Question

a circular rod is to be launched through 5 mm thick aluminium plate. if the ultimate shearing strength of aluminium is 120 mpa and safe compressive strength for the punch is 80 mpa, determine the maximum diameter of the hole that can be punched.

Answer 1

The maximum diameter of the hole that can be punched through the 5 mm thick aluminum plate is approximately 12.2 mm.

Step-by-step explanation:

To determine the maximum diameter of the hole that can be punched through the 5 mm thick aluminum plate, we need to consider the ultimate shearing strength of aluminum and the safe compressive strength of the punch.

Ultimate shearing strength is the maximum stress a material can withstand before it fractures. In this case, the ultimate shearing strength of aluminum is 120 MPa.

Safe compressive strength is the maximum stress a material can withstand before it deforms. For the punch, the safe compressive strength is 80 MPa.

Given that the plate thickness is 5 mm, the maximum diameter that can be punched can be calculated using the formula:

d = 2 * t * (√(ultimate shearing strength / safe compressive strength))

Substituting the values, we get:

d = 2 * 5 * (√(120 / 80))

= 2 * 5 * (√1.5)

≈ 2 * 5 * 1.22

≈ 12.2 mm

Therefore, the maximum diameter of the hole that can be punched is approximately 12.2 mm.