column effective length calculation
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DESIGN SHEETJob No. 8500 Building No. Rev. No. 0
Customer ADCO-Abu Dhabi Company for Onshore Oil Operations Date 13/5/09
Project SAS Full Field Development Designed By FBFjr.
Description Calculation of kz Checked By
START BY SELECTING YOUR TYPE OF STRUCTURE
BIENVENIDO! THIS SPREADSHEET WAS DESIGNED TO CALCULATE FOR THE EFFECTIVE LENGTH FACTOR, kz AS PER REQUIREMENT OF BS 5950-PART 1:2000. THE AIM IS TO MAKE THIS NOTE USER-FRIENDLY AND THE APPROACH WAS MADE AS SIMPLE AS POSSIBLE. ANY SUGGESTIONS/INPUTS
FOR FURTHER IMPROVEMENT WILL BE GREATLY APPRECIATED.------FBFernandezJr.
FIVE LEVELS-Exterior Columns FIVE LEVELS-Interior Columns
FOUR LEVELS-Exterior Columns FOUR LEVELS-Interior Columns
THREE LEVELS-Exterior Columns THREE LEVELS-Interior Columns
TWO/ONE LEVEL-Exterior Columns TWO/ONE LEVEL-Interior Columns
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK 5
Level 5
5.00 mLevel 4
4.00 m
Level 3
3.00 m
Level 2
2.00 m
Level 1
1.00 m
3.00 m
COLUMN LOCATION INTERIOR COLUMNS
NUMBER OF LEVELS 5
ENTER SECTION PROFILES
COLUMN PROFILE LENGTH, mk FACTORS
UC 203x203x60 6125 1.00 0.8182 1.0000 3.85
UC 203x203x60 6125 2.00 0.7143 0.8182 2.24
UC 203x203x60 6125 3.00 0.6364 0.7143 1.87
UC 203x203x60 6125 4.00 0.5745 0.6364 1.69
UC 203x203x60 6125 5.00 0.3750 0.5745 1.46
LEFT SECTION LENGTH, m
UC 203x203x60 6125 3.00
UC 203x203x60 6125 3.00
UC 203x203x60 6125 3.00
UC 203x203x60 6125 3.00
UC 203x203x60 6125 3.00
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
Segment 3
Segment 4
Segment 5
BEAM PROFILE
MOMENT OF INERTIA, cm4
Level 1
Level 2
Level 3
Level 4
Level 5
Back to HOME
1
2
3
4
5
Back to HOME
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK 5
Level 5
5.00 metersLevel 4
4.00 meters
Level 3
3.00 meters
Level 2
2.00 meters
Level 1
1.00 meters
3.00 meters
WARNING: PLEASE DO NO CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 UC 203x203x60 6125 1.00
Segment 2 UC 203x203x60 6125 2.00
Level 1 UC 203x203x60 6125 3.00
61.25
30.625
0.00
0.00
20.416666667
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.8182 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
1
2
3
4
5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 3.85
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 UC 203x203x60 6125 1.00
Segment 2 UC 203x203x60 6125 2.00
Segment 3 UC 203x203x60 6125 3.00
Level 1 UC 203x203x60 6125 3.00
Level 2 UC 203x203x60 6125 3.00
30.63
20.42
61.25
0.00
20.416666667
20.42
0.00
Figure E-3-Distribution factors for continuous columns
0.7143 from Figure E.3 page 182
0.8182
from Figure E.2 page 181
thus, kz = 2.24
C. FOR SEGMENT 3
ELEMENT SECTION PROFILE
Segment 2 UC 203x203x60 6125 2.00
Segment 3 UC 203x203x60 6125 3.00
Segment 4 UC 203x203x60 6125 4.00
Level 2 UC 203x203x60 6125 3.00
Level 3 UC 203x203x60 6125 3.00
20.42
15.31
30.63
0.00
20.416666667
20.42
0.00
Figure E-3-Distribution factors for continuous columns
0.6364 from Figure E.3 page 182
0.7143
from Figure E.2 page 181
thus, kz = 1.87
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
D. FOR SEGMENT 4
ELEMENT SECTION PROFILE
Segment 3 UC 203x203x60 6125 3.00
Segment 4 UC 203x203x60 6125 4.00
Segment 5 UC 203x203x60 6125 5.00
Level 3 UC 203x203x60 6125 3.00
Level 4 UC 203x203x60 6125 3.00
15.31
12.25
20.42
0.00
20.416666667
20.42
0.00
Figure E-3-Distribution factors for continuous columns
0.5745 from Figure E.3 page 182
0.6364
from Figure E.2 page 181
thus, kz = 1.69
E. FOR SEGMENT 5
ELEMENT SECTION PROFILE
Segment 4 UC 203x203x60 6125 4.00Segment 5 UC 203x203x60 6125 5.00
Level 4 UC 203x203x60 6125 3.00Level 5 UC 203x203x60 6125 3.00
12.25
0.00
15.31
0.00
20.416666667
20.42
0.00
Figure E-3-Distribution factors for continuous columns
0.3750 from Figure E.3 page 182
0.5745
from Figure E.2 page 181
thus, kz = 1.46
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURESTEEL PIPERACK 5
Level 5
7.00 mLevel 4
6.00 m
Level 3
5.00 m
Level 2
4.00 m
Level 1
3.00 m
4.00 m 5.00 m
COLUMN LOCATION INTERIOR COLUMNS
NUMBER OF LEVELS 5
ENTER SECTION PROFILES (in red fonts)
COLUMN PROFILE LENGTH, mk FACTORS
IPE220 2770 3.00 0.9722 1.0000 9.38
IPE240 3890 4.00 0.9011 0.9722 4.39
IPE120 318 5.00 0.7729 0.9011 2.72
IPE200 1940 6.00 0.5669 0.7729 1.89
IPE180 1320 7.00 0.3253 0.5669 1.42
LEFT SECTION LENGTH, m RIGHT SECTION LENGTH, m
IPE80 80.1 4.00 IPE100 171 5.00
IPE120 318 4.00 IPE100 171 5.00
IPE120 318 4.00 IPE100 171 5.00
IPE160 869 4.00 IPE160 869 5.00
IPE160 869 4.00 IPE160 869 5.00
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
Segment 3
Segment 4
Segment 5
BEAM PROFILE
MOMENT OF INERTIA, cm4
MOMENT OF INERTIA, cm4
Level 1
Level 2
Level 3
Level 4
Level 5
1
2
3
4
5
Back to HOME
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK 5
Level 5
7.00 mLevel 4
6.00 m
Level 3
5.00 m
Level 2
4.00 m
Level 1
3.00 m
4.00 m 5.00 m
WARNING: PLEASE DO NOT CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 IPE220 2770 3.00
Segment 2 IPE240 3890 4.00
Level 1Left IPE80 80.1 4.00
Right IPE100 171 5.00
9.2333333333
9.725
0.00
0.20
0.342
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.9722 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
thus, kz = 9.38
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
1
2
3
4
5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 IPE220 2770 3.00
Segment 2 IPE240 3890 4.00
Segment 3 IPE120 318 5.00
Level 1Left IPE80 80.1 4.00
Right IPE100 171 5.00
Level 2Left IPE120 318 4.00
Right IPE100 171 5.00
9.73
0.64
9.23
0.80
0.342
0.34
0.20
Figure E-3-Distribution factors for continuous columns
0.9011 from Figure E.3 page 182
0.9722
from Figure E.2 page 181
thus, kz = 4.39
C. FOR SEGMENT 3
ELEMENT SECTION PROFILE
Segment 2 IPE240 3890 4.00
Segment 3 IPE120 318 5.00
Segment 4 IPE200 1940 6.00
Level 2Left IPE120 318 4.00
Right IPE100 171 5.00
Level 3Left IPE120 318 4.00
Right IPE100 171 5.00
0.64
3.23
9.73
0.80
0.342
0.34
0.80
Figure E-3-Distribution factors for continuous columns
0.7729 from Figure E.3 page 182
0.9011
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 2.72
D. FOR SEGMENT 4
ELEMENT SECTION PROFILE
Segment 3 IPE120 318 5.00
Segment 4 IPE200 1940 6.00
Segment 5 IPE180 1320 7.00
Level 3Left IPE120 318 4.00
Right IPE100 171 5.00
Level 4Left IPE160 869 4.00
Right IPE160 869 5.00
3.23
1.89
0.64
2.17
1.738
0.34
0.80
Figure E-3-Distribution factors for continuous columns
0.5669 from Figure E.3 page 182
0.7729
from Figure E.2 page 181
thus, kz = 1.89
E. FOR SEGMENT 5
ELEMENT SECTION PROFILE
Segment 4 IPE200 1940 6.00
Segment 5 IPE180 1320 7.00
Level 4Left IPE160 869 4.00
Right IPE160 869 5.00
Level 5Left IPE160 869 4.00
Right IPE160 869 5.00
1.89
0.00
3.23
2.17
1.738
1.74
2.17
Figure E-3-Distribution factors for continuous columns
0.3253 from Figure E.3 page 182
0.5669
from Figure E.2 page 181
thus, kz = 1.42
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK 5
Level 4
3.00 m
Level 3
2.50 m
Level 2
2.10 m
Level 1
4.25 m
2.50 m
COLUMN LOCATION INTERIOR COLUMNS
NUMBER OF LEVELS 4BASE CONNETION PINNEDENTER SECTION PROFILES
COLUMN PROFILE LENGTH, mk FACTORS
UC 356x368x129 40250 4.25 0.8182 1.0000 3.21
UC 356x368x129 40250 2.10 0.7143 0.8182 2.49
UC 356x368x129 40250 2.50 0.6364 0.7143 3.08
UC 356x368x129 40250 3.00 0.5745 0.6364 2.60
LEFT SECTION LENGTH, m
UB 406x178x74 27310 2.50
UC 254x254x73 11410 2.50
UC 254x254x73 11410 2.50
UB 305x165x40 8503 2.50
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
Segment 3
Segment 4
BEAM PROFILE
MOMENT OF INERTIA, cm4
Level 1
Level 2
Level 3
Level 4
Back to HOME
1
2
3
4
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK 5
Level 4
3.00 meters
Level 3
2.50 meters
Level 2
2.10 meters
Level 1
4.25 meters
2.50 meters
WARNING: PLEASE DO NO CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 UC 356x368x129 40250 4.25
Segment 2 UC 356x368x129 40250 2.10
Level 1 UB 406x178x74 27310 2.50
94.70588235
191.6666667
0.00
0.00
109.24
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.7239 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
1
2
3
4
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 3.21
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 UC 356x368x129 40250 4.25
Segment 2 UC 356x368x129 40250 2.10
Segment 3 UC 356x368x129 40250 2.50
Level 1 UB 406x178x74 27310 2.50
Level 2 UC 254x254x73 11410 2.50
191.67
161.00
94.71
0.00
45.64
109.24
0.00
Figure E-3-Distribution factors for continuous columns
0.8854 from Figure E.3 page 182
0.7239
from Figure E.2 page 181
thus, kz = 2.49
C. FOR SEGMENT 3
ELEMENT SECTION PROFILE
Segment 2 UC 356x368x129 40250 2.10
Segment 3 UC 356x368x129 40250 2.50
Segment 4 UC 356x368x129 40250 3.00
Level 2 UC 254x254x73 11410 2.50
Level 3 UC 254x254x73 11410 2.50
161.00
134.17
191.67
0.00
45.64
45.64
0.00
Figure E-3-Distribution factors for continuous columns
0.8661 from Figure E.3 page 182
0.8854
from Figure E.2 page 181
thus, kz = 3.08
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
D. FOR SEGMENT 4
ELEMENT SECTION PROFILE
Segment 3 UC 356x368x129 40250 3.00
Segment 4 UC 356x368x129 40250 3.00
Level 3 UC 254x254x73 11410 2.50
Level 4 UB 305x165x40 8503 2.50
134.17
0.00
134.17
0.00
34.012
45.64
0.00
Figure E-3-Distribution factors for continuous columns
0.7978 from Figure E.3 page 182
0.8546
from Figure E.2 page 181
thus, kz = 2.60
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURESTEEL PIPERACK 5
Level 4
6.00 m
Level 3
5.00 m
Level 2
3.00 m
Level 1
4.00 m
7.00 m 4.00 m
COLUMN LOCATION INTERIOR COLUMNS
NUMBER OF LEVELS 4
ENTER SECTION PROFILES (in red fonts)
COLUMN PROFILE LENGTH, mk FACTORS
W14X370 226429.12 4.00 0.9722 1.0000 3.05
W14X370 226429.12 3.00 0.9011 0.9722 1.91
W14X211 110717.18 5.00 0.7729 0.9011 1.70
W14X257 141518.2 6.00 0.5669 0.7729 1.39
LEFT SECTION LENGTH, m RIGHT SECTION LENGTH, m
W14X370 226429.12 7.00 W14X211 110717.18 4.00
W14X211 110717.18 7.00 W14X211 110717.18 4.00
W14X211 110717.18 7.00 W14X211 110717.18 4.00
W14X211 110717.18 7.00 W14X211 110717.18 4.00
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
Segment 3
Segment 4
BEAM PROFILE
MOMENT OF INERTIA, cm4
MOMENT OF INERTIA, cm4
Level 1
Level 2
Level 3
Level 4
Back to HOME
1
2
3
4
Back to HOME
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK
Level 4
6.00 m
Level 3
5.00 m
Level 2
3.00 mLevel 1
4.00 m
7.00 m 4.00 m
WARNING: PLEASE DO NOT CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 W14X370 226429.12 4.00
Segment 2 W14X370 226429.12 3.00
Level 1Left W14X370 226429.12 7.00
Right W14X211 110717.18 4.00
566.0728
754.76373333
0.00
323.47
276.79295
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.6875 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
thus, kz = 3.05
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
4
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
1
2
3
4
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 W14X370 226429.12 4.00
Segment 2 W14X370 226429.12 3.00
Segment 3 W14X211 110717.18 5.00
Level 1Left W14X370 226429.12 7.00
Right W14X211 110717.18 4.00
Level 2Left W14X211 110717.18 7.00
Right W14X211 110717.18 4.00
754.76
221.43
566.07
158.17
276.79295
276.79
323.47
Figure E-3-Distribution factors for continuous columns
0.6918 from Figure E.3 page 182
0.6875
from Figure E.2 page 181
thus, kz = 1.91
C. FOR SEGMENT 3
ELEMENT SECTION PROFILE
Segment 2 W14X370 226429.12 3.00
Segment 3 W14X211 110717.18 5.00
Segment 4 W14X257 141518.2 6.00
Level 2Left W14X211 110717.18 7.00
Right W14X211 110717.18 4.00
Level 3Left W14X211 110717.18 7.00
Right W14X211 110717.18 4.00
221.43
235.86
754.76
158.17
276.79295
276.79
158.17
Figure E-3-Distribution factors for continuous columns
0.5125 from Figure E.3 page 182
0.6918
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 1.70
D. FOR SEGMENT 4
ELEMENT SECTION PROFILE
Segment 3 W14X211 110717.18 5.00
Segment 4 W14X257 141518.2 6.00
Level 3Left W14X211 110717.18 7.00
Right W14X211 110717.18 4.00
Level 4Left W14X211 110717.18 7.00
Right W14X211 110717.18 4.00
235.86
0.00
221.43
158.17
276.79295
276.79
158.17
Figure E-3-Distribution factors for continuous columns
0.3516 from Figure E.3 page 182
0.5125
from Figure E.2 page 181
thus, kz = 1.39
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)STRUCTURE: STEEL PIPERACK
Level 3
2.50 m
Level 2
2.10 m
Level 1
5.25 m
4.00 m
COLUMN LOCATION EXTERIOR COLUMNS
NUMBER OF LEVELS 3BASE CONNETION PINNEDENTER SECTION PROFILES
COLUMN PROFILE LENGTH, mk FACTORS
UC 356x368x129 40250 5.25 0.8182 1.0000 3.99
UC 356x368x129 40250 2.10 0.7143 0.8182 3.15
UC 356x368x129 40250 2.50 0.6364 0.7143 3.26
LEFT SECTION LENGTH, m
UB 406x178x60 21600 4.00
UC 254x254x73 11410 4.00
UC 254x254x73 11410 4.00
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
Segment 3
BEAM PROFILE
MOMENT OF INERTIA, cm4
Level 1
Level 2
Level 3
Back to HOME
1
2
3
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK 5
Level 3
2.50 meters
Level 2
2.10 meters
Level 1
5.25 meters
4.00 meters
WARNING: PLEASE DO NO CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 UC 356x368x129 40250 5.25
Segment 2 UC 356x368x129 40250 2.10
Level 1 UB 406x178x60 21600 4.00
76.66666667
191.6666667
0.00
0.00
54
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.8325 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
3
1
2
3
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 3.99
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 UC 356x368x129 40250 5.25
Segment 2 UC 356x368x129 40250 2.10
Segment 3 UC 356x368x129 40250 2.50
Level 1 UB 406x178x60 21600 4.00
Level 2 UC 254x254x73 11410 4.00
191.67
161.00
76.67
0.00
28.525
54.00
0.00
Figure E-3-Distribution factors for continuous columns
0.9252 from Figure E.3 page 182
0.8325
from Figure E.2 page 181
thus, kz = 3.15
C. FOR SEGMENT 3
ELEMENT SECTION PROFILE
Segment 2 UC 356x368x129 40250 2.10
Segment 3 UC 356x368x129 40250 2.50
Level 2 UC 254x254x73 11410 4.00
Level 3 UC 254x254x73 11410 4.00
161.00
0.00
191.67
0.00
28.525
28.53
0.00
Figure E-3-Distribution factors for continuous columns
0.8495 from Figure E.3 page 182
0.9252
from Figure E.2 page 181
thus, kz = 3.26
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT:
LOCATION:
CLIENT:
STRUCTURE:
Level 2
2.03 m
Level 1
6.50 m
1.60 m
COLUMN LOCATION EXTERIOR COLUMNS
NUMBER OF LEVELS 2BASE CONNETION FIXED
ENTER SECTION PROFILES
COLUMN PROFILE LENGTH, mk FACTORS
UC 356x368x129 40250 6.50 0.0000 0.0000 1.72
UC 356x368x129 40250 2.03 0.9043 0.0000 3.29
LEFT SECTION LENGTH, m
UB 254x146x31 4413 1.60
UB 254x146x31 4413 1.60
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
BEAM PROFILE
MOMENT OF INERTIA, cm4
Level 1
Level 2
Back to HOME
1
2
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURESTEEL PIPERACK 5
Level 3
4.00 m
Level 2
3.00 m
Level 1
6.00 m
4.50 m 5.00 m
COLUMN LOCATION INTERIOR COLUMNS
NUMBER OF LEVELS 3
ENTER SECTION PROFILES (in red fonts)
COLUMN PROFILE LENGTH, mk FACTORS
HEA240 7760 6.00 0.9722 1.0000 2.25
HEA240 7760 3.00 0.9011 0.9722 1.25
HEA240 7760 4.00 0.7729 0.9011 1.27
LEFT SECTION LENGTH, m RIGHT SECTION LENGTH, m
HEA360 33090 4.50 HEA240 7760 5.00
HEA360 33090 4.50 HEA240 7760 5.00
HEA240 7760 4.50 HEA240 7760 5.00
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
Segment 3
BEAM PROFILE
MOMENT OF INERTIA, cm4
MOMENT OF INERTIA, cm4
Level 1
Level 2
Level 3
Back to HOME
1
2
3
Back to HOME
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK
Level 3
4.00 m
Level 2
3.00 m
Level 1
6.00 m
4.50 m 5.00 m
WARNING: PLEASE DO NOT CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 HEA240 7760 6.00
Segment 2 HEA240 7760 3.00
Level 1Left HEA360 33090 4.50
Right HEA240 7760 5.00
12.933333333
25.866666667
0.00
73.53
15.52
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.3035 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
thus, kz = 2.25
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
1
2
3
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 HEA240 7760 6.00
Segment 2 HEA240 7760 3.00
Segment 3 HEA240 7760 4.00
Level 1Left HEA360 33090 4.50
Right HEA240 7760 5.00
Level 2Left HEA360 33090 4.50
Right HEA240 7760 5.00
25.87
19.40
12.93
73.53
15.52
15.52
73.53
Figure E-3-Distribution factors for continuous columns
0.3370 from Figure E.3 page 182
0.3035
from Figure E.2 page 181
thus, kz = 1.25
C. FOR SEGMENT 3
ELEMENT SECTION PROFILE
Segment 2 HEA240 7760 4.00
Segment 3 HEA240 7760 4.00
Level 2Left HEA360 33090 4.50
Right HEA240 7760 5.00
Level 3Left HEA240 7760 4.50
Right HEA240 7760 5.00
19.40
0.00
19.40
17.24
15.52
15.52
73.53
Figure E-3-Distribution factors for continuous columns
0.3719 from Figure E.3 page 182
0.3035
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 1.27
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT:LOCATION:CLIENT:DESIGNER:STRUCTURE:
Level 2
2.03 meters
Level 1
6.50 meters
1.60 meters
WARNING: PLEASE DO NO CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 UC 356x368x129 40250 6.50
Segment 2 UC 356x368x129 40250 2.03
Level 1 UB 254x146x31 4413 1.60
61.92307692
198.7654321
0.00
0.00
27.58125
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.9043 from Figure E.3 page 182
0.0000
from Figure E.2 page 181
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
for pinned base connection k2 =1
for fixed base connection k2 =0
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
1
2
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
thus, kz = 1.72
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 UC 356x368x129 40250 6.50
Segment 2 UC 356x368x129 40250 2.03
Level 1 UB 254x146x31 4413 1.60
Level 2 UB 254x146x31 4413 1.60
198.77
0.00
61.92
0.00
27.58125
27.58
0.00
Figure E-3-Distribution factors for continuous columns
0.8781 from Figure E.3 page 182
0.9043
from Figure E.2 page 181
thus, kz = 3.29
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURESTEEL PIPERACK 5
Level 2
0.00 m
Level 1
5.80 m
12.00 m 12.00 m
COLUMN LOCATION INTERIOR COLUMNS
NUMBER OF LEVELS 2
ENTER SECTION PROFILES (in red fonts)
COLUMN PROFILE LENGTH, mk FACTORS
UC 356x406x551 226900 5.80 0.9722 1.0000 7.80
none 0 0.00 0.9011 0.9722 N/A
LEFT SECTION LENGTH, m RIGHT SECTION LENGTH, m
UB 305x165x46 9899 12.00 UB 305x165x46 9899 12.00
none 0 0.00 none 0 0.00
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
Segment 2
BEAM PROFILE
MOMENT OF INERTIA, cm4
MOMENT OF INERTIA, cm4
Level 1
Level 2
Back to HOME
1
2
CALCULATION OF EFFECTIVE LENGTH RATIO-BS 5950 METHODReference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)STRUCTURE: STEEL PIPERACK
Level 1 1.09 m
1.20 m
COLUMN LOCATION EXTERIOR COLUMNS
NUMBER OF LEVELS 1BASE CONNETION FIXED
ENTER SECTION PROFILES
COLUMN PROFILE LENGTH, mk FACTORS
UC 152x152x30 1748 1.09 0.9094 0.0000 1.73
LEFT SECTION LENGTH, m
UC 152x152x30 1748 1.20
MOMENT OF INERTIA, cm4
COLUMN EFFECTIVE LENGTH FACTOR, KZk1 k2
Segment 1
BEAM PROFILE
MOMENT OF INERTIA, cm4
Level 1
Back to HOME
1
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)STRUCTURE: STEEL PIPERACK
1.09 meters
1.20 meters
A. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 UC 152x152x30 40250 1.09
Level 1 UC 152x152x30 4413 1.20
369.266055
0.00
0.00
0.00
36.775
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.9094 from Figure E.3 page 182
0.0000
from Figure E.2 page 181
thus, kz = 1.73
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
for pinned base connection k2 =1
for fixed base connection k2 =0
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
1
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 UC 152x152x30 40250 1.09
Segment 2 #REF! #REF! #REF!
Level 1 UC 152x152x30 4413 1.20
Level 2 UB 254x146x31 4413 1.60
#REF!
0.00
369.27
0.00
27.58125
36.78
0.00
Figure E-3-Distribution factors for continuous columns
#REF! from Figure E.3 page 182
#REF!
from Figure E.2 page 181
thus, kz = #REF!
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
CALCULATION SHEETCALCULATION OF STIFFNESS COEFFICIENT-BS 5950 METHOD
Reference: BS 5950 Part 1:2000 Annex E(normative) pages 178-182
PROJECT: SAS FULL FIELDS DEVELOPMENTLOCATION: ABU DHABI, UNITED ARAB EMIRATESCLIENT: ABU DHABI COMPANY FOR OIL OPERATIONS (ADCO)DESIGNER: FEDERICO FERNANDEZ, JR.STRUCTURE: STEEL PIPERACK
Level 2
0.00 m
Level 1
5.80 m
12.00 m 12.00 m
WARNING: PLEASE DO NOT CHANGE ANY ENTRYA. FOR SEGMENT 1
ELEMENT SECTION PROFILE
Segment 1 UC 356x406x551 226900 5.80
Segment 2 none 0 0.00
Level 1Left UB 305x165x46 9899 12.00
Right UB 305x165x46 9899 12.00
391.20689655
0.0
0.00
8.25
8.2491666667
0.00
0.00
Figure E-3-Distribution factors for continuous columns
0.9595 from Figure E.3 page 182
1.0000
from Figure E.2 page 181
thus, kz = 7.80
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
1
2
B. FOR SEGMENT 2
ELEMENT SECTION PROFILE
Segment 1 UC 356x406x551 226900 5.80
Segment 2 none 0 0.00
Level 1Left UB 305x165x46 9899 12.00
Right UB 305x165x46 9899 12.00
Level 2Left none 0 0.00
Right none 0 0.00
0.00
0.00
391.21
0.00
0
8.25
8.25
Figure E-3-Distribution factors for continuous columns
N/A from Figure E.3 page 182
N/A
from Figure E.2 page 181
thus, kz = N/A
MOMENT OF INERTIA, IY (cm4)
LENGTH (meter)
KC= cm3
KU= cm3
KL= cm3
KTL= cm3
KTR= cm3
KBR= cm3
KBL= cm3
k1 = [kC+kU]/[kC+kU+kTL+kTR] =
k2 = [kC+kL]/[kC+kL+kBL+kBR] =
kc=I/Lcolumn to be designed
KL
KBRKBL
Ku
KTRKTL K1
K2
kz=LEL
=[ 1−0 .2 (k1+k 2)−0 .12k1 k21−0 .8 (k1+k2)+0.6 k1k2 ]
0 .5
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