Derivation of ASME Joint Efficiency for fillet welds

[jrcan]

Folks,
Anyone know the history and/or derivation of the 0.45 joint efficiency used for single side fillet welds? Is it as simple as a slight reduction under the max shear stress or distortion energy criteria (0.5 or 0.577)?

 

[IdanPV]

Hi,

Refer to Interpretations VIII-80-29 and VIII-80-100 see also the note under UW-15(c)
EDIT:
I have now noticed that you are referring to the 0.45 joint eff. in Table UW-12, I don't know the answer but this requirement has been there at least since the 2004 Edition.

 

[TGS4]

Don't look too deep to try to figure out any derivation or inherent meaning in the weld joint efficiencies. The numbers/values were determined by committee consensus as a means for "penalizing" non-fully-inspected butt welds. Joint Efficiency = Penalty Factor.

 

[JStephen]

It's possible that's based on actual testing in the Olden Days, too.
I'm sure they had similar questions with riveted joints as well.

 

[DriveMeNuts]

Check out UW-15(c) for possible clues to your answer:

(c) The allowable stress values for groove and fillet welds in percentages of stress values for the vessel material, which are used with UG-41 calculations, are as follows:

(1) groove‐weld tension, 74%
(2) groove‐weld shear, 60%
(3) fillet‐weld shear, 49%

NOTE:These values are obtained by combining the following factors: 87.5% for combined end and side loading, 80% for shear strength, and the applicable joint efficiency factors.

I can't derive any of these with the instructions in the note. 74% seems impossible.

The 0.45 that you seek equals roughly 80% x 0.577 = 0.46, but that is unlikely correct. I reckon that they are derived from experiment plus perhaps a 70%ish arbitrary penalty factor.

 

[jrcan]

@ DriveMeNuts
I assumed something similar to your 0.8*0.577, but thought there may be something better.

@IdanPV
Yup. Gotcha on the committee stuff. I sometimes attend code meetings now and they are getting better at writing up the basis for all we do now...this is an oldy that I couldn't find definitive answers. I hadn't looked at those interpretations though...so I will have a look.

Thanks folks!

 

[DriveMeNuts]

Question: In UW-18(d), what does the 55% joint efficiency factor represent when calculating the allowable load on a fillet weld?

Reply: The 55% joint efficiency factor includes the conversion of the leg length of the fillet weld to the throat length (0.707), and the conversion of the tensile strength to the shear strength (0.8); therefore, no additional reduction need be taken in calculating the allowable load by computing the weld area.

That would suggest that the double lapped joint efficiency of 0.55 is calculated using this derivation, without any additional reduction. Does this number essentially mean 0.577 across the weld throat? This 0.55 efficiency appears to not have the 0.7 no-NDT factor applied to it.

Question(2): If the weld metal stress values specified in UW-15 apply to nozzles and reinforcement, in what cases does the stress value specified in UW-18(d) apply?

Reply(2): UW-15 is strictly for the reinforcing load-carrying paths of nozzles where UW-18 is for fillet welds which may not necessarily be around a nozzle. The values in UW-15(b) are 7/8 times the values in UW-18(d) to account for the combined end and side loadings. That is, 0.55 x 7/8 = 0.49.

This explains a lot. Perhaps the awkward asymmetric stress in the single side lap joint requires an efficiency reduction like 0.55 * 9/11 = 0.45. But still no application of the 0.7 No-NDT factor.

Would be interesting to see how the 0.8 tensile to shear strength factor is derived. I assume that it will relate to the geometry having the long weld leg being 1.3 times longer than the short thickness leg.

 

[IdanPV]

Some more interesting Interpretations regarding that issue:
Interpretation Number : VIII-1-95-42, VIII-77-102, VIII-80-26

 

[DriveMeNuts]

Question: The Note in UW-15(c) of Section VIII, Division 1 says the 49% factor for fillet welds in shear is the product of 87.5% times 80% for shear strength times the "joint efficiency factor." Is this "joint efficiency factor" the same as the joint efficiency factor given in Table UW-12 (0.45 for single fillet welds)?

Reply: No, the joint efficiencies given in Table UW-12 apply only to the weld types defined in Table UW-12.

Question: For UCS-23 material, General Note (b) to Table IA in Section II, Part D allows the shear stress in restricted members to be 0.8 times the value permitted in the Table. Does Section VIII, Division 1 specify the allowable shear stress in a member that is not restricted?

Reply: No, see U-2(g).

The Note 2 has been shifted to UG-23(g) in the latest codes. So, where you don't have ideal constrained shear, the efficiency needs to be less than 0.80, hence 0.45 for the lap joint. Still no answer as to how 0.80 is reduced to 0.45.

Question (2): In Section VIII, Division 1, construction, for a shell and head attached by a single, full fillet lap joint without plug welds for attaching shells and heads not over 5/8 in. thick, how is the joint efficiency of 0.45 from Table UW-12, column (c), applied?

Reply (2): For calculating the circumferential stress in the seamless shell, UG-27(c)(1) is used with E = 1.0, but with stress used in the formula equal to 80% of the allowable stress from the tables.

For calculating the longitudinal stress in the seamless shell, UG-27(c)(2) is used with E = 0.45, but with the stress used in the formula equal to the allowable stress from the tables.

As clear as mud.

 

[IdanPV]

Assumption only:
The closet I could get to the 0.45 joint efficiency in Table UW-12, is 0.7*0.65 (=0.455).
0.7 for conversion of the leg length of the fillet weld to the throat length, and 0.65 for the fact that the base metal acting as backing.
But, I probably be wrong, and as TSG4 suggested I think it is better to don't look too deep to try to figure out any derivation or inherent meaning in the weld joint efficiencies.

As for the 0.49 in UW-15, Interpretation VIII-80-29 explain that very good in my opinion.

 

[DriveMeNuts]

What baffles me is how Tension strength is converts to pure constrained shear strength with a factor of 0.8. Von Mises Yield Criterion shows that this should be 0.577.

Tension allowable stress S equals 2/3 x S_{yield,Tension}
Shear Allowable Stress equals 0.8 x 2/3 x S_{yield,Tension} = 1/2 S_{yield,Tension}

But S_yield,Shear = 0.577 x S_{yield,Tension}

Therefore Allowable shear stress S_Shear = 0.5/0.577 x S_yield,Shear = 0.924 x S_yield,Shear

So tension has a 1.5x design margin for protection against yielding. While shear has a slim 1/0.924 = 1.082 shear design margin, for pure shear of an unwelded component.

 

[DriveMeNuts]

I suppose that a shear efficiency of 0.8 'may' be able to be applied to pure constrained shear, because the shear modulus elastic distortion associated with the tensor cube doesn't apply. There is only a clean shear plane. More associated with a guillotine shear. This involves getting into the world of material science, and atomic dislocations.

A factor of 0.577 must apply when there is tensor cube which can distort elastically. The lap joint shear stress will do this and therefore be based on a shear efficiency of 0.577 (or 0.5 if the efficiencies are based on Tresca).

 

[IdanPV]

As far as I know, ASME VIII-1 is not based on von-Mises or Tresca.

 

[r6155]

Be careful
Fillet weld can be concave and machined with a radius, or convex (not recommended). Both have the same efficiency.
But for fatigue there are one more factor: Weld Surface Fatigue-Strength-Reduction Factors. See table 5.11 and 5.12 in ASME VIII Div.2

Regards

 

[IdanPV]

We are trying to understand the origin of the joint eff. of Table UW-12, especially 0.45.

 

[r6155]

Why not see the TGS4 post?
I agree.

Regards