Civil Engineering Reference
In-Depth Information
The completely stiff structure is shown in Fig. 16.11(b) in which we see that
M
=
17
and
N
8. However, since the truss is pin-jointed, we can obtain the internal statical
indeterminacy directly from Eq. (16.5) in which
M
=
=
16, the actual number of truss
members. Thus
n
s
=
16
−
16
+
3
=
3
and since, as can be seen from inspection, the support system is statically determinate,
the complete structure is three times statically indeterminate.
Alternatively, considering the completely stiff structure in Fig. 16.11(b) in which
M
8, we can use Eq. (16.3). The number of internal releases is found
from Eq. (16.4) and is
r
=
17 and
N
=
24. There are three additional releases from
the support system giving a total of 27 releases. Thus, from Eq. (16.3)
=
2
×
16
−
8
=
n
s
=
−
+
−
=
3(17
8
1)
27
3
as before.
The kinematic indeterminacy is found as before by examining the total degrees of
freedom of the nodes and the constraints, which in this case are provided solely by
the support system. There are eight nodes each having 2 translational degrees of
freedom. The rotation at a node does not result in a stress resultant and is therefore
irrelevant. There are therefore 2 degrees of freedom at a node in a plane truss and
3 in a space truss. In this example there are then 8
16 degrees of freedom and
three translational constraints from the support system. Thus
×
2
=
n
k
=
16
−
3
=
13
E
XAMPLE
16.3
Calculate the degree of statical and kinematic indeterminacy of
the frame shown in Fig. 16.12(a).
F
IGURE
16.12
Statical and
kinematic
indeterminacies
of the frame
of Ex. 16.3
(a)
(b)
In the completely stiff structure shown in Fig. 16.12(b),
M
6. The number
of releases,
r
, required to return the completely stiff structure to its original state is 3.
Thus, from Eq. (16.3)
=
7 and
N
=
n
s
=
3(7
−
6
+
1)
−
3
=
3
