(b) Moment diagrams
FIGURE 8
- Compute the shearing stress associated with the assumed
beam size
From the Code for an interior span, Vu = V 2 W 14 L' = !/2(390)(16) = 3120 Ib (13,877.8 N); d =
6 - 1 = 5 in (127 mm); vu = 3120/[12(5)] = 52 lb/in
2
(358.54 kPa); vc = 93 lb/in
2
(641.2
kPa). This is acceptable.
- Compute the two critical moments
Apply the appropriate moment equations. Compare the computed moments with the mo-
ment capacity of the assumed beam size to ascertain whether the size is adequate. Thus,
M,,neg = (VWw 11 Z/^2 = (^1 /n)(390)(16)^2 (12) = 108,900 in-lb (12,305.5 N-m), where the value
12 converts the dimension to inches. Then Mupos =^1 AtW 11 L'^2 = 74,900 in-lb (8462.2 N-m).
By Eq. 10, qmax = 0.6375(0.85)(87/137) = 0.344. By Eq. 6, MMallow = 0.90(12)
(5)^2 (3000)(0.344)(0.797) = 222,000 in-lb (25,081.5 N-m). This is acceptable. The slab
thickness will therefore be made 6 in (152.4 mm).
5. Compute the area of reinforcement associated with each
critical moment
By Eq. 7, bdfc = 12(5)(2.55) = 153.0 kips (680.54 kN); then 2bfcMunes/(l> =
2(12)(2.55)(108.9)/0.90 = 7405 kips^2 (146,505.7 kN^2 ); Asnes = [153.0 - (153.0^2 -
7405)°-^5 ]/50 = 0.530 in^2 (3.4196 cm^2 ). Similarly, Aspos = 0.353 in^2 (2.278 cm^2 ).
(a) Arrangement of reinforcing bars
Straight bar
•Trussed bar
Straight bar
-Trussed bar from
adjoining span
L
1
= clear span = 16' (4.9 m)
