Structural Concrete - Hassoun

17.12 Equivalent Frame Method 685
Therefore, slab stiffness is
K s = 4.1E cs × 10,240
= 140E
25 × 12
cs
4. Determine the column stiffness, K c :
( )
Ecb
K c = k ′ I c
× 2
l c
for columns above and below the slab.
k ′ = column stiffness factor
l c = 12 ft I c = (20)4 = 13,333 in.4
12
The stiffness factor, k ′ , can be determined as follows:
(
k ′ 1
= l c + Mc
)
A a I a
For the column, c = l c /2 and M = 1.0(l c /2) = l c /2.
A a = l c − h = 12 − 8
12 = 11.33
I a = (l c − h)3
12
= (11.33)3
12
= 121.2
( )( )
⎡
k ′ ⎢
1 × 12 12
1
= 12 ⎢
⎢
11.33 + 2 2
121.2
⎣
⎤
⎥
⎥ = 4.62
⎥
⎦
K c = 4.62E cb × 13,333
12 × 12 × 2 = 856E cb
In a flat-plate floor system, the column stiffness, K c , can be calculated directly as follows:
K c
E cb
=
I c
l c − h +
3I c l2 c
(l c − h) 3 (17.33)
5. Calculate the torsional stiffness, K t , of the slab at the side of the column:
∑
9Ecs C
K t =
and C = ∑ ( 1 − 0.63 x ) x 3 y
l 2 (1 − c 2 ∕l 2 ) 3 y 3
In this example, x = 8.0 in. (slab thickness) and y = 20 in. (column width). See Fig. 17.17.
(
c = 1 − 0.63 × 8 ) ( (8) 3 )
× 20
= 2553 in. 4
20 3
9E
K t =
cs × 2553
(20 × 12)[1 − 20∕(20 − 12)] = 124E 3 cs
For two adjacent slabs, K t = 2 × 124E cs = 248E cs .
6. Calculate the equivalent column stiffness, K ec :
or K ec = 192E cs .(E cb = E cs in this problem).
1
= ∑ 1 + 1 1 1
= +
K ec Kc K t 856E cb 248E cs