- Perform the second cycle of moment balancing and distribution
Thus MBA = -9.9(15/31) = -4.8; MBC = -9.9(16/31) - -5.1; MCB = +1.6(16/25) = + 1.0;
MO, = +1.6(9/25) = +0.6. - Continue the foregoing procedure until the carry-over moments
become negligible
Total the results to obtain the following bending moments: MA = -58.2 ft-kips (-78.91 kN
m); MB = -45.7 ft-kips (-61.96 kN-m); Mc = -66.1 ft-kips (-89.63 kN-m).
ANALYSIS OFA STATICALLY
INDETERMINATE TRUSS
Determine the internal forces of the truss in Fig. 46a. The cross-sectional areas of the
members are given in Table 5.
Calculation Procedure:
- Test the structure for static determinateness
Apply the following criterion. Let j = number of joints; m = number of members; r =
number of reactions. Then if 2/ = m + r, the truss is statically determinate; if 2/ < m + r,
the truss is statically indeterminate and
the deficiency represents the degree of in-
determinateness.
In this truss, j = 6, m = 10, r = 3, con-
sisting of a vertical reaction at A and D
and a horizontal reaction at D. Thus 2/ =
12; m + r = 13. The truss is therefore stat-
ically indeterminate to the first degree;
i.e., there is one redundant member.
The method of analysis comprises the
following steps: Assume a value for the
internal force in a particular member, and
calculate the relative displacement A,, of
the two ends of that member caused sole-
ly by this force. Now remove this member
to secure a determinate truss, and calcu-
late the relative displacement Aa caused
solely by the applied loads. The true inter-
nal force is of such magnitude that A 1 - =
-Aa. - Assume a unit force for one
member
Assume for convenience that the force in
BF is 1-kip (4.45-kN) tension. Remove
this member, and replace it with the as-
sumed 1-kip (4.45-kN) force that it exerts FIGURE 46. Statically indeterminate
at joints B and F, as shown in Fig. 46b. truss.