why should different material create problems ? E is just one property of the truss member stiffness (L/EA) ...
each element has L/EA as a stiffness
then at each node assemble the member stiffness in global co-ords

suggest you start with a very simple one first, one you can hand calc.

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

Element (ii) stiffness in local coordinates is given by


You need to transform the element stiffnesses from local coordinates to global and then assemble in global stiffness matrix K.

E(ii)= Element Young's modulus
A(ii)=element area
L(ii)=element length

rb1957 , goutam_freelance : Thank you for the replies. Here as you can see I have 17 beams and each beam has it's matrix, so my problem is : how should write each matrix as they are different , I can't use iteration so should I write each 17 lines for each Ke of a single beam ??!!!!
PS: goutam_freelance , I won't neglevt flexion in this example, so the matrix still like the one you showed me ?

ok, start with a simple example. You can model beams (with bending), just more complicated (axial and bending stiffness, two internal loads).

you are asking very basic questions, so get some texts ... there are plenty on-line.

you assemble the global matrix from the grids/nodes/points.

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

Thank you. I saw some examples and I did simple problems. If you see the picture of my problem you will notice that it's not the same thing , as we don't have same materials , same angle of the beams ... We will have different stiffness matrix that's why I'm asking

It shouldn’t be a problem to loop through all elements, create their local stiffness matrices and assemble them into global stiffness matrix. Which programming language do you use ?

That's my problem... It's in the creation of the Ke and the assemly .... It's not similar... I'm using python

if you can build a simple truss, then you should be able to do a more complicated truss. each truss element has it's own stiffness matrix (in local co-ordinates). assembling into the global matrix is "just" combining the elements based on which nodes they affect and what angle between the local axis and the global axis.

different material is "only" an impact to local stiffness.

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Ok I will see. Thank you brother

There

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Thank you brother

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or beyond

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Quote (Amine A)

That's my problem... It's in the creation of the Ke and the assemly .... It's not similar... I'm using python

Once you have seen and understood the basic procedure for combining the beam matrices into a global stiffness matrix your question will be answered, including why the beams having different lengths, angles and material properties is not a problem.

Regardless of the language you intend using in the end, I suggest setting up a simple problem on a spreadsheet and getting that to work.

The link below has links to a series of blog posts that work through that exercise, ending up with VBA and compiled code. The 3rd post addresses your question in this thread, but it is worth looking at the first two as well.

https://newtonexcelbach.com/2014/01/22/frame-analy...

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

Thank you so much

Hello,
Can you please tell me what is the size of stifness matrix of an inclined element if we consider traction , compression and flexion ? is it 4*4 or 6*6 ?
I have never saw a truss considering flexion problem....

Quote (Can you please tell me what is the size of stifness matrix of an inclined element if we consider traction , compression and flexion ? is it 4*4 or 6*6 ?)

I assume when you use the term "truss", you are using the standard interpretation that all members carry axial forces only, no shear forces or moments; hence node displacements are translations only, no rotations. (2 DOF per node in a 2D model, 3 DOF per node for a 3D model)

"Frame" members can carry axial forces, shear force, and bending moments, and node displacements include translations and rotations. (3 DOF per node in a 2D model, 6 DOF per node for a 3D model)

Hence a 2D truss element has a 4x4 element stiffness matrix; while a 3D frame element has a 12x12 element stiffness matrix.

http://julianh72.blogspot.com

The clearest way to understand how to form the global stiffness matrix is to consider the freedoms at each node, and how they interact. For your example, and ignoring bending moments, each node has two freedoms, deflection in the x and y direction.

There are 6 unrestrained nodes, so you need to find the magnitude of 12 deflections, so your stiffness matrix will be 12 x 12.

Numbering the freedoms in alphabetical order you will have 1,2 at Node E through to 11,12 at Node J. Each freedom number represents a row and column in the stiffness matrix, so for instance row and column 1 and 2 will combine the stiffness values from end 2 of elements AE and BE and End 1 of EF and EH.

If you want to include bending moments (assuming the nodes are fixed rather than hinged) there will be an additional freedom at each node, so your stiffness matrix will be 18 x 18.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

Yes , the problem is that in the exercice he didn't specify this... He said a structure composed of beams ( the figure I showed you on the first comment) and we will consider traction , compression and flexion. So I don't know with which case I will work based on that statement

Yes , thank you, so if I include bending...each local (for an element of 2 nodes) stifness matrix will be 6*6 no ?

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

• https://files.engineering.com/getfile.aspx?folder=991c7d8e-a459-461e-824a-0

Quote:

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/

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 😊

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

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).

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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

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

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

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

Quote:

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/

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 ?

Quote (rb1957)

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/

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).

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