Structural Concrete - Hassoun

352 Chapter 10 Axially Loaded Columns
Assume no. 3 spiral, a s = 0.11 in. 2 ,andd s = 0.375 in.
ρ s = 0.01152 = 4a s(D ch − d b ) 4(0.11)(17 − 0.375)
=
SD 2 S(17) 2
ch
Spacing s is equal to 2.2 in; use no. 3 spiral at s = 2 in. (as shown in Table 10.2).
Example 10.5
Design a rectangular tied short column to carry a factored axial load of 1765 kN. Use f c ′ = 30 MPa,
f y = 400 MPa, column width (b) = 300 mm, and a steel ratio of about 2%.
Solution SI Units
1. Using Eq. 10.9,
P u = 0.8φA g [0.85f ′ c + ρ g (f y − 0.85f ′ c )]
Assuming a steel percentage of 2%,
1765 × 10 3 = 0.8 × 0.65A g [0.85 × 30 + 0.02(400 − 0.85 × 30)]
A g = 102, 887 mm 2
For b = 300 mm, the other side of the rectangular column is 343 mm. Therefore, use a section
of 300 × 350 mm (Ag = 105,000 mm 2 ).
2. A s = 0.02 × 102,887 = 2057 mm 2 . Choose six bars, 22 mm in diameter (A s = 2280 mm 2 ).
3. Check the axial load strength of the section using Eq. 10.6:
φP n = 0.8φ[0.85f c ′(A g − A st )+A st f y ]
= 0.8 × 0.65[0.85 × 30(105,000 − 2280)+2280 × 400]×10 −3
= 1836 kN
This meets the required P u of 1765 kN.
4. Choose ties 10 mm in diameter. Spacing is the least of (1) 16 × 22 = 352 mm, (2) 48 × 10 =
480 mm, or (3) 300 mm. Choose 10-mm ties spaced at 300 mm. Note: Clear spacing of ties
should be at least 4/3 the nominal maximum size of the aggregate.
SUMMARY
Sections 10.1–10.4
Columns may be tied or spirally reinforced:
{ 0.65 for tied columns
φ =
0.75 for spirally reinforced columns
Also, ρ g must be ≤ 8% and ≥1%.