Simplified Design for Columns - Simplified Design of Reinforced Concrete Buildings - page 164

1 1 1 1

φ P ni = P nx + P ny − φ φ φ P o

where P ni = nominal axial load strength for a column subjected to an axial load P u at eccentricities e x and e y

P nx = nominal axial load strength for a column subjected to an axial load P u at eccentricity of e x only

(e y = 0)

P ny = nominal axial load strength for a column subjected to an axial load P u at eccentricity e y only

(e x = 0)

P o = nominal axial load strength for a column subjected to an axial load P u at eccentricity of zero (i.e.,
e x = e y = 0)

= 0.85 › (A g -A st ) + f y A st

The above equation can be rearranged into the following form:

1

φ P ni =

1 1 1

φ P nx + P ny −

φ φ P o

In design, P u < φ P ni where P u is the factored axial load acting at eccentricities e x and e y . This method is most

P ny are greater than the corresponding balanced axial loads; this is usually the case for
typical building columns.

suitable when φ P nx and φ

An iterative design process will be required when using this approximate equation for columns subjected to
biaxial loading. A trial section can be obtained from Figs. 5-18 through 5-25 with the factored axial load P u and
the total factored moment taken as M u = M ux + M uy where M ux = P u e x and M uy = P u e y . The expression for P ni

can then be used to check if the section is adequate or not. Usually, only an adjustment in the amount of rein-
forcement will be required to obtain an adequate or more economical section.

φ

5.5.3.1 Example: Simplified Design of a Column Subjected to Biaxial Loading

Determine the size and reinforcement for a corner column subjected to P u = 360 kips, M ux = 50 ft-kips, and M uy
= 25 ft-kips.

(1) Trial section

From Fig. 5-20 with P u = 360 kips and M u = 50 + 25 = 75 ft-kips, select a 14 × 14 in. column with 4-

No.9 bars.