No idea why this was in my recommendations, but I watched it and it was interesting. Props for Anneke Reinsperger for speaking English without even the hint of a German accent.
I dont work in material science just needed to learn something for work so I landed on your video. Not only the content of the video is informative but I also find how the entire process is explained is very neat! Thank you for sharing!
Mechanical stress sigma is always defined as Force F divided by cross sectional area S (or area A). Or more precise: NOMINAL stress sigma equals force F divided by ORIGINAL cross sectional area S zero. In the video TWO physical quantities are plotted on the vertical axis: Force F on the right side and stress sigma on the left side. Sorry for this, it may be confusing. You get sigma on the left side by simply taking F and dividing it by the original cross sectional area S zero.
"Why does the machine not apply any additional force in the 'luda' (Lüder) area of the curve?": This is an important question, and a research topic of mine, please see our publ. "On the nature of the yield point phenomenon" in Acta Materialia. "And why does the machine apply less and less force right before final fracture?" This is due to "material instability": Here the strain hardening effect is not strong enough to compensate the strength decrease due to the decrease in cross sectional area.
The term "elongation" in ASTM 615 indeed is not specified clearly. To my knowledge it can only mean "percentage elongation after fracture". So one has to put the fragments together and then determine the percentage plastic deformation after fracture has occurred.
"If this rod was a hollow tube or pipe how would the inner diameter be effected? Am I to assume that the point at which it would neck the inner diameter would increase roughly as much as the outer diameter decreases?" We do not have much experience on tensile tests of tubes, but as far as I know, the inner diameter at the neck will decrease, and the outer diameter as well, a bit more than the inner one. So from the side, the broken tube looks similar to the massive cylindrical specimen.
In materials of medium or low ductility, the fracture may take place anywhere along the prismatic part of the specimen, mostly at a weak point. In very ductile materials, the fracture tends to form in the middle region, because there is a certain influence of the grip regions.
"what would be if we take off tension on the middle of experiment for example on 18mm elongation. and apply it again?" Then an elastic spring-back parallel to the elastic straight line would take place, not to the original length, but to the plastically strained length of about 17.8 mm. On reloading, after elastic behaviour the original curve will be resumed, as if nothing had happened.
Please explain to me this. We are increasing the force gradually and observing the extension, right? Then how can the force reduce in the diagram? Now I see: the machine is not increasing the force gradually. But, the length is increased continuously and the force needed is plotted in the graph.
Answer to "Then how can the force reduce in the diagram? ....": Yes, this is the correct explanation. The test is mainly "strain controlled". In a "force controlled test" a drop of force would not be possible.
Abarajithan Gnaneswaran for what I understood the machine is design to increase and reduse the force automatic as long as it keep elongation at costant manner.So that why it can detect that at some point elongation may still proceed at low force input.That is what I think.
Thank you for your interest. We are sorry to state, that there is no Spanish text available, and we would not like to hand out the English or German text.
+ImJuStoOGoOd Answer to "how do you ensure that it fractures right at the centre? what are the parameters to be considered?" In tensile tests with very ductile and the same time very homogeneous materials the fracture tends to be in the central region of the specimen. This is caused by the influence of the specimen ends with a larger diameter. In less ductile materials, the fracture will be at a place with lower strength and/or a place with a defect. This may be at any point within the cylindrical part of the specimen.
Can u please explain why the neck formation occurs?? and if we carried out the same test twice(keeping the material and it's dimensions same) will the neck form at the same point in both the cases?
Answer to "Can u please explain why the neck formation occurs?? and if we carried out the same test twice(keeping the material and it's dimensions same) will the neck form at the same point in both the cases?" An important question. The neck may occur at any point along the cylindrical part of the specimen where the strength is lower than in the rest of the specimen. However, in very ductile materials (such as the steel in our video), the neck tends to be somewhere near the central region of the specimen. Necking takes place, when the increase of force by strain hardening is lower then the decrease of force by the reduction of the cross-sectional area. Sorry, this is very short, the proper explanation is more extensive - and very famous!
i have a problem..........just cant imagine how the speed is constant through out the test where at the beginning the grips dont seem to be moving but at the end they are moving somehow fast
Answer to "i have a problem..........just cant imagine how the speed is constant through out the test where at the beginning the grips dont seem to be moving but at the end they are moving somehow fast": Congratulations for your clear observation! This is one of several "simplifications" in the video. Actually, the whole tensile test should be carried out with a constant speed (constant strain rate). In order to get good results for the elastic part and the upper yield point, the strain rate is low and constant in the first part of the tensile test. If the hole tensile test were carried out with this low strain rate, the test would take a long time. So after a certain strain, it is allowed (even according to the international standards) to "speed up" the test. This has a certain influence on the results, but it is generally accepted. In addition to that, we used a flexible time scaling in the video ... Sorry for the late reply, too much work.
nice video.this video is much more suitable for professionals than students. this is widely used in industries. well we had this experiment done during my engg time.
Answer to "I like so much this video. Congratutions! May I use it for my classes?": Thanks for the praise. I am afraid, that all our videos are under the UA-cam standard license. This means, that no download and no use of downloaded material whatsoever is allowed. However, you can always set links to our videos or show them directly via UA-cam in your classes. Good luck!
Answer to "I am trying to calculate the Stress of this experiment using the formula shown at 8:33. I presume the tube is hollow, if so what is its thickness?" The specimen is completely solid, no tube. The original cross sectional are is the area of the circle, (pi*d^2)/4.
It seems like this would depend on how quickly the force is increased. How would the graph look different if the machine were applying force at 1/4 the rate. Would the neck snap?
kowalityjesus the test machines can be of two types stress controlled and strain controlled, latter being more accurate, and also the rate of application of strain is very less in these type of tests
Hi I had a question. Sometimes the standard will have a limited ruling section listed in it but the dimensions of the actual product can be larger than that. What happens in that case ? How does one comply with the standard.
Answer to: "Sometimes the standard will have a limited ruling section listed in it but the dimensions of the actual product can be larger than that. What happens in that case ? How does one comply with the standard." Different cross sections between the test piece and the actual product are very typical. This problems is solved by calculating the stresses (force divided by cross sectional area) in the test piece and then calculating the forces that the actual product can support.
Excellent video really. So wonderful to see real results. Another question. Did the second material retain permanent plastic deformation near the fracture point? It seems to spring back and not retain necking after break which seems like a cleaner cut. Is this like a somewhat brittle material then?
Answer to "Did the second material retain permanent plastic deformation near the fracture point? It seems to spring back and not retain necking after break which seems like a cleaner cut. Is this like a somewhat brittle material then?": The second material (Al alloy) retains all of the plastic deformation, including the neck. However, it is not as ductile as the first material and the neck is barely noticeable.
Did the testing machine gradually increase the strain/extension in a steady controlled fashion at a pre-determined rate (and hence measure what corresponding force developed), or did it gradually steadily increase the applied force at a steady rate (and hence measure what corresponding strain was developed)? Explain
Answer to "Why we should mark distance with regular intervals?": The marks at regular intervals are not absolutely necessary. Their main purpose in the video is to show the inhomogeneous plastic deformation of the sample. They also make it easier to determine the exact plastic properties. For professional or scientific purposes, the marks are very fine and precise, not as broad as in the video.
Answer to "In some books Elastic limit was given before upper yeild point. So when does the plastic deformation start? After upper yeild point?": This is an interesting question, and to the best of my knowledge not yet perfectly solved. There are some researchers who point out, that there is a very small amount of plastic deformation (which is difficult to measure) already before the upper yield point. It certainly depends on the material.
Answer to "Thank you for the sharing, to get the elongation rate, how do we measure the post-test length?": The post-test length (permanent gauge length after fracture) is commonly evaluated with two methods: (i) By calculating the elastic spring-back after fracture, and then applying it to the elongation immediately before fracture. This is done by the software of most machine manufacturers. (ii) By reassembling the broken test piece and then measuring the permanent gauge length.
Was millileter the force used or was it milipascals? On the second sample of Al did not show the yield strength. I would like to see 7068 (tennalum) test at 99,000 ksi tensil.
If this rod was a hollow tube or pipe how would the inner diameter be effected? Am I to assume that the point at which it would neck the inner diameter would increase roughly as much as the outer diameter decreases?
I googled "luda strain" and "luda effect" and couldn't find any information on it. Did I misspell it or is it a different word from what I heard on the video?
Answer to "I have a question though, is it going to make a difference when changing the rate of loading to the specimen?": Yes, definitely. As a general rule, increasing the strain rate results in slightly higher stresses during the tensile test.
So the same test could be done by suspending weight from the object till failure. Is you were to test rope you could suspend X amount of weight to the rope and gradually add more till failure ?
Answer to "So the same test could be done by suspending weight from the object till failure. Is you were to test rope you could suspend X amount of weight to the rope and gradually add more till failure ?": Loading a specimen with a suspended weight would be a test (without taking dynamic effects into account) with a slowly increasing force, also referred to as force-controlled. A force-controlled test could reasonably be performed on a specimen with a continuously increasing force-strain diagram, such as a brittle material. However, most materials exhibit an up-down or even an up-down-up-down behaviour in the force-elongation diagram, so a force-controlled test would not be reasonable. Instead, the typical tensile test is displacement controlled. So imagine a very strong tensile testing machine that slowly moves the test ends apart until failure occurs. During the test, the force that the test piece resists the extension is measured. Most ropes could actually be tested with a (gradually increased) suspended weight; nevertheless, displacement-controlled testing is better.
Answer to "What is the standard "mark leght value" to mark the gauge length for each steel different sizes?? How to calculate the mark leght?" Internationally, the most common gauge length is 5 times the diameter of cylindrical specimens. As an alternative, also 10 times the diameter may be used; this is what we have used in our video. In some cases, also quite different values are selected.
@@MaterialsScience2000 can you show me an example? I can't understand. Let's say Y16 with 650mm of length. Then what is the mark length from center to center?
Principally, you can use any length to diameter ratio. Most common, however, is the ratio of 5. If your length is 650 mm, then the diameter has to be 650/5 = 130 mm! Obviously, this would be a large sample. So just take a smaller diameter, let us say 10 mm, and a gauge length of 10x5 = 50 mm.
@@MaterialsScience2000 is it the formula is the diameter * 5??? Why must *5?? After the sample fracture, how many gauge length to measure? Is it measure 5 too??
Answer to "What do you mean by yield point phenomenon? Do ductile materials not show yield point for all materials?" Materials in which the yield point phenomenon occurs exhibit a characteristic drop in the stress-strain curve at the end of the elastic region. In contrast, in materials without yield point phenomenon, the stress continually rises during the tensile test (until necking occurs ...). It is the DROP of stress that makes the distinction. However, there are some people who define a yield point as an abrupt change of slope at the end of the elastic region.
thanks you very much for you effort .....I would like to ask you about young's modulus....does its value change after heat treatment for medium carbon steel or in other carbon steel ...and is that change is wide or not... ......because i had .. that changing when i did my research for aisi1030 with hardening, tempering,annealing,and normalizing ....thank you
Answer to "I would like to ask you about young's modulus....does its value change after heat treatment for medium carbon steel or in other carbon steel...?": In the case of steels, young's modulus does not change much with a heat treatment. There probably is a small influence, but this influence is so small, that it is almost impossible to detect it.
Question just to clarify, I had previously understood that a stress vs. strain graph, the stress was the Force DIVIDED BY THE AREA, which here seems to be the yield points... Is stress just the force measure or are these two different graphing methods?
Answer to "Good introduction, however you should have also explained how the extensometer is also used to calculate ductility properties.": You are absolutely right. The main idea of the video is to show the testing procedure. The calculation of the material properties from the "raw" data is very important, and much more extensive than most people think. So we just explain the most important strength properties.
Answer to "Is it possible to have the final cross sectional area? As it wasn't mentioned, it is possible to calculate from any other measure already given? Thx" We have not explained the calculation of the material properties very much and concentrated on the basics. If you wish to, you can measure the final cross sectional area yourself: Stop the video at the appropriate point (when the specimen is placed on the table) and then measure the diameter d at the point of fracture (should be around 5 mm in the case of the steel, and 9 mm in the case of the aluminum alloy). You can then calculate the reduction in area. It is not possible to calculate it from any other value.
Love this process. Crazy how edge and screw dislocations travel in the material. To imagine metal of that thickness being able to do necking and break is so cool. Interesting how material goes from elastic deformation to plastic deformation. It was nice being able to see when the yield point occured on the monitor. What was you factor of safety percentage?
Answer to "What was you factor of safety percentage?" There is no concept of "safety factor" in the tensile test itself. You need safety factors when you use the results of the tensile test to design a mechanically loaded component.
Here is something I'm really struggling to understand, please if someone can answer, I'd appreciate it: How can the displacement of the machine be larger than the gauge measured by the extensometer? I mean, the increase in the gauge... I realize the extensometer starts at 100 and the displacement at 0, but I would expect that gauge would increase at the same rate as the upper clamp moves, rigth? But I see in the video that at a point where the displacement is 25.02 mm, for example, the extensometer indicates, not 125.02, but 121.93 (at 3:40). How can this be possible? Is the answer to this question also the reason why I need an extensometer and I can't just rely on the displacement?
Answer to "Here is something I'm really struggling to understand, please if someone can answer, I'd appreciate it: How can the displacement of the machine be larger than the gauge measured by the extensometer? …": I try to do my best: the displacement of the machine is always greater than the elongation measured by the extensometer because the elongation of the machine frame and the elastic elongation of the specimen grips are included in the machine displacement. Let me explain it (in a simplified way) like this. Imagine a 100 mm long package string attached to a 200 mm long, strong elastic band that is itself attached to another 100 mm long package string. In total you have a rope of 100 + 200 + 100 = 400 mm in length. This is a simple model for the specimen grips and the specimen. One package string is the top grip and the other package string is the bottom grip, the elastic band is the specimen. Now imagine you are lengthening your entire rope from 400 mm to 420 mm, which means an elongation of 20 mm. This is a model for your tensile test. The "machine displacement" is then 420 - 400 = 20 mm. The extensometer measures the length of the prismatic test area, which is in or model the length of the elastic band. The length of the elastic band will be a little less than 220 mm, as the two package cords have their own (albeit small) elastic stretch. This is very simplified, the extension of the machine frame must also be included, and some other effects ...
Answer to "Does cross section area decrease at the point of fracture only throughout the whole experiment (even when sample is in elastic limit)?" The cross sectional area decreases from the beginning of the test. In the elastic region, this decrease is very small and barely noticeable. In the plastic region the decrease is important. The volume of the material here stays constant. The decrease of cross sectional area after fracture, an important material property, does not include the elastic part.
Answer to "In the first test, could you explain why the field "Position [mm]" differs from "Elongation/Weg[mm]" showed in the plot?" The "Position [mm]" is the (relative) position of the crosshead of the machine, measured from the start of the test. The "Weg_F [mm]" is the gauge length in the cylindrical part of the specimen. Delta L in the diagram is the increase of the gauge length in the cylindrical part of the specimen. The difference between "Position [mm]" and "Elongation Delta L" in the diagram is due - to a small gap between specimen and grips at the beginning of the test - to elastic deformation of the test machine and - to elastic and plastic deformation of the specimen outside of the gauge length Sorry for the late reply, I am a "scientific junkie", and very busy and fascinated at the moment by a new research project of mine.
7:33 Do you know guys why the specimen broke at 45 degrees? Because in a uniaxial state of stress (like tensile test) the shear stresses are maximum at 45 degrees w.r.t. the section of the specimen!!! Yeeeaah Science Bitch!
Metals are malleable and ductile materials with high electrical conductivity, high thermal conductivity, and high density. They are widely used in buildings, infrastructures, ships and automobile industry.
after yield point, the video said "the piece is strained uniformly ", what does that mean exactly? (2) since here we use force, but if we consider the stress, do we just take the original area? shouldn't the area decrease with extension?
Answer to "after yield point, the video said "the piece is strained uniformly ", what does that mean exactly? (2) since here we use force, but if we consider the stress, do we just take the original area? shouldn't the area decrease with extension?": (1) Strained uniformly means that the strain within the specimen is the same at any point of the specimen, just as in the case of a rubber band being strained. This is only true in the region between the end of the Lüders strain and the maximum load point. (2) Yes, the cross sectional area decreases during the tensile test. The stress mentioned in the video is the nominal stress or engineering stress. It is always calculated by dividing the force F by the original cross sectional area A_sub_0, so it is not the true stress. The true stress is the force F divided by the true cross sectional area A.
Is aluminium alloy close to Hight tensile steel properties? because i have noticed it's curve like it! ... and is there a difference between Aluminium and Aluminium alloy?
+Hesham Atwa As far as I know some aluminium alloys (aluminium together with other elements in a material) can be created to withstand high forces and have a high tensile strength. However, hard martensitic steels still have the highest tensile strength.
Answer to "what can you conclude about the yield strength of the two specimen tested here?": This is actually the most important question in connection with the tensile test. However, since the video was intended to show the tensile test in practice, we did not cover this important aspect. Both tensile tests have shown that the specimens meet all the requirements, not only the corresponding yield strength, but also the tensile strength and the required ductility. The requirements are specified in the relevant technical standards, in our case in the European standard EN 10025.
It would be valuable to add to the charpy energy testing video to include %shear area determination and may be include a drop weight testing video with %shear area determination.
At final stage of fracture if a straight line occurs with no elongation but drop of force and again increaseslittle bit and fractures means what can we decide from that
That 0.2 % seems arbitrary .. wouldnt it be better to just use that sudden curved point? or similar. Also it would be really great to plot diameter at the same time so we can see if it is contracting. Also the diameters and the stretch may be varying somewhat along the cylinder as the very ends of the cylinder are not free to thin like in the centre.
Answer to "That 0.2 % seems arbitrary .. wouldnt it be better to just use that sudden curved point? or similar. Also it would be really great to plot diameter at the same time so we can see if it is contracting. Also the diameters and the stretch may be varying somewhat along the cylinder as the very ends of the cylinder are not free to thin like in the centre." Absolutely correct, thank you. 0.2 % is the most common plastic strain for the proof stress, and it is arbitrary. If 0.2 % is not suitable for the application of the material, 0.1, 0.05, or 0.01 may be used instead. The "curved point" is not well defined, so this definition is not precise enough and not used in science or practice. And the comment on the diameter: yes, this is correct!
+Walkot2 A bolt generally has an equal area along its length, whereas this test specimen has been made thinner to ensure that fracture will occur there. Bolts will still fail at their threads.
Hi,thx for the video,we are a die casting company,we want to change our micro structure to increase our tensile strength from 300 to say 400,can u tell us what we have to add to A356 aluminium to get increased tensile strength
It''s very interesting to see how things work practically instead of just reading it from books.
Exactly
Totally agree.
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No idea why this was in my recommendations, but I watched it and it was interesting. Props for Anneke Reinsperger for speaking English without even the hint of a German accent.
Very clear speaking.
I searched it to see with my eyes what I was studying (for an exam about polymers)
I dont work in material science just needed to learn something for work so I landed on your video. Not only the content of the video is informative but I also find how the entire process is explained is very neat! Thank you for sharing!
I'm studying for the metallurgy exam at university and this video helps a lot to visualize and really understand how the complete process works.
Grande bro
good job i have the same thing to.
This proved to be very helpful in my engineering practical exams .
🙏🏻
this is the best video on tensile testing and strain on UA-cam
Never quite understood why we always had two different graphs but thanks to this video it's very clear. Great stuff
Mechanical stress sigma is always defined as Force F divided by cross sectional area S (or area A). Or more precise: NOMINAL stress sigma equals force F divided by ORIGINAL cross sectional area S zero.
In the video TWO physical quantities are plotted on the vertical axis: Force F on the right side and stress sigma on the left side. Sorry for this, it may be confusing. You get sigma on the left side by simply taking F and dividing it by the original cross sectional area S zero.
"Why does the machine not apply any additional force in the 'luda' (Lüder) area of the curve?": This is an important question, and a research topic of mine, please see our publ. "On the nature of the yield point phenomenon" in Acta Materialia. "And why does the machine apply less and less force right before final fracture?" This is due to "material instability": Here the strain hardening effect is not strong enough to compensate the strength decrease due to the decrease in cross sectional area.
what can you conclude about the yield strength of the two specimen tested here?
The best video I ever saw on this, keep it up! thank you for showing the curve as the test proceeds..
thank you so much for this video, while my professor has excellent knowledge we learn little so these videos are a lifesaver.
The term "elongation" in ASTM 615 indeed is not specified clearly. To my knowledge it can only mean "percentage elongation after fracture". So one has to put the fragments together and then determine the percentage plastic deformation after fracture has occurred.
I'm a sample machinist for the local steel mill and this is exactly what I machine every day.
This is an insteresting and informative video. I'm a Mechanical Engineering and this information clarifies some concepts.
This essay is great because it exemplifies that not all materials work in the same way.
"If this rod was a hollow tube or pipe how would the inner diameter be effected? Am I to assume that the point at which it would neck the inner diameter would increase roughly as much as the outer diameter decreases?" We do not have much experience on tensile tests of tubes, but as far as I know, the inner diameter at the neck will decrease, and the outer diameter as well, a bit more than the inner one. So from the side, the broken tube looks similar to the massive cylindrical specimen.
In materials of medium or low ductility, the fracture may take place anywhere along the prismatic part of the specimen, mostly at a weak point. In very ductile materials, the fracture tends to form in the middle region, because there is a certain influence of the grip regions.
"what would be if we take off tension on the middle of experiment for example on 18mm elongation. and apply it again?" Then an elastic spring-back parallel to the elastic straight line would take place, not to the original length, but to the plastically strained length of about 17.8 mm. On reloading, after elastic behaviour the original curve will be resumed, as if nothing had happened.
This video is really well made, thank you for making it. Finally some really good videos on materials testing. Thanks!!.
I’m in 6th grade and this was the first advanced thing I’ve learned like this so thx
What happy as same the company in your industry, I am a China company, good video!
Finally some really good videos on materials testing. Thanks!!
Please explain to me this. We are increasing the force gradually and observing the extension, right? Then how can the force reduce in the diagram?
Now I see: the machine is not increasing the force gradually. But, the length is increased continuously and the force needed is plotted in the graph.
Answer to "Then how can the force reduce in the diagram? ....": Yes, this is the correct explanation. The test is mainly "strain controlled". In a "force controlled test" a drop of force would not be possible.
Thanks. :-D
How the needed force is calculated in the region after elastic limit?
Abarajithan Gnaneswaran for what I understood the machine is design to increase and reduse the force automatic as long as it keep elongation at costant manner.So that why it can detect that at some point elongation may still proceed at low force input.That is what I think.
Abarajithan Gnaneswaran it is the resisting force offered by material. Not the external force
This is extremely fascinating. I'm super motivated to finish my remaining education. Working two full time jobs for school has been very tiresome.
well done *claps*
how old were u a year ago
@@philipranjit4586 ok
@@pan4909 wow i was a massive dick a year ago
@@philipranjit4586 lol 1 year ago m8
Best Tensile Test video in internet.
my mind just exploded. This is gonna be a fun semester
cipolla
Thank you for your interest. We are sorry to state, that there is no Spanish text available, and we would not like to hand out the English or German text.
how do you ensure that it fractures right at the centre? what are the parameters to be considered?
+ImJuStoOGoOd Answer to "how do you ensure that it fractures right at the centre? what are the parameters to be considered?" In tensile tests with very ductile and the same time very homogeneous materials the fracture tends to be in the central region of the specimen. This is caused by the influence of the specimen ends with a larger diameter. In less ductile materials, the fracture will be at a place with lower strength and/or a place with a defect. This may be at any point within the cylindrical part of the specimen.
Congratulations for your work, this is very helpful for beginners/students and experts. :)
Can u please explain why the neck formation occurs?? and if we carried out the same test twice(keeping the material and it's dimensions same) will the neck form at the same point in both the cases?
Answer to "Can u please explain why the neck formation occurs?? and if we carried out the same test twice(keeping the material and it's dimensions same) will the neck form at the same point in both the cases?" An important question. The neck may occur at any point along the cylindrical part of the specimen where the strength is lower than in the rest of the specimen. However, in very ductile materials (such as the steel in our video), the neck tends to be somewhere near the central region of the specimen. Necking takes place, when the increase of force by strain hardening is lower then the decrease of force by the reduction of the cross-sectional area. Sorry, this is very short, the proper explanation is more extensive - and very famous!
Too much helpfull practical knowledge
Thanks for your video, it was helped me to finish my final project
Why did I feel kinda bad about the piece? Like, it's gone through a lot
Very informative without any fuzz!
اللهم صلِّ على محمد و آل محمد كما صليت على إبراهيم و على آل إبراهيم انك حميدٌ مجيد.
It is interesting to see how steel can deform under load applications in a test. Thank you for made it
It is a good video, it explains the procedure very well.
Hello friends!!!
Very good video
Big Laik from me.
onion
why the fracture doesn't take place in the middle of the workpiece of the material but happen in one side of gauge length ?
i have a problem..........just cant imagine how the speed is constant through out the test where at the beginning the grips dont seem to be moving but at the end they are moving somehow fast
+samar fouad The speed at which Force is added is constant I guess
Answer to "i have a problem..........just cant imagine how the speed is constant through out the test where at the beginning the grips dont seem to be moving but at the end they are moving somehow fast": Congratulations for your clear observation! This is one of several "simplifications" in the video. Actually, the whole tensile test should be carried out with a constant speed (constant strain rate). In order to get good results for the elastic part and the upper yield point, the strain rate is low and constant in the first part of the tensile test. If the hole tensile test were carried out with this low strain rate, the test would take a long time. So after a certain strain, it is allowed (even according to the international standards) to "speed up" the test. This has a certain influence on the results, but it is generally accepted. In addition to that, we used a flexible time scaling in the video ... Sorry for the late reply, too much work.
Thanks for your care and illustration
tensile test live on utm machine
ua-cam.com/video/h2uIhAifD3Q/v-deo.html
@@MaterialsScience2000 wow thanks for the reply
nice video.this video is much more suitable for professionals than students. this is widely used in industries. well we had this experiment done during my engg time.
I like so much this video. Congratutions! May I use it for my classes?
Answer to "I like so much this video. Congratutions! May I use it for my classes?": Thanks for the praise. I am afraid, that all our videos are under the UA-cam standard license. This means, that no download and no use of downloaded material whatsoever is allowed. However, you can always set links to our videos or show them directly via UA-cam in your classes. Good luck!
I am trying to calculate the Stress of this experiment using the formula shown at 8:33. I presume the tube is hollow, if so what is its thickness?
Answer to "I am trying to calculate the Stress of this experiment using the formula shown at 8:33. I presume the tube is hollow, if so what is its thickness?" The specimen is completely solid, no tube. The original cross sectional are is the area of the circle, (pi*d^2)/4.
It seems like this would depend on how quickly the force is increased. How would the graph look different if the machine were applying force at 1/4 the rate. Would the neck snap?
kowalityjesus the test machines can be of two types stress controlled and strain controlled, latter being more accurate, and also the rate of application of strain is very less in these type of tests
kowalityjesus
thank you so much ,your voice is great ; peace from morocco
This video is really well made, thank you for making it
I'm not even sure how i ended up here but this is fascinating
somebody can explain me why i have that in my suggestions
It is really interesting. Good presentation.
Very good. The video is brilliant
Now thats what u call, THE BEST..!!
This is very well done, it helped my studies.
ma va a cagare
Hi I had a question. Sometimes the standard will have a limited ruling section listed in it but the dimensions of the actual product can be larger than that. What happens in that case ? How does one comply with the standard.
Answer to: "Sometimes the standard will have a limited ruling section listed in it but the dimensions of the actual product can be larger than that. What happens in that case ? How does one comply with the standard." Different cross sections between the test piece and the actual product are very typical. This problems is solved by calculating the stresses (force divided by cross sectional area) in the test piece and then calculating the forces that the actual product can support.
@@MaterialsScience2000 Incorrect answer Bro
@@samm7411 Use higher grade steel
Amazing video helped me a lot with my project.
Very neat! We just went over these concepts in Solid Mechanics.
Thank You for u practical demonstration.
Where i can find the aluminium , copper , Mild steel & brass round shoulder threaded pieces?
onion
Need to make it. (far too late?)
Excellent video really. So wonderful to see real results. Another question. Did the second material retain permanent plastic deformation near the fracture point? It seems to spring back and not retain necking after break which seems like a cleaner cut. Is this like a somewhat brittle material then?
Answer to "Did the second material retain permanent plastic deformation near the fracture point? It seems to spring back and not retain necking after break which seems like a cleaner cut. Is this like a somewhat brittle material then?": The second material (Al alloy) retains all of the plastic deformation, including the neck. However, it is not as ductile as the first material and the neck is barely noticeable.
Gajab nice work 🤗🤗❣️
Did the testing machine gradually increase the strain/extension in a steady controlled fashion at a pre-determined rate (and hence measure what corresponding force developed), or did it gradually steadily increase the applied force at a steady rate (and hence measure what corresponding strain was developed)? Explain
This video is very informative and has good explanation. Thank you
This is so cool, do you have a video with temperature instead of force aswell?
play.google.com/store/apps/details?id=highway.materialtesting&hl=en
must check my android app for highway material app, it may be helpful for you :)
Why we should mark distance with regular intervals?
Answer to "Why we should mark distance with regular intervals?": The marks at regular intervals are not absolutely necessary. Their main purpose in the video is to show the inhomogeneous plastic deformation of the sample. They also make it easier to determine the exact plastic properties. For professional or scientific purposes, the marks are very fine and precise, not as broad as in the video.
In some books Elastic limit was given before upper yeild point
So when does the plastic deformation start?
After upper yeild point?
Answer to "In some books Elastic limit was given before upper yeild point. So when does the plastic deformation start? After upper yeild point?": This is an interesting question, and to the best of my knowledge not yet perfectly solved. There are some researchers who point out, that there is a very small amount of plastic deformation (which is difficult to measure) already before the upper yield point. It certainly depends on the material.
wow!! thx guys. now i got the practical knowledge from your video.
Very informative and very proffesional. Thank you
Thank you for the sharing, to get the elongation rate, how do we measure the post-test length?
Answer to "Thank you for the sharing, to get the elongation rate, how do we measure the post-test length?":
The post-test length (permanent gauge length after fracture) is commonly evaluated with two methods:
(i) By calculating the elastic spring-back after fracture, and then applying it to the elongation immediately before fracture. This is done by the software of most machine manufacturers.
(ii) By reassembling the broken test piece and then measuring the permanent gauge length.
This was highly informative, thank you.
onion
Was millileter the force used or was it milipascals? On the second sample of Al did not show the yield strength. I would like to see 7068 (tennalum) test at 99,000 ksi tensil.
If this rod was a hollow tube or pipe how would the inner diameter be effected? Am I to assume that the point at which it would neck the inner diameter would increase roughly as much as the outer diameter decreases?
More information about yield phenomenon, please. That was very interesting, informative and helpful
I googled "luda strain" and "luda effect" and couldn't find any information on it. Did I misspell it or is it a different word from what I heard on the video?
It's Lüder's effect
Great video with excellent explanation
I have a question though, is it going to make a difference when changing the rate of loading to the specimen?
Answer to "I have a question though, is it going to make a difference when changing the rate of loading to the specimen?": Yes, definitely. As a general rule, increasing the strain rate results in slightly higher stresses during the tensile test.
@@MaterialsScience2000
Oh, so when the strain rate increase we should expect a faster failure in the specimen
Thank you
السلام عليكم
اخوي عندي سؤال ممكن تساعدني ؟
So the same test could be done by suspending weight from the object till failure. Is you were to test rope you could suspend X amount of weight to the rope and gradually add more till failure ?
Answer to "So the same test could be done by suspending weight from the object till failure. Is you were to test rope you could suspend X amount of weight to the rope and gradually add more till failure ?": Loading a specimen with a suspended weight would be a test (without taking dynamic effects into account) with a slowly increasing force, also referred to as force-controlled. A force-controlled test could reasonably be performed on a specimen with a continuously increasing force-strain diagram, such as a brittle material. However, most materials exhibit an up-down or even an up-down-up-down behaviour in the force-elongation diagram, so a force-controlled test would not be reasonable. Instead, the typical tensile test is displacement controlled. So imagine a very strong tensile testing machine that slowly moves the test ends apart until failure occurs. During the test, the force that the test piece resists the extension is measured.
Most ropes could actually be tested with a (gradually increased) suspended weight; nevertheless, displacement-controlled testing is better.
What is the standard "mark leght value" to mark the gauge length for each steel different sizes?? How to calculate the mark leght?
Answer to "What is the standard "mark leght value" to mark the gauge length for each steel different sizes?? How to calculate the mark leght?" Internationally, the most common gauge length is 5 times the diameter of cylindrical specimens. As an alternative, also 10 times the diameter may be used; this is what we have used in our video. In some cases, also quite different values are selected.
@@MaterialsScience2000 can you show me an example? I can't understand. Let's say Y16 with 650mm of length. Then what is the mark length from center to center?
Principally, you can use any length to diameter ratio. Most common, however, is the ratio of 5. If your length is 650 mm, then the diameter has to be 650/5 = 130 mm! Obviously, this would be a large sample. So just take a smaller diameter, let us say 10 mm, and a gauge length of 10x5 = 50 mm.
@@MaterialsScience2000 is it the formula is the diameter * 5??? Why must *5?? After the sample fracture, how many gauge length to measure? Is it measure 5 too??
What do you mean by yield point phenomenon? Do ductile materials not show yield point for all materials?
Answer to "What do you mean by yield point phenomenon? Do ductile materials not show yield point for all materials?" Materials in which the yield point phenomenon occurs exhibit a characteristic drop in the stress-strain curve at the end of the elastic region. In contrast, in materials without yield point phenomenon, the stress continually rises during the tensile test (until necking occurs ...). It is the DROP of stress that makes the distinction. However, there are some people who define a yield point as an abrupt change of slope at the end of the elastic region.
thanks you very much for you effort .....I would like to ask you about young's modulus....does its value change after heat treatment for medium carbon steel or in other carbon steel ...and is that change is wide or not... ......because i had .. that changing when i did my research for aisi1030 with hardening, tempering,annealing,and normalizing ....thank you
Answer to "I would like to ask you about young's modulus....does its value change after heat treatment for medium carbon steel or in other carbon steel...?": In the case of steels, young's modulus does not change much with a heat treatment. There probably is a small influence, but this influence is so small, that it is almost impossible to detect it.
Thank you so much for this video and for your explanation
Question just to clarify, I had previously understood that a stress vs. strain graph, the stress was the Force DIVIDED BY THE AREA, which here seems to be the yield points... Is stress just the force measure or are these two different graphing methods?
Good introduction, however you should have also explained how the extensometer is also used to calculate ductility properties.
Answer to "Good introduction, however you should have also explained how the extensometer is also used to calculate ductility properties.": You are absolutely right. The main idea of the video is to show the testing procedure. The calculation of the material properties from the "raw" data is very important, and much more extensive than most people think. So we just explain the most important strength properties.
Explained in a simple way
Excellent explanation
Is it possible to have the final cross sectional area? As it wasn't mentioned, it is possible to calculate from any other measure already given? Thx
Answer to "Is it possible to have the final cross sectional area? As it wasn't mentioned, it is possible to calculate from any other measure already given? Thx" We have not explained the calculation of the material properties very much and concentrated on the basics. If you wish to, you can measure the final cross sectional area yourself: Stop the video at the appropriate point (when the specimen is placed on the table) and then measure the diameter d at the point of fracture (should be around 5 mm in the case of the steel, and 9 mm in the case of the aluminum alloy). You can then calculate the reduction in area. It is not possible to calculate it from any other value.
Love this process. Crazy how edge and screw dislocations travel in the material. To imagine metal of that thickness being able to do necking and break is so cool. Interesting how material goes from elastic deformation to plastic deformation. It was nice being able to see when the yield point occured on the monitor.
What was you factor of safety percentage?
Answer to "What was you factor of safety percentage?" There is no concept of "safety factor" in the tensile test itself. You need safety factors when you use the results of the tensile test to design a mechanically loaded component.
Good presentation as well as explanation.
Here is something I'm really struggling to understand, please if someone can answer, I'd appreciate it: How can the displacement of the machine be larger than the gauge measured by the extensometer? I mean, the increase in the gauge... I realize the extensometer starts at 100 and the displacement at 0, but I would expect that gauge would increase at the same rate as the upper clamp moves, rigth? But I see in the video that at a point where the displacement is 25.02 mm, for example, the extensometer indicates, not 125.02, but 121.93 (at 3:40). How can this be possible?
Is the answer to this question also the reason why I need an extensometer and I can't just rely on the displacement?
Answer to "Here is something I'm really struggling to understand, please if someone can answer, I'd appreciate it: How can the displacement of the machine be larger than the gauge measured by the extensometer? …": I try to do my best: the displacement of the machine is always greater than the elongation measured by the extensometer because the elongation of the machine frame and the elastic elongation of the specimen grips are included in the machine displacement.
Let me explain it (in a simplified way) like this. Imagine a 100 mm long package string attached to a 200 mm long, strong elastic band that is itself attached to another 100 mm long package string. In total you have a rope of 100 + 200 + 100 = 400 mm in length. This is a simple model for the specimen grips and the specimen. One package string is the top grip and the other package string is the bottom grip, the elastic band is the specimen. Now imagine you are lengthening your entire rope from 400 mm to 420 mm, which means an elongation of 20 mm. This is a model for your tensile test. The "machine displacement" is then 420 - 400 = 20 mm. The extensometer measures the length of the prismatic test area, which is in or model the length of the elastic band. The length of the elastic band will be a little less than 220 mm, as the two package cords have their own (albeit small) elastic stretch.
This is very simplified, the extension of the machine frame must also be included, and some other effects ...
@@MaterialsScience2000 Thank you, thank you, thank you very much for your answer!! It help a lot :)
Does cross section area decrease at the point of fracture only throughout the whole experiment(even when sample is in elastic limit)?
Answer to "Does cross section area decrease at the point of fracture only throughout the whole experiment (even when sample is in elastic limit)?" The cross sectional area decreases from the beginning of the test. In the elastic region, this decrease is very small and barely noticeable. In the plastic region the decrease is important. The volume of the material here stays constant. The decrease of cross sectional area after fracture, an important material property, does not include the elastic part.
@@MaterialsScience2000 thank you for the answer
So, does it pull the metal apart or push it down until it breaks?
In the first test, could you explain why the field "Position [mm]" differs from "Elongation/Weg[mm]" showed in the plot?
Answer to "In the first test, could you explain why the field "Position [mm]" differs from "Elongation/Weg[mm]" showed in the plot?" The "Position [mm]" is the (relative) position of the crosshead of the machine, measured from the start of the test. The "Weg_F [mm]" is the gauge length in the cylindrical part of the specimen. Delta L in the diagram is the increase of the gauge length in the cylindrical part of the specimen. The difference between "Position [mm]" and "Elongation Delta L" in the diagram is due
- to a small gap between specimen and grips at the beginning of the test
- to elastic deformation of the test machine and
- to elastic and plastic deformation of the specimen outside of the gauge length
Sorry for the late reply, I am a "scientific junkie", and very busy and fascinated at the moment by a new research project of mine.
7:33
Do you know guys why the specimen broke at 45 degrees? Because in a uniaxial state of stress (like tensile test) the shear stresses are maximum at 45 degrees w.r.t. the section of the specimen!!! Yeeeaah Science Bitch!
Metals are malleable and ductile materials with high electrical conductivity, high thermal conductivity, and high density. They are widely used in buildings, infrastructures, ships and automobile industry.
after yield point, the video said "the piece is strained uniformly ", what does that mean exactly? (2) since here we use force, but if we consider the stress, do we just take the original area? shouldn't the area decrease with extension?
Answer to "after yield point, the video said "the piece is strained uniformly ", what does that mean exactly? (2) since here we use force, but if we consider the stress, do we just take the original area? shouldn't the area decrease with extension?": (1) Strained uniformly means that the strain within the specimen is the same at any point of the specimen, just as in the case of a rubber band being strained. This is only true in the region between the end of the Lüders strain and the maximum load point. (2) Yes, the cross sectional area decreases during the tensile test. The stress mentioned in the video is the nominal stress or engineering stress. It is always calculated by dividing the force F by the original cross sectional area A_sub_0, so it is not the true stress. The true stress is the force F divided by the true cross sectional area A.
Is aluminium alloy close to Hight tensile steel properties? because i have noticed it's curve like it! ... and is there a difference between Aluminium and Aluminium alloy?
+Hesham Atwa As far as I know some aluminium alloys (aluminium together with other elements in a material) can be created to withstand high forces and have a high tensile strength. However, hard martensitic steels still have the highest tensile strength.
Thanks, tim.
what can you conclude about the yield strength of the two specimen tested here?
Answer to "what can you conclude about the yield strength of the two specimen tested here?": This is actually the most important question in connection with the tensile test. However, since the video was intended to show the tensile test in practice, we did not cover this important aspect.
Both tensile tests have shown that the specimens meet all the requirements, not only the corresponding yield strength, but also the tensile strength and the required ductility. The requirements are specified in the relevant technical standards, in our case in the European standard EN 10025.
It would be valuable to add to the charpy energy testing video to include %shear area determination and may be include a drop weight testing video with %shear area determination.
Thank you. It really helps me a lot for preparing my exam
At final stage of fracture if a straight line occurs with no elongation but drop of force and again increaseslittle bit and fractures means what can we decide from that
That 0.2 % seems arbitrary .. wouldnt it be better to just use that sudden curved point? or similar.
Also it would be really great to plot diameter at the same time so we can see if it is contracting. Also the diameters and the stretch may be varying somewhat along the cylinder as the very ends of the cylinder are not free to thin like in the centre.
Answer to "That 0.2 % seems arbitrary .. wouldnt it be better to just use that sudden curved point? or similar.
Also it would be really great to plot diameter at the same time so we can see if it is contracting. Also the diameters and the stretch may be varying somewhat along the cylinder as the very ends of the cylinder are not free to thin like in the centre." Absolutely correct, thank you. 0.2 % is the most common plastic strain for the proof stress, and it is arbitrary. If 0.2 % is not suitable for the application of the material, 0.1, 0.05, or 0.01 may be used instead. The "curved point" is not well defined, so this definition is not precise enough and not used in science or practice. And the comment on the diameter: yes, this is correct!
It’s a very helpful video. Thanks alot
how can the threaded ends resist that much traction force? will a bolt behave the same?
+Walkot2 A bolt generally has an equal area along its length, whereas this test specimen has been made thinner to ensure that fracture will occur there. Bolts will still fail at their threads.
Hi,thx for the video,we are a die casting company,we want to change our micro structure to increase our tensile strength from 300 to say 400,can u tell us what we have to add to A356 aluminium to get increased tensile strength