Representing Beam as a Spring and positioning it, to achieve same deflection value. 3

[Housexn]

Hallo Engineers,
I really need your help urgently.
I have this problem as part of my project work.
I have a beam fixed on the wall and on-top of that beam, another beam with more less 1/3 or the total length of the initial beam. The two are attached together. with this, i will have two different stiffness xEI for the (attached beam + the mother beam) and the remaining 2/3 of the mother beam with EI. I used the method of super-positioning to achieve the total deflection. BUT my aim is to represent the beam on the top using a Spring and position it withing the 1/3 of the mother beam, to achieve the total deflection as when the two beams were attached
I have attached a file on this thread to give you a clear view of the problem.
Any tips with highly be appreciated as soon as you can. Thank you in advance

 

[BAretired]

It depends how the two beams are attached together. If they are attached at a point and the lower beam is free to flex between that point and the wall, then you need a spring whose stretch is equal to the deflection of the upper beam at the point of attachment under an assumed load.

If the two beams are continuously welded together such that the combined EI is equal to xEI, the problem has no solution because it is not possible to generate a bending moment diagram which satisfies the correct end deflection for all values of P.

BA

 

[Housexn]

@ BAretired, thank you very much sir, for your reply, its painting some clear picture now.
So, how do you go about it to find the Spring which will represent the upper beam for the CASE 1 that you just described. My problem is how to generate that spring and to position it. guide me on the steps further if you have some time. any attachments if POSSIBLE. Thank you.

 

[IDS]

Why not model what is there, rather than trying to get an approximation with a spring?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Housexn]

@IDS, Sir, the initial problem is that, i have an Hat(Omega) stringer. and with this, am trying to model it as a double/triple beam so that the mechanical properties of the Modeled Beam, will represent or be almost equal to that of the full Omega stringer. and to do this, i need to create the Beams and position them to the point that i can get some realist values which can match. If you check my attachment, you can see the upper beam which represents the lower flange of the omega stringer, with this lower flange, i have to model it as a beam and position it(beam) within the range distance of the flange.
N/B the most important result is that the deflection achieved when i have the full model, SHOULD correspond to the deflection achieved when i model the flange as a beam. As per now, am trying to model the beam as a spring to enable me to position it TO achieve the goal.

That's my goal Sir.Thank you.any reply will be appreciated highly.

 

[bpstruct]

It seems like, from what you said in your initial post, that you have solved the problem. But in case I'm missing something....

If it is attached at a point like you suggest. You equate the two deflections.

Delta(top beam) = function of P1 at attachment point
Delta (bottom beam) = function of P1 at attachment and P at end

Delta (top beam) = Delta (bottom beam)

Solve for P1

Spring stiffness is P1/Delta
Spring location is at attachment point

 

[rb1957]

the top beam is cantilevered at the LH end, and attaches to the lower beam at it's RH end, yes? sounds like a cantilever loaded with a tip load. standard references should tell you the displacement of the tip of the cantilever due to a load P.

then you can iterate ... assume P, how much deflection for the upper beam ? solve the lower beam (a cantilever with two loads), how much deflection at the desired pt ?

or you can think about setting up the equations for the lower beam, knowing that the force and the deflection at the desired pt are related.

 

[Housexn]

@rb1957 the top beam is cantilevered at the LH end?, RIGHT..BUT its also attached to the Lower beam entirely to its RH. Think of a fuselage stringer attached to the skin (Actually, the upper beam represent a stringer flange and the lower beam , the skin)...Both fixed at the far LH.

 

[rb1957]

as i understand it, the upper beam is cantilevered at the LH end and attached to the lower beam at the RH end. I'm assuming that this is just a shear attachment. so it'll behave like a cantilever, yes? the load it reacts is dependent upon the deflection of the common point.

i'd solve this iteratively ... assume P = 100 (1000?) lbs, calc the upper cantilever deflection. for the lower beam (a cantilever with two loads), calc the deflection at the common pt). then re-guess P and continue untill the deflection of the upper beam matches the lower beam.

knowing the upper beam results, a simple cantilever, you could derive a general result for the lower beam and get deflection as a function of P for the lower beam and solve the two beams.

 

[paddingtongreen]

Question. Is the right hand end of the upper piece, fixed to the lower in the vertical direction? the horizontal? or both directions? It makes a difference, because the top of the lower member wants to get longer and the bottom of the top member wants to get shorter.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[msquared48]

Equate the two deflections for the two beams at the same point and solve for P for a specific deflection.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[BAretired]

After all of these posts, we are still no closer to understanding the problem.

We know that the upper and lower beams are fixed at the left end, but we do not know how the two beams are attached to each other. If they are attached continuously, the problem is not soluble because no combination of spring location and stiffness in Setup 2 can simulate Setup 1.

If (a)the two beams are attached at a point by a tension member which can carry only vertical load and (b) the lower beam is free to flex between the left end and the attachment, the problem is easily solved by substituting a spring with proper stiffness at the same location in Setup 2.

Please tell us clearly how the two beams are fastened together.

BA

 

[Housexn]

it is actually attached using a tension member and my problem is to calculate the stiffness of this substitute Spring.

 

[structSU10]

Then take the deflection formula appropriate for the beam. if it is a cantilever with a point load at the end, then it is delta = P*L^3/3*E*I. you want your spring constant in kip/in so rearranging 3*E*I/L^3 = P/ Delta. your spring constant is 3*E*I/L^3. If this is not your situation, just take the same approach with whatever deflection formula of the upper beam is appropriate.

 

[BAretired]

Hooke's law states that
F = kx
where
x is the displacement of the spring's end from its equilibrium position;
F is the force exerted by the spring; and
k is the spring constant
From the above equation, k = F/x.
As stated by structSU10, the spring constant,
k = 3EI/L³
where I is the moment of inertia of the upper beam and L is the distance from the fixed end to the tension member connecting the upper and lower beams. The spring must be located at distance L from the fixed end.

The gap between upper and lower beam must be sufficient to allow the lower beam to flex freely. Otherwise, Setup 1 will behave differently than Setup 2.


BA

 

[Housexn]

Thank you all, it was really helpful, I got some ideas from all this post and everything was highly appreciated.