
In materials like aluminum and copper, the correct order of magnitude of various elastic moduli is
a)Young’s moduli< Shear moduli< Bulk moduli
b) Bulk moduli < Shear moduli< Young’s moduli
c) Shear moduli < Young’s moduli < Bulk moduli
d) Bulk moduli < Young’s moduli < Shear moduli
Answer
590.1k+ views
Hint: In the question it is asked to arrange the modulus of elasticity for metals like aluminum and copper. The poisson's ratio for most of the metals including aluminum and copper is equal to 0.35. Hence we will use the relation between young’s modulus, shear modulus and bulk modulus in terms of the poisson's ratio and accordingly compare which one of them is greater with respect to each other.
Formula used:
$B=\dfrac{Y}{3(1-2v)}$
$S=\dfrac{Y}{2(1+v)}$
Complete answer:
Let us say we have a material with Young’s modulus (Y), shear modulus(S) and bulk modulus (B). Let the poison ratio for the above material be ‘v’. Then the relation between young’s and bulk modulus of elasticity is given by,
$B=\dfrac{Y}{3(1-2v)}$
Similarly the relation between the young’s and shear modulus is given by,
$S=\dfrac{Y}{2(1+v)}$
For metals like aluminum and copper the poisson's ratio is found to be 0.35.
Hence from the above relation between the bulk and the young’s modulus we get as
$\begin{align}
& B=\dfrac{Y}{3(1-2v)} \\
& \Rightarrow B=\dfrac{Y}{3(1-2\times 0.35)} \\
& \Rightarrow B=\dfrac{Y}{3(1-0.7)}=\dfrac{Y}{0.9} \\
& \Rightarrow B>Y...(1) \\
\end{align}$
Similarly, using the above relation shear and young’s modulus we get,
$\begin{align}
& S=\dfrac{Y}{2(1+v)} \\
& \Rightarrow S=\dfrac{Y}{2(1+0.35)}=\dfrac{Y}{2.7} \\
& \Rightarrow Y>S...(2) \\
\end{align}$
Therefore from equation 1 and 2 we can conclude that, $S < Y < B$
Hence the correct answer of the above answer is option c.
Note:
Poisson's ratio is nothing but the lateral strain to that of longitudinal strain. For metal when they are subjected to longitudinal stress, the lateral strain is very small in comparison to that of longitudinal strain. Hence we get the poisson's ratio for most of the meats as less than 1.
Formula used:
$B=\dfrac{Y}{3(1-2v)}$
$S=\dfrac{Y}{2(1+v)}$
Complete answer:
Let us say we have a material with Young’s modulus (Y), shear modulus(S) and bulk modulus (B). Let the poison ratio for the above material be ‘v’. Then the relation between young’s and bulk modulus of elasticity is given by,
$B=\dfrac{Y}{3(1-2v)}$
Similarly the relation between the young’s and shear modulus is given by,
$S=\dfrac{Y}{2(1+v)}$
For metals like aluminum and copper the poisson's ratio is found to be 0.35.
Hence from the above relation between the bulk and the young’s modulus we get as
$\begin{align}
& B=\dfrac{Y}{3(1-2v)} \\
& \Rightarrow B=\dfrac{Y}{3(1-2\times 0.35)} \\
& \Rightarrow B=\dfrac{Y}{3(1-0.7)}=\dfrac{Y}{0.9} \\
& \Rightarrow B>Y...(1) \\
\end{align}$
Similarly, using the above relation shear and young’s modulus we get,
$\begin{align}
& S=\dfrac{Y}{2(1+v)} \\
& \Rightarrow S=\dfrac{Y}{2(1+0.35)}=\dfrac{Y}{2.7} \\
& \Rightarrow Y>S...(2) \\
\end{align}$
Therefore from equation 1 and 2 we can conclude that, $S < Y < B$
Hence the correct answer of the above answer is option c.
Note:
Poisson's ratio is nothing but the lateral strain to that of longitudinal strain. For metal when they are subjected to longitudinal stress, the lateral strain is very small in comparison to that of longitudinal strain. Hence we get the poisson's ratio for most of the meats as less than 1.
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