A circular hole of 150 mm diameter is to be punch out of a 2 mm thick plate. the shear stress needed to cause fracture is 400 Mpa. calculate the minimum force to be applied to the punch.
force=400 E6 * PI(.150/2)^2 Newtons
D=150mm
=0.15m
t=2mm
=0.002m
T=400MPa
=400E6Pa
T=F/Al
F=TAl
=T*Pi*D*t
=400E6*3.1416*0.15*0.002
=376911.12N
=376.9KN
To calculate the minimum force required to punch out a circular hole from a plate, you need to determine the area of the hole and then multiply it by the shear stress required to cause fracture.
First, calculate the area of the circular hole using its diameter. The diameter of the hole is given as 150 mm, so the radius (r) can be calculated as half of the diameter: r = 150 mm / 2 = 75 mm = 0.075 m.
The area of a circle is given by the formula: A = π r^2, where π (pi) is a mathematical constant approximately equal to 3.14159. Substituting the value of the radius into the formula:
A = π (0.075 m)^2 = 0.01767 m^2.
Next, multiply the area of the hole (A) by the required shear stress (σ) to calculate the force (F) using the formula: F = A * σ.
The required shear stress is given as 400 MPa. However, the shear stress needs to be converted to Newton per square meter (N/m^2) to match the SI units of the force:
1 MPa = 1 million N/m^2.
So, 400 MPa = 400 million N/m^2.
Substituting the values into the formula:
F = 0.01767 m^2 * 400 million N/m^2 = 7.068 million N.
Therefore, the minimum force required to punch out the circular hole is approximately 7.068 million Newtons.