Calculate the Poisson’s ratio when a rubber string is stretched and in that situation the change in the volume is negligible in comparison to the change in shape.
$ \left[ {{\text{Hint: }}\sigma = \dfrac{1}{2}\left\{ {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right\}} \right] $
Answer
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Hint
To solve this question, we have to use the formula of the Poisson’s ratio given as a hint. Then we need to apply the condition given in the question to get the value of the required Poisson’s ratio.
Formula Used: The formula used to solve this question is
$ \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right) $, where $ \sigma $ is the Poisson’s ratio, $ A $ is the area of cross section, $ V $ is the volume and $ L $ is the length.
Complete step by step answer
We know that the Poisson’s ratio is the ratio of the strain in the lateral direction to the strain in the longitudinal direction of a material. As given in the hint in the question, we have
$ \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right) $ (i)
Now, according to the question the change in the volume is negligible in comparison to the change in the shape. So, interpreting this information mathematically, we have
$ \dfrac{{dV}}{{dL}} \to 0 $
From (i) we have
$ \Rightarrow \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right) $
$ \Rightarrow \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\left( 0 \right)} \right) $
On solving we get
$ \Rightarrow \sigma = \dfrac{1}{2} $
$ \Rightarrow \sigma = 0.5 $
Hence, the value of the required Poisson’s ratio is equal to $ 0.5 $.
Note
The Poisson’s ratio is a very important ratio in the field of material engineering. It is used for finding out the deflection and the stress properties of the materials. It is used in the case of structures, such as beams, shells, discs, shells etc to determine their stress and deflection properties. Its value always ranges from $ 0 $ to $ 0.5 $.
A very important engineering material is cork. It has zero Poisson’s ratio. According to the definition of the Poisson’s ratio, zero value means that this material will not show any change in its lateral dimensions, when a longitudinal stress is applied to it. For this reason, cork finds its application in the slippers and the stoppers for wine bottles.
To solve this question, we have to use the formula of the Poisson’s ratio given as a hint. Then we need to apply the condition given in the question to get the value of the required Poisson’s ratio.
Formula Used: The formula used to solve this question is
$ \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right) $, where $ \sigma $ is the Poisson’s ratio, $ A $ is the area of cross section, $ V $ is the volume and $ L $ is the length.
Complete step by step answer
We know that the Poisson’s ratio is the ratio of the strain in the lateral direction to the strain in the longitudinal direction of a material. As given in the hint in the question, we have
$ \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right) $ (i)
Now, according to the question the change in the volume is negligible in comparison to the change in the shape. So, interpreting this information mathematically, we have
$ \dfrac{{dV}}{{dL}} \to 0 $
From (i) we have
$ \Rightarrow \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\dfrac{{dV}}{{dL}}} \right) $
$ \Rightarrow \sigma = \dfrac{1}{2}\left( {1 - \dfrac{1}{A}\left( 0 \right)} \right) $
On solving we get
$ \Rightarrow \sigma = \dfrac{1}{2} $
$ \Rightarrow \sigma = 0.5 $
Hence, the value of the required Poisson’s ratio is equal to $ 0.5 $.
Note
The Poisson’s ratio is a very important ratio in the field of material engineering. It is used for finding out the deflection and the stress properties of the materials. It is used in the case of structures, such as beams, shells, discs, shells etc to determine their stress and deflection properties. Its value always ranges from $ 0 $ to $ 0.5 $.
A very important engineering material is cork. It has zero Poisson’s ratio. According to the definition of the Poisson’s ratio, zero value means that this material will not show any change in its lateral dimensions, when a longitudinal stress is applied to it. For this reason, cork finds its application in the slippers and the stoppers for wine bottles.
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