In the stress-strain relationship for a stretched wire, we say that $\frac{F}{A}=Y\frac{β}{l}$; where β is the change in length and the rest have their usual meaning. We also say that there is a relative change in diameter every time there is a relative change in the length, namely given by the Poisson's ratio. My question is that for a given force, as the width of the wire also changes with length, cross-sectional area must also change. Thus, as the force starts acting, area would start decreasing continually. Therefore, area cannot be taken to be constant as in the equation for young's modulus. Also, if diameter were to reduce proportionally to the increase in length, every time the length increased by an increment there would be lesser cross-sectional area and this would permit more elongation. After sometime, the wire would have become so thin that it would break. Thus for any small a force, the wire would break. Please explain why this is not true and why my approach is incorrect.

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    $\begingroup$ Did you actually do the math of this, or are you coming to your conclusion by intuition only? If you write down the change in length and area for a given force, and from this compute the strain, you will see that the situation is stable for all valid values of Poisson ratio. Try writing down the equations and see if you have any questions after you did so. $\endgroup$
    – Floris
    Commented Aug 11, 2014 at 18:00
  • $\begingroup$ Engineering strain is defined as deflection relative to unstressed conditions, as opposed to real strain. This applies to both length and section area. What you describe is the unstable point beyond which the wire is going to accelerate the elongation and break. $\endgroup$ Commented Aug 11, 2014 at 19:07
  • $\begingroup$ @Floris: Yes, I did do the math and checked my derivations quite a few times over. If I take into account the width contraction, this is what I get (forgive me for the equations, I don't know latex): $\endgroup$
    – Gaurav
    Commented Aug 12, 2014 at 5:33
  • $\begingroup$ (F/A)=(-Y/2α){[(L/l)^2α] - 1}, where A is initial cross-sectional area, L is initial length & α is Poisson's ratio. $\endgroup$
    – Gaurav
    Commented Aug 12, 2014 at 5:44
  • $\begingroup$ @ja72:Yes, I do understand. But what stops further elongation? And in my equation in the previous comment, l is final length. $\endgroup$
    – Gaurav
    Commented Aug 12, 2014 at 5:58

2 Answers 2


The stress that we take into consideration is the Engineering stress i.e $\frac{F}{A_{original}}$. The actual stress is $\frac{F}{A_{actual}}$. We consider the engineering stress only because upto the yield point, the original area is approximately equal to the actual area. However above the yield point, plastic deformation takes place and the actual area becomes considerably smaller than the original area. So the stress increases much more rapidly and failure takes place.


I think I have arrived at a answer to my own question. It hit me that the answer is so simple after a brief inspection of the situation. Anybody else with the same doubt may find this helpful. The molecules of the solid are bound by spring-like bonds to each other. That is, the more you pull one molecule away from another, the more it tries to pull it back. Therefore, the wire will elongate until the force exerted by the spring-like bonds can hold back the external stress. However, the external stress will try to stretch the wire as much as it can. Thus, by decreasing cross-sectional area the wire can increase length as it will try to keep the total volume constant. However, area will not keep decreasing beyond a limit, as inter-atomic repulsion's will prevent atoms coming together any closer. So it will reach a steady state at this point. As for the area remaining constant, it is because the area remains practically constant; as for most materials, the width contraction is very small compared to length.

  • $\begingroup$ however, the further apart the molecules, the smaller the attraction is (In my opinion). So why should the force bahave like spring which the further apart, the greater the force? $\endgroup$
    – Bruce M
    Commented Jan 22, 2022 at 10:21
  • $\begingroup$ Your guess that beyond a limit, the area remain constant is not correct for most materials. Indeed the decrease in area can be a non-linear function of stretch in the direction perpendicular to the cross-section. $\endgroup$
    – Davius
    Commented Jul 5, 2022 at 12:40

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