Civil Engineering Reference
In-Depth Information
B
B
C
B
C
C
A
E
F
IGURE
16.9
Number of releases
for a plane truss
A
D
A
D
E
E
(a)
(b)
(c)
so that if
r
is the total number of releases required, the degree of statical indeterminacy,
n
s
, of the actual structure is
n
s
−
n
s
=
r
or, substituting for
n
s
from Eq. (16.1)
n
s
=
6(
M
−
N
+
1)
−
r
(16.2)
Note that in Fig. 16.8 no releases are required to return the completely stiff structure of
Fig. 16.8(b) to its original state in Fig. 16.8(a) so that its degree of indeterminacy is 18.
In the case of two-dimensional structures in which a ring is three times statically
indeterminate, Eq. (16.2) becomes
n
s
=
3(
M
−
N
+
1)
−
r
(16.3)
TRUSSES
A difficulty arises in determining the number of releases required to return the
completely stiff equivalent of a truss to its original state.
Consider the completely stiff equivalent of a plane truss shown in Fig. 16.9(a); we are
not concerned here with the indeterminacy or otherwise of the support system which
is therefore omitted. In the actual truss each member is assumed to be capable of
resisting axial load only so that there are two releases for each member, one of shear
and one of moment, a total of 2
M
releases. Thus, if we insert a hinge at the end of
each member as shown in Fig. 16.9(b) we have achieved the required number, 2
M
,
of releases. However, in this configuration, each joint would be free to rotate as a
mechanism through an infinitesimally small angle, independently of the members; the
truss is then excessively pin-jointed. This situation can be prevented by removing one
hinge at each joint as shown, for example at joint B in Fig. 16.9(c). The member BC
then prevents rotation of the joint at B. Furthermore, the presence of a hinge at B in
BA and at B in BE ensures that there is no moment at B in BC so that the conditions
for a truss are satisfied.
