 In this problem, we have this axial AB, which has a diameter D equal to 18 millimeters. We have a load P equal to 8.9 kN, which is applied here at point C and at point G. The axial is supported by two wheels with a spacing L equal to 1.45 meters. So these are the reaction forces. And the load and support are offset by B equal to 0.2 meters. So this distance B is equal to 0.2 meters. And we have to determine the maximum bending stress in the axial. So we know that the bending stress is equal to the moment times y, the distance from the neutral axis, divided by the moment of inertia. In this case, the moment of inertia is constant because the cross-section of the beam is constant. Then the maximum bending stress occurs where m is maximum and y is maximum. Then we need to look for the location of the maximum moment and also the location of the maximum distance y. Then we're going to start calculating first the reaction forces. This is our structure. And we have here the applied force P and the applied force P here as well. And these are the reaction forces R. And in this case it's very simple because from the vertical equilibrium we have that to R minus 2P is equal to 0. And since we have symmetry we directly find that R is equal to P. Then now we are able to calculate what are the expressions for the moment. Then we can divide our structure in different segments. Then we define the section or segment AC, CD, DF, FG, GP. For section AC we have, as you can see here we have no moment, then m is equal to 0. For section CD this force P is creating a moment which is counterclockwise. Then the internal reaction moment is clockwise. Then according to our sign convention we have that this is positive. Then this force P is creating an internal moment distribution different from this one. So we are creating here a distribution like this. So this is negative and it is equal to minus P times X. So we define X from C to the right. Now for section DF we have the same contribution of this force P minus PX. Now we have the contribution of this reaction force which is P. In this case this is positive and this is located at X minus 0.2. Then this is equal to minus PB. And for this section FG we have the same contribution as before minus PB. And also the contribution of this reaction force P. So this is positive again plus P and the distance from P to any point of this section is equal to X minus L plus B. So this is equal to minus P L plus 2B plus PX. And finally for GB we have again that this is free. We don't have moment M is equal to 0. We can now draw this moment distribution. From A to C we don't have moment. From C to D we have minus PX. So it's a straight line. And the value here at D is PB minus PB. So here is constant. Of course we know that this beam is symmetry. Then the maximum bend the moment is equal to minus PB. This is equal to minus 8.9 to the power of times 10 to the power of 3 times this distance B 0.2. And this is in unosmeter. So this is equal to minus 1.78 times 10 to the power of 3 nm. And of course you see that it occurs along this span DF. Now we go back to the flexural formula because we remember the sigma max. It was equal to M max times Y max divided by the moment of inertia. Then we can now calculate what is the moment of inertia. We have a circular section. And we have that the diameter is equal to 18 millimeters. Then we have a standard formula for this. The moment of inertia is equal to pi divided by 64 times D to the power of 4. And this is equal to 2.011 times 10 to the power of negative 6. So this is the moment of inertia. And we just need to know what is the Y max. So in this case we have this circular section. We have symmetry. This is the neutral axis. So Y is this distance. So Y max is this distance. Then we have that Y max is equal to the radius of the cross section. And then this is equal to diameter divided by 2. So of course we have to calculate the maximum and bend the moment. But the problem does not specify if it should be the maximum compressive or tensile. So we need to take into account also this distance which is also Y max. Then this is equal to plus minus this diameter divided by 2 is equal to 40 millimeters. Then now we have all the ingredients to calculate the maximum bending stress. So the maximum bend the moment it was equal to minus 1.78. The maximum Y is equal to plus minus 40 millimeters. And the moment of inertia is equal to 2.011 times 10 to the negative 6. Then this is equal to plus minus 35.4 megapascals. Then we have that the maximum bending stress is equal to 35.4 megapascals in tension and 35.4 megapascals in compression.