Simplified Frame Analysis - Simplified Design of Reinforced Concrete Buildings - page 43

(2) Determine the axial loads in the columns:

For the end columns, the axial loads can be obtained by summing moments about the column
inflection points at each level. For example, for the 2nd story columns:

= 0 : 12(13 + 6.5) + 23.1 (6.5) - P (90) = 0
P = 4.27 kips

∑ M

For this frame, the axial forces in the interior columns are zero.

(3) Determine the moments in the columns:

The moments can be obtained by multiplying the column shear force by one-half of the column length.

For example, for an exterior column in the 2nd story:

M = 5.85(13/2) = 38.03 ft-kips

(4) Determine the shears and the moments in the beams:

These quantities can be obtained by satisfying equilibrium at each joint. Free-body diagrams for the
2nd story are shown in Fig. 2-14.

As a final check, sum moments about the base of the frame:

= 0: 12.0(39) + 23.1(26) + 21.7(13) - 10.91(90) - 2(61.53 + 123.07) = 0 (checks)

∑ M

In a similar manner, the wind load analyses for an interior frame of Building #2 (5-story flat plate), in both the
N-S and E-W directions are shown in Figs. 2-15 and 2-16, respectively. The wind loads are determined in
Section 2.2.2.1.