Mounted Vessel Statics Problem 1

[pdiculous963]

Could anybody help me resolve a statics problem? What I have is a piece of equipment that is mounted to a vessel by two flanges (at points (a) and (b) in the drawing). I know the weight of the mounted equipment, and am trying to calculate the resultant moment and reaction forces at the two vessel flanges. I've summed the vertical forces, but whenever I try to sum the moments, nothing seems to make sense; I believe the system is statically indeterminate. What would be the best method for solving this problem?

Thanks.

 

[rb1957]

at first look, it looks redundant (statically indeterminate, hyper-static) but i don't think it is. considering the equipment to be rigid (reasonable?) then the deflection of the upper beam is the same as the lower beam. then it's a "simple" matter to solve both of the beams for an applied load P, then load one beam with x lbs and the other with W-x and iterate untill the two displacements are the same.

Quando Omni Flunkus Moritati

 

[pdiculous963]

Thanks for the quick reply. It is reasonable to consider the equipment to be rigid. So to solve this problem, I need to set cantilever beam deflection equations equal to each other with the loads of x and W-x, and solve for equal deflection?

 

[rb1957]

yep

Quando Omni Flunkus Moritati

 

[Cockroach]

Thank you for the clarity of your problem.

The problem is only statically indeterminate when l1=l2. Make them the same and the issue becomes statically determinate assuming the vertical reactions of the support carry the pressure vessel weight equally.

Otherwise, for differing l1 and l2 you must make an assumption of the system or define a stress mechanics issue at the support. RB1957 had a good point, assume a minimal deflection in the system for example. This is a classical assumption, there are others, look at the support and make a note on the factor of safety to axial loading, etc.

Otherwise, here is the solution based on my comments.

Regards,
Cockroach

 

[pdiculous963]

Thanks cockroach, I appreciate the alternate solution. My last question (hopefully) is when following rb1957's method, is exactly how to determine the reaction moments. If I resolve the reaction force at each point, do I then simply multiply that by the horizontal distance to the centroid of the system?

 

[rb1957]

the top beam is a cantilever, loaded by x, length l1

the bottom beam is a frame, loaded by W-x, length l2

but looking again (again !?) at the problem, there is another redundant couple of Fx forces (not shown in the original sketch), helping to react the offset Fy load ...
or is there ? i can see that there's the possiblity of a couple, that the weight would want to put the upper beam in tension ... but then i look at the weight in isolation and this couple wouldn't be balanced ??

Quando Omni Flunkus Moritati

 

[zekeman]

The problem is indeterminate and neither Rb nor Cockroach "solutions" adequately address the problem.
You can't assume a vertical virtual motion as in Rb's view.
And Cockroach, I don't see how you accounted for M1 and M2 in your equations.
I think it can be done by assuming a motion vector at the CM,dx,dy,d@ which translates into motion at the upper and lower positions on the beam connections to the mass,dx1,dy1,d@1 and dx2,dy2,d@2. Now from each of the stiffness matrices of the shafts, you can get the vector forces at 1 and 2.
Now write the three static equilibrium equations on the mass.
3 equations 3 unknowns.Not having done this problem in eons, i t also has to satisfy the energy equation, -Wdy=strain energy 1 +strain energy2; so I now have an additional equation. What to do??
Still thinking..........

 

[zekeman]

The problem is indeterminate and neither Rb nor Cockroach "solutions" adequately address the problem.
You can't assume a vertical virtual motion as in Rb's view.
And Cockroach, I don't see how you accounted for M1 and M2 in your equations.
I think it can be done by assuming a motion vector at the CM,dx,dy,d@ which translates into motion at the upper and lower positions on the beam connections to the mass,dx1,dy1,d@1 and dx2,dy2,d@2. Now from each of the stiffness matrices of the shafts, you can get the vector forces at 1 and 2.
Now write the three static equilibrium equations on the mass.
3 equations 3 unknowns.Not having done this problem in eons, i t also has to satisfy the energy equation, -Wdy=strain energy 1 +strain energy2; so I now have an additional equation. What to do??
Still thinking..........

 

[renderu]

As rb1957 suggested, but skip the iterations. Take the canned solution for bending of this case scenerio and make the top beam bending deflection equal to the bottom.

 

[zekeman]

"Now write the three static equilibrium equations on the mass.
3 equations 3 unknowns.Not having done this problem in eons, i t also has to satisfy the energy equation, -Wdy=strain energy 1 +strain energy2; so I now have an additional equation. What to do??
Still thinking.........."

The 3x3 set should yield a solution, dx,dy,d@ and the energy equation should therefore be redundant. Will try to prove that when I get some time.

 

[rb1957]

"3 equations 3 unknowns" = statically determinate, no?

i agree the stiffness of the vertical leg at pt2 is going to affect the lateral translation of the lower attmt pt of the weight. if the vertical leg was rigid then the lower support would essentially be a cantilever, l2 long; the weight would translate down, sharing the vertical reaction between the two beams, so that their displacement was the same. but if the lower vertical leg was a wimpy little section then it'd deflect a lot more for an applied moment and more weight would be reacted at the upper beam; in the limit all the weight would be on the upper beam.

this shows that you need to consider both deflections at the lower point (which whilst not expressly stated, could have been assumedreasoned) ... the vertical leg will deflect under the constant moment, (W-x)*l2, and axial load, (W-x), the horizontal moment will deflect under transverse load; i think it's reasonable to neglect the shortening of the upper baam as it deflects downward. and the distance between the two beams ends points has to remain the same.

Quando Omni Flunkus Moritati

 

[zekeman]

"The 3x3 set should yield a solution, dx,dy,d@ and the energy equation should therefore be redundant. Will try to prove that when I get some time. "

Just got a clearing. It's obviously redundant. Just the 3x3 set is sufficient. The proof of redundancy would be similar to having a weight at the end of of cantilevered shaft
would be
W*dy/2= strain energy=W^2*L^3/(6EI)
(corrected originally posted Wdy=strain energy )
which verifies that
dy=W*L^3/(3EI) classic formula

 

[zekeman]

pdiculous963 ,

If this is a real problem, post dimensions so we can see what assumptions we can take to get a reasonable answer.

Of course a general solution is available by the method I outlined, but involves significant labor.

 

[IDS]

Of course a general solution is available by the method I outlined, but involves significant labor.

I suppose it would be nice to have a "hand" solution, but it's a couple of minutes to put it into a frame analysis, then you can play around with the effect of varying dimensions and member stiffness values to your hearts content with hardly any labor at all.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[zekeman]

"I suppose it would be nice to have a "hand" solution, but it's a couple of minutes to put it into a frame analysis, then you can play around with the effect of varying dimensions and member stiffness values to your hearts content with hardly any labor at all. "


But many people don't have the access or resources to afford to do it that way.

 

[Cockroach]

Simple statics problem, gets amazing complicated by the addition of redundant moments and such. Yup, just a cantilever problem. I stand by my solution.

Regards,
Cockroach

 

[MJCronin]

I am guessing a little here..........but.......

Is this configuration really a vertical axis reboiler supported from a distillation column ??

If so, I would solve it as two seperate cantilever beam problems.

Most of the load of the reboiler would be supported by the cantilever pipe with the most stiffness.

You can divide the load up by the ratio of the stiffnesses.....the elbows complicate thingsas they ovalize under load

Or you can use CAESAR-II for a more exact analysis......

 

[IDS]

But many people don't have the access or resources to afford to do it that way.

There are plenty of free frame analysis programs available, including:

http://newtonexcelbach.wordpress.com/2012/12/06/frame4-version-3-05-with-support-displacements/

Simple statics problem, gets amazing complicated by the addition of redundant moments and such. Yup, just a cantilever problem. I stand by my solution.

But your solution requires that L1 = L2, but the sketch shows that L1 <> L2. Also it seems to ignore the effect of the vertical leg at the lower support.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Cockroach]

Actually Doug, if you read the solution provided near the bottom of the page, I clearly state the problem as statically indeterminate for the case of L1 unequal to L2. To show that there is a solution for the statically determinate case, it is a necessary evil that L1=L2.

In mathematics, we refer to the problem as a "closed solution set" for L1=L2 and open otherwise. Which is what I have been stating. Making use of redundant moments as M1 and M2 is pointless, the fixed foundation may suggest such a case for structures, but I impose the requirement of no deflection, M1 = M2 = 0 and the solution can be described as shown. That is obvious from the mathematics, for others, maybe not.

Regards,
Cockroach