Handbook of Civil Engineering Calculations

FIGURE 21. Interaction diagram.

1,596,000/193,400 = 8.25 in (209.55 mm). This result discloses that an eccentricity of 9.2

in (233.68 mm) corresponds to a point on AB and an eccentricity of 6 in (152.4 mm) cor-

responds to a point on BC.

2. Evaluate Pu when e = 9.2 in (233.68 mm)
   It was found that c = 9 in (228.6 mm) is a significant value. The corresponding value of e
   is 1,577,000/163,900 = 9.62 in (244.348 mm). This result discloses that in the present in-
   stance O 9 in (228.6 mm) and consequently fA =fy; FA = 80,000 Ib (355,840 N); FB =
   -80,000 Ib (-355,840 N); Fc = 30,600«; Pw/0.70 = 30,60Oa; MM/0.70 = 30,600a(18 - a)/2

• 160,000(6.5); e = MJP 14 = 9.2 in (233.68 mm). Solving gives a = 8.05 in (204.47 mm),
  P 14 = 172,400 Ib (766,835.2 N).

3. Evaluate Pu when e = 6 in (152.4 mm)
   To simplify this calculation, the ACI Code permits replacement of curve BC in the inter-
   action diagram with a straight line through B and C. The equation of this line is

M 14

Pu = P 0 -(Po-Bd- (39)

Mb

By replacing Mu with Pue, the following relation is obtained:

f"-i

+(P.-W

(39fl)

In the present instance, P 0 = 497,600 Ib (2,213,324.8 N); pb = 193,400 Ib (860,243.2 N);
Mb = 1,596,000 in-lb (180,316.1 N-m). Thus Pu = 232,100 Ib (1,032,380 N).

Compressionfailure

Tensionfailure