Page 422 - Mechanics of Asphalt Microstructure and Micromechanics

T
414 Ch a p t e r w e l v e
Σ
Based on the balance of force at the cross-section, i.e., F = 0, it can be written as:
0 b
c ∫
t ∫
σ dA + σ dA=0 (12-12a)
'
'
− a 0
Where s' t and s' c are tensile and compressive stresses distributed along the height
respectively. Solving Equation 12-12, it is obtained that:
E b 2 1
t = = (12-12b)
E a 2 r 2
c
Considering the resultant moment from the stress acting over the cross-section:
0 b
M= σ t ∫ ' ydA+ σ c ∫ ' ydA (12-13)
− a 0
With some simple integration one can obtain:
w
M= ( Ea ⋅ + E b ⋅ ) (12-14)
⋅
ε
ε
⋅
2
2
3 t t c c
Where E t and E c are the tensile and compressive moduli, respectively. With Equa-
tions 12-12b and 12-14, one can obtain:
M= w Eb 2 ⎛ ⎜ 1 + a⎞ ⎟ ε = w Ea 2 ⎛ ⎜ 1 + b⎞ ⎟ ε (12-15)
3 c ⎝ b⎠ c 3 t ⎝ a⎠ t

In their study, two extensometers are used to measure the deformation of the out-
side fiber of the beam. One extensometer is attached on the top and the other at the
bottom of the beam and along the longitudinal fiber. The center of the extensometer is
aligned with the mid-span of the beam. Next, the deformation of gauges amounted on
the beam is estimated. With self-weight, pure bending does not exist between the two
concentrated loads. The deformation measured by the extensometers should be calcu-
lated by:
t ∫
Δ = ε t dx (12-16)
GL
c ∫
Δ = ε c dx (12-17)
GL
Where GL is the gauge length. With the relationship in Equation 12-15, Equation
12-16 can be expressed as:
t ∫
Δ = a ε c dx= a Δ c (12-18)
GL b b

Therefore:
Δ ε a
t = t = = r (12-19)
Δ ε b
c c