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If a steel rod of length 1 m rests on a smooth horizontal base is heated from $T_1$ to $T_2$, what is the longitudinal strain developed?

I think that due to thermal expansion/contraction, the length of the rod should change, so strain should be developed. However, after googling my doubt it turns out the strain developed is zero.

Can I get an explanation as to why it is zero?

Here's a link to one of the pdfs I found on the Internet about this topic:

https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-20-structural-mechanics-fall-2002/lecture-notes/unit9.pdf

See page no. 5, figure 9.1.

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  • $\begingroup$ Hint: What is the definition of strain? What is the applied force here? $\endgroup$ Commented Feb 10, 2021 at 7:18
  • $\begingroup$ @GuyInchbald I think the restoring force is zero $\endgroup$
    – user287374
    Commented Feb 10, 2021 at 7:29
  • $\begingroup$ Well, even your lnked document says right where you cited (p5) "... thermal strain is equal to the total strain...", and on p3 it says "...The total strain of a material is the sum of the mechanical strain and the thermal strain...." and on p4 "(“actual”) total strain: that which you actually measure; the physical deformation of the part.." and on p2 "$\epsilon^T= \alpha (\Delta T)$" $\endgroup$
    – Toffomat
    Commented Feb 10, 2021 at 12:47

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It says that mechanical strain is zero. The author considers mechanical strain and thermal strain separately. The mechanical one is zero if there is no external force. He never say that the thermal strain, due to thermal expansion, is zero.

Edit According to the article: Thermal strain is due to the change in temperature of the bar. Mechanical strain is due to external forces acting on the bar. Both strains measure the change in dimensions, here focus being on length of the bar. If the temerature is fixed and you pull the ends of the bar, the bar expands and you have a mechanical strain. If the temperature is changed but no forces act on the bar, the bar again expands and this is thermal strain. This is what free expansion means. The bar may be on a frictionless table and the ends are free from any obstacle. So there are no horizontal forces on the bar. In general, you can hae both strains at the same time and they can cancel out in the right conditions. If you heat the bar but you aplly some compression at the ends, you may have no change in length. You can imagine this as having the two strains with opposite signs and cancelling each other. But the net effect is no change in length so no strain. But there is a stress in the bar. You can consider this situation in two steps, if you want. You heat the bar, it expands. Then you apply some compression force and you bring it back to the initial length.

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  • $\begingroup$ can u explain why the thermal strain in free expansion is zero $\endgroup$
    – user287374
    Commented Feb 10, 2021 at 13:44
  • $\begingroup$ I never said such a thing. And is not. The mechanical strain is zero in this case. By mechanical strain he means the strain produced by external forces. So if there are no external forces... $\endgroup$
    – nasu
    Commented Feb 10, 2021 at 13:48
  • $\begingroup$ im really sorry but i am really confused rn $\endgroup$
    – user287374
    Commented Feb 10, 2021 at 14:03
  • $\begingroup$ can u make it a little more elaborate $\endgroup$
    – user287374
    Commented Feb 10, 2021 at 14:03
  • $\begingroup$ See my edit to the answer. $\endgroup$
    – nasu
    Commented Feb 10, 2021 at 15:09
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Thermal strain is strain that develops when a material is heated or cooled. So the strain should not be zero.

In addition to the site you provided, thermal strain for a non-constrained object is not zero as long as the coefficient of thermal expansion is not zero, per the following:

http://emweb.unl.edu/NEGAHBAN/Em325/05-Thermal-strain/Thermal%20strain.htm

http://www.ce.memphis.edu/3322/Pdfs/PaulsPDFs/Thermal%20Strain.pdf

https://www.engineeringtoolbox.com/stress-restricting-thermal-expansion-d_1756.html

https://en.wikipedia.org/wiki/Thermal_expansion#Effects_on_strain

https://www.britannica.com/science/thermal-strain

Thermal stress (though a misnomer) is the internal stress that occurs if the object being heated or cooled is constrained. That would be the case if the steel rod was wedged between two rigid walls prior to heating.

When the heated rod is constrained from increasing in length, its thermal strain is cancelled by the compressive mechanical stress of the walls. If it was fixed to the rigid walls (i.e., unable to separate from them) and cooled so that it is constrained from decreasing in length, its thermal strain is cancelled by the mechanical tensile stress of the walls.

Hope this helps.

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    $\begingroup$ No. Thermal strain is the strain induced by the temperature change when the length is held constant. See for example mechanicaljournal.com/2019/11/22/thermal-stress-and-strain $\endgroup$ Commented Feb 10, 2021 at 12:40
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    $\begingroup$ @GuyInchbald you and the linked page seem to confuse stress and strain. When the length is held constent, there is no (total) strain, while in an unconfined situation, there is no stress $\endgroup$
    – Toffomat
    Commented Feb 10, 2021 at 12:49
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    $\begingroup$ @Toffomat. Thermal strain is subtly different. If you don't believe a respected mechanical journal, go find us a more reliable source to the contrary. $\endgroup$ Commented Feb 10, 2021 at 12:53
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    $\begingroup$ @GuyInchbald I guess it's clear from the question that OP is interested in the actual measurable longitudinal strain, which is $\Delta L/L$, and the "smooth surface" means there is no logitudinal force resisting thermal expansion, so $\epsilon=\alpha \Delta T$ (what your source calls "expansion of bar"$/l$).... $\endgroup$
    – Toffomat
    Commented Feb 10, 2021 at 13:03
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    $\begingroup$ @GuyInchbald Where in the article linked did you see the text quotedin your comment? I can see that the author calculate the STRESS when the length is held constant. But not the strain, which will be zero is the length is held constant. At least the longitudinal strain. There will be transversal strain but he does not seem to discuss this. $\endgroup$
    – nasu
    Commented Feb 10, 2021 at 13:31

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