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chandra_324_hw5_soln

# chandra_324_hw5_soln - Solution for ME324 HW#5 1 SOLUTION...

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Solution for ME324 HW #5 1 SOLUTION If too large of a billet is placed into the dies in a closed-die operation, presses will jam, not com- plete their stoke, subject press structures to high loads, and of course increase downtime.Numer- ous catastrophic failures have been attributed to such loads. If, on the other hand, the blank is too small, the desired shape will not be completely imparted onto the workpiece. 2 SOLUTION The answer is apparent, after balancing the horizontial forces on the element, we have an equation with two unknowns, and . In order to be able to solve for these equations, we need a second equation, i.e., Eq. of distorsion-energy criterion for plain strain. Note also that the element is in a state of yielding, hence it has to obey yield criterion. 3 SOLUTION (1) For annealed copper we have K=315 MPa = 46,000 psi and n=0.54. The force required for forging at any stage is the product of the flow stress of the material and the cross-sectional area.

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chandra_324_hw5_soln - Solution for ME324 HW#5 1 SOLUTION...

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