Page 237 - Handbook of Civil Engineering Calculations, Second Edition

2.22 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

CAPACITY OF A RECTANGULAR BEAM

The beam in Fig. 14a is made of 2500-lb/sq.in. (17,237.5-kPa) concrete. Determine the
flexural capacity of the member (a) without applying the basic equations of reinforced-
concrete beam design; (b) by applying the basic equations.

Calculation Procedure:
1. Record the pertinent beam data
Thus, f c
2500 lb/sq.in. (17,237.5 kPa); f c,allow 1125 lb/sq.in. (7756.9 kPa); n 10;
2
2
A s 3.95 sq.in. (25.485 cm ); nA s 39.5 sq.in. (254.85 cm ).
2. Locate the centroidal axis of the transformed section
2
Thus, 16(kd) /2 39.5(23.5 kd) 0; kd 8.58 in. (217.93 mm); d kd 14.92 in.
(378.968 mm).
3. Ascertain which of the two allowable stresses governs the
capacity of the member
For this purpose, assume that f c 1125 lb/sq.in. (7756.9 kPa). By proportion, f s
10(1125)(14.92/8.58) 19,560 lb/sq.in. (134,866 kPa) < 20,000 lb/sq.in. (137,900 kPa).
Therefore, concrete stress governs.
4. Calculate the allowable bending moment
1
Thus, jd 23.5 8.58/3 20.64 in. (524.256 mm); M Cjd /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.
5. Compute p and compare with p b to identify
the controlling stress
Thus, from Table 1, p b 0.0101; then p A s /(bd) 3.95/[16(23.5)] 0.0105 > p b .
Therefore, concrete stress governs.
Applying the basic equations in the proper sequence yields pn 0.1050; by Eq. 30, k
2 0.5
2
1
[0.210 0.105 ] 0.105 0.365; by Eq. 24, M ( /6)(1125)(0.365)(2.635)(16)(23.5)
1,593,000 in.·lb (179,977.1 N·m). This agrees closely with the previously computed
value of M.

FIGURE 14