Structural Concrete - Hassoun

434 Chapter 12 Slender Columns
9. Check the assumed φ:
a = 10.6in. c = 12.47in. d t = 19.5in.
( )
dt − c
ε t = 0.003
c
= 0.00169 < 0.002
φ = 0.65
Example 12.2
Check the adequacy of the column in Example 12.1 if the unsupported length is l u = 10 ft. Determine
the maximum nominal load on the column.
Solution
1. Applied loads are P n = 522 K and M n = 443.1 K.
2. Check if the column is long: l u = 10 ft, r = 0.3h = 0.3 × 22 = 6.6 in., and K = 1.0 (frame is braced
against sidesway).
Kl u 1 ×(10 × 12)
= = 18.2
r 6.6
Check if Kl u /r ≤ 34 − 12M 1b /M 2b ≤ 40
Right-hand side = 34 − 12 × 1 = 22 ≤ 40
Kl u
= 18.2 ≤ 22
r
Therefore, the slenderness effect can be neglected.
3. Determine the nominal load capacity of the short column, as explained in Example 11.4. From
Example 11.4, the nominal compressive strength is P n = 612.1 K (for e = 10 in.), which is greater
than the required load of 522 K, because the column is short with e = 10.2 in. (Example 12.1).
Example 12.3
Check the adequacy of the column in Example 12.1 if the frame is unbraced (sway) against sidesway,
the end-restraint factors are ψ A = 0.8 and ψ B = 2.0, and the unsupported length is l u = 16 ft, assume a
sway moment M s of 64 K ⋅ ft.
Solution
1. Determine the value of K from the alignment chart (Fig. 12.3) for unbraced frames. Connect the
values of ψ A = 0.8 and ψ B = 2.0, to intersect the K line at K = 1.4.
Kl u 1.4 ×(16 × 12)
= = 40.7
r 6.6
2. For unbraced frames, if Kl u /r ≤ 22, the column can be designed as a short column. Because actual
Kl u r = 40.7 > 22, the slenderness effect must be considered.
3. Calculate the moment magnifier δ s ,givenK = 1.4, EI = 17.40 × 10 6 K ⋅ in. 2 (from Example 12.1),
and