Page 242 - Handbook of Civil Engineering Calculations, Second Edition

REINFORCED CONCRETE 2.27

5. Alternatively, calculate the allowable bending moment by
assuming that the flange extends to the neutral axis
Then apply the necessary correction. Let C 1 resultant compressive force if the flange
extended to the neutral axis, lb (N); C 2 resultant compressive force in the imaginary ex-
1
tension of the flange, lb (N). Then C 1 /2(1350)(16)(8.40) 90,720 lb (403,522.6 N);
2
C 2 90,720(3.40/8.40) 14,860 lb (66,097.3 N); M 90,720(21.5 8.40/3)
14,860(21.5 5 3.40/3) 1,468,000 in.·lb (165,854.7 N·m).

DESIGN OF A T BEAM HAVING CONCRETE
STRESSED TO CAPACITY

A concrete girder of 2500-lb/sq.in. (17,237.5-kPa) concrete has a simple span of 22 ft (6.7
m) and is built integrally with a 5-in. (127-mm) slab. The girders are spaced 8 ft (2.4 m)
on centers; the overall depth is restricted to 20 in. (508 mm) by headroom requirements.
The member carries a load of 4200 lb/lin ft (61,294.4 N/m), exclusive of the weight of its
web. Design the section, using tension reinforcement only.

Calculation Procedure:
1. Establish a tentative width of web
Since the girder is built integrally with the slab that it supports, the girder and slab consti-
tute a structural entity in the form of a T beam. The effective flange width is established
by applying the criteria given in the ACI Code, and the bending stress in the flange is as-
sumed to be uniform across a line parallel to the neutral axis. Let A f area of flange
2
sq.in. (cm ); b width of flange, in. (mm); b
width of web, in. (mm); t thickness of
flange, in. (mm); s center-to-center spacing of girders.
To establish a tentative width of web, try b
14 in. (355.6 mm). Then the weight of
web 14(15)(150)/144 219, say 220 lb/lin ft (3210.7 N/m); w 4200 220 4420
lb/lin ft (64,505.0 N/m).
Since two rows of bars are probably required, d 20 3.5 16.5 in. (419.1 mm).
The critical shear value is V w(0.5L d) 4420(11 1.4) 42,430 lb (188,728.7 N);
v V/b
d 42,430/[14(16.5)] 184 lb/sq.in. (1268.7 kPa). From the Code, v max
5( f c
) 0.5 250 lb/sq.in. (1723.8 kPa). This is acceptable.
Upon designing the reinforcement, consider whether it is possible to reduce the width
of the web.
2. Establish the effective width of the flange according
to the Code
Thus, /4L /4(22)(12) 66 in. (1676.4 mm); 16t b
16(5) 14 94 in.
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1
(2387.6 mm); s 8(12) 96 in. (2438.4 mm); therefore b 66 in. (1676.4 mm).
3. Compute the moment capacity of the member
at balanced design
Compare the result with the moment in the present instance to identify the controlling stress.
With Fig. 16 as a guide, k b d 0.360(16.5) 5.94 in. (150.876 mm); A f 5(66) 330 sq.in.
2
1
(2129.2 cm ); f c1 1125(0.94)/5.94 178 lb/sq.in. (1227.3 kPa); C b T b /2(330)
(1125 178) 215,000 lb (956,320 N); z b ( /3)(5.94 2 0.94)/(5.94 0.94)
5
1.89 in. (48.0 mm); jd 14.61 in. (371.094 mm); M b 215,000(14.61) 3,141,000 in.·lb
2
(354,870.2 N·m); M ( /8)(4420)(22) (12) 3,209,000 in.·lb (362,552.8 N·m).
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