Beam deflection and curvature radius formula doubts

[FEAnalyst]

TL;DR Summary

What is the source of this formula for beam deflection based on curvature radius?

Hi,

I am working with leaf springs and studying the derivation of the formula for the deflection of such a structure. The derivation is shown here:



My only doubt is how to obtain the following formula: $$\delta=\frac{L^{2}}{8R}$$ where: $\delta$ - deflection, $L$ - length of the beam, $R$ - curvature radius. The beam under consideration is simply-supported with force applied in the middle. I haven't found such a simple equation anywhere. How is it derived ?

 

[Frabjous]

It‘s geometry. There are two equations:
Rθ=L/2
(R-δ)/R=1-δ/R=cosθ≈1-θ²/2

 

[FEAnalyst]

Thank you for quick reply. Those equations indeed give the formula from my post: $$1- \frac{\delta}{R}=1-\frac{(\frac{L/2}{R})^{2}}{2}$$ But how to formulate them? The geometry is just this, right? https://wikipedia.org/wiki/Circular_segment

 

[Frabjous]

I added a picture.

 

[FEAnalyst]

So $L$ in your equations is arc length (length of a bent beam)? In my case, it’s chord length (length of a straight beam). How does it affect the formula from my first post?

 

[Frabjous]

I haven’t done theory for a while, so it took me a while to figure it out.

Using your geometry
R²= (L/2)² + (R-δ)²
neglect terms of δ²

 

[FEAnalyst]

I’ve found a nice derivation in another YouTube video:

I just don’t get why $\delta^{2}=0$.

 

[Frabjous]

Another way is to make the assumption that R is large compared to L,δ (as opposed to δ is small compared to R, L)
If you do not neglect δ², you get

δ/R=1±√(1-L²/4R²)
≈1±(1-L²/8R²)
The sum violates the assumption, the difference gives the answer.

I am not familiar enough with the system to figure out which assumption.

 

[Lnewqban]

These may help:

https://www.hkdivedi.com/2017/10/leaf-spring-deflection-calculation.html

https://what-when-how.com/automobile/laminated-or-leaf-springs-automobile/

 

[FEAnalyst]

All right, so $\delta^{2}$ is just neglected as a small value (since we assume small deformations). Thank you very much for your help.

 

[erobz]

Unless I'm missing something, I don't think they can do what has been done for the entire leaf spring. I'm thinking it is for a single leaf, and it's not going to be a good predictor of the deflection entire leaf spring system.

Take for example the simple 2 leaf system:

We can apply Castigliano's Theorem to find the deflection at $C$, which says in general:

$$ \delta = \int \frac{M}{EI}~\frac{\partial M}{ \partial P} ~dx \tag{1} $$

$$ \delta_C = \delta_{AC}+\delta_{BC} \tag{2} $$

Due to symmetry, we can say that the contributions to the deflection at C integrating from $A \to C$ and $B \to C$ are equal, hence:

$$ \delta_C = 2 \delta_{AC} $$

Let the spatial coordinate $x$ start from $A$:

The moment as a function of $x$ is given by:

$$ M(x) = \frac{P}{2}x \tag{3} $$

$$ \implies \frac{\partial M}{ \partial P} = \frac{1}{2}x \tag{4} $$

Substitute (2),(3) and (4) into (1):

$$\delta_C = 2 \delta_{AC} = 2 \left[ \int_0^a \frac{ \frac{P}{2}x}{EI} ~\frac{1}{2}x ~dx + \int_a^{2a} \frac{ \frac{P}{2}x}{2EI} ~\frac{1}{2}x ~dx \right] $$

Simplifies to:

$$ \begin{aligned} \delta_C &= \frac{2P}{EI} \left[ \frac{1}{4} \int_0^a x^2 ~dx + \frac{1}{8} \int_a^{2a} x^2 ~dx \right] \\ \quad \\ &= \frac{2P}{EI} \left[ \frac{1}{12} \Bigg. x^3 \Bigg|_0^a + \frac{1}{24} \Bigg. x^3 \Bigg|_a^{2a} \right] \\ \quad \\ &= \frac{3Pa^3}{4EI} \end{aligned}$$

I'm sure there is a pattern to be found for $n$ leaves, but I didn't get that far.

 

[FEAnalyst]

I also wasn’t sure about the correctness of this formula but it gives good results when compared with FEA and even experiment (described in a Polish book about leaf springs). The derivation also makes sense to me. I think that the difference between it and other common approaches is that here initial curvature of the spring and force applied to "unbend" it is assumed.

 

[erobz]

I also wasn’t sure about the correctness of this formula but it gives good results when compared with FEA and even experiment (described in a Polish book about leaf springs). The derivation also makes sense to me. I think that the difference between it and other common approaches is that here initial curvature of the spring and force applied to "unbend" it is assumed.

It’s hard to say why it works. In my opinion the theory being applied in those videos should not be sound for a generalization. It’s hard to tell what they are saying though as it seems to mostly in a language other than English.

For the following generalization with 3 leaf's I get (applying the theory above):

$$\delta_C = \frac{P}{2EI} \left( \frac{11}{18}a^3+ \frac{1}{72}l^3 \right) $$

I'm not seeing how they could be equivalent.

 

[erobz]

Do over.

I was little hasty in defining the moments of inertia for each section. The above analysis isn't wrong per-se ( as far as I can tell ), its just they don't represent leaf plates of the same height in each layer.

This should be more like what's encountered with the addition of the same rectangular plate in each layer:

$$\delta_C = \frac{P}{2EI} \left( \frac{37}{72}a^3+ \frac{1}{648}l^3 \right) $$

with the condition $a < \frac{l}{4}$

Sorry if that caused any confusion.