Symmetrical Hingeless Arch - 2 Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : Symmetrical Hingeless Arch - 2 Civil Engineering (CE) Notes | EduRev

 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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