Les structures en treillis, en poutre, en dalles et en cadre sont essentielles pour une grande partie des constructions modernes : immeubles pour l'habitation ou de bureaux, halles et usines, ponts, ou passerelles, voies de transport de l'énergie ou des télécommunications, stades et grandes toitures. Le cours L'Art des Structures 2 vous propose d'en découvrir le fonctionnement et les bases de leur dimensionnement. Sur la base de la statique graphique, une discipline plus que centenaire qui allie rigueur et compréhension globale du fonctionnement, tout en limitant au maximum les aspects mathématiques, les efforts dans les structures sont déterminés, ce qui permet d'en choisir les dimensions. Après avoir suivi ce cours, vous serez en mesure d'identifier les structures en treillis, en poutres, en dalles et en cadre, ainsi que leurs déclinaisons en trois dimensions. Vous saurez comment déterminer les efforts qui agissent dans ces structures et de les prédimensionner. Vous serez à même de faire des propositions pour la réalisation d'une structure utilisant ces diverses formes et de comparer l'efficacité des diverses solutions possibles.
1120 Treillis à 5 noeuds
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Do curso por École Polytechnique Fédérale de Lausanne
L'Art des Structures 2 : treillis, poutres, dalles et cadres
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École Polytechnique Fédérale de Lausanne
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Treillis à 5, 7 et 9 noeuds
Vous allez maintenant résoudre vos premiers treillis, en commençant par déterminer les forces aux appuis puis en obtenant les efforts dans toutes les barres de proche en proche.
Conheça os instrutores
Olivier L. BurdetDr
Laboratoire de Construction en Béton
Hello, in this video, we are going to deal with a truss with 5 nodes.
As you can see in this video, we can move from a truss with 4 nodes,
the one which had a diagonal in compression,
to a truss with 5 nodes.
Adding 2 bars and 1 node,
we thus get a new structure.
We first want to see if this structure is statically determinate.
We have 2 support conditions on the right and 1 on the left, which means 3 in total.
Thereafter, we have (counts till 7) 7 bars.
We have (counts till 5) 5, 2 x 5 = 10 = 3 + 7.
Then, we are going to see
how to solve this structure,
what are the internal forces acting inside
and then we will compare these results
to the results we previously got
for the arch-cable and for the truss with 4 nodes.
Afterwards, we will perform a study of variations
to figure out what is the influence of height
on all the internal forces acting in this truss.
Let's start by solving the equilibrium.
We have got a small problem here because there is no node
like the one before. There are 3 bars intersecting on this node,
as well as on this one; there are 4 bars intersecting on this node,
here there are 2 bars, here there are 2 bars.
We will be able to solve this problem
if we first get the support reactions
on the left or on the right.
Here, I have already got them ; maybe in your exercise,
it will first be necessary to determine the support reactions,
from which we can solve, for example, the node E.
We are going to start by the node E, here.
The support reaction RE = 12.5N acts on this node.
Afterwards, rotating counter-clockwise
around this node
we meet the bar D-E
Be careful, the support reaction is drawn on a staggered way,
but we are going to make the Cremona construction
using the axis which is given by the forces, so here. We have the parallel to D-E,
then, we will arrive on C-D here, at this level.
Here are the internal forces.
What we can see here,
is that this force C-D is going to compress the node E,
so this bar is in compression.
Respectively, this vector is a vector in compression.
This is the internal force inside D-E.
Afterwards, the internal force inside C-D pulls on the free-body
so this is a tensile internal force, then this bar here is in tension
and then the polygon of forces is closed thanks to this support force on the right.
The contribution of the node E to the Cremona diagram
is highlighted in green.
We are now going to move to node D.
This node is subjected to the internal force in D-E, the force of 10 Newton,
the force in B-D and the force in C-D.
We take the internal force D-E in the other direction, we add the force of 10 Newtons,
and then the internal force in B-D, horizontal,
then the internal force in C-D, parallel to the segment [CD]
to finally reach again the beginning of the internal force D-E.
The internal force in B-D is pushing towards the right,
this is thus a compressive internal force acting on node D,
so this bar B-D, here, is in compression.
However, the internal force in C-D pulls on the free-body
and then bar C-D is in tension.
Here is the contribution of the node D to the Cremona diagram,
so a triangle over and another under.
We are now going to move to node C which is not loaded.
There are internal forces anyway. I had not noted it,
here is the internal force C-E
along which we are going to travel on the other direction,
then we are going to follow C-D also on the other direction,
and then we are going to travel along B-C
until here.
Then, I am going to shift to draw the internal force in A-C
till, again, the beginning of the internal force in C-E.
The internal force force in B-C, here, pushes on the free-body,
it is thus a compressive internal force,
however the internal force in A-C pulls on the free-body,
it is a tensile internal force.
The contribution of node C to the Cremeona diagram
is a bit complicated
since here, there are two lines one over each other
and there is this little triangle at the end.
We are now going to move to node B.
This node B is subjected to the internal force of B-C in the other direction,
then to the internal force of B-D also in the other direction, then to the force of 20 Newtons,
and to the internal force in A-B which comes back to the beginning of the internal force in B-C.
The internal force in A-B is thus in compression here.
I should have drawn before that the internal force in A-C is in tension.
The contribution of the node B to the Cremona Diagram, here.
Finally, we are going to move to node A.
We are simply going to check that the reaction at the support is right.
We have the internal force A-C in the other direction, followed by the internal force A-B in the other direction,
followed by the reaction force on the left RA = 17,5N.
The contribution of this gray subsystem to the Cremona diagram, here.
We have solved here the equilibrium of the entire structure,
finding the internal forces in all the nodes.
And here, like for a truss with 4 nodes, you have
all the resolution process which I encourage you to attentively follow
to understand well what happens.
To better understand, you could watch the video again to see what I have done.
Now, as we did before, we want to compare
the internal forces in the arch-cable, the truss with 4 nodes
and the truss with 5 nodes.
For this, we are going to concentrate on certain elements.
If we look at the internal force in the left diagonal,
we can see that it is quite similar in these calculations,
actually it is identical in 3 of the 4 calculations
and a little bit bigger in the fourth one.
What we can see as well is that this internal force is given
by this part of the Cremona diagram.
Likewise, the internal force in the lower bar
is given by this same part of the graphical construction,
sometimes with additional elements, but most of the time directly.
We can see that the order of magnitude of these internal forces is again the same.
If we look at the internal force in the right part of the truss,
it is given on the upper part
of the construction.
And again, the values are quite similar.
There is a big similarity
between the results obtain for the arch-cable
and the various types of trusses which we have seen till now.
Don't forget that these structures obviously have the same span,
the same height, and they are subjected to the same loads.
In this diagram,
I have gathered together the resolution for three trusses :
the truss with 5 nodes which we have already seen together,
a truss which is similar to the first one but with half the height
and a truss which is also similar to the first one but with twice the height.
As before, let's look at
what happens in the Cremona diagram.
Here, we get the internal forces in the left diagonal
which is the most loaded.
When we divide the height by 2,
we can see that the horizontal scale turns out to be the double.
However, the vertical scale is defined by Newtons.
The loads being the same, this vertical scale remains identical.
Horizontaly, the drawing lengthens, which means that the triangle becomes flatter.
What we can see is that these elements here,
the horizontal elements, got twice longer
so the internal forces double from 10.1 to 20.2N.
However, in the diagonal, we go from -20 to -26N
because, trigonometrically,
the hypotenuse increases less than the cathetus which doubles.
In a similar way, if we take an interest in the right part,
we have doubled the height of the truss and we have kept the same vertical forces,
this triangle becomes much smaller, actually half size,
it is classical, we go from 10.1 to 5.1N, it is a rounded 5.05,
then we decrease the internal forces in the horizontal elements
by a factor 2.
However, for the diagonal elements,
the internal force goes from 20.2 to 18.2, in compression, reciprocally.
As we had already seen it for the arch-cable
and for other types of structures,
the height that we give to a truss has a big influence.
If the figure becomes smaller, less high, the internal forces increase.
That means that we will use more matter.
If, conversely, the structures becomes higher,
the internal forces decrease and we will be able to save matter.
In this lecture,
we have seen that the truss with 5 nodes is statically determinate,
we have solved it
starting by a node on which there was support forces.
These support forces are independantely determined.
We have determined the internal forces for each of the bars of this truss
and we have compared them, on the one hand, to the ones
of the other types of structures which we had seen till here,
the arch-cable and the various types of trusses with 4 nodes,
and then we have carried out a variations study,
changing the height.
We have seen how this change of height
had repercussions on the Cremona diagram
and thus on the internal forces which act on the structure.
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