Effect of Reinforcement

To counteract brittle failure, one may use densely distributed reinforcement such as shear (or torsional) stirrups in beams. Together with the longitudinal reinforcement of beams, the shear reinforcement alone, at its yield limit, can resist a certain load, characterized by a nominal stress (J%u. When the structure is sufficiently large, there may be enough strain energy stored in the structure to drive a crack through the entire cross section at a load that only slightly exceeds the load carried by plastic reinforcement. However, there can be no size effect if that load is not exceeded. So, the size effect law tftust be applied only to the portion of the load capacity or nominal stress that is in excess of asN that is

Effect of Reinforcement

T CP ‘Nu





riNu — °Nu + aNu-,








where СТдСц is a possible contribution of concrete to the overall strength at the plastic limit, D0(p) is a function with the dimension of length, and p is the steel reinforcement ratio. Note the explicit dependence of the brittleness number on the steel ratio.

The procedure to determine oc£u is analogous to that sketched in the previous section, and requires no more than using the classical formulas of the code in (10.1.15). The determination of Do(p) is more difficult. For plain concrete, Eq. (10.1.16) shows that Do is a constant that is determined by the material fracture property 0/ and the structure geometry. Now, the structure geometry is altered by the presence of the reinforcement. Therefore, the geometrical factor must depend on the steel ratio as well as on other dimensionless ratios defining the layout of the reinforcement.

Let us now sketch a possible unified framework for the influence of the steel ratio on the value of Dq. We write /? in the form (10.1.8) and substitute the general form (5.3.11) for D:

Подпись: I<i CjldaKl Effect of Reinforcement(10.1.18)

where da denotes partial differentiation with respect to a. Next, we use the superposition principle and write the condition that the stress intensity factor is the sum of the stress intensity without reinforcement, minus the stress intensity factor caused by the steel-concrete interaction. The negative sign is due to the fact that the steel forces tend to close the crack. The resulting equation for Ki can always be written in the form

Kj — crK/~Dk(a) – pcrsfDks(a) (10.1.19)

where as is the stress in the steel bar and ks(a) a shape factor taking into account the steel distribution.

The simplest behavior one can encounter is that in which the effect of the reinforcement is exactly equivalent to decreasing the externally applied load (a pure parallel coupling). For such cases, ks(a) oc k(ct) and

P{p) = /3(0) -/? , Dq(p) — Dq(0) — D0 (10.1.20)

Pure cases of this kind are difficult to find in practice, but the results of Bazant and Kim (1984) seem to indicate that this is approximately valid for longitudinal reinforcement in diagonal shear.

For a densely distributed reinforcement, the behavior is quite different and one cannot assume that ks ос к. The result for Dg(p) is then obtained as

_ crsks(a0) afjk(ao)





1 – m, p n

c———– LA)

1 – mop








For small values of p, we can take a Macl. aurin expansion and write

Подпись: (10.1.22)Do(p) ~ (1 T m p) Do, rn — mo – mi

This is the form postulated by Bazant and Sun (1987) for the influence of the stirrups on the size effect in diagonal shear. As it transpires from the foregoing derivation, m is a geometrical factor that can be, in principle, determined either from experiment or from numerical simulation. As an example, for

Effect of Reinforcement

h———– s————– і

Figure 10.1.3 Longitudinally reinforced beam subjected to constant shear.

the diagonal shear of beams, Bazant and Sun (1987) proposed the following formula, determined by optimization of data fits and the condition that m —> 0 for short spans and m —> constant for large spans:

m — 400 j^l + tanh ^2-^ – 5.6^ J (10.1.23)

where s is the shear span and D the effective depth of the reinforcement (see Fig. 10.1.3).