Dear All,
During a seismic analysis of a storage tank,D=8.7 H=11.5, I cannot understand the way the software calculate the stress of the bolt from the Uplift force due to the seismic OPE.
The software has a strange behaviour:
1) If I select the "Design" method,it always says that the diameter is non sufficient,and it starts a loop in which eventhought I increase the bolt size at the step one,after a new check, this size will not be enough.and that again and again.
2) It's not clear the mathematical formula that the software uses to calculate the stress from the force. And that should be preatty simple and clear.

In my example, how from the load 0.14592E+06 N(total uplift load divided by 16 bolts,so it is the load/bolt)you can find the bolt stress of 160987.KPa with 16 anchor bolts?
It should be :

Load/area of bolt = 0.14592*10^6 N/(1140mm^2)=127000KPa that is NOT
160987.KPa .
Then,why,if Allowable stress is 320000 KPa and the stress is LESS of 320000KPa the software says that the size is not sufficient?!

Just for your info I show you a tipical output.


Case Uplift Allowable Load/Bolt
(N. ) Bolt Stress (N. )
(KPa )
Seismic OPE : 0.23348E+07 320000. 0.14592E+06



ANCHOR BOLT RESULTS - Section 5.12

Case Number Bolt Bolt Stress
of Bolts Diameter (KPa )
(cm. )

Seismic OPE : 16 3.810 160987.

Final Anchor Bolt Spacing .............. (m. )1.77
Specified number of user bolts is sufficient.
Specified user bolt diameter is insufficient.

You consider 1140mm^2 as bolt area and the figure corresponds to the geometrical area associated to 38.1 mm (1.5in) DIA of bolt.
You need to consider "root DIA", "minor DIA" or (better) directly the "tensile stress area" corresponding to UNC 1 1/2" (I suppose this is your bolt size).
So
0.14592*10^6 N/(1140mm^2)*(3.1415/4*1.5^2)/1.405=160.987 MPa
because the tensile stress area is 1.405 in^2
(or, if you prefer,
0.14592*10^6 N/(906.4498 mm^2)=160.979 MPa)

Note that the calculated area should be reduced by corrosion (at least it is what I understand API 650 intends to ask for, whereas 5.3.2.4 of API 650 is specifically written for design procedure of selecting nominal bolt size) and normally the CA specified in Data sheets refers to diameter, not to radius.

About the loop calculation in "Design" I haven't an answer because I don't know the problem.
"Open a Support Request on eCustomer and send in your job files" it is the right approach.

Best regards.

Dear Mariog,
you are right and I thank you for helping me.
But I still have a doubt.

In my case,with 16 anchor bolts and D=38.1 mm the software successfully verifies the analysis.
But, Why If I choose 24 number with a section of (for example 37 mm) it doesn't verifies and still needs section 38.1mm.
By Hand calculation, with 24 bolts and a section 37 mm diameter, It is more than enough to have a stress minus than the allowable.
Which one could be the problem?
Does the software make 2 different verification?
In the Configuration I can choose different anchorage,as TEMA Metric,TEMA Imperial, BS 3643, Does these standard refere to different dimension of the bolt that the software uses as standard dimension ?


Thank you for your help

In the API 650 last edition, E.7.1.2 requires that "When anchor bolts are used, they shall have a corroded shank diameter of no less than 25 mm (1 in.)".

Maybe the bolt size limit of 38mm was in the past somewhere (possibly in API or BS? I don't remember...).
BTW, as I remember, the bolt size limit was indeed 38 mm in old editions of SNAM/ AGIP specs, possibly to be the same in more recent editions?

Dear Alessandro,

I need to recognize I was wrong. In the old AGIP spec is required 27mm as minimum bolt size. In the past BS required 30 mm (adding specific CA). API 650 required that the minimum anchor bolt diameter should not be less than 25
mm, plus a corrosion allowance of at least 6 mm.

As a conclusion I have no idea from where is coming up 1 1/2" as minimum. In fact I don't know if TANK tries to force minimum at this size, I was just trying to understand the basis of such value.

Best regards.