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Re: Derivative recurrence
*To*: mathgroup at smc.vnet.net
*Subject*: [mg72718] Re: Derivative recurrence
*From*: dh <dh at metrohm.ch>
*Date*: Tue, 16 Jan 2007 03:27:40 -0500 (EST)
*References*: <eofikq$gc1$1@smc.vnet.net>
Hi,
Eij is obvioulsy the sum of a simplier tensors, say eij, and its
transpose. eij can be written as an outer product. Assume you defined
your functions in funs and the variables in vars
funs={z f1[x,y,t],z f2[x,y,t],z f3[x,y,t]};
vars={x,y,z}
eij=Outer[D[#1,#2]&,funs,vars];
Eij=(1/2)(eij+Transpose[eij])
Daniel
KFUPM wrote:
> Dear All
>
> I have the following
>
> u1(x,x,z,t) = z f1(x,y,t)
> u2(x,y,z,t) = z f2(x,y,t)
> u3(x,y,z,t) = z f3(x,y,t)
>
> which is basically the displacement fied. I want to compute the strain
> which is given in indecial notation as:
>
> Eij = 1/2( dui/dxj + duj/dxj ) where i, j, k go from 1 to 3,,, x1=x, x2
> =y, x3=z,
>
> I want mathematica to calculate the strain Eij autmoatically,
>
> for example
>
> when i = 1 and j= 2
>
> E12 =1/2 (du1/dx2+du2/dx1 ) but x1 =x and x2 =y
>
> therefore
>
> E12 =1/2 (du1/dy + du2/dx) = 1/2 ( z df1/dx + z df2/dy),,, where f1,
> f2 and f3 are arbitrary functions.
>
> Eij should generate 3 by 3 matrix containing the strain components.
> Frankly, i don't know how to do it using mathematica and i would
> appreciate any help in this regard.
>
>
> Thanks in anticipation
>
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