Stress - Cross-sectional and Inclined planes

[Atomic_Sheep]

Stress -- Cross-sectional and Inclined planes

As per attachment...

"On the cross-sectional plane mm the uniform stress is given
by P/A, while on the inclined plane mm the stress is of magnitude P/A'. In both cases
the stresses are parallel to the direction of P."

The parallel part makes sense... what doesn't to me is why even bother with the inclinded plane stress measured from area A'... shouldn't we just use A because if you're looking parallel to the force P, area A' will look like area A? After all, if you pick a point (say right in the middle of our solid spar or rectangular thing as per image) then you can draw an infinite amount of planes around that point at their correspondingly infinite numbers of angles... however no matter what plane you select, you'll end up with the same force being applied. Why even bother with non perpendicular planes?

 

[Angry Citizen]

however no matter what plane you select, you'll end up with the same force being applied. Why even bother with non perpendicular planes?

Because force is not stress. You're interested in the stress in the plane. Since stress is a tensor quantity, it does not obey the same rules that vector algebra gives us (i.e. the ones that forces use). You have to apply a tensor coordinate shift in order to get a traction vector that can be used to describe the state of stress along A'. This becomes important if, for instance, you have a weld along a surface that is at an angle. You would want to design the weld such that it can withstand stresses along the weld (shear) and across it (normal).