MECHANICS OF COMPOSİTE MATERIALS
(Last Update: July 16, 2026)
Professor Mehmet Zor / Dokuz Eylul University
Lecture Notes
(Dowload pptx)
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For Communication and Feedback
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Statement for Faculty Members
You can access the following materials related to this course from my website:
General Information:
More comprehensive general information about types, classifications, matrix and fibers of Composite Materials
Turkish exam questions and examples:
Some Exams and Answer Keys Taken in the Department of Mechanical Eng. and examples
Supporting Documents :
Standard Formula Paper used in exams, Literature Review Sample and other documents
Lecture Notes:
You can freely download the Turkish and English pdf and pptx files of this lecture notes.
Note: You can access the same type of materials regarding Statics, Dynamics, Strength and CAE courses on my website.
My personal WebSite:
My Youtube Channel for all turkish course videos: https://www.youtube.com/@mehmetzor
(tvid : turkish video)
Preface
Dear Student and Researcher Friends
Composite materials are used in many different sectors today due to the advantages they provide in terms of strength, lightness and economy. In parallel with technological developments, classical isotropic structural elements can be replaced by alternatives made of composites over time, and this situation is becoming more widespread day by day.
Today, it is possible to encounter such alternative composite elements, parts or equipment in many mechanisms, machines, structures or systems used in industry. R&D and innovation studies based on composite material alternatives are frequently carried out in many universities or private institutions.
For all these reasons, you, as future engineers, can only take an active part in such activities related to composites if you have sufficient knowledge of the basic principles, theories and approaches of composite material mechanics. Otherwise, you will not be able to provide the required level of information, guidance and evaluations expected from you in the design and analysis of a composite alternative structure that is very likely to be encountered in an R&D unit. This will undoubtedly affect your career negatively.
With these lecture notes, we aimed to convey to you the basic issues that an engineer should know in terms of mechanical calculations and measurements of composite materials. There is no doubt that an engineer who thoroughly understands these course notes will gain a privilege and a strong reason for being preferred in terms of basic knowledge and skills.
At the beginning, general information about composites is summarized in these notes, and each topic is associated with course training videos, which are my own explanations and can be accessed on mehmetzor.com. Tips that play a key role in understanding the course topics are specifically stated in each chapter. Lecture notes are updated over time, new topics or examples may be added, and last update dates are specifically stated. You can follow these updates on my website.
I hope my notes will be useful to all students and researchers.
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Mechanics of Composite Materials- Lecture Notes
January 2024
Mehmet Zor
Contents
1. Summary General Information
2. Anisotropic Material Types and Elastic Mechanical Properties (Anistropic Elasticity)
A- LAMINA (Unidirectional continuous fiber reinforced)
B- LAMINATED COMPOSİTES
8. Classical Lamination Theory (CLT) : Stress-Strain Calculations
9. Homegenization in Laminated Structures (Theoretical Approaches)
9.1 -9.3 Voigt, Reuss and CLT Methods
9.4 Zor Model
10. Strength and Failure Analysis of Laminated Structures Using the Zor Model
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Summary of �General Information� About �Composite Materials
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
You can access more comprehensive and detailed general information in the "General Information" document on en.mehmetzor.com.
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1.1 What is Composite Material?
A new material created by combining at least two different materials at the macro level (in such a way that they do not dissolve in each other) is called composite material.
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The aim is to develop and bring together some features (lightness, strength, flexibility, etc.) that are not available in the components alone.
1-Summary of General Information About Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1.2 Basic Properties of Composite Materials
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1-Summary of General Information About Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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1-Summary of General Information About Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1.3 Application Areas of Composites
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1-Summary of General Information About Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
The main sectors where composite materials are widely used and the product types used in these sectors are briefly summarized below:
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1.3.1 Some Areas Where Composites Are Used:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
1.3.2 Composite Product Examples
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1-Summary of General Information About Composites
Various Composite Products
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1-Summary of General Information About Composites
Concrete columns are created by combining iron and concrete and are actually a composite structure.
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1-Summary of General Information About Composites
iron bar (fiber)
Concrete (matrix)
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1-Summary of General Information About Composites
Composites used in an aircraft fuselage and their proportions
1.4 Advantages of Composite Materials.
Composite materials, which have many advantages over other materials with their characteristic features, are preferred due to their many superior properties such as,
1- long life,
2- lightness,
3-high chemical and mechanical resistance.
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1-Summary of General Information About Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Other Advantages Achievable with Composites:
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When the material properties in the table below are examined, it is seen that composite structures are both much lighter and much more durable than classical metals.
1-Summary of General Information About Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Isotropic Materials
Composites
Materials | Density ρ (gr/cm3) | Tensile Strength σç (MPa) | Modulus of Elasticiy GPa | Specific Tensile Strength σç / ρ | Specific Modulus of Elasticity E/ρ |
Non – Alloy Steel | 7.9 | 459 | 203 | 58 | 26 |
Alüminium | 2.8 | 84 | 71 | 30 | 25 |
Aluminium Alloy-2024 | 2.8 | 247 | 69 | 88 | 25 |
Brass | 8.5 | 320 | 97 | 38 | 11 |
Boron epoxy | 1.8 | 1600 | 224 | 889 | 124 |
Carbon-Epoxy-1 | 1.6 | 1260 | 218 | 788 | 136 |
Carbon-Epoxy-2 | 1.5 | 1650 | 140 | 1100 | 93 |
Kevlar-Epoxy | 1.4 | 1400 | 77 | 1000 | 55 |
S Glass-Epoxy | 1.8 | 1400 | 56 | 824 | 33 |
E Glass-Epoxy | 1.8 | 1150 | 42 | 639 | 23 |
1.5 Some Disadvantages of Composites
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
Delaminaiton
1.6 Composites and Engineering Activities
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1-Summary of General Information About Composites
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The fact that the mechanical behavior of composites is different from isotropic materials and can vary depending on the direction requires that they be examined with different mechanical approaches and criteria.
Sir, let's make the shaft material composite. Let's look at von-mises stresses again.
If we make the shaft composite, it would be more accurate to evaluate it according to the Tsai-Hill criterion, not Von-Mises. Yield-fracture criteria for composites are different.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
The options for creating composite materials are almost endless. Therefore, they are very difficult to classify.
However, common classifications will be emphasized here.
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1.7 Classification of Composite Materials
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
1.7.1 Classification According to Matrix Material Type :
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According to the type of matrix material, composites can be divided into 3 groups:
Classification Scheme by Matrix Material Type
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
1a) Thermosets: They are thermosetting plastics. One of the most well-known is Epoxy.
1b) Thermoplastics: Plastics that soften with heat.
1c) Elastomers: These are plastics that can stretch a lot (show large elastic deformation).The most well-known elastomer is rubber.
3. Ceramic Matrix Composites: They are high temperature composites and ceramic materials are used as matrix.
1. Plastic (Polymer) Matrix Composites
2. Metal Matrix Composites: These are composites in which light metals such as Aluminum and Zinc are used as matrix.
Since approximately 90% of composites are produced from polymer (plastic) based matrices, composite materials are also called reinforced plastics. They are divided into 3 groups.
1.7.2 Classification According to Shape and Placement of Reinforcement Elements :
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1-Summary of General Information About Composites
Since the composites in the first two groups are isotropic at the macro level, they are also called Quasi-Isotropic composites. , Group 3 composites show orthotropic properties.
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4 – Laminated (or Layered) Composites:
1.7.2 Classification According to Shape and Placement of Reinforcement Elements -Continue:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
fiber
lamina
matrix
laminated
composites
plates
θ: fiber orientation angle
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Sandwich composite structures fall into the laminated composite material class. Sandwich structures are obtained by gluing higher strength plates to the upper and lower surfaces of a low-density core material that does not carry load and has only insulation properties.
5- Sandwich Composites
The bottom and top layers can each be an isotropic material or a fiber-reinforced layer.
Sandwich composite panels used in exterior cladding
6- Hibrid Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
1.7.2 Classification According to Shape and Placement of Reinforcement Elements -Continue:
7- Natural Composites
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1-Summary of General Information About Composites
1.7.2 Classification According to Shape and Placement of Reinforcement Elements -Continue:
Composite structure of the tree
Composite structure of bone
Materials such as wood and bone are natural composite materials.
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1.8.1 Epoxy:
is an adhesive chemical resin from the thermoset group.
It is produced by polymerization of the epoxide group and its properties can be changed with different formulas. Depending on the type of hardener used, the properties of the composite material vary.
1.8.1.1 Some Superior Features of Epoxy :
1.8.1.2 Some Disadvantages of Epoxy
1.8 Some Important Materials Used in Composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
1.8.2 Glass Fiber:
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1-Summary of General Information About Composites
1.8.3 Carbon Fiber– Carbon Matrix (Carbon/carbon)
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1.8.4 Aramid Fiber (Kevlar)
Kevlar (Aramid) is a polymer fiber and a lightweight reinforcement material that provides high strength and rigidity to the composite structure. Aramid is aromatic polyamide, a type of nylon.
1.8.4.1 Some Outstanding Features
1.8.5 Boron Fiber :
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1.9.1 Mechanical and Thermal Properties of Thermoplastic Resins
1.9 Some Material Properties:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
Material | Specific Mass g/cm3 | Tensile Strength MPa | Modulus of Elasticity MPa | Temperature Limit oC |
Poly-Ethylene (PE) (low density) | 0.92-0.93 | 7-17 | 105-280 | 80 |
Poly-Ethylene (PE) (high density) | 0.95-0.96 | 20-37 | 420-1260 | 100 |
Poly-Vinyl-Chloride (PVC) | 1.50-1.58 | 40-60 | 2800-4200 | 110 |
Poly-Propylene (PP) | 0.90-0.91 | 50-70 | 1120-1500 | 105 |
Poly-Styrene (PS) | 1.08-1.10 | 35-68 | 2660-3150 | 85 |
Acronitrile-Butadiene-Strain(ABS) | 1.05-1.07 | 42-50 | - | 75 |
Poly-Meth-Metha-Archylic (PMMA) | 1.11-1.20 | 50-90 | 2450-3150 | 125 |
Poly-Tetra-Fluorine-Ethylene(PTFE)(Teflon) | 2.10-2.30 | 17-28 | 420-560 | 120 |
Polyamide (PA) Nylon 6.6 | 1.06-1.15 | 60-100 | 2000-3500 | 82 |
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1.9.2 Some Properties of Reinforcement Elements (Fibers)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites
Properties | E-Glass | S-Glass | Boron | Carbon | Kevlar49 |
Density (gr/cm3) | 2.54 | 2.49 | 2.68 | 1.85 | 1.44 |
Tensile Strength.(MPa) | 2000 | 4750 | 3450 | 2900 | 3750 |
Modulus of Elasticity(GPa) | 80 | 89 | 414 | 525 | 136 |
Fiber Diameter (µm) | 3-200 | 3-13 | 100-1000 | 5-13 | 12 |
Coeff. of Thermal Exp. (1/oC) | 5x10-6 | 2.9x10-6 | 3xx10-6 | -1x10-6 | -2x10-6 |
Kopma Uz. (%) | 2.75 | - | 0.7 | 0.5-1.3 | 2.5 |
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1.9.3 Tensile Curves of Fiber Materials
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1-Summary of General Information About Composites
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1.10.1 Major Composite Manufacturing Methods
Hand Lay-Up
Spray - up
Wet Filament Winding
RTM
Pultrusion
1.10 Manufacturing Technologies in Composites :
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1-Summary of General Information About Composites
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1.10.2 General Features of Manufacturing Methods:
Compression Molding
Vakum Bonding / Vakum Bagging
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1.10.3 Grouping of Manufacturing Methods According to Composite Material Type:
Manufacturing Methods
Thermoset Materials:
Thermoplastic Materials:
Short fiber Composites
Continius fiber Composites
Short fiber Composites
Continius fiber Composites
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1-Summary of General Information About Composites
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1.11 Let's remember the important concepts in strength :
1.11.1 Isotropic Material:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites / important concepts in stregth
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1.11.2 Separation Principle and Concept of Stress:
If a system subject to the influence of external forces is in balance, each of its parts, which we have separated on an imaginary basis, is also in balance separately. This is called the separation principle.
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1.11.3 Stress Types
1.11.3.1 Normal Stress : If the stress is parallel to the normal of the section plane, in other words, if it is perpendicular to the plane, it is called normal stress. It is denoted by σ . It occurs in tensile-compression and Bending loading.
1.11.3.2 Shear Stress : If the stress is perpendicular to the plane normal, in other words, if it is parallel to the plane, it is called shear stress. It is denoted by τ .
τyx
τxy
Meaning of shear stress indices:
plane normal
Stress Direction
τi j = τj i
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1.11.4 Deformation
L 0
Δ L/2
Δ L/2
1.11.4.1 Unit Elongation : Strain (ε)
1.11.4.2 Shearing Strain: Angle (γ):
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites / important concepts in stregth
1.11.5 Tensile test and stress-strain diagram:
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that best characterizes the mechanical behavior of materials .
Rupture
The part that connects to the chin
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Poisson's ratio:
Poisson's ratio: Varies between 0 and 0.5
Attention : Formula 1.1 is valid only in case of x-axis (uniaxial) loading . If there were forces in the y or z directions as well as P, this formula could not be used.
(Hooke's equation for uniaxial loading .)
Remember: Always,
Strain= Total elongation / initial length in that direction
If the rod in the figure is subjected to P load in the x direction, stress occurs in only x direction and no stress occurs in the y and z directions.
,
,
Poisson's ratio is the material property that gives the deformation effect of a load in other directions .
1.11.5.1 Poisson Ratio ( ν ):
(1.1 )
(1.2 )
(1.3ac )
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1.11.5.2 Modulus of Rigidity ( or Shear modulus) G
with The elastic relationship
(1.7 )
(1.4 )
(1.5 )
(1.6 )
Hooke’s equations between shear stresses and strains in other planes :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1-Summary of General Information About Composites / important concepts in stregth
1.11.6 Hooke’s Equations in the Most General Case:
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Temperature Effect
Shearing Strain
isotropic materials, these are the relations between stress-strain at any point Q:
(1.8.a )
(1.8.b )
(1.8.c )
(1.8.d )
(1.8.e)
(1.8.f )
Axial Strain
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2-3 plane
1-2 plane
1-3 plane
ANISOTROPIC MATERIAL TYPES
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2.
(tvid - 2)
(tvid: turkish course video number)
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2- Anisotropic Material Types
2.1 What is an Anisotropic Material?
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2- Anisotropic Material Types
2.2 Our Aims in This Chapter:
In article 1.11.6, it was shown that there are 2 independent material constants (E and ν) in the elastic region for isotropic materials and the stress-strain relations (Hooke’s equations) were summarized.
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Stiffness matrix: [C]
(9x9= contains 81 independent material constants)
2.3 Elastic Material Constants for Most Generally Anisotropic Materials
1st subscript: indicates the plane normal.
2nd subscript: shows the direction of stress or deformation.
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2- Anisotropic Material Types
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Compliance Matrix: [S]
It is the inverse of the stiffness matrix.
[S]=[C]-1
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types/ Elastic Material Constants for Most Generally Anisotropic Materials
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If the same analysis is made for planes 1-3 and 2-3:
it is found as
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types/ Elastic Material Constants for Most Generally Anisotropic Materials
Abbreviated Notations
From this last equation it is found as
t: thickness
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In this case, in an anisotropic material, the independent elastic material constants in the stiffness or compliance matrices are reduced to 6x6 = 36.
Thus, using shortened notations for an anisotropic object, stress-strain or strain-stress relations can be expressed in 2 different matrix formats as follows:
Or with index notation:
or
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types / Elastic Material Constants for Most Generally Anisotropic Materials
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The energy density stored in a nonlinear elastic material is dW,
In this case, the energy density can also be written as:
Stress-strain relationship can be writen as:
(I)
(II)
Since equations (I) and (II) are equal;
In this case, the stiffness (C) and compliance ( S ) matrices must be symmetric about the diagonal.
The independent, elastic material constants in these matrices are 21. In terms of material properties, such an anisotropic material that does not have planes of symmetry is called a triclinic material.
2.5 Triclinic Material
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types
The reciprocity condition holds for all elastic materials for which Hooke’s law is valid. Namely:
Finaly,
Reciprocity condition:
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In the figure, monoclinic material that is symmetrical with respect to the (x-y) plane is symbolized. (Note that each little cube is symmetrical with respect to this plane.)
Potassium Feldspar
(It is an example of a monoclinic material found in nature.)
2.6 Monoclinic Material
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types
Symbolic form of monoclinic structure
plane of symmetry
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We think that there is material symmetry with respect to the 1-2 plane at some point in an anisotropic, elastic and monoclinic structure.
Now we will determine the material constant number of the monoclinic material:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types/ Monoclinic Material
Similarly; In the following cases, there will be no distortion in the 1-3 and 2-3 directions:
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In this case, for a monoclinic and elastic material, the independent material constants in the compliance and stiffness matrices are reduced to 13.
,
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2- Anisotropic Material Types/ Monoclinic Material
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Below is shown the explicit expression of each term of the [S] compliance matrix for a monoclinic material.
(2.4)
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2- Anisotropic Material Types/ Monoclinic Material
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2.6.1. Hooke’s Equations for Monoclinic Material – (3D Case)
(2.5)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(2.6.a-f)
(a)
(b)
(c)
(e)
(f)
(d)
Explicit versions of the stress-strain (Hooke) equations in 3D for a material showing monoclinic behavior:
2- Anisotropic Material Types/ Monoclinic Material
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2.7 Orthotropic Material
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2- Anisotropic Material Types/ Ortotropic Material
2-3 plane
1-2 plane
1-3 plane
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2- Anisotropic Material Types/ Ortotropic Material
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Compliance [S] matrix for orthotropic material:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
In 2D Case:
In 3D Case
As a result, there are 9 independent material constants for an orthotropic material:
(2.7a)
(2.7b)
2- Anisotropic Material Types/ Ortotropic Material
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Stiffness [C] matrix for orthotropic material :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
In 2D Case
In 3D Case
Attention: The [C] matrix is symbolized as [Q] for orthotropic materials.
(2.7c)
(2.7d)
2- Anisotropic Material Types/ Ortotropic Material
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1
3
2
2.7.1 Industrial examples for orthotropic composites.
1
2
3
a-) Unidirectional continuous fiber reinforced composite layers: These are structures formed by arranging fibers, each of which is a single wire (without discontinuity), in a matrix in a way that they are parallel to each other in one direction.
(a)
(b)
Since the following composite types are symmetrical with respect to three planes, they exhibit orthotropic character and have widespread applications in industry.
b-) Bidirectional woven fabric composite layers: : These are structures formed by arranging fibers, each of which is a single wire (without discontinuity), in a matrix in a way that they are parallel and perpendicular to each other in two directions.
c-) Sandwich Composites: These are composite structures obtained by gluing two more durable plates to the upper and lower surfaces of a core material.
(c)
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Example 2.1: The plane stress state at a point of the unidirectionally reinforced Graphite/Epoxy Composite is shown in the figure.
Properties | Symbol | Unit | Glass/Epoxy | Boron/Epoxy | Graphite/Epoxy |
Fiber Volume Fraction | | | 0,45 | 0,50 | 0,70 |
Modulus of Elasticity in Fibers Direction (1) | | GPa | 38,6 | 204 | 181 |
Modulus of Elasticity Perpendicular to Fibers(2) | | GPa | 8,27 | 18,50 | 10,30 |
Major Poisson Ratio | | | 0,26 | 0,23 | 0,28 |
Shear Modulus | | GPa | 4,14 | 5,59 | 7,17 |
Mechanical Properties of a Unidirectional Layer for Different Materials
Accordingly, find the following values:
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a-)
Solution:
From the table on the previous page, the material properties for graphite/epoxy read as follows:
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2- Anisotropic Material Types/ Ortotropic Material
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b-)
c-)
d-)
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2- Anisotropic Material Types/ Ortotropic Material
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2.7.1 Hooke’s Equations for Orthotropic Material – (3D Case)
(2.8)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(2.9.a-f)
(a)
(b)
(c)
(e)
(f)
(d)
Stress-strain relations (Hooke’s equations) in 3D for a material showing orthotropic behavior.(valid for elastic loading)
2- Anisotropic Material Types/ Ortotropic Material
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If there are isotropy planes parallel to each other at every point of the orthotropic material, these materials are called transversely isotropic materials.
Although the material properties are the same in all directions on that plane, they differ in the direction perpendicular to the plane.
The structures obtained by placing the fibers parallel to each other and irregularly in the matrix show transverse isotropic character. The properties are almost the same in every direction in the 2-3 plane. The properties in the 1 direction are different.
2.8 Transversely Isotropic Material
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
An Isotropy plane
There are infinite planes of symmetry, the normal of which is direction 1 and parallel to plane 2-3.
2- Anisotropic Material Types/ Transversely Isotropic Material
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In composite materials with rotational symmetry with respect to fiber axis 1, material properties can be considered the same in all directions in sections parallel to plane 2-3.
Thus, the independent material constant is reduced from 9 to 5:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(2.10.a-e)
(2.11)
Since the 2-3 plane is isotropic, equation 2.11 is also valid for this plane.
While E2 and ν23 are known G23 can be calculated from equation 2.11 valid for isotropic materials:
2- Anisotropic Material Types/ Transversely Isotropic Material
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False
True
The tensile speciment transverse dimensions must be large enough and be representative of the entire structure.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Material constants matrices for transversely isotropic materials:
Test speciment
2- Anisotropic Material Types/ Transversely Isotropic Material
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2.8.1 Hooke’s Equations for Transversely Isotropic Materials
(2.12)
(3d Case)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(2.13.a-f)
(a)
(b)
(c)
(e)
(f)
(d)
Stress-strain relations (Hooke’s equations) in 3D for a material showing transverse isotropic behavior. (valid for elastic loading)
2- Anisotropic Material Types/ Transversely Isotropic Material
In fact, transversely isotropic materials are a special case of orthotropic materials, and the Hooke relations valid for orthotropic materials are also valid for transversely isotropic materials (See topic 2.7.1).
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a- Reinforced Concrete Columns: Reinforced concrete columns, which are mostly exposed to tension or compression, are transversely isotropic materials. Iron bars are placed in a mixed manner in the concrete matrix so that they are parallel to the column axis, which is the loading direction. The aim is to strengthen the matrix in the direction of the tension/compression load.
2.8.2 Natural and industrial examples of Transverse Isotropic composites :
b-) Sedimentary Rocks :The property of transverse isotropy is seen in nature in layered sedimentary rocks with long wavelengths. Each layer has approximately the same properties in its own plane (each layer is parallel to the 2-3 planes). However, there are different properties along the thickness (from bottom to top, as axis 1). The plane of each layer is the isotropy plane and the vertical axis is the axis of symmetry.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types/ Transversely Isotropic Material
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2.9 Isotropic Material
2
3
1
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types/ Isotropic Material
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2.9.1 Hooke’s Equations for Isotropic Materials (3D Case)
(2.14b)
2- Anisotropic Material Types/ Isotropic Material
Stress-strain relations (Hooke’s Equations) for a material showing isotropic behavior. (valid for elastic loading)
(a)
(b)
(c)
(e)
(f)
(d)
(2.15.a-f)
As can be seen, there are only two independent elastic material constants (E, ν) in isotropic materials.
ve
(2.14a)
Shear Modulus
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Or if we calculate the stresses from equation 2.14 b, we obtain the following equation for isotropic materials:
2- Anisotropic Material Types/ Isotropic Material
(2.14.c)
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2.9.2 Composites Having Isotropic Character:
b-) Chopped Fiber (Whiskers) Reinforced Composites
c-) Multidirectional, Continuous Fiber Reinforced Composites
a-) Particle reinforced composites and nano composites
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types/ Isotropic Material
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2.10 Hooke’s Equations in Plane Stress : (For Orthotropic, Transversely Isotropic and Isotropic Materials)
(2.16b)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
temperature effect *
(2.17)
Important points:
For orthotropic materials, in the case of plane stress we can write the Hooke equations in matrix format as follows:
(a)
(b)
(c)
(2.16a)
2- Anisotropic Material Types
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Material Character | Reinforcement Direction | Composite type | 3. Bölüm | 4. Bölüm | 5. Bölüm | 6. Bölüm | 7. Bölüm | 8. Bölüm | 9. Bölüm | 10. Bölüm | 11. Bölüm |
Theo. Calc. of Mat. Properties | Strength and yield limits | Experi-mental Measur. | Hooke and Transfor-mation Equations | Failure Criteria | Laminated Composite Calculations | Thermal Loadings | Calc.of Mat.Pro.& Strength Limits | Calc.of Mat.Pro. & Strength Limits | |||
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quasi Isotropic | Multidirectionally Reinforced Composites | Continuous Fiber Reinforced Composites
| Since these composites exhibit isotropic character, all mechanical calculations can be done with the approaches and equations in the topics shown in the Strength of Materials course. Therefore, they are not included in the scope of these lecture notes. | ||||||||
Discontinuous Fiber Reinforced Composites (whiskers) | |||||||||||
Composites Reinforced with Nano or Macro Particles | |||||||||||
Tablo 2.1 Classification of Composites According to Reinforcement Direction and Material, and Sections Explaining Mechanical Calculations
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
2- Anisotropic Material Types
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THEORETICAL CALCULATION OF ORTHOTROPIC MATERIAL PROPERTIES
3.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
in a Unidirectional and Continuous Fiber Reinforced Composites
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3.1 Our Purpose in This Section
While the elastic mechanical properties (Ef, νf, Em, νm) of the isotropic components (matrix and fiber) that make up the composite are known;
It is to theoretically calculate the properties (E1, E2, ν12, G12) of a one-way reinforced orthotropic composite layer obtained by combining these. However, it should be noted that these calculations will also be valid for transversely isotropic composites.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
3.2 Local Axes in the Orthotropic Layer
1: Axis parallel to the fiber direction in the layer plane,
2: axis perpendicular to the fiber direction in the layer plane
3: Axis perpendicular to the layer plane and in the thickness direction
3.3 Representative Volume Element in an Orthotropic Layer
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Unidirectional transverse isotropic composite structures (see: 2.11.3): The representative volume element is the same for these composites. This proves that the calculations made and the equations to be derived for composite plates reinforced with unidirectional continuous fibers are also valid for this type of composite.
1
2
3
3.4 Relationship Between Fiber and Matrix Volume Ratios�
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Fiber volumetric ratio
Matrix volumetric ratio
Theoretical Calculation of Composite Density:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(3.1)
(3.2)
Cross-sectional areas with normal in 1 direction :
A1
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Attention: There are only matrix and fiber volumes in the structure. Also, there is no independent volume called Composite. The material or volume called composite is theoretical and represents the entire structure. (Or we can think that since it is the only orthotropic material, the entire structure is called composite.)
3.5 Theoretical Calculation of E1 (Modulus of Elasticity in the Fiber Direction 1)
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We consider the representative volume element to which a pulling force P1 is applied in direction 1. Since the fiber and matrix are completely adhered to each other, They affect each other and extend by the same ΔL in the 1 direction. Since their initial lengths are equal, their unit elongation (ΔL/L) will also be equal.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Tip-1 :
In Direction 1, the strains of fiber, matrix and composite are always equal to each other.
From the static equilibrium of the left part of the Ι-Ι section,
(3.3)
(3.4)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Effect of Fiber Ratio on E1 Value
Additionally, from the equation:
(3.5)
This equation 3.5 will appear in future calculations.
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
the relationship between the stresses in direction 1 is obtained as:
3.6. Theoretical Calculation of E2 (Modulus of Elasticity Perpendicular to the Fibers)
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Tip 2: The stresses in both directions are equal to each other.
(3.6)
When we take Ι-Ι and II-II sections, respectively, in the representative volume element to which the P2 draft force is applied in the 2 direction; The internal forces and cross-sectional areas in the fiber and matrix are equal to those in the composite; Therefore, we can understand from the figures below that the stresses are the same as the stress in the composite.
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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Total elongation in direction 2;
t
L
σ2
σ2
(3.9a)
(3.8)
(3.7a-c)
When the Poisson effect is neglected, the strains in the 2nd direction from Hooke's relations are:
or
(3.9b)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Effect of Fiber Ratio on E2 Value
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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The general expression of Poisson's ratio for a layer showing orthotropic character is:
Major Poisson Ratio:
According to this;
Minor Poisson Ratio:
Transverse Poisson Ratio:
There are three other non-independent Poisson ratios:
There is also a general relationship between Poisson ratios and Elasticity Modules in an orthotropic material, as in equation 3.13:..>>
Notes: 1-) Equations 3.11, 3.12 and 3.13 can be used for all composite types with orthotropic properties.
2-) Poisson ratio cannot be negative except for some special materials. It cannot be greater than 0.5 in isotropic materials.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(3.11)
(3.12a-c)
(a)
(b)
(c)
(3.13)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
3.8 Theoretical Calculation of ν12 (Major Poisson Ratio)
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
In the representative volume element subjected to tension in direction 1, the total strain (ΔW) in direction 2 is equal to the sum of the strains in the fiber and matrix.
If we write the Poisson ratios of the matrix and fiber :
, Similarly for fiber..>>
(3.14)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Effect of Fiber Ratio on ν12
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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3.9 Theoretical Calculation of ν12 (Shear Modulus in plane 1-2)
shear strain angle in composite (γ12);
For the whole composite material;
We can also write this relation for fiber and matrix :
Due to static equilbrium, the internal shear forces must be equal. Since the A2 cross-sectional areas are also equal, the shear stresses in the fiber, matrix and composite are also equal.
or
,
Since the fiber and matrix are isotropic,
(3.15a)
(3.15b)
(3.16a-b)
,
(a)
(b)
(We thought of the entire structure as a single orthotropic material and named it composite.)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
15, Agust 2025
Shear stress (τ12) occurring in 1-2 plane in representative volume element creates different deformation angles (γf , γm) in fiber and matrix.
Example 3.1
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
| Glass Fiber | Epoxy |
Modulus of Elasticity | Ef =110 GPa | Em = 3,5 GPa |
Poisson Ratio | νf =0,27 | νm = 0,3 |
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Find the elastic properties of the Glass fiber-Epoxy composite to be obtained by combining the materials whose E, ν values are given in the table above.
(Fiber Volume Ratio = Vf = 0,3)
E1 = ? , E2= ?, G12 = ? , ν12=?
Solution:
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From equation (3.4):..>>
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
From equation (3.9):..>>
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
From equation (3.16a):..>>
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
From equation (3.16b):..>>
From equation (3.15b):..>>
From equation (3.14):..>>
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Materials | diameter (μm) | density ρ (kg/m3) | Modulus of Elasticiy E (GPa) | Poission Ratio ν | Tensile Strength σult (MPa) |
E-glass | 10 | 2600 | 74 | 0,25 | 2500 |
S-glass | 10 | 2500 | 86,9 | 0,22 | 2850 |
Kevlar 49 | 12 | 1450 | 130 | 0,4 | 2900 |
“HT«High Strength | 7 | 1750 | 230 | 0,3 | 3200 |
“HM” High Modulus | 6,5 | 1800 | 390 | 0,35 | 2500 |
Boron | 100 | 2600 | 200 | | 3400 |
| | | | | |
| | | | | |
3.11 Mechanical properties of some fiber materials
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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3.12 Mechanical properties of some matrix materials
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
Malzeme | Density ρ (kg/m3) | Modulus of Elasticiy E (GPa) | Poission Ratio ν | Tensile Stress σult (MPa) |
Epoxy | 1200 | 4,5 | 0,4 | 130 |
Phenolic | 1300 | 3 | 0,4 | 70 |
Polyester | 1200 | 4 | 0,4 | 80 |
Polycarbonate | 1200 | 2,4 | 0,35 | 60 |
Vinylester | 1150 | 3,3 | | 75 |
Silicone | 1100 | 2,2 | 0,5 | 35 |
Urethane | 1100 | 0,7-70 | | 30 |
Polyimide | 1400 | 4-19 | 0,35 | 70 |
PolyPropylene (PP) | 900 | 1,2 | 0,4 | 30 |
PolyPropylene Sulfone (PPS) | 1300 | 4 | | 65 |
PolyAmide (PA) | 1100 | 2 | 0,35 | 70 |
PolyEther Sulfone (PES) | 1350 | 3 | | 85 |
PolyEtherImide (PEI) | 1250 | 3,5 | | 105 |
PolyEther-Ether-Ketone (PEEK) | 1300 | 4 | | 90 |
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An aluminum sheet is reinforced unidirectionally with continuous boron fibers. Fibers constitute 36% of the total volume. The properties of the materials are given in the table below. A composite layer is produced by combining these materials. Calculate the following properties of this composite layer: a-) density, b-) Elasticity Modules in the 1 and 2 directions, c-) Poisson ratios in the 1-2 plane (major and minor), d-) Rigidity module in the 1-2 plane.
| Density (gr/cm3) | Modulus of Elasticity E (GPa) | |
Boron Fiber | 2,6 | 379 | 0,2 |
Aluminum | 2,7 | 70 | 0,33 |
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
d-)
Solution:
a-)
b-)
c-)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
ρ= 2,66 gr/cm3
Örnek 3.2
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3.13 ) Poisson Effects (p):
F1
F1
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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(3.17a)
(3.17b)
The stresses in 2 directions are equal:
(3.18a)
(3.18b)
(3.6)
If we substitute equation 3.6 into equations 3.17:
Strains in the fiber and matrix in the 1st direction: We substitute equation 3.6 into equations 2.17 (for isotropic materials).
Strains in direction 2:
(3.7a)
(3.19a)
(3.19b)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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(3.20)
(3.8)
(3.18b)
(3.18a)
(3.7a)
If the last equation is rearranged :
(3.5)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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(From equa. 3.19a):
(3.3)
(From equa. 3.12b and equa.3.7a):
(3.21)
(3.14)
(3.4 )
(3.13 )
(3.22)
(3.23)
(If equation 3.22 is substituted into 3.21 and rearranged,):
If we substitute equation 3.23 into equation 3.20:
..>>
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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The result after editing the last term is::
Poisson interaction term :
(3.24a)
(3.25)
or
(3.24b)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Example 3.3
What difference in E2 value occurs when you take into account the Poisson's ratio for the composite layer in Example 3.2? Calculate.
Poisson effect term :
From equa. 3.24 :
When Poisson effect is neglected :
was found
The difference is:
Solution:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
If the temperature of the fiber-reinforced orthotropic composite plate is increased by ΔT while it is free, changes in the dimensions of the plate occur. These size changes are called thermal deformations. We calculate them as follows:
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
, α1
, α2
Total Elongations:
Unit Elongatios (Strains):
(3.26)
(3.27)
Let's remember from equation 3.3 that the strains in the 1-direction in the fiber and matrix are the same as the strain of the composite (Tip 1):
No local shear deformation occurs due to ΔT :
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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3.14.1.a If we consider the structure as a single orthotropic material: Free deformation occurs in all directions. For this reason, the total internal forces, and therefore the stresses, that will arise in all axes within the material will be zero.
Now, we consider that the temperature of a unidirectionally reinforced orthotropic composite layer, which is not limited in any part, is increased by ΔT.
3.14.1 Thermal Stresses in Free State
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
representative volume element
We will examine the representative volume element for calculations.
(3.28)
(3.29)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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(Eq. (3.311) will be used in the α2 calculation)
3.14.1.b If we consider the structure as 2 different isotropic materials (matrix and fiber):
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(3.31)
(3.30)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Kompozit Malzeme Mekaniği-Ders Notları-Prof.Dr.Mehmet Zor
Eq. 3.26 :
(3.32)
(3.33)
3.14.2 Theoretical Calculation of α1
We think that the temperature of a composite layer that is unconstrained on any surface (i.e, free layer) is changed by the amount ΔT.
Similarly:
Eq. 3.30 :
Eq. 3.3 :
(3.34)
(from eq. 3.4)
The total strain in the 1 direction caused by ΔT in the fibers. :
The total strain in the 1 direction caused by ΔT in the matrix. :
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Kompozit Malzeme Mekaniği-Ders Notları-Prof.Dr.Mehmet Zor
3.14.3 Theoretical Calculation of α2
Again, we consider that the temperature of an unconstrained composite plate is changed by ΔT.
If we substitute equation 3.26 into equations 3.32 and 3.33;
(3.35)
(3.36)
Hooke relations in Equation 2.15 for fiber and matrix;
(a)
(b)
,
(3.26)
(3.3)
(3.32)
(3.33)
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Similarly for the matrix :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(3.37)
(3.38)
If we use equations 9.12 and 9.13 in equation 3.8:
If this equation is arranged..>>
(3.8)
In that case;
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
When the equations are arranged:
If we open the equation:
Then the equation takes the form:
When last edited:
(3.39)
(from eq. 3.14):
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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3.14.4 How are the theoretical calculations of α1 and α2 values of other types of composites made?
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Discontinuous fiber (whicker)reinforced composite
Particle reinforced composite
Double woven fabric reinforced composite
Sandwich Composite
multidirectional continuous fiber reinforced composites
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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Material | | |
Boron/Epoxy | 5 | |
Graphite/Epoxy | 0,88 | 31 |
E-glass/Epoxy | 6,3 | 20 |
Aluminum | 22 | 22 |
Copper | 16 | 16 |
Steel | 12 | 12 |
3.14.5 Thermal expansion coefficients of some materials
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
In the next section, we will further reinforce the subject with examples including formula deductions for a free or limited monolayer.
3. Unidirectional - Continuous Fiber Reinforced Composites / Theoretical Calculation of Orthotropic Material Properties
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THEORETICAL STRENGTH AND EFFICIENCY LIMITS
OF COMPOSITE
4.
(STRESS-STRAIN CALCULATION EXAMPLES RESULTING FROM STRUCTURAL LOADING)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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4. Theoretical Strength And Efficiency Limits of Composite
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4.1 Our Aims in This Section are:
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4.2 Load Distribution fraction Calculation in Composite Structure:
Tip for direction 1: Unit elongations are equal to:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(4.1)
4. Theoretical Strength And Efficiency Limits of Composite
Total Force on composite
Force carried by fibers
Force carried by matrix
;
Representative volume element :
How much of the tensile load in direction 1 do the fibers carry? First we are looking for an answer to this question :
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Stage 1: Elastic deformation occurs in the fibers and matrix.
Stage 2: While elastic deformation continues to occur in the fibers, the matrix undergoes plastic deformation.
Stage 3: Plastic deformation occurs in both fibers and matrix.
Stage 4: First the fibers and then the matrix are damaged.
(such as brittleness-ductility).
4.3 Deformation Stages of Composite Structure
When a unidirectionally reinforced composite layer with continuous fibers is subjected to tension in the 1-direction, it deforms in four stages as the load increases.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
(It is accepted that the composite is damaged when the fibers break.)
Fiber
damage
damage
matrix
composite
Stage 1
Stage 2
Stage 3
Stage 4
Stress
Strain
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(3.5)
koma
fiber
composite
matrix
rupture
rupture
a
b
c
d
rupture
Stages of deformation in a brittle structure
Figure 4.1
(4.5)
(4.2)
(4.3)
(4.4)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(It was explained in the calculation of E1.)
4. Theoretical Strength And Efficiency Limits of Composite
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Because the stresses in the fiber and matrix in the 2 direction will be equal, the matrix with lower strength will break first.
Under industrial working conditions, the direction of loading should coincide with the fiber direction (1 direction) in the composite. Because the 1st direction of the composite is more durable than the 2nd direction. Coinciding the load direction with the direction perpendicular to the fibers (direction 2) would be a wrong practice for unidirectional fiber reinforced composites. If there is loading in both 1st and 2nd directions, bidirectional fiber reinforced (cross-ply or woven fabric) composites must be used.
4.5 What is the Composite Strength in the (2) direction perpendicular to the fibers?�
(4.6)
4. Theoretical Strength And Efficiency Limits of Composite
For a unidirectional, continuous fiber reinforced layer,
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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4.6 Theoratical Compressive Strengths �
It is assumed that when the fibers in a layer subjected to compressive loading in the 1st direction are damaged, the composite is also damaged. Accordingly;
kpma
fiber
composite
matrix
damage
a
c
damage
Compression
Tension Region
(4.7)
(4.8)
(4.11.a)
Equation 3.5, which gives the compressive stress in the composite at any moment in direction 1, is written for the moment when the fibers are subjected to compressive damage.
(4.9)
(4.10)
4. Theoretical Strength And Efficiency Limits of Composite
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As stated in the theoretical calculation of G12 , which is the subject of 3.9, the shear stresses occurring in the composite, fiber and matrix are equal in case of shear loading.
Accordingly, the first component to be damaged due to shear loading will be the matrix with the lowest shear strength. (Because the matrix will reach its first strength limit.)
As a result, the shear strength of the composite is:
(4.11.c)
Any shear loading instant
(4.11.b)
Note: In some special cases, the shear strength of the fiber may be lower than that of the matrix. In this case, the theoretical shear strength of the composite should be taken as the shear strength of the fiber.
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Example 4.1
A unidirectional and continuous fiber reinforced orthotropic composite layer will be produced from the matrix and fiber materials whose properties are given in the table. In this layer with dimensions of 400mm x 400mm x 4mm, 25% volumetric fiber will be used. According to this,
a-) When pulling in direction 1, find the stresses in the fibers and matrix at the moment when the composite will be damaged.
b-) If tensile forces of F1 = 80kN, F2 = 48kN are applied to this layer simultaneously in directions 1 and 2, calculate the stresses and total deformations (extensions) that will occur in the composite, fiber and matrix in directions 1 and 2.
c-) If we had not neglected the Poisson effect, how much would the total elongation in the 2 direction we found in option c change?
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Solution:
a)
From equation (4.3):
To solve another options of this problem, we must first calculate the orthotropic properties:
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Stresses in the composites:
F1=80kN
F1
A1
F2
A2
F2=48kN
b-)
Unit elongation (strain) in the composite in direction 1:
Total elongation in the composite in direction 1:
From equation (2.17.a)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Total elongation in the composite in direction 2:
From equation (2.17.b)
4. Theoretical Strength And Efficiency Limits of Composite
Strain in the composite in direction 2:
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0
0
0
0
That is, from equation 3.3, strains in the 1 direction in the fiber and matrix:
Strains in direction 1 are equal (from tip-1)
Stress in direction 2 are equal (from tip-2)
That is, from equation 3.6, stresses in the 2 direction in the fiber and matrix:
We can apply Hooke's equations, which are valid for isotropic materials, separately for fiber and matrix. (Because fiber and matrix materials are isotropic.)
From equation (2.15.a):
From equation (2.15.b):
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c-)
Poisson effect factor :
Strain in direction 2 in the composite :
When we take the Poisson effect into consideration, the changing values are E2 and ν21.
If the Poisson effect is not neglected, we put (*) above the affected values.
Total elongation in composite in direction 2:
If the poisson effect is not neglected :
If the poisson effect is neglected :
From equation (3.10b):
From equation (3.13):
From equation (3.10.a):
From equation (2.17.b):
Difference between elongations:
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Example 4.2
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
The layer in Example 4.1 is placed in a fixed cavity suitable for its dimensions, as shown in the figure, and is subjected to a compression force of F = 60kN in the 1 direction. According to this; Calculate the changes in the side lengths of this layer in directions 1 and 2. (The bottom, back and side surfaces of the layer are in contact with the cavity.)
Çözüm:
stress in direction 1 :
Due to the constraints, the unit and total strains in the 2 direction are zero:
Total shortening in direction 1 :
From equation (2.13.b):
From equation (2.13.a):
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An aluminum plate with an initial temperature of 23 C is placed between two fixed walls as shown in the figure. According to this,
a-) Can the aluminum plate be used at 120 oC under these boundary conditions? Calculate.
Example 4.3* (video 3)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Material | E �Modulus of Elasticity | ν�Poisson ratio | α�Coefficient of Thermal Expansion (CTE) | Tensile/Yield Strength |
Steel - fiber (ductule) | 210GPa | 0,28 | 10x10-6 1/ oC | 800 / 400 MPa |
Aluminum - matrix (ductule) | 70 GPa | 0,27 | 23x10-6 1/ oC | 200/150 MPa |
(* Attention: This example also includes formula inferences regarding thermal loads.)
b-)Can this composite structure be used at 120 oC?
c-) When the right wall is removed, to what temperature can the composite structure be heated within its strength limits?
d-) When the right wall is removed, what will be the stress value that will occur in the matrix component of the composite structure at the maximum allowable (safe) temperature?
The same aluminum plate will be unidirectionally reinforced with 25% steel fibers to create a composite structure and this structure will be placed between two fixed walls. Accordingly, for this composite structure, answer the following questions with calculations.
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Aluminum
Aluminum
Aluminum
a-) Can the aluminum plate be used at 120 oC ?
P
As a result, when the temperature exceeds 116.1 °C, the aluminum material will flow. This means that Aluminum plate cannot be used at 120 oC. (We can make this calculation with our strength information)
= 0
Aluminum can be heated to its yield limit. In the limit case,σ = σakma.
According to the superposition principle, we first lift the right wall and increase the temperature, then we apply the reaction force P coming from the wall. We start from the fact that the total extension (δ) is zero.
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P
b-) Can the plate be used at the same operating temperature (120 oC) if it is unidirectionally reinforced with 25% steel fibers?
(4.12)
In an orthotropic layer constrained in direction 1, the temperature difference at any moment is:
This time we will use the same solution as in part a for the orthotropic composite structure. Note that the only thing that changes are the material constants in Hooke's relations.
From the last equality …>>
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Steel (fiber)
Aluminum (matrix)
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Then from the Equation 4.12:
:
The maximum temperature difference that can be applied to an orthotropic layer limited (constrained) in the 1 direction, within the strength limits:
(4.13)
The composite can be used up to this temperature in a constrained condition.
Note: In the constrained state, deformation is completely prevented. For this, it is necessary to have both walls.
Kompozit Malzeme Mekaniği-Ders Notları-Prof.Dr.Mehmet Zor
9. Kompozitlerde Termal Yüklemeler
From equ.4.4..>>
From equ. 4.3..>>
(See: chapter 4)
From eq. 3.4..>>
From equ. 4.13...>>
From equ. 3.34..>>
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c-) When the right wall is removed, to what temperature can the composite structure be heated within its strength limits?
As explained in article 3.14, even if the composite is allowed to expand freely, stresses in the 1 direction will occur in the fiber and matrix. In addition to thermal elongation, the fibers also lengthen a little more due to the pull of the matrix itself.
Total elongation in composite
Total elongation in fiber:
Thermal elongation
Elongation caused by the matrix pulling the fiber
=
=
Temperature difference at any instant in free state: :
: Stress caused by the matrix pulling the fibers:
(4.14)
Maximum allowable temperature difference in free state:
(4.15)
From the above equation
For option c of the example we are examining, there is an unconstrained situation since the right wall is removed. Maximum temperature difference from equation 4.15:
(Maximum operating temperature that can be reached within endurance limits when the right wall is removed)
Note: If one or both of the right or left walls are removed, extension is allowed and the Free state is obtained.
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=
Thermal elongation
Shortening caused by fibers preventing the matrix from elongating
: Stress caused by the fibers working to prevent the matrix from elongating
(4.16)
From the above equation, the stress in the matrix for any ∆𝑇 :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Total elongation in composite
Total elongation in matrix
=
d-) We will calculate the stress in the matrix at the maximum allowable temperature in the unconstrained case:
The maximum temperature difference was found in option c. At this instant the stress in the matrix:
(4.17)
or 2nd way
For free-state thermal loading:
From eq. 3.5:
If you pay attention, at maximum temperature, the yield strength of the matrix (150 MPa) is not exceeded and no damage occurs to the matrix.
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Sample Question 4.4
A cylindrical reinforcement sample with a diameter of 12cm and a length of 24cm, obtained by placing iron rods in concrete, is placed between two cylindrical rigid plates at the bottom and top and is subjected to a compression test. During the test, a 0.4mm collapse (shortening) in the length of the sample was measured when the compressive force P1 = 700kN. According to this,
P1
φ D = 12cm
24cm
| E(GPa) | ν | σmax (MPa) |
Iron | 200 | 0,3 | 400 |
Concrete | 32 | 0,2 | 65 |
a-) Which special type of anisotropic material does this material fall into? (Answer: Transversely Isotropic Material)
c-) Calculate the stresses in the concrete and iron bars for a compressive force of 700kN in this sample.(Answer: stress in concrete: -53.34MPa, stress in iron: -333.4MPa)
d-) Calculate theoretically the maximum compression force that the sample can withstand. (Answer: 839.66kN)
e-) How many 24mm diameter iron rods should be used in a 1m x 0.5m rectangular cross-section column with the same material properties? (Answer: 34)
b-) Find the narrowing in the sample diameter.(Answer: 40,56x10-3mm)
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4.8 Limits of Efficiency from Composite
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0
Strength of composite at break:
(4.18)
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4. Theoretical Strength And Efficiency Limits of Composite
kompozit
matris
kopma
kopma
d
e
(Remember: Equation 4.18 is valid for the case where the fibers do not carry any load and break immediately.)
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or
(4.18)
=
(4.4)
the fibers are said to carry load.
Considering equations 4.2, 4.3 and 4.5,
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
koma
fiber
composite
matrix
rupture
rupture
rupture
a
b
c
d
When
in terms of strains :
(4.19a)
(4.19b)
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or
in terms of strains
It is always valid.
From the above equaiton,
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
Equation (4.4)
(4.20a)
(4.20b)
4.8.3 Diagram: Fiber Volume fraction – Composite Strength
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A graphical summary of what is explained in this 4th topic can be seen on the side.
Important points :
The fibers break immediately and the matrix carries all the load. Composite strength is considered when the matrix is damaged.
2-) In the region Vfmin< Vf < Vfcr :
fibers carry the load, but the composite is not efficient.
fibers carry load and efficiency is obtained from the composite.
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4. Theoretical Strength And Efficiency Limits of Composite
Matrix dominant
Fiber dominant
Try to interpret the diagram for different situations.
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Example 4.5
It was produced in a unidirectionally reinforced orthotropic layer from matrix and fiber materials, both of which exhibit brittle character. The dimensions of the layer are 400mm x 400mm x 8mm and the material properties are given in the table above. 25% fiber was used in the structure. According to this;
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
4. Theoretical Strength And Efficiency Limits of Composite
Matrix Fiber
Modulus of Elasticity: Em=16GPa, Ef=82 GPa,
Maximum Strain( at breaking) εm-max=2,5x10-3 εf-max=1,71 x 10-3
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Solution:
b) Critical fiber volume fraction:
a) Minimum fiber fraction:
From equation(4.19b):
From equ. (4.20b):
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4. Theoretical Strength And Efficiency Limits of Composite
fibers carry load.
efficiency is obtained from the composite.
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Fiber strength:
c-)
F1max
F1max
A1
= 55,575 x 400 x 8
As explained in article 4.5, the strength of the composite in the 2 direction is equal to the strength of the matrix:
Matris strength:
Stress in the matrix when fibers break :
Composite Strength in direction 1:
Maximum tensile force that can be applied in direction 1:
Maximum tensile force that can be applied in direction 2:
F2max
F2max
A2
Composite Strength in direction 2 :
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5
EXPERIMENTAL DETERMINATION OF ORTOTROPİC PROPERTIES
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5.1 Purpose and Scope
Although any property of an isotropic material can only be determined by experimental measurements, the mechanical properties of an orthotropic composite material can be calculated theoretically (while the properties of its components are known) as explained in the 4th topic. However, the effects of factors such as internal material defects that may occur during composite production are ignored in theoretical calculations. In experimental measurements, the effects of these factors are reflected in the results, and therefore these results are much closer to reality. For this reason, as in other materials, the essential thing in composites is to determine the material properties experimentally
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
5. Experimental Determination of Ortotropic Properties
Our aim in this section
is to explain in detail how the mechanical properties (E1, E2, E3, G12 , ν12 ) of a unidirectional, continuous fiber reinforced composite exhibiting orthotropic properties are obtained through experimental measurements.
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5.2 Let's Remember the Experimental Determination of Mechanical Properties in Isotropic Materials:
Thanks to tensile testing in isotropic materials,
Modulus of elasticity (E),
Poisson's ratio (ν),
Plastic zone curve,
Yield ((σa) and tensile (σT) strengths can be obtained experimentally. (After reaching the tensile strength, the sample elongates very quickly and breaks suddenly. Therefore, the stress σr at the moment of rupture does not matter.)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
5. Experimental Determination of Ortotropic Properties
(First, check out 1.11.5.)
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Extensometers can be used instead of strain gauges in tensile tests. In this case, the ∆𝐿 elongation amounts on the sample are measured for different instants and unit elongations can be obtained for those instants from the formula 𝜀=∆𝐿/𝐿𝑜. Extonsometers provide significant convenience in this respect. Video extonsometers are also among the commonly used types.
Extensometer
Alternative to strain gauge :
5.2.1 Straingauges:
These are sensors used to measure strains.
Strain gauges
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While applying the tensile test in the y direction; for different instants such as b, c,… n, the following operations are performed respectively:
6) Yield Strength:
Py
Py
Py
Pyb
Pyc
Pk
Pa
Pç
5.2.2 Testing Stages for Isotropic Materials:
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5.2.3 If the 3rd strain gauge is used in the tensile test;
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Strain
Transformation Formula
(5.1b)
For Cartesian coordinates (x-y):
For local coordinates (1-2):
(5.1a)
5. Experimental Determination of Ortotropic Properties
Mohr Circle
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1: Axis parallel to the fiber direction in the layer plane
2: Axis perpendicular to the fiber direction in the layer plane
3: Axis in the direction of layer thickness
5.3.1 Local Axes:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
In a unidirectional and continuous fiber reinforced composite layer
5.3.2. Global axes: These are the Cartesian (x, y, z) axes that represent the entire structure in layered composites.
5. Experimental Determination of Ortotropic Properties
Each layer of layered composites has its own specific local axes (1,2,3).
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5.4.1- A sample is removed from the layer in direction 1.
5.4.2-) Strain gauges are glued in direction 1 and 2.
5.4.3-) The sample is connected to the tensile testing machine.
5.4.4-) Tensile Test is Performed, P-ΔL1 Diagram is measured from the device
Tensile Test is performed in Direction 1. The following steps should be followed in order:
(The sample thickness should be equal to the layer thickness.)
5. Experimental Determination of Ortotropic Properties
step 5.4.5..>>
L1
* P1 force values read from the tensile device are correct for the sample, but ΔL1 elongation values read from the device are not correct. Because the elastic extension effect of the device's jaws is also included in the ΔL1a , ΔL1b ,… values read from the device, and thus, it would be wrong to use the equation ε1= ΔL1 /L1 . For this reason, strain-gauges are attached to the sample and (ε1, ε2) values are measured directly. Or ΔL1a , ΔL1b ,.. values should be read accurately on the sample by using extonsometers instead of strain gauges.
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5.4.5) Stress – Strain Diagram and obtaining E1, ν12, XT values from there
For different instant such as a, b,.. n, the following operations are performed respectively
5. Experimental Determination of Ortotropic Properties
L1
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5.5.1-) A sample is removed from the layer in direction 2.
5.5.2-) Strain gauges are glued in direction 1 and 2.
5.5.3-) The sample is connected to the tensile testing machine.
5.5.4-) Tensile Test is performed, P-ΔL2 Diagram is measured from the device
Tensile Test is performed in 2 Direction Perpendicular to the Fibers. The following steps should be followed in order:
(The sample thickness should be equal to the layer thickness.)
5. Experimental Determination of Ortotropic Properties
Step 5.5.5..>>
L2
* P2 force values read from the tensile device are correct for the sample, but ΔL2 elongation values read from the device are not correct. Because the elastic extension effect of the device's jaws is also included in the ΔL2a , ΔL2b ,.. values read from the device, and thus, it would be wrong to use the equation ε2= ΔL2 /L2 . For this reason, strain-gauges are attached to the sample and (ε1, ε2) values are measured directly. Or ΔL2a , ΔL2b ,.. values should be read accurately on the sample by using extonsometers instead of strain gauges.
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5.5.5-) Stress – Strain Diagram and obtaining E2, ν21, YT values from there
5. Experimental Determination of Ortotropic Properties
L2
For different instant such as a, b,.. n, the following operations are performed respectively
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L2
L1
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By taking i=1, j=2 in this formula, experimentally obtained E1, E2, ν12, ν21 values can be verified.
If the experiments have been done correctly, this equation should also satisfy, even if approximately. If this equation is not satisfied, errors may have been made in the measurements or calculations in the experiments and these should be checked again.
Equation (3.13) valid for an orthotropic composite layer:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
5.6 Verification of Experimental Measurements for Tensile Tests
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5.7 Experimental Determination Methods for Shear Modulus (G12) and Shear Strength (S)
45°
5.7.1.1) A sample is removed from the layer at a 45° angle with the fibers.
5.7.1. Off – Axis Method (G12 is found.):
5.7.1.2) A strain gauge is glued in the x direction.
5.7.1.3) The sample is connected to the tensile testing machine.
5.7.1.4) Tensile Test is performed, Px -ΔLx Diagram is measured from the device
(The sample thickness should be equal to the layer thickness.)
(5.2)
In this method, the S value is not calculated. G12 is calculated from the transformation equation number 5.2 above. For this, the other values in equation 5 must have been determined beforehand. We accept that the values of E1, E2, ν12 have been found experimentally before. We choose θ = 45o. Elasticity Modulus (Ex) in the x direction is found by the experimental method whose steps are explained below.
5. Experimental Determination of Ortotropic Properties
As explained in section 5.4, since the jaws of the device also have an effect on the ΔLx values read from the device, direct measurement is made on the sample using a strain gauge and εx values are obtained for different instant.
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5.7.1.5) Stress – Strain Diagram and obtaining the Ex value from there.
*The explanations (important points) made in articles 5.4 and 5.5 also apply to this test.
5. G12 is drawn from equation 5.2 and calculated
(5.3)
5. Experimental Determination of Ortotropic Properties
For different instant such as a, b,.. n, the following operations are performed respectively:
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r
t
T
T
45o
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
5.7.2.a Torsional Tube Test Method (G12 and S are obtained)
(5.4)
Average cross-sectional area:
(5.5)
How these formulas are derived is explained on the next page...>>
5. Experimental Determination of Ortotropic Properties
Torsion Testing Machine
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Tiç = T
T
dA
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
5.7.2.b Derivation of the formulas used in the torsion tube method:
I-I section – left part
T
T
45o
I
B
a
b
I
1
B
I-I section
(avarage shear stress)
Strain Transformation equation from (5.1.b):
From the Hooke equations in equation (2.17);
(Average cross-sectional area)
5. Experimental Determination of Ortotropic Properties
(shear strength)
Shear Modulus in plane 1-2 :
(5.4)
(5.5)
Tinternal = T
r
ds=rdα
dα
dA=t.ds=trdα
t
Total internal moment in the section :
B
dA=tds
t
ds
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a
a
Section a-a
cross sectional area of notch:
A=c.t
Σ Fy =0
5. Experimental Determination of Ortotropic Properties
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Σ Fy =0
5.7.3 Iosipescu Method (obtains G12 and S):
load fixture apparatus
t:sample thickness
c:notch width
test sample
Shear Force
Bending Moment
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5) If the balance of the left part of the a-a section of the sample is examined; It is understood that the cutting force in the middle region of the sample, including the notch section, is V = P.
From equation (2.17);
From equation (5.1.b) :
(5.6)
(5.7)
5. Experimental Determination of Ortotropic Properties
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Section a-a
A=c.t
a
a
6) Calculations: Since the bending moment in the notch section is M=0, normal stresses will not occur r (σ1=σ2=0).
;
Situation at point D in the notch section:
D
D
Shear Force
Bending Moment
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Iosipescu Apparatus and Test Setup
5. Experimental Determination of Ortotropic Properties
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Example 5.1: A 2cm wide, 0.4cm thick sample is removed from a newly produced, unidirectional fiber reinforced layer as shown in the figure and is subjected to a tensile test. a and b strain gauges were glued on the sample in the directions shown in the figure. It is clear from the tensile diagram that the material is brittle and linear elastic. The values measured at different moments during the test are as shown in the table. Accordingly, by taking into account only the measurements in the table, determine the possible E1,E2,ν12, ν21, G12, XT, YT, S values of the material.
P (kN) | ε a | ε b |
3,2 | 6,1x10-4 | -1,1x10-5 |
6,6 | 12,7x10-4 | -2,4x10-5 |
8,2 | 16,3x10-4 | -3,1x10-5 |
11,2 | 29,2x10-4 | -4,3x10-5 |
18,9 (rupture) | 35,8x10-4 | -7,8x10-5 |
Solution:
Since a tensile test is performed perpendicular to the fibers, that is, in the 2 direction, E2, ν21 and YT values are obtained as a result of the test. (Explained in article 5.5)
a: direction 2 , b: direcition 1
From Equation ( 2.15b ) :
5. Experimental Determination of Ortotropic Properties
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
From Equation ( 3.12.b ) :
rupture
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| | |
| | |
| | |
| | |
| | |
| | |
Ölçüm No | P (kN) | ε2 | ε1 |
| 3,2 | 6,1x10-4 | -1,1x10-5 |
| 6,6 | 12,7x10-4 | -2,4x10-5 |
| 8,2 | 16,3x10-4 | -3,1x10-5 |
X 4 | 11,2 | 29,2x10-4 | -4,3x10-5 |
| Pult = 18,9 (break) | 35,8x10-4 | -7,8x10-5 |
Measurements (given in the question)
Calculatios
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Example 5.2
In the experiment to be carried out with the Iosipescu Method, the notch spacing of the sample taken from a brittle, unidirectional continuous fiber reinforced layer is c = 1cm and the sample thickness is t = 1mm. The measurements taken for different instants from this sample during the experiment are given in the table below. Accordingly, determine the stiffness modulus (G12) and shear strength (S) values of this sample.
P (N)= | 180 | 272 | 534 | Pult = 725 (instant of damage) |
ε45 = | 2x10-4 | 3,4x10-4 | 6,3x10-4 | 8,5x10-4 |
Instant Measurements
| | | | |
| | | | |
Solution:
(Calculation)
5. Experimental Determination of Ortotropic Properties
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(MPa)
(MPa)
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Example 5.3: With the off-axis experiment, the shear modulus (G12) of a composite material in the 1-2 plane will be determined.
Previously found properties of the material : E1 =26GPa, E2 = 6GPa, ν12 = 0.25
Values read during the test:
15kN
measurement number | 1 | 2 | 3 | 4 | 5 |
P(N) = | 2000 | 4600 | 8400 | 12800 | 18000 |
ΔL(mm) = | 0,21 | 0,49 | 0,87 | 1,3 | 2,2 |
For this purpose, a tensile test in the x direction was applied to a test sample with a cross section of: A = 100mm2, length: L = 10cm, and a fiber orientation of 45o. Instant values read from the tensile testing machine during the test are given in the table below. Accepting that ΔL total elongation values belong only to the sample; Obtain the G12 value from these measurements.
5. Experimental Determination of Ortotropic Properties
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HOOKE AND TRANSFORMATION EQUATIONS IN COMPOSITES
(STRESS-STRAIN CALCULATIONS IN AN ORTHOTROPIC LAYER)
6.
(tvid : 6.)
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6.1 Our aim in this section is to obtain Hooke’s and transformation equations that help us calculate the stresses and strains in global axes in a composite plate with orthotropic properties. The examination will be carried out for the plane stress condition, and the mechanical properties of the composite layer according to the local axes (E1 , E2 , ν12 , G12 ) will be assumed to be known. Examples will also be solved within the subject.
1-2: local axes, x-y: global axes
6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
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θ: Fiber orientation angle,
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For an orthotropic layer, the relations between stresses and strains according to the 1-2 (local) axes (i.e. Hooke’s Laws) are rewritten on the side (It was explained in the second topic).
[𝑄] stiffness matrix and [𝑆] compliance matrix depend only on the material properties in 1-2 directions and are also seen in the equations on the side. (Theoretical calculations of material properties were explained in topic 3; experimental calculations were explained in topic 5.)
or
6.2 Let's Remember Hooke’s Equations in Local Coordinates:
,
,
,
,
,
,
,
From Equation 2.17;
Note: The [C] matrix mentioned in the 2nd topic is expressed as [Q] here. [C] = [Q]
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Hooke equations between stresses and strains according to the x-y global axis set can be expressed as follows:
6.3 Deriving Hooke Equations in Global Coordinates :
(6.1)
(6.2)
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6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
Tip-4: Although Hooke equations want to be found in x-y global axes, material properties are known according to 1-2 local axes.
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Strain Transformation Equations:
Stress
Transformation Equaitions :
Now we continue our calculations to achieve our goal:
(These equations are obtained from static equilibrium and do not depend on material properties. Applies to all material types. It is shown in the Strength of Materials course.)
(6.3a-c)
(6.4a-c)
(a)
(b)
(c)
(a)
(b)
(c)
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where
In this case, if the stress transformation relations from equations 6.3 are written for directions 1-2:
Now we will write these equations in matrix format::
We can now write equations 6.5 in matrix format as follows:
(6.5a-c)
(6.6)
(6.7)
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6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
We define a transformation matrix :
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or;
(6.8)
(6.9)
Similarly, the strain transformation equations can be written in the same format.
(6.10a-c)
(6.11)
(6.12)
or;
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From equation 6.11
,
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6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
If we consider Equation 6.8 again;
Reduced Stiffness Matrix :
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When the multiple matrix multiplication process on the previous page is performed, the reduced stiffness matrix and its terms are found as follows:
(6.13)
(6.14a-f)
(a)
(b)
(c)
(d)
(e)
(f)
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(6.15)
(6.16a-f)
(a)
(b)
(c)
(d)
(e)
(f)
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6.5 ABSTRACT
Hooke’s Equations
Transformation Equations
Tip-5:
In single layer calculations, alternative solutions that do not use reduced matrices [𝑄 ̅ , 𝑆 ̅ ] should be preferred as much as possible, so that there are not too many operations.
Note:
It is inevitable to use reduced matrices in the calculations of layered composites that will be explained later.
In Local Coordinates :
In Global Coordinates:
For Stresses
For Strains
(6.1)
(6.2)
(6.7)
(6.8)
(6.11)
(6.12)
(2.16a)
(2.16b)
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Example 6.1 At a point q with a fiber orientation angle of 45o, the plane stress condition shown in the figure emerges. Calculate the global strains accordingly.
σ1 | = 30MPa |
σ2 | = -10MPa |
τ12 | = -30MPa |
ε1 | = 2,21x10-4 |
ε2 | = -5,64x10-4 |
γ12 | = -75x10-4 |
εx | = 35,7x10-4 |
εy | = -39,2x10-4 |
γxy | = 7,85x10-4 |
1st Solution Way: Using only [𝑸] 𝒗𝒆 [𝑺] Normal matrices:
First, let's find the local stresses from Equation 6.7:
Local strains from Equation 2.16b:
Global stresses from Equation 6.12:
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6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
q
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Global strains are calculated from Equation 6.2::
First, the terms of the Normal [S] matrix are calculated:
,
,
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6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
Similarly :
,
,
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E1 | E2 | ν12 | G12 |
140 GPa | 36 GPa | 0,28 | 14 GPa |
Mechanical Material Properties of the Composite Material
Example 6.2
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Let's first calculate the Minor Poisson Ratio :
When we look at the placement directions of strain gauges;
Local Strains
If we remember Equation 5.1b:
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Local Stresses
Let's calculate the local stress values from Equation 2.16a:
From equation 2.7d, the terms of the matrix [Q] are:
Global Stresses
Global strains from Equation 6.12 :
Global stresses from Equation 6.8 :
Global Strains
(Using reduced matrices)
or 2nd Solution
(From Equation 6.2 )
(From Equation 6.1 )
Due to the reduced matrices, Solution 2. The path is longer.
c: cos50o, s=sin50o
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6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)
6.6 Hooke's Relations Including the Temperature Effect
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ΔT
Let's assume that in a composite layer, in plane stress case, while there are tensile forces in the 1 and 2 directions, we increase the temperature of the layer by the amount ΔT. In this case, we calculate the resulting strain values using the superposition method and Hooke's relations as follows:
ΔT
P2
P1
P2
P1
P1
P1
P2
P2
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Hooke's equation including temperature for the plane stress case:
If we write it in matrix format:
For orthotropic or more specifically transversely isotropic materials; In local axes, α12 = 0. In other words, temperature change has no effect on shear deformation and shear strain.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Local strains:
Local stresses:
Global strains:
Global stresses:
(6.17.a)
(6.18.a)
(6.17.b)
(6.18.b)
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The transformation relation is also valid between local and global thermal expansion coefficients. Namely:
We can write the global strains that occur only due to the temperature difference ΔT as follows:
9.4.2 Global Thermal expansion coefficients (αx, αy, αxy) :
From equation 6.12 :
Recall the transformation matrix from Eq. 6.9:
From eq. 6.17.a
(While there is only ΔT)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(6.19.b)
(6.19.a)
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FAILURE CRITERIA
FOR COMPOSITES
7
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7.1 Failure Criteria and Importance
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7. Failure Criteria for Composites
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7.3 Main Failure Criteria in Orthotropic Composites
Unidirectional or bidirectional continuous fiber reinforced composites and sandwich composites show orthotropic properties and are examined according to the above criteria. These criteria are also used in damage detection of layered composites, which will be explained in the next topic. However, it is possible to find much different criteria developed specifically for composites in the literature.
1- Maximum Stress Criterion
2-Maximum Strain Criterion
3- Tsai – Hill Criterion
4- Modified Tsai – Hill Criterion
5- Tsai-Wu Criterion
6- Hoffman Criterion
7.2 Our aim in this chapter is to explain the main failure criteria used for composite materials with orthotropic properties. Criteria and examples will be explained according to the plane stress situation.
Now we will discuss the 6 criteria mentioned above one by one and understand with examples how damage (failure) detection is done for a single layer....>>
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7. Failure Criteria for Composites
Note: The term Theory can also be used instead of Criterion
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7.4. Let's remember the symbols of experimental material strengths according to Local Axes:
Direction 1
Direction 2
Tip 6: In order to detect failure (damage) at a point, the local stresses or local strains at that point must be known or calculated. Because all failure criteria have been developed according to these.
7. Failure Criteria for Composites
Experimental Strength Value
7.5.1 Maximum Stress Criterion
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According to this criterion, if all of the inequalities 7.1a-c shown on the side, for local stresses at a point of the composite are met, no damage will occur at that point.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
In other words, if at least one of these three inequalities is not met according to this criterion, damage will occur at that point.
This criterion gives much better and more realistic results in detecting the fracture of brittle materials, especially at points where tensile stresses exist.
(7.1a-c)
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7.5.2 Maximum Strain Criterion
In other words, if at least one of these three inequalities is not met according to this criterion, damage will occur at that point.
This criterion gives results much closer to reality in detecting the fracture of brittle materials and especially at points where tensile stresses occur.
(7.2a-c)
Strains at the time of damage are calculated from Hooke's relations (equations 7.3.a-e).
(7.3a-e)
7. Failure Criteria for Composites
(a)
(b)
(c)
(d)
(e)
(a)
(b)
(c)
According to this criterion, if all of the inequalities 7.2a-c shown on the side for local strains at a point of the composite are met, no damage (failure) will occur at that point.
7.5.3 TSAI – HILL Criterion
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This criterion is an adaptation of the Von Mises distortion energy damage theory, which is valid for isotropic materials, to anisotropic materials.
According to this criterion, for damage to occur in a layer, the following condition 7.4 must be violated:
How to calculate the constants G1, G2, G3, and G6? …we will examine this now…>>
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Since σ3 = τ31 = τ23 = 0 in the case of 2-Dimensional (Plane) stress, equation 7.4 turns into equation 7.5 below:
This criterion gives very good results
in damage detection in composites with the same tensile and compressive strengths,
in composites with ductile behavior and
in points subject to tensile stress.
In case of 3D Stress :
(7.4)
(7.5)
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Final forms of Equation 7.5:
In case of plane stress;
7. Failure Criteria for Composites
(7.5)
Let's rewrite equation 75 on this page:..>>
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If these values are substituted into Equation 7.5:
According to the Tsai-Hill Damage Criterion, if inequality 7.6 is violated in the case of plane stress, damage occurs.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(7.6)
,
,
It is obtained from these four equations as
,
7. Failure Criteria for Composites
If you pay attention to inequality 7.6, only tension strengths are taken into account in the Tsai-Hill criterion. However, in brittle materials, the compressive strength may differ from the tensile strength. In this case, this criterion needs to be modified to suit brittle materials…>>
7.5.4 Modified TSAI – HILL Criterion
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This criterion is a modified version of the Tsai-Hill criterion, taking into account compressive strength.
According to this criterion, if inequality 7.7 on side is violated for the plane stress condition, damage (failure) occurs at the examined point.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(7.7)
This criterion is a general criterion and can be used not only for brittle composite materials, but also for all orthotropic material types and all point stress states.
7. Failure Criteria for Composites
The meaning of the values in the denominator :
7.5.5 TSAI – WU Failure Criterion
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According to this criterion, if the following inequality is violated, the material will be damaged:
The constants H1, H2, H6, H11, H22, and H66 will be found for 5 different strength values of the layer. H12 can only be calculated experimentally.
This failure criterion is also a general criterion. It takes into account both the compressive and tensile strength of the materials. It can be used for all orthotropic material types and all point stress situations.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(7.8)
Now the determination of these constants will be explained…>>
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Equation 7.8 at the time of damage :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(7.9a)
(7.9b)
(7.9c)
(7.9d)
at the time of damage :
at the time of damage:
at the time of damage :
at the time of damage:
at the time of damage :
(7.9e)
(7.9f)
7. Failure Criteria for Composites
Let's write equation 7.8 here again:
(7.8)
Only the 𝐻12 calculation remains..>>
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7.5.5.1) Methods of Calculating the H12 constant
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
a.) 1st Experimental Method :
If there is a loading that will create equal tensile stress in the 1 and 2 directions at the examined point;
(σ1 =σ2=σ , τ12= 0)
At the time of damage, equation 7.8:
(7.10a)
Or any of the following combinations can be used in this method:
There may be different alternatives for the experimental setup. The important thing is that only normal stresses occur at the same intensity in directions 1 and 2 at the examined point.
7. Failure Criteria for Composites
2
1
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b-) 2nd Experimental Method :
For the instant of damage, from the stress transformation equations (Equation 6.5a-c)
(Or these values can also be seen from the Mohr Circle on the side..)
At the time of damage, equation7.8:
(7.10b)
Mohr Circle
7. Failure Criteria for Composites
are found.
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Per Tsai-Hill failure Criterion:
Per Hoffman Criterion:
Per Mises-Hencky Criterion:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
c.-) Method 3: Empirical Equations
Equations that generalize experimental measurement results or observational data are called empirical equations.
(7.10c)
(7.10d)
(7.10e)
7. Failure Criteria for Composites
7.5.6 Hoffman Criterion�
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According to this criterion, if the following inequality is violated, the material will be damaged:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(7.11)
This criterion gives better results
7. Failure Criteria for Composites
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Example 7.1
The plane stress state at one point of a composite layer reinforced with continuous fibers with an orientation angle of 45o is as shown in the figure. The properties of the composite layer are given in the table below.
At this point, whether damage will occur, check according to
a-Maximum Stress Criterion,
b- Maximum Strain Criterion,
c- Tsai-Hill Criterion,
d- Modified Tsai-Hill Criterion.
E1 | E2 | ν12 | G12 |
150GPa | 32GPa | 0,3 | 8 GPa |
| | | | |
100MPa | 200 MPa | 25MPa | 45MPa | 18MPa |
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a-)
According to Maximum Stress Theory (Criterion)
Local stresses and local deformations must be obtained to detect damage (Tip Point 6)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
c=cos45o , s=sincos45o
-200 < 5,5 < 100
-45 < -12,5 < 25
-18 < 16,5 < 18
No failure occurs.
Solution:
From equation 6.7,
Calculation of local stresses:
All 3 inequalities are satisfied.
(Equation 7.1 inequalities are checked. The values must be adjusted so that the left side of these equations must be negative..)
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b-)
According to Maximum Strain Criterion:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
-13,3x10-4 < 0,617x10-4 < 6,67x10-4
-14,1x10-4 < -4,01x10-4 < 7,81x10-4
-22,5x10-4 < 20,63x10-4 < 22,5x10-4
Calculation of Local Strains from Equation 2.16b:
From Equation7.3a-e,
Strains at time of damage:
We check the inequalities in Equation 7.2. (The left side of these inequalities must be negative.)
7. Failure Criteria for Composites
No failure occurs.
All 3 inequalities are satisfied.
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c-) According to Tsai-Hill Criterion,
d-) According to Modifiye Tsai-Hill Criterion:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
1,1 > 1
0,92 < 1
Damage occurs.
No Damage occurs.
The no-damage condition specified in Equation 7.6:
The no-damage condition specified in Equation 7.7:
(It has been found before.)
7. Failure Criteria for Composites
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Example 7.2
The forces shown in the figure act on a unidirectional and continuous fiber reinforced composite plate manufactured with a 60o fiber orientation angle and the plate remains in static balance. In the triple straingauge rosette attached to a Q point on the plate, a is placed horizontally, b is placed vertically, and c is placed at a 45o angle with the –x axis. Values measured with strain gauges:
E1 | E2 | ν12 | G12 |
140 GPa | 28 GPa | 0,35 | 10 GPa |
XT | XC | YT | YC | S |
130 MPa | 180 MPa | 30 MPa | 50 MPa | 20 MPa |
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
7. Failure Criteria for Composites
-,
Obtain the results for the criteria listed below yourself with the calculations.
εa = -12.49x10-4, εb = 6.747x10-4 , εc = -1.111x10-4
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In terms of all criteria: Local stresses or local strains must be obtained for damage detection. (Tip point 6)
Solution:
From Equation 5.1b, the strain (unit elongation) in the c strain-gauge is:
From Equation 6.11, local strains are calculated:
Looking at the directions of the Strain Gauges :
From this equation 𝛾xy is found:
Local Strains
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From equation 2.16a, local stresses are calculated:
We calculate each term of the [Q] matrix as follows:
From Equa. 3.13, Minor poisson ratio:
Local Stresses:
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e-) According to Tsai-Wu Criterion:
If we choose the Mises-Hencky empirical equation among the alternatives for H12 (Eq. 7.10 e);
Then, according to the modified Tsai-Wu criterion, no damage occurs at the Q point.
Let's rewrite the calculated local stresses:
| | | | |
130 MPa | 180 MPa | 30 MPa | 50MPa | 20 MPa |
Strength Values of Composite Material (Given in the Question)
From Equation 7.8, No Damage Condition:
Constants are found from Equation 7.9a-e:
If we substitute all numerical values into Equation 7.8:
(is zero in all cases)
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f-) According to Hoffman Criterion:
If numeric values are placed :
Therefore, no damage occurs at point Q according to the Hoffman criterion.
Local Stresses Found:
| | | | |
130 MPa | 180 MPa | 30 MPa | 50MPa | 20 MPa |
Strength Values of Composite Material (Given in the Question)
(No Damage Condition)
The inequality of Equation 7.11 is checked.
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E1 (GPa) | E2 (GPa) | ν12 | G12 (GPa) | | | | | | | |
130 | 30 | 0,28 | 12 | 85 | 170 | 40 | 80 | 15 | 10x10-6 | 24x10-6 |
Example 7.3*
A composite layer with a fiber orientation angle of 0o is placed in a rigid mold. The inner surfaces of the mold and the outer surfaces of the layer are in frictionless contact, and there is no compression between the surfaces. The required material properties of the layer are given in the table below. According to this,
a-) How much can the temperature of this layer be increased within the strength limits? Determine according to the Maximum Stress Criterion.
b-) Check for damage according to Tsai Hill and Modified Tsai-Hill criteria for fiber routing angle θ = 300 and ΔT = 50 0C
(This example includes formula inferences regarding thermal loads.)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Solution…>>
7. Failure Criteria for Composites
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Stresses at any temperature difference ΔT :
(7.12)
(7.13)
a-)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Solution:
1
2
There is no limitation in the thickness direction perpendicular to the plane (3 direction). The deformation is free and hence no stress occurs in this direction. Therefore, this problem is a plane problem.
If we first calculate the minor poisson ratio:
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If ΔT is found from equation 7.12 :
Review for direction 1:
(7.14)
If we examine it according to the Maximum Stress criterion:
Then, at the exact moment of damage :
No-damage condition for direction 1 (Eq. 7.1a):
Review for direction 2:
Similarly
Maximum temperature rise for direction 1, in the constrained layer in directions 1 and 2 :
Maximum temperature rise for direction 2, in the constrained layer in directions 1 and 2 :
Eq. 7.1.b:
at the time of damage:
From eq. 7.13 :
(7.15)
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When the given values are substituted:
(Temperature difference that will cause yield in direction 1)
Or the condition that no damage occurs in this layer :
(Temperature difference that will cause yield in direction 2)
Therefore, the temperature increase limit value in this layer is :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
b-) Check for damage according to Tsai Hill and Modified Tsai-Hill criteria for fiber routing angle θ = 300 and ΔT = 50 0C
300
According to Modified Tsai-Hill
no damage occur
According to for Tsai-Hill
damage occur
Answers:
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Example 7.3 (2021 - 2nd Visa Question)
The global stresses at point a of a unidirectionally reinforced composite layer with an orientation angle of θ= 450 were determined as follows:
σx = 27 MPa, σy = 26 MPa, τxy = 15MPa
A triple strain-gauge rosette with a 120o angle between them is glued to point a.
Material Properties
E1 (GPa) | ν12 | E2 (GPa) | G12 (GPa) | XT =(σ1T)ULT (MPa) | XC = (σ1C )ULT (MPa) | YT = (σ2T )ULT (MPa) | YC = (σ2C )ULT (MPa) | S = (τ12)ULT (MPa) |
84 | 0,35 | 33 | 9 | 50 | 104 | 30 | 60 | 10 |
At this point, determine whether damage will occur according to
a-) Tsai-Hill criterion and
b-) Maximum Strain Criterion.
c-) Calculate the unit elongation (strains ε ) values read from each of the strain-gauges.
Answers:
c-)
εd=3,38x10-4
εc=4,14x10-4
εb=1,79x10-4
a-)
0,64<1
No damage ocur.
b-)
-12,38x10-4< 4,461x10-4<5,95x10-4
-18,18x10-4< 1,756x10-4<9,09x10-4
-11,11x10-4<-0,556x10-4<11,11x10-4
No damage occcur.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
7. Failure Criteria for Composites
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CLASSİCAL LAMİNATİON THEORY (CLT)
8.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
STRESS–STRAİN CALCULATİONS İN LAMİNATED COMPOSİTES
(Note: The Zor Equivalent Volume Model, which is an alternative to CLT, will be explained in Sections 9.4 and Chapter 10.)
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8.1 Our aim in this chapter is to develop equations by which the stresses and strains occurring at any point in a laminated structure obtained by combining orthotropic layers reinforced with continuous fibers can be calculated theoretically.
We call the structures formed as a result of gluing more than one composite layer on top of each other layered or laminated composite structure.
8. Classical Lamination Theory (CLT)
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P1
P2
P3
P4
M1
M2
P5
1mm
1mm
Q
whole structure
(in static equilibrium)
Q element
(in static equilibrium)
Q
b
1mm
1mm
Point b
(plane stress state)
b
Meaning of Internal Force and internal moments in element Q:
Nx: Normal (tension or compression) internal force per unit length in the x direction,
Ny: Normal (tension or compression) internal force per unit length in the y direction,
Nxy = Nyx : Internal shearing forces per unit length,
Mx: Bending internal moment per unit length in a section whose normal is x
My: Bending internal moment per unit length in a section with normal y
Mxy: Torsional internal moment per unit length in a section whose normal is x
Myx: Torsional moment per unit length in a section whose normal is y
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
8.2 Internal forces and internal moments occurring per unit length:
We are considering a laminated composite structure that is subject to the
influence of external loads (P1, P2,.. M1, M2:..) and is in static equilibrium.
From this structure, we imaginary separate a Q element with the same
thickness as the lamina (layer) and side lengths of 1 mm.
According to the separation principle, this Q element is still in static equilibrium,
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on the midplane passing through the middle of the thickness of the
layered composite plate remains vertical and straight at the end of the
loading (after deformation) (line C'A’). That is, out-of-plane shear deformations do not ocur (γxz = γ yz =0)
selected in a downward direction.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
8.3 Classical Lamination Theory and Assumptions:
We consider the deformed state of the Q element of the layered (or laminated) structure. When we look at this element from the thickness side and from the x-z plane, we will make the following assumptions:
In the following stages, we will try to obtain the stresses and strains at any point of the layered structure in terms of internal forces and internal moments, which are known values…>>
8. Classical Lamination Theory (CLT)
5. Displacements remain very small compared to the plate thickness.
4. The plate (laminated structure) is assumed to be thin and only subjected to plane stress (σz = τxz = τ yz =0)
(center of curvature)
mid-plane
7. Although a layer tends to deform freely in the lateral direction due to the Poisson effect, this lateral deformation is constrained because
of the perfect bonding with the other layers. However, this lateral interaction between the layers (Poisson interaction) is neglected.
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: Vertical (z) displacement of point C, that is, the amount of collapse caused by bending (the index '0' is used for the middle plane)
: Horizontal (x) displacement of point C (total displacement in the x direction due to deformation)
u
: Horizontal (x) displacement of point A
α
: The angle made by the tangent passing through point C' with the horizontal
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
The slope of point C in the midplane is:
(8.1)
8.4 Calculation of Global Stress and Strains According to Mid-plane Values
8. Classical Lamination Theory (CLT)
C
A
z
C'
z
u
α
α
dx
dw0
.
.
A'
mid-plane
z
x
Now we enlarge the part within the oval frame in the figure and examine it in more detail:
C: a point on the midplane
A: a point on any layer k, at a distance z from C.
C' and A': are the final positions of the points C and A after changing shape.
According to the 3rd assumption, CA=C'A'=z can be written.
: The final deformed shape of the midplane due to bending
α
mid-plane
(center of curvature)
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A
C
z
C'
A'
z
u
α
α
dx
dw0
Similarly, when we look at it from the y-z plane, the displacement of point A in the y direction is:
For small angles;
If we look at the displacement (u) in the x direction of a point A on any layer such as k;
Here, the unit elongation (strain) value in the x direction:
and here the unit elongation (strain) value in the y direction:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(8.2)
(8.3)
(8.4)
(8.5)
(8.6)
(8.7)
8. Classical Lamination Theory (CLT)
mid-plane
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Shear strain angle in x-y plane :
is found
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
If Equations 8.4 and 8.6 we obtained before are substituted into equation 8.8 above:
(8.8)
(8.9)
8. Classical Lamination Theory (CLT)
(8.4)
(8.6)
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Meanings of the terms on the right side of equation 8.10:
Curvatures of the midplane:
If we write equations 8.5, 8.7 and 8.9 in matrix format:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(8.10)
=
Strains in the midplane:
(8.11)
(8.12)
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Global Strains:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(8.13)
In this case, at a point on layer k, at a distance z from the midplane:
(If equations 8.11 and 8.12 are substituted into 8.10,)
(If we remember equation 6.1)
Global Stresses:
(8.14)
From now on, we will obtain the strains and curvatures in the mid-plane in terms of internal forces and internal moments... >>
8. Classical Lamination Theory (CLT)
8.5 Coding of a layered structure
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The following points should be taken into consideration when coding:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Example 2: Coded version of a three-layer structure
Example 1: Coded version of a two-layer structure
8. Classical Lamination Theory (CLT)
8.6 Calculation of Mid-Plane Strains
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1mm
hk
1mm
1mm
Q
We examine the internal forces in a Q element with a side length of 1 mm in a layered structure:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Stress and internal force in x direction in layer k:
Internal force in the x direction in the area of a 1mm wide, dz thick dA differential strip:
(8.15)
Internal normal force in x direction in
layered Q element:
Internal normal force in y direction in layered Q element:
Internal shear force in layered Q element:
Similarly, if the same operations are performed for the y direction;
(8.16)
(8.17)
Nx, Ny, Nxy : These are the internal forces per unit length. Its units are N/mm.
Internal force in x direction in layer k:
If we write equations 8.15, 8.16 and 8.17 in a matrix format:
(8.18)
k
8. Classical Lamination Theory (CLT)
(midplane)
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Q
1mm
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Now we examine the internal moments in the Q element.
Stresses in x direction and internal bending moment (Mx-k) in layer k.
Internal force x
perpendicular distance
Similarly, if the same operations are performed for the y direction;
(8.19)
(8.20)
(8.21)
(8.22)
Internal bending moments per unit length are Mx, My; Internal torsional moment: Mxy. Units are Nmm/mm.
8. Classical Lamination Theory (CLT)
(midplane)
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In equation 8.14, we expressed the global stresses in a k layer in terms of the strains of the middle plane as follows:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Now if we substitute Equation 8.14 into Equations 8.18 and 8.22;
The deformations and curvatures of the middle plane do not depend on z.
(8.23)
(8.24)
(8.25)
Stifness Matrices
of the Whole Layered Structure
Since these stiffness matrices belong to the entire layered structure, there is 1 of each.
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(8.26)
(8.27)
Each of the Stifness Matrices is a 3x3 matrix and has 9 terms.
If the stiffness matrices are replaced:
Internal Forces:
Internal Moments:
(8.28)
(8.29)
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Strains and curvatures of the middle plane from equation 8.31:
If we combine Equations 8.28 and 8.29 into a matrix:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
(8.30)
(8.31)
Symbolically :
The explicit expression of the matrix is:
(8.32)
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8.7 Special Cases
1st Special Case: In structures that are symmetrical with respect to the midplane in terms of material, load, geometry and fiber arrangement, [B] = 0.
2nd Special Case: In symmetrical structures, if all internal moments are zero, the curvatures of the middle plane are also zero.
In this case, In this case, there is no need to calculate the [D] matrix since it will not be involved in the operations.
3rd Special Case:
In this case, there is no need to calculate the [D] matrix again.
However, in the following special cases we do not need to calculate some of these matrices:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
That is, if
then
since z = 0
In symmetrical or non-symmetrical structures, If the stresses or deformations in the mid-plane are asked;
In the stress or strain calculations to be made from Equations 8.13 and 8.14,
For this reason, before proceeding with the calculations of a layered structure, it should be checked whether it falls into the special conditions above.
8. Classical Lamination Theory (CLT)
4th Special Case: In symmetrical structures, if all internal forces are zero, the strains of the middle plane are also zero.
That is, if
then
In this case, there is no need to calculate the [D] matrix since it will not enter into the calculations.
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Example 8.1:
The material properties found through experimental measurements for a newly produced composite layer (lamina) with dimensions of 400mm x 400mm x 8mm are given in the table below.
E1 (GPa) | E2 (GPa) | | G12 (GPa) | XT (MPa) | XC (MPa) | YT (MPa) | YC (MPa) | S (MPa) |
126 | 78 | 0,33 | 29 | 35 | 68 | 18 | 38 | 8 |
A laminated (layered) structure was created by gluing two of these layers on top of each other with 45o and 0o fiber orientation arrangements, respectively. The internal forces occurring in an element with dimensions of 1 mm x 1 mm x 8 mm at the Q point of the laminated structure are shown in the figure.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
c-) Check whether damage will occur on the upper surface of the top layer according to the Hoffman criterion.
a-) Calculate the local stresses and local strains occurring on the upper and lower surfaces of the upper layer and on the upper, middle and lower surfaces of the lower layer.b-) Determine the proportions in which the Nx force is distributed among the layers
Accordingly, at point Q,
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Solution:
Step 3: Does the Structure Enter Special Cases?
Does this structure fall into any of the special cases described in Article 8.7 ? First of all, this is detected.
1st Special Case?
Since there is no symmetry with respect to the middle plane in terms of fiber arrangement, It does not fall into the 1st special case. Then the [B] matrix will be non-zero.
Since the structure is not symmetrical, it does not fall into the 2nd Special Case.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Since only the stresses or strains in the mid-plane are not asked, it does not fall into the 4th Special Case.
a-)
Step 1: Internal forces and internal moments are calculated.
Internal forces (Nx , Ny , Nxy) and internal moments (Mx , My , Mxy) per unit length from static equilibrium are calculated. But in this example, these values are given directly.
Step 2:
The layered structure is coded.
midplane
By following the steps below, we will obtain local stresses and local strains at the asked points.
8. Classical Lamination Theory (CLT)
each external surface and interface from top to bottom.
2nd Special Case?
3rd Special Case?
4th Special Case?
Since the structure is not symmetric, it does not fall into the 3rd Special Case.
As a results, since the structure does not fall into any special cases, all matrices [A], [B] and [D] must be calculated..>>
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Step 4:
The [Q] matrices of each layer are calculated:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Since the materials of both layers are the same, their [Q] matrices are equal:
[Q] matrices depend only on material properties.
(They do not depend on the fiber routing angleθ. )
First, the minor Poisson ratio is calculated;
[Q] matrix of layer 1
[Q] matrix of layer 2
From Equation 2.7:
E1 (GPa) | E2 (GPa) | | G12 (GPa) |
126 | 78 | 0,33 | 29 |
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Step 5:
The reduced matrices of each layer [𝑸 ̅ ] are calculated:
From equations 6.14;
For 1st layer : θ =45o
For 2nd layer: θ =0o
8. Classical Lamination Theory (CLT)
(Don't forget this observation... It brings practicality when solving problems..)
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Step 6:
The necessary ones are calculated from the [A], [B] and [D] matrices belonging to the entire structure.
Since the 2-layer structure in this example does not enter any special states, all 3 matrices will be calculated:
8. Classical Lamination Theory (CLT)
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Let's calculate other stiffness matrices :
8. Classical Lamination Theory (CLT)
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The strains and curvatures of the midplane are calculated:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Step 7:
From Equation 8.32,
Helpful Videos for Matrix Inversion:
In Excell:
In Matlab:
If we take the inverse of the 6x6 matrix on the right with the help of a program and substitute it;
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Similar calculations were made for other asked points and the results are given in the table below.
| | | | | |
| -4 | 0 | 0 | 2 | 4 |
| | | | | |
| | | | | |
| | | | | |
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Global strains are calculated:
Step 8:
From 8.13 :
For the upper surface of the 1st (Top) layer; (z=-4)
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Calculations have been made for other asked points and all global stresses are shown in the table below.
| | | | | |
| -4 | 0 | 0 | 2 | 4 |
| | | | | |
| | | | | |
| | | | | |
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Global stresses are calculated :
Step 9:
If equation 6.1 is applied;
8. Classical Lamination Theory (CLT)
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c:cos45o , s:sin45o
| | | | | |
| -4 | 0 | 0 | 2 | 4 |
| | | | | |
| | | | | |
| | | | | |
As can be noticed, the global stresses (in the table on the previous page) and local stresses occurring in the 2nd layer (layer with 0o fiber orientation) are equal.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Local stresses are calculated:
Step 10:
From Equation 6.7:
Calculations have been made for other asked points and all local stresses are shown in the table below.
8. Classical Lamination Theory (CLT)
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When calculations are made for other desired points;
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Local strains are calculated:
Step 11:
If equation 6.11 is applied;
c:cos45o , s:sin45o
8. Classical Lamination Theory (CLT)
| | | | | |
| -4 | 0 | 0 | 2 | 4 |
| | | | | |
| | | | | |
| | | | | |
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Alternative solution for option a :
Step 9-) Local strains are obtained from transformation equations:
Step 10-) Afterwards, local stresses are calculated from Hooke's relations :
or Global stresses could be calculated from local stresses:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
By doing the first 8 steps exactly the same, global strains are obtained. After this, the following steps are followed.
8. Classical Lamination Theory (CLT)
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On average, we can calculate the intensity and proportions in which the Nx force is distributed to the layers as follows:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Since
The ratio at which the 1st Layer carries the Nx force:
b )
8. Classical Lamination Theory (CLT)
The ratio at which the 2nd Layer carries the Nx force:
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The local stresses on the upper surface of the upper layer were calculated in step 10 of a as follows:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
c )
XT (MPa) | XC (MPa) | YT (MPa) | YC (MPa) | S (MPa) |
35 | 68 | 18 | 38 | 8 |
Material strength properties given in the question
According to Hoffman criterion;
If the numerical values are substituted into the equation:
(No Damage Condition)
Equation 7.11
… no damage occurs.
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E1 | E2 | ν12 | G12 | XT | XC | YT | YC | S |
(GPa) | (GPa) |
| (GPa) | (MPa) | (MPa) | (MPa) | (MPa) | (MPa) |
81 | 30 | 0,35 | 15 | 101 | 180 | 25 | 50 | 12 |
The material properties found through experimental measurements for a newly produced composite lamina with dimensions of 200mm x 200mm x 2mm are given in the table below.
Example 8.2:
480
480
00
6mm
Fx
x
y
Fy
Fx
Fy
200mm
200mm
8. Classical Lamination Theory (CLT)
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We will obtain local stresses and local strains by following the steps below. We'll check for damage later.
Calculations can be made on a Q element with unit edge lengths.
For this reason, first the total internal forces and then the internal forces per unit length are calculated from the separation principle and static equilubrium.
K1
Fx=30kN
x
y
Fy
Fx
Fy
200 mm
200 mm
Q
1mm
1mm
Q
1mm
We cut from the K1 plane and examine the equilbrium of the right side in the x direction:
Step 1: Internal forces and internal moments are calculated.
Solution:
8. Classical Lamination Theory (CLT)
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Fiç-x=Fx=30kN
200mm
t=8mm
Total internal force in x direction
Q
1mm
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Similarly, for the y direction;
Equilibrium in the y direction of the back side of the cut made from the K2 plane:
Fiç-y=Fy=-23kN
200mm
Total internal force in y direction
Fy
Q
1mm
Fx
x
y
Fy
Fx
Fy
200 mm
200 mm
K2
Q
1mm
1mm
Q
1mm
Normal internal force in y direction per unit length:
Q
Internal forces in the unit element Q:
1mm
1mm
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
8. Classical Lamination Theory (CLT)
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Step 3: Control os special cases
Does this structure fall into any of the special cases described in clause 8.7?
Does it fall under Special Case 1?
Does it fall under Special Case 2?
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Step 2:
The layered structure is coded.
If the structure is symmetrical, it enters the 1st Special Case.
Yes
Yes,
except option d
If the structure is symmetrical and internal moments are zero, it enters Special Case 2.
No, except option d
It will be explained while solving option d.
As explained in article 8.5
8. Classical Lamination Theory (CLT)
Does it fall under Special Case 4?
Does it fall under Special Case 3?
No
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Step 4:
The [Q] matrices of each layer are calculated:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Since their materials are the same, the [Q] matrices of all layers are the same.
From equation 2.7:
E1 (GPa) | E2 (GPa) | | G12 (GPa) |
81 | 30 | 0,35 | 15 |
8. Classical Lamination Theory (CLT)
(Recall that the [Q] and [S] matrices do not depend on the fiber orientation angle.)
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Step 5:
From equatios 6.14;
For layers 1 and 3 : θ =48o
For the 2nd layer: Since θ =0o
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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Step 6:
The necessary ones are calculated from the A], [B] and [D] matrices belonging to the entire structure.
For the reasons explained in Step 3: [B] = 0. The matrix [D] is non-zero, but there is no need to calculate it since it will not enter into operations. Then it is sufficient to just calculate the [A] matrix. (For option d, a separate explanation will be made.)
It is found as
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The strains and curvatures of the midplane are calculated:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Step 7:
From Equation 8.32
Or, since [B] = 0 and [D] will not enter into operations, the strains of the middle plane can be reduced to a 3x3 matrix multiplication as follows:
(It can also be found using a program such as Matlab, Excel etc.)
As explained in Step 3, the structure enters the 2nd special case and the curvatures of the middle plane become zero:
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Global strains are calculated:
Step 8:
From equation 8.13
Step 9:
Since they are independent of z, the global strains of all points are the same and equal to the strains of the midplane
Local strains are calculated:
The local stresses requested in the question on the lower surface of the upper layer are:
c:cos48o , s:sin48o
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Local stresses are calculated:
Step 10:
Since we have obtained local stresses and local deformations, we can move on to damage control..>>
If we apply Equation 6.1 for the bottom surface of the top layer;
As an alternative solution,
After doing the first 8 steps exactly ,
we could find,
global stresses in step 9,
local stresses in step 10,
the local strains n Step 11.
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No Damage Conditions :
a-) According to the Maximum Stress Criterion:
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Control for the bottom surface of the top layer (z=-1):
X
Note: Even if the shear stress (𝛕𝟏𝟐) is negative, it is taken as positive (+) in the equations. Because the direction does not matter in the shear strength of the material..
Previously calculated values
Material Properties Given in the Question
| | | | |
(MPa) | (MPa) | (MPa) | (MPa) | (MPa) |
101 | 180 | 25 | 50 | 12 |
Since all inequalities cannot be provided
on the lower surface of the upper layer.
according to this criterion damage occurs
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b-) According to the Maximum Strains Criterion,
No Damage Conditions :
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Previously calculated values
Additionally, the strength limits for strains are:
Control for the bottom surface of the top layer (z=-1):
X
Since all inequalities cannot be provided
Note: Even if the shear deformation angle (𝛄_𝟏𝟐) is negative, it is taken as positive (+) in the equations. Because the direction does not matter in the shear strength of the material.
on the lower surface of the upper layer.
according to this criterion damage occurs
8. Classical Lamination Theory (CLT)
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No Damage Condition:
c-) According to the Tsai-Hill Criterion
Control for the bottom surface of the top layer (z=-1):
Previously calculated values
8. Classical Lamination Theory (CLT)
on the lower surface of the upper layer.
according to this criterion damage occurs
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
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First of all, local stresses in the middle of the middle layer in the Q element must be determined.
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
Fx
x
y
Fy
Fx
Fy
Q
in the middle plane (z = 0) are asked, it enters the 4th special case and
there is no need to calculate the [D] matrix. (See article 8.7)
the intensities of the moments are not important. (We can see this situation by examining Equation 8.32.)
Q
1mm
1mm
Which of the special cases does option d fall into?
In this case, the first 8 steps done before are also valid for option d.
Otherr Steps..>>
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Alternative to Step 9 :
Local Strains are calculated:
Local strains in the middle of the middle layer:
c:cos0o , s:sin0o
Local Stresses are calculated:
Alternative to Step 10 :
Local stresses in the middle of the middle layer:
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According to the Modified Tsai-Hill Criterion (damage control in the middle of the middle layer):
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
No Damage Condition :
Previously calculated values
No damage occurs on the middle surface (midplane) of the Middle Layer.
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Example 8.3: A layered composite elbow was obtained by bending 3 of the continuous fiber reinforced composite layers, the properties of which are given in the table, in an L shape and gluing them on top of each other with a 450 / 00 / 450 fiber arrangement. The upper end of this elbow was connected to a fixed wall, a horizontal tensile force of F = 320kN was applied from a distance of h/2 from the upper surface and a torsional moment of T = 0.96kNm was applied to the lower free surface. Each of the layers is t = 2cm thick and h = 8cm wide. Accordingly, check whether damage will occur on the upper and lower surfaces of the middle layer under these loading conditions in this layered structure, according to the Tsai-Hill criterion.
| | | | |
96 MPa | 200 MPa | 48 MPa | 110MPa | 36 MPa |
E1 (GPa) | E2 (GPa) | | G12 (GPa) |
81 | 30 | 0,35 | 15 |
F
h/2
h
450
t
T
t
t
8. Classical Lamination Theory (CLT)
On the lower surface of the Middle Layer:
On the upper surface of the Middle Layer:
Answers:
1.46 > 1
(according to the Tsai-Hill criterion.)
1 = 1
Damage occur
Damage occur
inequality 7.6
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600
450
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Example 8.4:
BCDE beam will be manufactured by placing a bidirectional woven-fabric material no. 2 between two unidirectional and continuous fiber reinforced layers no. 1 and 3. The beam will be connected to a bar AB as shown in the figure and supported by the H element. The beam will also be subjected to uniformly distributed load in its middle region (between C-D). Material properties are given in the table below. Layer 1 has a fiber orientation angle of 600, layer 3 has a fiber orientation angle of 450 and the thickness of both is t1=t3=10mm. Material number 2 will have a thickness of t2 = 12mm and a fiber pair orientation of 00 / 900. In order to avoid damage to the structure under these conditions, determine the minimum value of b width according to the Tsa-Hill criterion.
Layer No | E1 (GPa) | E2 (GPa) | | G12 (GPa) |
| 84 | 30 | 0,35 | 12 |
| 40 | 40 | 0,32 | 26,4 |
1 and 3
2
| | | | | t (mm) | θ |
90 | 180 | 30 | 60 | 22 | 10 | 60o, 45o |
40 | 80 | 40 | 80 | 22 | 12 | 00 / 900 |
8. Classical Lamination Theory (CLT)
Answer:
A
H
1
3
2
B
C
2m
t2
q=240N/m
E
2m
2m
D
b
t1
t3
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8.8 Sandwich Composites and Calculation Methods
8. Classical Lamination Theory (CLT)
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8.9 Thermal Loading in Layered (laminated) Composites
(8.33)
We think that a layered composite structure is placed in a cavity that will prevent deformation in both x and y directions, and its temperature is increased by the amount ΔT.
Eq. (6.18.b)
Eq. (8.14)
Eq. (8.18)
Internal forces per unit length:
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Internal Forces per Unit Length due to temperature difference:
Internal Forces Per Unit Length Resulting from Structural Loads:
Total Internal Forces Per Unit Length Resulting from
Temperature Difference + Structural Loads:
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(8.34)
(8.36)
(8.37)
(8.35)
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Similarly, let's calculate the total internal moments :
Eq. (8.22)..>>
Internal moments per unit length:
If we substitute Equation (8.33) into Equation 8.22, we get
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(8.38)
Internal Moments per Unit Length due to temperature difference:
Internal Moments Per Unit Length Resulting from Structural Loads:
Total Internal Moments Per Unit Length Resulting from
Temperature Difference + Structural Loads:
(8.40)
(8.41)
(8.39)
(It is zero in symmetrical structures.)
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If Equations 8.34 and 8.41 are combined;
(8.42)
(8.43)
In this case, the strains and deformations of the middle plane are:
Equations 8.42 and 8.43 are general equations for layered composites, including the temperature effect.
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8.9.1 Steps for calculating Stresses and Strains (including temperature effect):
Other Steps are the same as described in topic 8. Only instead of equation 8.32, the operations start with equation 8.43.
The only difference with the calculations in topic 8 is that instead of structural internal loads, total internal loads (including temperature) will be used.
Now we will try to understand the subject better by solving an example..>>
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Example 8.6
3 of the layers in Example 7.3 are glued on top of each other with a 300 / 00 / 300 fiber arrangement and placed in the same mold. Each layer is 200mm x 200mm x 2mm in size. By tightening the bolts a little, a compression force of -20kN was created on the lateral surfaces. Additionally, the temperature of the system was increased by 50 0C. Calculate the local stresses arising on the lower surface of the middle layer, ignoring all friction.
Solution:
Material properties will be taken from example 7.3
a-) First of all, if we calculate the structural loads per unit length:
Nx = Ny = -20x103 N / 200mm = -100N/mm
Nxy = 0 (Because frictions are neglected)
200mm
200mm
6mm
Step 1-) Total internal forces and moments per unit length are calculated.
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b-) Calculation of internal forces and moments due to temperature difference:
300
300
00
For the 1st and 3rd layers with θ =300
[Q] matrix terms for all layers:
(It was calculated in Example 7.3)
Since θ =00 for the middle layer:
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Let's code the structure..>>
Calculation of global thermal expansion coefficients for each layer :
Eq. 6.19.b
Eq. 6.9
for the 2nd (middle) layer, θ =00
for the 1st and 3rd layer, θ =300
Values given in Example 7.3 :
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We will calculate Thermal Stifness Matrices:
From eq. 8.35
Because the structure is symmetrical :
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From eq. 8.38
Internal Forces
per Unit Length
due to temperature difference:
Internal Moments
per Unit Length
due to temperature difference:
From eq. 8.34
Total Internal Forces Per Unit Length Resulting from
Temperature Difference and Structural Loads:
Total Internal Moments Per Unit Length Resulting from
Temperature Difference and Structural Loads:
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Other Steps:
Then only matrix [A] will be calculated.
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- We will calculate the strains and curvatures in the midplane from equation 8.43
Or, since [B] = 0 and [D] have no effect in the calculations, the deformations of the middle plane can be reduced to a 3x3 matrix multiplication as follows:
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- Global strains are calculated:
From eq. 8.13
Since the global strains of all points are independent of z, they are the same and equal to the strains of the middle plane.
- Local strains are calculated:
The local stresses on the lower surface of the middle layer, requested in the question are:
c:cos0o , s:sin0o
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- Local stresses are calculated :
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HOMOGENİZATİON OF LAMİNATED COMPOSİTES
Theoretical Methods: Voigt, Reuss, CLT Approaches, and the Zor Model
9.
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For a homogenization approach to provide physically consistent results, the reciprocity condition must be satisfied. This relationship is one of the fundamental consequences of energy consistency in linear elasticity theory.
9.1 Scope and Objective
n-layered structure
Equivalent volume
Many different homogenization methods have been developed in the literature. Among the theoretical approaches used to calculate the properties of an equivalent volume, the most important are the Voigt method, the Reuss method, the Classical Lamination Theory (CLT) approach, and the newly developed Zor Model. We will now examine and compare these approaches one by one.
Reciprocity Condition :
(9.1)
Figure 9.1
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The Voigt approach assumes that, under in-plane tensile or compressive loading, all layers in a laminated structure undergo the same amount of elongation due to perfect bonding (iso-strain condition). Accordingly, the properties of the equivalent volume are obtained as the volume-weighted average of the layer properties.
9.1 Voigt Approach
z
x
For a laminated structure subjected to the external force Fₓ, the total elongation in the x-direction is Δl for every layer as well as for the entire structure. Since the lengths in the x-direction are identical (l), the corresponding normal strains are also equal:
The external force Fₓ is distributed among the layers in different proportions. From static equilibrium:
Fx
1
2
n
i
(9.1.1)
(9.1.2)
Figure9.2.a
Figure 9.2.b
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1
2
i
n
Elastic modulus of the equivalent volume in the x-direction:
Similarly, if an external force Fᵧ is applied to the entire structure, the elastic modulus of the equivalent volume in the y-direction is obtained as:
If the force equilibrium given in Eq. (9.1.2) is expressed in terms of normal stresses:
The volume fraction of the i-th layer is defined as:
(9.1.3)
(9.1.4)
(9.1.5)
(9.1.6)
(9.1.7)
Figure 9.3
From Hooke’s law:
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Thus,
or
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y
x
The strain in the y-direction for the i-th layer is
The overall strain in the y-direction becomes:
The volume fraction of the i-th layer is :
Similarly, if a force Fᵧ is applied, the equivalent Poisson’s ratio is obtained as
(9.1.8)
(9.1.9)
(9.1.10)
(9.1.11)
(9.1.12)
(9.1.13)
(9.1.14)
Figure 9.4 View of the Laminated Structure in the x–y Plane
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Therefore,
which gives
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1
i
2
n
z
y
x
Fxy
Fxy
1
i
2
n
z
y
x
(9.1.15)
(9.1.16)
(9.1.17)
(9.1.18)
(9.1.19)
(9.1.20)
Figure 9.5.a
Figure 9.5.b
Figure 9.5.c
1
i
2
n
z
y
x
From
Hooke’s law :
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Therefore,
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9.2 Reuss Approach : This approach assumes that, under a single loading condition (simple tension/compression or pure shear), the stresses in the loading direction are equal in all layers (iso-stress condition). The overall deformations of the equivalent volume are then defined as the volume-weighted averages of the deformations of the individual layers.
z
x
1
2
n
i
For a tensile load applied in the x-direction:
(9.2.1)
(9.2.2)
(9.2.3)
(9.2.5)
(9.2.4)
(9.2.6)
Figure 9.6.a
Figure 9.6.b
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For a tensile load applied in the x-direction:
(9.2.9)
(9.2.7)
(9.2.8)
(9.2.10)
(9.2.11)
(9.2.12)
y
x
Note: Although the Reuss approach satisfies the reciprocity condition, the iso-stress assumption cannot be considered a fully realistic representation of the actual mechanical behavior of laminated composite structures.
Figure 9.7
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y
x
y
x
Hooke’s laws
the volume-weighted average of the shear strains of the layers:
(9.2.14)
(9.2.15)
(9.2.16)
(9.2.17)
(9.2.18)
Figure 9.8.a
Figure 9.8.b
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Substituting these relations into Eq. (9.2.15) gives..
and therefore
which is the expression for the equivalent shear modulus according to the Reuss approach.
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9.3 Classical Lamination Theory (CLT) Approach :
(9.3.1)
(9.3.2)
(9.3.3)
(9.3.4)
(9.3.5)
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THE ZOR MODEL
[This new approach was published as a research article in the journal Composite Structures (Elsevier).]
( DOI: 10.1016/j.compstruct.2025.120025 )
A New Equivalent Volume Approach for Laminated Structures
9.4
9. Homogenization of Laminated Composites / The Zor Model
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9.4.1 Fundamental Characteristics of the Zor Model:
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z
x
This condition and Eq. (9.4.1) are directly assumed in the Voigt approach. In the Reuss approach, however, an iso-stress assumption is adopted, and therefore the strains are generally different. In the CLT approach, this condition is satisfied for symmetric laminates, whereas different results may be obtained for asymmetric laminates.
(9.4.1)
9.4.2 Consequences of Perfect Bonding in the Zor Model: Perfect bonding implies that common points between adjacent layers continue to move together after loading and that no separation occurs at the interfaces. As a result of perfect bonding, the following three conditions arise, which form the basis of the governing equations of the Zor Model.
Figure 9.9
A comparison of the different approaches with respect to the interpretation of perfect bonding is presented separately in Section 9.4.9.4.
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Case 2, which represents the lateral interactions between the layers, is a distinctive assumption of the Zor Model and is not included in the Voigt, Reuss, or CLT approaches.
y
x
(9.4.2)
Figure 9.10
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9.4.2.3) Case 3: Preservation of the Rectangular Shape of the Layer Plane
Consider a tensile load applied in the x-direction. A layer plane that is rectangular before loading remains rectangular after loading. In other words, lines that are initially perpendicular to each other remain perpendicular after deformation. Therefore, no shear strain develops in either the layers or the equivalent volume.
y
x
Note: These three conditions, which form the foundation of the Zor Model, are equally valid for tensile or compressive loading applied in either one or both directions within the x–y plane.
Since asymmetric laminates may produce different results in the CLT approach, this conclusion may appear unexpected at first glance. However, it should be remembered that the Zor Model is based on the macroscopic behavior of the equivalent volume and therefore follows a different framework from CLT.
(9.4.3)
Figure 9.11
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k
Fx
1
2
i
n
1
i
2
n
Fx
z
y
x
k
Fx
y
x
i
From static equilibrium,
(9.4.4)
(9.4.5)
Equation (9.4.4) is also present in the Voigt and CLT approaches. In contrast, Eq. (9.4.5) is a distinctive equation of the Zor Model.
Figure 9.12.a
Figure 9.12.b
Figure 9.12.c
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1
2
i
n
k
Expressing the force equilibrium equations, Eqs. (9.4.4) and (9.4.5), in terms of stresses gives
(9.4.4)
(9.4.6)
(9.4.7)
y
x
i
Equation (9.4.6) is also present in the Voigt and CLT approaches. In contrast, Eq. (9.4.7) is a distinctive result that emerges from the Zor Model.
Figure 9.13.a
Figure 9.13.b
Mechanics of Composite Materials- Lecture Notes / Mehmet Zor
9.4.4 Stress Distribution in the Zor Model:
Similarly, from Eq. (9.4.5)
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9.4.5 Application of Hooke’s Relations in the Zor Model
(9.4.8)
(9.4.9)
9.4.5.2) For the Equivalent Volume : The equivalent volume of a structure composed of monoclinic layers also exhibits monoclinic behavior. Therefore, the Hooke relations for the equivalent volume can be written in a similar form as
(9.4.10)
(9.4.11)
y
x
i
Continuous fiber-reinforced layers are orthotropic with respect to their local 1–2 coordinate system, but generally exhibit monoclinic behavior with respect to the global x–y coordinate system.
Figure 9.14
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9.4.6 Calculation of the Equivalent Properties in the Zor Model
Using the equations derived so far, the elastic properties of the equivalent volume representing the mechanical behavior of symmetric or asymmetric laminated structures composed of layers that generally exhibit monoclinic behavior with respect to the global x–y coordinate system will now be determined. In the Zor Model, the order of calculation is important and should follow the sequence presented below.
Substituting Eqs. (9.4.8) and (9.4.10) into Eq. (9.4.1) yields
(9.4.12)
Substituting Eqs. (9.4.9) and (9.4.11) into Eq. (9.4.2) gives
(9.4.13)
Solving Eqs. (9.4.12) and (9.4.13) for the normal stresses in the i-th layer, we obtain
(9.4.14)
(9.4.15)
We consider only the equations derived for the case of tensile loading in the x-direction.
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Substituting Eq. (9.4.15) into Eq. (9.4.7):
(9.4.16)
(9.4.17)
From Eq. (9.4.17):
Let us rewrite Eq. (9.4.18) in terms of these coefficients
(9.4.19)
(9.4.18)
(9.4.22)
(9.4.20)
To observe the Poisson effect more clearly in Eq. (9.4.18), we define layer-specific coefficients for each layer:
(9.4.21)
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Substituting Eq. (9.4.14) into Eq. (9.4.6):
(9.4.23)
(9.4.24)
(9.4.25)
(9.4.26)
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1
i
2
n
Fx
z
y
x
Fy
Fy
..>>
Figure 9.15
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(9.4.26)
(9.4.27)
(9.4.28)
(9.4.29)
(9.4.30)
(9.4.31)
(9.4.32)
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(9.4.33)
As a result,
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9.4.7) Natural Satisfaction of the Reciprocity Condition in the Zor Model
(9.4.34.a)
(9.4.34.b)
(9.4.34.c)
(9.4.34.d)
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Let us express the final equations of the Zor Model in terms of these variables:
Eq.(9.4.18):
Eq.(9.4.26):
Expansion of Eq. (9.4.24):
Expansion of Eq. (9.4.30):
(9.4.35)
(9.4.36)
(9.4.37.a)
(9.4.37.b)
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We know that reciprocity is satisfied for each i-th layer:
Let us express the reciprocity terms for the entire structure (equivalent volume) in terms of these variables:
or
(9.4.38)
(9.4.39)
(9.4.40)
(9.4.41)
(9.4.42)
Eq. (9.4.42) shows that the reciprocity condition is satisfied for the equivalent volume in the Zor Model.
Eqs. (9.4.40) and (9.4.41) are found to be equal.
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In this section, we will obtain the general stiffness matrix that gives the stress–strain relations in the Zor Model.
(9.4.44)
(9.4.43)
(9.4.45)
(9.4.3)
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Similarly,
(9.4.46)
(9.4.47)
(9.4.3)
(9.4.48)
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The equivalent stiffness matrix is obtained by taking the inverse of the compliance matrix and again takes an orthotropic form:
(9.4.49)
It is stated in the literature how the terms of the stiffness matrix under plane stress conditions for an orthotropic material can be written in terms of the engineering constants. Accordingly, the equivalent stiffness matrix of the Zor Model can be expressed as follows:
(9.4.50)
As can be seen, the Zor Model enables a structure composed of layers that are generally monoclinic in the global axes to be represented by an orthotropic equivalent volume. This makes other mechanical calculations much easier.
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Example 9.1
θ1
θ3
θ2
1
3
2
8mm
8mm
8mm
z
x
a-) Voigt
The layers are cut from a large continuous fiber-reinforced lamina, and the properties of this lamina with respect to the local 1–2 axes are as follows:
b-) Reuss
c-) CLT
d-) Zor Model
600
00
00
1
2
3
z
x
y
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Solution::
The cross Poisson’s ratios are calculated from Eq. (9.1):
Properties of the second layer: >>..>>
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(9.4.51.a)
(9.4.51.b)
(9.4.51.c)
(9.4.51.d)
Material Property Transformation Equations
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The cross Poisson’s ratio is calculated from Eq. (9.1):
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a-) According to the Voigt Approach
Step 3) Calculation of the Equivalent Volume Properties:
From Eq. (9.1.6):
From Eq. (9.1.7):
From Eq. (9.1.13):
From Eq. (9.1.14):
From Eq. (9.1.20):
From Eq. (9.1): Reciprocity Check
Reciprocity condition not satisfied
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b-) According to Reuss approach
From Eq. (9.2.5) :
From Eq. (9.2.6):
From Eq. (9.2.11):
From Eq. (9.2.12):
From Eq. (9.2.18):
From Eq. (9.1):
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Reciprocity condition is satisfied
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c-) According to the CLT Approach
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From Eq. (6.14) ;
where
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c3) The laminated structure is coded:
1
3
2
8mm
8mm
8mm
z
x
c4) The [A] matrix is calculated.
Eq. (8.23):
c5) The [A]-1 matrix is calculated.
(The result is given directly here.)
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From Eq. (9.3.1) :
From Eq. (9.3.2) :
From Eq. (9.3.3):
From Eq. (9.3.4):
From Eq. (9.3.5):
From (9.1) Reciprocity Check :
Reciprocity condition is satisfied
c6 ) Calculation of the Equivalent Properties
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d ) According to the Zor Model
In the Zor Model, the order of calculation is also important and should be followed as given below:
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The Zor Model adopts the Voigt approach for the calculation of the shear modulus.
From Eq. (9.4.33):
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Equivalent Properties | Voigt | Reuss | CLT | Zor Model |
Ex (GPa) | 87,14 | 82,69 | 88,09 | 87,82 |
Ey (GPa) | 75,51 | 74,94 | 77,73 | 76,79 |
νxy | 0,26 | 0,29 | 0,26 | 0,26 |
νyx | 0,25 | 0,26 | 0,23 | 0,23 |
Gxy (GPa) | 23,63 | 22,66 | 24,31 | 23,63 |
0o/ 60o / 0o (Symmetric)
Equivalent Properties | Voigt | Reuss | CLT | Zor Model |
Ex (GPa) | 82,45 | 79,12 | 85,28 | 82,96 |
Ey (GPa) | 73,00 | 72,58 | 76,5 | 74,02 |
νxy | 0,30 | 0,31 | 0,27 | 0,30 |
νyx | 0,28 | 0,29 | 0,24 | 0,26 |
Gxy (GPa) | 24,52 | 23,72 | 25,97 | 24,52 |
15o/ 60o / 0o (Asymmetric)
(calculated in parts (a)–(d))
(Only final results are presented)
Results Tables and Comparisons
From the tables above, it can be seen that the equivalent properties vary depending on the homogenization approach, and that these differences become somewhat more pronounced for asymmetric stacking sequences. In addition, factors such as the use of different materials in the layers and the number of layers are also important parameters that can affect the differences among the results obtained by the various approaches.
Table 9.1.a
Table 9.1.b
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Example 9.2 :
15o
1
3
2
8mm
8mm
8mm
z
x
60o
0o
x
y
The equivalent properties obtained for this structure as a result of the calculations were given previously in Table 9.1.b.
Calculate the stress and strain values at the mid-planes of the layers using the Zor Model and compare the results with those obtained from CLT. (The CLT results are given directly in the tables at the end of the solution.)
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Solution:
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According to the Zor Model, for plane tension/compression loading conditions, the strains are equal throughout the entire structure and in every layer; the shear strain is zero.
From Eq. (9.4.1):
From Eq. (9.4.2):
From Eq. (9.4.3):
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If the stresses are solved from Eqs. (9.4.8) and (9.4.9):
Eq. (9.4.8)
Eq. (9.4.9)
(9.4.52.a)
(9.4.52.b)
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Layer Strains..>> | | | | ||||||
Layer Number (i)..>> | 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |
Zor | 2,87 | 2,87 | 2,87 | -2,16 | -2,16 | -2,16 | 0 | 0 | 0 |
CLT | 3,021 | 2,825 | 2,629 | -2,243 | -2,067 | -1,890 | -0,057 | -0,057 | 0,790 |
Layer Stresses..>> | | | | ||||||
Layer Number (i)..>> | 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |
Zor | 21,62 | 14,38 | 26,08 | -10,21 | -8,25 | -13,11 | 0 | 0 | 0 |
CLT | 24,63 | 14,73 | 24,04 | -11,43 | -9,00 | -11,28 | 4,21 | -3,71 | 1,58 |
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The Voigt, Reuss, CLT, and Zor Model approaches, which theoretically calculate the mechanical elastic properties of an equivalent volume representing a laminated structure, are compared below from different perspectives.
9.5 Comparison of Theoretical Homogenization Approaches
9.5.1 In Terms of Interlayer Poisson Interactions
In order for the actual mechanical behavior of laminated composite structures to be represented accurately by an equivalent volume, it is extremely important to take into account the interlayer (system-level) Poisson interactions. These interactions arise from the restriction of the lateral deformations of the layers by one another and directly affect the elastic properties of the equivalent volume.
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9.5.2 In Terms of the Reciprocity Condition
The reciprocity condition is a fundamental relationship that must exist among the elastic properties of a material. Satisfaction of this condition means that the calculated equivalent elastic properties are mutually compatible and physically consistent.
9.5.3 In Terms of Computational Simplicity
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1
2
3
A
B
Lx
Fx
Fx
2
3
A'
B'
ΔLx/2
ΔLx/2
x
y
A'
B'
Şekil 9.16
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Strength and Failure Analysis
10.
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y
z
Equivalent Volume
of Laminated Structures Using the Zor Model
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10.1 Objectives of This Chapter: To determine the tensile, compressive and shear strength limits of the Zor equivalent volume described in Section 9.4, and to perform stress-strain calculations and failure evaluations for different loading conditions.
n layered structure
Zor equivalent volume
1
i
2
n
z
y
x
?
3. While determining the strength limit of the equivalent volume for a loading type, the weakest layer that will fail first under that loading is taken as the basis. The equivalent stress that brings this weakest layer to its own strength limit is determined. This equivalent stress is accepted as the strength-limit value of the equivalent volume.
4. . In the Zor approach, failure evaluation is performed in terms of the equivalent volume of the structure. In this way, rather than local failures that may occur in the layers beforehand, it is determined whether the structure as a whole has failed or not. (In CLT, failure evaluation is performed for each layer.)
10.2 General Framework
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y
x
i
y
x
i
1
i
2
n
Fx
z
y
x
k
Fx
1
2
i
n
k
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(9.4.14)
(9.4.15)
;
y
x
i
(10.1.a)
(10.1.b)
(10.1.c)
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Substituting Equations (9.4.14 and 9.4.15) into Equations (10.1 a–c) and rearranging, we obtain:
(10.2.a)
(10.2.b)
(10.2.c)
Layer stress coefficients:
(10.3.a)
(10.3.b)
(10.3.c)
(10.3.d)
(10.3.e)
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Local stresses before failure in terms of the stress coefficients:
(10.4.a)
(10.4.b)
(10.4.c)
(10.5)
(10.6.a)
(10.6.b)
(10.6.c)
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10.3.2 Application of the Failure Criterion to the Layers
(10.7.a)
Substituting the local stresses in Equation (10.6), which are valid at the moment of failure, into Equation (10.7.a), we obtain:
(10.7.b)
(10.7.c)
(10.7.d)
(i =1,2, …n )
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(10.8.a)
Substituting the local stresses in Equation (10.6), which are valid at the moment of failure, into Equation (10.8.a), we obtain:..>>
(i =1,2, …n )
(10.8.b)
(10.8.c)
(10.8.d)
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(10.9)
From Equation (7.9), the coefficients are:
According to the Tsai–Hill failure criterion: :
According to the Hoffman failure criterion:
According to the Mises-Hencky failure criterion:
(10.10.a)
(10.10.b)
(10.10.c)
(10.10.d)
(10.11.a)
(10.11.b)
(10.11.c)
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(10.13)
(10.12)
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(10.15)
(10.14)
(10.16)
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y
x
i
(10.1.a)
(10.1.b)
(10.1.c)
(10.17.a)
(10.17.b)
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Substituting Equations (10.17.a and 10.17.b) into Equations (10.1 a–c) and rearranging, we obtain:
Layer stress coefficients:
(10.18.a)
(10.18.b)
(10.18.c)
(10.19.a)
(10.19.b)
(10.19.c)
(10.19.d)
(10.19.e)
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Local stresses before failure in terms of the stress coefficients:
(10.20.a)
(10.20.b)
(10.20.c)
(10.21)
(10.22.a)
(10.22.b)
(10.22.c)
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10.4.2.a According to the Tsai–Hill Failure Criterion:
10.4.2.b According to the Modified Tsai–Hill Failure Criterion :
(10.24.a)
(10.24.b)
(10.23)
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(10.25)
(10.26)
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(10.27)
(10.28)
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y
x
(10.29)
(10.30)
(10.31)
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(10.34)
(10.32)
(10.33)
From the transformation Equations 6.3 a–c:
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(10.35)
(10.7.a)
At the moment of failure
(10.36)
After calculations are performed for all layers, from Equation (10.29) :
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(10.37)
At the moment of failure
(10.38)
(10.8.a)
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(10.9)
In the case of pure shear, at the moment of failure:
(10.39)
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(10.14)
In the case of pure shear,
at the moment of failure:
(10.40)
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10.8 Stress–Strain Calculations in the Zor Equivalent Volume :
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Example 10.1
480
480
00
6mm
Fx
x
y
Fy
Fx
Fy
200mm
200mm
(The structure, load, and material properties in Example 8.2 were used.)
E1 | E2 | ν12 | G12 | | | | | |
(GPa) | (GPa) |
| (GPa) | (MPa) | (MPa) | (MPa) | (MPa) | (MPa) |
81 | 30 | 0,35 | 15 | 101 | 180 | 25 | 50 | 12 |
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Solution:
Step 2. Calculation of the Elastic Properties of the Zor Equivalent Volume:
First, the elastic properties of each layer with respect to the global axes are determined:
The cross-Poisson ratios are calculated from Equation (9.1):
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From eqs. (9.4.51):
The cross-Poisson ratios are calculated from Equation (9.1):
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In the Zor Model, for the case of pure shear, the shear modulus is obtained by the Voigt-type volumetric average.
From eq. (9.4.33):
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Step 3: The Strength Limits of the Zor Equivalent Volume Are Calculated (According to the Modified Tsai–Hill Criterion):
a. Calculation of the Layer Stress Coefficients:
From eq. (10.3)
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From Eqs. (10.3)
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From Eqs. (10.19)
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From eq. (10.8.c), (i=1 , 3)
The equivalent stress that will cause failure of layers 1 and 3
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From eq. (10.8.c) , (i=2)
According to Equation (10.8.d), the tensile strength of the equivalent volume in the x direction:
The equivalent stress that will cause failure of layer 2
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Writing Equation (10.8.c) for the compression case :
The equivalent stress that will cause failure of layers 1 and 3
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(i=1 , 3)
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the compressive strength of the equivalent volume in the x direction:
The equivalent compressive stress that will cause failure of layer 2:
Writing Equation (10.8.c) for the compression case : (i=2)
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From Eq. (10.24) , (i=1 , 3)
The equivalent tensile stress that will cause failure of layers 1 and 3
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The equivalent tensile stress that will cause failure of layer 2
the tensile strength of the equivalent volume in the y direction:
From Eq. (10.24) , (i=2)
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The equivalent compressive stress that will cause failure of layers 1 and 3
Writing Equation (10.24.b) for the compression case (i=1 , 3)
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Writing Equation (10.24.b) for the compression case (i=2)
The equivalent compressive stress that will cause failure of layers 2
the compressive strength of the equivalent volume in the y direction
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The equivalent shear stress that will cause failure of layers 1 and 3
From eq. (10.38)
From eq (10.31):
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The equivalent shear stress that will cause failure of layer 2.
From eq. (10.38)
From eq. (10.31) :
From eq.(10.29):
Shear strength of the entire structure (equivalent volume)
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Step 4 – Calculation of the Stresses in the Equivalent Volume
6mm
x
y
200mm
200mm
200mm
6mm
200mm
x
y
Laminated Structure
Equivalent Volume and Forces
200mm
6mm
200mm
x
y
Equivalent Volume and Stresses
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Step 5 – Failure Check of the Equivalent Volume According to the Modified Tsai–Hill Criterion
For a single orthotropic layer, the no-failure condition of this criterion is written in the local 1–2 axes as given in Equations 7.7 or 10.8.a. The equivalent volume of the Zor model exhibits orthotropic behavior in the global axes. Therefore, for the Zor equivalent volume, this condition can be adapted to the global x–y coordinate system as shown in Equation (10.41).
Meaning of the values in the denominator:
(10.41)
Equivalent stresses calculated for this example:
In all cases:
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6mm
200mm
x
y
200mm
6mm
x
y
b-) In order to compare the local stress results with the CLT results in Example 8.2, we must also calculate the local stresses occurring in the layers in the Zor Model solution. This is because CLT gives results on a layer basis. However, it should also be remembered that, in strength calculations or failure checks in the Zor Model, the equivalent volume is taken as the basis, and there is no need for the layer stress or strain. According to the principle of superposition, we can apply the external forces successively and calculate the local stresses occurring in each layer.
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Total Local Stresses in the Layers and Comparisons with the CLT Results:
Local normal stress in the 1 direction:
Local normal stress in the 2 direction:
Total local stresses in layers 1 and 3
Shear stress in the 1–2 plane:
Total local stresses in layer 2
Local normal stress in the 1 direction:
Local normal stress in the 2 direction:
Shear stress in the 1–2 plane :
CLT
Zor
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200mm
6mm
200mm
x
y
z
h
b
x
z
At the instant of failure :
(total bending moment that will bring the entire structure to the strength limit)
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Example 10.2-)
x
y
200mm
200mm
6mm
Solution)
Stresses occurring in the equivalent volume
200mm
6mm
200mm
x
y
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Calculation of the Stiffness Matrix of the Equivalent Volume:
(from Equation 9.4.50):
The material properties calculated in Example 10.1 are substituted into the matrix:
Compliance Matrix of the Equivalent Volume:
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The strains in Equivalent Volume:
The total elongation/shortening occurring in the equivalent volume :
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200mm
6mm
200mm
x
y
z
b
h
From Eq. (10.41)
Failure occurs.
b-)
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