Beam loading problem

[310toumad]

Looking for some guidance to see if my methodology is correct in setting up this problem. Lets say you have a cylinder assembly that raises a fork/carriage structure with an overhanging load. The forks rest on the carriage via a formed strap on the top, and butting up against the bottom carriage tube. The overhanging load introduces a force couple at this location:

M = Fd1
M = 750*36.53 = 27398 in-lbs

Reactions in the force couple are equal to:

27398/d2
27398/15.5 = 1,767 lbs

So, approximating the long beam as fixed at one end at the top (bolted connection) with a roller support at the bottom (wheels tracking on cylinder casing), can I assume that a moment load (27,398 in-lbs) exists as shown? Or do I draw the FBD using the actual concentrated loads of the force couple that I solved for (1,767 lbs)?

If I did the latter, then you basically have one of the applied forces pressing down directly on the roller, so the reaction is at the same exact location. So the other force of 1767 at a distance of 75.79 would result in a moment at the fixed location of 133,920 in-lbs. That doesn't seem right though, because that's basically assuming that load is cantilevered with no support.

I could not find much info on beam formulas for the scenario of considering it as a moment load except this:

http://www.mathalino.com/reviewer/s...pped-beam-moment-load-span-area-moment-method

Following that method of R = 3M/2L yields R = (3*27398)/(2*91.29) = 450.2 lbs.
The moment at the wall would be M = M - RL = 27398 - (450.2*91.29) = -13,699 in-lbs

So unless I'm doing something wrong, considering the load as a moment load and using those equations would give me a much lower stress level than if I just modeled the forces as point loads. Am I missing something? Any help would be appreciated.

 

[Shu Jiang PEng SE]

the structure will be an indeterminate structure if the top end of the column being treated as fix. you can treat it as pin support

 

[WARose]

I ran it myself......using a variety of scenarios (one load case with it inputted as a couple, the other as just a pure moment)......and in all of them, the biggest reaction/force I got anywhere was 27.4 in-kips.

Keep in mind: I didn't consider any moment from vertical eccentricity (adding to the moment caused by the couple; I assume that would be fairly minimal anyway). And also: I only considered the forces/moment from the 750# force with the 36.5" eccentricity. (I.e. I disregarded that load you have at the top.....I assume that was just a reaction you drew in.)

 

[rb1957]

I can live with your couple load, but also add the 750 lbs axial load to the beam.

one 1/2 of the couple is reacted by the support below it, the other 1/2 is a point load on a propped cantilever (your 1767 lbs reaction at the far end is "wrong", it is more like 1767*15.5/91.29 about 1/6*1767 or about 300 lbs).

as noted above, this is an indeterminate problem (a single redundancy) but a sanity check for the reactions is Mapp/91.29 ... the applied moment is coming out as a couple (true if pinned support). The real reaction is (Mapp-Mfe)/91.29 where Mfe is the fixed end moment.

Google "propped cantilever" and you should get links to help you.

another day in paradise, or is paradise one day closer ?

 

[jec67]

The loads that comprise the couple is the proper way to go. In fact, that is the situation you have. Since you have a fixed support at one end and a roller at the other, it is an indeterminate structure. You could treat the top end as a pin; however, the connect will have some fixity (based on your sketch). You will need to analyze using a software application or find a classical solution (or derive one). Roark's may be a source - sorry, I don't have one available at the moment.

To echo another response, the reaction you calculated at the bottom is not correct. The load that is 15.5" above the support will pull the beam to the left, thereby reducing the reaction.

You may want to give some thought to fatigue concerns at the fixed end. It might be better to design the connection in such a way to accommodate rotation. Analysis will then be easy as it will be pin-roller beam. Making the analysis easy is not the goal of reconfiguring the connection (if you decide fatigue is an issue) - just an added benefit.

Regards

 

[310toumad]

So being that the beam is indeterminate, can it not be solved by simple superposition and using the deflections to solve for the reaction at the roller? See attached. Is the solution really any different if you consider the force couple as a moment load rather than the two concentrated loads?

 

[rb1957]

"can it not be solved by simple superposition and using the deflections to solve for the reaction at the roller?" ... no, it cannot.
"Is the solution really any different if you consider the force couple as a moment load rather than the two concentrated loads?" ... no it isn't, particularly in your case were the moment really is a couple; ie the moment is carried into the beam at two discrete points (a couple) rather than one concentrated point (a moment).

redundant structures cannot be solved by the standard equations of equilibrium, and so additional equations are needed. a simple method is unit force ... in your case remove the fixed end moment, solve as a SS beam. now calculate the rotation that would occur at the fixed end. then apply a unit moment as a load and calculate the rotation. now you can calculate the fixed end moment (the moment required to drive the end rotation to zero) and now you can solve the beam reactions.

another day in paradise, or is paradise one day closer ?

 

[canwesteng]

rb1957 - you just describer the method of superposition at a different location than the OP.

OP - you can calc deflection at the roller, then determine the force at that location required to bring deflection back to 0, and then solve via equilibrium. That is a valid method to determine forces.

 

[dhengr]

310toumad:
I’m not sure I really understand your structure and system, or that I understand your problem, or that you have defined your problem very well. Isn’t it true that the forks, both the horiz. tines which carry the load and their vert. legs which support the cantilevered tines, act as one unit and move up and down the two larger vert. beam/column members. As such, for a given load and load position the calcs. M = (750)(36.53") = 27398 “#’s and 27398/15.5" = 1767 #’s don’t change. They are the moment on the horiz. tines and reactions on the vert. fork legs to keep the system in equilibrium, and this fork lift tine system can move up and down the two larger vert. beam/column members. Of course, the DL of the fork tine system should be included in your calcs.

The reactions don’t change in magnitude, but they can change vert. position on the two larger vert. beam/column members. The left beam acts basically like a vert. simple beam with the 1767 # load acting to the left, but at varying vert. locations, so you can calc. the beam bending moment, shears and reactions. The right column in your sketch is a round pipe section and contains some means for lifting the fork lift tine system (a lifting chain or hydraulic cylinder or some such). At the moment, I wouldn’t consider either of the two larger vert. beam/column members fixed, certainly the four bolt top connection won’t provide much fixity. The bottom of the right column must be fixed or held in some way so it acts as a vert. cantilever beam/column. The right column has a right acting varying vert. location point load of 1767 #’s, a left acting reaction load at its top and a vert. column load equal to (2)(750 #’s, plus the weight of the fork tine sub-system). Is this a stationary lifting system or is it actually on a traveling fork lift? Fill in some of the details, do some free body diagrams (FBD’s) of the various components, in their proper orientations, and come back with corrections to my word picture, or with other questions.

 

[BAretired]

Like dhengr, I am not at all sure that you have stated the problem accurately. I doubt that you have a fixed condition at the top or a roller support at the bottom. However, assuming that you have properly stated the boundary conditions, you could make the structure determinant by removing the bottom reaction, calculate the lateral deflection as a result of the two loads, then calculate the bottom reaction required to reduce that deflection to zero.

Placing an applied moment halfway between the two applied forces is not exactly the same as applying the two forces as it results in a different moment diagram between the two forces.

BA

 

[310toumad]

The forks do not move up and down the vertical members, they are welded to the vertical members. The entire carriage/fork assembly moves up/down along the cylinder casing as it is raised as one unit. The wheels are also attached to the long vertical members. BA I will try to solve the way you have described.

 

[BAretired]

Unless I slipped up, the required reaction is 0.74776 P or about 1321#.
Edit: I slipped up. The reaction is 0.25224P or about 446#.
BA

 

[BAretired]

Applying a moment halfway between the two forces should produce a very good approximation of the correct answer.

Solving using the same method as before, gives:

RL³/3EI = (Ma/EI)(a/2+b)
where a = 83.54"; b = 7.75"; L = 91.29"; M = 27,398"#

So R = 446.9# which agrees very well with the previous post.

As expected, the two equal and opposite forces can be replaced with a moment in this instance. Agreement would not be as good if the forces were separated by a greater distance.


BA

 

[jec67]

In my original post on this subject, I recommended applying the load as a couple. My reason is, presumably, the internal shear forces will also be required. Applying the moment as a couple will allow accurate calculation of the shear diagram for the beam.