Structural Concrete - Hassoun

Review Problems on Concrete Building Components 965
c. Design the Column at C
1. Calculate effective length factors (assume column is fixed at both ends):
ΣE c I c
L
Ψ A = Ψ B = c
= 0, since fixed at both ends
ΣE b I b
L b
2. Find k:
Using the Alignment Charts (Figure 12.3), k = 0.5
3. Check Slenderness of Column West/East side (unbraced frame): (ACI Code, Section 6.2.5)
kL u
r
L u
≤ 22
)
− 4ft= 9.92 ft
( 25
= (16 ft)−6
12
( ) 12
r = 0.3h = 0.3 = 0.3 ft
12
kL u
r
16.53 ≤ 22
= 0.50(9.92)
0.3
= 16.53
∴ Slenderness effects permitted to be neglected → short column.
4. Column design calculations—west/east side direction
M u (K.ft) 11.65
P u (K) 27.45
h (in.) 12
A g =(12 in.)(12 in.) =144 in. 2
γh = 12 – (2in. cover) – (2in. cover) =8in.
γ = γh h = 8
12 = 0.67
φ = 0.65 (ACI 21.2.1)
(11.65 K ⋅ ft) (12)
M n = = 215.08 K ⋅ in.
0.65
R n =
P n e
f c ′ A g h → P n e = M n
R n = 215.08
4(144)(12) = 0.03
P n = P u
φ = 27.45
0.65 = 42.23 K
K n =
P n
f ′ c A g
= 42.23
4(144) = 0.07
From Figure 11.16 charts (c) and (d) in the textbook, ρ g = 0.01
A s = ρ g A g = 0.01(144 in. 2 )=1.44 in. 2 use 4 no.6 bars; d b = 0.75 in.
0.01A g ≤ A s ≤ 0.08A g (ACI Code, Section 10.6.1.1)