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I use this model to get the tangential and radial stress formulas but I want to examine a system also undergoes a torsion under constant speed. I'm a little confused at this point, do I need to point out the torsional stress in this picture and calculate all the formulas according to it, or do I need to just sum the results?

enter image description here

$ \sigma _ {1} =( \rho \omega ^ {2} /8)[(3+v)( R_ {1}^{2} + R_{2}^{2} + R_ {1}^ {2} R_{2}^{2} /r^ {2} )-(1+3v) r^ {2} $ ]

$ \sigma _ {2} =(3+v)( \rho \omega ^ {2} /8)( R_ {1}^ {2} + R_ {2}^ {2} - R_ {1}^ {2} R_ {2}^ {2} /r^ {2} - r^ {2}) $

Enes
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1 Answers1

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Unless you have direct transversal shear in the section, the torsion stress indicated on your differential element as $ \ \sigma_1 \ $ is the shear stress.

If the section is subject to transversal shear as well, however, that stress has to be added too, but then it may not be convenient to use polar coordinates, because transversal shear is easier to handle in cartesian coordinates!

Otherwise, if we use polar coordinates and have a transversal shear stress of v3, say parallel to the x-axis, we add

$$\sigma_{v3}*sin\theta^2(\delta r) \ $$ to the differential element stresses. and add

$$\sigma_{v3}*cos\theta^2(\delta r) \ $$ as compression or tension stress to the element. We then add the shear stress due to transverse shear to the differential element.

$$ \tau= \sigma v_3 (cos \theta)(sin \theta)$$

kamran
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