Structural Concrete - Hassoun

350 Chapter 10 Axially Loaded Columns
Example 10.1
Determine the allowable design axial load on a 12-in. square, short tied column reinforced with
four no. 9 bars. Ties are no. 3 spaced at 12 in. Use f c ′ = 4ksiandf y = 60 ksi.
Solution
1. Using Eq. 10.9,
P u = φP n = φK[0.85f ′ c A g + A st (f y − 0.85f ′ c )]
For a tied column, φ = 0.65, K = 0.8, and A st = 4.0in. 2
P u = φP u = 0.65(0.8)[0.85(4)(12 × 12)+4(60 − 0.85 × 4)] = 372 K
2. Check steel percentage: ρ g = 4 = 0.02778 = 2.778%. This is less than 8% and greater than 1%.
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3. Check tie spacings: Minimum tie diameter is no. ( 3. ) Spacing is the smallest ( ) of the 48-tie diameter,
16-bar diameter, or least column side. S 1 = 16 9
= 18 in., S
8 2 = 48 3
= 18 in., S
8 3 = 12.0in.
Ties are adequate (Table 10.1). Note: Clear spacing of ties should be at least 4/3 the nominal
maximum size of the aggregate.
Example 10.2
Design a square tied column to support an axial dead load of 400 K and a live load of 232 K using
f ′ c = 5ksi,f y = 60 ksi, and a steel ratio of about 5%. Design the necessary ties.
Solution
1. Calculate P u = 1.2P D + 1.6 P L = 1.2(400) +1.6(232) =851 K. Using Eq. 10.10, P u = 851 =
0.65(0.8) A g [0.85 × 5 + 0.05(60–0.8 × 5)], A g = 232.5in. 2 , and column side = 15.25 in., so use
16 in. (Actual A g = 256 in. 2 .)
2. Because a larger section is adopted, the steel percentage may be reduced by using A g = 256 in. 2
in Eq. 10.8:
851 = 0.65(0.8)[0.85 × 5 × 256 + A st (60 − 0.85 × 5)]
A st = 9.84 in. 2
Use eight no. 10 bars (A st = 10.16 in. 2 ). See Fig. 10.5.
Figure 10.5 Example 10.2.