Solid mechanics challenge question

[Mapes]

I just don't get why you're not proofreading your equations before you post them, or checking to see whether the values make physical sense. You're asking for comments on your work, but I can't tell which errors are simply typos vs. which indicate conceptual problems.

Once you have the stress in the wall written in terms of the rectangular coordinate system \bold{e_1}, \bold{e_2}, \bold{e_3}, there are several methods for doing a stress transformation. Once is using a direction cosine matrix \bold{a} like the one you wrote, and calculating the new stress \bold{\sigma^\prime}=\bold{a\sigma a^\mathrm{T}}. Another is using Mohr's circle to do it graphically or by equation.

 

[dislect]

thats the thing, its now written in cylindrical r,theta,z coordinates and not in rectangular. I don't know how to change it to rectangular!

 

[Mapes]

The trick here is to pick one angle on the structure, look at which directions the cylindrical axes point, and match these up to the rectangular axes. For example, let's work with the point midway between A and B. The radial axis \bold{r} is collinear with \bold{e_1}, so \sigma_\mathrm{rr}=\sigma_{11} at this point. Does this make sense?

 

[dislect]

yes. and e_3=e_z but what about e_2? you can also say that its the same direction as r on two locations..

 

[Mapes]

So let's only work with a single point. (Any point along the weld is equivalent, right?)

 

[dislect]

sorry i don't understand..

 

[Mapes]

At the top of the cylinder, the radial direction is also the \bold{e_2} direction. At the side of the cylinder, the radial direction is also the \bold{e_1} direction. So you have to commit to a single point before you begin to do the stress transformation part of the problem; otherwise, the axes will be coupled in an ambiguous way. I'm suggesting you pick a point at the weld and on the side of the cylinder closest to the viewer. Does this help clarify?

 

[Mapes]

so on any point on the e₁e₂ plane the stress is \sigma_rr and \sigma₁₁=\sigma₂₂=\sigma_rr?

No, I wouldn't say that. Rather, we have \sigma_{11}=\sigma_\mathrm{rr}\cos^2\theta+\sigma_{\theta\theta}\sin^2\theta within the tank material, where \theta is measured from \bold{e_1}.

 

[dislect]

will be edited later *

 

[dislect]

I'm going to start from the beginning now the make sure its all allright

1. finding the constants:
\sigma_{rr}(r=R)=a+b=-10P, \sigma_{rr}(r=2R)=4a+b=-4P
we comapre these exprassions to \ -10P,-P and not to \+10P,-P because the sign (-) says that the tank is under compression from the inner and outer surface?
This gives us a=2P, b=-12P

2.a The stress tensor in cylindrical coordinates (r,\theta,z):
\sigma_{(r,\theta,z)}=\begin{pmatrix}<br /> 2P-\frac{12P}{(r/R)^{2}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 2P+\frac{12P}{(r/R)^{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 2P \end{pmatrix}

2.b Switching from cylindrical coordinates (r,\theta,z) to cartesian (x,y,z), z remains the same:

R - transform matrix from cylinder to cartesian

\sigma_{(x,y,z)}=R^{T}\cdot \sigma_{(r,\theta,z)}\cdot R,
R=\begin{pmatrix}cos(\theta) &amp; -sin(\theta) &amp; 0 \\ sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix}

\sigma_{(x,y,z)}=\begin{pmatrix}\sigma_{xx} &amp; \sigma_{xy} &amp; \sigma_{xz} \\ \sigma_{yx} &amp; \sigma_{yy} &amp; \sigma_{yz} \\ \sigma_{zx} &amp; \sigma_{zy} &amp; \sigma_{zz}\end{pmatrix}=\begin{pmatrix}cos(\theta) &amp; sin(\theta) &amp; 0 \\ -sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix}\begin{pmatrix}\sigma_{rr} &amp; \sigma_{r\theta} &amp; \sigma_{rz} \\ \sigma_{\theta r} &amp; \sigma_{\theta\theta} &amp; \sigma_{\theta z} \\ \sigma_{zr} &amp; \sigma_{z\theta} &amp; \sigma_{zz}\end{pmatrix}\begin{pmatrix}cos(\theta) &amp; -sin(\theta) &amp; 0 \\ sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix}=
=\begin{pmatrix}cos(\theta) &amp; sin(\theta) &amp; 0 \\ -sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix}\begin{pmatrix}2P-\frac{12P}{(r/R)^{2}} &amp; 0 &amp; 0 \\ 0 &amp; 2P+\frac{12P}{(r/R)^{2}} &amp; 0 \\ 0 &amp; 0 &amp; 2P \end{pmatrix}\begin{pmatrix}cos(\theta) &amp; -sin(\theta) &amp; 0 \\ sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix}

2.c Transforming the cartesian coordinate system by 30 degrees - \sigma&#039;:

A - transformation matrix in the cartesian coordinate system by 30degrees

\sigma&#039;_{(x,y,z)}=A^{T}\cdot \sigma_{(x,y,z)}\cdot A=A^{T}R^{T}\sigma_{(r,\theta,z)}RA,
A=\begin{pmatrix}1 &amp; 0 &amp; 0 \\0 &amp; cos(60) &amp; sin(60) \\0 &amp; -sin(60) &amp; cos(60) \end{pmatrix}

The stress wall on AB is then:
\sigma&#039;_{(x,y,z)}= \begin{pmatrix}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; cos(60) &amp; -sin(60) \\<br /> 0 &amp; sin(60) &amp; cos(60) \end{pmatrix} \begin{pmatrix}cos(\theta) &amp; sin(\theta) &amp; 0 \\ -sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix}\begin{pmatrix}2P-\frac{12P}{(r/R)^{2}} &amp; 0 &amp; 0 \\ 0 &amp; 2P+\frac{12P}{(r/R)^{2}} &amp; 0 \\ 0 &amp; 0 &amp; 2P \end{pmatrix}\begin{pmatrix}cos(\theta) &amp; -sin(\theta) &amp; 0 \\ sin(\theta) &amp; cos(\theta) &amp; 0 \\ 0 &amp; 0 &amp;1\end{pmatrix} \begin{pmatrix}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; cos(60) &amp; sin(60) \\<br /> 0 &amp; -sin(60) &amp; cos(60) \end{pmatrix}

I'm waiting with the multiplication to see if you think I wrote it correctly this far, if so I can find the shear stress on AB by using the following connection:
\sigma_{ns}=s_i \cdot \sigma_{ij} \cdot n_i when \hat{s} and \hat{n} are unit vectors pointing in the direction of the shear and normal stress.

 

[dislect]

Mapes can you please evaluate my work?
I'm guessing I complicated things but its because I just don't understand the clues you gave me. I am stuck in the same place for almost a week and I'm just about the quit the all thing..
All I want is to understand in the clearest way how to change to cartesian coordinates and if after that change I am still suppose to have components in my tensor that depend on r or theta ?

 

[Mapes]

This approach looks fine.

 

[dislect]

so can you explain how come the shear stress expression has theta in it? i mean, am i suppose to leave it like that?
and, based on your intuition, where would the failure first occure? (based on that ill try to prove it using the stress expression)

 

[dislect]

the stress turns out to be an insanely huge matrix because of theta from the R transformation matrix. am i really suppose to leave it like that?

 

[Mapes]

You'll want to work at \theta=\pi so the outer stress transformation rotates around the correct axis. That should simplify things. And don't you want 30^\circ instead of 60^\circ?

EDIT: Corrected \theta typo.

 

[dislect]

can you expand more on the theat=pi/2 ? i catually tried working with pi/4 because i figured that way i my matrix components are maximal
what does it mean outer stress transformation rotates around the correct axis? and is it correct that the failure will start at the middle of AB?

i know you are probably sick of me by now but the submition is tommorow morning :)

 

[Mapes]

Whoops, I meant to say \theta=\pi (corrected above). As we discussed earlier, if you're going to rotate around \bold{e_1} to determine the shear stress at the weld, then \bold{e_1} needs to coincide with \sigma_\mathrm{rr}. This only occurs in the above equations if \theta=\pi or 0.

 

[dislect]

well that makes things a lot simpler :) now all i have to do is put it inside the multiplication above and look at the components A1,2
though its weird that the transformaition matrix from cylindrical to cartesian base turns into I, meaning does nothing to it..

and about where it first fails, middle of AB right?

 

[Mapes]

and about where it first fails, middle of AB right?

I believe that question is asking where in the wall the weld first yields: inside, middle, outside, etc. You'll get this from the transformed shear stress.