#### Installation — business terrible - 1 part

September 8th, 2015

Another possibility is to define the cracking load as the load that produces cracks of a given critical width wcr. Consider first the beams with stirrups. Under a certain load, a number of parallel diagonal cracks may initiate. The cracks are cohesive. This means that crack-bridging stresses are transmitted across the cracks (due to aggregate pullout and other phenomena). Reduction of the crack-bridging stress to zero requires a considerable opening displacement of the crack, as is clear from the typical stress-displacement diagram used in the cohesive (fictitious) crack model; see Chapter 7. Furthermore, it is known that when many parallel cracks form, only one of them may open widely, while the others unload and close. In fact, such a localization of crack openings into one among many parallel cracks is a necessity unless there is enough reinforcement to ensure a stiffening rather than softening behavior (see Chapter 8; also Chapter 12 in Bazant and Cedolin 1991). Thus, unless the stirrups are extremely strong, the situation as shown in Fig. 10.3.7b must be expected.

Since the reduction of the crack-bridging stress to zero requires a very large opening, we consider that the stress is reduced only to a certain small but finite fraction kt of the tensile strength // of concrete.

Consider now the relative displacement between points 5 and 6 at the bottom and top of the beam, lying on a line normal to the cracks after one large crack forms. This displacement may be approximately expressed as Ди/ = (D/ cos 9){ktfl/Ec) + wcr, in which D/cosO is the length of the line segment 56, and War a critical crack opening displacement at which the crack bridging stress is reduced from f[ to ktfl (Fig. 10.3.7c). Dividing this by the length of segment 56, we obtain the average normal strain in the direction orthogonal to the diagonal cracks:

Ай; ktf’t wcrcos 9 Dj cos 0 Ec D

Displacement Диі or strain £icr must be compatible with the overall deformation of the truss. Imagining the nodes of the truss to be attached to a homogeneously deforming continuum, this condition means that strain £/cr must be tensorially compatible with the normal strains £c in the inclined struts and ev in the vertical stirrups, as well as with the principal direction angle 0. This strain compatibility condition may be easily deduced from the Mohr circle in Fig. 10.3.7d. Noting that 14 — (ev —ffc)cot0, R = 05 = 01 = 14/ sin 2(? ~ (ev – £c) cot 9/ sin 29, gj — єс -1- 27?, we obtain the following expression for the average strain in the direction orthogonal to the diagonal cracks:

Єї = £c + —2~- = – £c cot2 0 (10.3.38)

sin 9 sin 9

In terms of the stresses, ev — <rv/E„ £c ~ crcJ Ec, in which Es elastic modulus of steel and Ec = secant modulus for the compression strut at the moment the diagonal cracks form, which is less than the initial elastic modulus but larger than the secant modulus for the peak stress point of the compression stress-strain diagram. Here the stresses may be expressed from the equilibrium conditions of the truss: от = vcrsvb tan 0/Av, (jc — — 2vcrj sin 29, where Av = cross section area of one stirrup, and vcr = Vcr/bD = nominal stress corresponding to the shear force at the moment of formation of large diagonal cracks. Substituting these expressions into (10.3.38), we obtain:

2 / svb cot2 9

sin 29 AVES + E™ ) V’

Setting this expression equal to (10.3.37), we obtain an equation for vcr, the solution of which furnishes the result:

Wrr

Vcr = Voo + V0 ~ (10.3.40)

Here we introduced the notations:

Equation (10.3.40) describes a size effect which is an alternative to (10.3.35). The asymptotic constant value foo exists because we assume that the critical crack opening wcr corresponds to nonzero crack bridging stress fct/f’; if this stress were neglected, we would obtain Voa = 0.

Consider now a beam without stirrups. This problem is more complicated because there is no truss model that could give the value of the average strain along the line 23 in Fig. 10.3.7c. Other simplifications are, therefore, needed to obtain a simple result. We will assume that the normal strains along the line segment 23 in Fig. 10.3.7e may be approximated according to the beam theory. The shear stress in the vertical plane is distributed parabolically, and so, at point 1 at mid depth of the beam (neutral axis), it has the value т = 1,5vcr. From the Mohr circle in Fig. 10.3.7f, we then obtain the normal stress <j in the direction 23 at point 1 and the corresponding strain: = 1.5vcrsin29/Ec. The normal strain in

the direction 23 may also be assumed distributed parabolically, in which case the average normal strain along this line is £i — vcr sin 29/Ec. Multiplying this by the length of segment 23, wc obtain the relative displacement between points 2 and 3 in the direction 23:

At the same time, in analogy to (10.3.37):

liquating the last two expressions, we obtain the same equation as (10.3.41), that is, vcr — Voo + vq(wc/D), in which we now make the notations:

2fct/t _ Ecwcr

3 sin 26′ 0 3 sin (?

/. The fracture modification (Bazant 1996b) of the classical widely used truss model (or the strut – and-tie model) for the shear failure of reinforced concrete beams describes the energy release and localization of damage into a band of compression splitting cracks (or a compression-shear crack) within a portion of the compressed concrete strut.

2. If the analysis of the maximum load based on the truss model is valid (and if the stirrups are designed sufficiently strong), the concrete strut must undergo compression softening (with progressive fracture) during the portion of loading history in which the maximum load is reached.

3. Analysis of the energy release into the crack band shows that a size effect on the nominal strength at shear failure of a reinforced concrete beam must occur and that it should approximately follow Bazant’s size effect law. Conversely, the fracture behavior of the truss model (slrut-and-lie model), particularly the damage localization with energy release, provides an explanation of the size effect widely observed in many tests as shown in the previous section.

4. The applied nominal shear stress that causes the initial large diagonal cracks also exhibits a size effect. The law of this siz. e effect depends on how the large diagonal cracks are defined.

5. The foregoing size effect formulae have not yet been calibrated and verified by the available test results for beams. The expressions for the coeflicients in these formulas need to be studied further in order to develop a design procedure incorporating the size effect.