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Page 1
The distance of the elastic centre from the crown is calculated by equation, d C
Page 2
The distance of the elastic centre from the crown is calculated by equation, d C
?
?
=
s
s
EI
ds
ds
EI
y
d
0
0
(1)
From Fig.34.7b, the ordinate at , is given by d
() ? cos 1 50 - = y
( )
?
?
-
=
6 /
0
6 /
0
50
50
cos 1 50
p
p
?
?
?
EI
d
d
EI
d
1
50
62
2.2535 m.
6
d
p
p
??
-
??
??
==
(2)
The elastic centre O lies at a distance of from the crown. The moment at
the elastic centre may be calculated by equation (34.12). Now the bending
moment at any section of the arch due to applied loading at a distance
m 2535 . 2
x from
elastic centre is
?
?
-
=
s
s
O
EI
ds
ds
EI
x
M
0
0
2
5
~
(3)
In the present case, ? sin 50 = x and ? d ds 50 = , constant = EI
/6
32
0
/6
0
550 sin
50
O
d
M
d
p
p
? ?
?
-×
=
?
?
%
2
1
sin
62 3 550
1081.29 kN.m
2
6
CC
MNd
pp
p
?? ??
-
?? ??
×
?? ??
+=- =- (4)
Page 3
The distance of the elastic centre from the crown is calculated by equation, d C
?
?
=
s
s
EI
ds
ds
EI
y
d
0
0
(1)
From Fig.34.7b, the ordinate at , is given by d
() ? cos 1 50 - = y
( )
?
?
-
=
6 /
0
6 /
0
50
50
cos 1 50
p
p
?
?
?
EI
d
d
EI
d
1
50
62
2.2535 m.
6
d
p
p
??
-
??
??
==
(2)
The elastic centre O lies at a distance of from the crown. The moment at
the elastic centre may be calculated by equation (34.12). Now the bending
moment at any section of the arch due to applied loading at a distance
m 2535 . 2
x from
elastic centre is
?
?
-
=
s
s
O
EI
ds
ds
EI
x
M
0
0
2
5
~
(3)
In the present case, ? sin 50 = x and ? d ds 50 = , constant = EI
/6
32
0
/6
0
550 sin
50
O
d
M
d
p
p
? ?
?
-×
=
?
?
%
2
1
sin
62 3 550
1081.29 kN.m
2
6
CC
MNd
pp
p
?? ??
-
?? ??
×
?? ??
+=- =- (4)
10 cos
2
O
L
Nx ?
??
=-
??
??
And.
1
y yd =-
Page 4
The distance of the elastic centre from the crown is calculated by equation, d C
?
?
=
s
s
EI
ds
ds
EI
y
d
0
0
(1)
From Fig.34.7b, the ordinate at , is given by d
() ? cos 1 50 - = y
( )
?
?
-
=
6 /
0
6 /
0
50
50
cos 1 50
p
p
?
?
?
EI
d
d
EI
d
1
50
62
2.2535 m.
6
d
p
p
??
-
??
??
==
(2)
The elastic centre O lies at a distance of from the crown. The moment at
the elastic centre may be calculated by equation (34.12). Now the bending
moment at any section of the arch due to applied loading at a distance
m 2535 . 2
x from
elastic centre is
?
?
-
=
s
s
O
EI
ds
ds
EI
x
M
0
0
2
5
~
(3)
In the present case, ? sin 50 = x and ? d ds 50 = , constant = EI
/6
32
0
/6
0
550 sin
50
O
d
M
d
p
p
? ?
?
-×
=
?
?
%
2
1
sin
62 3 550
1081.29 kN.m
2
6
CC
MNd
pp
p
?? ??
-
?? ??
×
?? ??
+=- =- (4)
10 cos
2
O
L
Nx ?
??
=-
??
??
And.
1
y yd =-
()
1
50 1 cos 2.25 y ? =- -
1
47.75 50cos y ? =-
Now
O
H
~
is given by equation (34.14). Thus
??
??
+
+
- = =
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
~
?
?
(5)
()
/6
2 01
00
1
5 47.75 50cos 50
s
My
ds x d
EI EI
p
? ? =-
??
()( )
/6
2
0
250
50sin 47.75 50cos d
EI
p
? ?? =-
?
() ()
/6
2
0
625000
23.875 1 cos 2 50cos sin d
EI
p
? ?? =--
?
?
() ()
/6
0
625000 1
23.875 1 cos 2 25 cos cos3 cos
2
d
EI
p
? ?? ?
?? ??
=---+
?? ??
?? ??
?
?
49630.735
EI
= (6)
()
/6
2 0
00
cos 1
10 25 cos
s
N
ds x d
EA EA
p
?
? ? =-
??
/6
2
0
10 1 cos 2
25 50sin cos
2
d
EA
p
?
? ??
?+ ? ??
=-
?? ??
?? ??
?
() ( ) ()
/6
0
10
12.5 1 cos 2 25 sin sin cos 2 d
EA
p
? ?? ? =+ - +
?
?
() ()
/6
/6
/6
0
0
0
10 1 1
12.5 sin 2 25 (cos ) cos3 cos
23 EA
p
p
p
?? ? ? ?
??
??
=+ -- +- -
??
??
??
??
??
Page 5
The distance of the elastic centre from the crown is calculated by equation, d C
?
?
=
s
s
EI
ds
ds
EI
y
d
0
0
(1)
From Fig.34.7b, the ordinate at , is given by d
() ? cos 1 50 - = y
( )
?
?
-
=
6 /
0
6 /
0
50
50
cos 1 50
p
p
?
?
?
EI
d
d
EI
d
1
50
62
2.2535 m.
6
d
p
p
??
-
??
??
==
(2)
The elastic centre O lies at a distance of from the crown. The moment at
the elastic centre may be calculated by equation (34.12). Now the bending
moment at any section of the arch due to applied loading at a distance
m 2535 . 2
x from
elastic centre is
?
?
-
=
s
s
O
EI
ds
ds
EI
x
M
0
0
2
5
~
(3)
In the present case, ? sin 50 = x and ? d ds 50 = , constant = EI
/6
32
0
/6
0
550 sin
50
O
d
M
d
p
p
? ?
?
-×
=
?
?
%
2
1
sin
62 3 550
1081.29 kN.m
2
6
CC
MNd
pp
p
?? ??
-
?? ??
×
?? ??
+=- =- (4)
10 cos
2
O
L
Nx ?
??
=-
??
??
And.
1
y yd =-
()
1
50 1 cos 2.25 y ? =- -
1
47.75 50cos y ? =-
Now
O
H
~
is given by equation (34.14). Thus
??
??
+
+
- = =
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
~
?
?
(5)
()
/6
2 01
00
1
5 47.75 50cos 50
s
My
ds x d
EI EI
p
? ? =-
??
()( )
/6
2
0
250
50sin 47.75 50cos d
EI
p
? ?? =-
?
() ()
/6
2
0
625000
23.875 1 cos 2 50cos sin d
EI
p
? ?? =--
?
?
() ()
/6
0
625000 1
23.875 1 cos 2 25 cos cos3 cos
2
d
EI
p
? ?? ?
?? ??
=---+
?? ??
?? ??
?
?
49630.735
EI
= (6)
()
/6
2 0
00
cos 1
10 25 cos
s
N
ds x d
EA EA
p
?
? ? =-
??
/6
2
0
10 1 cos 2
25 50sin cos
2
d
EA
p
?
? ??
?+ ? ??
=-
?? ??
?? ??
?
() ( ) ()
/6
0
10
12.5 1 cos 2 25 sin sin cos 2 d
EA
p
? ?? ? =+ - +
?
?
() ()
/6
/6
/6
0
0
0
10 1 1
12.5 sin 2 25 (cos ) cos3 cos
23 EA
p
p
p
?? ? ? ?
??
??
=+ -- +- -
??
??
??
??
??
81.795
EA
= (7)
()
/6 2
2
1
00
1
47.75 50cos 50
s
y
ds d
EI EI
p
? ? =-
??
()
/6
2
0
50
2280.06 2500cos 4775cos d
EI
p
? ?? =+ -
?
50 1
2280.06 1250 sin 4775sin
6623 EI 6
p pp ?? ?? ? ?
=++ -
?? ? ? ??
?? ? ? ??
p
105.046
EI
= (8)
/6 2
00
cos 50
(1 cos 2 )
2
s
ds d
EA EA
p
?
? ? =+
??
25 1
sin 23.915
62 3 EA
pp ??
=+ =
??
??
(9)
?
?
?
?
?
?
+
?
?
?
?
?
?
+ -
- =
EA EI
EA EI
H
915 . 23 046 . 105
795 . 81 735 . 49630
~
0 (10)
Consider an arch cross section of 300 500 mm × ; and
34
3.125 10 m I
-
=×
2
0.15 m A = . Then,
()
()
0
15881835.2 545.3
470.25 kN
33614.72 159.43
H
-+
=- =-
+
%
(11)
In equation (5), if the second term in the numerator and the second term in the
denominator were neglected then, we get,
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FAQs on Symmetrical Hingeless Arch - 3 - Structural Analysis - Civil Engineering (CE)
| 1. What is a symmetrical hingeless arch? |  |
Ans. A symmetrical hingeless arch is a type of structural arch that is designed to distribute loads evenly and symmetrically along its curved shape. Unlike traditional arches that rely on hinges for support, the symmetrical hingeless arch is self-supporting and does not require any external hinges or joints.
| 2. What are the advantages of using a symmetrical hingeless arch? | |
Ans. The use of a symmetrical hingeless arch offers several advantages. Firstly, it provides a more aesthetically pleasing design as there are no visible hinges or joints. Secondly, it allows for a continuous and uninterrupted load transfer, resulting in a more efficient structural system. Lastly, it provides greater flexibility in architectural design, allowing for wider spans and increased structural stability.
| 3. How is a symmetrical hingeless arch constructed? | |
Ans. The construction of a symmetrical hingeless arch involves a careful design process and the use of specialized materials. Typically, a curved formwork is used to shape the arch, and reinforced concrete or steel is used as the primary construction material. The arch is then carefully erected and secured in place, ensuring that it is structurally stable and capable of supporting the intended loads.
| 4. What are some common applications of symmetrical hingeless arches? | |
Ans. Symmetrical hingeless arches are commonly used in a variety of civil engineering applications. They are often employed in the construction of bridges, providing a visually appealing and structurally efficient solution for spanning rivers, highways, and other obstacles. Additionally, symmetrical hingeless arches can also be used in the design of architectural structures such as galleries, stadiums, and exhibition halls.
| 5. What are the key considerations when designing a symmetrical hingeless arch? | |
Ans. When designing a symmetrical hingeless arch, several key considerations must be taken into account. These include the intended span of the arch, the anticipated loads it will need to support, and the environmental conditions it will be exposed to. Additionally, factors such as material selection, construction techniques, and maintenance requirements should also be carefully considered to ensure the long-term durability and safety of the structure.
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