Three hinged arches – Concepts, types of arches, analysis of parabolic arch with supports at same levels, determination of horizontal thrust, radial shear and normal thrust for parabolic arch

1. Unit No.5:-Analysis of Arches
(Three Hinged Parabolic Arches. )
Subject: Structural Analysis –I
By,
Prof. S. R. Nawale
Assistant Professor
Sanjivani College of Engineering, Kopargaon

2. Unit No.5:-Analysis of Arches((Three Hinged Parabolic Arches. )
 Any section in arch will be subjected
to normal thrust(N), radial shear(Q)
& bending moment(M).
 In arches load get transferred by
axial compression & by flexure.
 Reactions:- -At both supports
reactions will be horizontal &
vertical
 Since arch is subjected to vertical
load only horizontal reactions are
equal &opposite.
 H is called as Horizontal Thrust
 Rise of arch(y) at any distance ‘x’
 Slope of tangent at any distance x
4ycy= ×x×(L-2x)
2L4yctanθ= ×(L-2x)
2L

3. Unit No.5:-Analysis of Arches(Three Hinged Parabolic Arch)

Normal Thrust (N) Normal Thrust (N)=
Radial shear(Q)=
Consider a three hinged parabolic arch subjected to load as shown in fig. since there
will be two reaction components at each end.
1)calculation of reaction-:
There are four unknowns
Additional Equilibrium equation at Hinge C :
Four unknown can be obtain.
2)calculation of vertical shear, radial shear, horizontal thrust and normal thrust.
At section ‘M’ let, V=vertical shear.
h=horizontal thrust.
Q=Radial shear.
N=Normal thrust.

4. Unit No.5:-Analysis of Arches(Three Hinged Parabolic Arch)
Que. A three hinged parabolic arch hinged at the supports and at the crown has a
span of 24m and a central rise of 4m.It carries a concentrated load of 50KN at 18m
from left supports and a udl of 30 KN/m over the left half portion .Determine the
moment, thrust and radial shear at a section 6m from the left support.
Calculation of supports reaction:-
B30×12×16+50×18 -24V = 0

5. To calculate ‘H’ consider portion CB
50x6-127.5x12+Hx4=0
H=307.5KN.
Bending moment at 6m from A
=
Find
BM at D =282.50x6-307.5x3-30x6x3
BM at D=232.50 kNm.
Unit No.5:-Analysis of Arches(Three Hinged Parabolic Arch)

6. Radial Shear;
Calculation of horizontal thrust and normal shear at 6m from support A:
H=horizontal thrust=307.5 KN.
Vertical shear at D=V.
S.F. at D=-30x6+282.50
V=102.5KN.
Calculate:-
Y=
Normal thrust
N=324.13KN
Q=0 KN
Unit No.5:-Analysis of Arches(Three Hinged Parabolic Arch)

7. . Show that for a three hinged parabolic arch carrying UDL over entire span,
Bending moment at any section is equal to zero.
Consider a Parabolic Arch ACB
VA=VB=WL/2
To calculate H, consider Arch CB.
∑M@C=0 Clockwise +ve &Anticlockwise –ve.
Unit No.5:-Analysis of Arches(Three Hinged Parabolic Arch)

8. Unit No.5:-Analysis of Arches(Three Hinged Parabolic Arch)
.
Hence, proved.