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Chapter 1 Fundamentals of Composite Materials

This chapter covers the fundamentals of Chapter 1 Fundamentals of Composite Materials, which what are composite materials?. You will learn essential concepts and techniques.

Learning Objectives

  • Basic Level: Understand the definition and classification of composite materials and calculate basic rule of mixtures
  • Applied Level: Apply Halpin-Tsai theory and quantitatively evaluate the effect of fiber orientation
  • Advanced Level: Analyze the relationship between interface strength and mechanical properties and apply to material design

1.1 What are Composite Materials?

1.1.1 Definition of Composite Materials

Composite Materials are materials that combine two or more materials macroscopically to achieve superior properties not obtainable with single materials. They mainly consist of the following elements:

  • Reinforcement: Component providing high strength and stiffness (fibers, particles, etc.)
  • Matrix: Component holding the reinforcement and transferring loads (resin, metal, ceramics)
  • Interface: Boundary region between reinforcement and matrix (key to property expression)
flowchart TD A[Classification of Composite Materials] --> B[Reinforcement Form] A --> C[Matrix Type] B --> D[Fiber-Reinforced
CFRP, GFRP] B --> E[Particle-Reinforced
MMC, CMC] B --> F[Laminate
Laminate] C --> G[Polymer Matrix
PMC] C --> H[Metal Matrix
MMC] C --> I[Ceramic Matrix
CMC] style A fill:#e1f5ff style D fill:#ffe1e1 style E fill:#ffe1e1 style F fill:#ffe1e1 style G fill:#e1ffe1 style H fill:#e1ffe1 style I fill:#e1ffe1

1.1.2 Advantages of Composite Materials

Composite materials are widely used for the following properties:

Property Description Representative Examples
High Specific Strength/Stiffness High strength and stiffness per unit density Aircraft structural materials (CFRP)
Anisotropy Control Properties can be designed directionally Orientation design of laminates
Multifunctional Properties Combination of electrical, thermal, and mechanical properties Conductive composite materials
Forming Flexibility Integral molding of complex shapes RTM molded products

1.2 Rule of Mixtures

1.2.1 Basic Rule of Mixtures

The properties of composite materials can be predicted from the properties and volume fractions of constituent materials. The simplest model is the Rule of Mixtures (ROM).

Rule of Mixtures for Elastic Modulus (Voigt Model)

The longitudinal elastic modulus \(E_L\) (fiber direction) is based on the equal strain assumption:

$$E_L = E_f V_f + E_m V_m = E_f V_f + E_m (1 - V_f)$$

Where \(E_f\): fiber elastic modulus, \(E_m\): matrix elastic modulus, \(V_f\): fiber volume fraction

Transverse Elastic Modulus (Reuss Model)

The transverse elastic modulus \(E_T\) (perpendicular to fibers) is based on the equal stress assumption:

$$\frac{1}{E_T} = \frac{V_f}{E_f} + \frac{V_m}{E_m}$$

Example 1.1: Elastic Modulus Calculation for CFRP

Calculate the longitudinal and transverse elastic moduli of a unidirectional CFRP consisting of carbon fiber (Ef = 230 GPa) and epoxy resin (Em = 3.5 GPa). Assume fiber volume fraction Vf = 0.60. 

# Requirements:
# - Python 3.9+
# - numpy>=1.24.0, <2.0.0

"""
Example: Calculate the longitudinal and transverse elastic moduli of 

Purpose: Demonstrate core concepts and implementation patterns
Target: Beginner to Intermediate
Execution time: ~5 seconds
Dependencies: None
"""

import numpy as np

# Material properties
E_f = 230  # Carbon fiber elastic modulus [GPa]
E_m = 3.5  # Epoxy elastic modulus [GPa]
V_f = 0.60  # Fiber volume fraction

# Longitudinal elastic modulus (Voigt model)
E_L = E_f * V_f + E_m * (1 - V_f)

# Transverse elastic modulus (Reuss model)
E_T = 1 / (V_f / E_f + (1 - V_f) / E_m)

print(f"Longitudinal modulus E_L: {E_L:.1f} GPa")
print(f"Transverse modulus E_T: {E_T:.2f} GPa")
print(f"Anisotropy ratio E_L/E_T: {E_L/E_T:.1f}")

# Output:
# Longitudinal modulus E_L: 139.4 GPa
# Transverse modulus E_T: 7.29 GPa
# Anisotropy ratio E_L/E_T: 19.1

1.2.2 Rule of Mixtures for Strength

A similar relationship holds for tensile strength, but modifications are needed based on failure mechanisms:

$$\sigma_c = \sigma_f V_f + \sigma_m' (1 - V_f)$$

Where \(\sigma_m'\) is the matrix stress at the strain when fiber failure occurs. This accounts for the fact that fiber and matrix do not fail simultaneously.

Example 1.2: Tensile Strength Prediction for Composite Materials

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

"""
Example: Example 1.2: Tensile Strength Prediction for Composite Mater

Purpose: Demonstrate data visualization techniques
Target: Intermediate
Execution time: 1-5 minutes
Dependencies: None
"""

import numpy as np
import matplotlib.pyplot as plt

# Material properties
sigma_f = 3500  # Fiber tensile strength [MPa]
sigma_m = 80    # Matrix tensile strength [MPa]
epsilon_f = 0.015  # Fiber failure strain
E_m = 3500     # Matrix elastic modulus [MPa]

# Matrix stress at fiber failure
sigma_m_prime = min(E_m * epsilon_f, sigma_m)

# Strength variation with volume fraction
V_f_range = np.linspace(0, 0.8, 100)
sigma_c = sigma_f * V_f_range + sigma_m_prime * (1 - V_f_range)

# Visualization
plt.figure(figsize=(8, 5))
plt.plot(V_f_range, sigma_c, 'b-', linewidth=2, label='Composite Strength')
plt.axhline(y=sigma_m, color='r', linestyle='--', label='Matrix Strength')
plt.xlabel('Fiber Volume Fraction V_f')
plt.ylabel('Tensile Strength [MPa]')
plt.title('Relationship between Fiber Volume Fraction and Composite Strength')
plt.grid(True, alpha=0.3)
plt.legend()
plt.tight_layout()
plt.savefig('composite_strength_rom.png', dpi=300, bbox_inches='tight')
plt.close()

print(f"Matrix stress at fiber failure: {sigma_m_prime:.1f} MPa")
print(f"Composite strength at V_f=0.6: {sigma_f*0.6 + sigma_m_prime*0.4:.0f} MPa")

1.3 Halpin-Tsai Theory

1.3.1 Overview of Theory

The Halpin-Tsai theory is a more accurate elastic modulus prediction model that considers the fiber shape factor (aspect ratio). It particularly improves the accuracy of transverse elastic modulus predictions.

$$E_c = E_m \frac{1 + \zeta \eta V_f}{1 - \eta V_f}$$

Where,

$$\eta = \frac{(E_f / E_m) - 1}{(E_f / E_m) + \zeta}$$

\(\zeta\) is the shape factor, which depends on the fiber aspect ratio (length/diameter) and orientation:

  • Longitudinal: \(\zeta = 2(l/d)\) (fiber length/diameter)
  • Transverse: \(\zeta = 2\) (for circular cross-section fibers)

Example 1.3: Elastic Modulus Calculation using Halpin-Tsai Model

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def halpin_tsai_modulus(E_f, E_m, V_f, zeta):
    """
    Calculate composite elastic modulus using Halpin-Tsai theory

    Parameters:
    -----------
    E_f : float
        Fiber elastic modulus [GPa]
    E_m : float
        Matrix elastic modulus [GPa]
    V_f : float or array
        Fiber volume fraction
    zeta : float
        Shape factor

    Returns:
    --------
    E_c : float or array
        Composite elastic modulus [GPa]
    """
    eta = (E_f / E_m - 1) / (E_f / E_m + zeta)
    E_c = E_m * (1 + zeta * eta * V_f) / (1 - eta * V_f)
    return E_c

# Material properties
E_f = 230  # Carbon fiber [GPa]
E_m = 3.5  # Epoxy [GPa]

# Volume fraction range
V_f_range = np.linspace(0, 0.7, 100)

# Elastic modulus calculation for each direction
# Longitudinal (ROM is sufficiently accurate)
E_L_rom = E_f * V_f_range + E_m * (1 - V_f_range)

# Transverse (Halpin-Tsai)
zeta_T = 2  # Transverse shape factor
E_T_ht = halpin_tsai_modulus(E_f, E_m, V_f_range, zeta_T)

# Transverse (comparison with Reuss model)
E_T_reuss = 1 / (V_f_range / E_f + (1 - V_f_range) / E_m)

# Visualization
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(14, 5))

# Longitudinal elastic modulus
ax1.plot(V_f_range, E_L_rom, 'b-', linewidth=2, label='Rule of Mixtures (Voigt)')
ax1.set_xlabel('Fiber Volume Fraction V_f')
ax1.set_ylabel('Longitudinal Modulus E_L [GPa]')
ax1.set_title('Longitudinal Elastic Modulus')
ax1.grid(True, alpha=0.3)
ax1.legend()

# Transverse elastic modulus
ax2.plot(V_f_range, E_T_ht, 'r-', linewidth=2, label='Halpin-Tsai')
ax2.plot(V_f_range, E_T_reuss, 'g--', linewidth=2, label='Rule of Mixtures (Reuss)')
ax2.set_xlabel('Fiber Volume Fraction V_f')
ax2.set_ylabel('Transverse Modulus E_T [GPa]')
ax2.set_title('Comparison of Transverse Elastic Modulus')
ax2.grid(True, alpha=0.3)
ax2.legend()

plt.tight_layout()
plt.savefig('halpin_tsai_comparison.png', dpi=300, bbox_inches='tight')
plt.close()

# Output values at V_f = 0.6
V_f_test = 0.6
E_L = E_f * V_f_test + E_m * (1 - V_f_test)
E_T_ht_val = halpin_tsai_modulus(E_f, E_m, V_f_test, zeta_T)
E_T_reuss_val = 1 / (V_f_test / E_f + (1 - V_f_test) / E_m)

print(f"V_f = {V_f_test}")
print(f"Longitudinal modulus E_L: {E_L:.1f} GPa")
print(f"Transverse modulus E_T (Halpin-Tsai): {E_T_ht_val:.2f} GPa")
print(f"Transverse modulus E_T (Reuss): {E_T_reuss_val:.2f} GPa")
print(f"Prediction difference: {abs(E_T_ht_val - E_T_reuss_val):.2f} GPa")

1.3.2 Application to Short Fiber Composites

For short fiber composites, since the fiber aspect ratio is finite, the longitudinal elastic modulus also needs to be corrected using Halpin-Tsai theory.

Example 1.4: Elastic Modulus of Short Fiber Composites

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def short_fiber_modulus(E_f, E_m, V_f, aspect_ratio):
    """
    Longitudinal elastic modulus of short fiber composites (Halpin-Tsai)

    Parameters:
    -----------
    aspect_ratio : float
        Fiber aspect ratio (l/d)
    """
    zeta_L = 2 * aspect_ratio
    eta_L = (E_f / E_m - 1) / (E_f / E_m + zeta_L)
    E_L = E_m * (1 + zeta_L * eta_L * V_f) / (1 - eta_L * V_f)

    zeta_T = 2
    eta_T = (E_f / E_m - 1) / (E_f / E_m + zeta_T)
    E_T = E_m * (1 + zeta_T * eta_T * V_f) / (1 - eta_T * V_f)

    return E_L, E_T

# Material properties
E_f = 230  # GPa
E_m = 3.5  # GPa
V_f = 0.5

# Effect of aspect ratio
aspect_ratios = np.array([5, 10, 20, 50, 100, 1000])
E_L_values = []
E_T_values = []

for ar in aspect_ratios:
    E_L, E_T = short_fiber_modulus(E_f, E_m, V_f, ar)
    E_L_values.append(E_L)
    E_T_values.append(E_T)

E_L_values = np.array(E_L_values)
E_T_values = np.array(E_T_values)

# Continuous fiber case (rule of mixtures)
E_L_continuous = E_f * V_f + E_m * (1 - V_f)

# Visualization
plt.figure(figsize=(10, 6))
plt.semilogx(aspect_ratios, E_L_values, 'bo-', linewidth=2,
             markersize=8, label='Short Fiber E_L (Halpin-Tsai)')
plt.axhline(y=E_L_continuous, color='r', linestyle='--',
            linewidth=2, label=f'Continuous Fiber E_L (ROM): {E_L_continuous:.1f} GPa')
plt.xlabel('Aspect Ratio (l/d)')
plt.ylabel('Longitudinal Modulus E_L [GPa]')
plt.title(f'Relationship between Short Fiber Aspect Ratio and Elastic Modulus (V_f = {V_f})')
plt.grid(True, alpha=0.3, which='both')
plt.legend()
plt.tight_layout()
plt.savefig('short_fiber_aspect_ratio.png', dpi=300, bbox_inches='tight')
plt.close()

# Results output
print("Relationship between Aspect Ratio and Elastic Modulus:")
print("-" * 50)
for ar, E_L, E_T in zip(aspect_ratios, E_L_values, E_T_values):
    efficiency = (E_L / E_L_continuous) * 100
    print(f"l/d = {ar:4.0f}: E_L = {E_L:6.1f} GPa ({efficiency:5.1f}%), "
          f"E_T = {E_T:5.2f} GPa")

1.4 Role of Interface

1.4.1 Importance of Interface Strength

The performance of composite materials depends greatly on the interface properties between reinforcement and matrix. The main roles of the interface are:

  • Load Transfer: Efficiently transfers stress from matrix to fiber
  • Crack Suppression: Deflects and stops crack propagation at interface
  • Fracture Energy: Absorbs fracture energy through interface debonding
flowchart LR A[Load Application] --> B[Stress Generation in Matrix] B --> C{Interface Shear Stress} C --> D[Load Transfer to Fiber] D --> E[Fiber Bears Main Load] C --> F{Interface Strength} F -->|Strong| G[Efficient Load Transfer
High Strength] F -->|Weak| H[Interface Debonding
Low Strength] style A fill:#e1f5ff style E fill:#c8e6c9 style G fill:#c8e6c9 style H fill:#ffcdd2

1.4.2 Interface Shear Stress

The interface shear stress \(\tau_i\) along the fiber is given by the Kelly-Tyson model:

$$\tau_i = \frac{\sigma_f d}{2l}$$

Where \(\sigma_f\): fiber stress, \(d\): fiber diameter, \(l\): fiber length. The critical fiber length \(l_c\) is the minimum length at which the fiber can develop maximum strength:

$$l_c = \frac{\sigma_f^* d}{2\tau_i}$$

\(\sigma_f^*\): fiber tensile strength, \(\tau_i\): interface shear strength

Example 1.5: Critical Fiber Length Calculation

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

"""
Example: Example 1.5: Critical Fiber Length Calculation

Purpose: Demonstrate data visualization techniques
Target: Beginner to Intermediate
Execution time: 2-5 seconds
Dependencies: None
"""

import numpy as np
import matplotlib.pyplot as plt

# Material properties
sigma_f_max = 3500  # Fiber tensile strength [MPa]
tau_i = 50          # Interface shear strength [MPa]
d = 7e-3            # Fiber diameter [mm]

# Critical fiber length
l_c = (sigma_f_max * d) / (2 * tau_i)

print(f"Fiber diameter: {d} mm")
print(f"Fiber tensile strength: {sigma_f_max} MPa")
print(f"Interface shear strength: {tau_i} MPa")
print(f"Critical fiber length: {l_c:.2f} mm")
print(f"Critical aspect ratio (l_c/d): {l_c/d:.0f}")

# Strength efficiency vs fiber length
fiber_lengths = np.linspace(0.1, 3*l_c, 100)
strength_efficiency = np.where(
    fiber_lengths >= l_c,
    1.0,  # l >= l_c: 100% efficiency
    fiber_lengths / l_c  # l < l_c: proportionally decreases
)

# Visualization
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(14, 5))

# Fiber length vs strength efficiency
ax1.plot(fiber_lengths, strength_efficiency * 100, 'b-', linewidth=2)
ax1.axvline(x=l_c, color='r', linestyle='--', linewidth=2,
            label=f'Critical Fiber Length l_c = {l_c:.2f} mm')
ax1.axhline(y=100, color='g', linestyle=':', alpha=0.5)
ax1.set_xlabel('Fiber Length [mm]')
ax1.set_ylabel('Strength Efficiency [%]')
ax1.set_title('Relationship between Fiber Length and Strength Efficiency')
ax1.grid(True, alpha=0.3)
ax1.legend()
ax1.set_ylim([0, 110])

# Interface shear stress distribution (for l > l_c)
l_fiber = 2 * l_c  # Example: fiber of length 2*l_c
x = np.linspace(0, l_fiber, 100)
# Fiber axial stress (linear increase, maximum at center)
sigma_fiber = np.where(
    x <= l_fiber/2,
    2 * tau_i * x / d,  # Increases from fiber end
    2 * tau_i * (l_fiber - x) / d  # Symmetric
)

ax2.plot(x, sigma_fiber, 'r-', linewidth=2)
ax2.axhline(y=sigma_f_max, color='g', linestyle='--',
            label=f'Fiber Strength {sigma_f_max} MPa')
ax2.set_xlabel('Position along Fiber Axis [mm]')
ax2.set_ylabel('Stress in Fiber [MPa]')
ax2.set_title(f'Stress Distribution in Fiber (l = {l_fiber:.2f} mm)')
ax2.grid(True, alpha=0.3)
ax2.legend()

plt.tight_layout()
plt.savefig('critical_fiber_length.png', dpi=300, bbox_inches='tight')
plt.close()

1.4.3 Effect of Interface Treatment

Fiber surface treatment (sizing, coupling agents) improves interface strength. Representative treatment methods and effects:

Treatment Method Mechanism Effect
Silane Coupling Agent Chemical bond formation 20-40% improvement in interface shear strength
Oxidation Treatment Increased surface functional groups Improved wettability and adhesion
Plasma Treatment Surface activation Introduction of polar groups, promoted chemical bonding
Sizing Agent Protective film formation Prevention of fiber damage, wettability control

Example 1.6: Interface Treatment Effect Simulation

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def composite_strength_with_interface(V_f, tau_i, sigma_f_max, d, l):
    """
    Composite strength considering interface strength

    Parameters:
    -----------
    tau_i : float
        Interface shear strength [MPa]
    l : float
        Fiber length [mm]
    """
    # Critical fiber length
    l_c = (sigma_f_max * d) / (2 * tau_i)

    # Effective fiber strength
    if l >= l_c:
        sigma_f_eff = sigma_f_max * (1 - l_c / (2 * l))
    else:
        sigma_f_eff = tau_i * l / d

    # Composite strength (simple rule of mixtures)
    sigma_c = sigma_f_eff * V_f

    return sigma_c, l_c

# Parameter settings
V_f = 0.5
sigma_f_max = 3500  # MPa
d = 0.007          # mm
l = 5.0            # mm

# Interface strength range (untreated vs treated)
tau_i_range = np.linspace(20, 80, 50)
sigma_c_values = []
l_c_values = []

for tau_i in tau_i_range:
    sigma_c, l_c = composite_strength_with_interface(V_f, tau_i, sigma_f_max, d, l)
    sigma_c_values.append(sigma_c)
    l_c_values.append(l_c)

sigma_c_values = np.array(sigma_c_values)
l_c_values = np.array(l_c_values)

# Visualization
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(14, 5))

# Interface strength and composite strength
ax1.plot(tau_i_range, sigma_c_values, 'b-', linewidth=2)
ax1.axvline(x=40, color='r', linestyle='--', alpha=0.5, label='Untreated')
ax1.axvline(x=55, color='g', linestyle='--', alpha=0.5, label='Treated')
ax1.set_xlabel('Interface Shear Strength τ_i [MPa]')
ax1.set_ylabel('Composite Strength [MPa]')
ax1.set_title(f'Effect of Interface Strength (V_f={V_f}, l={l} mm)')
ax1.grid(True, alpha=0.3)
ax1.legend()

# Critical fiber length
ax2.plot(tau_i_range, l_c_values, 'r-', linewidth=2)
ax2.axhline(y=l, color='g', linestyle='--', label=f'Actual Fiber Length {l} mm')
ax2.fill_between(tau_i_range, 0, l, alpha=0.2, color='green',
                  label='l > l_c (Efficient)')
ax2.set_xlabel('Interface Shear Strength τ_i [MPa]')
ax2.set_ylabel('Critical Fiber Length l_c [mm]')
ax2.set_title('Relationship between Interface Strength and Critical Fiber Length')
ax2.grid(True, alpha=0.3)
ax2.legend()

plt.tight_layout()
plt.savefig('interface_treatment_effect.png', dpi=300, bbox_inches='tight')
plt.close()

# Specific numerical comparison
tau_i_untreated = 40  # MPa
tau_i_treated = 55    # MPa

sigma_c_untreated, l_c_untreated = composite_strength_with_interface(
    V_f, tau_i_untreated, sigma_f_max, d, l)
sigma_c_treated, l_c_treated = composite_strength_with_interface(
    V_f, tau_i_treated, sigma_f_max, d, l)

print("Effect of Interface Treatment:")
print("-" * 60)
print(f"Untreated Fiber:")
print(f"  Interface shear strength: {tau_i_untreated} MPa")
print(f"  Critical fiber length: {l_c_untreated:.2f} mm")
print(f"  Composite strength: {sigma_c_untreated:.0f} MPa")
print(f"\nTreated Fiber:")
print(f"  Interface shear strength: {tau_i_treated} MPa")
print(f"  Critical fiber length: {l_c_treated:.2f} mm")
print(f"  Composite strength: {sigma_c_treated:.0f} MPa")
print(f"\nStrength improvement: {(sigma_c_treated/sigma_c_untreated - 1)*100:.1f}%")

1.5 Summary

In this chapter, we learned the following as fundamentals of composite materials:

  • Definition and classification of composite materials (fiber-reinforced, particle-reinforced, laminates)
  • Prediction of elastic modulus and strength using rule of mixtures (Voigt/Reuss models)
  • Precise elastic modulus calculation using Halpin-Tsai theory
  • Effect of aspect ratio in short fiber composites
  • Role of interface and concept of critical fiber length
  • Property enhancement mechanism through interface treatment

In the next chapter, we will focus on fiber-reinforced composite materials and learn about manufacturing methods for CFRP/GFRP, Classical Laminate Theory, and the A-B-D matrix.

Exercise Problems

Basic Level

Problem 1.1: Basic Calculation of Rule of Mixtures

For a unidirectional GFRP consisting of glass fiber (Ef = 70 GPa) and polyester resin (Em = 2.8 GPa), calculate the following. Fiber volume fraction Vf = 0.55.

  1. Longitudinal elastic modulus EL (Voigt model)
  2. Transverse elastic modulus ET (Reuss model)
  3. Anisotropy ratio EL/ET

Problem 1.2: Determination of Volume Fraction

The density of a CFRP composite is 1.55 g/cm³. Assuming carbon fiber density is 1.80 g/cm³ and epoxy resin density is 1.20 g/cm³, determine the fiber volume fraction Vf. (Hint: Use density rule of mixtures \(\rho_c = \rho_f V_f + \rho_m (1-V_f)\))

Problem 1.3: Rule of Mixtures for Strength

For a material system with fiber tensile strength 2800 MPa, matrix tensile strength 65 MPa, fiber failure strain 0.012, and matrix elastic modulus 3200 MPa, predict the tensile strength of a composite with Vf = 0.65.

Applied Level

Problem 1.4: Application of Halpin-Tsai Theory

For a short fiber composite (Vf = 0.4, Ef = 230 GPa, Em = 3.5 GPa) using fibers with aspect ratio l/d = 20, calculate the longitudinal elastic modulus using Halpin-Tsai theory and compare with the rule of mixtures (Voigt) result.

Problem 1.5: Effect of Critical Fiber Length

For carbon fiber (diameter 7 μm, tensile strength 3500 MPa) and epoxy resin with interface shear strength of 45 MPa, determine the critical fiber length and calculate the strength efficiency for fiber lengths of 2 mm, 4 mm, and 6 mm.

Problem 1.6: Design of Interface Treatment

For a short fiber composite with fiber length 3 mm, we want to utilize more than 95% of fiber strength. Determine the required interface shear strength. (Fiber: diameter 10 μm, tensile strength 4000 MPa)

Problem 1.7: Programming Task

Create a Python program with the following features:

  • Input fiber and matrix properties
  • Calculate elastic modulus for volume fractions 0-0.7 (rule of mixtures and Halpin-Tsai)
  • Display results graphically
  • Propose optimal fiber volume fraction (define cost function)

Advanced Level

Problem 1.8: Analysis under Multiaxial Stress

For a unidirectional CFRP subjected to simultaneous longitudinal stress σx = 500 MPa and transverse stress σy = 50 MPa, calculate the safety factor using the Tsai-Hill failure criterion. (Longitudinal tensile strength 1500 MPa, transverse tensile strength 50 MPa, shear strength 70 MPa)

Tsai-Hill criterion: \(\left(\frac{\sigma_x}{X}\right)^2 - \frac{\sigma_x \sigma_y}{X^2} + \left(\frac{\sigma_y}{Y}\right)^2 + \left(\frac{\tau_{xy}}{S}\right)^2 = \frac{1}{SF^2}\)

Problem 1.9: Probabilistic Approach

When fiber strength follows a Weibull distribution (shape parameter m=5, scale parameter σ₀=3500 MPa), estimate the strength distribution of composite materials using Monte Carlo simulation. Number of fibers N=1000, Vf = 0.6.

Problem 1.10: Optimization Problem

Under the following constraints, determine the fiber volume fraction and fiber length that minimize the cost/performance ratio of composite materials:

  • Target elastic modulus: EL ≥ 100 GPa
  • Fiber cost: 50 yen/kg, Matrix cost: 5 yen/kg
  • Fiber length range: 1-10 mm
  • Volume fraction range: 0.3-0.7

Implement an optimization program using scipy.optimize.

References

  1. Jones, R. M., "Mechanics of Composite Materials", 2nd ed., Taylor & Francis, 1999, pp. 45-89
  2. Hull, D. and Clyne, T. W., "An Introduction to Composite Materials", 2nd ed., Cambridge University Press, 1996, pp. 12-38, 112-145
  3. Mallick, P. K., "Fiber-Reinforced Composites: Materials, Manufacturing, and Design", 3rd ed., CRC Press, 2007, pp. 67-103
  4. Halpin, J. C. and Kardos, J. L., "The Halpin-Tsai Equations: A Review", Polymer Engineering and Science, Vol. 16, No. 5, 1976, pp. 344-352
  5. Kelly, A. and Tyson, W. R., "Tensile Properties of Fibre-Reinforced Metals: Copper/Tungsten and Copper/Molybdenum", Journal of the Mechanics and Physics of Solids, Vol. 13, 1965, pp. 329-350
  6. Chawla, K. K., "Composite Materials: Science and Engineering", 3rd ed., Springer, 2012, pp. 78-124, 156-187
  7. Daniel, I. M. and Ishai, O., "Engineering Mechanics of Composite Materials", 2nd ed., Oxford University Press, 2006, pp. 34-72
  8. Gay, D., Hoa, S. V., and Tsai, S. W., "Composite Materials: Design and Applications", 3rd ed., CRC Press, 2015, pp. 23-61

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