Question

What force will be required to punch a hole of 10 mm dia in a 1 mm thick plate, if the allowable shear stress is 50N/mm²? (π = 22/7) *

Answer 1

Answer:  3928.5 N

                       

Step-by-step explanation:

                                 we know that stress = F/A=50 N/mm^2

                                                 Area = pi r^2

                                      so, 50 = F/pi 5^2

                                         F = 3928.5 N

Answer 2

The given data is a 10 mm dia in a 1 mm thick plate.

we have to punch a hole the question is that what force is required to punch a hole on it.

The allowable shear stress is given as

$\frac{50 \: newton}{ { mm}^{2} }$

The formula of the stress is

$\frac{force}{area} = \frac{50newton}{ {mm}^{2} }$

area =π*r*r

so,

$50 = \frac{force}{\pi \times r \times r}$

π=22/7

$\frac{50 \times 5 \times 5 \times22 }{7} = force$

$\frac{27500}{7} = 3928.57$

Therefore, the final answer is F=3928.57N

# spj2

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