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QuantumATK => General Questions and Answers => Topic started by: fangyongxinxi on July 3, 2013, 15:25

Title: shear percentage "e12" in interface building
Post by: fangyongxinxi on July 3, 2013, 15:25

Dear Sir,

I have a problem on understanding the shear percentage "e12" in interface building.

I use atk-12.8, to build a interface structure.When "select surface cells", there are three "strain" (e11,e22,e12) for each cell match. I know the meanings of e11 and e22, but do not understand what does the "e12" mean.

could you give an example what the e12 means ?

Thanks~

Yong Fang

Title: Re: shear percentage "e12" in interface building
Post by: zh on July 4, 2013, 02:41

Because a surface cell consists of two unit vectors, it may be easy to understand them as below:
1--> x, 2-->y, so  11---> xx, 22-->yy;  12--> xy, 21--> yx.

Title: Re: shear percentage "e12" in interface building
Post by: fangyongxinxi on July 4, 2013, 03:20

Thanks for your reply.

I still confuse about how to get the shear percentage. I draw a picture, pink----blue. so e11=(a1-a2)/a2; e22=(b1-b2')/b2'. so what e12 mean?

hope to get your reply again.

Thanks~

Quote from: zh on July  4, 2013, 02:41

Because a surface cell consists of two unit vectors, it may be easy to understand them as below:
1--> x, 2-->y, so  11---> xx, 22-->yy;  12--> xy, 21--> yx.

Title: Re: shear percentage "e12" in interface building
Post by: zh on July 4, 2013, 14:18

|R11'   R12'|       |1+e11       0.5*e12|        |R11    R12|
|R21'   R21'|   = |0.5*e21        1+e22|  *   |R21    R22|

Rij'---> the unit vectors of a surface unit cell A;
Rij-->  the unit vectors of a surface unit cell B;

Title: Re: shear percentage "e12" in interface building
Post by: Anders Blom on July 5, 2013, 00:20

Explicitly, if we denote the first (left) surface lattice vectors as A and B, and the second (right) ones as A' and B', then in the Interface Builder these vectors are aligned such that both A and A' are parallel to the x-axis, i.e. A=ax*x and A'=ax'*x, and B=bx*x+by*y and B'=bx'*x+by'*y, and

e11 = (ax'-ax)/ax
e22 = (by'-by)/by
e12 = 1/2*(bx' - ax'/ax*bx)/by

Title: Re: shear percentage "e12" in interface building
Post by: fangyongxinxi on July 5, 2013, 03:05

Quote from: Anders Blom on July  5, 2013, 00:20

Explicitly, if we denote the first (left) surface lattice vectors as A and B, and the second (right) ones as A' and B', then in the Interface Builder these vectors are aligned such that both A and A' are parallel to the x-axis, i.e. A=ax*x and A'=ax'*x, and B=bx*x+by*y and B'=bx'*x+by'*y, and

e11 = (ax'-ax)/ax
e22 = (by'-by)/by
e12 = 1/2*(bx' - ax'/ax*bx)/by

Thanks for your patient.
I got it.