Buttress thread fatigue failures

[scottlel]

Well the tile sums it up we are having a fatigue failure that involves a buttress thread. I have a cylinder of 6061-T6 Aluminum we have cut an internal buttress thread in. After 7-10,000 cycles to pressure, 3,000psi, the tube fails at the last thread of the sealing cap. I've been through two designs. One I inherited that had a very high axial stress in the wall- 26,000psi. I thickend up the wall and have reduced the axial stress to 9,000psi. This did nothing for the life of the tube. After about the same number of cycles I still end up with the same failure.

Any ideas on what else I should be doing to fix this problem?

 

[rb1957]

ok, i forgot that nasty 2 ...
vM = sqrt((s1^2+s2^2+(s1-s2)^2)/2) = s1*sqrt((1+.25+.25)/2) = 0.87*s1
which is what i thought happened with vM when the stress state was bi-axial tension (the secondary tension stress "relieves" the primary tension)

 

[rizwa1n]

to find axial stress generated in the cylinder is it ok to use the following equation:

? = (?*(ri^2)*pi)/(?(ro^2 - ri^2))
where ri=inner radius
ro=outer radius
pi=pressure

i used these values
ro=2inches
ri=1.625inches
pi=3000psi
we get the following axial stress = 5824psi.
this axial stress can be used for calculating fatigue failure for buttress threads
the above equation is for thick walled cylinders exposed subjected to internal stresses.

 

[rizwa1n]

sorry or the typo ? is acutually the stress symbol

 

[rizwa1n]

just wanted to know if you are considering the following factors in your design calculations for fatigue strength which are as follows; endurance limit,surface finish,size,reliability,temperature,miscellaneous factors.
one more thing is 3000psi the mean stress or the maximum stress?

 

[rb1957]

back to your 1st post ... explain the "?"

hoop stress = pr/t
axial stress = hoop stress /2

now r = mean radius = ro+ri/2
and t = ro-r1

but no pi ...

 

[rizwa1n]

the ? on RHS of the equation is 3.14 and ? on the LHS is stress symbol i typed this equation in Word but it came out wrong. anyways if we are dealing with cyclic loadings and if we consider 3000 to be the maximum pressure this equation can be used to find maximum stress. but we would need minimum pressure to find minimum stress. both of these stress can be used to find mean stress which in turn can be used for fatigue calculations. i'm imagining thick-walled cylinder here. this equation was taken from fundamentals of machine element chapter 10 dealing with cylinders.