- #1

- 103

- 4

## n_d = \frac{1+\sqrt{1-(1-z^2c_s^2)(1-z^2c_\sigma^2)}}{(1-z^2c_s^2)} ##

where

z = the z score, this is determined from reliability which is given in the problem. A high z-score means we demand a very high reliability.

##c_s = \frac{\sigma_S}{\mu_S}## that is the standard deviation over the mean of the material strength.

##c_\Sigma = \frac{\sigma_\Sigma}{\mu_\Sigma}##

**Note:**I'm using capital sigma for stress and lower case sigma for standard deviation to hopefully help with confusion.

So my problem with this idea is that design factor is independent of the geometry we are working with. For example, if we have a cylinder that is under a tension load and we know the mean and standard deviations of the load, we can use that to calculate coefficient of stress ##c_\Sigma## by dividing by area: ##c_\Sigma = \frac{\sigma_F/area}{\mu_f/area}## areas cancel out and we are left with the coefficient of stress.

So since area cancels out, it looks like geometry is not even a factor in determining the design factor, and I know intuitively that this cannot be true. That's essentially saying that a cylinder the size of a tooth pick will have the same design factor as a cylinder the size as my arm. And from the equation above, they will both have the same reliability.

Seriously would appreciate any help here. My professor says it's independent of geometry but I can't see how that's true. If you need any further elaboration or anything, just let me know and I'll post them up.