Structural Concrete - Hassoun

12.6 Moment-Magnifier Design Method 435
P c = π2 EI
(Kl u ) 2 = π2 × 17.40 × 10 6
(1.4 × 16 × 12) 2 = 2377K
Assume factored loads are the same on all columns in the story level
δ s =
1.0
1 − ∑ P u ∕ ( 0.75 × ∑ 1.0
) =
= 1.24 ≥ 1.0
P c
1 − 339.2∕(0.75 × 2377)
4. From Example 12.1, the applied loads are P u = 339.2 K and M u = 288 K ⋅ ft, or
The design moment M c = M ns + δ s M s hence:
P n = 522K and M n = 443.1K⋅ ft
M c = 288 + 1.24(64) =367.4 K⋅ ft
e = M c 367.4 × 12
= = 13in.
P u 339.2
5. The requirement now is to check the adequacy of a column for P n = 522 K, M c = 307.6 K ⋅ ft, and
e = 13 in. The procedure is explained in Example 11.4.
6. From Example 11.4,
P n = 47.6a + 226.4 − 4f s
e ′ = e + d − h 22
= 13 + 19.5 −
2 2 = 21.5in.
P n = 1 [ (
47.6a 19.5 − a )
]
+ 226.4(19.5 − 2.5)
21.5
2
= 43.16a − 1.1a 2 + 179 a = 10.4in.
Thus, c = 12.24 in. and P n = 508 K. This load capacity of the column is less than the required
P n of 522 K. Therefore, the section is not adequate.
7. Increase steel reinforcement to four no. 10 bars on each side and repeat the calculations to get
P n = 568 K, ε t < 0.002, and φ = 0.65.
Example 12.4
Design an interior square column for the first story of an 8-story office building. The clear height of the
first floor is 16 ft, and the height of all other floors is 11 ft. The building layout is in 24 bays (Fig. 12.6),
and the columns are not braced against sidesway. The loads acting on a first-floor interior column due
to gravity and wind are as follows:
Axial dead load = 300K
Axial live load = 100K
Axial wind load = 0K
Dead − load moments = 32K ⋅ ft(top) and54K⋅ ft(bottom)
Live − load moments = 20K ⋅ ft(top) and36K⋅ ft(bottom)
Wind − load moments = 50K ⋅ ft(top)and50K ⋅ ft(bottom)
EI∕lfor beams = 360 × 10 3 K ⋅ in.