Solid mechanics challenge question

[dislect]

Hi all! Good evening

Here is the Question:

This question is given as a challenge extra credit task, and is a part of a course called solid mechanics 2 that deals more with shear strain and strain matrices.
While trying to translate the question I used the term "strength" instead of "strain" by mistake :)
I don't really have a lead how to begin, I would really appreciate if someone could guide me or write me some guidelines on how to find the constants and the max pressure

Thanks in advance and a good day to all :shy:

 

[Mapes]

Can you relate the stress tensor components to the known pressure on the inside and the outside of the tank? Can you also relate these components to the shear stress on the weld?

 

[dislect]

I know that the stress tensor components for a cylinder under internal pressure 10P are:

\sigma_{rr}=-10P,\sigma_{\theta\theta}=\frac{10PR}{R},\sigma_{zz}=\frac{10PR}{2R}
but I am confused as to how to refer to the external pressure?
about the shear stress, not so much :) i think that first we need to create the stress tensor for a cut shown on the right picture and then use a transformation matrice to move it by 30 degrees?

 

[Mapes]

Isn't the outside pressure P? What happens when you plug those two boundary conditions into \sigma_\mathrm{rr}?

I agree with the transformation strategy to find the shear stress along with weld.

 

[dislect]

I just get -9P (by the way am i right about the minus sign)? and the outside pressure dosent affect the other components?

 

[Mapes]

What do you get for a and b when you plug in a compressive stress of P when r corresponds to the outer surface, and a compressive stress of 10P when r corresponds to the inner surface?

 

[dislect]

you mean

while r<R (inner surface):

\sigma_{\theta\theta}=10P=a-\frac{b}{r/R^{2}},\sigma_{rr}=-10P=a+\frac{b}{r/R^{2}}
for r=R we get that a= 0 b= -10P

while r>2R (outside surface):

\sigma_{\theta\theta}=P=a-\frac{b}{r/R^{2}},\sigma_{rr}=-P=a+\frac{b}{r/R^{2}}
for r=2R we get that a= 0 b= -4P

im sure i got it wrong though.. and what about what happens between R to 2R

it all seems a bit odd because the formulas for the stress tensor components relay on a Thin-Walled Pressure Vessel and here its not that thin because t=R. We usually get that \sigma_{rr}&lt;&lt;\sigma_{\theta\theta but not here so are the formulas still valid?

 

[Mapes]

I'm not sure why you're using \sigma_{\theta\theta}; the pressures act radially. Also, a and b are the same constants throughout the problem.

These equations don't rely on the thin-wall assumption; if that were the case, the stresses would be constant.

 

[dislect]

right.. second try:

r=R
\sigma_{rr}=-10P=a+\frac{b}{(R/R)^{2}}\rightarrow a+b=-10P
r=2R
\sigma_{rr}=P=a+\frac{b}{(2R/R)^{2}}\rightarrow 4a+b=4P

a=3P, b=-13P or am i wrong about the signs of the pressures (+ / -) ?

 

[Mapes]

\sigma_\mathrm{rr}=10P is saying that the material is under tensile stress on the inner surface. This doesn't make much sense.

 

[dislect]

yeah i fixed that when i wrote the equation a+b=-10P and forget to correct it at the beginning. so these are the constants?
now i need to write the transformation matrice?

by the way, dosent the "attachments" (dont know the right word for it in english) at the bottom of the cylinder affect the pressure in some way? (what we see marked as a triangles)

 

[pongo38]

In your question, what is called external radius = 4R is at variance with the figure, where 4R is the external diameter

 

[dislect]

pongo i don't understand.. can you show me using the equations i wrote? the language gaps make it a bit difficult in a subject that is already hard for me to understand in my mother language :)

and btw, can i say that the shear strain is 0 at all components?

 

[Mapes]

\sigma_\mathrm{rr}=P or -P at r=2R?

Agreed that there is no shear stress.

 

[dislect]

i think that P at 2R and its what i used on the second equation *fingers crossed*

So in a cylindrical coordinate system i get:
\sigma=\begin{pmatrix}<br /> 3P-\frac{-13P}{(r/R)^{2}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 3P+\frac{-13P}{(r/R)^{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 2P \end{pmatrix}

Given that my matrix is correct I'm going to save some time and start working on the transformation. Now since we are currently looking at a cylindrical coordinate system I think need to translate it to a cartesian system in order to move it by 30 degrees so:
x=r\cdot cos(\theta), y=r\cdot sin(\theta) but I am unsure on how to apply the conversion in a tensor. Is the following conversion correct?

\begin{pmatrix}<br /> \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}

and later on the transformation matrix A to move the axis e2,e3 is:

A=\begin{pmatrix}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{1}{2} &amp; \frac{\sqrt{3}}{2} \\<br /> 0 &amp; -\frac{\sqrt{3}}{2} &amp; \frac{1}{2} \end{pmatrix}

How am I doing so far? :~)

 

[Mapes]

I'm still not seeing how you're setting the stress at the outer surface to be a positive value P, which represents a tensile stress. The outer surface is under compression.

Once you get this straightened out, try expressing the wall stress in terms of the rectangular coordinate system, apply the transformation, and check whether your answer makes sense.

 

[dislect]

But I did set it as possitive on the outside, its written in the second equation here below when r=2R:

right.. second try:

r=R
\sigma_{rr}=-10P=a+\frac{b}{(R/R)^{2}}\rightarrow a+b=-10P
r=2R
\sigma_{rr}=P=a+\frac{b}{(2R/R)^{2}}\rightarrow 4a+b=4P

a=3P, b=-13P or am i wrong about the signs of the pressures (+ / -) ?

I'm not familiar with the term "wall stress", isn't moving to a rectangular coordinate system is what I just wrote in the last reply?

 

[dislect]

ok correction, a=13P/4 and b=-44P/3 I solved it wrong somehow..
now I figured I could mark a vector \hat{n} facing out in a 90degree angle from line AB (which marks the direction of the normal) and a vector \hat{s} parallel to line AB (which marks the shear direction). Now since I found constants a,b and the shear strains \sigma_{\theta z},\sigma_{rz},\sigma_{\theta r} and so are equal to 0 I have the strain matrix \sigma_{\theta rz} in a cylindrical coordinate system.
Can I multiply \sigma_{\theta rz}\cdot \hat{n} \cdot \hat{s} to find the size of the shear strain on line AB? I connect that strain to what they marked as failure strain \sigma_{0} by saying that \sigma_{\theta rz}\cdot \hat{n} \cdot \hat{s} &lt; \sigma_{0} and then I isolate the pressure P?

thanks for your help!

 

[Mapes]

ok correction, a=13P/4 and b=-44P/3 I solved it wrong somehow..

But plugging in these values gives \sigma_\mathrm{rr}=-11.4P at r=R and \sigma_\mathrm{rr}=-0.4P at r=2R...

 

[dislect]

I seriously don't know what's wrong with me these days..
a=14p/3 b=-44p/3

whats about the other stuff I wrote? :)

 

[Top Cat]

Hi there, I was just looking for problems with cylindrical coordinates for practice, and I found this one, so I'll join in, I hope you don't mind.
Dislect, in section c, were you asked to find the angle of the initial failure or is it the distance r? BTW if you have other solved examples or if you know where can I find them, I'll be forever grateful :), Thanks.

 

[dislect]

The more the merrier!
I'm not sure what you just asked.. but the 30degree angle is given data. I need to find the max pressure P value before the welding starts to break but I'm not sure how to do it. I could probably find some cylindrical coordinates problems but its all in hebrew :)

 

[Top Cat]

The first failure will occur on the plane AB at the point of the maximum shear stress. Since this plane isn't symmetric about the Z axis, the maximum stress will be only for 2 specific angles, which I think we need to find in section c.

Hebrew wouldn't be very efficient for me lol... English or Russian would be better. Thanks anyway :)

 

[dislect]

the failure will accure somewhere in the AB plane, i don't see how the angles you are talking about get in. As I see it, the failure is suppose to occur either in point A, B or in the middle.

as to finding max pressure did you try solving that part? what result did you get?

 

[Mapes]

I seriously don't know what's wrong with me these days..
a=14p/3 b=-44p/3

This gives \sigma_\mathrm{rr}=P at r=2R, but \sigma_\mathrm{rr}=-P there (a compressive load).

After this is cleared up, you can write the stress tensor in terms of the \bold{e_1}, \bold{e_2}, \bold{e_3} coordinate system and perform the stress transformation.

 

[dislect]

you mean the pressures are -10p and -p? then i get that a=2p, b=-12p and hopefully that's it for these two equations :)

So in a cylindrical coordinate system i get:
\sigma=\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}

I hope i get the conversion to e1,e2,e3 right please correct me if i dont:
\begin{pmatrix}<br /> \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}
so:
\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(30) &amp; -sin(30) &amp; 0 \\ sin(30) &amp; cos(30) &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(30) &amp; sin(30) &amp; 0 \\ -sin(30) &amp; cos(30) &amp; 0 \\ 0 &amp; 0 &amp; 1\end{pmatrix}
which result in:

is that how i convert cylindrical coordinates to x,y,z moved by 30degrees?

 

[Mapes]

(1) Check the \sigma_\mathrm{rr} and \sigma_{\theta\theta} values in your first matrix.

(2) Check which axis you're rotating around. The stress value for that axis will be unchanged after the stress transformation.

 

[dislect]

correction:
cylindrical coordinate system:
\sigma=\begin{pmatrix}<br /> 2P-\frac{44P}{3(r/R)^{2}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 2P+\frac{-44P}{3(r/R)^{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 2P \end{pmatrix}

I'm rotatin around e_1 or \hat{r} so the value \sigma_{rr}=2P-\frac{44P}{3(r/R)^{2}} is suppose to remain the same as \sigma_{xx}?
then the matrix i am multiplying on both sides is changed to what?

\begin{pmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; cos(\theta) &amp; sin(\theta) \\ 0 &amp; -sin(\theta) &amp;cos(\theta)\end{pmatrix}

its just some sort of a formula i found on the internet, i don't really undertstand this transformation because no one knows/explains me how to change cylindrical coordinates to cartesian!
i know how to move a cartesian coordinate system by 30degrees, but i didnt find anything explaining how to move cylindrical to cartesian.

 

[Mapes]

Back to 44/3?

 

[dislect]

once and for all,
a=2p
b=-12p ..

\sigma=\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}

anyway.. its the conversion I am worried about i would have corrected and solved it from the beginning.