Structural Concrete - Hassoun

676 Chapter 17 Design of Two-Way Slabs
b. Voided volume ( of 14-in. ) rib = 6.54 ft 3 on 3 × 3 −ft 2 area. Total weight of 9-ft 2 area is
1.2(150) 9 × 17 − 6.54 = 1118 lb. Weight per square foot is 1118 = 125 psf.
12 9
c. Factored additional dead plus live load is 1.2(50) + 1.6(100) = 220 psf. Uniform q u at solid
(head) = 255 + 220 ≈ 500 psf. Uniform q u (at ribbed area) = 125 + 220 = 345 psf.
d. Loads on one panel (refer to Fig. 17.34): At the solid head, q u = 0.5(12) + 0.345(21) = 13.22
K/ft. At the ribbed area, q u = 0.345(33) = 11.39 K/ft.
3. Calculate shear and total static moment:
V u (at face of column) =13.22(5.17)+ (11.39)(21) = 188 K
2
M 0 (at midspan) =188(15.67)−13.22(5.17)(13.09)− 11.39(10.5)2 = 1424 K ⋅ ft
2
4. Check punching shear (refer to Fig. 17.35):
a. In the solid head at d/2 from the column face, h = 17 in., d = 17 − 1.25 = 15.75 in.,
c (column) = 20 in., b 0 = 4(20 + 15.75) = 143 in., V u = 11.39(21 ft) + 13.22(12 ft) − 0.5(37.75/
12) 2 = 393.4 K and
(
φV c = φ 2 + 4 ) √f
′
β c b 0 d
β = 1.0
(
φV c = 0.75 2 + 4 ) √ 5000
× 143 × 15.75
1.0 1000
= 716 K
( )
αs d √f
′
φV c = φ + 2
b c b 0 d
0
( ) √ 40 × 15.75 5000
= 0.75
+ 2 × 143 × 15.75
143 1000
= 765 K
φV c = φ4λ √ (√ )
5000
f c ′ b 0 d = 0.75(4)(1.0)
(143)(15.75) =478 K (controls) > V
1000
u
Waffle slab (looking upward).