 In this problem, we have this small structure here, in which we have three elements. This rod with circular section. This bar with a squared section. And these two elements are joined together through this pin with diameter equal to 25 mm. Then we need to find what is the average normal stress in the rod? What is the average normal stress in the bar? And what is the average shear stress in the pin? Then we can start with the average stress in the rod. So then we can cut here, for example. And we have here this applied load of 100 kN. Then if we apply the equations of equilibrium, we know that the sum of forces in the x direction is equal to zero. Then the internal reaction force is equal to 100 kN. Then from here we know that the average stress in the rod is equal to the force divided by the area. So we know that this force is equal to 100 kN. And the area is equal to 1000 mm squared. So this is equal to 100 MPa. We can do the same for the bar. We can cut here. And we have this. We have here this point load of 100 kN. And of course then the internal reaction force is equal to 100 kN as well. Then we know that the average normal stress in the bar is equal to the force divided by the area. So this is equal to 50 MPa. And finally we can calculate the shear stress in the pin. We can consider first the forces acting on the pin. So this is the piece. And we have here 50 kN. We have here 50 kN as well. And we have here this point load of 100 kN. Then for example if we cut this structure here at this plane and here at this plane. We have here something like this. So 100 kN acting in this direction. And we have here our shear stress and in this surface as well. Then we know that the average shear stress is equal to the force. 100 kN divided by these areas. This one. And of course this one. So then this is 2 times the area of the pin. Then this is equal to 102 MPa.