# Stress and Displacement of Cantilever Plate with a Uniform Load

Does any one happen to know of any resources that describe the stress and displacement of a uniformly distributed load in the direction perpendicular to the length direction of a plate analytically? I have checked both Roarks Formulas for Stress and Strain and Plates and Shells Theory and Analysis by Ansel Ugural and I do not see anything quite what I am looking for. Hopefully the images below help.   • Are you worried about the exact details of the stress distribution near the free edges? If not then use the normal beam equation to get the deflection at x1, and you know the angle it has bent so the deflection at L is deflection at x1+angle_in_radians*(L-x1) Feb 3 at 5:29
• This might be a start: engineering.stackexchange.com/q/3541/10902 Feb 3 at 6:43
• I am more interested in knowing the displacement roll off near the edges. We have some location sensitive parts. I am sure I could simply run a lot of FEA's to get my answer, but I thought it would be better to see if there was an analytical solution first, as we have a few different cases. Thank you Solar Mike, I will have to take a think on that and see if it helps. The post most similar to what I was looking for was like this: physicsforums.com/threads/bending-of-a-cantilever-plate.994320 Feb 3 at 20:47

If you are after the deflection of the beam at any point along the beam (but also for most intends and purposes - see below for exceptions) the problem you are after is equivalent to the following 2D case. The concentrated load at distance $$a$$, can be considered as applied to the entire width of the beam, so no asymmetry is present in the system. Then you can go to a table from a resource like Design Aid 6 and get the equations Notable exceptions would include the following:

• (heavily) anisotropic materials are employed (in some of those cases it is expected to have warping and twisting which would render the application to every point non possible.
• if the material is too soft, and you can expect the Hertzian pressure to affect the displacement locally at the force application point.

There might be others cases, but I already consider the above cases too exotic and unusual to be encountered, in engineering calculations that would be expected to be solved with closed form solutions.

• It is a complicated case if you need an exact answer. depending on the support condition if you don't have lateral shear restrain on the support the stresses vary. the deflection is a curve under the load, not a straight line. The same phrase as your question if googled brings some sources up. If you need to compare to FEM there are sources! Feb 3 at 17:46
• Hey NMech, I appreciate the write up, however this is not what I am looking for. The key assumption that I am breaking is that the beam remains plane before and after the bending, so an aspect ratio where L is ~10+ times greater then other dimensions. Out of curiosity I ran some super general FEA compared to beam bending of a plate, the difference is ~5-15% out of line of the FEA. While this might not be a problem for most applications this surface has some high sensitivity to displacement. You might be interested in looking at Roark's text book as they have analytical solutions to many cases. Feb 3 at 21:01
• @3BeaversInATrenchCoat So if I understood what you are saying is that the beam has a very low aspect ratio and as a result you are getting something like this. If that's the case, I'd suggest you put that infromation into the question, so that other people also can find it without digging into the comments. There are improved models that include shear induced rotation (e.g. Timoshenko beam).
– NMech
Feb 4 at 7:55

For bending of wide plates, we typically use the so-called plane-strain modulus $$\frac{E}{1-\nu^2}$$, where $$E$$ is Young's modulus and $$\nu$$ is Poisson's ratio. For $$\nu=0.3$$, for example—typical for ductile metals—the plate is predicted to be 10% stiffer than a narrow beam.

The reason is that the narrow beam is free to expand laterally on the compressive side and contract laterally on the tensile side, both due to Poisson effects. Because of the nearby free surfaces on either side of the width, the lengthwise cross-section is considered to be in a state of plane stress; lateral deformation is unencumbered. (This can be a point of confusion: Recall the Poisson effects aren't driven by external lateral stress; they occur freely because of the lack of an external lateral stress.)

This isn't the case with the wide plate, whose cross-section is instead in a state of plane strain (minimal strain allowed in the width direction). Now those Poisson effects are suppressed by an internal strain (in the width direction) acting on most of the lengthwise cross-sections (the $$L-t$$ plane in your diagram). As is typical, any extra constraint makes the material stiffer, and correspondingly $$\frac{E}{1-\nu^2}>E$$.

I derive various moduli in a page on generalized Hooke's Law (in progress) here. Briefly, generalized Hooke's Law

$$\varepsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}-\frac{\nu}{E}\delta_{ij}\sigma_{kk}$$

couples the 3-D strain $$\varepsilon_{ij}$$ and stress $$\sigma_{ij}$$ in isotropic materials. (Simple Hooke's Law $$\sigma=E\varepsilon$$ strictly applies only to axial loading of long, thin bars.) Here, $$\delta_{ij}$$ is 1 if $$i=j$$ and 0 otherwise, and $$\sigma_{kk}=\sigma_{11}+\sigma_{22}+\sigma_{33}$$. With $$\varepsilon_{33}=0$$ for plane strain and $$\sigma_{11}$$ loading, we have $$\sigma_{33}=-\nu\sigma_{11}$$ (the widthwise stress that arises because the wide plate can't easily shrink on one side and expand on the other), and thus

$$\varepsilon_{11}=\frac{1-\nu^2}{E}\sigma_{11},$$

whereas the plane-stress case would instead yield $$\varepsilon=\frac{1}{E}\sigma_{11}$$.