I would like to take a step back from this moment of clarity and thank you for taking the time to explain physics in a way that anyone could understand. So with my utmost sincerity "Thank you".
Sir,I am so glad that i came across your channel.Engineering never felt so easy before i watched you teaching.I never saw such a an amazing n enthusiastic professor in my 22 years of education.You the best.
I'm 30, been out of school for 10+ years and have just started an engineering course (online). Your videos are well paced and super clear and I will be using them as a resource throughout my studies. Thank you!!
Why are most physics teachers in college nowadays hopeless. My teacher just runs his own race on the blackboard while everyone are trying to catch up writing it all down. I have to give it it to you Michel, without your videos and the videos of other teachers on UA-cam, my understand of physics would have been way less. Thank you, greetings from Sweden, Örebro
Great, please could you work on this question and explain; Some physical property of a hypothetical material is as follows; Young's Modulus = 2.0 x 10^11 Pa , Poisson's ratio = 0.30 and electrical resistivity = 8.0 x 10^-4 Ωm.Find the change in resistance of a wire made of the material of the given dimensions when a 2.0m length of wire of cross sectional area 5.0mm squared is stretched by a 0.40 KN tensile force
Sir, in case a steel wire and a copper wire are of equal length and equal area of cross section are joined end to end and the combination is subjected to a tension,the tension in both the wires are equal.Why is it so ?
Professor, will the young's module change if the object was heated? For example, Iron would be easier to deform but I don't know if it would necessarily be easier to compress because the atoms would be moving faster so the object would have a greater resistance to compression, right?
Can this calculation also be used for compressive strength, ? Ps: since it’s actually 211Gpa instead of 210Gpa which you used the answer should be 0.02073459 mm.
Compressive strength refers to the maximum load per unit area that can be applied before the material breaks down. The units are force per unit area, which is the same as the units for pressure.
@@MichelvanBiezen I’m trying to work out what the compression (size ) of SiO2 for 1CM3 block at 1.1Gpa (The Compressive strength ) at (4000 Newtons ) My own number gives me (0.36mm (using this Young’s modulus @ the same 1.1Gpa ( the actual young’s is 71.1 Gpa) I get 0.0363636mm , (10mm > 9.64mm or 9.9636364mm ) Edited to give the correct number.
Young's modulus = (F/A) / (delta L / Lo) = P Lo / delta L Therefore delta L = P Lo / Y Is that the equation you used? What are the values you plugged in?
@@MichelvanBiezen 🤦🏼 ignore the 0.011 that’s the Bulk modulus, the correct number is closer to 0.36mm for 1.1Gpa Own - 1.1Gpa M2 > 110,000 cm2 (4000 N) ~ approximately 0.36mm Young (@ 1.1Gpa > 4000 x 0.01. /0.01 / 1,100,000,000 = 0.00000363636, Young(@71.1 Gpa> 4000 x 0.01 /0.01. / 71,000,000,000 = 5.63380282e-8 (approx 56.33 nanometers)
Im not sure about the way you calculate the area, for a 3D rectangle (which is ideally a cuboid), you can get the surface area (= 2(l*h + b*h + l*b)). So how it comes you only multiplay 2cm*2cm? pls help me :) And btw, great videos :)
Michel van Biezen Hey Michael I have a question for you why did you ignore the normal force and didn't calacuate the dL of at then add it to the main force dL
@@MichelvanBiezen I just completed the entire Stress and Strain playlist, thank you so much!! Looking forward to another successful physics semester taught primarily by your lectures haha
+Hein MT Young's modulus is the ratio of the stress to the strain. That is independent on the shape or dimensions of an object, and only depends on the material. Tensile strength is the ultimate "strength" and object has against damage or destruction. It depends on the size and shape of the object, as well as the material it is made of. It answers the question: " How much force can you apply before it breaks?"
Great teaching, thanks! But isn't the unit of the force required to deform a 1 by 1 meter iron beam supposed to be in Newton and not in Pascal? (The last calculation you're doing)
As A= 2cm × 2cm which is = 4 cm^2 and converting into meters area (4×10^-2)^2 then that would be = 0.0016; That would be the area right sir? and P:S 4×10^-2 for converting into meters and square for area
Awesome example on a percentange change, but what i know is that a 1% percentage change of 1.75 is 0.0175 not 0.01 am i wrong ?, so i think the force required was going to be 84 000 000N,I think Sir just calculated a change of a strain as 0.01% if i'm wrong please explain.
sizwe mbokazi The equation and answer is correct on the video. 1% = 0.01 F required = 840,000N Take a close look at the equation and see what 0.01 represents in the equation.
PLEASSSSSSSSSSSSEEEEE TEACH AT AUSTIN PEAY STATE UNIVERSITY........ cause you are my savior!!!!! I wouldn't be lost in class if you were my teacher hell. I may even like physics!!!!!
I don't understand squaring numbers when they are below 0. If you square 0.2m (20cm) it makes 0.04 (4cm) when really it would be 4 metres? Can you explain why you left the decimals as decimals and whether this creates inaccuracies
I would like to take a step back from this moment of clarity and thank you for taking the time to explain physics in a way that anyone could understand. So with my utmost sincerity "Thank you".
Sir,I am so glad that i came across your channel.Engineering never felt so easy before i watched you teaching.I never saw such a an amazing n enthusiastic professor in my 22 years of education.You the best.
In the portion of the problem where I use the blue color, (delta L / Lo = 0.01) so you don't have to divide by 1.75.
I'm 30, been out of school for 10+ years and have just started an engineering course (online). Your videos are well paced and super clear and I will be using them as a resource throughout my studies. Thank you!!
how is your engineering course going on?
Online? Where? I am 33
Every time I need to brush up on something, or fill in a gap, there you are. Thank you very much!
You are amazing, thank you so much :). You're the reason why I currently have an A in my general physics class!
I finally have a perfect understanding of stress and strain. Awesome explanations as usual.
Thank you. I have been at my homework for hours and I finally figured it out with the help of this.
Tell me again why I need my professor xD Thank you Michel!! You are awesome.
Thank you professor! No words to say, just you are the Man!
I believe the Force at the end of the video should be in N, not Pa. Thanks a lot for these videos. Michael is an awesome teacher.
You are correct. That should have been "N"
Was over-thinking it. Thanks for breaking it down Barney-style.
Why are most physics teachers in college nowadays hopeless. My teacher just runs his own race on the blackboard while everyone are trying to catch up writing it all down. I have to give it it to you Michel, without your videos and the videos of other teachers on UA-cam, my understand of physics would have been way less.
Thank you, greetings from Sweden, Örebro
I remember similar experiences when I went to school. Glad to be of help and welcome to the channel!
glad to know that , someone who knows how we suffer haha
very clear and understandable
best video of YOUNG MODULUS
Thanks for sharing these educational stuff ,helps me alot in my work. Thanks again. (Embossing on mild steel )
Vishal (mechanical engg,India)
Great, please could you work on this question and explain; Some physical property of a hypothetical material is as follows; Young's Modulus = 2.0 x 10^11 Pa , Poisson's ratio = 0.30 and electrical resistivity = 8.0 x 10^-4 Ωm.Find the change in resistance of a wire made of the material of the given dimensions when a 2.0m length of wire of cross sectional area 5.0mm squared is stretched by a 0.40 KN tensile force
Thanks! My lecturer gives confusing Symbols to represent change in length, force and Young's modulus..i understood yours clearly, thanks again.
Sir, in case a steel wire and a copper wire are of equal length and equal area of cross section are joined end to end and the combination is subjected to a tension,the tension in both the wires are equal.Why is it so ?
Great lecture videos. Very useful.
the video was very use full it made me to realy understand
Professor, will the young's module change if the object was heated? For example, Iron would be easier to deform but I don't know if it would necessarily be easier to compress because the atoms would be moving faster so the object would have a greater resistance to compression, right?
Can this calculation also be used for compressive strength, ?
Ps: since it’s actually 211Gpa instead of 210Gpa which you used the answer should be 0.02073459 mm.
Compressive strength refers to the maximum load per unit area that can be applied before the material breaks down. The units are force per unit area, which is the same as the units for pressure.
@@MichelvanBiezen I’m trying to work out what the compression (size ) of SiO2 for 1CM3 block at 1.1Gpa (The Compressive strength ) at (4000 Newtons ) My own number gives me (0.36mm (using this Young’s modulus @ the same 1.1Gpa ( the actual young’s is 71.1 Gpa) I get 0.0363636mm , (10mm > 9.64mm or 9.9636364mm )
Edited to give the correct number.
@@MichelvanBiezen ps: if I used the actual number for young’s I get approximately 56.33 nanometers.
Young's modulus = (F/A) / (delta L / Lo) = P Lo / delta L Therefore delta L = P Lo / Y Is that the equation you used? What are the values you plugged in?
@@MichelvanBiezen 🤦🏼 ignore the 0.011 that’s the Bulk modulus, the correct number is closer to 0.36mm for 1.1Gpa
Own - 1.1Gpa M2 > 110,000 cm2 (4000 N) ~ approximately 0.36mm
Young (@ 1.1Gpa > 4000 x 0.01. /0.01 / 1,100,000,000 = 0.00000363636,
Young(@71.1 Gpa> 4000 x 0.01 /0.01. / 71,000,000,000 = 5.63380282e-8 (approx 56.33 nanometers)
how find young modulus from stress-strain diagram
Is there any modulus which I can apply for no solid materials?Let say yarns from fibers?
There are for some materials, but the modulus is typically used for solid materials.
shouldn't the area be 0.04m instead of 0.02?
Neither. The area is (0.02m)^2 = 0.0004 m^2 as indicated in the video.
Yes, that is what I thought because area= b*h= 2*2=4 4cm = 4*10^-2m = 0.04m not (0.02)^2
Mr. Can ask some question realted problem if you don't mind????🙏
We try to answer all questions as time permits.
The last example is 2,100,000,000N = 21 x 10^8 Newtons right?
+Ace Chester Bernabe
Yes, I forgot to cancel out the m^2.
thank you very much!
i thought i was wrong hahaha!!!
at least i am learning XD
the last example .. F= 21*10^8 (Pa.m^2) = 21*10^8 N
Im not sure about the way you calculate the area, for a 3D rectangle (which is ideally a cuboid), you can get the surface
area (= 2(l*h + b*h + l*b)). So how it comes you only multiplay 2cm*2cm? pls help me :) And btw, great videos :)
We are calculating the cross sectional area.
Thank you for helpful explanation,,but i have a dot!!! why did you divided 0.02 with (21x10) 10 ??
9:50 i guess there a mistake in the unit at this time , it must be N
That is correct. The proper unit for force is Newton. Thanks for bringing that to my attention.
Michel van Biezen Hey Michael I have a question for you why did you ignore the normal force and didn't calacuate the dL of at then add it to the main force dL
Hi Thanks for the help.
im still having trouble getting the same answer as you. Can you explain how to enter it into a calculator please?
thanks
Let me know what you entered into your calculator.
Hi Thanks for your reply. i figured it out i didnt use brackets.
thanks
(referencing the first problem) I was a bit confused by A = 2cm x 2cm, wouldn't that make the area 4cm?
Yes, it is written as: (2 cm)^2 = 4 cm^2
@@MichelvanBiezen I just completed the entire Stress and Strain playlist, thank you so much!! Looking forward to another successful physics semester taught primarily by your lectures haha
keep it going
so what is the difference between tensile strength and young's modulus?
+Hein MT Young's modulus is the ratio of the stress to the strain. That is independent on the shape or dimensions of an object, and only depends on the material. Tensile strength is the ultimate "strength" and object has against damage or destruction. It depends on the size and shape of the object, as well as the material it is made of. It answers the question: " How much force can you apply before it breaks?"
+Michel van Biezen Thank you :)
+Michel van Biezen how do we get those Young's modulus value?
is it necessary to make the answer to be mm?
+paco gor
It doesn't matter. You can express it in any units.
Great teaching, thanks!
But isn't the unit of the force required to deform a 1 by 1 meter iron beam supposed to be in Newton and not in Pascal? (The last calculation you're doing)
hybenbabs,
You are correct. That should indeed be Newtons not Pascals.
Ok. And once again, thanks for the great lectures!
As A= 2cm × 2cm which is = 4 cm^2 and converting into meters area (4×10^-2)^2 then that would be = 0.0016; That would be the area right sir? and P:S 4×10^-2 for converting into meters and square for area
+nikhil monarch
4 cm^2 = 0.0004 m^2 (move the decimal 4 spaces)
Why not use n/mm2 instead of n/m2 for easy to see a whole number instead of decimal or with exponential form..
Because Young's modulus is expressed in standard units.
please make more mechanics videos,very helpful thank you
Fantastic video!!!
Thank you so much :)
Awesome example on a percentange change, but what i know is that a 1% percentage change of 1.75 is 0.0175 not 0.01 am i wrong ?, so i think the force required was going to be 84 000 000N,I think Sir just calculated a change of a strain as 0.01% if i'm wrong please explain.
sizwe mbokazi The equation and answer is correct on the video.
1% = 0.01
F required = 840,000N
Take a close look at the equation and see what 0.01 represents in the equation.
Michel van Biezen I saw my confusion thaks!
Thanks for the video, the last calculation involved working out force so unit should have been Newtons not pascal.
Sajid Alam
Oops. You are correct.
thanks hehe your vids are really helpful! :)
Thanks a lot for your lecture which can help me a lot
Great Video!!
man , i love you ;D
Isnt it 20.83 mm?
It depends on how many significant figures you have. If they are not specified, we usually keep it at 3.
Michel van Biezen i mean your answer is 0.028 mm.isnt that equivalent to 2.08x10^-3?
2.083 x 10^-5 m = 0.02083 mm
PLEASSSSSSSSSSSSEEEEE TEACH AT AUSTIN PEAY STATE UNIVERSITY........ cause you are my savior!!!!! I wouldn't be lost in class if you were my teacher hell. I may even like physics!!!!!
I don't understand squaring numbers when they are below 0.
If you square 0.2m (20cm) it makes 0.04 (4cm) when really it would be 4 metres? Can you explain why you left the decimals as decimals and whether this creates inaccuracies
Scott Latter Scott,
check your calculations, there are a few errors.
(0.2m)^2 = 0.04m^2 (20cm)^2 = 400 cm^2 they are both the same
i don't mean to sound dumb but can you explain how they are the same please?
doesn't 400cm = 4m? not 0.04m?
Scott Latter cm^2 = cm x cm = cm squared
there are 10,000 cm^2 in 1 m^2
therefor 400 cm^2 = 0.04 m^2
thank you for clarifying, i appreciate it
Very good no words
For the second F you computed you wrote Pa as its unit instead of Newton. Apart from that, very good video
Theo Bernier
You are correct.
keep the good work
the young modulus is wrong for Al, Brass and Glass!! it must be divided on 10
There are decimal points there, thus the numbers as written are correct. The dots are small and cannot be seen clearly. Thanks for checking.
Explained so well here
VOW EXPLAINATION CONTINUE
Glad you liked it. 🙂
Tq sir
oh wow no dislikes
rukhsana faiz there are no dislikes because here come only people who like to learn