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Calculate Pb (bending stress) linearization

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olgamsp

Mechanical
Aug 18, 2016
14
thread292-95338

Hi, I need some help. I have to calculate the stress bending tensor in a SCL (stress classification line), I have the stress tensor at the SCL but now I have to apply this:

σib=(∫σi(x-t/2)dx)/(t2/6)

How can I do this?.

I have 6 σi, the distance between Node 1 and Node 2 is the same for all the nodes (0.0018). So, I don't know what is t and x terms at the formule.
 
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As the word linearization implies, you calculate a linear stress distribution that has the same bending moment as the actual stress distribution. The result is a bending stress value Pb that, in your graph above, equals some 60 (MPa?). You have that value at one wall face, minus the same value at the other face and zero at mid thickness. It is hence irrelevant the local value of this (idealized) linear distribution, only the single bending stress Pb at the face has an interest.
My advice is to deepen your knowledge of the theory of the linearization procedure by reading some related texts.

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Hi again, I used the formule σcb=(t2/6)*Σ1n(σc+σc[i-1])/2*(x+x[i-1]-t)/2*(x-x[i-1]), but what equation I should used for the node 0 and the node n?
 
Hi, I made the model in ANSYS and the linearization bending results are not the same. I think this is because Ansys doesn't with the ASME Code Linearization. But Do you know what code ansys uses?
 
Yes, I am doing exactly what the code says, because I am doing an excel tool to linearized stresses, But I compared the results of bending stress using ansys with the integral at the code, and are very different.

Also I saw a post from you TGS4:

"When I said that ANSYS linearization doesn't follow the Code methodology, I mean that it does not follow the procedure in 5.A.4.1.2., which is a mandatory part of the new 2007 Edition of Division 2"

So I wish know what is the difference.

thanks

 
Without further details, I'm not sure what I can do to assist. You need to upload whatever calculations you are doing.
 
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