stability of steel structure chapter 2
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2. INSTABILITY OF BARS
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Chapter 2
INSTABILITY OF BARS
2.1. TORSION
Generally, torsion is avoided in structural metal (steel or aluminium alloy) members.
There are basically two types of torsion:
• St. Venant torsion (torsiunea cu deplanare liber ă);
• warping torsion (torsiunea cu deplanare împiedicat ă).
As a simplification, in the case of a member with a closed hollow cross-section, such
as a structural hollow section, it may be assumed that the effects of torsional warping
can be neglected; similarly, in the case of a member with open cross section, such
as I or H, it may be assumed that the effects of St. Venant torsion can be neglected.
2.1.1. St. Venant torsion
It occurs when all the following assumptions are accomplished (Fig. 2.1):
• the torsion moment is constant along the bar;
• the area of the cross-section is constant along the bar;
• there are no connections at the ends or along the bar that could prevent
warping.
Fig. 2.1. St. Venant torsion
the flanges remain rectangles
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2.1.1.1. Stress and strain state
The following aspects can be noticed:
• there is no increase or reduction of the length of the fibres (as there is nolongitudinal force):
εx = 0 → σx = 0 (2.1)
• warping (deplanarea ) of the cross-section is a result of the assumption εx = 0
(in order to keep the geometry);
∫ ⋅×τ=A
Ed dArT (2.2)
Fig. 2.2. St. Venant torsion – stress state
• each cross-section rotates like a rigid disk (it goes out of plane but the shape
does not change);
• the rotation between neighbour cross-section is the same along the bar.
.constdx
d=
ϕ=θ (2.3)
2.1.2. Warping torsion
It occurs anytime when at least one of the St. Venant assumptions is not fulfilled
(Fig. 2.3).
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Fig. 2.3. Warping torsion
2.1.2.1. Stress and strain state
The following aspects can be noticed:
• there are longitudinal stresses and strains (Fig. 2.4):
εx ≠ 0 → σx ≠ 0 → σw; τw (2.4)
• the rotation between neighbour cross-section is variable along the bar.
.constdx
d≠
ϕ=θ (2.5)
Fig. 2.4. Warping torsion – stress state
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2.1.2.2. Equilibrium equations
The following aspects can be noticed:
• there is no axial force acting on the bar:
∫∑ =σ⇒=⇒=A
wEdi,Ed 0dA0N0X (2.6)
• there are no bending moments acting on the bar:
∫∑ =⋅σ⇒=⇒=A
wEd,yi,Ed,y 0zdA0M0M (2.7)
∫∑ =⋅σ⇒=⇒=A
wEd,zi,Ed,z 0ydA0M0M (2.8)
• in each cross-section, the torsion moment is the sum of the St. Venant
component and the warping component (Fig. 2.5):
0hVdArT ewA
Ed =⋅+⋅⋅τ= ∫ (2.9)
Ed,wEd,tEd TTT += (2.10)
where:
Tt,Ed – the internal St. Venant torsion;
Tw,Ed – the internal warping torsion.
Fig. 2.5. St. Venant torsion and warping torsion
2.1.3. Torsion and bending
2.1.3.1. Bi-symmetrical cross-section subject to bending moment and shear force
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The force F, acting in the plane xOz, generates only bending moment about the y – y
axis (and shear force) and no torsion moment, as the resultant forces V w on the
flanges are balanced (Fig. 2.6).
Fig. 2.6. Shear stresses in a bisymmetrical cross-section in bending
2.1.3.2. Mono-symmetrical cross-section subject to bending moment and shear force
A force F, acting in the plane xOz in the centre of gravity of a mono-symmetrical
cross-section, generates not only bending moment about the y – y axis (and shear
force) but torsion moment too (Fig. 2.7).
Fig. 2.7. Shear stresses for force acting in the centre of gravity
eFhFT wef Ed ⋅+⋅= (2.11)
The shear centre (centrul de t ăiere, centrul de încovoiere-r ăsucire ) is the point
through which the applied loads must pass to produce bending without twisting. A
force F, acting in the plane xOz in the shear centre of a mono-symmetrical cross-
section, generates only bending moment about the y – y axis (and shear force) and
no torsion moment (Fig. 2.8).
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Fig. 2.8. Shear stresses for force acting in the shear centre
cVT EdEd ⋅= (2.12)
eFhFcV wef Ed ⋅+⋅=⋅ (2.13)
Ed
wef
V
eFhFc
⋅+⋅= (2.14)
Notations: EdwEd
f VF;V
F==α (2.15)
Ed
EdeEd
V
eVhVc
⋅+⋅⋅α= (2.16)
ehc e +⋅α= (2.17)
F acting in the centre of gravity F acting in the shear centre
Fig. 2.9. Effects of a force acting in or outside of the shear centre
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2.1.4. Torsion – calculation
2.1.4.1. St. Venant torsion
The case of open cross-sections
a) Rectangular cross-section
T
Edmax
I
tT ⋅=τ t = minimum edge (2.18)
3
T tb3
1I ⋅⋅= (2.19)
.constIG
T
dx
d
T
Ed =⋅
=ϕ′=ϕ
=θ (2.20)
ϕ′⋅⋅= TEd IGT (2.21)
b) Cross-section made of several rectangles
Rigid disk assumptions (simplifying assumptions):
1. each cross-section rotates one about the other;
2. the rotation varies from one cross-section to the other but it is constant
for all the points on the same cross-section; the cross-section does not
change its shape in plane but it can go out of plane;
3. the rotation occurs around an axis parallel to the axis of the bar.
As a result of assumption 2,
T
Ed
n
1
i,T
n
i
i,Ed
n,T
n,Ed
1,T
1,Ed
IG
T
IG
T
IG
T...
IG
T
⋅=
⋅
=⋅
==⋅
=θ
∑
∑ (2.22)
∑ ⋅⋅=n
1
3
iiT tb3
1I (2.23)
Remark: For hot-rolled shapes,
∑ ⋅⋅α
=n
1
3
iiT tb3
I α = 1,1 … 1,3 (2.24)
T
maxEdmax
I
tT ⋅=τ tmax = maximum thickness (2.25)
ϕ′⋅⋅= TEd IGT (2.26)
t
b
1
i
n
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The case of hollow sections (Fig. 2.10)
aVbVT baEd ⋅+⋅= (2.27)
Fig. 2.10. Torsion of hollow sections
It is accepted that: (Bredt relation)
2
TaVbV Edba =⋅=⋅ (2.28)
b2
TV Eda
⋅= ;
a2
TV Edb
⋅= (2.29)
a
Ed
a
aa
tab2
T
ta
V
⋅⋅⋅
=
⋅
=τ (2.30)
b
Ed
b
bb
tba2
T
tb
V
⋅⋅⋅=
⋅=τ (2.31)
min
Edmax
tA2
T
⋅⋅=τ (2.32)
2.1.4.2. Warping torsion
An exact calculation would consider the bar as a sum of shells (Fig. 2.11).
Fig. 2.11. Shell modelling of a bar in torsion
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In daily practice a simplified approach is used, based on the Vlasov theory. The
simplifying assumptions are the following ones:
1. rigid disk behaviour:
• each cross-section rotates one about the other;
• the rotation varies from one cross-section to the other but it is constant
for all the points on the same cross-section;
• the rotation occurs around an axis parallel to the axis of the bar (Fig.
2.12);
Fig. 2.12. Axis of rotation of the bar
2. the shear deformations are zero in the mid-line of the cross-section (Fig.
2.13);
Fig. 2.13. Mid-line of the cross-section
3. σw and τw are constant on the thickness of the cross-section, because it is
thin (the mid-line is representative for the cross-section);
4. when calculating σw, it is assumed that τw = 0.
mid-line
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Based on these assumptions, the cross-section of the bar is reduced to its mid-line
(Fig. 2.14) and the following relations can be written between in-plane strains and
longitudinal ones (Fig. 2.15), considering rotation around point C:
dv'nn = (2.33)
dx
dv
ds
du= (2.34)
α⋅′′= cosnn'nn (2.35)
α⋅′′== cosnn'nndv (2.36)
Fig. 2.14. Mid-surface of the member
ϕ⋅=′′ dCnnn (2.37)
α⋅ϕ⋅== cosdCn'nndv (2.38)
α⋅= cosCnr (2.39)
ϕ⋅= drdv (2.40)
dx
ddsrdu
dx
dr
ds
du ϕ⋅⋅=⇒
ϕ⋅= (2.41)
ϕ ′′⋅ω=ε⇒=εdx
du (2.45)
Fig. 2.15. Geometric relations
du
mid-line
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By definition (Fig. 2.15),
( )triangletheof area22
dsr2ddsr ×
⋅⋅=ω=⋅ (2.42)
Notation (Fig. 2.15):
[ ]2s
0
s
0
Lddsr ∫∫ ω=⋅=ω normalised warping function (coordonat ă sectorial ă) (2.43)
it is also known as sectorial area
ϕ′⋅ω=⇒ϕ′⋅ω=ϕ′⋅⋅= uddsrdu (2.44)
Expressing σw and τw
ϕ ′′⋅ω⋅=ε⋅=σ=σ EEwx (2.46)
dAEdA 2
w ⋅ω⋅ϕ ′′⋅=⋅ω⋅σ (2.47)
∫∫ ⋅ω⋅ϕ ′′⋅=⋅ω⋅σ=A
2
A
w dAEdAB (bimoment) (2.48)
(bimoment de încovoiere-r ăsucire )
∫ ⋅ω=A
2
w dAI (warping constant [L6]) (2.49)
(moment de iner ţ ie sectorial )
Parallel between bending moment and warping torsion
zI
M
y
Ed,y
x ⋅=σ ω⋅=σw
wI
B (2.50)
y
yEd,z
zIt
SV
⋅
⋅=τ
w
wEd,w
wIt
SM
⋅
⋅=τ (2.51)
∫ ⋅ω=A
w dAS (warping static moment [L4]) (2.52)
Sw = … [L4] (moment static sectorial )
The coordinates of the shear centre about the centre of gravity are:
y
AC
I
dAz
y∫ ⋅⋅ω
= (2.53)
z
AC
I
dAy
z∫ ⋅⋅ω
= (2.54)
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2.1.5. Cross-section characteristics associated to torsion
Considering a mono-symmetrical cross-section (Fig. 2.16), the following can be
calculated:
Fig. 2.16. Mono-symmetrical cross-section (SN030a-EN-EU [14])
• the position of the shear centre S from the bottom fibre of the cross-section:
1312
32
1
3
1
s
2
SC tbtb
tb
h2
t
z ⋅+⋅
⋅
⋅+= (2.55)
• the St. Venant torsional constant:
3
thtbtbI
3
ww
3
22
3
11T
⋅+⋅+⋅= (2.56)
• the warping constant (SN030a-EN-EU [14]):
( )223
21
3
1
2
3
21
3
1z
2
sw
tbtb
tbtbIhI
⋅+⋅
⋅⋅⋅⋅⋅= (2.57)
2.2. BUCKLING LENGTH
The first known theoretical approach for solving a bar in compression belongs to
Euler (1744) [1]. He started by writing the following equilibrium equation (Fig. 2.17)
for a pin connected bar axially loaded in compression:
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ρ=−=
+
1
EI
M
dx
dv1
dx
vd
232
2
2
(2.58)
where:
vFM ⋅= (2.59)
Fig. 2.17. The equilibrium of a pin connected bar in compression
The solution he obtained is the very well known:
2
2
crL
EIF
⋅π= (2.60)
for the critical force that generates buckling of the bar and:
L
xsinez 0
⋅π⋅= (2.61)
for the deformed shape of the bar.
This relation was then extended to other types of restraints at the ends, by inscribing
the bar on an equivalent pin-connected bar (Fig. 2.18). To allow this, the buckling
length was defined as a concept. All these theoretical approaches are based on the
theory of bifurcation of equilibrium.
Definition
The system length (EN 1993-1-1 [6] def. 1.5.5) is the distance in a
given plane between two adjacent points at which a member is
braced against lateral displacement in this plane, or between one
such point and the end of the member.
Definition
The buckling length (Lcr) (EN 1993-1-1 [6] def. 1.5.6) is the system
length of an otherwise similar member with pinned ends, which has
the same buckling resistance as a given member or segment of
member.
L
e0
F x
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It is also defined as the distance between two consecutive inflection
points along the deformed shape of a bar. Sometimes, in practice, it
is replaced by the system length.
Euler’s relation is then expressed as:
2
cr
2
crL
EIF
⋅π= (2.62)
where Lcr = kL is the buckling length (Fig. 2.18).
k – end fixity condition.
k = 1,0 k = 0,7 k = 2,0 k = 0,5 k = 1,0
Fig. 2.18. Different values of the buckling length factor
2.2.1. Buckling length of columns
In everyday situations, bars are part of a structure, they are connected to other bars
and so the joints are not purely fixed or purely pinned. As a result, the buckling
length of an element depends on its loading state and on the stiffness of the
neighbour bars. Relations for calculating it are given in different books and were
given in Annexe E (informative) of the previous version of Eurocode 3 – ENV 1993-
1-1 [111]. For defining the buckling length of a column, (parts of) structures are
separated in sway and non-sway, depending whether the (lateral) displacements of
the joints at the end of the bar are permitted or not. This separation is done by
means of stiffness criteria that will be presented later. Usually, the non-sway
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behaviour is guaranteed by means of bracings. The distribution factors used in figure
2.19 – 2.22 are calculated using the following relations:
1211C
C1
KKK
K =
++η (ENV 1993-1-1 [3], rel. (E.1)) (2.63)
2221C
C2
KKK
K =
++η (ENV 1993-1-1 [3], rel. (E.2)) (2.64)
where:
KC – stiffness of the column (I/L);
Kij – stiffness of the beam ij.
Remark: A more precise formulation for Kij would be stiffness of the connection
between beam ij and column, as semi-rigid connections could be used. In this case a
more careful analysis should be carried out.
The buckling length for non-sway buckling mode is presented in figure 2.19 [111].
Fig. 2.19. Non-sway buckling mode (ENV 1993-1-1 [111] Fig. E.2.3)
The buckling length for sway buckling mode is presented in figure 2.20 [111].
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Fig. 2.20. Sway buckling mode (ENV 1993-1-1 [111] Fig. E.2.3)
Fig. 2.21. End fixity condition, k, for non-sway buckling (ENV 1993-1-1 [111] Fig.
E.2.1)
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Fig. 2.22. End fixity condition, k, for sway buckling (ENV 1993-1-1 [111] Fig. E.2.2)
This model can be expanded to continuous columns, presuming the loading factor
N/Ncr is constant on their entire length. If this does not happen (which is the actual
case) the procedure is conservative for the most critical part of the column [111]. In
this case, the distribution factors are calculated using the following relations:
12111C
1C1
KKKKKK =
++++η (ENV 1993-1-1 [3], rel. (E.3)) (2.65)
22212C
2C2
KKKK
KK =
+++
+η (ENV 1993-1-1 [3], rel. (E.4)) (2.66)
where K1 and K2 are the values of the stiffness of the neighbour columns (Fig. 2.23).
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Fig. 2.23. Distribution factors for continuous columns (ENV 1993-1-1 [111] Fig.
E.2.4)
2.2.2. Buckling length of beams
Presuming the beams are not subject to axial forces, their stiffness can be taken
from table 2.1, as long as they remain in the elastic range [111].
Table 2.1. Stiffness of a beam in the elastic range (ENV 1993-1-1 [111] Tab. E.1)
Connection at the other end of the beam Stiffness K of the beam
Fixed 1,0 × I/L
Pinned 0,75 × I/L
Rotation equal to the adjacent one (double curvature) 1,5 × I/L
Rotation equal and opposite to the adjacent one
(simple curvature)
0,5 × I/L
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General case: θa rotation at the adjacent end and θb rotation at the opposite end
(1,0 + 0,5 × θa / θb) × I/L
For regular buildings with rectangular frames and reinforced concrete floors, subject
to uniform loads, it is accepted to consider the stiffness of the beams given in table
2.2.
Table 2.2. Stiffness K of beams – structures with reinforced concrete floors ([111]
Tab. E.2)
Loading condition of the beam Non-sway bucklingmode
Sway buckling mode
Beams supporting directly the
reinforced concrete slabs
1,0 × I/L 1,0 × I/L
Other beams under direct loads 0,75 × I/L 1,0 × I/L
Beams subjected only tobending moments at the ends
0,50 × I/L 1,5 × I/L
When the beams are subject to axial forces, stability functions must be used for
expressing their stiffness. A simplified conservative approach is proposed in ENV
1993-1-1 [111], neglecting the increase of stiffness generated by tension and
considering only compression in the beams. Based on these assumptions, the
values in table 2.3 can be considered.
Table 2.3. Stiffness of beams in compression (ENV 1993-1-1 [111] Tab. E.3)
Connection at the other end of the beam Stiffness K of the beam
Fixed 1,0 × I/L × (1,0 – 0,4 × N/NE)
Pinned 0,75 × I/L × (1,0 – 1,0 × N/NE)
Rotation equal to the adjacent one (doublecurvature)
1,5 × I/L × (1,0 – 0,2 × N/NE)
Rotation equal and opposite to the adjacent one(simple curvature)
0,5 × I/L × (1,0 – 1,0 × N/NE)
where:
2
2
EL
EIN
⋅π= (2.67)
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2.2.3. Empirical relations for the buckling length of columns
ENV 1993-1-1 [111] provides empirical expressions as safe approximations that can
be used as an alternative to the values from figures 2.21 and 2.22. The k coefficient
for the buckling length can be calculated by the following relations:
a. for non-sway buckling mode (Fig. 2.21)
( ) ( )22121 055,014,05,0=k η+η⋅+η+η⋅+ ([111], rel. (E.5)) (2.68)
or, alternatively,
( )
( ) 21212121
247,0364,00,2
265,0145,00,1 =k
η⋅η⋅−η+η⋅−
η⋅η⋅−η+η⋅+ ([111], rel. (E.6)) (2.69)
b. for sway buckling mode (Fig. 2.22)
( )( )
5,0
2121
2121
60,08,00,1
12,02,00,1 =k
η⋅η⋅+η+η⋅−
η⋅η⋅−η+η⋅− ([111], rel. (E.7)) (2.70)
2.2.4. Comments on the buckling length of beams
If the buckling length is generally easy to identify for members subject to axial
compression forces, the effective lateral buckling length is a more delicate subject,
given the complexity of the deformed shape (at the same time buckling and torsion).
This leads to a temptation to simplified approaches, like considering the effective
lateral buckling length as equal to the distance between points of zero (Fig. 2.24) in
the bending moment diagram, or between inflection points of the strong axis
deformed shape [8].
Important
In order to prevent this, the American code ANSI/AISC 360-10 [7]
states in the 6.3 commentary: “In members subjected to double
curvature bending, the inflection point shall not be considered a
brace point.”
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