Hi all,
I have a question about mohr's circles, especially the circle related to uniaxial traction stress.
I don't understand the reason why the maximum shear stress is found on a direction rotated 45° from the tensile axis (90° in the mohr's circle).
In my opinion, the shear component (which is obtained with the projection of the traction stress on the rotated element surface) increase more and more when rotating the element over 45° from the tensile axis..
Please, I would be grateful if someone clarifies my doubt.
thank you very much
The Mohr circle is drawing over an two axis diagram. Vertical axis is shear stress (Tau), and horizontal axis tensile or axial strees (Sigma).
If you have an uniaxial tensile stress, you have a point (Sigma 1) over the horizontal axis, and the two others pincipals axials stress (Sigma 2 and Sigma 3) are zero and coincident with the origin. In this case, you can trace a circle with a diameter = Sigma 1.
What sucess? You obtain a circle with a Radius =Tau maximun = (Sigma 1 ) / 2, and for this reason, a maximun Tau sucess in a plane rotated 45° in relation with axial axis (arises from the geometric construction).
Dear MATTEO
Dear Matteo, begin from transformation equations for stresses on inclined planes and try to get the equation of corresponding Mohr cycle. Its shape will help you in analysis of max shear stress plane orientation
thanks for the replies, but I need to know the physical expanalation about the fact for which the tau reaches is max value at 45° of rotation.
Why the tau in not max at 90° of rotation?
I hope my question is now more clear.
We see by means of Mohr circle that in an uniaxial tensile test, Tau maximun is 0,5 times the value of principal axial stres. Then, if we make an tensile test, the surface fracture area is always normal to the axial direction. If in some case, Tau will be greater than tensile strees, the fracture area will be inclined in relation with tensile stress direction.
Hi Matteo Ubertin
Let two planes make an angle of θ between them. On the Mohr circle the angle between these two planes = 2θ. Now on the Mohr circle the angle between plane of uniaxial tension and max shear stress is 2θ = 90 deg. Hence θ = 45 deg in the specimen. Hope this answer is very useful.
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