Handbook of Civil Engineering Calculations

Mb

— = QA3pgmDs + O. I4t (47 a)

Pb

For symmetric tied columns:

Mb

— = d(Q.61Pgm + 0.17) (4Ib)

Pb

For unsymmetric tied columns:

Mb _ p'm(d-d') + Q.ld

~P~b ^ (X-P)IH+ 0.6 (4?C)

where/?' = ratio of area of compression reinforcement to effective area of concrete. The

value ofPa is taken as

Pa = 034f;Ag(l+Pgm) (48)

The value of Mf is found by applying the section modulus of the transformed un-
cracked section, using a modular ratio of 2n to account for stress transfer between steel
and concrete engendered by plastic flow. (If the steel area is multiplied by 2n - 1, al-
lowance is made for the reduction of the concrete area.)
Computing Pa and Mf yields Ag = 260 in^2 (1677.5 cm^2 ); Ast = 7.62 in^2 (49.164 cm^2 );
pg = 7.62/260 - 0.0293; m = 50/[0.85(4)] = 14.7; p/n = 0.431; n = 8; Pa = 0.34(4)(260)
(1.431) = 506 kips (2250.7 kN).
The section modulus to be applied in evaluating Mf is found thus: /= (^1 Xw)(13)(20)^3 +
7.62(15)(7.5)^2 = 15,100 in^4 (62.85 dm^4 ); S = Vc= 15,100/10 = 1510 in^3 (24,748.9 cm^3 );
Mf= Sf 0 = 1510(1.8) = 2720 in-kips (307.3 kN-m).

2. Compute Pb and Mb
   By Eq. 476, M 1 JP 13 = 17.5(0.67 0.431 + 0.17) = 8.03 in (203.962 mm). By Eq. 446, Pb =
   PaMf/(Mj + 8.03PJ = 506 2720/(2720 + 8.03 x 506) = 203 kips (902.9 kN); Mb =
   8.03(203) = 1630 in-kips (184.2 kN-m).


3. Compute M 0
   By Eq. 466, M 0 = 0.40(3.81)(50)(15) = 1140 in-kips (128.8 kN-m).


4. Compute the limiting value of P
   As established by Eq. 41, Pmax = 0.2125(4)(260) + 0.85(20)(7.62) = 351 kips (1561.2
   kN).


5. Construct the interaction diagram
   The complete diagram is shown in Fig. 23.

AXIAL-LOAD CAPACITY OF A

RECTANGULAR MEMBER

The member analyzed in the previous calculation procedure is to carry an eccentric longi-

tudinal load. Determine the allowable load if the eccentricity as measured from TV is (a) 10

in (254 mm); (b) 6 in (152.4 mm).