Design of Short Columns Subject to Axial Load and Bending - Design of Reinforced Concrete - page 281

of the column cross section, while for nonsymmetrical sections it can be located by taking
moments.
Example 10.1 illustrates the calculations involved in locating the plastic centroid for
a nonsymmetrical cross section. The ultimate load P n is determined by computing the
total compressive forces in the concrete and the steel and addi ng them together. Then P n
is assumed to act downward at the plastic centroid at a distance from one side of the col-
umn, and moments are taken on that side of the column of the upward compression forces
acting at their centroids and the downward P n .

x

EXAMPLE 10.1

c Determine the plastic centroid of the T-shaped column shown in Figure 10.2 if f 4000 psi and

f y 60,000 psi.

SOLUTION The plastic centroid falls on the x axis as shown in Figure 10.2 due to symmetry. The column is di-

vided into two rectangles, the left one being 16 6 and the right one 8 8 . C 1 is assumed to be
the total compression in the left concrete rectangle, C 2 the total compression in the right rectangle,
and C s the total compression in the reinforcing bars.

C 1 (16)(6)(0.85)(4) 326.4 k

C 2 (8)(8)(0.85)(4) 217.6 k

C s In computing , the concrete where the bars are located is subtracted; that is,

s (4.00)(60 0.85 4) 226.4 k C

Total compression P n 326.4 217.6 226.4 770.4 k

Figure 10.2