1. Properties and Applications of Composites & Nanocomposites
Arya Ebrahimpour
Professor, Civil and Environmental Engineering
PSCI 640 Elements of Nanosciences, November 9, 2009

2. Outline of the Lecture Introduction
Engineering Properties of Materials
Composite Materials
Carbon Molecules
Nanocomposites
Applications of Nanocomposites & Smart Materials
Will use PowerPoint and the board

3. Introduction
Predictions involving applications of nanotechnology (Booker & Boysen):
By 2012 significant products will be available using nanotechnology (medical applications including cancer therapy and diagnosis, high density computer memory, …)
By 2015 advances in computer processing
By 2020 new materials and composites
By 2025 significant changes related to energy

4. Engineering Properties of Materials
Normal stress is the state leading to expansion or contraction. The formula for computing normal stress is:
Where, s is the stress, P is the applied force; and A is the cross-sectional area. The units of stress are Newtons per square meter (N/m2 or Pascal, Pa). Tension is positive and compression is negative.
Normal strain is related to the deformation of a body under stress. The normal strain, e, is defined as the change in length of a line, DL, over it’s original length, L. P L DL A

5. Engineering Properties, cont.
Young's modulus of elasticity (E) is a measure of the stiffness of the material. It is defined as the slope of the linear portion of the normal stress-strain curve of a tensile test conducted on a sample of the material.
Yield strength, sy, and ultimate strength, su, are points shown on the stress-strain curve below.
For uniaxial loading (e.g., tension in one direction only): s = E e s 1 E
Stress, s
Strain, e su sy
Rupture

6. Engineering Properties, cont.
Shear stress, t, is the state leading to distortion of the material (i.e., the 90o angle changes). The corresponding change in angle, in Radians, is called shear strain, g. The slope of the linear portion of the t-g is called shear modulus of elasticity, G. 1 G
Stress, t
Strain, g g t

7. Engineering Properties, cont.
Poisson’s ratio, n, is another property defined by the negative of the ratio of transverse strain, e2, over the longitudinal strain, e1, due to stress in the longitudical direction, s1. 1 2 s1
Original shape e2 e1

8. Engineering Properties, cont.
Isotropic Materials have properties that do not depend on the orientation of the coordinate system (xyz). That is,
E1 = E2 = E3, G23 = G31= G12, & n12 =n21 =n13 =n31 …
Isotropic materials can be fully described with only two (2) of the three material constants (E, G, and n).
Examples of isotropic materials: steel, aluminum, … 1 2 3

9. Engineering Properties, cont.
Anisotropic materials have different properties in different directions. In the most general case, they are defined by 21 independent constants. Special cases include:
Orthotropic: wood and some composites
Transversely isotropic: some continuous fiber reinforced composites
Fibers

10. Engineering Properties, cont.
A 3D stress element
Stresses and Strains in 3D:
Knowing that tij = tji, we have six
independent stresses:
s1, s2, s3, t23, t31 , and t12
For shear stresses, the first index is the
plane number and the second is the direction.
Stresses in terms of strains or vice versa are given by:
Where, [C] is the stiffness matrix and the [S] is the compliance. matrix. Vectors {s} and {e} represent both normal and shear stresses and strains, respectively.

11. Engineering Properties, cont.
Stresses in terms of strains:
Strains in terms of stresses:

12. Composite Materials
Composites consist of two or more materials in a structural unit. There are four types:
Fibrous composites (fibers in a matrix)
Continuous fibers, woven fibers, chopped fibers
Laminated composites (layers of various materials)
Particulate composites
Combinations of some or all of the above

13. Composite Materials, cont.
Engineering Applications: Composite materials have been used in aerospace, automobile, and marine applications (see Figs. 1-3). Recently, composite materials have been increasingly considered in civil engineering structures. The latter applications include seismic retrofit of bridge columns (Fig. 4), replacements of deteriorated bridge decks (Fig. 5), and new bridge structures (Fig. 6).
Figure Figure Figure 3
Figure Figure Figure 6

14. Composite Materials, cont.
Medical Applications: Stents are made with steel and more recently with polymers with shape memory effects (Wache, et al.).
The material is deformed within a temperature range of glass transition temperature (Tg) of amorphous phase and melting temperature (Tm) of crystalline phase, then was cooled below Tg. After the material was reheated between Tg and Tm, the original structural shape was recovered. High dosage (up to 35% by weight) and at a high rate of release of medication were noted in this study.

15. Composite Materials, cont.
Fabrication Process
Open mold, spray-up
Open mold, hand lay-up
Roll-forming process
Pultrusion process

16. Composite Materials, cont.
Fabrication Process
Sheet-molding compounds (SMCs) are used extensively in the automobile industry.
Machine for producing SMCs
Compression molding process

17. Composite Materials, cont.
Lamina: Basic building block of a laminate consisting of fibers in a thin layer of matrix.
Laminate: Bonded stack of laminae (plural of lamina) with various orientations.
Note: Unlike metals, with composites we can design the structure and the material that goes with it.

18. Composite Materials, cont.
Glass fiber versus bulk glass:
Strength Ratio = 3400/170 = 20
3400 MPa
170 MPa
< 1mm
Griffith’s measurement of tensile strength as a function of fiber thickness (Gordon, J.E., The New Science of Strong Materials, 1976)

19. Composite Materials, cont.
Behavior of orthotropic vs. anisotropic materials:
In orthotropic (and special case of isotropic) materials, shear-extension coupling (SEC) and shear-shear coupling (SSC) terms are zero. That is, if you pull on the material, it will not distort.
For example, for an orthotropic material, if we let all stresses other than s1 be zero, then we have no shear strain, g12, as shown below:
g12 = S16s1 + S26 s2 + S36 s3 + S46t23 + S56t31 + S66t12 = S16s1 = 0

20. Composite Materials, cont.
Taking advantage of coupling in composites: In the forward-swept wings of Grumman X-29 aircraft, bending and twisting coupling was used to eliminate the aerodynamic divergence (gross wing flapping that tears off the wings).

21. Composite Materials, cont.
Orthotropic material compliance matrix can be expresses in terms of the previously defined materials properties Ei, Gij, and nij .
Note that the SEC and SSC terms are zero.
SEC
SSC
Because of symmetry, of the matrix, we have:

22. Composite Materials, cont.
Example 1:
Given: The unidirectionally-reinforced glass-epoxy lamina shown has the following properties: E1 = 53 GPa, E2 = 18 GPa, n12 = 0.25, G12 = 9 GPa. The load P is applied in the 1-direction. Note: This lamina is orthotropic.
Find:
a. Determine strains e1 and e2 under the force P.
b. What are reasonable values for E3 and n13?
c. Based on the values in Part (b), find e3.
d. What are the final dimensions of the lamina?

23. Composite Materials, cont.
Predicting stiffness E1 using Rule of Mixtures

24. Composite Materials, cont.
Predicting stiffness E1
Load sharing is analogous to a set of springs in parallel (see figure on the left)
Figure on the right shows the predicted vs. measured values

25. Composite Materials, cont.
Predicting stiffness E2

26. Composite Materials, cont.
Predicting stiffness n12 and G12

27. Composite Materials, cont.
Example 2:
Given: A unidirectional carbon/epoxy has the following properties:
Ef = 220 GPa, Em = 4 GPa, and Vf = 0.55
Find: a. Estimate the value of the composite longitudinal modulus E1
b. Estimate the value of the composite transverse modulus E2
c. If fiber Poisson’s ratio nf = 0.25 and nm = 0.35, find the lamina n12
d. Assuming that the fiber and the matrix behave individually as
isotropic materials, estimate G12
e. What Vf is needed to obtain composite E1 that matches stiffness
of aluminum (E = 69 GPa)?

28. Composite Materials, cont.
Predicting Composite strength
Function of individual stiffness, strength, and strain values at the points of failure
Will go over an example, if time permits.

29. Carbon Molecules Graphite versus Diamond
Graphite: Used as lubricant and pencil lead is composed of sheets of carbon atoms in a large molecule. Only weak van der Waals’ forces hold the sheets together. They slide easily over each other.
Diamond: Carbon atoms stacked in a three-dimensional array (or lattice), giving a very large molecule. This gives diamond its strength.
Graphite sheets
Diamond structure

30. Carbon Molecules, cont.
Graphite sheet is a molecule of interlocking hexagonal carbon rings. Each carbon bonds covalently with three others, leaving one electron unused. The orbital for these “extra” electrons overlap, allowing electrons to freely move throughout the sheet. This is why graphite conducts electricity.
Structure of a sheet of graphite

31. Carbon Molecules, cont.
Buckyballs were discovered by Smalley (Rice University), Kroto and Curl in 1985 by vaporizing carbon with a laser and allowing carbon atoms to condense.
A buckyball is short for buckmisterfullerene after Buckminster Fuller, an American architect and engineer, who proposed an arrangement of pentagons and hexagons for geodesic dome structures.
It has 60 carbon atoms in a ball shaped with 20 hexagons and 12 pentagons and has a diameter of about one nanometer.
A buckyball

32. Carbon Molecules, cont.
In 1991, carbon nanotubes (CNTs) were discovered by Sumio Iijima of NEC Research Lab. After taking pictures of buckyballs in an electron microscope, he noticed needle shaped structures (i.e., cylindrical carbon molecules).
Single-wall carbon nanotubes (SWNTs) versus multiwalled carbon nanotubes (MWNTs)
The length of CNTs vary, but the smallest diameter seen in SWNTs is about one nm.
A single-walled carbon nanotube

33. Carbon Molecules, cont. CNT
A scanning electron microscope (SEM) image of a CNT hanging off the tip of an atomic force microscope (AFM) cantilever.

34. Tensile Strength (MPa)
Carbon Molecules, cont.
Strength (su), stiffness (E modulus), and density of common materials
Material
Tensile Strength (MPa)
Tensile Modulus (GPa)
Density (g/cm3)
6061 Aluminum (bulk)
310 69
2.71
4340 Steel (bulk)
1,030
200
7.83
Nylon 6/6 (polymer) 75
2.8
1.14
Polycarbonate (polymer) 65
2.4
1.20
E-glass fiber
3,448 72
2.54
S-2 glass fiber
4,830 87
2.49
Kevlar 49 aramid fiber
3,792
131
1.44
T-1000G carbon fiber
6,370
294
1.80
Carbon nanotubes
30,000
1,000
1.90
From: Gibson, R.F., 2007

35. Nanocomposites, cont.
Nanofibers and MWNTs: hollow tubular geometries with aspect ratios (L/d) ranging in the thousands.
Material
Diameter
(nm)
Length
Young’s Modulus
(GPa)
Tensile Strength
Vapor-grown carbon nanofibers
10-200
30, ,000
SWNT
~ 1.3
500-40,000
13-52
From: Gibson, R.F., 2007
10 mm
Scanning electron microscope image of vapor-grown carbon nanofibers in a polypropylene matrix
300 nm
Image of MWNTs in a polystyrene matrix

36. Nanocomposites, cont.
Challenge: Unlike fibers in conventional laminates, waviness of the nanotubes and nanofiber reinforced materials complicates the material property calculations. Representative volume elements (RVEs) may be modeled as shown below:
Waviness is defined by the waviness factor,

37. Nanocomposites, cont.
Predictions of the Young’s modulus of elasticity:
The modulus of the RVE2 (the right diagram in the previous page), Ex= ERVE2, and the effective modulus for randomly oriented nanotubes, E3D-RVE2, have complex formulas, but are both are functions of the waviness factor. E3D-RVE2 as functions of nanotube volume fraction and w, is shown below.

38. Nanocomposites, cont.
Combinations of nanoparticles and conventional continuous fibers:
Nanoparticle reinforced matrix
Conventional continuous fiber

39. Nanocomposites, cont. Example 3:
Given: A unidirectional carbon/epoxy lamina with Ef = 220 GPa, Em = 2 GPa, and Vf = 0.55 is also reinforced with randomly placed carbon nanotubes with volume fraction, VNT, equal to 25% of the matrix. Assume nanotube waviness factor of 0.05.
Find:
a. Estimate the value of the composite longitudinal modulus E1
b. Estimate the value of the composite transverse modulus E2
Hint: Use the given graph of E3D-RVE2 in place of the complicated formulas.

40. Nanocomposites, cont.
Strength prediction: In general, relations for predicting strength are complex. However, for randomly oriented fibers, an approximate equation may be used to estimate the tensile strength, as follows (Gibson, 2007):

41. Applications of Nanocomposites & Smart Materials
Shape Memory Alloys (SMAs) are used in reconstructive surgery where sustained pressure is needed for faster healing process. Nickel and Titanium alloy developed by Naval Ordinance Laboratory (named Nitinol).
Shape memory phase changes
SMA plate connected to a jaw bone

42. Applications of Nanocomposites & Smart Materials
Tether between two space outposts for providing artificial gravity (Scientific American)

43. Applications of Nanocomposites & Smart Materials
Carbon nanotube reinforced polymer composites for structural damping
Application: large amplitude vibrations of space structures
Damping loss modulus values for polycarbonate with and without nanotube fillers (Ajayan, et al, 2006)

44. References
University of Alberta’s Smart Material and Micromachines web site (November 6, 2007), available from:
Ajayan, P. M., Suhr, J., and Koratkar, N., (2006). “Utilizing interfaces in carbon nanotube reinforced polymer composites for structural damping, Journal of Material Science, 41,
Suhr, J., Koratkar, N., Keblinski, P., and Ajayan, P. (2005). "Viscoelasticity in carbon nanotube composites,” Nature Materials Vol. 4.
Gibson, R. F., Principles of Composite Material Mechanics, Second Edition, CRC Press, 2007.
Jones, R. M., Mechanics of Composite Materials, Taylor and Francis, 1999.
Advani, S. G., Processing and Properties of Nanocomposites, World Sciences, 2007.
Booker, R. and Boysen, E., Nanotechnology, Wiley Publishing, 2005.
Scientific American, Understanding Nanotechnology, 2002.