An engineer has been given the assignment of finding how much money can be saved

Question

An engineer has been given the assignment of finding how much money can be saved over a year's time by redesigning the press plates from the pattern shown in Fig. 4.18a to the pattern shown in Fig 4.18b for stamping out 2.400 in diameter disks. The stamping material is 14 gauge sheet metal and can be purchased in 100 ft rolls in varyig widths in 0.5 in increments. One square forot of metal weighs 3.20 lb. The metal is sold for $0.20/lb. Do not consider the ends of the rolls. The company expects to produce 38000 parts this year. How much can be saved?

Explanation / Answer

First we have to calculate width and then material required

See for the first die,width required=2.4*4=9.6 inch=10 inch(approx)

For second die=Width of first die+radius=10+1.2=11.2=11(approx)

Material required= Total parts/number of columns=38000/4*diameter=38000/4*2.4=22800 inch=1900 feet

Similarly for second=1645 feet=1700(approx)

So

cost for first,as it is 10 inch

so

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