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Is deformation gradient always a two point tensor?

Mac_Mechanics's picture

I am trying to derive the Green Lagrange  strain tensor and encountered the use of material displacement gradient where I need to define the tensor F as the difference of the material displacement gradient U and the identity tensor I.  In the book of Malvern, He mentioned that the displacement vector can be defined using the same indices (Big letter) to the current and reference configuration, thus if I take the material gradient of the displacement vector, Grad(U), IJ = xsub(I)/Xsub(J) - Kroneckerdelta,IJ, can I then define the value xsub(I)/Xsub(J) as F even if their indices are both capital?  I hope someone can enlighten me on this one.  

 

rajan_prithivi's picture

Yes, the Deformation gradient is a two point tensor.

F is defined as

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F_iJ = partial (x_i) / partial (X_J) , this is in indicail notation

In tensor notation, its F = partial (x) / partial (X)

We use small letters to indicate the quantities in the eulerian and capital letters for thse in Lagrangian.

we have the displacement defined as ,

 U(X,t) = x(X,t) -X (which is with respect to lagrangian)

Differentiating the above with respect to X will give you,

partial (U)/partial(X) = partial (x)/partial (X) - I

or

partial (x)/partial (X)= I + partial (U)/partial(X)

so, F =  I + partial (U)/partial(X)

This is deformation gradient expressed in terms of diaplacement

Now, the green tensor is given by

E = 0.5 *(FTF - I)

Hope this helps to some extent!

 

 

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