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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

  • Faculty members who want to follow these lecture notes in their own department's course are sufficient to send me an information e-mail. Solutions to the questions marked (*) in the notes are sent to these faculty members, if they wish.

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)

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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

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Contents

1. Summary General Information

2. Anisotropic Material Types and Elastic Mechanical Properties (Anistropic Elasticity)

A- LAMINA (Unidirectional continuous fiber reinforced)

  1. Theoretical Calculation of Material Properties of Unidirectional Continuous Fiber Composites
  2. Theoretical Strength and Efficiency Limits of Composites (Structural Loading Examples)
  3. Experimental Determination of Mechanical Properties
  4. Hooke and Transformation Equations in Composites (Stress-strain calculations for single layer)
  5. Yield and Failure Criteria in Composites

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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Summary of �General InformationAbout �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

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1.2 Basic Properties of Composite Materials

  1. Although composite actually means mixture, it does not consist of soluble and dissolving components.
  2. There is no exchange of atoms between the components.
  3. Composite components do not chemically affect each other.
  4. If the materials dissolve in each other and there is a mixture at the atomic level, such materials are not composites but alloys.
  5. If the mixture is at the level of nanometer particles, these types of composites are called nano composites.

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  1. Composites are generally created from a main material called "matrix" and a more durable material called "reinforcement element (fiber)".
  2. Of these two groups of materials, the reinforcement material increases the strength and load-bearing ability of the composite material.
  3. The matrix material plays a role in preventing crack propagation that may occur during the transition to plastic deformation and delays the rupture of the composite material.

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1-Summary of General Information About Composites

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

  1. By placing continuous fibers, especially in directions where strength is important, the composite structure is provided with higher strength in those directions.
  2. Since there is no need to place fibers in other directions, the composite structure is both lighter than classical metallic materials, has higher strength in the desired direction, and is more resistant to the same external loads. In fact, this is one of the most important purposes of composite manufacturing.

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1.3 Application Areas of Composites

  • Composite materials are used in a wide variety of areas thanks to their structure and properties. Since each sector has different needs and expectations, the product flexibility of composite materials appears as an important advantage.
  • Composites are used as raw materials in different sectors as well as as auxiliary equipment in manufacturing.

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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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  • Space technology,
  • Maritime industry,
  • In the field of medicine (Manufacture of medical devices),
  • Robotics technology,
  • Chemical industry,
  • Electrical-Electronic technology,
  • Musical instruments industry,

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1.3.1 Some Areas Where Composites Are Used:

  • Construction and building industry,
  • Automotive industry,
  • Defense Industry and Aviation Sector,
  • Food and Agriculture Sector
  • Manufacturing of Sports Equipment (high jump poles, tennis rackets, surfing, racing boats, skis, etc.).

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1.3.2 Composite Product Examples

  • Building Industry

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  • Exterior and interior coverings
  • Decorative Applications
  • Roofing Sheets
  • Carrier profiles
  • Rainwater Transport Systems
  • Concrete molds
  • prefabricated buildings
  • Bridges, Water tanks,
  • Food Section Coverings
  • Automobile bodies
  • Bus side panels
  • Truck trailer side slats
  • Truck dashboards
  • Container manufacturing
  • Highway signs
  • Highway edge posts
  • Pedals, Rear View Mirrors
  • Air Intake Manifold
  • Automotive industry
  • Defense industry
  • Aircraft and helicopter body parts,
  • Airplane nose and wing parts
  • Mortar hulls and chests
  • Bulletproof panel manufacturing
  • Helmets
  • Mine and assault boat parts and hulls
  • Shelters etc...
  • Food and Agriculture Sector
  • Silos
  • food storage tanks
  • brine tanks
  • Aquaculture equipment
  • Greenhouses, Grain warehouses
  • Irrigation channels etc...
  • Maritime industry
  • Sailboats
  • Motor boats,
  • Lifeboats,
  • Buoys,
  • Pontoons-piers,
  • Marine motorcycle,
  • Canoes,
  • Surfboards,
  • Marina equipment etc...
  • Chemical Industry
  • Pipes for various purposes
  • Chemical plant floor grates
  • Acid tanks and coatings
  • Purification equipment
  • Industrial platform and railings
  • Ventilation ducts etc...
  • Energy Sector
  • Insulators,
  • Antennas,
  • Circuit Breakers,
  • Fuse-panel Boxes,
  • Lighting Bodies,
  • Insulated Platforms,
  • Lighting Poles,
  • Circuit Breaker Boxes,
  • Cable Carriers,

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Various Composite Products

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Concrete columns are created by combining iron and concrete and are actually a composite structure.

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iron bar (fiber)

Concrete (matrix)

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Composites used in an aircraft fuselage and their proportions

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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:

  1. High Dielectric Resistance
  2. Sealing
  3. Color variety
  4. U.V. resistance to rays
  1. Repairability
  2. Machinability
  3. Ease of Assembly, Design and Molding
  4. Can be applied to concrete, metal and wooden surfaces

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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

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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

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Alüminium

2.8

84

71

30

25

Aluminium Alloy-2024

2.8

247

69

88

25

Brass

8.5

320

97

38

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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

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1.5 Some Disadvantages of Composites

  1. The properties of the manufactured composite may not always be ideal. The quality of the material depends on the quality of the production method, there is no standardized quality.

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Delaminaiton

  1. Since laminated composites are sensitive to interlayer shear stresses, delaminations (separation between layers) may occur.
  2. Since some composites are brittle, they are easily damaged, and their repair may create new problems.
  3. They need to be cleaned very well and dried hot before they can be repaired. Some drying techniques can take a long time and be difficult.

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1.6 Composites and Engineering Activities

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1-Summary of General Information About Composites

  • As can be seen, composite materials can be used in many different sectors, they are constantly being developed and their usage rate is increasing day by day.
  • The probability of encountering composites in any business is quite high.
  • R&D activities for composite materials are frequently carried out not only in composite producing companies, but also in other companies depending on the usage situation.
  • In these activities, solutions and development alternatives with composite materials are possible.
  • For this reason, it is an important privilege and reason for preference for engineers to be able to carry out activities such as design, analysis, theoretical calculations, experimental measurements, developments and manufacturing for composite materials that they can do for isotropic (metal, ceramic, etc.) materials. This is only possible with a good basic knowledge of composite material mechanics.

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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.

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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

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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

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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.

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1.7.2 Classification According to Shape and Placement of Reinforcement Elements :

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  1. Particulate Reinforced Composite
  2. Discontinous Fibers or Whiskers Reinforced Composites
  3. Continous Fibers Reinforced Composites

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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:

  • Composite layers (lamina) are obtained by placing more durable rod or woven fiber (rod or woven) materials into the matrix material.
  • Then, multiple layers are bonded on top of each other and a laminated composite plate is obtained.
  • In these layered structures, the orientation angle (θ) of the fibers may differ from layer to layer.
  • The orientation angle of the fibers is selected in directions where the stresses will be higher under operating conditions, and in this case, both lightness and strength are achieved at the same time.

1.7.2 Classification According to Shape and Placement of Reinforcement Elements -Continue:

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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

  • It is possible to have two or more fiber types in the same composite structure.
  • These types of composites are called hybrid composites.
  • This field is very suitable for the development of new types of composites.

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1.7.2 Classification According to Shape and Placement of Reinforcement Elements -Continue:

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7- Natural Composites

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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. Its resistance to water, acid, oil and chemicals is very good and does not lose its resistance over time.
  2. Epoxies, which are generally two-component, change from liquid to solid after a certain period of time.
  1. It has excellent mechanical durability. It can withstand temperatures up to 140oC when wet and 220 oC when dry.
  1. It creates surfaces that are resistant to friction and wear.
  1. Low shrinkage occurs during hardening.
  1. Their costs are high.
  2. They are harmful to the skin.

1.8.1.2 Some Disadvantages of Epoxy

1.8 Some Important Materials Used in Composites

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1.8.2 Glass Fiber:

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  1. Glass fiber is produced from materials such as silica, colemanite, aluminum oxide and soda.
  2. It is the most commonly used fiber type in the production of fiber reinforced composites.
  1. Glass fiber is produced by passing molten glass under pressure through a specially designed furnace with small holes at its base.
  1. There are types with different properties: A, C, E, S and R glass.

1.8.3 Carbon Fiber– Carbon Matrix (Carbon/carbon)

  1. Composites made of carbon matrix and carbon fiber can withstand up to 4000oC.
  2. These composites have very good thermal and mechanical properties at high temperatures.
  3. Carbon fiber is lighter and has better mechanical properties than glass fiber. However, production costs are high.

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  1. The type of fiber that is stronger and also more expensive than carbon fiber is boron fiber.
  2. Boron is the second lightest element that is solid at room temperature. It is manufactured by coating boron on a thin wire called the core (usually Tungsten/Wolfram). Therefore, boron fiber is a composite in itself.

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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. Low density,
  2. High strength and fatigue resistance,
  3. High impact resistance and wear resistance,
  4. High chemical resistance,
  5. E-Compressive strength close to glass fiber,
  6. Kevlar fiber composites are 35% lighter than glass fiber composites.

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:

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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)

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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. Hand Lay-Up
  2. Spray-Up
  3. Wet Filament Winding
  4. Resin Transfer Molding (RTM)
  5. Pultrusion
  6. Compression molding
    1. Sheet Moulding Composites (SMC)
    2. Bulk Moulding Composites (BMC)
  7. Vakum Bonding / Vakum Bagging
  8. Autoclave bonding

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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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1.10.2 General Features of Manufacturing Methods:

Compression Molding

Vakum Bonding / Vakum Bagging

  1. Manufacturing methods of the composite may vary depending on the reinforcement and matrix material, part shape, and the properties targeted from the composite.
  2. Raw materials, mold, heat and pressure are generally needed to produce a part.
  3. During manufacturing, care should be taken to ensure
  4. that the fiber has an evenly spaced and homogeneous distribution,
  5. that the fiber materials are thoroughly wetted by the matrix since they are sensitive to mechanical contact,
  6. and that a strong interface is created between the fiber and the matrix.
  1. In addition to the resin and reinforcement material used, the manufacturing method also plays an important role in determining the final properties of a composite structure.

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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

  1. SMC molding
  2. SRIM
  3. BMC molding
  4. Spray -Up
  5. Injection molding
  1. Flament Winding
  2. Pultrusion
  3. Resin Trans. Molding
  4. Hand Lay-Up
  5. Autoclave
  6. SCRIMP,RIFT,VARTM..
  1. Injection Molding
  2. Blow Molding
  1. Thermal Shaping
  2. Tape Wrapping
  3. Press Molding
  4. Autoclave

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1.11 Let's remember the important concepts in strength :

1.11.1 Isotropic Material:

  1. They are materials that show the same mechanical behavior in all directions and directions against thermal or mechanical loading.
  2. They show a homogeneous grain distribution in terms of internal structure distribution. Like all pure metals , alloys are isotropic materials.
  1. In addition, particles whiskers or discontinuous fiber reinforced composites are considered isotropic materials at the macro level if they have a homogeneous distribution . Nano-composites also fall into this class. All mechanical calculations valid for isotropic materials also apply to them. ( They are also called quisa-isotropic materials because they contain more than one material . )

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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.

 

 

  • In addition to external forces, internal forces and internal moments are also applied to each part that we separate imaginary from the separation part. Internal and external forces provide static balance in each part.
  • Internal forces and internal moments: These are the reactions that occur on the separation surface. It is the response of the system to external forces in that part.
  • Stress: It can be defined as the force per unit area resulting from internal force and moment reactions.
  • Stress is not applied, it occurs inside the object under the influence of external loads.

 

 

 

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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

  • The most important concept in deformation is the magnitude we call strain.
  • In its simplest definition, it is the amount of extension (or shortening) per unit length of an object.

 

  • The subscript e also indicates its direction.( could be ε x , ε y or ε z ).

L 0

Δ L/2

Δ L/2

1.11.4.1 Unit Elongation : Strain (ε)

1.11.4.2 Shearing Strain: Angle (γ):

  • At a point, in the most general case, there may be 3 different shear stresses (τxy , τyz , τxz )
  • These stresses cause shear strain angles (γxy , γyz , γxz )
  • It is the deformation angle caused by shear stress at a point. The angle γ is in the same plane as the shear stress and has the same indices. Try to understand this angle by carefully examining the figures below .

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1-Summary of General Information About Composites / important concepts in stregth

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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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

1-Summary of General Information About Composites / important concepts in stregth

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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

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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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1-Summary of General Information About Composites / important concepts in stregth

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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?

  • Materials whose properties vary with direction are called anisotropic materials.
  • Anisotropic materials show different mechanical behavior (stressing, deformation, etc.) depending on the direction, against thermal or mechanical loading.
  • Composite materials are anisotropic materials and are more specialized versions of them.
  • In order to understand the mechanics of composites, it is important to first know the more general mechanics of anisotropic materials.
  • In fact, isotropic materials are the most special form of anisotropic material.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

2- Anisotropic Material Types

2.2 Our Aims in This Chapter:

  1. To explain the types of anisotropic materials, starting from the most general to the specific,
  2. for each type, it is to derive the stress-strain relations (Hooke’s equations) in the region where the loading is elastic and
  3. the material constants that provide these equations.

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.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

2- Anisotropic Material Types

 

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Compliance Matrix: [S]

It is the inverse of the stiffness matrix.

[S]=[C]-1

 

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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:

 

 

  • (i , j , k , l = 1, 2, 3,…,6)

or

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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.)

  • Materials that, in addition to showing triclinic material properties, also have one plane of symmetry at the crystal or larger macro level are called monoclinic materials.
  • Physical material properties are symmetrical with respect to this plane.
  • In homogeneous structures, these symmetry planes at all points are parallel to each other.
  • An example of these is Feldspar, which is the raw material of ceramic materials.
  • Layered (laminated)composite materials can have planes of symmetry or planes of symmetry parallel to each other at much larger sizes than the crystal level (at the macro level that can be seen with the eye).

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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2.6.1. 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

  • In other words, in the orthotropic material type, there are 2 more symmetry planes in addition to the monoclinic material.
  • At each point of the orthotropic material, there is symmetry with respect to all three planes 1-2, 2-3 and 1-3.
  • Continuous fiber reinforced composite layers show orthotropic properties.
  • In addition to the triclinic material feature, materials that have three planes of symmetry at the crystal or larger macro level are called orthotropic materials.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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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.

  1. The Compliance [S] matrix

  • The Minor Poisson’s ratio

  • The Stiffness [C ] matrix

  • The strain values in directions 1 and 2.

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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2- Anisotropic Material Types/ Ortotropic Material

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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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2.7.1 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.

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2- Anisotropic Material Types/ Transversely Isotropic Material

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2.9 Isotropic Material

  • If there are an infinite number of isotropy planes parallel or angled to the axes in the structure, these types of materials are called isotropic materials.
  • In other words, the properties are the same in all directions parallel or at an angle to the axes in the material, so there are infinite directions.
  • In this case, the elastic material constants are reduced from 5 to 2 (E,ν):

2

3

1

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

2- Anisotropic Material Types/ Isotropic Material

  • All metallic materials such as steel, aluminum are isotropic.
  • No matter which direction the sample is removed from the material, the tensile test curve and E, n values do not change.

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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

  • These types of composites show isotropic properties at the macro level.
  • Since they are isotropic materials consisting of at least two different materials, they are also called quasi-isotropic materials.
  • If the size of the particles is at the nano level, these types of composites are called nano-composites.
  • There are 2 independent material constants: elasticity modulus (E ) and poisson's ratio (n).
  • The shear modulus (G) value depends on E and n and is calculated from equation 1.5.
  • The values ​​of E and n cannot be found theoretically; they can only be obtained experimentally.
  • Since they are isotropic, all calculations shown in the strength course are valid for these composites.
  • Therefore, for calculations regarding these composites, it is necessary to look at the sources where strength topics are explained.

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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:

  1. The temperature effect will be explained in more detail in the 9th topic. Its effect on strains has been added here.
  2. These formulas are also valid for orthotropic, transversely isotropic and isotropic materials.
  3. For isotropic materials; It is taken as (E1=E2=E, ν12=ν21=ν, G12 =G, α1=α2=α )

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

Orthotropic

Unidirectionally Reinforced Composites

Continuous Fiber

Reinforced Composites

 

 

 

 

 

 

 

 

 

 

Transvesly Isotropic

Composites

 

 

 

 

 

 

 

 

 

 

Bidirectionally Reinforced Composites

Continuous Fiber Reinforced

Woven Composites

 

 

 

 

 

 

 

 

 

Sandwich Composites

 

 

 

 

 

 

 

 

 

 

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

(tvid. 3a and 3b.)

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

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3.3 Representative Volume Element in an Orthotropic Layer

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  • Calculations will be made on the representative volume element extracted from the unidirectional fiber reinforced composite plate.
  • This element should be chosen in such a way that it represents the composite plate at a minimum level.
  • Since a representative element is examined at the micro level, this subject is also called «micromechanics of composite materials».
  • The mechanical properties of the composite will be determined by elementary strength calculations.

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

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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.)

 

 

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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:

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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

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Effect of Fiber Ratio on E2 Value

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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)

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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)

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Effect of Fiber Ratio on ν12

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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.

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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=?

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Solution:

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  • Calculation of E1 :
  • Calculation of E2 :

 

 

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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  • Calculation of G12:

 

 

  • Calculation of ν12 (major poisson ratio) :

 

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

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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

 

 

 

 

 

 

 

 

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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)

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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

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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

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The result after editing the last term is::

 

Poisson interaction term :

 

(3.24a)

  • When the Poisson effect term (𝑝) is neglected, equations 3.24 and 3.9 will be the same.
  • Although "p" can be at negligible levels in single-direction fiber-reinforced composites, "p" occurs at higher values ​​in bidirectionally woven composites and must be taken into account.

 

(3.25)

or

 

(3.24b)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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:

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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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, 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 :

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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)

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  • from static equilibrium in direction 2:
  • from static equilibrium in direction 1:

 

(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)

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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. :

 

 

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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)

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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;

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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):

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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

  • Particle reinforced, discontinuous fiber (whicker) reinforced, multidirectional continuous fiber reinforced composites are considered isotropic (called quasi-isotropic) and have 1 α value, and this value can only be found experimentally.
  • The α1 and α2 values of double woven fabric reinforced composites or sandwich composites are equal to each other and can be calculated with one of the equations 3.34 or 3.39. (The same result should come from both equations.)
  • Theoretical calculation methods of other mechanical properties (E1,E2,ν12,G12) of other types of composites are explained in section 3.10
  • Equations 3.34 and 3.39 are used for unidirectional, continuous fiber reinforced composites. Other types of composites;

Discontinuous fiber (whicker)reinforced composite

Particle reinforced composite

Double woven fabric reinforced composite

Sandwich Composite

multidirectional continuous fiber reinforced composites

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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.

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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

(tvids : 4a and 4b)

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  1. While the strength limit values (breaking stresses) of the fiber and matrix and the fiber volumetric fraction (ratio) are known, theoretically calculating the tensile and compressive strengths of the composite in the 1st and 2nd directions,
  2. Explaining the stresses and deformations that occur as a result of structural loading with examples,

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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:

  1. To formulate, using examples, the temperature increases that will bring the strength limits to their peak in thermal loading.
  2. and are to determine the lower limit values of the fiber volumetric fraction so that the fibers can carry loads and the composite can be efficient in terms of strength.

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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.

  • Some of these stages may not occur depending on the properties of the composite components

(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

  • Attention: Even if the fibers have higher strength, the stress on them will be higher and the fibers will break first. (For a correctly and efficiently designed lamina).

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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  • If the fibers and matrix have a brittle structure, there will be no plastic zone during the deformation stages described in article 4.3.
  • Tip 3: As soon as the fibers break, the composite is considered damaged
  • In a strength-efficient composite, fibers break before the matrix.

 

 

  • Therefore, at the moment the fibers break, the composite is at point b and the matrix is at point c.
  • When the fibers break, no break occurs in the matrix. In order for the matrix to break, it must reach point d.
  • The stress in the composite at any given moment is:

 

  • Breaking stress (or tensile strength) of fiber :

 

  • Stress in the matrix when fibers break :

 

 

 

(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

  • Breaking stress (or tensile strength) of matrix :

 

  • Theoretical tensile strength of the composite in direction 1:

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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 �

  • Compressive strength of fibers :
  • Stress in the matrix at the time of compressive damage to the fibers:

 

 

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.

 

 

 

  • Compressive Strength of Matrix :

 

(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.

4. Theoretical Strength And Efficiency Limits of Composite

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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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4. Theoretical Strength And Efficiency Limits of Composite

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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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4. Theoretical Strength And Efficiency Limits of Composite

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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):

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

4. Theoretical Strength And Efficiency Limits of Composite

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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:

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

4. Theoretical Strength And Efficiency Limits of Composite

 

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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):

 

 

 

4. Theoretical Strength And Efficiency Limits of Composite

 

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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

νPoisson ratio

αCoefficient 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.

4. Theoretical Strength And Efficiency Limits of Composite

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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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4. Theoretical Strength And Efficiency Limits of Composite

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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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4. Theoretical Strength And Efficiency Limits of Composite

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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.

 

4. Theoretical Strength And Efficiency Limits of Composite

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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. Theoretical Strength And Efficiency Limits of Composite

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  • When we think about strength, our goal in reinforcing the matrix with fibers is to obtain a more durable structure from the matrix.
  • For this, the fiber ratio must remain above certain limit values.
  • Otherwise, composite production would have no benefit or meaning.

4.8 Limits of Efficiency from Composite

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

4. Theoretical Strength And Efficiency Limits of Composite

  • Now we will see how these limit values are calculated theoretically:..>>

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0

 

Strength of composite at break:

(4.18)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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)

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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 :

  1. In the region 0 < Vf < Vfmin :

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.

  1. In the region Vfcr < Vf <1 :

fibers carry load and efficiency is obtained from the composite.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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;

  1. Whether the fibers can carry load,
  2. Determine whether efficiency can be obtained from the composite.
  3. Calculate the maximum tensile forces that this structure can carry in directions 1 and 2.

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 :

 

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

4. Theoretical Strength And Efficiency Limits of Composite

 

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5

EXPERIMENTAL DETERMINATION OF ORTOTROPİC PROPERTIES

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

(tvid: 5a and 5b)

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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.)

  • The tensile sample can be removed from the material in any direction. Because the material is isotropic, the properties are the same in all directions.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

  • In the same experiment, measurements are taken for more than one moment in the elastic region.
  • Measurements that are too different from the others are eliminated and not included in the calculation.
  • The average of compatible values obtained from measurements in the same experiment is taken.
  • Experiments are performed for different samples and the general average of the averages of all samples is taken.

5. Experimental Determination of Ortotropic Properties

(First, check out 1.11.5.)

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  • At least 2 strain gauges should be attached to the tensile test sample.
  • A strain gauge measures the unit elongation (ε) in its bonding direction.
  • In this way, (εx, εy ) values are measured with these two strain gauges.
  • The shearing strain ( angle γ) cannot be measured with strain gauges.

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 :

  • Since the elastic extension of the device jaws is included in the ∆𝐿 values read from the tensile test device's own indicator, these values mislead us. For this reason, extonmeters that take measurements directly on the sample are used.
  • Strain gauges are also used to experimentally find the stresses at a point on the surface of an object under any loading condition.

5.2.1 Straingauges:

These are sensors used to measure strains.

Strain gauges

5. Experimental Determination of Ortotropic Properties

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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:

5. Experimental Determination of Ortotropic Properties

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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5.2.3 If the 3rd strain gauge is used in the tensile test;

 

  • γxy is calculated from equation 5.1a.

 

  • Because the Mohr circle is tangent to the y axis from the left. R= τ45y / 2

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

 

Strain

Transformation Formula

  • Generally θ' = 45o is chosen

(5.1b)

For Cartesian coordinates (x-y):

For local coordinates (1-2):

(5.1a)

5. Experimental Determination of Ortotropic Properties

Mohr Circle

  • The G value found as a result of the tensile test should be equal to or close enough to the value to be obtained from this theoretical formula.

 

  • During the test, G values are found for different moments and averaged.

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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

5. Experimental Determination of Ortotropic Properties

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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

5. Experimental Determination of Ortotropic Properties

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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)

 

  1. By substituting the T and ε45o values measured for different instants into equation 5.4, the G12 value is determined for each instant and the shear modulus (G12) value of the material is obtained by taking the general average.

(5.4)

 

Average cross-sectional area:

  1. By reading the torsional moment (Tult) at the time of damage from the device, the shear strength of the material is calculated from equation 5.5:

(5.5)

How these formulas are derived is explained on the next page...>>

  1. In this method, first of all, a hollow, cylindrical (tube or pipe-shaped) sample with an average radius of "r" and a wall thickness of "t" must be prepared.
  2. The sample must be reinforced with continuous fibers along the axis of the cylinder.
  3. A strain gauge is placed on the outer surface of the cylinder at an angle of 45o with the horizontal.
  4. The sample is connected to the torsion tester.

5. Experimental Determination of Ortotropic Properties

 

Torsion Testing Machine

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T = 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

  1. The apparatus and the sample are connected together to the tensile device and a compressive load P is applied. P compressive loads must be equidistant from the notch axis and in opposite directions. In this case, equal P1 forces will occur between the contact parts of the apparatus to the sample close to the notch, and P2 forces will occur in the distant contact parts.

Σ Fy =0

5.7.3 Iosipescu Method (obtains G12 and S):

  1. In this method, a special sample with a V-notch is prepared in the direction of the fibers, as shown in the figure.
  1. The sample is connected to the load fixture apparatus consisting of 2 parts. The most important point is that the V notch is right in the middle. Because only in this case, the bending moment in the notch section becomes zero and the shear stress in the notch section is found with the formula 𝜏12=V/A. (V: shear force in section)
  1. A strain-gauge is glued to the sample at an angle of 45° with the fiber axis, as shown in the figure.

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)

  1. During the compression test performed on the tensile testing machine, P and 𝜀45o values are read for different instants.
  2. For these recorded instants, S values are calculated from equation (5.6) and G12 values are calculated from equation (5.7).
  3. These instantaneous values are averaged and the S and G12 values of the material are obtained.
  4. Results much closer to reality can be obtained by taking the general average of the tests performed for different samples.

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 (σ12=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

  • 1

3,2

6,1x10-4

-1,1x10-5

  • 2

6,6

12,7x10-4

-2,4x10-5

  • 3

8,2

16,3x10-4

-3,1x10-5

X 4

11,2

29,2x10-4

-4,3x10-5

  • 5

Pult = 18,9

(break)

35,8x10-4

-7,8x10-5

 

Measurements (given in the question)

Calculatios

 

 

  • Measurements 4 are not taken into account as they are too different from the others.

5. Experimental Determination of Ortotropic Properties

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

  • The tensile value at break is also equal to the breaking strength (YT) value.

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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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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]

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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)

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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;

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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From equation 6.11

 

,

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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(6.15)

(6.16a-f)

(a)

(b)

(c)

(d)

(e)

(f)

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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:

 

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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:

 

 

 

,

,

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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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:

 

 

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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6.6 Hooke's Relations Including the Temperature Effect

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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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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)

6. Hooke's and Transformatıon Equations In Composıtes (Stress-Strain Calculations In An Orthotropic Layer)

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FAILURE CRITERIA

FOR COMPOSITES

7

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

(tvid : 7a and 7b.)

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7.1 Failure Criteria and Importance

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  • Although increasing the size of the parts that make up a solid system allows us to obtain a safer structure in terms of strength, it will result in a more expensive system.
  • Reducing the dimensions will allow us to obtain a more economical (cheaper) system, but this time it will risk durability.
  • Therefore, obtaining an optimum design in terms of durability and economy and producing a problem-free system is only possible by calculating the strength limits, that is, the minimum dimensions of the parts. This is the fundamental subject of strength science and is a complete engineering study.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

  • Calculation of these minimum dimensions is only possible by knowing and applying the failure (damage) criteria very well.
  • However, when changing part dimensions, it should be ensured that there is no loss of functionality of the system.
  • This is also related to deformation calculations, which are the subject of strength, and appears as a third criterion.

7. Failure Criteria for Composites

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  • As the main failure criteria in isotropic materials, we can list Tresca and Von-Mises Yield Criteria, Rankine, Coulmb and Mohr Fracture criteria. These criteria are explained in «Strength of Materials» lessons.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

  • Nano- or macro-sized particle reinforced, chopped fiber (Whiskers) reinforced or multi-directional continuous fiber reinforced composites show isotropic behavior and are evaluated according to the above-mentioned criteria.
  • Failure criteria tell us whether yielding or fracture will occur at the examined point of the object, depending on the stress or strain values.
  • Failure or damage can be defined as deformity and loss of functionality that occurs as a result of overload in a structure. Failure occurs when yielding occurs for ductile (formable) materials, and when rupture or fracture occurs for brittle materials.
  • It is necessary to choose the appropriate criterion for damage detection according to the type of material and mechanical behavior of the object. It is extremely wrong to use a suitable damage criterion to determine the yielding of ductile materials in determining the fracture of a brittle material, or to use a criterion used for isotropic materials for a composite orthotropic material. Such mistakes also have negative effects on our engineering careers.
  • For this reason, it is very important that we have full knowledge of the damage criteria.

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....>>

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

7. Failure Criteria for Composites

Note: The term Theory can also be used instead of Criterion

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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

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

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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)

7. Failure Criteria for Composites

 

 

 

 

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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

 

 

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.

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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)

7. Failure Criteria for Composites

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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

 

 

 

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…>>

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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 :

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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…>>

7. Failure Criteria for Composites

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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;

1 2=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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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

 

 

 

 

 

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

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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

  • in composites with different tensile and compressive strengths,
  • especially in composites that exhibit brittle behavior.

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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

7. Failure Criteria for Composites

 

 

 

 

 

 

 

 

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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..)

7. Failure Criteria for Composites

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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.)

 

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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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

7. Failure Criteria for Composites

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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:

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

7. Failure Criteria for Composites

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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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

7. Failure Criteria for Composites

 

 

 

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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.

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

7. Failure Criteria for Composites

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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:

7. Failure Criteria for Composites

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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)

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

  • Since ΔT has no effect on shear stress and shear deformation, shear stress is zero and there is no need to examine inequality 7.1.c.

7. Failure Criteria for Composites

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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:

7. Failure Criteria for Composites

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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.

(tvids: 8a and 8b)

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)

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

  • The equations will be developed within the scope of the Classical Lamination Theory (CLT) and will also be valid for structures consisting of layers with different fiber orientation angles, different thicknesses, and different material properties.
  • In the examples to be solved, damage assessments will also be made according to the criteria explained in section 7.

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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

  • Now we consider that all internal forces and moments in element Q have been previously calculated from static equilibrium and each of them is known.

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,

8. Classical Lamination Theory (CLT)

  • We want to calculate the stresses and deformations at a point such as "b" belonging to the Q element.
  • We accept that there is a plane stress state in element "b".

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  1. Each layer is elastic, orthotropic and homogeneous within itself.
  2. It is accepted that the layers adhere to each other very well.
  3. A vertical and straight line (for examp: line CA) drawn before loading

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)

  1. The Cartesian coordinate system used is placed on the middle plane of the layered composite and the z axis is necessarily

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)

 

 

 

 

8. Classical Lamination Theory (CLT)

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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)

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8.5 Coding of a layered structure

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  • When solving problems related to layered composites, the layers must be numbered correctly and the interfaces must be coded. (The reason for this will become clearer later.)
  • The figure on the side shows a composite structure consisting of n layers, shown in the x-z plane.

The following points should be taken into consideration when coding:

  1. First, the layers are numbered. The top one is the 1st layer, the other layer numbers are given downwards.
  2. The layer interfaces are then coded. The outer surface of the top layer is h0 (or z0), and as you go down, each interface is given code numbers as h1, h2, ...
  3. h values are actually the z coordinates of that interface and can be negative. The symbol z can also be used instead of h.
  4. The midplane may not correspond to an interface. In this case, the middle plane is not given a separate code number.

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)

  • The origin of the cartesian coordinate system is placed at any point in the midplane passing through the middle of the thickness. The middle plane is coincident with the x-y plane. The z axis is selected in the direction of the thickness and downwards.

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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.

  • Moment of the stresses 𝜎𝑥 on the strip area dA about the midplane:

Internal force x

perpendicular distance

  • Moment of 𝜎𝑥 stresses in layer k relative to the mid-plane:

 

 

  • Internal bending moment in the section with normal x in the layered Q element :

 

 

Similarly, if the same operations are performed for the y direction;

 

  • Total bending moment in the section with normal y in the layered Q element:

 

  • Torsional internal moments in the layered Q element:

 

  • If we write equations 8.19, 8.20 and 8.21 in a matrix format:

(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.

8. Classical Lamination Theory (CLT)

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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)

8. Classical Lamination Theory (CLT)

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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)

8. Classical Lamination Theory (CLT)

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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,

8. Classical Lamination Theory (CLT)

 

 

 

 

 

 

 

 

 

 

 

 

 

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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)

  • The midplane (the middle of the total thickness) is the interface of the two layers. The x-y plane coincides with the middle plane. z is selected in the downward direction.
  • The layers are numbered from top to bottom(1 and 2).
  • As explained in article 8.5, h values are given sequentially to

each external surface and interface from top to bottom.

  • In fact, the h value is the z coordinate of that interface.

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

8. Classical Lamination Theory (CLT)

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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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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.

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;

8. Classical Lamination Theory (CLT)

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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)

 

 

8. Classical Lamination Theory (CLT)

 

 

 

 

 

 

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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:

 

  • For different solid systems and different boundary conditions, the internal forces at the most critical Q point must first be found from static calculations.

1mm

1mm

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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.

  • According to the figure above,
  • Since a single material is used in our construction, it is symmetrical with respect to the middle plane in terms of material.
  • Geometrically, it is symmetrical with respect to the middle plane.
  • Since the forces acting on the structure are distributed homogeneously, they are symmetrical with respect to the middle plane.
  • In terms of fiber arrangement, our material is again symmetrical with respect to the middle plane.
  • Therefore, this structure is a symmetrical structure. The 1st Special Case is satisfied and [B] = 0

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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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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.

8. Classical Lamination Theory (CLT)

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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

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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

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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

 

 

 

 

  • Since the structure is symmetrical, it is provided in the first special case and [B] = 0.
  • Although there are internal moments in the Q element, since only the stresses

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)

  • Since the internal moments will not be taken into account only for the middle plane,

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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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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:

8. Classical Lamination Theory (CLT)

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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.

8. Classical Lamination Theory (CLT)

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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

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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  • Sandwich composites are actually a special case of laminated structures, consisting of a total of three layers, and all calculations applicable to laminated composites can also be directly applied to sandwich composites.
  • Sandwich composites are a type of laminated composite formed by bonding two stronger face sheets to the top and bottom surfaces of a core material.
  • The components of a sandwich composite — the core and the reinforcing top and bottom layers — can each be made of isotropic materials, or they may themselves be different types of composite structures.�The core can be a solid block or have partially hollowed-out regions, such as honeycomb configurations.�For this reason, sandwich composites offer a wide range of possible configurations.

8.8 Sandwich Composites and Calculation Methods

  • In sandwich composites, stress-strain calculations can be performed using the [A], [B], and [D] matrices explained in Topic 8, or by employing alternative approaches developed for laminated composites.

 

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:

 

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

8. Classical Lamination Theory (CLT)

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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:

 

 

 

 

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

(8.34)

(8.36)

(8.37)

(8.35)

8. Classical Lamination Theory (CLT)

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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. Classical Lamination Theory (CLT)

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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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8. Classical Lamination Theory (CLT)

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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.

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

8. Classical Lamination Theory (CLT)

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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 :

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

8. Classical Lamination Theory (CLT)

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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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8. Classical Lamination Theory (CLT)

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  • Since the structure is symmetrical, it enters the 1st special case, therefore [B] =0
  • Does the structure fall into the special situations in article 8.7?

Other Steps:

  • The matrix [D] is non-zero. However, since the total internal moments per unit length are zero, the 2nd special case is achieved. For this reason, there is no need to calculate the [D] matrix as it will not enter into the calculations.

Then only matrix [A] will be calculated.

  • Since the stresses in the mid-plane are not asked, the structure does not enter the 3rd special case.

 

 

 

 

 

 

 

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8. Classical Lamination Theory (CLT)

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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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8. Classical Lamination Theory (CLT)

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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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8. Classical Lamination Theory (CLT)

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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

9. Homogenization of Laminated Composites

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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

9. Homogenization of Laminated Composites

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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:

9. Homogenization of Laminated Composites

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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

9. Homogenization of Laminated Composites

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

 

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:

  • The normal stresses in all layers are assumed to be equal in the loading direction:
  • The elongation of the equivalent volume in the x-direction is assumed to be the volume-weighted average of the elongations of the layers: :

 

(9.2.1)

(9.2.2)

 

 

(9.2.3)

 

 

 

  • From Hooke’s law: :

 

  • If a tensile load is applied in the y-direction, a similar procedure yields :..>>

(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)

  • The contraction of the equivalent volume in the y-direction is assumed to be the volume-weighted average of the contractions of the individual layers:

 

 

 

 

 

  • The lateral strain in the y-direction for the i-th layer is

 

 

  • For the equivalent volume, the Poisson behavior is given by

 

 

 

(9.2.7)

(9.2.8)

(9.2.10)

(9.2.11)

  • Similarly, for a tensile load applied in the y-direction, :…>>

 

(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 shear stresses in the entire structure and in all layers are assumed to be equal: :
  • The shear strain of the equivalent volume is assumed to be

the volume-weighted average of the shear strains of the layers:

 

  • For the i-th layer: :
  • For the equivalent volume:

(9.2.14)

(9.2.15)

(9.2.16)

(9.2.17)

(9.2.18)

Figure 9.8.a

Figure 9.8.b

9. Homogenization of Laminated Composites

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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 :

  • The fundamental equations of CLT were presented in detail in Chapter 8, where several examples involving stress–strain calculations for laminated structures were solved.
  • In this approach, each layer is evaluated individually, and the contributions of all layers are combined to determine the equivalent elastic properties of the laminated structure.
  • Perfect bonding between the layers is assumed. One of the most important limitations of the CLT approach is that it does not account for the lateral interactions arising from poisson effects between the layers at the structural level.
  • In the CLT approach, the reciprocity condition is embedded in the fundamental equations of the theory and is satisfied as a mathematical consequence of its formulation.

 

 

 

 

 

 

(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

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  1. The Zor Model (or Zor Approach), a new homogenization method for laminated structures, assumes perfect bonding between the layers, similar to the Voigt and CLT approaches.
  2. Under any loading condition, the layers cannot deform independently of one another in any direction.
  3. The model aims to represent the equivalent and overall macroscopic behavior of the entire structure.
  4. In the general case, a layer may exhibit monoclinic behavior with respect to the global x–y coordinate system.
  5. The reciprocity condition is not imposed as an initial assumption. Instead, it emerges naturally from the equivalent properties derived using the laws of static equilibrium and Hooke’s law.
  6. The resulting equations are valid for both symmetric and asymmetric stacking sequences.

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.

  • Case 3 is not included in the Voigt and Reuss approaches. In the CLT approach, it is valid only for symmetric laminates, whereas in the Zor approach (model), it is applied to both symmetric and asymmetric laminates.

 

 

 

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

 

 

 

 

 

 

 

  • Since no external force is applied to the structure in the y-direction : Fy= 0

 

 

 

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)

  • In this case, the third row of the matrix equation numbered (9.4.43) is:

 

(9.4.43)

  • The stress state in the equivalent volume (at the macro level) is:
  • From Eq. (9.4.3), the shear strain is:

 

 

  • As a necessary consequence of Eq. (9.4.44):

(9.4.45)

(9.4.3)

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Similarly,

 

 

(9.4.46)

  • In this case, the third row of the matrix equation given in Eq. (9.4.43) becomes:

 

  • The stress state at the macro level in the equivalent volume is:
  • Eq. (9.4.3) is also valid for this loading condition:

 

 

  • As a necessary consequence of Eq. (9.4.46):

(9.4.47)

(9.4.3)

  • Thus, the equivalent compliance matrix of the Zor Model takes the following orthotropic form:

 

(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::

 

  • Step 2) Determination of the Material Properties of the Layers with Respect to the Global x–y Axes:

 

 

The cross Poisson’s ratios are calculated from Eq. (9.1):

  • Since the layers are made of the same material, their properties with respect to the local 1–2 axes are the same as the properties of the large lamina from which they were manufactured.

Properties of the second layer: >>..>>

  • Step 1) Determination of the Volume Fractions:

 

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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:

  • Equivalent volume properties calculated for the Zor Model from Table 9.1.b:

 

 

 

 

  • Stresses developed in the equivalent volume

 

 

  • Strains developed in the equivalent volume

 

 

 

 

 

 

 

 

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  • Global strains in each layer

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):

 

  • Global properties of each layer:

 

 

 

 

 

 

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  • Global stresses in the 1st layer:

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)

  • Global stresses in each layer::

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  • Global stresses of 2nd layer:

 

 

 

 

 

 

  • Global stresses of 3th layer:

 

 

 

 

 

 

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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

  • The CLT values for the mid-plane of each layer were obtained following the procedures described in Chapter 8 and were entered directly into the tables.

 

  • The differences between the Zor and CLT results may vary depending on the stacking sequence, layer material properties, and the applied loads.

 

  • While all layers share the same global strains in the Zor Model, the strains in CLT may vary from one layer to another.
  • Although failure assessment in CLT is performed on a layer-by-layer basis, in the Zor Model the failure assessment is performed on the equivalent volume. This subject will be discussed in greater detail in Chapter 10. In Example 10.1, the CLT and Zor Models are compared separately for a symmetric structure..

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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.

  • Voigt, Reuss, and CLT approaches do not take interlayer Poisson interactions into account.
  • The Zor Model takes interlayer Poisson interactions into account and uses them in the calculation of the equivalent properties. This constitutes the most important and distinctive feature of the Zor Model.

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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.

  • The Voigt approach cannot satisfy the reciprocity condition at the equivalent volume level.
  • In the Reuss and CLT approaches, the reciprocity condition is inherently satisfied because it is embedded in the equations used from the very beginning.
  • In the Zor Model, the reciprocity condition is never used as an initial or intermediate equation. All final equations are derived from static equilibrium, Hooke’s law, and the equalities resulting from the perfect bonding condition. Nevertheless, the final equations of the Zor Model automatically satisfy the reciprocity condition in all cases.

9.5.3 In Terms of Computational Simplicity

  • The Voigt, Reuss, and Zor approaches reduce the system to an equivalent volume and provide simple and fast calculations.
  • CLT, on the other hand, requires extensive matrix operations for mechanical calculations.

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  • CLT, although it assumes perfect bonding between layers, may predict different strains and different shear strains in the upper and lower layers adjacent to an interface under uniaxial plane tension/compression loading in asymmetric stacking sequences. This result is not consistent with the interpretation of perfect bonding described above.
  • The Voigt and Reuss approaches, on the other hand, do not discuss the physical consequences of perfect bonding between layers and do not provide any assessment of the deformation behavior of common points at the interfaces.

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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x

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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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

 

 

 

 

 

 

 

  • The Zor Model and CLT results are very close to each other. This is mainly because the structure in the example is symmetric and therefore the Poisson interactions remain at a low level. In asymmetric stacking sequences, the differences between the results of the methods become somewhat more pronounced because the Poisson interactions are greater. However, it should not be forgotten that the Zor Model and CLT are different approaches. In Example 9.2, the CLT and Zor Model results were also compared for an asymmetric stacking sequence.

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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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Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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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:

10. Strength and Failure Analysis of Laminated Structures Using the Zor Model

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The strains in Equivalent Volume:

The total elongation/shortening occurring in the equivalent volume :

10. Strength and Failure Analysis of Laminated Structures Using the Zor Model

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor

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200mm

 

 

6mm

200mm

x

y

 

 

 

z

b

h

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

From Eq. (10.41)

 

 

Failure occurs.

 

b-)

 

10. Strength and Failure Analysis of Laminated Structures Using the Zor Model

Mechanics of Composite Materials- Lecture Notes / Mehmet Zor