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A steel rod of some length rests on a SMOOTH horizontal base. If it is heated for a considerable temperature difference, say $0^\circ C$ to $100^\circ C$, what is the strain produced?

Strain is difference length/original length. New length is $L(1+\alpha \Delta T)$ and original length is L. Therefore strain comes out to be $\alpha \Delta T$. But if you apply Young's modulus which says $Y=\text{Stress/Strain}$. On rearranging we get $\text{Strain=stress/Y}$. As its a smooth surface now force of resistance is applied by surface. Therefore stress is zero and subsequently strain should also be zero.

The question in the book says strain is zero.

Which one of these is correct and why doe we consider that ?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jun 21 at 6:15
  • $\begingroup$ @DavidZ so can you please answer it then? $\endgroup$ – Shaurya Goyal Jun 21 at 8:27
  • $\begingroup$ Do you have a question about my Answer? Is there something about it that you disagree with? Note the similarity of my answer to that of @Claudio Saspinski, and that we disagree with the answer that you accepted. $\endgroup$ – Chet Miller Jun 22 at 2:27
  • $\begingroup$ @ChetMiller the answer that i accepted is the theory that is being taught to everyone else at this level of education (pre university) thus i accepted it. The answers that you and Claudio added make it clearer. Thank you $\endgroup$ – Shaurya Goyal Jun 22 at 14:19
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For a rod experiencing both thermal strain and stress loading, the formula is not $$\sigma=Y\epsilon$$. The correct formula is $$\sigma=Y(\epsilon-\alpha \Delta T)$$In this example, since the surface is smooth, there is no friction to allow tensile stress to build up in the rod when it is heated. So it experiences unconstrained expansion, in which case, $$\sigma=0$$ and $$\epsilon = \alpha \Delta T$$If your book says that the strain is zero, it is incorrect.

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I suppose it is a one dimensional problem, and the rod is just lying on the smooth surface, so that it is free to elongate as being heated. We can not expect any stress from a possible restriction to deformation caused by the surface.

Even if the heating process is such that one end of the rod reaches a temperature much greater than the other end, again there is no stress because the heated portions of the rod are free to expand.

Thermal stresses require some sort of constraining for displacements. It will be the case if we think of a rod suddenly placed inside a furnace at 1000 K for example. A gradient will develop between surface and interior before the temperature reaches an equilibrium. During that period there will be thermal stresses inside the rod.

Edit Jun,22th: My answer is all about stresses, and I realize now that the OP question is about strain. The elastic strain is zero, but the total strain is not zero.

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  • $\begingroup$ Both are answers are consistent with one another and in conflict with the answer that the OP accepted. What are your thoughts on this, and how can she be properly set straight. $\endgroup$ – Chet Miller Jun 22 at 11:49
  • $\begingroup$ @Chet Miller After reading your comment, I read again the OP question and updated my answer. The strain is not zero as you said, unless the book specifies elastic strain. $\endgroup$ – Claudio Saspinski Jun 22 at 15:29
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First you have to know the proper definition of stress. By stress what immediately comes to our mind is $\frac{\textit {Force}}{\textit {Area}}$. But which force? Many mistakenly take it as the deforming force which is being applied on the body; just for example, you take it here as the force exerted by the surface on the body.

This is quite wrong. Actually, when a body is under the application of a deforming force, internal force of reaction comes into play within it, which tends to resist the applied force so as to maintain the original configuration of the body. This internal reaction force per unit area of the body is defined as stress.
However, according to Newton's third law of motion, the force of action is equal and opposite to the force of reaction. So, in case of mechanical stress, the stress is measured by considering the magnitude of the deforming force (obviously, that is easier than measuring the internal reaction force).

But, thermal stress is measured from the strain due to thermal expansion of the body which is not zero (why should it be because there is non-zero change in its length if you consider longitudinal strain). As evident from your equations, thermal stress $$\sigma=Y\times \textit {thermal strain}=Y\alpha\Delta T$$ So, as you can clearly see, stress is not zero.

This is as far as I made out of this although I do not understand why your book says that the strain should be zero.
You can also check out this answer : How would one derive the equation of thermal stress?

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  • $\begingroup$ this is exactly what i was thinking. it must be a mistake by the author then. $\endgroup$ – Shaurya Goyal Jun 21 at 9:45

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