- #1

JulienB

- 408

- 12

##\mu = \frac{E}{2G} - 1##

I imagine that ##Cov(X,Y)## refers to ##Cov(E,G)## in our case. The problem is that we have only one value for E and for G (determined experimentally), which is probably why we were asked to estimate the covariance. I've seen that ##0 < Cov(X,Y) < 1## when the two values grow together, and that's the result we got:

##G = \frac{E}{2(1 + \mu)} = kE = 0.3708 E## (that's the value we got for the Poisson ratio)

Since ##\mu## is supposed to be a constant, a graph of ##G## in function of ##E## should be linear. Can I then estimate ##Cov(E,G) \approx 1##? Or did I completely misunderstand it?

Thanks a lot in advance for your answers.

Julien.