Issue 41

Y. Yang et alii, Frattura ed Integrità Strutturale, 41 (2017) 339-349; DOI: 10.3221/IGF-ESIS.41.45 343 Fig. 4(a) shows the simplified force diagram of the specimen. (b) shows the equivalent force diagram for the right part of the specimen. The approximate differential line of deflection equation for the beam can be expressed by section as follows: When 0≤x≤L/6: ω = '' 6 f PL E I (1) When L/6≤x≤L/2: ω = − '' 4 2 f PL Px E I (2) Where, E f is the traditional flexural modulus; I is the inertia moment; w is the mid-span deflection; P is the load applied; L is the specimen span. By integrating Formula (1) and (2) twice respectively, based on the boundary conditions and continuity conditions, we will know that when L/6≤x≤L/2, ω = − − + 2 3 2 3 8 12 144 2592 f PLx Px PL x PL E I (3) When x=L/2, by substituting the inertia moment I=bh 3 /12 into Formla (3), we obtain the flexural modulus calculation formula: ω = 3 3 23 108 f PL E bh (4) Where, h is the mid-span sectional height; b is the cross section width of the specimen; other symbols are the same meanings as above. Flexural modulus calculation correction formula considering the shear effect When the span height of the beam is greater than 5, the positive stress formula for pure bending can be applied to the positive stress calculation for transverse bending. However, for a deep beam with a span height of less than 5, if we use the pure bending theory and assumptions for slender beam in material mechanics, the error will increase rapidly with the decreasing depth-span ratio [15]. In the current inorganic binder test regulation for flexural modulus, the span height ratio L/h used is 3, but it did not consider the shear effect on the beam deflection, so the result is inconsistent with the actual situation [14]. Therefore, this paper deduces the flexural modulus calculation correction formula considering the shear effect. (the flexural modulus mentioned later is the result considering the shear effect). ¦ ¤¦ Ø γ τ P/2 P/2 τ P/2 P/2 P/2 P/2 Figure 5: Force Analysis Diagram of Specimen Considering the Shear Effect As shown in Fig. 5, let the additional deflection caused by the shear effect be ω ∆ . Then: ω γ ∆ = ⋅ 3 L (5)