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• # Bending stress due to 2 different, perpendicular bending moments

Discussion in 'Calculations' started by julian89, Nov 11, 2013.

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3. ### KevinKNew Member

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I don't think superposition applies in this case, because it would involve calculating the stress due to each moment independent of the effect of the other moment. If you think about it, bending a beam that is already bent about a separate axis is different from bending a straight beam. I'll try to find the formulae and get back to you.

4. ### PierArgWell-Known Member

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Hi Julian and welcome to the forum.
I think your question is just to have a theoretical approach to the problem and not to solve a real case. Right?
(I'm finding hard to associate your problem to a real case/component )

I'd consider two separated cases and then I'd add together their effects.

What Kevink says is correct, especially if you want to consider the deformation of the shape of the section (it would be difficult even with a FEM analysis).
But if you want to calculate just the global stresses and the total displacement of the entire beam, you could apply the superposition.

Last edited: Nov 18, 2013
5. ### delboyNew Member

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bending with horiz & vert bending

hi,
if you have bending in two planes i.e. horizontal bending on
shaft from tension in pulley/belt and vertical load
from.gears or self weight of component on shaft then
you calculate the resultant moment by squaring both,
adding result and get square root ( pythagorous)

6. ### Paul.RMember

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Moments are vectors - they can be added as such (as in your diagram). If the cross-section is circular, that makes things much easier. Bending stress is Mc/I, where M is the bending moment, c is the distance from the neutral axis to the extreme fibre (which in your case is just the radius of the circle), and I is the moment of inertia about the neutral axis of the bend.
For your specific case of a circular cross-section, the bending stress would be ((Mx^2+My^2)^0.5)*c/I.

If your cross section were not circular, you would need to find the moment of inertia about the neutral axis of the combined moment (i.e. the moment of inertia of a rotated section). c would be the maximum perpendicular distance measured from the centroid on the neutral axis to the edge of the cross-section. For example, if the two moments were equal and the cross section were square, then the neutral axis would be at 45 degrees from horizontal - which means that the moment of inertia would be that of a diamond cross section and c would be the distance from the centre of the square to the corner.

7. ### sarnathNew Member

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just find resultant of two moments and since its a circular C/S so MI remains same along any axis, so that it.

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