Yeah, welcome, good afternoon to our tutorial today and I would like to come back to my

inquiry about the possible shifting of lecture and tutorial and in this regard

I double-checked with the people responsible for the video recording and

they told me that Friday at 8.15 it's not possible to take a recording so if we do

it on Friday then we would do this in 8, 8, 8, lecture hall 8 from 10.15 to 11.45

and after a lunch break till 1.45. May I ask you once again after the tutorial

which of these dates are possible and which not? I did it already in the

lecture but just to make sure that those of you who did not attend the lecture

also have the opportunity to vote here. So the background is I won't be available

on June the 4th so I'm looking for alternative dates to give a lecture and

tutorial and of course we will also take the video recordings so even if you

cannot attend you will be able to find that online. Okay and this morning the

question was whether we do it on Wednesday on Thursday or on Friday. I

will ask you after the tutorial about your opinion. Okay so this is our today's

task, it's exercise 6 and this subsumes more or less all the elements we

considered so far, elements in terms of the ingredients of the finite element

method and we have given here a truss which is clamped on the left-hand side

and it is subjected to a distributed load and a single force on the right-hand

side. First of all we should discretize system into two equidistant linear

truss elements and set up the connectivity matrix. This is what we

start with and then there are a couple of other questions posed here. So when it

comes to discretization then if there is nothing specified about the ordering of

the nodes and so on then it's up to you. In the examination of course I would

give some some constraints that you are not completely free to choose that but

this shouldn't be a problem. So let me just copy the first question here and

let us start.

So

if there are no specifications given about the node numbering, ordering and so

on, then what would be a very obvious way to discretize and to label nodes and

elements in this case just element 1, 2 and the nodes accordingly. So let us do

this simple thing here. So then it is specified that these elements should be

equidistant so if the entire length is L then the length of a single element is L

by 2.

And here we have element 1, here we have element 2 and the node numbers are just

1, 2 and 3. So this is the first part. The second part we have to do is here to

set up the connectivity matrix and closely linked to the connectivity matrix

is the question of node, nodal positions and this we should provide here as well.

Let me just give you the number of elements which is just 2, then the number

of element nodes in linear truss elements it's also 2 and the number of

nodes is 3. So maybe this is the number of elements, this is the number of

element nodes and this is the number of nodes. Just to clarify this, then the

length of element 1 is the same like length of element 2, this is L by 2 and

then let us have a look to the coordinates.

So X, E, eta means this notation, this is the eta node of element E, maybe we can

just add here that's the local, the eta local node of element E and Xi, this is

the i-th global node and with that we can set up the nodal positions. So X, 1, 1, this is

the same like X, 1 and this is just position 1. We have a coordinate system

given in the sketch, I just put it here for clarification and then we have here

the second local node of element 1. What would be another labeling for the same

node? This is from the point of view of element 1. What would be the labeling here