Page 231 - Handbook of Civil Engineering Calculations, Second Edition

2.16 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

6. Select the reinforcing bars, and locate the bend points
For positive reinforcement, use no. 4 trussed bars 13 in. (330.2 mm) on centers, alternat-
ing with no. 4 straight bars 13 in. (330.2 mm) on centers, thus obtaining A s 0.362 sq.in.
2
(2.336 cm ).
For negative reinforcement, supplement the trussed bars over the support with no. 4
2
straight bars 13 in. (330.2 mm) on centers, thus obtaining A s 0.543 sq.in. (3.502 cm ).
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).”
7. Investigate the adequacy of the reinforcement beyond
the bend points
In accordance with the Code, A min A t 0.0020bt 0.0020(12)(6) 0.144 sq.in.
2
(0.929 cm ).
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 re-
inforcement at section A (Fig. 8). Then L
/5 3.2 ft (0.98 m); AM 8 3.2 0.5 4.3
ft (1.31 m).
2
The conditions immediately to the left of A are M u M u,pos /2w u (AM) 74,900
1
2
2
1 /2(390)(4.3) (12) 31,630 in.·lb (3573.56 N·m); A s,pos 0.181 sq.in. (1.168 cm ); q
0.181(50)/[12(5)(3)] 0.0503. By Eq. 5, M u,allow 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: M u,allow 74,900(0.181)/0.353 38,400 in.·lb (4338.43 N·m).
The trussed bars in effect terminate as negative reinforcement at B, where O B
3.2 0.33 0.5 2.37 ft (72.23 m). The conditions immediately to the right of B are
2
|M u | M u,neg 12(3120 2.37 /2 390 2.37 ) 33,300 in.·lb (3762.23 N·m).
1
Then A s,neg 0.362 sq.in. (2.336 cm ). As a conservative approximation, M u,allow
2
108,900(0.362)/0.530 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, U u,allow 800 lb/sq.in. (5516 kPa).
If CDE in Fig. 8b 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 0.354L
; V u at N/V u at support
0.354L
/(0.5L
) 0.71; V u at N 0.71(3120) 2215 lb (9852.3 N). By Eq. 18, u u
V u /( ojd) 2215/[0.85(1.45)(0.875)(5)] 411 lb/sq.in. (2833.8 kPa). This is accept-
able. It is apparent that the maximum bond stress in the top bars has a smaller value.

ANALYSIS OF A TWO-WAY SLAB
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 w u lb/sq.ft. (kPa). Calculate the
ultimate unit moment m u for which the slab must be designed.