This laboratory will explore the basic stress analysis techniques used to determined the material properties such as the modulus of elasticity and the Poisson's ratio. This could also be extended to determining other material properties such as proportional limit, yield strength, ultimate strength, or fracture strength of a material. It will be demonstrated in the lab that what so call the material properties are not constant under different type of load configuration. At the same time, it will be demonstrated that these material properties could be determined from different type of testing. The students should also learn that the same material properties in different type of loading, i.e., bending, tension, or compression, could be approximated if some of these properties are known.

In addition to basic stress analysis techniques, students will be exposed to a material not considered to be traditional engineering material, i.e., composite. The use of uniaxial composite will show a more general case of material model that goes beyond the isotropic Hooke's law. The most general case of constitutive equations will be discussed then simplified down to the case of transversely isotropic material that will be the material used in this laboratory.

A Composite in engineering sense is any materials that have been physically assembled to form one single bulk without physical blending to foam a homogeneous material. The resulting material would still 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 characteristics of each of the materials.

Usually, composite materials will consist of two separate components, the matrix and the filler. The matrix 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 may be 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 fillers can be of any material such as carbon fiber, glass bead, sand, or ceramic.

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

• Particulate
• Short fiber
• long fiber
• laminate

Particulate composite consists of the composite material in which the filler materials are roughly round. An example of this type of composite would be the unreinforced concrete where the cement is the matrix 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 metal matrix with ceramic filler.

Short and long fiber composites are composites in which the filler material has a length to diameter ratio, l/d, greater than one. Short fiber composites are generally 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. Carbon fiber, 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 particles or fibers. Formica countertop is a good example of this type of composite. The matrix material is usually phenolic type thermoset polymer. The filler could be any 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 different type of loading may call on different component of the composite to take the load. This implied that the material properties of composite materials may be different in tension and in compression as well as in bending. Throughout the lab, the subscript t, c, and b will be used to designate the properties in tension, compression, and bending respectively.

Advantages of composite materials
The main advantage of most composites materials are in the weight savings. A quick way to illustrate this advantage 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 same volume (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. But doing 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 fiber
Carbon fiber can be manufactured from 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 to form honey combs networks similar to graphite. ,



Designation of carbon fiber composites consists of the fiber material follows by the matrix material. For example, AS4/3501-5 means that the carbon fiber used 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 lab will likely be IM7/3501-6.

Layup
It can easily be seen that the long fiber composite will have directionality depending on the direction in which 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 in layers 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 the relative 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 global coordinate of a composite piece. Each layer in that composite may have different fiber orientation that will require a separate coordinate system. The direction 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 coordinate systems coincide in the case of uniaxial composite.


Stress-strain relationships
Material reactions under stresses can be described by a set of constitutive equations. For isotropic material, this is known as Hooke's law or sometimes, in an inverse form, Lamé [la-may] equations. The 3-D Hooke's law in matrix form is:

This Hooke's law is in the compliance form where the strains 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 S is used for compliance and the C is used for the stiffness.

In the most general case, the stress or strain with subscript ij is not the same as the one with supscript ji. 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 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 be symmetric. 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.

The last simplification will give the inverse of the Hooke's law. That is the material is isotropic and has infinite planes of symmetry. The number of constants reduces down to two. They are normally express in term of three interdependent constants E, ν, and G.

or using more familiar constants:

The isotropic case of course can be reduced even further using plane stress or plane strain assumption where the Z direction is ignored. Similar plane strain and plane stress assumption can also be made for the orthotropic and transversely isotropic materials. In this case they would reduce to exactly the same 2-D form.

It should be note that the coordinate system in isotropic material is defined by the loading situation, that is not the case for other materials. In all other cases, the coordinate system is defined by the directionality of the material.

Transversely isotropic material
For the composite specimen used in this lab, the plane containing the 2 direction is the isotropic plane. Therefore, the uniaxial composite is a transversely isotropic material. Five constants are needed to describe this composite. In our coordinate system they are:

Inaddition,

It is necessary to keep track of the subscript of the material properties. The subscript 1 denotes the 1 direction 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, ν₁₂ is the Poisson's ratio in the 2 direction due to the load being applied 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, and bending are different. For example, the modulus in tension, compression, and bending for the 1 direction will all have different values. They will be denoted as E_1t, E_1c, E_1b.

Strain transformation equation
In the case where the stresses or strains are know in a different direction, transformation equations could be used to calculated the stresses or strains in the direction desired. The elementary strength of materials books usually derive the transformation equation using the wedge method. In order to demonstrate that the transformation equation and Mohr's circle are one and 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.

Suppose that strain readings were taken off a strain rosette with three gages labeled a, b, and c. Strains ε_x, ε_y, γ_xy can be found from the three gage readings and three gage positionings.

where, i = a, b, and c.

Rule of mixtures
One quick way to estimate the material properties, i.e., the moduli in 1 and 2 direction of a composite is by using 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.

Composite (laminated) beam
Not 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 will result 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 basic idea 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, M₁ and M₂ with moduli of elasticity E₁<E₂, the expansion factor for replacing M₂ with M₁ is

.

The new beam is now made out of one material and can be solved using the usual method. The first step is to determine the centroid of the cross section.

The neutral axis of this beam goes through the centroid of its cross section. All measurements and calculations 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 of each 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. The stress in the original section is then the calculated stress divide by a factor of n. For the unexpanded section, the stress is as calculated.

The above outline also works when E1>E2. In this case the expanded section is actually contracted. Think of it as an expansion by a factor of less than one and continue.

1. Three strain readings were recorded from a tensile specimen with the corresponding position of the strain gages with respect to X axis. They are:
   
   

   

   Find ε_x, ε_y, γ_xy using the strain transformation equations.

   

   Rewrite strain transformation equation in matrix form:

   

   The three unknowns in the equation above can be solved in any manner. The solution to a matrix equation

   

   is

   

   Hence,

   

   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 the results 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 Π is "pi()".
   
2. Young's modulus in bending is given by,
   

   E = Mc/I ε

   
   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 ε is the strain. For the case of three-point bending beam as shown in Figure 4, express the E in term of P, L, c, I, and ε.
   
   
   
   Figure 4 Three-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 distance x from the end will give the bending moment in term of x. Letting x goes to L/2 will give the expression for the maximum bending moment.
   
   
   
   
   
   Substitute the moment into the equation above:
   

   

3. Derive an equation, in terms of P, L₁, L₂, E, and I, for the deflection, δ_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 of P,L₁, and/or L₂ for this loading condition. (Assume the beam is made from an isotropic and homogeneous material.) You may want to solve the three-point bending case first.
   
   
   
   Figure 5 Four-point bending beam
   
   
   
   Beam equations.
   Region (1) x < L1

   

   Region (2) L₁< x < (L₁+L₂)

   

   There four equations θ₁, θ₂, δ₁, and δ₂ with four unknowns C₁, C₂, C₃, and C₄, altogether. We need four boundary equations (BC) to solve the four equations. They are:
   

   

   
   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 rosettes mounted near but not necessary at the center.
   
5. For the beam shown in Problem 2 solve for the maximum stress using the beam cross-section shown below and knowing that h₁ = h₂ = 0.125 in., b = 1.25 in., L = 10.0 in., and P = 1000 lb. E₁ is that of aluminum (10,000 ksi), E₂ is that of steel (30,000 ksi).
   
   
   
   
   

The term "composite material" can be broadly defined as the resultant of combining two or more materials, each of which has their own unique properties, to form one new material. In a way, we studied composite materials on a microscopic scale when we investigated multicomponent structures in metals, ceramics, and polymers. However, when we speak of engineering composite materials, we generally mean that 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 to nature, such as combining cellulose fibers and lignin to form wood. The advantages of composite materials are that they can be constructed to exhibit the best qualities from their constituents that neither constituent 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 be used 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 combined in certain configurations, composite materials can be manufactured that behave in an orthotropic nature (nine independent material properties) or even in a transversely isotropic nature (five independent material constants). The samples we are going to test have been manufactured so they behave in a transversely isotropic nature. Before we can start the experiment, we need to gain an understanding of some of the terminology used for 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, the horizontal lines represent fibers while the shaded region represents the matrix material. Figure 2 shows a schematic of a typical specimen used in this laboratory. 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.



Besides the fact that fibrous composite materials have different material properties in the 1 and 2 directions, the material can also have different material properties in 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 reasoned by realizing that, in bending, part of the beam cross-section will be in tension while the other part will be in compression. As a result, the neutral axis will shift and will no longer be in the center. Figure 3 shows a representation of the stress and strain distribution for a sample in bending with and without equal moduli in tension and compression. The fact that the moduli can be different leads to several initial observations. The moment of inertia changes due to the neutral axis shift. The correct moment of inertia can be determined using the parallel axis theorem. In addition, the distance from the neutral axis to the outside fibers is no longer equal to half the beam thickness. The last problem of the pre-lab explores these observations. ASTM D790, entitled "Standard Methods of Test for Flexural Properties of Plastics, " outlines methods for testing samples in bending. We will use this standard as a guideline for part of our experiment.

Prelab

1. Three strain readings were recorded from a tensile specimen with the corresponding position of the strain gages with respect to X axis. They are:

      (1)

   Find ε_x, ε_y,γ_xy using the strain transformation equations.

      (2)

2. Young's modulus in bending is given by,

   E = Mc/I ε   (3)

   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 ε is the strain. For the case of three-point bending beam as shown in Figure 4, express the E in term of P, L, c, I, and ε.
   
   
   
   Figure 4 Three-point bending beam.
   
   
3. Derive an equation, in terms of P, L₁, L₂, E, and I, for the deflection, δ_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 of P, L₁, and/or L₂ for this loading condition. (Assume the beam is made from an isotropic and homogeneous material.) You may want to solve the three-point bending case first.
   
   
   
   Figure 5 Four-point bending beam
   
   
4. What advantages or disadvantages does 4-point bending have over 3-point bending knowing our samples have strain gage rosettes mounted near but not necessary at the center.
   
5. For the beam shown in Problem 2 solve for the maximum stress using the beam cross-section shown below and knowing that h₁ = h₂ = 0.125 in., b = 1.25 in., L = 10.0 in., and P = 1000 lb. E₁ is that of aluminum (10,000 ksi), E₂ is that of steel (30,000 ksi).
   
   
   
   
   

Tasks

1. Load a composite specimen (fiber direction) in tension. Record the strains from all the strain gages for 3-10 different applied loads. Report these values in tabular form.
   
2. Calculate E_1t and ν₁₂ from 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 matrix direction. We will refer to this modulus as E_2t. 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. E_1t and E_1c can be calculated using Equation 4 and 5.^[2] M is the maximum moment at the current strain reading. ε_c and ε_t are the absolute value of strains in compression and tension, respectively. w is the specimen width and h is the specimen thickness.

      (4)
      (5)

   Comment on any differences between the tension modulus, E_1t, and compression modulus, E_1c, in the fiber direction. Why might they be different? Is the difference significant? How does this value of E_1t compare with the value calculated in Task 2? Explain the difference in E_2t and E_1t.
   
7. Load the same composite specimen again. This time, record the load deflection history on the strip chart. Using beam theory from the prelab, calculate the bending (flexural) modulus, E_1b, in the fiber direction. Use Figure 3 to determine the moment of inertia. From Figure 3:

      (6)
      (7)
      (8)

   Does your value of E_1b seem reasonable? Explain your answer in detail. It should be noted that you just determine E_1t, E_1c, and E_1b from a single bending test.
   
8. It has been suggested that the flexural modulus can also be determined for a beam of unequal tensile and compressive moduli from the following 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 this result compare to E_1b you calculated previously? Will this equation be applicable if your E_1t and E_1c values are very different?
   
9. It can be shown analytically that the following relationship has to be true for a transversely isotropic material.^[4]

      (10)

   Calculate the ν₂₁ using the tension moduli from Task 4 and 6. How would you experimentally determine the value of ν₂₁?
   
10. The rule of mixtures if often employed as an analytical method of determining the moduli of a uniaxial composite plate just from knowing the properties and amounts of constituent materials used.^[5] Determine E₁ and E₂ using the rule of mixtures. Compare these values to the experimentally determined values.

       (11)
       (12)

    Values and definitions of E_m, E_f, V_m, and V_f will be supplied in the laboratory. Will the rule of mixtures be able to predict the different in the properties in tension and compression?
    
11. Summarize the material properties you determined in the result section of your report.
    

References

1. Jones, R.M., Mechanics of Composite Materials, Hemisphere Publishing Corporation, New York, 1975, p.1.
2. Yu, M., Tarnopolskii, T. Kincis, Handbook of Composites, vol.3, Institute of Polymer Mechanics, Vatvian S.S.R., p. 266.
3. Carlsson ,L.A., Pipes, R.B., Experimental Characterization of Advanced Composite Materials, Prentice-Hall, Inc., New Jersey, 1987, pp. 91-93.
4. Jones, R.M., Mechanics of Composite Materials, Hemisphere Publishing Corporation, New York, 1975, p.38.
5. Jones, R.M., Mechanics of Composite Materials, Hemisphere Publishing Corporation, New York, 1975, pp.90-92.