Structural Concrete - Hassoun

686 Chapter 17 Design of Two-Way Slabs
Figure 17.39
Example 17.13: Analysis by moment distribution. All moments are in K ⋅ ft.
7. Moment distribution factors (DF): For the exterior joint,
DF (slab) =
DF (columns) = K ec
∑ K
= 0.58
K s 140
=
K s + K ec 140 + 192 = 0.42
The columns above and below the slab have the same stiffness; therefore, the distribution factor
of 0.58 is divided equally between both columns, and each takes a DF of 0.58/2 = 0.29. For the
interior joint,
DF (slab) =
K s
140
=
2K s + K ec 2 × 140 + 192 = 0.295
DF (columns) = K ec 192
∑ = K 2 × 140 + 192 = 0.41
Each column will have a DF of 0.41/2 = 0.205.
8. Fixed-end moments: Because the actual LL/DL is less than 0.75, the full-factored load is assumed
to act on all spans.
Fixed − end moment = k ′′ q u l 2 (L 1 ) 2
The factor k ′′ can be determined by the column analogy method: For a unit load w = 1.0 K/ft
over the longitudinal span of 25 ft, the simple moment diagram is shown in Fig. 17.38b. The area