HYDRO MECHANICAL SHEAR

[kiddiecorner]

If anybody please help me telling how to determine the Shear Load (if its the right term) required to cut a strip 0.53 inch thick and has a yield strength of 80,000 psi. Im just confused as others use the shear strength and others use the yield strength when using F=A x Ss. The blade angle is 2.2 degrees.

Thanks

 

[CoryPad]

Shear force is the correct term, but load is (mis)used frequently.

You need to provide the width of the strip in addition to the thickness.

You need to use the shear strength of the material, not the yield strength. If you don't know the shear strength, you can estimate it as ~ 0.6 [·] ultimate tensile strength.

Blade angle is irrelevant for macro-scale force calculation.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[kiddiecorner]

Thanks Cory,

The width of the material is 92 inches, the blade angle as i understand will also affect the Area and the force eventually.
And what do you think would be the proper penetration to fracture?

 

[CoryPad]

No comment regarding penetration.

Assuming the material ultimate tensile strength is 100 ksi (should be in the ballpark for 80 ksi yield), then the shear strength is ~ 60 ksi. Then the force would be 92 in [·] 0.53 in [·] 60 ksi = 2.9 million pounds-force.

That is a stout shear you have in mind.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[bvanhiel]

Cory,

Your assuming the whole thing gets chomped at once. I would think for this application that the shear blade would be angled and progressively shear the material.

-b

 

[MikeHalloran]

Shearing SY80 is an unusual requirement. Stuff that hard is normally cut by waterjet, plasma, or oxyfuel.



Mike Halloran
Pembroke Pines, FL, USA

 

[kiddiecorner]

Yes thats the reason that the blade is angled so that only a portion of the material will be in contact with the blade at any given one time.