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
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.
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.
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
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
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.
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.
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.
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
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
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.
@@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.
@@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.
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)
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
It's very yummy lecture for me. My sir
Such a wonderful teacher. Salute to you sir. 🙏
Wow sir!!! You changed my whole perspective towards physics.
last portion was always my bdout ....now it is clear...thanks srr
Finally i came to know the difference between Engineering stress and True Stress
Dear Sir,
One video on proof stress. Humble request.
Sir, you are extremely good.
you saved my mme exam! thnks sir!
Which Class ??
Thank for you presentation and please review engineering strain e =deltaL/L0 in example final
This was helpful...keep up the good work sir
Very nice explanation sir
Thanks so much sir
Thank you sir ❤
Thank you sir for your excellent explanation
IIT D jindabad❤️❤️❤️❤️❤️❤️❤️❤️
Thank you Rajesh 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 in Hook's law which stress and strain is used.
Engineering or True or Both can be used.
Please reply.
Engg stress
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.
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.
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
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
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.
@@rajeshprasadlectures thanks for your detailed reply
Sir In the entire Strength of materials we study is Engineering stress or True Stress or both?
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.
Veru useful sir
Love this guy
Sir, why volume constancy is valid in plastic deformation ?
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.
L=100
After tension
L=120
Then after unloading
L=110
There where ,which strain we apply??
Sir is power law Stress=K (strain)^n valid for engg. values after yield point?
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
Sorry, I am not able to get your question. Could you please express it differently.
almost fell asleep watching this, but thanks! It was really easy to understand
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
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
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.
@@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.
@@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.
If we superimpose the true and engg stress strain curves, upto what point will they be same?
Mathematically they would be same only at zero strain. Beyond that, they are rather close upto yield point in metals.
Tq sir
@ 3:47 where did A0/A come from in (F/A0) * (A0/A)????
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)
just multiply and divide by A0
Sir, can we consider only initial and final condition for engg strain ? Plz reply
That's right. Both engineering strain and true strain depend only on the Initial and final lengths.
Hindi please
Sir at UTS why is the true strain equal to strain hardening exponent
L=100
After tension
L=120
Then after unloading
L=110
There where ,which strain we apply??