I think Young’s modulus is used for elasticity (i.e. a rubber band stretching); How does one quantify how floppy or bendable a rod is?
If I knew the name of it, I’d look it up myself.
You guys are both right. Young’s modulus, is a property of the material, while the second moment of area is a property of the shape. The general relationship is 1/rho=M/EI, where rho is the radius of curvature, M is the moment applied, E is Youngs Modulus, and I is the second moment of area.
If you want a more and/or less technical explination, or something more specific, just let us know.
Thanks. I knew I was missing something.
So if I were to say something is more bendable then something else, I would say it has a lower Young’s modulus?
How is this different than saying something is more “stretchy” than something else?
To me; “stretchy” and “bendable” are two different properties. Are they the same?
Yes and no. “Stretchy” refers to a sample being put in tension or compression. “Bendable” refers to a sample with a moment applied to it.
The formula for bending I mentioned above. The formula for stetching is delta=F*l/AE, where F is force, l is length, A is area, and E is Young’s modulus.
You can see that moment and force are on the top on both sides of their respective equations. Area and second moment of area are also on the bottom. These values are analagous to each other. Notice E is in the same spot in both equations.
While you didn’t ask, the formula for torsion is theta = T l / G J, where theta is deflection in radians, T is torque, l is length, G is modulus of rigidity (I think), and J is polar moment of inerta. Again, this formula is completely analagous to the other two. So, basically, you can change the “bendyness”, “strechyness”, and “twistyness” by varying the material type or its shape.
They are not the same, per se(sp?), but they are very closely related.
Here’s an example that may help you relate them a little better.
First, immagine a 2x4 piece of pine lumber, say, 10 feet long, supported at the ends. Now immagine a piece of steel the same size next to it. It’s pretty intuitive that the wood will bend alot more.
What if we turn the 2x4 on its edge? Its momment of inertia (another name for second moment of area, hence the “I” term) is bigger in that direction, so it will bend less than it did before.
Don’t reed on if you’re satisfied, the next example is kind of wordy and possibly too complex.
Second, imagine the same piece of 2x4 steel. Lets say that we make a piece of box tubing that is 4"4" outside diameter, hollow, with the same ammount of total material and the same length. If we compress them from their ends, with the same amount of force, they will “shrink” the same amount. Now, if we put both of these pieces supported on their ends, and stand on them, the 2x4 piece will deform more than the HSS (hollow structural steel) if you lay it flat. This is because more of the material is further away from the center of bending (defined as the point of zero stress, as the top of the piece will be in compression, and the bottom will be in tension. This must mean that there’s some imaginary plane that has no stress). I’m not quite sure what the comparison would be to the HSS if you laid the 2x4 steel on its edge, I’d have to run the calc, but it will bend much less than the same steel laid flat.
Wouldn’t the location of this plane, and hence the various moments of the area, depend on the material in question? If, for instance, you have a material which stretches very easily, but which is very difficult to compress, the zero-strain plane would be very close to the side which is in compression.
Absolutely (for the second part of your post). A common example of a situation where the center of bending is not in the center of the material would be a member whose cross-section is not symmetric, with respect to the applied force, or with respect to the plane perpendicular to the applied force. Channel iron, for example. I can’t think of a homogeneous material, off the top of my head, that is more difficult to compress than to put into tension. (In its elastic regime (concrete doesn’t count, it’s a composite)).
Edit: As for your first question, the center of bending can be calculated without knowing the material’s E, or the magnitude of the force. The direction of the force(s), however, must be known.
In normal engineering practice with common structural materials and laminar composites, materials are assumed to be homogeneous with respect to bending or compression, and thus the second moment of area (sometimes stilly anachronistically referred to as “area moment of inertia” even though it has nothing to do with intertia, strictly speaking) can be determined geometrically without accounting for materials. This holds true within engineering tolerances and linear material properties range. Ceramics, matrix composites, nonhomogenous materials, et cetera will have different (sometimes dramatically different) properties in compression, tension, bending, buckling, and torsion, and this has to be taken into account with a variety of different analytical and modeling techniques, or sometimes just cheat factors based upon semi-empirical data and testing of said materials under the anticipated range of conditions.
Don’t even get me started on materials that undergo a state change during use. Explaing to people why they can’t use corrosion-resistant duplex blah blah superalloy of steel for a piece of equipment that will see artic-like conditions 'cause it’ll turn into peanut brittle at those temps is an exercise in frustration, especially when they’ve already had purchasing buy a mill run of it.
Well, you can calculate I[sub]1[/sub] and I[sub]2[/sub] about the primary axes, and then transform to whatever axis the imposed bending load is about. That’s pretty easy, and you can find the I values for standard structural shapes in any mechanics or steel construction handbook.
Stranger