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2.5 INDETERMINATE ARCHES
An arch may be defined as a plane-curved bar or rib
supported and loaded in a way that makes it act in direct
compression. For example, the structure in Fig. 2.8a is a
parabolic symmetrical arch that is loaded with a distributed
load that varies linearly over the span L of the arch. It is fixed
at end A and supported by an immovable hinge at end B. The
horizontal distance between the end supports of the arch is the
span of the arch, and the line AB joining its points of supports
is the springing line.
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Fig. 2.8 (a) Parabolic arch of variable thickness loaded as
shown. (b) Parabolic arch loaded with redundant force XA,
redundant moment MA, and applied distributed loading. (c)
Free-body diagram of a segment of the arch.
For the arch in Fig. 2.8a, the span and springing lines of the
arch are the same because points A and B are at the same level
with respect to the y axis. The highest point C of the arch is the
crown, and an arch may be a symmetrical arch or an
unsymmetrical arch. If, for example, one end of the arch is
lower than the other, then the arch is unsymmetrical. Various
types of arches are shown in Figs. 2.9 and 2.10.
Fig. 2.9 Circular arches: (a) Fixed at one end and hinged at the
other. (b) Fixed at both ends. (c) Three-hinge arch. (d) Twohinge arch with supports at different elevation. (e) Two-hinge
arch.
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It is assumed here that the plane of curvature of the arch rib is
also a plane of symmetry for each cross section of the arch, and
the externally applied loads are assumed to act only in this
plane. On this basis, we have a two-dimensional problem and
the deformation of the arch will take place in the plane of
symmetry. The maximum vertical distance from the springing
line to the arch axis, denoted as H in Fig. 2.8a, is the rise of the
arch.
2.4.1 Parabolic Arch of Variable Thickness
We consider the linearly elastic variable-thickness parabolic
arch in Fig. 2.8a that is loaded by a distributed load of
maximum intensity w0 and varying linearly over the span of
the arch. At any arc length s from support A, the moment of
inertia Is is assumed to vary as follows:
I
dx
ds I s
(2.108)
I s  I c sec , cos  c 
 sec 

Is
ds
dx
Ic
where Ic is the moment of inertia at the crown C of the arch,
and
 dy 
  tan 1  
(2.109)
 dx 
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Fig. 2.10 (a) Two-hinge parabolic arch. (b) Two-hinge elliptical
arch. (c) Two-hinge hollow arch. (d) Two-hinge with horizontal
tie. (e) Twin circular arch
Since the parabolic arch in Fig. 2.8a is statically indeterminate
to the second degree, the horizontal force XA and bending
moment MA at support A are taken as the redundants. On this
basis, the arch is reduced to one that is hinged at end B,
supported by roller at end A, and loaded as shown in Fig. 2.8b.
By considering a segment A0 of the arch as shown in Fig.
2.8c, the normal force Ns, shear force Vs, and bending moment
Ms at the end 0 of the segment may be determined by using the
three static equilibrium equations. For example, by taking
moments about the end 0 of the segment and assuming
counterclockwise moments as positive, we have
w0 x 3
 w0 L M A 
M



x

M

X
y

M

0


 0  6 L 
A
A
s
6L
(2.110)
Note that in Fig. 2.8c is the vertical reaction at end A in Fig.
2.8b, and it may be determined from this figure by using
statics. By solving Eq. (2.110), we find
w Lx
x

M s  M A 1    X A y  0 (1  x 2 )
(2.111)

L
6
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By setting equal to zero the sum of the forces in the
vertical direction, Fig. 2.8c, and assuming upward forces as
positive, we have
w0 L w0 x 2 M A
 Fy  6  2L  L  Vs cos  N s sin   0
(2.112)
By setting equal to zero the sum of the forces in the horizontal
direction of the segment, we find
(8.113)
 Fx  X A  Vs sin   N s cos  0
The simultaneous solution of Eqs. (2.112) and (2.113) for Vs
and Ns yields
 w L w x2 M 
N s   X A cos   0  0  A  sin 
2L
L 
 6
 w L w x2 M 
Vs   X A sin    0  0  A  cos
2L
L 
 6
(2.114)
(2.115)
From Example 1.3 of Section 1.5, an equation analogous
to Eq. (10) may be written for the complementary strain
energy U of the parabolic arch. This equation is
U 
s
0
2
2
s M
s KV
N s2
s
s
ds  
ds  
ds
0 2 EI
0 2GA
2 AE
s
(2.116)
where S is the arc length of the parabolic arch, A is its crosssectional area at any coordinate s, E is the modulus of
elasticity, G is the shear modulus, and K is the shear factor.
The shear factor K may be determined as shown in Example
1.3. For rectangular cross sections K is 1.2.
When the rise of the arch is large compared to its
thickness, say a ratio of 10 or larger, then the complementary
strain energy due to Ns and Vs would be small compared to the
one produced by Ms and it can be neglected. On this basis, Eq.
(8.116) yields
U 
s
0
M s2
ds
2 EI s
(2.117)
The integrations in Eq. (2.116), or Eq. (2.117), may be
simplified by using the expression
I
(2.118)
ds  s dx
Ic
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Thus, by substituting Eqs. (2.111) and (2.118) into Eq. (2.117),
we find
U 
L
0
1
2 EI c
2
w0 Lx

x

2 
M A 1  L   X A y  6 (1  x ) dx




(2.119)
2.4.2 Additional Subjects and Methods
The values of the redundant force XA and redundant
moment MA may be obtained from the minimizing conditions
U
0
X A
U
0
M A
(2.120)
(2.121)
Application of Eqs. (2.120) and (2.121) yields
w Lx


x

M A 1    X A y  0 (1  x 2 ) ( y )dx  0

0
6
 L


L
w0 Lx
1 
x
x

2 
0 2EI c M A 1  L   X A y  6 (1  x )(1  L )dx  0

L
1
2 EI c
(2.122)
(2.123)
By considering the geometry of the arch in Fig. 2.8a, we
find
y
4 Hx
( L  x)
L2
(2.124)
By substituting Eq. (2.124) into Eqs. (2.122) and (2.123) and
performing the required integrations, we obtain the following
two equations, which are in terms of the redundants XA and
MA:
16 HX A  10 M A  w0 L2  0
 120 HX A  120M A  7 w0 L2  0
(2.125)
(2.126)
Simultaneous solution of Eqs. (2.125) and (2.126) yields
1
5w0 L2
72 H
w0 L2
MA 
90
XA 
(2.127)
(2.128)
With known XA and MA, the values of Ms, Ns, and Vs may
be determined from Eqs. (2.111), (2.114), and (2.115),
respectively. They are as follows:
x  5w0 L2 y w0 Lx

1


(1  x 2 )


72 H
6
 L
2
2
w L w x
5w L
w L2 
N s   0 cos   0  0  0  sin 
72 H
2L
90 L 
 6
Ms 
w0 L2
90
 w0 L w0 x 2 w0 L2 
5w0 L2
Vs  
sin   


 cos
72 H
2L
90 L 
 6
(2.129)
(2.130)
(2.131)
2.4.3 Semicircular Arch with Hinged Ends
We consider now the uniform semicircular arch in Fig. 2.11a
that is hinged at the support points A and B and loaded by a
uniformly distributed load w as shown. The moment of inertia
I is uniform throughout the arch. Since the arch is statically
indeterminate to the first degree, the horizontal reaction XA at
the end A is taken as the redundant. On this basis, we have an
arch that is hinged at end B, supported by roller at end A, and
loaded by the reactive force XA and applied load w as shown in
Fig. 2.11b.
By considering the free-body diagram of the arch
segment AC in Fig. 2.11c and applying the three static
equilibrium equations, the expressions for the normal force
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Fig. 2.11 (a) Semicircular arch hinged at the end supports. (b)
Semicircular arch loaded with redundant force XA and applied
distributed load w. (c) Free-body diagram of a segment of the
arch.
2.4.4 Additional Subjects and Methods
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Ns, shear force Vs, and bending moment Ms, may be
determined. For example, by setting equal to the sum of the
moments about point C of the segment, we find
M s  wr 2 (1  cos )  X A r sin  
wr 2
(1  cos ) 2
2
(2.132)
The static equilibrium equations in the horizontal and vertical
directions of the arch segment yield
wr  wr (1  cos )  Vs sin   N s cos  0
(2.133)
(2.134)
X A  Vs cos  N s sin   0
Simultaneous solution of Eqs. (2.133) and (2.134) yields
N s   X A sin   wr cos 2 
(2.135)
(2.136)
Vs  wr cos sin   X A cos
By considering only the complementary strain energy due to
bending and using Eq. (2.132), we find
 /2
U  2
0
1

EI
M s2
d
2 EI
 /2

0
2
wr
 2
2
 wr (1  cos  )  X A r sin   2 (1  cos  )  d
The redundant reaction XA may be determined by using the
equation
U
0
X A
(2.137)
Equation (2.137) yields
2
EI
 /2

0
 2

wr 2
wr
(
1

cos

)

X
r
sin


(1  cos ) 2 (r sin  )d  0
A

2


(2.138)
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By integrating Eq. (2.138) and solving for XA, we find
XA=0.4099wr
(2.138a)
By substituting Eq. (2.138a) into Eqs. (2.132), (2.135), and
(8.136), we find
wr 2
M s  wr (1  cos ) 
(1  cos ) 2  0.4099wr 2 sin 
2
N s  0.4099 wr sin   wr cos 2 
2
Vs  wr cos sin   0.4099wr cos
(2.139)
(2.140)
(2.141)