Torsion in a Basically Shaft Which is Circular Cross Section

Defination a problem of torsion belongs to shaft:

Above, a shaft is with fixed supported which is under the force and this force 3kN. Firstly, we found the torque on the shaft,it is T=2100Nm. Secondly, we calculated the moment of inertia, it is J=40,8677 * e-6 m**4. Finally, we calculated maximum shear stress (128,46 MPa) and minumum shear stress (97,63 MPa). Now we will create a model of this problem and we will see impact a moment which is 2100 Nm on the shaft. We will get von Mises stress.

Primarily, we have to choose a material for our shaft. Material should be steel which have high yield strenght. It has about 350 MPa tensile yield strenght, 200 GPa modulus of elasticity, density 7.8 g/cc and 0.25 poisson ratio. So we determined our material properties, now we can create a model to do analysis. for moment of 2100 Nm:

This is our model
Plane of YZ

For 2100 N m moment we have maximum von Mises stress which is 304 MPa. This stress is lower than yield strenght of our steel (350 MPa). So this shaft does not plastic deformed.

Viewing maximum (304 MPa) and minum (138 MPa) stress on shaft
Displacements above the shaft

The maximum displacement is 2.841 mm.

Result of Analysis

We applied 2100 N m moment this shaft, we obtained 304 MPa stress. This stress is enough for our material to resume. If we obtain more than 350 MPa stress, our material would be deformed. Also maximum displacement occurs out surface.

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