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If the shear stress in steel excecds about 4.00 x 1O'Nlm"it muptures: Detemmine the shear force necessary to break & bolt /.50 inch in diamete....

Question

If the shear stress in steel excecds about 4.00 x 1O'Nlm"it muptures: Detemmine the shear force necessary to break & bolt /.50 inch in diamete.

If the shear stress in steel excecds about 4.00 x 1O'Nlm"it muptures: Detemmine the shear force necessary to break & bolt /.50 inch in diamete.



Answers

Assume if the shear stress in steel exceeds about $4.00 \times 10^{8} \mathrm{N} / \mathrm{m}^{2}$ the steel ruptures. Determine the shearing force necessary to (a) shear a steel bolt 1.00 $\mathrm{cm}$ in diameter and (b) punch a 1.00 -cm-diameter hole in a steel plate 0.500 $\mathrm{cm}$ thick.

So for party, we know that the force needed to share the bolt of through its cross sectional area would be f equaling the cross sexual area most supplied by the stress. So this would be the cross sectional area. Pie times are squared 5.0 times 10 to the negative third meters quantity squared, multiplied by the stress of 4.0 times 10 to the eighth Newtons per square meter. And we find that the Forest Office equaling 3.14 times 10 to the fourth Newtons. This would be our final answer for Party four party for part B. Rather, we know the area over which the sheer occurs is equal to the circumference of the whole times the thickness and so here a the cross sexual areas equaling two pi r The circumference times t the thickness. This would be equaling two pi times 5.0 times 10 to the negative third meters multiplied by the thickness of 5.0 times 10 to the negative third meters. So this is equaling 1.57 times 10 to the negative fourth meters squared and so their force required to punch the hole would be f equaling the cross sectional area eight times again the stress. So this would be equaling 1.57 times 10 to the negative forth square meters multiplied by the stress of 4.0 times Ted to the Eighth Newtons per square meter. And we find that the force required is to punch a hole is 6.28 times 10 to the fourth Newton's. This would be our final answer for Part B. That is the end of the solution. Thank you for watching.

For still to be failed. The shear stress that is F panel upon a must be equivalent to the limiting she justice and represent this way. How limited. Okay, so from this you can buy it for a circular projection. The first pallet with those offices by these were into the limiting cheer stress upon. For now it's obscured the value so apparently liquids buy into the diameter. D is uh 0.5 cm square into the limiting shear stress that is full in to tenders eight newton four m square, about four to convert centimeter centimeter, multiply satisfying. one m 400 cm sq on the force needed to break the balls is 7.9 into tenders three notice and two significant is it?

Hello. Everyone here has given if the CIA spare in the speed XY the maximum Sierra Strip, she got a spreading. It still may be four in 10 to the power eight Newton per meter square that still rupture. Return my sharing force necessary. We have the required sharing ports necessary for for bullet off diameter. One centimeter B but through a bunch off one centimeter own diameter hole. You nice. Beautiful it off. Why fight sentimental? But it course it's called toe area into spreads Area is why did square by foot into a stress substitute Developed 3.1 ft diameter is given bunch intimidated in tow The stress pouring toe tend to devour it on solving it You will get 3.1 ft in tow tend to the pervert for nuclear So this is the answer up part it No, we are solving the second part the area over which if you see that agape, we have to open chariot So the area over which the the awkward is equal toe that circumference of the whole time If Cygnus you haven't we by D in Tokyo 3.1 ft The meeting of the whole one tent image. It on sickness is but you will get. And yes, will be. Has fourth Bilby area into its threats and its president wouldn't do tend to devour it. So it will be six point to it in tow. Tend to the power for Newton. This is the answer off, people. Thanks for watching it.

My friends, This is a problem based on elasticity. Here it is given I still wear having that I am eater 1.25 inch. That is 1.25 and took point 0883 ft. It is under the force off five G I. P. That is £5000. Do you have to fight maximums here? Stress maximums here expressed. Then it is bad little the face. So it would be have upon fighting square by foot Like this. My home? Yeah. Yeah. So for maximum, the force will be in this direction. That is full f upon by the script substituting the volume full in tow. £5000 3.14 and 1.25 and 2.0. Hated three square. So on solving it maximum. It's serious press. You will get 5.5 ft 10 20 power, £5 per feet square. Hertzel. Thanks for watching it


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