psyclones

Sure. It's an awkward shaped bracket. Concerns regarding cabling & access is not an issue. Thanks for your help. your right you can't do much without the drawings, and I can't provide them (again apologies). CCF_000006.pdf

Good point. My apologies, I can't give more info, Yes. I agree the code(s) goes into great detail about how to deal with different wall/ roof pressures and local area pressures on corners and around eaves of buildings. Sway becomes a significant problem for multi level buildings also. What information would you need?

As far as I know, there was no (physical) test model created. But, the code is very robust if applied properly. Structural Engineer's for domestic buildings (I know) rely on the Wind code not CFD modelling. This maybe of little importance, but I know once all loads were determined the structural analysis would have been designed using (Siemens) FEMAP. I have the (3D) CAD model. Yes. I reviewed the ASCE 7-16 for Wind loadings, and the American Standard (in some aspects) is very similar to the Australian. And Yes. "multiple fatalities if the structure failed" is correct. Maybe your right, due to where this structure will be mounted, Winds loads become a primary concern, so conservative loading would be ok, before applying SF. But; on second thought, if we take conservative loading as a given, could we/I just treat the whole structure as a Rectangular Prism (of which it would occupy <50% Vol). Then trying the calculate wind loads becomes much less cumbersome?

Hello, An Engineer I know determined wind loads for a small but rather odd shaped structure assembled from various standard steel sections. It's a frame attached to the topside of a stadium roof to suspend speakers. We have AS/NZS 1170.2 (Australian) Wind Code, which gives detailed tables to calculate drag co-efficient’s for standard members. But my problem is this; Even if the sum of all areas and co-efficient’s were calculated, the wind loading (on any given face) would be incorrect, or very conservative! Due to the standard only applying to individual members. *Wouldn’t be easier to use a CFD package to model the wind loading, the wind velocities (per direction) can be easily found in Aust. Standard? O/A approx. dimensions; 1300mm x 370mm x 670 mm, Steel sections include; 100mm sq hollow section, 50x50x5 equal angles. 12mm and 16mm thick plates to mount speakers & structure existing rafters. I’ve very hesitate to provide an image of model due to confidentially. Which I realise without it could make it difficult to get an idea of the structures geometry.

The lecturer has posted a solution to solving the problem using the above method with only a vertical force. Problem solved. But upon reflection, I looked at a textbook & the only problems (MDM) which include frames & horizonal forces took sway (or horizontal deflection) & principle of superposition into account to solve the problem.

Hello, I have question is with applying Moment distribution method in calculating moments in a frame with two forces; one vertical & one horizonal. (Please see attached) No sway is assumed. It's a frame i.e; joints B & C are fixed. My lecturer has included an example with moments created from both vertical & horizontal forces applied to frame. Then he goes on to say (in an email) that the Moment distribution method can't be used to calculate horizontal forces, only vertical forces I have spent time trying to email the Lecturer & Tutors about this problem and they won't even take the time to read my email(s) properly and look at attachments & answer my questions to give me any real clarity. I would like to know (if anyone can tell me) if it's possible to solve the attached problem with two forces in a single Moment distribution method table? Otherwise how can I solve it using this method?CCF_000013.pdf I've created an excel spreadsheet and can't get the numbers to work.