Understanding How an I Beam Supports Loads

[retrac]

Homework Statement

Why is an 'I' beam good at supporting loads? (Bear in mind that I'm only 16, so I haven't met moments of inertia or anything like that, but have met basic calculus)

The Attempt at a Solution

Here's what I think so far. When a weight is placed on an I beam, the vertical part of the 'I' is very good at bearing the vertically downward force, and the flanges stops the beam from rotating.

However if this is true, surely all you'd need is a thick solid beam which is relatively wide which could support the load itself and because of its width, wouldn't need to worry about twisting.

However somewhere I read it has something to do with shear and bending forces, but I don't really understand what a shear force is and how it comes into it. If someone could explain it like this it would be really helpful

Thanks.

 

[mathmate]

Yes, the web (the vertical part) of the I-beam resists shear force, which you called the vertically downward force. The flanges stop the beam from rotating, both in the plane of the web, which is caused by bending moments, as well as rotating around the axis of the beam (twisting).
The reason I-beams are used instead of a rectangular beam is simply economy. Steel being relatively strong, does not need to be a rectangle to resist the shear forces and bending moments, but it needs to be deep enough to reduce the deflection under load, as well as wide enough to resist lateral buckling. One of the best shapes is the I-beam, which allows other things to be tucked between the flanges (wood joists, for example). On the other hand, a hollow steel section (HSS) which is tubular in shape is also very economical, and is even more robust when it comes to buckling and twisting.

If wood is used as the building material, it is usually done in rectangular sections because it is relatively less strong, and also more expensive to shape.

It sounds like you will be a very good civil or mechanical engineer, because you seem to see how things work intuitively.

 

[minger]

To clarify a little more, when we load a beam like that, there are essentially two stresses. The shear stress acts to tear the beam into two "steps", with one above the other. The bending stress acts to bend the beam into a "smiley" or "frown" face.

In the equations, only area affects pure shear stress. This means that if I apply a pure shear stress on a beam, it doesn't matter what the shape is, as long as the cross sectional area is the same.

However, bending stress is affected by geometry. More specifically, it's determined by what we call the Area Moment of Inertia, or the Second Moment of Area.
http://en.wikipedia.org/wiki/Second_moment_of_area
The larger the Area Moment of Inertia, the less the beam will deflect, and the more loading it can take.

To make this value larger, you try to put areas further away from the centerline as possible. Take your I-beam for example. In normal operation, it stands tall, with the top and bottom flange far away from its centerline. This acts to increase its area moment of inertia.

However, if the I-beam is loaded sideways, then all of a sudden, those flanges are now on top of the centerline (draw a picture if needed, it now looks like a short H)! Now, the I-beam can actually become quite weak.

 

[Dr.D]

Actually, I think it is a bit premature to try to address this question without first considering how beams bend in general. This fundamental insight is essential to the understanding of how beams work, and that is what we spend a good bit of time introducing in a mechanics of materials course. Without that, I think the only answer that can properly be given to this question is to simply say, "I-beams are effective in resisting bending because they put the material in the position where it is most effective." This is not a very satisfying answer, but the real answer cannot be given in 25 words or less, even for someone who knows a little bit of calculus.