design of steel frames

Design of steel frames using SAP2000 – Illustrative examples

CSI Portugal & Spain

Contents

• Introduction

• Example 1 – Column• Example 2 – Beam• Example 3 – Beam-column

2

• Example 4 – Planar frame

• Conclusion

• Example 5 – Spatial frame• Example 6 – Short class 4 column• Example 7 – Long class 4 column

Introduction

Objective:• Present illustrative examples concerning the safety check and design of steel members and structures using (i) EC 3 design formula and (ii) different SAP2000 design tools (based on frame or shell FE)

• SAP2000 provides tools to both (i) check the safety of steel frame structures according to Eurocode 3 and (ii) optimise their design

Scope:

3

• In order to fully exploit the potential of SAP2000 tools, it is necessary to know how to apply different EC 3 design methods in SAP2000

4

Example 1 – Column (1/3)• Spatial column (flexural buckling):

SAP2000 frame FE model SAP2000 shell FE model

• Simply supported for major and minor bending • Laterally unbraced• Torsion prevented at both extremities • S 235 steel, IPE 200 profile (class

1)IPE 200

5

Example 1 – Column (2/3)

kN

NN RdzRdb

02.19275.6692867.0

.

MPaNNf RdbEdyx 53.232.max.

kNLEIN zzcr

25.2405.3420.1210 2222

.

670.1

25.24075.669.

zcrRkz NN

2867.0, zz

Buckling curve bImperf.:

Model 3(SAP2000 shell)



EC 3 formulae/SAP2000 frame design

34.0mm

Le142500

mmAWe zel

98.4)2.0( .0

mkNLeNp Ed

/74.18 2

00

mkNLeNp Ed

/618.08 2

00

mmAWe zel

98.4)2.0( .0

Method: • Ayrton-Perry formula• equiv. lateral forces with imperf. according to Table 5.1 (EC 3)

• equiv. lateral forces with imperf. equiv. to buckling curves

• 2nd order shell FEM • 2nd order shell FEM • 2nd order shell FEM

• imperf. factor from buckling curves

• geometric imperf. equiv. to buckling curves

• Column design according to EC 3 formulae and thin-walled rectangular shell FE models:

kNAfN yRk 75.669235850.2

kNLpP 08.600 kNLpP 163.200

6

Example 1 – Column (3/3)

EC 3 formulae/SAP2000 frame design

Model 1(SAP2000 shell) Diff. Model 2

(SAP2000 shell) Diff. Model 3(SAP 2000 shell) Diff.

Ncr.z [kN] 240.25 239.68 -0.2% 239.68 -0.2% 239.68 -0.2%

x.max [MPa] 232.53 538.64 +132% 236.12 +1.5% 232.29 -0.1%

Nb.Rd [kN] 192.02 144.03 -25.0% 189.69 -1.2% 190.59 -0.8%

• Shell models considers shear flexibility (more accurate)• Model 1 is too conserv. due to high imperf. values of Table 5.1 (EC3) • Shell models 2 and 3 are accurate when compared to EC 3 formulae • Differences in buckling resistance are usually lower than differences in stresses

• Column flexural buckling load may be determined using frame model (e.g., for arbitrary support conditions)

Nr FE Ncr.z [kN] Diff. (vs Ncr.shell)

1 292.11 +21.9%

3 240.17 +0.2%

6 239.79 +0.05%

• Shell models don’t consider the exact cross-section but a reduced one (conservative)

Longitudinal normal stress

(Model 3)

• Flexural buckling analysis (using frame FE):

• Discretisation in at least 3 FE is recommended (e.g., using SAP2000 automatic mesh)

• Column resistance results:

7

Example 2 – Beam (1/3)

kNmLpM EdEdy 62.30

8

2

max..

IPE 200

SAP2000 frame FE model SAP2000 shell FE model

• Spatial beam (lateral torsional buckling):

• Simply supported for major and minor bending • Laterally unbraced• Torsion prevented and warping free at supports • S 235 steel, IPE 200 profile (class

1)• Loaded in major bending plane

kNLpP EdEd 70

8


EC 3 formulae

Buckling curve a: 21.0LT

971.0 crRdLT MM

Elastic design Plastic designkNmfWM yelyRdely 66.45... kNmM Rdply 94.51..

kNm

EIGIL

II

LEIM

z

t

z

wzcr

25.4342.16.2

0692.05.342.101299.0

5.342.1210

2

2

2

2

2

2

2

2

0.

035.1LT

640.0LT 686.0, LTLTLT

kNmMM RdLTRdb 32.31. kNmM Rdb 26.33.

Model A(SAP2000 shell)

mmLe 75000

Imperfection in minor axis bending:

Model B(SAP2000 shell)

mmee column 98.4.00

kNmMCM crcr 44.4825.4312.10.1

kNmM cr 26.430.

SAP2000 frame design

kNmM Rdply 94.51..

Always considers C1=1

by default

096.1LT

599.0LT

kNmM Rdb 11.31. kNmM Rdb 01.28.

MPa

MMf RdbEdyx

7.229.max.

MPax 5.279max.

21.0LT

kNmM Rdb 86.29.

MPax 3.246max.

kNmM cr 03.43

kNmM Rdely 17.44..

Shell models consider a reduced cross-section

(+14%) (-3.3%)

kNmMM crcr 26.430.

9


EC 3 – elastic design

EC 3 – plastic design Diff. SAP2000

frame designDiff.

(plast.)Model A

(SAP2000 shell)Diff.

(elast.)Model B

(SAP 2000 shell)Diff.

(elast.)

Mcr [kN] 48.44 48.44 0% 43.26 -10.7% 43.03 -11.2% 43.03 -11.2%

x.max [MPa] 229.7 - - - - 279.5 +21.7% 246.3 +7.2%

Mb.Rd [kN] 31.32 33.26 +6.2% 31.11 -6.5% 28.01 -10.6% 29.86 -4.7%

• Shell models consider shear and local/distortional deformation when determining buckling loads (more accurate)

• Models A (and B) give accurate (reasonable) resistances when compared to EC 3 elastic results• A 14% increase from the elastic to the plastic moment resistance only results in a 6% increase in the member resistance. When instability plays an important role, plastic strength reserve cannot be fully exploited

In-plane deformation(Model A)

3D deformation (Model A)

• Beam resistance results:

• SAP2000 frame design yields conservative results (-6.5% in bending resistance) by considering the most unfavourable bending moment distribution (uniform)• Shell models consider a reduced cross-section (lower buckling loads and resistances)

10

Example 3 – Beam-column (1/5)• Spatial beam-column (flexural and lateral torsional buckling):

• Simply supported for major and minor bending • Laterally unbraced• Torsion prevented and warping free at supports • S 235 steel, IPE 500 profile (class

1)• Loaded axially and in major and minor bending planes

SAP 2000 shell FE model

kNmLpM EdzEdy 8.198100

8

2.

max.. Maximum major axis

bending moment:

11

Example 3 – Beam-column (2/5)

EC3 design formulae

SAP2000 frame design Diff. Shell model

(SAP2000 shell) Diff.

Ncr.z [kN] 3157 3157 0% 3085 -2.1%

Ncr.y [kN] 71040 71040 0% 57711 -16.9%

Mcr.0 [kN] 900.4 900.5 0% 861.2 -4.4%

Mcr [kN] 1080.5 900.5 -16.7% 913.3 -15.5%

Buckling loads

kNLEIN zzcr 315722.

kNLEIN yycr 7104022.

kNmEIGIL

II

LEIM

z

t

z

wzcr 4.9002

2

2

2

0.

kNmMCM crcr 5.10804.9002.10.1

• C1 factor from tables is unconservative when compared with numerical results (1.2 vs 1.06)

• EC3 design formulae and SAP2000 frame design considers exact web-flange joint geometry and neglects shear deformability, resulting in higher buckling loads when compared to the shell model

12


EC 3 - elastic

Flexural buckling resistance

Lateral torsional buckling resistance

EC 3 design formulae

EC 3 - plastic

929.031572726. zcrRkz NN

642.0, zz

kNNN RdzRdzb 1751..

2.0196.0710402726 y

1y

kNNN RdyRdyb 2726..

(buckling curve b)

648.05.10801.453.. crRdyelLT MM

kNmfWM yyelRdyel 1.453...

kNmMM RdLTRdb 1.368.

812.0, LTLT


34.0LT

691.05.10806.515.. crRdyplLT MM

kNmM Rdypl 6.515..

kNmMM RdLTRdb 8.406.

789.0, LTLT

+ 14%

+ 11%

929.0z 196.0y

642.0z 1y

kNN Rdzb 1751.. kNN Rdyb 2726..

kNmM Rdb 1.387.


kNmM Rdypl 6.515..

757.0LT

751.0LT

34.0 21.0 34.0 21.0

34.0LT

minor axis: major axis: major axis:minor axis:

13

Example 3 – Beam-column (4/5)Beam-column resistance (Method 2)

925.0myC 924.0myC

6.0mzC 6.0mzC

925.0 mymLT CC 924.0 mymLT CC

EC 3 design formulae SAP2000 frame design

945.027265002.06.01925.06.01

..

Rdyb

Edymyyy NNCk 924.0yyk

816.017515006.0929.0216.06.021

..

Rdzb

Edzmzzz N

NCk 816.0zzk

489.0816.06.06.0 zzyz kk 489.0yzk

961.01751500

25.0925.0929.01.01

25.01.01

..

Rdzb

Ed

mLT

zzy N

NC

k 961.0zyk

1800.096.7825489.0

8.4068.198945.0

2726500

.

.

.

.

..

Rdz

Edzyz

Rdb

Edyyy

Rdyb

Ed

MMk

MM

kNN

eq. (6.61):

eq. (6.62): 1013.196.7825816.0

8.4068.198961.0

1751500

.

.

.

.

..

Rdz

Edzzz

Rdb

Edyzy

Rdzb

Ed

MMk

MM

kNN

1028.1

14


EC 3 – plastic(method 2)

SAP2000 frame design (method 2) Diff. Model 1 - elastic

(SAP2000 shell) Diff.

Failure parameter 1.013 1.028 +1.5% 1.169 +15.4%

3D deformation (shell model)

mmAWe zel 58.42.0 .0 • Imperfection (minor axis):

SAP2000 shell FE model

• EC3 design formulae and SAP2000 frame design yield very similar results

• SAP2000 shell model yields moderately conservative results because it (i) is based on elastic design and (ii) considers a reduced cross-section

Beam-column resistance results:

• Failure parameter:yx fFP max.

15

Example 4 – Frame (1/7)

• Planar frame:

• Load combinations:

Dead load Wind load Live load

HE

A 18

0

• Laterally braced at joints

• ‘Dead + Wind’ and ‘Dead + Life’ (1.35Gk + 1.5Qk)

• Major axis bending in the frame plane

HE

A 18

0

IPE 220 IPE 220

6

1

12 [m]

• Pinned to the ground

• Lateral and lateral torsional buckling not prevented!• S 355 steel

16


radmh 003536.08660.08165.02001

0

• Global imperfection (life load combination):

mh 6 8165.0622

hh

Height:

2mNr columns:

8660.02115.0115.0

mm

Imperfection angle:

Equiv. lateral force:

kNNH Ed 1718.059.48003536.0

Global imperf. as equiv. lateral forces

(live load comb.)

• Buckling analysis:

• Wind combination • Live load combination

1028.55 cr No P-D effects to consider

1056.5 cr P-D effects must be considered

17

Example 4 – Frame (3/7)• P-D analysis (live load combination):

Deformed config.[m]

N[kN]

My

[kN.m]Vz

[kN.m]

18

Example 4 – Frame (4/7)• EC3 design check (life load combination):

• All members satisfy EC3 design formulae (FP<1)

19

Example 4 – Frame (5/7)• 1st order analysis (wind load combination):

Deformed config.[m]

N[kN]

My

[kN.m]Vz

[kN.m]

20

Example 4 – Frame (6/7)• EC3 design check (wind load combination):

• All members satisfy EC3 design formulae (FP<1)

21

Example 4 – Frame (7/7)• EC3 automatic design (wind and live load combinations):

Initial sections estimate

Run analyses(all load comb.)

Run

analyses

Modified sections

(automatic)

Sections to be modified by user due to symmetry

Columns: HEA160Beams: IPE 200

Columns: HEA180Beams: IPE 220

Final sections

Not safe ! Safe

Safe

22

Example 5 – Frame (1/4)• Spatial frame:

• Longitudinally braced

6

1

12[m]

• Pinned to the ground• S 355 steel• HEA 180 (columns), IPE 220 (transv. beams), IPE 100 (long. beams), 4 mm cable (bracing)

SAP2000 frame FE model

4

4

4

4

4

• Load combination:

• ‘Dead + Live’ (1.35Gk + 1.5Qk)• Load values and configuration equal to example 4

• Two cross cables may be substituted by one rod with the same diameter that resists tension and compression

• Note:

23

Example 5 – Frame (2/4)• Buckling analysis:

Torsion

Mode 1 Mode 3Mode 2

Longitudinal sway Transversal sway

37.21. b 82.22. b 37.53. b

1037.21. bcr Option 1: increase bracing stiffness until 2nd order analysis is no

longer necessary for torsion and longitudinal sway (cr>10)

Option 2: perform the spatial frame 2nd order analysis with imperf.

24


Option 1

• 10 mm cable

Torsion

1058.136. b

Buckling analysis

1037.51. b

Longitudinal sway

1016.1731. b

• Transversal sway 2nd order effects and imperfections already checked in Example 4

• No torsion or longitudinal 2nd order effects and imperfections to consider

Option 2

radmh 003118.07638.08165.02001

0

• Global imperfection:

mh 6 8165.0622

hh

Height:

6mNr columns:

7638.06115.0115.0

mm

Imperfection angle:

Equiv. lateral force:

kNNH Ed 1525.091.48003536.0

Transversal sway

• 4 mm cable

25


Option 2 (cont.)Torsion Longitudinal sway

Members resistance:

Cable resistance:

OK

Imperfection:

OK

RdEd NkNN 67.2max.

kNAfN yRd 88.2735507854.0

OK RdEd NkNN 56.2max. OK

26

Example 6 – Short class 4 column (1/4)• Square hollow section short column:


• Simply supported • S 355 steel, welded SHS profile, class 4 (compression)

SHS 300

300

300[mm]

6


• Objective: determine the column buckling resistance

27

Example 6 – Short class 4 column (2/4)• Effective cross-section:

Gross section:

300

300[mm]

6

486/288 tc 3481.04242

344248 tc Class 4 walls

043.1481.04.28

62884.28

k

tbf

cr

yp

7565.0043.1

13055.0043.13055.022

p

p

Plate slenderness:

Reduction factor:

115

115[mm]

Effective section (pure compression):

2310056.7 mA 4410017.1 mIII dzy

44

.. 10764.6 mWW zelyel

2310616.5 mAeff

44. 108586.0 mI yeff

34

. 10689.5 mW yeff

44. 10783.4 mW del

34. 10085.4 mW deff

Classification (pure compression):

109

mmbbeff 10922887565.0

22

Effective width:

28

Example 6 – Short class 4 column (3/4)

kN

NN RdeffzRdb

189219949491.0

..

kNLEIN zzcr

172075.37.101210 2222

.

3404.0

172071994..

zcrRkeffz NN

19491.0, zz




EC 3 formulae

34.0

mmae47.12000

mm

te pp

032.066)8.0043.1(13.0

6)8.0(0

Method: • Ayrton-Perry formula

• local geometric imperf. according to Table 3.1 (EC 3-1-5)

• 2nd order shell FEM • 2nd order shell FEM



kNfAN yeffRkeff 1994355616.5.

SAP2000 design (SAP2000 frame)

kNN Rkpl 2505. kNAfN yRkpl 2505355056.7.

kNNcr 17205

337.0z

95.0z

kNN Rdb 1857.

34.0

• local geometric imperf. equiv. to local buckling curves

Flexural buckling of minute importance

• no global imperf. • no global imperf.

12 longitudinal half-waves

29

Example 6 – Short class 4 column (4/4)

EC 3 formulae SAP2000 design(SAP2000 frame) Diff. Model 1


Ncr.local [kN] - - - 2258 - 2258 -

Nb.Rd [kN]Lower /upper

bound1892 1857 -1.8% 2135/

2303+12.8%+17.8%

1316/1918

-30.4%+1.4%

• Model 2 is conserv. when compared to Model 1 because it considers a higher imperfection

• Shell models 1 and 2 are reasonably accurate when compared to EC 3 formulae

• SAP2000 design is very accurate when compared to EC 3 formulae

Deformation (Model 1)


• Shell models lower and upper bounds correspond to first yielding due to plate bending and corner yielding due to membrane normal stress resultant (the real resistance is between the two)

Upper bound analysis (Model 1)

Lower bound analysis (Model 1)

30

Example 7 – Long class 4 column (1/5)• Square hollow section long column:


• Simply supported • S 355 steel, welded SHS profile, class 4 (compression)

SHS 300

300

300[mm]

6


• Objective: determine the column buckling resistance

31

Example 7 – Long class 4 column (2/5)

kN

NN RdeffzRdb

158419947944.0

..

kNLEIN zzcr

430277.101210 2222

.

6808.0

43021994..

zcrRkeffz NN

17944.0, zz




EC 3 formulae

34.0

mmLe282500

mm

AWeeffdeff

89.11616.55.408)2.06808.0(34.0

)2.0( .0

Method: • Ayrton-Perry formula • 2nd order shell FEM • 2nd order shell FEM




SAP2000 design (SAP2000 frame)

kNN Rkpl 2505. kNAfN yRkpl 2505355056.7.

kNNcr 4301

674.0z

798.0z

kNN Rdb 1559.

34.0

Flexural buckling of significant importance(local-global buckling

interaction occurs)

mmae47.12000

mm

te pp

032.066)8.0043.1(13.0

6)8.0(0

• local geometric imperf. according to Table 3.1 (EC 3-1-5)

• local imperf. equiv. to local buckling curves

• global imperf. from buckling curves

12 longitudinal half-waves

• global imperf. from Table 5.1 (EC 3-1-1)

32


• Re-determine effective cross-section for load NEd=1584 kN:

a) Determine bending moment in the critical cross-section:

b) Determine stress distribution in the gross cross-section:

MPaANEd

mean 5.224056.71584

MPaWM

Rdd

Ed 5.1304783.043.62

.

D

MPamean 0.355max D

c) Walls reduction factors:

632.03555.224

835.0

3055.02

p

p

88.405.12.8

K

945.088.481.04.28

62884.28

k

tbp

Walls AB & BD:

MPamean 0.94min D

419.05.2240.94

1043.1

3055.02

p

p

58.505.12.8

K 883.0

58.581.04.286288

4.28

ktb

pWalls AC & CD:

702.03555.224883.0.

. yd

Edcompredp f

kNmMMWfM

NN

EdEd

dely

Ed

Rdpl

Ed 43.6214783.03552505

15841..

33


• Re-determine effective cross-section for load NEd=1584 kN:

[mm]

Effective section (NEd + MEd):

2310480.6 mAeff

44. 109388.0 mI deff

34

. 10689.5 mW yeff

34. 10265.4 mW deff

mmbbb

mmbb

mmbb

eeffe

effe

eff

130110240

110240632.052

52

240288835.0

12

1

110130

• Column design according to EC 3-1-1 formulae:

kN

NN RdeffzRdb

176123007658.0

..

7312.0

43022300..

zcrRkeffz NN

17658.0, zz

Buckling curve b

Imperf.:

34.0


34


• Shell model 1 is reasonably accurate when compared to analytical calculations

• SAP2000 design is slightly conservative when compared to EC 3 procedure because it does not iterate to find effective cross-section (considers the unfavourable case of pure compression)


• Shell models lower and upper bounds correspond to first yielding due to plate bending and corner yielding due to membrane normal stress resultant (the real resistance is between the two)

EC 3 formulae SAP2000 design(SAP2000 frame) Diff. Model 1


Ncr.local [kN] - - - 2253 - 2253 -

Nb.Rd [kN]Lower /upper

bound1761 1559 -11.5% 1928/

2001+9.5%

+13.6%1016/1349

-42.3%-23.4%

• Shell model 2 is too conservative due to considering too large imperfections

35

Conclusion

• It is not possible to fully exploit the plastic strength reserve of members prone to instability. An elastic design (e.g., using shell FE) is usually not too conservative, even for members with class 1 cross-sections

• SAP2000 design tools for steel frame structures are practical, fast and on the safe side. It is possible not only to (i) check if the members satisfy the EC3 resistance requirements, but also (ii) optimise their sections

• SAP2000 shell design is valid for arbitrary thin-walled members (e.g., tapered, with non-symmetrical cross-sections, etc) and support conditions, while EC3 design formulae are limited to bisymmetrical simply supported uniform members

36

References

• ECCS Technical Committee 8, Rules for Members Stability in EN 1993 – 1 – 1, Background documentation and design guidelines

design of steel frames

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