OpenLLM-France/wikimedia huggingface dataset:
# Instructional design/Force field analysis/What is FFA?

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## Harry Potter vs. Voldemort in the Goblet of Fire

The principles of force field analysis are all around us, even though the forms they show might vary. For example, forces can be seen in the following clip from "Harry Potter Vs. Voldemort in the Graveyard", when Harry and Voldemort are dueling. In this video, Harry’s goal is to escape Voldemort. You will observe people, places, and magic powers that help or hinder Harry to escape. Please pay more attention on the section of 1:44 - 2:55 and think about how he can use these different forces to accomplish his goal.
At the beginning of the fighting, Voldemort got the upper hand because his power was stronger. In other words, if the magic power made by Harry was stronger than the one from Voldemort, Harry would be able to control and defeat the enemy.

## Similar Examples

Some sports also apply the concepts related to force field analysis, such as tug of war and sumo. You can see the pictures and the descriptions provided below.
| There is a kind of sport called tug of war or rope pulling in which two teams pull at opposite ends of a rope to test strength. How do you think how a team is able to defeat the other? | Tug of war |
| Sumo is a Japanese style of wrestling in which each competitor attempts to force another one out of the circular ring. What should a sumo wrestler do to push the opponent out of the ring? | Sumo |

## What is Force Field Analysis?

The most common thing in the video and pictures is that there are always two forces against each other within these events.
Now, come back to your workplace. Have you ever thought why your supervisor hinders your effort in launching a new reformation? Why your good suggestions are never implemented? While you may or may not be able to observe the reasons, you can feel that if the supporting forces are not with you, you cannot reach to a result as desired as you thought.
Force field analysis is a management tool to analyze the driving forces and restraining forces which affect change initiatives in organizations. Kurt Lewin introduced force field analysis in the 1950’s. Organizations widely use force field analysis as a technique for
   * assessing and identifying effects,
   * planning feasible implementations,
   * implementing organizational changes, and
   * evaluating strategies.
In this lesson, we place the focus on the use of FFA as a tool in the area of needs analysis to help you identify your client’s problems and then plan feasible recommendations.
Click the Next button to continue.
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# Instructional design/Generate PBL Problems

← Go back to ID Page
| 1. ID PBL Problems | 2. Characteristics of PBL Problems | 3. Generate PBL Problems | 4. Practice Now | 5.Summary and References |
Knowing the characteristics of PBL problems helps you generate PBL problems that stimulate learning. The table below lists the three features of PBL with examples of problems that represent each feature. For each problem example, you will:
   * Review the example.
   * Answer some questions about the example.
Features of PBL problems
Click on each sub-features to learn more.
| Feature 1 Enhance Students' Motivation | 1.1 Problems reflect students' learning interests |
| Feature 1 Enhance Students' Motivation | 1.2 Problems should challenge students |
| Feature 1 Enhance Students' Motivation | 1.3 Real-life problems |
| Feature 2 Enhance Students' Critical Thinking Skills | 2.1 Ill-structured problems |
| Feature 2 Enhance Students' Critical Thinking Skills | 2.2 Require higher level thinking activities |
| Feature 3 Enhance Students' Collaboration | 3.1 Group collaboration rules |
When you finish reading all the features of PBL problems,please click on forth session of this class "4. Practice Now."
| 1. ID PBL Problems | 2. Characteristics of PBL Problems | 3. Generate PBL Problems | 4. Practice Now | 5. Summary and References |
# Instructional design/Force field analysis/How FFA is presented in a figure?

Back to ID Homepage > ID force field analysis > What is FFA? > How FFA is presented in a figure? > A case study for practicing FFA > Other FFA figure > Making change successful > How FFA is used?

## Driving Force VS. Restraining Force

As you have seen in the Harry Potter video, there were two different forces helping and hindering Harry at the same time. Here lists these “forces” shown in the video. Could you identify which ones helped Harry and which ones hindered Harry?
From the forces you have identified, we can say:
Those forces helping Harry escape Voldemort can be viewed as driving forces which led him to achieve his goal. On the other hand, those forces hindering Harry can be viewed as restraining forces.
The same concept can be applied to force field analysis which has two parts - driving force and restraining force.
   * Driving force: the force favoring the proposed change or keeping it going
   * Restraining force: the force resisting change or keeping the status quo

## How is FFA Presented?

This graphic is one of the representations of force field analysis. Let’s relate the graphic to the Harry Potter video.
* Harry is currently at ‘a’ - captured by Voldemort. He wants to move to the situation of ‘b’ - escaped from Voldemort
* The arrows pointing right represent driving forces.
* The arrows pointing left represent restraining forces.
The other quality the diagram indicates is the strength of each force.
* Longer arrow = greater strength to drive or restrain change
* Shorter arrow = lower strength to drive or restrain change
If we see this figure from the viewpoint of organization change, according to Lewin’s point, organizations are systems in which the current situation is not static but dynamic with forces working in opposite directions. There are always both driving forces that push for change and restraining forces that act against change in any situation. When one does an analysis, these two forces are shown in balance, like Harry and Voldemort, in a visual depiction that looks above.

## Questions

Now you know what driving forces and restraining forces are and how they are presented in a graphic. Before moving forward, let’s review a case study for practice and see how force field analysis is applied in a bicycle manufacturer.
Click the Next button to practice and review the case.
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# Instructional design/Force field analysis/How force field analysis is used?

Back to ID Homepage > ID force field analysis > What is force field analysis? > How force field analysis is presented in a figure? > How force field analysis is used?
Now you have basic concept of force field analysis and its representative figures, and you might start wondering about how force field analysis can be used in your organization. Generally force field analysis provides functions of
   * Reducing insufficient specification of variables,
   * Identifying strengths and weaknesses, and
   * Recognizing relation to larger system.

## An Example of Using FFA as Motivation Techniques

This fragment of the interview conducted by Paul Grantham will provide you an example of how force field analysis is used in the field of psychological intervention.
Context: Talia is a 30 years old asylum seeker presenting with employment problems and low mood. Her motivation for addressing her problems is affected by repeated failure to obtain work. Paul Grantham, a consultant and a clinical psychologist, demonstrates how to bypass issues of resistance to help the client.
In the clip, after having talked about factors driving Talia to seek more works, Paul Grantham tried to find out those factors decreasing Talia’s motivation of working. Obviously, he was using force field analysis to help create personal change on motivation. Now, try to relate this case to your working situation and think about:
   * Have you ever used any strategy similar to force field analysis before? If you have, what was the situation or problem you were encountering?
   * Is there any change happening in your organization that you think force field analysis might be an effective tool?

## Summary

Now you are almost done with the lesson. The last task for you is to watch a summary video below which will help you review what you have learned during the lesson. In next lesson, you will learn a set of procedures of conducting a force field analysis.
# Instructional design/Force field analysis/What is force field analysis?

Back to ID Homepage > ID force field analysis > What is force field analysis? > How force field analysis is presented in a figure? > How force field analysis is used?

## Harry Potter vs. Voldemort in the Goblet of Fire

The principles of force field analysis are all around us, even though the forms they show might vary. For example, forces can be seen in the following clip from "Harry Potter Vs. Voldemort in the Graveyard", when Harry and Voldemort are dueling. Let's watch the video first and please pay more attention on the section of 1:44 - 2:55. What might you call these forces?
At the beginning of the fighting, Voldemort got the upper hand because his power was stronger. In other words, if the magic power made by Harry was stronger than the one from Voldemort, Harry was able to control and defeat the enemy.

## Similar examples

Some sports also apply the concepts related to force field analysis, such as tug of war and sumo. You can see the pictures and the descriptions provided below.
| There is a kind of sport called tug of war or rope pulling in which two teams pull at opposite ends of a rope to test strength. How do you think how a team is able to defeat the other? | Tug of war |
| In Japan,sumo is a Japanese style of wrestling in which each competitor attempts to force another one out of the circular ring. What should a sumo wrestler do to push the opponent out of the ring? | Sumo |

## What is Force Field Analysis?

The most common thing in the video and pictures is that there are always two forces against each other within these events.
Now, come back to your workplace. Have you ever thought why your supervisor hinders your effort in launching a new reformation? Why your good suggestions are never implemented? While you may or may not be able to observe the reasons, you can feel that if the supporting forces are not with you, you cannot reach to a result as desired as you thought.
Force field analysis is a management tool to analyze the driving forces and restraining forces which affect change initiatives in organizations. Kurt Lewin introduced force field analysis in the 1950’s. Organizations widely use force field analysis as a technique for
   * assessing and identifying effects,
   * planning feasible implementations,
   * implementing organizational changes, and
   * evaluating strategies.
In this lesson, we place the focus on the use of FFA as a tool in the area of needs analysis to help you identify your client’s problems and then plan feasible recommendations.
Click the Next button to continue.
| ID Homepage | ID force field analysis | < Back | Next > |
# Elasticity/Flat punch indentation
## Indentation due to a frictionless rigid flat punch

* Indentation by a plat rigid punch
* Start with uneven surface profile $u_{0}(x_{1})\,$.
* Unsymmetric load $F\,$, but sufficient for complete contact over the area $A\,$.
Displacement in $x_{2}\,$ direction is
    $u_{2}=-u_{0}(x_{1})+C_{1}x_{1}+C_{0}\,$
where $C_{0}\,$ is a rigid body translation and $C_{1}x_{1}\,$ is a rigid body rotation.
Rigid body motions can be determined using a statically equivalent set of forces and moments
    ${\begin{aligned}\int _{A}p(\xi )~d\xi &=-F\\\int _{A}p(\xi )\xi ~d\xi &=-F~d\end{aligned}}$
The displacement gradient is
    $-{\frac {du_{0}}{dx_{1}}}+C_{1}=-{\frac {(\kappa +1)}{4\pi \mu }}\int _{-a}^{a}{\frac {p(\xi )}{x-\xi }}~d\xi ~;~~-a<x<a$
Integral is a Cauchy Singular Integral that appears often and very naturally when the problem is solved using complex variable methods.
Note that the only thing we are interested in is the distribution of contact forces $p(\xi )\,$.If we change the variables so that
$x=a\cos \phi \,$ and $\xi =a\cos \theta \,$, then
    ${\frac {1}{a\sin \phi }}{\frac {du_{0}}{d\phi }}+C_{1}=-{\frac {(\kappa +1)}{4\pi \mu }}\int _{0}^{\pi }{\frac {p(\theta )\sin \theta }{\cos \phi -\cos \theta }}~d\theta ~;~~0<\phi <\pi$
If we write $p(\theta )\,$ and $du_{0}/d\phi \,$ as
    ${\begin{aligned}p(\theta )&=\sum _{0}^{\infty }{\frac {p_{n}\cos(n\theta )}{\sin \theta }}\\{\frac {du_{0}}{d\phi }}&=\sum _{1}^{\infty }u_{n}\sin(n\phi )\end{aligned}}$
and do some algebra, we get
    ${\begin{aligned}p_{0}&=-{\frac {F}{\pi a}}\\p_{1}&=-{\frac {Fd}{\pi a^{2}}}\\p_{n}&=-{\frac {4\mu u_{n}}{(\kappa +1)a}}~;~~n>1\end{aligned}}$

### Flat punch with symmetric load

    $u_{0}=C\,$
In this case,
    ${\frac {du_{0}}{d\phi }}=0\Rightarrow u_{n}=0~;~~n=1{\infty }$
Also, $d=0\,$ (origin at the center of $A\,$), hence $p_{1}=0\,$. Therefore,
    $p(x)={\frac {p_{0}}{\sin \phi }}=-{\frac {F}{\pi {\sqrt {a^{2}-x^{2}}}}}$
At $x=\pm a$, the load is infinite, i.e. there is a singularity.
# Elasticity/Equilibrium example 1
## Example 1

Given:
Euler's second law for the conservation of angular momentum
    ${\text{(1)}}\qquad \int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)$
The divergence theorem
    ${\text{(2)}}\qquad \int _{\partial B}n_{i}~\sigma _{ij}~dS=\int _{B}{\frac {\partial \sigma _{ij}}{\partial x_{i}}}~dV$
The equilibrium equation (Cauchy's first law)
    ${\text{(3)}}\qquad {\frac {\partial \sigma _{ij}}{\partial x_{i}}}+\rho ~b_{j}={\frac {d}{dt}}\left(\rho ~v_{j}\right)$
Show:
    ${\text{(4)}}\qquad \sigma _{ij}=\sigma _{ji}$

### Solution

Let us first look at the first term of equation~(1) and apply the divergence theorem (2). Thus,
    ${\begin{aligned}\int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS&=\int _{\partial B}n_{l}~(e_{ijk}~x_{j}~\sigma _{lk})~dS\\&=\int _{B}{\frac {\partial {(e_{ijk}~x_{j}~\sigma _{lk})}}{\partial x_{l}}}~dV\\&=\int _{B}\left(e_{ijk}~{\frac {\partial x_{j}}{\partial x_{l}}}~\sigma _{lk}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ijk}~\delta _{jl}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\end{aligned}}$
Plugging this back into equation~(1) gives
    $\int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)$
Therefore, bringing all terms to the left hand side,
    ${\text{(5)}}\qquad \int _{B}\left[e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}\left({\frac {\partial \sigma _{lk}}{\partial x_{l}}}+\rho ~b_{k}-{\frac {d}{dt}}\left(\rho ~v_{k}\right)\right)\right]~dV=0$
Using the equilibrium equations~(3), equation~(5) reduces to
    ${\text{(6)}}\qquad \int _{B}e_{ilk}~\sigma _{lk}~dV=0$
Since this holds for any $B$, we have
    ${\text{(7)}}\qquad e_{ilk}~\sigma _{lk}=0$
If you work this expression out, you will see that $\sigma _{ij}=\sigma _{ji}$. Hence, the stress tensor is symmetric.
# Elasticity/Energy methods example 2
## Example 2

Given:
The potential energy functional for a membrane stretched over a simply connected region ${\mathcal {S}}$ of the $x_{1}-x_{2}$ plane can be expressed as
    $\Pi [w(x_{1},x_{2})]={\frac {1}{2}}\int _{\mathcal {S}}\eta \left[(w_{,1})^{2}+(w_{,2})^{2}\right]~dA-\int _{\mathcal {S}}pw~dA$
where $w(x_{1},x_{2})$ is the deflection of the membrane, $p(x_{1},x_{2})$ is the prescribed transverse pressure distribution, and $\eta$ is the membrane stiffness.
Find:
* The governing differential equation (Euler equation) for $w(x_{1},x_{2})$ on ${\mathcal {S}}$.
* The permissible boundary conditions at the boundary $\partial {\mathcal {S}}$ of ${\mathcal {S}}$.

### Solution

The principle of minimum potential energy requires that the functional $\Pi$ be stationary for the actual displacement field $w(x_{1},x_{2})$. Taking the first variation of $\Pi$, we get
    $\delta \Pi ={\frac {\eta }{2}}\int _{\mathcal {S}}\left[(2w_{,1})\delta w_{,1}+(2w_{,2})\delta w_{,2}\right]~dA-\int _{\mathcal {S}}p~\delta w~dA$
or,
    $\delta \Pi =\eta \int _{\mathcal {S}}\left[w_{,1}~\delta w_{,1}+w_{,2}~\delta w_{,2}\right]~dA-\int _{\mathcal {S}}p~\delta w~dA$
Now,
    ${\begin{aligned}(w_{,1}~\delta w)_{,1}&=w_{,11}~\delta w+w_{,1}~\delta w_{,1}\\(w_{,2}~\delta w)_{,2}&=w_{,22}~\delta w+w_{,2}~\delta w_{,2}\end{aligned}}$
Therefore,
    $w_{,1}~\delta w_{,1}+w_{,2}~\delta w_{,2}=(w_{,1}~\delta w)_{,1}-w_{,11}~\delta w+(w_{,2}~\delta w)_{,2}-w_{,22}~\delta w$
Plugging into the expression for $\delta \Pi$,
    $\delta \Pi =\eta \int _{\mathcal {S}}\left[(w_{,1}~\delta w)_{,1}+(w_{,2}~\delta w)_{,2}-(w_{,11}+w_{,22})~\delta w\right]~dA-\int _{\mathcal {S}}p~\delta w~dA$
or,
    $\delta \Pi =\eta \int _{\mathcal {S}}\left[(w_{,1}~\delta w)_{,1}+(w_{,2}~\delta w)_{,2}\right]~dA-\eta \int _{\mathcal {S}}\nabla ^{2}{w}~\delta w~dA-\int _{\mathcal {S}}p~\delta w~dA$
Now, the Green-Riemann theorem states that
    $\int _{\mathcal {S}}(Q_{,1}-P_{,2})~dA=\oint _{\partial {\mathcal {S}}}(P~dx_{1}+Q~dx_{2})$
Therefore,
    $\delta \Pi =\eta \oint _{\partial {\mathcal {S}}}\left[(w_{,1}~\delta w)~dx_{2}-(w_{,2}~\delta w)~dx_{1}\right]-\int _{\mathcal {S}}\left[\eta \nabla ^{2}{w}+p\right]~\delta w~dA$
or,
    $\delta \Pi =\eta \oint _{\partial {\mathcal {S}}}\left[w_{,1}~{\frac {dx_{2}}{ds}}-w_{,2}~{\frac {dx_{1}}{ds}}\right]\delta w~ds-\int _{\mathcal {S}}\left[\eta \nabla ^{2}{w}+p\right]~\delta w~dA$
where $s$ is the arc length around $\partial {\mathcal {S}}$.
The potential energy function is rendered stationary if $\delta \Pi =0$. Since $\delta w$ is arbitrary, the condition of stationarity is satisfied only if the governing differential equation for $w(x_{1},x_{2})$ on ${\mathcal {S}}$ is
    ${\eta \nabla ^{2}{w}+p=0~~~~~\forall ~~(x_{1},x_{2})~\in ~{\mathcal {S}}}$
The associated boundary conditions are
    ${w_{,1}~{\frac {dx_{2}}{ds}}-w_{,2}~{\frac {dx_{1}}{ds}}=0~~~~~\forall ~~(x_{1},x_{2})~\in ~\partial {\mathcal {S}}}$
# Elasticity/Energy methods example 3
## Example 3 : Torsion of a cylinder

Suppose that the cross-section of the cylinder is bounded by the curves:
    ${\begin{aligned}x_{1}&=c\\x_{1}&=d\\x_{2}&=g(x_{1})\\x_{2}&=-g(x_{1})\end{aligned}}$
A statically admissible Prandtl stress function for this cross-section is
    $\phi =[x_{2}^{2}-g^{2}(x_{1})]f(x_{1})$
with the restrictions that $f(x_{1})\,$ is twice continuously differentiable and $f(c)=f(d)=0\,$.
We seek to derive the best approximate Prandtl stress function of this form by minimizing the complementary energy.
You can show that the complementary energy per unit length of the cylinder can be expressed as
    $\Pi ^{c}[\phi (x_{1},x_{2})]=\int _{\mathcal {S}}\left[{\frac {1}{2\mu }}(\phi _{1}^{2}+\phi _{,2}^{2})-2\alpha \phi \right]~dA$
Plugging in the given form of $\phi$ and after some algebra, we get
    $\Pi ^{c}[f(x_{1})]={\frac {15}{16\mu }}\int _{c}^{d}\left[{\frac {1}{2}}g^{5}(f^{'})^{2}+{\frac {5}{2}}g^{4}g^{'}ff^{'}+{\frac {15}{4}}g^{3}(g^{'}f)^{2}+{\frac {5}{4}}g^{3}f^{2}+{\frac {5}{2}}\mu \alpha g^{3}f\right]~dx_{1}$
Taking the variation of $\Pi$, and after considerable manipulation, we get
    $\delta \Pi ^{c}={\frac {15}{16\mu }}\int _{c}^{d}{\frac {g^{5}}{4}}\left[f^{''}-{\frac {5}{g^{2}}}(g^{2}f)^{''}+{\frac {10}{g^{2}}}f+{\frac {10\mu \alpha }{g^{2}}}\right]\delta f~dx_{1}$
Now, if we consider the cross-section to be rectangular, then we have $c=-a$, $d=a$, and $g=b$. Therefore, the above equation reduces to
    $\delta \Pi ^{c}={\frac {15}{16\mu }}\int _{-a}^{a}{\frac {b^{5}}{4}}\left[f^{''}-5f^{''}+{\frac {10}{b^{2}}}f+{\frac {10\mu \alpha }{b^{2}}}\right]\delta f~dx_{1}$
Therefore, the function $f$ that minimizes $\Pi ^{c}$ satisfies the equation
    $-4f^{''}+{\frac {10}{b^{2}}}f+{\frac {10\mu \alpha }{b^{2}}}=0$
or,
    $f^{''}-{\frac {5}{2b^{2}}}f-{\frac {5\mu \alpha }{2b^{2}}}=0$
with the static admissibility conditions $f(-a)=f(a)=0\,$.
The general solution of the above equation is
    $f=A\sinh \left({\sqrt {2.5}}~{\frac {x_{1}}{b}}\right)+B\cosh \left({\sqrt {2.5}}~{\frac {x_{1}}{b}}\right)-\mu \alpha$
Therefore, from the BCs,
    $A=0~~;~~B={\frac {\mu \alpha }{\cosh {{\sqrt {2.5}}~a/b}}}$
We can substitute back into $\phi$ to get the approximate Prandtl stress function for this problem. The error between the exact and this approximate solution is generally less than 1%.
# Elasticity/Flamant solution
## The Flamant Solution

* Elastic wedge loaded by two forces at the tip
* This problem is also self-similar (no inherent length scale).
* All quantities can be expressed in the separated-variable form $\sigma =f(r)g(\theta )$.
* The stresses vary as $(1/r)$ (the area of action of the force decreases with increasing $r$). How about a conical wedge ?
From Michell's solution, pick terms containing $1/r$ in the stresses. Then,
    $\varphi =C_{1}r\theta \sin \theta +C_{2}r\ln r\cos \theta +C_{3}r\theta \cos \theta +C_{4}r\ln r\sin \theta$
Therefore, from Tables,
    ${\begin{aligned}\sigma _{rr}&=C_{1}\left({\frac {2\cos \theta }{r}}\right)+C_{2}\left({\frac {\cos \theta }{r}}\right)+C_{3}\left({\frac {2\sin \theta }{r}}\right)+C_{4}\left({\frac {\sin \theta }{r}}\right)\\\sigma _{r\theta }&=C_{2}\left({\frac {\sin \theta }{r}}\right)+C_{4}\left({\frac {-\cos \theta }{r}}\right)\\\sigma _{\theta \theta }&=C_{2}\left({\frac {\cos \theta }{r}}\right)+C_{4}\left({\frac {\sin \theta }{r}}\right)\end{aligned}}$
From traction BCs, $C_{2}=C_{4}=0$. From equilibrium,
    ${\begin{aligned}F_{1}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\cos \theta d\theta &=0\\F_{2}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\sin \theta d\theta &=0\end{aligned}}$
After algebra,
    $\sigma _{rr}={\frac {2C_{1}\cos \theta }{r}}+{\frac {2C_{3}\sin \theta }{r}}~;~~\sigma _{r\theta }=0~;~~\sigma {\theta \theta }=0$

## Special Case : α = -π, β = 0

    $C_{1}=-{\frac {F_{1}}{\pi }}~;~~C_{2}={\frac {F_{2}}{\pi }}$
The displacements are
    ${\begin{aligned}u_{1}&=-{\frac {F_{1}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}+{\frac {F_{2}(\kappa +1){\text{sign}}(x_{1})}{8\mu }}\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}-{\frac {F_{1}(\kappa +1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}$
where
    ${\begin{aligned}\kappa =3-4\nu &&{\text{plane strain}}\\\kappa ={\frac {3-\nu }{1+\nu }}&&{\text{plane stress}}\\\end{aligned}}$
and
    ${\text{sign}}(x)={\begin{cases}+1&x>0\\-1&x<0\end{cases}}$
# English-Chinese/Some experts refer to long sitting as "the new smoking" because of its negative effects on overall health

Some experts refer to long sitting as "the new smoking" because of its negative effects on overall health.
一些专家将长期坐着称为“新吸烟”，因为它对整体健康有负面影响。
# English-Chinese/Some organizations have promoted active travel

As a response to concerns over health and environmental issues, some organizations have promoted active travel, which seeks to promote walking and cycling as safe and attractive alternatives to motorized transport.
作为对健康和环境问题的关注的回应，一些组织促进了积极的交通方案，旨在促进步行和骑自行车作为机动交通的安全和有吸引力的替代方案。
# English-Chinese/Some parts of the report were blacked out.

Some parts of the report were blacked out.
报告的某些部分被涂黑了。
# English-Chinese/Some risk factors are immutable

Some of the risk factors, such as age, gender or family history/genetic predisposition, are immutable; however, many important cardiovascular risk factors are modifiable by lifestyle change, social change, drug treatment (for example prevention of hypertension, hyperlipidemia, and diabetes).
一些风险因素，如年龄，性别或家族史/遗传倾向，是不可改变的; 然而，许多重要的心血管危险因素可通过改变生活方式，社会变化，药物治疗（例如预防高血压，高脂血症和糖尿病）来改变。