Mechanical Properties - Science topic

You mean the zone axes? This is quite surprizing since as far as I know this was one reason to introduce the Hough transform and go away from the originally used zone axes for orientation determination by EBSD. For TEM it seems to be different since the Kossel lines are much sharper than Band edges and deviations from straight lines are even smaller. I guess finally it doesn't change that much whether one uses lattice or reciprocal lattice directions since (hkl)_1 x (hkl)_2 = [uvw] and vice versa. It is more a question what can be better experimentally extracted. Nevertheless, thanks for the reference. Maybe, I am overlooking something...

Pls find attached some references which may be of interest for you

Maybe, you can try the commonly used matching materials , such as epoxy resin, vinylic/acrylic copolymer, cellulose ester etc. These materials have the lower acoustic impedance to be easily used as one-layer or multi-layers matching for achieving the optimum one-fourth or half wavelength acoustic impedance design of your air-coupled transducer

First, you should know the relation between the strength and the grain size is inverse relation.

Sure no, because the ultimate tensile strength increase and the ductility decrease after the cold rolling of the material. The increase of the ultimate tensile strength due to the decrease of the grain size (the relation between the strength and the grain size is inverse relation). However, in order to increase the ductility, reduce the strength and eliminate any stress concentration, the material should be heat treatment such as annealing. This annealing increase the grain size.

Thanks Robert for the article. Now I am comparing the FCC Fe-C binary system to the role of C in the Fe-Mn-C system where Mn-C complexes form and how they influence the climb rate of dislocation. Thanks for your helpful response.

High. you can use nanoindentation to calculate fracture toughness by simply using information like a pdf in below link:

then have in mind that:

Toughness is given by the critical strain energy release rate (Gc) while fracture toughness is given by the critical stress intensity factor (Kc). The strain energy originates from the elastic stresses around the crack. Toughness and strain energy are thus closely related. The formula relating the two is:

Kc = (E'Gc)^0.5

where E' depends on the stress state and is close to or equal to the Young’s modulus.

In your case "composite material" you should define what kind of composite you have . If it is Fibrous Composite or some other kind it is acceptable to use indentation test result as bulk specimen result. If your specimen is inhomogeneous you may repeat the test for different area and samples and use average value as your final result. I hope it help u.

Hope this paper will be helpful for you.

You can try autoignition route. Where chemicals are taken in nitrate and chloride and added some fuel like citric acid. After ignition you get some powder. These powder if you calcine at around 1000 C you will get very fine grained even nano sized powder

The structure of any compound is decided by inter-particle interaction represented basically by two body central potential between every pair of particles; here I presume that three body interactions if present are weak. The two body interaction has two component: (i) repulsive part and attractive part. The inter-particle distance (bond length) is decided by the equilibrium between repulsive and attractive forces. When we replace any atom substitutionally (as you have stated), it is natural that interaction energy of bonds with substituted atom is different than that with host atom that has been replaced. There are two possibilities: (a) the net cohesive energy of the doped system is lower than host or (b) it is higher. In the former case the doped system is more stable than host, while in the latter case it is less stable. In what follows your example corresponds to the former. While increase in bond length corresponds increase in the bond energy, yet the net change in cohesive energy could be negative if new bonds are stronger in their binding in comparison to old bonds in the host. So what is found to be in your example need not be true with every such case. However, if it is always true, it will be a useful information for my perusal.

The austenitic stainless steel grades (the 300 series) are more difficult to machine that the ferritic and martensitic stainless steel grade families. Austenitic stainless steels are more highly alloyed and more prone to higher work-hardening rates. This leads to the need for higher horsepower and more rigid setup. Milling of stainless steel involve generally high cutting forces and powerful machines are necessary. As for turning, it is recommended to use coated carbide tools or inserts. For carbide end milling tools, cutting speeds between 90 and 200 m/min are possible for free-machining grades (like X8CrNiS18-9/1.4305 – AISI 304 grade with high sulphur content) and between 50 and 150 m/min for regular materials (like 1.4301 – AISI 304). Feed rates vary between 0.012 and 0.125 mm/rev/tooth, depending on the cutter diameter. For face milling equipped with carbide inserts, cutting speeds range from 80 to 400 m/min for free-machining grades and from about 50 to 300 m/min for standard alloys. Feed rates are between 0.05 and 0.20 mm/rev/tooth.

Machining setup recommendation for turning austenitic stainless steel AISI 304, depth of cut, in: 0.3;

- High-speed steel tool: speed (fpm): 60, feed (irp): 0.020, tool material AISI M2 and M3

- Coated tool: speed (fpm): 325, feed (irp): 0.020, tool material CC-2

References:

1. Michael McGuire, Stainless Steels for Design Engineers, ASM International 2008

2. P.J. Cunat, Working with stainless steel, Euro-Inox, 1998

I do not have experience in mechanical testing of hollow spheres. However, I found the attached paper interesting in response to your question. The authors utilized the nano-indentation set-up, described above by others, and made it appropriate to test the microballoon. I hope this would help.

First of all the relation between tensile strength(TS) and shear strength(SS) is 0.50 via Tresca and 0.577 vis von Mises. (Not as in the quastion)

The answer is yes for linear elastic analysis

A very good publication to explain the initial question is this:

Ductile–Brittle Transition Temperature of Ultrafine Ferrite/Cementite Microstructure in a Low Carbon Steel Controlled by Effective Grain Size, T. HANAMURA, F. YIN and K. NAGAI, ISIJ International, Vol. 44 (2004), No. 3, pp. 610–617

This paper is a very good read and explaines in terms of the Yoffee diagram, how DBTT is decreased with decreasing effctive grain size (or unit crack path).

If somebody can not access this publication and is interested in more about this, please contact me.

Maybe EBSD could be a easier to use technique in order to get a first impression of your microstructure than TEM. You probably would not find the small martensite needels within your austenitic grains, but the degree of defomation of the grains yould be easily estimated from.

Thank you Sadaf

Do you know which kind of spray is used in making mirrors?

Which are the dimensions of the sheet/plate? If you check the standard E-23 (Notched bar impact testing of metallic materials), in the figure Fig A3.1 you will find the required dimensions for the impact test (Charpy). These dimensions can be from 2.5 mm to 20 mm thickness

Hello Shahram,

I usually use the solver in excel for this kind of problems. I formulate the problem as a least squares minimization problem. And then I use the solver to find the values of the unknowns that minimize the residual. The only problem is to find a good-enough starting point. In my opinion the solver in excel is very robust and easy to use.

You can find Wöhler curves and other mechanical properties of engineering plastics (including PP) in the handbook "Characterization and failure analysis of plastics, ASM International, 2010"

Hi Kateryna, you needs check the quality of the surface, the contact and mainly review again the setup of the indenter,parameters, the apparent contact is up to 1000nm?. also review quality of the tip, and I recommend you start using similar parameters from others researh works.

Niko:

That is true. This can be understood from the Ms: Start Temperature of the Martensitic Transformation [°C] of steel. As per Kunitake,

Ms = 560.5 – 407.3 C – 37.8 Mn – 14.8 Cr – 19.5 Ni – 4.5 Mo – 7.3 Si – 20.5 Cu

As you can see the combined effect of Ni and Mo is to reduce the Ms by a larger amount than Mo alone. This increases the hardenability.

KUNITAKE, T. Prediction of Ac1, Ac3 and Ms Temperatures by Empirical Formulas. Heat Treating (Japan), 41, 2001, p. 164-168.

It is traditionally made in Zinc. The main reason for the choice is it is low cost, corrosion resistant, and melts at low temperature. Because of the complex geometry it is normally made by Hot chamber die casting.

Huang...thanks a lot for offering help. problem is solved.

Paul, just as Daria wrote, temperature will increase the dissociation process (obviously at the initial stages), but your concern would be how the ions and radicals produced after a few machining process influence the subsequent chemical reactions between the material/solution interface (i.e. influence of products of reactions on subsequent chemical reactions) and also, the influence of temperature on the solution/material/reaction products interfaces. This will all depend on the material being machined, the etchant and the by-product of the reactions after a couple of machining. If the by-products of the reactions (ions and radicals) do not impede the reaction process in the presence of temperature, then you can expect a continues increase in machining rate.

I think this book will be helpful: "Atlas of Stress-strain Curves, ASM International".

Hi Ramesh Naidu, thanks a lot for your interest in this question. I copy here a very nice comment from Dr. John Guerra (President and CEO, Nanoptek Corporation):

"We apply strain to titanium dioxide and other semiconductors like silicon and germanium by growing them onto nano-structured substrates . The nanostructures cause a geometrical lattice mismatch, something like epitaxial growth but with just one semiconductor instead of two. And this approach allows much higher strain levels, and therefore bandgap shifts, with good adhesion at the same time. You can see some of our work at our website www.nanoptek.com, and it has also been published in Phys. Rev. B.. We produce real photoanodes in this way for hydrogen production, and we have also extensively modeled the process. The strain is very sensitive to nanostructure size, film thickness, and crystal morphology. "

The material properties of the martensite phase and the Poisson ratio can be easily declared within ANSYS (APDL). However, in the martensite phase the transformation strain upon heating (the SMA) is regained. How can the transformation strain at the martensite phase can be inserted in that case?

My humble submission is that the elastic modulus measurements by tensile/compression testing are not so accurate. The tests often lead to erroneous results. If you are trying to get a good accurate value of elastic modulus of your material for modelling purpose, it is worth trying ultrasonic methods. its much easier method to obtain accurate modulus values. There are several papers are available on this.....the one comes to my mind first is the following:

Kumar A, Jayakumar T, Raj B, Ray KK. Acta Mater 2003;51:2417.

As natural fibers have varying dimensions which are not uniform, at least 10 specimens have to be tested and average taken. Further, the statistical parameter like SD has to be considered for each set of results.

I also agree with Neeraj and Rajashekhara. For our configuration Cu ETP and AA5754 in buttjoint, thickness 0,8mm, the optimum in tensile strength was obtained to the following FSW paramters: 4000 rpm, 200mm/min, 2 degrees, tool displacement of 0,2mm in aluminum. But all parameters depends on your command variables e.g. tensile strength, thermal / eletrical conductivity, thickness of intermetallics.

I did some measurement work with steel shots (round and sharp), brass chips and sand on wheel blasting machine. From my point of view the minimum diameter of steel particles is approx. 0.5 mm. The wheel blades are not very smooth and the acceleration of smaller spheres is poor. Sharp shots accelerate even worse. Glass particles are used in pneumatic driven devices for cleaning of hardly reachable spots. From my point of view glass particles might be used in wheel machines instead of sand, but their field of use is limited to soft material (aluminum) and mostly for making surface less light reflecting.

See the paper "A short and versatile finite element multiscale code for homogenization problems" http://www.sciencedirect.com/science/article/pii/S0045782509001273

and the book "Multiscale Finite Element Methods"

Prof. Bouhala, thank you for your response!

I think that i found the reason, it was just due to the combination of crosslinking and filler effects of the stiff crystalline phase.

I am using Electrochemical discharge machining, as part of my research work for channels and holes with its characterization. So I cannot change the machining process.

Hi Yasin. When Graphene is added to a thermoset, the curing reaction is afected by this addition. The 3 D network that is created by the thermoset will have a higher molecular weight between croslinkings (not entanglements, it is not a thermoplastic). This means that when a tensile stress experiment is performed, the young modulus will be lower and also the tensile strengh will be lower. You will need less force to achive the same strain. All this explanation takes into account only the matrix, not the interface that can drive the mechanical propeties (but this depends on the graphene concentration)

Dimple structure is an important characteristic of the ductile failure, which you can get from Fractography. Recently, the 'damage' variable are quantifies by measuring the void (coalescence) fraction that are formed during ductile deformation [see the following references http://dx.doi.org/10.1016/j.actamat.2012.03.017, http://dx.doi.org/10.1016/j.scriptamat.2009.02.049].

Shear Fracture Area (SFA) is another important parameter often used to define the ductile to brittle transition temperature (DBTT) of a material from a conventional Charpy impact test. This is often used in complementary with the fracture energy measurements for DBTT (ASTM E 23).

The strength of cement is a function of its porosity. As said above, you need a certain amount of water for the formation of the hydration products. But if you add more and more water (increase the w/c ratio), you dilute the cement paste and create more water-filled pore space between the grains, i.e. you find less nuclei for the hydrates in each volume unit. Hydrates have to grow larger and larger to cover the spatial gap (the water) between them and to interact and to develop strength - either physically (interlocked growth) or chemically (e.g. van-der-Waals attraction).

I did something very similar with microseismic waves, in order to estimate the damage parameter in brittle solids. You can take a look to these papers by me (all available on my ResearchGate profile):

E. Ferretti, A. Di Leo, Modelling of Compressive Tests on FRP Wrapped Concrete Cylinders Through a Novel Triaxial Concrete Constitutive Law, «SCIENTIFIC ISRAEL – TECHNOLOGICAL ADVANTAGES», 2003, 5, pp. 20 – 43.

E. Ferretti, A Discussion of Strain-Softening in Concrete, «INTERNATIONAL JOURNAL OF FRACTURE» (Letters section), 2004, 126, pp. L3 – L10.

E. Ferretti, Experimental Procedure for Verifying Strain-Softening in Concrete, «INTERNATIONAL JOURNAL OF FRACTURE» (Letters section), 2004, 126, pp. L27 – L34.

E. Ferretti, A. Di Leo, Cracking and Creep Role in Displacements at Constant Load: Concrete Solids in Compression, «COMPUTERS, MATERIALS & CONTINUA», 2008, 7, pp. 59 – 80.

E. Ferretti, Shape-effect in the effective laws of Plain and Rubberized Concrete, «COMPUTERS, MATERIALS & CONTINUA», 2012, 30, pp. 237 – 284.

Dear Amin.

tank you for your kind suggestion.

Please let me know, why you Consolidated materials in 500MPa at 380C. Do you follow any the article or standard? How can you control grain growth at the mentioned temperature?

What is the question here?

It is straightforward to show that the slip systems you mention exist, simply by considering the number of possible permutations of the indices. For the {110}<1-11> systems, for example:

(110)[1-10] (101)[10-1] (011)[01-1]

(110)[-110] (101)[-101] (011)[0-11]

(1-10)[110] (10-1)[101] (01-1)[011]

(1-10)[-1-10] (10-1)[-10-1] (01-1)[0-1-1]

Note that six of these are simply the other six running in the reverse direction. Whether you consider there to be twelve {110}<1-11> systems or six is a matter of convention.

(Note also that while these systems may be physically distinct, only five of them can be truly independent. For proof of this statement, see Kelly & Knowles, 'Crystallography and Crystal Defects'.)

If you are asking whether all 48 systems can be activated in any given bcc material, that is a very different question. It can only be answered by looking for papers on the material in question.

I suggest you the PEN substrate (polyethylene naphthalate) or PET

Thanks Joachim .....

If you have a high energy laser available, laser shock processing is a well proven method to introduce shock waves inside steels and other methods. You need a few GW/sqcm of energy density introduced on the steel surface, under restricted plasma expansion conditions (that is, for instance, underwater), and you will have a shock wave due to the strong pressure of the ablation plume, which expands at ultrasonic speed against the .

One advantage of this method is that the "projectile" used is just a bunch of photons, so you can monitor the workpiece through it and study precisely the behaviour of front and bottom surfaces. Normally high power Q-Switched nanosecond lasers are used.

I agree with Amin Ezazi, Kevlar has good strength to weight ratio. It is good in low temperatures. Kevlar fabric has wide applications where strength to weight ratio plays major role. Example: Kevlar finds importance in aviation industries as it is good impact resistant and in tailored composites with directional properties.

Van Der Voort lists several "Modified" Keller's Reagents. This one seemed the most appropriate for 6061.

Strain to failure gives the measure of how much the specimen is elongated to failure. By this it means that, it you have strain to failure of 3% measured in specimen of length 100 mm, the material will fail when it it elongated 3 mm, as experimented in tensile test. Therefore, for a material with higher strain to failure rate, the ductility is higher and tensile strength is lower.

I do not see really the question in your statement, but if you want to compute the properties of the aggregate based on the properties of the precipitates/constituents there are many ways to do it the simplest of which is the Rule/law of Mixtures as described here: http://en.wikipedia.org/wiki/Rule_of_mixtures

Also, other alternatives are here:

You should cover your gauges by layer of resin and then by metal foil to have 'closed metal box' with strain gauge inside. Closed metal box will not have magnetic field inside.

Two to four passes are needed to achieve homogeneous distribution without any agglomerations.

Hi, About the powder, SiC has more hardness than Al2O3, I think add SiC to aluminum alloys on FSW process that can increase weld strength. Because add SiC powder, which making dislocation of weld motion difficult. Among to compare of the structure and melting point, Al2O3 is similarly with specimen, relatively the powder of SiC is compete not.

here is an article, I hope that it help you youhttp://www.sciencedirect.com/science/article/pii/S0022460X73803736

Generally, there is. for example, the relation between Young’s modulus and Porosity can be expressed by Nielsen Eq.:

E = (E0 * (1 - P) ^ 2) / (1 + P / (f - 1))

where E is Young’s modulus, E0 is Young’s modulus for pore-less sample, P is volume fraction of porosities, and f is shape factor.

I think you can find the " f " value by fitting the best curve for your results based on mentioned equation.

If the SF appear on each second {111} plane (ABABAB, CACACA or BCBCBC arrangements), this defect structure in austenite forms a hexagonal atomic package of lattice planes (hcp), which is recognised as ε- martensite. The arrangement of the SF on each next {111} lattice plane leads to the formation of the inverse (mirror symmetric) arrangement of the lattice planes CBACBA to the austenitic matrix, that have been detected as (coherent) twin boundaries. Increasing plastic deformation leads to the formation of ’-martensite with the body centred cubic structure (bcc) in the areas of the highest density of microstructure defects. Important factors for the defect interaction are for example, the stacking fault energy, the temperature, the degree of deformation.

Another guess is intensive oxidation of copper by the grain boundary diffusion but I don't know what scheme is used for measuring the heat resitance with respect to FSP trace.

You cannot measure the fracture energy from the results of bend testing alone. This is because there is no information on the size of the flaw that is causing fracture. Griffith's criteria relates the flaw size and the applied stress to the fracture energy. But, you do not know either the flaw size or the fracture energy from a bending test. In order to obtain he fracture energy you must start with a known crack size and measure the critical applied "stress" that results in crack extension.

If you have limited equipment available you could try the indentation toughness method. This requires only a Vicker's microhardness test machine. See papers by B. R. Lawn from the early 1980s. Also note that this will not usually ive an entirely accurate number due to two main issues: 1) the indentation process establishes a residual stress field and 2) many ceramics display R-curve behavior (increase in toughness with increase in crack length) via a number of mechanisms (Marco's answer above alludes to this behavior by invoking a non-negligible process zone).

Note that it is possible to use indentation in initiate a crack of known (measured) size and then perform a bending test to find the critical applied loading that causes material failure. If one assumes this indentation crack is an elliptical crack there are solutions for the KI under uniform far-field tension. Note that the comments on residual stress field around the indent and R-curve behavior are still applicable.

I do agree with Thomas, I have had a lot of data dispersion just by changing a bit the "pre-shearing time", from sample to sample, it would be right consider leaving for some time (e.g. 10 min), the sample on the fixed plate before starting any test, to disipate any previous strain caused by putting the sample onto the plate

@E. Garcia-Sanchez, i had down loaded the software, but not getting properly the working procedure. can you help me in getting the tutorial for this software..pls

Are you working on large scale (macro, >micron) or small scale (nano)? If small scale, it will be anisotropic; if large scale, it could be tricky depending on the grain. You mentioned the grain is big, so it is probably anisotropic as well.

There are two reasons for creation of twin boundaries: phase transformation when cooling from a high-symmetry phase at high temperature and mechanical twinning due to stresses (see the previous article “twinning_mechanisms” by Su Leen Wong). Thermal fatigue will increase the twin density because of the inhomogeneous stress distribution.

If you are sure that there are no other defects, like dislocations or large angle grain boundaries, then twin boundaries in general are considered as not so disturbing for electric conductivity rather than disturbing thermal conductivity. So overall ZT should increase a little, but I don’t know a paper with such evidence.

If you want to avoid twin boundaries at all, you need to switch to a material with higher symmetry, like TiNiSn half-Heusler thin films.

High strength does not necessarily correlate with high wear resistance. Take a look at tribological research on the materials you think are suitable.

First of all, a little of back ground for the answer is needed as:

There are five systems (Vickers, Brinell, Rockwell, Knoop, Scleroscope ) of measuring the hardness in terms of Mohs scale from Hardness Number 1(softest-Talc) to 10(hardest-Diamond). In most of the alloys, the Brinell HardnessNumber [BHN] (predecessor of Rockwell) is calculated from their diameter as:

BHN = 2 P / [pi. D (D - (D^2 – d^2) ^1/2)].

P = load on the indenting tool (kg); D = diameter of steel ball (mm) d = depression diameter at the rim (mm) in the range of 2.5-4.75 mm.

Rockwell Hardness Number (RHN) is measured by indenting test material with a diamond cone penetrator having an angle of 120° with a spherical tip of radius 0.2 mm. The indenter is forced into test material by a preliminary minor load, followed by incremental increase in major load.When a permanent increase in depth of penetration is reached,the RHN =E-e where E is a constant depending on form of indenter and e is permanent increase in depth of penetration by major load measured in units of 0.002 mm

Rockwell testing is of two categories-Regular (C&D Scales); Superficial – (30N&30T).

High RHN represent Hard and the low RHN represent Soft materials.

In regular Rockwell testing, the minor load is apways 10 kgf and the major load can be 60,100,150kgf. RHN is marked as:

Blck Divisions (Represent Brale Tester) and Red divisions (Represent any other Tester)

100(Hardeston Diamond penetrator scale) or130 (Hardest on ball penetrator scale).

HRC 60 represents [H (Hardness); R (Rockwell); C (Type of Scale); 60(kgf)].

So penetration from a minor to major load of 100 to 60 kgf= 40.0.002 = 0.080mm.

Measuring Hardness Number of alloys (Al, Pb,Cu) in terms of their thickness became a common trend because in The International Alloy System, the alloys were marked in many different thicknesses primarily because of their Roof flashing properties where their particle size would not matter much.With the invension of numereous types of steels used for different purposes where the hardness was largely related to their particle size which varies on the surface and in its bulk of the steel, Rockwell invented his method measuring the hardness number in terms the depth (penetration) of the sheets of the steel.Later, he found it to be more advantageous because:

Rockwell testing is rapid, manually operated, takes 5-10sec , identination extremely small which need not be removed by machining ( non-destructive).A Rockwell C scale pentrates up to depth of appxo 0.0035 inches with identination dia. 0.019 inches which is barely visible.So sheets of size as small as .006 inch thickness can be tested.

I think, you need some time to read and learn:

B. E. Warren: X-Ray Diffraction

(http://www.amazon.com/X-Ray-Diffraction-Dover-Books-Physics/dp/0486663175/ref=sr_1_1?ie=UTF8&qid=1380616308&sr=8-1&keywords=B.+E.+Warren%3A+X-Ray+Diffraction)

If you know the basics, you can go into detail, for instance:

Corrosion is a chemical process*. It involves chemical reactions between the material itself and the atmosphere around it. This usually means oxygen (either gaseous or in water), although sulfur can be very corrosive if present. Most corrosion mechanisms are readily characterised and easily understood, but in terms of the chemical properties of the material, NOT its mechanical properties.

Abrasive wear is a much more complex phenomenon. The wear mechanism(s) involved may involve small-scale plastic flow (which may correlate with hardness), localised fracture of asperities (which may correlate with fracture toughness), pullout of grains from the material (which may itself involve plastic processes, or it may involve decohesion at the grain boundaries), or one of several other possibilities. Which of these is seen in any particular case depends on the material itself, the counter-surface it is wearing against, whether abrasion is two- or three-body, the rate of wear, the normal force on the surface and more.

I recommend you find a good textbook on abrasion and wear and try to get some understanding of the basic processes involved and how they might apply to your material. I'm sorry this answer isn't as simple as you might have wanted!

Paresh, the COF can not be a material constant since it depends on many factors such as load, speed, rougness, temperature, adhesion, transfer film and so on.

Also in spite of low COF there may be intense wear due to intensive transfer of PTFE onto metal surface. Therefore, I believe that only tests carried out at conditions close to the real one may be helpful.

There are different test methods that you can use, they are available in ASME and ASM handbooks, for example ASTM B769 tends to obtain shear strength of aluminum alloys, ASTM F606 also obtains the shear strength of bolts and ASTM D732 shows the procedure of obtaining Shear strength of Plastics by Punch tool.As Sylvain told you can also obtain the tensile strength and use the von Mises or Tresca approximation to achieve the Shear strength.

Hi Sarvesh,

The absorption coefficient at the 400-800nm range of nitrate and chloride salts can be found in this dissertation http://dspace.mit.edu/handle/1721.1/62706

Hope it helps,

Georgios

There is a LOT of literature which examines the indentation size effect in great detail. As a starting point, I recommend Pharr, Herbert & Gao in Annual Review of Materials Science, 40, pp271-292 (2010) and the references contained therein.

If you know that TYPE of slip direction, then Sergei's answer is correct. For example, fcc metals slip on {111}<1-10>, but on the (111) plane the directions [1-10], [-110], [10-1], [-101], [01-1] and [0-11] are all possible. You can deduce which will occur in any given loading situation using Schmid's law, an introduction to which is given here:

I believe, however, the question is asking about how to work out the TYPE of slip direction if this is not known. My first suggestion would be to search in the literature for information on similar compounds with the same crystal structure.

If no information is available, the most reliable method involves viewing dislocations under dark field transmission electron microscopy (TEM). Two diffracting conditions in which the dislocation completely disappears are found, and from these the Burgers vector of the dislocation can be calculated. Details of this method may be found in any good textbook on microscopy or dislocations; I can recommend Hull & Bacon, 'Introduction to Dislocations' or Williams & Carter, 'Transmission Electron Microscopy'.

... for cryogenic temperatures:

Central difference for convection does not work unless the cell Peclet no.is less than 2.0

maybe be appropriate.

Considering the three point bend test of ceramics and composites, it is the same.

Dear Akeem

you can describe your PDMS films as a four parameter mechanical model(combination of Maxwell and kelvin model) i,e the parallel or series combination of spring and dash-pot. for your model you can derive the constitutive equation, using this constitutive you can get the relaxation curve and creep curve equation for your described model. simultaneously you need to perform nano indentation test on films and then compare the relaxation curve and creep curve which you obtain from test and mechanical model. by comparison you can find the value of modulus of elasticity and Poisson ratio of the film. for more information you can refer the attachment.

Actually I want to estimate a life of twist drill tool not a drill bit..

And i want to create a machine that can be estimate the life of the twist drill tool....OK

Thanks and regards,

Bipin K. Boghara

Kindly find nptel.iitm.ac.in/courses/.../IISc.../Composite%20Materials/.../LNm11.pdf‎

Assuming linear elastic / SSY conditions at the fatigue crack tip, the analytic expression for the stress field ahead of the crack at the minimum of the load cycle is the well known K-field equation (see e.g. Broek). However you are probably interested in the stress field within the (reversed) plastic zone, for which available solutions are mostly based on FE.

I'm afraid I can't give you a simple answer - I'm not a creep testing expert - but as nobody else has answered this question, let me suggest a couple of things you might want to consider:

Creep occurs by a number of different mechanisms, depending on the stress and the temperature. Do a Google image search for 'creep mechanism map' to get some idea. You tested your material at up to half its yield stress, which puts you firmly in the dislocation creep regime. If the material is to be used at lower stresses, you need to make sure you test in the appropriate regime.

Secondly, a creep test will ideally show you primary and secondary creep, before tertiary creep sets in and the material ruptures. If you test at too high a stress, you may see failure before true secondary creep is established.

Hope that's useful!

i agree with Sergey Bogdanov. but you should also keep in mind that what indentation load you are applying, what indenter you are using, how does plastic deformation take place, what is the young's modulus of your material, what is the crystal size for certain plane, indentation depth, Thermal drift correction of your instrument and etc

Hi Frederik,

thank you for your suggestion. The approach to surface is identical for all three strain rates (10 nm/s). Only after surface contact the loading rate differs for the three strain rates. If there was cracking during indentaion, I should see it on the indentation curve. Indeed, one measurement of 1.25 s-1 strain rate showed a typicall crack progression and was excluded from analysis.

Hello Hai,

Since polyimide is a polymer, you can bend or stretch it. Now, if you use it as a substrate for thin metal films, it would be much easier to study its mechanical properties.

Thanks for the help.

Hello Justin Robbins, the best practical way to measure would be to install resistance strain gauges, I think. However if you only need a rough number and your vessel has a rather big circumference, low wall thickness and you use it at higher pressures you might be able to measure the increase of the circumference before and after loading the vessel. Each 0.1 MPa (1 bar) pressure would contribute for a steel vessel to about 0.6 mm increase in circumference (Radius 1m, wall thickness 5 mm). If you know pressure, size and wall thickness you can, of course, calculate the stresses (since Sigma,hub=pressure*radius/wall thickness; Sigma,cyl= 0.5*Sigma, hub).