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Understanding the Solution to ε = xe-x2
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[QUOTE="Chestermiller, post: 4715949, member: 345636"] 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 [/QUOTE]
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Introductory Physics Homework Help
Understanding the Solution to ε = xe-x2
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