Handbook of Civil Engineering Calculations

ALLOWABLE ECCENTRICITY OF A MEMBER

The member analyzed in the previous two calculation procedures is to carry an ultimate

longitudinal load of 150 kips (667.2 kN) that is eccentric with respect to axis N. Deter-

mine the maximum eccentricity with which the load may be applied.

Calculation Procedure:

1. Express Pu in terms of c, and solve for c
   From the preceding calculation procedures, it is seen that the value of c corresponding to
   the maximum eccentricity lies between 8 and 9 in (203.2 and 228.6 mm), and therefore fA
   <fy. Thus fB = -40,000 lb/in^2 (-275,800 kPa); fA = 40,000(c - 2.5)7(15.5 - c); Fc =
   30,600(0.85c) = 26,00Oc; 150,000 = 0.70{26,000c + 80,000[(c - 2.5)7(15.5 - c) - 1]}; c =
   8.60 in (218.44 mm).


2. Compute Mu and evaluate the eccentricity
   Thus, a = 7.31 in (185.674 mm); Fc = 223,700 Ib (995,017.6 N);/, = 35,360 lb/in^2
   (243,807.2 kPa); Mu = 0.70(223,700 x 5.35 + 150,700 x 6.5) = 1,523,000 in-lb (172,068.5
   N-m); e = Mu/Pu = 10.15 in (257.81 mm).

Design of Compression Members

by Working-Stress Method

The notational system is as follows: Ag = gross area of section, in^2 (cm^2 ); A 5 = area of ten-
sion reinforcement, in^2 (cm^2 ); Ast = total area of longitudinal reinforcement, in^2 (cm^2 ); D
= diameter of circular section, in (mm); pg = Ast/Ag; P = axial load on member, Ib (N);/ =
allowable stress in longitudinal reinforcement, lb/in^2 (kPa); m =/,/(0.85//).
The working-stress method of designing a compression member is essentially an
adaptation of the ultimate-strength method. The allowable ultimate loads and bending
moments are reduced by applying an appropriate factor of safety, and certain simplifica-
tions in computing the ultimate values are introduced.
The allowable concentric load on a short spirally reinforced column is P = Ag(Q.25f +
fsPg\ °r

P = 0.25/^+/5^r (40)

where/ = 0.40/, but not to exceed 30,000 lb/in

2

(206,850 kPa).

The allowable concentric load on a short tied column is P = 0.85^(0.25/' +//?g), or

P = 0.2125/'^ + 0.85/^ (41)

A section of the ACI Code provides that Pg may range from 0.01 to 0.08. However, in

the case of a circular column in which the bars are to be placed in a single circular row,

the upper limit of Pg is often governed by clearance. This section of the Code also stipu-

lates that the minimum bar size to be used is no. 5 and requires a minimum of six bars for

a spirally reinforced column and four bars for a tied column.