# Beam deflection and curvature radius formula doubts

• FEAnalyst
In summary, the equation for deflection of a simple-supported beam with force applied in the middle is derived using two equations: the first is the curvature radius multiplied by the length of the beam, and the second is the cosine of the angle between the force vector and the beam's original direction.
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 ?

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

Last edited by a moderator:
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

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?

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

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

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

I just don’t get why ##\delta^{2}=0##.

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.

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.

Lnewqban
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.

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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.

FEAnalyst said:
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.

Last edited:
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.

FEAnalyst

## What is beam deflection?

Beam deflection is the bending of a beam under an applied load. It is a measure of how much the beam has moved from its original position.

## What is the formula for beam deflection?

The formula for beam deflection depends on the type of beam and the type of load applied. For example, the formula for a simply supported beam under a point load is (wL^3)/(48EI), where w is the load, L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia.

## What is the curvature radius formula?

The curvature radius formula is used to calculate the radius of curvature of a beam at a specific point. It is given by the formula R = (1 + (y')^2)^(3/2)/y'', where y' is the first derivative of the beam deflection equation and y'' is the second derivative.

## How do I calculate the deflection of a cantilever beam?

To calculate the deflection of a cantilever beam, you can use the formula d = (wL^3)/(3EI), where d is the deflection, w is the load, L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia.

## What factors affect beam deflection?

The factors that affect beam deflection include the type of beam, the type and magnitude of the load applied, the material properties of the beam, and the length and support conditions of the beam. Other factors such as temperature and moisture can also affect beam deflection.

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