# Design Factor with known reliability

1. Oct 23, 2015

In my textbook we are given the equation for design factor as:

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

2. Oct 23, 2015

### jackwhirl

What is a design factor, exactly?
It looks to me a lot like a safety factor, just accounting for statistical variations in your material and load. Am I on the right track?

3. Oct 23, 2015

They are very similar concepts. From what I understand $n_d = \frac{what- the- part- can -handle}{what- the -part- experiences}$ the main difference between the two is that design factor includes all modes of failure (buckling, vibrations, etc...). So I don't think the differences is statistics, in this example we're just trying to use statistics to get a more accurate idea of when our part could fail. ^Not 100% confident on my definition there, but yeah, I know they are very similar concepts.

4. Oct 23, 2015

### jackwhirl

Ok, with that semi-settled:
All other things being equal, they would only have the same design factor if they have the same reliability.
You're not getting reliability from the equation. You're getting it from the problem statement.
In a real world scenario(within a certain range of loading), arm cylinder is likely to have a higher reliability than toothpick cylinder and, therefore, a different design factor.

5. Oct 24, 2015