Simplified Frame Analysis - Simplified Design of Reinforced Concrete Buildings - page 41

columns. Under the assumptions of this method, however, a frame configuration with unequal spans will have axial
load in those columns between the unequal spans, as well as in the end columns. The general term for axial load in
the end columns in a frame of n bays with unequal spans is:

Wh
2n

and Wh
2n

, n

= length of bay n

1 n

The axial load in the first interior column is:

Wh
2n

Wh
2n 1 −

2

and, in the second interior column:

Wh
2n

Wh
2n 2 −

3

Column moments are determined by multiplying the column shear with one-half the column height. Thus, for
joint B in Fig. 2-11, the column moment is (W/3) (h/2) = Wh/6. The column moment Wh/6 must be balanced
by equal moments in beams BA and BC, as shown in Fig. 2-12.

Note that the balancing moment is divided equally between the horizontal members without considering their
relative stiffnesses. The shear in beam AB or BC is determined by dividing the beam end moment by one-half
the beam length, (Wh/12)(˜/2) = Wh/6˜ .

The process is continued throughout the frame taking into account the story shear at each floor level.

2.6.2 Examples: Wind Load Analyses for Buildings #1 and #2

For Building #1, determine the moments, shears, and axial forces using the Portal Method for an interior frame
resulting from wind loads acting in the N-S direction. The wind loads are determined in Section 2.2.1.2.

Moments, shears, and axial forces are shown directly on the frame diagram in Fig. 2-13. The values can be
easily determined by using the following procedure:

(1) Determine the shear forces in the columns:

For the end columns:
3rd story: V = 12.0 kips/6 = 2.0 kips
2nd story V = (12.0 kips + 23.1 kips)/6 = 5.85 kips
1st story: V = (12.0 kips + 23.1 kips + 21.7 kips)/6 = 9.50 kips

The shear forces in the interior columns are twice those in the end columns.