Structural Concrete - Hassoun

658 Chapter 17 Design of Two-Way Slabs
and y 2 = 15 in.
(
C =
1 − 0.63 × 12
22
= 9528 in. 4
β t = E cb C
2E cs I s
= 9528
2 × 6860 = 0.69
) ( )
12 3 (
× 22
+ 1 − 0.63 × 7
3
15
) ( )
7 3 × 15
3
a. Distribute the interior negative moment, M ni : Referring to Table 17.5 and by interpolation, the
percentage of moment assigned to the column strip (for l 2 /l 1 = 0.83 and α f1 l 2 /l 1 > 1.0 is 80%).
Column strip = 0.8 × 254.8 =−203.8K⋅ ft
Middle strip = 0.2 × 254.8 =−51.0K⋅ ft
Because α f1
l 2 ∕l 1 > 1.0, 85% of the moment in the column strip is assigned to the beam. Therefore,
Beam = 0.85 × 203.8 =−173.3K⋅ ft
Column strip = 0.15 × 203.18 =−30.6K⋅ ft
Middle strip =−51.0K⋅ ft
b. Distribute the positive moment, M p : Referring to Table 17.5 and by interpolation, the percentage
of moment assigned to the column strip is 80% (85% of this value is assigned to the beam).
Therefore,
Beam =(0.85)(0.8 × 207.5) =+141.1K⋅ ft
Column strip =(0.15)(0.8 × 207.5) =24.9K⋅ ft
Middle strip = 0.2 × 207.5 =+41.5K⋅ ft
c. Distribute the exterior negative moment, M ne : Referring to Table 17.5 and by interpolation,
the percentage of moment assigned to the column strip (for l 2 /l 1 = 0.83, α f1 l 2 /l 1 > 1.0, and
β t = 0.69) is 94%, and 85% of the moment is assigned to the beam. Therefore,
Beam =(0.85)(0.94 × 58.2) =−46.5K⋅ ft
Column strip =(0.15)(0.94 × 58.2) =−8.2K⋅ ft
Middle strip = 0.06 × 58.2 =−3.5K⋅ ft
17.9 DESIGN MOMENTS IN COLUMNS
When the analysis of the equivalent frames is carried out by the direct design method, the moments
in columns due to the unbalanced loads on adjacent panels are obtained from the following equation,
which is specified by the ACI Code, Section 8.10.7.2:
M u = 0.07[(q Du + 0.5q Lu )l 2 l 2 n − q ′ Du l′ 2 (l′ n) 2 ]
(17.22a)
If the modified stiffness method using K ec and α ec is used, then the moment M u is computed as
follows:
M u = 0.08[(q Du + 0.5q Lu )l 2 l 2 n − q ′ Du l′ 2 (l′ n) 2 ]
1 + 1∕α ec
(17.22b)