Force required to reduce the diameter

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Discussion Overview

The discussion revolves around calculating the force required to reduce the diameter of a mild steel bar from 40 mm to 39.99 mm under tensile stress. It includes aspects of material properties, such as Young's modulus and Poisson's ratio, and involves mathematical reasoning related to strain and stress calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial calculation of transverse strain is presented as -0.25 x 10^-3 based on the diameter change.
  • Axial strain is derived from the transverse strain using Poisson's ratio, resulting in an axial strain of 833.333 x 10^-6.
  • Axial stress is calculated as 166.666 x 10^6 N/m² using the derived axial strain and Young's modulus.
  • The force required is computed as 209439.51 N based on the calculated axial stress and the cross-sectional area of the bar.
  • One participant suggests rounding the answer to 200 kN, while another questions the rationale behind this rounding.
  • Discrepancies in answers arise, with one participant obtaining 209.36 kN due to differences in unit conversion and strain calculations.
  • Discussion includes a point about significant figures, suggesting that the final answer should reflect the lowest number of significant figures from the given values.

Areas of Agreement / Disagreement

Participants express varying opinions on the rounding of the final answer and the accuracy of the calculations, indicating that there is no consensus on the correctness of the initial calculations or the rounding approach.

Contextual Notes

Participants highlight potential issues with unit conversions and significant figures, which may affect the final results, but do not resolve these concerns.

oxon88
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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
 
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can anyone check my workings please?
 
Looks good! Just round off your answer to 200 kN.
 
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Many Thanks.
 
Was this correct Oxon88 as I get a answer which is slightly different from yours?
 
yes. what answer did you get?
 
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
 
looks acceptable, its pretty close to what i got. As PhanthomJay states, just round it off to 200kN
 
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
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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