Total design moment in both x and y axis for concrete columns - Eurocode2 3

[Pretty Girl]

I'm designing a short column in accordance with Eurocode 2. This column is situated at a corner with bi-axial bending as it has beams fixed to it.
I want to calculate the total moment (suppose I have already calculated the moments of both axes), so I can feed to the design moment equation "M02 + (Ned ✕ ei)" to get the design moment.
As I see the steel bar at the corner gets both "x" and "y" moments in both axes (example below).
So, I feel we need to combine both moments (x and y axes) before feeding to the above equation.

So, M02 = (Mx + My) = (100 + 150 kNm) = 250 kNm total for both axes?

My questions are,
1. Do we need to sum up both the moments (100 + 150 kNm = 250 kNm) as I did above before feeding it to the equation?
2. Do we feed total 250 kNm to the above equation to get the design moment or just the 150kNm (bigger moment only)?

The calculation example books are bit unclear of the above.

 

[HTURKAK]

Q1. Do we need to sum up both the moments (100 + 150 kNm = 250 kNm) as I did above before feeding it to the equation?

A= NO!!


2. Do we feed total 250 kNm to the above equation to get the design moment or just the 150kNm (bigger moment only)?

A= In this case NO!!!

If we assume Medy= 150 kN-m and Medx= 100 kN-m ( Assuming these are design moments at orthogonal directions )

Medy/Medx= 150 /100 kN-m = 1.5 that is; > 0.2 and < 5.0 ; you can not ignore, Biaxial check is required.

You can use the formula (5.39) suggested at EC-2 ,( BRESLER'S FORMULA)

I have copy and pasted the relevant page of the code. ( This is law resource and free )

Don't underestimate a nail. A nail saves a horseshoe, a horseshoe saves a horse, a horse saves a commander, a commander saves an army, an army saves a whole country.. GENGHIS KHAN

 

[Pretty Girl]

@HTURKAK
Thank you for your detailed reply.
After I went through the pages you provided, I understood what to do.

Further, as I see, the equations you provided alone can't be used, after the bi-axial check you provided, we need to use the BS 8110 equations to determine the moments. I didn't notice anything similar in Eurocode2. So I had to use BS8110 equations below. Hope these are ok to use.

 

[IDS]

Further, as I see, the equations you provided alone can't be used, after the bi-axial check you provided, we need to use the BS 8110 equations to determine the moments

It's really not a good idea mixing codes, unless the code you are using specifically says alternative procedures are acceptable for a specific calculation.

In the case of biaxial bending, Equation 5.39 at the bottom of page 74 tells you how to combine the moments, if you are not going to do a detailed analysis:

You also may find the blog post below useful, which includes a comparison of Eurocode 2 and AS 3600 results with a more detailed analysis, and a spreadsheet for doing the detailed analysis.

https://newtonexcelbach.com/2019/11/21/biaxial-bending-update/

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[HTURKAK]

MR. IDS (Civil/Environmental) has a valid point and a PBS for his explanation. I disagree with your opinion. Follow one of the codes ( relevant at your region) . You do not need to use the BS 8110 equations for biaxial bending .









Don't underestimate a nail. A nail saves a horseshoe, a horseshoe saves a horse, a horse saves a commander, a commander saves an army, an army saves a whole country.. GENGHIS KHAN

 

[Pretty Girl]

@IDS

Isn't that the final check to see if the r/f and section we designed is safe/not?
How do you combine X and Z moments with the equation you have given? that's just a "resistance vs effects" check isn't it?

What I previously planned to do
1. Combine moments using: BS8110
2. Calculate the area of r/f and other section to that moment and requirements: (Eurocode)
3. Check the resistance vs effects the equation you have supplied, to check if the design is safe (Eurocode)

Why doesn't that create a safe design, I'm trying to learn.
I have seen almost all the books out there used the BS8110 to combine moments.

And even the manual published by the institute of the structural engineers (United Kingdom) shows that the BS8110 equations can be used.

 

[Pretty Girl]

@HTURKAK
Thank you for your response.

 

[IDS]

How do you combine X and Z moments with the equation you have given? that's just a "resistance vs effects" check isn't it?

I don't. I just choose a cross section and reinforcement based on the larger of the two moments then check it using Equn. 5.39, or the equivalent in whichever code I am using, and adjust the design if necessary. If the capacity under the combined moments is close to the limit I will also check it for the resultant moment, rotating the section so the resultant of the bending capacity is in the same direction as the resultant applied moment.

Why doesn't that create a safe design, I'm trying to learn.

If you are checking the final section to the Eurocode requirements then that's OK. I don't see the point of using the BS 8110 equation to get an initial design moment myself, but if you find it useful there is no problem using it as long as you do the final check to Eurocode.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[dik]

I don't know if I have the source code anymore, but about 25 or 30 years back I wrote a biaxial bending program for concrete rectangular columns, and it is actually quite involved. Far more complicated than uniaxial bending.

-----*****-----
So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[IDS]

Regarding the need to check the resultant capacity of the section in the direction of the resultant applied moment, even though it isn't required by the code, see the graphs below from the link given in my first post:

https://newtonexcelbach.com/2019/11/21/biaxial-bending-update/

The blue line is the capacity of the section in the direction of the applied resultant moment. The red and green lines are the capacities to the Australian and Eurocode formulas respectively.

Section with very high axial load:

Non-rectangular section:

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Pretty Girl]

@IDS

don't. I just choose a cross section and reinforcement based on the larger of the two moments

You're just designing for the larger moment. I feel like your method is unsafe.
The combination always grabs some portion of the other moment to the moment you're considering.

Example moments: Z = 400 kNm, Y = 300 kNm

Your method (choosing larger moment)

Design moment = max (400;300) = 400 kNm

Combination method

Design moment = 400 kNm + some percentage from 300 kNm = probably would end up around 550 kNm or something. (so this is higher than the method you used)

Combination method always ends up as a bigger moment than considering a single axis moment
So, isn't your method is not safe than the combination method?

I don't see the point of using the BS 8110 equation to get an initial design moment myself

Because there aren't no other way in Eurocode to see what is the correct combined moment would be. I feel like you're basically designing for the wrong moment (by just selecting and designing only for the larger moment). So at the end it will show like you have a sufficient design when you compare the "effects (design moment) vs the resistance". But basically you designed it for the wrong moment (so the design will be wrong as I feel). So isn't the entire calculation of yours inaccurate and not safe?

 

[IDS]

I feel like you're basically designing for the wrong moment (by just selecting and designing only for the larger moment).

But that isn't what I said you should do.

Whatever method you use to derive the initial moment, you have to check that the section is strong enough for the combined moment.

Using Eurocode, you can either use Equation 5.39, or you can check the section capacity under the resultant moment, analysing the section as asymmetric.

If you use the BS8110 formula as the starting point, you still have to do the same checks to make sure your final section is OK.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

Further to my previous post, it's actually quite easy to derive an equation for the minimum resistance moment, similar to the BS8110 one.

For applied moment Me and moment resistance Mr, if Me/Mr is equal about the two axes then Equation 5.39 reduces to:

2(Me/Mr)^a <= 1.0

So for minimum Mr: (Me/Mr)^a = 0.5

Me/Mr = 0.5^(1/a)

Mr = Me/(0.5^(1/a))

In the graph below I have plotted 1/(0.5^(1/a}} with 1 + Beta from BS8110, and an equivalent factor from the Australian code (AS 3600), for a range of axial loads:

The higher factors are more conservative, but in practice any of them would be OK to use, as long as you check the resulting section against Equation 5.39 (if using Eurocode 2), or against a detailed biaxial analysis.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/