Stability
Both methods 2.1 and 2.2 are extensively used in practice. When verifying
members according to EN 1993-1-1 section 6.3, system length should be
used as the buckling length.
In Method 3, the designer must determine an appropriate effective
length that allows for the consideration of P-Δ effects while performing
member checks according to section 6.3 of EN 1993-1-1. As the design
is based on first order internal forces, the complexity of the analysis is
removed, but the effective length needs to be specified for each column.
The concept of effective length was introduced in Part 1 of the current
article for isolated struts, where the horizontal or rotational restraints of the
strut ends were assumed as infinitely rigid. This does not represent reality:
(i) rotational stiffness of the nodes is related to the flexural stiffness of the
elements that are connected to the nodes, resulting in a rotational spring
on each node – kr,i (Figure 2); (ii) if a structure is susceptible to second order
global effects, the complexity is increased, as the structure is horizontally
flexible (assessed by the value of αcr ), resulting in horizontal springs on each
node – kh,i (Figure 2).
When a column is integrated in a frame, the concept of effective length
may be described as the fictional pin-ended strut length that buckles at the
same time as the frame for a specified load case6. Based on the value of αcr
for the entire frame, the critical load Ncr for each column can be calculated
by multiplying the design axial load on each column by the value of αcr . The
effective lengths can then be obtained by a back calculation, knowing that
Ncr=(π2 EI) ⁄ (leff)2. Thus, the effective length of a column is dependent on the
applied load and spring stiffness at the nodes. The values of leff obtained are
only appropriate within the load arrangement assumed to calculate αcr . This
method is described in Annex E.6 of BS 5950-17.
b) Sway frame8
In practice, while using Method 3, the definition of the effective buckling
length is often obtained indirectly by a simplified analysis where each
column is considered individually, with no dependency on the applied load.
There are several resources to assess the problem, such as the well-known
Wood method⁹, which provides effective buckling lengths for sway or
non-sway frames. These approximate methods are intended to provide an
Boundary conditions Section
12 NSC
Technical Digest 2019
answer for the problem shown Figure 2c. The Wood method can be found
in Annex E of BS 5950-1 as well as in NCCI SN008a10. Based on the model in
Figure 2c, simplified methods usually assume that kh,L = ∞ and kh,U = 09.
The approximate methods provide exact results if every member has the
same rigidity parameter Ør = EI/NEdl where EI is the flexural stiffness of the
column, NEd is the design axial load on the column, and l is the system
length of the column⁶. This means that all columns would buckle at the
same time. The columns with low values of Ør are the critical members
(members which induce frame instability), for which the method gives
conservative values of the buckling length. For members with high values of
Ør , buckling lengths are unconservative. For the critical members, the
method can be seen as a conservative approximation for the critical load of
the frame⁶.
The approximate methods provide an efficient and systematic procedure
to assess the problem. However, the following effects/simplifications are
usually disregarded/considered in the process6,8,9,11,12: (i) only columns are
affected by P-Δ effects, while internal forces to design other elements
(beams, connections) will be always based on first order theory; (ii) for
frames sensitive to second order effects, the effective lengths calculated are
the same for any value of αcr ; (iii) there is no influence of the applied load;
(iv) for columns in non-sway frames, the rotation at opposite ends of the
restraining elements are equal in magnitude and opposite in direction,
producing single curvature bending; (v) for columns in sway frames, the
rotation at opposite ends of the restraining elements are equal in
magnitude and opposite in direction, producing double curvature bending;
(vi) all columns buckle simultaneously; (vii) stiffness parameter Ør is the
same for all columns; (viii) no significant axial force exists in the beams;
(ix) all joints are rigid; (x) joint restraint is distributed to the column above
and below the joint in proportion to EI/l for the two columns. Further
information about approximate methods can be found in reference 11.
Two worked examples follow, where the results obtained from the
application of methods 2.1, 2.2 and 3 are compared.
Worked example 1: simple column
Influence of the number of finite elements on simple struts (Table 1):
The results support the conclusions from Part 1: for low values of NEd ⁄ Ncr the
errors in using an approximate stiffness matrix are less significant than for
cases where NEd ⁄ Ncr is close to 4. The consideration of 3 finite elements for
the strut gives reasonable results for the four cases.
The design of the column based on Method 2 (2.1 by a numerical P-Δ
or 2.2 considering ksw) and Method 3 will be undertaken for the structure
in Figure 3. Two examples are considered for different levels of horizontal
load. A comparison of the Unity factor (UF) for relevant checks according to
EN 1993-1-1 is presented in Table 2.
From Worked Example 1, it can be noted that there is very close
agreement in the utilization factor between methods 2.1 and 2.2. Method 3
is conservative for NEd = 75 kN and H ⁄ 2 = 10 kN. If the horizontal load H ⁄ 2 is
increased to 20 kN, Method 3 becomes unconservative.
a) Non-sway frame8
αcr ≥ 10 for elastic global analysis
αcr < 10 for elastic global analysis
Figure 2: Effective length concept in sway and non-sway frames
c) Equivalent isolated
column model
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);