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Chapter 3: Problem 4

Show that \(Y=9 \mu B /(\mu+3 B)\) connects the three elastic moduli \(Y, \mu\), and \(B\). By requiring that the three moduli be positive, show that Poisson's ratio is constrained to the range \(-1<\sigma<\frac{1}{2}\). Does this result imply that a rod increases its volume on being stretched?

Short Answer

Expert verified
Answer: The given equation is derived using the definitions of Poisson's ratio and its relation to the elastic moduli. The equation holds for -1 < σ < 1/2, which determines that all elastic moduli are positive for this range. A rod increases its volume when stretched if its Poisson's ratio is between -1 and 0. Therefore, depending on the rod's material and its Poisson's ratio, the volume of a rod can either increase or decrease when stretched.

Step by step solution

01

Write down given equation and conditions for positive moduli

The given equation is \(Y=9 \mu B /(\mu + 3 B)\). We need to show that the three moduli (Y, μ, and B) are positive.
02

Derive a relation between Poisson's ratio and elastic moduli

To obtain a relation between Poisson's ratio and the elastic moduli, we need to know their definitions. Poisson's ratio (σ) is defined as the negative ratio of the transverse strain to the axial strain, and it relates to the moduli as: $$\sigma = \frac{1}{2}\left( \frac{Y}{\mu} - 1\right)$$ Also, we have the relation: $$Y = 3 B(1-2\sigma)$$
03

Derive Poisson's ratio from the given equation

We will derive Poisson's ratio from the given equation, which is \(Y = 9 \mu B / (\mu + 3 B)\). Firstly, write the given equation in terms of μ: $$\mu = \frac{3YB}{(3 - Y)}$$ Now, substitute this expression into the relation between Poisson's ratio and elastic moduli: $$\sigma = \frac{1}{2}\left(\frac{Y}{\frac{3YB}{(3-Y)}}-1\right) = \frac{1}{2}\left(\frac{3-Y}{3B}-1\right)$$
04

Show that moduli are positive if Poisson's ratio is between -1 and 1/2

By plugging the given equation \(Y = 9 \mu B /(\mu + 3 B)\) into the relation \(Y = 3 B(1-2\sigma)\), we get: $$3 B(1 - 2\sigma) = \frac{9 \mu B}{(\mu + 3 B)}$$ The equation holds if: 1. Y > 0: \(\sigma < \frac{1}{2}\), and 2. Y < 3B, μ > 0: \(\sigma > -\frac{1}{2}\) Thus, we have demonstrated that, for -1 < σ < 1/2, all moduli are positive.
05

Determine whether a rod increases its volume when stretched based on Poisson's ratio

The change in volume of a rod when stretched depends on Poisson's ratio. If σ > 0, the rod's volume decreases, as there is a positive transverse contraction when it is stretched. For -1 < σ < 0, the volume of a rod increases when it is stretched, as there is a negative (expansive) transverse strain when subjected to axial stress. Since our derived result implies that Poisson's ratio (σ) can vary from -1 to 1/2, it means that a rod can increase its volume when stretched depending on the material it is made of and its Poisson's ratio.

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