It this post I am going to try to tie together all of the recent posts about the forces on beams, moment of inertia, and the stress-strain curve. This is an overview and not number crunching like the last post. Since we are dealing with unknowns with our popsicle sticks we really don’t have any numbers to crunch, but as always if we have an understanding of what is important and what isn’t we know how to improve things.

If you think about it, the purpose of a column is to support a weight above the surface of something. The simplest answer to that problem is to bring in more dirt and get the ground up to the height we want to have the weight. Obviously that is not a good answer, we might like to live or work in that area if it happens to be a building. The opposite extreme is to completely bridge over the area so we have a large unobstructed space. (Columns are kind of bad in the middle of an ice skating rink.) In that last case we would still probably want columns around the edge of the building, unless it is a very very big “A frame” building. So the problem is: How big do the columns have to be to support the weight?

In the post “Stress, Strain and finding center” we talked about the stress on a material and that was defined as force divided by cross sectional area. The force could either be tension or compression as our columns will experience. The stress will deform the material and up to a certain point the material will “spring back”. This deformation was called strain and the one term I forgot to mention was the **modulus of elasticity **which is defined as the unit stress divided by the unit deformation. In other words, in the simple terms I like to think in, the spring constant. The higher the modulus of elasticity is the harder it will be to pull a given amount of stretch out of the material, or in the image I have in my head, the stronger the spring is.

A few posts later in “How a beam resists bending. (Moment of Inertia)”, we applied a bending force to the beam and it experienced both compression and tension at the same time. The analogy here was that we have two springs, one on either side of the center of the beam and one spring is being compressed and the other is being stretched. At that point I started talking about the moment of inertia and probably bored you to tears with a bunch of math. The whole point of that was that when it comes to resisting bending, not only is the area of the material important, but how far it is from the center of the beam is even more important.

Now imagine we have a chair and and three structures, lets say each is 40 ft. tall and you have your choice on which structure you want to place the chair on top of and sit in it. One is a rod of metal. Another is a plate of metal, fairly wide, but very narrow. The third is a tube or the same material or in other words, a large diameter pipe. All three have the same cross sectional area, but it is obvious all three have very different moments of inertia. Since all three have the same area and the force would be the same, the compressive stress would be the same in all three cases. However, common sense says the pole and the plate would probably bend and buckle while we might have a chance with the large diameter tube.

Of course people wanted a way to quantify the difference. The first number that is used is called the radius of gyration. This number allows us to compare different shapes and does so by calculating a distance that produces the same moment of inertia if all the area was at that distance. If the structure has a different values in the x direction than it does in the y direction there would be two different radius of gyrations. Radius of gyration is given the value k and k = the square root of (I / A ) where I is the moment of inertia and A is the cross section area. So what does this do for us? Again this shows that the more of the area we can put the farther from the center of the structure the better off we will be because the radius of gyration will be higher.

In the late 1700’s a Swiss mathematician named Euler was struggling with the problem of buckling beams and came up with the formula. shown in the picture to the left. Since we are not going to calculate anything using this formula, the important concepts for us is to understand the relationships. As the slenderness ratio becomes larger (L gets longer or k becomes smaller) the acceptable compressive force per square inch becomes smaller. Likewise, if a material that has a a smaller modulus of elasticity is used the acceptable pounds per square inch load is much less. That is why my rubber eraser column in the first picture is not performing very well.

Euler’s formula is not used very often except to explain the principles. There are different formulas used for different materials and different authorities also have different formulas, charts, and tables for their areas. Many of those have been developed from experience and testing. So although we saw a very interesting formula, we really can’t use it except in the way we have which is to understand what is happening and what is important. This is especially the case in our popsicle construction where our columns are not uniform.

Timoshenko is = L/A10

I am glad you were able to determine the answer for yourself. I would have had to try to look that up because I am not familiar with the term. Although I am an engineer, I am not a civil engineer and am self-taught in that area. That is why I limit my design to popsicle sticks mostly.

Dear Mr/Mrs

Thank you for you.

What is main concept Timoshenko and Euler-Bernoulli Beam?

Is important the Radius of Gyration in both Beams?

Rg=sqrt(I/A)

and,

Rg=r/2

and,

Can you help me?

Timoshenko is = L/A10

or

Timoshenko is = h/L>20

and

Euler-Bernoulli = h/L<20

Which is correct?

I will be pleased to hear from you.

Best Regards,

Mostafa keshavarz Emami

Emami2435@yahoo.com

989191894158