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A hole of diameter d is to be punched in a plate of thickness t. For the plate material, the maximum crushing stress is 4 times the maximum allowable shearing stress. For punching the biggest hole, the ratio of diameter of hole to plate thickness should be equal to
1. 1/4
2. 1/2
3. 1
4. 2

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Correct Answer - Option 3 : 1

Concept:

Punching:

in sheet metal operation, punching operation the process in which sheared part is considered as scrap and the remaining part is desired part.

For punching opreration,

⇒ Punch tool size = Hole size = d

The maximum crushing stress induced in the punching tool,

\( ⇒ {{\bf{\sigma }}_{{\bf{cmax}}}} = \frac{{4{\bf{F}}}}{{{\bf{\pi }}{{\bf{d}}^2}}}\;\;\;\;\; \ldots \left( 1 \right)\;\)

The maximum allowable shearing stress,

\( ⇒ {{\bf{\tau }}_{{\bf{smax}}}} = \frac{{\bf{F}}}{{{\bf{\pi dt}}}}\;\;\;\;\;\; \ldots \left( 2 \right)\;\)

where, F = Punching force, d = diameter of the hole (punching tool diameter), t = thickness of the metal sheet

Calculation:

Given:

σcmax = 4τsmax

\( ⇒ \frac{{4{\bf{F}}}}{{{\bf{\pi }}{{\bf{d}}^2}}} = 4 \times \frac{{\bf{F}}}{{{\bf{\pi dt}}}}\;\)

⇒ d = t

⇒ The ratio of the diameter of the hole to plate thickness = 1

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