Press Tonnage (No Flywheel)

[Bob Warren]

TL;DR

Looking for the resulting tonnage of press ram.

2.00" of Stroke and 300 strokes per minute.

 

[Bob Warren]

1/32" from BDC

 

[jrmichler]

Is this homework? If so, we can move it.

To calculate forces, you start with a free body diagram (search the term). Every calculation in this problem requires getting the free body diagram correct. There are other considerations that will enter the problem at a later stage, but nothing will make sense until the free body diagram is correct. So start there, and we will assist as needed.

 

[Baluncore]

Since you specify torque as inch pounds, and the crank radius conveniently as one inch, I assume you are using US customary units, so want an answer in short tons.
The crank torque of 24,000 pound⋅inch = 12 ton⋅inch.
My estimate is; 12 / Sin( Acos( -31/32 ) ) = 48.37 ton(short).
But I certainly would not trust it without a diagram.

 

[Ranger Mike]

ram, dimensions, die opening width, material thickness, type of material, bending rate to name a few

 

[Baluncore]

Define the problem. You offer no specs on the ram. No data.

What more do you need that is not provided in the diagram or text ?

 

[Baluncore]

ram, dimensions, die opening width, material thickness, type of material, bending rate to name a few

Ram dimensions: determine pressure, not total tonnage.
Die opening width: As above, irrelevant, the throw on the crank is ±1", so the opening will be a maximum of 2" high.
Material thickness: is not important as; 1/32" above BDC has been specified.
Type of material: Bending rate: Irrelevant.

The torque of 24,000 pound·inch is 12.0 short ton·inch.
The crank throw is 2” total, so the crank follows; y = 1” * Cos(theta) .
TDC will be at 0°, with BDC at 180° .
The work will first be done at; y = 1/32” above BDC, then ;
Theta = Acos( -31/32 ) = Acos( -0.96875 ) = 165.638°.
Mechanical advantage is the reciprocal of the slope of the Cos() function, = Sin() ;
1 / Sin( theta ) = 1 / 0.248 = 4.0316 advantage.
So the ram force will be 12 * 4.0316 = 48.38 ton(short).

That ignores the connecting rod diagonal length of 13” when offset 0.248” which will lower the tonnage slightly. A crude estimate is; (13” - 1/32”) / 13” = 99.76% .
So the; 48.38 becomes 48.25 ton(short).

The die, below the work, should be sprung to protect the crank and bearings. That cushion must begin to compress before 48.25 ton, which will set the actual limit to the press tonnage applied to the work.