复合材料力学(宏观力学,微观力学)

vf vm 1 = + E2 E f Em
Generalized equation for n – constituent composite:
n vi 1 = ∑ E2 i =1 Ei 1 E2 = n (transverse modulus) vi ∑ E i =1 i
(∆ mW = ε 2 mWm = −υmε1mWm ) = −(υmε1Wm + υ f ε1W f ) / W
ε 2 = −υmν mε1 − υ f ν f ε1 υ12 = −ε 2 / ε1 = υm vm + υ f v f
modulus : E1 = Emvm + E f v f
Determination of transverse Modulus
Longitudinal Modulus
• σ 1 / ε1 = σ m vm / εm + σ f vf / εf • For linear fiber and matrix (Hooke’s Law) • E1 = Em vm + E f vf
Generalized equation for composites with n constituents:
Determination of transverse Modulus(E2)
ε2 = ∆W 2 / W = ∆Wf / W + ∆Wm / W
(ε f = ∆Wf / Wf ; ε m = ∆Wm / Wm ) ⇒ ε 2 = ε f Vf + ε mVm
(ε f =
σ2
Ef
; εm = + Vm
Volume Fractions
Composites Volume V= V f + V m + V v
Fiber Matrix Void
Composites Mass M= Mf + Mm
Volume Fractions: vf + v m + vv = 1 Mass Fractions: mf + mm = 1
(Vv / M c = vvVc / M c = vv ρc ) ⇒ 1 = m f ρc / ρ f + mm ρc / ρ m + vv ⇒ vv = 1 − ρc (m f / ρ f + mm / ρ m )
How to determine the ρc , m f , mm ?
Burnout test of glass/epoxy composite
Void content determination
M c M f + M m ρ f V f + ρ mVm ρc = = = = ρ f v f + ρ m vm Vc Vc Vc
Mc Mc Mc ρc = = = Vc V f + Vm + Vv M f / ρ f + M m / ρ m + Vv ⇒ ρc = 1 m f / ρ f + mm / ρ m + vv ρc
Weight of empty crucible = 47.6504 g Weight of crucible +composite = 50.1817 g Weight of crucible +glass fibers = 49.4476 g ρ f = 2.5 g 3 , ρ m = 1.2 g 3 cm cm Find vv if ρ c = 1.86 g 3
Generalized equations for n – constituent composite n 1 ρ c = ∑ ρi vi = n wi i =1 ∑ ρ i =1 i
Void content determination
• The void content of a composite is detrimental to its mechanical properties. • These detriments include lower: • Shear stiffness and strength • Compressive strengths • Transverse tensile strengths • Fatigue resistance • Moisture resistance
Micromechanics of composites
Assumptions
1.The bond between fibers and matrix perfect. 2.The elastic moduli, diameters, and space between fibers are uniform. 3.The fibers are continuous and parallel. 4.The fibers and matrix follow Hooke’s law(linearly elastic). 5.The fibers possess uniform strength. 6.The composite is free of voids.
or
Biblioteka Baidu
Transverse Strength
⇒ σ2 < σm
(σ 2 )
Due to stress (strain) concentration
Factors influence • properties of fiber and matrix • the interface bond strength • the presence and distribution of voids (flaws) • internal stress and strain distribution (shape of fiber, arrangement of fibers)
Alternative solution
Practical situation
1. Perfect packing unlikely 2. Fiber Waviness 3. Fibers shouldn’t touch Commonly vf is 0.4 to 0.7 What effects on the composites if vf too small or too large?
Density of composite
Mc = M f + Mm + 0 ⇒ Mf Mc + Mm = 1 ⇒ m f + mm = 1 Mc
M c M f + M m ρ f V f + ρ mVm ρc = = = Vc Vc Vc ∴
ρc = ρ f v f + ρ m vm
or : ρc = ρ f v f + ρ m (1 − vm )
cm
Sol:
mf =
Mf Mc
=
49.4476 − 47.6504 = 0.71 50.1817 − 47.6504
mm = 1 − m f = 1 − 0.71 = 0.29
vv = 1 − ρc (m f / ρ f + mm / ρm ) = 1 −1.86(0.71/ 2.5 + 0.29 /1.2) = 2.22%
σ2
Em
)
E2
ε2 = Vf
σ2
Ef
σ2
Em
1 V f Vm ⇒ = + (reciprocal theory) E2 E f Em
⇒
E2 1 = Em Vm + V f ( Em / E f )
Determination of transverse Modulus
Assume all constituents are in linear elastic range:
E1 = ∑ Ei vi L
i =1
n
Rule-of-mixture
Factors influencing E1 and σ1
• • mis-orientation of fibers fibers of non-uniform strength due to variations in diameter, handling and surface treatment, fiber length stress concentration at fiber ends (discontinuous fibers) interfacial conditions residual stresses
Micromechanics of Unidirectional Composites
Including: • Void volume fraction • Density of composite • Composite stresses and strains • Elastic modulus of composite • Thermal expansion properties of composite
Determination of Longitudinal Modulus
under a Longitudinal stress Parallel model
Longitudinal direction
• σ1 Ac = σm Am + σf Af • σ1 = σm Am / Ac + σf Af / Ac • σ1 = σm V m/ V c + σf / V f Longitudinal composite stress: σ1 =σm vm + σf v f= σm vm + σf (1-v m) Longitudinal composite strain: ∆L ε1 = ε f = ε m = L
• Under a transverse stress
• the stress same in composite, fiber and matrix.
• Series model
• (transverse direction) (σ = F / AT ) • the stress on composite: σ 2 = σ m = σ f • The strain on composites: ∆ W 2 = ∆ W f + ∆ W m
• • •
Determination of Poisson ratio(υ12)
• Transverse strain due to σ1:
ε 2 = ∆W / W
• Transverse poisson’s ratio
υ12 = −ε 2 / ε1
ε 2 = ∆W / W = (∆ mW + ∆ fW ) / W