Steel ruptures when a shear of 3.5 * 10 ^8 Nm^-2 is applied. The force neede to punch a 1cm diameter hole in a steel sheet 0.3cm thick is nearly?

Question

priyar · Answer

@ganeshie8 @samigupta8 @parthkohli @mayankdevnani

irishboy123 · Answer

isn't this just $\tau = \dfrac{F}{A}$  ?

priyar · Answer

oh its that simple? shear means stress..? but i have learnt only categories in strain..not stress i am bit confused.. @IrishBoy123 ..

irishboy123 · Answer

yes, the A is the cylindrical area that is being sheared, and that, if recall these things correctly, is that.  bish bosh.

priyar · Answer

so would it be 3.5 * 10 ^8 * 3*10^-5 = 1.05 * 10^4N?

irishboy123 · Answer

not too sure i follow all that but in letters $ F = \tau \times 2 \pi r t$ where t is metal thickness and r = radius of hole

priyar · Answer

i missed out pi in the abv exp. !

priyar · Answer

so F = 3.3 *10^4 N.. thats correct!

priyar · Answer

Thank You @IrishBoy123!!

samigupta8 · Answer

BT shear is applied tangentially if we really want to punch the steel sheet then how can one do that without applying  a normal force on it...?

samigupta8 · Answer

@priyar do u know this?

samigupta8 · Answer

Did you get my doubt?