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.
Let A 5 - area of tension reinforcement, in^2 (cm^2 ); A's = area of compression reinforce-
ment, in^2 (cm^2 ); fs = stress in tension reinforcement, lb/in^2 (kPa);/' = stress in compres-
sion reinforcement, lb/in^2 (kPa);<C" = resultant force in compression reinforcement, Ib
(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 compression reinforce-
ment.
The data recorded for the beam are/c = 1800 lb/in
2
(12.411 kPa); n = S;Kb = 324 lb/in
2
(2234.0 kPa); kb = 0.419; jb = 0.860; M= 230,000(12) = 2,760,000 in-lb (311,824.8 N-m).
- 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/ Then// = 0.860(21.5) = 18.49 in (469.646 mm); A 3 = Ml(fjd) =
2,760,000/[20,000( 18.49)] = 7.46 in^2 (48.132 cm^2 ). This area of steel cannot be accom-
modated in the 15-in (3 81-mm) beam width, and two rows of bars are therefore required. - Evaluate the moments M 1 and M 2
Thus, d = 24 - 3.5 = 20.5 in (520.7 mm); M 1 = KJ>f = 324(15)(20.5)^2 = 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). - 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 =jyd = 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 Ib (692,553.6 N); C = 40,000 Ib (177,920 N). - Compute the areas of reinforcement and select the bars
Thus A 3 = TIf 8 = 155,700/20,000 = 7.79 in^2 (50.261 cm^2 ); kd = 0.419(20.5) = 8.59 in
(218.186 mm); d - kd = 11.91 in (302.514 mm). By proportion, fs' = 2(20,000)
(6.09)/11.91 = 20,500 lb/in^2 (141,347.5 kPa); therefore, set// - 20,000 lb/in^2 (137,900
kPa). Then, A 3 = CIf 8 ' = 40,000/20,000 = 2.00 in^2 (12.904 cm^2 ). Thus tension steel: five
no. 11 bars, As = 7.80 in^2 (50.326 cm^2 ); compression steel: two no. 9 bars, As = 2.00 in^2
(12.904 cm^2 ).
DEFLECTION OFA CONTINUOUS BEAM
The continuous beam in Fig. I9a 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// = 2500 lb/in^2 (17,237.5 kPa) and/, = 40,000 lb/in^2 (275,800 kPa).
Calculation Procedure:- Record the areas of reinforcement
At support: A 3 = 4.43 in^2 (28.582 cm^2 ) (top); A 3 = 1.58 in^2 (10.194 cm^2 ) (bottom). At cen-
ter: A 5 = 3.16 in
2
(20.388 cm
2
) (bottom).