3

Shear areas have been obtained for a few different cross-sections. I-sections usually define their shear areas as equal to the product of the height of the section and the thickness of the web (assuming they fall under the thin-walled category).

However, what would the shear area for these sections be (assuming thin-walled theory applies)?

enter image description here

Should we adopt the smallest width which gives a continuous rectangle? But then the section to the right would have zero shear area.

enter image description here

Or should we adopt the width which gives the largest continuous rectangle by area?

enter image description here

Or is there some other method?

Wasabi
  • 13,077
  • 8
  • 36
  • 61
  • 1
    assuming thin-walled theory applies Are these thin-walled sections or solid sections? Also, "inverted I" is a little misleading, I feel. Maybe just "Shear area of atypical cross-section"? – grfrazee Mar 16 '16 at 19:08
  • Yeah, I couldn't find a good title. It's an "inverted I" in that instead of having horizontal protrusions above and below, they actually reduce their dimensions at those points. But yes, bad title. Actually, I just realized a better description would be as convex sections, while I-sections are concave. Or something. And these are solid sections, but very thin (the images themselves are symbolic, assume the height is magnitudes larger than the thickness). This is only relevant in that thin "sheet-like" sections have $A_s = A_w$, while wide rectangular sections have $A_s = \frac{5}{6}A$. – Wasabi Mar 16 '16 at 19:25
  • It's kind of an "inverse" I in that sense, I think. Also, if the height is magnitudes larger than the thickness, then the shear area approaches the depth times the thickness anyway. – grfrazee Mar 16 '16 at 19:28
  • @grfrazee, "inverse" would have totally been better than "inverted." And I agree, but my question is basically which thickness to adopt. So you're suggesting the web, right? That's what I was thinking as well. – Wasabi Mar 16 '16 at 19:35
  • 1
    For weird shapes, shear area has always been an "engineering judgment" type of thing, in my experience. It's roughly defined as the continuous rectangle by area. Your second shape is interesting, as you've pointed out, and I think would depend greatly on how it's loaded and how it's supported. It's worth noting that for anchor bolts and other dowel type fasteners, the entire shaft area is used as the shear area. Not 100% analogous to this situation, but something worth considering. – William S. Godfrey- S.E. Mar 18 '16 at 00:52
  • If your beams have high aspect ratio and no support flanges buckling becomes a major risk factor. – Rick May 30 '17 at 17:59

1 Answers1

2

You can calculate the shear stiffness directly and then get an effective area after the fact if you'd like.

$$A_{shear}=\frac{V}{\gamma_{ave}\,G}$$

You can calculate average shear by averaging over the height (not area):

$$\gamma_{ave}=\frac1h \int \gamma\;dz$$

You can calculate shear strain from the shear stress:

$$\gamma=\frac{\tau}G$$

And shear stress from shear flow:

$$\tau=\frac{g}t$$

$$g(z)=\frac{V\,Q(z)}{I}$$

Now a lot of the variables cancel if you plug in at this point:

$$A_{shear}=\frac{h\,I}{\int \frac{Q}{t}\;dz}$$

Assuming a beam with shear center in the middle (for example symmetric):

$$A_{shear}=\frac{h\,\int^\frac{h}2_{-\frac{h}2}t\,z^2 dz}{\int^\frac{h}2_{-\frac{h}2} \frac{\int^\frac{h}2_zt\,z\; dz}{t}\;dz}$$

That will give you the effective shear area for any symmetric beam given the thickness/width as a function of z.

Rick
  • 1,335
  • 8
  • 24