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# Symmetrical Hingeless Arch - 1 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : Symmetrical Hingeless Arch - 1 Civil Engineering (CE) Notes | EduRev

``` 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
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
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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## Structural Analysis

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