Longitudinal Modulus Equal strain assumption: εc = εf = εm 2

Matrix Fiber

σ1

1

σ1

Matrix L Representative Volume Element (RVE) ΔL

Static equilibrium requires that the total resultant force on the element must equal the sum of the forces acting on the fiber and matrix.

σ C1 Ac = σ F 1 AF + σ M 1 AM σ C1 = σ F 1

AF A + σ M1 M AC AC

Where, AC, AF, AM are composite, fiber, and matrix cross sections. Then, we can also say that: VF = AF/AC and VM = AM/AC where VF and VM are volume fractions and not volumes of fiber and matrix.

Invoking Hooke’s Law, we get EC1εC1 = EF1

εF1VF + EM1 εM1VM

Then, to have strain compatibility, we have to assume that the average strains in the composite, fiber, and matrix along the 1-direction are equal. Therefore, EC1 = EF1 VF + EM1 VM = EF1 VF + EM1 (1-VF) -- Parallel combination rule of mixtures

Therefore, the fraction of load carried by fibers in a unidirectional continuous fiber lamina is EF VF EM

σ F VF PF E F VF = = = PC σ F V F + σ M (1 − V F ) E F V F + E M (1 − V F ) E F VF + (1 − VF ) EM If you assume wood-plastic composite and that EF is 1,000,000 psi and Em is 175,000 psi (HDPE for example), then PF 6V F = PC 6V F + (1 − V F )

Then, PF/PC = 0.6 or 60% Strength – Unidirectional Continuous Fiber Lamina: In general, fiber failure strain is lower than the matrix failure strain. Assuming all fibers have the same strength, the tensile rupture of fibers will determine the rupture in the composite. Therefore, estimation of longitudinal tensile strength could be calculated as ' σ LTU = σ FU VF + σ M (1 − VF )

' Where σFU is the fiber tensile strength and σ M is matrix stress at fiber failure strain. For effective reinforcement of the matrix (i.e., σ LTU ≥ σ MU ), the VF must be greater than the critical value defined as: ' σ MU − σ M ' σ FU − σ M

CriticalV F =

V. Yadama, CE 537, Fall 2007

2

Transverse Modulus

2

σ2

δ Matrix W Fiber Matrix 1

σ2

Representative Volume Element (RVE)

Geometric compatibility requires that the total transverse composite displacement in the 2-direction, δC2, must equal the sum of the corresponding transverse displacements in the fiber, δF2, and the matrix, δM2. Therefore, δC2 = δF2 + δM2 But, εF2 = σ2/EF2 and

εM2 = σM2/EM2 δC2 = ε2W = VFWεF2 +VMWεM2 or

Invoking Hooke’s Law:

ε2 = VFεF2 + VMεM2

σ C2

EC 2

=

σ F2

EF 2

VF +

σM2

EM 2

VM

But , σ C 2 = σ F 2 = σ M 2 Therefore, 1 1 1 = VF + (1 − VF ) EC 2 E F 2 EM 2 or EC 2 = EF 2 EM 2 E F 2VM + E M 2VF

Series combination rule of mixtures

V. Yadama, CE 537, Fall 2007

3

Assumptions are not so good: • Transverse strain mismatch exists at the boundary between fiber and matrix • Transverse stresses may not be same as VF ≠ VM • Poisson’s Ratio is not the same which would cause shear stress between fiber and matrix Fibers do not contribute significantly, matrix dominated.

Poisson’s Ratio

2 ΔW Matrix W Fiber Matrix L Representative Volume Element (RVE) ΔL

σ1

1

σ1

εF = εM

Macroscopically, ΔW = −Wε 2 = Wν 12 ε 1 But , microscopically ΔW = Δ MW + Δ FW Δ FW = WVFν F ε 1 Δ MW = WVMν M ε 1 Therefore, ν 12 = ν M VM + ν F VF and

ν 21 =

E 22 ν 12 E11

Neither fiber nor matrix dominate the composite ν12 V. Yadama, CE 537, Fall 2007 4

Shear Modulus

Presume that the shearing stresses on the fiber and on the matrix are the same (which is clearly not the case).

2

Matrix W Fiber Matrix

τ

1

Matrix Fiber Matrix ΔM/2 ΔF Δ

τ

Then, γ M =...