0
$\begingroup$

The book I am referring to states the following -

The ratio of shearing stress to the corresponding shearing strain is called shear modulus of the material and is represented by G. It is also called modulus of rigidity.


After surfing on the internet for a while, I found one answer stating that shear modulus was nothing but the resistance of a material against shearing.

If the above is true, why don't we consider Young's modulus to be called modulus of rigidity. Doesn't it also resist longitudinal displacement?


PS - Moreover, my textbook states that stretching of a string is determined by the shear modulus. But I feel like it should be Young's modulus.

$\endgroup$

2 Answers 2

3
$\begingroup$

The modulus of elasticity (Young's Modulus), $E$, and the shear modulus, $G$, are related by the equation:

$$G=\frac{E}{2(1+ν)}$$

Where $ν$ is Poisson's ration = -(lateral strain)/longitudinal strain).

As you can see, the two are proportional to one another. I personally never heard of the shear modulus being called modulus of rigidity and I agree with you it doesn't seem to make sense to call one "rigidity" and the other "elasticity" when the are linearly related. You'd think they would both be called rigidity or elasticity, but not the opposite.

Young's modulus involves longitudinal stress/strain (tension/compression). The shear modulus involves transverse or lateral stress/strain (shear), so it is logical they are related to each other by Poisson's ratio (ratio of lateral to longitudinal strain). You can also see this because when you longitudinally compress or stretch something it laterally expands and contracts, respectively, as well.

My question was, 'Why is it called modulus of rigidity?'. The above answer does not answer that question

The only reason I can think of is to avoid confusion in the use of terms. If both the shear modulus and Young's modulus were referred to as "modulus of elasticity", or, for that matter, "modulus of rigidity" how would we know which modulus was being referred to? What I was trying to say is there should be no technical reason for the difference in terms for $G$ and $E$, since they both refer to resistance to deformation (lateral and longitudinal).

Hope this helps.

$\endgroup$
4
  • 1
    $\begingroup$ Note this holds true for an isotropic solid, not a more complicated solid. $\endgroup$
    – Jon Custer
    Commented Sep 17, 2019 at 14:37
  • $\begingroup$ @JonCuster Yes Jon, that's correct. $\endgroup$
    – Bob D
    Commented Sep 17, 2019 at 14:46
  • $\begingroup$ @BobD I don't see the above answer answering the question, 'Why is it called modulus of rigidity?' . Could you share your thoughts on it? $\endgroup$
    – Kaushik
    Commented Sep 17, 2019 at 14:53
  • 1
    $\begingroup$ @Kaushik See my revision in response to this follow up question. $\endgroup$
    – Bob D
    Commented Sep 17, 2019 at 15:00
3
$\begingroup$

Shear strain does not change the volume of the material. Therefore, the shear modulus measures how easy it is to change the shape of an object (i.e. how rigid it is) while Young's modulus or the elastic modular measures how easy it is to stretch the object (i.e. how elastic it is).

For isotropic materials the two moduli are related by $E/G = 2(1 + \nu)$ but for anisotropic materials they are completely independent. It is quite possible for anisotropic materials like woven cloth to have the $G < 10^{-3}E$, and taking $G = 0$ can be a useful approximation

$\endgroup$
2
  • 1
    $\begingroup$ Thank you for this answer. I knew the reasons behind the naming was something along these lines, but I couldn't think of the right way to word it. It was bugging me all morning that I couldn't think of the clear way to express this. $\endgroup$
    – JMac
    Commented Sep 17, 2019 at 16:55
  • $\begingroup$ Stretching the object is nothing but changing the shape longitudinally. Isn't? So even Young's modulus is measuring how rigid a body is. $\endgroup$
    – Kaushik
    Commented Sep 17, 2019 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.