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Page 1
Solution:
Since, the arch is symmetrical and the loading is also symmetrical,
kN 40 = =
by ay
R R (1)
Now the strain energy of the arch is given by,
Page 2
Solution:
Since, the arch is symmetrical and the loading is also symmetrical,
kN 40 = =
by ay
R R (1)
Now the strain energy of the arch is given by,
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(2)
Let us choose and as redundants. Then we have,
a
H
a
M
0 =
?
?
a
M
U
and 0 =
?
?
a
H
U
(3)
The bending moment at any cross section is given by,
D a a ay
y H M x R M ? ? = = - - = 0 (4)
2 / ) 10 ( 40 p ? ? = = - - - - =
D a a ay
x y H M x R M
? ? cos ) 90 cos(
a a
R H N + - =
D a a
R H N ? ? ? ? = = + = 0 cos sin (5)
sin ( 40)cos
aa
NH R ? ? =+- /2 ? ?p = = (6)
? sin R y =
) cos 1 ( ? - =R x
And ? Rd ds =
See Fig 34.5.
0 ) 0 ( ) 1 (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds
EI
M
M
U
0
0
=
?
s
ds
EI
M
Since the arch is symmetrical, integration need to be carried out
between limits 2 / 0 p to and the result is multiplied by two.
0 2
2 /
0
=
?
p
ds
EI
M
0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40
2 /
552 . 2 /
2 /
0
2 /
0
2 /
0
= - - - - - -
? ? ? ?
? ? ? ? ? ? ?
p
p
p p p
Rd R Rd R H Rd M Rd R
a a
Page 3
Solution:
Since, the arch is symmetrical and the loading is also symmetrical,
kN 40 = =
by ay
R R (1)
Now the strain energy of the arch is given by,
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(2)
Let us choose and as redundants. Then we have,
a
H
a
M
0 =
?
?
a
M
U
and 0 =
?
?
a
H
U
(3)
The bending moment at any cross section is given by,
D a a ay
y H M x R M ? ? = = - - = 0 (4)
2 / ) 10 ( 40 p ? ? = = - - - - =
D a a ay
x y H M x R M
? ? cos ) 90 cos(
a a
R H N + - =
D a a
R H N ? ? ? ? = = + = 0 cos sin (5)
sin ( 40)cos
aa
NH R ? ? =+- /2 ? ?p = = (6)
? sin R y =
) cos 1 ( ? - =R x
And ? Rd ds =
See Fig 34.5.
0 ) 0 ( ) 1 (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds
EI
M
M
U
0
0
=
?
s
ds
EI
M
Since the arch is symmetrical, integration need to be carried out
between limits 2 / 0 p to and the result is multiplied by two.
0 2
2 /
0
=
?
p
ds
EI
M
0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40
2 /
552 . 2 /
2 /
0
2 /
0
2 /
0
= - - - - - -
? ? ? ?
? ? ? ? ? ? ?
p
p
p p p
Rd R Rd R H Rd M Rd R
a a
0 92 . 135 304 . 41 571 . 1 8310 . 22
2 2 2
= + - - - R R R H R M R
a a
0 92 . 135 56 . 169 15 571 . 1 477 . 342 = + - - -
a a
H M
0 837 . 308 15 571 . 1 = - +
a a
H M
0 ) (sin ) (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds y
EI
M
H
U
?
0 cos 40 ) (sin
1
) (sin
) cos sin (
]]} 10 ) cos 1 ( [ 40 ){[ sin (
1
)} sin ( )] cos 1 ( 40 ){[ sin (
1
2 /
552 . 2 /
2 /
0
2 /
552 . 2 /
2 /
0
= -
+
+ - - - - - - - -
? ?
? ?
p
p
p
p
p
p
? ? ? ? ?
? ?
? ? ? ? ? ? ?
Rd
EA
Rd
EA
R H
Rd R R
EI
Rd R H M R R
EI
a a
a a
?
?
= - - -
+ - + + + + -
2 /
552 . 2 /
2 3 3
2
2 /
0
2
3 2 3 3
0 } cos sin
40
sin
400
cos sin
40
sin
40
{
}
cos sin ) (
sin sin sin cos sin
40
sin
40
{
p
p
p
? ? ? ? ? ? ?
?
? ?
? ? ? ? ? ?
d
EA
R
EI
R
EI
R
EI
R
d
EA
R R
EA
R H
EI
R H
EI
R M
EI
R
EI
R
ay
a a a
0 ) 0555 . 0 (
40
) 333 . 0 (
400
) 0554 . 0 (
40
) 333 . 0 (
40
)
2
1
(
40
) 785 . 0 ( ) 785 . 0 ( ) 1 ( )
2
1
(
40
) 1 (
40
2
2 2
= - - -
+ - + + + +
-
A R RI I I
A R A R
H
I
H
IR
M
I I
a a a
0 2 58 . 23 266 = + + -
a a
M H (7)
Solving equations (6) and (7), and are evaluated. Thus,
a
H
a
M
kN 28 . 28 =
a
H
kN 42 . 466 - =
a
M (8)
Page 4
Solution:
Since, the arch is symmetrical and the loading is also symmetrical,
kN 40 = =
by ay
R R (1)
Now the strain energy of the arch is given by,
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(2)
Let us choose and as redundants. Then we have,
a
H
a
M
0 =
?
?
a
M
U
and 0 =
?
?
a
H
U
(3)
The bending moment at any cross section is given by,
D a a ay
y H M x R M ? ? = = - - = 0 (4)
2 / ) 10 ( 40 p ? ? = = - - - - =
D a a ay
x y H M x R M
? ? cos ) 90 cos(
a a
R H N + - =
D a a
R H N ? ? ? ? = = + = 0 cos sin (5)
sin ( 40)cos
aa
NH R ? ? =+- /2 ? ?p = = (6)
? sin R y =
) cos 1 ( ? - =R x
And ? Rd ds =
See Fig 34.5.
0 ) 0 ( ) 1 (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds
EI
M
M
U
0
0
=
?
s
ds
EI
M
Since the arch is symmetrical, integration need to be carried out
between limits 2 / 0 p to and the result is multiplied by two.
0 2
2 /
0
=
?
p
ds
EI
M
0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40
2 /
552 . 2 /
2 /
0
2 /
0
2 /
0
= - - - - - -
? ? ? ?
? ? ? ? ? ? ?
p
p
p p p
Rd R Rd R H Rd M Rd R
a a
0 92 . 135 304 . 41 571 . 1 8310 . 22
2 2 2
= + - - - R R R H R M R
a a
0 92 . 135 56 . 169 15 571 . 1 477 . 342 = + - - -
a a
H M
0 837 . 308 15 571 . 1 = - +
a a
H M
0 ) (sin ) (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds y
EI
M
H
U
?
0 cos 40 ) (sin
1
) (sin
) cos sin (
]]} 10 ) cos 1 ( [ 40 ){[ sin (
1
)} sin ( )] cos 1 ( 40 ){[ sin (
1
2 /
552 . 2 /
2 /
0
2 /
552 . 2 /
2 /
0
= -
+
+ - - - - - - - -
? ?
? ?
p
p
p
p
p
p
? ? ? ? ?
? ?
? ? ? ? ? ? ?
Rd
EA
Rd
EA
R H
Rd R R
EI
Rd R H M R R
EI
a a
a a
?
?
= - - -
+ - + + + + -
2 /
552 . 2 /
2 3 3
2
2 /
0
2
3 2 3 3
0 } cos sin
40
sin
400
cos sin
40
sin
40
{
}
cos sin ) (
sin sin sin cos sin
40
sin
40
{
p
p
p
? ? ? ? ? ? ?
?
? ?
? ? ? ? ? ?
d
EA
R
EI
R
EI
R
EI
R
d
EA
R R
EA
R H
EI
R H
EI
R M
EI
R
EI
R
ay
a a a
0 ) 0555 . 0 (
40
) 333 . 0 (
400
) 0554 . 0 (
40
) 333 . 0 (
40
)
2
1
(
40
) 785 . 0 ( ) 785 . 0 ( ) 1 ( )
2
1
(
40
) 1 (
40
2
2 2
= - - -
+ - + + + +
-
A R RI I I
A R A R
H
I
H
IR
M
I I
a a a
0 2 58 . 23 266 = + + -
a a
M H (7)
Solving equations (6) and (7), and are evaluated. Thus,
a
H
a
M
kN 28 . 28 =
a
H
kN 42 . 466 - =
a
M (8)
Equations (34.7a) and (34.7b) are quite difficult to solve. However, they can be
further simplified if the origin of co-ordinates is moved from to in Fig. 34.3.
The distance d is chosen such that
C O
1
( ) y yd = - satisfies the following condition.
( )
0
0 0
1
=
-
=
? ?
s s
ds
EI
d y
ds
EI
y
(34.10a)
Solving which, the distance d may be computed as
?
?
=
s
s
EI
ds
ds
EI
y
d
0
0
34.10b)
The point is known as the elastic centre of the arch. Now equation (34.7a) can
be written with respect to new origin . Towards this, substitute
O
O d y y + =
1
in
equation (34.7a).
34.4 Elastic centre method
Page 5
Solution:
Since, the arch is symmetrical and the loading is also symmetrical,
kN 40 = =
by ay
R R (1)
Now the strain energy of the arch is given by,
? ?
+ =
s s
EA
ds N
EI
ds M
U
0
2
0
2
2 2
(2)
Let us choose and as redundants. Then we have,
a
H
a
M
0 =
?
?
a
M
U
and 0 =
?
?
a
H
U
(3)
The bending moment at any cross section is given by,
D a a ay
y H M x R M ? ? = = - - = 0 (4)
2 / ) 10 ( 40 p ? ? = = - - - - =
D a a ay
x y H M x R M
? ? cos ) 90 cos(
a a
R H N + - =
D a a
R H N ? ? ? ? = = + = 0 cos sin (5)
sin ( 40)cos
aa
NH R ? ? =+- /2 ? ?p = = (6)
? sin R y =
) cos 1 ( ? - =R x
And ? Rd ds =
See Fig 34.5.
0 ) 0 ( ) 1 (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds
EI
M
M
U
0
0
=
?
s
ds
EI
M
Since the arch is symmetrical, integration need to be carried out
between limits 2 / 0 p to and the result is multiplied by two.
0 2
2 /
0
=
?
p
ds
EI
M
0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40
2 /
552 . 2 /
2 /
0
2 /
0
2 /
0
= - - - - - -
? ? ? ?
? ? ? ? ? ? ?
p
p
p p p
Rd R Rd R H Rd M Rd R
a a
0 92 . 135 304 . 41 571 . 1 8310 . 22
2 2 2
= + - - - R R R H R M R
a a
0 92 . 135 56 . 169 15 571 . 1 477 . 342 = + - - -
a a
H M
0 837 . 308 15 571 . 1 = - +
a a
H M
0 ) (sin ) (
0 0
= + - =
?
?
? ?
s s
a
ds
EA
N
ds y
EI
M
H
U
?
0 cos 40 ) (sin
1
) (sin
) cos sin (
]]} 10 ) cos 1 ( [ 40 ){[ sin (
1
)} sin ( )] cos 1 ( 40 ){[ sin (
1
2 /
552 . 2 /
2 /
0
2 /
552 . 2 /
2 /
0
= -
+
+ - - - - - - - -
? ?
? ?
p
p
p
p
p
p
? ? ? ? ?
? ?
? ? ? ? ? ? ?
Rd
EA
Rd
EA
R H
Rd R R
EI
Rd R H M R R
EI
a a
a a
?
?
= - - -
+ - + + + + -
2 /
552 . 2 /
2 3 3
2
2 /
0
2
3 2 3 3
0 } cos sin
40
sin
400
cos sin
40
sin
40
{
}
cos sin ) (
sin sin sin cos sin
40
sin
40
{
p
p
p
? ? ? ? ? ? ?
?
? ?
? ? ? ? ? ?
d
EA
R
EI
R
EI
R
EI
R
d
EA
R R
EA
R H
EI
R H
EI
R M
EI
R
EI
R
ay
a a a
0 ) 0555 . 0 (
40
) 333 . 0 (
400
) 0554 . 0 (
40
) 333 . 0 (
40
)
2
1
(
40
) 785 . 0 ( ) 785 . 0 ( ) 1 ( )
2
1
(
40
) 1 (
40
2
2 2
= - - -
+ - + + + +
-
A R RI I I
A R A R
H
I
H
IR
M
I I
a a a
0 2 58 . 23 266 = + + -
a a
M H (7)
Solving equations (6) and (7), and are evaluated. Thus,
a
H
a
M
kN 28 . 28 =
a
H
kN 42 . 466 - =
a
M (8)
Equations (34.7a) and (34.7b) are quite difficult to solve. However, they can be
further simplified if the origin of co-ordinates is moved from to in Fig. 34.3.
The distance d is chosen such that
C O
1
( ) y yd = - satisfies the following condition.
( )
0
0 0
1
=
-
=
? ?
s s
ds
EI
d y
ds
EI
y
(34.10a)
Solving which, the distance d may be computed as
?
?
=
s
s
EI
ds
ds
EI
y
d
0
0
34.10b)
The point is known as the elastic centre of the arch. Now equation (34.7a) can
be written with respect to new origin . Towards this, substitute
O
O d y y + =
1
in
equation (34.7a).
34.4 Elastic centre method
? ? ?
- =
+
+
s s
c
s
c
ds
EI
M
ds
EI
d y
N
EI
ds
M
0
0
0
1
0
) (
(34.11)
In the above equation,
?
s
ds
EI
y
0
1
is zero. Hence the above equation is rewritten as
?
?
- = +
s
s
c c
EI
ds
ds
EI
M
d N M
0
0
0
(34.12)
Now, is the moment ) ( d N M
c c
+
0
~
M at (see Fig. 34.3). Similarly the equation
(34.7b) is also simplified. Thus we obtain,
O
?
?
- = + =
s
s
c c
EI
ds
ds
EI
M
d N M M
0
0
0
0
~
(34.13)
and,
??
??
+
+
- = =
ss
ss
c
ds
EA
ds
EI
y
ds
EA
N
ds
EI
y M
N H
00
2
1
2
00
0 1 0
0
cos
cos
~
?
?
(34.14)
34.4.1Temperature stresses
Consider a symmetrical hinge less arch of spanL, subjected to a temperature
rise of . Let elastic centre Obe the origin of co-ordinates and be the
redundants. The magnitude of horizontal force be such as to counteract the
increase in the span
C T °
0 0
~
,
~
M H
0
~
H
2
LT a
due to rise in temperature Also from Symmetry,
there must not be any rotation at the crown. Hence,
. T
?
=
?
?
= =
?
?
s
O
O
ds
M
M
EI
M
M
U
0
0
~
0 (34.15)
2
~ ~ ~
0 0
T L
ds
H
N
EA
N
ds
H
M
EI
M
H
U
s
O
s
O O
a
? ?
=
?
?
+
?
?
=
?
?
(34.16)
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FAQs on Symmetrical Hingeless Arch - 2 - Structural Analysis - Civil Engineering (CE)
| 1. What is a symmetrical hingeless arch? |  |
Ans. A symmetrical hingeless arch is a structural element in civil engineering that is designed to support loads by transferring them to the ground through arch action. It is called "hingeless" because it does not have any hinges or joints along its length, which allows for a continuous and uninterrupted load transfer. The arch is symmetrical in shape, meaning that the profile on one side is a mirror image of the other side.
| 2. How does a symmetrical hingeless arch work? | |
Ans. A symmetrical hingeless arch works by utilizing the inherent strength of the arch shape to carry and distribute loads. When a load is applied to the arch, it generates compressive forces along its curved profile. These forces are transferred to the abutments or supports at the ends of the arch, which then transfer the load to the ground. The arch shape helps to distribute the load evenly along its length, resulting in a stable and efficient load-bearing structure.
| 3. What are the advantages of using a symmetrical hingeless arch? | |
Ans. Some advantages of using a symmetrical hingeless arch include:
- Increased load-carrying capacity: The arch shape allows for efficient load distribution, enabling the structure to carry heavier loads compared to other structural forms.
- Simplified construction: The absence of hinges or joints simplifies the construction process and reduces the need for maintenance.
- Aesthetic appeal: The symmetrical arch shape is visually pleasing and can enhance the overall architectural design.
- Durability: The continuous and uninterrupted load transfer in a symmetrical hingeless arch reduces the chances of structural failures and increases its longevity.
- Flexibility: The arch shape allows for flexibility in design, making it suitable for various applications such as bridges, tunnels, and architectural structures.
| 4. What are some examples of structures that use symmetrical hingeless arches? | |
Ans. Symmetrical hingeless arches are commonly used in various civil engineering structures. Some examples include:
- Bridges: Arch bridges, such as the Sydney Harbour Bridge in Australia, often utilize symmetrical hingeless arches for their load-bearing capabilities and aesthetic appeal.
- Tunnels: Tunnel structures can incorporate symmetrical hingeless arches to provide support and stability to the tunnel walls and roof.
- Roofing systems: Architectural structures like stadiums, exhibition halls, and auditoriums may use symmetrical hingeless arches as part of their roofing systems to create large open spaces without the need for intermediate supports.
- Pedestrian walkways: Footbridges and pedestrian walkways can incorporate symmetrical hingeless arches to span over obstacles or water bodies, providing a safe and visually appealing passage for pedestrians.
| 5. How are symmetrical hingeless arches designed and analyzed in civil engineering? | |
Ans. The design and analysis of symmetrical hingeless arches involve several considerations. Firstly, the structural engineers determine the required load-carrying capacity and select appropriate materials, such as steel or reinforced concrete, based on the specific project requirements. They then analyze the arch structure using mathematical models and computer simulations to ensure its stability and structural integrity under various load scenarios. Factors such as geometry, material properties, and support conditions are taken into account during the design process. Additionally, structural engineers may also consider factors such as seismic loads, wind loads, and temperature effects to ensure the arch's performance and safety throughout its lifespan.
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Symmetrical Hingeless Arch - 2 Civil Engineering (CE) Questions
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