SA-516-70 at Elevated Temperatures 8

[mdoppler]

I am looking for the Young's Modulus and Yield Stress for SA-516 Gr. 70 at elevated temperatures up to or over 1200F. I can't seem to find it online.
Thank you for any help you can give.

 

[JoeTank]

You will find that information in the ASME Boiler and Pressure Vessel Code, Section II, Subsection D.

Steve Braune
Tank Industry Consultants

http://www.tankindustry.com

 

[metengr]

There is some additional data that might be of use from Dieter,"Mechanical Metallurgy" using the assumption that SA 516 Grade 70 falls under "carbon steel" see page 335;

The Modulus of Elasticity at different temperature for carbon steel is as follws -

room temp 30 X 10*6 psi
400 deg F 27.0
800 deg F 22.5
1000 deg F 19.5
1200 deg F 18.0

I was able to locate actual yield strength data as a function of test temperature for SA 516 Gr 70 plate material, 3/4" - 2" in thickness. This yield strength data is from a Bethlehem Lukens publication on SA 516 and SA 387 steel plates;

@ 100 deg F 55 ksi
200 deg F 50
400 deg F 40
600 deg F ~37
800 deg F ~34
1000 deg F ~30
1200 deg F ~20

 

[metman]

Hi metengr,

I am intereted to see the related info in Dieter but I have an old edition, 1961, so maybe the info is not there. At least it is not on page 335. In what chapter is it located?

Jesus is THE life,
Leonard

 

[metengr]

metman;
It is the 2nd Edition, Part III, in Chapter 9 - The Tension Test. PS; I hope having the 1st Edition is not giving away your age (haha).

 

[metman]

metengr,
Thanks. Would you believe that my anchient edition even has "parts?"
OK you whippersnapper - help me out here a bit: I have been trying to convince a young coworker draftsman that stiffness is largely independent of alloy/temper/strength and Dieter touches on it in the article you referenced.

I showed Mr. M the list of physical/mechanical properties of elements in the front of metals handbook where E is a property of the element (at std temperature)with NO regard to condition (alloy/temper/strength).

The article in Dieter might be more convincing but I need to be clear about the terminology that Dieter uses. "...it follows that the modulus of elasticity is one of the most structure-insensitive of the mechanical properties." Apparently Dieter is saying LATTICE structure as in BCC/FCC/TETRAGONAL etc? OK - just before this, Dieter says, "The modulus of elasticity is determined by the binding forces between atoms. Since these forces cannot be changed without changing the basic nature of the material, it follows that the modulus..." Is Dieter referring to the theoretical cohesive strength of the material in this sentence? And if so, how does this relate to UTS vs E?

For example; E for steel is largely independent of tetragonal structure vs BCC yet UTS is very much dependent upon one or the other or both of these structures. Let's not get sidetracked with YS or plastic instability right now because plastic slip is a very separate mechanism from elastic deformation on the one hand (E) and total separation (UTS)of the "..binding forces.." that Dieter alludes to if in fact Dieter means cohesive strength and maybe that is where my confusiono is.

Probably this old coger needs to completely re-read Dieter
plus Richard's Engineering Materials Science plus Reed-Hill's Physical Metallurgy Principles but I probably won't live that long.

Sorry guys as this probably deserves a separarte thread but then it would lose some flavor and I might get busted as a student disguised as a has-been.

Jesus is THE life,
Leonard

 

[NickE]

metman- you are exactly right, the binding forces btw atoms give the elastic modulus. And likely you wont be able to convince the young guy that you are right.

If I remember MY101 correctly, there is a 1st principals derivation of modulus and why it doesnt change significantly with changes in heat-treat/alloying/strength. I'll see if Ive got it in any of my notes. I think its directly related to sub atomic forces and atom-atom bonding w/in the lattice independant of the shape.

The reason strength is affected by temper/alloy/etc.. is that strength relates directly to dislocation movement. After the yield point the material's properties are now controlled by different mechanisms.

I am also interested in having a good logical proof of this available. I am continually confronted by people who mistake stiffness for strength.

I think we could all think really hard and work back to when we were in school and come up with a good derivation we can make it a FAQ.

nick

 

[Maui]

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 41x106 psi and 19x106 psi, depending on 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, three-dimensional 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 two-dimensional example this type of plane is called a close-packed plane, and the directions along which the balls touch are called close-packed 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. If the atoms are pulled apart to a separation of (r - ro), a resisting force appears. For small displacements (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. I can walk you through a simple example of this type of calculation if you like to show you the relationship between the modulus, the atomic microstructure, and the bonding.


Maui

 

[Maui]

"For example, the elastic modulus for a single crystal of iron varies between 41x106 psi and 19x106 psi, depending on the direction of measurement." Please note that these values are 41,000,000 psi and 19,000,000 psi, respectively.



Maui

 

[NickE]

WOW thanks a bunch Maui... Now I'm going to propose the next part:

The reason that alloying doesnt significantly change modulus is that for most metals the alloying element do not change the bulk poly x-tal structure. And dont alter the overall atomic bonds. (or maybe enough of them)

I'm not sure why heat treat condition doesn't affect modulus thou.

nick

 

[CoryPad]

NickE,

Because you still have a polycrystal composed primarily of iron atoms. The polycrystalline nature provides an "averaging" effect so that the bulk modulus is not dependent on the different modulus tensor values. The heat treatment doesn't change the fact that iron atoms bound to each other have similar atom-atom bond strengths.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[NickE]

Thanks guys thats sorta what I was thinkin.

It's been a long (not as long as some of you thou.) time since these concepts were introduced and derived.

nick

 

[metman]

NickE,
I'm giving you a big star especially for the following:

"The reason strength is affected by temper/alloy/etc.. is that strength relates directly to dislocation movement. After the yield point the material's properties are now controlled by different mechanisms."

OK this makes sense and maui/CoryPad the bulk idea fits well with this picture accounting for minimal affect of processing/structure on E.

Thanks for this insightfull input.

maui---I followed most of your detailed explanation and even though I am weak in differential equations would like you to do the walk-thru:

maui said:
"I can walk you through a simple example of this type of calculation if you like to show you the relationship between the modulus, the atomic microstructure, and the bonding."

This should help to explain maui's opening comment:

"There are two main things that influence the value of the modulus:

1.) The atomic microstructure
2.) The interatomic bonds."

because no. 2.) does not seem to agree with Dieter's comment "...it follows that the modulus of elasticity is one of the most structure-insensitive of the mechanical properties."


Jesus is THE life,
Leonard

 

[metman]

OOPS,

I meant to say because no. 1.)((The atomic microstructure))does not seem to agree with Dieter's comment "...it follows that the modulus of elasticity is one of the most structure-insensitive of the mechanical properties."



Jesus is THE life,
Leonard

 

[Maui]

Metman, I will attempt to do the walk through. As I stated above, there are two main things that influence the value of the modulus:

1.) The interatomic bonds.
2.) The atomic microstructure.

The atoms that comprise a crystalline structure are held together by bonds that behave like linear springs, so long as these bonds are not stretched too far apart. For relatively small displacements from the equilibrium spacing, the stiffness S of the bond is approximately constant and is given by

So = (d^2U/dr^2) evaluated at r = ro

where U(r) is the interatomic potential function and ro is the equilibrium separation between two bonded atoms. 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 is

F = So(r-ro)

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 to a distance (r-ro), then the total force per unit area, defined as the stress, is given by

Stress = NSo(r-ro)

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 (r-ro) into strain by dividing by the initial separation distance ro to obtain

strain = (r-ro)/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 and setting the resulting expression equal to zero we can solve for B. We find

B = [(q^2)*ro^(n-1)]/[4*pi*n*epsilon]

So the stiffness of the bond is given by

So = [alpha*q^2]/[4*pi*epsilon*ro^3]

where alpha=n-1. The coulombic attraction in an 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. This results in a value for E of 34.2 GPa. Not bad. Does this answer your questions?


Maui

 

[NickE]

Put this into a FAQ - or make it really big--->


"Why modulus is relatively insensitive to changes in chemistry/temper/coldwork"

nick
(thanks bunches maui- good work)