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Methods to Reduce Punch Force - IV

A hole of 40 ...

Question

A hole of $40 m m$ diameter is pierced in a steel sheet of $4 m m$ thickness without shear on the tool. Shear strength of steel is $400 N / m m^{2}$ and penetration is $25$ %. What is the expected percentage load reduction if a shear of $1 m m$ is provided on the punch?

25.00

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33.33

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50.00

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66.67

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Solution

The correct option is C $50.00$
As $s = p t$
$F = \frac{F_{m a x} K t}{K t + S} = \frac{F_{m a x} \times 0.25 \times 4}{0.25 \times 4 + 1} = \frac{F_{m a x}}{2}$
$\Rightarrow$ Hence, $50$ % reduction in punching load.

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A hole of 40mm diameter is pierced in a steel sheet of 4mm thickness without shear on the tool. Shear strength of steel is 400N/mm2 and penetration is 25%. What is the expected percentage load reduction if a shear of 1mm is provided on the punch?

The correct option is C 50.00As s=pt F=FmaxKtKt+S=Fmax×0.25×40.25×4+1=Fmax2 ⇒ Hence, 50% reduction in punching load.

A hole of $40 m m$ diameter is pierced in a steel sheet of $4 m m$ thickness without shear on the tool. Shear strength of steel is $400 N / m m^{2}$ and penetration is $25$ %. What is the expected percentage load reduction if a shear of $1 m m$ is provided on the punch?

The correct option is C $50.00$
As $s = p t$
$F = \frac{F_{m a x} K t}{K t + S} = \frac{F_{m a x} \times 0.25 \times 4}{0.25 \times 4 + 1} = \frac{F_{m a x}}{2}$
$\Rightarrow$ Hence, $50$ % reduction in punching load.