DESIGN OFA RECTANGULAR BEAMA beam on a simple span of 13 ft (3.9 m) is to carry a uniformly distributed load, exclu-
sive of its own weight, of 3600 Ib/lin ft (52,538.0 N/m) and a concentrated load of 17,000
Ib (75,616 N) applied at midspan. Design the section, using/; = 3000 lb/in^2 (20,685 kPa).
Calculation Procedure:
- Record the basic values associated with balanced design
There are two methods of allowing for the beam weight: (a) to determine the bending mo-
ment with an estimated beam weight included; (b) to determine the beam size required to
resist the external loads alone and then increase the size slightly. The latter method is used
here.
From Table 1, Kb = 223 lb/in
2
(1537.6 kPa);/^ = 0.0128;^ = 0.874. - Calculate the maximum moment caused by the external loads
Thus, the maximum moment Me =^1 APL + Y*wL^2 =^1 X 4 (17,00O)(13)(12) + !/8(360O)
(13)^2 (12) = 1,576,000 in-lb (178,056.4 N-m). - Establish a trial beam size
Thus, bd^2 = M/Kb = 1,576,000/223 = 7067 in^3 (115,828.1 cm^3 ). Setting b = (^2 A)d, we find
b - 14.7 in (373.38 mm), d = 22.0 in (558.8 mm). Try b = 15 in (381 mm) and d= 22.5 in
(571.5 mm), producing an overall depth of 25 in (635 mm) if the reinforcing bars may be
placed in one row. - Calculate the maximum bending moment with the beam weight
included; determine whether the trial section is adequate
Thus, beam weight = 15(25)(150)/144 = 391 Ib/lin ft (5706.2 N/m); Mw = (%)(391)
(13)^2 (12) = 99,000 in-lb (11,185.0 N-m); M = 1,576,000 + 99,000 = 1,675,000 in-lb
(189,241.5 N-m); Mb = Kbbcf- = 223(15)(22.5)^2 = 1,693,000 in-lb (191,275.1 N-m). The
trial section is therefore satisfactory because it has adequate capacity. - Design the reinforcement
Since the beam size is slightly excessive with respect to balanced design, the steel will be
stressed to capacity under the design load. Equation 25 is therefore suitable for this calcu-
lation. Thus, A 3 = MI(JJd) - 1,675,000/[20,000(0.874)(22.5)] = 4.26 in^2 (27.485 cm^2 ).
An alternative method of calculating A 8 is to apply the value of pb while setting the
beam width equal to the dimension actually required to produce balanced design. Thus,
A 3 = 0.0128(15)(1675)(22.5)/1693 = 4.27 in^2 (27.550 cm^2 ).
Use one no. 10 and three no. 9 bars, for which A 3 = 4.27 in^2 (27.550 cm^2 ) and (^6) min =
12.0 in (304.8 mm).
- Summarize the design
Thus, beam size is 15 * 25 in (381 x 635 mm); reinforcement is with one no. 10 and three
no. 9 bars.
DESIGN OF WEB REINFORCEMENTA beam 14 in (355.6 mm) wide with an 18.5-in (469.9-mm) effective depth carries a uni-
form load of 3.8 kips/lin ft (55.46 N/m) and a concentrated midspan load of 2 kips (8.896
kN). The beam is simply supported, and the clear distance between supports is 13 ft (3.9