Beam Deflection 1

[hydromech]

Can anyone help with a the equation for calculating simple beam deflection.

I have a beam with a point load and a UDL, when I calculate the deflection using the moments to left of x I get a different answer than when I consider the whole beam with the point load.

I differentiate both equations twice using the same value for x and also the constants for each equation and the value when I consider the whole beam is always approx half that of when I consider only half of the beam.

what is the equation that considers the point load and the UDL..?

Thanks

Adrian

 

[rb1957]

what's the difference between "McCauley's Method" and double integration ?

looking at desertfox's post (5:55, 4/Apr) it looks to be the same ?

and whilst i like someone going back to 1st principles to solve a problem, doesn't anyone use Roark (or any of the other standard methods) ? with a central popint load and a UDL, the answer is two very short sums ... WL^3/(48*EI) + 5wL^4/(384*EI) = (8W+5wL)*L^3/(384*EI)

 

[desertfox]

Hi rb1957

The only difference is that brackets with negative numbers are ignored and it makes I believe the intergration easier.
You can solve it by doing the double intergration method but if you try to do it with both loadings combined I think the intergration gets long winded.
There are other ways as have been suggested Strain Energy and the variation Castiglano's, Superposition etc. which are all valid.

http://www.freestudy.co.uk/statics/beams/beam tut3.pdf

Here is the link posted in my original post have a look it contains several methods.

desertfox

 

[hydromech]

It's not working out for me...


We all agree that the max deflection occurs in the middle of the 5mtr beam.

Following desertfox's example

Macaulay's method = -5.089x10^3 mtr

Using the roy mech method, the point load and UDL should equal -5.089x10^3 mtr.

The point load uses lengths a+b to calculate the load in the middle and gives 3.04x10^3 mtr

The UDL equation gives the maximum 2.0413x10^3 mtr. The problem that I see is that this equation uses the length at 5m. The maximum deflection occurs at 2.5m.

Using 2.5 for the length instead of 5 gives a whole different answer.

Any ideas..?

Adrian

 

[desertfox]

Hydromech

Quote:-

"The point load uses lengths a+b to calculate the load in the middle and gives 3.04x10^3 mtr"

"The UDL equation gives the maximum 2.0413x10^3 mtr. The problem that I see is that this equation uses the length at 5m. The maximum deflection occurs at 2.5m."

If you add the two values 3.04 and 2.04 X10^-3 together you get the 5.089mm answer.

The Roy Mech link I gave you gives the maximum deflection for the beam but what it doesn't say and I suppose it should is that those deflections occur at the centre of the beam. The formula on the link are correct you are supposed to use the length of the beam (L) in the formula.

desertfox

 

[desertfox]

hi hydromech

Go to pages 8 and 10 of the link you will find examples of a single point load central in the beam span and a UDL load for both beams simply supported and how to calculate each deflection.

http://www.freestudy.co.uk/statics/beams/beam tut3.pdf

desertfox

 

[hydromech]

Got it now...

thanks again

 

[desertfox]

your welcome

desertfox

 

[desertfox]

Hi hydromech

I presume now your all sorted with this beam.
I just wondered whether it was just methods of solving it you were interested in rather then the actual beam deflection?
the reason I ask is I can show you other ways of solving it if thats what your looking for.

desertfox

 

[hydromech]

Desertfox...

It started with me trying to work out the deflection of the beam to understand the level of compensation I would need for a control system for a hydraulic cylinder that would push again the beam.

I could calculate the deflection using standard methods and got the answer...no problem!

The issue was in the checking...the engineer in me wanted everything to check out and be equal and balance. That is where I got stuck.

I also wanted an equation that could be put into Excel to plot the deflection at various point across the beam.

With your help, I got what I needed so thanks once again.

Regards

Adrian

 

[desertfox]

Hi hydromech

Well it sounds an interesting application I did wonder what an hydraulics engineer was doing getting involved with beam deflections (GRIN).
Now you have told me the application it does bring two questions to mind:-


1/ Was the UDL the mass of the beam ?

2/ I assumed that the point load and UDL were acting in the
same direction is that correct? if the point load you
gave us was the cylinder then should it have been acting
in the opposite direction.

If the UDL was not the mass of the beam then you would need to include it in the calculations.

regards

desertfox

 

[hydromech]

Yes the UDL is the mass of the beam + a load suspended from the underside of the beam. All equal to -15KN/m.

The point load is the cylinder that is push against the cylinder. This load being equal to -70KN.

Both load are in the same direction.

Cheers

Adrian

 

[desertfox]

hi hydromech

Well that clears that up then.

regards

desertfox

 

[hydromech]

Having re-read my last post, I see that it makes no sense at all.

When the cylinder is extended, it pushes against a resistive load thus deflecting the beam that the cylinder is mounted on.

The stated loads and their direction were correct, the rest was nonsense...sorry.

Cheers

Adrian

 

[desertfox]

hi hydromech

I didn't se the error but once you confirmed the directions of the load that was good enough for me.

desertfox

 

[LSThill]

desertfox (Mechanical)

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