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Part-3a Material Type: Notes; Professor: Sottos; Class: Mechanics of Composites; Subject: Theoretical and Appl Mechanics; University: University of Illinois - Urbana-Champaign; Term: Fall 2010;
Typology: Study notes
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REINFORCEMENT + MATRIX + INTERFACE = COMPOSITE
The properties and performance of a composite depend on:
**- properties of the reinforcement and the matrix
Aspect Ratio : l/d
Volume Fraction:
Mass Fraction:
“Rule of Mixtures”
Let P be any property of the composite ….
“Rule of Mixtures”
“Inverse Rule of Mixtures”
Need to consider the microstructure
Mechanics of Materials Approach Case 1. Modulus in Fiber Direction = Longitudinal Modulus
Parallel Response
F 1
Assumptions: 1. Fiber/matrix bond is perfect (no slippage) 2. Uniform stress & strain (linear elastic) 3. Iso-strain: f = m= 1
Applied force is distributed between the fibers and matrix: F 1 = Ff + Fm 1 Ac = f Af + m Am 1 v = f v f + m v m 1 = f Vf + m Vm
1 = f Vf + m Vm
E 1 1 = Vf Ef1 f + Vm Em m
Recall: 1 = f = m
E 1 = Ef1 Vf + Em Vm
Rule of Mixtures
An expression for Poisson's ratio can be derived in a similar fashion
12 = f12 Vf + m Vm
A similar expression can also be derived for the shear modulus of the composite:
For any property P of the unidirectional composite: Parallel Reaction:
Series Reaction:
Drawbacks: - Series model is not very accurate. - Assumptions of uniform stress and strain are not valid. - Strain is magnified between fibers. Need more realistic assumptions about microstructure.
P 2
Vfs
Vms
V (^) mp
Vfs
Vms
V (^) mp
= 0 Series Reaction = Parallel Reaction
- **can be interpreted as "reinforcing efficiency”
**- Mechanics of Materials
z = ( +2 G ) z + r + r = z + ( +2 G ) r + (^) = z + r + ( +2 G ) rz = r = (^) z = 0
Matrix: Linear elastic, isotropic
where:
In radial coordinates:
Fiber: linear elastic, transversely isotropic
z = n (^) Af z + Af r + Af (^)
r = Af z + ( kTf + μ) r + ( kTf μ Tf )
(^) = Af z + ( kTf μ Tf ) r + ( kTf + μ Tf )
rz = r = (^) z = 0
C 11 = n (^) a C 12 = a C 22 = kT + μ C 23 = kT - μT C 44 = μT C 66 = μa
Where:
In the fiber:
In the matrix:
@ r=r (^) f ; u (^) rf^ = u (^) rm, rf= rm @ r=r (^) m ; rm^ = 0
@ r=0; u (^) r is finite
To determine constants Ao, A 1 , A 2 and A 3 the boundary conditions must be satisfied
Small … can neglect
Solving for the constants and back substituting yields:
Place RVE under uniform shear loading in 1-2 (r-z) plane: @ r=r m
Place RVE under uniform shear loading in 2-3 (r- ) plane:
@ r=r m
Obtained from properties calculated above:
Composite Properties!
Self-consistent field model:
Same holds for
Types of Short Fiber Composites:
Aligned Partially Aligned Random
Prediction of short fiber composite properties requires:
**- Properties of fiber and matrix
Predictions for carbon fiber / epoxy matrix, Vf = 50%
Halpin-Tsai Approach:
where:
where: = 2
