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]

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

 

[zekeman]

"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.

 

[rb1957]

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

 

[zekeman]

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.

 

[renderu]

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

 

[IDS]

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/

 

[zekeman]

"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.

 

[pdiculous963]

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.

 

[rb1957]

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

 

[renderu]

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.

 

[rb1957]

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

 

[zekeman]

"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

 

[rb1957]

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

 

[renderu]

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).

 

[rb1957]

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

 

[renderu]

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

 

[rb1957]

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

 

[renderu]

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).

 

[IDS]

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/

 

[zekeman]

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.