Structural Concrete - Hassoun

17.8 Analysis of Two-Way Slabs by the Direct Design Method 657
3. a. Calculate α f (α f = EI b /EI s ):
α l (long direction) = 22,453
6860 = 3.27
α s (short direction) = 22,453
8232 = 2.72
α (edge beam) = 15,302
4401 = 3.48
Average α = α fm =
b. β = ratio of long to short clear span.
c. Calculate h:
Min.h =
3.27 + 2.72 × 2 + 3.48
4
22.33
18.33 = 1.22
(22.33 × 12)(0.8 + 60,000∕200,000)
36 + 9(1.22)
Use h = 7in.> 3.5 in. (minimum code limitations).
4. Calculate factored loads:
5. Calculate total static moments:
q u = 292 psf (from Example 17.6)
= 3.05
= 6.3in.
M 0l = 364.0K⋅ ft M 0s = 294.3K⋅ ft (from previous example)
6. Calculate the design moments in the short direction (span = 20 ft): Because the slab is continuous
in this direction, the moments are the same as those calculated in Example 17.23 and shown in
Fig. 17.23 for an interior panel.
7. Calculate the moments in one panel using the coefficients given in Table 17.2 or Fig. 17.14
(Case 3):
Interior negative moment (M nt )=0.7M 0 = 0.7 × 364 =−254.8K⋅ ft
Positive moment within span (M p )=0.57M 0 = 0.57 × 364 =+207.5K⋅ ft
Exterior negtive moment (M ne )=0.16M 0 = 0.16 × 364 =−58.2K⋅ ft
Note: If the modified stiffness method is used, then C = 9528, K t = 1520 E c , K c = 370 E c , K b = 312
E c , K s = 95 E c , K ec = 498 E c ,andα ec = 1.22. The interior negative moment becomes −253.13 K⋅
ft (same as before). The positive moment becomes −173.19 K ⋅ ft (16% decrease) and the exterior
moment becomes −128.16 K ⋅ ft (220% increase).
8. Distribute the panel moments to beam, column, and middle strips:
l 2
= 20
l 1 24 = 0.83 α f 1
= α s = 3.27
Calculate C:
α f1
l 2
l 1
= 3.27 × 0.83 = 2.71 > 1.0
C = ∑ ( 1 − 0.63 x y
) x 3 y
3
Divide the section of the edge beam into two rectangles in such a way as to obtain maximum C.
Use for a beam section 12 by 22 in., x 1 = 12 in., y 1 = 22 in., and a slab section 7 × 15 in., x 2 = 7in.,