Surgery For Facial Flushing

COMBINATIONS for optimum digestion

Photo by Alessandra Cenci
Sometimes even after a meal of wholesome food may feel a sense of heaviness due to digestive problems. This can also occur in the absence of food allergies because not all foods are digested with the same times and in the same way. Inside of mouth, stomach and intestines are many enzymes. Enzymes are the proteins that act as biological catalysts allowing all biochemical reactions occur, for example, during digestion. Enzymes are very specific in the sense that not only some are present only in certain specialized cells, but are specific to certain foods: amylases such as are delegated the digestion of starch, lipase for digestion of fats and protease for protein digestion. Each of these enzymes to function at optimum level, it needs a particular environment that may be acidic or alkaline. The first digestion occurs in the mouth where the enzyme is present ptyalin for an initial carbohydrate digestion, so you should eat slowly and chew longer. Other enzymes critical for digestion of carbohydrates are found in the gut. Both in the mouth in the gut is an alkaline environment. In the stomach gastric juices will provide to create an acidic environment for the protein component of grains. The protein foods are digested in a rather acidic, so when chewing, which must always be accurate, there will be no digestion in the stomach but a specific enzyme pepsinogen converted to the active pepsin by the gastric juices will start the digestion of proteins. From all this it follows that for optimal digestion and assimilation of food perfect, it would be advisable not to combine with each other protein foods (meat, fish, legumes, eggs and cheese) with carbohydrate foods (grains, flour, pasta, bread and potatoes). By contrast, the combination of too many foods in addition to straining the digestive system would give rise to different toxins to eliminate a waste of energy. Even different kinds of protein foods should not be combined with each other. Le verdure sia cotte che crude possono essere associate a qualsiasi cibo. L’ideale sarebbe poter iniziare il pasto o la cena con un piatto di insalata fresca per fornire al corpo tutti gli oligoelementi e le vitamine necessarie per una digestione ottimale. La frutta non andrebbe mangiata a fine pasto perché potrebbe indurre fermentazione nell’intestino, mentre mangiata lontano dai pasti si digerisce rapidamente. I dolci sarebbe meglio mangiarli dopo un pasto di carboidrati. Alcuni abbinamenti possono essere più tollerati come farro e fagioli, riso e lenticchie, frutta acida al termine di un pasto proteico. Questa teoria del Dott. Shelton, pur conservando la sua validità è stata ultimamente anche un po’ rivista, nel senso che il succo di lemon over the meat and fish would favor the absorption of iron due to vitamin C. In conclusion then these rules are only indicative, the food must also be a pleasure, a moment of sharing with others, and serenity ... like I said in my initial post naturally healthy ... We need to nourish and nurture our next positive thoughts.

From: H. Shelton, "The simple combination of foods" Ed G. Manca
From the Street "food combinations" Red Ed
http:// www.vitalfitness.it / comb-ali.html
http://www.cibo360.it/alimentazione/dietologia/diete/combinazioni_alimentari.html

Cubefield 2 Descargar

hyperstatic: Antoni Gaudí i Cornet TRUSSES

  TRAVATURE IPERSTATICHE

Le travi continue rientrano nella categoria delle travi inflesse e sono caratterizzate dall'essere sostenute da più di due appoggi e dall'essere prive di sconnessioni interne. Si noti che gli appoggi sono tutti costituiti da carrelli, idonei quindi a consentire il libero esplicarsi delle dilatazioni assiali, tranne uno, in genere intermedio, che è fisso, per impedire gli spostamenti rigidi paralleli all'asse. Un esempio di trave continua è riportato nella figura 8.8.

Fig. 8.8. Trave continua

The continuous beam has a lot of uncertainty as there are intermediate supports. The degree of hyperstaticity is then given by n - 2 with n number of supports. It may be associated with any main system obtained eg. removing intermediate supports and replacing them with the reactions, as shown in Figure 8.9 where both n = 7 .

Figure 8.9. Mainline obtained by removing the supports abundant

This choice, however, may be adopted when the degree of hyperstaticity is low, while it is generally more convenient to introduce the kind of internal incoherence M = 0, ie the hinges.

Eg. In the case of continuous beam on four supports (see Fig. 8.10) can take the main structures shown in Figure 8.11. These patterns are linked to the name of the engineer. Bavarian GH Gerber so that he named.

Figure 8.10. Continuous beam on four supports

Figure 8.11. Examples of isostatic continuous beams, beams Gerber

8.6.2. L'equazione dei tre momenti

La scelta di struttura principale più conveniente risulta in genere quella ottenuta introducendo delle cerniere in corrispondenza degli appoggi. A questa scelta restano associati, come incognite iperstatiche, i momenti flettenti sugli appoggi, come indicato nella figura 8.12. Questa scelta risponde alla regola generale che consiglia di adottare delle strutture principali alle quali corrispondano delle configurazioni deformate le più vicine possibili a quella della struttura iperstatica alla quale sono associate.

Fig. 8.12. Esempio the main structure of the continuous beam on four supports.

It may be noted that in a system as the main transmission has chosen not to share a field to another so that, in the case of a continuous beam on n supports, (see fig. 8.8), to write the general equation of Mueller-Breslau, we can examine a field consists of two adjacent fields, as shown in Figure 8.13.

Figure 8.13. This main structure

, referring to the support m-th, where he made the disconnection M m = 0 , esprimerà la condizione di continuità della trave ossia la condizione che la linea elastica abbia una tangente unica e quindi una rotazione relativa nulla fra le due sezioni facenti capo alla sconnessione, cioè:

(8.14)

dove rappresenta la rotazione della sezione C m considerata appartenente alla campata di sinistra e la rotazione della medesima sezione pensata appartenente alla campata di destra. Naturalmente si tratta di rotazioni indotte dai carichi, dai cedimenti vincolari, dalle variazioni termiche e dalle incognite iperstatiche sulla struttura principale.
Trattandosi di travi inflesse, ed possono più agevolmente calcolarsi mediante il teorema di Mohr illustrato al paragrafo 7.2.5.
Supponendo che la trave continua abbia rigidezza costante per ogni campata, è facile vedere che ed possono calcolarsi come qui di seguito indicato.
Rotazioni indotte dai carichi:

(8.15)

Le reazioni e rappresentano le reazioni vincolari della trave ausiliaria sottoposta al carico fittizio p* costituito dal diagramma dei momenti flettenti. Le rotazioni ed sono perciò ottenute dividendo e rispettivamente per ed .

Le reazioni e , note come termini di carico , si possono calcolare, una volte per tutte, per le condizioni di carico più frequenti. Nella tabella seguente sono riportati i valori di e per alcune condizioni di carico.

Rotazioni indotte dalle incognite iperstatiche:

(8.16)

(8.17)

Rotazioni indotte dai cedimenti vincolari:


Se indichiamo con , ed i cedimenti, supposti anelastici, degli appoggi , e rispettivamente, posto





is:

(8.18)

Taking into account all the contributions, the matching equation (8.14) gives rise to the following expression:

(8.19)

The (8.15) assumes a particularly simple form when the beam is a stiffness constant that is when we can ask:

In this case, the fact (8.19) is simplified in the following:

(8.20)

The general equation (8.18), as well as (8.19 ) contains the unknown moments on three consecutive supports. It is therefore known as the equation of three moments. In the form (8.20) it is also known as equation Bertot - Clapeyron. From these equations we can write as many as the number of intermediate supports or in the degree of hyperstaticity. The first and the last equation contains only two unknown moments as is

They therefore take the form:

first equation:

(8.21)

last equation:

(8.22)

But if the details A and B are stuck, and the moments are statically indeterminate. In this case requires two additional equations that can be easily write by observing that if the joints A and B settlements are home to a corner and b respectively, for the two bays of the shore , you can write

(8.23)

would lead to the two additional equations are searching

(8.24)

is interesting to note that the equations (8.24) can also be obtained instead of thinking that each of the joints A and B there is a pair of supports are very close to each other, comprising a light field that is infinitesimal. It is thus possible to write an equation analogous to (8.20) namely:

putting in (8.20) can be arrived at:

while putting , we arrive at: both equations obtained coincide with (8.24).

We are thus able, in any case, writing a number of congruence equations equal to the number of unknowns. Solving this system of algebraic equations is reached values \u200b\u200bof bending moments at the supports. We are then able to calculate the support reactions and draw the diagrams of the stress characteristics M and T .

8.6.3. Joint reactions

The study of the continuous beam is completed with the calculation of reactions, easily obtained once we know the bending moments at the supports. In fact, with reference to Figure 8.14, where the load applied on the m-th span is, for simplicity, limited to a single concentrated force , the bending moment at the end of the span, that is:

Figure 8.14. Calculation of reaction forces

(8.25)

which implies the cut on the left of that span

(8.26 )

is clear that if the first is the support of the continuous beam then constrain the reaction coincides with . Otherwise it is easy to see that in a generic section S of abscissa z , cutting and bending moment are valid:

• If the section S is right load

(8.27)

• If the section S is to the left of the load


The binding reaction is easily obtained by noting that it follows from the first of the cut to the right of the span is given by:

(8.28)

While it can be deduced from Figure 8.14: