Issue 41

Y. Yang et alii, Frattura ed Integrità Strutturale, 41 (2017) 339-349; DOI: 10.3221/IGF-ESIS.41.45

τ G is the shear strain; is the shear stress; τ = T A

2 P T is the sectional area;

γ =

=

Where,

is the shear force;

µ + 2(1 ) f E

µ 2(1 ) 3 f E bh +

PL

=

ω ∆ =

is the shear modulus; μ is Poisson’s ratio. We substitute this into Formula (5) and obtain

.

G

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:

3

µ 2(1 ) +

PL

23 ' PL

=

+

ω

(6)

3

E bh

3

E bh

108

f

f

' f E considering the shear effect can be expressed as:

Then the flexural modulus

3

µ 2(1 ) +

PL

23 ' PL

=

+

E

(7)

f

3

ω 3 ' bh

ω 108 ' bh

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

¦ Ò P

x dx

E p

h 1

h 1

h

y

E t

h 2

h 2

b

¦ Ò t

(a)

(b)

Figure 6: Stress Distribution of Specimen Mid-Span Section

= dA bdx . For the mid-span section, the bending

Let the micro-area unit parallel to the neutral axis (as shown in Fig. 6(a)) 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)

is the compressive stress of the upper surface; σ t

Where, M is the bending moment; σ p

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:

2

2

σ M bh =

+

σ t

bh

/ 3

/ 3

(9)

p

1

2

344