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It is known that Shear Modulus is generally less than Young's modulus for most materials. What does this mean? Does this mean that it is easier to change shape of a fixed body by applying force than stretching or compressing it?

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    $\begingroup$ Hi Mohak. Can you clarify your question? Are you asking why the shear modulus is usually less than the bulk modulus, or are you asking what it means for the shear modulus to be less than the bulk modulus? $\endgroup$ Oct 19, 2015 at 5:59
  • $\begingroup$ @JohnRennie I'm asking the what it means for shear modulus to be lesser than Young's modulus, which would be the latter of your statement. $\endgroup$ Oct 19, 2015 at 6:23

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For simplicity let's consider a cubic shape of side $\ell$, so the volume is $\ell^3$.

Suppose we compress the cube so its volume decreases from $V$ to $V - \delta V$, and suppose this needs a pressure $P$.

Bulk modulus

Then the bulk modulus is:

$$ K = -\delta P \frac{V}{\delta V} $$

in the limit of $\delta V \rightarrow 0$. Now suppose we deform the cube slightly by pushing the top of it to one side. Suppose this takes a force $F$:

Shear modulus

Then the shear modulus is:

$$ G = \frac{F}{A} \frac{\ell}{x} $$

where $A$ is the area of the top face of the cube.

The first process involves compression only and no shear, while the second process involves shear only and no compression, so there is no reason why the bulk and shear moduli should necessarily be related. Of course in practice they are related because both originate from the interatomic forces in the material.

You ask:

Does this mean that it is easier to change shape of a fixed body by applying force than stretching or compressing it?

but this is not a well defined question. Remember that compression and shear are unrelated processes. You could ask:

Does this mean that it is easier to shear a body by $x/l = 1\%$ than to compress it by $\delta V/V = 1\%$?

and this we can answer because if $G < K$ then the answer is yes, provide we equate the pressure in compression with the force per unit area in shear. However even though both have the units of pressure I'm not sure I would really say they were equivalent.

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  • $\begingroup$ whoa sir, that was a well defined answer, thank you for your support! $\endgroup$ Oct 19, 2015 at 7:59

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