Question 1.9: A punch for making holes in steel plates is shown in Fig. 1-...
A punch for making holes in steel plates is shown in Fig. 1-51a. 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. 1-51b).
If a force P = 110 kN is required to create the hole, what is the aver-age shear stress in the plate and the average compressive stress in the punch?
The average shear stress in the plate is obtained by dividing the force P by the shear area of the plate. The shear area A_{s} is equal to the circumference of the hole times the thickness of the plate, or
A_{s}=\pi d t=\pi(20 mm )(8.0 mm )=502.7 mm ^{2}in which d is the diameter of the punch and t is the thickness of the plate.
Therefore, the average shear stress in the plate is
The average compressive stress in the punch is
\sigma_{c}=\frac{P}{A_{\text {punch }}}=\frac{P}{\pi d^{2} / 4}=\frac{110 kN }{\pi(20 mm )^{2} / 4}=350 MPain which A_{\text {punch }} 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.)