Lifting-lug.xls

& Boyce Mfg. Co. Ltd., process Equipment Division, Plant - 15, Vikhroli, Mumbai - 400079 MDC No. : 851

Client : Reliance Petroleum Ltd.

Page No.

170

JOB NO. : 110120

PMC : Bechtel UK  

Rev.

:3

Made By : SBS

Equipment No. : MV-RF414-C-113

Date

: 02/02/04

LIFTING LUG STRESSES IN BENDING, TENSION & SHEAR 

F1 Ø

F1=W/2

F1

F2=F1.tan(Ø)

F2

[ [ [ [ [

a

[

cm

]

29.000

[

--

]

IS2062 Gr A

[

kg / cm²

]

2345.00

Bending = 0.66Y

[

kg / cm²

]

1547.70

Shear = 0.4 x Y

[

kg / cm²

]

938.00

Tension = 0.6Y

[

kg / cm²

]

1407.00

Combined=0.9Y

[

kg / cm²

]

2110.50

W

[

cm

]

61.00

T

[

cm

]

10.00

Section Modulus=WT²/6

Z

[

cm³

]

1016.67

Bending Stress=F2 X a / Z

Sb

[

kg / cm²

]

1003.16

Shear Stress=F2 / (W X T)

Ss

[

kg / cm²

]

5 7 .6 5

Tension Stress=F1 / (W.T)

St

[

kg / cm²

]

215.16

Net Normal Stress=Sb+St

Sn

[

kg / cm²

]

1218.33

Sc [

kg / cm²

]

1221.05

Half Lifting Angle

Ø

Erection Weight

W

Impact Factor

i

Arm for Bending

Degrees kgf -kgf kgf

] ] ] ] ]

15.000 175,000 1.500 131,250 35,168.332

F2 a

Material of lifting Lug Yield Strength

Y

Allowable Stresses for

Lug Width Lug Thickness

Combined Stress={Sb+St+[(Sb+St)^2+4Ss^2]^0.5} / 2

SINCE ALL THE STRESSES ARE < ALLOWABLE STRESSES, THE DESIGN IS SAFE.

Combined Max Shear=S's=[(Sb/2)²+Ss²]^0.5 Combined Maximum Tension=(Sb/2)+S's

S's S

[

kg / cm²

]

[

kg / cm²

]

504.884 1,006.465

---

[ [

---

] ]

2.787 2.330

Safety Factors In Shear = 0.6 X Yield / S's In Tension=Yield/S

GODREJ & BOYCE MFG. CO. LTD., DESIGN DEPT., PLANT 15, MUMBAI, INDIA.

DESIGN OF INTERMEDIATE LIFTING LUG (TYPE "C") MDC No.: 216

Client

Job No. : 97025

Consultant : TOYO

Rev. : 0

Made By : NSL

Eqpt. No. : 2C - 450

Date : 26 / 03 / 97

Area of cross section

: TOYO

Page :

1

A

[

cm ^ 2

]

#REF!

Z

[

cm ^ 3

]

#REF!

M

[

kgf - cm ]

#VALUE!

Bending Stress = M / Z

Sb

[ kgf/cm^2 ]

#VALUE!

Shear Stress = 1.25 X F / A

Ss

[ kgf/cm^2 ]

#VALUE!

Combined maximum shear :

S's

[ kgf/cm^2 ]

#VALUE!

S

[ kgf/cm^2 ]

#VALUE!

= b.d - k.h

Section Modulus of Cross Section = (b.d^3 - h.k^3) / ( 6 x d )

Moment at column O/D = 1.25 X We X L

= [(Sb/2)^2+Ss^2]^0.5

Combined maximum tension : = (Sb / 2) + S's

Safety factors : in shear = 0.6 x Yield strength / S's

S.F.sh

[

-

]

#VALUE!

in tension = 0.6 x Yield strength / S

S.F.ten [

-

]

#VALUE!