Structural Concrete - Hassoun

400 Chapter 11 Members in Compression and Bending
bending about the x and y axes, the equivalent uniaxial M u moment can be calculated using the
following equations:
√
M u = (M ux ) 2 +(M uy ) 2 = P u e (11.29)
and
where
e =
√
(e x ) 2 +(e y ) 2 = M u
P u
(11.30)
M ux = P u e y = factored moment about the x-axis
M uy = P u e x = factored moment about the y-axis
M u = P u e = equivalent uniaxial factored moment of the section due to M ux and M uy
In circular columns, a minimum of six bars should be used, and these should be uniformly
distributed in the section.
Example 11.17 Circular Column
Determine the load capacity P n of a 20-in.-diameter column reinforced with 10 no. 10 bars when e x =
4in.ande y = 6in.Usef c ′ = 4ksiandf y = 60 ksi.
Solution
1. Calculate the eccentricity that is equivalent to uniaxial loading by using Eq. 11.30.
√
e(for uniaxial loading) = e 2 x + e 2 y = √ (4) 2 +(6) 2 = 7.211 in.
2. Determine the load capacity of the column based on e = 7.211 in. Proceed as in Example 11.9:
3. For a balanced condition,
d = 17.12 in. a = 9.81 in. c = 11.54 in. (by trial)
C c = 521.2 K ΣC s = 269.8K ΣT = 132.1 K
P n = C c + ΣC s − ΣT = 650 K
c b =
( ) ( )
87 87
d
87 + f t = 17.12 = 10.13 in.
y 147
c = 11.54 in. >c b
which is a compression failure case.
Example 11.18 Circular Column
Design a 16-in. circular column subject to biaxial bending using the equivalent uniaxial moment method.
Given P u = 200 K, M ux = 1000 K ⋅ in, M uy = 700 K ⋅ in, f c ′ = 4ksi,f y = 60 ksi.
Solution
1. Determine nominal load:
For spiral column φ = 0.75
Nominal load = 200∕0.75 = 266.67 K
Nominal moment capacity about x axis M nx = 1000∕0.75 = 1333.33 K ⋅ in.
Nominal moment capacity about y axis M ny = 700∕0.75 = 933.33 K ⋅ in.