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Page 1
Instructional Objectives:
After reading this chapter the student will be able to
1. Analyse hingeless arch by the method of least work.
2. Analyse the fixed-fixed arch by the elastic-centre method.
3. Compute reactions and stresses in hingeless arch due to temperature
change.
34.1 Introduction
As stated in the previous lesson, two-hinged and three-hinged arches are
commonly used in practice. The deflection and the moment at the center of the
hingeless arch are somewhat smaller than that of the two-hinged arch. However,
the hingeless arch has to be designed for support moment. A hingeless arch
(fixed–fixed arch) is a statically redundant structure having three redundant
reactions. In the case of fixed–fixed arch there are six reaction components;
three at each fixed end. Apart from three equilibrium equations three more
equations are required to calculate bending moment, shear force and horizontal
thrust at any cross section of the arch. These three extra equations may be set
up from the geometry deformation of the arch.
34.2 Analysis of Symmetrical Hingeless Arch
Consider a symmetrical arch of span L and central rise of Let the loading on
the arch is also symmetrical as shown in Fig 34.1. Consider reaction components
c
h
Page 2
Instructional Objectives:
After reading this chapter the student will be able to
1. Analyse hingeless arch by the method of least work.
2. Analyse the fixed-fixed arch by the elastic-centre method.
3. Compute reactions and stresses in hingeless arch due to temperature
change.
34.1 Introduction
As stated in the previous lesson, two-hinged and three-hinged arches are
commonly used in practice. The deflection and the moment at the center of the
hingeless arch are somewhat smaller than that of the two-hinged arch. However,
the hingeless arch has to be designed for support moment. A hingeless arch
(fixed–fixed arch) is a statically redundant structure having three redundant
reactions. In the case of fixed–fixed arch there are six reaction components;
three at each fixed end. Apart from three equilibrium equations three more
equations are required to calculate bending moment, shear force and horizontal
thrust at any cross section of the arch. These three extra equations may be set
up from the geometry deformation of the arch.
34.2 Analysis of Symmetrical Hingeless Arch
Consider a symmetrical arch of span L and central rise of Let the loading on
the arch is also symmetrical as shown in Fig 34.1. Consider reaction components
c
h
at the left support A i.e., bending moment , vertical reaction and horizontal
thrust as redundants.
a
M
ay
R
a
H
Considering only the strain energy due to axial compression and bending, the
strain energy U of the arch may be written as
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(34.1)
where M and are respectively the bending moment and axial force of the
arch rib. Since the support
N
A is fixed, one could write following three equations at
that point.
0 =
?
?
a
M
U
(34.2a)
0 =
?
?
a
H
U
(34.2b)
0 =
?
?
ay
R
U
(34.2c)
Knowing dimensions of the arch and loading, using the above three equations,
the unknown redundant reactions and may be evaluated.
a a
H M ,
ay
R
Since the arch and the loading are symmetrical, the shear force at the crown is
zero. Hence, at the crown we have only two unknowns. Hence, if we take the
internal forces at the crown as the redundant, the problem gets simplified.
Page 3
Instructional Objectives:
After reading this chapter the student will be able to
1. Analyse hingeless arch by the method of least work.
2. Analyse the fixed-fixed arch by the elastic-centre method.
3. Compute reactions and stresses in hingeless arch due to temperature
change.
34.1 Introduction
As stated in the previous lesson, two-hinged and three-hinged arches are
commonly used in practice. The deflection and the moment at the center of the
hingeless arch are somewhat smaller than that of the two-hinged arch. However,
the hingeless arch has to be designed for support moment. A hingeless arch
(fixed–fixed arch) is a statically redundant structure having three redundant
reactions. In the case of fixed–fixed arch there are six reaction components;
three at each fixed end. Apart from three equilibrium equations three more
equations are required to calculate bending moment, shear force and horizontal
thrust at any cross section of the arch. These three extra equations may be set
up from the geometry deformation of the arch.
34.2 Analysis of Symmetrical Hingeless Arch
Consider a symmetrical arch of span L and central rise of Let the loading on
the arch is also symmetrical as shown in Fig 34.1. Consider reaction components
c
h
at the left support A i.e., bending moment , vertical reaction and horizontal
thrust as redundants.
a
M
ay
R
a
H
Considering only the strain energy due to axial compression and bending, the
strain energy U of the arch may be written as
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(34.1)
where M and are respectively the bending moment and axial force of the
arch rib. Since the support
N
A is fixed, one could write following three equations at
that point.
0 =
?
?
a
M
U
(34.2a)
0 =
?
?
a
H
U
(34.2b)
0 =
?
?
ay
R
U
(34.2c)
Knowing dimensions of the arch and loading, using the above three equations,
the unknown redundant reactions and may be evaluated.
a a
H M ,
ay
R
Since the arch and the loading are symmetrical, the shear force at the crown is
zero. Hence, at the crown we have only two unknowns. Hence, if we take the
internal forces at the crown as the redundant, the problem gets simplified.
Hence, consider bending moment and the axial force at the crown as the
redundant. Since the arch and the loading is symmetrical, we can write from the
principle of least work
c
M
c
N
0 =
?
?
c
M
U
(34.3a)
0 =
?
?
c
N
U
(34.3b)
0
0 0
=
?
?
+
?
?
=
?
?
? ?
s
c
s
c c
ds
M
N
EA
N
ds
M
M
EI
M
M
U
(34.4a)
0
0 0
=
?
?
+
?
?
=
?
?
? ?
s
c
s
c c
ds
N
N
EA
N
ds
N
M
EI
M
N
U
(34.4b)
Where, is the length of centerline of the arch, s I is the moment of inertia of the
cross section and A is the area of the cross section of the arch. Let and
be the bending moment and the axial force at any cross section due to external
loading. Now the bending moment and the axial force at any section is given by
0
M
0
N
Page 4
Instructional Objectives:
After reading this chapter the student will be able to
1. Analyse hingeless arch by the method of least work.
2. Analyse the fixed-fixed arch by the elastic-centre method.
3. Compute reactions and stresses in hingeless arch due to temperature
change.
34.1 Introduction
As stated in the previous lesson, two-hinged and three-hinged arches are
commonly used in practice. The deflection and the moment at the center of the
hingeless arch are somewhat smaller than that of the two-hinged arch. However,
the hingeless arch has to be designed for support moment. A hingeless arch
(fixed–fixed arch) is a statically redundant structure having three redundant
reactions. In the case of fixed–fixed arch there are six reaction components;
three at each fixed end. Apart from three equilibrium equations three more
equations are required to calculate bending moment, shear force and horizontal
thrust at any cross section of the arch. These three extra equations may be set
up from the geometry deformation of the arch.
34.2 Analysis of Symmetrical Hingeless Arch
Consider a symmetrical arch of span L and central rise of Let the loading on
the arch is also symmetrical as shown in Fig 34.1. Consider reaction components
c
h
at the left support A i.e., bending moment , vertical reaction and horizontal
thrust as redundants.
a
M
ay
R
a
H
Considering only the strain energy due to axial compression and bending, the
strain energy U of the arch may be written as
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(34.1)
where M and are respectively the bending moment and axial force of the
arch rib. Since the support
N
A is fixed, one could write following three equations at
that point.
0 =
?
?
a
M
U
(34.2a)
0 =
?
?
a
H
U
(34.2b)
0 =
?
?
ay
R
U
(34.2c)
Knowing dimensions of the arch and loading, using the above three equations,
the unknown redundant reactions and may be evaluated.
a a
H M ,
ay
R
Since the arch and the loading are symmetrical, the shear force at the crown is
zero. Hence, at the crown we have only two unknowns. Hence, if we take the
internal forces at the crown as the redundant, the problem gets simplified.
Hence, consider bending moment and the axial force at the crown as the
redundant. Since the arch and the loading is symmetrical, we can write from the
principle of least work
c
M
c
N
0 =
?
?
c
M
U
(34.3a)
0 =
?
?
c
N
U
(34.3b)
0
0 0
=
?
?
+
?
?
=
?
?
? ?
s
c
s
c c
ds
M
N
EA
N
ds
M
M
EI
M
M
U
(34.4a)
0
0 0
=
?
?
+
?
?
=
?
?
? ?
s
c
s
c c
ds
N
N
EA
N
ds
N
M
EI
M
N
U
(34.4b)
Where, is the length of centerline of the arch, s I is the moment of inertia of the
cross section and A is the area of the cross section of the arch. Let and
be the bending moment and the axial force at any cross section due to external
loading. Now the bending moment and the axial force at any section is given by
0
M
0
N
0
M y N M M
c c
+ + = (34.5a)
0
cos N N N
c
+ = ? (34.5b)
1 =
?
?
c
M
M
; y
N
M
c
=
?
?
; ? cos =
?
?
c
N
N
; 0 =
?
?
c
M
N
. (34.6)
Equation (34.4a) and (34.4b) may be simplified as,
0 ) 0 ( ) 1 (
0 0
= +
? ?
s s
ds
EA
N
ds
EI
M
0
00 0
ss s
cc
M ds y ds
MN ds
EIEI EI
+=-
?? ?
(34.7a)
0 cos
0 0
= +
? ?
s s
ds
EA
N
yds
EI
M
?
2
2 00
00 0 00
cos cos
ss s ss
cc c
My Ny N My N
ds ds ds ds ds
EI EI EA EI EA
?? ++ =- -
?? ? ??
(34.7b)
From equations 34.7a and 34.7b, the redundant and may be calculated
provided arch geometry and loading are defined. If the loading is unsymmetrical
or the arch is unsymmetrical, then the problem becomes more complex. For such
problems either column analogy or elastic center method must be adopted.
However, one could still get the answer from the method of least work with little
more effort.
c
M
c
N
Page 5
Instructional Objectives:
After reading this chapter the student will be able to
1. Analyse hingeless arch by the method of least work.
2. Analyse the fixed-fixed arch by the elastic-centre method.
3. Compute reactions and stresses in hingeless arch due to temperature
change.
34.1 Introduction
As stated in the previous lesson, two-hinged and three-hinged arches are
commonly used in practice. The deflection and the moment at the center of the
hingeless arch are somewhat smaller than that of the two-hinged arch. However,
the hingeless arch has to be designed for support moment. A hingeless arch
(fixed–fixed arch) is a statically redundant structure having three redundant
reactions. In the case of fixed–fixed arch there are six reaction components;
three at each fixed end. Apart from three equilibrium equations three more
equations are required to calculate bending moment, shear force and horizontal
thrust at any cross section of the arch. These three extra equations may be set
up from the geometry deformation of the arch.
34.2 Analysis of Symmetrical Hingeless Arch
Consider a symmetrical arch of span L and central rise of Let the loading on
the arch is also symmetrical as shown in Fig 34.1. Consider reaction components
c
h
at the left support A i.e., bending moment , vertical reaction and horizontal
thrust as redundants.
a
M
ay
R
a
H
Considering only the strain energy due to axial compression and bending, the
strain energy U of the arch may be written as
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(34.1)
where M and are respectively the bending moment and axial force of the
arch rib. Since the support
N
A is fixed, one could write following three equations at
that point.
0 =
?
?
a
M
U
(34.2a)
0 =
?
?
a
H
U
(34.2b)
0 =
?
?
ay
R
U
(34.2c)
Knowing dimensions of the arch and loading, using the above three equations,
the unknown redundant reactions and may be evaluated.
a a
H M ,
ay
R
Since the arch and the loading are symmetrical, the shear force at the crown is
zero. Hence, at the crown we have only two unknowns. Hence, if we take the
internal forces at the crown as the redundant, the problem gets simplified.
Hence, consider bending moment and the axial force at the crown as the
redundant. Since the arch and the loading is symmetrical, we can write from the
principle of least work
c
M
c
N
0 =
?
?
c
M
U
(34.3a)
0 =
?
?
c
N
U
(34.3b)
0
0 0
=
?
?
+
?
?
=
?
?
? ?
s
c
s
c c
ds
M
N
EA
N
ds
M
M
EI
M
M
U
(34.4a)
0
0 0
=
?
?
+
?
?
=
?
?
? ?
s
c
s
c c
ds
N
N
EA
N
ds
N
M
EI
M
N
U
(34.4b)
Where, is the length of centerline of the arch, s I is the moment of inertia of the
cross section and A is the area of the cross section of the arch. Let and
be the bending moment and the axial force at any cross section due to external
loading. Now the bending moment and the axial force at any section is given by
0
M
0
N
0
M y N M M
c c
+ + = (34.5a)
0
cos N N N
c
+ = ? (34.5b)
1 =
?
?
c
M
M
; y
N
M
c
=
?
?
; ? cos =
?
?
c
N
N
; 0 =
?
?
c
M
N
. (34.6)
Equation (34.4a) and (34.4b) may be simplified as,
0 ) 0 ( ) 1 (
0 0
= +
? ?
s s
ds
EA
N
ds
EI
M
0
00 0
ss s
cc
M ds y ds
MN ds
EIEI EI
+=-
?? ?
(34.7a)
0 cos
0 0
= +
? ?
s s
ds
EA
N
yds
EI
M
?
2
2 00
00 0 00
cos cos
ss s ss
cc c
My Ny N My N
ds ds ds ds ds
EI EI EA EI EA
?? ++ =- -
?? ? ??
(34.7b)
From equations 34.7a and 34.7b, the redundant and may be calculated
provided arch geometry and loading are defined. If the loading is unsymmetrical
or the arch is unsymmetrical, then the problem becomes more complex. For such
problems either column analogy or elastic center method must be adopted.
However, one could still get the answer from the method of least work with little
more effort.
c
M
c
N
Consider an unloaded fixed-fixed arch of spanL . The rise in temperature, would
introduce a horizontal thrust and a moment at the supports. Now due to
rise in temperature, the moment at any cross-section of the arch
t
H
t
M
t H M M
t t
- = (34.8)
Now strain energy stored in the arch
?
=
s
EI
ds M
U
0
2
2
Now applying the Castigliano’s first theorem,
0
s
tt
UM
LT ds
HEI
a
?
==
??
?
M
H
?
2
00
ss
t
t
My y
LT ds H ds
EIE
a == -
??
I
(34.9)
Also,
ds
M
M
EI
M
M
U
s
t t
?
?
?
= =
?
?
0
0
0
) (
0
=
-
?
s
t t
ds
EI
y H M
??
= -
ss
t t
EI
yds
H
EI
ds
M
00
0 (34.10)
34.3 Temperature stresses
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FAQs on Symmetrical Hingeless Arch - 1 - Structural Analysis - Civil Engineering (CE)
| 1. What is a symmetrical hingeless arch? |  |
Ans. A symmetrical hingeless arch is a type of architectural structure that consists of two identical halves, which are connected at the top without the use of hinges. This design allows for even distribution of forces and provides stability to the arch.
| 2. How does a symmetrical hingeless arch differ from a traditional arch? | |
Ans. Unlike a traditional arch, which relies on hinges at the base to transfer loads, a symmetrical hingeless arch does not require any hinges. Instead, the arch is designed in a way that the load is evenly distributed across the two halves, providing greater stability and eliminating the need for hinges.
| 3. What are the advantages of using a symmetrical hingeless arch in civil engineering? | |
Ans. There are several advantages of using a symmetrical hingeless arch in civil engineering. Firstly, it offers improved structural stability and can withstand greater loads compared to traditional arches. Additionally, its symmetrical design provides aesthetic appeal and architectural flexibility. Moreover, the absence of hinges reduces maintenance requirements and the risk of failure due to hinge deterioration.
| 4. In what applications can symmetrical hingeless arches be used? | |
Ans. Symmetrical hingeless arches find application in various civil engineering projects. They can be used in the construction of bridges, tunnels, and even large-scale architectural structures like stadiums and exhibition halls. The design versatility of these arches allows for their adaptation to different spans and load requirements.
| 5. Are symmetrical hingeless arches cost-effective compared to traditional arches? | |
Ans. The cost-effectiveness of symmetrical hingeless arches compared to traditional arches depends on several factors, including the span length, design complexity, and construction materials. While the initial construction cost may be higher for symmetrical hingeless arches due to their specialized design, their long-term maintenance costs are typically lower due to the absence of hinge maintenance. Additionally, the durability and load-bearing capacity of these arches often result in a longer service life, making them a cost-effective choice in the long run.
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