Unity check for biaxial bending

[Polecat]

In AISC's LRFD Manual, when you have combined shear and tension on bolts, a unity check is made by ratios of (applied shear/ strength)^2 + (applied tension/strength)^2 <= 1.0

Similar checks are made when you have combined shear and moment in beams or columns.

My question is: when you do a unity check on a beam subject to biaxial bending with , are the individual ratios squared or are they added together on a straight linear basis?

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[WillisV]

See 2005 Specification - Chapter H, Equation H1-1b - added linearly.

 

[Taro]

I'm not sure where you are getting your information. For bolts, AISC LRFD has always used a linear approximation to the elliptical interaction curve as given in Table J3.5.

For beam-columns, there are two equations in Chapter H to check depending on the amount of axial load.

 

[Polecat]

Taro:

I was referring to the equation shown in the chapter on A390 bolts; specifically Section 5.2, Combined
Shear and Tension (page 16.4-32 of the LRFD 3rd Edition), equation 5.2 describing the ultimate limit state interaction.

This equation shows the two ratios of shear and tension being squared and then added together to compare to unity.

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[Taro]

Thanks for clarifying. I hadn't noticed that the RCSC specification was revised. The RCSC equation is more accurate, but the AISC specification supersedes the RCSC specification so you are technically supposed to be using the linear approximation. Maybe after a few revision cycles the elliptical unity equation will work its way into the AISC specification.

 

[WillisV]

Taro,

AISC already took this into account - see the commentary to J3.7 of the 2005 Specification. The equations presented in the specification are a tri-linear approximation of the ellipse (see Fig. C-J3.1).

 

[Taro]

WillisV, that is exactly my point! AISC uses a tri-linear approximation that supersedes the RCSC elliptical unity equation. So technically you are supposed to use the AISC equations.

Of course, the RCSC equation is theoretically more correct, so I wouldn't have a problem with Polecat using it. But some building officials are sticklers for enforcing the code exactly as written and won't listen to logic.

 

[Polecat]

Taro & WillisV:

Thanks for your thoughtful inputs. I would, however, like to pursue this subject a little further.

The following references are from AISC's LRFD 3rd Ed.
The 3 line approximation you mentioned is shown on Fig C-J3.3, page 16.1-244.

I'm having a problem reconciling the equation shown there for ft to the results obtained from the shear-tension interaction equations shown on Table J3.5 (as in 117-2.5*fv <90). Since the latter was presumably derived from the former, should they not be equal?

Perhaps I'm not interpreting the nomenclature properly for the ft equation. I'm taking fv as the applied shear stress, Fnt as the limiting nominal tensile stress (which in the case of A325 material would be 90 ksi), and Fnv as the limiting shear stress (which would be 48 ksi according to Table J3.2)

Any ideas?

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[WillisV]

That table was refined for the 13th edition manual and no longer includes those equations. I recommend using Section J3.7 of the 2005 Spec (13th ed. manual) - the tri-linear approach.

 

[Taro]

The equations remain the same in the 13th edition. They just changed the presentation a little to fit with the dual ASD/LRFD format. It's still the same tri-linear approximation it always was in previous LRFD specifications.

Polecat, I think your interpretation is correct. What is the confusion?

 

[Polecat]

Taro:

The confusion is that I am getting apples and oranges. To be specific, let us use the following numbers as an example: (again, using the 3rd edition of LRFD)
bolt diam d = 0.625 nominal area Ab = 0.307 phi = 0.75
Fnt = 90 Fnv = 48 (from Table J3.2)

(As an aside, I don't quite understand why the full shaft area, Ab, is use to compute shear stress instead of the root area if the threads are in the shear plane. But, that's what the equations call for)

NOTE: instead of following these numbers here, you can also download the attached MathCad sheet and see the work.

Factored tension load Tu = 7k, shear Vu = 6k
Tensile stress = ft1 = Tu/Ab = 22.82 ksi
Shear stress = fv = Vu/Ab = 19.56 ksi
From the interaction equation:
ft = phi*sqrt[Fnt^2-(Fnt/phi*Fnv)^2*fv^2] (Fig C-J3.3 from Page 16.1-244)
The result of this is ft = 56.67 ksi

Now, going to Table J5.3, Page 16.1-65:
Ft = 117-2.5*fv <90 (with the threads in the shear plane)
Ft = 68.1 ksi, and phi*Ft = 51.1 ksi
If this difference is the result of the Ft equation being a 3-line approximation of the elliptical equation, then that would explain it. That being the case, then using this lower Ft value would be a bit more conservative.
A Unity check of this would be (ft1/phi*Ft)+(fv/phi*Fnv)<1
which resolves to 0.990

Now, let us fast forward to the specifications for A325 bolts, Section 5.2 on Page 16.4-32 where we have a combined shear and tension unity expression for ulitmate limit state (Equation 5.2).
Here, Tu & Vu are the applied loads (same as above) and Rn is based on the same relative Fnt and Fnv values.
phi*Rnt = phi*76.56*Ab = 17.63k phi*Rnv = phi*48*Ab = 11.05
The ratios become:
Tu/phi*Rnt = 0.403 Vu/phi*Rnv = 0.543
Now if these are added linearly, we have 0.946, which would be just slightly less conservative than the 0.99 figure as above.

HOWEVER, Equation 5.2 says that each of these quantities is to be SQUARED, then added. That produces a sum of 0.457, about half of the other sum.

So, to conclude this rather long story, my confusion is the inconsistency between two interaction equations that should yield the same result. Unless it is related to the fact that A325's are supposed to be pretensioned, it would almost lead one to believe that Eq 5.2 is a misprint.

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[Taro]

The "unity" check for the tri-linear approximation is not compared to a value of 1.0. That would be extremely conservative. That would result in a straight line between the end points of the elliptical segment on Fig. C-J3.3. The limiting value is 1.3 instead of 1.0. Even this value of 1.3 is quite conservative in the middle portion of the curve, but it is a trade-off so that we aren't too nonconservative at the ends of the curve.

You also appear to have a numerical error in your application of Eq. 5.2. The tensile strength should be 90 ksi, not 76.56.