An Exact Theory for Curved
Beams with Any Thin-walled
Open Sections
Genshu Tong* and Qiang Xu
Department of Civil Engineering, Zhejiang University, Hangzhou, 310027, China
Abstract: Currently available theories for thin-walled curved beams lack rigorous
theoretical development. This paper provides a detailed derivation of an exact theory
for biaxial bending and torsion of thin-walled circularly curved beams with any open
profile. The derivation is based on two well-accepted assumptions in the theory of
thin-walled members. Exact expressions for longitudinal displacement, longitudinal
normal stress and shear stress and their resultants are presented. Simplified theories are
also given for practical applications.
Key words: curved beam, thin-walled beam, bending and torsion
1. INTRODUCTION
Curved girders having a streamlined shape find wide
applications in highway bridges and city transit
services, in public buildings, in industrial buildings and
in oil-sealed dry gas-holders, etc. The cross-sections of
these curved girders have different shapes (P , I, C, box
and H with no axis of symmetry ).
A curved beam usually bends and twists
simultaneously under practical load conditions. The
earliest investigator of curved girders is St. Venant
(McManus 1969). Vlasov (1961) proposed a systematic
bending and torsion theory for curved beams.
Dabrowski (1964) studied bending and torsion of
curved girders with P -shaped sections that were used
mainly in bridges. Timoshenko (1965) investigated the
buckling of I-shaped curved beams without considering
warping resistance. Kristek (1979) introduced the
theories of curved box girders both with and without the
assumption of a rigid contour of the cross-section.
Most of the research work has the common feature
that the internal force-curvature relationships of a
curved beam were obtained by modifying the
counterparts of the straight beam considering the effect
of initial curvature. Substituting these relations into the
force equilibrium equations of curved beams (Vlasov
1961, Timoshenko 1965) or into the total potential
energy expression (Yoo 1982), the differential
equilibrium equations in terms of the displacements and
twist rotation were obtained. Dabrowski (1964) studied
curved beams by assuming strain distributions on the
cross-section. Kristek (1979) assumed the warping
displacement in the curved beam. Yoo’s (1982) paper
aroused further researches on curved beams and a series
of discussions were made on Vlasov’s and Yoo’s
theories. New theories were also put forward by
Rajaskaran and Padmanabhan (1989), Yang and Kuo
(1987), Kang and Yoo (1994). However, all these
studies are applicable only for curved beams with
doubly symmetrical cross-sections.
Rajasekran and Padmanabhan (1989) and Kang and
Yoo (1994) adopted the longitudinal displacement
presented by Usami and Koh (1980). To determine the
longitudinal displacement of the centroid of a curved
girder, Usami and Koh (1980) introduced a fictitious thin-
*Corresponding author
Advances in Structural Engineering Vol. 5 No. 4 2002
195
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
coordinate systems are set up: ¬r 2 f 2 Y is a
cylindrical coordinate system whose positive direction
is shown in Figure 1a; ­x 2 y 2 z is a Cartesian
coordinate system set up on the cross-section. x axis has
the same direction as the radial r-coordinate and z axis
is directed in the direction of f increasing; ®n 2 s 2 z
is a right-handed coordinate system on the middle
surface of the thin-walled section where n is a normal
positive when outwards directed, s is tangent to the
middle surface positive when counterclockwise relative
to the shear centre S.
The radius of the centroid line is R. The following
two widely accepted assumptions are used to derive the
general theory of curved beams with thin-walled
sections: (1) the shape of the cross-section does not
deform in its own plane during the deformation; (2)
shear strain on the middle surface may be neglected.
Other assumptions are that the material is linearly
elastic and deformation is small. The displacements of
any point on the middle surface in x and y directions are
– v– respectively. The displacements in the n
denoted as u,
and s directions are j and h and the longitudinal
– The deformations
displacement in the z direction is w.
of the shear centre S (xs, ys) are u and v and the twist
angle is u , positive when the x-axis rotates towards the
y-axis (counter-clockwise). a is the angle between the
normal n and x-axis.
walled branch connecting the centroid and any point on
the middle surface of the thin-walled section (when the
centroid is not on the middle surface of the cross section),
and defined a warping coordinate on this fictitious branch.
After introducing a further simplification, Usami obtained
an approximate longitudinal displacement for the curved
beam. Although Usami and Koh derived their theory
starting from the two fundamental assumptions of thinwalled members (i.e. the shear strain on the middle
surface may be neglected and the cross-section does not
deform in its own plane), his longitudinal displacement
would not satisfy the zero shear strain assumption on the
middle surface and introducing a fictitious thin-walled
branch is hardly justifiable.
To the authors’ knowledge, Yang (1987) was the
only researcher who derived the correct longitudinal
displacements of curved beams based on the two
fundamental hypotheses of thin-walled members, but
his research was restricted to beams with doubly
symmetrical H-shaped sections. Pi and Trahair (1997)
also analyzed a doubly symmetrical I-section curved
beam. Tong (1996) presented theories of curved beams
with mono-symmetrical H-sections where the effect of
mono-symmetry was calculated. Tong (1997) proposed
a general theory of curved beams with any open foldedplate cross-section.
This paper presents a new general theory for curved
beams with arbitrary thin-walled open cross-sections.
The theory is exact within the framework of classical
thin-walled members. Simplified theories with good
accuracy are also proposed for practical applications.
2.2. Longitudinal Displacement
A thin-walled circularly curved beam can also be
viewed as a shell generated by the revolution of section
contour B0E about Y axis. Then the curvilinear
coordinate s is along the meridian of the shell and the
coordinate z is in the circumferential direction. The
shear strain on the middle surface of curved beam is the
same as in a shell of revolution (Flugge 1973):
2. THEORETICAL DEVELOPMENT
2.1. Fundamental Assumptions
Shown in Figure 1 is a thin-walled circularly curved
beam with arbitrary open section. The following
My
y
Qy
Qx
Mz
S
y
Mx
S ( x s , ys )
s
x
Js
n
a
E
z
P(x,y)
N
r
C
s
f
B
x
q
O
0 ( x0 , y 0 )
Y
(a)
(b)
Figure 1. A thin-walled curved beam with arbitrary open cross section
196
Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
g
sz
=
¶h
–
¶w
+
r¶f
¶s
–
w
+
sin a
r
(1)
s_2
_ s_2
in which Dy = e0 Rr2 cos a ds 2 A1eA Rr e0 Rr2 cosa dsdA,
s_2
s
_ _2
D v = e0 Rr2 r sds 2 A1 eA Rr e0 Rr2r sdsdA, they satisfy the
following equations:
The first assumption leads to the following
equations:
u– = u 2
ys)u and v– = v + (x 2
(y 2
xs)u
h = 2 u– sin a + v– cos a = 2 u sin a + v cos a + r su
(3)
where r s = (y 2 ys)sin a + (x 2 xs) cos a is the distance of
shear centre S to the tangent of point P (x , y) on the middle
surface, positive when s is counter-clockwise relative to
the shear centre. Substituting Eqn 3 into Eqn 1 and using
the second assumption, the following equation
+
¶s
–
w
r
1
sin a = 2
r
(r su 9 + v9 cos a 2
r
r0
w0 +
x0 2
r0
x
e
s
u9 2
r
cos a
r2
0
sr
e r dsu 9 (5)
dsv9 2
r
0
s
2
where w0 is the longitudinal displacement of a reference
point 0 on the middle surface of the thin-walled section
and x0 is its x coordinate, r0 = R + x0.
It is noted that point 0 may be any point on the
middle surface and is not necessarily the centroid.
When the centroid is not on the middle surface, the
longitudinal displacement of the centroid cannot be
determined because the integration route is not through
the centroid. Here the average longitudinal
– is
displacement of the whole cross-section w = A1 eA wdA
introduced as a basic unknown (A is the cross section
area). Then the longitudinal displacement can be
expressed as:
r
R
w2
x
R
u9 2
r
Dy
R2
v9 2
r
R
R
A
dA = 0
(7)
When the cross section has multiple branches, taking
the point 0 as a single reference point, the longitudinal
displacement may be calculated for each of the branches
from Eqn 5 and Eqn 6.
As a comparison, the longitudinal displacement
presented by Usami and Koh (1980) is reproduced here
in a linearized form:
w– = w 2
x
R
1
u9 2
ys
w+
Rs
2
y
v9 2
Rs
v
v9 2
Rs
1
u 9 2
v9
Rs
2
(8)
where Rs = R + xs is the radius of the shear centre line,
v is the warping function of the thin-walled crosssection. As a general expression, Eqn 8 doesn’t satisfy
Eqn 4.
2.3. Normal Stress and Its Resultants
The longitudinal normal stress on the cross-section may
be calculated from the circumferential strain in the shell
of revolution (Flugge 1973):
s = Ee
e=
Dv
R2
u 9
Advances in Structural Engineering Vol. 5 No. 4 2002
(6)
u–
r
+
–
¶w
r¶f
=
(9a)
w9 2
u0
R
u2
+
u0
r
(9b)
2
y2
ys
Dy
r
u 2
R2
v0 2
Dv
R2
u 0
Its resultants are:
Axial force
Bending moments
e s dA
M = e s ydA,
M = 2 e s xdA
B = e s v dA
N=
x
A
A
y
Bimoment
w– =
y
r
e
dA = 0, Dv
u9 sin a ) (4)
is obtained. Here r = R + x, ( )9 = d( )/df . This is a
partial differential equation with variable coefficients,
with the curvilinear coordinate s as a variable. Unlike
the straight beam, the longitudinal displacement in a
curved beam cannot be obtained by direct integration.
Although a , r, r s are functions of the curvilinear
coordinate s which is varied for different cross sections,
it may be verified that the exact solution of Eqn 4 is:
w– =
A
r
(2a,b)
So the displacement along curvilinear coordinate s is:
–
¶w
eD
v
(10a)
(10b,c)
A
A
(10d)
Substituting Eqn 9a,b into above equations and
denoting
197
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
C0 =
Cv =
e
r
A
v
e
A
r
e
dA, Cy =
dA, Cyy =
A
y
r
y2
e
x
e
¶f
u N
y
A
A
e yD dA, C = e v
A
v
vv
A
v
vx
DydA, Cu
y
A
=
v
ev
EA
R
(w9 2
r
¶q
¶s
2
2q sin a = 0
2
A
Et
2q sin a = 2
Et(y 2
Dv dA,
the following expressions are obtained:
N = EC0(u + u0 ) +
¶s
(12)
dA,
r
A
¶q
+r
Substituting Eqn 9a,b into Eqn 12:
yv
e
¶s t
dA,
r
A
dA, Cyv =
r
A
dA, Cx =
e D dA, C = e D dA, C = e yD dA,
CvN =
Cu x =
1
ys)
r
R
(w0 2
u 9 + Et
Et
u- ) 2
r
Dy
Dv
R
R2
v- + Et
2
(u9 + u- ) +
u -
Again this is a partial differential equation with
variable coefficients and q cannot be found by direct
integration. But it can be solved exactly by taking a free
edge B as the start point of integration and the exact
solution is expressed as follows:
u0 ) 2
(11a)
E(Cy 2
ysC0)u 2
E
CvN
R2
v0 2
E
Cu
N
u
2
R
0
E(w0 2
q=2
u- )
Rr
2
EAs(u9 + u- )
(RAs + Sys) 2
r2
+
(13)
My = 2 ECx(u + u0 ) + E(2 RCy 2
E(Sxs 2
ysCx)u 2
(11b)
E
CvN
R
v0 2
Mx = ECy(u + u0 ) 2
E
Cu
N
R
u 0
ysCy)u 2
E(Cyy 2
(11c)
E
Cvx
R2
v0 2
Bv = ECv (u + u0 ) 2
E
Cu
x
2
R
u 0
Cvv
R2
v0 2
E
Cu
v
R2
+
R2r2
EFv u R 2r 2
where A
= eB tds, Sys = eB xtds, Sxs = eB ytds, Fy = eB rDytds,
s s
Fv = eB rDv tds. The resultants of the shear flow q are:
s
s
s
s
shear force in x direction Qx = 2
e q sin a
shear force in y direction Qyl = 2
e q cos a
E
B
E
B
e qr
E
B
sds
ds
(14a)
ds (14b)
(14c)
Hv =
es
A
Dy
r
R
dA and Hu =
es
A
Dv
r
R
dA
(15a,b)
u 0
2.4. Shear Stress and Its Resultants
The shear flow q on the middle surface is shown in
Figure 1b, positive when it has the same direction as s.
The equilibrium condition of the revolutionary shell
element in the circumferential direction is (Flugge 1973)
198
EFyv-
where Qyl is the shear force in y direction corresponding to
the shear flow q. Here two new notations are introduced:
(11d)
E
u 9 +
warping torsion moment Mv =
y s C v )u 2
E(Cyv 2
ysAs)
r2
Since Eqn 12 may be rewritten as:
r
¶s t
¶f
+
¶
¶s
(r2q) = 0
(16)
we obtain:
Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
e
Qx = 2
E
B
e (r q)d12
E
2
x
2
rR
B
=
e
R
B
e
Qy1 = 2
1
e
E
E
B
1
er
R
E
Mv =
1
R2
e
E
B
R
e qr
E
B
(r2q)dDv =
1
R2
sds
=
R2
R
2
2 (r q)Dv
R2
2
2
2 B
es
E
R ¶f
e
B
(r2q)
u
B
2
Dy
B
E
¶
R2
r2
1
R2
e
y
¶s
r
R
B
Dv
tds
Tsv = GJ
where
1
¶
¶s
(r2q)ds
J=
B
r
R
tds
(17a,b,c)
1
3
v9
R2
e
A
R2
r2
(18)
t2dA.
du
_
1
drdu
g 1’
r
(a)
e
s Dv
1
_ drdu
2
dr
df
(R + xs)u 9 2
g 2’
dr
r
R ¶f
E
Thus the total torsion moment of a curved beam is Mz
= Tsv + Mv . It should be noted that there is another shear
force related to the St.Venant torsion Qy2 = 2 Tsv/Rs
required by the equilibrium condition of torsion moment.
The total shear force in the y direction is Qy = Qy1 + Qy2.
dvv-
2
g
(r2q)ds
r sds =
E
¶f
1 ¶
D v ds =
2.5. Free Torsional Moment
Unlike the straight beam, the displacement out of the
plane of curvature v– produces a shear strain that is similar
to the St.Venant torsion, as shown in Figure 2. The total
–
twisting rate of any point is u r9 2 rv29 which is also derivable
from Vlasov’s general theory of shells if the
displacements of a shell are represented by Eqn 2a,b and
Eqn 6. If the shear modulus of the material is G, one has
cos a ds =
E
B
¶(s t)
Based on these equations, we find that Hv and H u
may be seen as modified definitions of bending moment
and bimoment in a curved beam. The boundary
conditions are also expressed in terms of these two
quantities, see Table 1.
s xtds
eD
R
E
1
e
R2
r
Qx = 2 M 9y/R, Qy1 = H9v/R, Mv = Hu 9 /R
E
B
1 ¶
(r2 q)ds
rR ¶s
1
B
1
which means:
B
u
Dyds =
¶
x
e (r q) r
E
e
E
=
B
R ¶f
1
sin a ds =
2
E
1 ¶
E
g
v-
u2 e
rR
(r2q)Dy 2
2
¶f
r
E
x
1
¶(s t)
2 B
(r2q)
B
ds =
q cos a ds =
(r2q)dDy =
=
rR
1
E
B
¶f
E
2 B
2
= (r2 q)
¶(s t) x
r
e
q sin a ds = 2
2
df
(b)
Figure 2. Twisting rate in curved beams
Advances in Structural Engineering Vol. 5 No. 4 2002
199
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
moment about the x axis caused by a component of
shear force Qx (which is through the shear centre) on a
infinitesimal segment of the curved beam (Figure 3).
Eqn 19f is a equilibrium equation about the shear centre
axis zs, ysNdf is a torsion moment about zs axis
generated by a component of N on the segment (Figure
3). These terms may be verified using virtual
displacement principle (Usami and Koh 1980).
Eliminating the shear forces leads to
Qx
x
Qx d f
Q x+dQ x
N
Ndf
(s
N+dN
xs
(ce
df
zs
ntr
he
oid
R
ar
ce
nt e
r)
)
M0 y + RN = qx(R + xs)R
RN9 2
RMx0 2
Figure 3. Equilibrium of a curved beam segment
2.6. Differential Equilibrium Equations
Assuming the transversely distributed loads are qx, qy
whose action lines are through the shear centre, the
distributed torsion moment about the shear centre is mz,
the axial load qz is through the centroid, then the
equilibrium equations in terms of the internal forces are:
N2
M 9x 2
Qx9 = qx(R + xs)
ysM0y = 2 qyR(R + xs)2
(20c)
ysN = 2 mz(R + xs)
(20d)
Substituting the force-displacement relations into
these equations, four equilibrium equations with
constant coefficients are obtained in terms of four
displacements u, v, w, u . They are omitted here because
of their lengthiness. The boundary conditions are given
in Table 1.
Table 1. Boundary conditions for curved beams
(19b)
N9 + Qx = 2 qzR
(19c)
Mz 2 QyRs + Qxys = 0
(19d)
M9 y + Qx R = 0
(19e)
ysN = 2 mz(R + xs)
M9 z + M x 2
(20b)
(19a)
Q9 y = 2 qy(R + xs)
M9 z + Mx 2
RMz9 2
M9 y = 2 qzR2
(20a)
(19f)
It is noted that the term Qxys in Eqn 19d and ysN in
Eqn 19f were not included by Vlasov (1961) and Tong
(1996,1997). Eqn 19d is a moment equilibrium
condition about the principal x axis, Q xysdf is a bending
Displacement
w
u
u9 /R
Force
N 2 M y/R
Qx = 2 M9 y/R
My
Displacement
Force
v
Qy = H9 v/R 2 Tsv/Rs
v9 /R
2 Hv
Mz = Hu9 + Tsv
u
u 9 /R
2 Hu
3. COMPARISON WITH CURRENT
AVAILABLE THEORY
Because the equations presented above are lengthy,
simplified theories are needed for practical applications.
This section gives simplified theories for curved beams
with specific types of cross-sections and comparisons with
current available theories are given whenever possible.
Figure 4. A curved beam with mono-symmetrical H-section, web perpendicular to the curvature plane
200
Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
3.1. Mono-symmetrical H Section with Web
Perpendicular to the Curvature Plane (Figure 4)
For a mono-symmetrical H section whose web is
perpendicular to the curvature plane, one obtains
R
D y = y, Dv = v
r
, v = 2 x(y 2
ys)
M x = ys
EIy
R3
x
R
1
u9 2
w + ys
v9
R
2
R
v
v9 2
R
1
v9
u 9 2
R
e=
r
5
x
2
R
R2
1
R
1
u +
v0
R
2
v
w9 + ys
2
R2
1
v0
u 0 2
R
My
R
2
v0
R
2
2
26
R
EIy
R3
Bv
My =
R2
R
(25a)
EIy(uIV + 2u0 + u) + EAR2(w9 + u + ysu ) = qxR4 (25b)
EIv (u
IV
v0 ) = qyR4
+ 2u 0 + u ) + EIxR(v0 + Ru ) 2
GJR(Ru 0 2
ysEAR2(w9 + u + ysu ) = m zR4
e=
(25c)
v0 ) +
(25d)
R
w9 + u
x
2
R
R2
(u0 + u) 2
(26)
(23)
1
y v0
+u
2
v
2
R2
1
u 0 2
v0
R
2
and the equilibrium equations are:
EIv
EIx(vIV + Ru 0 ) 2
R2
(Ru
IV
2
vIV) + GJ(Ru 0 2
(u + u0 ),
(u + u0 )
Advances in Structural Engineering Vol. 5 No. 4 2002
v0 ) = qyR4
(27a)
(24a,b)
EIy
R
2
v9
u 9 2
The longitudinal normal strain given by Vlasov is:
It is found from Eqn 17c that the relation Mv =
the theory of straight thin-walled member is also valid for
curved members with this type of cross section because
of Hu = Bv . The force-displacement relations are:
(w9 + u + ysu ) +
1
GJ
EA(w0 + u9 + ysu 9 ) = 2 qzR2
R R
EA
(u 9 + u - ) ,
R3
EIx(vIV + Ru 0 ) + GJ(Ru 0 2
dB
_v
dz in
N=
(u + u 0 ),
R3
where A = eAdA, Iy = eAx2dA, Ix = eAy2dA, Iv = eAv 2dA.
Substituting Eqns 24a~f into Eqns 20a~d, one obtains:
Two quantities defined in Eqns 15a,b are simplified
ys
EIv
Tsv =
as:
Hu = Bv , Hv = Mx 2
2
(24e,f)
(22)
y
R
EIv
2
2
2
u0 2
1
v0
(u + u 0 )
R2
Mv = 2
Usami and Koh (1980) gave the same longitudinal
displacement for this of type section and therefore gave
the same results as the present paper. Eqn 21b becomes
identical to Yang and Kuo (1987) when the section is
doubly symmetrical. The longitudinal normal strain is:
R w9 + u + ysu
EIv
Bv = 2
(21a)
(21b)
y
R
u +
(24c,d)
The longitudinal displacement is:
–=w2
w
EIx
(u + u0 ) 2
1
EIv u
IV
2
vIV
R
2
+ EIxR(v0 + Ru ) 2
GJR(Ru 0 2
v0 ) = mzR4
(27b)
201
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
Vlasov’s (1961) theory required that both x and y axes
are the principal axes although the cross-section may be
non-symmetrical.
1. Beams with doubly symmetrical H sections
Loaded by mid-point concentrated vertical forces or
uniformly distributed loads, curved beams with its ends
simply supported or fixed are analyzed by using
Vlasov’s equations and Eqns 25c,d (setting ys = 0).
Comparison of this paper and Vlasov’s theory shows
that, for simply supported beams, the results of two
theories are almost identical. For beams with fixed ends,
both theories also give the same deformations and
bending moment as well as torsion moments, but
difference exists between bimoments, especially when
flanges are wide, the subtended angle is large and the
radius is small. In most cases the maximum discrepancy
is still less than 6% (the present paper gives smaller
bimoment). For beams with hot-rolled I-section or
narrow-flanged H section, the differences are negligible.
2. Beams with mono-symmetrical H sections
In this case Eqns 25a~d show a coupling between inplane and out-of-plane deformations while Vlasov’s
equations are uncoupled.
Usually the axial load qz = 0, one obtains w9 + u + ysu
= C by integrating Eqn 25a. For simply supported
beams whose end may move freely along
circumferential axis, the mechanics boundary
conditions are: N = 0 and My = 0. Using Eqns 24a,b, C
= 0 is obtained. Substituting w9 + u + ysu = 0 into Eqns
25c,d, we get the decoupled equations governing out-ofplane deformations of curved beams, which are
identical to those equations when the cross section is
doubly-symmetrical. So the conclusions for simplesupported curved beams with doubly symmetrical H
sections are also applicable for curved beams with
mono-symmetrical H sections when only out-of-plane
behaviors are concerned.
For curved beams with fixed ends, the problem is
more complex because C cannot be determined easily.
But it is still possible to find an exact solution to Eqns
25a~d. A curved beam with fixed ends under mid-span
vertical load P = 2 1.0 kN is solved. Beam length is 3.2m
and the subtended angle varies from 10° to 180°. The
dimensions of cross section are: -3003 12(upper flange),
-1503 12(lower flange) and -3763 8(web) and
corresponding section properties are A = 84.08cm2, Ix =
22424cm4, Iy = 3039cm4, J = 32.34cm4, Iv = 451632cm6
and ys = 10.93cm. The material properties used are E =
206GPa, G = 79GPa. The results are tabulated in table 2.
The locations where the maximum deflection and
twist rotation occur are always at the mid-span. For the
curved beam discussed, the maximum bending moment
occurs at the end. The location of maximum bimoment
may appear at the mid-span or at the end, depending on
the subtended angle. So the deformations listed in table
2 are at the mid-span while the forces are at the end if
no additional statement is made.
It can be seen from Table 2 that, the difference in the
internal forces is not great, but the twist rotation of the
present theory is 10% smaller when the subtended angle
is small. Another difference is that the present theory
predicts an axial force (see Table 2) and an in-plane
moment although these forces are small.
It is interesting to note that the vertical deflection
increases with the value of subtended angle, while twist
rotation increases first until b = 30° then decreases until
b = 60° and then increases again. Figure 5 shows the
deformations along half of the curved beam with the
subtended angle varying. One can find that the
deflection mode is similar but the twist mode is
changing when the subtended angle increases. When b
reaches 40°, inverse twist rotations near the ends begin
Table 2. Beam with fixed ends (b is the subtended angle )
b
10°
30°
50°
70°
90°
120°
150°
180°
v (102 5 m)
-0.399
-0.495
-0.569
-0.647
-0.742
-0.931
-1.186
-1.519
Vlasov
m) Mx (N · m)
-0.204
425.5
-0.289
497.7
-0.268
525.1
-0.260
533.6
-0.267
535.0
-0.296
530.8
-0.339
522.0
-0.396
509.3
u (102
4
Bv (N · m2)
-3.808*
-6.788*
-8.233*
-12.54
-18.39
-27.0
-35.76
-44.93
v (102 5 m)
-0.396
-0.481
-0.555
-0.635
-0.732
-0.921
-1.174
-1.505
u (102
4
m)
-0.190
-0.258
-0.251
-0.251
-0.261
-0.291
-0.336
-0.392
This paper
N (N)
-86.22*
-168.5*
-118.6*
-77.48*
-52.41*
-31.68*
-20.70*
-14.24*
Mx (N · m)
423.7
485.5
514.3
526.3
530.3
528.5
521.1
509.3
Bv (N · m2)
-3.643*
-6.399*
-8.067*
-12.31
-17.91
-26.23
-34.76
-43.73
Note: The asterisk * bimoments or axial forces at the mid-span.
202
Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
Figure 5. Deformation of the curved beams with I-shape section
Figure 6. A curved beam with mono-symmetrical H-section, web on the curvature plane
to occur and the maximum twist rotation decreases
slightly. Thereafter the maximum inverse twist rotation
increase with the value of subtended angle while the
maximum twist at mid-span decreases first until b = 60°
and then increases again.
The longitudinal displacement presented by Usami
and Koh (1980) is slightly different from Eqn 28b for
this type of sections. Eqn 28b becomes identical to
Yang and Kuo (1987) if the section is doubly
symmetrical. The longitudinal normal strain is
3.2. Mono-symmetrical H Section with Its Web
Lying on the Curvature Plane (Figure 6)
For a mono-symmetrical H section whose web is lying
on the curvature plane, one has
r
y
Dy = y
R2
r2
R2
, Dv = v
r2
, v = y(x 2
xs )
R
(28a)
5
R w9 + u
e=
R
1
u +
v0
Rs
2
x
2
2
(u0 2
R2
v
rR
w9 ) 2
1
v0
u 0 2
Rs
26
(29)
Here the quantities defined in Eqns 15a,b are reduced
The longitudinal displacement is:
– =w2
w
x
R
(u9 2
w) 2
y
Rs
v9 2
v
r
to:
1
u 9 2
v9
Rs
Advances in Structural Engineering Vol. 5 No. 4 2002
2
(28b)
Hu =
e
s v
A
R
r
dA ž Bv , Hv =
R
R2
1
Mx 2
Hu
R
2
(30)
203
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
dHv
u
It can be seen that the relation M v = dH
—
dz ž —
dz in this
case. So it is much more convenient to use Hu in curved
beams instead of Bv . The force-displacement relations
are:
N=
EA
(w9 + u) +
R
EIy
R3
EIy(uIV + 2u0 + u) + EAR2(w9 + u) = qxR3Rs (32b)
1
EIx u 0 +
EIy
EIx
R
1
Rs
1
ESyv
2
R2
1
u 0 2
v0
Rs
IV
vIV
2
Rs
ESyv
R
1
u +
v0
Rs
u
2
2
Rs
+
v0
Rs
2
= qyRRs2
2
2
EIv
R2
1
u 0 2
v0
Rs
2
1
v0
+ EIxR2 u +
Rs
2
2
(32d)
2
1
GJR2 u 0 2
(31c,d)
Hu = 2
1
R2
GJ u 0 2
1
2
v0
2
vIV
(u + u0 )
R2
u +
2
IV
(32c)
EIv u
Mx = 2
R
EIv
(u + u0 ),
(31a,b)
My =
vIV
2
v0
Rs
2
= mzR3Rs
Here Syv < 0 is used for convenience.
3.3. Channel Section (Figure 7a)
For channel-section, we can obtain:
Mv = 2
ESyv
R2
1
u +
vRs
2
2
EIv
R3
1
u - 2
vRs
2
(31e,f)
Tsv =
GJ
R2
(Rsu 9 2
1
R2
R2 R12
R1
R12
y, Dv = y
2
r
2
2
Rs ,
(33)
v9 )
v
where Syv = eAyv R–
r2 dA. The corresponding differential
equilibrium equations are:
2
EA(w0 + u9 ) = 2 qzR2
Dy =
(32a)
= 2 y(x + xs 2
2x1)
where R1, Rs are radius of the web and the shear centre
axis, and x1, xs are their x coordinates, respectively. Eqn
33 is also valid when the channel is open to the right.
The normal strain is:
Figure 7. A curved beam with channel section or nonsymmetrical H-section
204
Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
e=
w9 + u
R
x
2
rR
Substituting Eqns 36a~d into Eqns 20c~d, one
obtains:
(u0 + u) 2
(34)
y
r
yRs
(u + u 0 ) +
R12
1
2
v0
u 0 2
Rs
EIx(vIV + Rsu 0 ) 2
Here the quantities defined in Eqns 15a,b are:
RRs
R12
3
Bv + (xs 2
x1)2
Mx
RS
4
,
(35)
Hv =
R
Rs
1
Mx 2
Hu
R
Mv = 2
EIx
R12
(v0 + Rsu ),
EIv
R12R
Hu = 2
2
(u 9 + u - ), Tsv =
EIv
R12
GJ
R2
(u 0 + u
IV
)+
R12
2
R
(Rsu 0 2
v0 ) = qyRs2R12
EIv (u 9 + u - ) + EIxR(Rsu 2
v0 ) 2
(37b)
GJ
The force-displacement relations for out-of- plane
problems are:
Mx = 2
R
(37a)
GJ
Hu =
EIv
(u + u 0 ) (36a,b)
(Rsu 9 2
v9 ) (36c,d)
R12
R2
(Rsu 0 2
v0 ) = mzRsR12R
Figure 8 shows a comparison of solutions of a fixed
curved beam whose section is given in Figure 7a. The
section properties used are A = 78.08cm2, I x =
21615cm4, Iy = 3306cm4, J = 29.46cm4, Iv =
873321cm6, xs = 13.93cm. Beam length is 2.8m and the
subtended angle is 50°. The beam is subjected to a midspan vertical load P = 2 1.0 kN which acts through the
shear centre.
Figure 8. Deformations and internal forces of a curved beam with channel section
Advances in Structural Engineering Vol. 5 No. 4 2002
205
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
Here Hu and Bv defined above are also given in Figure
8. It is seen from Figure 8 that the bending moments are
essentially the same for both solutions but differences
between the deformations and the bimoments are nonnegligible. This paper predicts greater bimoment and
vertical deflection for this example. When the subtended
angle becomes greater and the radius smaller,
differences will be greater. It is also noted that the
maximum positive bimoment is not at the fixed support,
but at a section away from the support in this example.
3.4. H Section without any Symmetrical Axis
(Figure 7b)
For curved beams of H section without any symmetrical
axis, which find applications in dry gasholders, the finite
element method is proposed for analysis. Figure 7b shows
an H-section without any symmetrical axis, the following
expressions are used in the finite element analysis:
e cos a
s
R2
0
r2
ds = y
R2
R12
,
er
R2
s
0
s
ds =
r2
(38)
2 y
R2
(Rs 2
2
R1
R1) + (y 2
ys)R2
1
1
r
2
1
R1
2
in which R1 and Rs are radii of the web and the shear
centre axis respectively. In this case, Vlasov’s theory
cannot be used without great modification.
Shape functions used in the finite element analysis
are second order polynomials for the longitudinal
displacement and third order polynomial for the radial
displacement, vertical displacement and the twist
rotation. Because the expressions for Dy and Dv are very
lengthy, the cross sections of curved beams are
subdivided into many small area elements and
numerical integration is used to calculate the section
properties effectively. Gaussian integration is used to
calculate the stiffness matrix.
A curved beam with this type of cross section is
analyzed here. The section dimensions are given in
Figure 7b and the section properties are A = 87.68cm2,
Ix = 24332cm4, Iy = 3869cm4, Ixy = 2 2927cm4, J =
34.07cm4, Iv = 941388cm6, xs = 2 5.08, ys = 7.21cm.
The length of the curved beam is 3.2m and the
subtended angle is 30°. The beam is fixed at both ends,
subjected to a uniformly distributed load qy =
2 400N/m, uniform torsional moment m z =
2 20.33N·m/m and a vertical load P = 2 1.0kN acting
through the shear centre at mid-span. The deformations
and internal forces along half of beam length are shown
in Figure 9. Here also Hu is used instead of Bv .
Figure 9. Deformations and internal forces of a curved beam with nonsymmetrical H section
206
Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
3.5. Simplified Equations
For those beams with non-symmetrical cross sections,
simplified equations may also be needed. Because x and
r coordinates have the same direction, as a general
theory, it is not required for the x and y axes to be the
principal axes, but they must be the centroidal axes. is
the normalized warping function calculated as in the
theory for a straight member.
Additional to section properties defined above, we
adopt the following approximations:
e
0
s
R
2
r2
cos a ds = y 2
y0,
e
s
0
R
2
r sds = v
r2
2
v
0,
Hu = 2
EIv
R2
u 0,
Ixy
AR
A
R
+
Iy
Ixy
R
R
R2
Ix
Cyy =
R
, Cy = 2
2
, Cv = 0,
, Cyv = 0, CvN = Ixy, Cvx = Ix,
N
= 0, Cu x = 0, Cu
= Iv
v
e xydA. The force-displacement relations are:
A
N=
EA
R
(w9 + u + ys u ) +
(39a)
EIy
R3
My =
Mx = 2
(u0 + u + ys u ) +
EIy
R2
R2
EIxy
R3
(u0 + u + ys u ) +
EIx
(39d,e)
(40a)
EIy(uIV + 2u0 + u) + E(Iyys + RIxy)(u 0 + u ) +
EIxy(v
IV
(40b)
+ v0 ) + EAR2(w9 + u + ysu ) = qx(R + xs)R3
GJRv0 2
EIÃ u
IV
+
(40c)
(EIxyR + EIyys)(uIV + u0 + ysu 0 ) = qyR3(R + xs)2
EIv u
IV
2
GJR(R + xs)u 0 + E(IxR + Ixyys)Ru +
(GJR + EIxR + ysEIxy)v0 +
(40d)
(EIxyR + ysEIy)(u0 + u + ysu ) = mzR3(R + xs)
Cvv = 0, Cu
where Ixy =
u -
[GJ(R + xs) + EIxR + EIxyys]Ru 0 +
Iy
, Cx = 2
3
R3
EA(w0 + u9 + ys u 9 ) = 2 qzR2
, Dv = v
then the coefficients in Eqns 11a-d are simplified as
following:
C0 =
EIv
The equilibrium equations are:
(EIxR + EIxyys)vIV 2
Dy = y 2
Mv = 2
(v0 + Ru ) 2
EIxy
R2
(v0 + Ru )
EIxy
R2
(v0 + Ru ) (39b)
(u0 + u + ys u ) (39c)
Advances in Structural Engineering Vol. 5 No. 4 2002
The simplification made above achieves high
accuracy for in-plane deformations, but has less
accuracy for Dy and D v . If a higher accuracy is required
for the out-of-the-curvature-plane deformation, some
new section properties must be introduced and the
equations obtained will be lengthier.
A curved beam is calculated using Eqns 25c,d and
Eqns 40c,d to examine the accuracy of simplified
equations (40c,d). The beam has two ends fixed and is
loaded by a mid-span concentrated force. The section is
a welded H-section with flanges –4003 20 and web
–4603 12. Length of the beam L = 4.8m remains
constant for all beams and the subtended angle of the
curved beams vary from 10° to 180°. Comparisons
show that, within the applicable range of the theory of
thin-walled members, the two sets of equations give
similar results in deformations, the difference in forces
related to the restrained torsion is greater but still less
than 5% for the majority of the beams. Only when the
subtended angle is large (greater than 120°) and the
radius is small, does the difference become larger, but is
still less than 10%. This indicates that Eqns 40c,d
achieve good accuracy in most situations.
207
An Exact Theory for Curved Beams with Any Thin-walled Open Sections
4. CONCLUSIONS
This paper presents a new theory for the bending and
torsion of thin-walled circularly curved beams with
open profiles. Compared with previous researches, the
following progresses have been made:
(1) An exact longitudinal displacement was lacking
in the current general theory of thin-walled
curved members. This paper gives an exact
longitudinal displacement[Eqn 6] that conforms
exactly to the two basic assumptions commonly
accepted in the theory of thin-walled members.
The longitudinal displacement is applicable for
any thin-walled cross-sections with open
profile.
(2) Based on the exact longitudinal displacement, a
new general theory is presented for the linear
elastic analysis of circularly curved thin-walled
members.
(3) Two quantities Hv and Hu [Eqns 15a,b] are
introduced from which the resultants of the
shear flow may be derived conveniently [see
Eqns 17a,b,c]. It is also found that the boundary
conditions are simpler if they are related to these
two quantities (see Table 1).
dBv
(4) The relation M v = —
dz for a straight member is
not always valid for curved beams. By adopting
dHu
the new quantity Hu , we have Mv = —
dz for
curved beams with arbitrary open sections. So it
is more convenient to use Hu than B v in curved
member theory.
(5) Because of the complexity of the exact theory,
simplified theories are also presented for curved
beams with sections frequently encountered.
Comparisons show that discrepancies between
present theory and Vlasov’s become significant
when the subtended angle of the curved member
is large, the radius is small and the flange is
wide. In these situations, the more rigorous
theory presented in this paper may be used for
specific type of cross-sections.
(6) For curved beams with non-symmetrical cross
sections, FEM is recommended for use in
analysing the internal forces and deformations.
An example is given of the use of FEM to show
the application of the theory presented to this
kind of beam. A simplified theory is also
provided although it is not so easy to solve it
analytically.
208
ACKNOWLEDGMENTS
This research work (Grant No.59778037) is financially
supported by the Chinese National Science Foundation.
The authors are grateful to the reviewers for their
discerning comments on this paper.
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Advances in Structural Engineering Vol. 5 No. 4 2002
Genshu Tong and Qiang Xu
Tong Genshu is a Professor of Civil Engineering at Zhejiang University. He was born in
Oct. 1963 and received his B.E. at Zhejiang University in 1983. In 1988 he obtained the
degree of Doctor of Engineering at Xi’an Institute of Metallurgy & Architecture. His
research work is concentrated on stability of steel structures, including the stability of
braced steel structures, curved beams and portal frames and design of dry gasholders. He
has authored about 50 papers in refereed archival journals and conference proceedings.
Xu Qiang is a Ph.D. candidate in the Department of Civil Engineering & Architecture at
Zhejiang University. He received his B.E. from Zhejiang University in 1996. His research
interests are in the behaviour of curved beams and nonlinear finite element analysis.
Advances in Structural Engineering Vol. 5 No. 4 2002
209