Understanding the Solution to ε = xe-x2

[lecammm]

Homework Statement

ε = xe^-x2
L = L

Homework Equations

ε = ΔL/L

The Attempt at a Solution

xe^-x2 = (ΔL - L)/L

∴ ΔL = Lxe^-x2 + LI really don't understand how the solution got "dx" on the first step, also the delta L. I want to understand the first step, the rest of the solution I understand. Thanks!

 

[Chestermiller]

I've been around stress and strain all my professional life, and I can tell you that this is an unbelievably poorly worded problem. I think what they are trying to say is that the axial strain ε is a function of the distance x from the left end of the bar. If u is the axial displacement of a material cross section from its initial location before the bar was strained, then the displacement is related to the strain by du/dx = ε. You integrate this equation from x = 0 to x = L to get the axial displacement of the right end of the bar.

Chet

 

[lecammm]

Thank you for your response,
I'm sorry, I still don't understand how you are getting du/dx = ε, I understand up to the point where you say, "is related to the strain". The rest I understand, thanks!

Is it just that du is the extension and dx is the original length of a segment of length?

But then what is δL? Is that du?

 

[Chestermiller]

Thank you for your response,
I'm sorry, I still don't understand how you are getting du/dx = ε, I understand up to the point where you say, "is related to the strain". The rest I understand, thanks!

Is it just that du is the extension and dx is the original length of a segment of length?

But then what is δL? Is that du?

What this problem is trying to do is to show you what to do in cases where the local strain in the bar is non-uniform.

Let x represent the axial location of a material cross section before the non-uniform strain is imposed, and let x + u(x) represent the location of the same material cross section after the strain is imposed. The parameter u is called the local displacement. Suppose we focus on the short segment of material that was originally between x and x + dx in the unstrained bar. The original length of this short segment of material was dx. If we now focus on this same short segment of material after the non-uniform strain is imposed, the new length of this same short segment of material is now (dx + du). The change in length of the short segment of material is du, and its original length was dx, so the local strain of the material is du/dx (the change in length divided by the original length). So,

ε(x) = du/dx

This equation is called the "strain-displacement" relationship for the non-uniform deformation.

Chet

 

[HalfLostAndSearching]

The first step is using the definition of epsilon (ε) in terms of delta L (ΔL) and L. This definition states that ε is equal to the change in length (ΔL) divided by the original length (L). In this case, we are given that ε is equal to xe-x2, so we can set that equal to ΔL/L. This is the first step in trying to solve for ΔL.

To get from this step to the next step (ΔL = Lxe-x2 + L), we can rearrange the equation by multiplying both sides by L. This will give us ΔL = xe-x2L, and then we can add L to both sides to get the final solution.

I hope that helps! Let me know if you have any other questions.