design of steel frames
TRANSCRIPT
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)
Model 2(SAP2000 shell)
Model 1(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
Example 2 – Beam (2/3)
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
Example 2 – Beam (3/3)
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
Example 3 – Beam-column (3/5)
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
SAP2000 frame design
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.
SAP2000 frame design
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
Example 3 – Beam-column (5/5)
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
Example 4 – Frame (2/7)
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
Example 5 – Frame (3/4)
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
Example 5 – Frame (4/4)
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:
SAP2000 frame FE model
• Simply supported • S 355 steel, welded SHS profile, class 4 (compression)
SHS 300
300
300[mm]
6
SAP2000 shell FE model
• 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
Buckling curve bImperf.:
Model 2(SAP2000 shell)
Model 1(SAP2000 shell)
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
• imperf. factor from buckling curves
• Column design according to EC 3 formulae and thin-walled rectangular shell FE models:
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
(SAP2000 shell) Diff. Model 2(SAP 2000 shell) Diff.
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)
• Column resistance results:
• 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:
SAP2000 frame FE model
• Simply supported • S 355 steel, welded SHS profile, class 4 (compression)
SHS 300
300
300[mm]
6
SAP2000 shell FE model
• 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
Buckling curve bImperf.:
Model 2(SAP2000 shell)
Model 1(SAP2000 shell)
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
• imperf. factor from buckling curves
• Column design according to EC 3 formulae and thin-walled rectangular shell FE models:
kNfAN yeffRkeff 1994355616.5.
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
Example 7 – Long class 4 column (3/5)
• 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
Example 7 – Long class 4 column (4/5)
• 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
kNfAN yeffRkeff 2300355480.6.
34
Example 7 – Long class 4 column (5/5)
• 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)
• Column resistance results:
• 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
(SAP2000 shell) Diff. Model 2(SAP 2000 shell) Diff.
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