Beam Deflection

[_MechEng95]

Hi there,

I am struggling with working out maximum deflection in a beam.

It has two point loads, at L/4 on each side and a distributed load at L/2.

I know max deflection for a distributed load but I am not sure how I calculate both at the same time.

Do I have to treat this problem as three seperate beams? Calculate for each 'cross section' and add deflections? Really haven't got a clue.

The beam is simply supported with pin and roller boundary constraints.

Ty in advance.

 

[rb1957]

research "superposition"

another day in paradise, or is paradise one day closer ?

 

[XR250]

Just put both point loads in the middle and calc. that deflection. It will be conservative but fine for most purposes.

 

[DaveAtkins]

5ML^2/48EI is another quick way to approximate deflection.

Superposition will get you a better answer.

DaveAtkins

 

[CBSE]

rb1957 has it right. Use superposition. Setup a spreadsheet that calculates deflections for point loads, distributed loads, etc. and many different locations. Then add the deflections at each location together. This will give you your deflection.

 

[_MechEng95]

Cheers, I will look in to superposition.

The beam is 6m long, Point loads at 1.5m & 4.5m and a distributed load across the length.

Should I calculate the deflection at 1.5m with the point load and multiply it by 2 or calculate deflection at 1.5m and 4.5m seperately?

 

[BAretired]

Use the Conjugate Beam Method for an exact result.

Properties of Conjugate Beam
The length of a conjugate beam is equal to the length of the actual beam.
The load on the conjugate beam is the M/EI diagram of the loads on the actual beam.
A simple support for the real beam remains simple support for the conjugate beam.
The point of zero shear for the conjugate beam corresponds to a point of zero slope for the real beam.
The point of maximum moment for the conjugate beam corresponds to a point of maximum deflection for the real beam.

BA

 

[BAretired]

With symmetric loading, the maximum deflection occurs at midspan and is equal to:

Pa(3l[sup]2[/sup] - 4a[sup]2[/sup])/24EI + 5wl[sup]4[/sup]/384EI

BA

 

[rb1957]

the problem is symmetric so that makes sense. the trick is to calculate the deflection at 3m due to the 1.5m point load, double this for the other point load, and add the distributed load deflection (max deflection at L/2)

another day in paradise, or is paradise one day closer ?

 

[_MechEng95]

@BARetired, is that the overall equation or the deflection at 1.5/3/4.5?

I worked out deflection to be ~6mm however, ANSYS modelling suggests it should be arround 14. Of course there may be error in my model on ANSYS.

 

[BAretired]

@ME95,
It is the deflection at midspan for two loads 'P', each at a distance 'a' from each support plus a uniform load 'w' across the entire span. It is also the maximum deflection for that combination of loads.



BA

 

[_MechEng95]

https://gyazo.com/166343a6b10ccf380eefd975b1565a56

So that is the beam. It has a cross section of (0.3x0.35m)

Am I correct in thinking deflection is as follows:

2{Pa(3l[sup]2[/sup] - 4a[sup]2[/sup])/24EI} + (5wl[sup]4[/sup]/384EI)

or do I just make P the value of the sum of the point loads?

This help is really appreciated. Helping my understanding a lot.

 

[BAretired]

No, the factor '2' should be removed from the first term and the value of P is the value of just one point load.

Δ = Pa(3l[sup]2[/sup] - 4a[sup]2[/sup])/24EI + 5wl[sup]4[/sup]/384EI

where Δ is the deflection at midspan (also the maximum deflection in the beam)

The first term takes into account the fact that there are symmetrical loads P placed 'a' away from each support. The second term is the deflection due to a uniformly distributed load.

BA

 

[_MechEng95]

On a seperate note, is this method of superposition applied to finding the natural frequencies of a beam such as this as I can only find formulae for beams with point loads or uniform loads, not a combination.

 

[BAretired]

I don't remember. Calculating natural frequencies is not something I do as a rule.

BA

 

[_MechEng95]

Thanks for clarifying that. Much appreciated.

 

[BAretired]

You're welcome.

BA

 

[_MechEng95]

The current beam has loads at 1.5m from each support. Is a = 1.5m, or the mid-point of the beam?

 

[rb1957]

"The first term takes into account the fact that there are symmetrical loads P placed 'a' away from each support." a = 1.5m (and the calculation is for the mid-point deflection given the two loads, symmetric loading).

another day in paradise, or is paradise one day closer ?

 

[Tmoose]

Calculations/modeling of Natural frequencies of a beam must consider masses, not loads per se.