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Dear all I have this simple structure in SAP2000: I want the self-weight of the slab to transfer to the beam at the slab edges. The self-weight = 25 kN/m^3*0.2m = 5 kN/m^2. I have done the appropriate meshing, but the self-weight of the slab does no seem to be transferring to the beams...

Thanks rb1957.

Thanks for you reply, rb1957. After much staring at the screen and wondering why (lol) I have realised that you have to add on the fixed end forces/moments to the vector A in the code. Thus, for member 2 we get Member2_forces = [ A(1) A(2) + 600 A(3) + 4000 A(4) A(5) + 600 A(6) - 4000]...

Hi rb1957, 1) So, these are the fixed end moments. the UDL is 30 kN/m. Thus, for member 1 that is 10m long, the force at the node is 30*10/2 = 150 kN, and the moment is 30*10^2/12 250 kNm. For member 2 that is 40m long, the force at the node is 30*40/2 = 600 kN, and the moment is 30*40^/12 =...

Dear all, I have the following structure with its loading that I am trying to solve I am using the stiffness method, and my MATLAB code is below: clear, clc, close all % Structural properties E = 2.1E+08; I = 1022000/100^4; A = 620/100^2; EI = E*I EA = E*A alpha = atand(30/40) L1 = 10; L2...

Its just another phrase to emphasize that He will never stop taking care of you.

Thank you for your answer, BulbTheBuilder. And its great to meet another Christian on here. Just in closing If you love that verse in Deuteronomy, look at it when it is quoted in Hebrews 13:5 in the Amplified Bible: "Let your character [your moral essence, your inner nature] be free from the...

When I click the link it goes to 'page not found'. I will have look for that. Unfortunately, my student trial expires in about 10 days. Shame, I was starting to like Tekla Structural Designer. At Uni I used SAP2000 and was accustomed to using that. It took me a while to enjoy Tekla. Yes...

I didn't know why the loading was changing on the beam along its length (the one I labelled), and also all the beams at the same level that were parallel to it - members B to F. So, it has been answered somewhat. Wow, that's incredible. Praise God for your long life. I guessed you were retired...

Yes, BAretired, for those areas (not in red) I would use the loading for roof space, which on my loading sheet is 1.35*0.25 + 1.5*0.25 = 0.71 kN/m^2. I have learned that its good to group members, so I would use the same beam size for B-F. In regard to labelling the joists and wall etc I would...

Yes, sure, but I can tackle it as I go along. There are many other members in the house other than this one, and I am starting with the roof rafters and working down. My first thoughts were that the beam above the wall will take 14.5 kN/m and it will transmit 14.5 kN/m to the wall. This is...

Can I ask what made you use 20%? I didn't post it, but strangely some of the other walls the loading was different. The middle wall for example went down to 6.6 kN/m from 14.75 kN/m. EDIT: would it be the relevant stiffness that would determine how much load share would be between the wall and...

Well, most load paths I get. This was one load path amongst many that I was stuck on. Even some of the helpers/members here said they didn't know. Just because there is one case where I don't know should I not try to design the cases that I do know? This project of mine is all about practise...

Thanks for your explanation, Celt, and I will do what you say.

Its okay, guys, I just removed the steel beam from the top of the wall and I got a constant load of 14.75 kN/m.

It's been a long time since I first posted and there's no-one?

Dear all I am currently working on a hypothetical project for the following structure: Firstly, I am tackling the load distrubtion on beams in kN/m from an applied area load in kN/m^2. I am analysing the load that comes from the floors including the imposed load which is 2.95 kN/m^2. You can...

Thanks, Celt. I think I will definitely buy the book.

Okay, I made a 4 storey symmetric (so no torsion) structure consisting of two shear walls, and using G*A/(1.2*H) I obtained accurate results. I tried again for our more complex structure with two shear walls instead of the core, but didn't get accurate results. Perhaps there is an error in my...

Yes, its not matching the computer software. To test the accuracy I compute the maximum deflection under the calculated load shares. When I have the load shares I use the equation delatMax = w*H^2/(2*K). Where w is the calculated load share, H is the building height, and K is the stiffness of...