# Stress in a rod clamped between two rigid walls when the temperature is increased

The usual approach to calculate stress is to equate thermal expansion in the unclamped condition to the magnitude of contraction caused by strain produced due to the walls. I have some questions about this approach:

1. Wouldn't Young's modulus of the rod change with temperature?

2. Moreover, in this method, $$L$$ is the length in the unclamped condition. After increasing the temperature, the 'original length' of the rod should be $$L(1+\alpha \Delta T)$$, which should be used in the relation (i) below because the original length of the rod is taken with reference to a particular temperature.

$$\Delta L=\frac{FL}{AY}=L\alpha \Delta T \cdots (i)$$ $$\frac{F}{A}=Y \alpha\Delta T= \text{stress} \cdots (ii)$$

Am I correct in both? This is really bothering me because I cannot find a discussion of these points in any books, but I strongly feel their validity.

After heating it, the length is $$L_0(1+\alpha \Delta T)$$ and its length has been increased by $$\Delta L=L_0\alpha \Delta T$$. To get it back to its original length, we have to compress it so that the second $$\Delta L$$ is minus the first $$\Delta L$$: $$\sigma=Y\frac{(L_0\alpha \Delta T)}{L_0(1+\alpha \Delta T)}=Y\frac{\alpha \Delta T}{(1+\alpha \Delta T)}\tag{1}$$But, from the equation for an infinite geometric progression, we have $$\frac{1}{1+\alpha \Delta T}=1-(\alpha \Delta T)+(\alpha \Delta T)^2-(\alpha \Delta T)^3+...$$If we substitute that back into Eqn. 1 and recognize that, in the approximations associated with Hooke's law, we are retaining only linear terms in the strains, we obtain (for the compressional stress): $$\sigma=Y\alpha \Delta T$$

• That means neglecting the denominator is only a practical approximation , unfortunately my doubt sprung up from a conceptual setback in various books I have seen where the natural length of the rod in the heated condition is taken as L itself and not L(1+α∆T) , which inspite of its incorrectness gives the same value for stress (the approx.) which you got in the end. Jan 29, 2019 at 16:17
• If you are saying that sometimes books don't explain things adequately, I agree. They should have indicated, in this case, that the analysis is confined to small (linearized) strains. Jan 29, 2019 at 16:23

The first point. We usually neglect the fact that Young's modulus changes with temperature. Point 2 Now you are right the original length is just as you wrote it. So now if we rise the temperature in the absence of any compressive force length will be L(1+@T). So in order to keep it at the previous length strain =L@T. Now so now we will put in the young modulus formula.

• In real-life engineering you don't ignore the change in E with temperature, and if you need to use logarithmic strain instead of engineering strain (because the strains are large) and a nonlinear stress-strain relationship (depending on the material) you can do that. The expansion coefficient $\alpha$ also changes with temperature. The OP's equations are just the simplest "toy model" to demonstrate the principle of how to do the analysis, and not necessarily accurate in any particular real-world situation. Jan 28, 2019 at 17:52
• Hmm Yess if given Y is given as a particular function then sure why not . It can be calculated. I said that for what he sees in his book it has been neglected. Jan 28, 2019 at 18:09

@dawood mansoor is correct, we normally neglect temperature dependence of Young's Modulus. This is reasonable as long as the temperature range is smal and maximum temperature does not affect other relevant material properties as well (softening, in particular).

That being said, the equation for thermal stress can be derived from the equation for mechanical stress. The main difference is that thermal stress is the result of restricting thermal expansion. See the figure below. If the right side of the bar were free, heating would result in the increase in length, $$dl$$, shown. The existence of a restraint (the wall shown on the right side) preventing the elongation creates thermal stress in the bar.

For mechanical stress involving uniaxial loading and deformation the stress-strain relationship is given by

$$σ=\frac{F}{A}=Eε$$

Where $$σ$$ is the normal stress $$E$$ is Young’s modulus and $$ε$$ is the mechanical strain in units per unit (e.g., m/m).

Linear expansion, $$dl$$ due to temperature increase, is given by

$$dl=αldt$$

Where $$α$$ is the temperature coefficient of thermal expansion in units of $$\frac{m}{m^0K}$$. $$l$$ is the original length of the bar, and $$dt$$ is the temperature change.

Thermal strain for unrestricted expansion is given by

$$ε_t=\frac{dl}{l}$$

Substituting in the previous equation we have

$$ε_t=αdt$$

With the expansion restricted, thermal stress, $$ε_t$$ is converted to mechanical stress $$ε$$ of the first equation. So setting $$ε= ε_t$$ in the first equation we get:
$$αdT$$

$$\frac{F}{A}=EαdT$$

Hope this helps.

• The relation '(i)' in my question is wrong ∆L=Lα∆T=(FL(1+α∆T))/AY , This is correct because natural length of rod has changed with rise in temperature and therefore , F/A=STRESS=Yα∆T/(1+α∆T), got confused because in your answer you have written STRESS=Yα∆T , and nothing in the denominator , am I correct with this ? Jan 29, 2019 at 6:30
• @ADITYAPRAKASH"STRESS" in my final equation is $\frac{F}{A}$ which makes my equation identical to your equation $(ii)$. As far as equation $(i)$ I don't know what "Y" is in the middle part, but since $\frac{\Delta L}{L}=ε$ where $ε$ is thel strain the equality between the first and last term is correct and is the same as the third to last of my equations. Hope this helps. Jan 29, 2019 at 13:52
• I will put an answer to my question which I think is correct , and I request you to check it. Jan 29, 2019 at 15:23

The original length of the rod is L (when the temperature is not raised), when the temperature is raised by ∆T the NATURAL TENDENCY of the rod is to stay in a length of L(1+α∆T), where α is the coefficient of linear expansion.Due to the fixed walls , the rod is being kept at a length of L , although at that temperature it should be in a length of L(α∆T) , so there is a case of compression in the rod , the compression being Lα∆T
Listing the components of the analysis:-
natural length=L(1+α∆T)
Young's modulus of elasticity of rod=Y
..and other components having their usual meanings
$$Lα∆T=\frac{FL(1+α∆T)}{AY}\Rightarrow F=\frac{YAα∆T}{1+α∆T}$$which is the compressive force provided by the walls from which we can get the stress diving F by A.