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For 50% penetration of work material, a punch with single shear equal to thickness will
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- reduce the punch load to half the value
- increase the punch load by half the value
- maintain the same punch load
- reduce the punch load to quarter load
Correct Option: A
Punch load = π Dt τ | ![]() | ![]() | |
t1 |
t1 = shear on punch.
P = 50%
Punch load = π Dt τ × | |
t |
⇒ 0.5 π Dt τ