Cylinder Thickness Equation - why does it have an extra term?

[roseda]

According to PD5500, the min cylinder thickness, e = p x Di / (2 x f - p)
where p is pressure, Di is inner diameter and f is design strength

Re-arranged to give stress:
f = p x Di / (2 x e) + p / 2

However, according to standard thin cylinder theory, the stress is:
f = p x Di / (2 x e)

Anyone know why the extra p / 2 factor is added in PD5500?

Is it an additional safety factor? (I thought the reductions applied to obtain the design strength were sufficient to not require extra safety factors)

 

[TGS4]

It's a term to adjust for non-thin-shell applications.

 

[OSIRIS7]

Re-arranged to give stress:
f = p x(Di+d)/ (2 x d)

 

[roseda]

Thanks osiris, so it simply uses the outer diameter rather than the inner for the calc.

I note that it gets more conservative as e/Di increases.

 

[prex]

p/2 is the membrane component of the through thickness stress: p at inner wall and zero outside.
As this stress is negative (compressive), it adds up to the hoop stress to form the stress intensity, or the difference between the maximum principal stress and the minimum one. So the Tresca failure criterion is used (as expected).

prex
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[OSIRIS7]

No, it uses the mean diameter

You can also use the outside diameter:
= p x(Do-d)/ (2 x d)

Hence:
e = p x Do / (2 x f + p)

And it is for relatively thin walled cylinders. The design code will mention in what thickness/diameter ratio this formula is applicable.

 

[roseda]

Thanks prex, that makes sense.

However, I'm not sure why the average (-p/2) of the radial stress is used rather than the maximum (-p).

 

[prex]

Because the check is based onto the membrane (through thickness average) components of stresses: also the hoop stress varies (slightly) along the thickness. The linearly varying portion of stresses gives rise to the bending components of stresses, that need not be checked for thin cylinders, as minor with respect to the allowable (=1.5S for membrane+bending)

prex
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[OSIRIS7]

Just an extra: the stress (cross section) in the pipe caused by thrust is 1/2x stress in longitudinal direction. That is why a pipe will (normally) tear in longitudinal direction.