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Elastic deformation problem

  • Thread starter savva
  • Start date
  • #1
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I am not sure which formula to apply, can anyone help me out?

Homework Statement



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

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
SteamKing
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What is the formula for stress?
 
  • #3
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Exams and leaving study to the last minute
 
  • #4
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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)
 
  • #5
235
1
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.
 
  • #6
235
1
@ 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.
 

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