(a) Calculate the force required to punch out a rectangular hole 100m x 50mm in a 2mm thick aluminium sheet. The ultimate strength in shear is 125 MN/m^2.
(b) If the maximum force which can be exerted by the press is 45kN, what is the greatest thickness that can be punched out?
Two co-axial shafts are to be connected by a flanged coupling having six 6mm diameter bolts equally spaced on a PCD of 200mm. If the maximum permissible shear stress in the bolt material is 180 MN/m^2, find the factor of safety for the bolts if the coupling transmits a torque of 2KNm.
In the case of hole punching the shear area is taken to be the surface area of the plug to be punched the shear stress is calculated from the Force divided by the shear area.
The shear area is given from: 2 x (50 mm + 100 mm) x 2mm = 600 mm^2
The shear stress given is 125 MN/m^2 =125 x10^6 N/ (1000mm)^2 = 125 N/mm^2
The force required is then 125 N/mm^2 x 600 mm^2 = 75 kN
Now a maximum force of 45 kN is all we ...
The solution talks the reader through each step of calculations carefully to promote a fuller understanding of shear stress and force.