# Roark's formulas for stress and strain

1. Dec 1, 2007

### matthias_t

Hi all,

i hope that i am in the right forum.

As you see in my topic i have a problem with Roark's formula especially for elastic stability of plates and shells.
The formula i got a problem with is the following:
If you have the book:
sixth edition: page 689 table 35 formula 15.
or
seventh edition: page 735 table 15.2 formula 15.

if you dont have the book:
"thin walled circular tube under uniform longitudinal compression" (ends not constrained)
formula:
stress = (1/root(3))*(Young's Modu / (1-poissons ratio²) *( t / r )
where
r = radius of tube
t = wall thickness (i hope...)

"Tests indicate an indicate buckling stress of 40 - 60% of this theoretical value"

So on i got an indicate stress of 1363 N/mm² by using a wall thickness equal 1.5 mm,
radius 139.82, steel with E = 210000 N/mm² and Poissons ratio = 0.3.
Even with 40% of this stress i got approx. 545 N/mm² anyway.
So I need 714 kN to buckle this tube.
With: force = stress * area
area = (radius² - (radius - t)²)PI
It seems a bit too much for me.
Could anyone help me with my doubts? I helpful link would be very nice too.

What other material properties will influence my buckling tube in a relevant way?

I thank you in advance

Matt

2. Dec 3, 2007

### Q_Goest

Hi Matt,
I agree, this looks rather peculiar. I have the 7'th edition. Note however, that the stress they are calculating is what they are calling the "critical unit compressive stress", not simply the critical compressive stress. I can't seem to locate the definition of this value, but I suspect it has to do with how this critical unit compressive stress compares to the stress calculated from F/A. I'll look through this a bit more when I get a chance, and if you or anyone else comes up with the answer, great! There are obviously other formulas for calculating this such as the Euler, Johnson parabola or secant formula. You might want to try that first.
Dave.

3. Dec 3, 2007

### Q_Goest

ah wait... just found it. Table 15.1 (pg 718 of 7'th edition) talks about critical unit load which is force per unit length. By analogy, I'm assuming the critical unit compressive stress is the critical stress divided by the length. If I do that, it all seems to work nicely.