240 7 The Force Method
Ta b l e 7. 7 The chord length of each portion of the arch
Portion 0 11 22 33 44 55 66 77 8
Length (m) 3.9863 3.5377 3.2040 3.0233 3.0233 3.2040 3.5377 3.9863
Internal forces in the unit state The arch is subjected to unit primary unknown
X 1 D 1 (Fig.7.15b). Horizontal reactionHD 1 and the positive directions of inter-
nal forces are shown in Fig.7.15c.
MN 1 D 1 y
QN 1 D 1 sin' (7.11)
NN 1 D 1 cos':Internal forces at specified points in the unit state according to (7.11) are presented
in Table7.8.
Ta b l e 7. 8 Internal forces of
the arch in the unit state Points
MN 1 QN 1 NN 1
0 0.0 0:7070 0:7070
1 2:625 0:6000 0:8000
2 4:50 0:4472 0:8944
3 5:625 0:2425 0:9701
4 6:00 0.0 1:0000
5 5:625 0.2425 0:9701
6 4:50 0.4472 0:8944
7 2:625 0.6000 0:8000
8 0.0 0.7070 0:7070
The unit displacement caused by primary unknownXD 1 equalsı 11 DZ.s/MN 1 MN 1
EIds: (a)Thus theonlycolumnMN 1 (Table7.8) will be used for calculation of unit displace-
ment; the columnsQN 1 andNN 1 will be used for computation of final shear and axial
forces as indicated at step 5 of procedure.
Internal forces in the loaded state Displacement in the primary system caused by
applied load equals
1PDZ.s/MN 1 M^0
P
EIds; (b)whereMP^0 is bending moments in the arch in the primary system due to given load
q. Thus, as in case of unit displacement, for computation of loaded displacement we
will take into account only bending moment.
The reactions of supports of the primary system in the loaded state areR^0 AD
R^0 BD 24 kN; this state is not shown. Expressions for internal forces are follows
( 0 x 24 )