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Deflection of beam check...
4

Deflection of beam check...

Deflection of beam check...

(OP)

Hi,

Its been a while since ive needed to calculate the defelection of a beam, in doing so i have came up with two quite different answers, using Macauleys and the direct deflection equation. calcs attached, any pointers in my errors would be soundly appreciated.

Thanks!

RE: Deflection of beam check...

2
C1 has negative sign. Also there are problems with signs in the integrations, but they cancel out.
If you take this simple problem in symbolic form, recognizing, as an example, that RA=RB=wL/2, it will be much simpler to avoid trivial mistakes.
Look also in the first site below under Beams -> Single beam -> Simply supported -> Unif.load (or here) to check your result.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Deflection of beam check...

yes, C1 negative.

also deflection units (N/mm*mm^4/(N/mm^2*mm^4) = mm
and index ... i expect something closer to E-17.

but good on you for trying 1st principles ...

RE: Deflection of beam check...

(OP)
Thanks prex for comments i noticed my C1 error and now corrected.

also took a look at the link - nice tool thanks, it was showing a maximum deflection of 83x10^3 mm defelction . ..?

@ rb1957 - also got myself muddled up on my units thanks.

however just using the single equation i get -8.34x10^-15 mm.

just one last question is my Ixx value ok it seems quite large.....?

thanks bert.

RE: Deflection of beam check...

it's a pretty darn'd big beam, 2m deep, 1"thick, 2' wide ...

quick calc, 0.8*78^3/12 + 2*24*.8*39.4^2 = 91247 in4 = 3.8E10 mm4

RE: Deflection of beam check...

Hi Bert,
Firstly your Ixx value is correct. I calculate it to be 3.61*10^10 mm^4.
Regarding the deflection, I agree with your value on the standard equation ie 8.34mm*10^-12.
I am going to look at the Macauley method in more detail. It's good to get the old grey matter working, however, I think if it were me I would have just gone with the standard equation.
Regards,
desertfox

RE: Deflection of beam check...

I typically start with EI*d^4y/dx^4 = -w, integrate four times, solve for the boundary conditions, and get the answer that Bert2 ultimately got.  For a uniformly distributed beam with simple supports, you would end up at 5wL^4/384EI anyway...so one could start there.

Seems the only thing at issue is the computation of I.

Regards,

SNORGY.

RE: Deflection of beam check...

(OP)
@ rb1957 - yeah the beam is large; which in turn i guess gives a large Ixx value.


@Desterfox - Thanks for the confirmation, await your response to the macauley's.

To conclude the beam has a very small amount of deflection at 8.34mm*10^-12.?

Thanks for the help guys!
 

RE: Deflection of beam check...

Have you considered it might buckle first?

RE: Deflection of beam check...

20mm thick caps, deflecting nothing ... doesn't sound like a buckling issue

RE: Deflection of beam check...

it's a pretty darn'd big beam, 2m deep, 1"thick, 2' wide ...

quick calc, 0.8*78^3/12 + 2*24*.8*39.4^2 = 91247 in4 = 3.8E10 mm4


Seems like a bigger problem (if in real life) will be the depth of the member: At nearly 2 meters depth and 15.7 meters long (51-1/2 feet) combined a relatively "thin" but very wide caps, won't the failure mechanism be the web twisting?  (Unless you could always ensure you have a perfectly symmetric and even load exactly centered on the mid-plane of the beam every day with no induced moment being ever transferred to the beam.....)

Seems several intermediate stiffeners are essential if he wants little deflection.   

RE: Deflection of beam check...

(OP)
@ prex - could you elaborate how you came to the conclusion on 8.3mm deflection / what method did you use? - thanks.

@ racookpe1978 - i can see your point but with the small amount of deflection i dont see this as a problem.

bert.

RE: Deflection of beam check...

Hi Bert,
I initially agreed with your calculation on the deflection and I got the same value, however we both made the same error, which was that we mixed our units. The 'I' value that we calculated we both used in the standard equation and the units being mm^4. However the rest of the units were in metres. If you change the units of 'I' to metres m^4 you will find the deflection comes out at 8.34mm as stated by prex in his earlier post.
I also calculated the deflection by direct integration and got 8.34mm deflection again. I cannot post my solution yet but will do on Friday when I return home.
Still looking at Macauleys method and will post later.
Regards,
desertfox

RE: Deflection of beam check...

(OP)

Ahh of course mm^4 for 'Ixx' thanks desertfox !

RE: Deflection of beam check...

Hi Bert2

Your welcome!
I'm believe Macauley's method is applicable for the for the standard case of beam loading which is the type your beam falls into.
Use the direct intergration method its much easier.

desertfox  

RE: Deflection of beam check...

sorry I meant not applicable

RE: Deflection of beam check...

Other than refreshing your memory, which I will admit we all need to do at times, you are taking the long hard road to the solution.

The AISC steel book, and many other handbooks, have the solution for the deflection for any point "x" along a uniformly loaded simply supported beam, to include the maximum point of deflection, the center.    

Mike McCann
MMC Engineering
http://mmcengineering.tripod.com
 

RE: Deflection of beam check...

(OP)
thanks desertfox! great help as always!!!

RE: Deflection of beam check...

Hi Bert2

your welcome!

RE: Deflection of beam check...

Some belated comments on this problem:

- You really need to have an idea of the expected magnitude of deflections to pick up gross errors straight away, but even without that it's obvious that a deflection approaching 1 billion light years is a little on the high side.

- A few minutes entering the deflection formula in a spreadsheet would pick up the error with the sign of the exponents, but that then gives a deflection of 8.3E-18 mm, which is also obviously much too small, being much less than the diameter of an iron atom.

- The E value for steel used in the original calc was in Pa (N/m^2), whereas the load was in KN, so one or the other needs to be adjusted.  This was obviously picked up in later iterations, but no-one mentioned it. Correcting this brings the deflection up to 8.3E-15 mm, but this is still obviously much too small.

- I don't know how the units of mm^4 for deflection in the original calc were arrived at, but this should have rung alarm bells, because it should have been in m (as the span length) and not raised to the fourth power.

- I don't agree that Macaulay's Method is inapplicable to this problem.  With a uniform load over the full length it reduces to the integration method.  It doesn't provide any advantage for this case, but checking this sort of simple example does provide a quick check that the method is being applied properly.

- I have recently posted a spreadsheet for analysing continuous beams using Macaulay's Method on my blog.  It now allows for single or multi-span beams with cantilevers and spring supports (translation and/or rotation) and optionally shear deflection.  There are six posts on the subject, the first one (including a link to the latest version of the spreadsheet) being:
http://newtonexcelbach.wordpress.com/2011/11/12/beam-actions-and-deflections-by-macaulays-method/

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
 

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