Structural Concrete - Hassoun

468 Chapter 13 Footings
9. Development length of the main reinforcement: l d = 29 in. for no. 6 bars and 54 in. for no. 9 bars:
( L
Provided l d (long direction) =
2 − c )
2 − 3in. = 60 in.
Provided l d (short direction) =39 in. >29 in.
Example 13.4
Determine transfer of force between column and footing for a 14 × 14−in. tied column with four
no. 11 longitudinal bars: f c ′ = 4000 psi, f y = 60,000 psi, axial dead load = 250 kips, axial live load
P L = 150 kips, and footing size is 10 × 10 ft.
Solution
1. Determine the factored load P u = 1.2 × 250 + 1.6 × 150 = 540 K
2. Bearing strength of column concrete:
φP nb = φ(0.85f c ′ A 1 )=0.65(0.85 × 4 × 14 × 14) =433 K < P u = 540 K
The bearing strength of the column is less than the axial load. So concrete only cannot transfer the
load to the footing. The excess load (540 − 433 = 107 kips) must be transferred by reinforcement.
3. Bearing strength of footing concrete:
√
A
φP nb = 2
φ(0.85f c ′ A A 1 )
1
√
10 × 10 × 144
A 2 =
= 8.6 > 2
14 × 14
Use 2:
φP nb = 2(433) =866 K > P u = 540 K safe
4. Required area of dowel bar:
A s = P u − φP nb 540 − 433
= = 2.74 in.2
φf y 0.65 × 60
A s,min = 0.005(14 × 14) =0.98 in. 2
Provide four no. 8 bars (A s = 3.16 in. 2 ).
5. Development of dowel reinforcement
a. For development into the column (Fig. 13.21) no. 11 bar may be lap spliced with no. 8 dowel
bars. Dowels must extend into the column:
{ }
Devlopment length of no. 11 bars
Greater of →
Lap splice length of no. 8 footing dowels
For no. 11 bars:
[ ] [
]
0.02fy
l dc =
λ √ 0.02 (60,000)
d b =
f c
′ 1.0 √ 1.41 = 26.75 in.
4000
l dc,min =(0.0003f y )d b = 0.0003 × 60,000 × 1.41 = 25.38 in.
Forno.8bars:
Lap splice length =(0.0005)f y d b = 0.0005(60,000)(1.0) =30 in.(governs)