INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS
Log In
Come Join Us!
Are you an Engineering professional? Join EngTips Forums!
 Talk With Other Members
 Be Notified Of Responses
To Your Posts
 Keyword Search
 OneClick Access To Your
Favorite Forums
 Automated Signatures
On Your Posts
 Best Of All, It's Free!
 Students Click Here
*EngTips's functionality depends on members receiving email. By joining you are opting in to receive email.
Posting Guidelines
Promoting, selling, recruiting, coursework and thesis posting is forbidden. Students Click Here

Metal and Metallurgy engineering FAQ
Material Properties
Why is the elastic modulus relatively insensitive to changes in chemistry/heat treatment/coldwork? by Maui
Posted: 5 Jun 08 (Edited 6 Jun 08)

The physical basis of material properties like Young's modulus can be understood by examining materials on the atomic scale. There are two main things that influence the value of the modulus:
1.) The atomic microstructure 2.) The interatomic bonds.
Different values are obtained for the elastic modulus depending upon the crystallographic direction in which we measure E. This directional variation in properties is known as anisotrophy. For example, the elastic modulus for a single crystal of iron varies between 41,000 ksi and 19,000 ksi, depending upon the direction of measurement. Tabulated values of E are usually average values taken from polycrystalline materials with a random orientation of the individual grains.
ATOMIC MICROSTRUCTURE All solid materials may be classified as either crystalline or amorphous based upon the way in which the atoms arrange themselves. Crystalline materials are characterized by long range order. This means that the atoms arrange themselves into regular, repeating, threedimensional patterns. The crystals formed by these rather large groups of atoms are called grains. An example of a crystalline structure would be the zinc coating on a galvanized steel sheet. Amorphous solids do not possess any long range order, although they may have short range order. Glass is a good example of this type of material.
The fact that crystalline solids have long range order means that the atom or group of atoms that make up the basic unit of the material must have identical surroundings. If we model the atoms as hard spheres, then we can think of packing them together in a plane as though we were racking a set of billiard balls for a game of pool. The balls are arranged so that they take up the least amount of space. In this twodimensional example this type of plane is called a closepacked plane, and the directions along which the balls touch are called closepacked directions. We could extend this pattern by adding balls until it completely covers the pool table. The important thing to notice is that the balls are arranged in a regular repeating two dimensional pattern.
Now suppose that we start adding balls on top of the first plane that we already arranged. How we position the second plane of atoms is important, because it will determine the type of three dimensional structure that will be produced. The depressions that are formed in the first plane of atoms where three atoms touch are ideal locations for the atoms in the second layer to sit. By dropping atoms into these convenient "seats" we can build a second close packed plane on top of the first one. By adding more planes on top of the previous ones in this way, we find that we can produce a three dimensional structure where the atoms take up the least amount of space. This is an example of a close packed structure. FCC is one microstructure that can be formed using this type of construction.
ATOMIC BONDS The strength of an interatomic bond depends upon the forces that exist between the bonding atoms. From a theoretical standpoint we can determine the force F between two atoms for any separation distance r from the relationship
F = dU/dr
where U(r) is the interatomic potential function. F is zero at the equilibrium point r = ro. Suppose that the atoms are pulled apart to a separation distance r which displaces them from the equilibrium separation distance ro by an amount (r  ro). In this case a resisting force appears. For small displacements from equilibrium (r  ro) the resisting force is proportional to the displacement for all materials in both tension and compression.
The stiffness S of the resulting bond is given by
S = dF/dr = d^2U/dr^2
If the bond is not stretched too far, S is approximately constant and is given by
So = (d^2U/dr^2) evaluated at r = ro
So the bond behaves in a linear elastic manner. This is the physical origin of Hooke's Law. A narrow, steep potential well corresponds to a stiff material with a high modulus. A broad, shallow potential well represents a material with a low modulus. You can think of So as the "spring constant" of this tiny atomic spring. So the force between a pair of atoms stretched apart to a distance r, where r is slightly greater than (or less than) ro, is
F = So(rro)
Now imagine a solid held together by these linear springs, joining two adjacent planes of atoms together. The number of bonds that are formed between the atoms in these two adjacent planes will directly impact the mechanical response that the material has to an applied stress. The greater the number of bonds that are formed per atom (which is a direct result of the atomic microstructure), the more resistant the material is to deforming under load. The stronger the forces that exist between the atoms (which is a direct result of the bonding), the more resistant the material is to deformation. So each of these parameters influences the response of the material to an applied stress. For simplicity, imagine that we are dealing with a material that has ionic bonds, with a simple cubic crystal structure. If our adjacent planes are stretched apart so that they are displaced from their equilibrium separation distance by an amount (rro), then the total force per unit area, defined as the stress, is given by
Stress = NSo(rro)
where N is the number of bonds per unit area. If we draw a simple cubic structure, we find that the atoms are spaced a distance ro apart, with each atom at the corner of a cube. So the average area per atom is equal to ro^2. It follows that N = 1/ro^2. We can convert displacement from equilibrium (rro) into strain by dividing by the initial separation distance ro to obtain
strain = (rro)/ro
so that
stress = (So/ro)*strain
Young's modulus is therefore
E = stress/strain = So/ro
If you know the interatomic potential function, then you can calculate a theoretical value for the modulus based upon this simple model. Keep in mind that the crystal structure that was assumed is simple cubic, which most materials do not possess. And we have also ignored the effects of secondary, or long range bonding as well. Still, the results that are obtained from this calculation are surprisingly accurate when compared to actual measured values of real materials. For the ionic bond, the interatomic potential function can be expressed in the form
U(r) = constant  q^2/[4*pi*epsilon]+ B/r^n
where q is the electron charge, pi=3.1412, epsilon is the permittivity of free space, B is a constant for a particular material, and n is an integer. Differentiating this expression once, setting r=ro, and setting the resulting expression equal to zero we can solve for B. We find
B = [(q^2)*ro^(n1)]/[4*pi*n*epsilon]
So the stiffness of the bond is given by
So = [alpha*q^2]/[4*pi*epsilon*ro^3]
where alpha = n1. Consider the case of NaCl. The coulombic attraction in this ionic bond is a long range interaction that varies as 1/r. Because of this, an Na+ ion not only interacts attractively with its neighboring Cl ions, but it also interacts repulsively with its slightly more distant Na+ neighbors. To calculate So properly, we must sum over all of these bonds, taking both attractions and repulsions into account. Doing this, we find that the result is the above expression for So with a value of alpha=0.58. Substituting in values for the physical constants as well as using a value for the atomic spacing of 2.5 angstroms, we find that So = 8.54 N/m. In the case of NaCl this results in a value for E of 34.2 GPa. The measured value of E for NaCl is 39.96 GPa. So the calculated result is within 15% of the actual value. 
Back to Metal and Metallurgy engineering FAQ Index
Back to Metal and Metallurgy engineering Forum 

Resources
Manufacturing professionals are beginning to explore digital manufacturing technologies in their efforts to reduce cost and improve quality and efficiency of parts and processes. Additive manufacturing is one digital technology which is changing the way we think about the design and lifecycle of manufactured parts. Download Now
The adoption of 3D printing into major companiesâ€™ product development life cycles is a testament to the technologyâ€™s incredible benefits to consumers, designers, engineers and manufacturers. While traditional production methods have limitations in manufacturability, 3D printing provides unparalleled design freedom due to the additive method of building parts layer by layer. Download Now
In this ebook, we discuss five stages of the product development life cycle and key services that help engineers and designers move forward effectively. Stratasys Direct Manufacturingâ€™s advanced manufacturing solutions can help your team accomplish its goals quickly and with ease. Download Now
Keeping your application and its intended use in
mind is key when prototyping. For example, early in
the design process, focusing too greatly on creating
components or products specifically for the additive
manufacturing (3D printing) process used for
prototyping may cause challenges down the road. Download Now

