Lifting-lug.xls
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& 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
Degrees kgf -kgf kgf cm
] ] ] ] ] ]
15.000 175,000 1.500 131,250 35,168.332 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
Half Lifting Angle
Ø
Erection Weight
W
Impact Factor
i
F1=W/2
F1
F2=F1.tan(Ø)
F2
Arm for Bending
a
[ [ [ [ [ [
F2 a
Material of lifting Lug Yield Strength
Y
Allowable Stresses for
Lug Width
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
]
57.65
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
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. As per Page 264 of Bedner : Combined Max Shear=S's=[(Sb/2)²+Ss²]^0.5
S's S
[
kg / cm²
]
Combined Maximum Tension=(Sb/2)+S's
[
kg / cm²
]
504.884 1,006.465
Safety Factors In Shear = 0.6 X Yield / S's In Tension=Yield/S
---
[ [
---
] ]
2.787 2.330
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!
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
Degrees kgf -kgf kgf cm
] ] ] ] ] ]
15.000 175,000 1.500 131,250 35,168.332 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
Half Lifting Angle
Ø
Erection Weight
W
Impact Factor
i
F1=W/2
F1
F2=F1.tan(Ø)
F2
Arm for Bending
a
[ [ [ [ [ [
F2 a
Material of lifting Lug Yield Strength
Y
Allowable Stresses for
Lug Width
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
]
57.65
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
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. As per Page 264 of Bedner : Combined Max Shear=S's=[(Sb/2)²+Ss²]^0.5
S's S
[
kg / cm²
]
Combined Maximum Tension=(Sb/2)+S's
[
kg / cm²
]
504.884 1,006.465
Safety Factors In Shear = 0.6 X Yield / S's In Tension=Yield/S
---
[ [
---
] ]
2.787 2.330
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!
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