1
$\begingroup$

The equation of thermal stress is:

Stress =$\frac{F}{A}$ = -$E$ $a$ $dT$, where $E$ is Young's Modulus, $a$ is the coefficient of linear thermal expansion, and $dT$ is the change in temperature.

I can't think of an intuitive reason for which $E$, $a$, and $dT$ would be multiplied together, and I haven't been able to find anything online. In this case, I would assume that a derivation would explain the properties of this formula. Does anyone have an idea as to what this derivation would look like?

$\endgroup$
1
  • $\begingroup$ I can think of an intuitive reason for this: there is a relation between expansion of a material, be it mechanical (pulling it apart) or thermic. The fact that both types of stresses (mechanical and thermic) can be linked to expansion via Young modulus, speaks of the common underlying physics: molecular/ionic forces. As for dL = a*dT it speaks of the linear relation between thermal expansion and increase of temperature. So I think all this give sense to multiplying E, a and dT. $\endgroup$
    – rmhleo
    Sep 14, 2015 at 8:45

4 Answers 4

2
$\begingroup$

From the definition of Young's Modulus, we have the following expression: $$ Y = \frac{Stress}{Strain}=\frac{\frac {F}{A}}{\frac {\Delta l}{l_0}} $$

Also from the definition of coefficient of thermal expansion, we have $$ \alpha =\frac{l-l_0}{l_0t} $$ or, $$ \Delta l = l - l_0 = l_0 \alpha t $$

Substituting the value of $\Delta l$ in the first expression, we have $$ Y = \frac{\frac {F}{A}}{\frac { l_0 \alpha t}{l_0}} =\frac {F}{\alpha A t}$$ or $$ \frac{F}{A} = Y \alpha t$$

$\endgroup$
1
  • $\begingroup$ The reason the minus sign of the original question doesn't appear is that this approach calculates the external load required to mimic a certain amount of thermal expansion. What is more often sought is the stress that arises from a strain constraint (e.g., of zero) under a temperature change. See also this discussion. $\endgroup$ Apr 26, 2022 at 18:07
1
$\begingroup$

The stress in an element (rod, beam, etc) is:

$$ \frac{F}{A} = \frac{E \Delta l}{L_0}$$

where $\Delta l$ is the length change of the element. That really just stems from the definition of Young's Modulus.

From here, it's easy to see where the equation for thermal stress comes from -- the linear coefficient of thermal expansion multiplied by the change in temperature gives the change in length of the element. This is just substituted into the expression for stress.

$\endgroup$
0
$\begingroup$

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

$$ε_t=αdt$$

Restricted expansion is converted to thermal stress. Setting $ε= ε_t$ in the first equation we get:

$$\frac{F}{A}=EαdT$$

Not sure where you got the negative sign in your equation. It could be based on the fact that the reactions at the walls in the figure below are equal and opposite to the expansion forces shown. The reactions reflect compressive stress, which is considered negative vs. tensile stress, which is considered positive.

Hope this helps. enter image description here

$\endgroup$
-2
$\begingroup$

you have to know something here

thermal didn't add stress by it self , the part have to be fixed in direction of expansion in order to convert the change of length to stress

so that you can understand that the heat is not add any stress if the parts are free because nothing stops change in length

I hope this point is clear because some times the part is free or partially free which mean it can change its length freely for example 1 mm but the heat expansion is 3mm then the stress will calculate on 2mm strain jut

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.