Elastic deformation problem

[savva]

I am not sure which formula to apply, can anyone help me out?
1. The problem statement, all variables and given/known data

28. A square copper bar experiences only elastic
deformation if it is stressed less than 95MPa.
To support a load of 1340kg without exceeding this stress,
the minimum square cross-section ( i.e. width of one side
of the square cross section) required is

a) 1.8mm square
b) 3.6 mm square
c) 23.6mm square
d) 11.8mm square

2. Relevant equations



3. The attempt at a solution

[SteamKing]

What is the formula for stress?

[savva]

Exams and leaving study to the last minute

[savva]

Although the scientific formula for stress is τ = F / A
Rearranging this gives A = F / τ

Thus:
F = 1340kg x 9.8
τ = 95x10^6 Pa (as stated in the question)

Subbing these into A = F / τ gives 1.38x10^-4 m^2

Convert this to mm^2 by multiplying by 10^6 which gives 138.2 mm^2

Asquare = L^2, therefore to find the width of one side square root the area,\

so (138.2^0.5) = 11.8mm (Making the answer D)

[chikis]

Rember we have tensil stress and we have tensil strain.
Stress is the ratio of the force or load, F on the elastic material eg spring or string to the cross sectional area, A. It can be summarized as: stress= F/A.
While tensil strain is the ratio of the extention, e of the elastic material to the lenght, l. It can be given as strain=e/l . They are two different things. Always differenciate them for a better understanding.
The unit of stress is Nm-2 while strain has no unit since e and l are in metres.

[chikis]

@ Savva,
are you satisfied with the answer you got concerning your question on elastic deformation.
If you are not, then what does the unit MPa (95MPa) stand for; let me see how I can come in.
I know that v, is velocity which is ms-1. m is for mass which is kg. a is for acceleration which ms-2 e.t.c. But that MPa I don't know.