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ELASTIC FLEXURALTORSIONAL BUCKLING OF FIXEDARCHES
by YONG-LIN PI and MARK ANDREW BRADFORD
(School of Civil and Environmental Engineering, University of
New South Wales, Sydney,NSW 2052, Australia)
[Received 6 May 2003. Revise 22 June 2004]
Summary
This paper is concerned with the elastic flexuraltorsional
buckling of laterally fixed circulararches that are subjected to
uniform axial compression, and to uniform bending. The
finitestrains and the energy equation for the flexuraltorsional
buckling of arches have been derivedbased on an accurate orthogonal
rotation matrix. Closed form solutions for the elastic
flexuraltorsional buckling resistance of laterally fixed arches in
uniform compression and in uniformbending have been obtained, which
are quite different from those of pin-ended and simplysupported
arches. The results demonstrate that the routine effective length
approach, which isoften useful for determining the buckling
response of straight members, is not suitable for theelastic
flexuraltorsional buckling response of fixed arches.
1. Introduction
Arches that are loaded in-plane may suddenly displace laterally
and twist out of their plane ofloading and fail in a
flexuraltorsional buckling mode as shown in Fig. 1. The elastic
flexuraltorsional buckling loads of arches that are subjected to
uniform axial compression and to uniformbending play an important
role in the investigation of the mechanics of the buckling response
ofarches under general loading. They are also often used as the
reference load (or moment) in thedesign of steel arches (1 to 3).
This paper is concerned with the elastic flexuraltorsional
bucklingof laterally fixed circular arches with a doubly symmetric
cross-section that are subjected to uniformcompression and to
uniform bending.
Elastic flexuraltorsional buckling of arches that are subjected
to uniform compression or touniform bending has been investigated
by a number of researchers (4 to 14). Closed form solutionsfor the
elastic flexuraltorsional buckling load of pin-ended arches in
uniform compression andfor the elastic flexuraltorsional buckling
moment of simply supported arches in uniform bendinghave been
obtained (5,8,10 to 12); however, the elastic flexuraltorsional
buckling of laterally fixedarches that are subjected to uniform
compression and to uniform bending does not appear to havebeen
reported. It might be thought that the solutions for the elastic
flexuraltorsional buckling loadof pin-ended arches in uniform
compression and for the elastic flexuraltorsional buckling momentof
simply supported arches in uniform bending can be extended
routinely to fixed arches by usingthe effective length concept
(15). The method that adopts the effective length approach has
beenused in the elastic buckling analysis of columns and beams (4,
5, 15). For example, the closed formsolution for the flexural
buckling load Ny of a column about its minor principal axis is
given by
Ny = 2 E Iy/2,Q. Jl Mech. Appl. Math. (2004) 57 (4), 551569 c
Oxford University Press 2004; all rights reserved.
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552 Y.-L. PI AND M. A. BRADFORD
Fig. 1 Geometry and loading. (a) Arch in uniform compression;
(b) arch in uniform bending; (c)flexuraltorsional buckling; (d)
lateral restraints (plan)
where E is Youngs modulus, Iy is the second moment of area of
the cross-section, and is theeffective length of the column,
related to the actual length S of the column by = kS in which k
isthe effective length factor and whose value depends on the end
support conditions. For a laterallypin-ended column, it is well
known that k = 1 and = S; for laterally fixed column, k = 05and =
S/2. Thus, the first mode flexural buckling load of a laterally
fixed column in uniformcompression is equal to the second mode
flexural buckling load of a laterally pin-end column inuniform
compression. This method can also be used for the elastic
flexuraltorsional buckling ofbeams, for which the first mode
flexuraltorsional buckling moment of a laterally fixed beam
inuniform bending is equal to the second mode flexural buckling
moment of a laterally pin-endedbeam in uniform bending. However,
because the lateral deformations primarily couple with thetwist
rotational deformations during the flexuraltorsional buckling of
arches, the method of usingthe concept of an effective length may
be unsuitable for the elastic flexuraltorsional buckling
ofarches.
An analysis of the cross-section of an arch has usually been
undertaken in order to derive thestrains for determining its
flexuraltorsional buckling response (6, 8 to 10). Most of these
types ofanalysis can be considered to be equivalent to using a
rotation matrix R that does not satisfy thedesired proper
orthogonality conditions that RRT = RT R = I and that det R = +1 in
the strainderivations, and so some terms in the strains that are
significant for modelling the flexuraltorsionalbuckling of arches
may be lost. Therefore, an accurate rotation matrix that satisfies
the orthogonalityconditions needs to be sought for the strain
derivations that are to be used in the buckling analysis,in order
to model properly the mechanics of the buckling response.
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BUCKLING OF FIXED ARCHES 553
Fig. 2 Axes system, basis and position vectors
The purpose of this paper is to investigate analytically the
elastic flexuraltorsional buckling oflaterally fixed circular
arches with a doubly symmetric open thin-walled cross-section using
anenergy approach that is based on an accurate rotation matrix, and
to obtain the analytical solutionsfor the elastic flexuraltorsional
buckling load of fixed arches in uniform compression and for
theelastic flexuraltorsional buckling moment of laterally fixed
arches in uniform bending.
2. Rotation and curvatures
In general, the centroidal axis os of a circular arch has an
initial curvature x0 about the majorprincipal axis ox (that is, in
the direction of the minor principal axis oy of the cross-section)
asshown in Fig. 1. To describe the deformation of the arch, a
body-attached curvilinear orthogonalaxis system oxys is defined as
follows. The axis os passes through the locus of the centroids of
thecross-section of the undeformed arch and the axes ox and oy
coincide with the principal axes of thecross-section, as shown in
Figs 1 and 2. After the deformation, the origin o displaces u, v, w
to oand the cross-sections (that are assumed to remain rigid in
their plane and so do not distort) rotatethrough an angle , and so
the body attached curvilinear orthogonal axis system oxys moves
androtates to a new position oxys as shown in Fig. 2.
A unit vector ps in the tangent direction of os, and unit
vectors px and py in the direction of oxand oy form a right-handed
orthonormal basis as shown in Fig. 2. The unit vectors px , py, ps
areused as the fixed reference basis. They do not change with the
deformation, but their directionschange from point to point along
the arch axis os. In the deformed configuration, a unit vector qs
isdefined along the tangent direction of the axis os of the axis
system oxys, and unit vectors qxand qy are defined along the
principal axes ox, oy of the rotated cross-section at o as
shown
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554 Y.-L. PI AND M. A. BRADFORD
in Fig. 2. The unit vectors qx , qy , qs also form an
orthonormal basis. They attach to the arch andmove with the arch
during the deformation with the vector qs being normal to the
cross-section atall times.
The rotation matrix that describes the rotation from the basis
vectors px , py , ps in the undeformedconfiguration to the basis
vectors qx , qy , qs in the deformed configuration can be obtained
in matrixform as (see the Appendix)
[qx , qy, qs] = [px , py, ps]R, (2.1)
where the rotation matrix R is given by
R = Rxx Rxy RxsRyx Ryy Rys
Rsx Rsy Rss
(2.2)
with
Rxx = (1 u2)C uvS, Rxy = (1 u2)S uvC, Rxs = u, (2.3)Ryx = (1
v2)S uvC, Ryy = (1 v2)C + uvS, Rys = v, (2.4)Rsx = uC vS, Rsy = uS
vC, Rss = w, (2.5)
where C cos , S sin , u = u/(1+), v = v/(1+), w = (1+w)/(1+), v
= vwx0,w = w +vx0, ( ) d( )/ds, (1+ ) =
(u)2 + (v)2 + (1 + w)2, is the longitudinal normal
strain at the centroid and = 1/(1 + w).The rotation matrix R in
(2.2) belongs to a special orthogonal rotation group denoted by
SO(3)
because it satisfies the proper orthogonality and unimodular
conditions (16)
RRT = RT R = I and det R = +1. (2.6)
For SO(3), the curvatures and strains are invariant during
rigid-body rotations.A fixed (space) right-handed rectangular
coordinate system O XY Z is defined in space as shown
also in Fig. 2. The position of the undeformed and deformed arch
can be defined in the axissystem O XY Z . Unit vectors PX , PY , PZ
in directions O X , OY and O Z also form a right-handedorthonormal
basis.
In the undeformed configuration, the position vector of the
centroid o in the fixed axes O XY Z isr0 (Fig. 2), and so the unit
vector ps tangential to the centroidal axis os can be expressed in
termsof the position vector r0 as
ps = dr0/ds. (2.7)
The FrenetSerret formulae in terms of basis vectors px , py , ps
in the undeformed configurationcan then be written as
dpids
= K0pi (i = x, y, s), (2.8)
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BUCKLING OF FIXED ARCHES 555
Mx Mx Mx Mx
(d)(c)
s
sy
yx
x
(b)(a)
Fig. 3 Curvature and bending moment conventions. (a) Positive
initial curvature; (b) negative initialcurvature; (c) positive
bending; (d) negative bending
where K0 is the skew-symmetric matrix for the initial curvature
x0 and is given by
K0 = 0 0 00 0 x0
0 x0 0
(2.9)
and the positive and negative curvatures are defined as shown in
Fig. 3.In the deformed configuration, the position vector of the
centroid o in the fixed axis system
O XY Z is r as shown in Fig. 2, and so the vector qs tangential
to the deformed centroidal axis oscan be expressed in terms of the
position vector r of the centroid o as
qs = drds =1
1 + drds
, (2.10)
where the relationship ds = (1 + )ds is used. Because the
differentiation of the position vectoris taken with respect to the
deformed length s, qs is a unit vector.
The position vector r of the centroid o can be expressed as
(Fig. 2)r = r0 + upx + vpy + wps . (2.11)
Substituting (2.11) into (2.10) and considering (2.7)
produces
qs = 11 + drds
= 11 + [u
px + vpy + (1 + w)ps] = upx + vpy + wps . (2.12)In the deformed
configuration, according to the FrenetSerret formulae, the
relationship between
the derivatives of the basis vectors and the curvatures and
twist can be written asdqids
= 11 +
dqids
,= Kqi (i = x, y, s), (2.13)
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556 Y.-L. PI AND M. A. BRADFORD
where the matrix K for the curvatures and twist in the deformed
configuration is given by
K = 0 s ys 0 x
y x 0
(2.14)
in which x and y are the curvatures about the unit vectors qx
and qy (that is, about ox and oy),respectively, and s is the twist
about the unit vector qs (that is, about os) after the
deformation.
Differentiating (2.1) with respect to s yields[dqxds
dqyds
dqsds
]= [px py ps] dRds +
[dpxds
dpyds
dpsds
]R (2.15)
and substituting (2.8) and (2.13) into (2.15) and considering ds
= (1+)ds leads to the curvaturesin the deformed configuration
as
(1 + )K = RT dRds
+ RT K0R. (2.16)
Substituting (2.2) and (2.9) into (2.16) leads to the curvatures
x and y and the twist s in thedeformed configuration expressed
as
x = {uS vC w(uS vC) + [(1 u2 w2)C uvS + wC]x0}/(1 + ),y = {uC +
vS w(uC + vS) [(1 u2 w2)S + uvC + wS]x0}/(1 + ),s = [ + (uv uv) +
ux0]/(1 + ),
where S sin and C cos .
3. Position vectors and strains
The position vector a0 of an arbitrary point P(x, y) on the
cross-section of the arch in theundeformed configuration can be
expressed as (Fig. 2)
a0 = r0 + xpx + ypy, (3.1)where x and y are the coordinates of
the point P in the principal axes ox and oy (Fig. 1(c)).
The position of the point P(x, y) in the deformed configuration
is determined based on thefollowing two assumptions. First, the
cross-sectional plane remains both plane and perpendicularto the
member axis during the deformation (the EulerBernoulli hypothesis).
Secondly, the totaldeformation of the point P results from two
successive motions: translation and finite rotationof the
cross-section, and a superimposed warping displacement along the
unit vector qs in thedeformed configuration. Under these two
assumptions, the position vector a of the point P1, whichis the
position of the point P after the deformation, can be expressed in
terms of the basis vectorsqx , qy, qs as (Fig. 2)
a = r + xqx + yqy (x, y)sqs (3.2)in which (x, y) is the
normalized section warping function.
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BUCKLING OF FIXED ARCHES 557
The warping function (x, y) in (3.2) can be obtained by
considering the Saint-Venant uniformtorsion problem for a prismatic
bar, which results in the Laplace equation
2 = 2
x2+
2
y2= 0 (3.3)
and which may be solved by considering that the shear stresses
sx and sy conjugate to the shearstrains sy and sx satisfy a
traction-free boundary condition on the lateral surface. The
solutionfor the warping function (x, y) can be uniquely specified
by using the following three additionalorthogonality
conditions:
A(x, y) d A =
A
x(x, y) d A =
Ay(x, y) d A = 0. (3.4)
In order to obtain the expressions for the finite strains, the
following simplifications are made. Itis assumed that the effects
of third and higher order terms of the strains are very small and
can beignored and that w + u2/2 + v2/2 + w2/2. The longitudinal
normal strain ss at the pointP(x, y) on the cross-section can then
be obtained as
ss = 12(
a
s
a
s a0
s
a0s
)
w + 12 u2 + 12 v2 + 12 w2 x{uC + vS x0S}+ y{uS vC + x0C 12 u2x0
x0} { + ux0} + 12 (x2 + y2){ + ux0}2, (3.5)
where the approximation = 1/(1 + w) = 1/(2 + w) 1/2 is used.
Similarly,
sx = 12(
a
s
a
x a0
s
a0x
)
(y +
x
)s (3.6)
and
sy = 12(
a
s
a
y a0
s
a0y
)
(x
y
)s . (3.7)
4. Buckling analysis
4.1 Stress resultants
In classical flexuraltorsional buckling analysis, the following
assumptions are usually made. First,there are no lateral and twist
deformations before buckling. Secondly, the conservative loads that
actas well as the in-plane stress resultants are constant during
the flexuraltorsional buckling. Hence,during flexuraltorsional
buckling, there are no changes of in- plane deformations. Thirdly,
theprebuckling strains are small so that linearized prebuckling
strains can be used in the bucklinganalysis, while fourthly, the
effects of prebuckling deformations on the flexuraltorsional
bucklingcan be ignored. According these assumtions, prior to
buckling, u = = 0 and the prebuckling
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558 Y.-L. PI AND M. A. BRADFORD
shear strains sx , sy and shear stresses sx sy are equal to
zero. Hence, only the in-plane stressresultants prior to buckling
exist and they are given by
N =
Ass d A and M =
A
ss y d A,
where A is the area of the cross-section, N is the axial force,
M is the bending moment aboutthe axis ox and ss is the longitudinal
normal stress given by ss = Ess with E being Youngsmodulus. During
the buckling, N and M remain constant.
4.2 Arches in uniform compressionA laterally fixed circular arch
(u = = u = = 0 at s = 0 and s = S) with a radial loadq uniformly
distributed around its centroidal axis (Fig. 1(a)) can be
considered to be primarilysubjected to the uniform compression Q =
N = q R, because the bending moments in the archare very small and
can be ignored (M = 0). Under the uniform compression, an arch may
bifurcatein a flexuraltorsional buckling mode.
Because there are no changes of in-plane displacements during
flexuraltorsional buckling, thepotential energy of in-plane
uniformly distributed conservative loads q due to the lateral
andtorsional buckling deformations is equal to zero. The potential
energy of the arch due to lateraland torsional buckling
deformations can then be written as
= 12
V(E2ss + G 2sx + G 2sy) dV, (4.1)
where V indicates the volume of the arch and G is the shear
modulus of elasticity.Substituting (3.5) to (3.7) and M = 0 into
(4.1), and considering the potential energy given by
(4.1) is produced by the lateral and torsional buckling
deformations only, leads to
= 12
S0
{E Iy
(u +
R
)2+ E Iw
( u
R
)2+ G J
( u
R
)2
+ N[(u)2 + r20
( u
R
)2]}ds, (4.2)
where R 1/x0 is the radius of the circular arch because the
curvature x0 of an arch is in thenegative direction of the axis oy,
S is the length of the arch, Iy is the second moment of area of
thecross-section about its minor principal axis, Iw is the warping
constant of the cross-section, J is theSaint-Venant torsional
constant of the cross-section, and Iy , Iw and J are defined by
Iy =
Ax2 d A, Iw =
A
2 d A and J =
A
[(x
y
)2+
(y +
x
)2]d A.
From symmetry, the first mode possible buckled shapes of a
laterally fixed arch can be defined by
u
uC=
C= 1
2
(1 cos s
S/2
), (4.3)
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BUCKLING OF FIXED ARCHES 559
where uC is the central lateral displacement and C is the
central angle of twist during flexuraltorsional buckling. The mode
shapes of (4.3) satisfy the kinematic boundary conditions u = =u =
= 0 at s = 0 and s = S and the symmetric conditions u = = 0 at s =
S/2.
By substituting (4.3) and N = Q into (4.2) and integrating
(4.2), the potential energy due tolateral and torsional buckling
deformations becomes
= 18
S2
(
S/2
)2 {E Iy
[u2C
(
S/2
)2 2uCC
R+ 3
2C
R2
(S/2
)2]
+[
G J + E Iw(
S/2
)2] [u2CR2
2uCCR
+ 2C]
u2C Q(
1 + r20
R2
)+ 2uCC Q r
20R
2Cr20 Q}
. (4.4)
According to the principle of stationary potential energy (17),
the buckling equilibriumconfiguration is defined by the
equations
uC= 0 and
C= 0
which leads to [k11 k12k21 k22
] {uCC
}=
{00
}, (4.5)
where
k11 =[
1 + a2f b2f Q
Ny f
(1 + r
20
R2
)]Ny f ,
k12 = k21 =(a f
b f a f b f + QNy f
r20R
Ny fMys f
)Mys f ,
k22 =(
1 + 3a2f
b2f Q
Ny fNy fNs f
)r20 Ns f ,
in which Ny f and Ns f are the first mode flexural buckling load
and the torsional buckling load of afixed column of length S that
is subjected to uniform axial compression respectively, and are
givenby
Ny f = 2 E Iy
(S/2)2and Ns f = 1
r20
[G J +
2 E Iw(S/2)2
], (4.6)
Mys f =
r20 Ny f Ns f is the first mode flexuraltorsional buckling
moment of a laterally fixed beamof length S that is subjected to
uniform bending, and the parameters a f and b f are defined by
a f = S/2 R
and b f = Mys fNy f S/2 . (4.7)
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560 Y.-L. PI AND M. A. BRADFORD
Alternatively, the potential energy given by (4.4) is stationary
when = 0 (17),
uCuC +
CC = 0 (4.8)
which can be written as {uCC
}T [ k11 k12k21 k22
] {uCC
}= 0 (4.9)
which must hold for any variations {uC , C }T and leads to
(4.5).Equation (4.5) has a non-trivial solution for uC and C
when
k11k22 k12k21 = 0 (4.10)which can be written as the quadratic
equation
A1(Qays f /Ny f )2 + A2 (Qays f /Ny f ) + A3 = 0 (4.11)
in which A1 = 1,A2 =
(2a2f 1 3a4f
)
(1 + 3a2f /b2f
)Ns f /Ny f ,
and
A3 =(
1 2a2f + 3a4f + 2a2f /b2f)
Ns f /Ny f .
The first mode flexuraltorsional buckling load of a fixed arch
in uniform compression is thenobtained by solving (4.11) as
Qays fNy f
= 12
Ns fNy f
{(1 + 3a
2f
b2f
)+
(1 2a2f + 3a4f
) Ny fNs f
[1 (1 2a2f + 3a4f ) Ny fNs f
]2
(1 + 6a2f 9a4f
) 2a2fb2f
Ny fNs f
+ 6a2f
b2f+ 9a
4f
b4f
. (4.12)
By using the I-section shown in Fig. 4 with typical material
properties for steel (E = 2 105MPa and G = 8 104 MPa; these values
are also used throughout the paper), the first
modeflexuraltorsional buckling load Qays f of fixed arches in
uniform compression given by (4.12) withn = 1 are compared in Fig.
5 with the first and second mode flexuraltorsional buckling loads
Qaysn(n = 1 and 2) of pin-ended arches in uniform compression given
by (13)
QaysnNyn
= 12
NsnNyn
[(1 + a
2n
b2n
)+
(1 a2n
)2 NynNsn
(
1 + a2n
b2n
)2+ 2
(a2nb2n
1) (
1 a2n)2 Nyn
Nsn+ (1 a2n)4
(NynNsn
)2 , (4.13)
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BUCKLING OF FIXED ARCHES 561
Fig. 4 (a) Cross-section, and (b) dimensions
Fig. 5 Flexuraltorsional buckling of fixed arches in uniform
compression
where the parameters an and bn are defined by
an = Sn R
and bn = n MysnNyn S (4.14)
in which Mysn =
r20 Nyn Nsn is the nth mode flexuraltorsional buckling moment of
a simply
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562 Y.-L. PI AND M. A. BRADFORD
supported beam of length S that is subjected to uniform bending.
Here Nyn and Nsn are the nthmode flexural buckling load and the
torsional buckling load of a pin-ended column of length S thatis
subjected to uniform axial compression respectively, and are given
by
Nyn = (n)2 E Iy
S2and Nsn = 1
r20
[G J + (n)
2 E IwS2
]. (4.15)
It is noted in Fig. 5 that the reference second mode flexural
buckling load Nys2 of a pin-endedcolumn is equal to the reference
first mode flexural buckling load Nys f of a fixed column (4, 5,
15).It can be seen from Fig. 5 that when the included angle is
small, the difference between the firstmode buckling load Qays f of
a fixed arch and the second mode buckling load Qays2 of a
pin-endedarch is very small. The difference increases rapidly with
an increase of the included angle .The first mode buckling load
ratio Qays f /Nys f of a fixed arch decreases slowly with an
increaseof the included angle and still has a substantial value
when = 180, whereas the first andsecond mode buckling load ratios
Qays/Nys and Qays2/Nys2 of the corresponding pin-ended archdecrease
rapidly with an increase of the included angle and the first mode
buckling load becomeszero and the second mode buckling load becomes
very small when = 180. Hence, using thesecond mode buckling load of
a pin-ended arch as the first mode buckling load of a fixed
archwill significantly underestimate the flexuraltorsional buckling
resistance of the fixed arch. This isquite different from columns
that are subjected to uniform axial compression. The second
modeflexural or torsional buckling load of a pin-ended column is
equal to the first mode counterpart ofthe corresponding fixed
column (that is, when = 0 in Fig. 5) (15).
The variations of the dimensionless first mode flexuraltorsional
buckling load ratio Qays f /Nys fof fixed arches ( = 90) given by
(4.12) with the out-of-plane slenderness ratio S/ry are shownin
Fig. 6, where ry =
Iy/A is the radius of gyration of the cross-section about its
minor
principal axis. The variations of dimensionless first and second
mode flexuraltorsional bucklingload ratios Qays/Nys and Qays2/Nys2
of the corresponding pin-ended arches given by (4.13) withthe
slenderness ratio S/ry are also shown in Fig. 6. It can be seen
that the difference between thefirst mode buckling load of fixed
arches and the second mode buckling load of pin-ended
archesincreases with an increase of S/ry .
4.3 Arches in uniform bendingWhen an in-plane simply supported
and out-of-plane fixed arch is subjected to equal and
oppositemoments M , the arch is under uniform bending and the axial
force N = 0. Under the uniformbending, an arch may also bifurcate
in a flexuraltorsional mode. Because there are no changesof
in-plane displacements during flexuraltorsional buckling, the
potential energy of external in-plane end moments M due to the
lateral and torsional buckling deformations is equal to zero.
Thepotential energy of the arch due to flexuraltorsional buckling
deformations can also be given by(4.1).
Substituting N = 0 and (3.5) to (3.7) into (4.1) and considering
the potential energy given in(4.1) is produced by the lateral and
torsional buckling deformations only leads to the
followingexpression for the potential energy due to these
deformations:
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BUCKLING OF FIXED ARCHES 563
Fig. 6 Effects of the slenderness ratio S/ry on the buckling of
fixed arches in uniform compression
= S
0
12
{E Iy
(u +
R
)2+ E Iw
( u
R
)2+ G J
( u
R
)2
+M(
2u + 2
R+ u
2
R
)}ds. (4.16)
Substituting (4.3) into (4.16) and then integrating gives
= 18
S2
(
S/2
)2 {[u2C
E Iy2
(S/2)2 2uCC E IyR + 3
2C
E IyR2
(S/2
)2]
+[
G J + E Iw2
(S/2)2
] [u2CR2
2uCCR
+ 2C]
2uCC M + 32CMR
(S/2
)2+ u2C
MR
}.
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564 Y.-L. PI AND M. A. BRADFORD
As in section 4.2, we obtain (4.10), where
k11 =(
1 + a2f b2f + a f b fM
Mys f
)Ny f ,
k12 = k21 =(a f
b f a f b f MMys f
)Mys f ,
k22 =(
1 + 3a2f
b2f+ 3a f
b fM
Mys f
)r20 Ns f .
This leads to the quadratic equation for the buckling moment
Mays f
B1(Mays f /Mys f
)2 + B2 (Mays f /Mys f ) + B3 = 0 (4.17)in which B1 = 1 3a2f
,
B2 = a f b f 3a3f b f a fb f
3a3f
b fand B3 = 2a2f 3a4f
2a2fb2f
1.
Equation (4.17) has two solutions, corresponding to positive
bending (Fig. 3(c)) and negativebending (Fig. 3(d)).
Typical relationships between the first mode flexuraltorsional
buckling moment Mays f of alaterally fixed arch given by (4.17) and
the included angle are shown in Fig. 7. The first andsecond mode
flexuraltorsional buckling moment Maysn (n = 1 and 2) of a simply
supported archgiven by (5)
MaysnMysn
= anbn2
an2bn
(
anbn2
an2bn
)2+ (1 a2n)2 (4.18)
are also shown in Fig. 7. It is noted in Fig. 7 that the
reference second mode flexuraltorsionalbuckling moment Mys2 of a
simply supported beam is equal to the reference first mode
flexuraltorsional buckling moment Mys f of a laterally fixed beam
(4, 5, 15). It can be seen fromFig. 7 that under positive uniform
bending (Fig. 3(c)), the dimensionless buckling moment ratioMays f
/Mys f of laterally fixed arches increases with an increase of the
included angle , whereasthe dimensionless first and second mode
buckling moment ratios Mays/Mys and Mays2/Mys2 ofsimply supported
arches decreases with an increase of the included angle (the first
mode bucklingmoment of simply supported arches becomes zero when =
180). The differences betweenthe buckling moments of laterally
fixed arches and the second mode buckling moments of
simplysupported arches are significant. This is quite different
from beams that are subjected to uniformbending. Under negative
uniform bending (Fig. 3(d)), however, the absolute values of the
bucklingmoments of both laterally fixed arches and simply supported
arches increase with an increase of theincluded angle . The
buckling moments of laterally fixed arches in negative uniform
bending alsodiffer from the second mode buckling moments of simply
supported arches.
Typical relationships between the dimensionless first mode
flexuraltorsional buckling momentratio Mays f /Mys f of laterally
fixed arches ( = 90) and the slenderness ratio S/ry are shown
in
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BUCKLING OF FIXED ARCHES 565
Fig. 7 Flexuraltorsional buckling of laterally fixed arches in
uniform bending
Fig. 8. The relationships of the dimensionless first and second
mode flexuraltorsional bucklingmoment ratios Mays/Mys and
Mays2/Mys2 of the corresponding laterally pin-ended arches withthe
slenderness ratio S/ry are also shown in Fig. 8. It can be seen
that for positive bending, thedifference between the first mode
buckling moment ratio Mays f /Mys f of laterally fixed arches
andthe second mode buckling moment ratio Mays2/Mys2 of laterally
pin-ended arches is significant andincreases with an increase of
the slenderness ratio S/ry . For negative bending, the difference
is lesssignificant.
5. Conclusions
This paper has investigated the elastic flexuraltorsional
buckling of laterally fixed circulararches in uniform compression
and in uniform bending. The nonlinear relationship between
thedisplacements and strains during the flexuraltorsional buckling
has been obtained based on anaccurate orthogonal rotation matrix. A
classical energy approach for investigating the buckling
oflaterally fixed arches was then formulated. Closed form solutions
for the elastic flexural- torsionalbuckling resistances of
laterally fixed arches in uniform compression and in uniform
bending havebeen obtained in the paper.
It was found that for arches that are subjected to uniform axial
compression, when the includedangle is small, the difference
between the first mode flexuraltorsional buckling load of a
fixed
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566 Y.-L. PI AND M. A. BRADFORD
Fig. 8 Effects of the slenderness ratio S/ry on the buckling of
laterally fixed arches in uniform bending
arch and the second mode flexuraltorsional buckling load of a
pin-ended arch is very small. Thedifference increases rapidly with
an increase of the included angle and becomes very large whenthe
included angle exceeds a certain value. The difference also
increases with an increase of theslenderness ratio S/ry .
For a laterally fixed arch that is subjected to positive uniform
bending, as the included angleincreases, the first mode buckling
moment of a laterally fixed arch increases whereas the secondmode
buckling moment of a laterally pin-ended arch decreases and the
difference between themgrows rapidly with the included angle. The
difference also grows with S/ry . For a laterally fixedarch that is
subjected to negative uniform bending, as the included angle
increases, the absolutevalue of both the first mode buckling moment
of a laterally fixed arch and the second mode bucklingmoment of a
laterally pin-ended arch increases. The absolute value of the
second mode bucklingmoment of a laterally pin-ended arch is
somewhat higher that of the first mode buckling moment ofa
laterally fixed arch.
Acknowledgement
This work has been supported by an Australian Professorial
Fellowship and a Discovery Projectawarded to the second author by
the Australian Research Council.
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BUCKLING OF FIXED ARCHES 567
References
1. Standards Australia, AS4100-1998, Steel structures (SA,
Sydney, Australia 1998).2. Y.-L. Pi and N. S. Trahair, Out-of-plane
inelastic buckling and strength of steel arches, J. Struct.
Engng 124 (1998) 174183.3. and , Inelastic lateral buckling
strength and design of steel arches, Engng Struct. 22
(2000) 9931005.4. S. P. Timoshenko and J. M. Gere, Theory of
Elastic Stability, 2nd edition (McGrawHill, New
York 1961).5. V. Z. Vlasov, Thin-Walled Elastic Beams, 2nd
edition (Israel Program for Scientific
Translation, Jerusalem 1961).6. C. H. Yoo, Flexuraltorsional
stability of curved beams, J. Engng Mech. 108 (1982) 13511369.7.
SSRC, Guide to Stability Design, Criteria for Metal Structures, 4th
edition (ed. T. V. Galambos;
Wiley, New York 1984).8. J. P. Papangelis and N. S. Trahair,
Flexuraltorsional buckling of arches, J. Struct. Engng 113
(1987) 889906.9. Y.-B. Yang and S.-R. Kuo, Effects of curvature
on stability of curved beams, ibid. 113 (1987)
821841.10. S. Rajasekaran and S. Padmanabhan, Equations of
curved beams, J. Engng Mech. 115 (1989)
10941111.11. N. S. Trahair, FlexuralTorsional Buckling of
Structures (E & FN Spon, London 1993).12. Y.-L. Pi, J. P.
Papangelis and N. S. Trahair, Prebuckling deformations and
flexuraltorsional
buckling of arches, J. Struct. Engng 121 (1995) 13131322.13. M.
A. Bradford and Y.-L. Pi, Elastic flexuraltorsional buckling of
discretely restrained arches,
ibid. 128 (2002) 719727.14. Y.-L. Pi and M. A. Bradford, Elastic
flexuraltorsional buckling of continuously restrained
arches, Int. J. Solids Struct. 39 (2002) 22992322.15. N. S.
Trahair and M. A. Bradford, The Behaviour and Design of Steel
Structures to AS4100,
3rd editionAustralian (E & FN Spon, London 1998).16. R. P.
Burn, Groups. A Path to Geometry (Cambridge University Press,
2001).17. G. J. Simitses, An Introduction to the Elastic Stability
of Structures (PrenticeHall, Englewood
Cliffs 1976).18. J. Argyris, An excursion into large rotations,
Comput. Meth. in Appl. Mech. and Engng 32
(1982) 85155.
APPENDIXRotationsThe rotation matrix R that rotates a vector pi
(i = x, y, s) to a new position q j ( j = x, y, s) about a
rotationvector can be expressed as (18)
R = I + sin
S() + 12
sin2(/2)(/2)2
S2(), (A.1)
where = p = x px + ypy + sps and the auxiliary matrix S() is
skew-symmetric and given by
S() = 0 s ys 0 x
y x 0
. (A.2)
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568 Y.-L. PI AND M. A. BRADFORD
The scaled rotation vector given by
= 2p tan (/2) (A.3)is used to construct the rotation matrix
R.
Substituting the rotation vector of (A.3) for the rotation
vector of (A.1) gives
R = I + 11 + /4 S() +
12
1(1 + /4)S
2(). (A.4)
The vector can be expressed in component form by using basis
vectors px , py , ps as
= x px + ypy + sps , (A.5)and S() is given by (A.2), where x , y
, s replace x , y , s , respectively.
By using (A.2) and (A.5), the components of the matrix R of
(A.4) can be written as
Rxx = [1 14 (1 + cos )(2y + 2s )], Rxy = 12 (1 + cos )(s +
12xy), (A.6)Rxs = 12 (1 + cos )(y + 12xs), Ryx = 12 (1 + cos )(s +
12yx ), (A.7)
Ryy = [1 14 (1 + cos )(2s + 2x )], Rys = 12 (1 + cos )(x +
12ys), (A.8)Rsx = 12 (1 + cos )(y + 12sx ), Rsy = 12 (1 + cos )(x +
12ys), (A.9)Rss = [1 14 (1 + cos )(2x + 2y)]. (A.10)
It can be derived from (A.5) that
2x + 2y + 2s = 4 tan2
2= 4(1 cos )
1 + cos (A.11)
and from (2.1) and (A.6) to (A.10), the unit vector qs can then
be expressed in terms of the basis vectors px ,py , ps as
qs = Rxspx + Ryspy + Rssps = 12 (1 + cos )(y + 12xs)px+ 12 (1 +
cos )(x + 12ys)py + [1 14 (1 + cos )(2x + 2y)]ps . (A.12)
On the other hand, the unit vector qs can also be expressed by
(2.12). Comparison of (2.12) and (A.12)leads to
12 (1 + cos )(y + 12xs) = u, (A.13)
12 (1 + cos )(x + 12ys) = v, (A.14)
1 14 (1 + cos )(2x + 2y) = w. (A.15)In order to express the
rotation matrix in terms of general displacements, in addition to
the displacements
u, v, w, a fourth displacement parameter s associated with the
twist rotation of the cross-section isintroduced such that
s = 2 tan (/2). (A.16)Substituting (A.16) into (A.11) and then
solving (A.11) and (A.15) simultaneously gives
cos = 1 + (1 + cos )(1 + w)/2. (A.17)
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BUCKLING OF FIXED ARCHES 569
Substituting (A.16) and (A.17) into (A.13) and (A.14) yields
x = 2{u tan (/2) v} and y = 2 {u + v tan (/2)} , (A.18)
where = 1/(1 + w). Substituting (A.16) to (A.18) into (A.6) to
(A.10) leads to the following relationshipbetween basis vectors px
, py , ps in the undeformed configuration and basis vectors qx , qy
, qs in the deformedconfiguration being stated in tensor form
as
qi = R ji p j , i, j = x, y, s, (A.19)or in a matrix form as
[qx , qy , qs ] = [px , py , ps ]R, (A.20)where the matrix R is
given by (2.2) with components given by (2.3) to (2.5). These
components are expressedin terms of the four conventional
displacement parameters u, v, w, and their derivatives, and they
becomeinfinite only when 1 + w = 0 and/or 1 + = 0. These situations
cannot occur during the deformation of areal structure.
The rotation matrix R of (2.2) belongs to the special orthogonal
rotation group SO(3) because it satisfiesthe orthogonality
conditions of (2.6). For the SO(3) group, the invariant requirement
needed for the rigid bodyrotation is satisfied, namely, the
curvatures and strains are invariant during the rigid body
rotations.
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