Elastic energy of a dislocation line

Elastic energy of a dislocation line

[@abhisheksoni1115]

congratulations for a half century sir

[@ravindarsingh3002]

Sir, is it the energy stored in the material during elastic deformation (called as resilience) or energy stored during plastic deformation (in the form of lattice strains) ?

[@introductiontomaterialsscience]

Sorry, somehow missed your question in time. It is the elastic energy of the dislocation line itself due to internal strains caused by the dislocation. It is not resilience which is elastic energy due to stains caused by external forces.

[@akithjabed553]

The dimensionsal analysis at the end of the video... How E/L came at last.. It was clueless.. Please check once

[@TheSourav77]

[A] represents the dimensions of area i.e [L]^2. So the [L]^2 term in the denominator cancels the one in the numerator, leaving us with just the dimensions of Force. Now, multiplication of the dimension of length [L] in both numerator and denominator gives us units of (Energy/Length). Hence, the elastic strain energy per unit length.

[@introductiontomaterialsscience]

@@TheSourav77 Thanks Sourav for your clarification.

[@karlampraneeth972]

@@TheSourav77 Bro,are we considering the change in work?

[@blackheart6897]

Sir, how line energy of a dislocation line can be associated with Burgers vector? Shouldn't it be associated with tangent vector ( because tangent vector is along the direction of dislocation line ) ?

[@introductiontomaterialsscience]

The energy of the dislocation line depends on the deformation or distortion caused by it. Since the line is a boundary between slipped and unslipped regions the deformation associated with it depends upon the magnitude of the slip associated with the line. This is the magnitude of the Burgers vector. The tangent vector only gives the orientation of the line. Thus in the isotropic assumption it has no influence on the energy. But real crystals are anisotropic, so the tangent vector, i.e., orientation of the line, also has some influence on the energy.

[@blackheart6897]

@@introductiontomaterialsscience Thank you so much sir, I got it now.

[@Bhola_Sonai]

As per the drawing, Above the slip plain will be tension and below the slip plane will be compression....please Sir clear my doubt......

[@rajeshprasad101]

There is one more plane above the slip plane than below. Thus the average spacing between the planes is less above the slip plane and more below it. Therefore, it is compression above and tension below.

[@pailasaisravan5423]

why there will be a tension in the bottom part

[@introductiontomaterialsscience]

In the same distance, there is one less plane below the slip plane. Thus interplanar spacings are slightly larger than the ideal interplanar spacing. This is equivalent to tension perpendicular to the plane.

[@shivamgoyal3888]

Sir please check again the dimensional analysis at the end...

[@introductiontomaterialsscience]

Thanks for pointing this out. There is no mistake but I think I could have explained it a bit better. Sourav Sahoo has given a nice explanation in his answer to a similar query by Akith Jabed Barlaskar. You may like to have a look at it.

[@introductiontomaterialsscience]

Thanks, Sourav for explaining it better than I have done it.

[@shivamgoyal3888]

Got it sir. I read it as F by mistake instead of E.

[@jarjarbinks9937]

sir isn't slip dislocation a plastic deformation