4.2 Internal Forces 83
Analysis of three-hinged arch subjected to fixed loads is presented below. This anal-
ysis implies determination of reactions of supports and construction of internal force
diagrams.
Design diagram of the three-hinged circular arch subjected to fixed loads is pre-
sented in Fig.4.5. The forcesP 1 D 10 kN;P 2 D 8 kN;qD 2 kN=m. It is necessary
to construct the internal force diagramsM,Q,N.
P (^1) q P 2
C
A B
0 0
Reference
beam
11.5
Q^0 (kN)
14.5 4.5
19.5
- −
154
M^0 (kNm)
116
152
78
125 124
58
Design diagram
RA
RA
RB
RB
q=2kN/m
P 2 =8kN
P 1 =10kN
A
l=32m
C
f=8m
B
HH
1
3 4 5
2
k
7
6
n
4m 4m
y
x
Circle
Fig. 4.5 Three-hinged circular arch. Design diagram, reference beam, and corresponding internal
force diagrams
Solution.
Reference beam The reactions are determined from the equilibrium equations of all
the external forces acting on the reference beam
R^0 A!
X
MBD 0 W
R^0 A 32 CP 1 24 Cq 8 12 CP 2 4 D 0 !RA^0 D14:5kN;
R^0 B!
X
MAD 0 W
R^0 B^32 P 1 ^8 q^8 ^20 P 2 ^28 D^0 !R^0 BD19:5kN:
The values of the two reactions just foundshould be checked using the equilibrium
equation
P
YDR^0 ACR^0 BP 1 q 8 P 2 D14:5C19:5 10 2 8 8 D
34 34 D 0.