The document describes 7 examples of designing steel frames using SAP2000. Example 1 analyzes a column using both frame and shell finite element models in SAP2000, finding the shell models more accurately consider shear flexibility. Example 2 analyzes a beam, with shell models again more accurate by including shear and local/distortional effects. Example 3 analyzes a beam-column, showing shell models provide more accurate buckling loads than formulas alone by including shear and joint geometry. Overall, the examples illustrate how SAP2000 tools can be used to both check and optimize steel frame designs according to Eurocode 3, and that shell models tend to provide more accurate analyses and results compared to frame models or formulas alone.

1. Design of steel frames using SAP2000 –
Illustrative examples
CSI Portugal & Spain

2. 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

3. 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. 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. 5
Example 1 – Column (2/3)
kN
NN RdzRdb
02.192
75.6692867.0
.


 
MPaNNf RdbEdyx 53.232.max. 
kN
LEIN zzcr
25.240
5.3420.1210 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
34.0
mm
Le
14
2500


mm
AWe zel
98.4
)2.0( .0

 
mkN
LeNp Ed
/74.1
8 2
00


mkN
LeNp Ed
/618.0
8 2
00


mm
AWe 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. 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. 7
Example 2 – Beam (1/3)
kNm
Lp
M Ed
Edy 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. 8
Example 2 – Beam (2/3)
EC 3
formulae
Buckling curve a: 21.0LT
971.0 crRdLT MM
Elastic design Plastic design
kNmfWM yelyRdely 66.45...  kNmM Rdply 94.51.. 
kNm
EI
GIL
I
I
L
EI
M
z
t
z
wz
cr
25.43
42.16.2
0692.05.3
42.1
01299.0
5.3
42.1210
2
2
2
2
2
2
2
2
0.











035.1LT
640.0LT  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 
kNmMcr 26.430. 
SAP2000
frame design
kNmM Rdply 94.51.. 
Always
considers C1=1
by default
096.1LT
599.0LT
kNmM Rdb 11.31.  kNmM Rdb 01.28. 
MPa
MMf RdbEdyx
7.229
.max.

 MPax 5.279max. 
21.0LT
kNmM Rdb 86.29. 
MPax 3.246max. 
kNmMcr 03.43
kNmM Rdely 17.44.. 
Shell models consider a
reduced cross-section
(+14%) (-3.3%)
kNmMM crcr 26.430. 

9. 9
Example 2 – Beam (3/3)
EC 3 – elastic
design
EC 3 – plastic
design
Diff.
SAP2000
frame design
Diff.
(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. 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
kNm
Lp
M Edz
Edy 8.198100
8
2
.
max.. 
Maximum major axis
bending moment:

11. 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
. 
kNm
EI
GIL
I
I
L
EI
M
z
t
z
wz
cr 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. 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
1y
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
34.0LT
691.05.10806.515..  crRdyplLT MM
kNmM Rdypl 6.515.. 
kNmMM RdLTRdb 8.406.  
  789.0,  LTLT 
+ 14%
+ 11%
929.0z 196.0y
642.0z 1y
kNN Rdzb 1751..  kNN Rdyb 2726.. 
kNmM Rdb 1.387. 
SAP2000 frame design
kNmM Rdypl 6.515.. 
757.0LT
751.0LT
34.0 21.0 34.0 21.0
34.0LT
minor axis: major axis: major axis:minor axis:

13. 13
Example 3 – Beam-column (4/5)
Beam-column resistance (Method 2)
925.0myC 924.0myC
6.0mzC 6.0mzC
925.0 mymLT CC 924.0 mymLT CC
EC 3 design formulae SAP2000 frame design
945.0
2726
500
2.06.01925.06.01
..
















Rdyb
Ed
ymyyy
N
N
Ck  924.0yyk
    816.0
1751
500
6.0929.0216.06.021
..













Rdzb
Ed
zmzzz
N
N
Ck  816.0zzk
489.0816.06.06.0  zzyz kk 489.0yzk
   
961.0
1751
500
25.0925.0
929.01.0
1
25.0
1.0
1
..






Rdzb
Ed
mLT
z
zy
N
N
C
k

961.0zyk
1800.0
96.78
25
489.0
8.406
8.198
945.0
2726
500
.
.
.
.
..

Rdz
Edz
yz
Rdb
Edy
yy
Rdyb
Ed
M
M
k
M
M
k
N
N
eq. (6.61):
eq. (6.62): 1013.1
96.78
25
816.0
8.406
8.198
961.0
1751
500
.
.
.
.
..

Rdz
Edz
zz
Rdb
Edy
zy
Rdzb
Ed
M
M
k
M
M
k
N
N
1028.1 

14. 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. 15
Example 4 – Frame (1/7)
• Planar frame:
• Load combinations:
Dead load Wind load Live load
HEA180
• Laterally braced at joints
• ‘Dead + Wind’ and ‘Dead + Life’ (1.35Gk + 1.5Qk)
• Major axis bending in the frame plane
HEA180
6
1
12 [m]
• Pinned to the ground
• Lateral and lateral torsional
buckling not prevented!
• S 355 steel

16. 16
Example 4 – Frame (2/7)
radmh 003536.08660.08165.0
200
1
0  
• Global imperfection (life load combination):
mh 6 8165.0
6
22

h
h
Height:
2m
Nr columns:
8660.0
2
1
15.0
1
15.0 












m
m
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. 17
Example 4 – Frame (3/7)
• P-D analysis (live load combination):
Deformed config.
[m]
N
[kN]
My
[kN.m]
Vz
[kN.m]

18. 18
Example 4 – Frame (4/7)
• EC3 design check (life load combination):
• All members satisfy EC3 design formulae (FP<1)

19. 19
Example 4 – Frame (5/7)
• 1st order analysis (wind load combination):
Deformed config.
[m]
N
[kN]
My
[kN.m]
Vz
[kN.m]

20. 20
Example 4 – Frame (6/7)
• EC3 design check (wind load combination):
• All members satisfy EC3 design formulae (FP<1)

21. 21
Example 4 – Frame (7/7)
• EC3 automatic design (wind and live load combinations):
Initial sections
estimate
Run analyses
(all load comb.)
Modified
sections
(automatic)
Sections to be
modified by user
due to symmetry
Columns: HEA160
Beams: IPE 200
Columns: HEA180
Beams: IPE 220
Final sections
Not safe ! Safe
Safe

22. 22
Example 5 – Frame (1/4)
• Spatial frame:
• Longitudinally braced
6
1
[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
• 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. 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. 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 003118.07638.08165.0
200
1
0  
• Global imperfection:
mh 6 8165.0
6
22

h
h
Height:
6m
Nr columns:
7638.0
6
1
15.0
1
15.0 












m
m
Imperfection angle:
Equiv. lateral force:
kNNH Ed 1525.091.48003536.0 
Transversal sway
• 4 mm cable

25. 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. 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. 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
043.1
481.04.28
6288
4.28





k
tbf
cr
y
p
    7565.0
043.1
13055.0043.13055.0
22





p
p



Plate slenderness:
Reduction factor:
115
115
[mm]
Effective section (pure compression):
23
10056.7 mA 

44
10017.1 mIII dzy


44
.. 10764.6 mWW zelyel


23
10616.5 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
mm
bbeff
109
2
2887565.0
22




Effective width:

28. 28
Example 6 – Short class 4 column (3/4)
kN
NN RdeffzRdb
1892
19949491.0
..


 
kN
LEIN zzcr
17207
5.37.101210 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
mm
ae
47.1
2000


mm
te pp
032.0
66)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
337.0z
95.0z
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. 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
bound
1892 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. 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. 31
Example 7 – Long class 4 column (2/5)
kN
NN RdeffzRdb
1584
19947944.0
..


 
kN
LEIN zzcr
4302
77.101210 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
mm
Le
28
2500


mm
AWe effdeff
89.11
616.55.408)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
674.0z
798.0z
kNN Rdb 1559. 
34.0
Flexural buckling of
significant importance
(local-global buckling
interaction occurs)
mm
ae
47.1
2000


mm
te pp
032.0
66)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. 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:
MPa
A
NEd
mean 5.224
056.7
1584
 MPa
W
M
Rdd
Ed
5.130
4783.0
43.62
.
D
MPamean 0.355max D 
c) Walls reduction factors:
632.0
355
5.224

 
835.0
3055.0
2



p
p



88.4
05.1
2.8




K
945.0
88.481.04.28
6288
4.28





k
tb
p
Walls AB & BD:
MPamean 0.94min D 
419.0
5.224
0.94

 
1043.1
3055.0
2


 



p
p
58.5
05.1
2.8




K 883.0
58.581.04.28
6288
4.28





k
tb
pWalls AC & CD:
702.0
355
5.224
883.0.
. 
yd
Edcom
predp
f


kNmM
M
Wf
M
N
N
Ed
Ed
dely
Ed
Rdpl
Ed
43.621
4783.03552505
1584
1
..




33. 33
Example 7 – Long class 4 column (4/5)
• Re-determine effective cross-section for load NEd=1584 kN:
[mm]
Effective section (NEd + MEd):
23
10480.6 mAeff


44
. 109388.0 mI deff


34
. 10689.5 mW yeff


34
. 10265.4 mW deff


mmbbb
mmbb
mmbb
eeffe
effe
eff
130110240
110240
632.05
2
5
2
240288835.0
12
1









110130
• Column design according to EC 3-1-1 formulae:
kN
NN RdeffzRdb
1761
23007658.0
..


 
7312.0
43022300..

 zcrRkeffz NN
  17658.0,  zz 
Buckling curve b
Imperf.:
34.0
kNfAN yeffRkeff 2300355480.6. 

34. 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
bound
1761 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. 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. 36
References
• ECCS Technical Committee 8, Rules for Members Stability in EN 1993 – 1 – 1,
Background documentation and design guidelines