1- Loading:W = 1,700 kg Weight of component to be lifted

FS = 2 Impact factor of Safety

SWL = 35 ton shackle safe working Load

N = 1 No.s No. of Lugs

a = 90 deg. in Plane Force Angle

b = 0 deg. Out of Plane Force Angle

Wfactored = W X FS = 1700 X 2 = 3,400 kg

= 33.34 kN factored Load

Ts = W / N = 1700 / 1 = 1,700 kg Max Tension on Padeye

= 16.671 kN

P = Ts X FS = 16.671 X 2 = 33.34 kN < SWL OK

Fz = P Sin (a) = 33.343 kN Vertical Force on Padeye

Fy = P Sin (b) = 0.000 kN Out of Plan Force on Padeye

Fx = P Cos (a) = 0.000 kN Horizontal Force on Padeye

2- Dimensioning:

a- Skid Lug Sizing: Profle: W10X49

k = 70 mm Distance from lug hole to edge of beam

rL = 70 mm Lug radius

tL = 35 mm Lug thickness, tL

tcp = 12 mm Collar plate thickness, tcp

d = 62 mm Diameter of hole, d

hL = 193 mm Distance from lug hole to base, hL

wL = 253 mm Lug base width, wL

Dcp = 100 mm Collar ring diameter, Dcp

H = ( 253 )/2 - 70 = 56.50

a- Shakle Sizing:

B = 57 mm Pin Dia, Dp

A = 83 mm Shackle Clearance

C = 197 mm Shackle Depth Clearance

c- Clearance between shackle & Lug :

59 mm

A = 83 mm

tLt < A OK

rL = 70 mm

C = 197 mm

rL < C OK

OK so Lug is Acceptable

d- Lug Hole Diameter Check:

- Max of : B X 1.05 = 57 X 1.05 = 59.85 mm

or B+3 = 57 + 3.00 = 60.00 mm

Max = 60.00 mm

- Also d< (B+6) Shall be Satisfied

57 + 6.00 = 63.00 mm

60.00 < 62 < 63.00

So d is: Acceptable

3- Lifting Lug Material & Properties:

ASTM A36

Fy = 248.21 MPa Yield Strength

fby all = 0.66 Fy = 163.82 MPa Allowable bending Stress in Plane

fbz all = 0.75 Fy = 186.16 MPa Allowable bending Stress out of Plane

ft all = 0.60 Fy = 148.93 MPa Allowable tensile Stress

fbr all = 0.90 Fy = 223.39 MPa Allowable Bearing Stress

fs all = 0.40 Fy = 99.28 MPa Allowable Shear Stress

For Skid VESSEL

33.343 X Sin (90) =

33.343 X Sin (0) =

33.343 X Cos (90) =

tLT = tL + 2 X tcp = 35 + 2 X12 =

Hence Dimensions are

Material Used :

H

4- Stress Check at Base of Lug:

a- Bending Stresses:

My = in Plane Moment

= -6,425,120.95 N.mm

fby = in Plane Stress

= -17.21 MPa

fby < fby all

17.21 < 163.82 in Plane is: OK

Mz = (Fy X hL) Out of Plane Moment

= 0.00 N.mm

fbz = Out of Plane Stress

= 0.00 MPa

fbz < fbz all

0.00 < 186.16 out of Plane is: OK

b- Tensile Stresses:

ft =

= 0.00 MPa

ft < ft all

0.00 < 148.93 tensile is: OK

c- Combined Stresses:

U =

= 0.11 < 1.00

OK

d- Shear Stresses:

fsz = in Plane Stress

= 3.77 MPa

fsz < fs all

3.77 < 99.28 In Shear is: OK

fsy = Out of Plane Stress

= 0.00 MPa

fsy < fs all

0.00 < 99.28 Out Shear is: OK

e- Von-Mises Stresses:

fz = ft + fby = 0 + 17.21 = 17.21 MPa

fy = ft + fbz = 0 + 0 = 0.00 MPa

fxy = fsz2 + fsy

2 Average Shear Stresses

= = 3.77 MPa

fcomb =

=

=

fcomb all = 0.66 Fy = 163.82 MPa

fcomb < fcomb all

15.37 < 163.82 So: OK

( 0 X 192.7 ) =

(Fx X H)-(Fz X hL)

( 0 X 56.5 ) - ( 33342.61 X 192.7 ) =

6 My / (tL X wL2

)

6 X -6425120.95 / ( 35 X 253 ^2 ) =

fz2 + fy

2 - (fz + fy+3fxy2)

6 My / (wL X tL2 )

6 X 0 / ( 253 X 35 ^2 ) =

Fx / (tL X wL )

0/ ( 35 X 253 ) =

(17.21 / 163.82) + (0 / 186.16) + (0 / 148.93)

So Combined Stressess are:

Fz / (tL X wL )

33342.61/ ( 35 X 253 ) =

Fy / (tL X wL )

0/ ( 35 X 253 ) =

3.77^2 + 0^2

17.21^2 + 0^2 - ( 17.21 + 0 + 3X[3.77]^2 )

15.37 MPa

5- Stress Check at Pin Hole:

a- Tensile Stresses:

P = 33,342.61 N

At =

= [ 2 X ( 35 X (70 - 62/2) ) ] + 2 X [ 2 X ( 12 X ( (100/2) - 62/2 ) ) ]

= 3642.00 mm2

ft = P/At = 9.16 MPa

ft < ft all

9.16 < 148.93 Padeye T is: OK

b- Bearing Stresses:

Abr =

= 57 X (35 + 2 X 12 ) t cp

= 3363.00 mm2

fbr = P/Abr = 9.91 MPa t L

fbr < fbr all

9.91 < 223.39 Padeye Br is: OK t cp

B , Shackle Bolt Dia

c- Shear Stresses:

As = At & fs = ft

fs < fs all

9.16 < 99.28 Padeye S is: OK

33342.61 / 3363 = Lug Thickness

Collar Plate Thickness

[ 2 X ( tL X (rL - d/2) ) ] + 2 X [ 2 X (tcp X ( (Dcp / 2) - d/2) ) ]

33342.61 / 3642 =

B X (tL + 2tcp )

Collar Plate Thickness

At

At