I don't understand 2007 HSC Question 10. (1 Viewer)

[CMCDragonkai]

A mild steel test piece was made into 1/4 scale. It was stretched to 200kN tension until destroyed.

At what minimum force would you expect the full size component to fail?

The answer is 3200kN.

How do you do this?

 

[philkurts]

stress=force/area 1/4 size is 1/16 of the area. So full size is 16 times more area, 16 times more load for the same stress.

 

[mahdi]

stress = force / area

say the normal test piece has a raidus of 1 (it doesnt matter what you give it)
its area will be, pi x 1 sqaured
which equals pi

if the area is 1/4, then the new radius would be 0.25
so area would equal, pi x 0.25 sqaured
and 0.25 squared equals 1/16
so area now equals pi/16

therefore the force will be 16 times greater
200Kn x 16 = 3200Kn

 

[gurmies]

stress = force / area

say the normal test piece has a raidus of 1 (it doesnt matter what you give it)
its area will be, pi x 1 sqaured
which equals pi

if the area is 1/4, then the new radius would be 0.25
so area would equal, pi x 0.25 sqaured
and 0.25 squared equals 1/16
so area now equals pi/16

therefore the force will be 16 times greater
200Kn x 16 = 3200Kn

Solid stuff, Mah-Deh xD

 

[Kaos1]

mahdi is exactly right in his maths. good work

I would have just said its inversely proportional (ie. 1 over root 2), but your method is alot simpler

 

[CMCDragonkai]

Nice answer, thanks.