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Consider a taut wire or rod lying along the x axis. To
Chapter 16, Problem 16.27(choose chapter or problem)
Consider a taut wire or rod lying along the x axis. To define Young's modulus YM we apply a pure tension along the axis; that is, a stress with \(\sigma_{11}>0\) and all other \(\sigma_{i j}=0\).
(a) Use Equation (16.95) to write down the corresponding strain tensor E.
(b) Argue from the definition (16.55) of Young's modulus to show that \(\mathrm{YM}=\sigma_{11} / \epsilon_{11}\).
(c) Combine these two results to verify the expression (16.100) for YM, showing in particular that \(\mathrm{YM}=9 \mathrm{BM} \cdot \mathrm{SM} /(3 \mathrm{BM}+\mathrm{SM})\).
Questions & Answers
QUESTION:
Consider a taut wire or rod lying along the x axis. To define Young's modulus YM we apply a pure tension along the axis; that is, a stress with \(\sigma_{11}>0\) and all other \(\sigma_{i j}=0\).
(a) Use Equation (16.95) to write down the corresponding strain tensor E.
(b) Argue from the definition (16.55) of Young's modulus to show that \(\mathrm{YM}=\sigma_{11} / \epsilon_{11}\).
(c) Combine these two results to verify the expression (16.100) for YM, showing in particular that \(\mathrm{YM}=9 \mathrm{BM} \cdot \mathrm{SM} /(3 \mathrm{BM}+\mathrm{SM})\).
ANSWER:Step 1 of 4
The given equation of the tensor, which gives the strain tensor, is
\(E=\frac{1}{3 \alpha \beta}[3 \alpha \Sigma-(\alpha-\beta)(\operatorname{tr} \Sigma) I] \rightarrow(i)\)
Given that, \(\sigma_{11}>0\) and the rest of the \(\sigma_{i j}=0\).
Thus the stress tensor is
\(\begin{array}{l}\Sigma=\left[\begin{array}{ccc}\sigma_{11} & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0\end{array}\right] \rightarrow\\
\text{(ii)}\end{array}\)
Then,
\(\operatorname{tr} \Sigma=\sigma_{11} \rightarrow(\text { iii })\)