Sir you are extremely good. I just worship you. Excellent explanation.

watched almost all of your videos I love your teaching style The way you implant interest before starting the lecture is amazing

Wow sir!!! You changed my whole perspective towards physics.

Such a wonderful teacher. Salute to you sir. 🙏

last portion was always my bdout ....now it is clear...thanks srr

It's very yummy lecture for me. My sir

Finally i came to know the difference between Engineering stress and True Stress

Thank you Rajesh Sir !!!!

Sir, you are extremely good.

Very nice explanation sir

Thank for you presentation and please review engineering strain e =deltaL/L0 in example final

This was helpful...keep up the good work sir

Thank you sir for your excellent explanation

Dear Sir, One video on proof stress. Humble request.

Veru useful sir

IIT D jindabad❤️❤️❤️❤️❤️❤️❤️❤️

you saved my mme exam! thnks sir!

@itsonyou

5 жыл бұрын

Which Class ??

Love this guy

Thanks so much sir

sir while defining true incremental strain does your definition imply that strain is additive in nature? and if so, then how do you know its additive? and secondly in the example why did you take L0 as 2L while calculating compression in the definition you said that L0 is initial length (natural length? ) from the formula you derived for true strain. Thankyou

Sir is power law Stress=K (strain)^n valid for engg. values after yield point?

Tq sir

SIR in Hook's law which stress and strain is used. Engineering or True or Both can be used. Please reply.

@lidyaangelin8927

5 жыл бұрын

Engg stress

@introductiontomaterialsscience

5 жыл бұрын

The difference between engineering and true stresses and strains becomes negligible for small elastic deformation. Thus Hooke's law is valid for both and will give the same modulus.

Sir at UTS why is the true strain equal to strain hardening exponent

almost fell asleep watching this, but thanks! It was really easy to understand

@introductiontomaterialsscience

5 жыл бұрын

I agree with you. I am rather slow. If ever, I do a second edition, I will try to be faster. thanks for the feedback rajesh prasad

Thank you! 12:21 L0 and L in equations of eng.strain and true strain: what is dieeference? L0 must be initial length of sample, but it was shown as L... is it mistake? or i lost completely the idea And how a true strain can be equal to =0 after full iteration of deformation cycle to 100% and back??? True - means it is always related to previous L so it can only grow undependantly from direction of loading.

@rajeshprasad101

4 жыл бұрын

Your first confusion: Lo vs. L I have indeed made it very confusing by mixing symbols L (instantaneous or final length) and Lo (initial length). Engineering strain eE should be defined eE=DL/Lo where DL=L-Lo is the change in length. However, in the slide I have written it as DL/L. I have added to the confusion by labelling the initial length in my sketch as L and not as Lo. Thanks for pointing this out and sorry for the confusion it has created. Your second doubt: And how a true strain can be equal to =0 after full iteration... During the compression stage from 2Lo to Lo the true strain eT=ln L/Lo is negative as L

Sir is it mean that in the engineering strain will be greater 0.5 than initial strain ? And a little deformation will be done or not?? Please tell me sir

@introductiontomaterialsscience

4 жыл бұрын

Sorry, I am not able to get your question. Could you please express it differently.

Sir In the entire Strength of materials we study is Engineering stress or True Stress or both?

@introductiontomaterialsscience

2 ай бұрын

In the strength of materials we usually confine ourselves to elastic regime. In the small strains corresponding to elasticity there is not much difference between true and engineering stress. The difference becomes significant only when the strains are large, like in plastic deformation.

L=100 After tension L=120 Then after unloading L=110 There where ,which strain we apply??

If we superimpose the true and engg stress strain curves, upto what point will they be same?

@introductiontomaterialsscience

4 жыл бұрын

Mathematically they would be same only at zero strain. Beyond that, they are rather close upto yield point in metals.

Sir, why volume constancy is valid in plastic deformation ?

@introductiontomaterialsscience

Ай бұрын

This is good question. I think we can say that this is an experimental observation. The atomic mechanisms of plastic deformation are consistent with this fact. So in slip one part of the crystal slides over the other part and so the volume remains constant. In twinning, the crystal reorients itself, again keeping the volume constant.

@TimeEnergyConsciousness

Ай бұрын

@@introductiontomaterialsscience thank you Sir, got the idea. (it's a question asked in BARC interview)

Stress ( or one of the component ) is Force by Area. Both Area and Force are vectors and stress is a tensor. A vector by vector (division of vectors ) is a tensor?. How can we explain this thing from equation sigma = F/A

@rajeshprasadlectures

3 жыл бұрын

Stress can be a scalar, a vector, or a tensor depending on the context. When we discuss the results of a uniaxial tensile test we are thinking of a single tensile force acting along the longitudinal axis of the sample on the transverse plane. Here we simply divide the magnitude of the force by the transverse area and treat it as a scalar. The scalar stress here is the division of two scalars. This is the context in which I have used stress in this course. But when you think of a vector force acting on a plane of a specific area then you can divide the force vector by the scalar area. This is a well-defined vector quantity and is called a TRACTION vector. Here we are dividing the vector by a scalar to obtain another vector, which is fine. So in the example of the uniaxial tensile test mentioned in the previous paragraph, you can treat the tensile stress also as a tensile traction vector of the magnitude force/ area acting in the direction of the longitudinal axis. If one wishes to describe stress at a point in material then one uses stress tensor, a symmetric second-rank tensor having nine components of which six are independent. If you know this tensor sigma then you can find the traction vector T on any plane passing through the point with an outward plane normal vector n. This relation is given as T= sigma n So sigma can be thought of as a second rank tensor that relates the plane normal vector n to the traction vector T acting on that plane. Since the above is a tensor relation, you CANNOT invert it to write sigma=T/n and think of sigma as a division of two vectors, an operation that is not defined in vector algebra.

@sanammamba2933

3 жыл бұрын

@@rajeshprasadlectures thanks for your detailed reply

in last comparison elongation compression example, i dont get why you take 2l as lo in final compression state, shouldnt it be l in case of engineering strain

@introductiontomaterialsscience

4 жыл бұрын

I extend from L to 2L and then compress it back to L again. Obviously, the net strain is zero. It comes out to be zero if we calculate true strain. But it is not so if we use engineering strain.

@bs143

4 жыл бұрын

@@introductiontomaterialsscience Sir, here the initial condition starts with Lo. "so engineering strain denominator should be always Lo and not 2Lo". please Correct me Sir,if the statement is wrong.

@rajeshprasad101

4 жыл бұрын

@@bs143The idea is that if you consider tension and compression steps separately. So during tension, the initial length is Lo but during compression the initial length is 2Lo.

Sir, can we consider only initial and final condition for engg strain ? Plz reply

@introductiontomaterialsscience

4 жыл бұрын

That's right. Both engineering strain and true strain depend only on the Initial and final lengths.

@ 3:47 where did A0/A come from in (F/A0) * (A0/A)????

@MI-gs1qo

5 жыл бұрын

So this is how you can do it. you know e(true)=F/A and e(eng) = F/A0...................................................... Divide e(true) by e(eng)............ u should get the following....... e(true)/e(eng= F*Ao/A*F............ both forces will cancel each other and you will have will end up with e(true)=e(eng)*(Ao/A)

@saurabhpatil7045

5 жыл бұрын

just multiply and divide by A0

Hindi please

L=100 After tension L=120 Then after unloading L=110 There where ,which strain we apply??