Simply Supported Deam Beam Deflection

Variation in one digit is only 1/10 as significant as in the preceding digit.

Stress in the beam is dependend on where the load enters the beam. You have three scenarios so you would expect changes.

Post some pics. Always good to see what one is talking about.

Thanks for the response IdeaSmith,

See attached two of the scenarios I'm looking at - could you explain why I'm getting a big difference in deflection between the two scenarios?

Appreciate the help

Regards

Dan

There could (and should) be a small difference, probably very small. Perhaps you could give the relevant dimensions and the software's calculations.

Post 1 was entirely on topic, regardless of any particulars in the discussion. I don't know which fool thought otherwise.

First figure, bigger deflection. Moment radius from pin point are larger compared to bottom figure, also when Length is much longer deflection is directly proportional.

when Length is much longer deflection is directly proportional

No. Deflection is proportional to the square of the length,

Cube of length.

Fourth power of length, for a given load per metre. Max y = 5*w*L⁴/(384*E*I). You can check the dimensions

You weren't the only one!

I'm not by any stretch of the imagination proficient in mechanical stress calculations however;

In the top scenario you are using only half the width of the beam whereas in the bottom scenario you are using the full width of the beam to disperse the load weight.

Hence; The bottom scenario will allow almost twice the loading factor.

No

Danny, when you remove the support of the beam, as in the case with the "support", or lack of, at the top of the beam, what do you think will happen?

I think it will split.

Homework makes me crancky.

Just a quick guess on this one, but a beam pinned at the midpoints at both ends would have less "pivot" at its ends then a beam pinned at the top or bottom of its end points. More pivot range in the end points will result in more beam movement. Minimizing pivot range in the end points will limit beam movement. End point pin placement will alter tensile, shear and compression forces by limiting or maximizing movement of the beam.

Thanks for the responses gents,

My beam is 6m long, 3m depth and my load is 8kN/m.My deflection is around 40% greater when the beam is pinned at bottom compared to when its pinned in mid height. Please note that I do have a pin at one end of the beam and a roller at the other end

Thanks

Regards

Daniel

You haven't mentioned its cross-section, which might be significant.

sorry thickness is 0.5m...

.5 meters??? That's around 19.5 inches thick!

Okay, I read "deep beam" as a general description in the OP.

Deep beams behave differently from normal depth beams, but I am not sufficiently familiar with them to have an opinion on whether the position of the pin would have that much effect.

I suggest that when the pin is low, there is vertical compression of the material above it; when placed high, there would be vertical tension in the material below it. These would add to the flexural deflection.

I hate these impossible hypothetical questions, a beam such as this could not be supported over such small sections of its depth. I wonder how you applied the force to the beam without failing the local material.

For a beam under a vertical uniform distributed load would there not always be compression in the beam material immediately below the load?...I fail to see how the support point would make a difference here in relation to tens/comp. For instance if a beam is vertically loaded then surely the half of beam above the neutral axis is always in compression with the half of beam below the neutral axis always in tenion

I agree, unless the ends are constrained to reduce strain (as others have suggested) I don't see how the support point makes any difference. But if the ends are so constrained it's no longer simply supported, the ends must remain horizontal (I think) and it becomes a fixed end case. That gives max deflection (at centre) 5 x less than simply supported, if I remember my Roark. I would re-check the calcs.

But a beam 3m deep x 0.5m thick is a piece of work, specially as it's only 6m long. What is it made of? Presumably not concrete or you wouldn't be talking about tensile loads. What is the applied load?

Its just a generic coursweork problem with a modulus of elasticity of 30kN/mm2. applied load is 6kN/m. The beam is just simply supported, not fixed...

That's about the right elastic modulus for concrete (without reinforcement). Are you sure it's not a trick question?

You're right there is local compression under the load, but that is at right angles to the beam axis so does not contribute to deflection. There is also local compression at the supports if these are underneath. In practice the is much more likely to be under than at midpoint or on top.

Also you give a load 6kN/m, but the dead weight (concrete at 2500kg/m³) is about 37 kN/m. But still with a beam that size I make tensile/comp stress only 0.257MPa, and max deflection 0.02mm.

You need to give more information on your calcs to comment on why defections are different for the 3 scenarios.

I've given all the information I can..what am I missing?

I think the q I ask is more of a general question and I'd imagine the behaviour that I'm seeing for the different support points is fairly similiar for all deep beams no matter what the specific dimensions and loads i have. I seem to be going around in circles....

You are missing experience that comes with trying to solve difficult engineering problems.

At least you're asking good questions.

Try to find some studies done by designers and see how your numbers look, compared to theirs.

Do you belong to any organizations that may help students?

I know very little about FEA, but I'm sure it must start from initial conditions, and presumably you entered different data for the 3 cases, giving rise to different outputs. If you can give some information on inputs we might be able to comment further.

"For a beam under a vertical uniform distributed load would there not always be compression in the beam material immediately below the load?"

Absolutely. If you then place the reaction under, say, the last 0.5m at each end. This puts a vertical force into the beam only at those points, the material is under high compression vertically. This force would eventually be distributed as you get higher, you would have to design for that compression if this were a real thing. Think of them as columns at each end that picks up the reacting force and distributed it evenly up the end of the beam. Look at how you put those reactions into the beam and where the forces/pressures go.

You have an FEA model, check out how the load distributed itself through the beam ends.

I've never had to deal with beams where the support point varied vertically. Yes, above NA is compression and below is tension; however, I suspect the NA shifts with the different support conditions. That is, with the support at the bottom, wouldn't the NA be along the bottom and the entire beam then is in compression? Then you must consider buckling--it may control. I think that is part of the criteria for "compact" beams. Also consider that the whole beam can roll over by lateral buckling. Support at the top may also be controlled by lateral buckling. I remember being told that a bar joist should NOT be loaded until the lateral bracing is installed or it may roll over.

I think you made either an error of 1 order of magnitude (only) or your FEA soft is not correct. In such "wall type" beams shear plays a much more important role as in slender beams (L/h>20) where the flexion is the major.

I was amazed by the 40% so that I made the test with same dimensions and the differences are ONLY around 5% (the Young module does play a role only for absolute values of deflexion but not for the ratio between the 2 cases.

You will best understand what happens if you consider the "stress flow" through the beam to the supports and compare both cases.

If your FEA allows it make the simulation if not let me know and I shall make it for you.

Can you get a printout of the deformed shapes and compare them.

You have the FEA program and three solutions, what is the difference, not just in deflection, but in the deflected shapes of the elements. (I assume you used plate elements).

I still cannot understand why you, and some others, cannot see that the bottom of the beam would be crushed when it is seated on the support. Under that high load, there will be strain, in the vertical direction to be added to the flexural deflection.

I still don't know how you input the supports, were the single points or what?

Lest you think I am being tiresome, I am a retired professional engineer and my discipline is structural. The fact that you can't see what I am trying to tell you comes down to you, not me. You appear to want an answer without doing any more thinking.

When you calculate the properties,of the beam at the center, you will find that the deflected beam is shorter at the top and longer at the bottom (because of shear stresses). When you calculate the simply supported cases you must take the new beam lengths into account. This is covered by "Roark and Young" under "Beams of relatively great depth".

I suspect that if you try your FEA with a variety of beam thicknesses, you will find that the solutions converge as the beam thickness is reduced. This should describe and explain the the properties of relatively thick beams.

I agree with the structural engineers in saying that for your scenario, the beam could and would crush at the simple support points due to excessive stress accumulation in the vertical, and I am just a chemist. Like a wire cutting through cheese.

The answer to your homework (shame on you), is EΨ=HΨ, and you need to find the eigenvectors of green cheese on the moon.

You have a deep bean, with two tiny points of support under a huge load, obvious trick question, something has to give.

If I don't get an off-topic on this I will be shocked, but I never mark myself off-topic after reading some of the drivel that presents itself here.

brilliant!

In theory a simply-supported beam rests on points or maybe knife-edges, giving theoretically very high local stress. I'm not strictly a mechanical engineer, but I'm sure in practice the supports would have enough area to give realistic bearing stress, while still counting as simple supports for calculation of bending stresses, deflection etc.

With deep beams come shear deflections. Is this something you have covered in your recent course work ?

The position of the supports will change the shear forces through your FEA model. I am not sure, but maybe this has some measurable effect.

The rectangular beam section has been chosen in this problem because you can work it out by hand calcs too. Do a web search for deep-beams-shear-deflections.

Is your FEA model linear or non linear ?

How are your results outputted ? Are you showing the outer fibres of the beam in each case ? Sometimes it is the output that is not the same and so not comparable.

Hi

Would you please be able to tell me how to manually calculate the bending and shear stresses in a simply supported deep beam (with uniform distributed load) so that i can try to verify my results? They are different to a normal thin beam i believe. I can't find anything on it in Timoshenko book

Any guidance is much appreciated

The best thing is to do the web search. I did it and found some very good source material. For instance : http://www.aboutcivil.org/deep-beams-concept-applications-assumptions.html

Timoshenko is excellent. But try Den Hartog too. He explains things a bit better for a first approach. With worked examples.

For a beam length L, depth D, breadth B, with UDL w/m.

Max shear stress is the load on each support (1/2 the total load w*L) divided by beam area B*D. Zero at the centre, opposite sign each end if you're into the details.

Max bending moment M, at centre = w*L²/8. Max tensile/compressive stress = M/Z where section modulus Z = B*D²/6.

Max deflection y, at centre = 5*w*L⁴/(384*E*I) where second moment of area I = B*D³/12.

Formulas from Roark

Above assumes the supports do only that and put no longitudinal force on the beam. In that case IMHO it doesn't matter where the support is placed vertically, and the neutral axis is along the beam centre irrespective of where the supports are.

But it could be different if the supports exert force along the beam, while still allowing the beam ends to pivot in a vertical plane. E.g. by putting a hole through the beam, with a shaft constrained horizontally. The beam then bends into an arc, and the vertical force is partly taken by the vertical component of the tension in the beam at the supports. Tension constant through the depth of the beam. Think of the beam as a wire, which has negligible stiffness, but can still take a load via that mechanism. However, doing some calcs, the tensile stress comes to 3.3MPa, and the deflection 39mm i.e. much higher than treating it as a beam (see my #18) so my conclusion is restraining the ends makes negligible difference, it still behaves as simply supported. It would be different for a beam of much smaller section. Also tensile stress would not be acceptable in a plain concrete beam.

In any case I still think you need to check the inputs to your FEA. If you told it the ends are restrained as above, it might come to a conclusion different from mine, accounting for the outputs you're getting.

I don't think this applies to deep beams as its doesn't account for the effect of the shear stresses on deflection etc

Replying to Danny740 #41, passingtongreen #44 and nick name #48

That's a good point about deflection due to shear. It's not something I've thought about before, and I'm fairly sure it's not included in Roark's deflection formula. I did a calculation, a bit of integration but quite straightforward and got max shear deflection (at beam centre, obviously) = f_s*L/(4*G) where f_s = max shear stress, G = shear modulus. I couldn't find a figure for shear modulus for concrete so assumed it is E/3, by comparison with steel data. With total load 43kN/m (including dead load) I get shear deflection 13mm, vs 22mm a la Roark.

I assume total deflection = sum of the two, and the shear figure is a significant part, if my sums are right. Maybe for a more typical length/depth ratio shear deflection could be smaller, I might try it sometime. For an I-beam, I think the shear fraction would be greater, not less, as 2^nd MoA/beam area is greater than in a solid beam.

I don't see how the support details affect it. Any beam has to be supported, and the local bearing stress must be limited to a design figure. Also it doesn't explain the difference in FEA results for different support arrangements (though for middle or top support as I suggested it would not be simply a case of resting the end of the beam on something).

I don't understand why you don't see how the support affects it.

In this thread we are discussing a model. As far as I can tell, the support is a single node, The force must spread out from this point into the local material. If it causes stress in that material, it also causes strain. When the point is at the bottom of the beam, the strain must be in the material above the point. The distance from the support node to the centerline is reduced by this strain.

When the support is at the top of the beam, tension must spread out in the material below, causing strain. The distance between the support and the centerline is increased.

When the support is in the center, the is compression above and tension below.

Each of these causes different effects on the distribution but adds to the deflection.

I can understand you difficulty but please use the suggestion I made look at the "stress flow" and how it "flows" through the beam from the loaded side to the support. If you draw the "flow lines" you immediately see what is the reason for the difference.

In your thinking you are still "slender beams" where the stress is allover the section and its distribution correlates with the principle that a section stays plane after deformation which is NOT any more valid for L/H ratios as for the beam object of this discussion.

The deflection formula you quote is from the fundamental bending equations. This distribution of stresses is disrupted at the supports, this disruption is negligible in normally proportioned beams, but it is not negligible in "deep beams", The disruptions cover a large part of the total length and therefor the formulae do not apply in the OP's case.

The support forces at the ends of beams are not introduced in a shape matching the classic stress distribution, therefor the behavior is different.

Think about a beam, sitting on a wall for support. There will be a high bearing stress between them. Think about the distribution of this stress as it spreads out in the beam. Compare this with the classic bending stress diagram.

Valid for slender beams not for beams with a low ratio L/H

See:

Formulas for Stress and Strain by

R. Roark and W. Young

McGraw Hill Book Company

note the chapter on Beams of Great Depth

Roark only gives formula for deflections..can't find anything on stresses

Roark doesn't give stress directly. It gives bending moments, which are used to find stress via the section properties. See my #36.

"Roark only gives formula for deflections..can't find anything on stresses"

Remember that stress and strain are related by the proportionality of elastic modulus.