OpenLLM-France/wikimedia huggingface dataset:
# DevOps/Ansible/Modules

Modules add new capabilities to Ansible and is a core component of the software, as of September 2019 there are more than 2.800 modules available: ansible-doc --list. For example you can manage AWS infrastructure using Ansible, execute commands with shell module or collect information using setup module.
* ansible -m setup SERVER_NAME will collect information from your system
* ansible -m setup SERVER_NAME -a 'filter=ansible_distribution'. Collect just OS.
Other common useful modules include:
* Execution modules:
** command, default for ad-hoc commands, bypasses the shell so won’t be impacted by local shell variables.
** shell, allow to use shell functionalities such as redirection.
** raw, do not require python on remote system
* gather_facts, lineinfile, user, group, mount, git, docker, docker_compose and mail modules, see all of them by category: https://docs.ansible.com/ansible/latest/modules/modules_by_category.html

## Modules

* apt_repository: DevOps/Ansible/Add a repository
* docker compose: New in version 2.1 and was call docker_service before Ansible 2.8
* file, for managing files and directories: https://docs.ansible.com/ansible/latest/modules/file_module.html
* ovirt_vm (KVM)

## Module raw example

* ansible -m raw DESTINATION_HOSTNAME -a uptime
# DevOps/Ansible/Playbooks

Ansible playbooks are written in YAML text files and define actions to execute against host/devices, check official Ansible playbook introduction: https://docs.ansible.com/ansible/devel/user_guide/playbooks_intro.html

### Task definitions or Playbooks

Task definitions are defined in text files, called playbooks in Ansible terminology, in YAML format. Execution of task definitions or playbooks are make using ansible-playbook command followed by your desired playbook to execute. Playbooks can also use Jinja templates to enable dynamic expressions and access to variables.
Ansible Playbook execution:
* ansible-playbook my_new_created_playbook.yml
* ansible-playbook my_new_created_playbook.yml -f 10 -f 10 parallelize 10 executions
       -f forks

### Create "Executable" Playbooks

Add the following line #!/usr/bin/env ansible-playbook at the beginning of your playbooks and make it executable chmod +x /path/to/your/playbook.yml to execute without calling ansible-playbook binary.

### Ansible Playbooks Examples

Create a new file touch /tmp/test_file.txt, REMOTE_SERVER should be already present in your Ansible inventory:/etc/ansible/host
Playbook using module file:
Important Note: dash symbol (-) and indentation has to be properly formatted.

### Variables

You can use variables in your playbooks, check official documentation for further information: https://docs.ansible.com/ansible/devel/user_guide/playbooks_variables.html

## Activities

* Read Ansible introduction to playbooks: https://docs.ansible.com/ansible/devel/user_guide/playbooks_intro.html
# DevOps/Ansible/Ansible changelog and versioning

w:Ansible (software) Changelog: https://github.com/ansible/ansible/releases
    ansible --version

### Releases

* v2.9: 31/10/2019 https://github.com/ansible/ansible/blob/stable-2.9/changelogs/CHANGELOG-v2.9.rst
** v2.9.1 13/11/2019
* v2.8: 16/05/2019 https://github.com/ansible/ansible/blob/stable-2.8/changelogs/CHANGELOG-v2.8.rst, https://www.ansible.com/blog/whats-new-in-red-hat-ansible-engine-2.8
** Ansible Collections and content namespacing
** BECOME is the default Ansible word to prompt for a password
** gather_facts module: https://docs.ansible.com/ansible/latest/modules/gather_facts_module.html
** python interpreter discovery
* v2.7 09/10/2018: https://github.com/ansible/ansible/blob/stable-2.7/changelogs/CHANGELOG-v2.7.rst, https://www.ansible.com/blog/the-release-of-red-hat-ansible-engine-2-7
** New reboot module
** ignore_unreachable added
* v2.6 02/07/2018 https://www.ansible.com/blog/your-time-has-come-ansible-2-6
** Inventory Plugins: foreman, generator, gcp_compute, nmap
* v2.5 23/03/2018 https://groups.google.com/forum/#!topic/ansible-announce/V2HLb1lKKoI
* v2.4 19/09/2017
** import_playbook keywords https://docs.ansible.com/ansible/latest/modules/import_playbook_module.html
** import_tasks
** import_role https://docs.ansible.com/ansible/latest/modules/import_role_module.html#import-role-module
* v2.0 12/01/2016 https://www.ansible.com/blog/ansible-2.0-launch https://raw.githubusercontent.com/ansible/ansible/stable-2.0/CHANGELOG.md
** block introduce the concept of exception handling to playbooks, and were modeled after the try/except/finally structure of Python (and many other languages)
* v1.5 28 February 2014
* 1.4.3 https://github.com/ansible/ansible/releases/tag/v1.4.3
* 1.4.2 December 18, 2013 https://github.com/ansible/ansible/releases/tag/v1.4.2
** ansible-galaxy command added for retrieving roles from the Galaxy
* 1.2 June 10, 2013
** Ansible Roles introduced
# DevOps/Ansible/User ssh access

Solution:
touch create_user_with_public_key_access.yml
If you are using this playbook for the first time and ssh key authentication is not configured modify first line to: #!/usr/bin/env ansible-playbook --ask-pass
Execute: ansible-playbook create_user_with_public_key_access.yml References:
* User Module doc: https://docs.ansible.com/ansible/latest/modules/user_module.html#user-module
* SSH authorized key module: https://docs.ansible.com/ansible/latest/modules/authorized_key_module.html#authorized-key-module
You can also loop over Subelements, check https://docs.ansible.com/ansible/2.4/playbooks_loops.html#looping-over-subelements.
# DevOps/Ansible/ansible-vault

ansible-vault encrypt passwords.yml
ansible-vault [create|decrypt|edit|encrypt|rekey] [--help] [options] file_name

## Commands

* ansible-playbook --vault-id @prompt create-users.yml

## Activities

* Learn about ansible-vault common uses cases
# DevOps/Ansible/gather facts

Ansible gather_facts module, new in Ansible 2.8, allows to gather useful variables about remote hosts that can be used in playbooks
# DevOps/Ansible/Configure user to be able to use sudo with no password

Example with insertafter
# DevOps/Ansible/Use loops in task

Ansible offers two keywords for creating loops: loop and with_<lookup>.
# English-Chinese/In manufacturing wheat products, gluten is valuable to impart viscoelastic functional qualities in dough, enabling the preparation of diverse processed foods

In manufacturing wheat products, gluten is valuable to impart viscoelastic functional qualities in dough, enabling the preparation of diverse processed foods such as breads, noodles, and pasta that facilitate wheat consumption.
在制造小麦产品时，面筋有助于在面团中赋予粘弹性功能品质，使得能够制备促进小麦消费的各种加工食品，例如面包，面条和面食。
# English-Chinese/In many countries including the United States, United Kingdom, Philippines, and Brazil, Japanese restaurants have become popular

In many countries including the United States, United Kingdom, Philippines, and Brazil, Japanese restaurants have become popular.
在许多国家，包括美国，英国，菲律宾和巴西，日本餐馆已经变得流行。
# English-Chinese/In most Quorn products, the fungus culture is dried and mixed with egg albumen, which acts as a binder, and then is adjusted in texture and pressed into various forms

In most Quorn products, the fungus culture is dried and mixed with egg albumen, which acts as a binder, and then is adjusted in texture and pressed into various forms.
在大多数Quorn产品中，将真菌培养物干燥并与蛋清混合，蛋清作为粘合剂，然后调整质地并压制成各种形状。
# English-Chinese/In observational studies of people with severe obesity, weight loss following bariatric surgery is associated with a 46% reduction in cardiovascular risk

In observational studies of people with severe obesity, weight loss following bariatric surgery is associated with a 46% reduction in cardiovascular risk.
在对严重肥胖人群进行的观察性研究中，减肥手术后体重减轻与心血管风险降低46％相关。
# English-Chinese/In mammals, the ear is usually described as having three parts—the outer ear, middle ear and the inner ear

In mammals, the ear is usually described as having three parts—the outer ear, middle ear and the inner ear.
在哺乳动物中，耳朵通常被描述为具有三个部分 - 外耳，中耳和内耳。
# English-Chinese/In many contexts, potato refers to the edible tuber, but it can also refer to the plant itself

In many contexts, potato refers to the edible tuber, but it can also refer to the plant itself.
在 许多 情况下，马铃薯 是指 可食用的 块茎，但 它 也可以 指 植物本身。
# English-Chinese/In order to stem the tide, education and awareness that cardiovascular disease poses the greatest threat, and measures to prevent or reverse this disease must be taken

In order to stem the tide, education and awareness that cardiovascular disease poses the greatest threat, and measures to prevent or reverse this disease must be taken.
为了阻止心血管疾病构成最大威胁的潮流，教育和意识，必须采取预防或逆转这种疾病的措施。
# English-Chinese/In manufacturing wheat products, gluten is valuable to impart viscoelastic functional qualities in dough, enabling the preparation of diverse processed foods such as breads, noodles, and pasta that facilitate wheat consumption

In manufacturing wheat products, gluten is valuable to impart viscoelastic functional qualities in dough, enabling the preparation of diverse processed foods such as breads, noodles, and pasta that facilitate wheat consumption.
在制造小麦产品时，麸质对于赋予面团粘弹性功能品质是有价值的，使得能够制备促进小麦消费的各种加工食品，例如面包，面条和面食。
# English-Chinese/In modern-day wars, civilians are often targets

In modern-day wars, civilians are often targets. An example of this is the nuclear bombs dropped on Hiroshima and Nagasaki at the end of World War II.
在现代战争中，平民往往是目标。 这方面的一个例子是第二次世界大战结束时在广岛和长崎投下的核弹。
# English-Chinese/In nutrition, diet is the sum of food consumed by a person or other organism

In nutrition, diet is the sum of food consumed by a person or other organism. The word diet often implies the use of specific intake of nutrition for health or weight-management reasons (with the two often being related).
在 营养 方面，饮食 是 人 或 其他 生物 消耗 的 食物 的 总和。 饮食 这个 词 通常 意味着 为了 健康 或 体重 管理 的 原因 而 使用 特定 的 营养 摄入量（这 两者 通常 是 相关 的）。
# Fermat cubic/Syz (x^2,y^2,z^2)/Strongly semistable/Example

Let $R=K[x,y,z]/{\left(x^{3}+y^{3}+z^{3}\right)}$, where $K$ is a field of positive characteristic $p\neq 3$, $I={\left(x^{2},y^{2},z^{2}\right)}$, and
    $C=\operatorname {Proj} {\left(R\right)}\,.$
The equation $x^{3}+y^{3}+z^{3}=0$ yields the short exact sequence
    $0\longrightarrow {\mathcal {O}}_{C}\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}(3)\longrightarrow {\mathcal {O}}_{C}\longrightarrow 0.$
This shows that $\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}$ is strongly semistable.
# Fermat quartic/Hilbert-Kunz/Dependence on characteristic/Example

The Fermat quartic $x^{4}+y^{4}+z^{4}=0$ is the easiest example where the Hilbert-Kunz multiplicity of the maximal ideal $(x,y,z)$ fluctuates with the characteristic. We have
    $e_{HK}(R_{p})={\begin{cases}3{\text{ for }}p=\pm 1\,\operatorname {mod} \,8\\3+{\frac {1}{p^{2}}}{\text{ for }}p=\pm 3\,\operatorname {mod} \,8\,.\end{cases}}\,$
The limit is of course $3$, which corresponds to the fact that the syzygy bundle is semistable in characteristic zero. The syzygy bundle is semistable for all prime characteristics $p\geq 3$, but not strongly semistable for the prime numbers $p=\pm 3\,\operatorname {mod} \,8$.
# Fermat quartic/Syz (x,y,z)/Characteristic 3/Semistable, not strongly semistable/Example

Let $C$ be the smooth Fermat quartic given by $x^{4}+y^{4}+z^{4}$, and consider on it the syzygy bundle $\operatorname {Syz} {\left(x,y,z\right)}$ (which is also the restricted cotangent bundle from the projective plane). This bundle is semistable. Suppose that the characteristic is $3$. Then its Frobenius pull-back is $\operatorname {Syz} {\left(x^{3},y^{3},z^{3}\right)}$. The curve equation gives a global non-trivial section of this bundle of total degree $4$. But the degree of $\operatorname {Syz} {\left(x^{3},y^{3},z^{3}\right)}(4)$ is negative, hence it can not be semistable anymore.
# Fermat cubic/z^2 in tight closure of (x,y)/Example

We consider the Fermat cubic $R=K[X,Y,Z]/{\left(X^{3}+Y^{3}+Z^{3}\right)}$, the ideal $I=(X,Y)$ and the element $Z^{2}$. We claim that in positive characteristic $\neq 3$ the element $Z^{2}$ does belong to the tight closure of $I$. Equivalently, the open subset
    $D(X,Y)\subseteq \operatorname {Spec} {\left(R[S,T]/(XS+YT+Z^{2})\right)}\,$
is not affine. The element $Z^{2}$ defines the cohomology class
    $c={\frac {Z^{2}}{XY}}\in H^{1}(D(X,Y),{\mathcal {O}}_{X})\,$
and its Frobenius pull-backs are
    $F^{e*}(c)={\frac {Z^{2q}}{X^{q}Y^{q}}}\in H^{1}(D(X,Y),{\mathcal {O}}_{X})\,.$
This cohomology module has a $\mathbb {Z}$-graded structure (the degree is given by the difference of the degree of the numerator and the degree of the denominator) and, moreover, it is $0$ in positive degree (this is related to the fact that the corresponding projective curve is elliptic). Therefore for any homogeneous element $t\in R$ of positive degree we have $tF^{e*}(c)=0$ and so $Z^{2}$ belongs to the tight closure.
From this it also follows that in characteristic $0$ the element $Z^{2}$ belongs to the solid closure, because affineness is an open property in an arithmetic family.
# Fermat equation/x^2 in (y,z)^*/Forcing algebra/Several characteristics/Example

Let $K$ be a field and consider the Fermat ring
    $R=K[X,Y,Z]/{\left(X^{d}+Y^{d}+Z^{d}\right)}\,$
together with the ideal $I=(X,Y)$ and $f=Z^{2}$. For $d\geq 3$ we have $Z^{2}\notin (X,Y)$. This element is however in the tight closure $(X,Y)^{*}$ of the ideal in positive characteristic (assume that the characteristic $p$ does not divide $d$) and is therefore also in characteristic $0$ inside the tight closure and inside the solid closure. Hence the open subset
    $D(X,Y)\subseteq \operatorname {Spec} {\left(K[X,Y,Z,S,T]/{\left(X^{d}+Y^{d}+Z^{d},SX+TY-Z^{2}\right)}\right)}\,$
is not an affine scheme. In positive characteristic, $Z^{2}$ is also contained in the plus closure $(X,Y)^{+}$ and therefore this open subset contains punctured surfaces (the spectrum of the forcing algebra contains two-dimensional closed subschemes which meet the exceptional fiber $V(X,Y)$ in only one point; the ideal $(X,Y)$ has superheight two in the forcing algebra). In characteristic zero however, due to remark the superheight is one and therefore by fact the algebra $\Gamma ({D(X,Y),{\mathcal {O}}}_{B})$ is not finitely generated. For $K=\mathbb {C}$ and $d=3$ one can also show that $D(X,Y)_{\mathbb {C} }$ is, considered as a complex space, a Stein space.
# Fermat cubic/xyz in (x^2,y^2,z^2)^*/Cohomological proof/Example

Let $R=K[x,y,z]/{\left(x^{3}+y^{3}+z^{3}\right)}$, where $K$ is a field of positive characteristic $p\neq 3$, $I={\left(x^{2},y^{2},z^{2}\right)}$ and $f=xyz$. We consider the short exact sequence
    $0\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\longrightarrow {\mathcal {O}}_{U}^{3}{\stackrel {x^{2},y^{2},z^{2}}{\longrightarrow }}{\mathcal {O}}_{U}\longrightarrow 0$
and the cohomology class
    $c=\delta (xyz)=H^{1}{\left(U,\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}\,.$
We want to show that $zF^{e*}(c)=0$ for all $e\geq 0$ (here the test element $z$ equals the element $z$ in the ring). It is helpful to work with the graded structure on this syzygy sheaf (or to work on the corresponding elliptic curve $\operatorname {Proj} R$ directly). Now the equation $x^{3}+y^{3}+z^{3}=0$ can be considered as a syzygy (of total degree $3$) for $x^{2},y^{2},z^{2}$, yielding an inclusion
    $0\longrightarrow {\mathcal {O}}_{U}\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}.$
Since this syzygy does not vanish anywhere on $U$ the quotient sheaf is invertible and in fact isomorphic to the structure sheaf. Hence we have
    $0\longrightarrow {\mathcal {O}}_{U}\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\longrightarrow {\mathcal {O}}_{U}\longrightarrow 0$
and the cohomology sequence
    $\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{s}\longrightarrow H^{1}{\left(U,\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{s+3}\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{s}\longrightarrow 0,$
where $s$ denotes the degree-$s$th piece. Our cohomology class $c$ lives in $H^{1}{\left(U,\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3}$, so its Frobenius pull-backs live in $H^{1}{\left(U,F^{e*}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3q}$, and we can have a look at the cohomology of the pull-backs of the sequence, i.e.
    $\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{0}\longrightarrow H^{1}{\left(U,F^{e*}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3q}\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{0}\longrightarrow 0.$
The class $zF^{e*}(c)$ lives in $H^{1}{\left(U,F^{e*}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3q+1}$. It is mapped on the right to $H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{1}$, which is $0$ (because we are working over an elliptic curve), hence it comes from the left, which is
    $H^{1}(U,{\mathcal {O}}_{U})_{1}=0\,.$
So $zF^{e*}(c)=0$ and $f\in {\left(x^{2},y^{2},z^{2}\right)}^{*}$.
# Fermat cubic/z not in tight closure of (x,y)/Example

We consider the Fermat cubic $R=K[X,Y,Z]/{\left(X^{3}+Y^{3}+Z^{3}\right)}$, the ideal $I=(X,Y)$ and the element $Z$. We claim that for characteristic $\neq 3$ the element $Z$ does not belong to the solid closure of $I$. Equivalently, the open subset
    $D(X,Y)\subseteq \operatorname {Spec} {\left(R[S,T]/(XS+YT-Z)\right)}\,$
is affine. For this we show that the extended ideal inside the ring of global sections is the unit ideal. First of all we get the equation
    $X^{3}+Y^{3}=(XS+YT)^{3}=X^{3}S^{3}+3X^{2}S^{2}YT+3XSY^{2}T^{2}+Y^{3}T^{3}\,$
or, equivalently,
    $X^{3}{\left(S^{3}-1\right)}+3X^{2}YS^{2}T+3XY^{2}ST^{2}+Y^{3}{\left(T^{3}-1\right)}=0\,.$
We write this as
    ${\begin{aligned}X^{3}{\left(S^{3}-1\right)}&=-3X^{2}YS^{2}T-3XY^{2}ST^{2}-Y^{3}{\left(T^{3}-1\right)}\\&=Y{\left(-3X^{2}S^{2}T-3XYST^{2}-Y^{2}{\left(T^{3}-1\right)}\right)},\end{aligned}}$
which yields on $D(X,Y)$ the rational function
    $Q={\frac {S^{3}-1}{Y}}={\frac {-3X^{2}S^{2}T-3XYST^{2}-Y^{2}{\left(T^{3}-1\right)}}{X^{3}}}\,.$
This shows that $S^{3}-1=QY$ belongs to the extended ideal. Similarly, one can show that also the other coefficients $3S^{2}T,3ST^{2},T^{3}-1$ belong to the extended ideal. Therefore in characteristic different from $3$, the extended ideal is the unit ideal.
# Headache assessment
## HPI

* FIRST or WORST headache ever?
* DIFFERENT from usual headaches?
* Onset = when did it start?
** gradual, sudden (thunder-clap)
** prodrome, aura
* Palliation/Provocative
** stress, food, menstrual cycle, rest
* Quality
** unilateral, bilateral, band-like?
** does it spread?
** throbbing, stabbing, dull, pressure
* Radiation
** where does it spread?
* Severity
** Nausea/vomiting
** photo/phonophobia
** vision changes
** fever
** stiff neck
** confusion

## Review of systems:

* General systemic illness: weightloss
* CNS: change in mentation, personality, aura

## RED FLAGS OF SERIOUS HEADACHES
### New onset

* headaches beginning at age 40
** mass lesion, temporal arteritis

### More SEVERE and FREQUENT headaches (worst headache ever)

* mass lesion, subdural hematoma, medication overuse, post-coital headache/migraine

### SUDDEN onset (maximal at onset - no increase over time)

* SAH, mass lesion (especially in the posterior fossa)
* new onset in patient with risk factors for HIV or CANCER
** meningitis, brain abscess, metastasis

### headache ASSOCIATED with

* fever (meningitis, encephalitis, systemic infection)
* projectile vomiting
* impaired mental status
* focal neurological signs - weakness, paresthesia (mass lesion, stroke)
* recent head injury
* papilledema (mass lesion, pseudotumor, meningitis)
* meningismus (meningeal irritation)
* seizures
* trauma (intracranial hemorrhage, subdural/epidural hematoma, SAH)

## PMHx

* Hypertension
* HIV
* Cancer
* Trauma
* Recent procedures

## Medication

* Analgesic abuse
* Recreational drugs
* Birth control

## Family history

* Migraines
* Subarachnoid hemorrhage
* Stroke
# Health/Essays

* Medical disclaimer: This page is for educational and informational purposes only and may not be construed as medical advice. The information is not intended to replace medical advice offered by physicians. Please refer to the full text of the Wikiversity medical disclaimer.
* Health/Essays
* Health/Essays/Depression in older adults
* Health/Essays/Myths and Misconceptions about Type II Diabetes
* Health/Essays/Parental health
* Health/Essays/Second hand smoke policies on college campuses
* Health/Essays/Second hand smoking policies on college campuses
* Health/Essays/Secondhand Smoke
# Elasticity/Stress example 1
## Example 1

For a unit cube that is in statical equilibrium and has a homogeneous state of stress, show that there are only six possible components of stress.

### Solution

Since the stress in the cube is homogeneous, the forces on opposite faces of the cube should be equal and opposite. Hence, the possible components of stress are
    $\sigma _{xx},~\sigma _{yy},~\sigma _{zz},~\tau _{xy},~\tau _{yx},~\tau _{yz},~\tau _{zy},~\tau _{zx},~\tau _{xz}.$
The homogeneity of the stress also means that the three components of force on any face must pass through the center of the face.
Since the cube is in static equilibrium, the total moment about an axis passing through the center of the cube and parallel to the $x\,$ axis should be zero. (Recall that the moment of a force about an axis measures the tendency to rotate a rigid body about the axis. Also recall that the sign of a moment is positive or negative depending on the clockwise or anticlockwise direction of rotation due to the applied force).
Hence, $\tau _{yz}=\tau _{zy}\,$. Similarly, $\tau _{zx}=\tau _{xz}\,$, and $\tau _{xy}=\tau _{yx}\,$.
Therefore, the six possible components of stress are
    $\sigma _{xx},~\sigma _{yy},~\sigma _{zz},~\tau _{xy},~\tau _{yz},~\tau _{zx}.$
# Elasticity/Stress example 3
## Example 3

Given:
A stress field whose components in the basis $(\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3})\,$ are given by the matrix
    $[{\boldsymbol {\sigma }}]={\begin{bmatrix}6~x_{1}~x_{3}^{2}&0&-2~x_{3}^{3}\\0&1&2\\-2~x_{3}^{3}&2&3~x_{1}^{2}\end{bmatrix}}$
Find:
* Assuming negligible body forces, show whether this field satisfies equilibrium.
* Find the traction acting at a point P whose position vector is $\mathbf {x} =2~\mathbf {e} _{1}+3~\mathbf {e} _{2}+2~\mathbf {e} _{3}\,$ acting on a plane $2~x_{1}+x_{2}-x_{3}=5\,$.
* Determine the normal traction at this point on the plane.
* Determine the projected shear traction at this point on the plane.
* Determine the principal stresses at point P.
* Determine the principal directions of stress at point P.

### Solution

This problem is very similar to example 2.
The stress is a function of $x_{1}\,$ and $x_{3}\,$. We first plug the stress into the equilibrium equations (Cauchy's first law) and find that the sum is zero. Hence the stress satifies equilibrium.
Next, we find the stress at the given point byplugging in the values of $x_{1}\,$ and $x_{3}\,$ into the given stress.
Then we find the normal to the given plane $a_{i}~x_{i}=b\,$ using the relation
    ${\widehat {\mathbf {n} }}=\pm {\cfrac {a_{i}~\mathbf {e} _{i}}{\sqrt {a_{j}~a_{j}}}}$
and find the traction vector and thenormal and projected shear tractions.
We find the eigenvalues (principal stresses) by first forming the cubic characteristic equation and then solving for it. Then we plug the values of the principal stresses into the eigenfunction and solve for the direction of the principal strains.
Since the stress tensor is symmetric, we need an extra relation between $n_{1}\,$, $n_{2}\,$, and $n_{3}\,$ to reduce the system down to two equations and two unknowns. This relation is $n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1\,$, which means that each eigenvector is a unit vector.
The Maple output is shown below.