REINFORCED CONCRETE DESIGN
M,, for design
5.7 DESIGN O F DOUBLY REINFORCED RECTANGULAR SECTIONS As explained earlier, doubly reinforced sections are generally resorted to in situations
where the cross-sectional dimensions of the beam are restricted (by architectural or other considerations) and where singly reinforced sections (with are not adequate in terms of capacity. Doubly reinforced beams are also used in situations where reversal of moments is likely (as in multi-storeyed frames subjected to lateral loads). The presence of compression reinforcement reduces long- term deflections due to shrinkage (refer Section All compression reinforcement must be enclosed by closed stirrups [Fig. in'order to prevent their possible and to provide some ductility by confinement of concrete.
5.7.1 Design F o r m u l a s
As the dimensions of the beam section already are fixed, the design problem is one of determining the areas of reinforcement required in tension and compression As explained in Section from the design of view, it is necessary to limit the neutral axis depth to This can be done conveniently by
and resolving the factored moment = into components [Fig. 5.71:
(5.13)
BALANCED STRAINS SINGLY "STEEL BEAM" STRESSES
BEAM REINFORCED (balanced)
SECTION SECTION
5.7 Concept underlying the design of a 'balanced' doubly reinforced section where is the limiting moment capacity of a singly reinforced 'balanced' section (without compression steel) [Fig. given by Eq. 5.8, and A M , is the additional moment capacity desired from the compression steel A, and the corresponding additional tension steel
-
which may be visualised as the flanges of an equivalent 'steel beam' [Fig. As the contribution of concrete in compression is entirely accounted for in it does not contribute to AM,,. The distribution ofstrains, given by the condition, [Fig. is identical the and corresponding distributions of stresses are as shown in Fig. and
If denotes the total percentage of tension steel required for a 'balanced' doubly reinforced section, then corresponding to Eq. 5.13; it can also be resolved into two components.
=
where or (given by Eq 5.7) is the tension steel corresponding to
.--
and is that corresponding to A M , , . Evidently, the moment AM, is obtained from a couple comprising a (compressive) force -
and an equal and opposite (tensile) force with a arm The in the compression steel (at the ultimate limit state) depends on the strain (given by which is controlled by the linear strain distribution with the neutral axis located at x = [Fig. Values off,, different grades of steel and typical d ' l d ratios are listed in Table 4.5. (It may be noted from Table 4.5 that the full design yield stress is attained only in the case of mild steel.) Based on the above, following formulas are obtainable:
where R 2 and
Applying Eq. 5.16 and Eq. design aids can be generated to give values of and for a given 2 )- for various combinations of concrete and steel grades and different d'ld ratios. These have been developed in Table A.4 (placed in Appendix A of this book) for the commonly used combination of M 20 and M 25 grades of concrete with 415 steel. Four typical ratios of 0.05,
and 0.20) are covered in Table A.4. The Design Handbook, SP 16 [Ref. gives such Tables for other combinations of concrete and steel grades.
DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE 199
5.7.2 Procedure for Given Determining
For a given rectangular section (with given. b, and given the
moment capacity for a singly reinforced section should be first determined, using Eq. 5.8. If is greater than or equal to the factored moment the section should be as a singly reinforced section
-
as described i n Section Otherwise (for the section should be designed as doubly reinforced section.The value of is determined and the values of AA,, from Eq. assuming a suitable value for d'. The total is then obtained from The bars should be suitably selected such that the A,, actually provided is as close as possible in magnitude to, but not less than Tables and A.6 may be used for this purpose.
If the placement of bars results in a new value of the depth which is significantly different from the original value assumed, and
should be recalculated at this stage itself.
Determining
Using the value of provided, the value of may be calculated
from The value off, bc obtained from the [Fig. o r
and the stress-strain relation [Table 3.21, or alternatively from Table 4.5 by interpolating for the calculated value of
.
The compression bars should be suitably selected such that the provided is as close as possible (but not less than,
a design procedure will result invariably in an adequately 'safe' design with and however, a design check for strength is always desirable.Alternative: Using Design
Design aids (Table A.4, SP: may be used to determine the and for the calculated value of accordingly, and A,, are suitably This is most commonly adopted in practice. However, it should be noted that if the A , actually provided is well in excess of the there is a possibility of ending up with an section (with In order to avoid such a situation, (which is undesirable, and also not permitted by the Code), a correspondingly higher value of A, should by such that the resulting (given by Eq. 4.84).
200 REINFORCED CONCRETE
Check for Control
The limiting ratio for deflection control [Eq. is satisfied by doubly reinforced beams, on account of the modification factor for the compression steel [Table However, in the case of relatively shallow beams, a check for deflection control becomes necessary.
EXAMPLE 5.4
Design flexural reinforcement for the beam in Example 5.1, given its size limited to 250 mm x 400 and that it has to carry, in addition to the loads already mentioned, a concentrated dead load of 30 placed at the point. Assume that beam is subjected to moderate exposure conditions.
SOLUTION
for design
Given b 250 mm, D 400 25 415
Let d D-50 = 350
span 6.0 m (as in Example
Loads: 5.0 + (sclf-weight), at
Due to self-weight. = = 2.5 = 10.0
Factored Load: (5.0
+
2.5+
10.0) x 1.5 = 26.25and 30 45 (at
:.Factored Moment (maximum at midspan):
+ 45 185.6 186
Singly or reinforced section?
For Fe 415 with M 25)
106.3 x Nmm 106
As M the section has to be doubly reinforced, with p,
,
where 41.61- ---
1.201 for Fe 415 with M 25..
Considering a 'balanced section' = A,,1.201 2
where 1051
.
Assuming 20 100 bars for compression steel, 48 (30 mm clear cover+
8 stirrup+
DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE 201
Using 3 bars,
--
27.5Provide 3 nos 28 3 x 616 = 1848
.
Actual 30 mm clear cover and 8d 348 350 assumed earlier
Revising the above calculations with d = 348
1045 mm2,
748 mm2, = 1793 mm2,
= 1848
-
1045 803 mm2 A,,Assuming for = 481348 = 0.138, from Table 4.5;
- 344.7
0.15-0.10
[Alternatively. applying = 0.479 in 4.78,
= = 0.00249
344.7 (from Table 3.2 or design stress-strain curve)]
Using 3 b a s , 19.2
Provide 3 nos 20 [A, 3 x 314 942 869 The proposed section is in Fig. 5.8.
[As an exercise in analysis, the student may verify that section satisfies the design
conditions: and
Alternative method: using design aids Assuming d 350 nun, 50
Referrine to 25 concrete and Fe 415 steel). ,
.
for = 501350 = 0.143 and = 6.073 by interpolation,
-2.042 1787 mm2
REINFORCED CONCRETE DESIGN
Provide 3-28 for tension steel [A,, 3x616 1848 nun2 17871 and for compression steel [A,, 4x201 804 7991.
BEAM SECTION
Fig. 5.8 Doubly reinforced section design - Example 5.4 Design check
To ensure , it suffices to establish Actual d provided: d 400 - 30
-
8 - 2812 = 348d'= 46
345.8 [Table
[Alternatively, = = 0.00253
345.8 (Table
Actual provided: 2.124
Actual provided: = 100x8041 (250x348) = 0.924
Asp, is slightly less than , the section is slightly [This can also be verified by applying Eq. 4.81, which gives
0.505 =
Revised design
To ensure ductile failure,
bd 100 867
Provide for steel A,, 3x314 = 942 867 as
shown in Fig. 100 x x 348) = 1.083 OK.
DESIGN OF BEAMS AND ONE-WAY SLABS FOR FLEXURE 203
for
2.124 and 233
0.842 [Table 5.2 Fig. 4 of Code]
1.083 1.263 [Table 5.3 or Fig. 5 of Applying Eq. 5.5,
20 x 0.842 x 1.263 21.27
60001348 17.24 21.27 -Hence OK.