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A problem requires to find the stress in the portions AC and BC of the given figure when a given temperature change takes place.

It is mentioned in the solution that total deformation of the structure (A to B), and hence the strain is zero. It is also mentioned that it is wrong to assume that the the total deformation, and hence the strain is zero for the portions AC and BC.

Can anyone please clarify this ?

A verified answer to a similar query (Confused with stress, strain and linear thermal expansion) in this site explains that the strain would be calculated with considering the denominator as natural length at a specific temperature.

But then, wouldn't it be incorrect to say that 'deformation' rather than tendency to deform is necessary for strain ?

  • $\begingroup$ You are correct in thinking that just because the distance A to B is constant doesn't mean that strain is zero. The strain will be based on the difference in FREE LENGTH of the beam at the two temperatures. $\endgroup$
    – James
    Commented Aug 21, 2019 at 14:54
  • $\begingroup$ Related: physics.stackexchange.com/q/21154/226902 $\endgroup$
    – Quillo
    Commented Jul 22, 2023 at 12:41

2 Answers 2


The "average strain" for the complete component is zero, but that doesn't tell you anything useful. It's a bit like saying that if you drive somewhere and then return to where you started, your average velocity is zero!

The total strain at any point is the sum of the thermal strain and the elastic strain. The thermal strain is the same in both components, but the elastic strain is not. The axial force in each component is the same, but the cross section areas are different, so the axial stress and the axial strain is different in AC and CB.

If A and C are fixed and you heat up the object, then point C will move towards point A, because AC has a smaller elastic stiffness than CB.

  • $\begingroup$ That is helpful. What about my second question ?. How can stress exist if the average strain for the entire component is zero ? $\endgroup$
    – Zam
    Commented Aug 21, 2019 at 2:12
  • $\begingroup$ When you have both thermal expansion and axial compression present at the same time, the equation for the stress is $$\sigma=E(\epsilon -\alpha \Delta T)$$where $\epsilon$ is the actual strain, $\alpha$ is the coefficient of thermal expansion, and E is Young's modulus. $\endgroup$ Commented Aug 22, 2019 at 11:32

When your reference said that the total deformation is zero, what they meant is that the distance between A and B does not change.

In this system, the strain in AC is not equal to the strain in BC, so strain is not uniform throughout the object. If BC gets larger, then AC gets shorter, such that distance AB remains constant. Since BC and AC are initially equal, it means that the strain in AC is minus the strain in BC. Suppose that the initial lengths of AC and BC are $L_0$, and the initial length of AB is $2L_0$. Then if $\epsilon$ is the strain in BC, the new length of BC is $L_0(1+\epsilon)$ and the new length of AC is $L_0(1-\epsilon)$. So the new total length of AB is $$L_0(1+\epsilon)+L_0(1-\epsilon)=2L_0$$So there is no change in the length of AB, meaning that the total deformation is zero (even through the strain is different in different parts of the object).


Up to now, you appear to have been dealing only with systems in which the deformation is uniform and homogeneous. However, in systems like the present one where the deformation is not uniform (e.g., in one part there is extension and in another part there is compression), the definition of strain needs to be extended to include the concept of local strain. Here's how it works.

Consider a rod running along the x axis. Consider two arbitrary material points along the rod, 1 and 2, located prior to the deformation at coordinates $x_1$ and $x_2$. After the deformation has been applied, the new locations of the material points are now $x_1+u_2$ and $x_2+u_2$, where $u_1$ and $u_2$ are the "displacements" of the two points. The original length of the segment of the rod between points 1 and 2 is $$l_0=(x_2-x_1)$$and the final length of the segment (after deformation) is $$l=(x_2-x_1)+(u_2-u_1)$$So the change in length of the segment is just $$\Delta l=(u_2-u_1)$$And the overall (average) strain in the segment is just $$\epsilon_{1,2}=\frac{\Delta l}{l_0}=\frac{(u_2-u_1)}{(x_2-x_1)}=\frac{\Delta u}{\Delta x}$$As the two arbitrary material points are taken closer together and $\Delta x$ becomes smaller, we approach the local strain at position x: $$\epsilon=\frac{du}{dx}$$ Now, if the local strain is the same value at all positions along the rod (for example, if the rod is not tapered or the elastic modulus of the material does not vary along the rod), then the strain is said to be uniform, and the overall strain of the rod (determined by taking the two material points at the two ends of the rod) is the same as the local strain at all locations.

In this problem, the local strain is uniform from A to C and also from C to B, but not from A to B. The local strain from A to C is minus the local strain from C to B, such that the overall average strain from A to B is zero.

  • $\begingroup$ @That clarifies stuff with respect to my question. Please do answer my second question. Like how can we say that there exists strain in AB when its deformation is zero unless there is a change in the definition of strain. $\endgroup$
    – Zam
    Commented Aug 21, 2019 at 2:06
  • $\begingroup$ See my ADDENDUM $\endgroup$ Commented Aug 21, 2019 at 12:00

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