Checking of Interaction surface

[Anonymous]

Hi,

I am checking a RC section using EUROCODE. For the section moment resistance in Y / Z direction where I can use the function of interaction curve to plot the Ax vs My / Mz to see whether the load case exceeds the moment capacity.

When come to biaxial bending problem, we have to check the (Mz/MRdz)2+(My/MRdy)2 < 1 or not. And the interaction curve doesn't allow such 3-direction checking.Is the function of interaction surface,added in 2018,used for biaxial bending checking? 

Thanks a lot.

[dave_geeves]

Hi Matthew,

The interaction curves can be plotted for  My against Mz for a given axial load which will show if your combined applied My and Mz are within the curve indicating they are safe.  You can define various values of axial force by adding extra tabs and if these values coincide with the axial forces in your load case definitions then the  associated My-Mz point will be shown. 

This is an explicit solution rather then the applied rule of "Unity checking" that is allowed for in rectangular and circular columns within BS5400 part 4.  I wasn't aware that this was allowed for biaxial bending in Eurocode 2 as I was under the impression that the explicit approach should always be used.  I may be wrong in this and if so please let me know the clauses or documents I should be looking at.

The interaction surface is an extension of the interaction line but is calculated in a different and more robust way so that iterations failing to converge do not occur.  This has the added advantage that all load cases are seen in the same diagram and the factor of safety on either the total applied load or just the applied variable action are calculated and tabulated for all load cases.

I hope this helps - you can find a worked example on the interaction surfaces in the latest examples manual provided with the software (Example 3.9).

Kind regards

Dave Geeves

[Anonymous]

Thanks for your reply, Dave.

I take a simple case as an example for designing a column capacity. Data are listed below.

 Section: 3500mmx3500mm Rectangular Column

Axial load : N = 51635 kN

Shear along y: Vy = 9068kN

Shear along z: Vz = 1192kN

Torsion: T = 11902 kNm

Moment about y-axis: My = 215100 kNm

Moment about z-axis: Mz = 85075 kNm

I would try to calculate this column capacity in two ways. The first way I will input the axial load only in the SAM model to get the enlarged Mux and Muy for the above section with T50 reinforcement, then calculate the ultimate capacity follow the equation 16 in BS5400 Part 4.

It is found that the ultimate Muy and Muz are equal to 296820 kNm and 273803 kNm under axial load = 51635 kN.

And since N /Nuz <= 0.2, αn = 1.0. (table 12)

So, the equation 16 becomes (Mx/Mux) + (My/Muy) <= 1 for biaxial bending capacity.

If the value of αn = 1.0, it should be a linear equation and the capacity area should be a triangular shape.

Now, we input the value into this equation,

(My/Muy) + (Mz/Muz) = 215100/296820 + 85075/273803 = 0.724+ 0.310 = 1.03 > 1, NOT OK.

In the second way, I will directly input all the force into SAM model to get the final result using interaction surface.

As shown above, the My vs Mz is 0.87 under axial load = 51635kN. It is acceptable since <1.

In conclude, It is found that by hand calculation the result of the biaxial bending moment exceeds the ultimate capacity, however, in SAM model it does not. I don’t know why the calculation result is not equal, can you help me? Thanks a lot.

In addition, the interaction curves as shown in the SAM model, no matter the load is 0 or any numbers, the curve for biaxial bending moment would always become a parabolic shape instead of triangular shape for αn = 1.0.

[Anonymous]

SAM MODEL is attached for your reference.

[Anonymous]

Moreover, it is noted that SAM cannot consider the column as a slender column in calculating the ultimate moment capacity even i set the design code to be EUROCODE. There should be a additional moment in y and z direction respectively if the column is defined as a slender column following CL 5.2 (7), CL 6.1(4), BS EN 1992-1-1:2004. Is there any ways I can input the column height and the buckling mode for more accurate checking in the moment capacity. 

[dave_geeves]

Hi,

Thanks for all the clarification - I see now that you are referring to the bi-axial design of slender columns that have to include second order effects.  I have often wondered myself why eccentricity moments and slenderness  moments are not included as part of the load application for Eurocodes, as they were in the BS5400 code option.  Unfortunately I don't think there is currently a way in the software to calculate these effects and they need to be calculated manually each time and added onto the first order moments.

With regard to the simplified unity check, I will need a bit more time to think about that and get back to you.

Kind regards

Dave Geeves

[Anonymous]

Thanks for your reply. Can you also provide the formula how the number (0.87) comes from in the interaction surface?

You can also try another case without axial load. 

Pn = 0 kN

My = Mz = 500000 kNm

Under this situation without axial load, the Muz = 247924, Muy = 263841

the ultimate capacity =  My/Muy + Mz/Muz = 500000/263841 + 500000/247924 = 3.9117

In the interaction surface, it shows 3.072.

So, this problem does occur not only in the case with axial load but also without axial load. 

[dave_geeves · Accepted solution]

Hi Matthew,

I've now had a chance to review your comments - and thanks for attaching your data file.  I must admit I was a little confused as your data file is analysed using Eurocodes with UK NA but the clauses and tables you are quoting with regard to biaxial bending come from BS5400 part 4.  The "Unity Check" is also present in the Eurocodes in EN1992.1.1 clause 5.8.9(4) but is a little clearer in stating that this simplified approach should be use "In the absence of an accurate cross section design for bi-axial bending":

ASBD provides this accurate cross section design in it's iterative procedures so should be used in preference.  The unity check is a simplified approach and will always provide a more conservative solution.  You will also notice that the value of "a" given in the table above is a little different than in BS5400 in that it takes a value of 2 for circular columns, which should provide a circular interaction diagram (the same as the ASBD method).  For a rectangular section with NEd = 0.0 the interaction is tending towards the linear solution implied by the unity check and becomes more and more linear with an increase in reinforcement

So,  I believe that there will always be a difference between the two approaches but the ASBD solution is an explicit and more accurate solution than the simplified approach, so should be used in preference.  If your section was not rectangular or circular or the reinforcement was not symmetrical than you would have to use the explicit method.

I hope this has helped but please get back to me if I have missed something.  If this has answered your query then please mark my reply as a solution so that others may benefit.  Thanks

Kind regards

Dave Geeves

[dave_geeves · Accepted solution]

Hi Matthew,

I forgot to add a comment about the value of 0.87 in your previous post.

The performance factor 0.87 is the ratio between the length of the applied load vector My, Mz, Ax and the length of a vector if the applied load vector is extended to the interaction surface (in 3D).

If you require the performance vector for just the applied moment (with the axial force constant) then define the applied load with AX as a permanent action and the moments as variable actions as shown below.

This then shows a performance ration (ie 1/loadfactor) of 0.873

Hope you find this useful.

Kind regards

Dave Geeves

[Anonymous]

Dear Dave,

But the most confusing me thing is how the number of "performance" comes out, could you list out the equation?

[dave_geeves · Accepted solution]

Hi,

Ok, let's think about this another way.

If you carry out a "Bending Axial and Shear" analysis of the section, and apply your axial load only load case you can easily get your My resistance (with Mz=0.0) and your Mz resistance (with My = 0.0) ; but you can also get a combined My - Mz resistance with a certain ratio of Mz/My

In your example Mz = 85075 and My = 215100 which gives a ratio of 0.3955

The resulting resistance is then 247416 which is My giving the coexisting Mz moment of 97856.

The performance ratio is the applied load/resistance load which for My is 215100/247416 = 0.869.  which will be the same for Mz.  This is of course assuming that the coexisting axial load remains the same at 51635kN.  This is essentially the same as the interaction surface method when I have declared the axial load to be a permanent load.

I hope this helps to make things clearer and if it does please mark my post as a solution so that others can benefit.  Thanks

Kind regards

Dave Geeves