objectives of composite lab

8/13/2019 Objectives of Composite Lab

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Objective

Lecture

Lab work

Data Reduction

Handout

Miscellaneous

LectureCompositesA Composite in engineeringsense is any materials

that have been physically assembled to form onesingle bulk without physical blending to foam a

homogeneous material. The resulting material wouldstill have components identifiable as the constituent of

the different materials. One of the advantage of

composite is that two or more materials could be

combined to take advantage of the good

characteristicsof each of the materials.

Usually, composite materials will consist of two

separate components, the matrix and the filler. Thematrix is the component that holds the filler together

to form the bulk of the material. It usually consists of

various epoxy type polymers but other materials maybe used. Metal matrix composite and thermoplastic

matrix composite are some of the possibilities. The

filler is the material that has been impregnated in the

matrix to lend its advantage (usually strength) to the

composite. The fillerscan be of any material such

as carbon fiber, glass bead, sand, or ceramic.

Composites can be classified into roughly three orfour types according to the filler types:

Particulate Short fiber long fiber laminate

Particulate composite consists of the composite

materialin which the filler materials are roughly

round. An example of this type of composite would be

the unreinforced concrete where the cement is thematrix and the sand serves as the filler. Lead particles

in copper matrix is another example where both the

matrix and the filler are metals. Cermet is a metalmatrix with ceramic filler.

Short and long fiber composites are composites in
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which the filler material has a length to diameter

ratio, l/d, greater than one. Short fiber composites aregenerally taken to have l/d of ~100 while long fiber

type would have l/d~ . Fiber glass filler for boat

panel is an example of short fiber composite. Carbonfiber, aramid fiber (Kevlar) fiber are some of the

filler material used in the long fiber type composites.

Laminate is the type of composite that uses the filler

material in form of sheet instead of round particlesor

fibers. Formica countertop is a good example of this

type of composite. The matrix material is usually

phenolic type thermoset polymer. The filler could beany material from craft paper (Formica) to canvas

(canvas phenolic) to glass (glass filled phenolic).

Since the composites are non-homogeneous, the

resulting properties will be the combination of the

properties of the constituent materials. The differenttype of loading may call on different component of the

composite to take the load. This implied that

the material propertiesof composite materials may be

different in tension and in compression as well as in

bending. Throughout the lab, the subscript t, c,

and bwill be used to designate the properties

intension, compression, and bending respectively.

Advantages of composite materialsThe main advantage of most composites materials are

in the weight savings. A quick way to illustrate thisadvantage is in the strength to weight ratio. Different

materials has different strength, that is each material

can take different of amount of load for the samevolume (cross sectional area) of the material. For a

given design, the material used must be strong enough

to withstand the load that is to be applied. If a material

selected is not strong enough, the part must be

enlarged to increase the load bearing capacity. Butdoing so increases the bulk and weight of the part.

Another option is to change material to one that has

high enough strength to begin with.

Carbon fiberCarbon fiber can be manufactured from


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polyacrylonitrile (PAN) fiber or pitch fiber. Most

carbon fibers in use in high performance application,however, are made from PAN. The PAN fibers

undergo multi step process to drive off all the

hydrogen atoms and some of the nitrogen atoms toform honey combs networks similar to graphite. ,

Figure 2.1

Figure 2.1Molecular structure of PAN being

converted to carbon fiber.

Designation of carbon fiber composites consists of the

fiber material follows by the matrix material. For

example, AS4/3501-5 means that the carbon fiberused has the name AS4 with the epoxy name 3501

being used as the matrix material. The 5 signifies the

fifth reformulation of the 3501 epoxy. The composite

uses in this labwill likely be IM7/3501-6.

LayupIt can easily be seen that the long fiber composite will

have directionality depending on the direction inwhich the fibers are laid out in the composite.

Composites usually come in sheet called Prepreg. The

sheet will consist fibers preimpregnated with uncured

matrix material. The sheet can be cut and lay up inlayers to form the composite. The lay up could be

done in a manner where the fibers all line up in one

direction. This is called uniaxial composite. The lay

up used in this lab will be of this type due to therelative ease of analysis. Most of the time, the prepreg

sheets will be laid in different directions. The analysis

of this type of layup is beyond the scope of the lab.

In analyzingcomposite material, it is necessary to

designate some type of coordinate system.

Customarily, the X-Y will designate the globalcoordinate of a composite piece. Each layer in that

composite may have different fiber orientation that

will require a separate coordinate system. Thedirection along the fiber is designated 1 direction

while the direction transverse to it is designated 2

direction or the matrix direction. The X-Y and 1-2


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coordinate systems coincide in the caseof uniaxial

composite.

Figure 2.2Schematic of uniaxial composite material.

Stress-strain relationshipsMaterial reactions under stresses can be described by

a set of constitutive equations. For isotropicmaterial,

this is known as Hooke's law or sometimes, in an

inverse form, Lam [la-may] equations. The 3-DHooke's law in matrix form is:

This Hooke's law is in the compliance form where thestrains are expressed in term of stresses and a

compliance matrix. The inverse of this expresses the

stresses in term of strain and the stiffness matrix.

Compliance form

Stiffness form

The convention is that the symbol Sis used for

compliance and the Cis used for the stiffness.

In the most general case, the stress or strain with

subscript ijis not the same as the one with supscriptji.The [C] and [S] matrices would each be a [9x9]

matrix. This reduces to [6x6] matrices because of the

definition of the shear stresses and strains. For the

general [6x6] matrices, it will be totally populated
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with non equal terms inside. This would imply that

there is a need of 36 constants to describe the stress-strain behavior of any generic material. Invoking

compatibility condition where no two materials may

occupy the same space, the [C] and the [S] must besymmetric. This leads to the first useful set of

constitutive equations describing material behavior.

Anisotropic or triclinic material has no plane of

symmetry. A total of 21 material constants is needed

to describe the stress-strain behavior. In generic form:

If the material has one plane of symmetry,

monoclinic, some constants are zero and the stress-

strain behavior can be described with 13 constants.

Next simplification is when the material has 2 (3)planes of symmetry. This is called orthotropic

material. The number of constants reduces to nine.

Next simplification can be made when the material

has one plane of isotropy. That is the material has one

plane of symmetry or the transverse. This transverse

plane has infinite plane of symmetry. Another word,the material behaves in isotropic manner within that

plane. This is called transversely isotropic material.

The number of independent constants reduces to five.


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Inaddition,

It is necessary to keep track of the subscript of the

material properties. The subscript 1 denotes the 1direction and the subscript 2 denotes the 2 direction.

The subscript 12 denotes the reaction in 2 direction

due to the act in 1 direction. For example, 12is the

Poisson's ratio in the 2 direction due to the load beingapplied in the 1 direction.

Writing out the transversely isotropic Hooke's law that

will be required for our analysis:

Keep in mine that the material properties in tension,

compression, andbendingare different. For example,

the modulus in tension, compression, and bending for

the 1 direction will all have different values. They willbe denoted asE1t,E1c,E1b.

Strain transformation equationIn the case where the stresses or strains are know in adifferent direction, transformation equations could be

used to calculated the stresses or strains in the

direction desired. The elementary strength of materialsbooks usually derive the transformation equation

using the wedge method. In order to demonstrate that

the transformation equation and Mohr's circle are oneand the same, a Mohr's circle method will be used. It

is necessary that the students be made aware of the

equivalency of the transformation and the Mohr's

circle to prepare them for the Pressure Vessel lab.


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Suppose that strain readings were taken off a strain

rosette with three gages labeled a, b, and c.

Strains x, y, xycan be found from the three gagereadings and three gage positionings.

where, i= a, b, and c.

Rule of mixturesOne quick way to estimate the material properties, i.e.,

the moduli in 1 and 2 direction of a composite is byusing the rule of mixture. It assumes that the modulus

of a composite is the combination of the modulus of

the fiber and the matrix that are related by the volume

fraction of the constituent materials.


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Composite (laminated) beamNot to be confused with beam made up of composite

material, this composite (laminated) beam in this

context refers to a beam with layering material having

different Young's moduli. The different in moduli willresult in the beam having a shift in neutral axis under

bending load. One way to work a composite beam

problem is by using an equivalent beam. The basicidea is to make a beam out of one material but expand

or contract the substituted part laterally so that it has

the same functionality as the original beam.

For beam made up of two materials,M1andM2with

moduli of elasticityE1


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The neutral axis of this beam goes through the

centroid of its cross section. All measurements andcalculations must be done with respect to this neutral

axis.

The next step is to calculate the moment of inertia, I.

Since the neutral axis is not at the geometric center a

parallel axis theorem must be use to shift the I's ofeach area to the neutral axis.

Normal stress equation can then be used to calculate

the tensile and compressive stress in that equivalent

beam (as oppose to the stress in the original beam).Since the expanded section is enlarged by a factor

of n, the area is also increased by the same factor. Thestress in the original section is then the calculated

stress divide by a factor of n. For the unexpandedsection, the stress is as calculated.

The above outline also works whenE1>E2. In thiscase the expanded section is actually contracted.

Think of it as an expansion by a factor of less than one

and continue.


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Objective

Lecture

Lab work

Data Reduction

Handout

Miscellaneous

Lab WorkPrelab

1. Three strain readings were recorded from atensile specimen with the corresponding

position of the strain gages with respect to X

axis. They are:

Find x, y, xyusing the strain transformation

equations.

Rewrite strain transformation equation inmatrix form:

The three unknowns in the equation above can

be solved in any manner. The solution to a

matrix equation

is

Hence,
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Some of the commands that could prove to be

useful in MS Excel are "minverse" and"mmult". "minverse" inverts a matrix and

"mmult" multiplies two matrices. To use these

two commands, highlight an area that theresults will go then enter the formula:

=minverse(matrix1) follow by a[command]+[return] or [crtl]+[return]

=mmult(matrix1,matrix2) follow by a

[command]+[return] or [crtl]+[return].

The degrees must be in radian in MS Excel.

The formula for is "pi()".

2.

Young's modulus in bending is given by,

E = Mc/I

Where, M is the maximum bending moment, I

is the moment of inertia, c is the distance from

the neutral axis to the outer fiber, and is thestrain. For the case of three-point bending

beam as shown in Figure 4, express theEin

term ofP,L, c,I, and .

Figure 4Three-point bending beam.

Draw the shear and moment diagrams to

determine the maximum bending moment.Alternatively, summing the moment about an

arbitrary cut at a distancex from the end will

give the bending moment in term ofx.

Lettingxgoes toL/2 will give the expressionfor the maximum bending moment.
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Substitute the moment into the equation above:

3. Derive an equation, in terms ofP,L1,L2,E,andI, for the deflection, L1, of the beam

directly under either of the applied loads as

shown in Figure 5. Draw the shear and

moment diagrams for the beam shown below.What is the maximum bending moment in

terms ofP,L1, and/orL2for this loadingcondition. (Assume the beam is made from anisotropic and homogeneous material.) You

may want to solve the three-point bending case

first.

Figure 5Four-point bending beam
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BC 2:

BC 4:

BC 1: C3 = 0 hence,

BC 3:

4. What advantages or disadvantages does 4-point bending have over 3-point bending

knowing our samples have strain gage rosettesmounted near but not necessary at the center.

5. For the beam shown in Problem 2 solve for themaximum stress using the beam cross-section

shown below and knowing that h1= h2= 0.125in., b= 1.25 in.,L= 10.0 in., andP= 1000

lb.E1is that of aluminum (10,000 ksi),E2is

that of steel (30,000 ksi).


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Objective

Lecture

Lab work

Data Reduction

Handout

Miscellaneous

Data Reductionsss
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Objective

Lecture

Lab work

Data Reduction

Handout

Miscellaneous

HandoutThe term "composite material" can be broadly defined

as the resultant of combining two or more materials,

each of which has their own unique properties, toform one new material. In a way, we studied

composite materials on a microscopic scale when weinvestigated multicomponent structures in metals,

ceramics, and polymers. However, when we speak of

engineering composite materials, we generally meanthat two or more different materials are assembled

macroscopically in a mechanical way. One example

could be assembled by man, such as combining glass

fibers with epoxy. Another example could be due tonature, such as combining cellulose fibers and lignin

to form wood. The advantages of composite materialsare that they can be constructed to exhibit the best

qualities from their constituents that neitherconstituent possesses singularly.

[1] In this laboratory,

we are mainly interested in composite materials that

are manmade.

Although the objective of this laboratory is to become

more familiar with high strength composites, there are

other objectives that are an integral part of this lab.Since the fibrous composite is generally not isotropic,

a more general or alternative material model must beused to describe the composite. After identifying the"constants" needed to describe our composite, we will

perform various experiments to determine some of

these constants.

Composites are commonly classified as being

anisotropic materials (materials with 21 independent

material constants). However, if the constituent

materials are combinedin certain configurations,

composite materials can be manufactured that behave

in an orthotropic nature (nine independent materialproperties) or even in a transversely isotropic nature

(five independent material constants). The samples we

are going to test have been manufactured so theybehave in a transversely isotropic nature. Before we

can start the experiment, we need to gain an

understanding of some of the terminology used for
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these materials.

The samples used in this laboratory are constructed

from a uniaxial composite material. Figure 1 shows a

schematic of a uniaxial composite. In Figure 1, thehorizontal lines represent fibers while the shaded

region represents the matrix material. Figure 2 shows

a schematic of a typical specimen used in thislaboratory. Each specimen has a strain gage rosette

mounted at its center as shown in the figure. Gages A

and C make angles of -45 and +45, respectively,

with gage B and the X axis.

Figure 1Schematic of uniaxial composite material.

Figure 2Strain gages layout for the tensile specimen.

Figure 3Stress and strain distribution for sampleswith and without equal moduli in tension and

compression.

Besides the fact that fibrous composite materials have

different material properties in the 1 and 2 directions,the material can also have different material propertiesin tension than it has in compression. One of the

following tasks will investigate this phenomenon. In

addition, since the composite material has a different

modulus in tension than it does in compression, the"bending" modulus will also be different. This can be
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distance from the neutral axis to the outer

fiber, and is the strain. For the case of three-point bending beam as shown in Figure 4,

express theEin term ofP,L, c,I, and .

Figure 4Three-point bending beam.

3. Derive an equation, in terms ofP,L1,L2,E,andI, for the deflection, L1, of the beam

directly under either of the applied loads as

shown in Figure 5. Draw the shear andmoment diagrams for the beam shown below.

What is the maximum bending moment interms ofP,L1, and/orL2for this loading

condition. (Assume the beam is made from an

isotropic and homogeneous material.) Youmay want to solve the three-point bending case

first.

Figure 5Four-point bending beam

4. What advantages or disadvantages does 4-point bending have over 3-point bendingknowing our samples have strain gage rosettes

mounted near but not necessary at the center.

5. For the beam shown in Problem 2 solve for themaximum stress using the beam cross-section

shown below and knowing that h1= h2= 0.125in., b= 1.25 in.,L= 10.0 in., andP= 1000

lb.E1is that of aluminum (10,000 ksi),E2isthat of steel (30,000 ksi).
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Tasks

1. Load a composite specimen (fiber direction) intension. Record the strains from all the strain

gages for 3-10 different applied loads. Report

these values in tabular form.2. CalculateE1tand 12from 3-D Hooke's law.

Report these values in tabular form.

Remember to apply your statistics knowledge,i.e., mean, standard deviation, mode, median,

etc.

3. Load a composite specimen (matrix direction)in 3-point bending using the Instron load

frame. Record the strain for 3-5 different

applied load.

4. Determine the elastic modulus in the matrixdirection. We will refer to this modulus asE2t.Assume that the neutral axis is at the center of

the cross-section.5. Load another composite specimen (fiber

direction) in 4-point bending. Record the strain

from the top and bottom gages for 10-15

different applied loads.6. Determine a tension and compression

moduli.E1tandE1ccan be calculated using

Equation 4 and 5.[2]

Mis the maximummoment at the current strain

reading. cand tare the absolute value ofstrains in compression and tension,

respectively. w is the specimen width and histhe specimen thickness.

(4)


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(5)

Comment on any differences between the

tension modulus,E1t, and compressionmodulus,E1c, in the fiber direction. Why mightthey be different? Is the difference significant?

How does this value ofE1t compare with the

value calculated in Task 2? Explain thedifference inE2tandE1t.

7. Load the same composite specimen again. Thistime, record the load deflection history on thestrip chart. Using beam theory from the prelab,

calculate the bending (flexural) modulus,E1b,

in the fiber direction. Use Figure 3 todetermine the moment of inertia. From Figure

3:

(6)

(7)

(8)

Does your value ofE1bseem reasonable?

Explain your answer in detail. It should be

noted that you just determineE1t,E1c,andE1bfrom a single bending test.

8. It has been suggested that the flexural moduluscan also be determined for a beam of unequal

tensile and compressive moduli from thefollowing equation:

[3]

(9)

This equation was derived assuming the

neutral axis was at the center of the cross-

section when calculating the moment of inertia

and the maximum stresses. How does thisresult compare toE1byou calculated

previously? Will this equation be applicable if


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yourE1tandE1cvalues are very different?

9. It can be shown analytically that the followingrelationship has to be true for a transversely

isotropic material.[4]

(10)

Calculate the 21using the tension moduli

from Task 4 and 6. How would you

experimentally determine the value of 21?

10.The rule of mixtures if often employed as ananalytical method of determining the moduli

of a uniaxial composite plate just fromknowing the properties and amounts of

constituent materials

used.[5]

DetermineE1andE2using the rule ofmixtures. Compare these values to the

experimentally determined values.

(11)

(12)

Values and definitions ofEm,Ef, Vm,and Vfwill be supplied in the laboratory. Will

the rule of mixtures be able to predict thedifferent in the properties in tension and

compression?

11.Summarize the material properties youdetermined in the result section of your report.

References

1. Jones, R.M.,Mechanics of CompositeMaterials, Hemisphere PublishingCorporation, New York, 1975, p.1.

2. Yu, M., Tarnopolskii, T. Kincis,Handbook ofComposites, vol.3, Institute of Polymer

Mechanics, Vatvian S.S.R., p. 266.

3. Carlsson ,L.A., Pipes, R.B.,Experimental


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Characterization of Advanced Composite

Materials, Prentice-Hall, Inc., New Jersey,1987, pp. 91-93.

4. Jones, R.M.,Mechanics of CompositeMaterials, Hemisphere PublishingCorporation, New York, 1975, p.38.

5. Jones, R.M.,Mechanics of CompositeMaterials, Hemisphere PublishingCorporation, New York, 1975, pp.90-92.


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Objective

Lecture

Lab work

Data Reduction

Handout

Miscellaneous

Miscellaneous

Last Modified

Sep 2005
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objectives of composite lab

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