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Mounted Vessel Statics Problem

Mounted Vessel Statics Problem

Mounted Vessel Statics Problem

(OP)
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.

RE: Mounted Vessel Statics Problem

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

RE: Mounted Vessel Statics Problem

(OP)
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?

RE: Mounted Vessel Statics Problem

yep

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

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

RE: Mounted Vessel Statics Problem

(OP)
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?

RE: Mounted Vessel Statics Problem

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

RE: Mounted Vessel Statics Problem

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..........

RE: Mounted Vessel Statics Problem

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..........

RE: Mounted Vessel Statics Problem

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.

RE: Mounted Vessel Statics Problem

"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.

RE: Mounted Vessel Statics Problem

"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

RE: Mounted Vessel Statics Problem

"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

RE: Mounted Vessel Statics Problem

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.

RE: Mounted Vessel Statics Problem


Quote:

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/

RE: Mounted Vessel Statics Problem

"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.

RE: Mounted Vessel Statics Problem

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

RE: Mounted Vessel Statics Problem

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......

RE: Mounted Vessel Statics Problem

Quote (zekeman)

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/fr...

Quote (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.

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/

RE: Mounted Vessel Statics Problem

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

RE: Mounted Vessel Statics Problem

i wonder why the problem is determiinate if L1 = L2 ? the load won't 1/2 between the two beams even if they have the same I 'cause the vertical leg of the lower beam means that the deflection of the lower point won't be the same as the upper point. and even if it does 1/2, (same I), you're obtaining that result from other than static equilibrium ... you're saying the deflection of the two loadpaths is the same based on the geometry of the problem ... with different Is the deflection of the two points is still the same (from geometry, not from equilibrium).

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

"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. "

Oh really, so you are saying the pinned solution is the same as the built-in case. Seems like you are taking a lot of liberties here.

RE: Mounted Vessel Statics Problem

no deflection and M1 = M2 = 0 are incompatiable ... isn't it either no deflection and a fixed end moment or no moment and a pinned joint ?

i think your solution (11:51 14th March) is treating the problem as a three force body ... you're relying on the beams being fixed to the load, otherwise they can't carry transverse load. as a three force body, then sure your's is a (the) solution. if there was a direct loadpath for the vertical load, i'd be inclined (no pun intended) to agree with your approach. but i'm not so sure the original sketch is intended to work the way you propose, and it is more likely that the beams are pinned to the load; sure that is not specifically spelt out in the original sketch, but ...

just my 2c ...

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

I wonder if somebody with access to all those powerful programs would post a solution for a few different beam values so maybe we can put an end to this endless speculation .(Personally, I think this problem does not lend itself to a simple model.)

I propose:
L=10,L1=10,l2=3
L=10; L1=20, L2=10
L=10,L1=10, L2=0

I did these with the flexibility matrices and Excel and got some interesting results but I'm prone to making errors so I need verification before I present my results.
In general,however I was surprised to find how sensitive the system is to the relative length of L2.


RE: Mounted Vessel Statics Problem

Here's my solution inluding the bottom vertical beam. My goal was to find the sharing of the load between top beam and bottom beam. I only showed loads on my links that are necessary for this task so take it easy when you see Fx not shown in there.

http://imgur.com/a/JDGtC

RE: Mounted Vessel Statics Problem

Quote:

I wonder if somebody with access to all those powerful programs would post a solution for a few different beam values so maybe we can put an end to this endless speculation

Well anyone with access to the Internet can access my frame analysis spreadsheet, but since I know how it works I'll spend a few minutes doing the analyses.

The flexural stiffness of the supports is also significant, so I have done runs assuming concrete supports with a cross section of 0.2, 0.5, and 1.0 m and an elastic modulus of 20,000 MPa. I have assumed all dimensions are in m and applied a load of 1 kN. For the geometry the length of the vertical downstand also needs to be considered, so where L = height between supports, L1 = top horizontal, L2 = bottom horizontal, L3 = vertical, I have looked at 4 cases:

1: L=10,L2=10,L1=3, L3 = 2.5
2: L=10, L2=20, L1=10, L3=2.5
3: L=10, L2=10, L1=1, L3=2.5
4: L=10, L2=2, L1=1, L3=1.0

Note that I have reversed L1 nd L2 to be consistant with the original sketch.

Results are attached; FX1, FY1, M1 are reactions at the top support and FX2, FY2, M2 are at the bottom support.


Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Mounted Vessel Statics Problem

"Here's my solution inluding the bottom vertical beam. My goal was to find the sharing of the load between top beam and bottom beam. I only showed loads on my links that are necessary for this task so take it easy when you see Fx not shown in there. "

Good approach, but why are you assuming a pinned connections and the absence of the horizontal forces limits the analysis. In my analysis, for the example, I did, the horizontal forces played a dominant role.

RE: Mounted Vessel Statics Problem

(OP)
Yes, it is a vertical axis reboiler off a column. The reboiler has support lugs, however it is hovering off of its structural steel. Dimensions are as follows l1=29 inches, l2=42 inches. And the vessel empty weight is 3287#.

I agree that Caesar II would be the way to go for analysis, but I like to be able to verify the numbers that come out of programs such as Caesar II; it helps remind me why I went to college. When I treated it as a 2 beam cantilever problem (ignoring the vertical leg), I get a reaction force at (a) of 2972 pounds and a load at (b) of 304 pounds. Modeling this in caesar results in a load of about 3200# and 315# respectively.

RE: Mounted Vessel Statics Problem

something doesn't look right, between your hand calc and the ceasar results ... ceasar gave higher loads at both beams, and neither matches the applied load (your hand calc is close, the ceasar reaults might be vector sums ?).

also, it sounds like the reboiler is pinned to it's support beams, so we can't neglect the fixed end moments.

of course an easy (conservative) way to analyze the structure would be to put all the load through the top beam, a moment of 3287*29 = 8000 ft.lbs (empty, but loaded wt ?)

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

Zekeman, I think it's a good assumption to assume pinned at the vessel since this would yield a conservative solution at the supports 'a' and 'b'. But if pdiculous963 were to insist that it isn't pinned at the vessel then the problem can still be solved with the same approach.

Concerning the forces in 'x', I have to admit that I made a mistake on link 3. The angle of rotation of link 3 will be further increased by the horizontal load. Therefore I should have superimposed the solution for the angle of rotation of the cantilever beam scenario.

I honestly believe this is the best approach to take if solving by hand.

RE: Mounted Vessel Statics Problem

i'd've thought the horizontal forces were small and determinate ... consider the reboiler as a free body, with the load on the center-line and the (vertical) reactions offset (on the side of the vessel) making a couple (independent of how the load is distributed between the two beams) that needs to be balanced by a horizontal couple. now this'll change slightly as the vessel rotates slightly (due mostly to bending of the lower beam's vertical leg), thought i'd expect the rotation would reduce the offset.

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

"When I treated it as a 2 beam cantilever problem (ignoring the vertical leg), I get a reaction force at (a) of 2972 pounds and a load at (b) of 304 pounds. Modeling this in caesar results in a load of about 3200# and 315# respectively. "

Since vertical stiffness is inversely proportional to third power of lengths, I would expect a larger proportion than 10%, namely

(29/42)^3=0.329

RE: Mounted Vessel Statics Problem

but i think bending of the vertical leg at 2 create large displacements at the vessel end of the beam ... unloading the lower beam.

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

In the 2-beam cantilever assumption with pinned ends at the vessel, I believe the reaction at the top beam should be W L2^3/(L1^3 +L2^3) = 2472 Lbs (see my posted solution above).

RE: Mounted Vessel Statics Problem

same as zeke posted previously.

i think that assumption (that the lower beam is fixed at the end of the horizontal leg) is unconservative for the upper beam. the vertical leg is loaded by a moment, so the end will rotate (theta) and this'll cause theta*L2 deflection at the vessel end, which IMHO will be significant

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

Rb1957, my solution includes part of that rotation due to the L2 beam bending. I forgot the bending due to the horizontal load though.

RE: Mounted Vessel Statics Problem

i don't see how, given your previous post distributing the load by the ratio of the lengths cubed. that says (to me) that both beams are cantilevers, L1 and L2 respectively. the support for L1 is clearly much stiffer than L2's vertical leg.

Quando Omni Flunkus Moritati

RE: Mounted Vessel Statics Problem

I worked out both....the general solution taking into account the rotation then I presented the solution of the specific case where the vertical leg lenght is small/negligeable (as a check to the general solution).

RE: Mounted Vessel Statics Problem

This is getting frustrating.

To make the problem determinate possibilities are:
Two equal horizontal legs, both pinned at one end, in which case the vertical load is distributed 0.5:0.5 (top:bottom)
Two unequal horizontal legs, top pinned at one end, bottom pinned at both ends, load is distributed 1.0:0.0
Legs as shown on the sketch, top pinned at one end, bottom on a vertical or horizontal roller. For a vertical roller the load distribution is 1.0:00. For a horizontal roller it depends on the relative length of the horizontal legs. For the lengths given (29 inches top, 42 inches bottom) I get 3.231:-2.231.

But none of these are applicable to the problem presented. Even if connections at the boiler are pinned it is an indeterminate problem, so we need to know the length of all three legs and their flexural stiffness, and the distance between the connections at the boiler. With that information it is a two minute job to analyse it in any 2D frame analysis program, or if you prefer a 10 minute job to do by moment distribution, or your preferred "hand calculation" method. Treating it as a determinate problem is a gross over-simplification, and even designing both connections to take the entire vertical load isn't necessarily conservative, since the bottom leg could be in tension.

I just don't understand the desire to either over-simplify it or feed it into an industry specific (and no doubt very expensive) analysis package.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Mounted Vessel Statics Problem

IDS,
Thank you for the posting the 4 cases.

My spreadsheet solution is close in most of the cases, but I think the change of my moment equilibrium equation on the reboiler due to the rotation ( since I sum the moments about the top connecting point at the reboiler and the CM shifts a horizontal distance leftward of L@/2) may account for some of the differences.

In any case, your solution shows that the simplified hand calculation
is problematical on several levels as you pointed out.

I guess the message is that a simple looking indeterminate system is not easy to solve without computer power.



RE: Mounted Vessel Statics Problem

I need to say that moment distribution by hand is more than a 10-minute exercise. Closer to an hour or more to double check the tedius algebra.

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