Structural Detailing of Reinforcement for Economy - Simplified Design of Reinforced Concrete Buildings - page 253

Two legs of the No.3 ties are effective in each direction; thus, the required spacing s can be determined
as follows:

2 0.11 = 0.22 in.2 = 0.0015 16 s

or, s = 9.2 in.

Splice length = 0.83 30 = 24.9

Use a 2 ft-1 in. splice length with No.3 ties @ 9 in. throughout the splice length.

8.8.4 Example: Lap Splice Length for an Interior Column of Building #2, Alternate (1) Slab and

Column Framing Without Structural Walls (Sway Frame)

As in Example 8.8.3, the required lap splice length will be determined for an interior column in the 2nd
story, located just above the slab at the 1st level.

(1) Load Data

In Example 5.7.1, the following load combinations were obtained for the 2nd story interior columns:

Gravity loads: P u = 399 kips ACI Eq. (9-2)

M u = 13 ft-kips

Gravity + Wind loads: P u = 399 kips ACI Eq. (9-3)

M u = 47 ft-kips

P u = 371 kips ACI Eq. (9-4)

M u = 77 ft-kips

P u = 243 kips ACI Eq. (9-6)

M u = 68 ft-kips

(2) Column Size and Reinforcement

A 16 16 in. column size was established in Example 5.7.1 for the 1st story columns. This size is used
for the entire column stack; the amount of reinforcement can be decreased in the upper levels. It was
also shown that the 1st story columns were slender, and the 8-No.10 bars were required for the
factored axial loads and magnified moments at this level.

The reinforcement for the 2nd story columns can be determined using the procedure outlined in
Chapter 5. As was the case for the 1st story columns, the 2nd story columns are slender (use k = 1.2;
see Chapter 5):

( ) − 8.5

⎣ ⎦

1.2 12 × 12

k u

r =

= 34 > 22

0.3 × 16