Think of it this way. 1) The cutting blade applies the force. 2) The cross sectional area of the cut is the "area" 3) F/A is the shear stress 4) The distance from the support to the position of the blade is "h", the height. 5) The distance the material moves under the force of the blade is the "delta x". 6) delta x / h is the shear strain. 7) If the delta x becomes to large, the material will be cut.
Those cutting machines you referring to. The blade is in this lecture straight. But to reduce the F orce required to cut material the blade have an angle of 87 degrees for the upper blade the bottom blade is 90 degrees. How to calculate the real force required for the machine with a 3degree cutting blade? And also the F orce needed to cut material will ruin the machine itself due stress by keep forcing it to the limit of the max material itself. So is it good to consider a couple percentage more force to destress the machine?
Sir, i have a doubt,in all three cases, you take the Cross sectional Area PERPENDICULAR to the Force applied.... Is it always the case? When we need to find Shear stress, we take Force upon the Area perpendicular to it?
Basic question: The shear plane in the very first example is the top of the cuboid. So shouldn't the Strain be a function of the length of the top portion or it is meant that is a square section where the top is also h
When you make a cut, the inner surface marked "A" will deform before the knife actually cuts through it. If not enough force is applied the knife will not go through the material and only deform like in the example at the top. Thus the areas are the same in the equation. When the knife cuts through the material you exceed the shear strength of the material.
Thanks for the explanation. Maybe I was not clear. Stain = deformation / original length. In the first example, the deformation is dx, and strain is dx/ h. But the plan of deformation is perpendicular to h, so shouldn't it be dx/ b ? Or if I got this right, overall length of plane h is deformed and it increases by dx ?
Amazing lecture. However, I don't get the part of shear strain. How can it be delta x over height, since most of previous lectures considering strain of a string as delta Length over Length for example
you don't seem to mention the material strength. it seems as the shear modulus you define only involves the area and force applied , and lacks material strenth modulus per specific or how to determine this. tyvm for your efforts on this I look forward to viewing all your force/material ;-)
to find the shear strain you need delta x. So how could you find delta x if you were given the dimensions of the object and the force as well as the shear modulus
Loved every minute of it. I wished the shear modulus was calculated formthe last example just for us tomhave annidea but this was suoerbly informative nonetheless. Trillions of thanks,
thank you for your excellent lecture. i have a question about how to calculate shear strain of the application of knife cutting a piece of steel plate.
In the first video strain is given by deformation/original shape and I can see that both with the examples given by the change in length for the wire and the videos on bulk modulus. Should I view shear modulus as a different approach to strain or is there a link that I'm missing when saying the change in length/height represents the deformation/original shape as well?
i dont really understand why the blade that is cutting the sheet is doing a shearing force rather than a compressing force? The way I see it is that the blade compresses the sheet untill it breaks
+A. Kessler If the punch doesn't punch through the metal and dents it, then you would have a distance h where the metal is sloped and a distance delta x equal to the indentation of the dent. When punching through the metal the stress applied is greater than the breaking stress.
+Abdelrahman Wael Helaly The force required to cut a sheet of metal or cut off a bold, or anything like that depends on the shear strength of the material.
People who explain theories should just be as respected as People who invent it !
RESPECT for YOU !
Think of it this way. 1) The cutting blade applies the force. 2) The cross sectional area of the cut is the "area" 3) F/A is the shear stress 4) The distance from the support to the position of the blade is "h", the height. 5) The distance the material moves under the force of the blade is the "delta x". 6) delta x / h is the shear strain. 7) If the delta x becomes to large, the material will be cut.
Complimenting the well organized and illustrated features of this lecture are the clear explanations given by the instructor. Great video.
This is seriously an amazing video, so clear yet so concise MashaAllah, may Allah (God Almighty) bless you. Ameen
what if he is a scientific aethiest that doesnt believe in a God?
It is by definition. That is how the shear strain was defined.
Those cutting machines you referring to. The blade is in this lecture straight. But to reduce the F orce required to cut material the blade have an angle of 87 degrees for the upper blade the bottom blade is 90 degrees.
How to calculate the real force required for the machine with a 3degree cutting blade?
And also the F orce needed to cut material will ruin the machine itself due stress by keep forcing it to the limit of the max material itself. So is it good to consider a couple percentage more force to destress the machine?
thank you for the one of the most clear lectures I ever had
Great Video I understood it lot better than a regular class !!
Great video. Just have a question... I thought young's modulus was stress/strain. In this video Shear modulus is also stress/strain?
Yes, all three the Young's modulus, the bulk modulus, and the shear modules are defined as stress/strain
Sir, i have a doubt,in all three cases, you take the Cross sectional Area PERPENDICULAR to the Force applied....
Is it always the case? When we need to find Shear stress, we take Force upon the Area perpendicular to it?
+Vicky Da Vinci That is a good way to think about it.
Basic question: The shear plane in the very first example is the top of the cuboid. So shouldn't the Strain be a function of the length of the top portion or it is meant that is a square section where the top is also h
When you make a cut, the inner surface marked "A" will deform before the knife actually cuts through it. If not enough force is applied the knife will not go through the material and only deform like in the example at the top. Thus the areas are the same in the equation. When the knife cuts through the material you exceed the shear strength of the material.
Thanks for the explanation. Maybe I was not clear. Stain = deformation / original length. In the first example, the deformation is dx, and strain is dx/ h. But the plan of deformation is perpendicular to h, so shouldn't it be dx/ b ? Or if I got this right, overall length of plane h is deformed and it increases by dx ?
Amazing lecture. However, I don't get the part of shear strain. How can it be delta x over height, since most of previous lectures considering strain of a string as delta Length over Length for example
Shear strain is defined like that. You are talking about normal strain
HI SIR MICHEL VAN BIEZEN, CAN YOU PUT IN YOUR DISCRIPTION A LINK FOR PRACTICE PROBLEMS, PLZZZ
you don't seem to mention the material strength. it seems as the shear modulus you define only involves the area and force applied , and lacks material strenth modulus per specific or how to determine this. tyvm for your efforts on this I look forward to viewing all your force/material ;-)
The shear modules depends on the material
The stress and the strain don't
to find the shear strain you need delta x. So how could you find delta x if you were given the dimensions of the object and the force as well as the shear modulus
You can use tanTHETA if you have all the dimensions, Shear Strain also = tanTHETA which equalls = DX over height or Length
Loved every minute of it. I wished the shear modulus was calculated formthe last example just for us tomhave annidea but this was suoerbly informative nonetheless.
Trillions of thanks,
thank you for your excellent lecture.
i have a question about how to calculate shear strain of the application of knife cutting a piece of steel plate.
In the first video strain is given by deformation/original shape and I can see that both with the examples given by the change in length for the wire and the videos on bulk modulus. Should I view shear modulus as a different approach to strain or is there a link that I'm missing when saying the change in length/height represents the deformation/original shape as well?
Shear stress (even though it has the same units) is somewhat different as compared to the stress seen with Young's modules and the Bulk modulus.
Your drawing skills are great!
Your drawing is awesome, it really looks 3D :)
i dont really understand why the blade that is cutting the sheet is doing a shearing force rather than a compressing force? The way I see it is that the blade compresses the sheet untill it breaks
realy superb lecture. it helps me lot
Thank u sir......great....
U rocks sir best explanation...
Hello, Michel. In your hole punch example, what would be h?
+A. Kessler If the punch doesn't punch through the metal and dents it, then you would have a distance h where the metal is sloped and a distance delta x equal to the indentation of the dent. When punching through the metal the stress applied is greater than the breaking stress.
Ah, I see, thank you for the explanation! My question seems ill worded now, but you managed to get the gist. Thank you for setting me straight 👍
not at all bad in drawing
Thank you Very much! VERY GOOD teaching, really help me out
ALWAYS THE BEST
Hey there thanks sir :)
i didn't get the part where you cut that sheet in half.
can you further explain please ?!
+Abdelrahman Wael Helaly The force required to cut a sheet of metal or cut off a bold, or anything like that depends on the shear strength of the material.
an awesome video , thank you sir .
I really like all of your lecture:) thanks a lot that I could get this video
thank you . it helps me alot
Awesome!
10^10 thanks