Example 1-4 A punch for making holes in steel plates is shown in Fig. 1-29a. Assume that a punch having diameter d = 20 mm is used to punch a hole in an 8-mm plate, as shown in the cross-sectional view (Fig. 1-29b). If a force P= 110 kN is required to create the hole, what is the average shear stress in the plate and the average compressive stress in the punch? FIG. 1-29 Example 1-4. Punching a hole in a steel plate (a) P = 110 kN (b) -d = 20 mm 8.0 mm Solution The average shear stress in the plate is obtained by dividing the force P by the shear area of the plate. The shear area 4, is equal to the circumference of the hole times the thickness of the plate, or A₁ = dt = (20 mm) (8.0 mm) = 502.7 mm² in which d is the diameter of the punch and t is the thickness of the plate. There- fore, the average shear stress in the plate is P Taver A, 110 kN 502.7 mm² 219 MPa The average compressive stress in the punch is P σε Apunch P 110 kN 350 MPa πd²/4 77 (20 mm)²/4 in which Apunch is the cross-sectional area of the punch. Note: This analysis is highly idealized because we are disregarding impact effects that occur when a punch is rammed through a plate. (The inclusion of such effects requires advanced methods of analysis that are beyond the scope of mechanics of materials.)