As the title of this post says, how do the contraints increase the stiffness. For example, I was doing an exercise where I had to calculate the deflection on a rubber block with and without deflection and they mentioned that "the constraint increases stiffness by a factor of 1.33 (for v=0.5)" where v is the Poisson's ratio. So, where does the 1.33 come from and what are the effects of the constraints.

I understand that imposing constraints will make it harder to stretch/bend, but I do not understand how they came up with 1.33



1 Answer 1


Let me see if I understood it correctly:

You have a rubber block under a uniaxial load (compression or tension). That block may or may not be constrained on one pair of sides (the load is applied along the block's x-axis and the block can't deform in the y-axis but can in the z-axis, for example).

If this is the case, then the 1.33 is easy enough to explain.

Let's start by remembering that the Poisson ratio describes how big perpendicular deformations are when under axial loads.

Given that, let's think about the unconstrained case: you have a rubber block under some load. You calculate the axial deformation (along the x-axis, let's say) and it comes out as $\epsilon_x = \epsilon$. We also know the rubber's Poisson ratio, so we immediately know that $\epsilon_y = \epsilon_z = -\nu\epsilon$.

But, oh, sorry, actually we've been doing the constrained version this entire time, sorry! So we actually know that $\epsilon_y = 0$. So, how can we fix our mistake? Well, we can apply forces along the y-axis which will resist that $\nu\epsilon$ deformation. The value of that force doesn't matter to us, only the knowledge that it generated a "counter-deformation" of $\nu\epsilon$.

But wait, if we're applying an axial load along the y-axis which causes a deformation of $\nu\epsilon$, then we also have a simultaneous deformation along the x-axis! And the value of this deformation will be equal to $-\nu^2\epsilon$.

So, the force applied along the x-axis caused a deformation of $\epsilon$, which caused a transversal deformation of $-\nu\epsilon$ along the y-axis, which caused a reaction from the walls which caused a counter-deformation of $\nu\epsilon$ (so that the final deformation along the y-axis is zero, as expected), which then causes a deformation of $-\nu^2\epsilon$ along the x-axis. Fittingly for a rubber block, there was a lot of bouncing back and forth here.

So, at the end of the day, our deformation along the x-axis is $$\begin{align} \epsilon_x &= \epsilon - \nu^2\epsilon \\ &= \epsilon(1 - \nu^2) \\ \end{align}$$

And since Young's modulus is defined as $E = \dfrac{\sigma}{\epsilon}$, we end up getting that the "effective Young's modulus" in this case is

$$\begin{align} E_{x,\ \mathrm{nominal}} &= \dfrac{\sigma}{\epsilon} \\ E_{x,\ \mathrm{effective}} &= \dfrac{\sigma}{\epsilon_x} \\ &= \dfrac{\sigma}{\epsilon(1 - \nu^2)} \\ &= \dfrac{E_{x,\ \mathrm{nominal}}}{1 - \nu^2} \end{align}$$

And for $\nu = 0.5$, that gives us:

$$\begin{align} E_{x,\ \mathrm{effective}} &= \dfrac{E_{x,\ \mathrm{nominal}}}{1 - \left(\dfrac{1}{2}\right)^2} \\ &= \dfrac{E_{x,\ \mathrm{nominal}}}{\left(\dfrac{3}{4}\right)} \\ &= \dfrac{4E_{x,\ \mathrm{nominal}}}{3} \\ &\approx 1.33E_{x,\ \mathrm{nominal}} \end{align}$$

And since stiffness is proportional to Young's modulus, that 1.33 factor simply carries over to the stiffness.

  • 2
    $\begingroup$ good explanation. $\endgroup$
    – kamran
    Mar 12, 2021 at 2:34
  • 1
    $\begingroup$ My apologies for writing an unclear question. I didn't know how to formulate it appropriately. But the explanation that you provided was what I was looking for. Thank you! $\endgroup$
    – iZxCV
    Mar 12, 2021 at 9:33

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