Civil Engineering Reference
In-Depth Information
Figure 14.23
Practically speaking, however, it is unnecessary to go through such a lengthy process for
buildings of normal proportions because the values that would be obtained do not vary
significantly from the values given by the ACI Code, which are shown in Figure 14.23.
Equivalent Rigid-Frame Method
When continuous beams frame into and are supported by girders, the normal assumption
made is that the girders provide only vertical support. Thus they are analyzed purely as
continuous beams, as shown in Figure 14.24. The girders
d
o provide some torsional stiff-
ness, and if the calculated torsional moments exceed as specified in ACI
Section 11.6.1 they must be considered, as will be described in Chapter 15.
Where continuous beams frame into columns, the bending stiffnesses of the columns to-
gether with the torsional stiffnesses of the girders are of such magnitude that they must be
considered. An approximate method frequently used for analyzing such reinforced concrete
members is the
equivalent rigid-frame method
. In this method, which is applicable only to
gravity loads, the loads are assumed to be applied only to the floor or roof under consideration
and the far ends of the columns are assumed to be fixed, as shown in Figure 14.25. The sizes
of the members are estimated, and an analysis is frequently made with moment distribution.
For this type of analysis it is necessary to estimate the sizes of the members and com-
pute their relative stiffness or
I
/
(
A
cp
/
cp
)
c
f
values. From these values, distribution factors can be
computed and the method of moment distribution applied. The moments of inertia of both
columns and beams are normally calculated on the basis of gross concrete sections, with
no allowance made for reinforcing.
There is a problem involved in determining the moment of inertia to be used for contin-
uous T beams. The moment of inertia of a T beam is much greater where there is positive
moment with the flanges in compression than where there is negative moment with the
flanges cracked due to tension. Because the moment of inertia varies along the span, it is
necessary to use an equivalent value. A practice often used is to assume that the equivalent
moment of inertia equals twice the moment of inertia of the web, assuming that the web
Figure 14.24
