Page 245 - Handbook of Civil Engineering Calculations, Second Edition

2.30 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

Since plastic flow generates a transfer of compressive stress from the concrete to the
steel, the ACI Code provides that “in doubly reinforced beams and slabs, an effective
modular ratio of 2n shall be used to transform the compression reinforcement and com-
pute its stress, which shall not be taken as greater than the allowable tensile stress.” This
procedure is tantamount to considering that the true stress in the compression reinforce-
ment is twice the value obtained by assuming a linear stress distribution.
2
Let A s area of tension reinforcement, sq.in. (cm ); A s
area of compression rein-
2
forcement, sq.in. (cm ); f s stress in tension reinforcement, lb/sq.in. (kPa); f s
stress in
compression reinforcement, lb/sq.in. (kPa); C
resultant force in compression rein-
forcement, lb (N); M 1 moment capacity of member if reinforced solely in tension to
produce balanced design; M 2 incremental moment capacity resulting from use of com-
pression reinforcement.
The data recorded for the beam are f c 1800 lb/sq.in. (12.411 kPa); n 8; K b 324
lb/sq.in. (2234.0 kPa); k b 0.419; j b 0.860; M 230,000(12) 2,760,000 in.·lb
(311,824.8 N·m).
2. Ascertain whether one row of tension bars will suffice
Assume tentatively that the presence of the compression reinforcement does not apprecia-
bly alter the value of j. Then jd 0.860(21.5) 18.49 in. (469.646 mm); A s M/( f s jd)
2
2,760,000/[20,000(18.49)] 7.46 sq.in. (48.132 cm ). This area of steel cannot be accom-
modated in the 15-in. (381-mm) beam width, and two rows of bars are therefore required.
3. Evaluate the moments M 1 and M 2
2
2
Thus, d 24 3.5 20.5 in. (520.7 mm); M 1 K b bd 324(15)(20.5) 2,040,000
in.·lb (230,479.2 N·m); M 2 2,760,000 2,040,000 720,000 in.·lb (81,345.6 N·m).
4. Compute the forces in the reinforcing steel
For convenience, assume that the neutral axis occupies the same position as it would in
the absence of compression reinforcement. For M 1 , arm j b d 0.860(20.5) 17.63 in.
(447.802 mm); for M 2 , arm 20.5 2.5 18.0 in. (457.2 mm); T 2,040,000/17.63
720,000/18.0 155,700 lb (692,553.6 N); C
40,000 lb (177,920 N).
5. Compute the areas of reinforcement and select the bars
2
Thus A s T/f s 155,700/20,000 7.79 sq.in. (50.261 cm ); kd 0.419(20.5) 8.59 in.
(218.186 mm); d kd 11.91 in. (302.514 mm). By proportion, f s
2(20,000)
(6.09)/11.91 20,500 lb/sq.in. (141,347.5 kPa); therefore, set f s
20,000 lb/sq.in.
2
(137,900 kPa). Then, A s
C
/f s
40,000/20,000 2.00 sq.in. (12.904 cm ). Thus ten-
2
sion steel: five no. 11 bars, A s 7.80 sq.in. (50.326 cm ); compression steel: two no. 9
2
bars, A s 2.00 sq.in. (12.904 cm ).
DEFLECTION OF A CONTINUOUS BEAM

The continuous beam in Fig. 19a and b carries a total load of 3.3 kips/lin ft (48.16 kN/m).
When it is considered as a T beam, the member has an effective flange width of 68 in.
(1727.2 mm). Determine the deflection of the beam upon application of full live load, us-
ing f c
2500 lb/sq.in. (17,237.5 kPa) and f y 40,000 lb/sq.in. (275,800 kPa).

Calculation Procedure:

1. Record the areas of reinforcement
2
2
At support: A s 4.43 sq.in. (28.582 cm ) (top); A s
1.58 sq.in. (10.194 cm ) (bottom).
2
At center: A s 3.16 sq.in. (20.388 cm ) (bottom).