How to combine lateral wind load into the EQU equation? Eurocode 2 4

[Pretty Girl7]

I was checking the load combinations and then was trying to check for EQU. (Trying to learn how to do it in real-life situation)

My aim was to find out the loads acting on the beam and find the forces. Feed the forces into EQU equation stated below. Then feed the forces into bending moment equations to design the r/f needed etc.

The equation is from: clause 6.4.3.2, page 47, Eurocode 0.

I did the calculation to find the loads as below, but I don't know where to place the wind load in the equation as it's lateral load. I don't even know if the above equilibrium equation is supposed to have wind loads as well.

combination 1: (Self weight) + (1.5 x PL1) + (1.5 x 0.7 x PL2) + WIND LOAD

Ed = (1.35 x (2 kN/m x 6)) + (1.5 x 15 kN) + (1.5 x 0.7 x 20 kN) + WIND LOAD
Ed = 16.2 kN + 22.5 kN + 21 kN + WIND LOAD
Ed = 61.95 kN + WIND LOAD

Then I calculated total moments, but since I don't know where to put "wind load" in above equation, I didn't include it.
Total moments (MEd) = (Design load x length) / 8 + (Design load x length) / 4
= "(16.2kN x 6 m) / 8" + "(22.5kN x 6 m) / 4" + "(21kN x 6 m) / 4"
= 9 kN + 33.75 kN + 31.5 kN
= 77.40 kNm

I have mainly two questions.
1. How to include lateral wind load into that equation.
2. Is the calculation correct if we consider the wind-load didn't exist.

 

[Pretty Girl7]

Assuming the subject beam is an element of frame , the EQU combinations for the overall str.in this case ,
Ed1= 0.90 Gk,j + 1.5 Qk (wind) + 0.0* Qk (Imposed LL) FOR overturning,sliding, uplift..

I later realised I have more questions to ask about your post

1. Why did you used 0.90 for GK as we're trying to find out both EQU and overall STR. The Gk,j should be 1.35 isn't it?
2. The self-weight should be "favourable (supportive)" for the building as it resists the overturning from wind isn't it? so the fact that you used 0.90 doesn't make sense to me.

 

[Pretty Girl7]

Don't you have a more experienced colleague you can ask? If you had a sit-down with somebody that has some experience with load combinations in any code I think it would save you a lot of time.

You guys/ladies are my friends. Unfortunately I don't have anyone who knows about these things. Since you asked me several times, my purpose of asking questions is, I want to understand properly how to design/build safer structures regardless of the cost. Just think of me as an alien trying to gather accurate knowledge and engineering concepts from this world

 

[ThomasH]

You guys/ladies are my friends. Unfortunately I don't have anyone who knows about these things. Since you asked me several times, my purpose of asking questions is, I want to understand properly how to design/build safer structures regardless of the cost.

Is it fair to say that you are a student, with little or no background in structural engineering? And you are asking questions with the purpose of learning how to design structures according to the Eurocod?

The I suggest that you start with Eurocode 0 EN 1990 and read Section 3 "Principles of limit State Design", that defines the different limit states you must be familiar with.
Then you can read Section 6.4 "Ultimate Limit State". That describes the different ultimate limit states, including EQU and STR. That may explain why in your first post the question regarding EQU perhaps was not entirely relevant, it should probably have been STR instead.
Finally, you can go to Annex A.1, Table A1.2(B) where you will find the coefficients for the STR combination.

What I did now (I hope), was to pick the two chapters that gives the background to your specific question as I interpret it. And then I gave you the answer .

But to learn Eurocode with this approach and no background i structural engineering, I would say that it will take a lot of time, if at all feasible.

 

[HTURKAK]

A= NO !!!! Dead load (Gk) is favorable in this case . and the coeff. shall be 0.9 ..



A= It makes sense ..Consider a frame will be checked for EQUILIBRIUM , say for overturning. The resisting moment to overturning shall be calculated with min. dead load and zero imposed load . While the destabilizing load due to wind the coeff. for wind shall be 1.50 ( if leading action)

static equilibrium may be expressed as follows:
Ed,dst < Ed,stb

In this case ( checking for overturning due to wind ) the design effects of destabilising actions (Ed,dst= factored OT moment due to wind )shall be less than the design effects of the stabilising actions (Ed,stb= resisting moment due to minimum dead load ).

Literally this means , minimum overturning F.S.= 1.5/0.9=1.67 ( when we do not consider the lever arm effect)..




Tim was so learned that he could name a
horse in nine languages: so ignorant that he bought a cow to ride on.
(BENJAMIN FRANKLIN )

 

[Pretty Girl7]

Is it fair to say that you are a student, with little or no background in structural engineering? And you are asking questions with the purpose of learning how to design structures according to the Eurocod?

The I suggest that you start with Eurocode 0 EN 1990 and read Section 3 "Principles of limit State Design", that defines the different limit states you must be familiar with.
Then you can read Section 6.4 "Ultimate Limit State". That describes the different ultimate limit states, including EQU and STR. That may explain why in your first post the question regarding EQU perhaps was not entirely relevant, it should probably have been STR instead.
Finally, you can go to Annex A.1, Table A1.2(B) where you will find the coefficients for the STR combination.

What I did now (I hope), was to pick the two chapters that gives the background to your specific question as I interpret it. And then I gave you the answer smile.

But to learn Eurocode with this approach and no background i structural engineering, I would say that it will take a lot of time, if at all feasible.

Thank you for the comments and your insights. I am indeed going through the codes. Sometimes it's confusing, so I turn to this forum.

 

[Pretty Girl7]

A= NO !!!! Dead load (Gk) is favorable in this case . and the coeff. shall be 0.9 ..

A= It makes sense ..Consider a frame will be checked for EQUILIBRIUM , say for overturning. The resisting moment to overturning shall be calculated with min. dead load and zero imposed load . While the destabilizing load due to wind the coeff. for wind shall be 1.50 ( if leading action)

Yeah I understand it's favourable here and should use a lower value for dead load as well, however, I noticed a second set of factors should be used because we're checking for both EQU and STR? The following A1.2(A) mentions we can use "1.35" for favourable dead loads. Isn't that correct?