Two beams joined after bending

[abayerle]

TL;DR Summary

Take two beams and bend one to a certain radius of curvature (maybe with a uniform load). Then take the second beam that is not bend and glue it onto the first. The glue is super stiff! Now how do I calculate the resulting curvature of the glued beams? How does the stress profile before and after gluing look like?

Thinking about Bernoulli Beam Theory and how to calculate the curvature of a beam, that you first bend and then glue to another beam. The latter is flat initially. How do I calculate the resulting curvature of the composite beam? How do I get the stress diagrams right - my first idea is in the image below.
Second scenario: Imagine you have glued the one beam to a very stiff, flat surface. Secondly you bend a second beam and then glue it on top of the other beam. In the third step you dissolve the glue layer of the first beam and now the composite beam is free to relax. It will curve, right? Is the curvature of the composite beam now different from that of the first scenario.
Ultimately, I want to think of a model where I stack many beams on top of each other and release the stack after all beams are fused.
Maybe someone closer to bending physics can help me out?

 

[Lnewqban]

Welcome, @abayerle !

Anything can happen for both situations, according to many factors.
You will have initial opposite forces and moments, as well as temporary and permanent deformations.
The resulting curvature or final state will be one of balance.

 

[Baluncore]

You need to specify when a beam is being bent beyond the elastic limit, and when it is being released to take up the plastic deformation imparted by that initial bending.

You need to specify how the first bent beam (or assembly) is held during the next gluing process. Is the initial bent beam fixed or free to comply with the next laminate, while they are being clamped and glued onto it?

 

[jrmichler]

Ultimately, I want to think of a model where I stack many beams on top of each other and release the stack after all beams are fused.

This is a well known problem, especially in woodworking. Good search terms to learn more are springback laminated beams. The photo below shows 9 mm of springback in a wood beam made from three laminations clamped to a form 47.5" long with 106" bending radius. The portion of the laminated beam to the left of the bending form is straight because it is supposed to be straight.

 

[Juanda]

This seems like one of those problems that is fairly easy to explain and conceptually understand. However, the math to solve it is not as direct as it could be presumed at the beginning.
I have been thinking about it all evening but I couldn't find a proper answer. However, I will share some insights I had along the way.
For the problem I will be assuming that both beams are the same, beams have square sections, only elastic deformations, small displacements, pure bending before gluing, and infinitely strong glue. Basically as many simplifications as possible (and even then I couldn't solve it).

1. An instant before gluing:
   
   
   To get matching surfaces (same curvature at the contact points) the green beam is less loaded than the brown beam. This is because the curvature of its neutral surface is lower and the deformation at the section is inversely proportional to the curvature of the neutral line.
   $\varepsilon = \frac{y}{\rho }$
   Therefore, before gluing, the tensile diagram for those sections will look different and not equal as shown in OP's text.
   
   
2. After gluing and releasing the torque that initially deformed the beams.
   I don't think the neutral line will be at the point where the glue is keeping the beams together as OP initially proposed. In fact, that glue will be carrying the shear load from the tension caused by the beams trying to come back to their original shape.
   From the action/reaction principle, the tension the green beam causes on the brown one must be the same as the compression the brown beam does on the green beam.
   Green arrows: Force that the green beam causes on the brown beam due to the glue.
   Red arrows: Force that the brown/red beam causes on the green beam due to the glue.
   
   
   
   Each beam will retain a neutral line but it will be displaced when compared with the original pure bending case due to the forces from the glue. I think the normal tension diagram would be like this.
   
   
   
   
3. I believe the key to solving this problem is to think about the beams separately. For example, this is what the green beam would be experiencing (I will fix one end of the beam just for convenience).
   
   
   
   Where q is the force the red beam is causing on the green beam when trying to come back to its original shape while being held by the glue that links it to the green beam. So far I have not been able to obtain a value for q. Once knowing q, the problem should not be too hard to solve if the equivalency shown is true. The equivalent Moment due to the uncentered force would be $M = q*S_N*\frac{h}{2 }$ where $S_N$ is the neutral surface during pure bending. I used $\frac{h}{2}$ because I am assuming a symmetric beam as shown.

I'll keep my eyes on this post in case someone writes a solution for this. I consider it a pretty interesting problem but I feel I lack the knowledge to solve it on my own.

 

[erobz]

This seems like one of those problems that is fairly easy to explain and conceptually understand. However, the math to solve it is not as direct as it could be presumed at the beginning.
I have been thinking about it all evening but I couldn't find a proper answer. However, I will share some insights I had along the way.
For the problem I will be assuming that both beams are the same, beams have square sections, only elastic deformations, small displacements, pure bending before gluing, and infinitely strong glue. Basically as many simplifications as possible (and even then I couldn't solve it).

1. An instant before gluing: View attachment 327918
   To get matching surfaces (same curvature at the contact points) the green beam is less loaded than the brown beam. This is because the curvature of its neutral surface is lower and the deformation at the section is inversely proportional to the curvature of the neutral line.
   $\varepsilon = \frac{y}{\rho }$
   Therefore, before gluing, the tensile diagram for those sections will look different and not equal as shown in OP's text.
   View attachment 327919
2. After gluing and releasing the torque that initially deformed the beams.
   I don't think the neutral line will be at the point where the glue is keeping the beams together as OP initially proposed. In fact, that glue will be carrying the shear load from the tension caused by the beams trying to come back to their original shape.
   From the action/reaction principle, the tension the green beam causes on the brown one must be the same as the compression the brown beam does on the green beam.
   Green arrows: Force that the green beam causes on the brown beam due to the glue.
   Red arrows: Force that the brown/red beam causes on the green beam due to the glue.
   View attachment 327921
   Each beam will retain a neutral line but it will be displaced when compared with the original pure bending case due to the forces from the glue. I think the normal tension diagram would be like this.
   View attachment 327920
3. I believe the key to solving this problem is to think about the beams separately. For example, this is what the green beam would be experiencing (I will fix one end of the beam just for convenience).
   View attachment 327923
   Where q is the force the red beam is causing on the green beam when trying to come back to its original shape while being held by the glue that links it to the green beam. So far I have not been able to obtain a value for q. Once knowing q, the problem should not be too hard to solve if the equivalency shown is true. The equivalent Moment due to the uncentered force would be $M = q*S_N*\frac{h}{2 }$ where $S_N$ is the neutral surface during pure bending. I used $\frac{h}{2}$ because I am assuming a symmetric beam as shown.

I'll keep my eyes on this post in case someone writes a solution for this. I consider it a pretty interesting problem but I feel I lack the knowledge to solve it on my own.

I think you can use Castigliano’s Theorem? The strain energy of the first beam that is deflected under load is preserved in the composite beam when the load is released. I’m out of town, so I won’t be able to try it for a while. It might not be sufficient by itself though…

 

[Juanda]

I think you can use Castigliano’s Theorem? The strain energy of the first beam that is deflected under load is preserved in the composite beam when the load is released. I’m out of town, so I won’t be able to try it for a while. It might not be sufficient by itself though…

For sure the beams will find a new equilibrium when the external loads disappear and the glue is acting. Such equilibrium will have the minimum strain energy for the 2 beams but I don't know how to mathematically apply that knowledge to this problem.

As a simpler scenario, I solved the problem with axial springs. Imagine you have two springs with different lengths of equilibrium. Then, you glue them together (initial with initial and final with final nodes). The stiffness will be the sum of each spring because they will be working in parallel and the new equilibrium length will be in between the two previous equilibrium points.

The force in each spring will be
$F_{k_1}=-k_1x+k_1L_{1eq}$
$F_{k_2}=-k_2x+k_2L_{2eq}$

Then, the force of the combined spring will be
$F_{comb} = F_{k_1}+F_{k_2} = -k_1x+k_1L_{1eq} -k_2x+k_2L_{2eq}$

The new equilibrium point will happen when $F_{comb} = 0 \rightarrow F_{k_1}=-F_{k_2}$ which is the same as saying that the new equilibrium length will be at the point where the compression of the blue spring is the same as the tension in the red spring. As a result, we will have a new purple spring.
$x_{eq_{comb}}=\frac{k_1L_{eq_1}+k_2L_{eq_2}}{k_1+k_2}$

That's why I tried to solve the problem by finding the force that the red beam does on the green beam (and the other way around). That's the force I called {\color{Red} q}.

I believe the key to solving this problem is to think about the beams separately. For example, this is what the green beam would be experiencing (I will fix one end of the beam just for convenience).
View attachment 327923
Where q is the force the red beam is causing on the green beam when trying to come back to its original shape while being held by the glue that links it to the green beam. So far I have not been able to obtain a value for q. Once knowing q, the problem should not be too hard to solve if the equivalency shown is true. The equivalent Moment due to the uncentered force would be $M = q*S_N*\frac{h}{2 }$ where $S_N$ is the neutral surface during pure bending. I used $\frac{h}{2}$ because I am assuming a symmetric beam as shown.

Coming back to the simplification with the axial springs, to find the new equilibrium point it is also possible to obtain the potential energy of the combined spring $U_{comb}=\int_{0}^{+\infty }-F_{comb}dx$ and find the minimum value. However, that would require additional steps that add no interesting information.
I don't know if such a method could be more useful to solve this problem with the beams.

I'll keep watching this thread in case someone actually manages to crack the case. Although I find it interesting I will have to give it up for now.

 

[jrmichler]

TL;DR Summary: Take two beams and bend one to a certain radius of curvature (maybe with a uniform load). Then take the second beam that is not bend and glue it onto the first. The glue is super stiff! Now how do I calculate the resulting curvature of the glued beams? How does the stress profile before and after gluing look like?

Thinking about Bernoulli Beam Theory and how to calculate the curvature of a beam, that you first bend and then glue to another beam. The latter is flat initially. How do I calculate the resulting curvature of the composite beam? How do I get the stress diagrams right

Even though the OP is (apparently) long gone, this thread presents an interesting challenge. Here's how I would attack it. Start with some simplifying assumptions, because we can always add complication after getting a simple solution. Assumptions:

1) The two beams are thin relative to the radius of curvature, so both beams are bent to the same radius.
2) The radius of curvature is "large". Curved beam theory does not apply.
3) The beams are linear elastic.
4) After bonding, ignore shear at the joint. The shear is on the neutral axis, so does not affect the bending.

The initial state, Figure 1 below, has the two beams, one curved with radius R$_0$, the other straight. Both beams are stress free.

The second state, Figure 2 below, has the curved bent straight by equal moments at each end. That beam now has bending stress, the other beam is still stress free. Now bond the two beams together and remove the moments.

That results in the third state, Figure 3 below. That results in a composite curved beam that is curved to a larger radius than the initial state of the curved beam. The resulting stress state is the superposition of two different stress states.

The top beam is bent to a radius R$_f$ larger than its initial stress free curved shape. That, if done by a pair of external moments, would require a pair of moments less than M in Figure 2. The resulting bending stress is less than the stress in Figure 2, but larger than zero. That bending stress is sketched in Figure 3.

The composite beam is bent to a curve with radius R$_f$. Bending a straight beam to that radius would require a pair of moments. That pair of moments is equal to, and opposite to, the pair of moments acting to partially straighten the top beam. That bending stress is sketched in Figure 3.

The bending moment to straighten the top beam from radius R$_0$ to radius R$_f$ is equal to the bending moment to bend the composite beam from straight to radius R$_f$. Two equations, two unknowns - the moment and the final radius.

The final stress is the sum of those two stresses. The step change at the bond line is the shear stress in the glue.

I would stop here, but a more ambitious person would take it a step further and develop an analytical solution for N laminations.

 

[Juanda]

Even though the OP is (apparently) long gone, this thread presents an interesting challenge. Here's how I would attack it. Start with some simplifying assumptions, because we can always add complication after getting a simple solution.

I am all down for simplifications but I would like to better understand some points of what you propose here. I'll be writing in red commenting on your post copying only the bits that don't sit well in my mind.

2) The radius of curvature is "large". Curved beam theory does not apply.
I believe it's the other way around. When the radius of curvature is big compared with the section of the beam, curved beam theory can be applied.
https://roymech.org/Useful_Tables/Beams/Curved_beams.html

Did you mean the opposite? Meaning: The radius of curvature is "small". Implying the beams are bent a lot.
Also, if curved beam theory does not apply, what does apply then?

4) After bonding, ignore shear at the joint. The shear is on the neutral axis, so does not affect the bending.
It is necessary to have stress at the joint. Otherwise the joint would not be necessary in the first place because the elements would stay together by themselves. I believe this stress at the joint will mainly be shear because one beam wants to contract and the other wants to expand. I am not sure if even some normal stress would be necessary to keep the beams together too.
In fact, you later derive that the shear at the joint will be the responsible for the jump in the normal tension profile at the sections of the bonded beam so it has a non zero value disagreeing with what you assumed in this point.The initial state, Figure 1 below, has the two beams, one curved with radius R0, the other straight. Both beams are stress free.
The intial problem started with two straight beams. I'm not sure if by this simplifications we're solving a too different system. Besides, I don't think you can use beam theory with beams that aren't straight. And if you can, the related math is more complex than the typical math we are all used to. These days we just rely on FEA for such cases although I found this source talking a little about the analytical approach for a curved beam.
https://www.ques10.com/p/21894/curved-beams-cannot-be-designed-by-applying-the-si/#:~:text=a min ago.-,Curved beams cannot be designed by applying the simple bending,straight beams Justify this statement.&text=In straight beams, the neutral,in the beam is linear.

I guess it could be argued that if the stress free radius of curvature of the beam is very small, the linear analsysis won't be too far from the real solution. However, the main problem remains which is that the initial situation is that both beams are straight.

Do you think you can solve the problems from the starting point of two straight beams?

I am certain there must be quite some literature about this kind of problem because its got practical uses in the industry. We have mentioned carpentry in this thread (you mentioned springback laminated beams in post #4) but also a similar thing can be seen in old suspensions for heavy machinery (leaf spring suspension). I believe it's still in use but not so common nowadays.

I'll try to find more to read about it now that I finally remembered the keywords to search it online.

 

[Juanda]

I am certain there must be quite some literature about this kind of problem because its got practical uses in the industry. We have mentioned carpentry in this thread (you mentioned springback laminated beams in post #4) but also a similar thing can be seen in old suspensions for heavy machinery (leaf spring suspension). I believe it's still in use but not so common nowadays.
View attachment 328222

I'll try to find more to read about it now that I finally remembered the keywords to search it online.

Don't mind the leaf spring suspension. The leaves are allowed to slide between one another so it is not the same situation we're considering here. Besides, they don't start being straight.