1: Comparing elastic effects with thermal expansion (10 points) We\'re going to

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1: Comparing elastic effects with thermal expansion (10 points) We're going to figure out which effect is larger, in some sense. It may seem like a silly question, because you can't compare apples with oranges, and you can't compare things that have different units. However, Young's modulus has units of pressure, which is equivalent to energy density (since 1 Newton per square meter is equal to one Joule per cubic meter) One microscopic interpretation of Young's modulus is then that aY is the energy per cubic bond length (since the cube of the bond length is approximately the volume available to an atom) when AL/L-1 (i.e. the elastic energy density required for a very large compression or stretch) Likewise, the coefficient of thermal expansion (usually denoted o) has units of 1/tem perature, and ifAL/L 1 then ??~ 1/a. Moreover, the vibrational energy per atorn in a solid is approximately kBT, so we could think of kB/a as the thermal energy you'd need to add to each atom i order to get a very large expansion. Look up the thermal expansion coefficient a, lattice constant (bond length) a, and Young's modulus Y for copper, and use them to determine if the thermal or elastic effects

Explanation / Answer

thermal expansion coefficients and the elastic moduli of the material are important physical parameters to predict the shrinkage and structure. Effective elastic effect is more accurate than the effective thermal expansion coefficients of the composites, which results from the imperfection of interfacial bonding. In ductile materials As soon as the yield stress is reached and plastic deformation starts and the stress just won't go up much anymore.

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