Bending Moment of 2 parallel beams 4

[ogg22]

Hi,

I am currently designing a machine that has 3 pairs of parallel beams supporting a single central unit.

I am unaware of how to calculate the bending moment when there are two beams in parallel, as a single beam it can be modeled as a cantilever beam, but if the beams are connected at either end (a rectangular construction can be assumed), how does the bending moment change? This construction has been adopted to increase the system stiffness.

Any guidance would be great.

Ollie

 

[CELinOttawa]

Post a sketch, please.

Two beams have twice the strength, three have tripple. Likewise with stiffness. The devil, however, is in the details. If the beams are not bound composite they will each resist and deflect according to their proportionate loading.

If your unit has its centroid at the centre and a reasonably stiff frame with skid type supports, the centre beam is going to see half of the load and each of the litter beams one quarter. If the unit has legs on the putter corners, the centre beam will never see any load (don't laugh my friend, I has seen this mistake be built!). You also have to watch for an offset centroid (very common) which will change the apportionment of load to the individual beams.

Sketch; Anything else is conjecture and half measure in helping.

 

[ogg22]

The attached image gives a basic example of the configuration. Initial analysis is of the bending of the beams in one orientation, with the final analysis looking at beams mounted in 3 locations equally about the triangular center piece.

thanks,

 

[ztengguy]

What book is this from?

 

[ogg22]

It's not from a book, its a concept I am currently working on. The sketch was a solid edge model annotated as a JPEG.

 

[CELinOttawa]

What you've attached is an indeterminate beam-couple... This does not readily yield itself to hand solutions.

Quick questions:

- Can the left hand side rotate? I'm betting yes.
- Are the ends of the rods (beams) fixed to the supports (triangle and box)? I'm betting yes.
- Can the beams translate freely up and down? I'm betting yes.

If so, this is going to result in a beam-column for the top beam (ie: flexural and compressive forces) and an uplift condition beam column for the bottom (flexural and tension forces).

I'm betting this isn't going to work. Your top beam-column is going to be too slender to take the imparted compression loads.

In my opinion you need a 2D FE structural analysis software to work this one out.

 

[ogg22]

You are correct in your answers, for the sake of the basic model I'm looking at it in only one plane, i.e a complicated cantilever beam. Is it possible to work out the bending moment simply for the attached image? I have access to Ansys software, I was looking to validate results using hand calcs, and possibly produce a BM diagram.

Cheers,

 

[TehMightyEngineer]

You could come up with a rough (possibly unconservative) approximation by figuring out the capacity of one cantilevered beam in your given setup and then doubling the capacity. As CELinOttawa stated this wont be accurate as one beam will be in flexural-compression while one is in flexural-tension. Should work well enough for a hand-calc double check of software though.

Maine EIT, Civil/Structural.

 

[kingnero]

cannot you make a truss out of your proposed set-up?

 

[CELinOttawa]

Okay, the biggest problem you have is restraining the motion of the left side so that you actually get flexural-compression and flexural-tension. The beam is going to do everything it can to yield in only flexure, meaning you have a serious torsional problem if you cannot or have not restrained the out of plain motion of the left hand "triangle".

You need to make sure this can work in the real world. As detailed, it will not.

Here is an INACCURATE back-of-the-envelope checking procedure:

0) Solve for the fixed end moment on the Right Hand Side (RHS) as if this was one beam.
1) Split the end reaction into two equal reactions on the RHS's two beams (NB: NOT accurate, but realistic).
2) Solve for all forces in the system as if pinned.
3) Apply fixity to each end of beams on LHS in terms of 20 or 30% of the fixed moment calculated. Deduct same from RHS.
4) Rework with the moments in your sum of forces analysis from (2).
5) Verify you have a possible solution by summing and accounting for all force applied.

Dirty, and not quick, but it should give you results within 20% or so of the "correct" solution.

Your rig must be FORCED to behave this way as well as be constrained in order to be stable.

 

[ogg22]

Thanks

 

[CELinOttawa]

Hey ogg22, One more thing: When you deduct the moment from the RHS and solve for all forces, you should find that there is now a couple (compressive force at top, tension force at bottom) on your RHS supports which compensates for the deducted moment.

You cannot have a valid solution if the tension couple plus the two RHS fixed end moments do not add up to your original input moment of 29.5kN·m. That value is assuming your drawing is to scale, and the load is actually input at the horizontal centre of the triangular LHS. [Thus 590mm from ultimate support to point of load application, making force times distance = 50kN x 590mm]

 

[rb1957]

i think the structure will work as two independent beams in bending, 1/2 shear on each ...
where you have some uncertainity is does the outbd fttg fix that end of the beam, so you have a "guided" cantilever. then these end moments (at the outbd end) would be balanced by a couple

Quando Omni Flunkus Moritati

 

[paddingtongreen]

The triangle looks very stiff to me, much stiffer than the beams, so:
Move the force to the end of the beams, calculate the moment of the force multiplied by its distance from the ends of the beam and divide it by the beam spacing. Apply these as tension and compression loads on the beams. Now divide the force by two beams, multiply by the length of the beams, divide by two ends. These are the moments at the ends of the beams, there is zero moment in the center. If the triangle is not comparatively rigid, Cel's solution is better.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[rb1957]

i think it's important to recognise that the two beams are carrying shear loads (and have increasing moments in them)
and that they are not working like caps of a single beam (which'd have increasing endload).

if the beams weren't parallel it'd behave very differently (the beams would be predominantly axially loaded).

Quando Omni Flunkus Moritati

 

[CELinOttawa]

Lol... If the sum of forces analysis doesn't show shear in each beam, I'll chew on my own shoe...

 

[paddingtongreen]

rb1957, The only beam that I know of that has bending without exterior shear is one in circular bending.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[rb1957]

i was thinking that the OP might be looking at the two bars as caps of a beam.

and yes, i know that doesn't work ('cause there's no web to take the shear, to join the two caps together).

so then the problem is "merely" two cantilever beams, each carrying 1/2 the shear (that is if you assume the loaded end is pinned). if the loaded end is fixed then you have a guided cantilever, the outbd ends of both beams resists some moment, which is reacted as a couple between the beams; could become a large deflection problem (but if it does there are probably other issues with the structure).

Quando Omni Flunkus Moritati

 

[BAretired]

Assuming:
[li]triangular center piece is a rigid body[/li]
[li]both beams are fixed each end[/li]
[li]distance from applied load to beam ends is x[/li]
[li]c/c distance between beam is h[/li]

F = 50 x/h (compression in top beam, tension in bottom)
M = 50*500/4 = 6250 N-mm in each beam

In other words, I agree with PaddingtonGreen.

BA

 

[CELinOttawa]

Ah, I missed the fact that repeats on the two other sides of the triangle, thus simplifying the problem greatly.

My analysis applies to the problem as-drawn if the triangle is free to rotate. Go with the simplified assumption and you should be okay in this case as rotation and translation relative to the RHS are fixed in reference by the two other identical "mirrorring" beam sets.