Calculation Procedure:- Identify the controlling stress
Thus, M= (^1 / 8 )(4100)(22)^2 (12) = 2,977,000 in-lb (336,341.5 N-m). From the previous cal-
culation procedure, Mb = 3,141,000 in-lb (354,870.2 N-m). Since Mb > M, the beam size is
slightly excessive with respect to balanced design, and the steel will therefore be stressed
to capacity under the stipulated load. - Compute the area of reinforcement
As an approximation, this area may be found by applying the value ofjd associated with
balanced design, although it is actually slightly larger. From the previous calculation pro-
cedure,^ = 14.61 in (371.094 mm). Then A 5 = 2,977,000/[20,000(14.6I)] = 10.19 in^2
(65.746 cm^2 ). - Verify the design by computing the member capacity
Thus, nAs = 101.9 in^2 (657.46 cm^2 ); kd = (330 x 2.5 + 101.9 x 16.5)7(330 + 101.9) = 5.80
in (147.32 mm); z = (^5 / 3 )(5.80 + 2 x 0.80)7(5.80 + 0.80) = 1.87 in (47.498 mm); Jd= 14.63
in (371.602 mm); Mallow = 10.19(20,000)(14.63) = 2,982,000 in-lb (336,906.4 N-m). This
is acceptable.
REINFORCEMENT FOR DOUBLY
REINFORCED RECTANGULAR BEAMA beam of 4000-lb/in
2
(27,580-kPa) concrete that will carry a bending moment of 230
ft-kips (311.9 kN-m) is restricted to a 15-in (381-mm)'width and a 24-in (609.6-mm) total
depth. Design the reinforcement.
Calculation Procedure:- Record the pertinent beam data
In Fig. 18, where the imposed moment is substantially in excess of that corresponding to
balanced design, it is necessary to reinforce the member in compression as well as ten-
sion. The loss in concrete area caused by the presence of the compression reinforcement
may be disregarded.
Since plastic flow generates a transfer of compressive stress from the concrete to the
FIGURE 18. Doubly reinforced beam.