Dear All,

I am working on a Tanks Iquiry some of thank having flush type nozzle connection and facing problem when i meet the requirement as per clause API650 5.7.81b: (Same requirement exist in API620 also.)

b. The vertical or meridional membrane stress in the cylindrical shell at the top of the opening for the flush-type connection shall not exceed one-tenth of the circumferential design stress in the lowest shell course containing the opening.

API620 gives us the ciriteria of meridional or longitudinal stress calculation (5.10.2.5c of API620). when i calculate the stress at the top of flush type opening the required thickeness become extreamly high.

in my case the shell thickness is 6mm and 32mm insert plate is required due to meet the flush-type connection requirement.

I need your opinion and give a right path if i am going wrong way.

Thanks and best Regard's
Jamil

Dear Jamil,

Looking only to that API 650 paragraph (and accepting it "as is") you can consider the following logic.
Under pressure load, the longitudinal stress in one-half of the circumferential one. In case the first course thickness has been selected based on calculation, that means the circumferential there is about the allowable stress (or "the circumferential design stress" as they mentioned). Consequently, under normal circumstances, the longitudinal stress there is about 1/2 of the circumferential design stress.
Imposing that the longitudinal stress shall not exceed one-tenth of the circumferential design stress, means to have the lower course about 5 times than those calculated under normal circumstances.

The above "rule" is not an exact one, because the "longitudinal stress" is not only due to pressure and, in some cases, 6mm is considered not by calculation but because is the minimum thickness of the first course.

I would consider that your "trouble" (instead 6mm you have 32mm..) is "normal".

The main question is the reason of such rule. I don't know from where it was "imported".
As terminology, it is clear that such nozzle connection shall not exceed the first course (see height limits), so why that cryptic language as 1/10 of "circumferential design stress in the lowest shell course containing the opening"? It is clear that it is the same course, isn't it?

For me it appears an inconsistency in API 650; I can see the rule into API 650/ 5.7.8 Flush-Type Shell Connections but I cannot see the same rule for 5.7.7 Flush-Type Cleanout Fittings. Why, what exactly makes the difference in "the vertical or meridional membrane stress in the cylindrical shell at the top of the opening" for two similar cases?

Moreover, looking to 5.7.7.4 it appears that they want to calculate the reinforcement thickness and considering the graph of Figure 5.11 it is clear that the maximum thickness of reinforcement would be 1.4 of shell thickness. So following this criteria only, you may conclude that a total thickness of (1+1.4)*t=2.4*6=14.4mm is the maximum necessary.

I think it is not a problem that can be solved by us, discussing in this forum. It is a rule prohibitive for Flush-Type Shell Connections and I'm afraid I don't have enough knowledge to understand it.

Just a remark... in case you consider there an "insert" including reinforcement area, you may have another trouble with API inspectors. As it is written in API 650, it appears that they want to see an assembly with a piece of shell. It is true that in the Agenda Item: 650-771 they discussed to consider an insert there, but not an insert including the reinforcing area. Please see the attachment to understand what I'm referring to. Anyway this is only a proposal )not yet approved), so instead it please consider the last edition of API 650 about "insert" subject.

My best regards.

Dear Mariog,

Thanks for your reply and remarks on the issue.

as per API620 when calculating vertical Force using equation
T1=(Rc/2)*(P+(-W + -F/At))

-Negitive sign shall be considered with W and F force which is acting on the shell and shall be opposite to the pressure.

so calculated vertical stress is become very minor and negligible.

Jamil

Dear Jamil,

It is true that W and F are compressive forces whereas pressure gives tensile longitudinal stress.
It depends on the magnitude of the effects to say "is negligible" or "it is important".
So in your case the result is negligible?

Dear Jamil,

The main reason for which the vertical stress becomes "very minor and negligible" is API 620 which evaluates wrong (in my opinion) that vertical stress.

API 620 considers P=Pl+Pg and add/substract a term related with metal, liquid and "F".

Considering weight of liquid (lets note it as W_liquid) and dividing it by At, we get exactly the hydrostatic pressure which is Pl.

So for our case:
T1= (Rc/2)*[p-(W+F)/At]=(Rc/2)*[Pl+Pg-W_liquid/At-W_metal/At-F/At]=
=(Rc/2)*[Pl+Pg-Pl-W_metal/At-F/At]= (Rc/2)*[Pg-W_metal/At-F/At]
and we may note that Pl (hydrostatic term) disappeared!

So which is the effect of hydrostatic pressure?
Of course it enters in T2 calculation (and gives the thickness), but does not enter into T1 result?
Would I expect that the longitudinal stress is not depending on the hydrostatic pressure? So with tank full, half-full or near empty would I get the same result for longitudinal (vertical) stress in a particular position on the first course?

Dear Mariog,

I agree with you "T1" where Pl and W_liquid act twice which is wrong so we can assumed W_liquid suppress from "W" as below:

T1=(Rc/2)*[Pl+Pg-W_metal/At-F/At]

I have not actually check in the case with this scenario due to EID holidays.

Jamil

Dear Jamil,

Finally, I succeed to understand what is the problem and reconcile all the stuff.

API 620 considers a free-body approach which was rather confusing for me.

In fact, the free body diagram in this case would be the upper part of the tank, "floating in midair".
It has a pressure force acting up on the bottom of the fluid (which is the reaction force of bottom to liquid, bottom being subject of Pl+Pg pressure). It has weight of contents (liquid weight) and tank itself (shell, roof, stairs, platforms, appurtenances) acting down.
The vector resultant is the compressive or tensile force in shell.

I would prefer another approach, which is equivalent, based on cause- effect i.e. a correlation between load and longitudinal stress effect.
So:
- liquid hydrostatic pressure---> zero longitudinal stress;
- gas pressure---> Pg*Rc/2t longitudinal stress;
- W1+F load transferred to shell (anything but not the liquid content!)---> -(W1+F)/(2*π*Rc*t)= -[(W1+F)/At]*Rc/(2t)

As you can see, and surprising for me (but never too late for the truth... which is I need to educate myself more!), the longitudinal effect of pressure is not P*Rc/2t but Pg*Rc/2t!

Indeed, looking into the document attached (a Roark's extract), the longitudinal stress due to liquid hydrostatic pressure is zero, the effect of gas pressure is Pg*Rc/2t (due to the "cap effect") and the effect of weight transferred to shell is -(W1+F)/(2*π*Rc*t)= -[(W1+F)/At]*Rc/(2t) with W1 not including the weight of liquid.

Now everything makes sense:
T1=Pg*Rc/2-(W1+F)/(2*π*Rc)=Pg*Rc/2-[(W1+F)/At]*Rc/2=[Pg-(W1+F)/At]*Rc/2 and the result is identical with the final result of API 620 for cylindrical shells.

Dear Mariog,

checked Meridional stress with E-TANK calculation. In E-tank, pressure of liquid head is considered in P, but weight of liquid has not been used in W. (image Attached)

The reason in our case, which is cylinderical Tank and Flat bottom can not take effect of Liqued weight because of fully supported on Foundation and that supported area we shouldn't tacke the effect of Pl and W_liquid as per clause 5.11.1 of API620 (Shaped bottom).

Finally in Cylidrical Tanks with Flat or Fully rested Bottom we shall use Equation for Meridional Stress is:

T1=(Rc/2)*(P-(W_Metal+F/At))

Best Regard's
Jamil

Where:
P=Pl+Pg

Dear Jamil,

T1=(Rc/2)*(Pg-(W_Metal+F/At))
and not
T1=(Rc/2)*((Pl+Pg)-(W_Metal+F/At))

This may be confusing (and it was, in my case), because we are very familiar with the longitudinal stress as PD/4t (or PR/2t) in pressure vessels and piping, so we would expect there should be a term as (Pl+Pg)R/2t.

In fact it is very simple; that term exists IF there is a cap effect.
The hydrostatic pressure hasn't a cap effect whereas the gas pressure pg gives a cap effect as in every pressure vessel.

API 620 prefers to give an explanation (a free body equilibrium equation) which is correct, but still confusing because gives no figure associated. I'll try to do a sketch in the next days.

Seismic Hoop combined Stress Calculation for Shell Course:

Impulsive Hoop Membrane Force per E.6.1.4-2b [Ni]:
= (2.77 * Ai * G * D^2)
* ((Y / D * 0.75) - (0.5 (Y / D * 0.75)^2))
= (2.77 * 0.375 * 1.100 * 49.213^2)
* ((6.562 / 49.213 * 0.75) - (0.5 (6.562 / 49.213 * 0.75)^2))
= 448.234 lb. [1993.744 N.]

Convective Hoop Membrane Force per E.6.1.4-4b [Nc]:
= cosh ( 3.68 ( H - Y ) / D )
/ cosh ( 3.68 * H / D )
* 0.98 * Ac * G * D^2
= cosh ( 3.68 ( 72.178 - 6.562 ) / 49.213 )
/ cosh ( 3.68 * 72.178 / 49.213 )
* 0.98 * 0.093 * 1.100 * 49.213^2
= 148.879 lb. [662.214 N.]

Product Hydrostatic Membrane Force [Nh]:
= 0.0361 * G * Y * 12.0 * D * 12.0 / 2.0
= 0.0361 * 1.100 * 6.562 * 12.0 * 49.213 * 12.0 / 2.0
= 923.258 lb. [4106.653 N.]

Total Combined Hoop Stress per E.6.1.4-6:
= (sqrt( Ni^2 + Nc^2 + (AV * Nh)^2 ) + Nh) / Thk
= (sqrt( 0.000^2 + 0.000^2 + (0.168 * 0.000 )^2 ) + 0.000 ) / 0.197
= 7215.565 psi [49748.438 KPa]

please check the report

the values are not matching in this calculation
and the units are not matching as per API 650 STD

and please clarify TH Nh value calculation

when i had provide the anchor bolt dia in input - 42mm
but the generated out put report shows 82.0mm instead of 42mm

please clarify