A punch for making holes in steel plates is shown in Fig.1-54a. Assume that a punch having diameter d = 3/4 in. is used to punch a hole in an 3/10 in. plate, as shown in the cross-sectional view (Fig.1-54b).

If a force P = 24 kips is required to create the hole, what is the average shear stress in the plate and the average compressive stress in the punch?

Question

A punch for making holes in steel plates is shown in Fig.1-54a. Assume that a punch having diameter d = 3/4 in. is used to punch a hole in an 3/10 in. plate, as shown in the cross-sectional view (Fig.1-54b).

If a force P = 24 kips is required to create the hole, what is the average shear stress in the plate and the average compressive stress in the punch?

Accepted Answer

Use the four-step problem-solving approach to find average shear stress in the plate and average compressive stress in the punch.

1. Conceptualize: Assume that the shaft of the punch is in compression over its entire length due to applied load P. Force P acts downward on the plate and is applied as a uniformly distributed force along a circle of diameter d as the punch passes through the plate.

2. Categorize : The average shear stress in the plate is obtained by dividing the force P by the shear area of the plate. The shear area is the cylindrical area of the plate that is exposed when the punch passes through the plate. The compressive stress of interest is the one acting on a circular cross section through the lower segment of the punch (Fig.1-54).

3. Analyze : The shear area [latex]A_{s}[/latex] is equal to the circumference of the hole times the thickness of the plate, or

[latex]A_{s}=\pi d t=\pi(3 / 4 ext { in. })(3 / 10 ext { in. })=0.707 in ^{2}[/latex]

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

[latex] au_{ aver }=\frac{P}{A_{s}}=24 kips / 0.707 in ^{2}=34 ksi[/latex]

The average compressive stress in the punch is

[latex]\sigma_{c}=\frac{P}{A_{ ext {punch }}}=\frac{P}{\pi d^{2} / 4}=24 kips / \pi(0.75 in .)^{2} / 4=54.3 ksi[/latex]

in which [latex]A_{punch}[/latex] is the cross-sectional area of the lower segment of the punch.

4. Finalize: The normal and shear stress distributions are not uniform due to stress concentration effects; hence, the calculations result in “average” stresses. In addition, this analysis is highly idealized because impact effects that occur when a punch is rammed through a plate are not part of this analysis.

Question 1.9: A punch for making holes in steel plates is shown in Fig.1-5...

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