Area of an inclined surface with respect to the original surface

[tent]

TL;DR Summary

Relationship of inclined area with respect to original area

Hi, I have a problem with inclined planes. The idea is to calculate the stress in an inclined plane of a bar under tension for which you need the surface. I have no idea how this surface is derived, though. In the attached file, you can see what I mean. For a rectangular cross-section, it's straightforward, just applying the rectangle area with the new inclined length. Now, everywhere I see, everyone uses the same rectangular bar as an example.

However, in one single textbook, the exercise uses an elliptical cross-section to seemingly represent a random surface. They use the same formula for the area, but without any explanation, apparently trivially and immediately deriving, but I don't see why the area of an inclined elliptical surface with respect to the original surface is the same as the rectangular one.

My suspicion is that it has to do with the vector area which, being the same direction as the normal, is somehow projected onto the other's area vector, but I don't see it. Thanks for the help.

 

[mfb]

If you cut a cylinder you get an ellipse, which is just a stretched circle, so the area of an ellipse is simply pi*(semi-minor axis)*(semi-major axis). The first is the radius of the cylinder, and the second one you can find in the same way as the rectangular case.

 

[tent]

Okay, I see that now. It seems to me that for all common cross-sections this is true, at least the ones I can think of, even compound ones such as an H-beam.

But what about any cross-section? By any I mean, an area enclosed by a loop that doesn't cross itself such as a horseshoe, a star/asterisk, sickle, quarter-moon, etc. Could it be proven whether or not $A _\theta=\frac {A_0} {cos \theta}$ is valid for the area of a section resulting from an inclined plane cutting through a bar with cross-section as described previously, where $\theta$ is the angle of inclination?

 

[mfb]

As long as the bar stays the same along its axis that formula stays true - all you do is stretch the area in one direction.

 

[tent]

Alright, looking at it as a scaling factor in one direction does help. This clears it up, thanks.

 

[robphy]

The concept of "flux" might help
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c3

Here's a useful diagram from
https://www.ux1.eiu.edu/~cfadd/1360/24Gauss/Flux.html