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API 6A App. D RTJ Torque Values

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rickylee

Mechanical
Aug 16, 2002
38
How are API 6A App. D recommended torque values applied to an RTJ. Am I to perform a calculation similar to ASME Sec 8 Div 1 App2 to determine bolt stress? API 6A, APP D recommends 40 and 52.5 ksi bolt stress, but for what application?

Rick
 
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The values in App D are provided as acceptable values for flange make up, to provide sufficient gasket pre-load.

This is not the same as the design allowable stress of .83Sy which must be calculated based on the thread root area and take into account all load conditions.
 
Thanks, and why are 2 values presented (40 ksi and 52.5 ksi)?
 
Values are 50% of stud yield.

For B7 & L7 studs Sy=105 ksi, make up stress 52.5 ksi
For B7M & L7M studs Sy=80ksi, make up stress 40 ksi
 
Does API 6A mention that 50% of bolt yield stress should be targeted? If not, how is the reader to know that or have I missed something?

Also Sy changes with diameter as per ASME PV code. Therefore these tables would only apply to a range of sizes.

Rick
 
Sy is listed at the top of tables D.1 & D.2, Also in section 10. Also see last paragraph Section D.2 and section D.4.

The ASME B&PVC does not apply to API 6A except where it is specified by API 6A.
 
After bolting up at 50% of YS the flanges appear to contact on the outside diameter. Is this by design or have we over torqued?

Rick
 
What size flange and ring gasket?

There are three different styles of API flanges as well as some common proprietary styles. Some will bolt to contact most do not.
 
A 3" 2500 is not an API 6A flange, the roughly equivalent API 6A flange is 3-1/16 10,000 lb using a BX154 gasket. You should be following ASME bolting requirements not API. For a 3" 2500 with an R32 you should have approximately 1/8" gap between the faces at make up.
 
On this 3" 2500# ASME B16.5, using the Wm1/Wm2 (B&PV Code)I would bolt up to 14,000 psi bolt stress, and using API I would tension an equivalent API flange to 52,500 psi bolt stress. Why?

Weld neck and flange thickness dimensions are similar between the ASME and API flanges.ASME has a bit larger OD.

I think bolting this joint to 14,000 psi bolt stress is too low.

Rick

 
I'm not sure how you derived 14,000 psi. Its been a while since I did any Div. 8 stuff and I made a couple of assumptions on materials but a quick calc on a 3" 2500 rated for 6000 psi gives me a minimum bolt load of 23,560 and a max of 61,590 which puts it in the same neighborhood as 6A for bolt stress on a 1-1/4 bolt.

Also note the B16.5 and 6A dimensions for some flanges are similar up to class 1500 (6A 5000 psi) but not above that. Also material requirements for the two specs are considerably different.
 
What is the nature of your quick calc?

Rick
 
Keep in mind this is the internet so for all you know I'm some anonymous flake.

Minimum required bolt load at operating condition = pressure end load + gasket load

Wm1 = H + Hp = (.785*G*G*P) + (2*b*3.14*G*m*P)

G gasket pitch = 5.000"
P operating pressure = 6000 psi (assumed pressure rating, pressure ratings vary between 2000 psi and 6250 psi at room Temp and decrease with temp increase)
b gasket seating width = w/8
w gasket width = .500
m gasket factor = 6 (assumed value for rough calculation mild steel would be 5.5 stainless would be 6.5)

Maximum bolt load

W = (Am + Ab)* Sa/2

Am = Wm1/Sb
Sb = Sa for room temp but will decrease with temp increase
Ab = Area of bolts under root = 7.59 in*in
Sa = yield stress of bolt at room temp = 105 ksi

Divide Wm1 & W by 8 to get load per bolt.

I didn't bother to check the minimum load for gasket seating because it rarely is the controlling factor for RTJs
but if you want to

Wm2 = 3.14*b*G*y

y = gasket seating stress = 18 ksi to 26 ksi


6A can get away with using 1/2 the bolt yield as bolt load because the minimum yield stress for 6A flanges is 45 ksi for weld necks and 60 ksi for everything else (75 ksi for 15,000 psi and above). ASME flanges are made from every material you could think of and bolt loads should be calculated according to the material and temperature conditions.
 
Thanks, Hush.

Your advice has been helpful.

I agree with your internet comment, and it's up to all of us to verify the opinions and advice we are receiving.

Rick
 
ASME VIII div.1 , has also an appendix S , which recommends
an impirical value for bolt tensioninig
( I think it is 25000 psi / (sqrt(bolt diam.) ).

API 6A - Appendix D gives also the bolt tensioning force based on 50% of the yield.

In both cases the bolt tensioning is translated into a torque and this is function of the lubricant, diameters, pitch...
---------------------

The question which bother me is what is the relation between this 50 % yield = 360 N/mm2 and Sa =172 N/mm2 for A-193 B7 bolt < 2".

This Sa , as explained above is used in
W = (Am + Ab)* Sa/2

Why are we talking about 2 bolt stresses??
API 6A Appendix D has nothing to do (in my opinion, I maight be wrong!) with ASME!!!

 
Comment on Hush presentation:
quote : Sa = yield stress of bolt at room temp = 105 ksi
End of quote.


Sa: design stress value taken from, for instance B31.3 table A-2

For A-193 B7 or A 320 L7 :
Sa 25 ksi = 172 N/mm2 for d< 2 1/2"
for up to 400 Fahr.
Sa = Tensile strength ( and not yield ) divided by FIVE



I believe derivation of bolt stresss is given in ASME VIII div.1 appendix P
 
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