text:
Interesting social sciences/Jurisprudence

https://www.youtube.com/watch?v=AMCYrKAh8-I&ab_channel=%D0%92%D0%B0%D0%BB%D0%B5%D1%80%D0%B8%D0%B9%D0%A1%D1%82%D0%B0%D1%80%D0%B8%D0%BA%D0%BE%D0%B2

The causes of criminality.

English subtitles exist in this video. On the cover is a thief named Yaponchik (Vyacheslav Ivankov).

https://www.youtube.com/watch?v=gHjD4w2Msps&ab_channel=%D0%92%D0%B0%D0%BB%D0%B5%D1%80%D0%B8%D0%B9%D0%A1%D1%82%D0%B0%D1%80%D0%B8%D0%BA%D0%BE%D0%B2

Michel Foucault on prison reform.

English subtitles exist in this video. Valery Starikov Interesting social studies. On the cover is a still from the film "Gentlemen of Fortune."

The law is a system established by the State of norms and rules of behavior, fulfillment of which is ensured by force of State’s coercion. State-enforced laws are norms and the rules of behavior. Branches of criminal justice system (police, court, prison) must apply by sanctions to the culprit in case of violation of law. Corporal punishments (tortures and executions) served such legal sanctions in ancient times usually. The execution were replaced by life imprisonment in prison in modern democratic society often, and tortures were replaced by imprisonment for the term limited to the law, tortures were replaced by fine, confiscation of property, compensation of damage to the injured party and so on. But, nevertheless, law were built at the armed control of State over their citizens with antiquity before our days most frequently. Legal norms make behavior of people to a predictable, this is the function and the benefits of the law for society.

Legal norms should not be contradictory each to other; otherwise, citizens may find themselves in a difficult situation and citizens will not know what law they should perform. The State must to keep public order, to protect right to private property and free competition. The state has to stop rigidly all attempts of criminals to disturb public order. The state has to stop rigidly all attempts of criminals to seize somebody else’s property unlawfully. The state has to stop rigidly all attempts of criminals to prevent  to other citizens to be engaged freely in business activity. People enter the relations in society inevitably, people sign economic contracts with each other, at the same time the state is a guarantor of the fact that both parties have to satisfy conditions of this contract that the violator of terms of the contract will be punished and this violator have to compensate losses to the injured party. All citizens shall have the identical rights. All citizens shall to keep law irrespectively of social status of these citizens. Even high-ranking officials and rulers must keep the law and even rulers shouldn't allow of cases of an arbitrariness and lawlessness in own activity. Only published laws should apply, otherwise, citizens will not know the laws for which they may be punished and the power will receive the right to arbitrariness. If a constitution exists in the country that even the ruler is obliged to keep of the norms of this constitution.

The state-enforced laws as norms of behavior.
The state-enforced laws are adopted by the supreme bodies of the government. Laws are adopted by the king or the monarch with help of individual decision under tyranny, the kingdom or a hereditary monarchy, in such a way, for example, the king Hammurabi has accepted for the first time the code of laws known in the history. The state-enforced laws are adopted under the aristocracy or oligarchy by the senate, a Boyar Duma, the Central Committee or the Politburo of the ruling Communist Party with help of vote of members of this ruling privileged meeting. The state-enforced laws are adopted by vote in people's assembly under direct democracy. The state-enforced laws are adopted by parliament in the conditions of representative democracy. A hierarchy of the state-enforced laws and codes exist today. The code is a great number of the the state-enforced laws describing norms of behavior in some sphere of activity - in economy, the family sphere and so on. The most important the state-enforced laws are united in the constitution. The state-enforced laws from other codes shouldn't contradict to constitutional norms. The state-enforced laws from other codes become invalid according to the decision of the constitutional court in that case when the state-enforced law from the code is contrary to the norms of the constitution. Along with the state-enforced laws, there are also other legal acts subordinated to laws: decrees of the president, government regulations, instruction, rule, orders and so on. That to live in society, that to make success in affairs, the person has to study the state-enforced laws.

'''The causes of criminality. '''    Many viewpoints exist on the causes of criminality in science.

Neil Smelser has allocated three approaches in decision of this difficult problem:

 * biological approach,

 * psychological approach,

The essence of biological approach consist in the following:  the born criminal has the inclination to criminal behavior, the born criminal was born by thief and savage, the born criminal is able to do nothing else, a born criminal can only steal, rob and kill. Caesare Lombroso claimed within this approach that criminals are a result of degradation to earlier stages of human evolution. Cesare Lombroso named such external features of the born criminal, as the protruding lower jaw, rare beard and reduced sensitivity to pain. By the way, in my opinion, these features are most characteristic for speakers on personality type. Similarity of criminal with primitive savage is shown in the passion to tattoo, the weakness of mind and even in degradation of the handwriting similar to hieroglyphic writing. Caesar Lombroso has allocated four types of criminals: murderer, thief, rapist and swindler. And this typology remains to this day.
 * sociological approach.

According to Henry H. Goddard, criminals are mentally retarded men, occurring from hereditarily burdened families. My experience of school teaching of criminals on lessons of history in women's prison shows that criminals have different levels of development of intellect like ordinary people from very low level of development of intellect before very high level of development of intellect. Very high level of development of intellect was discovered at one thief‐women which received trust of men. she drank of men by vodka with sleeping pills and she robbed of men when men were sleeping. This thief-women has remember each my word in my lesson on history, each date, each name of hero of history, it is possible, the best memory is necessary feature of character of “thief on trust”. Level of development of intellect affects to degree of the success of growth for thief’s qualifications.

Stages for such growth for thief’s qualifications exist:

• Specialization on apartment's burglaries.

• Specialization on pocket thefts.

• Theft is based on trust of victim to criminal.

Thus, the opinion of Henry H. Goddard seems not entirely truth.

From the point of view of the theory of "criminal psychodynamics", the criminal in own acts is guided not by mind but only emotions, the criminal is a psychopathic personality who has deviates from the norm, but the criminal isn't neither mentally sick, nor mentally retarded personality.

Ernst Kretschmer have given the description of three main body types of man:

• asthenic (tall and thin);

• pyknic (stocky, fat),

• athletic (with advanced musculature);

An athlete is person with the developed musculature, inclined to the organization of conflict.

A pyknic is person, inclined to obesity, which attempts to settle conflict with help of peaceful way.

Asthenic is thin person, which is potential victim in the conflict. William Herbert Sheldon considered that the athletics are most inclined to the crimes.

The genetic explanation of the reasons of crime consists that some criminals have chromosomes like XYY whereas ordinary men have chromosomes like XY, and ordinary women have chromosomes like XX. These people with chromosomes like XYY have frightening appearance, high growth and this people are heavy psychopaths sometimes.

Sigmund Freud has entered concept "criminals with sense of guilt" within the framework of psychological approach, this criminals wish that they were caught and punished because this criminals feel own guilty because of their "inclination to destruction," this criminals believe that imprisonment will help them overcome this inclination.

The essence of sociological approach consists that the main reason of crime is that thieves are brought up in criminal group purposefully. Edwin Sutherland claimed that the prison corrects nobody, and prison is school of training in criminal behavior often.

Émile Durkheim considered that the social anomie or, in other words, lawlessness, is the main reason for criminality in the period of social crisis. A similar period was in Russia during government of Mikhail Gorbachev and Boris Eltsin. The a main goal of Vladimir Putin became the establishing of order in the country, that to drive of criminal world into an underground. The condition of an anomy is observed in big cities among teenagers where there is a mixture of various ethnic groups to the different systems of values.

Robert King Merton considered that the gap between the purposes and the approved means of their achievement is the reason of crime, for example, the purpose is financial success, and means is receiving the higher education, but paid education is often inaccessible today therefore some people can use illegal ways of enrichment.

According to the theory of stigmatization of Howard S. Becker, influential groups are capable to put stigma of criminal on members of less influential groups and thus influential groups train of members of less influential groups to consider themselves by criminals. For example, whites did with African Americans in the United States this way.

From the point of view of Marxism and radical criminologists, criminals are rebels against capitalist exploitation. The example of such "rebel" is Joseph Stalin who had 6 criminal records for extortion and 5 escapes from exile. It is possible to draw a conclusion that all these theories are partly true and crime is a consequence of many causes.

An empirical study of the causes of criminality.
My preliminary forecast before the empirical researches consisted that a potential criminal is the speaker. This forecast has found partial confirmation, but it has turned out that speakers are capable besides that to show resourcefulness, speakers know how to engage other personality types in a criminal group, speakers are able to use of other personality types as performers of own criminal plans, then speakers are capable to avoid criminal punishment and speakers are capable to shift all fault for crimes on other personality types. There is a specialization in the criminal world in various roles, such as organizer, executor, helper and instigator. Thief in law in Russia is an example of the organizer of the high qualification. Skilled in thieves' affairs speaker is capable to play a role of the organizer. The inexperienced speaker or the technician approach for a role of executor.

A principle: "You mustn't to be by a executor!" exists in criminal world, because executors are arrested by first of all then the executor is obliged according to thieves' laws to take all fault on himself and executor mustn't betray of other members of criminal group, otherwise executor can be killed in prison. The role of the executor is the most unprofitable in comparison with other roles.

The psychologist and the speaker can undertake role of the instigator, at the same time the speaker forces of executor to commit to crime by means of threats, and the psychologist persuades by executor to commit to crime by means of deception and cunning, thus, the speaker and psychologist use different coercion methods. The role of the helper is carried out different the personality types, for example the technician is capable to make high-quality weapon, the speaker is capable to sell stolen property. Psychologist and speaker are able to provide information about candidates of potential victims of robberies and about their property status at apartment's burglaries. The psychologist knows how to rub into the trust to potential victims.

By the way, my research has shown that, unlike rules in men's prisons, women having the status of thiefs in law do not exist in women's corrective labor colony of strict regimen and high hierarchy doesn't exist among the prisoner-women. The speaker and the psychologist approach for a role of the drug dealer. The technician can play a role of the cruel killer-executioner or killer-sniper because of inclination of the technician to insensibility and low emotionality.

The speaker has the highest inclination to be by leader of criminal world, to make all kinds of serious crimes and to improve thieves' skills, speaker has inclination to adventurism and frivolous behavior, readiness to submit to pressure of criminal group and inability to foresee heavy consequences of own crimes. This inclination of the speaker to criminal behavior is shown at young age and in state of intoxication especially brightly. The psychologist is the potential victim of crime often because the psychologist provoke of criminals involuntarily thanks to peculiar for psychologists weakness of character. A woman-psychologist provokes involuntarily of a rapist to rape due to the fact that the  woman-psychologist has the highest degree of sexuality.

This my research was conducted on the basis of women's corrective labor colony of strict regimen of UT 389/28 in town Berezniki during the period from November 2003 to March 2004. The status of colony was changed at the end of this period – the strict regimen was changed on a ordinary regimen. The informal unstructured interview (35 respondents) and the included observation in the course of training of prisoner in 8 and 9 classes of middle school (15 respondents) became forms of carrying out a research. Total, 50 prisoner-women were interviewed and subjected to observation by means of casual selection. 40% were by psychologists, 20% were by technicians, 40% were speakers from this prisoner-women, theorists were not found among prisoner-women. That to appear in this colony and to conduct a research, I had to get a job of the teacher of history in this prison's school. The generalized results of my research are given in the table given below:

It is possible to draw the following conclusions on the basis of these results of a research.

Speakers have the highest status in colony. Speakers have the biggest inclination to group crimes though all prisoner-women have noted that prisoner-women executed orders of the man-recidivist in criminal group. Re-education of the speaker is the most difficult task that finds manifestation that the speaker has the most large average amount of criminal records on one prisoner among the personality types. The task of re-education of the speaker is complicated by those circumstances that speaker has the worst marital status – speaker marries more rare, speaker has of children less often than other personality types, speaker has high inclination to the use of drugs, to alcoholism and high probability to ache by HIV.

Psychologists specialize in crimes on sale of drugs and apartment's burglaries. Psychologists have high inclination to fall into a state of passion and psychologists have high inclination to finish off victim in the course of causing grievous bodily harm, but psychologists have low inclination to robbery. Psychologists are most of all inclined to commit of crimes alone rather than in a group. The task of re-education of the psychologist is complicated by his inclination to the use of drugs, high probability to ache by HIV. Task of re-education of the psychologist is facilitated by a relatively good marital situation – psychologist marries more often and psychologist has children more often than others personality types.

Technicians are inclined to commission of crimes much less than other personality types. Technicians make only 20% of prisoners. But technicians have high inclination to commission of deliberate and unintentional domestic murders (malice aforethought murders and manslaughter), technicians are capable to kill as the executioner and technicians are capable to be cold insensitive to the death suffering of the victim. Process of re-education of technicians is facilitated by low inclination to the use of drugs and quite good marital status. 1 technician and 1 psychologist have two criminal records for murder each that is an example of unique cruelty.

All women of Romani people (5 women) were psychologists and women of Romani people specialize in sale of drugs – 60%, in thefts - 20% and in premeditated murders (malice aforethought) - 20%.

A knife (84% of cases), the axe (8% of cases), the noose of (4% of cases}, a rolling pin (4% of cases) have served as the crime tool in 24 cases of women crimes.

The question about necessity of the capital punishment

Punishment for the crimes connected with use of violence became much more strict under the new Russian criminal code. The top limits of punishment are raised from 15 to 20 years and to life imprisonment. Russian Federation keeps of conditions of the moratorium on the capital punishment, though such punishment as the capital punishment is prescribed in the criminal code for the most serious crimes, but courts replace of capital punishment to life imprisonment. It is connected with fear to make fatal mistake of justice when a theoretical opportunity remains to sentence of innocent person to death penalty. It is impossible already to return to life of the innocent person in case of detection of such mistake of justice after enforcement of sentence about capital punishment. For example, it has become clear during the investigation of criminal case of the serial killer Andrei Chikatilo who has killed about 50 people that the court has sentenced to capital punishment for one of these murders of innocent person who was executed already.

According to sociological polls, most of residents of Russia are sure of need of the capital punishment, especially in connection with threat of the Chechen terrorism. Most of residents of Russia are sure of need of application of the capital punishment to those terrorists who blew up houses in Moscow, Volgodonsk and Buynaksk. Most of residents of Russia are sure of need of application of the capital punishment to those terrorists who cut off heads of russian prisoner of war in Chechnya. To abolish of capital punishment is absurdity during war.

Region of Upper river Kama (the cities of Solikamsk, Berezniki, Krasnovishersk, Kizel) was turned into the prison region at the Soviet power. One of leaders of the Chechen terrorists – Salman Raduyev – was sentenced to life imprisonment and Salman Raduyev served sentence in Solikamsk supermax prison "White Swan". Salman Raduyev said in one of an interview that Salman Raduyev hopes to serve sentence in supermax prison 25 years, to receive amnesty and to be released, but Salman Raduyev has died a year later on December 14, 2002 in hospital from old wounds. To survive in prison is very difficult for the former terrorist. Prisoners and security guards put pressure upon terrorist also.

Almost all squad of the Perm OMON (Special Purpose Police Unit) – 43 militiamen (from them 32 were killed in battle, and 11 were captured and executed with special cruelty) has died in one of the ambushes made by the Chechen terrorists. This event has taken place on March 29, 2000. It is possible to understand hostility of security guards to the condemned terrorists in these conditions. Many leaders of the Chechen terrorists – Dzhokhar Dudayev, Shamil Basayev, Ibn al-Khattab, Aslan Maskhadov was killed with help of special operations. It is possible to abolish of capital punishment in civilized Europe, but it is impossible to abolish of capital punishment in criminal Russia. The capital punishment remains even in the USA. When the prisoner was sentenced to capital punishment, the prisoner can choose a method of execution in the USA – a lethal injection, an electrocution, firing squad, hanging or gas inhalation, and relatives of the victim have an opportunity to watch for this procedure of an execution with help of cable television.

Michel Foucault about the prison’s reform
Prison’s reform

Michel Foucault has described a cruel execution of Robert-François Damiens on March 2, 1757 at the beginning of own book "Discipline and Punish: The Birth of the Prison". Prisons in modern understanding have appeared eighty years later after this execution, and a daily routine was drawn up in prison for young prisoners already. Prisoners were obliged to work 9 hours in day in workshops, prisoners were obliged to study 2 hours in day in prison school. So, public execution were complemented by imprisonment and a prison daily routine. Public execution and imprisonment became different types of punishment in Europe and America. The prison was just the place earlier where the criminal was interrogated and tortured, and then criminal waited an execution here. The prison became the place of life and work for the prisoner after reform according to strict routine where the prisoner spent a long term of imprisonment. These punishments have become applicable for different kinds of crime and these punishments have become applicable to different types of criminals. This reform was carried out in period just under 100 years. During this time, new codes were drawn up that provided for a new types of punishment: in Russia – in 1769, in Prussia – in 1780, In Pennsylvania and Tuscany – in 1786, in Austria – 1788, in France – in 1791, 1808 and 1810.

The essence of prison’s reform (since the late 18th century until the early 19th century) consisted in following:

 * Public executions and tortures were cancelled,

 * Accurate codes, uniform rules of legal proceedings, jury trial were introduced,

 * The corrective character of the penalty were introduced, assignment of various terms of imprisonment were introduced, kinds of punishment depending on weight of crime were introduced. Only flagellation  remain still some time in Russia, Austria and Prussia.

 * The human branding disappeared on the criminal's face or shoulder. A certain caution has arisen in the art of inflicting pain. The body disappeared as the major target of punishment.

 * Amende honorable and pillory were cancelled.

 * Use of convicts at public works was cancelled – at repair of roads when convicts were chained in shackles and iron collars and convicts exchanged with crowd by curses in response on contempt from this crowd.

 * Chain-gang were cancelled. While the chain-gang, which had dragged convicts across the whole of France, as far as Brest and Toulon was replaced in 1837 by inconspicuous black-painted cell-carts. This cell-carts was prison on wheels. The central corridor divides of cell-carts on all length, six single cells exist on each side of the corridor where prisoners sit facing a corridor. Legs and hands of criminals are chained by a chain.  An iron door with a small window was between a corridor and cell that prevents to communication between prisoners on the way.  There are two security guards armed by clubs in a corridor. Punishment had gradually ceased to be a spectacle, a rite and a show for gapers gradually.

Reasons of the abolition of public executions:

 * The execution exceed in savagery the crime itself.

 * The execution accustomed of spectators to cruelty whereas execution should wean of spectators from cruelty.

 * The execution makes of the executioner resemble a criminal. The execution makes judges resemble murderers.

 * The execution caused of admiration and pity to the executed criminal instead of contempt and hatred.

Conclusion: Punishment becomes the most hidden part of the penal process gradually. Effectiveness of punishment is seen as resulting from its inevitability, not from its visible intensity.

Powers began to apply imprisonment, confinement, forced labour, penal servitude, prohibition from entering certain areas, deportation instead of tortures and executions during modern times. Imprisonment is connected somewhat with physical suffering too unlike penalties, but intolerable corporal pain is not connected with criminal penalty any more already. The whole army of warders, doctors, chaplains, psychiatrists, psychologists, educationalists comes to replace the executioner as to the specialist in causing pain.

The shortcomings of modern imprisonment:

 * Imprisonment has no effect on the public.

 * Imprisonment is incapable of corresponding to the specificity of crimes.

 * Imprisonment is costly.

 * Imprisonment maintains of convicts in idleness. Imprisonment multiplies their vices.

 * Work of the jailer is an exercise in tyranny.

Panopticon
Possibility of the round-the-clock supervision of prisoners is here a main goal of construction of Panopticon prison according to the plan of Jeremy Bentham. This prison represents the round or crosswise building, for example, Kresty Prison in St. Petersburg represents several crosswise buildings. The central tower is in the center of such building where a supervisor settles down. A supervisor has an opportunity to observe through wide windows for life of prisoners in cells which are located around the circumference of the building as in an aquarium or in a zoo. Two windows exist in each cell, one window comes to light, the second big window leaves inside therefore the cell is looked through. Only one criminal has to sit in each cell that to avoid by bad influence from prisoners at each other. Walls between cell prevent to communication between prisoners. Therefore there is no of danger of a plot, planning of collective escape and the planning of new crimes for the future. This structure of the building of prison is opposite to the principle of a dungeon. Constant supervision eliminates the need for corporal punishment to some extent too. Modern technical means allow to provide constant supervision over prisoners in cells not by means of a special structure of the building of prison, and by means of video supervision. Such projects of prisons as Panopticon are still too expensive for Russia. The available prisons are overcrowded beyond any measure, closeness and insanitary conditions exist in cells in Russia today.

Two systems of keeping of criminals in prison in the West
Powers in the West wanted to interfere with association of criminals in prison in criminal community.

Two systems of keeping of criminals in prison were invented for this purpose in the USA:

 * Auburn system in  the 19th century which prescribes keeping  of the prisoners in  individual cell at night,  work and meals in common in the afternoon, as in the monastery, but on condition of absolute silence. Prisoners can speak only with warders.

 * Pennsylvania system (Separate system) provides keeping of prisoners in the conditions of absolute isolation from each other in the solitary cell that to leave of criminal alone with own conscience.

There were discussions about an opportunity to pay of work of prisoners and a possibility of the prisoner to refuse from some kinds of works and to organize of strike.

Arguments of opponents of prison reform and of replacement of executions and torture to imprisonment:

 * The prison organizes training of inexperienced criminals to "thieves' professions" and the prison organizes transformation of inexperienced criminals into professional criminals.

 * Criminals plan future crimes in prison, сriminals plan organization of escapes from prison and сriminals plan collective prison revolts.

 * Prisons don't diminish the crime rate. Prisons can be extended, multiplied or transformed, but the quantity of crime and criminals remains stable or, worse, increases.

 * Prison generates of recidivists, those criminals leaving prison have more chance than before of going back to prison. Existence in prison is lifestyle for recidivists. Recidivists get used to not taking care about food and housing, job and livelihood in prison.

 * The prison cannot fail to produce of delinquents. The prison produces more and more criminals, rather than correcting of criminals, whether criminals are in solitary cell or criminals perform useless work for them. Between 1,000 and 1,500 convicts live under the surveillance of between thirty and forty supervisors, who can preserve some kind of security only by depending on informers.

 * The prison makes possible, even encourages, the organization of a milieu of delinquents, loyal to one another, hierarchized, ready to aid and abet any future criminal act. The prison gives to "thieves' profession and the prison learns to be the informer. Сriminals find themselves after release from prison in such conditions which doom of criminals to crime repetition. Criminals lose residence during the inprisonment and criminals become homeless tramps after an exit from prison. Imprisonment dooms of family of the criminal to poverty after detention of the only breadwinner. As a result the son of the thief becomes by thief too frequently  that leads to new crimes eventually.

Conclusions: In my opinion, the prison has many shortcomings, but the prison can't be cancelled. It is impossible to return to tortures and executions because such return contradicts to the principles of humanity. Besides, corporal punishments remain in prison system today to some extent. Corporal punishments are the great invention of mankind too and corporal punishments can't be cancelled completely too because such people exist on which only corporal punishments act. It is impossible to replace corporal punishments on administrative punishments completely, that is it is impossible to replace corporal punishments on conversion of the person from the status  "free person" to status of "prisoner". The attempt to replace corporal punishments and an executions on administrative punishments (imprisonment) was not quite successful because this attempt has led to emergence of new problems which weren't earlier. If Gleb Zheglov from the movie "The Meeting Place Cannot Be Changed" was guided the principle that "the thief has to be imprisoned" at any cost, then the judge was guided earlier by the principle: the place of recidivists – on the gallows at any cost, because it was possible to reduce number of recidivists with help of executions, it was possible to destroy criminal hierarchy with help of an executions of leaders of criminal world,  it was possible to interfere to process of training of the inexperienced criminals to thieves' professions because executions reduced number  of similar "teachers" really. The execution of notorious villains allows to reduce costs on keeping of criminals to a minimum. Besides, the execution of terrorists and maniacs allows to satisfy sense of justice at relatives of the victims on the ancient law of retaliation – "Eye for an eye", these relatives are forced to commit murder of the criminal from revenge and  these relatives are forced to be exposed to criminal prosecution today sometimes. This example shows that cancellation of any social invention, even corporal punishments and the capital punishment leads to emergence of big problems always.


text: Category:Book:Overview of Elasticity of Materials

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Overview of Elasticity of Materials/Example Problems

Problem Statement:
Given a point on an elastic body with stress state $\boldsymbol{\sigma}$, determine the principal stresses and the angle between the original orientation and the orientation with maximum shear stress. Solve this problem both using the analytical equations and using Mohr's circle construction. Stresses are given in MPa.

\boldsymbol{\sigma}= \begin{pmatrix} 72 & 21 & 0 \\ 21 & 32 & 0 \\ 0 & 0  & 0 \end{pmatrix} $$

Solution:
We're going to solve this by two methods. First using the analytic solution then using Mohr's circle.

Analytic Solution
We have the equations
 * $$\begin{align}

\sigma'_{11}=\frac{\sigma_{11}+\sigma_{22}}{2}+\frac{\sigma_{11}-\sigma_{22}}{2}cos\left ( 2\theta \right )+\sigma_{12}sin\left ( 2\theta \right ) \\

\sigma'_{22}=\frac{\sigma_{11}+\sigma_{22}}{2}-\frac{\sigma_{11}-\sigma_{22}}{2}cos\left ( 2\theta \right )-\sigma_{12}sin\left ( 2\theta \right ) \end{align}$$ and
 * $$\begin{align}

\sigma'_{12}=\frac{\sigma_{22}-\sigma_{11}}{2}sin\left ( 2\theta \right )+\sigma_{12}cos\left ( 2\theta \right ) \end{align}$$

where the prime stresses are in the rotated reference frame. When $$\sigma'_{12}=0$$ the system is in its principal orientation and $$\sigma'_{11}=\sigma_{p1}$$ and $$\sigma'_{22}=\sigma_{p2}$$. Setting $$\sigma'_{12}=0$$ and solving yields

tan\left ( 2\theta \right )=\frac{2\sigma_{12}}{\sigma_{22}-\sigma_{11}} $$ and upon substituting the stresses given in this problem we find $$\theta=23.2$$°. Substituting this value of $$\theta$$ and the given values for $$\boldsymbol{\sigma}$$ yields $$\sigma'_{11}=\sigma_{p1}=81$$ MPa $$\sigma'_{22}=\sigma_{p2}=23$$ MPa and $$\sigma'_{12}=0$$ MPa, which verifies that this is indeed the principal axis.

If we are given a stress state that is the principal orientation then in our rotated reference frame above $$\boldsymbol{\sigma'}$$ the $$\sigma_{12}$$ term is zero. We find the direction with maximum shear by taking the derivative of $$\sigma'_{12}$$, setting it to zero, and solving for $$\theta$$.

\frac{d \sigma'_{12}}{d\theta}=\frac{d}{d\theta}\left(\frac{\sigma_{22}-\sigma_{11}}{2}sin\left ( 2\theta \right )\right)=\left ( \sigma_{22}-\sigma_{11} \right ) cos\left ( 2\theta \right ) $$ and the maximum occurs when $$\theta=45$$°. This means the maximum or minimum occurs at $$23.2 + 45 = 68.2$$°. Substituting this into our equations along with $$\boldsymbol{\sigma}$$ yields $$\sigma'_{11}=\sigma_{p1}=52$$ MPa, $$\sigma'_{22}=\sigma_{p2}=52$$ MPa and $$\sigma'_{12}=-29$$ MPa, the minimum. By the symmetry of the system rotating an additional $$90$$° yields the maximum, $$\sigma'_{11}=\sigma_{p1}=52$$ MPa, $$\sigma'_{22}=\sigma_{p2}=52$$ MPa, and $$\sigma'_{12}=29$$ MPa.

Mohr's Circle Solution
Following the diagram given here,

Radius=AB=BC=\sqrt{BD^2+CD^2} $$ and

tan\left ( 2\theta \right )=\frac{CD}{BD} $$ Given $$CD=21$$, $$OD=72$$, and $$OE=32$$, we concluded that $$AD=40$$ and $$EB=BD=20$$. Therefore, $$Radius=29$$ and $$\theta=23.2$$°.From the principal orientation$$P1=OB+Radius=OE+EB+Radius=81$$ MPa and $$P2=P1-2 Radius=23$$ MPa. The minimum and maximum shear stress is $$\pm Radius=\pm 29$$ MPa and are found by rotating the system $$23.2+45=68.2$$° and $$23.2 + 45 + 90 = 158.2 = -21.8^\circ$$.

Problem Statement:
Given a point on an elastic body with stress state $\boldsymbol{\sigma}$, determine the principal stresses. Stresses are given in MPa. [Hint: There is only one way to draw the Mohr's circle representation. Use this to simplify your work. You'll find that there is almost no math once you draw the picture.]

\boldsymbol{\sigma}= \begin{pmatrix} 0 & 6  & 8 \\ 6 &  0  & 10 \\ 8 &  10 & 0 \end{pmatrix} $$

Problem Statement:
Given a point on an elastic body with stress state $\boldsymbol{\sigma}$, and that rotating about $x_{3}$ yields stress state $\boldsymbol{\sigma'}$ , fully determine both stress states and the unknown parameter $q$. Stresses are given in MPa.

\boldsymbol{\sigma}= \begin{pmatrix} 25q & \sigma_{12} & 0 \\ \sigma_{21} & 5q & 0 \\ 0 & 0  & 0 \end{pmatrix} ~\boldsymbol{\sigma'}= \begin{pmatrix} 50 & 5 & 0 \\ 5 & \sigma'_{22} & 0 \\ 0 & 0  & 0 \end{pmatrix} $$

Solution:
We have the equations
 * $$\begin{align}

\sigma'_{11}=\frac{\sigma_{11}+\sigma_{22}}{2}+\frac{\sigma_{11}-\sigma_{22}}{2}cos\left ( 2\theta \right )+\sigma_{12}sin\left ( 2\theta \right ) \\

\sigma'_{22}=\frac{\sigma_{11}+\sigma_{22}}{2}-\frac{\sigma_{11}-\sigma_{22}}{2}cos\left ( 2\theta \right )-\sigma_{12}sin\left ( 2\theta \right ) \end{align}$$ and
 * $$\begin{align}

\sigma'_{12}=\frac{\sigma_{22}-\sigma_{11}}{2}sin\left ( 2\theta \right )+\sigma_{12}cos\left ( 2\theta \right ) \end{align}$$ where the prime stresses are in the rotated reference frame.

This immediately allows us to substitute for $$\sigma'_{12}$$ and $$\theta$$ to find $$q=-1/2$$. This give $$\sigma_{11}=-12.5$$ MPa and $$\sigma_{22}=-2.5$$. The invariant relation

\sigma_{11}+\sigma_{22}+\sigma_{33}=I_1 $$ allows

\sigma_{11}+\sigma_{22}+\sigma_{33}=\sigma'_{11}+\sigma'_{22}+\sigma'_{33} $$ with substitutions determines $$\sigma'_{22}=-65$$ MPa. At this point all the parameters are determined except $$\sigma_{12}$$ and simply substituting into either the $$\sigma'_{11}$$ or $$\sigma'_{22}$$ allows us to find $$\sigma_{12}=57.5$$ MPa. The resulting stress tensors, in units of MPa, are

\boldsymbol{\sigma}= \begin{pmatrix} -12.5 & 57.5 & 0 \\ 57.5 & -2.5 & 0 \\ 0 & 0  & 0 \end{pmatrix} ~\boldsymbol{\sigma'}= \begin{pmatrix} 50 & 5 & 0 \\ 5 & -65 & 0 \\ 0 & 0  & 0 \end{pmatrix} $$

The Mohr's circle representation of this solution is shown here. Note that the shaded triangles are similar and therefore can be used to simplify the solution if sought graphically.

Problem Statement:
Given a point on an elastic body with stress state $\boldsymbol{\sigma}$, determine the principal stresses and the angle between the original orientation and the orientation with maximum shear stress. Stresses are given in MPa.

\boldsymbol{\sigma}= \begin{pmatrix} 18 & 12 & 9 \\ 12 & 12 & -6 \\ 9 & -6  & 6 \end{pmatrix} $$

Problem Statement:
Both mathematically and with words explain the impact of applying the transformation tensors $\boldsymbol{T_1}$, $\boldsymbol{T_2}$ ,$\boldsymbol{T_3}$ , and $\boldsymbol{T_4}$ on vector $\boldsymbol{a}$. Apply the same transformation to rank two tensor $\boldsymbol{Z}$.

\boldsymbol{T_1}= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0  & 1 \end{pmatrix} ~ \boldsymbol{T_2}= \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0  & -1 \end{pmatrix} ~ \boldsymbol{T_3}= \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0  & -1 \end{pmatrix} ~ \boldsymbol{T_4}= \begin{pmatrix} cos\left(\theta\right) & 0 & sin\left(\theta\right) \\ 0 & 1 & 0 \\ -sin\left(\theta\right) & 0  & cos\left(\theta\right) \end{pmatrix} $$

\boldsymbol{a}= \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} ~ \boldsymbol{Z}= \begin{pmatrix} Z_{11} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{pmatrix} $$

Solution:
Remembering the definition of tensor multiplication and the implicit summations used in Einstein notation, we know that the rank two $$\boldsymbol{T}$$ tensor acting on the vector $$\boldsymbol{a}$$ will necessarily result in terms that look like

a'_2=a_1T_{11}+a_2T_{12}+a_3T_{13} $$ which is similar to "normal" matrix multiplication as you've seen before. In contrast applying the rank two transformation tensor on the rank two $$\boldsymbol{Z}$$ tensor requires a double sum which results in terms that look similar to

Z'_{12}=T_{11} T_{21} Z_{11} + T_{11} T_{22} Z_{12} + T_{11} T_{23} Z_{13} + T_{12} T_{21} Z_{21} + T_{12} T_{22} Z_{22} + T_{12} T_{23} Z_{23} + T_{13} T_{21} Z_{31} + T_{13} T_{22} Z_{32} + T_{13} T_{23} Z_{33} $$ When working with these type of 9-term sums it is usually advisable to use a software package to simply the work. The results are given here.

Transformation tensor 1 results in

\boldsymbol{a'}= \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} ~ \boldsymbol{Z'}= \begin{pmatrix} Z_{11} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{pmatrix} $$ which corresponds to the identity transformation, i.e., leave the tensors unmodified.

Transformation tensor 2 results in

\boldsymbol{a'}= \begin{pmatrix} -a_1 \\ -a_2 \\ -a_3 \end{pmatrix} ~ \boldsymbol{Z'}= \begin{pmatrix} Z_{11} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{pmatrix} $$ This transformation is the inversion transformation. This can be seen in the behavior of the vector. Interestingly enough it leaves $$\boldsymbol{Z}$$ unmodified.

Transformation tensor 3 results in

\boldsymbol{a'}= \begin{pmatrix} -a_1 \\ a_2 \\ -a_3 \end{pmatrix} ~ \boldsymbol{Z'}= \begin{pmatrix} Z_{11} & -Z_{12} & Z_{13} \\ -Z_{21} & Z_{22} & -Z_{23} \\ Z_{31} & -Z_{32} & Z_{33} \end{pmatrix} $$ This transformation involves mirroring across the $$x_1$$ and $$x_3$$ directions. This is equivalent to rotation around the $$x_2$$-axis by $$180$$°.

Transformation tensor 4 results in $$ \boldsymbol{a'}= \begin{pmatrix} a_1 cos\left(\theta\right) + a_3 sin\left(\theta\right) \\ a_2 \\ a3 cos\left(\theta\right) - a_1 sin\left(\theta\right) \end{pmatrix} ~ \boldsymbol{Z'}= \begin{pmatrix} Z_{11} cos\left(\theta\right)^2 + Z_{13} cos\left(\theta\right) sin\left(\theta\right) + Z_{31} cos\left(\theta\right) sin\left(\theta\right) + Z_{33} sin\left(\theta\right)^2 & Z_{12} cos\left(\theta\right) + Z_{32} sin\left(\theta\right) & Z_{13} cos\left(\theta\right)^2 - Z_{11} cos\left(\theta\right) sin\left(\theta\right) + Z_{33} cos\left(\theta\right) sin\left(\theta\right) - Z_{31} sin\left(\theta\right)^2 \\ Z_{21} cos\left(\theta\right) + Z_{23} sin\left(\theta\right) & Z_{22} & Z_{23} cos\left(\theta\right) - Z_{21} sin\left(\theta\right) \\ Z_{31} cos\left(\theta\right)^2 - Z_{11} cos\left(\theta\right) sin\left(\theta\right) + Z_{33} cos\left(\theta\right) sin\left(\theta\right) - Z_{13} sin\left(\theta\right)^2 & Z_{32} cos\left(\theta\right) - Z_{12} sin\left(\theta\right) & Z_{33} cos\left(\theta\right)^2 - Z_{13} cos\left(\theta\right) sin\left(\theta\right) - Z_{31} cos\left(\theta\right) sin\left(\theta\right) + Z_{11} sin\left(\theta\right)^2 \end{pmatrix} $$ This transformation involves rotation around the $$x_2$$-axis. Substituting $$180$$° for $$\theta$$° yields identical results to transformation tensor 3.

Problem Statement:
Consider the stress state $\boldsymbol{\sigma}$. Write a transformation tensor that rotates the reference frame into the principal directions. Stresses are given in MPa.

\boldsymbol{\sigma}= \begin{pmatrix} 72 & 21 & 0 \\ 21 & 32 & 0 \\ 0 & 0  & 0 \end{pmatrix} $$

Solution:
We know from Example 1 that the solution is to rotate by $$\theta=23.2$$ around the $$x_3$$ axis and we know the form of the rotation transformation matrix from Example 5, therefore the solution is

\boldsymbol{\sigma}= \begin{pmatrix} cos\left(\theta\right) & sin\left(\theta\right) & 0 \\ sin\left(\theta\right) & cos\left(\theta\right) & 0 \\ 0 & 0  & 0 \end{pmatrix} $$ where $$\theta=23.2$$.

Problem Statement:
Given the displacement tensor $\boldsymbol{e}$, identify the rotation tensor and strain tensor.

\boldsymbol{e}= \begin{pmatrix} 0.001 & 0.001 & 0.002 \\ -0.001 & 0.002 & 0.002 \\ 0.001 & 0.003  & 0.002 \end{pmatrix} $$

Solution:
We can break any tensor into a fully symmetric and fully anti-symmetric tensor resulting in an anti-symmetric rotation tensor

\boldsymbol{\omega}= \begin{pmatrix} 0 & \frac{e_{12}-e_{21}}{2} & \frac{e_{13}-e_{31}}{2} \\ \frac{e_{21}-e_{12}}{2} & 0 & \frac{e_{23}-e_{32}}{2} \\ \frac{e_{31}-e_{13}}{2} & \frac{e_{32}-e_{23}}{2}  & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0.001 & 0.0005 \\ -0.001 & 0 & -0.0005 \\ -0.0005 & 0.0005  & 0 \end{pmatrix} $$ and symmetric strain tensor

\boldsymbol{\varepsilon}= \begin{pmatrix} e_{11} & \frac{e_{12}+e_{21}}{2} & \frac{e_{13}+e_{31}}{2} \\ \frac{e_{21}+e_{12}}{2} & e_{22} & \frac{e_{23}+e_{32}}{2} \\ \frac{e_{31}+e_{13}}{2} & \frac{e_{32}+e_{23}}{2}  & e_{33} \end{pmatrix} = \begin{pmatrix} 0.001 & 0 & 0.0015 \\ 0 & 0 & 0.0025 \\ 0.0015 & 0.0025  & 0 \end{pmatrix} $$ resulting ultimately in $$\boldsymbol{e}=\boldsymbol{\omega}+\boldsymbol{\varepsilon}$$.

Problem Statement:
Within linear, homogeneous, isotropic elasticity theory, for a given stress state (units in MPa) determine the strain state given the Poisson ratio is 0.40 and the shear modulus is 50 GPa. Identify the hydrostatic stress, deviatoric stresses, strain dilatation, and strain deviator.

\boldsymbol{\sigma}= \begin{pmatrix} 10 & 2 & 2 \\ 2 & 5 & -3 \\ 2 & -3  & -5 \end{pmatrix} $$

Solution:
We're given the stress, so returning a solution requires determining the strain, which can be determined by

\varepsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}-\frac{\nu}{E}\sigma_{kk}\delta_{ij} $$ We're given the Poisson ratio, $$\nu$$ and shear modulus $$G$$ but to use this equation we need the elastic modulus $$E$$. This can be found from

G=\frac{E}{2\left(1 - \nu \right)} $$ which can be inverted to

E=2 G \left(1 - \nu \right) $$ Upon substitution this yields $$E=60$$ GPa. The resulting strain tensor is

\boldsymbol{\varepsilon}= \begin{pmatrix} \frac{1+\nu}{E}\sigma_{11}-\frac{\nu}{E}\left(\sigma_{22}+\sigma_{33}\right) & \frac{\sigma_{12}}{2G} & \frac{\sigma_{13}}{2G} \\ \frac{\sigma_{21}}{2G} & \frac{1+\nu}{E}\sigma_{22}-\frac{\nu}{E}\left(\sigma_{11}+\sigma_{33}\right) & \frac{\sigma_{23}}{2G} \\ \frac{\sigma_{31}}{2G} & \frac{\sigma_{32}}{2G}  & \frac{1+\nu}{E}\sigma_{33}-\frac{\nu}{E}\left(\sigma_{11}+\sigma_{22}\right) \end{pmatrix} = \begin{pmatrix} 0.1667 & 0.0467 & 0.0467 \\ 0.0467 & 0.0500 & -0.0700 \\ 0.0467 & -0.0700  & -0.1833 \end{pmatrix}. $$

The hydrostatic stress is

\sigma_m=\frac{\sigma_{kk}}{3}=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3}=\frac{10+5-5}{3}=3.33 \text{ GPa} $$ The strain dilatation is

\Delta=\left(1+\varepsilon_{11}\right)\left(1+\varepsilon_{22}\right)\left(1+\varepsilon_{33}\right)-1=\left(1+0.1667\right)\left(1+0.0500\right)\left(1+0.1833\right)-1=0.4496 $$ which results in a mean strain of

\varepsilon_m=\frac{\Delta}{3}=\frac{0.4496}{3}=0.1499 $$ (Note that when strain is small $$\Delta \approx \varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}=0.400$$, which yields $$\varepsilon_m \approx 0.1333$$. Unfortunately, in this example the strain is relatively large.) The deviatoric stress and strain are then determined by subtracting the mean stress and strain from their respective tensors diagonal.

\boldsymbol{\sigma'}= \begin{pmatrix} 10-3.33 & 2 & 2 \\ 2 & 5-3.33 & -3 \\ 2 & -3  & -5-3.33 \end{pmatrix} = \begin{pmatrix} 6.67 & 2 & 2 \\ 2 & 1.67 & -3 \\ 2 & -3  & -8.33 \end{pmatrix} $$ and

\boldsymbol{\varepsilon'}= \begin{pmatrix} 0.1667-0.1499 & 0.0467 & 0.0467 \\ 0.0467 & 0.0500-0.1499 & -0.0700 \\ 0.0467 & -0.0700  & -0.1833-0.1499 \end{pmatrix} = \begin{pmatrix} 0.0168 & 0.0467 & 0.0467 \\ 0.0467 & 0.0999 & -0.0700 \\ 0.0467 & -0.0700  & -0.3332 \end{pmatrix} $$

Problem Statement:
Within linear, homogeneous, isotropic elasticity theory, for a given strain state determine the stress state given the bulk modulus of 100 GPa and Lam$$\acute{e}$$ parameter of 50 GPa. Identify the hydrostatic stress, deviatoric stresses, strain dilatation, and strain deviator.

\boldsymbol{\varepsilon}= \begin{pmatrix} 0.002 & 0.002 & 0.001 \\ 0.002 & 0.002 & 0.001 \\ 0.001 & 0.001  & -0.001 \end{pmatrix} $$

Solution:
We are given the strain, so returning a solution requires determining the stress, which can be determined by

\sigma_{ij}=\frac{E}{1+\nu}\varepsilon_{ij}-\frac{\nu E}{\left(1 + \nu \right)\left(1 - 2 \nu \right)}\varepsilon_{kk}\delta_{ij} $$ where $$\frac{\nu E}{\left(1 + \nu \right)\left(1 - 2 \nu \right)}$$ is the Lamé parameter $$\lambda$$. Since

G=\frac{E}{2\left(1+\nu\right)} $$ and

G=\frac{3\left(K-\lambda\right)}{2} $$ we know

\frac{E}{\left(1+\nu\right)}=3\left(K-\lambda\right) $$ The stress expression becomes

\sigma_{ij}=3\left(K-\lambda\right)\varepsilon_{ij}+\lambda\varepsilon_{kk}\delta_{ij} $$ Substituting and solving yields

\boldsymbol{\sigma}= \begin{pmatrix} 0.45 & 0.3 & 0.15 \\ 0.3 & 0.45 & 0.15 \\ 0.15 & 0.15  & 0.0 \end{pmatrix} $$GPa

The hydrostatic stress is

\sigma_m=\frac{\sigma_{kk}}{3}=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3}=\frac{0.45+0.45+0.0}{3}=0.30 \text{ GPa} $$ The strain dilatation is

\Delta=\left(1+\varepsilon_{11}\right)\left(1+\varepsilon_{22}\right)\left(1+\varepsilon_{33}\right)-1=\left(1+0.002\right)\left(1+0.002\right)\left(1-0.001\right)-1=0.003 $$ which results in a mean strain of

\varepsilon_m=\frac{\Delta}{3}=\frac{0.003}{3}=0.001 $$ An interesting observation is that the strain dilation and hydrostatic (mean) stress are related by the bulk modulus $$\sigma_m=K\Delta$$, which in this case is $$0.3=100\times0.003$$. When performing calculations it is important to use known checkpoints such as these to validate your work.

The deviatoric stress and strain are then determined by subtracting the mean stress and strain from their respective tensors diagonal.

\boldsymbol{\sigma'}= \begin{pmatrix} 0.45-0.30 & 0.30 & 0.15 \\ 0.30 & 0.45-0.30 & 0.15 \\ 0.15 & 0.15  & 0.00-0.30 \end{pmatrix} = \begin{pmatrix} 0.15 & 0.30 & 0.15 \\ 0.30 & 0.15 & 0.15 \\ 0.15 & 0.15  & -0.30 \end{pmatrix} $$ and

\boldsymbol{\varepsilon'}= \begin{pmatrix} 0.002-0.001 & 0.002 & 0.001 \\ 0.002 & 0.002-0.001 & 0.001 \\ 0.001 & 0.001  & -0.001-0.001 \end{pmatrix} = \begin{pmatrix} 0.001 & 0.002 & 0.001 \\ 0.002 & 0.001 & 0.001 \\ 0.001 & 0.001  & -0.002 \end{pmatrix} $$

Problem Statement:
For a single crystal of cubic zirconia, which has a elastic constants approximately $$c_{11}=575$$, $$c_{12}=115$$, and $$c_{44}=75$$ GPa, determine the elastic energy required to apply a uniaxial strain of 0.001 in the $$\left[ 100 \right]$$ direction. Determine the elastic energy required to apply a uniaxial strain of 0.001 in the $$\left[ 110 \right]$$ direction. Compute the Zener anisotropy ratio. Using isotropic elasticity theory, the elastic modulus of 200 GPa, and Poisson ratio of 0.3 the elastic modulus of $$200 \ GPa $$ and the Poisson's ratio of $$0.3$$ to compute the elastic energy to apply a uniaxial strain of 0.001.

Solution:
Say that the $$\left[100\right]$$ is in the $$x_1$$ direction. The elastic energy, $$U_o$$, is

U_o=\frac{1}{2}\sigma_{ij}\varepsilon_{ij} $$ so

dU_o=\frac{1}{2}\sigma_{ij}d\varepsilon_{ij} $$ In the case of anisotropic elasticity theory

\sigma_{ij}=c_{ijkl}\varepsilon_{kl}. $$ In this case all $$\varepsilon_{ij}=0$$ except $$\varepsilon_{11}=0.001$$. Therefore all $$\sigma_{ij}=0$$ except $$\sigma_{11}=c_{1111}\varepsilon_{11}$$, $$\sigma_{22}=c_{1122}\varepsilon_{22}$$, and $$\sigma_{33}=c_{1122}\varepsilon_{33}$$.

Substituting into the above equationsː
 * $$\begin{align}

dU_0 &= \varepsilon_{11} c_{1111} d\varepsilon_{11} + \underbrace{\varepsilon_{11} c_{1122} \ d\varepsilon_{22} + \varepsilon_{11} c_{1122} \ d\varepsilon_{33}}_{= \ 0 \ upon \ integration} \\ &= \varepsilon_{11} c_{1111} d\varepsilon_{11} \\ \\ U_0 &= \int_0^{\varepsilon_{11} \ = \ 0.001} \varepsilon_{11} c_{11} d\varepsilon_{11} \\ &= \left. c_{11} {\varepsilon_{11}^2 \over 2} \ \right |_{0}^{0.001} \\ &= (575 *10^9 \ Pa){0.001^2 \over 2} \\ &= 288 \ kJ \end{align}$$

Problem Statement:
For a single crystal of cubic zirconia, which has a elastic constants approximately $$c_{11}=575$$, $$c_{12}=115$$, and $$c_{44}=75$$ GPa, determine the elastic energy required to apply a uniaxial strain of 0.001 in the $$\left[ 100 \right]$$ direction followed by a shear strain of 0.001 on the $$\left( 001 \right)$$ face in the $$\left[ 010 \right]$$ direction.

Solution:
Obeying the given order of operation, first we apply a uniaxial load of $$0.001$$ in the $$[100]$$ direction. (Also covered in Example 10)

Note that our general equations are $dU = \sigma_{ij} \ d\varepsilon_{ij}$, and $\sigma_{ij} = c_{ijkl} \varepsilon_{kl}$.

Restating the strain tensorː

\varepsilon = \left ( \begin{matrix} \varepsilon_{11} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right )$$

From this we get the non-zero stress tensorː

\sigma = \left ( \begin{matrix} \varepsilon_{11} c_{11} & 0 & 0 \\ 0 & \varepsilon_{11} c_{12} & 0 \\ 0 & 0 & \varepsilon_{11} c_{12} \end{matrix} \right )$$

Thus the only non-zero $$d\varepsilon_{ij}$$ is $$d\varepsilon_{11}$$, and by integrating our equation for $dU$ we getː $$U_1 = 288 \ kJ$$

Now let's apply the shear $$0.001$$ on the $$(100)$$ face in the $$[010]$$ direction.

Our new stress tensor isː

\sigma = \left ( \begin{matrix} \varepsilon_{11} c_{11} & \varepsilon_{12} c_{44} & 0 \\ \varepsilon_{12} c_{44} & \varepsilon_{11} c_{12} & 0 \\ 0 & 0 & \varepsilon_{11} c_{12} \end{matrix} \right )$$

Here, the only non-zero terms are $$d\varepsilon_{12}$$, and $$d\varepsilon_{21}$$. As these are equivalent terms, due to symmetry, we can solve for one and multiply the answer by two.

Once again utilizing our basic energy equation we getː

\begin{align} dU &= \sigma_{ij} \ d\varepsilon_{ij} = \varepsilon_{12} c_{44} \ d\varepsilon_{12} \\ \\ U_2 &= 2\int_0^{\varepsilon_{12}} \varepsilon_{12} c_{44} \ d\varepsilon_{12} \\ &= 2 \left [c_{44} {\varepsilon_{12}^2 \over 2} \right ]_0^{\varepsilon_{12}} \\ &= (75*10^9 \ Pa) 0.001^2 \\ U_2 &= 75 \ kJ \end{align}$$

Finally, adding $$U_1$$ and $$U_2$$ together to get the total energy of this combined transformation gives us a final answer of $$363 \ kJ$$.

Problem Statement:
For a polycrystal specimen of cubic zirconia, which has a elastic constants approximately $$c_{11}=575$$, $$c_{12}=115$$, and $$c_{44}=75$$ GPa, use isotropic elasticity theory and the elastic modulus of $$200 \ GPa $$ and the Poisson's ratio of $$0.3$$ to compute the elastic energy to apply a strain state

\boldsymbol{\varepsilon}= \begin{pmatrix} 0.001 & 0.001 & 0.000 \\ 0.001 & 0.000 & 0.000 \\ 0.000 & 0.000 & 0.000 \end{pmatrix} $$ If the polycrystal material has a porosity of 2% approximately how much will this change the elastic modulus? Approximately how much will this change the elastic energy for this applied strain?

Solution:
Note that our general equations are $dU = \sigma_{ij} \ d\varepsilon_{ij}$, and $\sigma_{ij} = c_{ijkl} \varepsilon_{kl}$.

Here, we can put the stress in terms of the elastic modulus and the Poisson's ratioː

 * $$\sigma_{ij} = {E \over 1 + \nu} \varepsilon_{ij} +

{\nu E \over (1 + \nu) (1 - 2 \nu)} \varepsilon_{kk} \partial_{ij} $$

Where the latter constant is equivalent to the Lamé Constant ($$\lambda $$)ː $$\lambda = {\nu E \over (1 + \nu)(1 - 2\nu)} $$

Briefly solving for the Lamé Constant yields usː

\begin{align} \lambda &= {\nu E \over (1 + \nu)(1 - 2\nu)} \\ &= {0.3 * 200 \ GPa \over (1 + 0.3) (1 - 2(0.3))} = 115.3846 \ GPa \end{align} $$

Keeping in mind that $$\varepsilon_{12}$$ equals $$\varepsilon_{21}$$ thanks to symmetry, the non-zero stresses for this problem areː

\sigma = \left ( \begin{matrix} {E \over 1+ \nu} \varepsilon_{11} + \lambda \varepsilon_{11} & {E \over 1 + \nu} \varepsilon_{12} & 0 \\ {E \over 1 + \nu} \varepsilon_{12} & \lambda \varepsilon_{11} & 0 \\ 0 & 0 & \lambda \varepsilon_{11} \end{matrix} \right )$$

Therefore, we can write out the non-zero energy terms asː

 * $$\begin{align}

dU &= \sigma_{ij} \ d\varepsilon_{ij} \\ &= \left ( {E \over 1 + \nu} +\lambda \right ) \varepsilon_{11} \operatorname{d} \! \varepsilon_{11} + \left ( {E \over 1 + \nu} +\lambda \right ) \varepsilon_{12} \operatorname{d} \! \varepsilon_{12} + \left ( {E \over 1 + \nu} +\lambda \right ) \varepsilon_{21} \operatorname{d} \! \varepsilon_{21} + \lambda \varepsilon_{11} \operatorname{d} \varepsilon_{11} + \lambda \varepsilon_{11} \operatorname{d} \varepsilon_{11} \\ &= \left ( {E \over 1 + \nu} +\lambda \right ) \varepsilon_{11} \operatorname{d} \! \varepsilon_{11} + 2 \left ( {E \over 1 + \nu} +\lambda \right ) \varepsilon_{12} \operatorname{d} \! \varepsilon_{12} \\ \\ U &= \left. \left (    {E \over 1+\nu} + \lambda \right ) {1 \over 2} \varepsilon_{11}^2 \right |_{0}^{\varepsilon_{11}} + \left. \left ( {E \over 1 + \nu} \right ) \varepsilon_{12}^2 \right |_0^{\varepsilon_{12}} \\ &= 288 \ kJ \end{align}$$


text:
Overview of Elasticity of Materials/Introducing Stress

Introduction
We will begin by developing the constitutive equations that describe the relationship between stress, $$\sigma$$, and strain, $$\varepsilon$$. This is the subset of continuum mechanics that focuses on the purely elastic regime, and in particular, will focus on linear elasticity where Hooke's Law holds true. The concepts of stress and strain originate by considering the forces applied to a body and its displacement. Beginning with forces, there are two types of forces that can be applied. First, there is surface force which can either be point forces or distributed forces that are applied over a surface. Second, there is body force which is applied to every element of a body, not just a surface (i.e., gravity, electric fields, etc.).

The body of interest has numerous forces acting on it and these are transmitted through the material. At any point inside the body, you can imagine slicing it to observe the forces present on the imagined cut surface, as pictured in Figure 1. These forces are the interactions between the material on either side of the imagined cut. We define the stress at a point in the body as the forces acting on the surface of such an imagined cut. As you recall, the stress is defined as the force over the area which it is applied. The force, $$\mathbf{P}$$, is a vector quantity, allowing the components to be projected into the normal and tangential directions. As shown in Figure 1, the normal component is defined according to the angle $$\theta$$, yielding a normal stress $$\sigma_{33} = \frac{P \cos \! \theta}{A}$$. The tangential component of the force, $$P \sin \! \theta$$, can be further projected into the two orthogonal directions identified in Figure 1 as $$x_1$$ and $$x_2$$, yielding two orthogonal shear stresses. This is performed according to the angle $$\phi$$, giving $$\sigma_{31} = \frac{P \sin \! \theta \cos \! \phi}{A}$$ and $$\sigma_{32} = \frac{P \sin \! \theta \sin \! \phi}{A}$$.

Note here that we've defined the coordinate system such that the $$x_3$$ direction is the direction normal to the cut surface. It is convenient to use $$(x_1, x_2, x_3)$$ instead of $$(x, y, z)$$ because it allows us to pass the indexes to the stress and strain quantities. In this example, the normal stress is given by $$\sigma_{33}$$ to specify that the normal stress is applied to the surface with a normal in the $$x_3$$ direction with a force projected in the $$x_3$$ direction. The tangential components $$\sigma_{31}$$ and $$\sigma_{32}$$ specify the surface having a normal $$x_3$$ with forces projected in the $$x_1$$ and $$x_2$$ directions, respectively. Cutting an infinitesimal cuboid, the stresses are defined in all three directions as shown in Figure 2. For comparison, the notation used in some textbooks will write normal stresses $$\sigma_{x}$$, whereas here we will use $$\sigma_{11}$$. These textbooks also use $$\tau$$ to denote shear stress, such as $$\tau_{xy}$$ whereas here we will use $$\sigma_{12}$$. This allows the stress state to be succinctly written in matrix (tensor) form

 * $$\sigma = \left ( \begin{align}

\sigma_{11} \quad \sigma_{12} \quad \sigma_{13} \\ \sigma_{21} \quad \sigma_{22} \quad \sigma_{23} \\ \sigma_{31} \quad \sigma_{32} \quad \sigma_{33} \end{align} \right ) $$

The imaginary slice taken through point in the body in Figure 1 could have been any plane, but the force would remain the same. This would result in a new definition of the surface normal, and potentially a new expression for the stress. The physical presence of the stress does not change, but the description does, i.e., the coordinate system is modified. The remainder of this section is devoted to expressing the coordinate transformation and analysis of the stresses.

Plane Stress
We will begin by simplifying the picture we are working with. The plane stress condition is observed for a thin 2D object, e.g., a piece of paper, which has no stress out of the plane. This allows us to write $$\sigma_{33}=0$$. Further, there is no shear in the $$x_3$$ direction such that $$\sigma_{13} = \sigma_{23} = 0$$. For an object in the plane stress condition, our goal is to determine the state of stress at some point for any orientation of the axis. For this object, the direction with zero force is $$x_3$$ coming out of the page and the non-zero stress state in the $$x_1$$ and $$x_2$$ directions have components $$\sigma_{11}$$, $$\sigma_{22}$$, and $$\sigma_{12}=\sigma_{21}$$.

Imagine a new area defined on a plane rotated about $$x_3$$ such that the normal, defined $$x_1'$$, is related to $$x_1$$ by $$\theta$$ as shown in Figure 3.

The components of force on the area is determined by the application of the original stresses to the projection of the new area:
 * $$\begin{align}

F_{1} = \sigma_{11} A \cos \! \theta + \sigma_{12} A \sin \! \theta = S_{1}A \\ F_{2} = \sigma_{22} A \sin \! \theta + \sigma_{12} A \cos \! \theta = S_{2}A \end{align}$$    [1 & 2]

where the elements $$ A \cos \! \theta $$ and $$ A \sin \! \theta $$ are the projection of the A in the original orientation, shown in Figure 3 (a), and $$S_{1}$$ and $$S_{2}$$ are the total stresses in the $$x_1$$ and $$x_2$$ directions, where $$\sigma_{12}=\sigma_{21}$$. Then, dividing by A yields:
 * $$\begin{align}

S_{1} = \sigma_{11} \cos \! \theta + \sigma_{12} \sin \! \theta \\ S_{2} = \sigma_{22} \sin \! \theta + \sigma_{12} \cos \! \theta \end{align}$$    [3 & 4]

Projecting the total stresses shown in Figure 3 (b) into the normal direction in the $$x_1'$$ coordinate yields
 * $$\sigma_{11}' = S_{1} \! \cos \! \theta + S_{2} \! \sin \! \theta $$    [5]

In a similar fashion, we project tangential to the plane and yield
 * $$\sigma_{12}' = S_{2} \! \cos \! \theta - S_{1} \! \sin \! \theta$$    [6]

Resulting in

 * $$\begin{align}

\sigma_{11}' &= ( \sigma_{11} \cos \! \theta + \sigma_{12} \sin \! \theta ) \cos \! \theta + ( \sigma_{22} \sin \! \theta + \sigma_{12} \cos \! \theta ) \sin \! \theta \\ &= \sigma_{11} \cos^2 \! \theta + \sigma_{22} \sin^2 \! \theta + 2 \ \sigma_{12} \sin \! \theta \cos \! \theta \end{align}$$    [7] and
 * $$\begin{align}

\sigma_{12}' &= ( \sigma_{22} \sin \! \theta + \sigma_{12} \cos \! \theta) \cos \! \theta - (\sigma_{11} \cos \! \theta + \sigma_{12} \sin \! \theta ) \sin \! \theta \\ &= \sigma_{12} (cos^2 \! \theta - \sin^2 \! \theta ) + (\sigma_{22} - \sigma_{11}) \sin \! \theta \cos \! \theta \end{align}$$    [8] It is known that $$\sigma_{12}'=\sigma_{21}'$$ and therefore only $$\sigma_{22}'$$ needs determining. To do so, we define a new area that is rotated by $\pi$ /2 relative to our original plane as shown in Figure 4. In this new orientation,
 * $$\begin{align}

S_{1} A &= \sigma_{11} A \cos (\theta + {\pi \over 2}) + \sigma_{12} A \sin (\theta + {\pi \over 2}) \\ S_{1} &= -\sigma_{11} \sin \! \theta + \sigma_{12} \cos \! \theta \end{align}$$    [9] and
 * $$\begin{align}

S_{2} A &= \sigma_{22} A \sin (\theta + {\pi \over 2}) + \sigma_{12} A \cos (\theta + {\pi \over 2}) \\ S_{2} &= \sigma_{22} \cos \! \theta - \sigma_{12} \sin \! \theta \end{align}$$    [10]

Projecting the total stress in the normal direction yields
 * $$\begin{align}

\sigma_{22}' &= S_{1} \cos (\theta + {\pi \over 2}) + S_{2} \sin (\theta + {\pi \over \theta}) \\ &= -S_{x_1} \sin \! \theta + S_{x_2} \cos \! \theta \end{align}$$    [11]

Substituting Equations 9 and 10 for $$S_{1}$$ and $$S_{2}$$ into Equation 11 for $$\sigma_{22}'$$ yields
 * $$\begin{align}

\sigma_{22}' &= - (-\sigma_{11} \sin \! \theta + \sigma_{12} \cos \! \theta) \sin \! \theta + (\sigma_{22} \cos \! \theta - \sigma_{12} \sin \! \theta) \cos \! \theta \\ &= \sigma_{11} \sin^2 \! \theta + \sigma_{22} \cos^2 \! \theta - 2 \ \sigma_{12} \sin \! \theta \cos \! \theta \end{align}$$    [12]

The well-known trigonometric identities

 * $$\begin{array}{lcl}

\cos^2 \! \theta = {\cos 2\theta +1 \over 2} & \quad & \sin^2 \! \theta = {1-\cos 2\theta \over 2} \\ 2\sin \! \theta \cos \! \theta = \sin 2\theta & & \cos^2 \! \theta - \sin^2 \theta = \cos 2 \theta \end{array}$$

are applied to Equations 7, 8, and 12 for $$\sigma_{11}'$$, $$\sigma_{12}'$$, and $$\sigma_{22}'$$ respectively, resulting in
 * $$\begin{align}

\sigma_{11}' &= {\sigma_{11} + \sigma_{22} \over 2}+ {\sigma_{11} - \sigma_{22} \over 2} \cos {2\theta} + \sigma_{12} \sin {2\theta} \end{align}$$    [13]
 * $$\begin{align}

\sigma_{22}' &= {\sigma_{11} + \sigma_{22} \over 2} - {\sigma_{11} - \sigma_{22} \over 2} \cos {2\theta} - \sigma_{12} \sin{2 \theta} \end{align}$$    [14] and
 * $$\begin{align}

\sigma_{12}' &= {\sigma_{22} - \sigma_{11} \over 2} \sin 2\theta + \sigma_{12} \cos 2\theta \end{align}$$    [15]

Principal Stress
There are numerous immediate results that come from this derivation, from which we can gain greater insights. One result that comes from the equations for $$\sigma'$$ is $$\sigma_{11}' + \sigma_{22}' = \sigma_{11} + \sigma_{22}$$, for all $$\theta$$. This means that the trace of the stress tensor $$\boldsymbol{\sigma}$$ is invariant.

A second result is that the maximum normal stresses and shear stresses vary as a sine wave with period $$\pi$$. Within this oscillation, the normal and shear stresses are shifted by a phase factor that results in (1) the maximum and minimum normal stresses occur when the shear is zero, (2) the maximum and minimum shear stresses are shifted from each other by $$\pi$$/4, (3) the maximum and minimum normal stresses are shifted from each other by $$\pi$$/2, and (4) the maximum and minimum shear stresses are shifted by $$\pi$$/4 from the minimum and maximum normal stresses.

Any stress state can be rotated to yield $$\sigma_{12}=\sigma_{21}=0$$. This diagonalizes the stress tensor and gives normal stresses that are extreme. In this orientation, the planes are called the principal planes and the normal stresses are called the principal stresses. The directions that give these principal stresses are called the principal axis. As a matter of convention, we define the first principal stress $$\sigma_{p1}$$ to be the largest and the sequentially smaller principal stresses to be $$\sigma_{p2}$$ and $$\sigma_{p3}$$, although here we have limited ourselves to 2D plane stress and only enumerate $$\sigma_{p1}$$ and $$\sigma_{p2}$$.

We know $$\sigma_{12}=0$$ in the principal orientation, which means we can use Equation 8 for $$\sigma_{12}'$$ to determine the angle ($$\theta$$) needed to rotate the tensor $$\boldsymbol{\sigma}$$ into $$\sigma'$$ which is principal,

 * $$\begin{align}

0 &= \sigma_{12}' = \sigma_{21}' \\ 0 &= \sigma_{12} (\cos^2 \! \theta - \sin^2 \! \theta ) + (\sigma_{22} - \sigma_{11}) \sin \! \theta \cos \! \theta \\ -(\sigma_{22} - \sigma_{11}) \sin \! \theta \cos \! \theta &= \sigma_{12} (\cos^2 \! \theta - \sin^2 \! \theta ) \\ {\sin \! \theta \cos \! \theta \over \cos^2 \! \theta - \sin^2 \! \theta} &= {\sigma_{12} \over \sigma_{11} - \sigma_{22}} \end{align}$$ Resulting in
 * $$\begin{align}

\tan 2\theta &= {2 \ \sigma_{12} \over \sigma_{11} - \sigma_{22}} \end{align}$$    [16]

It is observed graphically by plotting $$\tan 2\theta$$ in Figure 5 that adjacent roots are each separated by $\pi$ /2. Furthermore, we can now utilize the Pythagorean Theorem to solve for our principal stresses. For a simple right triangle with hypotenuse $$c$$ and sides $$a$$ and $$b$$ we know

 * $$\begin{align}

\sin \xi = {b \over c} \\ \cos \xi = {a \over c} \\ \tan \xi = {b \over a} \end{align}$$ which can be combined with the Pythagorean Theorem, $$a^2+b^2=c^2$$ and Equation 16,

 * $$\begin{align}

a &= \sigma_{12} \\ b &= {1 \over 2} (\sigma_{11} - \sigma_{22}) \\ c &= \plusmn \left ( {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \right ) ^ {1 \over 2} \end{align}$$

These can be further combined which yields
 * $$\begin{align}

\sin 2 \theta &= \plusmn {\sigma_{12} \over \left ( {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \right ) ^ {1 \over 2} } \end{align}$$    [17] and
 * $$\begin{align}

\\ \cos 2 \theta &= \plusmn { {1 \over 2} (\sigma_{11} - \sigma_{22}) \over \left ( {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \right ) ^ {1 \over 2} } \end{align}$$    [18]

These equations tell us for a given stress state, $$\sigma$$, what rotation is needed to align $$\sigma$$ with the principal axis.

Substituting these equations into Equation 13 for $$\sigma_{11}'$$, determines the principal stresses

 * $$\begin{align}

\sigma_p &= {\sigma_{11} + \sigma_{22} \over 2} + {\sigma_{11} - \sigma_{22} \over 2} \left ( \plusmn {{1 \over 2} (\sigma_{11} - \sigma_{22}) \over        \left ({1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \right )^{1 \over 2}}    \right ) + \sigma_{12} \left ( {\sigma_{12} \over \plusmn       \left ( {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \right )^{ 1 \over 2}}   \right ) \\ &= {\sigma_{11} + \sigma_{22} \over 2} + {{1 \over 4} (\sigma_{11} - \sigma_{22})^2 \over \plusmn \left ( {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2       \right )^{1 \over 2} } + {{\sigma_{12}}^2 \over \plusmn \left (        {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2    \right )^{1 \over 2}} \\ &= {\sigma_{11} + \sigma_{22} \over 2} \plusmn \left (    {{1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \over         \left ( {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2 \right )^{1 \over 2}}   \right ) \end{align}$$ Resulting in
 * $$\begin{align}

&= {\sigma_{11} + \sigma_{22} \over 2} \plusmn \left (   {1 \over 4} (\sigma_{11} - \sigma_{22})^2 + {\sigma_{12}}^2     \right )^ {1 \over 2} \end{align}$$    [19]

Use Equation 19 in Equation 16 to find $$\theta$$ for $$0<\theta<{\pi \over 4}$$.

To find the maximum shear stress, we take the derivative with respect to theta of our simplified Equation 15 for $$\sigma_{12}'$$ and set it equal to $$0$$.

 * $$\begin{align}

0 &= {\operatorname{d} \over \operatorname{d} \! \theta} \left (    {\sigma_{22} - \sigma_{11} \over 2} \sin 2\theta + \sigma_{12} \cos 2\theta \right ) \\ &= 2{\sigma_{22} - \sigma_{11} \over 2} \cos 2\theta - 2 \sigma_{12} \sin 2 \theta \\ &= (\sigma_{22} - \sigma_{11}) \cos 2\theta - 2 \sigma_{12} \sin 2\theta \end{align}$$ Resulting in an expression for $$2\theta$$:
 * $$\begin{align}

\\ {\sigma_{22} - \sigma_{11}\over 2 \ \sigma_{12}} &= {\sin 2\theta \over \cos 2\theta} = \tan 2\theta \end{align}$$    [20]

Notice that Equation 20 and Equation 16 are negative reciprocals which means that $$2\theta_p$$ and $$2\theta_{MAX} $$ are shifted by $$\pi$$/2. This is indicative of

 * $$\begin{align}

\tan \phi = \frac{a}{b} \\ \tan \phi + \frac{\pi}{2} = \frac{-b}{a} \end{align}$$

which implies that $$2\theta_p$$ and $$2\theta_{MAX} $$ are separated by $$\pi$$/4. Through substitution of Equation 20 into Equation 15, we arrive at an expression for $$\sigma_{12MAX}$$:

 * $$\sigma_{12MAX} = \plusmn \left [

\left ({\sigma_{11} - \sigma_{22} \over 2} \right )^2 + {\sigma_{12}}^2 \right]^{1 \over 2}$$    [21]

Mohr's Circle
A convenient means of visualizing angular relationships is through Mohr's circle, which we derive here. Rearrange Equation 13 for $$\sigma_{11}'$$ and Equation 15 for $$\sigma_{12}'$$,

 * $$\begin{align}

\sigma_{11}' - {\sigma_{11} + \sigma_{22} \over 2} &= {\sigma_{11} - \sigma_{22} \over 2} \cos {2\theta} + \sigma_{12} \sin {2\theta} \end{align}$$    [22]
 * $$\begin{align}

\sigma_{12}' = {\sigma_{22} - \sigma_{11} \over 2} \sin {2\theta} + \sigma_{12} \cos {2\theta} \end{align}$$    [23]

Square both expressions,

 * $$\begin{align}

\left (\sigma_{11}' - {\sigma_{11} + \sigma_{22} \over 2} \right )^2 &= \left ( {\sigma_{11} - \sigma_{22} \over 2} \cos {2\theta} + \sigma_{12} \sin {2\theta}    \right )^2 \end{align}$$

 * $$\begin{align}

\left ( \sigma_{12}' \right )^2 &= \left ( {\sigma_{22} - \sigma_{11} \over 2} \sin 2\theta + \sigma_{12} \cos 2\theta        \right )^2 \end{align}$$ Next, add them together to yield

 * $$\begin{align}

\left ( \sigma_{11}' - {\sigma_{11} + \sigma_{22} \over 2} \right )^2 + {\sigma_{12}'}^2 &= \left ( {\sigma_{11} - \sigma_{22} \over 2} \right )^2 (\cos^2 \! 2\theta + \sin^2 \! 2\theta) + {\sigma_{12}}^2 (\cos^2 \! 2\theta + \sin^2 \! 2\theta) \\ {\left ( \sigma_{11}' - {\sigma_{11} + \sigma_{22} \over 2} \right )^2} + {\sigma_{12}'}^2 &= \left ( {\sigma_{11} - \sigma_{22} \over 2} \right )^2 + {\sigma_{12}}^2 \end{align}$$

The resulting expression is the equation for a circle: $$(x - h)^2 + y^2 = r^2$$

 * $$\underbrace{ \left ( \sigma_{11}' - {\sigma_{11} + \sigma_{22} \over 2} \right )^2 }_{(x-h)^2}

+ \underbrace{{\sigma_{12}'}^2}_{y^2} = \underbrace{ \left ( {\sigma_{11} - \sigma_{22} \over 2} \right )^2 + {\sigma_{12}}^2 }_{r^2}$$    [24]

From this expression, Mohr's circle is drawn in Figure 6. For a given stress state, $$\sigma$$, the center of the circle is $$h=\frac{\sigma_{11}-\sigma_{22}}{2}$$ and the radius $$r = \left ( \left ( \frac{\sigma_{11}-\sigma_{11}}{2} \right )^{2} + \sigma_{12}^{2} \right )^\frac{1}{2}$$. A bisecting line intercepts the circle such that the projection onto the x-axis identifies $$\sigma_{11}$$ and $$\sigma_{22}$$. The projection onto the y-axis identifies $$\sigma_{12}$$. Rotating the bisection is equivalent to transforming the stress state by $$2 \pi$$, i.e., a rotation by $$\phi$$ on the diagram is equivalent to rotating by $$2 \pi$$ in our equations. This allows the new stress state to be read from the diagram. When the bisector is horizontal, the principal orientation is identified. Rotating the bisection on the diagram by $$\pi$$ is equivalent to rotating the system by $$\theta=\pi$$/2, which can be imagined as rotating the cuboid faces until the system is back in registry, i.e., it returns to the original stress state. Further, rotating the bisection on the diagram by $$\pi$$/2 is equivalent to rotating by $$\theta=\pi$$/4, which is known to be the orientation with maximum shear stress. Thus, from a given initial stress state, $$\sigma$$, all stress states that can be achieved through rotation are visualized on the circle.

Generalizing from 2D to 3D
Generalizing from 2D to 3D, we move from a biaxial plane stress system to a triaxial system. Determining the principal axis and angular relations is similar to the case of 2D and will be shown below. Note as a matter of convention, when two of the three principal stresses are equal, we call the system "cylindrical", and if all three principal stresses are equal, we call the system "hydrostatic" or "spherical".

As in the case of the biaxial system, we begin by defining a plane with area $$A$$ that passes through our $$x_1$$, $$x_2$$, and $$x_3$$ coordinate system, as shown in Figure 7. The plane intercepts the axis at ($$J$$, $$K$$, and $$L$$) as demonstrated in the figure. To simplify the problem and allow us to make progress toward our derivation, we will say that the plane is one of the principal planes so that the shear stress components are zero. Thus, we only need to consider our principal stress that is normal to the plane.

Define $$\ell$$, $$m$$, and $$n$$ to be the direction cosine between $$x_1$$, $$x_2$$, and $$x_3$$ and the normal to the stress. Using the unit vectors $$\hat{i}$$, $$\hat{j}$$, and $$\hat{k}$$ parallel to $$x_1$$, $$x_2$$, and $$x_3$$, we have

 * Sec1 Fig7.png$$\ell = \cos \theta_{1} = {\hat{i} \cdot \boldsymbol{\sigma} \over |\boldsymbol{\sigma|}}

\quad m = \cos \theta_{2} = {\hat{j} \cdot \boldsymbol{\sigma} \over |\boldsymbol{\sigma|}}; \quad n = \cos \theta_{3} = {\hat{k} \cdot \boldsymbol{\sigma} \over |\boldsymbol{\sigma|}} $$

The projection of stress along $$x_1$$, $$x_2$$, and $$x_3$$ direction give the total stresses $$S_{1}$$, $$S_{2}$$, and $$S_{3}$$:

S_{1} = \sigma_p \ell ; \quad S_{2} = \sigma_p m ; \quad S_{3} = \sigma_p n $$

In the biaxial derivation, the area is projected into three directions, producing the triangles in Figure 7 which have areas $$LOK = A \ell$$, $$JOL = Am$$ and $$JOK = An$$. We can now equate the forces in the two reference frames:

 * $$\begin{align}

S_{1} A = \sigma_{p} \ell A = F_x &= LOK \sigma_{11} + JOL \sigma_{21} + JOK \sigma_{31} + \\ &= A \ell \sigma_{11} + A m \sigma_{21} + A n \sigma_{31} \end{align}$$ So,
 * $$\begin{align}

\sigma_p \ell &= \sigma_{11} \ell + \sigma_{21} m + \sigma_{31} n \end{align}$$    [25]

By a similar process, the $$F_{x_2}$$ and $$F_{x_3}$$ components yield

 * $$\begin{align}

\sigma_p m &= \sigma_{12} \ell + \sigma_{22} m + \sigma_{32} n \end{align}$$    [26]
 * $$\begin{align}

\sigma_p n &= \sigma_{13} \ell + \sigma_{23} m + \sigma_{33} n \end{align}$$    [27]

These equations rearrange to

 * $$\begin{align}

0 &= (\sigma_{11} - \sigma_p) \ell + \sigma_{12} m + \sigma_{13} n \end{align}$$    [28]
 * $$\begin{align}

0 &= \sigma_{12} \ell + (\sigma_{22} - \sigma_p) m + \sigma_{23} n \end{align}$$    [29]
 * $$\begin{align}

0 &= \sigma_{13} \ell + \sigma_{23} m + (\sigma _{33} - \sigma_p)n \end{align}$$    [30]

This set of equations can be solved for $$\left [ \ell, m, n \right ]$$ for a particular value of $$\sigma_{p}$$. This set of secular equations can be solved for eigenvalues $$\sigma_{p}$$ and eigenvectors $$\left [ \ell, m, n \right ]$$. The non-trivial solutions, when $$\ell, m,$$ and $$n$$ are non-zero, involves setting the determinant

 * $$det \ \left |

\begin{matrix} \sigma_{11} - \sigma_p & \sigma_{12} & \sigma_{13} \\ \sigma_{12} & \sigma_{22} - \sigma_p & \sigma_{23} \\ \sigma_{13} & \sigma_{23} & \sigma_{33} - \sigma_p \end{matrix} \right |$$

to zero and solving for the eigenvalues and subsequent eigenvectors.

Upon rearranging, we get

 * $$\begin{align}

0 = {\sigma_p}^3 - & (\sigma_{11} +\sigma_{22} + \sigma_{33}) {\sigma_p}^2 \\ & \quad + (\sigma_{11} \sigma_{22} + \sigma_{22} \sigma_{33} + + \sigma_{33} \sigma_{11} -        {\sigma_{12}}^2 - {\sigma_{23}}^2 - {\sigma_{31}}^2) \sigma_p \\ & \qquad \qquad - (\sigma_{11} \sigma_{22} \sigma_{33} + 2\sigma_{12} \sigma_{23} \sigma_{31}        - \sigma_{11} {\sigma_{23}}^2 - \sigma_{22} {\sigma_{13}}^2 - \sigma_{33} {\sigma_{12}}^2 ) \end{align}$$    [31]

The three roots of this cubic equation give the principal stresses, $$\sigma_{p1}$$, $$\sigma_{p2}$$, and $$\sigma_{p3}$$. The principle stresses, once determined, are substituted back into the secular Equations 28-30 to determine the eigenvectors corresponding to $$\left [ \ell, m, n \right ]$$, also recognizing that $$\ell^2 + m^2 + n^2 = 1 $$.

Solving the cubic equation is not the focus of this text, but Equation 31 is important because the coefficients in front of the principal stress must be invariant, i.e., the same principal coordinates must exist no matter the orientation of the coordinate system. From the cubic equation, the three invariants are

 * $$\begin{align}

I_1 = (\sigma_{11} +\sigma_{22} + \sigma_{33}) \end{align}$$    [32]
 * $$\begin{align}

I_2 = (\sigma_{11} \sigma_{22} + \sigma_{22} \sigma_{33} + \sigma_{33} \sigma_{11}        -{\sigma_{12}}^2 - {\sigma_{23}}^2 - {\sigma_{31}}^2) \end{align}$$    [33]
 * $$\begin{align}

I_3 = (\sigma_{11} \sigma_{22} \sigma_{33} + 2\sigma_{12} \sigma_{23} \sigma_{31}        - \sigma_{11} {\sigma_{23}}^2 - \sigma_{22} {\sigma_{13}}^2 - \sigma_{33} {\sigma_{12}}^2 ) \end{align}$$    [34]

This is useful because these invariant relations determine the relationship between stresses in different orientations, i.e. given $$\sigma$$, you can now directly determine $$\sigma_{1}$$, $$\sigma_{2}$$, and $$\sigma_{3}$$.

Now, we will generalize our solution to include not only the principal stresses. Just as we did earlier, we can write out the total forces:

 * $$\begin{align}

S_{x_1} A = F_{x_1} = \sigma_{11} \ell A + \sigma_{12} m A + \sigma_{31} n A \end{align}$$
 * $$\begin{align}

\qquad S_{x_1} = \sigma_{11} \ell + \sigma_{12} m + \sigma_{33} n \end{align}$$    [35]
 * $$\begin{align}

\qquad S_{x_2} = \sigma_{12} \ell + \sigma_{22} m + \sigma_{23} n \end{align}$$    [36]
 * $$\begin{align}

\qquad S_{x_3} = \sigma_{13} \ell + \sigma_{23} m + \sigma_{33} n \end{align}$$    [37]

Which gives the total stress:
 * $$S^2 = {S_{x_1}}^2 + {S_{x_2}}^2 + {S_{x_3}}^2$$    [38]

From this, the projection onto the normal component is:
 * $$\sigma' = S_{x_1} \ell + S_{x_2} m + S_{x_3} n$$    [39]

Substituting Equations 34-36 into Equation 38 gives us:

 * $$\begin{align}

\sigma' &= (\sigma_{11} \ell + \sigma_{12} m + \sigma_{31} n) \ \ell + (\sigma_{12} \ell + \sigma_{22} m + \sigma_{31} n) \ m   + (\sigma_{13} \ell + \sigma_{31} m + \sigma_{33} n) \ n \end{align}$$ Which simplifies to,

\sigma' = \sigma_{11} \ell^2 + \sigma_{22} m^2 + \sigma_{33} n^2 + 2\sigma_{12} \ell m + 2\sigma_{23} m n + 2\sigma_{31} n \ell $$    [40]

The magnitude of the shear component can be determined utilizing $$S^2 = {\sigma'}^2 + \tau^2$$, but we cannot easily decompose our shear stress into its constituent elements. Fortunately, we are primarily interested in the maximum shear stress. We know that the plane containing the maximum shear stress is located midway between the planes of principal normal stresses. Starting by setting our known stress state as the principal axis such that $$\sigma_{11} = \sigma_{p_1}$$, $$\sigma_{22} = \sigma_{p_2}$$, and $$\sigma_{33} = \sigma_{p_3}$$, our direction cosine is between the principal axis and the normal of the plane with the maximum shear stress. This means that Equation 39 for projection is rewritten as:

 * $$\sigma' = \sigma_{p_1} \ell^2 + \sigma_{p_2} m^2 + \sigma_{p_3} n^2 $$    [41]

Squaring this equation gives us:

 * $${\sigma'}^2 = {\sigma_{p_1}}^2 \ell^4 + {\sigma_{p_2}}^2 m^4 + {\sigma_{p_3}}^2 n^4

+ 2\sigma_{p_1} \sigma_{p_2} \ell^2 m^2 + 2\sigma_{p_1} \sigma_{p_3} \ell^2 n^2 + 2\sigma_{p_2} \sigma{p_3} m^2 n^p_2 $$    [42]

We can then use the principal components and substitute Equations 34-36 into Equation 37 to get:

 * $$S^2 = {\sigma_{11}}^2 \ell^2 + {\sigma_{22}}^2 m^2 + {\sigma_{33}}^2 n^2 $$    [43]

After much algebra and putting Equations 41 & 42 into Equation 40, we get:

 * $$\tau_{MAX}^2 =

(\sigma_{11}-\sigma_{22})^2 \ell^2 m^2 + (\sigma_{11} - \sigma_{33})^2 \ell^2 n^2 + (\sigma_{22} - \sigma_{33})^2 m^2 n^2 $$    [44]

With this solution, we now have three possible planes. One plane bisects $$\sigma_{11} $$ and $$\sigma_{22} $$, another plane bisects $$\sigma_{11} $$ and $$\sigma_{33} $$, and the final plane bisects $$\sigma_{22} $$ and $$\sigma_{33} $$. (Bisecting means $\theta={\pi \over 4} $, and $\cos {\pi \over 4} = {\sqrt{2} \over 2} $ ). Here are the values of $$\ell,m,n$$, and $$\tau$$ for these three planes:

 * {| class="wikitable"
 * {| class="wikitable"

!$$\boldsymbol{\ell}$$ !$$\boldsymbol{m}$$ !$$\boldsymbol{n}$$ !$$\boldsymbol{\tau}$$
 * $$0$$
 * $$\frac{\sqrt{2}}{2}$$
 * $$\frac{\sqrt{2}}{2}$$
 * $$   \frac{1}{2}(\sigma_{22}-\sigma_{33})$$
 * $$\frac{\sqrt{2}}{2}$$
 * $$0$$
 * $$\frac{\sqrt{2}}{2}$$
 * $$\frac{1}{2}(\sigma_{11}-\sigma_{33})$$
 * $$\frac{\sqrt{2}}{2}$$
 * $$\frac{\sqrt{2}}{2}$$
 * $$0$$
 * $$\frac{1}{2}(\sigma_{11}-\sigma_{22})$$
 * }
 * $$0$$
 * $$\frac{1}{2}(\sigma_{11}-\sigma_{22})$$
 * }

By convention, $$\sigma_{11} > \sigma_{22} > \sigma_{33}$$, and therefore our maximum shear stress is:

 * $$\tau_{MAX}={\sigma_{11} - \sigma_{33} \over 2}$$

Note that we know there are two planes of maximum shear stress, rotated $\pi$ /2 from each other. Thus, the direction cosine above are actually $\plusmn {\sqrt{2} \over 2} $.

Because these axial rotations are decoupled, we can represent 3D stress states using Mohr's Circles as seen in Figure 8.


text:
Talk:Overview of Elasticity of Materials/Representing Stress: Rotating the Reference Frame Using Mohr's Circle

I think this pages should be deleted. I have moved it to be inside the the book instead of outside.


text:
Modern Greek/A1/Exercise/Lesson 03.1

Greek fascination is the forever the philosophical views of as much as imortally possible to be understood in order to be broken down into works work and comprehend studies in to the Greek language for permanent basis of the philosophical views and comprehend studies have found that it is not to be a bit more than the other people who are in place permanent deal of money and comprehend studies have found a few days of future and the philosophical views on the way.

title: Modern Greek/A1/Exercise/Lesson 03.1


text:
Exercise as it relates to Disease/The effect of lumbar stabilization and walking exercises on chronic low back pain

This article is an original critical analysis of; "The effect of lumbar stabilisation and walking exercises on chronic low back pain - a randomised controlled trial" by JH Suh, H Kim, GP Jung, JY Ko, JS Ryu - Medicine, 2019 - ncbi.nlm.nih.gov.

What is the background to this research?

 * In modern society, lower back pain (LBP) is one of the most prevalent physical disorders that affects performance at work as well as general well-being across all ages.

 * Chronic pain is closely associated with structural and histomorphological changes in spinal muscles. These back muscles, become fatigued through over-exertion. Over time poor coordination of the spinal muscles leads to chronic LBP.

 * Exercise can be a proactive treatment for improving muscle strength, endurance and functional impairment amongst low back pain patients.
 * Many avenues of physical activity have been referred to and introduced in the study, but no superior exercise mode was suggested.

 * This study aims to compare the effectiveness of two exercises that can help reduce LBP, namely individualised graded lumbar stabilisation exercises (IGLSE) and walking exercises (WE).

Where is the research from?

 * This research highlights the comparison between two successful exercise modes that are integrated into the treatment of LBP.
 * This study can be found on PubMed(https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6616307/).
 * This research is supported by Research Resettlement Fund for the new faculty of Seoul National University, and basic science research program via the National research Foundation of Korea (NRF), which is funded by the Ministry of Education.
 * Commercial parties that have directly contributed to the research financially do not stand to profit from the results.
 * According to the authors, there were no identifiable conflicts of interest.
 * Authors
 * Ju Seok Ryu
 * Dr. Ryu belongs to the Seoul National University College of Medicine, and has been actively involved in research since 2004 to now. (895 of citations, 16 of h-index and 58 article, 1 letter, and 1 review).

What kind of research was this?

 * The study format was a RCT with a total of 48 participants who all experienced chronic LBP. Within the study there were 4 experimental groups: flexibility exercise, walking exercise, stabilisation exercise, and stabilisation with walking exercise. The study duration was 6 weeks.
 * Lower back pain is a widespread common chronic condition, exercise is one of the most effective treatment strategies and there is a pleather of available research on PubMed.
 * In comparison to similar studies this research involved a smaller testing population and was shorter in duration.

What did the research involve?

 * Summary
 * The participants were randomly sorted into 4 different groups, and performed the appropriate exercises.
 * Dosage: 30-60mins sessions, completed 5 times/week, for 6 weeks in total
 * Education: All participants received education surrounding correct and healthy posture for managing LBP as well as how to engage and brace their abdominal muscles.
 * Primary outcome measurement:
 * Visual analog scale(VAS) of their pain during physical activity and at rest.
 * Secondary outcome measurement:
 * VAS of radiating pain measured at rest and during physical activity
 * Individual medication dosage (number of times/day)
 * Oswestry disability index
 * Beck depression inventory
 * Endurances of patient in specific postural positions
 * Strength of lumbar extensors
 * Methodology
 * This research used a randomised control trial format and the subjects were recruited from the universities outpatient rehab clinic.
 * The RCT structure of this study makes for reliable and accurate comparisons between the exercise groups.

 * Limitation
 * The causes or severity of each patient's lower back pain was not disclosed during the study.
 * The duration of research could have been extended to assess the long term effects of the exercises.
 * The results from this study only reflect the short term benefits as opposed to positive changes overtime.
 * The types and effects of medications participants were using during study were not considered as a contributing factor to the results.

What were the basic results?

 * Results
 * LBP reduced during physical activity across all four experimental groups
 * LBP reduced at rest: FE group and SE group
 * Frequency of medication use decreased in the FE group
 * Exercise frequency increased: SE, WE
 * The highest compliance: SE group
 * Overall there were no clinically significant differences between the 4 experimental groups

 * Interpret
 * Clinicians from the study confidently recommended SE and WE as suitable treatments for patients experiencing chronic LBP. The exercises were found not only to relieve lower back pain but in addition, improved overall muscle endurance, core stability and prevented the progression of LBP.
 * The results from the WE group and the stabilisation training + WE group produced substantial improvements in the muscle endurance of the back muscles.
 * Researchers also identified the beneficial contribution posture related education and muscle bracing exercises produced within all experimental groups

What conclusions can we take from this research?

 * As a result of this study, there is no doubt that the exercise treatments were effective in the LBP patients.
 * As the research highlighted, the combination of targeted exercises, stabilisation and postural training as well as patient education are effective in reducing LBP related symptoms and altering the management of their pain through adjusting the patient's perspective of treatment options

Practical advice

 * Exercises including walking, lumbar stabilisation, postural training and stretching are low cost activities that do not require any additional equipment and can be completed in any setting at a time of convenience. In addition, these exercises are easy to self-pace and adapt to avoid high levels of fatigue and exhaustion, the sets and reps can be modified to increase the recovery period.
 * Maintaining compliance to the exercises outside of a clinical setting could be hard to continue due to barriers in patient motivation and the lack of persistent feedback.
 * According to additional research some strategies for increasing compliance and patient motivation include group sessions, involving a second person such as a a friend or family member and keeping an exercise log/diary.

Further information/resources

 * Causes of Lower Back Pain
 * Lower Back Pain Symptoms, Diagnosis, and Treatment
 * Lower Back Pain: Symptoms, Stretches, Exercise for Pain Relief
 * 11 Exercises for Lower Back Pain Relief


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History of wireless telegraphy and broadcasting in Australia/Topical/Biographies/Walter Francis Maxwell Howden/Graphics

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Cataloging and Classification/Controlled Vocabularies

The definition of Controlled vocabulary, as per Weihs & Intner (2017):
 * "A list of terms authorized for indexing, such as a subject heading list or thesaurus."


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Skaneateles Conservation Area/Invasive species/Ampelopsis glandulosa

Ampelopsis glandulosa var. brevipedunculata (porcelain berry)
Some sources, including iMapInvasives, treat porcelainberry or Amur peppervine as a separate species (Ampelopsis brevipedunculata), while others (e.g. the New York Flora Atlas) treat Ampelopsis brevipedunculata as a synonym of Ampelopsis glandulosa. For the purposes of this document, we'll generally refer to this plant as Ampelopsis glandulosa (or porcelainberry) except where references using Ampelopsis brevipedunculata are being quoted directly.

Identification
Porcelainberry (Ampelopsis glandulosa) is a deciduous, woody, perennial vine in the grape family, Vitaceae, and without its colorful fruit it looks quite similar to grape vines. Non-fruiting vines can be differentiated by the pith, which is white in porcelain berry stems but brown in grapevines. These vines also differ by the presence of lenticels on porcelainberry bark and the shaggy straight-peeling bark of grape vines.

The flowers and fruit of porcelainberry generally grow upright as opposed to grapes, which hang down. Porcelainberry fruit is also harder than grapes and is distinctive in the range of colors from white, aqua, pink, blue, purple, etc.

Southeastern-native Ampelopsis peppervines are occasionally planted as "native" and also have similar appearances.

Nearest infestations
Ampelopsis glandulosa var. brevipedunculata (porcelain berry or Amur peppervine) as been confirmed in iMapInvasives in Cayuga County near the south end of Skaneateles Lake and at Filmore Glen State Park. A larger infestation is present in Ithaca at Buttermilk Falls.

Invasiveness ranking for Ampelopsis glandulosa (porcelainberry)
In 2010, Ampelopsis brevipedunculata (Maxim.) Trautv. (Porcelain berry) was ranked as highly invasive with a score of 62/87 = 71.26%

Regulated by New York State law.

1. Ecological impact (17/30)
1.1. Impact on Natural Ecosystem Processes and System-Wide Parameters: Influences ecosystem processes to a minor degree (3/10)

 * Large accumulation of biomass affect organic matter, nutrient content of soil and fire regime.

 * Large stands impact light availability.

 * Can dominate the vegetation by forming a uniform blanket over shrubs, trees, and the ground, especially on forest edges.

1.2. Impact on Natural Community Structure: Significant impact in at least one layer (7/10)

 * Form a sprawling mat shading out other plants below it, creating a new vine layer.

 * May not completely cover the canopy or eradicate layers below.

1.3. Impact on Natural Community Composition:

1.4. Impact on other species or species groups:

2. Biological characteristics and dispersal ability (/)
2.1. Mode and rate of reproduction

2.2 Innate potential for long-distance dispersal:

2.3. Potential to be spread by human activities:

2.4. Characteristics that increase competitive advantage:

2.5. Growth vigor:

2.6. Germination/Regeneration:

2.7. Other species in the genus invasive in New York or elsewhere:

3. Ecological amplitude and distribution (/)
3.1. Density of stands in natural areas in the northeastern USA and eastern Canada:

3.2. Number of habitats the species may invade:

3.3. Role of disturbance in establishment:

3.4. Climate in native range:

3.5. Current introduced distribution in the northeastern USA and eastern Canada:

3.6. Current introduced distribution of the species in natural areas in the 8 NY PRISMs:

4. Difficulty of control (/)
4.1. Seed banks:

4.2. Vegetative regeneration:

4.3. Level of effort required:


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User talk:Scott.beckman

Welcome, Scott.beckman!

If you have any questions, you can ask in the assistance reading room or possibly contact me personally. JackPotte (discuss • contribs) 13:46, 4 September 2021 (UTC)

title: User talk:Scott.beckman