sir there is compression along one diagonal or elongation along another diagonal then how you take AC AS ORIGINAL LENGTH in both compression and elongation
Sir waise necessary nahi thha ki hum "Cube" hi consider kare yaha because humko bas ek square cross-section chahiye and we can get it even if we have a cuboid (rectangular prism) which has one set of parallel faces(at which shear deformation is happening) as squares. Amazing video Sir thank you so much 🙏
Perhaps I missed a crucial point in the explanation but it seems to me that it was assumed without proof that the tensile stress along the face diagonal AC is the same as the shear stress applied over the top face of the cube (both of these different types of stress denoted by T). I believe it is indeed the case that they are equal but I don't think it was shown or mentioned in the presentation.
The right way to explain is as follows: You need to consider cross-sectional area and calculate the force along the diagonal line. The force (and stress) in diagonal should be calculated by equating 2*tau*sin(45)*L=p*L*sqrt(2); where p is stress along the diagonal. Anyway this will give p=tau.
Hello Bharat ! Thank you for following us. You may click the following links to watch the related videos. (1) Relation among Young's modulus, Bulk Modulus and Modulus of rigidity (2) Relation among Young's modulus , Bulk modulus and Poisson's ratio And as always ... thanks for watching.
@@dopeTV007 if this is true then by trignometry (let the length of diagonal be D)Dcos45 and Dsin45 should give sides of rectangle which should be the same acc to you
Hello Ashok Kumar ! I am not an Engineer. I am a Physicist. I think before I teach engineering stuff I have to work very hard. And I shall work hard if it helps you. Leave a comment #ShouldIWorkHard Thank you for your support and as always ... thanks for watching.
Sir video ke thumbnail me thoda mistake hai, (1+sigma) ke jaga (1-sigma) likh diya hai apne... Video lekin ekdam zabardast hai...
Thank you so much sir❤️. I didn't understand it from my university but it is very clear now from your video
Sir why we add alpha T AC and bita TAC ,i donot understand please explain me soon
12:25 mein jo likha hei wo jara samajh nehi a rahi hei sir??
excelent video! Sir, what about Relation among Young's modulus , Lamé Constant(λ) and Poisson's ratio. Could you please solve it.
Thanks a lot sir.Your explanation way is best.
sir there is compression along one diagonal or elongation along another diagonal then how you take AC AS ORIGINAL LENGTH in both compression and elongation
Well। Done sir.।।।।। Salute to uuu.। 🙏🙏🙏🙏
Sir , excellent...i understood clearly🙏🙏
young modulus of a material is 1254 pa and the poision ratio is 0.25 the modulus of rigidity of a matterial is??
Thank you very much sir aapne bahut aachha explain kiya hai sir👌👌
sir bsc maths ka bhi channel banaye hme bhut jrurat hai thanks
Sir waise necessary nahi thha ki hum "Cube" hi consider kare yaha because humko bas ek square cross-section chahiye and we can get it even if we have a cuboid (rectangular prism) which has one set of parallel faces(at which shear deformation is happening) as squares.
Amazing video Sir thank you so much 🙏
thank you so much sir for your amazing explanation this helps me a lot
Perhaps I missed a crucial point in the explanation but it seems to me that it was assumed without proof that the tensile stress along the face diagonal AC is the same as the shear stress applied over the top face of the cube (both of these different types of stress denoted by T). I believe it is indeed the case that they are equal but I don't think it was shown or mentioned in the presentation.
Really helpful thank you soo much sir
Very well explained 😇
Thanks a lot sir💓💓
Thank your sir🙆♀️😊🤗🙆♀️
Thank u...for such...a simple derivation...
*Sir thumb nail me equations galti he. (-ve) ka place me (+ve) likha he*
Rc'=T*AC*alpha?????
Alpha is not along the diagonal..
We should take that stress whic is along the diagonal..
How u eqaute that equal to alpha?????
The right way to explain is as follows: You need to consider cross-sectional area and calculate the force along the diagonal line. The force (and stress) in diagonal should be calculated by equating 2*tau*sin(45)*L=p*L*sqrt(2); where p is stress along the diagonal. Anyway this will give p=tau.
it is very easily explained..... thank you sir....
Sir, what about bulk modulus......?
Hello Bharat !
Thank you for following us. You may click the following links to watch the related videos.
(1) Relation among Young's modulus, Bulk Modulus and Modulus of rigidity
(2) Relation among Young's modulus , Bulk modulus and Poisson's ratio
And as always ... thanks for watching.
Can you please make more videos of bsc physics??
amzing and more understandable video
in rectangle diagonals also gives 45 angle
Hello terra aliens !
Is it ?
I think 'no'.
And as always .... thanks for watching.
how all the angle of vertices of rectangle is 90. then daigonal devide it so it become 45. if no what is the angle then. brother
@@dopeTV007 if this is true then by trignometry (let the length of diagonal be D)Dcos45 and Dsin45 should give sides of rectangle which should be the same acc to you
sir it's very helpful .... keep uploading this types of topics on physics part 2 and also from mathematics
sir easily explain 👌👌
Thank you sir 😊
highly educative and useful .. :)
Tell me the value of modulus of rigidity for perfectly elastic body
Sir apne sin ka formula galat likha h
Apko sin45 ke jagah cos45 likhna chahiye tha
sir aap kamal hoo👏👏👏
sir I am confused between the formulas of lateral and longitudinal in Stress,strain and poisson ratio.
+Naveen Singh
everything is there.
Aap galat ho sin ke ghaga par cos hoga at 8:00 see.
Thank you so much sir
sir why are u Rc'=a.T.Ac +b.T.Ac I think it will be ( - ) insted of +
I requested sir, Young modulus, Bulk modulus and Modulus of rigidity. En teeno topics ko alg alg se define kijiye SIR, PLZ...........
Available. Par bahut purana video hai.
Hello just comment to make it century
Too good🤠
Please sir video upload on depression at the centre of beam loaded at middle and supported ends(lab method and light beam)
+prepation of Isc
Yes.
Thanks a lot sir ☺
more video strength of material ,plzzzzzz... plz..
Hello Ashok Kumar !
I am not an Engineer. I am a Physicist.
I think before I teach engineering stuff I have to work very hard. And I shall work hard if it helps you.
Leave a comment #ShouldIWorkHard
Thank you for your support and as always ... thanks for watching.
Mast sir
why did you multiplied Tangential stress(T) with alpha and beta when they are longitudinal and lateral strains.
*Sir thumb nail me equations galti he. (-ve) ka place me (+ve) likha he
plzz upload relation between E,Nand M
Super
At 13:50
nice video sir
Thanku so much sir
Hello !
You are welcome and as always .... thanks for watching.
Sir , thumbnail is wrong
thank u so much sir..
Sir why the shear angle is consider as very very small.
Thanks ..
Hello KAJAL GUPTA,
You are welcome.
Aap sab ki duao se hi ye channel hai.
thanku sir
welcome
Sir apne thumbnail me galt dala hai + ki jagah minus dala maine galt likh diya exam me 😔😔 plzz sahi thumbnail rakh karo 🙏🏻🙏🏻🙏🏻
Thank you sir
superb
Nice
thanks sir
Helpful video
+Gpi Pcmc
You are welcome.
Bas aap sab ka saath rahe.
thanks sir
Sir, 2(1-sigma) in thumbnail
tankyou sir
👍
sir please koi bhi chij btaie to ese ki ydi hm sir exam me likhe to achhe Max ae for Allahabad university bsc1 😌
👏👏
Thanks sir.
I
Hello !
You are welcome and as always .... thanks for watching.
Video ka title me 1-sigma diya he
English plz
English pls
1+sigma hona chahiye tha
Hello Rikima,
Yes, you are right.
I must rectify the thumbnail.
Thank you very much for your valuable feed.
..
My plz r sir
Ghatiya
Thank you Sir!
*Sir thumb nail me equations galti he. (-ve) ka place me (+ve) likha he
Thnks sir
Thank u sir ji
Hello Rabindranath Kumar !
You are welcome.
Bas aap sab ki dua bani rahe.
*Sir thumb nail me equations galti he. (-ve) ka place me (+ve) likha he
Abe +ve ki jgah -ve aaega
Thanks sir
thanks sir
Thank you sir
Thanks sir