Building Design and Construction Handbook - Merritt - Ventech!

5.94 SECTION FIVE
joints by the actual unbalanced moments at those joints divided by 1000, and summing
(see also Art. 5.11.9 and Table 5.6).
5.11.9 Procedure for Sidesway
Computations of moments due to sidesway, or drift, in rigid frames is conveniently
executed by the following method:
1. Apply forces to the structure to prevent sidesway while the fixed-end moments
due to loads are distributed.
2. Compute the moments due to these forces.
3. Combine the moments obtained in Steps 1 and 2 to eliminate the effect of the
forces that prevented sidesway.
Suppose the rigid frame in Fig. 5.77
is subjected to a 2000-lb horizontal load
acting to the right at the level of beam
BC. The first step is to compute the moment-influence
factors (Table 5.6) by
applying moments of �1000 at joints B
and C, assuming sidesway prevented.
Since there are no intermediate loads
on the beams and columns, the only
fixed-end moments that need be considered
are those in the columns resulting
from lateral deflection of the frame
caused by the horizontal load. This deflection,
however is not known initially.
FIGURE 5.77 Rigid frame.
So assume an arbitrary deflection, which
produces a fixed-end moment of
�1000M at the top of column CD. M is an unknown constant to be determined
from the fact that the sum of the shears in the deflected columns must be equal to
the 2000-lb load. The same deflection also produces a moment of �1000M at the
bottom of CD [see Eq. (5.126)].
From the geometry of the structure, furthermore, note that the deflection of B
relative to A is equal to the deflection of C relative to D. Then, according to Eq.
(5.126) the fixed-end moments in the columns are proportional to the stiffnesses of
TABLE 5.6 Moment-Influence Factors
for Fig. 5.77
Member �1000 at B �1000 at C
AB
BA
BC
CB
CD
DC
351
702
298
70
�70
�35
�105
�210
210
579
421
210