A problem about virtual work principle for continuous system

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SUMMARY

The discussion focuses on the application of the virtual work principle to derive equilibrium equations for continuous mechanical systems. The user encountered difficulties in obtaining simple equilibrium equations, suggesting a misunderstanding in the application of the principle. The equations presented involve stress, compression, and integration by parts, leading to the conclusion that the derivative of tension with respect to position, \(\frac{dT}{dx}\), is valid except at the open end of the system. The final relationship established is that tension \(T\) equals pressure \(P\) at the boundary \(x=L\).

PREREQUISITES
  • Understanding of the virtual work principle in mechanics
  • Familiarity with equilibrium equations in mechanical systems
  • Knowledge of calculus, specifically integration by parts
  • Basic concepts of stress and strain in continuous materials
NEXT STEPS
  • Study the application of the virtual work principle in continuous systems
  • Learn about equilibrium equations in mechanical engineering
  • Explore integration techniques in calculus, particularly integration by parts
  • Investigate the relationship between stress, strain, and tension in materials
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Mechanical engineers, students studying mechanics, and researchers focusing on continuous systems and the virtual work principle will benefit from this discussion.

athosanian
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dear all, the virtual work pinciple can be used to derive the equilibrium equations for the mechanical systems. however, when I want to apply it to a continuous system, I found it can not give out the simple equilibrium equations. there should be something wrong with my thinking. I expect some expert could give me some advice. thanks very much.

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The virtual work needed to compress (or elongate) the small element is the stress times the amount of the compression. So,
<br /> \begin{align}<br /> &amp;TA(\delta u+\frac{d \delta u}{dx} dx - \delta u)=TA\frac{d \delta u}{dx} dx,\\<br /> &amp;\delta W=\int_0^L TA\frac{d \delta u}{dx} dx=TA\delta u|_{x=0}^{x=L}-A \int_0^L \delta u\frac{dT}{dx}dx =TA\delta u|^{x=L}-A \int_0^L \delta u\frac{dT}{dx}dx,<br /> \end{align}<br />
where we use integration by parts.
As \delta u is arbitrary, we have \frac{dT}{dx} anywhere other than the open endx=L. Happily we know that T=P at x=L. We get equation (1).
 

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