column effective length calculation

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


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


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


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



0.8182


thus, kz = 2.24

C. FOR SEGMENT 3


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



0.7143


thus, kz = 1.87


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


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



0.6364


thus, kz = 1.69

E. FOR SEGMENT 5


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



0.5745


thus, kz = 1.46


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


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


NUMBER OF LEVELS 5

ENTER SECTION PROFILES (in red fonts)


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



Segment 1

Segment 2

Segment 3

Segment 4

Segment 5

BEAM PROFILE



Level 1

Level 2

Level 3

Level 4

Level 5

1

2

3

4

5

Back to HOME




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


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



1.0000


thus, kz = 9.38


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3



1

2

3

4

5


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


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



0.9722


thus, kz = 4.39

C. FOR SEGMENT 3


Segment 2 IPE240 3890 4.00


Segment 4 IPE200 1940 6.00


Right IPE100 171 5.00


Right IPE100 171 5.00

0.64

3.23

9.73

0.80

0.342

0.34

0.80



0.9011



LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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



Segment 4 IPE200 1940 6.00

Segment 5 IPE180 1320 7.00


Right IPE100 171 5.00


Right IPE160 869 5.00

3.23

1.89

0.64

2.17

1.738

0.34

0.80



0.7729


thus, kz = 1.89

E. FOR SEGMENT 5


Segment 4 IPE200 1940 6.00

Segment 5 IPE180 1320 7.00


Right IPE160 869 5.00


Right IPE160 869 5.00

1.89

0.00

3.23

2.17

1.738

1.74

2.17



0.5669


thus, kz = 1.42


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


Level 4

3.00 m

Level 3

2.50 m

Level 2

2.10 m

Level 1

4.25 m

2.50 m


NUMBER OF LEVELS 4BASE CONNETION PINNEDENTER SECTION PROFILES


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


UB 406x178x74 27310 2.50

UC 254x254x73 11410 2.50

UC 254x254x73 11410 2.50

UB 305x165x40 8503 2.50



Segment 1

Segment 2

Segment 3

Segment 4

BEAM PROFILE


Level 1

Level 2

Level 3

Level 4

Back to HOME

1

2

3

4




Level 4

3.00 meters

Level 3

2.50 meters

Level 2

2.10 meters

Level 1

4.25 meters

2.50 meters



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



1.0000



LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3



1

2

3

4


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


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



0.7239


thus, kz = 2.49

C. FOR SEGMENT 3


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



0.8854


thus, kz = 3.08


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


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



0.8546


thus, kz = 2.60


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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



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


NUMBER OF LEVELS 4



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


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



Segment 1

Segment 2

Segment 3

Segment 4

BEAM PROFILE



Level 1

Level 2

Level 3

Level 4

Back to HOME

1

2

3

4

Back to HOME



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



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



1.0000


thus, kz = 3.05


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3



4


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


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



0.6875


thus, kz = 1.91

C. FOR SEGMENT 3


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



0.6918



LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


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



0.5125


thus, kz = 1.39


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


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


UB 406x178x60 21600 4.00

UC 254x254x73 11410 4.00

UC 254x254x73 11410 4.00



Segment 1

Segment 2

Segment 3

BEAM PROFILE


Level 1

Level 2

Level 3

Back to HOME

1

2

3




Level 3

2.50 meters

Level 2

2.10 meters

Level 1

5.25 meters

4.00 meters



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



1.0000



LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3



3

1

2

3


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


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



0.8325


thus, kz = 3.15

C. FOR SEGMENT 3


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



0.9252


thus, kz = 3.26


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


PROJECT:

LOCATION:

CLIENT:

STRUCTURE:

Level 2

2.03 m

Level 1

6.50 m

1.60 m


NUMBER OF LEVELS 2BASE CONNETION FIXED



UC 356x368x129 40250 6.50 0.0000 0.0000 1.72

UC 356x368x129 40250 2.03 0.9043 0.0000 3.29


UB 254x146x31 4413 1.60

UB 254x146x31 4413 1.60



Segment 1

Segment 2

BEAM PROFILE


Level 1

Level 2

Back to HOME

1

2



Level 3

4.00 m

Level 2

3.00 m

Level 1

6.00 m

4.50 m 5.00 m


NUMBER OF LEVELS 3



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


HEA360 33090 4.50 HEA240 7760 5.00

HEA360 33090 4.50 HEA240 7760 5.00

HEA240 7760 4.50 HEA240 7760 5.00



Segment 1

Segment 2

Segment 3

BEAM PROFILE



Level 1

Level 2

Level 3

Back to HOME

1

2

3

Back to HOME




Level 3

4.00 m

Level 2

3.00 m

Level 1

6.00 m

4.50 m 5.00 m



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



1.0000


thus, kz = 2.25


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


Segment 1 HEA240 7760 6.00

Segment 2 HEA240 7760 3.00

Segment 3 HEA240 7760 4.00


Right HEA240 7760 5.00


Right HEA240 7760 5.00

25.87

19.40

12.93

73.53

15.52

15.52

73.53



0.3035


thus, kz = 1.25

C. FOR SEGMENT 3


Segment 2 HEA240 7760 4.00

Segment 3 HEA240 7760 4.00


Right HEA240 7760 5.00


Right HEA240 7760 5.00

19.40

0.00

19.40

17.24

15.52

15.52

73.53



0.3035



LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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



PROJECT:LOCATION:CLIENT:DESIGNER:STRUCTURE:

Level 2

2.03 meters

Level 1

6.50 meters

1.60 meters



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



0.0000



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



1

2


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


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



0.9043


thus, kz = 3.29


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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



Level 2

0.00 m

Level 1

5.80 m

12.00 m 12.00 m


NUMBER OF LEVELS 2



UC 356x406x551 226900 5.80 0.9722 1.0000 7.80

none 0 0.00 0.9011 0.9722 N/A


UB 305x165x46 9899 12.00 UB 305x165x46 9899 12.00

none 0 0.00 none 0 0.00



Segment 1

Segment 2

BEAM PROFILE



Level 1

Level 2

Back to HOME

1

2



Level 1 1.09 m

1.20 m


NUMBER OF LEVELS 1BASE CONNETION FIXED



UC 152x152x30 1748 1.09 0.9094 0.0000 1.73


UC 152x152x30 1748 1.20



Segment 1

BEAM PROFILE


Level 1

Back to HOME

1




1.09 meters

1.20 meters

A. FOR SEGMENT 1


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



0.0000


thus, kz = 1.73


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




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


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


#REF! from Figure E.3 page 182

#REF!


thus, kz = #REF!


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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




Level 2

0.00 m

Level 1

5.80 m

12.00 m 12.00 m



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



1.0000


thus, kz = 7.80


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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


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


N/A from Figure E.3 page 182

N/A


thus, kz = N/A


LENGTH (meter)

KC= cm3

KU= cm3

KL= cm3

KTL= cm3

KTR= cm3

KBR= cm3

KBL= cm3




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

column effective length calculation

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