Shearing Modulus/Shearing Strength

[Myung]

Homework Statement

How much force is required to punch a hole ½ in. in diameter from a 1/8 in. thick steel sheet of shearing strength 4x104 psi.

Homework Equations

Shearing Modulus = Pressure/Shearing Strain

Pressure = F/A

Shearing Strain = DistanceSheared/Length-to-be-sheared

The Attempt at a Solution

Can I first ask how can I use the Value of its Shearing Strength to the equation?

 

[PhanthomJay]

Homework Statement

How much force is required to punch a hole ½ in. in diameter from a 1/8 in. thick steel sheet of shearing strength 4x104 psi.

Homework Equations

Shearing Modulus = Pressure/Shearing Strain

Pressure = F/A

Shearing Strain = DistanceSheared/Length-to-be-sheared

The Attempt at a Solution

Can I first ask how can I use the Value of its Shearing Strength to the equation?

You don't need the shear modulus to calculate the force. You can use the 2nd equation where pressure is the shear strength. And where A = ___??___

 

[Myung]

You don't need the shear modulus to calculate the force. You can use the 2nd equation where pressure is the shear strength. And where A = ___??___

A = ??

should I use the diameter or the thickness? or both? will that matter?

 

[PhanthomJay]

A = ??

should I use the diameter or the thickness? or both? will that matter?

Yes, it will matter; shear stresses act parallel to the cross section area, in the plane in which they act. Visualize the punch tearing through the entire thickness of metal. Intuitively, the thicker the metal, the greater will be the force required to punch through it. So the thickness is important in determining the Area. The area is the thickness times what other term?

 

[Myung]

Yes, it will matter; shear stresses act parallel to the cross section area, in the plane in which they act. Visualize the punch tearing through the entire thickness of metal. Intuitively, the thicker the metal, the greater will be the force required to punch through it. So the thickness is important in determining the Area. The area is the thickness times what other term?

A = the cross-sectional area of material with area parallel to the applied force vector.
A = Area of Hole(∏(1/2^2)/4) + Area of Thickness ( W = 1/8 , L = 1/2)

F = 10353.98 N?

Cross-sectional area is perpendicular to the Applied Force Vector while the Area of the Thickness is parallel to the said Vector.

 

[PhanthomJay]

A = the cross-sectional area of material with area parallel to the applied force vector.
A = Area of Hole(∏(1/2^2)/4) + Area of Thickness ( W = 1/8 , L = 1/2)

F = 10353.98 N?

Cross-sectional area is perpendicular to the Applied Force Vector while the Area of the Thickness is parallel to the said Vector.

Since the area of the hole is perpendicular to the applied force, it should not be included in the calculation of the area parallel to the applied force. So you just want the Area of the 'thickness', which is the thickness times the length of the thickness measured along the hole's circumference. That length is not 1/2 inch. When you get the answer for force, the units of force are in pounds, not Newtons.