Intuitive explanation of expected value of sample standard deviation I am having trouble to intuitively understand why the expected value of the unbiased sample variance is not equal to the square of the expected value of the unbiased sample standard deviation. 
In other words, it's been shown that for 
$$s^2  = \frac{1}{N-1}\sum_{k=1}^N \left(x_i -\overline{x} \right)$$
where $\overline{x}=\frac{1}{N}\sum_{i=1}^Nx_i$ is the sample mean, we have : 
$$E\left[s^2 \right] = \sigma^2$$
and
$$E\left[ s \right] = \sigma \sqrt{\frac{2}{N-1}} \frac{\Gamma(N/2)}{\Gamma((N-1)/2)}$$
Now, how come, if I expect to measure a variance, for eg $v=2$, I can't expect to measure a standard deviation of $s = \sqrt{v} = \sqrt{2}$ ?
Again, I'm looking for an intuitive explanation (the math-side has been done already).
 A: Graphs for Comments.
Stripchart of values of $x^2$ for $x = 1, 2, \dots, 99,$ with sample mean.

Stripchart for values of $100x.$

Simulation illustrating your formula for $E(S)$ where
$S$ is the SD of a sample of size $n = 5$ from a
normal population with standard deviation $\sigma = 10.$
Based on a million normal samples of size five. [I chose $n = 5$ for
illustration because the discrepancy between $E(S)$ and $\sigma$ is especially large for small $n.]$
set.seed(626)
s = replicate(10^6, sd(rnorm(5, 100, 10)))
mean(s)
[1] 9.403078                      # aprx E(S) from simulation
10*sqrt(2/4)*gamma(5/2)/gamma(4/2)
[1] 9.399856                      # exact from formula


Illustrating the convergence of $E(S_n)$ to $\sigma = 10$ for normal data.
n = 2:100
E = 10*sqrt(2/(n-1))*gamma(n/2)/gamma((n-1)/2)
plot(n, E, pch=20)
 abline(h=10, col="green2")


To understand why, for $n=2,$ $E(S_2)$ is so small, it may help to
show that $S_2$ is the sample range divided by $\sqrt{2}:$
$$S_2^2 = (X_1 - \bar X)^2 + (X_2 - \bar X)^2 \\
= \left(\frac{2X_1-X_1-X_2}{2}\right)^2+\left(\frac{2X_2-X_1-X_2}{2}\right)^2\\
= \frac {1}{4} (X_1 - X_2)^2 + \frac {1}{4} (X_2 - X_1)^2
=\frac 12(X_1 - X_2)^2.$$
Thus $S_2 = |X_1-X_2|/\sqrt{2}.$
