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Properties of metals at Cryo temp 4

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DanStro

Mechanical
Dec 11, 2004
391
Hello all,

Can anyone recommend a site,book, aricle,... that could explain why some materials are better at cryo temps than others?

I guess I would just like to get a basic understanding of the effect of the very low temperatures. I have found some posts/sites that basically say this is good (mainly low carbon steels and aluminum) or this is bad (high carbon steels) but I haven't come up with anything that explains why. Is it based mainly on the ductile-brittle transfomation temperature of the material or is it much more complicated?

Any help is appreciated.
Dan


 
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It basically comes back to crystal structure. In steels there is a low temperature brittle phase (martensite) which forms from the austenite. In some steel groups, the austenite is converted to another phase structure, that prevents marteniste forming, so the are not susceptible to a ductile - brittle transformation. There is no equivalent low temperature phase formed in aluminum, so they are also not susceptible to a ductile - brittle transformation.
 
To add to the above post, the low temperature fracture properties of metals are governed by chemical composition, grain size, and heat treatment, to name just a few of the important contributors. Here is a basic explanation of this behavior for metals;

The single contributor that effects low temperature fracture behavior of metals is chemical composition. The element or elements that are added to a metal determines its crystal structure, which will effect fracture behavior at low and high temperatures. Heat treatment of certain metals will also effect crystal structure, and fracture behavior.

The crystal structure in metals that results in poor fracture toughness at low temperature is a body centered cubic (BCC) lattice structure. This lattice structure pertains to the arrangement of atoms in the metal. The reason why a BCC lattice structure has inherently poor fracture toughness at low temperature is that there are a limited number of slip systems in comparison to other metal crystal structures like FCC (face centered cubic) or HCP (hexagonal close packed). When a metal is subjected to stress, it can deform elastically and plastically (permanent) because planes of atoms that make-up the lattice structure can move. If a crystal structure can allow planes of atoms to move relative to each other (slip), you will not cleave or separate the planes of atoms. If the slip systems are limited in number (as with the BCC structure) or become ineffective as the temperature of the metal is lowered, the planes of atoms will cleave (break bonds and separate) instead of slip. As planes of metal atoms separate from cleavage, this results in cracks and brittle fracture behavior.

Metals like austenitic stainless steel, copper and aluminum that have FCC or HCP lattice structures have increased slip systems operative at even low temperatures - so you don't exhibit brittle fracture behavior.

With the above background, the two articles below explain low temperature fracture behavior;


and

 
A little more info is here:

thread330-66521

One minor correction to metengr's post - the HCP structure has few slip systems at low temperatures and is more like BCC than FCC regarding low temperature mechanical properties.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
>>To add to the above post, the low temperature fracture properties of metals are governed by chemical composition, grain size, and heat treatment, to name just a few of the important contributors. Here is a basic explanation of this behavior for metals;

While very true these factors are very important to low temperature behaviour, they are only suppressing the transformation temperature. To totally avoid such transformations you need to choose metals that have no martensite transformations, no matter how sluggish.
 
"To totally avoid such transformations you need to choose metals that have no martensite x-formations, no matter how sluggish."

What if they form soft martensite?

What if I have a structure composed completly of Ferrite and Pearlite, that contains a relatively large amount of carbon? (Such that the steel could if quenched rapidly form martensite.) Would I then see a Ms and Mf? If I cooled low enough would I begin to form martensite?

I just dont understand why you seem to be insistant that the ductile to brittle transition temp is an intrinsic property of materials. I thought that all parameters that have to be empirically measured are extrinsic? While properties that are inherent such as crystal structure, density, etc that can be derived from first principals are intrinsic.

 
Consider a perfectly elastic material, such as glass, which contains a sharp crack. The crack grows when the atomic bonds at the crack tip are broken. Work must be done to break these bonds and separate the atomic planes. The total energy required to form a crack of length 2a is

W = 4a*(gamma)

where 4a is the total surface area of the crack per unit thickness and gamma is the surface energy of the material. As the crack grows, more surface area is created and so more work is done by the applied forces. The total energy per unit thickness required to produce a crack of length 2a under an applied tensile stress is

Wtotal = 4a*(gamma) - (S^2)*pi*a^2/E

where S is the applied stress, pi=3.1416 and E is the elastic modulus. The condition for unstable crack growth is obtained by taking the derivative of Wtotal with respect to crack length a and setting the resulting expression equal to zero (taking the derivative with respect to either a or 2a results in the same answer). We find

dWtotal/da = 0

4*(gamma) - 2*pi*S^2*a/E = 0

S = [2*E*(gamma)/(pi*a)]^0.5

or
S*[(pi*a)^0.5] = (2*E*gamma)^0.5

As the length of a pre-existing crack increases, the stress required for fracture decreases. This equation is known as the Griffith criterion for fracture. It often appears in the form

S = [E*Gc/(pi*a)]^0.5
or
S*[(pi*a)^0.5] = (E*Gc)^0.5

where Gc is called the critical strain energy release rate, or the total work of fracture. This equation can be used to predict the critical values of stress and crack length that are required for a crack to grow in an unstable manner. When the term S*[(pi*a)^0.5] reaches the critical value (E*Gc)^0.5, the crack will begin to grow. In this context, it is convenient to treat S*[(pi*a)^0.5] as a measure of the driving force for crack propagation. It is common practice to define

K = S*[(pi*a)^0.5]

as the stress intensity factor. Fracture occurs when the stress intensity factor K equals or exceeds the critical stress intensity factor KIC where

KIC = (E*Gc)^0.5

KIC is usually referred to as the fracture toughness.

For most metals, the measured values of the critical strain energy release rate Gc are two to four orders of magnitude greater than the surface energy term. The reason for this difference is that the stresses at the crack tip cause localized yielding to occur before fracture. Yielding extends over a region called the plastic zone, within which tensile stresses are comparable to the yield stress. The local stress ahead of a sharp crack in an elastic material is given by

Slocal = S + S*[(a/2r)^0.5]

where r is the radial distance from the crack tip, and S is the applied stress. We can estimate the radius of the plastic zone ry by setting Slocal equal to the yield stress Sy. If r << a, then to a good approximation,

ry = (S^2)*a/[2*(Sy^2)]

ry = (K^2)/[2*pi*(Sy^2)]

Note that the radius of the plastic zone decreases rapidly as the yield strength increases. Cracks in relatively soft materials produce a large plastic zone compared to cracks in hard ceramics which result in a small plastic zone, or none at all.
Metals often contain alloying elements or impurities that form non-metallic inclusions. If inclusions are contained inside the plastic zone, then plastic flow will occur around them. As plastic deformation continues, elongated cavities form around these inclusions ahead of the crack tip. As these cavities link up, the crack tip advances by means of this ductile tearing. The plastic flow at the crack tip turns the initial sharp crack into a blunt crack, and a great deal of energy must be expended to make the crack grow. This is why the measured value of Gc for metals is much higher than the theoretical values. Since Gc is high, so is KIC. This is why metals are so tough.
The fracture surface of a ceramic or glass tends to be very flat and shiny as opposed to the dull, rough fracture surface of a ductile metal. This occurs because ceramics and glasses have very high yield strengths. Very little plastic deformation takes place at the crack tip, and even if a small amount of crack tip blunting is allowed for, the stress at the crack tip still exceeds the ideal strength of the material. The resulting crack grows between a pair of atomic planes giving rise to an atomically flat surface. This is called cleavage. The energy required to break the atomic bonds is much less than that absorbed by ductile tearing, and this is the reason why materials like glass are so brittle. It also explains why some steels become brittle and fail like glass at low temperatures.
At low temperatures, metals with BCC and HCP microstructures become brittle and fail by cleavage even though they may be tough at room temperature. Only metals with an FCC microstructure are unaffected by temperature in this way. In metals that do not have the FCC microstructure the motion of the dislocations is affected by the thermal agitation of the atoms. As the temperature drops, the thermal agitation decreases and dislocations become much less mobile than they are at room temperature. This increases the yield strength, which causes the plastic zone at the crack tip to shrink until it becomes so small that the failure mechanism changes from ductile tearing to cleavage. This effect is called the ductile-to-brittle transition. This is the reason why many of the Liberty ships that were constructed during WW II failed. When the ships were exposed to the cold water temperatures in the North Atlantic the welds in the ship’s hulls underwent a ductile-to-brittle transition. Many of the ships split in two under their own weight while they were docked. I hope that this gives you some insight into this type of phenomenon.


Maui
 
Maui -- you amaze me again (As in the modulus' insensitivity to alloying elements or heat treat) with your ability to concisely explain metallurgical phenomena with a great use of 1st principals and hard science. A star and a Thank You. I remember talking about this topic in several classes, It just become so hazy when it's not used.


nick
 
Thanks Nick! Your positive feedback is always welcome.


Maui
 
Maui - it is interesting that HCP is susceptible to ductile-brittle transition type fracture and FCC is not, when they are so much closer in arrangment than HCP is to BCC. Have you any explanation available?
 
SMF1964-- The answer to your question goes back to CoryPad's answer and reply. There are fewer slip systems available for dislocation movement in the HCP and BCC crystal systems. At low temperatures the lack of slip systems in BCC and HCP lattices causes a change from failure by ductile tearing to cleavage. Cleavage is normally a brittle fracture mechanism.

(I hope I summarized well.)
 
>>What if I have a structure composed completly of Ferrite and Pearlite, that contains a relatively large amount of carbon? (Such that the steel could if quenched rapidly form martensite.) Would I then see a Ms and Mf? If I cooled low enough would I begin to form martensite?

Discusssing in terms of steels, but also relevent to other allow systems. Martensite will not form from the pearlite or ferrite phase, ferrite and cementite and hence pearlite is the stable low temperature state of iron (with carbon).

However, martensite will form from the meta stable austenite steel phase, as martensite is just another meta stable phase with a lower free energy than the autenite, as the temperature is reduced, the crystal structure attempts to change to a lower free energy state. At low temperature the atom movement within the lattice becomes extremely limited so the large movements required to change from FCC to BBC won't occur, but shorter range movements and distortions can and do occur to form the Body Centered Tetragonal of martensite.
 
"However, martensite will form from the meta stable austenite steel phase, as martensite is just another meta stable phase with a lower free energy than the autenite, as the temperature is reduced, the crystal structure attempts to change to a lower free energy state. At low temperature the atom movement within the lattice becomes extremely limited so the large movements required to change from FCC to BBC won't occur, but shorter range movements and distortions can and do occur to form the Body Centered Tetragonal of martensite."

Right, I understand the mechanism for the phase change btw austenite/martensite.

What got my attention is that you seemed (and I am likely wrong) to imply that materials that undergo the ductile to brittle transformation is a result of an accompanying martensite transformation.

IE: If I have steel (only for example since its easy) with a structure composed entirely of ferrite and pearlite. No Retained Austenite, No tempered martensite, No bainite. This steel has a BCC crystal structure, hence it has a ductile to brittle transformation temp. From this: "In some steel groups, the austenite is converted to another phase structure, that prevents marteniste forming, so the are not susceptible to a ductile - brittle transformation." I thought you were implying that the ductile to brittle change is due to the martensite X-form, not the fact that BCC has few slip systems available so that the failure must occur by cleavage.

Nick
I love materials science!
 
I've probably said it poorly. Although other steel phases may be brittle or may lose their ductility with decreasing temperature, to the best of my knowledge they do not suffer a ductile to brittle transition which is temperature dependent. This phenomena is restricted to the austenite / martensite phase transformation.
 
AlanD.
All BCC metals and alloys, whether they be iron based, tungsten, molybdenun, niobium, etc. undergo a ductile-to-brittle transformation with decreasing temperature. The transformation temperature range is primarily dependent on chemistry, microstructure, grain size and inclusion morphology.

 
My apologies everyone, I've been barking up the wrong tree. Please disregard everything I've posted in this thread.
 
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