OK, back to the original topic, i.e., how to deal with an understrength shear web. I did another experiment, this time using maths instead of grabbing some stuff out of the garage and stretching it this way and that.
I drew a beam section, cantilevered it out from a wall 12", applied a concentrated vertical up load, and then increased it until the outer surfaces of the caps were at yield for the material, in this case 1144 steel @ 100,000 psi, and 58% of that for shear strength (fig."A"sub1). Then I used the vertical force (Fsubv) to calculate normal (comp/tensn) and shear at the middle (point "a") and at the top (point "b") of the shear web, see fig."B"sub1. The maximum shear at the middle ("a") is the calculated shear at that point because there is no normal stress there. So I concentrated on the top of the web (point "b") where there is significant normal stress as well as shear stress and these combine to create a greater amount of normal and shear stress at various angles to the beam axis. These stresses can be found with a graphical device called Mohr's circle (fig."C"sub1). In Mohr's circle, the calculated normal stress is laid along the horizontal axis (not the x-axis of the beam, though) and the shear stress is measured, to the same scale, from the normal stress endpoint vertically up for positive or down for negative. The center point of the normal stress line is found and then a circle drawn with that point as its center and the endpoint of the shear stress on the circumference. Then, a diameter is drawn through the shear stress endpoint and the midpoint of the normal stress. The maximum normal stress is on the horizontal axis from the origin ("0") to point "u" on the circumference; here there is no shear stress. The maximum shear stress is equal to the radius of Mohr's circle.
The maximum normal and shear stress at point "b" in the shear web, (fig."C"sub1) is seen to be less than the yield of the 1144 steel. But now I replace the web with a fictional material called "staluminite" (could have called it RCS, for really crappy steel) which has the weight by volume of steel, but the strength charactaristics of 6061-t6 aluminum, i.e., 40,000 psi tensile and 58% (or so) of that for shear. Now my web is severely understrength. I tried three ways to bring it back up to strength: increase the size of the caps, increase the size of the web, or, as per Billski's admonition, increase both the web and the caps in some balanced manner.
1) In fig."D"sub1, I tried increasing the size of the caps but the shear gradually went up and I could not strengthen the web this way (duh!).
2) In fig."E"sub1, I enlarged the web and was able to bring it back up to strength, but not without making the web actually wider than the caps, possibly because I chose a material with 40% of the strength of the caps. Makes for an odd looking "I" beam.
3) I increased both the web and the caps and was able to bring the beam back up to strength and still have a reasonable looking beam section. Also, the cross-sectional area is about half that of the one in fig."E"sub1, and so half the weight. I also found the angle at which the maximum stresses would act; not so important for steel and aluminum, but would be of interest when using FRP.
So, I have tried to replicate the "trend" of having to increase web and caps when the web is understrength as when using orthotropic materials such as FRP composites, but by using isotropic material so as to avoid doing all of that matrix algebra with pencil and paper. Did I succeed? I think I may have, but I thought I had it after the last experiment, too! Hopefully, Billski will be kind enough to evaluate this one too
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Edit: Normal stress is designated "sigma (lower case Greek symbol)", and shear stress is designated "tau (lower case Greek symbol)". "I" is moment of inertia of the beam cross-sectional area, "t" is thickness of the beam at the point in question, i.e., "t" = 1/2" in the shear web, or 4" in the caps, and "V", here, is equal to the vertical force "Fsubv" since it is the only vertical force to the left of the beams cross section at the root, and the "root" cross sectionis where we're calculating these forces to apply.
Also, in fig."A"sub1, in the equation for Fsubv = 56,075.8 (bottom right hand corner), the numerator of 100,000/1.7833 is supposed to have five zeros instead of four.