4.3 Short-term (immediate) deflection
4.3.1 Effects of cracking
If reinforced concrete beams are uncracked, deflection analysis is a simple matter and can be done readily using Equations 4.2(1), 4.2(2) or 4.2(3), as the case may be.
However, reinforced concrete beams crack even under service or working load
Chapter 4 Deflection of beams and crack control 77
Table 4.2(2) Values ofαPfor concentrated loads (in all cases, deflection is positive downwards) Support and
deflection at C
C
deflection at C aPfor
Table 4.2(3) Values ofαMfor moments
C
78 Part I Reinforced concrete
Table 4.2(4) Deflection limits
Type of member
Deflection to be considered
Deflection
limitation (D/Lef) for spans∗
Deflection
limitation (D/Lef) for cantilevers∗
All members Total deflection 1/250 1/125
1/500 1/250 Bridge members The live load and
impact deflection
1/800 1/400
a In Equation 4.5(8), use total g and total q.
∗ Note: See Table 2.3.2 in AS 3600-2009 for qualifications.
conditions and cracking occurs at discrete sections along the beam at quite
unpredictable positions. Thus, it may be futile to resort to rigorous analytical methods for deflection calculations. Instead, most researchers choose to solve these problems using one semi-empirical method or another. Loo and Wong (1984) studied the relative merits and accuracy of a group of nine such methods. They came to the conclusion that the so-called effective moment of inertia approach is a convenient and accurate one. It is convenient because the traditional deflection formulas (as given in Equations 4.2(1), 4.2(2) and 4.2(3)) are readily applicable with some modifications to the bending rigidity term or EI.
For a cracked reinforced concrete beam, the E term is replaced by Ecas defined in Equations 2.1(5) or 2.1(6), and for I an effective value Iefshould be used, where in general
Icr ≤ Ief ≤ Ig Equation 4.3(1)
in which Igis the gross moment of inertia of the uncracked beam section and Icris that of a fully cracked beam (see Appendix A).
4.3.2 Branson’s effective moment of inertia
The formula for calculating the effective moment of inertia (Ief) adopted in the Standard and several other major codes of practice (including the American Concrete Institute) is originally attributed to Branson (see Loo and Wong 1984). The empirical formula, which takes into consideration the stiffening effects of the concrete in tension between cracks (i.e. tension stiffening), is explicit and all-encompassing. That is
Ie f = Icr + (Ig− Icr) the Branson formula in its original form underestimates the deflection of very lightly
Chapter 4 Deflection of beams and crack control 79
reinforced beams (see Gilbert 2008). The quantities Mcr, Icrand Msin Equation 4.3(2) are discussed in detail below.
The quantity Msis the maximum bending moment at the section, based on the short-term serviceability load under consideration. For simply supported beams, Ms
may be taken as the mid-span moment; for cantilever beams it should be taken as the root moment.
The moment of inertia for a fully cracked section (Icr) can be determined in the usual manner, once the position of the neutral axis of the transformed section is known. A brief discussion on this topic may be found in Appendix A. For the specific case of a singly reinforced rectangular section
Icr = bd3 12
4k3+ 12pn(1 − k)2
Equation 4.3(3) in which the elastic neutral axis parameter
k=
( pn)2+ 2pn − pn Equation A(5)
Note that Equation 4.3(3) may be used for a conservative evaluation of Icrfor a doubly reinforced rectangular section.
Finally in Equation 4.3(2), the cracking moment Mcr = Ig
yt
fct. f − fcs
Equation 4.3(4)
in which ytis the distance between the neutral axis and the extreme fibre in tension of the uncracked section; fcsis the maximum shrinkage-induced tensile stress on the uncracked section, which may be computed using a rather tedious process as detailed in the Standard in Clause 8.5.3.1 in conjunction with Clause 3.1.7. By virtue of Equation 2.1(2)
This equation is applicable for beam sections of any given shape.
In Clause 3.1.7.2 of the Standard, the formula for shrinkage strains has an accuracy range of±30%. Relevant provisions in ACI 318-2008 ignore the shrinkage effects.
In view of the cumbersome computational process involved, especially in determining shrinkage-induced stress, Clause 8.5.3.1 of the Standard recommends the following approximate equations for flanged beams.
(a) For p= Ast
80 Part I Reinforced concrete
Note that Equations 4.3(6) and 4.3(7) are valid for flanged beams with the flange in compression. Otherwise, for example, for an inverted T or L beam, bef= bw. Similarly, for a rectangular beam bef = bw= b.
It is worth mentioning here that some writers opt for a conservative and easy solution and simply recommend Ief = Icr(see, for example, Warner et al (2006)).
4.3.3 Load combinations
After replacing EI with EcIef, Equations 4.2(1), 4.2(2) and 4.2(3), which were developed for homogeneous uncracked beams, may be used for cracked reinforced concrete beams. It may be assumed that, under service load, the principle of superposition holds.
The load combination formulas recommended in the Standard for serviceability design have been discussed in Section 1.3.2. Using Equation 1.3(6),1.3(7) or 1.3(8), as the case may be, the dead, live and wind loads can be combined accordingly and the bending moment (Ms) can be readily determined. With this value of Ms, the effective moment of inertia (Ief) is computed using Equation 4.3(2). Needless to say, the deflection under combined loads may be obtained as the sum of the individual effects, each of which is calculated using the same Ief.
4.3.4 Illustrative example
Problem
Given a simply supported beam (with Lef= 10 m, b = 350 mm, d = 580 mm, D = 650 mm and pt= 0.01), compute the mid-span deflection under a combination of dead load including self-weight (g= 8 kN/m) and live load (q = 8 kN/m). Take Ec= 26000 MPa, Es= 200000 MPa and fc= 32 MPa; assume that the beam forms part of a domestic floor system; ignore the shrinkage effects.
Solution
The gross moment of inertia (see Figure 4.3(1)) Ig = 350 ×6503
26000 = 7.69 and pt= 0.01, Equation A(5) (from Appendix A) yields k= 0.3227. In turn, Equation 4.3(3) gives Icr= 3174 × 106mm4.
Chapter 4 Deflection of beams and crack control 81
650
pt= 0.01 350
580
Figure 4.3(1) Cross-sectional details of the example simply supported beam Note: all dimensions are in mm
For domestic floor systems,ψsandψlare given in Table 1.3(1) as 0.7 and 0.4, respectively. For short-term deflection, Equation 1.3(8)(iii) governs and we have combined load= g + 0.7q = 8 + 0.7 × 8 = 13.6 kN/m
The moment at mid-span is Ms = 13.6 ×102
8 = 170 kNm Equation 4.3(2) thus gives
Ie f = 3174 × 106+ (8010 − 3174) × 106×
83.65 170
3
= 3750 × 106 mm4< Ig, which is acceptable.
Finally, Equation 4.2(1) in conjunction with Table 4.2(1) gives D= 5
384× 13.6 × 100004
26000× 3750 × 106 = 18.2 mm
It may be apparent in Table 4.2(4) that the immediate deflection under dead and live loads is not a criterion for serviceability design. Its calculation, however, is essential in the analysis of long-term deflection as discussed in Section 4.4 and the total
deflections under normal or repeated loading (Section 4.6).
4.3.5 Cantilever and continuous beams
The discussion so far in this section has centred mainly on simply supported beams.
For cantilever beams, the procedure remains basically unchanged, except that for Lef, Equation 4.2(5) should be used, and Msin Equation 4.3(2) should be the bending moment at the root of the cantilever, or at the face of the support girder, column or wall, as the case may be.
For continuous beams, the definition of Lef is the same as for simple beams.
However, the procedure for computing Ief may require some elaboration. Figure 4.3(2) illustrates typical spans of continuous beams and the corresponding moment diagrams. For the sections i, c and j, the bending moments are Mi, Mcand Mj,
82 Part I Reinforced concrete
C
Mj
i j C
W
Mj
i j
Mi
+ Mc
–Mj
Mo
+ Mc
–Mj
(Mo-Mc) –Mi
L/2 L/2 L/2 L/2
w
(a) End span (b) Interior span
Figure 4.3(2) Typical spans of continuous beams and the corresponding moment diagrams
respectively, where Miand Mjare the results from a statically indeterminate analysis of the continuous beam (or output from a computer analysis). For an end span
Mc= Mo− Mj
2 Equation 4.3(8)
and, for an interior span Mc= Mo−
Mi+ Mj
2 Equation 4.3(9)
in which Mo = w Le f2
8 Equation 4.3(10)
is the maximum moment of a simply supported span under uniformly distributed load (w).
With Mstaking the value of Mi, Mcand Mj, respectively, Equation 4.3(2) may be used to compute Ieffor the three nominated sections (i.e. Ief,i,Ief,cand Ief,j). Note that AS 3600-2009, like its predecessor AS 3600-2001, has no provisions for the deflection analysis of continuous beams per se. However, Clause 8.5.3 in AS 3600-1994 recommends that, for an end-span
Ie f =
Ie f,c+ Ie f, j
2 Equation 4.3(11)
and, for an interior span Ie f =
2Ie f,c + Ie f,i+ Ie f, j
4 Equation 4.3(12)
Chapter 4 Deflection of beams and crack control 83
For computing midspan deflections, these interpolated values of Iefshould be used.
Following the assumption that the principle of superposition holds for deflection analysis, the midspan deflection of a continuous span, as illustrated in Figure 4.3(2), can be determined by appropriately combining Equations 4.2(1) and/or 4.2(2) with 4.2(3).