Hi, I am not sure if this is the right forum for my question but here goes!
This is the problem:
The stress tensor in the Si coordinate system is given below:
σ’ij = {{-500, 0, 30}, {0, -400, 0}, {30, 0, 200}} MPa
Calculate the stress tensor in the L coordinate system if: cos-1a33=45o, and X’2 is in the plane defined by X1, X2 and is rotated 60o counterclockwise from X2.
The answer the book give is:
From the formula
σ’ij = aikajlσkl, where σ’ij is defined in L,
(σ’33) = a312σ11 + 2a31a32σ12 + 2a31a33σ13 + a322σ22 + 2a32a33σ23 + a332σ33
The direction cosine matrix is given by
aij = {{0.35,0.61,-0.71}, {-0.87,0.5,0}, {0.35,0.61,0.71}}
Thus, σ’11 = -150.0 MPa; σ’12 = 92.0 MPa; σ’13 = -225.0 MPa; σ’22 = -450.0 MPa; σ’23 = 31.0 MPa; σ’33 = -100.0 MPa.
Here is the answer with some explanations and guidance and I’m still not able to reach the same answer as the author of the book I’m following.
My idea was to just insert the given σij and aij from the answer in the σ’ij = aikajlσkl function but already here I encounter the problem that σ’33 = 0,35^2*(-500)+2*0,35*0,71*30+0,61^2*(-400)+0,71^2*200 = 97,5 MPa, which isn’t that much off, but when they express their answers with one decimal precision I assume it is wrong. I have tested to solve the other stresses too but the error is even larger for them.
I would very much appreciate if somebody can show how to solve this problem with the given information.
