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# Symmetrical Hingeless Arch - 3 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : Symmetrical Hingeless Arch - 3 Civil Engineering (CE) Notes | EduRev

``` 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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## Structural Analysis

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