standard deviation may fallows normal distribution, but nature rarely fallows normal distribution

Article TO DETERMINE SKEWNESS, MEAN AND DEVIATION WITH A NEW APPROAC...

1. The distribution of the standard deviation  \sqrt{s^2} as well as the variance s^2 is NEVER  normal, since they assume only POSITIVE values;

2. For simple sample of  n IID normally distributed observations with variance  v ,   the distribution of  s^2  is of  Gamma type, with mean   (n-1)v/n, with exponent (shape parameter) (n-1)/2.  For sufficiently large   n , the distribution of  s^2   can be approximated by  normal distribution with standard deviation  v \sqrt{2/n}.

3. The basic properties of distribution of the standard variation  \sqrt{s^2} is perfectly presented at the Wolfram page linked to  by Julie, and rather completely developed at the MsWiki pages linked to by Waldemar.

Regards

For a sample of size n and standard deviation s, (n-1)s^2/(sigma^2) follows a chi-square distribution with degree-of-freedom n-1 where sigma is the population standard deviation. Appropriate transformation may be used to find the distribution of s. This distribution cannot be Gaussian except perhaps asymptotically with the use of the Central Limit Theorem.

The Wikipedia article on Standard Deviation has pretty much all you would want to know along with references. As noted, the distribution is chi-sq, not normal, and the usual n-1 estimator is biased.

 You probably ask not about standard deviation distribution per se,  but about sample standard deviation.

The standard deviation unknown constant and has only defective distribution (if you don't leave in bias word, where you need to suggest prior distribution of st.dev)

The sample standard deviation distibution can be found as hier:

http://www.unc.edu/~hannig/STOR655/handouts/Handout-asymptotics.pdf

https://stats.stackexchange.com/questions/105337/asymptotic-distribution-of-sample-variance-of-non-normal-sample

Th answer by C. Mitchell Dayton can be misleadind, since the claim

" the distribution [of standard deviation] is chi-sq,"

is incorrect in two points: first, this holds for normal samples, only (additionaly: after suitable multiplication); secondly this is related to the sample variance not to the sample standard deviation. Regards, Joachim.