Composites

o http://composite.about.com/

o http://www.netcomposites.com/

o http://www.gurit.com/

o http://www.hexcel.com/

o http://www.toraycfa.com/

o http://www.e-composites.com/

o http://www.compositesone.com/basics.htm

o http://www.wwcomposites.com/ (World Wide Search Engine for Composites)

o http://jpsglass.com/

o http://www.eirecomposites.com/

o http://www.advanced-composites.co.uk/

o http://www.efunda.com/formulae/solid_mechanics/composites/comp_intro.cfm

o

 Introduction

There is an unabated quest for new materials which will satisfy the specific requirements for

various applications like structural, medical, house-hold, industrial, construction, transportation,

electrical; electronics, etc. The metals are the most commonly used materials in these

applications. In the yore of time, there have been specific requirements on the properties of these

materials. It is impossible of any material to fulfill all these properties. Hence, newer materials are

developed. In the course, we are going to learn more about composite materials. First, we will

deal with primary understanding of these materials and then we will learn the mechanics of these

materials.

In the following lectures, we will introduce the composite materials, their evolution; constituents;

fabrication; application; properties; forms, advantages-disadvantages etc.  In the present lecture

we will introduce the composite materials with a formal definition, need for these materials, their

constituents and forms of constituents.

Definition of a Composite Material

A composite material is defined as a material which is composed of two or more materials at a

microscopic scale and has chemically distinct phases.

Thus, a composite material is heterogeneous at a microscopic scale but statistically

homogeneous at macroscopic scale. The materials which form the composite are also called as

constituents or constituent materials. The constituent materials of a composite have significantly

different properties. Further, it should be noted that the properties of the composite formed may

not be obtained from these constituents.  However, a combination of two or more materials with

significant properties will not suffice to be called as a composite material. In general, the following

conditions must be satisfied to be called as a composite material:

1. The combination of materials should result in significant property changes. One can see

significant changes when one of the constituent material is in platelet or fibrous from.

2. The content of the constituents is generally more than 10% (by volume).

3. In general, property of one constituent is much greater   than the corresponding

property of the other constituent.

The composite materials can be natural or artificially made materials. In the following section we

will see the examples of these materials.

Why do we need these materials?

There is unabated thirst for new materials with improved desired properties. All the desired

properties are difficult to find in a single material. For example, a material which needs high

fatigue life may not be cost effective. The list of the desired properties depending upon the

requirement of the application is given below.

1. Strength

2. Stiffness

3. Toughness

4. High corrosion resistance

5. High wear resistance

6. High chemical resistance

7. High environmental degradation resistance

8. Reduced weight

9. High fatigue life

10. Thermal insulation or conductivity

11. Electrical insulation or conductivity

12. Acoustic insulation

13. Radar transparency

14. Energy dissipation

15. Reduced cost

16. Attractiveness

The list of desired properties is in-exhaustive. It should be noted that the most important

characteristics of composite materials is that their properties are tailorable, that is, one can design

the required properties.

History of Composites

The existence of composite is not new. The word “composite” has become very popular in recent four-five

decades due to the use of modern composite materials in various applications. The composites have

existed from 10000 BC. For example, one can see the article by Ashby [1].  The evolution of materials

and their relative importance over the years have been depicted the Figure 1 of this article. The common

composite was straw bricks, used as construction material.

Then the next composite material can be seen from Egypt around 4000 BC where fibrous composite

materials were used for preparing the writing material. These were the laminated writing materials

fabricated from the papyrus plant. Further, Egyptians made containers from coarse fibres that were drawn

from heat softened glass.

One more important application of composites can be seen around 1200 BC from Mongols. Mongols

invented the so called “modern” composite bow. The history shows that the early existing of composite

bows dates back to 3000 BC as predicted by Angara Dating. The bow used various materials like wood,

horn, sinew (tendon), leather, bamboo and antler. The horn and antler were used to make the main body

of the bow as it is very flexible and resilient. Sinews were used to join and cover the horn and antler

together. Glue was prepared from the bladder of fish which is used to glue all the things in place. The

string of the bow was made from sinew, horse hair and silk. The composite bow so prepared used to take

almost a year for fabrication. The bows were so powerful that one can throw the arrows almost 1.5 km

away. Until the discovery of gun-powder the composite bow used to be a very lethal weapon as it was a

short and handy weapon.

As said, “Need is the mother of all inventions”, the modern composites, that is, polymer composites came

into existence during the Second World War. During the Second World War due to constraint impositions

on various nations for crossing boundaries as well as importing and exporting the materials, there was

scarcity of materials, especially in the military applications. During this period the fighter planes were the

most advanced fighting means. The light weight yet strong materials were in high demand. Further, for

application like housing of electronic radar equipments require non-metallic materials. Hence, the Glass

Fibre Reinforced Plastics (GFRP) were first used in these applications. Phenolic resins were used as the

matrix material. The first use of composite laminates can be seen in the Havilland Mosquito Bomber of

the British Royal Air Force.

The composites exit in day to day life applications as well. The most common existence is in the form of

concrete. The concrete is a composite made from gravel, sand and cement. Further, when it is used

along with steel to form structural components in construction, it forms one further form of composite. The

other material is wood which is a composite made from cellulose and lignin. The advanced forms of wood

composites can be ply-woods. These can be particle bonded composites or mixture of wooden

planks/blocks with some binding agent. Now a days, these are widely used in the furniture and

construction materials.

An excellent example of natural composite is muscles of human body. The muscles are present in a

layered system consisting of fibers at different orientations and in different concentrations. These result in

a very strong, efficient, versatile and adaptable structure. The muscles impart strength to bones and vice

a versa. These two together form a structure that is unique. The bone itself is a composite structure. The

bone contains mineral matrix material which binds the collagen fibres together. 

The other examples include: wings of a bird, fins of a fish, trees and grass. A leaf of a tree is also an

excellent example of composite structure. The veins in the leaf not only transport the food and water but

also impart the strength to the leaf so that the leaf remains stretched with maximum surface area. This

helps the plant to extract more energy from sun during photo-synthesis.

What are the constituents in a typical composite?

In a composite, typically, there are two constituents. One of the constituent acts as a reinforcement and

other acts as a matrix. Sometimes, the constituents are also referred as phases.

What are the types of reinforcements?

The reinforcements in a composite material come in various forms. These are depicted through Figure

1.1.

           

1. Fibre: Fibre is an individual filament of the material. A filament with length to diameter ratio above

1000 is called as a fibre. The fibrous form of the reinforcement is widely used. The fibres can be

in the following two forms:

a. Continuous fibres: If the fibres used in a composite are very long and unbroken

or cut then it forms a continuous fibre composite. A composite, thus formed using

continuous fibres is called as fibrous composite. The fibrous composite is the

widely used form of composite.

b. Short/chopped fibres: The fibres are chopped into small pieces when used in

fabricating a composite. A composite with short fibres as reinforcements is called

as short fibre composite.

In the fibre reinforced composites, the fibre is the major load carrying constituent.

2. Particulate: The reinforcement is in the form of particles which are of the order of a few microns

in the diameter. The particles are generally added to increase the modulus and decrease the

ductility of the matrix materials. In this case, the load is shared by both particles and matrix

materials. However, the load shared by the particles is much larger than the matrix material. For,

example in an automobile type carbon black (as a particulate reinforcement) is added in rubber

(as matrix material). The composite with reinforcement in particle form is called as particulate composite.

3. Flake: Flake is a small, flat, thin piece or layer (or a chip) that is broken from a larger piece. Since

these are two dimensional in geometry, they impart almost equal strength in all directions of their

planes. Thus, these are very effective reinforcement components. The flakes can be packed

more densely when they are laid parallel, even denser than unidirectional fibres and spheres. For

example, aluminum flakes are used in paints. They align themselves parallel to the surface of the

coating which imparts the good properties.

4. Whiskers: These are nearly perfect single crystal fibres. These are short, discontinuous and

polygonal in cross-section.

5.

6. The classification of composites based on the form of reinforcement is shown in Figure 1.2. The

detailed classification further is given in Figure 1.3. The classification of particulate composites is

depicted further in Figure 1.4. Some of the terms used in these classifications will be explained in

the following paragraphs/lectures.

Figure 1.2: Classification of composites based on reinforcement type8.

Figure 1.3: Classification of fibre composite materials10.

Figure 1.4: Classification of particulate composites

Why is the reinforcement made in thin fibre form?

There are various reasons because of which the reinforcement is made in thin fibre form. These reasons

are given below.

a) An important experimental study by Leonardo da Vinci on the tensile strength of iron wires of various

lengths (see references in [2, 3]) is well known to us. In this study it was revealed that the wires of same

diameter with shorter length showed higher tensile strength than those longer lengths. The reason for this

is the fact that the number of flaws in a shorter length of wire is small as compared longer length. Further,

it is well known that the strength of a bulk material is very less than the strength of the same material in

wire form.

The same fact has been explored in the composites with reinforcement in fibre form. As the fibres are

made of thin diameter, the inherent flaws in the material decrease. Hence, the strength of the fibre

increases as the fibre diameter decreases. This kind of experimental study has revealed the similar

results [2, 3]. This has been shown in Figure 1.5 qualitatively.

Figure 1.5: Qualitative variation of fibre tensile strength with fibre diameter

b) The quality of load transfer between fibre and matrix depends upon the surface area between fibre and

matrix. If the surface area between fibre and matrix is more, better is the load transfer. It can be shown

that for given volume of fibres in a composite, the surface area between fibre and matrix increases if the

fibre diameter decreases.

 

Let    be the average diameter of the fibres,   be the length of the fibres and  be the number of

fibres for a given volume of fibres in a composite. Then the surface area available for load transfer is

(1.1)The volume of these fibres in a composite is

(1.2)

Now, let us replace the fibres with a smaller average diameter of  such that the volume of the fibres is

unchanged. Then the number of fibres required to maintain the same fibre volume is

(1.3)

The new surface area between fibre and matrix is

(1.4)

Thus, for a given volume of fibres in a composite, the area between fibre and matrix is inversely

proportional to the average diameter of the fibres.

c) The fibres should be flexible so that they can be bent easily without breaking. This property of the

fibres is very important for woven composites. In woven composites the flexibility of fibres plays an

important role. Ultra thin composites are used in deployable structures.

The flexibility is simply the inverse of the bending stiffness. From mechanics of solids study the

bending stiffness is  EI, where  is Young’s modulus of the material and  is the second moment of

area of the cross section of the fibre. For a cylindrical fibre, the second moment of area is

(1.5)

Thus,

Flexibility (1.6)

Thus, from the above equation it is clear that if a fibre is thin, that is, small in diameter it is more flexible

 

 

 

Boron Fiber

This fibre was first introduced by Talley in 1959 [15]. In commercial production of boron fibres, the method

of Chemical Vapour Deposition (CVD) is used. The CVD is a process in which one material is deposited

onto a substrate to produce near theoretical density and small grain size for the deposited material. In

CVD the material is deposited on a thin filament. The material grows on this substrate and produces a

thicker filament. The size of the final filament is such that it could not be produced by drawing or other

conventional methods of producing fibres. It is the fine and dense structure of the deposited material

which determines the strength and modulus of the fibre.

In the fabrication of boron fibre by CVD, the boron trichloride is mixed with hydrogen and boron is

deposited according to the reaction

In the process, the passage takes place for couple of minutes. During this process, the atoms diffuse into

tungsten core to produce the complete boridization and the production of  and . In the

beginning the tungsten fibre of 12  diameter is used, which increases to 12  . This step induces

significant residual stresses in the fibre. The core is subjected to compression and the neighbouring boron

mantle is subjected to tension.

The CVD method for boron fibres is shown in Figure 1.7.

Figure 1.7: Schematic of reactors for silicon carbide fibres by Chemical Vapour Deposition

The key features of this fibre are listed below:

These are ceramic monofilament fiber.

Fiber itself is a composite.

Circular cross section.

Fiber diameter ranges between 33-400  and typical diameter is 140  .

Boron is brittle hence large diameter results in lower flexibility.

Thermal coefficient mismatch between boron and tungsten results in thermal residual stresses

during fabrication cool down to room temperature.

Boron fibres are usually coated with SiC or   so that it protects the surface during contact with

molten metal when it is used to reinforce light alloys. Further, it avoids the chemical reaction

between the molten metal and fibre.

Strong in both tension and compression.

Exhibits linear axial stress-strain relationship up to 650 .

Since this fibre requires a specialized procedure for fabrication, the cost of production is relatively

high.

The boron fibre structure and its composite is elucidated in Figure 1.8.

Figure 1.8: Boron fibre structure and its composite

         

 

CarbonFiber:

The first carbon fibre for commercial use was fabricated by Thomas Edison.

Sixth lightest element and carbon- carbon covalent bond is the strongest in nature.

Edison made carbon fiber from bamboo fibers.

Bamboo fiber is made up of cellulose.

Precursor fiber is carbonized rather than melting.

Filaments are made by controlled pyrolysis (chemical deposition by heat) of a precursor material

in fiber form by heat treatment at temperature 1000-3000  

The carbon content in carbon fibers is about 80-90 % and in Graphite fibers the carbon content is

in excess of 99%. Carbon fibre is produced at about 1300 while the graphite fibre is produced

in excess of 1900 .

The carbon fibers become graphitized by heat treatment at temperature above 1800 .

“Carbon fibers” term is used for both carbon fibers and graphite fibers.

Different fibers have different morphology, origin, size and shape

The size of individual filament ranges from 3 to 147  .

Maximum use of temperature of the fibers ranges from 250 to 2000 .

The use temperature of a composite is controlled by the use temperature of the matrix.

Precursor materials: There are two types of precursor materials (i) Polyacrylonitrile (PAN) and (ii)

rayon pitch, that is, the residue of petroleum refining.

Fiber properties vary with varying temperature.

Fiber diameter ranges from 4 to10  .

A tow consists of about 3000 to 30000 filaments.

Small diameter results in very flexible fiber and can actually be tied in to a knot without breaking

the fiber.

Modulus and strength is controlled by the process. The procedure involves the thermal

decomposition of the organic precursor under well controlled conditions of temperature and

stress.

Cross section of fiber is non-circular, in general, it is kidney bean shape.

Heterogeneous microstructure consisting of numerous lamellar ribbons.

Morphology is very dependent on the manufacturing process.

PAN based carbon fibers typically have an onion skin appearance with the basal planes in more

or less circular arcs, whereas the morphology of pitch-based fiber in such that the basal planes lie

along radial planes. Thus, carbon fibers are anisotropic.

 

Glass Fibre

Fibers of glass are produced by extruding molten glass, at a temperature around 1200  

through holes in a spinneret with diameter of 1 or 2 mm and then drawing the filaments to

produce fibers having diameters usually between 5 to15 .

The fibres have low modulus but significantly higher stiffness

Individual filaments are small in diameters, isotropic and very flexible as the diameter is small.

The glass fibres come in variety of forms based on silica   which is combined with other

elements to create speciality glass.

What are the different types of glass fibres? What are their key features?

The types of glass fibres and their key features are as follows:

E glass - high strength and high resistivity

S2 glass - high strength, modulus and stability under extreme temperature and corrosive

Introduction

In this lecture we are going see some more advanced fibres. Further, we will see their key features,

applications and fabrication processes.

Alumina Fibre

These are ceramics fabricated by spinning a slurry mix of alumina particles and additives to form

a yarn which is then subjected to controlled heating.

Fibers retain strength at high temperature.

It also shows good electrical insulation at high temperatures.

It has good wear resistance and high hardness.

The upper continuous use temperature is about 1700  .

Fibers of glass, carbon and alumina are supplied in the form of tows (also called ravings or

strands) consisting of many individual continuous fiber filaments.

Du Pont has developed a commercial grade alumina fibre and known as Alumina FP

(polycrystalline alumina) fibre. Alumina FP fibres are compatible with both metal and resin

matrices. These fibres have a very high melting point of 2100 . They can withstand

temperatures up to 1000 without any loss of strength and stiffness properties at this elevated

temperature. They exhibit high compressive strengths, when they are set in a matrix.

The Alumina whiskers are available and they exhibit excellent properties. Alumina whiskers can

have the tensile strength of 20700 MPa and the tensile modulus of 427 GPa.

What are the applications of Alumina fibres?

The Alumina has a unique combination of low thermal expansion, high thermal conductivity and

high compressive strength. The combination of these properties gives good thermal shock

resistance. These properties make Alumina is suited for the applications in furnace use as

crucibles, tubes and thermocouple sheaths.

The good wear resistance and high hardness properties are harnessed in making the

components such as ball valves, piston pumps and deep drawing tools.

Aramid Fibre

These fibres are from Aromatic polyamide, that is, nylons family.

Aramid is derived from “Ar” of Aromatric and “amid” of polyamide.

Examples of fibres from nylon family: Polyamide 6, that is, nylon 6 and Polyamide 6.6, that is,

nylon 6.6

These are organic fibers.

Melt-spun from a liquid solution

Du Pont developed these fibers under the trade name Kevlar. From poly (p-phenylene

terephthalamide (PPTA) polymer.

Morphology – radially arranged crystalline sheets resulting into anisotropic properties.

Filament diameter about 12  and partially flexible

High tensile strength.

Intermediate modulus

Significantly lower strength in compression.

5 grades of Kevlar with varying engineering properties are available. Kevlar-29, Kevlar-49,

Kevlar-100, Kevlar-119 and Kevlar-129.

Silicon Carbide Fibre (SiC)

Silicon carbide fibres are ceramic fibers. These fibres are produced in similar fashion as boron fibres are

produced. The fibres are produced by two methods as follows:

CVD on Tungsten or Carbon Core

NICALON™ by NIPPON Carbon Japan

CVD on Tungsten or Carbon Core:

This fiber is similar in size and microstructure to boron.

The fibres are produced on both tungsten and carbon cores.

These fibres are relativity stiff due to thicker diameter of the fibres. The diameter of the fibres is

about 140  .

The fibres have strength in the range of 3.4 – 4.0 GPa.

Failure strain is in the range of 0.8 - 1%.

The Young’s modulus is about 430 GPa.

The fibres show high structural stability and strength retention even at temperatures above

1000  .

The CVD with  as the reactant, SiC is deposited on the core as follows:

The SiC fibres produced on a tungsten core with a diameter about 12  . It shows a thin

interfacial layer between the SiC mantle and the tungsten core. In case, when carbon fibre is

used the fibre diameter of the carbon fibres is about 33  .

Both type of SiC fibre have smoother surfaces than a boron fibre. This is because there is a

deposition of small columnar grains as compared to conical nodules in boron fibres.

NICALON™ by NIPPON Carbon Japan

The fibres are manufactured by the process of controlled pyrolysis (chemical deposition by heat)

of a polymeric precursor.

The fiber is homogeneously composed of ultrafine beta-SiC crystallites and carbon.

The filament is similar to carbon fiber in size.

The diameter of the fibre is about 14  

The fibres more flexible due to small diameter.

The fibres arranged in tows of 250 to 500 filaments per tow.

These fibres come in two grades:

a. Ceramic Grade: provides good high temperature performance and mechanical properties

b. High Volume Resistivity Grade: It is a low dielectric fibre. It has good electrical and

mechanical properties. These are used in dielectric structures.

Uses of the NICALON™ Fibres     

These fibres are used to form fibrous products such as high temperature insulation, filters, etc. These

fibres have high resistance to chemical attack. Hence, these can be used in harsh environments.

These are also used as a reinforcement in plastic, ceramic and metal matrix composites.

 

Cross Sectional Shapes of Fibres

The cross sectional shapes of fibre of various types we have studied above are different. The

cross sectional shape of the fibres, although is assumed to be circular, is not circular in general.

The various cross sectional shapes of the fibre are shown in Figure 1.10.  

Figure 1.10: Cross sectional shapes of fibres

Fiber Properties

The following are the important points regarding the fibre properties.

Density, axial modulus, axial Poisson’s ratio, axial tensile strength and coefficient of

thermal expansion are some of the important properties.

Advanced fibers exhibit a broad range of properties.

Properties of carbon fiber can vary significantly depending upon fabrication process.

For the advanced fibres studied above one can attain either high modulus (> 700 GPa)

or high strength (> 5 GPa) but not both attainable simultaneously.

 SCS-6, IM8, boron and sapphire fibers offer the best combination of stiffness and

strength but have large diameters and thus limited flexibility. However, IM8 fibers are

exception for flexibility.

The specific stiffness of some of these fibres is almost 13 times of structural metals.

Similarly, the specific strength of some of these fibres is almost 16 times of structural

metals.

Weight saving, when the composites of these fibres are used, is tremendous due to high

specific stiffness and strength.

Actual properties of composite (fiber + matrix) are reduced.

Specific properties are reduced even further when the loading is in a direction other than

the length direction of fibers.

Tailorable properties.

One can get the desired heat transfer or electrical conductivity with proper designing.

The increased fatigue resistance is attainable with the use of these fibre composites.

Introduction

In the previous lecture we have introduced various advanced fibres along with their fabrication processes,

precursor materials and key features. In the present lecture we will introduce some matrix materials, their

key features and applications.

What are the matrix materials used in composites? 

The matrix materials used in composites can be broadly categorized as: Polymers, Metals, Ceramics and

Carbon and Graphite.

The polymeric matrix materials are further divided into:

1. Thermoplastic – which soften upon heating and can be reshaped with heat and pressure.

2. Thermoset – which become cross linked during fabrication and does not soften upon reheating.

The metal matrix materials are: Aluminum, Copper and Titanium.

The ceramic materials are: Carbon, Silicon carbide, Silicon nitride.

The classification of matrix materials is shown in Figure 1.11.

Figure 1.11: Matrix materials

 

 

Comparison between Thermoplastics and Thermosets:

The comparison between the thermoplastic and thermoset matrix materials is given in Table 1

below:

Table 1.1: Comparison between thermoplastics and thermosets.

ThermoplasticsThermosets

Soften upon heat and pressure Decompose upon heating

Hence, can be repaired Difficult to repair

High strains are required for failureLow strains are required for

failure

Can be re-processed Can not be re-processed

Indefinite shelf life Limited shelf life

Short curing cycles Long curing cycles

Non tacky and easy to handleTacky and therefore, difficult to

handle

Excellent resistance to solvents Fair resistance to solvents

Higher processing temperature is required.

Hence, viscosities make the processing difficult.

Lower processing temperature

is required.

What are the common metals used as matrix materials? What are their advantages and disadvantages?

The common metals used as matrix materials are aluminum, titanium and copper.

Advantages:

1. Higher transfer strength,

2. High toughness (in contrast with brittle behavior of polymers and ceramics)

3. The absence of moisture and

4. High thermal conductivity (copper and aluminum).

Dis-advantages:

1. Heavier

2. More susceptible to interface degradation at the fiber/matrix interface and

3. Corrosion is a major problem for the metals

            The attractive feature of the metal matrix composites is the higher temperature use. The

aluminum matrix composite can be used in the temperature range upward of 300ºC while the titanium

matrix composites can be used above 800 .

           

What are the ceramic matrix materials? What are their advantages and disadvantages?

The carbon, silicon carbide and silicon nitride are ceramics and used as matrix materials.

Ceramic:           

The advantages of the ceramic matrix materials are:

1. The ceramic composites have very high temperature range of above 2000  .

2. High elastic modulus

3. Low density

The disadvantages of the ceramic matrix materials are:

1. The ceramics are very brittle in nature. 

2. Hence, they are susceptible to flows.

Carbon

The advantages of the carbon matrix materials are:

1. High temperature at 2200

2. Carbon/carbon bond is stronger at elevated temperature than room temperature.

The disadvantages of the carbon matrix materials are:

1. The fabrication is expensive

2. The multistage processing results in complexity and higher additional cost.

It should be noted that a composite with carbon fibres as reinforcement as well as matrix material is

known as carbon-carbon composite. The application of carbon-carbon composite is seen in leading

edge of the space shuttle where the high temperature resistance is required. The carbon-carbon

composites can resist the temperature upto 3000  .

The advantages of these composites are:

1. Very strong and light as compared to graphite fibre alone.

2. Low density

3. Excellent tensile and compressive strength

4. Low thermal conductivity

5. High fatigue resistance

6. High coefficient of friction

The disadvantages include:

1. Susceptible to oxidation at elevated temperatures

2. High material and production cost

3. Low shear strength

Figure 1.12 depicts the range of use temperature for matrix material in composites. It should be noted that

for the structural applications the maximum use temperature is a critical parameter. This maximum

temperature depends upon the maximum use temperature of the matrix materials

 

What are the different forms of composites?

1. Unidirectional lamina:

o It is basic form of continuous fiber composites.

o A lamina is also called by ply or layer.

o Fibers are in same direction.

o Orthotropic in nature with different properties in principal material directions.

o For sufficient number of filaments (or layers) in the thickness direction, the effective

properties in the transverse plane (perpendicular to the fibers) may be isotropic. Such

a composite is called as transversely isotropic.

 

 

2. Woven fabrics:

o Examples of woven fabric are clothes, baskets, hats, etc.

o Flexible fibers such as glass, carbon, aramid can be woven in to cloth fabric, can be

impregnated with a matrix material.

o Different patterns of weaving are shown in Figure 1.13.

Typical weaving patterns are shown in Figure 1.13.

Figure 1.13: Types of weave

3. Laminate:

1. Stacking of unidirectional or woven fabric layers at different fiber orientations.

2. Effective properties vary with:

1. orientation

2. thickness

3. stacking sequence

A symmetric laminate is shown in Figure 1.14.

Figure 1.14: A symmetric laminate

4. Hybrid composites:

The hybrid composite are composites in which two or more types of fibres are used.

Collectively, these are called as hybrids. The use of two or more fibres allows the combination

of desired properties from the fibres. For example, combination of aramid and carbon fibres What are the advantages of the composite materials?

The following are the advantages of composites:

1. Specific stiffness and specific strength:

The composite materials have high specific stiffness and strengths. Thus, these material offer

better properties at lesser weight as compared to conventional materials. Due to this, one gets

improved performance at reduced energy consumption.

2. Tailorable design:

A large set of design parameters are available to choose from. Thus, making the design

procedure more versatile. The available design parameters are:

1. Choice of materials (fiber/matrix), volume fraction of fiber and matrix, fabrication method,

layer orientation, no. of layer/laminae in a given direction, thickness of individual layers,

type of layers (fabric/unidirectional) stacking sequence.

2. A component can be designed to have desired properties in specific directions.

3. Fatigue Life:

   The composites can with stand more number of fatigue cycles than that of aluminum. The

critical structural components in aircraft require high fatigue life. The use of composites in

fabrication of such structural components is thus justified.

4. Dimensional Stability:

Strain due to temperature can change shape, size, increase friction, wear and thermal stresses.

The dimensional stability is very important in application like space antenna. For composites, with

proper design it is possible to achieve almost zero coefficient of thermal expansion.

5. Corrosion Resistance:

Polymer and ceramic matrix material used to make composites have high resistance to corrosion

from moisture, chemicals.

6. Cost Effective Fabrication:

The components fabricated from composite are cost effective with automated methods like

filament winding, pultrusion and tape laying. There is a lesser wastage of the raw materials as the

product is fabricated to the final product size unlike in metals.

7. Conductivity:

The conductivity of the composites can be achieved to make it a insulator or a highly conducting

material. For example, Glass/polyesters are non conducting materials. These materials can be

used in space ladders, booms etc. where one needs higher dimensional stability, whereas copper

matrix material gives a high thermal conductivity.

The list of advantages of composite is quite long. One can find more on advantages of composite in

reference books and open literature.

What are the disadvantages of Composites?

1. Some fabrics are very hard on tooling.

2. Hidden defects are difficult to locate.

3. Inspection may require special tools and processes.

4. Filament-wound parts may not be repairable. Repairing may introduce new problems.

5. High cost of raw materials.

6. High initial cost of tooling, production set-up, etc.

7. Labour intensive.

8. Health and safety concerns.

9. Training of the labour is essential.

10. Environmental issues like disposal and waste management.

11. Reuse of the materials is difficult.

12. Storage of frozen pre-pregs demands for additional equipments and adds to the cost of

production.

13. Extreme cleanliness required.

14. The composites, in general, are brittle in nature and hence easily damageable.

15. The matrix material is weak and hence the composite has low toughness.

16. The transverse properties of lamina or laminate are, in general, weak.

17. The analysis of the composites is difficult due to heterogeneity and orthotropy.

 IntroductionThis lecture is dedicated to the use of composites. The use of composites is almost ubiquitous! The

use of composites is an unexhaustive list. In the following we cite some important applications.

What are the applications of the composite materials?

The applications of the composites are given in the following as per the area of application.

Aerospace:

Aircraft, spacecraft, satellites, space telescopes, space shuttle, space station, missiles,

boosters rockets, helicopters (due to high specific strength and stiffness) fatigue life,

dimensional stability.

All composite voyager aircraft flew nonstop around the world with refueling.

Carbon/carbon composite is used on the leading edges nose cone of the shuttle.

B2 bomber - both fiber glass and graphite fibers are used with epoxy matrix and polyimide

matrix.

The indigenous Light Combat Aircraft (LCA - Tejas) has Kevlar composite in nose cone, Glass

composites in tail fin and carbon composites form almost all part of the fuselage and wings,

except the control surfaces of the wing.

Further, the indigenous Light Combat Helicopter (LCH – Dhruvh) has carbon composites for its

main rotor blades. The other composites are used in tail rotor, vertical fin, stabilizer, cowling,

radome, doors, cockpit, side shells, etc.

Missile:

Rocket motor cases

Nozzles

Igniter

Inter stage structure

Equipment section

Aerodynamic fairings

 

 

Launch Vehicle:

Rocket motor case

Interstage structure

Payload fairings and dispensers

High temperature Nozzle

Nose cone

Control surfaces

Composite Railway Carrier:

Composite railway auto carrier

Bodies of Railway Bogeys

Seats

Driver’s Cabin

Stabilization of Ballasted Rail Tracks

Doors

Sleepers for Railway Girder Bridges

Gear Case

Pantographs

Sports Equipments

Tennis rockets, golf clubs, base-ball bats, helmets, skis, hockey sticks, fishing rods, boat hulls,

wind surfing boards, water skis, sails, canoes and racing shells, paddles, yachting rope, speed

boat, scuba diving tanks, race cars reduced weight, maintenance, corrosion resistance.

Automotive

Lower weight and greater durability, corrosion resistance, fatigue life, wear and impact

resistance.

Drive shafts, fan blades and shrouds, springs, bumpers, interior panels, tires, brake shoes,

clutch plates, gaskets, hoses, belts and engine parts.

Carbon and glass fiber composites pultruted over on aluminum cylinder to create drive shaft.

Fuel saving –braking energy can be stored in to a carbon fiber super flywheels.

Other applications include: mirror housings, radiator end caps, air filter housing, accelerating

pedals, rear view mirrors, head-lamp housings, and intake manifolds, fuel tanks.

B. Spray Lay-Up:

Fibre is chopped in a hand-held gun and fed into a spray of catalyzed resin directed at the mould.

The deposited materials are left to cure under standard atmospheric conditions. The fabrication

method is depicted in Figure 1.16.

The polyester resins can be used with glass rovings is best suited for this process.

Figure 1.16: Wet or hand lay-up fabrication

Advantages:

The spray-up process offers the following advantages:

o It is suitable for small to medium-volume parts.

o It is a very economical process for making small to large parts.

o It utilizes low-cost tooling as well as low-cost material systems.

Limitations:

The following are some of the limitations of the spray-up process:

o It is not suitable for making parts that have high structural requirements.

o It is difficult to control the fiber volume fraction as well as the thickness. These

parameters highly depend on operator skill.

o Because of its open mold nature, styrene emission is a concern.

o The process offers a good surface finish on one side and a rough surface finish on the

other side.

o The process is not suitable for parts where dimensional accuracy and process

repeatability are prime concerns. The spray-up process does not provide a good surface

finish or dimensional control on both or all the sides of the product.

o Cores, when needed, have to be inserted manually.

o Only short fibres can be used in this process.

o Since, pressurized resin is used the laminates tend to be very resin-rich.

o Similar to wet/hand lay-up process, the resins need to be of low viscosity so that it can be

sprayed.

Applications:Simple enclosures, lightly loaded structural panels, e.g. caravan bodies, truck fairings, bathtubs,

shower trays, some small dinghies.

 

 

C. Autoclave Curing:

The key features of this process are as follows:

o An autoclave is a closed vessel for controlling temperature and pressure is used for

curing polymeric matrix composites.

o Composites to be cured is prepared either through hand lay up or machine placement

of individual laminae in the form of fibers tape which has been impregnated with resin.

o Components is then placed in an autoclave and subjected to a controlled cycle of

temperature and pressure.

o After curing, the composite is “solidified”.

o One can use the fibres like carbon, glass, aramid etc. along with any resin.

Advantages:

o Large components can be fabricated.

o Since, the curing of matrix material is carried out under controlled environment the

resin distribution is better as compared to hand or spay lay-up processes.

o Less possibility of dilution with foreign particles.

o Better surface finish.

Disadvantages:

o Initial cost of tooling is high.

D. Filament Winding:

This process is an automated process. This process is used in the fabrication of components or

structures made with flexible fibers. This process is primarily used for hollow, generally circular or

oval sectioned components. Fibre tows are passed through a resin bath before being wound onto

a mandrel in a variety of orientations, controlled by the fibre feeding mechanism, and rate of

rotation of the mandrel. The wound component is then cured in an oven or autoclave.

           

One can use resins like epoxy, polyester, vinylester and phenolic along with any fibre. The fibre

can be directly from creel, non-woven or stitched into a fabric form.

           

The filament winding process is shown in Figure 1.17.

Advantages:

o Resin content is controlled by nips or dies.

o The process can be very fast.

o The process is economic.

o Complex fibre patterns can be attained for better load bearing of the structure.

 Disadvantages:

o Resins with low viscosity are needed.

o The process is limited to convex shaped components.

o Fibre cannot easily be laid exactly along the length of a component.

o Mandrel costs for large components can be high.

o The external surface of the component is not smoothly finished.

Figure 1.17: Filament winding

Applications:

Pressure bottles, rocket motor casing, chemical storage tanks, pipelines, gas cylinders, fire-

fighters, breathing tanks etc.

E. Pultrusion:

It is a continuous process in which composites in the form of fibers and fabrics are pulled through

a bath of liquid resin. Then the fibres wetted with resin are pulled through a heated die. The die

plays important roles like completing the impregnation and controlling the resin. Further, the

material is cured to its final shape. The die shape used in this process is nothing the replica of the

final product. Finally, the finished product is cut to length.

In this process, the fabrics may also be introduced into the die. The fabrics provide a fibre

direction other  0°. Further, a variant of this method to produce a profile with some variation in the

cross-section is available. This is known as pulforming.

The resins like epoxy, polyester, vinylester and phenolic can be used with any fibre.

The pultrusion process is shown in Figure 1.18.

Advantages:

o The process is suitable for mass production.

o The process is fast and economic.

o Resin content can be accurately controlled.

o Fibre cost is minimized as it can be taken directly from a creel.

o The surface finish of the product is good.

o Structural properties of product can be very good as the profiles have very straight fibres.

 Disadvantages:

o Limited to constant or near constant cross-section components.

o Heated die costs can be high.

o Products with small cross-sections alone can be fabricated.

Applications:

Beams and girders used in roof structures, bridges, ladders, frameworks

Figure 1.18: Pultrusion

 IntroductionIn this lecture we will see some more composites fabrication processes. Further, as we have done in

the previous lecture, we will see the advantages, disadvantages and application of these processes.

A. Braiding:

This is an automatic fabrication process. The toes are interlaced together to the final form of

the product. Further, this interlacing can be over the mandrel which has the final shape of the

product. The toes can be impregnated with the resin. Then the product is cured at room

temperature or in autoclave.

Advantages:

o Cost effective automated technique for interlacing fibers into complex shapes.

o Final product is obtained. B. Vacuum Bagging:

This is basically an extension of the wet lay-up process described above where pressure is

applied to the laminate once laid-up in order to improve its consolidation. This is achieved by

sealing a plastic film over the wet laid-up laminate and onto the tool. The air under the bag is

extracted by a vacuum pump and thus up to one atmosphere of pressure can be applied to the

laminate to consolidate it.

Materials Options:

o Resins: Primarily epoxy and phenolic. Polyesters and vinylesters may have problems due

to excessive extraction of styrene from the resin by the vacuum pump.

o Fibres: The consolidation pressures mean that a variety of heavy fabrics can be wet-out.

o Cores: Any.

Advantages:

o Higher fibre content laminates can usually be achieved than with standard wet lay-up

techniques.

o Lower void contents are achieved than with wet lay-up.

o Better fibre wet-out due to pressure and resin flow throughout structural fibres, with

excess into bagging materials.

o Health and safety: The vacuum bag reduces the amount of volatiles emitted during cure.

Disadvantages:

o The extra process adds cost both in labour and in disposable bagging materials.

o A higher level of skill is required by the operators.

o Mixing and control of resin content still largely determined by operator skill.

Applications:

Large one-off cruising boats, race car components, core-bonding in production boats.

Figure 1.19: Vacuum Bagging

C. Resin Transfer Molding - RTM

The process consists of arranging the fibres or cloth fabrics in the desired configuration in a

preform. These fabrics are sometimes pre-pressed to the mould shape, and held together by a

binder. A second matching mould tool is then clamped over the first. Then pressurized resin is

injected into the cavity. Vacuum can also be applied to the mould cavity to assist resin in being

drawn into the fabrics. This is known as Vacuum Assisted Resin Transfer Moulding (VARTM) or

Vacuum Assisted Resin Injection (VARI). The laminate is then cured. Both injection and cure can

take place at either ambient or elevated temperature.

In this process, the resins like epoxy, polyester, vinylester and phenolic can be used. Further, one

use the high temperature resins such as bismaleimides can be used at elevated process

temperatures. The fibres of any type can be used. The stitched materials work well in this process

since the gaps allow rapid resin transport. Some specially developed fabrics can assist with resin

flow.

Advantages:

o The process is very efficient.

o Suitable for complex shapes.

o High fibre volume laminates can be obtained with very low void contents.

o Good health and safety, and environmental control due to enclosure of resin.

o Possible labour reductions.

o Both sides of the component have a moulded surface. Hence, the final product gets a

superior surface finish

o Better reproducibility.

o Relatively low clamping pressure and ability to induce inserts.

Disadvantages:

Matched tooling is expensive and heavy in order to withstand pressures.

Generally limited to smaller components.

Unimpregnated areas can occur resulting in very expensive scrap parts.

Applications:

The applications include the hollow cylindrical parts like motor casing, engine covers, etc.

Figure 1.21: Centrifugal casting

D. Centrifugal Casting:

In this process the chopped fibres and the resin is sent under pressure to the cylindrical

moulding. The moulding is rotating. Due to centrifugal action, the mixture of resin and chopped

fibres get deposited on wall of the moulding. Thus, the mixture gets the final form of the product.

Advantages:

1. Suitable for small hollow cylindrical products.

2. Economic for small production.

Disadvantages:

3. Complex shape can not be made.

4. Resin with low viscosity is needed.

5. The finish of the inner side of the product is not good.

6. The structural properties may not be good as the chopped fibres are used.

Figure 1.21: Centrifugal casting

Applications:The applications include the hollow cylindrical parts like motor casing, engine covers, etc.

 In this lecture, we are going to introduce some

concepts from solid mechanics which will be useful

for better understanding of this course. It is presumed

that the readers have some basic knowledge of linear

algebra and solid mechanics.

In solid mechanics, each phase of a material is

considered to be continuum, that is, there is no

discontinuity in the material. Thus, in this course

individual fibres and the matrix of a lamina/composite

are considered to be continuum. Further, this results

in saying that heterogeneous composite is also a

continuum.

In this lecture, we will introduce some of the notations

that will be followed for the rest of the course. Hence,

the readers are advised to understand them clearly

before they proceed to further lectures.

Concept of Tensors

Tensors are physical entities whose components are

the coefficients of a linear relationship between

vectors.

The list of some of the tensors used in this course is

given in Table 2.1.

Table 2.1 List of some of the tensor quantities

Quantity Live subscripts

Scalar (zeroth order tensor) 0

Vector (first order tensor) 1

ij Second order tensor 2

Fourth order tensor 4

It is often needed to transform a tensorial quantity

from one coordinate system to another coordinate

system. This transformation of a tensor is done using

direction cosines of the angle measured from initial

coordinate system to final coordinate system. Let us

use axes as the initial coordinate axes and  as

the final coordinate axes (denoted here by symbol

prime –  ). Now, we need to find the direction

cosines (denoted here by aij) for this transformation

relation. Let us use the convention for direction

From/To

 

  

Let us derive the direction cosines for a

transformation in a plane. Let the coordinate axes x1-

x2 (that is, plane 1-2) are rotated about the third axis

x3 by an angle as shown in Figure 2.2. Thus, from

the figure it is easy to see that . A

careful observation of the figure shows that the angle

between  is not the same as the angle

between . It means that the direction

cosines .

Now, we will find all the direction cosines. The list is given below.

(2.1)

The matrix of direction cosines given above in Eq.

(2.1) is also written using short forms for

. Then Equation (2.1)

becomes

(2.2)

Note: Some of the books and research articles also

use .

Note: This matrix is also called Rotation Matrix.

Note: The above direction cosine matrix can be

obtained from the relation between unrotated and

rotated coordinates. For the transformation shown in

Figure 2.2 (a) one can write this relation  using the

geometrical relations shown in Figure 2.2 (b) as

Now the direction cosines are given by the following

relation:

Now we will use the direction cosines to transform a

vector, a second order tensor and a fourth order

tensor from initial coordinate (unprimed) system to a  Let us assume that, the unprimed and primed coordinate systems are as shown in Figure 2.2. The

transformation matrix for this rotation is given in Equation (2.1). Then, the components  can be given

Note: In two dimensional case, the above

transformation is written as

Equation (2.3) can also be written in an inverted form

to give the components Pi in unrotated axes in terms

of components  in rotated axes system as

(2.5)

The rotation matrix aij in Equation (2.2) has a property

that

(2.6)

Now, we will extend the concept to transform a

second order tensor. Let us transform the stress

tensor  as follows

The transformation of a fourth order tensor  is

given as

(2.8)

The readers are suggested to write the final form of

Equation (2.8) using similar procedure used to get the

last of Equation (2.7).

 Deformation of a Body

When a deformable body is subjected to external forces, a body may translate, rotate and deform as well. Thus, after deformation the body occupies a new region. The initial region occupied by the body is called Reference Configuration and the new region occupied by the body after translation, rotation and deformation is called Deformed Configuration. Let us consider a point P in reference

configuration. Its position with respect to origin of a reference axes system  is shown in Figure 2.3.

The point P occupies a new position  and its position vector  is also given.

Figure 2.3 Reference and deformed configurations

The deformation map is defined as

(2.9)

Thus, deformation map is a vector valued function. Similarly, for deformation of a point Q to , we can

write

(2.10)

We can find the deformation  as

(2.11)

(2.7)

where  is called Deformation Gradient. In component form, one can write

(2.12)

Now, let us give the deformation map for the displacement of a point. Let us consider the point P in

reference configuration again. It goes a deformation  and occupies a new position . Thus, we

can write this deformation as follows

Now, we will define strain tensor. We are going to find . We know that

. Thus,

(2.16)

where E is Lagrangian Strain Tensor. Now using the last two of Equation (2.16) for  

we get,

(2.17)

This equation can be written in index form as

(2.18)

where  is given as . Thus, the strain components are nonlinear in . Here,

 are the displacement components in three directions. For example, let us

write the expanded form of strain components .

(2.19)

Similarly,

(2.20)

The readers should observe that from the definition of strain tensor in Equation (2.18), the strain tensor is

symmetric (that is, ). If the gradients of the displacements are very small the product terms in

Equation (2.18) can be neglected. Then, the resulting strain tensor (called Infinitesimal Strain Tensor) is

given as

(2.21)

The individual strain components are given as

(2.22)

The readers are very well versed with these definitions. This strain tensor can be written in matrix form as

(2.23)

Note: The shear strain components mentioned above are tensorial components. In actual practice,

engineering shear strains (which are measured from laboratory tests) are used. These are denoted by

. The relation between tensorial and engineering shear strain components is

(2.24)

 

The engineering shear strain components are given as follows:

(2.25)

Using the engineering shear strain components, the strain

tensor can be written in matrix form as

(2.26)

Stress

Now, we will introduce the concept of stress. The components

of stress at a point (also called State of Stress) are (in the limit)

the forces per unit area which are acting on three mutually

perpendicular planes passing through this point. This is

represented in Figure 2.4. Stress tensor is a second order

tensor and denoted as . In this notation, the first subscript

corresponds to the direction of the normal to the plane and the

second subscript corresponds to the direction of the stress. For

example,  denotes the stress component acting on a plane

which is perpendicular to direction 2 and stress is acting in

direction 3. The tensile normal stress components

 are positive. The shear stress components

 are defined to be positive when the normal to the plane

and the direction of the stress component are either both

positive or both negative.

The readers should note that the state of stress shown in

Figure 2.4 represents all stress components in positive sense.

In this figure, the stress components are shown on positive

faces only.

Figure 2.4 State of stress at a pointThe stress tensor can be written in matrix form as follows:

(2.27)

In general, instead of using global 1-2-3 coordinate system, x-y-

z global coordinate system is used. Further, the shear stress

components are shown using notation . Thus, the stress

tensor in this case can be written as

(2.28)

Note: The stress tensor will be symmetric, that is  only

when the there are no distributed moments in the body. The

readers are suggested to read more on this from any standard

solid mechanics book. In this entire course, we will deal with

symmetric stress-tensor.

 

Equilibrium Equations

The equilibrium equations for a body to be in static equilibrium at a point are given in index notations as

(2.29)

where,   are the body forces per unit volume. If the body forces are absent, then the equilibrium

equation becomes

(2.30)

The equilibrium equations without body forces are written using xyz coordinates as follows:

(2.31)

Boundary Conditions            

The boundary conditions are very essential to solve any problem in solid mechanics. The boundary

conditions are specified on the surface of the body in terms of components of displacement or stress.

However, the combination of displacement and stress components is also specified.

Figure 2.5 shows a body, where the displacement as well as stress components are used to specify the

boundary conditions.

We define traction vector  for any arbitrary point (for example, point P in Figure 2.5) on surface as a vector consisting of three stress components acting on the surface at same point. Here, the three stress

components are normal stress  and shear stress  and . The traction vector at this point is written as

(2.32)

where  is the unit normal to the surface at point P. For example, if this surface is perpendicular to axis

2, then  and the components of traction acting at a point on this surface are given as follows

(2.33)

Figure 2.5: (a) A body showing displacement and traction boundary conditions

                   (b) traction vector at any arbitrary point P on the surface of a body

  : Constitutive Equations

The relationship between stress and strain is known as constitutive equation. The general form of this equation is

(2.34)

Here,  are called elastic constants. This is also referred to as elastic moduli or elastic stiffnesses. This form of constitutive equation is known as generalized Hooke’s law. Very soon, we

will see this equation in detail for various material types.

The inverse of this equation can be written as

(2.35)

where  is known as compliance.

Plane Stress Problem

Plane stress problem corresponds to a situation where out of plane stress components are negligibly

small. Thus, we can say that the state of stress is planar. The planar state of stress in x-y plane is

shown in Figure 2.6. For the case shown in this figure, the normal and shear stress components in z

directions, that is  are zero. Please note that the state of stress shown in this figure

assumes the stress symmetry.

Note: A careful observation for strain components in z direction ( ) reveals that these

need not be zero. This is a common mistake made by many readers. The magnitude of these strain

 

components can be found with the help of constitutive equation given in Equation (2.34).

Homework

1. Verify the property given in Equation (2.6) for rotation matrix.

2. Using Equation (2.6), show that

3. where the term , called Kronecker delta, has the value 1 on the diagonal and 0 on the off

diagonal, that is, it represents an identity matrix when represented in matrix form.

4. Using relation for strain components (given in Equation (2.21)) write the expanded form of all

strain components and understand the physical significance of all strain components. (The

normal strain components denote the stretching of a line element, etc.)

5. Derive the principles of minimum of total potential and total complementary potential energy.

6. Derive the principle of virtual work.

 Introduction

In this lecture, we are going to develop the 3D constitutive equations. We will start with the the

generalized Hooke’s law for a material, that is, material is generally anisotropic in nature. Finally, we

will derive the constitutive equation for isotropic material, with which the readers are very familiar.

The journey for constitutive equation from anisotropic to isotropic material is very interesting and will

use most of the concepts that we have learnt in Lecture 5.

The generalized Hooke’s law for a material is given as

(3.1)

where, σij is a second order tensor known as stress tensor and its individual elements are the stress

components. is another second order tensor known as strain tensor and its individual elements

are the strain components. Cijkl is a fourth order tensor known as stiffness tensor. In the remaining

section we will call it as stiffness matrix, as popularly known. The individual elements of this tensor

are the stiffness coefficients for this linear stress-strain relationship. Thus, stress and strain tensor

has ( ) 9 components each and the stiffness tensor has (=) 81 independent elements.

The individual elements =) 81 are referred by various names as elastic constants, moduli and stiffness coefficients. The reduction in the number of these elastic constants can be sought with

the following symmetries.

Stress Symmetry:

The stress components are symmetric under this symmetry condition, that is, . Thus, there

are six independent stress components. Hence, from Equation. (3.1) we write

(3.2)

Subtracting Equation (3.2) from Equation (3.1) leads to the following equation

(3.3)

There are six independent ways to express i and j taken together and still nine independent ways to

express k and l taken together. Thus, with stress symmetry the number of independent elastic

constants reduce to ( ) 54 from 81.

Strain Symmetry:

The strain components are symmetric under this symmetry condition, that is, . Hence, from

Equation (3.1) we write

Subtracting Equation (3.3) from Equation (3.2) we get the following equation

(3.4)

It can be seen from Equation (3.4) that there are six independent ways of expressing i and j taken

together when k and l are fixed. Similarly, there are six independent ways of expressing k and l taken

together when i and j are fixed in Equation (3.4). Thus, there are  independent constants

for this linear elastic material with stress and strain symmetry.

With this reduced stress and strain components and reduced number of stiffness coefficients, we can

write Hooke’s law in a contracted form as

(3.5)

where

(3.6)

   

Strain Energy Density Function (W):

The strain energy density function W is given as

(3.7)

with the property that

(3.8)

It is seen that W is a quadratic function of strain. A material with

the existence of W with property in Equation (3.8) is called as

Hyperelastic Material.

                The W can also be written as

(3.9)

Subtracting Equation (3.9) from Equation (3.7) we get

(3.10)

which leads to the identity . Thus, the stiffness matrix

is symmetric. This symmetric matrix has 21 independent elastic

constants. The stiffness matrix is given as follows:

(3.11)

The existence of the function W is based upon the first and

second law of thermodynamics. Further, it should be noted that

this function is positive definite. Also, the function W is an

invariant (An invariant is a quantity which is independent of

change of reference).

The material with 21 independent elastic constants is called

Anisotropic or Aelotropic Material.

Further reduction in the number of independent elastic

constants can be obtained with the use of planes of material

symmetry as follows.

Material Symmetry:

It should be recalled that both the stress and strain tensor

follow the transformation rule and so does the stiffness tensor.

The transformation rule for these quantities (as given in

Equation (3.1)) is known as follows

(3.12)

where are the direction cosines from i to j coordinate

system. The prime indicates the quantity in new coordinate

system.

When the function W given in Equation (3.9) is expanded using

the contracted notations for strains and elastic constants given

in Equation (3.11) W has the following form:

(3.1

3)

(A) Symmetry with respect to a Plane:

Let us assume that the anisotropic material has only one plane of material symmetry. A material with one plane of material symmetry is called Monoclinic Material.Let us consider the x1- x2 ( x3= 0) plane as the plane of material symmetry. This is shown in Figure 3.1. This symmetry can be formulated with the change of axes as follows

(3.15)

With this change of axes,

(3.16)

This gives us along with the use of the second of Equation (3.12)

(3.17)

Figure 3.1: Material symmetry about x1-x2 plane

First Approach: Invariance Approach

Now, the function W can be expressed in terms of the strain components . If W is to be invariant, then

it must be of the form

(3.18

)

Comparing this with Equation (3.13) it is easy to conclude that

(3.19)

Thus, for the monoclinic materials the number of independent constants are 13. With this reduction of

number of independent elastic constants the stiffness matrix is given as

Second Approach: Stress Strain Equivalence Approach

The same reduction of number of elastic constants can be derived from the stress strain equivalence

approach. From Equation (3.12) and Equation (3.16) we have

(3.21)

The same can be seen from the stresses on a cube inside such a body with the coordinate systems

shown in Figure 3.1. Figure 3.2 (a) shows the stresses on a cube with the coordinate system x1, x2,x3

and Figure 3.2 (b) shows stresses on the same cube with the coordinate system .

Comparing the stresses we get the relation as in Equation (3.21).

Now using the stiffness matrix as given in Equation (3.11), strain term relations as given in Equation

(3.17) and comparing the stress terms in Equation (3.21) as follows:

Using the relations from Equation (3.17), the above equations reduce to

Noting that , this holds true only when

Similarly,

 

This gives us the  matrix as in Equation (3.20).

First Approach: Invariance Approach

We can get the function W simply by substituting  in place of and using contracted notations for the

strains in Equation (3.18). Noting that W is invariant, its form in Equation (3.18) must now be restricted to

functional form

(3.25)

From this it is easy to see that

Thus, the number of independent constants reduces to 9. The resulting stiffness matrix is given as

(3.26)

When a material has (any) two orthogonal planes as planes of material symmetry then that material is

known as Orthotropic Material. It is easy to see that when two orthogonal planes are planes of material

symmetry, the third mutually orthogonal plane is also plane of material symmetry and Equation (3.26)

holds true for this case also.

               

Note: Unidirectional fibrous composites are an example of orthotropic materials.

Second Approach: Stress Strain Equivalence Approach

The same reduction of number of elastic constants can be derived from the stress strain equivalence

approach. From the first of Equation (3.12) and Equation (3.23) we have

(3.27)

The same can be seen from the stresses on a cube inside such a body with the coordinate systems

shown in Figure 3.3. Figure 3.4 (a) shows the stresses on a cube with the coordinate system x1, x2, x3 and

Figure 3.4 (b) shows stresses on the same cube with the coordinate system . Comparing the

stresses we get the relation as in Equation (3.27).

Now using the stiffness matrix given in Equation (3.20) and comparing the stress equivalence of Equation

(3.27) we get the following:

This holds true when . Similarly,

This gives us the  matrix as in Equation (3.26).

Figure 3.4: State of stress (a) in x1, x2, x3 system

                                         (b) with x1-x2  andx2-x3 planes of symmetry

Alternately, if we consider x1-x3 as the second plane of material symmetry along with x1-x2 as shown in

Figure 3.5, then

(3.28)

And

(3.29)

This gives us the required strain relations as (from Equation (3.12))

or in contracted notations, we write

Substituting these in Equation (3.18) the function W reduces again to the form given in Equation (3.25) for

W to be invariant. Finally, we get the reduced stiffness matrix as given in Equation (3.26).

Figure 3.5: Material symmetry about x1-x2  andx1-x3 planes

  :

The stress transformations for this coordinate transformations are (from the first of Equation (3.12)

and Equation (3.29))

The same can be seen from the stresses shown on the same cube in x1, x2, x3 and  

coordinate systems in Figure 3.6 (a) and (b), respectively. The comparison of the stress terms leads to

the stiffness matrix as given in Equation (3.26).

Note: It is clear that if any two orthogonal planes are planes of material symmetry the third mutually

orthogonal plane has to be plane to material symmetry. We have got the same stiffness matrix when

we considered two sets of orthogonal planes. Further, if we proceed in this way considering three

mutually orthogonal planes of symmetry then it is not difficult to see that the stiffness matrix remains

the same as in Equation (3.26).

 

 

Homework:

1. Starting with hyperelastic material, first take x2-x3 plane as plane of material symmetry and obtain

the stiffness matrix. Is this matrix the same as in Equation (3.20) ? Justify your answer.

2. Starting with the stiffness matrix obtained in the above problem, take x1-x3as an additional plane

of symmetry and obtain the stiffness matrix. Is this matrix the same as in Equation (3.26)? Justify

your answer.

Transverse Isotropy:

Introduction:

In this lecture, we are going to see some more simplifications of constitutive equation and develop the

relation for isotropic materials.

First we will see the development of transverse isotropy and then we will reduce from it to isotropy.

First Approach: Invariance Approach

This is obtained from an orthotropic material. Here, we develop the constitutive relation for a material with

transverse isotropy in x2-x3 plane (this is used in lamina/laminae/laminate modeling). This is obtained with

the following form of the change of axes.

(3.30)

Now, we have

Figure 3.6: State of stress (a) in x1, x2, x3 system

                                         (b) with x1-x2  and x1-x3 planes of symmetry

From this, the strains in transformed coordinate system are given as:

(3.31)

Here, it is to be noted that the shear strains are the tensorial shear strain terms.

For any angle α,

(3.32)

and therefore, W must reduce to the form

(3.33)

Then, for W to be invariant we must have

Now, let us write the left hand side of the above equation using the matrix as given in Equation (3.26)

and engineering shear strains. In the following we do some rearrangement as

Similarly, we can write the right hand side of previous equation using rotated strain components. Now, for

W to be invariant it must be of the form as in Equation (3.33).

1. If we observe the terms containing and   in the first bracket, then we conclude that

  is unchanged.

2. Now compare the terms in the second bracket. If we have  then the first of Equation

(3.32) is satisfied.

3. Now compare the third bracket. If we have , then the third of Equation (3.32) is

satisfied.

4. Now for the fourth bracket we do the following manipulations. Let us assume that and

is unchanged. Then we write the terms in fourth bracket as

To have W to be invariant we need to have so that the third of Equation (3.32) is satisfied.

Thus, for transversely isotropic material (in plane x2-x3) the stiffness matrix becomes 

(3.34)

Thus, there are only 5 independent elastic constants for a transversely isotropic material.

 Second Approach: Comparison of Constants

This can also be verified from the elastic constants expressed in terms of engineering constants like

. Recall the constitutive equation for orthotropic material expressed in terms of engineering constants. For the transversely isotropic materials the following relations hold.

When these relations are used in the constitutive equation for orthotropic material expressed in terms

of engineering constants, the stiffness matrix relations in Equation (3.34) are verified.

Isotropic Bodies

If the function W remains unaltered in form under all possible changes to other rectangular Cartesian

systems of axes, the body is said to be Isotropic. In this case, W is a function of the strain invariants.

Alternatively, from the previous section, W must be unaltered in form under the transformations

(3.35)

and

(3.36)

In other words, W when expressed in terms of  must be obtained from Equation (3.33) simply by

replacing  by . By analogy with the previous section it is seen that for this to be true under the

transformation Equation (3.35). We can write

And the transformed strains are given as

(3.37)

Thus, for any angle α,

(3.38)

and therefore, W must reduce to the form

(3.39)

Then, for W to be invariant we must have

Now, let us write the left hand side of above equation using the  matrix as given in Equation (3.34)

and engineering shear strains. In the following we do some rearrangement as

(3.40

)

Similarly, we can write the right hand side of the previous equation using rotated strain components.

Now, for W to be invariant it must be of the form as in Equation (3.39)

1. From the second bracket, if we propose , then we can satisfy the first of Equation

(3.38).

2. From the third bracket, third of Equation (3.38) holds true when

. 3. The fourth bracket is manipulated as follows:

Thus, to satisfy the second of Equation (3.38) we must have . Further, we should have

. From our observation in 2, we can write . 

It follows automatically that W is unaltered in form under the transformation in Equation (3.36).

Thus, the stiffness matrix for isotropic material becomes as

(3.41)

Homework:

1. Starting with the stiffness matrix for transverse isotropic material, take the transformations about

x1 and x2 and show that you get the stiffness matrix as given in Equation (3.41).

The Lecture Contains:

Constitutive Relations for Orthotropic Materials and Stress-Strain Transformations

In the previous lecture we have seen the constitutive equations for various types of (that is, nature of)

materials. There are 81 independent elastic constants for generally anisotropic material and two for an

isotropic material. Let us summarize the reduction of elastic constants from generally anisotropic to

isotropic material.

1. For a generally anisotropic material there are 81 independent elastic constants.

2. With additional stress symmetry the number of independent elastic constants reduces to 54.

3. Further, with strain symmetry this number reduces to 36.

4. A hyperelastic material with stress and strain symmetry has 21 independent elastic constants.

The material with 21 independent elastic constants is also called as anisotropic or aelotropic

material.

5. Further reduction with one plane of material symmetry gives 13 independent elastic constants.

These materials are known as monoclinic materials.

6. Additional orthogonal plane of symmetry reduces the number of independent elastic constants to

9. These materials are known as orthotropic materials. Further, if a material has two orthogonal

planes of symmetry then it is also symmetric about third mutually perpendicular plane. A

unidirectional lamina is orthotropic in nature.

7. For a transversely isotropic material there are 5 independent elastic constants. Plane 2-3 is

transversely isotropic for the lamina shown in Figure 3.7.

8. For an isotropic material there are only 2 independent elastic constants.

Principal Material Directions:

The interest of this course is unidirectional lamina or laminae and laminate made from stacking of these

unidirectional laminae. Hence, we will introduce the principal material directions for a unidirectional fibrous

lamina. These are denoted by 1-2-3 directions. The direction 1 is along the fibre. The directions 2 and 3

are perpendicular to the direction 1 and mutually perpendicular to each other. The direction 3 is along the

thickness of lamina. The principal directions for a unidirectional lamina are shown in Figure 3.7.

Engineering Constants:

The elastic constants which form the stiffness matrix are not directly measured from laboratory tests on a

material. One can measure engineering constants like Young’s modulus, shear modulus and Poisson’s

ratio from laboratory tests. The relationship between engineering constants and elastic constants of

stiffness matrix is also not straight forward. This relationship can be developed with the help of

relationship between engineering constants and compliance matrix coefficients.

In order to establish the relationship between engineering constants and the compliance coefficients, we

consider an orthotropic material in the principal material directions. If this orthotropic material is subjected

to a 3D state of stress, the resulting strains can be expressed in terms of these stress components and

engineering constants as follows:

Figure 3.7: Unidirectional lamina with principal material directions

(3.42)

whereas the engineering shear strain components are given as

(3.43)

Here, are the Young’s moduli in 1, 2 and 3 directions, respectively. Thus,  represents

the axial modulus and  represent in-plane transverse and out-of-plane transverse moduli,

respectively. Note that axial direction is along the fibre direction. 

 represents the shear moduli. G12,G13 are the axial shear moduli in two orthogonal planes that contain

the fibers.G23 represents out-of-plane transverse shear modulus. Further, it should be noted that

.

The term represents the Poisson’s ratio. It is defined as follows

(3.44)

where  represents the strain in the direction of applied stress and  represents the strain the

associated lateral direction. It should be noted that, in general .We will mimic some (thought) experiments that we actually do in laboratory to extract these engineering constants. For example, we find engineering constants of a transversely isotropic lamina.

Experiment 1: The lamina is loaded in traction along the axial direction as shown in Figure 3.8 (a) and the strains in along three principal directions are recorded as the load is varied. The slope of the axial

stress versus axial strain curve yields the axial Young’s modulus . The ratios  give

the Poisson’s ratios respectively.

Experiment 2: The lamina is loaded in traction along direction 2. The two views of this loading case are

shown in Figure 3.8 (b). The slope of stress-strain curve in direction 2 gives the in-plane transverse

Young’s modulus . Since, the material is isotropic in 2-3 plane,  is also equal . The strains in all

three directions are measured. The ratios  give the Poisson’s ratios ,

respectively.

Experiment 3: The lamina is loaded in shear in plane 1-2 as shown in Figure 3.8 (c). The slope of the in-

plane shear stress and engineering shear strain curve gives the shear modulus . Please note that if

we load the lamina in 1-3 plane by shear then also we will get this modulus because the behaviour of

material in shear in these two planes is identical. Thus, by shear loading in plane 1-2 gives .

Experiment 4: The lamina is loaded in shear in 2-3 plane as shown in Figure 3.8(d). The corresponding

shear stress and engineering shear strain curve yields the shear modulus .

Note: We will see the experimental details to measure some of these engineering constants in a chapter on experimental characterization of lamina, laminates, fibres and matrix materials.

 

Constitutive Equation for an Orthotropic Material:

Now, let us assume that we have measured all the engineering constants of an orthotropic material along principal directions. With these engineering constants we know the relation between the strain and stress components as given in Equation (3.42) and Equation (3.43). Thus, it is easy to see that we can relate the strain components to stress components through compliance matrix. Let us recall from previous lecture the stiffness matrix for orthotropic material (Equation (3.26)). The inverse of this matrix (compliance) will have the same form as the stiffness matrix. Thus, we write the relationship between strain and stress components using compliance matrix as follows

             

Figure 3.8: Experiments to extract engineering constants for

a transversely isotropic material

(3.45)

Now compare Equation (3.42) and Equation (3.43) with Equation (3.45). This gives us the compliance

coefficients in terms of engineering constants. The coefficients are given in Equation (3.46).

(3.46)

It should be noted that like stiffness matrix, the compliance matrix is also symmetric. The compliance

matrix given in Equation (3.45) is shown symmetric.

Note: It is known from our elementary knowledge of linear algebra that inverse of a symmetric matrix is also a symmetric matrix. Since, the stiffness matrix, which is the inverse of compliance matrix, is symmetric; the compliance matrix has to be symmetric.

Now, let us derive some more useful relations using the symmetry of compliance matrix. If we compare

 and  we get . Similarly, comparison of  and  and comparison of  and  

give two more similar relations. All these relations are given in Equation (3.47). 

Constraints on Engineering Constants in Orthotropic Materials

For orthotropic materials there are constrains on engineering constants. These constrains arise due to

thermodynamic admissibility. For example, in case of isotropic materials it is well known that the Young’s

modulus and shear modulus are always positive. Further, the Poisson’s ratio lager than half are not

thermodynamically admissible. If these constrains are violated then it is possible to have a nonpositive

strain energy for certain load conditions. However, for isotropic materials the strain energy must be a

positive definite quantity.

In this section, based on the work done by Lempriere we are going to assess the implications of this

thermodynamic requirement (positive definiteness of strain energy) for orthotropic materials.

The sum of work done by all stress components must be positive, otherwise energy will be created. This

condition imposes a thermodynamic constraint on elastic constants. This condition requires that both

compliance and stiffness matrices must be positive definite. In other words, the invariants of these

matrices should be positive.  

Let us look at this condition with physical arguments. For example, consider that only one normal stress

component is applied. Then we can find the corresponding strain component from the corresponding

diagonal entry of the compliance matrix. Thus, we can say that for the strain energy to be positive definite

the diagonal entries of the compliance matrix must be positive. Thus,

(3.53)

In a similar way, it is possible under certain conditions to have a deformation which will give rise to only

one normal strain component. We can find the corresponding stress using the corresponding diagonal

entry in stiffness matrix. For the strain energy produced by this stress component to be positive the

diagonal entry of the stiffness matrix must be positive. Thus, this condition reduces to

(3.54)

and the determinant of the compliance matrix must also be positive. That is,

(3.55)

Now, using the reciprocal relations given in Equation (3.49), the condition in Equation (3.54) can be expressed as

(3.56)

This condition also justifies that the Poisson’s ratio greater than unity is feasible for orthotropic lamina.

Poisson’s ratios greater than unity is sometimes observed in experiments.

The condition in Equation (3.55) can be written as

The terms inside the brackets are positive. Thus, we can write

(3.57)

This condition shows that all three Poisson’s ratios cannot have large positive values and that their

product must be less than an half. However, if one of them is negative no restriction is applied to

remaining two ratios.

Let us consider the transversely isotropy as a special case. Let us consider transverse isotropy in 2-3

plane. Let

(3.58)

Then, the conditions in Equation (3.56) reduce as

(3.59)

and Equation (3.57) (using reciprocal relations in Eq. (3.49)) becomes as

(3.60)

The condition posed by above equation is more stringent than that posed in Equation (3.59). Note that the

quantities  and  are both positive. Thus, the limits on Poisson’s ratio in transverse plane are

(3.61)

Further, consider a special case of isotropic material where  and . This simplifies Equation

(3.61) to a well known condition

(3.62)

 Stress and Strain Transformation about an Axis

Often it is required to transform the stress or strain tensor from one

coordinate axes system to another. For example, if the fibres in a

lamina are not oriented along direction x, then we may need to

transform the stress and strain components from principal material

directions 1-2-3 to global directions xyz or vice-a-versa. It should be

noted that the stress and strain tensors are second order tensors.

Hence, they follow tensor transformation rules.

In this section we are going to introduce two notations. The

subscripts 123 will denote a quantity (like constitutive equation,

engineering constants, etc.) in principal material directions, while

subscripts xyz will denote the corresponding quantity in global

coordinate directions.

Figure 3.9: Unidirectional lamina with global xyz directions

and principal material 1-2-3 directions

Let us transform the stress and strain components for the case

shown in Figure 3.9. Here, xy plane is rotated about direction z to 1-

2 plane. Here, direction z and direction 3 are in same directions, that

is, along the thickness direction of lamina. The direction cosines for

this axes transformation are as given in Equation (2.2). However,

these are again given below.

Stress Transformation:

Let us do the stress transformation as given in Equation (2.7). In this

equation the primed stress components denotes the component in

123 coordinate system. Using the expanded form of Equation (2.7)

and stress symmetry, let us obtain  component of stress

Thus, substituting the values of direction cosines from above, we get

The remaining five stress terms (using stress symmetry) on the left

hand side are also obtained in a similar way. Let us write the final

form of the relation as

(3.63)

Here,  and

 and  is the stress

transformation matrix. Thus, comparing all the terms as in Equation

(3.63), we can write  as

(3.64)

 where . It should be noted that  is not

symmetric.

Strain Transformation:

In a similar way, we can transform the strain components from xy plane to 1-2 plane. In this

transformation we will use engineering shear strains. Let us find the  using the transformation

equation similar to stress transformation and using strain symmetry as

(3.65)

 

Substituting the direction cosines and rearranging, we get

(3.66)

We know from Equation (2.24) that the tensorial shear strains are half the engineering shear strains.

Thus, in Equation (3.53) we substitute

On simplification and putting , we get

(3.67)

The other five strain terms (using strain symmetry) on the left hand side are also obtained in a similar

way. Let us write the final form of the relation as

(3.68)

Here,  and  and  is the strain

transformation matrix. Thus, comparing all the terms as in Equation (3.68), we can write  as

(3.69)

Note: The transformation matrices,  and  differ by factors 2 in two terms.

Note: The transformation matrices,  and  are not symmetric.

Note: The order of stress and strain components in Equatjion (3.63) and Equation (3.68) is important.

Some books and research articles follow different orders. The readers are cautioned to take a note of it.

 

Examples:

Example 1: Calculate the stiffness and compliance coefficients for transversely isotropic material

AS4/3501 Epoxy. The properties are as given below for a fibre volume fraction of 60%.

Solution:

Unit of all compliance coefficients is 1/GPa.

The corresponding stiffness coefficients are calculated by inversion of the compliance matrix.

 

Unit of all stiffness coefficients is GPa.

Note: Both stiffness and compliance matrices are symmetric.

Homework:1. Write the number of independent elastic constants for 3D hyperelastic, monoclinic, orthotropic, transversely isotropic and isotropic materials.

2. Are the Poisson’s ratio  and  independent of each other for an orthotropic unidirectional lamina?

3. Take the form of stiffness matrix for an orthotropic material as given in Equation (3.26). Using any symbolic calculation software like Maple or Mathematica, obtain the inverse of this matrix and confirm that the form of compliance matrix written in Equation (3.42) is correct. Further, confirm that this matrix is symmetric. (One should be able to do this using the concepts of linear algebra alone.)

4. Extend the Problem 3 to get the stiffness matrix given in Equation (3.51).

Stiffness, Compliance Transformation and Hygro-thermo-elastic Constitutive Relation

 

Introduction

In the previous lecture we have derived the stiffness matrix in terms of engineering constants using the compliance relations. Further, we have seen the constraints over the engineering constants in orthotropic materials. Then we have visited the stress and strain transformation. In the present lecture we will see the stiffness and compliance transformation about an axis. Further, we will address the effect of thermal and hygroscopic actions on lamina constitutive equations.

Stiffness Transformation:

It is required to relate the stress components with strain components in global xyz directions. The stiffness matrix which relates the stress and strain components in global directions is called as transformed stiffness matrix. We will derive an expression for the transformed stiffness matrix as follows.

The constitutive equation in principal material coordinates, as given in Equation (3.11), is

(3.70)

Now, we express  using Equation (3.63) to transform stresses and

Equation (3.68) to transform strains. Substituting these equations, we get

(3.71)

Pre-multiplying both sides by , we get

where we define the transformed stiffness matrix as

(3.73)

The transformation matrices  and  can be inverted as follows

(3.74)

The final form of the transformed stiffness matrix is given in Equation (3.75).

The individual  terms of this matrix are determined using  and relation for

. The individual terms are given in Equation (3.76).

Note: The transformed stiffness matrix is symmetric in nature.

Note: The transformed stiffness matrix given in Equation (3.75) has exactly the same form

as a stiffness matrix for a monoclinic material. Thus, we can conclude that a transformation

through an arbitrary angle  about direction 3, leads to a monoclinic material behaviour.

The same can be seen from the plane of elastic symmetry considerations in xyz coordinate

system. The given lamina is symmetric only about xy plane. Thus, the transformed stiffness

matrix in Equation (3.75) is consistent with monoclinic material.

Note: Transformed stiffness coefficient terms are fourth order in the sine and cosine

functions. It is very important to use appropriate precision level while calculating (in

examinations and writing computer codes) these coefficients.

(3.76)

The constitutive equation  becomes

Compliance Transformation:

We are going to follow a procedure to transform compliance matrix similar to one used for transformation of a stiffness matrix. We have the constitutive equation in principal material direction as in Equation (3.70). We can write this in inverted form as

(3.77)

using Equation (3.63) and Equation (3.68) we get

(3.78)

Pre-multiplying both sides by , we get

(3.79)

where, we define the transformed compliance matrix as(3.80)

Alternately, we can find  by inverting the transformed stiffness matrix . Thus, inverting  

from Equation (3.73), we get

After carrying out the calculation for , it is easy to give its form as follows 

(3.81)

Note that  has the same symmetric form as the transformed stiffness matrix.

The individual terms of the compliance matrix are obtained by carrying out multiplication of matrices

as in Equation (3.82) and are given below. 

 

Thermal Effects:

Thermal effects (effects due to change in temperature) are very important in composite materials for

various reasons. The analysis of composites with thermal effects and effective thermal properties of

the composite are two of the main reasons.

Important issues from analysis point of view:

1. The composite materials are used in environment where thermal gradients are

unavoidable. For example, the helicopter containing composite fuselage operates at -50° C

during winter at Leh and same helicopter can operate at +50° C during summer in the

desert of Rajasthan. Thus, the effect of temperature gradient on the service performance of

the composite is very important. In such service conditions, the layers of composite material

tend to expand or contract but are restricted due to adjacent layers. Thus, it induces

thermal stresses.

2. Most of the fabrication processes of polymer matrix composites have thermal cycles for

matrix curing. A typical cycle involves raising the temperature to a certain level and holding

it there for specified time and bringing it back to room temperature. It is well known that the

fibre and matrix materials have different coefficients of thermal expansion (defined below).

This mismatch produces residual thermal stresses because the fibres and matrix material is

constrained in a composite.

Important issues from effective thermal properties point of view:

The second reason for the study of thermal effects is the effective properties of the composite

materials.

1. Finding effective thermal properties of the composite theoretically to get an estimate

requires sophisticated mathematical modeling when one considers:

a. Difference in coefficients of thermal expansion of fibre and matrix materials

b. The direction dependence of coefficients of thermal expansion in these materials

c. Curing cycle temperature variations. This point is important because for some of

the materials the coefficient of thermal expansion changes with temperature.

2. Finding the effective thermal properties for lamina in global direction with oriented fibres as

shown in Figure 3.9 requires a special attention.

Further, finding these effective properties by laboratory test is also a challenge. Thus, for the

various reasons mentioned above the study of thermal effect is very important. In the following, we

develop a systematic way to handle effective thermal properties of a lamina along global directions.

It is well known that when a material is subjected to thermal gradient, it undergoes a deformation.

The strain due to thermal changes is called thermal strain (denoted by superscript (T)). In general,

the thermal strain is proportional to the temperature change . The constant of proportionality is

called coefficient of thermal expansion. Thus, we can write the thermal strains in principal material

directions for an orthotropic material as

(3.83)

where  denote the coefficients of thermal expansion in principal

material directions. It should be noted that for an orthotropic material in principal directions there are

no shear strains due to thermal effects like in an isotropic material. For an isotropic material the

coefficient of thermal expansion is same in any direction. However, for an orthotropic material

. The thermal expansion of an elemental cube in principal directions for an isotropic

and orthotropic material is shown in Figure 3.10.

These thermal strains will not produce stresses unless these are constrained. The thermal strains

which do not produce stresses are known as free thermal strains. However, in case of composites

the fibres and matrix are constrained in a lamina and layers are constrained a laminate. Thus, in

composite the thermal strains produce the thermal stresses.

The thermal strains are given in principal material directions as given in Equation (3.83). Let us

consider that we need to find these strains in a global coordinate system (refer Figure 3.9). We

need to transform them from 123 coordinate system to xyz coordinate system by a rotation  about

3-axis. Thus, similar to Equation (3.68), we can write

(3.84)

Substituting Equation (3.83) in the above equation,

(3.85)

where

(3.86)

Figure 3.10: Thermal expansion in an isotropic and

orthotropic material

On substitution of  in the above equation, we get the following individual terms of coefficients

thermal expansion in xyz directions.

 Thermo-Elastic Constitutive Equation:

Let us assume that the total strain,  in a composite is a superposition of the free thermal strain

 and the strain due to mechanical loads (also known as mechanical strains) . Thus,

Now, for mechanical strains we use the constitutive equation as

Thus, we can write the total thermo-elastic strain as

(3.89)

Equation (3.89) can be written as

Premultiplying the above expression by , we get the stresses as

(3.90)

Equation (3.90) is the basic constitutive equation for thermo-elastic stress analysis.

Using Equation (3.90) and similar to Equation (3.72) we can find the stresses due to thermo-elastic

effects in global directions as,

(3.91)

where

Equation (3.91) is inverted to give the total strains in terms of the mechanical and free thermal

strains as

(3.92)

Effect of Moisture:

The polymer matrix composite materials, during their service can absorb moisture from the environment.

The effect of absorption of moisture is to degrade the various material properties of the composite.

 

 

 

 

 

 

 

 

 

 

 

 

 

Further, this results in an expansion. It is called hygroscopic expansion. However, this expansion is

again constrained as in thermal expansion. Hence, when dealing with the hygroscopic expansions, a

treatment similar to thermal expansion is used.

The hygroscopic strains  are assumed to be proportional to the percentage moisture absorbed,

This percentage is measured in terms of weight of the moisture. The constant of proportionality,  is

the coefficient of hygroscopic expansion.

Thus, in principal coordinates the hygroscopic strains are

(3.93)

 where

 

 

Hygro-Thermo-Elastic Constitutive Equation:

This is the most general formulation for the mechanical, thermal and hygral effects on stress

analysis in composites. Here, we superimpose the strains due to these three effects to give us the

total strain as

 

 

 

 

(3.96)

Using constitutive equation for mechanical strains, we get

(3.97)

The stresses in the composite can be given as

(3.98)

These stresses in global coordinates xyz can be written as

(3.99)

 

 

 

 

 

 

 

 

 

  

Examples:

Example 2: Transform the stiffness and compliance matrix of Example 1 about axis 3 by an

angle of

= 30°.

Solution:

Approach 1: One can find the transformation matrices  and do the matrix

multiplication as given in Equation (3.73) for transformed stiffness matrix and then inverse this

matrix or do the matrix multiplication as given in Equation (3.80) to get the transformed

compliance matrix. The use of Equation (3.73) and Equation (3.80) is suggested because

remembering  is not so difficult. Further, their inverse can be easily found with the

help of Equation (3.74).

            

 

For

Thus

Unit of all transformed stiffness coefficients is GPa.

Unit of all transformed compliance coefficients is 1/GPa.

Approach 2: You can write the expanded form for transformed stiffness and compliance

coefficients in Equation (3.76) and Equation (3.82). However, the readers are suggested to use

this approach only when they are confident of remembering these terms.

Example 3: The coefficients of moisture absorption for T300/5208 composite material are

. Plot the variation these coefficients between

.

Solution:

We have the expression for variation of the coefficients of moisture absorption as

where, . We plot the above variation using a computer code. The final plot is shown in Figure 3.11.

 Homework:

1. Verify the result given in Equation (3.74).

2. Using the invariance property of strain energy density function, show that

and 3.4. Obtain the individual terms of transformed stiffness and compliance matrices using Equation (3.73) and

Equation (3.80), respectively and verify it with Equation (3.76) and Equation (3.82), respectively.

5. Obtain the strain transformation matrix  using tensorial shear strains. Further, using this transformation matrix obtain the transformed stiffness and compliance matrix in the form similar to Equation (3.75) and Equation (3.81). Compare the new matrices and comment on the observations with justifications.

6. Calculate the stiffness and compliance coefficients for following transversely isotropic materials given in Table 3.1.

Table 3.1: Properties for unidirectional transversely isotropic lamina

Property\Material T300/BSL914C Epoxy E-glass/LY556/ HT907/DY063 EpoxyS-glass/MY750/ HY917/DY063 Epoxy

E1 (GPa) 138 53.48 45.6

E2=E3(GPa) 11 17.7 16.2

G12 = G13(GPa) 5.5 5.83 5.83

ν12= ν13 0.28 0.278 0.278

ν23 0.4 0.4 0.4

             

α1 (×10-6/°C) -1 8.6 8.6

α2 = α3 (×10-6/°C) 26 26.4 26.4

6. The stiffness matrix for an orthotropic material is given as

7. Find:

a. The compliance matrix

b. The engineering constants .

7. Why are the thermal effects important in composite materials? Explain in detail.

8. Plot variation of between   for T300/BSL914C Epoxy and E-glass/LY556/

Dimensional Lamina Analysis

Introduction

In the previous chapter, we have developed 3D constitutive equations. While analyzing composites, most

of the times a planar state of stress actually exists. It is noted that a typical unidirectional lamina has very

small thickness compared to its planar (xy) dimensions. Thus, it is appropriate to assume a planar state of

stress in a lamina. In this chapter, we are going to derive a constitutive equation for plane stress problem

in unidirectional laminar composite.

Plane Stress for Monoclinic (or Rotated Orthotropic) Material

3D constitutive equation for a single layer of a unidirectional composite with a fiber orientation  relative

to the global coordinate

(4.1)

For a state of plane stress, we have

(4.2)

Thus, it is easy to see that the two out of plane shear strains  are zero. We can write these

strains using Equation (4.1) as

(4.3)

The out of plane normal strain  is expressed using Equation (4.1) and Equation (4.2) as

(4.4)

Note that this strain component is not zero.

In plane components of strain for a plane stress state can be written using Equation (4.1) as

(4.5)

From 3D constitutive equation (Equation (3.72)) for the transformed stiffness, we can write  as

(4.6)

From this equation, we can get the out of plane transverse normal strain as

(4.7)

Thus, the out of plane normal strain is expressed in terms of in-plane strain components and known

stiffness coefficients.

 Reduced Transformed Stiffness Matrix

The equation for in-plane components of stress in terms of the transformed stiffness coefficients is

(4.8)

Substituting  from Equation (4.7) into Equation (4.8) and upon simplification, we get

             

(4.9)

The above equation is written in matrix form as

(4.10)

where the transformed reduced stiffness coefficients   are defined as

(4.11)

Note: Transformed reduced stiffness matrix is symmetric.

Note: It is very important to note that the transformed reduced stiffness terms for plane stress,

, are not simply the corresponding terms,  taken from the 3D stiffness matrix. This should be clear from the fact that the inverse of a  matrix is different from that of a . This can easily be seen from Equation (4.11). The readers should easily understand that when

 terms are used to define a constitutive equation, then it is a reduced transformed constitutive equation.

Plane Stress for Orthotropic Material

Let us recall the constitutive equation for orthotropic material in principal directions.We can write the constitutive equation using compliance matrix as (Equation (3.45))

(4.12)

We have planar state of stress for orthotropic lamina. Then we have out of plane transverse  

Thus, the out of plane transverse shear strains are zero. Now, let us write the out of plane

transverse normal strain using Equation (4.12) as

(4.15)

Using and Equation (4.13) in the above equation, we get

(4.16)

and the inplane strain components are given as

(4.17)

This equation is called reduced constitutive equation using compliance matrix.

We have the 3D constitutive equation using stiffness matrix in principal material directions as

(Equation . (3.26))

(4.18

We have from the condition of plane stress problem that . Thus, using Equation. (4.18),

we can write

(4.19)

This leads to non-zero transverse normal strain  as

(4.20)

Using Equation. (4.18), we can write the inplane stress components as

(4.21)

Putting the expression for  from Equation. (4.20) in above equation, we get

This equation is written in matrix form as

(4.23)

where, the  terms can be written using index notations as follows

(4.24)

 

Let us compare this equation with corresponding 3D equation (Eq. (4.12). It can easily be seen that the

compliance terms of the constitutive equation are identical for 3D and plane stress problems. Thus, we

can write for the plane stress problems as

(4.25)

It is easy to invert a  matrix. In fact, you need to invert a  matrix. Thus, we can write the

individual reduced stiffness entries in terms of compliance entries as

(4.26)

Compliance and Stiffness Coefficients Using Engineering Constants

              

Let us write the compliance and stiffness matrices using engineering constants. It is easy to see that

the individual entries of the compliance matrix in plane stress problem are the as 3D compliance. Thus,

we write for the plane stress problem the compliance entries as

(4.27)

Here, we have used the property that compliance matrix is symmetric, that is, . Using

, we can develop the reciprocal relationship for 2D case as

(4.28)

Note: It is easy to see for a plane stress problem of an orthotropic material that there are only four of

the five material constants independent.

We can write the individual terms of reduced stiffness matrix in principal material directions by using

Equation (4.26) and Equation (4.27) as

(4.29)

Note: For a transversely isotropic material there is no reduction of the number of independent

constants for plane stress problem.

 

One can write the constitutive equation in material coordinates, using Equation (3.42), Equation (3.43)

and introducing the corresponding reduction in out of plane direction as   

(4.30)

It is easy to write the compliance coefficients in Equation (4.27) from these relations. Further, we can

write the above relations in inverted form as

 2D Transformations about an Axis:

In planar stress condition we need to transform the stresses in plane. Let us write, similar to Equation

(3.63), the transformation equation for stresses as

(4.32)

where  is the transformation matrix for stress tensor. For the above equation, using Equation

(3.64), this matrix can be written as

(4.33)

Similarly, we can write the strain transformation equation in the following form.

(4.34)

         

where  is the transformation matrix for strain tensor. We can find this matrix using Equation (3.39)

and the above relations as

             

(4.35)

Note: The transformation matrices  and  are not symmetric. There is a difference of factor 2 in

two entries of these matrices.

Note: The transformation matrices  and  can be inverted using following relation

Lamina Constitutive Relations in Global Coordinates:

The plane stress constitutive equation in principal material coordinates is

(4.37)

Let us write the stresses and strains in terms of components in global directions using Equation (4.32)

and Equation (4.34). The above equation can be re-written to give stresses in global coordinates as

(4.38)

We define the plane stress transformed reduced stiffness matrix  as

(4.39)

Introducing this definition in Equation (4.38), we get

(4.40)

               

The above equation is written in expanded form as

(4.41)

Note:  is a symmetric matrix. Further, it is a fully populated matrix with non zero  

coefficients.

Thus, using Equation (4.33), Equation (4.35 in Equation (4.39), we can write the individual terms in

expanded form as

Note: Reduced stiffness coefficients are fourth order in the sine and cosine functions.

Note:  are very important in that they define the coupling both in-plane normal and shear

responses. Figure 4.1 shows response of an isotropic and orthotropic material under traction. The

behaviour of orthotropic lamina loaded along fibre direction and perpendicular to fibre direction is

essentially similar to an isotropic material. However, for an off axis lamina, the behaviour clearly shows

the coupling between normal and shear terms.

Figure 4.1: Normal shear coupling in orthotropic lamina

The transformed plane stress constitutive equation can also be given in inverted form of Equation

(4.40) as

(4.43)

where

(4.44)

Using  from Equation (4.33) and  from Equation (4.35) in the above equation, we get the

individual coefficients of transformed reduced compliance matrix as

  Thermal Effects:

Thermal strains in principal material coordinates are proportional to the temperature change . These are given using coefficient of thermal expansion as

(4.46)

where .

Transformation of the thermal strains  to the strains  in global coordinates gives

(4.47)

Let   . Thus, Equation (4.47) becomes

(4.48)

or

(4.49)

   Thermo-Elastic Constitutive Equation:

The total strain due to mechanical and thermal loading in principal material directions is given as

(4.50)

We can write for the mechanical strains as

(4.51)

Thus, Equation (4.50) becomes 

Moisture Effect:

The hygroscopic expansion in principal material direction is proportional to the amount of percentage

weight of moisture absorbed. Further, the hygroscopic expansion will be in principal normal directions

only. This expansion will not lead to any shear. Thus, we write the hygral strains in principal directions

for planar problem as

(4.54)

Here, denotes the coefficient of hygroscopic expansion in principal material directions for planar

              

problem and  denotes the amount by percentage weight of moisture absorbed.

Now let us transform the hygroscopic strains in global coordinate system as

(4.55)

Using Equation (4.54), we can write

(4.56)

where

(4.57)

It is clearly seen from Equation (4.48) and Equation (4.57) that  and behave in a similar

way.

Hygro-Thermo-Elastic Constitutive Equations:

Homework:

1. Verify the result given in Equation (4.36). 2. Using the invariance property of strain energy density function, show that:

and

3. Using the relation between  and  as given in Equation (4.24) and  in terms of

engineering constants, show that  are as given in Equation (4.29).

4. Write the compliance coefficients in Equation (4.45) in terms of engineering constants.

5. Using Equation (4.24) in Equation (4.42) obtain the individual terms of   in terms of .

6. For fibre orientation  and  obtain  matrix for materials given in Table 3.1.

7. The  matrix for a composite with fibre orientation of  is given as

8. Find all engineering constants for this material.

9. Write a computer code to calculate reduced transformed stiffness and compliance matrix for any

angle of fibre orientation with respect to global coordinate system.

10. Extend the code written for the above problem to plot the variation of  terms for orientation of

fibres between . Plot the variation for materials given in Table 3.1.

11. Write a computer code to plot the variation of thermal and hygroscopic expansion coefficients

with fibre orientation between  for T300/5208 composite.

Lamina Engineering Constants

Introduction

In the previous lecture we have derived constitutive equation for planar state of stress in a lamina. We

have derived these constitutive equations in principal material and global directions. In this lecture, we are

going to see a practical application of the planar constitutive equations in industry. Although, the

engineering constants in principal material directions are known, it is difficult to comment on the

engineering constant for off axis lamina, instantly. When laminae are used for designing a structure, the

engineering constants in global directions become very useful for a quick estimate of the behavior of the

structure under certain loads. Thus, for practical application purpose, the various lamina engineering

constants are obtained and their variation for fibre orientation between  for a range of

composite materials is given together. Thus, a designer can use the required lamina with appropriate fibre

orientation and material.

Here we are going to obtain engineering constants for any off axis lamina as a function of engineering

constants of that lamina in principal material directions and fibre orientation. This can be done with the

help of lamina constitutive equation with appropriate one dimensional state of stress.

We have constitutive equation in global directions as given in Equation (4.5)

Thus, for a given state of stress in global directions we can find the strains in global directions from this

equation.

Axial Modulus:

Consider a unidirectional off axis lamina. This lamina is subjected to the loading  and

 as shown in Figure 4.2.Thus, from Equation (4.5) for this state of stress we can write the axial strain as

(4.61)

The Young’s modulus in x-direction is now defined as

(4.62)

Thus, from Equation (4.61), we can write

(4.63)

            

 

Figure 4.2: Off axis lamina loaded in traction along x direction

In the above equation,  is written using compliance terms in principal material directions as

(4.64)

Further, it can be writing compliance terms in principal directions using engineering constants doing

some rearrangements as

(4.65)

From this expression it is easy to see that the modulus  when   and  when

. The variation of the modulus  with fibre orientation for AS4/3501-6 Epoxy material is

shown in Figure 4.3. The variation of the modulus  for both positive and negative fibre orientations is

identical in nature.

Figure 4.3: Variation of axial modulus with fibre orientation for AS4/3501-6 Epoxy

  

        In-plane Shear Modulus:

The in-plane or axial shear modulus for an off axis lamina can be obtained when it subjected to a pure

shear loading as shown in Figure 4.7. Thus, for this loading condition we have   and .

.

For this loading, we define the in-plane shear modulus  as

(4.75)

With the help of Equation (4.5) we rewrite this equation as

(4.76)

And in terms of engineering constants, it becomes

(4.77)

The variation of  with fibre orientation between  to  for AS4/3501-6 Epoxy material is

shown in Figure 4.8. From this figure it can be seen that shear modulus is maximum when .

At  the value of shear modulus is

(4.78)

             

Figure 4.7: Off axis lamina loaded in pure shear

Note: When the material is isotropic, that is  and , then the above expression

reduces to the familiar relation

(4.79)

The minimum value of shear modulus is seen when the lamina is loaded in shear in principal material

directions and its value becomes

(4.80)

Note: It is very important to note that the shear modulus of the lamina is a minimum when lamina is in

principal directions and a maximum when fibre orientation is  or . Further, the behavior of a

lamina under same pure shear for fibre orientation  is significantly different from that of lamina with

fibre orientation of . The physical significance of this phenomenon is explained in greater details

in the later section.