Technical
NSC 27
April 19
Boundary conditions Section
l
m
Theoretical value of Ncr,z
kN
Table 1: Buckling analysis of a strut considering different number of finite elements (FE)13.
Ncr,1
1 FE
Ncr,3
3 FE
Ncr,5
5 FE
Ncr,10
10 FE
Cantilever 254 UC 107 10 Ncr,z = EI
(2l)
= 307.16 309.47 307.19 307.17 307.16
Pinned
Pinned
254 UC 107 10 Ncr,z = EI
l
= 1228.65 1493.86 1230.59 1228.91 1228.66
Pinned
Fixed
254 UC 107 10 Ncr,z = EI
(0.6992l)
= 2513.18 3734.64 2528.93 2515.68 2513.64
Fixed
Fixed
254 UC 107 10 Ncr,z = EI
(0.5l)
= 4914.59 ∞* 5022.36 4930.35 4915.65
* - See Part 15, Figure 8; this example represents NEd ⁄ Ncr = 4);
h = 10 m;
Columns: 254 UC 107; lz = 5928 cm⁴; Steel: S355 JR
NEd = 75 kN; Example 1.1: H = 20 kN;
Example 1.2: H = 40 kN (factored loads)
EN 1993-1-1 section 5.3.2:
EHF =
1
200 * 75 = 0.375 kN
Ncr = EIz
(2l)
= 307.16 kN
If the system is represented by a single column:
cr =
Ncr
NEd
307.16
75
= = 4.10
As αcr > 4, local bow imperfections can be
disregarded in the analysis – EN 1993-1-1 5.3.2 (6).
ksw =
1
1-1/cr
=
1
1-1/4.10 = 1.32
Figure 3: Example of a slender simple column.
Worked example Method
leff
m
NEd
kN
H/2 + EHF
kN
First order bending
moment kNm
Second order bending
moment kNm
UF
1.1
2.1 10 75 10.375 103.75 131.29 0.53
2.2 10 75 10.375 103.75 103.75 * 1.32 = 136.95 0.55
3 20 75 10.375 103.75 - 0.63
1.2
2.1 10 75 20.375 203.75 257.77 1.04
2.2 10 75 20.375 203.75 203.75 * 1.32 = 268.95 1.09
3 20 75 20.375 203.75 - 0.96
Table 2: Results for two different load arrangements: simple column13.
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