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Force required to reduce the diameter

  • Thread starter oxon88
  • Start date
  • #1
176
1

Homework Statement



A mild steel bar 40 mm diameter and 100 mm long is subjected to a tensile force along its axis.
Young’s modulus of elasticity for mild steel = 200 GN m–2.
Poisson’s ratio is 0.3.

Calculate the force (F) required to reduce the diameter to 39.99 mm.

Use the x–y coordinate system as shown above.

Homework Equations



Poissson's Ratio = - (transverse strain / axial strain)

force = Stress x Area

The Attempt at a Solution



transverse strain = (39.99 - 40) / 40 = -0.25x10^-3

axial strain = - (-0.25x10^-3 / 0.3) = 833.333 x 10^-6

axial stress = (833.333 x 10^-6) x (200 x 10^9) = 166.666 x 10^6

force (F) = 166.666x10^6 x (0.25∏ x 0.04^2) = 209439.51 N
 

Answers and Replies

  • #2
176
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can anyone check my workings please?
 
  • #3
PhanthomJay
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Looks good! Just round off your answer to 200 kN.
 
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  • #4
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Many Thanks.
 
  • #5
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Was this correct Oxon88 as I get a answer which is slightly different from yours?
 
  • #6
176
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yes. what answer did you get?
 
  • #7
101
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I got 209.36kN that's why I was wondering. I changed the dimensions of the tube into m from the start and got a different answer to your transverse strain which then saw all my answers being different all the way through. Hence why I wanted to know if yours was correct a I thought it was slightly out
 
  • #8
176
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looks acceptable, its pretty close to what i got. As PhanthomJay states, just round it off to 200kN
 
  • #9
101
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why would you round it down to 200kN? Also I commented on another question regarding tubular column which is question 2 and was waiting to hear back from that thread if you could that would be great.
 
  • #10
PhanthomJay
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Values should be rounded to the lowest number of significant figures in the given values. All such values have only 1 significant figure. Therefore, the answer is good only to one significant figure.
 
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