Beams of Relatively Great Width Roark's Bending Theory 1

[sword26]

A Colleague of mine asked me a question and I couldn't answer it, so I was hoping someone else could. It's in reference to Roark's 7th edition Beams of Greater Width Analysis (Section 8.11) in which he discusses the stresses on very short wide cantilever beams. It mentions the beam bending formula can be reduced to 'stress= Km*6*P/t^2' where Km is a dimensionless factor based on a ratio of c/a, c being the moment arm of a point load, a being the depth of the beam.

Now the question is, assuming the moment arm is always half of the depth, if you increase the depth, no matter how big your moment arm is, Km stays the same since it's based on a ratio of the two dimensions, thus the stress is the same. Is there any explanation for this?

The only thing I can think of is since the analysis is assuming short beams, and Km in itself is a conservative replacement for a near zero value (c/b, b being the width and very wide would reduce that to zero), the difference in increasing the depth is negligible. But when using this to size the depth of a flange on a slat track as he's doing, issues of weight savings and cost come into play and any little bit helps, but how is one supposed to know when this analysis no longer applies, or the depth is too great? Should a simple cantilever beam bending computation with a short width be taken as well as a side check? Any thoughts from the community?

 

[GregLocock]

"assuming the moment arm is always half of the depth"

That seems an odd assumption.

Cheers

Greg Locock


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[sword26]

Thanks for the response Greg, if it's all the same, I'd rather not get into specifics as to why that assumption was made...so let's take it another way, assume c=a, signifying the moment arm is always at the edge of the flange...still c/a would be the same. My question is in regards to the theory itself, not any individual analysis or assumptions...thank you...

 

[GregLocock]

OK. As the cantilever gets wider the bending stresses local to the point load start to approach those seen in a plate encastre at one edge only.

Since I have no Roarke handy I can't check this today.



Cheers

Greg Locock


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[JStephen]

I don't have the book, so I can't see the context.

In your first paragraph, you're saying stress does not vary with width of the beam, so something is screwy there.

 

[zekeman]

"Now the question is, assuming the moment arm is always half of the depth, if you increase the depth, no matter how big your moment arm is, Km stays the same since it's based on a ratio of the two dimensions, thus the stress is the same. Is there any explanation for this?"

Isn't that statement true for a standard rectangular beam, where Km=1

s=M*6/bt^2
M=Pc
s=Pc*6/bt^2
if c/b constant, then
s proportional to P

So I think he is saying that Km being a function of c/b is constant when c/b is constant.

 

[GregLocock]

"'stress= Km*6*P/t^2' where Km is a dimensionless factor based on a ratio of c/a, c being the moment arm of a point load, a being the depth of the beam.

"

If a is the depth of the beam what is 't'?

Cheers

Greg Locock


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[sword26]

"So I think he is saying that Km being a function of c/b is constant when c/b is constant."

Initially I thought the same thing, but Roark's formula is not relating the moment arm to the width as b is infinite in a very wide flange. Thus c/b becomes near zero. There seems to be some confusion and I fully take responsibility for not putting a diagram of the problem.

a is not the width of the beam, b (not shown in the diagram) is in the z direction and as mentioned it is infinite...so the question is referring to Km being a factor of the ratio of the moment arm(c) to the depth of the beam(a)

Thanks again for the responses, look forward to reading more...

 

[zekeman]

Looks like he is saying the width of stress captured at the fixed end is equal to c times a function of c/t.

 

[sword26]

The width is essentially ignored in this analysis, so please disregard it, since it is a very wide short beam...only the moment arm (c) and the depth of the flange of the beam (a) is taken into consideration...attached is the accompanying table. Now it does offer different values for a different position on the fixed end (z) but that's not of interest to the question. Assuming one point z=0, if the moment arm (c) is always taken at a certain point [half of the depth, or full depth], increasing the depth of the flange (a), is not accounted for in the analysis, as the c/a stays exactly the same...what's the justification for that?

 

[GregLocock]

Take a normal cantilever width b

sigma/y=M/I
M=P*c
I=1/12*b*t^3
y=t/2
so sigma=P*c*t/2*12/b/t^3
sigma=P*(c/b)*6/t^2

which looks at least intriguingly similar to

'stress= Km*6*P/t^2' where Km is a dimensionless factor based on a ratio of c/a, c being the moment arm of a point load, a being the depth (I'd call it length) of the beam.

well I've found Roarke, i'll have a read



Cheers

Greg Locock


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[GregLocock]

OK, the reason is as follows.

Take the obvious case where c/a=1

The stress due to the bending moment at the root of the cantilever, in line with the load, is 51% of that of a normal cantilever. Seems reasonable, handy rule of thumb is that infinity seems to equal 2!

Now weld an additional strip of material along the infinite length of the beam, to the edge of the flange, so c/a=0.5. The stress drops to 37% of the simple case, as the additional material outboard of P helps distribute load along the infinite length.

Of course this is easily checked in FEA

Cheers

Greg Locock


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[dhengr]

GregL:
Do you have Roark’s Ed. 7? Which ever Ed. you have, would you please post those couple pages as a pdf file, that’s the format/file I can read and print best. I can’t read his png files. I have his 5th Ed. and got to the same intriguing spot you are at, yesterday, then got pulled of on something else. That is: stress = 6P(c/2z)/t^2, where b = 2z. There is a fair amount of confusion caused by mixed up terminology btwn. the OP’er. and Roark and me, in terms of width, depth, span length ‘a’, etc., etc. Look at Roark’s deflection equation, it has an “a^2" term in the numerator; and I suspect we should have a “c” (the lever arm) term in the numerator of Roark’s stress equation. But, I haven’t gotten quite that far yet.

For all the ranting I do about FEA software being the go-to solution method, means of analysis, for every conceivable problem, for young engineering people, this problem would be a good application for that software. There have also been several other threads on this general topic, here on E-Tips, do a search looking for point loads on beam flange tips, under hung trolley cranes and crane rail beams, etc. I sure we would all like to see the OP’er. do a FEA on this problem, I suspect it’s already been done many times, and show his results, so we could compare them with the above.

 

[sword26]

with all due respect, i find it a bit condescending that one assumes I have not searched the forum before posting...secondly, the terminology I'm using is straight from Roark, I dont understand why one would even respond and make such comments when they haven't grasped the scope of the problem, due to not being able to open a simple picture file(pretty much a jpg file, and almost positive pdf readers usually have the ability to open png files)...but I digress...

Greg, the similarity of roark's equation to the basic moment formula was alluded to in my first post when I spoke of c/b nearing zero at infinite width, and moreover it was flatly noted by zekeman and while I do appreciate you trying to help, I'm afraid you don't understand what I'm asking...I'm not wanting to weld any material after the fact, this is preliminary design, so we are investigating different theories...this particular one seems to have a flaw in that if c is always at the edge or c=a, no matter how big 'a' gets, you will always get c/a=1, thus km doesn't change, ergo stress doesn't either, there is something inherently wrong with this is what we suspect, however I decided to present the question to the community to see if someone with more experience and expertise could shed some light on why roark seemingly overlooked this detail...I have no desire to run a fea on this as it is a question of theory, not any specific design.

 

[GregLocock]

Shrugs, I thought my explanation made sense. I'm sorry it didn't to you. Have a good day.

Cheers

Greg Locock


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[GregLocock]

Ah OK, I didn't explain the scaling thing. If a typical cantilever's length (x in roarke) doubles, the stress at the wall due to a tip load doubles. If the cantilever is made twice as wide (z in roarke), the stress halves. As you make a cantilever twice as long and twice as wide the stress doesn't change.

With an infinite beam the stress pattern will be identical, but twice as long, at the root, if the x dimension in Roarke is doubled. The moment has doubled, but so has the available material at the root to react it.







Cheers

Greg Locock


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[sword26]

Greg, apologies if I seemed curt earlier, I assure you it was not directed at you...the part to you was simply trying to make myself a bit more clear...So it sort of sounds like we're cooking now...By your last post I gather that you've arrived at a similar conclusion as to one of my only explanations in my original post. That since b is very wide, in a normal beam formula, c/b goes to zero; however, this is obviously not reality. So Roark is basically stating a conservative assumption that a wide beam can at best reduce the stress by about half, and in his text he mentions that this is confirmed by testing.

I understand that, but it still just seems odd to me that if you start with a wide beam (with fixed width), again b is so wide it an be considered infinite even though it's not, and then are tasked with sizing the depth of the flange (a) with a point load always on the edge of the flange, by his theory, it makes no difference if 'a' is 1,2,3,4...Km stays the same...I should mention, that this analysis is for short wide beams, so I suspect he wasn't expecting anyone to size an extravagant depth with it, but still, if you are trying to save weight and what not, minute details like that count, and the small discrepancies in a close to zero margin could possibly make a difference. Then again, since Km in itself is conservative, perhaps it would not.

 

[IDS]

...secondly, the terminology I'm using is straight from Roark

Roark doesn't use the term "depth", and that is what caused a lot of the initial confusion.

this particular one seems to have a flaw in that if c is always at the edge or c=a, no matter how big 'a' gets, you will always get c/a=1, thus km doesn't change, ergo stress doesn't either, there is something inherently wrong with this is what we suspect

The expalnation given by Greg seems to have answered this point; the effective width of distribution of the load is proportional to the distance from the web, and the bending moment is proportional to the same value, so the flexural stress doesn't change. No doubt this is a simplification, but it seems a reasonable one, and it is supported by measurements.

Doug Jenkins
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