximum bending force,Fb

=

Bending stress, Sb = [(Fb/Aw)]

=

3.41 N/mm²


=

61.25 N/mm²

Since Sb

11/02/2012

<


46723 N

SATISFACTORY.

Page 20 of 20

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