In a blanking operation to produce steel washers- the maximum punch load used is 3-105 N- The plate thickness is 5 mm and percentage penetration is 25- What is the work done during this shearing operation-

Work done = F_max× pt nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;=300 kN×0.25×5=375 J ...

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Force and Work done in Punching and Blanking

In a blanking...

Question

In a blanking operation to produce steel washers, the maximum punch load used is $3 \times 10^{5} N$ . The plate thickness is 5 mm and percentage penetration is 25. What is the work done during this shearing operation?

325 J

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475 J

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375 J

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350 J

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Solution

The correct option is C 375 J
$Work done = F_{m a x} \times p t$
$= 300 k N \times 0.25 \times 5 = 375 J$

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In a blanking operation to produce steel washers, the maximum punch load used is 3×105N. The plate thickness is 5 mm and percentage penetration is 25. What is the work done during this shearing operation?

The correct option is C 375 JWork done=Fmax×pt =300kN×0.25×5=375J

In a blanking operation to produce steel washers, the maximum punch load used is $3 \times 10^{5} N$ . The plate thickness is 5 mm and percentage penetration is 25. What is the work done during this shearing operation?

The correct option is C 375 J
$Work done = F_{m a x} \times p t$
$= 300 k N \times 0.25 \times 5 = 375 J$