Problem 3 (Shear Stress) Let's solve the above problem with a different load and pin diameter as given below: Problem Case Load (N) Pin Diameter Shear Stress (mm) (Mpa) Guided Practice 3500 12 XX Top view we width H 5000 12 D 2 8000 15 3 8000 20 2 Retaining ring Shear planes Pictorial view Link Reaction Side virw The force on the link in the simple pin joint shown in the above figure is 3550 N. If the pin has a diameter of 10.0 mm, compute the shear stress in the pin. Solution Objective Compute the shear stress in the pin. Given F= 3550 N; D= 10.0 mm Analysis Pin is in direct shear with one cross section of the pin resisting all of the applied force (single shear). Use the below equations direct shear stress = r m applied force F shear area A, Results The shear area, As, is (10.0 mm) - 78.5 mm Then the shear stress is F 3530 N 78.5 mm - 45.2 N/mm = 45.2 MPa