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Structures

It is possible to be shelp that a structure is a set of elements had such form that can bear a burden with stability. This is important, because  of what it is worth to us to make a skyscraper S.A. arrive a blow from air, one falls to us? Examples of structures are, aside from the building, a chair, a bridge, the towers of electricity and even the shell of the turtle. The structures must be designed to bear the burdens that are going to contain, this is their own weight but the variable weight. In the case of the bridge, the own weight is the weight of the elements that compose it and the variable is the vehicles that circulate around the same. There are times that appear sudden forces as winds or earthworks.

## A good architect must consider these factors at the time of his design.

Types of efforts The forces that can appear in a structure can be:Tracció

n.  It happens when they appear forces that they have to extend to an element of structure. A clear example is the cable that supports the weight of a bridge. On the one hand it is the beam that supports the highway and on the other the pillar to which it is moored Compression. In this case, the force must to diminish the length of the element. Example, the leg of a chair Flexion. The force appears perpendicular to the element of the structure. On the road of the bridge, when happening a truck, exerts a force downwards, trying to double to him. Torsion. In this case, the force that appears is to circulate and lies down to twist the material. An example we have it in the key of a lock or a screwdriver  Shears. The forces are parallel, of opposite sense and perpendiculars to the element

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Ejercicio1 In the swing of the image, to say to what type of effort is put under the different metallic bars, the cords of the swing and the seat that the weight of the boy supports, Exercise 2: To name 2 examples where they appear each one of the studied efforts. Example, traction in the chain of the anchor of a boat.

## Exercise 3. In the poster of the figure, to make a study indicating the efforts which they are put under each element, whereas clause that can have important winds and that they can strike in the end of the poster

Elements of one structure

1. From a bridge to a skyscraper we found that all structure has a series of elements that are repeated almost always in all of them. We will happen to comment them
2. Foundations.  Where it falls all the weight. Usually they are formed by a framework of irons and concrete and serves basic to support all the weight of the structure. The form and the size depend on what is going to support. He is not the same for a bridge on the sea that stops a house.Columns
3. .  Vertical elements that starting off of the foundations lift the structure. In a building, those of down are but wide that those of above, because as we raised plants, the weight that it supports is smaller
4. Beams. Seemed to the columns, but in this case they are placed horizontally. The beams that there are between two beams joist is called
5. Arcs. Element very used in the antiquity, allows to support a weight of the structure “unloading it” on the ends. We see an example of our bridge of Alcantara

Braces. They are special steel cables that support a great tensile stress. Usually one is used much in the bridges

## Arcs. As we saw before, the arc is used to both support a horizontal element “unloading” this weight on lateral pillars. Up to here all good, the mystery is in which each stone that composes the arc is loose. Then, because one does not fall. The answer is in the form of the stones, that are not rectangular blocks, but has the sides of different size and of this form when positioning itself on the arc, the greater side is placed in the part superior and the minor in the part of under the arc. To the stone superior key stone is called to him. We will see in the following illustration this detail. The “key” of this key stone is in how it transmits the vertical force to other horizontals. We can see in this detail

Stable and Unstable structures.

Tenth that a structure is stable when before any sudden effort (it pushes) it recovers his original position. Cristina Sierra Moreno (1º Baccalaureate 2012) has done us a work with the GIMP to show this concept. We can see it next.

1.  In the doll of the left we see that when pushing one does not fall it. He is very stable and its stability is due to two things
2. The base allows to reclaim its initial state The center of gravity has it very low

## The center of gravity is a virtual point (imaginary) where this all the mass of the body assumes that. That is to say, from the point of view of the gravity, it is as if it threw of that point. If a car that has motor falls to the water, the entry point is by the hood, since most of the weight this in the front part (motor). If we found the center of gravity and “we held it” with a ball-point pen, then, the object does not fall (to see image)

Profiles

Some of the elements of the structures as the beams have very characteristic forms. To this form Profile is denominated to him. We can make this profile as massive squaring of iron so that it has supported the weight, but, There will be no another way to use less material and than it has supported the same weight. The mechanical engineers already considered this subject and discovered that Forma of the profile is determining, that is that to that massive bar we cleared the iron to him of the interior and have a hollow square bar with fabulous results. We show some of the profiles but used in the construction first of them she is square hollow one, the second double T (also call H), the third T and the fourth L. Are many others, perhaps but these but are used of all.

Forces.

In order to finish this subject we are going to study a little how the forces in the simple elements act but, as a beam, to determine as the effort in the same is distributed. We put an example. If we must in center support a weight of a bridge of 1000 kg, and the bridge is supported by two pillars, What effort supports each pillar. Answer: As the weight is the same of each pillar and since there are two, the effort in each is 500 kg We will see other complicated situations a “bit” but.

Exercise. To calculate that effort supports to support X and the support and if the force of 1000 Nw is to 1 meters of X. Solution: a) is fulfilled that the forces upwards are equal that the forces downwards and therefore, 1000 = Fx + Fy b) If we cleared the strongpoint X, then the bar would fall by the left. We must hold it so that one does not fall. If we held it right under the force of 1000 Nw, the force necessary to avoid that it falls would be also of 1000 Nw. If we moved away of the point to the left we see that the force that we must use decreases. That is to say, to but far, less force to avoid than falls. Therefore, and applying the law of the handle, we have: But since there are displaced 1 meter to the left, the point and also is going to work, and therefore we must again make the calculations for this point. Good, as we at the outset shelp that the sum of the forces is 1000 we have Fy is going to be 200 Nw, but would be good that you make the calculations to verify it. Exercise to 2.Calcular as the points green and red of the image are distributed to the weight. Solution: Force in Red point (right) = 2500 KgFuerza in the green point = 500 kg Exercise 3. To calculate the weight that falls on the supports if the forces that act on the bar are 100 and 200 kg. Solution: Force in the support of the right: 120+ 30 = 150 KGEn the other strongpoint fall the rest of load 300 - 150, that is to say, kg = 150 kg. Exercise 4.  We have made two supports with two different plastic types. One of the supports holds a maximum weight of 200 kg and the other of 100 kg. The length of the bar is of 10 meters. To solve the following assumptions. a) Compute the range to the left support if the applied force is of 400 kg. b) To calculate the maximum force that can support in case the distance to the left support is of 2 meters. Solution. If you do not find how to do it, I explain it to you in the section of Solutions (down)

Making better structures. Viva el Triángulo

As doing resistant to the structures? Good, better he would be to become the question  How to make resistant structures to the smaller price. All we can make a resistant building incorporating some enormous columns with some very deep foundations, but the price would be very high.

The idea is to make the structure so that with the possible material minor, it has supported to the weight for which have been designed. Here it takes part of determining form, the form of the structure. We will see one of them, the triangle.

If we do squaring with bars of metal and we united them some to others with screws, we can make the experiment apply forces in the vertices. Resultado-> the sides yield. If we do the same with a triangle (we used less bars and less material) the result is well different. We have a geometric form that supports the efforts very well that they have to deform it. In this image is a detail of the suspension bridge of portugalete (Biscay) where aside from several braces, we can observe that the structure is very triangularizada, that is to say, exist many triangular forms in the same. We can make the same observation in the Eiffel tower, buildings and other structures and will see the same. Triangles. Because not squared. We will see the reason. We will see it better in this example.

In both examples we have applied a force, first a squaring and soon to a pentagon. We have the same result, a deformation of the original form. If we happened to triangulizar and we put the green element, the result is two triangles where there was squaring and the same of 2º example (in this case 3 triangles). If we applied the external force again, the new structure will support the effort far better. Exercises. In the following elements, it comments if he is stable and as it is possible to be done

It is worth the trouble, to finish this section, to see this video where the constructive process of a bridge is on the Edge for the AVE. A wonder. http://vimeo.com/67119686

Solutions:

Section a). Simply it does not have solution, since when putting a load of 400 kg, some of the supports always is going to have a greater load for the one than they have been designed. The total of the supports is of 200 + 100 = 300 kg, and we want to him to put a load of 400.

b) To calculate the maximum force that can support in case the distance to the left support is of 2 meters. In this case, we have:

Force X Distance to the opposite strongpoint/total Distance = Force that supports the support;

F X 8/10 = 200; F = 200 Xs 10/8 = 250 kg

We will verify that the red support does not have a greater load of 100 kg;

250 x2/10 = 500/10 = 50 kg