- 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, Ib (N); C 2 = resultant compressive force in the imaginary ex-
tension of the flange, Ib (N). Then C 1 =^1 X 2 (1350)(16)(8.40) = 90,720 Ib (403,522.6 N); C 2
- 90,720(3.40/8.4O)^2 = 14,860 Ib (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 Nm).
DESIGNOFA TBEAM HAVING CONCRETE
STRESSED TO CAPACITY
A concrete girder of 2500-lb/in^2 (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 Ib/lin ft (61,294.4 N/m), exclusive of the weight of its web.
Design the section, using tension reinforcement only.
Calculation Procedure:
- 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 Af= area of flange in^2
(cm^2 ); 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 Ib/lin ft (3210.7 N/m); w = 4200 + 220 = 4420
Ib/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 Ib (188,728.7 N);
v = V/b'd = 42,430/[l4(l6.5)] = 184 lb/in^2 (1268.7 kPa). From the Code, vmax - 5(/c')°-^5 =
250 lb/in^2 (1723.8 kPa). This is acceptable.
Upon designing the reinforcement, consider whether it is possible to reduce the width
of the web. - Establish the effective width of the flange according
to the Code
Thus,^1 AL =^1 A(22)(\2) = 66 in (1676.4 mm); I6t + bf = 16(5) + 14 = 94 in (2387.6 mm);
s = 8(12) - 96 in (2438.4 mm); therefore b = 66 in (1676.4 mm). - 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, kbd = 0.360(16.5) - 5.94 in (150.876 mm); Af= 5(66) =
330 in^2 (2129.2 cm^2 ); /cl = 1125(0.94)75.94 = 178 lb/in^2 (1227.3 kPa); Cb = Tb =
(^1) X
2 (SSO)(1125 + 178) = 215,000 Ib (956,320 N); zb = (
(^5) X
3 )(5.94 + 2 x 0.94)7(5.94 + 0.94) -
1.89 in (48.0 mm);yW = 14.61 in (371.094 mm); Mb = 215,000(14.61) - 3,141,000 in-lb
(354,870.2 N-m); M= (^1 X 8 )(4420)(22)^2 (12) = 3,209,000 in-lb (362,552.8 N-m).