Strain-stress, cross sectional, elongation

[ruffryder]

Hey guys, I will admit that I am having extra troubles on this chapter and I don't understand most of it so be gentle :\. Anyways I realize these problems are porbably easy but here they are:

1) A Load of 140,000 N (31,500lbf) is applied to a cylindrical specimen of a steel alloy that has a cross-sectional diameter of 10mm. If the original specimen length is 500mm (20in.) how much will it increase in length when this load is applied?

so i think its load/original cross sectional area? 140,000/ 10,000 = 14.0

the real answer is 8.5mm. how do i get there?

2)A cylindrical rod 500mm long, having a diameter of 12.7mm is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 1.3mm when the applied load is 29,000N which of the four metals or alloys are possible candidates.

Aluminum alloy 70GPa, 255MPa yield strength, 420MPa tensile strength
Brass Alloy 100, 345 , 420
Copper 110, 210 , 275
Steel Alloy 207, 450 , 550

The answer is steel and brass. I have no clue how to do this one. So you don't have to answer it

 

[radou]

Hey guys, I will admit that I am having extra troubles on this chapter and I don't understand most of it so be gentle :\. Anyways I realize these problems are porbably easy but here they are:

1) A Load of 140,000 N (31,500lbf) is applied to a cylindrical specimen of a steel alloy that has a cross-sectional diameter of 10mm. If the original specimen length is 500mm (20in.) how much will it increase in length when this load is applied?

so i think its load/original cross sectional area? 140,000/ 10,000 = 14.0

the real answer is 8.5mm. how do i get there?

2)A cylindrical rod 500mm long, having a diameter of 12.7mm is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 1.3mm when the applied load is 29,000N which of the four metals or alloys are possible candidates.

Aluminum alloy 70GPa, 255MPa yield strength, 420MPa tensile strength
Brass Alloy 100, 345 , 420
Copper 110, 210 , 275
Steel Alloy 207, 450 , 550

The answer is steel and brass. I have no clue how to do this one. So you don't have to answer it

1) The expression for elongation is \Delta l = \frac{F\cdot l}{E \cdot A}, where l is the length before deformation, F if, of course, the applied load, E the elasticity module for steel, and A the area of the cross section. I hope this works.

2) A similar problem - after solving 1), you should have a clue.

 

[ruffryder]

ya thnx for the response, i think i can take it from here.