OnStrike

Ill be flooding the forum with political opinions now too. If others do it, why not me.

For example, if the pipes are long enough, you might need the tank higher to overcome friction and losses. Something to be aware of....

No worries. Also note that friction and flow in pipes (reference) might effect practical considerations.

That is how a manometer works

^^ What he said. Make the U-Bend higher than the tank

If the U-bend were higher than the water basin, water would not flow

@PatoSáinz Because the water basin is higher than the rest of the pipe system, hydrostatic pressure causes water to flow.

To clarify my answer before: $a(t)$ into bar --> $a(x)$ in the bar at a given value of $a(t)$ until $a$ is constant on all differential mass elements. What is the formal way to write this?

@ACuriousMind Noted, will do. Thanks- learning every day.

DavidZ is the man!

@ACuriousMind He deleted them. They werent bad, just "will you skype", "I dont get it", etc.

@DavidZ On a separate note, the comments here are outdated and not useful, as I edited my answer. Not sure how much you wish to micromanage deletion of comments...

Cool, thanks DavidZ.

@DavidZ Agreed. I thought Id alert you because my answer prevented the normal process from deleting it. I didnt want my name on a crap answer.

Thanks, enjoy.

I wasnt sure how to formally write it, even though the answer and logic are correct

@Ocelo7

@DavidZ It was closed and scheduled for deletion until my answer was upvoted, preventing its deletion

Within the bar, $a$ is a function of x until it is damped, assuming $a(t)$ does not change again.

The acceleration applied to the rod is strictly a function $a(t)$.

So the force on the bar is $a(t)$ but when $a$ changes, accelerations in the bar vary until the damping coefficient of the material forces the accelerations to be constant.

I wrote the partial $\frac{\partial \;a(x,t)}{\partial x}$ for completeness but in reality, if $a$ changes as a function of time, $a(x)$ will vary. What is the formal way to write this?

Another quick question if I may? Are the stresses the same in Euler and Timoshenko beams? I know Timoshenko beams include the deflection that Euler beams neglect but I believe the stress states are identical?

I have read that cover to cover and think I understand EB beams well.

Thanks @Wasabi

^vs only 10 quad elements

**sorry, wrong JPG.. 20 elements

^This is the 30 linear elements.

All plots shown use ANSYS Beam 188 element. All have the same geometry and material. Only the formulation is changed between linear and quad.

The quad designates the internal formulation of the shape functions and total DOF for the stiffness matrix

No. Im not talking about quadrilateral elements (like Quad 182). Im talking about quadrilateral beam elements (3 nodes, each with 6 DOF).

A standard beam stiffness matrix is based on Euler-Bernoulli formulation with 6 DOF per node. The Timoshenko stiffness matrix helps with deep sections where shear deformation becomes a factor.

The linear elements are terrible at reflecting the stress state (I get that) but I dont understand why the nodal deflection is different from theory.

Here is the comparison for 2 linear elements vs two quad elements.

That is the single Quad.

If I were to model it with a quadratic element I get very close to the exact solution with only a single element

For a slender beam I get the same stress results because it is only a single, linear element.

Hi, thanks for your help.

@joojaa Thanks. Im not entirely sure what you mean by 'does not rotate' but I will edit the question to reflect your points and better clarify.

@joojaa As I understand it: Euler-Bernoulli assumes the cross section is perpendicular to the N.A. (best for slender beams); Timoshenko assumes cross sections remain planar but does not assume they are perpendicular to the N.A.; My knowledge of Mindlin elements is minimal but I dont think they require the cross section be planar. Regardless, the difference between theories is negligible compared to the FE error.

@joojaa Thanks for your interest. It is true that ANSYS uses Timoshenko or Mindlin beam elements. While this does account for some error in comparison to Euler-Bernoulli, Im wondering why the result doesnt better reflect solutions obtained analytically. BTW, all beam theories do account for rotation (defined as $ROTZ=\frac{dy}{dx}$ etc.) .

I know open-ended topics arent well received in QA. Alright, thats the topic... fire away, destroy my dreams (if you must)?!

This made me wonder if anyone has suggested a means to slowly diffuse its energy (to prevent a catastrophic eruption in the very distant future), with the added bonus of harnessing its energy.

I recently read about the massive volcano under Yellowstone National Park.

Im posting in chat so y'all cant downvote..;)