# Help on stress

I'm having problem knowing how to work out this question, can anyone point me in the correct direction.

Question
14mm diameter steel rod carries a load of 3.08kN in tension. calculate the stress.?

I ended up with this:
Stress = Force/Area.
Stress = 3.08/154 = Answer = 0.02N/mm²

if the rod is 2.5m long and the modulus of elasticity is 200GN/m2
how much will it extend?

so far i have know idea of the formula or how to work this out?

I also have one other question related to stress that is

A brass rod measures 10mm in dia, and 150mm length. The rod extends 0.3mm when subjected to a load. given that brass E=85GN/m2
Calculate the stress in the material and the magnitude of the load:
if you could help with the formula that would be a big help.
I'm not able to find any thing on the net and the books i got from the library are over the top in calculations
many thanks

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Hi donkey131, welcome to PF. For a rod undergoing axial loading, stress $$\sigma[/itex] and strain [tex]\epsilon[/itex] are related by the equation [tex]\sigma=E\epsilon[/itex]. Make sure to check your units in your calculations. Another formula that you will need is [tex]\epsilon$$= [l(final)-l(initial)]/l(initial)

so how do i now find the Magnitude.

not sure about this as im trying to teach my self, it an NVQ course. NOt good teaching your self something you don't know.

Calculate the stress in the material and the magnitude.
10mm = Diameter steel rod
?? = Area of the rod
3.08kN = Tension
?? = Stress
?? = Strain
150mm = Length
0.3mm = Extends
85 GN/m² = Elasticity

Area:
Area = pi x radus² There for: Area = Pi (3.142) x 2.5² (6.25) = 19.637mm² (19.64)

Strain:
Strain = extension (change in Length)/Original Length
Strain = 0.3/150 = 0.002

There for we now can find the Stress:
Stress = Strain x Elasticity
Stress = 0.002 x 85 =0.17N/m²

so how do i find Magnitude:?????

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