Baffling Eccentricity 1

[engleprechaun]

Im am new to the design world and have not been quite understanding some things. Just when i think i have itt all figured out I run into something even more confusing.

In my down time i have been revisiting a design that to me from feild experince seemed like it was way over designed. Its a design for a concrete cassion foundation for a monopole. I thought i could develop an interaction diagram for the foundation and then see from the eccentricity where my actual loads would fall on the diagram. Well, all the capactiy numbers are coming out well above actual loads.

My problem is that in figured the eccentricity of a the capacity loads my large end eccentricity came out to 880 in. and the eccentricity for the actual loads came out to be 14,000 in.

They kinda make sense since the tower reactions are 20 kip axial and 2012 ft-kip moment (both factored). What does an inexperinced person do in this situation. It is to be built soon. Do i worry about ti that much.

Also, in developing my interaction diagram in the situation for a large eccentricity the load capacities came out to be 1600 kip axial and 9600 ft-kip moment.(e=880) Please help!!

Thank you in advance.

 

[Lion06]

I'm not quite following you - if you have M=2012 k-ft with P=20 K, I am coming up with e=1207.2" (my understanding is that this is your actual loading).

For M=9600 k-ft and P= 1600 k, I am coming up with d=72".

I don't know where you're coming up with 14000" and 880", respectively. Can you elaborate?

Additionally, if your actual load falls inside your interaction diagram then you are fine no matter what the eccentricity is.

 

[engleprechaun]

well now i feel a little dumb. You are right. The 2012 k and 20 k is the actual loading. And you have the right e. I was changing from ft-kip to in-kip in mathcad twice and didnt realize it until you came up with the correct e.

Also i discovered that i was setting up my compression block area way wrong. The actual d=60". I was taking it times a. Instead of the actual area creating by "a" in from the edge, making the compression capacity much smaller, and the e much larger. Making it become much more realistic.

So my apologizes for not looking closer before asking. Thank you though for your help!!!!

 

[miecz]

If you are using Mathcad, try carrying the units with your calculations. This is a great advantage that Mathcad has over spreadsheets. You can enter your moment in kip-ft or kip-in or kN-m and mathcad automaticlly makes the conversion for you, so you never have to worry about this kind of error.