 In this problem we consider this shaft with diameter equal to 200 mm, which is subjected to a torque t equal to 126 kNm and a bent moment m equal to 157 kNm, as you can see here in the picture below. And using more circle we need to determine where are the principal stresses and the maximum shear force at here point A. And we need also to estimate the orientation of the plane at which they occur. So first of all we need to determine what are the stresses due to the torque at this moment. Then according to the reference system that we have defined here, we have that at A this is my small differential element. So we have here normal stresses sigma y and sigma x and shear stress tau xy. Then we have normal stresses sigma x due to bending, right? So this bending moment is creating a shear stress in the x direction. So this is equal to m times y divided by the moment of inertia. And we know that for a cylinder with the circular cross-section the moment of inertia with respect to the neutral axis is equal to pi d to the power of 4, this is diameter divided by 64. Then this is equal to, and y is the distance from the neutral axis to this point. And this is equal to the diameter divided by 2. And of course we know that the moment is equal to 157 kNm. Then with this information we have that sigma x is equal to 200 MPa. We know that we don't have sigma y because the torsion is not creating any moment in the y direction. So sigma y equal to 0. Then these are the stresses due to bending. And now we can calculate the stresses due to torsion. This torque t is only creating shear stresses. Then we have the tau xy is equal to the torque times r divided by the polar moment of inertia. And the polar moment of inertia is equal to 2 times the moment of inertia i. Then we know the radius is d-halves. So then this is equal to, then now we have all the information that we need in order to draw the Mohr circle. Remember that we have the sigma x is equal to 200 MPa. Sigma y is equal to 0 and the shear stress is equal to 80 MPa. And we define this to be the plane A and this to be the plane B. So then first we can calculate what is the center of the circle which is equal to sigma average. So this is sigma x plus sigma y divided by 2. Then this is sigma x divided by 2 which is equal to 100 MPa. And the radius is equal to sigma x minus the average squared plus the shear stress squared. And the square root of this so this is equal to 128 MPa. Then now we can start drawing the circle. The average is located here, center around 100 MPa. So first for this plane A I know that the rotation due to the shear stress is counterclockwise. So according to my criteria, to the criteria that we are using this is a negative shear stress. Because it's creating a counterclockwise rotation. So this is located somewhere around here and corresponds with sigma x equal to 200 MPa. This is point A and the plane B we have of course a positive shear stress with correspond with 0. Then this is point B and this is point A. Now we can finish the circle more or less like this. Then this is the first principle plane and this is the second one. So the coordinates of sigma 1 is equal to the average plus radius. And sigma 2 is equal to the average minus the radius. So this is equal to 100 plus 128. So this is 228 and this is equal to 100 minus 128. Then this is negative 28 MPa. Of course this is the maximum shear stress which is equal to the radius. And now we need to define what is the orientation of the planes at which the principal stresses occur. So we know that this is our plane A. And we also know that the angles in the mercerical are two times the angles in a real problem. So this is the orientation of A with respect to the principal plane. So this is two times V1. Then I know that two times V1 is equal to the arc tangent of this distance here which is equal to 80 divided by this distance here which is equal to sigma x 200 minus the average. Then this is equal to 38.6 degrees. And from here I have the sigma 1 phi 1, sorry. This angle is equal to 19.3. So counterclockwise from A. And of course the orientation of this second plane with respect to A is 2 times phi 2 which is equal to 2 times phi 1 plus 180 degrees. So from here we obtain that phi 2 is equal to 109.3 degrees. Again counterclockwise from A.