CAPACITY OFA RECTANGULAR BEAM
The beam in Fig. I4a is made of 2500-lb/in^2 (17,237.5-kPa) concrete. Determine the flex-
ural capacity of the member (a) without applying the basic equations of reinforced-
concrete beam design; (b) by applying the basic equations.
Calculation Procedure:
- Record the pertinent beam data
Thus,// = 2500 lb/in
2
(17,237.5 kPa); /./callow = 1125 lb/in
2
(7756.9 kPa); n = 10; A 5 =
3.95 in
2
(25.485 cm
2
); nAs = 39.5 in
2
(254.85 cm
2
).
- Locate the centroidal axis of the transformed section
Thus, l6(kd)^2 /2 - 39.5(23.5 - kd) = O; kd = 8.58 in (217.93 mm); d - kd = 14.92 in
(378.968 mm).
- Ascertain which of the two allowable stresses governs the
capacity of the member
For this purpose, assume that/ = 1125 lb/in^2 (7756.9 kPa). By proportion, fs =
10(1125)(14.92/8.58) = 19,560 lb/in^2 (134,866 kPa) < 20,000 lb/in^2 (137,900 kPa).
Therefore, concrete stress governs.
- Calculate the allowable bending moment
TJms 9 Jd= 23.5 - 8.58/3 = 20.64 in (524.256 mm); M= Qd=^1 X 2 (1125)(16)(8.58)(20.64) =
1,594,000 in-lb (180,090.1 N-m); or M = Tjd = 3.95(19,560)(20.64) = 1,594,000 in-lb
(180,090.1 N-m). This concludes part a of the solution. The next step comprises part b.
- Compute p and compare with pb to identify
the controlling stress
Thus, from Table l,pb = 0.0101; then;? =Asl(bd) = 3.95/[16(23.5)] = 0.0105>/? 6 There-
fore, concrete stress governs.
Applying the basic equations in the proper sequence yieldspn = 0.1050; by Eq. 30, k =
[0.210 + 0.105^2 ]^05 - 0.105 = 0.365; by Eq. 24, M= (^1 /6)(1125)(0.365)(2.635)(16)(23.5)^2
= 1,593,000 in-lb (179,977.1 N-m). This agrees closely with the previously computed val-
ue of M.
(a) Section (b) Stresses and resultant forces
FIGURE 14
