PSI - Issue 64

Tom Molkens et al. / Procedia Structural Integrity 64 (2024) 1484–1491 Tom Molkens & Mona El-Hallak Author name / Structural Integrity Procedia 00 (2019) 000 – 000

1486

3

after redistribution. This is because the concrete is the limiting factor for reinforced cross-sections, although it should be noted that before the Eurocodes appeared, a value of 0.85 was often used (at least in Belgium), regardless of the application and even the material. To take advantage of moment redistribution in normal service conditions, the bond properties between the (G)FRP bar and the concrete matrix should be included in the design. This can be achieved by a short top reinforcement that slips in a controlled manner. In all previous studies shown in Table 1, continuous GFRP top reinforcement was always used, which is less economical (incidentally, this is rarely done with steel either) and also does not promote redistribution due to the lack of slip. In this study, each configuration was repeated twice to detect at least random results. Two configurations were tested, identical in the beam dimensions and reinforcement lengths, but different in the diameter of the rebars above the intermediate support. 3. Experimental setup 3.1. Theoretical background Simple structural mechanics can be used to calculate the effect of the redistribution factor on the internal force action and the support reactions. With a total load P distributed over two spans of length L , the following static diagram was obtained as illustrated in Figure 1:

M M M M M M M M

δ=1,00 δ=0,90 δ=0,80 δ=0,70 δ=0,60 δ=0,50 δ=0,40 δ=0,00

ELA

ELA

ELA

ELA

ELA

ELA

ELA

ELA

Figure 1. Static model of the experimental test setup.

A redistribution factor δ = 1 corresponds to a linear elastic distribution giving a reaction force of 5/32∙ P at the end supports and 11/16∙ P for the central support. A value of 0 reduces the system to a double isostatic arrangement, a lower limit of 0.44 should be respected for concrete with f ck <50 N/mm² in combination with steel reinforcement. The measurement of the reaction forces allows the calculation of the δ -factor based on the rotational equilibrium according to Eq. (2). = 2 2 − (136 2 ) → =83(1−4 ) (2) For the evaluation of deflections and curvatures (measured by the strains), a distinction should be made between the moment of inertia of the section in the uncracked state I nc and in the cracked state Ic . Where αe is the ratio between the elastic modulus of the reinforcement ( E g ) and the secant modulus of the concrete ( E cm ), the second moment of area I for a symmetrical reinforcement with n top and bottom bars of diameter D (16 mm) and coverage c (25 mm) is placed in a section of width b and height h : = 1 ℎ2³ + 2( − 1) 4 ² (ℎ2 − − 2 ) 2 (3)