- Select the reinforcing bars, and locate the bend points
For positive reinforcement, use no. 4 trussed bars 13 in (330.2 mm) on centers, alternating
with no. 4 straight bars 13 in (330.2 mm) on centers, thus obtaining A 5 = 0.362 in^2 (2.336
cm^2 ).
For negative reinforcement, supplement the trussed bars over the support with no. 4
straight bars 13 in (330.2 mm) on centers, thus obtaining A 5 = 0.543 in^2 (3.502 cm^2 ).
The trussed bars are usually bent upward at the fifth points, as shown in Fig. 8a. The
reinforcement satisfies a section of the ACI Code which requires that "at least... one-
fourth the positive moment reinforcement in continuous beams shall extend along the
same face of the beam into the support at least 6 in (152.4 mm)." - Investigate the adequacy of the reinforcement beyond
the bend points
In accordance with the Code, Amin = At = 0.0020^ = 0.0020(12)(6) = 0.144 in^2 (0.929
cm^2 ).
A section of the Code requires that reinforcing bars be extended beyond the point at
which they become superfluous with respect to flexure a distance equal to the effective
depth or 12 bar diameters, whichever is greater. In the present instance, extension =
12(0.5) = 6 in (152.4 mm). Therefore, the trussed bars in effect terminate as positive rein-
forcement at section A (Fig. 8). Then Z//5 = 3.2 ft (0.98 m); AM = 8 - 3.2 - 0.5 = 4.3 ft
(1.31m).
The conditions immediately to the left of A are Mu = M 14 pos - Vzwu(AM)^2 = 74,900 -
(^1) /
2 (390)(4.3)
(^2) (12) = 31,630 in-lb (3573.56 N-m); A
spos = 6.181 in
(^2) (1.168 cm (^2) ); q =
0.181(50)/[12(5)(3)] = 0.0503. By Eq. 5, MMallow = 0.90(0.181)(50,000)(5)(0.970) =
39,500 in-lb (4462.7 N-m). This is acceptable.
Alternatively, Eq. 11 may be applied to obtain the following conservative approxima-
tion: MM>allow = 74,900(0.181)70.353 = 38,400 in-lb (4338.43 N-m).
The trussed bars in effect terminate as negative reinforcement at B, where O'E = 3.2-
0.33 - 0.5 = 2.37 ft (72.23 m). The conditions immediately to the right of B are \MU\ =
M,,neg - 12(3120 x 2.37 - V 2 x 390 x 2.37
2
) = 33,300 in-lb (3762.23 N-m). Then^,neg =
0.362 in^2 (2.336 cm^2 ). As a conservative approximation, MM>allow = 108,900(0.362)70.530
- 74,400 in-lb (8405.71 N-m). This is acceptable.
8. Locate the point at which the straight bars at the top may
be discontinued
9. Investigate the bond stresses
In accordance with Eq. 19, t/M>allow = 800 lb/in^2 (5516 kPa).
If CDE in Fig. 86 represents the true moment diagram, the bottom bars are subjected
to bending stress in the interval NN'. Manifestly, the maximum bond stress along the bot-
tom occurs at these boundary points (points of contraflexure), where the shear is relative-
ly high and the straight bars alone are present. Thus MN= 0.3541'; Vu at Nl Vu at support
= 0.354Z//(0.5Z/) - 0.71; Vu at TV= 0.71(3120) = 2215 Ib (9852.3 N). By Eq. 18, uu =
VJ(JS 1 OJd) = 2215/[0.85(1.45)(0.875)(5)] = 411 lb/in^2 (2833.8 kPa). This is acceptable. It
is apparent that the maximum bond stress in the top bars has a smaller value.
ANALYSIS OFA TWO-WAYSLAB
BY THE YIELD-LINE THEORY
The slab in Fig. 9a is simply supported along all four edges and is isotropically rein-
forced. It supports a uniformly distributed ultimate load of wu Ib/ft^2 (kPa). Calculate the
ultimate unit moment mu for which the slab must be designed.