# Bending theory

1. Feb 18, 2009

### jet10

I would appreciate it if some one could help me with this.
I read from Timoshenko's book Theory of Elasticity that for prismatical beam bending (uniaxial), the assumption is
$$\sigma_x>>\sigma_z,\sigma_y=0$$
where $$\sigma_x$$is stress in the direction parallel to the deflection curve.
And for uniaxial plate bending the assumption is
$$\sigma_x,\sigma_y>>\sigma_z=0$$
and
$$\epsilon_y=0$$
where x is the direction parrallel to the deflection curve.
I don't understand the assumption for uniaxial plate bending. Why should it be different to the assumption for uniaxial beam bending? Why is $$\epsilon_y=0$$for plate bending?
What is the difference between uniaxial plate bending and uniaxial beam bending apart from the fact that the thickness of the plate compared to its width is much smaller than for a beam?

2. Feb 18, 2009

### Mapes

I don't have the book handy, but are you sure the strain assumption isn't $\epsilon_z=0$ for plates, where z is the direction of plate thickness? This would make more sense.

Alternately, we might assume $\epsilon_y=0$ for beams. Either way, the assumptions don't seem symmetric the way you've written them.

3. Feb 19, 2009

### jet10

I have checked with Timoshenko's Theory of Plates and Shells. There is no mistake in what I wrote. It even says "The lateral strain in the y-direction must be zero in order to maintain continuity in the plate during bending.."
This follows from Hooke's law that
$$\epsilon_x=\frac{\sigma_x}{E}-\frac{\nu\sigma_y}{E}$$
and
$$\epsilon_y=\frac{\sigma_y}{E}-\frac{\nu\sigma_x}{E}=0$$
and thus,
$$\epsilon_x=\frac{(1-\nu^2)\sigma_x}{E}$$.

This assumption would imply that that cross section of the plate (y-z slice) has to remain exactly rectangular and not deform into a trapezoid like in beam bending. This resistance against deformation leads to stress in the y-direction. But I don't understand why. There shouldn't be a difference between bending a beam and bending a plate, I think.

Last edited: Feb 19, 2009
4. Feb 19, 2009

### Mapes

OK, got it. Thanks for checking; I thought that Timoshenko was modeling simultaneous and identical bending moments in both axes, and couldn't understand why the y strain component alone was being selected to be zero. Instead, however, it looks like the only load is a y-axis bending moment (that is, a line in the x direction will curl up under load).

The reason why the difference exists between a beam and a plate is that the beam is free to contract or expand laterally along its width due to Poisson's effect. It will contract ($\epsilon_y<0$) where $\epsilon_x>0$ and expand where $\epsilon_x<0$. Because a beam's width is insignificant (by definition), the lateral contraction and expansion is also insignificant, and edges in the z direction remain nearly vertical. Accompanying this unopposed strain is an absence of stress in the y direction.

A plate's width, however, is not insignificant. There is no way that the top or bottom face of a plate could contract a fraction of its width while the other can expand. (Imagine what that would look like.) To keep vertical edges vertical, we instead assume that the strain in the y direction is zero; this results in nonzero stress in the y direction. The consequence is that plates are stiffer than beams, even for identical loading per unit width. And you've given the factor, $1/(1-\nu^2)$, in the equation you wrote above.

5. Feb 19, 2009

### jet10

Thanks very much. Yes, I think it does answer my question. So it is like two thin layers that are glued together. The top tries to expand and the bottom tries to contract. None of them succeeds because the other is doing the opposite.

6. Feb 19, 2009

### jet10

Just one more question: at which width does a beam become a plate?

7. Feb 19, 2009

### Mapes

You're welcome. You may be interested in some brief http://john.maloney.org/Papers/Generalized%20Hooke%27s%20Law%20(3-12-07).pdf" [Broken] that I wrote recently, including the increase in effective stiffness of plates.

Last edited by a moderator: May 4, 2017
8. Feb 19, 2009

### Mapes

There's a gradual transition as the width increases to become a nonnegligible fraction of the length. (For beams and plates we've already assumed that the thickness is negligible compared to the length.)

9. Feb 19, 2009

### jet10

What are the criteria for the fraction to be nonnegligible? Is it just a matter of definition?

10. Feb 19, 2009

### Mapes

The criterion is often taken to be an order of magnitude. It would be interesting to see the details of where the transition occurs; you might find it plotted in a solid mechanics test somewhere. You could also simulate the geometry if you have FEA capability. Perhaps another forum member has seen a plot of the transition region (bending stiffness as a function of w/L).

11. Feb 19, 2009

### jet10

Thanks a lot for the help. I will see if I can find such a plot in a paper somewhere. Your note is a nice summary. Thanks for that too

12. Mar 1, 2009

### Mapes

Prompted by your question, I decided to simulate beam/plate deflection by FEA and compare it to the analytical solutions,

$$\delta=\frac{4PL^3}{Ewt^3}$$

for cantilevered beams subject to a transverse end load and

$$\delta=\frac{4(1-\nu^2)PL^3}{Ewt^3}$$

for cantilevered plates, as derived in my link.

As shown in the attached image, the end deflection smoothly transitions from the plane stress to the plane strain solution as the beam width increases (that is, as the beam transitions into a plate). The thickness was 1% of the length for all simulations.

The results support the rule of thumb that the beam solution is accurate when beam width is at least an order of magnitude less than beam length.

#### Attached Files:

• ###### Cantilevered beam-plate FEM vs analytical deflection.png
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13. Mar 3, 2009

### jet10

I can see here that the stiffness really depends on the width to length ratio, and the factors derived analytically are only approximations. The diagram is illuminating.