and L-beams were in Section connection
between the slab and in cast construction makes the two act integrally, so that portion of functions as a flange of the beam. It should be noted that the is effective only when it is on the compression side, when the beam is in a 'sagging' mode of 'hogging') the on top. if the beam is (inverted T-beam) and it is subjected to 'hogging' moments (as in a cantilcvcr), the T-beam action is as flange is under compression.
Ideal flanged beam action occurs when the flange dimensions are small while the beam is deep in closely spaced long-span bridge girders in a T-beam bridge. beam is invariably heavily reinforced in such cases.
T r a n s v e r s e Reinforcement in F l a n g e
The integral action the and the web is usually by the transverse bars in the slab and the stirrups in the beam. In the case of isolated flanged (as in of staircases), the detailing of reinforcement depicted in Fig. may be adopted. The overhanging portions of the slab should be designed ascantilevers the provided accordingly.
.
Adequate transverse reinforcement must be near the of the flanne.
reinforcement is usually in the form of negative reinforcement in the continuous slabs which span across and the flanges of the T-beams.
is not the case (as in slabs the main bars to the beam), the Code specifies that transverse reinforcement should be provided in the flange of T-beam (or as shown in area of such steel should be not less than 60 percent of the main area of steel provided at the of the slab, and should extend on either side of the beam to a distance not less than fourth of the span of the
Where the slab are beam action cannot be
unless connectors are at the and the
slab.
204 REINFORCED CONCRETE DESIGN
Fig. 5.9 Detailing of flanged beams to ensure integral action of slab and beam.
Design Procedure
In the case of a continuous flanged beam, the negative moment at face of the support generally cxcecds maximum moment (at or near the midspan), and governs the proportioning of the beam cross-section. In such cases of negative moment, if the slab is located on top of the beam (as is usually the case), flange is under flexural tension and the concrete in the flange is rendered ineffectivc. section at the is therefore to bc designed as a rectangular section for the factored negative moment'
.
Towards the of the beam, however, the as a proper flanged beam (with the flange under flexural compression). As the width of web and the overall depth D are already fixed design considerations at the support, all that remains to be determined is the area of reinforcing steel; the effective is determined as suggested by the CodeIn simply supported flanged beams, the web dimensions must also b e designed (if not otherwise specified). The width of the web is generally fixed as 250 300 350 (as for a rectangular and overall depth to he approximately to An appmximate estimate of the area of tension steel A,, can be obtained as follows:
where the lever arm may be taken approximately as or whichever is larger. If convenient, the reinforcement should bc accommodated in one layer - although, often this may not be possible. When steel is provided in more than one layer, the effective depth gets reduced.
The determination of the actual reinforcenicnt in a flanged beam depends on the location of neutral axis which, of course, should be limited to If
In such it is desirable to the steel the top of the web 'across the effective width the flange, to protect the integral flange as recommended by the ACI Code. additional be provided in the flange region far this
DESIGN OF BEAMS AND ONE-WAY FOR FLEXURE 205
exceeds for a singly reinforced flanged section, the depth o f t h e section should be suitably increased: otherwise, a doubly reinforced Section is to b e designed.
Neutral Axis Flange
This is, by far, the most common situation encountered in building design. Because of very large compressive concrete area by the flange in T-beams and
of usual proportions, the neutral axis lies within the flange
whereby the behaves like a rectangular section having width effective depth
A simple way of first Dj is by verifying where
is the limiting ultimate moment of resistance for the condition =
It may be noted that the above equation is meaningful only if In rare situations involving very thick flanges and relatively shallow beams, may be less than In such cases, is obtained by substituting in place of in Eq. 5.19.
Neutral Axis within Web
When M,, , it follows that The accurate of
can be somewhat laborious'. As explained in Chapter 4, the contributions of compressive forces and in the and 'flange' may b e accounted for
as follows:
(5.20) where,
(5.21)
and the equivalent flange thickness is equal to or less than depending on whether exceeds or not.
For the value of the ultimate moment of resistance corresponding to = may be first computed. If
As an alternative to this procedure, a design based an the of A,, may be assumed, and the resulting section to determine design becomes acceptable M,,
2 0 6 CONCRETE DESIGN
the appropriate value or the for (given by Eq. 5 23)
-
5.20, the resulting (in terms of the unknown can be solved to yield the correct value of Corresponding to this value of , the values of and can be computed [Eq. 5.21, and the required A,, obtained by solving the forceEXAMPLE 5.5
Design the interior beam in the floor system in Example 5.3 [Fig. Assume that the beam is subjected to moderate exposure conditions. Use 415 steel.
SOLUTION
.
The slab is one-way, spanning between the beams, which are simply supportedand hence behave as T-beams = 8230 160 300
width (CI 23.1.2 Code):
2632
which is acceptable as it is less than span of 3700).
.
Assume overall depth 1 15 = 550 effective depth deffective span 8.0 0.23 m 8.23 m (less than 8.0 0.5 = 8.5 for design
.
Distributed loads from (refer 5.3):5.5 m 20.35
14.8
Additional dead load due to self weight of web:
2.93
.
:.Factored load =+
14.8+
2.93) 57.12.
Factored moment (maximum at= 484
Assuming a lever arm equal to the larger of 450 and - = 420 450
DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE 207
or, providing 4 bars, - - 30.8 [Alternatively, this is obtainable Table
It may be observed that the bars (either 3-364 or 4-324) can be accommodated in layer, given = 300 mm. Assuming 32 bars and 8 stirrups,
Actual d 494
(clear cover shall not be less than the diameter of the bar) A,,
= 237 160
.
Assuming axis to be located= [for M 25 concrete]
484
Hence, the neutral axis is located definitely within the flange
.
Accordingly, designing the as a singly reinforced rectangular sectionwith 2632 and 494 nun,
(which incidentally is about 6 percent less than the value calculated earlier).
Provide 2-32 2-28 bars
[A,, = (2x804)
+
(2x616) = 2840 2808 nun2]].The cross-section of the beam, showing the location of bars, is depicted in Fig. 5.10.
Fig. 5.10 T-beam of Example 5.5
208 REINFORCED CONCRETE DESIGN
Check for Control
Ignoring the contribution of flanges (conservative) Section 1.92;
494 2840
k, 0.844 [Table
= 82301494 16.66 16.88 -Hence. OK.
EXAMPLE 5.6
A continuous T-beam has the cross-sectional dimensions shown in Fig. The web dimensions been from the consideration of negative moment at support and shear strength requirements. The span is 10 m i n d moment at under loads is 800 Determine the flexural reinforcement at Consider Fe 415 steel. the beam is subjected to moderate exposure conditions.
SOLUTION
A,,
Actual flange width provided = 1500 mm; 100 b , 300 Maximum width (0.7 x 300
+
(6 x 100) 20671500 mm
Assuming d = 650 and a lever arm equal to the larger of 585 and
d - 600 600
34.3
Providing 4 bars,
-
As bars cannot be accommodated in one layer within the width 300 two layers are required.
Assumine a reduced d 625 625
-
10012 = 575 mm.Provide 5-324 [A,, 804 x 5 4020 mm2] with 3 bars in the lower layer plus 2 bars in the upper layer, with a clear vertical separation of 32 mm
-
as shown in Fig. 5.1 Assuming 8 stirrups and a clear 32 cover to stirrups,DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE 209
(a) given section (b) proposed reinforcement Fig. of Example 5.6
0.479 x 618
. .
Assuming M 25 concrete, 100 the 25 = 0 362 x 25 x 1500 x x -= 800
and
-
0.416+ -
0.362 25
0.447 25 (1500
-
296 100 mm.
Considering = 233-
= (2715 x +
(13410 x x (618
-
.
Evidently, 7 for which 3+
0.65 x 100) =+
871650) N.800 x = -
+ +
871650) x+
2790229.5+
Solving this quadratic equation,
109.3
x
= 296Applying =
=
+=
The reinforcement (5-32 A,, 4020 based on the approximate estimate of A,, [Fig. 5.1 is evidently adequate and appropriate.
REINFORCED CONCRETE