Strain at the Fibre in Tension of One of Two Beams Stacked 4

[Logan82]

Hi,

I want to follow up on this post (

https://www.eng-tips.com/viewthread.cfm?qid=498797&action=updated),

but with a slightly different question.

The assumptions are the following:
- Inertia_total = Inertia_{bottom_beam} + Inertia_{top_beam} (this was the conclusion of this thread (

https://www.eng-tips.com/viewthread.cfm?qid=498797&action=updated)

- The top beam has less inertia than the bottom beam
- The two beam shapes are different.
- The two beams are not linked to each other, they are just stacked.
- We can assume that there is no friction.
- The two loads are symmetrical.
- In reality there are bolts, but there are not enough bolts to make sure that the two beams are linked.
- The deflections of the two beams are equal (δ₁ = δ₂).

I have the following situation:

Questions:
- What would be the maximum strain at the tension fibre of one of those two stacked beams?
- More specifically, what should I take for the distance between the neutral axis and the fiber in tension "y"?

 

[driftLimiter]

Cant you just determine what proportion of your loading ends up in each member then analyze it as an isolated beam, Calculate the stress at the extreme fiber then use Hooke's law to determine strain? I think because your taking the assumption that there is no composite action then you can just can just find the relative stiffness of each one.

Pb1 = P total * (Ib1) / (Ib1 + Ib2)

 

[Just Some Nerd]

I'd probably write it out in terms of moments rather than P out of personal preference, but basically same solution as you Drift by proportioning out the load. Assuming fully elastic behaviour for this also

Calculate the combined moment M first
M1/Ib1 = M2/Ib2 if deflections are the same (similar solution for differing elastic modulus, I think it just becomes M1/(E1*Ib1) = M2 / (E2*Ib2))
Solve simultaneous equations M = M1 + M2 and M1/M2 = Ib1/Ib2 knowing M, Ib1, Ib2

Following that σ = My/I and ε = My/EI are easy to calculate for each beam

 

[Tomfh]

Agree with the above.

Note however that the maximum slip between between the beams will occur at the ends, where the beam moment and hence the strain is zero.

 

[rb1957]

won't the two beams behave as one if they are firmly attached together ? This is not summing the Is but I of the composite section.

this will tell you have much shear is happening at the interface, and so design the fasteners accordingly.

If you want to say the fasteners are elastic then there will be a limited slip at the interface. (and a much longer analysis)

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Logan82]

Thank you all, I used what you suggested.

won't the two beams behave as one if they are firmly attached together ? This is not summing the Is but I of the composite section.

this will tell you have much shear is happening at the interface, and so design the fasteners accordingly.

If you want to say the fasteners are elastic then there will be a limited slip at the interface. (and a much longer analysis)

This is a structural evaluation of existing structures, so it's either the fasteners pass or fail, I can't change them.

Good idea to check the max slip that they can endure.

 

[rb1957]

well surely you can add fasteners if needed, to avoid a "fail" ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[3DDave]

The top beam has two point loads and the bottom one has two point reactions, but the loads between the two won't be point loads. It's possible the beams don't maintain full contact.

Not sure what to do besides the previously suggested approximations which should be adequate.

 

[driftLimiter]

Winkler springs as the bottom beam ? I feel that the solution is bounded by the assumption that the point loads are distributed to each beam.