Coding by finite element method a truss

[Amine A]

Hello,
I am coding the truss problem by finite element (figure below).I found difficulties at the level of the assembly of the stiffness matrices . Since the beams do not have the same materials, we will not have the same stiffness matrices. In the program , how to do the loop of assembly ? Any ideas please ?

 

[Amine A]

Should I work with this matrix (figure below) in my case ? (considering traction / compression and flexion)

 

[IDS]

Should I work with this matrix (figure below) in my case ? (considering traction / compression and flexion)

Yes, so you will have 17 matrices, each 6x6. Do you see how these are combined into a single 18x18 global matrix?


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Amine A]

Yes yes the global will be 30*30 and considering displacement on the 4 nodes it will be reduced to 18*18.
Thank you so much

 

[Amine A]

Hello ,
Please ,should I find 17 or 18 forces in the postprocessing results ?

 

[rb1957]

you've got 14 elements, no ? (AF and FD are single elements, no?

are your reaction points fixed or pinned ?

this is school work, no? please comply with site rules (about student posts).


another day in paradise, or is paradise one day closer ?

 

[Amine A]

I didn't pay attention to that sorry...okay I will( because when I did this account I was working with a company...).
I have 17 elements like shown in the picture I've put on the first comment. Yes AF and FD are elements so the total elements is 17. What is the difference between pinned and fixed ?
Like shown all displacement for A,B,C and D are 0

 

[rb1957]

apologies, yes 17 elements (colour blind to the horizontal ones !)

do the supports (A thru D) react moment (fixed) or just axial load (pinned) ?

another day in paradise, or is paradise one day closer ?

 

[Amine A]

No colour blind and no horizentaol ones, it's 17 elements with AF and FD. The supports A thru D are all fixed

 

[IDS]

Please ,should I find 17 or 18 forces in the postprocessing results ?

Why do you think there will only be 17 or 18 forces?

You have 17 members, each with 2 ends, and each end has two forces and a moment.

You also have 17 6x6 member stiffness matrices, and the deflections and rotations at each end of each member, so you can find the 17x2x3 member actions by matrix multiplication.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

there will be 17 axial loads (why would you think 18 ?). this is the complete solution for a simple pin ended truss.

with fixed ends you can anticipate that end member will have different loads at each end, different moments balanced by different shear forces.

sure in this specific case, you can invoke symmetry with your simple load.

another day in paradise, or is paradise one day closer ?

 

[IDS]

there will be 17 axial loads (why would you think 18 ?). this is the complete solution for a simple pin ended truss.

He's said several times that all the connections and supports are fixed rather than pinned.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

yes, I know ... which is why I went on with the other loads (my point was to show how much simpler a simple truss would've been).

another day in paradise, or is paradise one day closer ?