Page 255 - Handbook of Civil Engineering Calculations, Second Edition

2.40 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

For symmetric spiral columns:

M b
0.43p g mD s 0.14t (47a)
P b
For symmetric tied columns:

M b
d(0.67p g m 0.17) (47b)
P b
For unsymmetric tied columns:
p
m(d d
) 0.1d
M b
(47c)
( p
p)m 0.6
P b
where p
ratio of area of compression reinforcement to effective area of concrete. The
value of P a is taken as

P a 0.34f c
A g (1 p g m) (48)
The value of M f 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.)
2
2
Computing P a and M f yields A g 260 sq.in. (1677.5 cm ); A st 7.62 sq.in. (49.164 cm );
p g 7.62/260 0.0293; m 50/[0.85(4)] 14.7; p g m 0.431; n 8; P a 0.34(4)(260)
(1.431) 506 kips (2250.7 kN).
3
The section modulus to be applied in evaluating M f is found thus: I ( /12)(13)(20)
1
4
2
3
3
4
7.62(15)(7.5) 15,100 in (62.85 dm ); S I/c 15,100/10 1510 in (24,748.9 cm );
M f Sf c 1510(1.8) 2720 in.·kips (307.3 kN·m).
2. Compute P b and M b
By Eq. 47b, M b /P b 17.5(0.67 0.431 0.17) 8.03 in. (203.962 mm). By Eq. 44b,
P b P a M f /(M j 8.03P a ) 506 2720/(2720 8.03 506) 203 kips (902.9 kN);
M b 8.03(203) 1630 in.·kips (184.2 kN·m).
3. Compute M o
By Eq. 46b, M o 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, P max 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 N is (a) 10
in. (254 mm); (b) 6 in. (152.4 mm).