Issue 41

Y. Yang et alii, Frattura ed Integrità Strutturale, 41 (2017) 339-349; DOI: 10.3221/IGF-ESIS.41.45 344 Where, τ γ = G is the shear strain; is the shear stress; τ = T A is the shear force; = 2 P T is the sectional area; µ = + 2(1 ) f E G is the shear modulus; μ is Poisson’s ratio. We substitute this into Formula (5) and obtain µ ω + ∆ = 2(1 ) 3 f PL E bh . The deflection ω ' under the mutual action of moment and shear can be expressed as: ω ω ω = + ∆ ' . By substituting the above equations, we obtain: µ ω + = + 3 3 2(1 ) 23 ' 3 108 f f PL PL E bh E bh (6) Then the flexural modulus ' f E considering the shear effect can be expressed as: µ ω ω + = + 3 3 2(1 ) 23 ' 3 ' 108 ' f PL PL E bh bh (7) If we compare the flexural modulus formula (7) considering the shear effect and formula (4) without considering the shear effect, the change ratio of the two is: µ − + = 2 2 ' 216(1 ) 69 f f f E E h E L . As the required span height ratio L/h of the cement stabilized macadam beam specimen is 3, we know that with the shear effect taken into account, the flexural modulus is increased by about 47% (Poisson’s ratio μ is 0.25). Derivation of tensile modulus calculation formula based on flexural test The tensile and compressive stress distribution of the mid-span section is shown in Fig. 6(b). x y x dx h 1 h h 2 h 1 h 2 ¦ Ò P ¦ Ò t (a) (b) E p E t b Figure 6: Stress Distribution of Specimen Mid-Span Section Let the micro-area unit parallel to the neutral axis (as shown in Fig. 6(a)) = dA bdx . For the mid-span section, the bending moment M caused by the internal force can be expressed as follows: σ σ − = + ∫ ∫ 0 1 2 2 0 2 1 2 h p t h M bx dx bx dx h h (8) Where, M is the bending moment; σ p is the compressive stress of the upper surface; σ t is the tensile stress of the lower surface; h 1 is the vertical distance between the upper surface of the specimen and the neutral axis moved up; h 2 is the vertical distance between the lower surface of the specimen and the neutral axis; other symbols have the same meanings as above. By integrating this formula, we obtain: σ σ = + 2 2 1 2 / 3 / 3 p t M bh bh (9)