Stress tensor transformation and rotation of coordinate system

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 S_i 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=45^o, and X’₂ is in the plane defined by X₁, X₂ and is rotated 60^o counterclockwise from X₂.

 

The answer the book give is:

 

From the formula

 

 

σ’_ij = a_ika_jlσ_kl, where σ’_ij is defined in L,

 

(σ’₃₃) = a₃₁²σ₁₁ + 2a₃₁a₃₂σ₁₂ + 2a₃₁a₃₃σ₁₃ + a₃₂²σ₂₂ + 2a₃₂a₃₃σ₂₃ + a₃₃²σ₃₃

 

 

The direction cosine matrix is given by

 

a_ij = {{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 σ_ij and a_ij from the answer in the σ’_ij = a_ika_jlσ_kl 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.

Edited January 9, 201411 yr by theade88