text:
'use strict';

const CALLABLE_SERVICE_FACTORY_MODULE = `${process.cwd()}/lib/helpers/factories/CallableServiceFactory`;
const RUNNABLE_SERVICE_FACTORY_MODULE = `${process.cwd()}/lib/helpers/factories/RunnableServiceFactory`;
const CONFIGURATION_LOADER_MODULE = `${process.cwd()}/lib/helpers/loaders/ConfigurationLoader`;
const SERVICE_CONTEXT_LOADER_MODULE = `${process.cwd()}/lib/helpers/loaders/ServiceContextLoader`;
const SERVICE_FUNCTIONALITY_LOADER_MODULE = `${process.cwd()}/lib/helpers/loaders/ServiceFunctionalityLoader`;
const LOGGER_MANAGER_MODULE = `${process.cwd()}/lib/helpers/logging/LoggerManager`;
const CHECK_SCHEMA_MODULE = `${process.cwd()}/lib/helpers/schemas/CheckSchema`;
const CREDENTIALS_GENERATOR_MODULE = `${process.cwd()}/lib/helpers/security/CredentialsGenerator`;
const CLIENT_MODULE = `${process.cwd()}/lib/platform/Client`;
const MICROREST_MODULE = `${process.cwd()}/lib/platform/Microrest`;
const SERVER_MODULE = `${process.cwd()}/lib/platform/Server`;
const SERVICE_MANAGER_MODULE = `${process.cwd()}/lib/platform/ServiceManager`;
const CALLABLE_SERVICE_MODULE = `${process.cwd()}/lib/services/CallableService`;
const RUNNABLE_SERVICE_MODULE = `${process.cwd()}/lib/services/RunnableService`;
const SERVICE_MODULE = `${process.cwd()}/lib/services/Service`;
const SERVICE_CONTEXT_MODULE = `${process.cwd()}/lib/services/ServiceContext`;

const CONFIGURATION_REAL_FILE = `${process.cwd()}/configuration.json`;
const CONFIGURATION_TEST_FILE = `${process.cwd()}/test/env/configuration.json`;
const CONFIGURATION_LOGGER_TEST_FILE = `${process.cwd()}/test/env/configurationLogger.json`;

/**
 * Microrest Framework Modules for testing
 *
 * @public
 * @static
 * @readonly
 * @constant {Object}
 */
module.exports = {
    callableServiceFactory: CALLABLE_SERVICE_FACTORY_MODULE,
    runnableServiceFactory: RUNNABLE_SERVICE_FACTORY_MODULE,
    configurationLoader: CONFIGURATION_LOADER_MODULE,
    serviceContextLoader: SERVICE_CONTEXT_LOADER_MODULE,
    serviceFunctionalityLoader: SERVICE_FUNCTIONALITY_LOADER_MODULE,
    loggerManager: LOGGER_MANAGER_MODULE,
    checkSchema: CHECK_SCHEMA_MODULE,
    credentialsGenerator: CREDENTIALS_GENERATOR_MODULE,
    client: CLIENT_MODULE,
    microrest: MICROREST_MODULE,
    server: SERVER_MODULE,
    serviceManager: SERVICE_MANAGER_MODULE,
    callableService: CALLABLE_SERVICE_MODULE,
    runnableService: RUNNABLE_SERVICE_MODULE,
    service: SERVICE_MODULE,
    serviceContext: SERVICE_CONTEXT_MODULE,

    configurationRealFile: CONFIGURATION_REAL_FILE,
    configurationTestFile: CONFIGURATION_TEST_FILE,
    configurationLoggerTestFile: CONFIGURATION_LOGGER_TEST_FILE
};

redpajama_set_name: RedPajamaGithub


text:
Med-Esthetics Magazine Feature Article for Esthetic Skin Institute
The Esthetic Skin Institute, a highly accredited resource for medical aesthetics training and certification for medical professionals, owes its success to its founder's unique combination of skills and experience.
Working as a registered nurse in Florida in the 1990s, Sasha Parker discovered that she also had a keen interest in skin care and holistic health. "I had a friend who was working as an esthetician," she explains, "and she was so happy working with healthy, appreciative clients. She was also doing very well." So Parker obtained licenses both as an esthetician and as an electrologist and started to establish a skin care practice. Soon she was working in both fields with a goal of eventually phasing out of nursing and working full time as a skincare specialist. Her plans changed however after an encounter with a physician at the facility where she worked. Observing his high degree of stress and fatigue, she suggested that he consider incorporating skin care into his practice, explaining the advantages and benefits of doing so. At the time, the concept of aesthetic medicine was so new that the MD was uncertain and declined. Later, after attending one of her educational presentations, he invited her to meet with him and his partner. Convinced of the potential for enhancing his practice with aesthetics, he invited her to join them.
That was the beginning of the Esthetics Skin Institute. As time passed and the medical aesthetics field developed, more and more doctors and nurses contracted with Parker for training. Because she was a gifted and passionate educator, training soon eclipsed all of her other activities and in 1997 she established the Institute.
Today, Parker provides courses for doctors, registered nurses, nurse practitioners, physician assistants, dentists and dental surgeons as well as select courses for licensed practical nurses and licensed vocational nurses. No matter what the subject, her primary goal is training practitioners to provide safe, high-quality care. "It's amazing how many of us feel that just because we have a license to provide medical care, we are knowledgeable enough to provide aesthetics care," she says. But this isn't always the case. "In my state," Parker reports, "from 1997 to 2003, 244 cases of adverse events were reported from laser treatments. Out of those 244 events, 202 treatments were administered by physicians." She believes it's imperative, therefore, for every individual in the medical aesthetics profession—from physicians to nurses to physician assistants—to obtain in-depth, hands-on training. "The key to the good delivery of care," says Parker, "is a good education."
Her high standards for Esthetic Skin Institute courses receive continuous praise from her students, and her return and referral rates are high. "My students return for training in all of the aesthetics services they plan to offer in their practices," she notes. "They also refer colleagues who wish to transition into aesthetics or add aesthetic services to their practices."
Another indication of the quality of her service is the caliber of physicians who comprise her board of advisers. "Such highly respected physicians do not become affiliated with an organization that is not high quality," she observes.
Parker has been referred to as "a visionary and a pioneer ahead of her time" for developing this training system. Her response? "I have always been a seeker and giver of knowledge," she notes, "and I have had some great teachers along the way who helped me, so now I strive to help others."

redpajama_set_name: RedPajamaCommonCrawl


text:
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Credit Ratings Set Up Sharjah for More Investments, Debt Sales
debt sales
Nicolas Parasie
AccessDeniedAccess Denied7608D3F8D3332448XKoyxFG3VjAuyd57ocEw8lj9MjuX9PIkLORt0ieBupNxjNiKX2KLvxAB51a/bJr3qQRro/YpG8Y=
Sharjah, one of seven semi-autonomous members that make up the United Arab Emirates, on Wednesday said it was assigned its first sovereign credit rating by Standard & Poor's and Moody's – a move that helps strengthen its investment profile and paves the way for future debt sales.
S&P rated the emirate A, while Moody's handed it an A3 rating – strong credit scores that are broadly in the same league as those of fellow emirates Abu Dhabi and Ras Al Khaimah.
S&P said its rating for Sharjah reflected "the emirate's comparatively wealthy and diverse economy, limited fiscal risks and low government debt, and the likelihood of support from the UAE if the need arises."
Moody's meanwhile referred to Sharjah's strong finances. "The government has recorded only small fiscal deficits of around 1%-2% of GDP on average since 2008. Moreover, government debt is very low at around 6% of GDP as of 2012 and Moody's expects it to remain below 10% in 2014."
Waleed Al Sayegh, director general of Sharjah's finance department
Waleed Al Sayegh, director general of Sharjah's finance department, said the credit rating would open the door to new investors from abroad and boost the confidence of local business. Combined with the benefit of potentially lower borrowing costs, the credit rating will help the emirate to further invest in infrastructure, he said.
Officials said the emirate isn't planning a conventional or Islamic bond issuance for now but said the credit rating ensures that is a realistic alternative to secure funds in the future.
"What we can say is that at some point in the future when we want to go to the debt capital markets, this marks an important step on our journey if we want to do that," said Tom Koczwara, director of Sharjah's debt management office. Despite Mr. Koczwara's cautious comments, bankers from both local and international financial institutions were swarming him after the press conference, possibly to get a head start should the emirate decide to tap advisors for a debt sale.
Sharjah is the U.A.E.'s third-largest emirate. It is often eclipsed by its fast-growing neighbor Dubai and oil-rich Abu Dhabi. Sharjah was named Capital of Islamic Culture for 2014 and Capital for Arab Tourism in 2015, underlining the emirate's ambitions to keep up with some of its neighbours as it continues to diversify its economy.
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Sarangra is a village in Chogawan-2 Tehsil in Amritsar District of Punjab State, India. It is located 33 km to the west of District headquarters Amritsar. 8 km from Chogawan-2. 275 km from State capital Chandigarh

Sarangra Pin code is 143110 and postal head office is Preet Nagar (Amritsar).

Odhar (3 km), Kakar (4 km), Manj (4 km), Bachiwind (4 km), Lopoke (5 km) are the nearby villages to Sarangra. Sarangra is surrounded by Harsha Chhina Tehsil to the east, Ajnala-1 Tehsil to the east, Gandiwind-9 Tehsil to the south, Amritsar Tehsil to the east.

Amritsar, Tarn Taran, Patti, Batala are the nearby cities to Sarangra.

Demographics
Punjabi is the local language.

References

Villages in Amritsar district

redpajama_set_name: RedPajamaWikipedia


text:
Live Monster Energy Nascar Cup Series Dover Online
AAA 400 Drive For Autism NASCAR Stream Live
at 1:00 P.M ET On 4 June 2017
You can watch Monster Energy NASCAR Cup Series stock car race AAA 400 Drive For Autism live stream on your PC, laptop, Mac, Tablet, Smart TV and other Android devices globally. The NASCAR AAA 400 Drive For Autism 2017 is a 400-mile (640 km) race which will be taking place every year at the Dover International Speedway in Dover, Delaware. Watch AAA 400 Drive For Autism Monster Energy NASCAR live online on Sunday 4th June 2017 at the time 1:00 P.M ET. In 1969 it was first established, before two years it was a 300 laps race and in 1971 it was 500 laps and from 1998 it has been 400-lap event till now. Don't miss out to watch 2017 AAA 400 Drive for Autism Monster Energy NASCAR Cup Series race live HD streaming on your modern devices across the world.
Event: NASCAR Monster Energy Cup Series – AAA 400 Drive For Autism
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Location: Dover International Speedway, Dover, DE
TV: Fox Sports 1
About Track:
Dover International Speedway is a one of the sport's iconic racetracks and it is nicknamed "The Monster Mile. In 1969 it was first opened, it is one mile long with banking 24 degrees in the corners and 9 degrees down straightaways.
Last Year's Winners:
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2015: Jimmie Johnson
2013: Tony Stewart
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Reindeers are a fun and exciting Christmas decoration for your children to make.
Begin by marking out in pencil where you want the 4 feet to go at the bottom of your toilet roll tube. Once you have done that, cut out an arch shape between the feet. Use this arch shape as a template to cut out the other legs.
Cut approximately the top third off the toilet roll, keeping a longer neck for your reindeer.
Draw an oval shape for the reindeer's head onto your piece of white card. Draw two fairly large ear shapes about a third of the way down. Eventually you will want to fold the head along the ears in order to attach the antlers so bear this in mind when drawing your head. If you are making more than one reindeer, then use the first head shape as a template.
Paint the toilet roll tube and head with brown paint. Leave to dry.
Draw half an antler shape along the fold.
Cut it out, then unfold the card to reveal the complete antler. Use this as a template for the second antler.
Add the eyes and nose using a black marker pen.
Fold the head shape along the line of the ears and cut two small slits for the antlers.
Slot the antlers into the slits and sellotape in place.
Using the sellotape, stick the head onto the reindeer's neck.
my toddler and I had a lot of fun making our reindeers and they make fantastic festive decorations. If we were making them again, I would use a lighter brown paint, as then the eyes and mouth would show up better. We hope you enjoy making these with your children. You could put them up around the house, or make a small hole in them, add a pretty ribbon and hang them on your Christmas tree.
This morning the Toddlebike2 is officially launching in the UK. Several toddlers were chosen to take part in the launch event and we are thrilled that my toddler is one of them! She was given the important task of testing out the Toddlebike2 in advance of the launch and we are excited to have the opportunity to review it.
Two weeks ago, a huge parcel arrived in the post for my toddler.
… and get her hands on her brand new Toddlebike2!
The Toddlebike2 is suitable for toddlers aged 18-36 months and is a pre-balance bike. It is available in three colours: Racing Red, Midnight Blue and Pinky Pink. We opted for Racing Red.
my toddler took to her Toddlebike2 straight away and immediately got the hang of moving forwards on the bike.
It is easy to steer and she was able to manoeuvre easily around the room. Initially, she preferred to sit on the cross-bar but soon learned that she could still grip the handlebars while sitting back in the seat.
She found it tricky to get on and off the bike by herself, but her confidence in her ability to do this has really improved.
One of the appealing features of the bike is that it is super lightweight, weighing approximately 0.8kg – around four times lighter than most balance bikes. my toddler immediately set about testing out this claim for herself and found that she could lift it with ease.
This means that the bike is easily transportable, and it was simple to pop it under the buggy/clip it on with a buggy hook to take to the park, although since the temperatures have dropped we have not been able to use the bike outside as much as we would have liked. my toddler toppled over a couple of times while getting used to it, but the bike is so light that she was able to get back up easily and lift the bike into an upright position by herself.
The Toddlebike2 is very hard-wearing and can be ridden on a range of terrains from mud and grass to carpets. We've taken ours out into the garden and my toddler has enjoyed steering it around.
Even better for mummies, it is also very easy to clean!
The Toddlebike2 has quickly become one of my toddler's favourite toys. It has really helped to develop her confidence on a bike, and she can move quite quickly on it, so it is great to take out on short trips. It's helping to build up her independence, as she can use it at times when she would usually go in the buggy.
The Toddlebike2 comes with a three year guarantee and can be bought at John Lewis or on Amazon.
We would highly recommend the Toddlebike2 to anyone looking for a sturdy, safe and fun first bike for their little ones. It would make a perfect Christmas present for your toddler!
Thank you to the wonderful Jo Hockley for letting us try out this fantastic product and we wish you every success with the launch!
Over the weekend, we had a fantastic outing to Waddesdon Manor, Buckinghamshire (https://www.nationaltrust.org.uk/waddesdon-manor/). The grounds were bright with autumnal colours; we were surrounded by splashes of red, gold and amber. Inspired by the beautiful scenery, we collected some pretty leaves, conkers, pine cones and twigs covered in lichen.
It was lovely to bring a little bit of autumn into our house so my toddler could explore these interesting things in more detail.
A sensory box is a safe and simple way for toddlers to explore new ideas and objects. This was extremely straightforward to set up. All you need is a plastic box, or a cardboard box would do. I filled the base of the box with bird seed, but you could also use rice, popcorn kernels or sand. I put in some of the leaves, pine cones, twigs and conkers that we had collected at Waddesdon.
Alongside these, I added some little toy containers, a funnel, a bucket and a little plastic spade.
You could also add a few cinnamon sticks to stimulate your child's sense of smell. Pop the box on the floor on top of an old sheet/splash mat and your toddler can have lots of fun exploring! my toddler began by sorting all the pine cones into the bucket.
But her favourite thing to do was to play with the bird seed and transfer it into the different containers.
I actually added three more little bowls of various sizes for her to use.
my toddler loved this activity and was happily absorbed for over an hour.
Next, we decided to use some of the leaves we had collected to create some foraging creatures of our own: hedgehogs! This is a quick activity to set up. Draw a hedgehog shape onto some brown card or sugar paper. Cut out the hedgehog. To make several identical ones, use the first as your template.
Your toddler (guided by you) can then stick some leaves onto the body of the hedgehog, pointy sides at the top to emulate prickles.
To finish, simply draw the eyes, nose and mouth onto the face of the hedgehog using a black marker pen.
my toddler enjoyed sticking the leaves down, but I think that an older toddler would probably get more out of this activity, as they would be able to do more of it themselves.
We were lucky enough to find some nice conkers at Waddesdon, which were perfect for our next craft activity. To create conker caterpillars, simply make a hole through the middle of the conker. You can use a skewer, scissors or even a drill. I used a skewer. Then, grab some pipe cleaners. You will need to twist the end of the pipe cleaner to stop the conkers from falling off and also to avoid the sharp end from hurting your toddler. Then your toddler can have lots of fun practising his or her fine motor skills, threading the conkers onto the pipe cleaner.
I put a little bit of sellotape around the top of the pipe cleaner to protect my toddler's fingers from the metal tip and also to make it easier for her to thread the conkers. When enough are on the pipe cleaner, you can remove the sellotape and twist the other end to secure the conkers. Then simply stick two goggly eyes onto the first conker to make your very own conker caterpillar!
my toddler found the finished product fascinating and enjoyed playing with it. She managed to pull the goggly eyes off very quickly though, so this is something to watch out for.
Next time you and your children are out and about enjoying the autumnal surroundings, don't forget to collect some treasures of your own. We hope you enjoy recreating these activities. Happy foraging!
One of my toddler's favourite books is What the Ladybird Heard by Julia Donaldson. For anyone unfamiliar with Julia Donaldson, she was the Children's Laureate from 2011-13 and has written a host of children's books, her most famous arguably being The Gruffalo. We discovered her books when my toddler was still a small baby and they were an instant hit, thanks to the wonderful rhyming text and imaginative storylines. my toddler particularly likes What the Ladybird Heard because of the different animal noises and the vibrant illustrations by Lydia Monks. On every page there is a small, sparkly ladybird, and my toddler enjoys finding and pointing to her while we are reading.
Inspired by the book, I created some ladybird-themed craft activities for my toddler.
To prepare, I cut a potato in half then scooped out several small circular holes to create the distinctive spots on a ladybird. I used the pointy end of a potato peeler to do this.
I also cut out a 'handle' to make it easier to hold the potato.
I poured some red paint onto a paper plate and my toddler had lots of fun dipping the potato into the paint and printing ladybird shapes onto the paper.
We used black sugar paper so that the red paint would really stand out. Also, it would have been tricky for my toddler to separately paint in the black spots at her age (17 months), so this solved the problem! my toddler loved this activity and I would highly recommend it for any toddlers of a similar age.
This activity requires parental help and takes a little longer to complete, as you have to wait for the paint to dry, but it is worth the effort to create your very own little ladybirds!
Cut the cups out of the egg cartons.
Paint each of the cups red and leave to dry. For a sparkly ladybird, you could add some red glitter or use red glitter paint.
Once the cups are dry, add black dots. You could use a marker pen, black circular stickers or black paint. Next stick or glue the goggly eyes onto the black pom pom.
Cut some black pipe cleaners into about 4 cm long strips and bend them into "L" shapes for the antennae. You will need two strips per ladybird.
Finally, glue two pipe cleaners and the pom pom onto the red body, and your ladybird is complete!
Next, we needed to create somewhere for our ladybirds to live!
For the base of the sensory box, I used white rice that I had dyed green to give the illusion of grass.
• Pour 500g uncooked rice into a plastic container which has a lid.
• Add green food colouring and 1 ½ – 2 tsp of white vinegar.
• I also added a few drops of lavender oil to eradicate the smell of vinegar.
• Pop the lid on and shake the container to evenly distribute the colour.
• Finally, tip the rice mixture onto a baking tray/into a roasting pan, and leave to dry for at least 24 hours. It actually took longer than this for my rice to dry, so I put it into the oven on a low heat for about an hour to speed things up.
• Once dry, the rice can be used and re-used. Just make sure you store it in an air-tight container.
Once the 'grass' was dry, I added the egg carton ladybirds, some pine cones, fir tree twigs and conkers to the box.
Next to the box I placed some containers, a mini muffin tray, a little plastic spade and a rake.
my toddler had lots of fun with this activity but it was rather messy, so I would definitely advise you to put down a large tablecloth/old sheet on the floor to catch the rice. my toddler spent most of her time picking up the rice and letting it trickle out of her hand into all the different containers.
I will do you no harm.
your colourful wings are my joy.
your children are crying so.
They will do you no harm.
and greet them both for me.
We hope you and your children have lots of fun creating some ladybirds of your own and you enjoy these activities as much as we have.
my toddler and I had lots of fun getting ready for Guy Fawkes Night, or Bonfire Night as it is also known, this year. At 17 months, my toddler is still a little young for an evening of bonfires, sparklers and fireworks, so I decided to create a toddler-friendly Bonfire Afternoon for her to enjoy with four different activities.
First, we made Foil Sparklers. These are incredibly easy to make. All you need is a rectangle of kitchen foil. Tear or cut horizontally in strips about 1cm apart from the left side into the middle until you have a fringe.
Then carefully roll the foil from the bottom upwards, taking care not to damage the fringe. Twist to keep it all together and then give it to your toddler to play with!
my toddler loved her foil sparkler, and particularly the exciting noise it made when shaken!
Now my toddler had a sparkler, it was time to recreate some fireworks! For this, we made sensory bottles.
Add equal quantities of water and oil (you can add a little more oil than water if desired), then add some food colouring. Finally, pop between 1 and 3 effervescent tablets into the water bottle and enjoy watching the reaction!
I only used 1 tablet in each bottle but if I were doing this activity again I would use more, as the reaction was not as powerful as I had anticipated. Nevertheless, my toddler found these bottles mesmerising and played with them for ages!
We finished with two painting activities.
my toddler began by painting fireworks using pipe cleaners. This was very simple to set up. I took 3-4 pipe cleaners, twisted them at the middle point and then fanned the ends out like a star. This creates the effect of fireworks exploding.
I also put out three different colours of paint, but you could use as many colours as you like. As my toddler is still quite young, I showed her what to do first but she got the hang of it quickly and was able to dip one of the pipe cleaner 'fans' in the paint and then press it down onto the sugar paper. We used navy sugar paper to create the effect of the night sky, so the bright colours of the fireworks would stand out on the paper.
To prepare, I cut strips along a toilet paper roll to about the middle and fanned the ends out. We were going to use these in the same way as we did the pipe cleaner 'fans'.
The original idea was to create more firework-like shapes but the overall effect ended up being more similar to that of a bonfire. my toddler had great fun splattering the paint over the page with the toilet roll tubes.
We used red, orange and yellow colours, which were perfect for creating flames.
Next time we do this activity, it would be good to try to create the effect of the wood burning at the bottom of the page. We could do this by painting logs, either with brown finger paint or dipping pipe cleaners into paint and pressing onto the paper, or possibly by sticking brown pipe cleaners onto the page. Let me know what you decided to do if you recreated this activity.
my toddler and I thoroughly enjoyed our Bonfire Afternoon and we hope you all enjoy Guy Fawkes Night, whatever you choose to do. Have a wonderful and safe time!


text:
Mr. Gilroy Putie, General Manager of the Nevis Electricity Company Limited, at the Department of Information's office
NIA CHARLESTOWN NEVIS (July 23, 2020) — Mr. Gilroy Putie, General Manager of the Nevis Electricity Company Limited (NEVLEC), says Nevisians are benefiting from a more reliable electricity supply.
Mr. Pultie, who recently spoke with the Department of Information on the strides the company has made in the past year, noted that moving forward the expectations are for an even higher rate of reliability.
"Reliability in my mind is at a fairly high-level right now. Yes, there is room for improvement but it is quite good…I believe the plans that we've had, plans that we have started to roll out over the next year, two years, at the most, I think the reliability of the supply in Nevis is going to be on par or better than a lot of islands in the region because we already getting close to that…and customers are already enjoying the benefits of our more reliable supply.
"They can expect it to be even more reliable, and even with a small storm the system is going to be a lot more resilient, the level of damage is going to be less and we would be able to restore much more quickly than maybe a few years ago. So, our customers have already begun to see the benefits of this but they can expect to see a lot more over the next year to two years," he said.
The NEVLEC GM stated that the workers had embraced the challenge to better the electricity supply on the island.
"The guys have responded well. The guys have worked hard and we have achieved a lot and you can see the results of it.
"The work that we have been doing to improve the reliability on the transmission and distribution side is in two different areas. We first of all decided to improve the redundancy in the grid, and what that means is to provide more possibilities of feeding an area from different sources," he said.
Prior to 2019 in most areas on Nevis there was only one possibility of feeding electricity to the area. Mr. Pultie explained that if they were required to work on the system in one area many customers would be without electricity. That situation proved to be a challenging one for the company and they set out to rectify the matter.
"We decided because the system needed to be reinforced, there is a lot of work that needed to be done on the system, we did not want to be taking outages that affect so many customers at one time, and so we built some redundancy into the system…We are putting a number of switches on the line so that we can isolate a much smaller area," he said.
The NEVLEC GM gave examples of the work done so far to improve reliability of the electricity supply.
"We were able to provide a connection to the Cotton Ground feeder from the Charlestown feeder so the line that goes into the town and feeds the town that ended in the town at one time, we were able to create a ring to tie it back to the Cotton Ground feeder, and that tie is done around the section of the by-pass road and the main road.
"If a vehicle were to crash into a pole, on the by-pass road in the vicinity of the Artisan Village, at one time everybody would have been out, from Prospect to Camps for six hours, to replace the pole. If that were to happen now, we would be able to isolate that problem to between [St. Kitts -Nevis] Observer and just before the Island Main Road, and we would be able to feed up to The Observer from the very Cotton Ground feeder, and the rest of the areas from the by-pass all the way up to Camps we would be able to import from the Charlestown 1 feeder which we have connected to it, and the upper part of it we would be able to take from Gingerland via the West Coast and so that was the first prong of our efforts at improving reliability," he said.
General Manager of the Nevis Electricity Company Limited
Mr. Gilroy Putie
NIA announces stimulus package for the construction sector in Nevis


text:
How do atoms form the ions in ionic compounds? | 16-18 years
Check students' understanding of why atoms form ions, and how ionic compounds form by ionic bonding, using this lesson plan with activities for 16–18 year olds
In this activity, students, in small groups, decide whether statements they are given about the formation of sodium and chloride ions are true or false. By thinking about and discussing the statements, students check their understanding about ionic compounds and ionic bonding.
This activity could be used to assess prior learning.
An alternative version of the 'Why do atoms form ions?' activity, featuring different questions, is available as a PDF book extract on 'Ionic bonding' from K. Taber's Chemical misconceptions.
Students will:
Understand how and why ionic compounds such as sodium chloride are formed.
Sequence of activities
Demonstration and introduction
Demonstrate the reaction of sodium with chlorine and the subsequent test for chloride ions using silver nitrate solution.
Issue a mini whiteboard to each student and ask them to write down:
What particles are present in sodium and chlorine before the reaction.
What particles are present in sodium chloride after the reaction.
Pose questions to stimulate discussion about the use of the words atoms, molecules and ions before sharing the purpose of the session. For example:
'What has changed during the chemical reaction to cause such an enormous change in the appearance of reactants and products?'
Activity: stage 1
Organise students into groups of three. Tell each group to number its members, one, two or three. Ask:
Number one's to explain what they mean by an atom
Number two's to explain what they mean by a molecule
Number three's what they mean by an ion
to the other two students in the group.
Allow time for discussions to come to agreed explanations. Support individual groups by using questions to focus their ideas.
Give each student a 'Why do atoms form ions?' sheet. Ask them to:
Work individually.
Decide whether they think that the 10 statements are true or false.
Record this in the table.
Invite students to:
Return to their groups of three.
Share their ideas.
Agree group responses.
If a data projector is available, show an electronic animation of the formation of the sodium chloride ionic lattice. In a plenary:
Select members of different groups to describe and explain their responses.
Encourage other groups to challenge or add to explanations.
Provide an opportunity for students to write down in what ways they have changed their original ideas, as a result of listening to others during the session.
Take in the sheets and comment on how students' ideas have developed. Draw attention to ways in which individuals might need to develop their explanations further.
The initial demonstration and discussion about the meaning of words such as atom, molecule and ion draw attention to the learning objectives.
Students compare their responses to statements about the formation of ions and have the opportunity to articulate their ideas. The electronic animation showing the formation of a sodium chloride lattice helps students recognise the standard they are aiming for.
They examine how their ideas have developed during the session and written feedback supports students should they need to develop their understanding further.
Practical notes
Demonstration equipment and procedure
A procedure for demonstrating the reaction of chlorine and sodium when the metal is heated is available as the second part of a practical series on heating group 1 metals in air and in chlorine. This resource includes a full kit list and safety instructions for heating sodium, lithium and potassium in air and in chlorine – only the steps required for heating sodium in chlorine are needed for this lesson plan.
Intructions for using silver nitrate solution to test for chloride ions are available as part of another resource on testing for negative ions.
Mini whiteboards, one for each student
Animation showing the formation of the sodium chloride lattice
Data projector
A sodium atom spontaneously loses an electron to get a full shell of electrons.
False: the ionisation energy has to be supplied to remove an electron. It is not a spontaneous process.
A Na7- ion is more stable than a sodium atom because it has a full shell of electrons.
False: the Na7- ion would be highly unstable because the repulsions between electrons would far out balance the attractions between protons and electrons. Having a full shell of electrons does not automatically mean that an ion is stable.
A Cl7+ ion is just as stable as a Cl- ion because they both have a full shell of electrons.
False: a great deal of energy would be required to ionise a chlorine atom to give a 7+ cation. Having a full shell of electrons does not automatically mean that an ion is stable.
Each proton in the nucleus of an atom attracts one specific electron.
False: all the protons in the nucleus attract all of the electrons (and vice versa).
Energy is required to remove an electron from an atom.
True: this is the ionisation enthalpy.
When an atom is ionised, it then requires even more energy to remove a second electron.
True: this is because there is less repulsion from other electrons counteracting the attraction from the nucleus and because the ionic radius decreases so that the second electron is closer to the nucleus.
Once you've removed an electron from a sodium atom you can never put it back.
False: the separated cation and electron are less stable which is why energy is needed to ionise an atom.
Once you have removed one electron from a sodium atom you can't remove another because that would mean it no longer had a full electron shell.
False: it is perfectly possible to remove another electron from the sodium ion, although the ionisation energy needed will be significantly higher than that needed to remove the first electron.
Solid sodium chloride contains pairs of sodium and chloride ions which are kept together by their opposite charges.
False: solid sodium chloride is made up of a three dimensional lattice of sodium and chloride ions.
When sodium chloride dissolves in water the solution contains molecules of sodium chloride.
False: when solid sodium chloride dissolves in water the sodium and chloride ions become hydrated and are dispersed through the solution.
Why do atoms form ions student sheet
Editable handout | Word, Size 66.5 kb
Handout | PDF, Size 46.48 kb
This lesson plan was originally part of the Assessment for Learning website, published in 2008.
Assessment for Learning is an effective way of actively involving students in their learning. Each session plan comes with suggestions about how to organise activities and worksheets that may be used with students.
T. Lister, Classic Chemistry Demonstrations. London: Royal Society of Chemistry, 1995.
K. Taber, Chemical misconceptions – prevention, diagnosis and cure, Volume 2: Classroom resources, Ionic bonding. London: Royal Society of Chemistry, 2002.
Higher-order thinking and metacognition
Structure and bonding
Atoms and bonds
AQA Chemistry
4.2 Bonding, structure and the properties of matter
4.2.1 Chemical bonds, ionic, covalent and metallic
4.2.1.2 Ionic bonding
Metal atoms lose electrons to become positively charged ions.
Non-metal atoms gain electrons to become negatively charged ions.
AQA Combined science: Synergy
4.6 Interactions over small and large distances
4.6.2 Structure and bonding
Explain chemical bonding in terms of electrostatic forces and the transfer of electrons. Construct dot and cross diagrams for simple ionic substances.
AQA Combined science: Trilogy
Edexcel Chemistry
Topic 1 - Key concepts in Chemistry
Ionic bonding
1.21 Explain how ionic bonds are formed by the transfer of electrons between atoms to produce cations and anions, including the use of dot and cross diagrams
Edexcel Combined science
OCR Chemistry B: 21st century
C2 Chemical patterns
C2.3 How do metals and non-metals combine to form compounds?
C2.3.6 explain ionic bonding in terms of electrostatic forces and transfer of electrons
OCR Combined science B: 21st Century
OCR Combined science A: Gateway
C2 Elements, compounds and mixtures
C2.2 Bonding
C2.2e explain chemical bonding in terms of electrostatic forces and the transfer or sharing of electrons
OCR Chemistry A: Gateway
SQA Chemistry
Chemical changes and structure
Atomic structure and bonding related to properties of materials
Ionic compounds
Ions are formed when atoms lose or gain electrons to obtain the stable electron arrangement of a noble gas.
In general, metal atoms lose electrons forming positive ions and non-metal atoms gain electrons forming negative ions.
A/AS level
CCEA Chemistry
Unit AS 1: Basic Concepts in Physical and Inorganic Chemistry
1.2 Atomic Structure
1.2.1 describe the properties of electrons, protons and neutrons in terms of their location in the atom and their relative masses and charges;
1.2.2 explain the terms atomic number and mass number and use them to deduce the numbers of protons, neutrons and electrons in an atom or ion;
1.2.4 define and demonstrate understanding of the term isotopes;
1.2.8 deduce the electronic configuration of atoms and ions up to krypton in terms of shells and subshells using the building up principle (s, p and d notation and electrons in boxes notation);
WJEC Chemistry
Unit 1: THE LANGUAGE OF CHEMISTRY, STRUCTURE OF MATTER AND SIMPLE REACTIONS
1.2 Basic ideas about atoms
(e) significance of standard molar ionisation energies of gaseous atoms and their variation from one element to another
(f) link between successive ionisation energy values and electronic structure
1. Periodic table and atomic structure
1.4 Electronic Structure of Atoms
Organisation of particles in atoms of elements nos 1 - 20 (numbers of electrons in each main energy level).
Matter is composed of particles, which may be atoms, molecules or ions.
2. Chemical bonding
2.1 Chemical Compounds
Compounds. Simple chemical formulas.
2.2 Ionic Bonding
Representation of ionic bonds using dot and cross diagrams.
Examination of a model of the NaCl crystal.
Positive and negative ions.
Sodium chloride crystal structure.
How to teach ionic bonding at 14–16
Steer students away from ionic bonding misconceptions with these ideas for your classroom
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text:
Jean-Marie-Théodore Baudouin anomenat «d'Aubigny» (París, 19 d'agost de 1786 - 1866) fou un dramaturg francès.

Fill gran de l'impressor Baudouin, Aubigny es va interessar aviat pel teatre. Es va iniciar l'abril 1815 amb l'assistència de Louis-Charles Caigniez amb el melodrama La Pie voleuse, que va merèixer el títol de comèdia per la seva forma regular i simple. Algun temps més tard, va estrenar a l'Odèon Les Petits Protecteurs, que va escriure sol i que li va assegurar una reputació duradora com a dramaturg.

Dramaturgs francesos en francès
Dramaturgs parisencs


text:
\section{Introduction}
Systems with a high density of sensors and actuators often lack centralized information and computing capability. Thus, structural constraints, e.g,  on information exchange among subsystems, have to be incorporated into the design procedure of the control system. However, imposing such constraints may greatly complicate controller synthesis.

We address \texorpdfstring{$H_{\infty}$}{H-infinity} structured static state feedback, a problem that is recognized as genuinely hard given arbitrary plant and controller structures. However, we give a simple form for an optimal control law applicable to linear time invariant (LTI) systems with symmetric and Hurwitz state matrix. Furthermore, if the system possess a compatible sparsity pattern the proposed controller is distributed. Consider the following LTI system
\begin{equation}
\dot{x} = \underbrace{-\textrm{diag}(1,3,2)}_Ax+\underbrace{\begin{bmatrix}
-1 & 0 &0\\
1 &1 &-1\\
0 &0 &1
\end{bmatrix}}_Bu + w \label{bufferSys}
\end{equation}
where the state $x$, the control input $u$ and the disturbance $w$ are real valued. The static state feedback controllers
\begin{displaymath}
L_1 =\begin{bmatrix}
1 &-\frac{1}{3} &0\\
0 &-\frac{1}{3} &0\\
0 &\frac{1}{3} &-\frac{1}{2}
\end{bmatrix} \textrm{ and }
L_2  =\begin{bmatrix}
   0.93 & -0.11&
  0.00 \\   -0.05 &
   -0.17 &   -0.01 \\
  0.04 &
  0.16 &
  -0.26
\end{bmatrix}
\end{displaymath}
both minimize the \texorpdfstring{$H_{\infty}$}{H-infinity} norm of the closed-loop system from disturbance $w$ to performance output $\left(x,u\right)$, i.e., when ${u = L_1x}$ and $u = L_2x$, respectively. However, they have different structural properties, e.g., $L_1$ is sparser than $L_2$. Furthermore, the feedback law $u=L_1x$ is distributed as the matrix $L_1$ has the same structure as the sparse matrix $B^T$. This is not the case for controller $L_2$. Controller $L_1$ can be given on the simple form we propose. More specifically $L_1$ can be written as $L_1=B^TA^{-1} $. Controller $L_2$ is derived by the algebraic riccati equation (ARE) approach. That is, iteration over an ARE-constraint until the minimal value of the norm is obtained, see \cite{zhou1996robust} for details. Controllers synthesized by the ARE method are often dense, as is the case for controller $L_2$. Moreover, as the control law we give, i.e., $u= B^TA^{-1}x$, is optimal, it is equal in performance to any centrally derived optimal controller. Additionally, it is transparent in its structure, easy to synthesize and scalable.

In the 1980's, synthesis of controllers that achieve $H_{\infty}$ norm specifications became a major research area and was formulated in \cite{zames1981feedback}. The solution to the synthesis problem was initially based on operator design but evolved to a state-space based design that paved the way for optimization tools to be used, e.g., see~\cite{doyle1989state}. The \texorpdfstring{$H_{\infty}$}{H-infinity} norm condition can be turned into a linear matrix inequality (LMI) by the Kalman-Yakubovich-Popov lemma~\cite{gahinet1994linear}, see Lemma~\ref{KYP} in Appendix for the version used in this paper. This reformulation made the synthesis computationally easier. As the theory on \texorpdfstring{$H_{\infty}$}{H-infinity} feedback control emerged, a decentralized version took form, e.g., see \cite{zhai2001decentralized}. Imposing general sparsity constraints on the controller might complicate the design procedure. However, design is simplified if the constrained set of controllers $K$ is quadratically invariant with respect to the given system \cite{rotkowitz2006characterization}. It is also simplified if the closed-loop system is constrained to be internally positive~\cite{tanaka2011bounded}. However, in contrast to our method, the methods previously mentioned do not result in controllers that are equal in performance to the central non-structured controller of the system.

The optimal control law $u=B^TA^{-1}x$ only requires some relatively inexpensive matrix calculations for its synthesis, especially for sparse systems. This is in relation to general \texorpdfstring{$H_{\infty}$}{H-infinity} controller synthesis where more expensive computational methods are required. Additionally, its structure is transparent, which is not often the case in \texorpdfstring{$H_{\infty}$}{H-infinity} controller synthesis.

The \texorpdfstring{$H_{\infty}$}{H-infinity} framework treats worst-case disturbance as opposed to stochastic disturbance in the $H_2$ framework. However, the transparent structure and simple synthesis of the derived optimal feedback law might motivate its use even when some characteristics of the disturbance are known. Moreover, it can be extended to incorporate coordination in a system of heterogeneous subsystems, given a linear coordination constraint. The coordinated control law is a superposition of a decentralized and a centralized part, where the latter is equal for all agents. This structure might be well suited for distributed control purposes as well. See~\cite{madjidian2014distributed} for a similar problem treated in the $H_2$ framework.

The outline of this paper is as follows. This section is ended with some notation. In Section II, the main results is stated and proved. Section III treats system sparsity patterns that result in a distributed control law. Section IV gives an extension of the control law that incorporates coordination. In Section V, the performance of our optimal control law is compared, by a numerical example, to an optimal controller synthesized by the ARE approach. Concluding remarks are given in Section VI.

The set of real numbers is denoted $\mathbb{R}$ and the space $n$-by-$m$ real-valued matrices is denoted $\mathbb{R}^{n \times m}$. The identity matrix is written as $I$ when its size is clear from context, otherwise $I_n$ to denote it is of size $n$-by-$n$.  Similarly, a column vector of all ones is written $\mathbf{1}$ if its length is clear form context, otherwise $\mathbf{1}_n$ to denote it is of length $n$.

For a matrix $M$, the inequality $M \geq 0$ means that $M$ is entry-wise non-negative and $M \in \mathbb{R}^{n \times n}$ is said to be Hurwitz if all eigenvalues have negative real part. The matrix $M$ is said to be Metzler if its off-diagonal entries are non-negative and the spectral norm of $M$ is denoted $\|M\|$. Furthermore, for a square symmetric matrix $M$, $M \prec 0$ ($M \preceq 0$) means that $M$ is negative (semi)definite while $M \succ 0$ ($M \succeq 0$) means $M$ is positive (semi)definite.

The $H_{\infty}$ norm of a transfer function $F(s)$ is written as $\|F(s)\|_{\infty}$. It is well known that this operator norm equals the induced 2-norm, that is
\begin{displaymath}
\|F\|_{\infty} = \textrm{sup}_{v \neq 0} \frac{\|Fv\|_2}{\|v\|_2}.
\end{displaymath}

\section{An optimal \texorpdfstring{$H_{\infty}$}{H-infinity} state feedback law}
Consider a LTI system of the following structure
\begin{equation}
\dot{x} = Ax+Bu+w \quad \label{G}
\end{equation}
where state matrix $A \in \mathbb{R}^{n \times n}$ is symmetric and Hurwitz and state $x \in \mathbb{R}^{n}$ can be measured. Moreover, control input ${u \in \mathbb{R}^m}$, disturbance $w \in \mathbb{R}^n$ and matrix ${B \in \mathbb{R}^{n \times m}}$.

Given (\ref{G}) and performance output $\left(x,\,u\right)$, consider a stabilizing static state feedback law $u \coloneqq Lx$, where $L \in \mathbb{R}^{m \times n}$. The transfer function of the closed-loop system, i.e., from disturbance $w$ to performance output $\left(x,\,u\right)$, is given by
\begin{equation}
G_L(s) = \begin{bmatrix}
I \\L
\end{bmatrix} (sI-(A+BL))^{-1}. \label{Gtransfer}
\end{equation}
For (\ref{G}) with $A$ symmetric and Hurwitz, an optimal \texorpdfstring{$H_{\infty}$}{H-infinity} static state feedback controller $L$, i.e., a matrix $L$ such that $\| G_L  \| _{\infty}$ is minimized, can be given explicitly in the matrices $A$ and $B$. This is the main result of this paper and it is stated in the following theorem, followed by a proof.

\begin{theo} \label{theo1}
Consider the system (\ref{G}) with $A$ symmetric and Hurwitz. Then, the norm $\| G_L\| _{\infty}$ is minimized by the static state feedback controller $L_* = B^TA^{-1}$. The minimal value of the norm is $\sqrt{\|(A^2 + B^TB)^{-1} \|}$.
\end{theo}
\begin{proof}
Given $\gamma>0$, the following statements are equivalent.
\begin{enumerate}[(i)]
\item There exists a stabilizing controller $L$ such that
\begin{displaymath}
\| G_L \|_{\infty} = \left \| \begin{bmatrix}
I \\ L
\end{bmatrix} (i \omega -A-BL)^{-1} \right \|_{\infty} < \gamma.
\end{displaymath} \label{th1}
\item \label{th2}There exists a matrix $P \succ 0$ such that
\begin{displaymath}
\begin{bmatrix}
(A+BL)^TP+P(A+BL) & P & [I \enskip L^T] \\ P & -\gamma^2I &0 \\
[I \enskip L^T]^T & 0 & -I \\
\end{bmatrix} \prec 0. \\
\end{displaymath}
\item \label{th3}There exist matrices $X\succ 0$ and $Y$ such that
\begin{displaymath}
\begin{bmatrix}
AX+XA+BY+Y^TB^T & I & [X \enskip Y^T] \\ I & -\gamma^2I &0 \\
[X \enskip Y^T]^T & 0 & -I \\
\end{bmatrix} \prec 0.
\end{displaymath}
\item \label{th4}There exist matrices $X\succ 0$ and $Y$ such that
\begin{align*}
\left(X+A\right)^2+ \left(Y+B^T\right)^T&\left(Y+B^T\right) \\ &-A^2 -BB^T + \gamma^{-2}I \prec 0.
\end{align*}
\item \label{th5} \begin{displaymath}
-A^2 -BB^T + \gamma^{-2}I \prec 0.
\end{displaymath}
\item \label{th6}\begin{displaymath}
\gamma > \sqrt{\|\left(A^2 + BB^T \right)^{-1}\|}.
\end{displaymath}
\end{enumerate}
The equivalence between (\ref{th1}) and (\ref{th2}) is given by the K-Y-P-lemma, Lemma~\ref{KYP} given in Appendix. Statement (\ref{th2}) can be equivalently written as (\ref{th3}) after right- and left-multiplication with $\textrm{diag}(P^{-1},I,I)$ and change of variables ${(P^{-1},LP^{-1}) \rightarrow (X,Y)}$. The equivalence between (\ref{th3}) and (\ref{th4}) is obtained by applying Schur's complement lemma and completion of squares to the inequality in (\ref{th3}). Choosing $X=-A$ and $Y=-B^T$ shows equivalence between (\ref{th4}) and (\ref{th5}). It is possible to choose $X=-A$ as $A$ is symmetric and Hurwitz, i.e., $A \prec 0$. Finally,  notice that $A^2+BB^T\succ 0$ and thus $\left(A^2+BB^T\right)^{-1} \succ 0$. Thus,
\begin{displaymath}
\textrm{(\ref{th5})} \iff \gamma^2I \succ  \left(A^2+BB^T\right)^{-1}
\iff \textrm{(\ref{th6})}.
\end{displaymath}
Given $X=-A$ and $Y=-B^T$, $\gamma$ is minimized. Thus, ${L_* = YX^{-1} = B^TA^{-1}}$ minimizes the norm in (\ref{th1}) and the minimal value of the norm, i.e,  $\|G_{L_*}\|$, is less than $\gamma$. Now, define $\gamma_* \coloneqq \sqrt{\|\left(A^2+BB^T\right)^{-1}\|}$ and assume that $\|G_{L_*}\|_{\infty} \neq \gamma_*$. Then $\|G_{L_*}\|_{\infty}$ has to be strictly larger than or strictly smaller than $\gamma_*$. Consider $\|G_{L_*}\|_{\infty}>\gamma_*$. This statement contradicts statement (\ref{th1}) and (\ref{th6}) and is therefore false. Now, consider instead $\|G_{L_*}\|_{\infty} <\gamma_*$. This statement contradicts that $\gamma$ is minimized and is therefore also false. Hence, the statement $\|G_{L_*}\|_{\infty} \neq \gamma_*$ is false and
\begin{displaymath}
\|G_{L_*}\|_{\infty} = \sqrt{\|\left(A^2+BB^T\right)^{-1}\|}.
\end{displaymath}
\end{proof}

\begin{remark}
Instead of performance output $ \left(x, \,u\right)$ consider $ \left(\tilde{x}, \,\tilde{u}\right)$ defined as
\begin{equation}
\begin{bmatrix} \tilde{x} \\ \tilde{u} \end{bmatrix} = \begin{bmatrix}
C & 0 \\ 0 & D
\end{bmatrix}\begin{bmatrix}
x \\ u
\end{bmatrix} \label{scaled}
\end{equation}
with square matrices $C \in \mathbb{R}^{n \times n}$ and $D \in \mathbb{R}^{m\times m}$. Define matrices $Q\coloneqq C^TC\succ 0$ and $R \coloneqq D^TD \succ 0$. If $-AQ^{-1}$ is symmetric and positive definite it is still possible to find an explicit expression for an optimal static state feedback controller, specifically $L_* = R^{-1}B^TQA^{-1}$. Hence, other performance objectives than $\left(x,\,u\right)$ can be included in this framework. However, cross terms between $x$ and $u$ are not possible for Theorem~\ref{theo1} to hold. Matrices $Q$ and $R$ can be seen as the cost matrices for state $x$ and control input $u$, respectively, and used as design parameters in the synthesis of the optimal $H_{\infty}$ static state feedback controller $L_*$. The restrictions on $Q$ and $R$, i.e., $Q \succ 0$ and $R \succ 0$, are not significantly more conservative than restrictions given on similar design matrices in static state feedback synthesis by linear quadratic control, see~\cite{athans1971role} for comparison.
\end{remark}

Synthesis of an optimal static state feedback controller $L$ that minimizes the \texorpdfstring{$H_{\infty}$}{H-infinity} norm of (\ref{Gtransfer}) generally requires additional computation beyond what is needed to compute $L_*$ given by Theorem~\ref{theo1}, i.e., some relatively simple matrix calculations. Moreover, optimal controllers generated by other methods than Theorem~\ref{theo1} are rarely as transparent as $L_*$. The transparency simplifies analysis of the structure of the optimal controller as well as enables scalability. This will be exploited in the following section.

In order for Theorem~\ref{theo1} to be applicable, the system of interest has to have a state space representation with symmetric and Hurwitz state matrix $A$. The symmetry property of $A$ demands that states that affect each other does so with equal rate coefficient. Such representations appear, for instance, in buffer networks and models of temperature dynamics in buildings. We will now give an example of the latter.

\begin{example}
Consider a building with three rooms as depicted in Fig.~\ref{fig:RoomTemp}. The average temperature $T_i$ in each room $i=$ 1, 2 and 3, around some steady state, is given by the following model
\begin{align}
&\dot{T}_1 = -r_1T_1+r_{12}\left(T_2-T_1\right)+u_1+w_1\nonumber \\
&\dot{T}_2 = -r_2T_2+r_{12}\left(T_1-T_2\right)+r_{23}\left(T_3-T_2\right)+u_2+w_2 \label{temp} \\
&\dot{T}_3 = -r_3T_3+r_{23}\left(T_2-T_3\right)+u_3+w_3 \nonumber
\end{align}
governed by heat balance. The parameters $r_\bullet $ are constant, real-valued and positive. They are the rate coefficients of the system. For instance, $r_{12}$ is the rate coefficient of the heat transfer through the wall between room 1 and 2. Changes in outdoor temperature and disturbances specific for each room, such as a window is opened, are modeled by disturbances $w_i$. The average temperatures can be measured as well as controlled through heating and cooling devices, given by control inputs $u_i$. If (\ref{temp}) is written on form (\ref{G}), it is easy to see that matrix $A$ is symmetric. Thus, Theorem~\ref{theo1} is applicable to (\ref{temp}), assuming that that parameters $r_\bullet$ are such that $A$ is also Hurwitz. Given a disturbance, the feedback law with $L_*$ from Theorem~\ref{theo1} tries to keep the average temperature as close to the steady state as possible while minimizing the cost that comes with heating and cooling.
\end{example}

\begin{figure}[t]
\centering
\vspace{0.3cm}
\includegraphics[width=0.5\linewidth]{temp.eps}
\caption{Schematic of a building with three rooms. The average temperature in each room $i=1,2$ and $3$ is denoted $T_i$ and given by (\ref{temp}).  \label{fig:RoomTemp}}
\end{figure}

\section{Distributiveness and scalability}
The structure of the optimal controller $L_*$ given in Theorem~\ref{theo1} is clearly dependent on the structure of matrices $A$ and $B$ in (\ref{G}). For instance, if $A$ is diagonal and $B$ is sparse, $L_*$ has the same sparsity pattern as $B^T$. Moreover, controller $L_*$ is distributed if (\ref{G}) possess a compatible sparsity pattern. It is worthwhile to point out that for some sparsity patterns of (\ref{G}) the representation $L_*^{-1}u=x$ instead of $u = L_*x$ might be beneficial for computation of $u$, that is if $B^T$ is invertible. We will begin this section with an example to demonstrate the distributiveness and scalability of the optimal controller given a system of structure (\ref{G}) with compatible sparsity pattern.
\begin{example}
Consider the following LTI system, containing three subsystems denoted $S_1$, $S_2$ and $S_3$,
\begin{align}
&S_1: \quad \dot{x}_1 = A_1x_1 + B_1u_1+w_1 \nonumber \\
&S_2: \quad \dot{x}_2 = A_2x_2 + B_2u_1+B_3u_2 +w_2 \label{Garea} \\
&S_3: \quad \dot{x}_3 = A_3x_3 + B_4u_2 +w_3 \nonumber
\end{align}
where each subsystem $S_i$, $i=$1, 2 and 3, has finite state dimension $n_i \geq 1$, each control input $u_i$, $i=$1, 2 and 3, is a vector of finite length $m_i \geq 1$ and the matrices are of suitable dimension. Furthermore, matrices $A_1$, $A_2$ and $A_3$ are assumed to be symmetric and Hurwitz. Then,  Theorem~\ref{theo1} is applicable to (\ref{Garea}) and results in the optimal controller
\begin{equation}
L_* = \begin{bmatrix}
B_1^TA_1^{-1} & B_2^TA_2^{-1} & 0 \\ 0 & B_3^TA_2^{-1} & B_4^TA_3^{-1}
\end{bmatrix}. \label{controlLaw}
\end{equation}\label{exampleArea}
\end{example}

Notice that,  if (\ref{Garea}) is written on form (\ref{G}) the optimal controller $L_*$ has the same sparsity pattern as $B^T$.  Thus, each control input vector $u_i$ is only constructed by the states it affects in (\ref{Garea}). If we consider each subsystem $S_i$ in (\ref{Garea}) to represent an area of the physical system it models, the optimal controller (\ref{controlLaw}) is distributed according to these areas. See Fig.~\ref{fig:areaSys} for a graphical representation of the system, drawn with solid lines. Each subsystem $S_i$ is depicted by a circular node while each control input $u_i$ is given by a link connecting the subsystems it affects in (\ref{Garea}). Each disturbance $w_i$ is drawn as arrows that points toward the subsystem it affects in (\ref{Garea}).

\begin{figure}[b]
\centering
\includegraphics[width=0.65\linewidth]{area.eps}
\caption{Graphical representation of (\ref{Garea}) in solid lines. Additional subsystem $S_4$ and control input $u_3$ in dashed lines.}
\label{fig:areaSys}
\end{figure}

Consider that a fourth subsystem denoted $S_4$, of finite dimension $n_4 \geq 1$, is connected to (\ref{Garea}) via a third control input denoted $u_3$, of finite length $m_3 \geq 1$, as depicted by dashed lines in Fig. \ref{fig:areaSys}. The dynamics of subsystem $S_4$ and the altered dynamics of subsystem $S_3$ are then given by
\begin{align*}
&S_3: \quad \dot{x}_3 = A_3x_3+B_4u_2+B_5u_3+w_3\\
&S_4: \quad \dot{x}_4 = A_4x_4+B_6u_3+w_4
\end{align*}
where matrix $A_4$ is also assumed to be symmetric and Hurwitz. Then, Theorem~\ref{theo1} is still applicable and the extended optimal controller becomes
\begin{displaymath}
L_* = \begin{bmatrix}
B_1^TA_1^{-1} & B_2^TA_2^{-1} & 0 & 0 \\ 0 & B_3^TA_2^{-1} & B_4^TA_3^{-1} & 0 \\ 0 & 0 & B_5^TA_3^{-1} & B_6^TA_4^{-1}
\end{bmatrix}.
\end{displaymath}
The expansion of the system does not alter the initial control inputs $u_1$ and $u_2$. Thus, for systems with this type of sparsity pattern, the control law $u = L_*x$ is easily scalable. Moreover, the control law is still distributed as the additional control input $u_3$ is only constructed by states $x_3$ and $x_4$.

We will now consider systems of structure (\ref{G}) with diagonal and Hurwitz matrix $A$. Then, the closed-loop system from disturbance $w$ to state $x$ with control law $u=L_*x$, from Theorem~\ref{theo1}, is internally positive by Lemma~\ref{lemIP}, given in Appendix, if and only if $-BB^T$ is Metzler. Consider the closed loop system from disturbance $w$ to output $y \coloneqq x$, with $L_* = B^TA^{-1}$, i.e.,
\begin{align*}
&\dot{x} = (A+BL_*)x+w \\
&y = x
\end{align*}
where $A+BL_* = A+BB^TA^{-1}$. In order to fulfil the requirements in Lemma~\ref{lemIP} we only need to check if $A+BL_*$ is Metzler as the other matrices are entry-wise non-negative. If $A$ is diagonal and Hurwitz, i.e, all diagonal elements are negative, it is necessary and sufficient that $-BB^T$ is Metzler for $A+BL_*$ to be Metzler.

\begin{remark}
Consider (\ref{scaled}), $A$ diagonal and $-BB^T$ Metzler. Then, diagonal $Q$ and $R$ would suffice in order for the closed-loop system from disturbance $w$ to state $x$ to be internally positive.
\end{remark}

\begin{example}
Consider three buffers of some quantity connected via links with flow $u_1$ and $u_2$ as depicted in Fig.~\ref{fig:buffersys}. The dynamics of the levels in the buffers, around some steady state depicted by the dashed lines in Fig.~\ref{fig:buffersys}, is given by
\begin{equation}
\begin{bmatrix} \dot{x}_1\\\dot{x}_2 \\ \dot{x}_3\end{bmatrix} = \underbrace{-\textrm{diag}\left(1,2,4\right)}_A\begin{bmatrix} x_1\\x_2 \\ x_3\end{bmatrix} +\underbrace{\begin{bmatrix} -1 & 0  \\ 1 & -1  \\ 0 & 1 \end{bmatrix}}_B\begin{bmatrix} u_1\\u_2\end{bmatrix} +w. \label{waterbuffer}
\end{equation}
State $x_i$ corresponds to the level in buffer $i=$1, 2 and 3, respectively. Each buffer has some internal dynamics dependent on its own state, as given by matrix $A$. However, with different rate coefficients for the different buffers. We want to construct a control law that minimizes the impact from disturbance $w$  on performance output $\left(x,u\right)$ in the $H_{\infty}$ norm sense. That is, we want to keep the system at its steady state, i.e, $x_i= 0$ for all $i$, while also keeping the cost down, i.e, the magnitude of the control input.

\begin{figure}[t]
\centering
\includegraphics[width=0.6\linewidth]{buffer.eps}
\caption{Three buffers denoted $1$, $2$ and $3$ connected via links with flow $u_1$ and $u_2$, respectively. The dashed lines represent some steady state of the system. \label{fig:buffersys}}
\end{figure}

Given the matrix $B$ in (\ref{waterbuffer}), $-BB^T$ is Metzler. Thus the closed-loop system with the optimal control law given by Theorem~\ref{theo1}, i.e.,
\begin{displaymath}
L_* = \begin{bmatrix}
1 & -1/2 & 0 \\ 0 &1/2&-1/4
\end{bmatrix},
\end{displaymath}
is internally positive. This implies that, in closed-loop with controller $L_*$, the states $x_i$ of (\ref{waterbuffer}) will always be non-negative, i.e., the buffer levels will never go below their steady state values, given non-negative disturbance.

To get some further intuition of what controller $L_*$ does consider control input $u_1$. It is given by $u_1= x_1-x_2/2$. Thus, $u_1$ is strictly positive if $x_1>x_2/2$ and the controller $L_*$ redistributes the quantity of buffer 1 and buffer 2 relative to their internal rate coefficients. Again, as in the previous example, $L_*$ has the same sparsity pattern as $B^T$. In this case, with one-dimensional subsystems, it means that each control input only considers local information, i.e., from the buffers it connects. \label{exampleBuff}
\end{example}

The system (\ref{waterbuffer}) can be depicted by a graph, much like the system in the previous example, Example \ref{exampleArea}. However, due to the structure of the matrix $B$ in (\ref{waterbuffer}), the links in this graph could be drawn as directed arrows. In other word, the matrix $B$ in (\ref{waterbuffer}) is the node-link incidence matrix of a directed graph. See~\cite{newman2010networks} for a formal definition of this notion. It is well known that for such matrices $B$,  the matrix product $-BB^T$ is Metzler.

\section{Coordination in the \texorpdfstring{$H_{\infty}$}{H-infinity} framework}
In this section we will extend the optimal control law given by Theorem~\ref{theo1} in order to include coordination. The problem formulation is as follows. Consider a LTI system of $\nu$ subsystems
\begin{equation}
\dot{x}_i = A_ix_i+B_iu_i+w_i, \quad i = 1,\dots, \nu  \label{Gcoord}
\end{equation}
where $A_i$, for $i = 1,\dots, \nu$, is symmetric and Hurwitz. Furthermore, the control inputs $u_i$ have to coordinate in order to fulfil the following constraint
\begin{equation}
u_1+u_2+\dots+u_{\nu}=0. \label{constraint}
\end{equation}
Given the performance objective $(x,u)$ and the coordination constraint in (\ref{constraint}) we want to construct an optimal \texorpdfstring{$H_{\infty}$}{H-infinity} static state feedback controller for (\ref{Gcoord}).

Our solution to the given problem is as follows. Rewrite control input $u_1$ in terms of the other control inputs given (\ref{constraint}), i.e.,
\begin{equation}
u_1=-u_2-u_3\,\dots \,-u_{\nu}, \label{u1}
\end{equation}
and define $\tilde{u} = [u_2, \,u_3,\,\dots ,\,u_{\nu}]^T$. Then,
\begin{displaymath}
u = \underbrace{\begin{bmatrix}
-\mathbf{1}^T_{\nu-1} \\ I_{\nu-1}
\end{bmatrix}}_{D} \tilde{u}
\end{displaymath}
and the overall system of (\ref{Gcoord}) can be written
\begin{displaymath}
\dot{x} = \underbrace{\textrm{diag}(A_1,\dots,A_{\nu})}_Ax+\underbrace{\textrm{diag}(B_1,\dots,B_{\nu})}_BD\tilde{u} +w
\end{displaymath}
with performance output $(x,u)=(x,D\tilde{u})$. Now, define $$R = D^TD = I+\mathbf{1} \mathbf{1}^T,$$ as described in Remark 1, and notice that $R^{-1}=  I-\frac{1}{\nu}\mathbf{1}\mathbf{1}^T$. The optimal control law by Theorem~\ref{theo1} is then
\begin{align*}
\tilde{u} &= R^{-1}D^TB^TA^{-1}x  \\&= \left(I_{\nu-1}-\frac{1}{\nu}\mathbf{1}_{\nu-1}\mathbf{1}_{\nu-1}^T \right)\begin{bmatrix}
-\mathbf{1}^T_{\nu-1} \\ I_{\nu-1}
\end{bmatrix}^TB^TA^{-1}x \\ &= \left(\begin{bmatrix}
0& I_{\nu-1}
\end{bmatrix}-\frac{1}{\nu}\mathbf{1}_{\nu-1}\mathbf{1}_{\nu}^T \right)B^TA^{-1}x.
\end{align*}
Thus, $u_i$ for $i=2,\,\dots,\,\nu$, i.e., the elements in $\tilde{u}$, is
\begin{equation}
u_i = B_i^TA_i^{-1}x_i-\frac{1}{\nu}\sum_{k=1}^{\nu}B_k^TA_k^{-1}x_k. \label{controllawutilde}
\end{equation}
Now, consider $u_1$ again,
\begin{align*}
u_1 \stackrel{\textrm{(\ref{u1})}}{=}-\sum_{i=2}^{\nu}u_i \stackrel{\textrm{(\ref{controllawutilde})}}{=} -\sum_{i=2}^{\nu}\left( B_i^TA_i^{-1}x_i-\frac{1}{\nu}\sum_{k=1}^{\nu}B_k^TA_k^{-1}x_k \right)\\
= -\left(\sum_{k=1}^{\nu}B_k^TA_k^{-1}x_k-B_1^TA_1^{-1}x_1-\frac{v-1}{v}\sum_{k=1}^{\nu}B_k^TA_k^{-1}x_k \right) \\
= B_1^TA_1^{-1}x_1-\frac{1}{\nu}\sum_{k=1}^{\nu}B_k^TA_k^{-1}x_k,
\end{align*}
i.e., it has the same structure as (\ref{controllawutilde}). Thus, the optimal control law can be written
\begin{equation}
u_i =  B_i^TA_i^{-1}x_i-\frac{1}{\nu}\sum_{k=1}^{\nu}B_k^TA_k^{-1}x_k\label{coordlaw}
\end{equation}
for each subsystem $i=1,\,\dots,\,\nu$ in (\ref{Gcoord}). The first term of $u_i$ in (\ref{coordlaw}) is a local term, only dependent upon the subsystem $i$,  while the second term is dependent on global information of the overall system. However, as this term is equal for all control inputs $u_i$, (\ref{coordlaw}) might still be appropriate for distributed control use.

In \cite{madjidian2014distributed}, a similar type of problem is considered, however in the $H_2$ framework with stochastic disturbances and the necessity of homogeneous subsystems. The optimal control law derived in \cite{madjidian2014distributed} and the one we suggest in (\ref{coordlaw}) are similar in structure. However, our approach can treat heterogeneous systems in addition to homogeneous ones. On the contrary, it is only applicable to systems with symmetric and Hurwitz state matrix, properties that are not necessary in \cite{madjidian2014distributed}.

\section{Numerical example}
Consider a system of the same structure as (\ref{bufferSys}) given in Section I, i.e., a system
\begin{align}  \label{numEx}
\begin{bmatrix} \dot{x}_1\\\dot{x}_2 \\ \dot{x}_3\end{bmatrix} = &\underbrace{-\begin{bmatrix} a_1 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_3\end{bmatrix}}_A\begin{bmatrix} x_1\\x_2 \\ x_3\end{bmatrix} \nonumber \\&+\underbrace{\begin{bmatrix} -b_1 & 0 & 0  \\ b_2 &b_3 & -b_4  \\ 0&0 & b_5 \end{bmatrix}}_B\begin{bmatrix} u_{12}\\u_2\\u_{23}\end{bmatrix} + \begin{bmatrix} w_1\\w_2 \\ w_3\end{bmatrix}
\end{align}
where $a_i>0$, for $i=$1, 2 and 3, and $b_j>0$, for $j = 1, \dots, \,5$, and the performance output is $\left(x,\,u\right)$. We will now compare the optimal controller given by Theorem~\ref{theo1}, i.e., $L_*$, and an optimal controller derived by the ARE-approach, see~\cite{zhou1996robust}, denoted $L_{G}$ for global. In the latter approach, we consider the minimal value of the \texorpdfstring{$H_{\infty}$}{H-infinity} norm of (\ref{Gtransfer}) given by Theorem~\ref{theo1} and iterate over the ARE-constraint until this minimal value is reached. See \cite{MATLAB} for the software used. Controllers $L_1$ and $L_2$ given in Section I are examples of controllers $L_*$ and $L_G$ treated here, respectively.

Controllers $L_*$ and $L_G$ are optimal and thus they both obtain the minimal value of the $H_{\infty}$ norm of (\ref{Gtransfer}). Now we want to compare how they affect the closed-loop dynamics more in detail. We randomly generate values of the parameters $a_i$ and $b_j$ in $(0.1,5]$ and compare the step-response of the states of (\ref{numEx}) in closed-loop with $L_*$ and $L_G$. In other words, given constant disturbance of value 1. The average dynamics over 50 such randomly generated systems is shown in Fig.~\ref{fig:numEx}. To clarify, we average over the absolute value of the step response in each time instance.

\begin{figure}[b]
\centering
\includegraphics[width=0.9\linewidth]{plot.eps}
\caption{Average step response for states $x_1$, $x_2$ and $x_3$ for closed-loop systems with controller $L_*$ (solid lines) and $L_G$ (dashed lines).} \label{fig:numEx}
\end{figure}

\begin{figure}[t]
\centering
\includegraphics[width=0.6\linewidth]{numex.eps}
\caption{Associated graph of~(\ref{numEx}).}
\label{fig:buffersystem}
\end{figure}

The system (\ref{numEx}) can be depicted by the graph given in Fig.~\ref{fig:buffersystem}, as described in Section III. If we compare the step responses shown in Figure~\ref{fig:numEx}, it seems as if controller $L_*$ is better at attenuating local disturbances than $L_G$ is. With local disturbances we mean the disturbance that points towards the state in  Fig.~(\ref{fig:buffersystem}). However, this is at the expense of larger impact on distance. For instance, consider  disturbance $w_1$. Its impacts on state $x_1$ is lower for controller $L_*$ than $L_G$ while its impact on the remaining states is the opposite for the given controllers. However, overall they are comparable in performance.

We will end this numerical example by commenting on controller $L_2$ given in Section I, that is an example of controller $L_G$ treated in this numerical example. Some entries of $L_2$ are small in magnitude compared to the other entries, i.e., entries (2,1), (2,3) and (3,1), where the first number in each parenthesis is the row and the second is the column. However, only entry (3,1) can be replaced with a zero for the controller to still achieve the optimal bound. Furthermore, for systems of much larger dimension than (\ref{bufferSys}), this type of reduction analysis might be difficult.

\section{CONCLUSIONS}

We give a simple form for an optimal  $H_{\infty}$ static state feedback law applicable to LTI systems with symmetric and Hurwitz state matrix. This simple form is given in the matrices of the system's state space representation which makes it transparent. The structure of the control law also simplifies synthesis and enables scalability, especially given sparse systems. Furthermore, given compatible system sparsity patterns the control law is distributed. The examples we give consider diagonal or block diagonal state matrices and somewhat more general sparsity patterns of the remaining system matrices. These types of system sparsity patterns are common among the systems for which distributed control methods are needed. Furthermore, we extend the optimal control law in order to incorporate coordination among subsystems. The resulting coordinated control law is similar for all subsystems. More specifically, for each subsystem, it is a superposition of a local term and an averaged centralized term where the latter is equal for all subsystems involved in the coordination. In conclusion, our control law is well suited for distributed control purposes. Future research directions include to consider saturation constraints on the optimal control law as such are common in the systems intended for its application. Furthermore, to investigate the existence of an analogous optimal control law given output feedback instead of state feedback.

\addtolength{\textheight}{-12cm}

\section*{APPENDIX}

\begin{lem} \textit{The Kalman-Yakubovich-Popov lemma} \\
Given $A \in \mathbb{R}^{n \times n}$, $B \in \mathbb{R}^{n \times m}$, $M=M^T \in \mathbb{R}^{(n+m)\times(n+m)}$, with $\textrm{det}(j\omega I-A) \neq 0$ and $(A,B)$ controllable, the following two statements are equivalent:\\
(i)
\begin{displaymath}
\begin{bmatrix}
\left(j \omega I-A \right)^{-1}B \\ I
\end{bmatrix}^*M \begin{bmatrix}
\left( j \omega I-A \right)^{-1} B \\ I
\end{bmatrix} \preceq 0
\end{displaymath}
$\forall \omega \in \mathbb{R}\cup\{\infty \} $.\\
(ii) There exists a matrix $P \in \mathbb{R}^{n \times n}$ such that $P=P^T$ and
\begin{displaymath}
M+ \begin{bmatrix}
A^TP + PA & PB \\ B^TP & 0
\end{bmatrix} \preceq 0
\end{displaymath}
The corresponding equivalence for strict inequalities holds even if $(A,B)$ is not controllable. \label{KYP}
\end{lem}
\begin{proof}
See ~\cite{rantzer1996kalman}.
\end{proof}
\textit{Remark.} If the upper left corner of $M$ is positive semidefinite, it follows from (\ref{KYP}) and Hurwitz stability of $A$ that $P \succeq 0$ ~\cite{rantzer1996kalman}.

\begin{lem}
The LTI system
\begin{align*}
\dot{x} = Ax + Bv\\
y = Cx + Dv
\end{align*}
is internally positive if and only if
\begin{enumerate}
\item[i] $A$ is Metzler, and
\item[ii] $B \geq 0$, $C \geq 0$ and $D \geq 0$.
\end{enumerate} \label{lemIP}
\end{lem}
\begin{proof}
See \cite{kaczorek2001externally}.
\end{proof}

\bibliographystyle{IEEEtran}

redpajama_set_name: RedPajamaArXiv


text:
PHOBIA United States
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Few bands represent the US grindcore scene better than PHOBIA, and that's why there's no better band for the first ever OBSCENE EXTREME AMERICA. The Orange County grind punks have worked with labels as defining as SLAP-A-HAM, RELAPSE and WILLOWTIP, and released splits with bands as diverse as CORRUPTED, PLUTOCRACY, GADGET and EXTINCTION OF MANKIND. Furiously political and unflinchingly hard working, these DIY tour dogs are rarely off the road, and fifth full-length, 2012's 'Remnants Of Filth', isn't some music school bullshit that lives in the studio ? it's raw and unpolished, punk-as-fuck grindcore as it's meant to be.
Formed in Orange County California in 1990, Phobia has been punishing ears with their murderous socio political grinding noise assault for two decades, Dealing out their savage combination of the energy and ideals of hardcore punk mixing it with the power and brutality of early grind!!
2012 Remnants of Filth CD, LP
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I know we are incredibly behind on gallery updates, but here are some new candids for you from this month. Added candids of Aly & AJ Signing Autographs At Airport in LA on June 6th & Arriving in LA on June 10th. More updates to come..
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text:
La teoría de vigas de Timoshenko fue desarrollada por el ingeniero ucraniano-estadounidense Stephen Timoshenko, estableciéndose como un modelo matemático riguroso ampliamente utilizado para describir la vibración transversal de vigas, postulado en la década de 1920. También denominada la teoría de vigas gruesas. Históricamente el primer modelo de viga importante fue la Teoría de Vigas de Euler-Bernoulli o teoría clásica de vigas como consecuencia de las obras de Bernoulli (Jacob y Daniel) y Euler, creado en el año de 1744. En 1921 y 1922, Timoshenko propuso una mejora al añadir el efecto de deformación de corte. Mostró, a través del ejemplo de una viga apoyada sencilla, que la corrección frente a cortante es cuatro veces más importante debido a la inercia de rotación en comparación con la teoría de Euler-Bernoulli.

Suposiciones básicas
Las suposiciones básicas de la teoría de vigas de Timoshenko están dadas por.
 El material del elemento es homogéneo, isotrópico y lineal elástico; cumpliendo la ley de elasticidad de Hooke. Coeficiente de Poisson despreciable.
 El desplazamiento vertical  de los puntos que se encuentran sobre la sección transversal del elemento en una posición , son pequeños e iguales a los desplazamientos del eje de la viga.
 El desplazamiento lateral  a lo largo del eje de la viga es nulo.
 La sección transversal del elemento en una posición  normal al eje de la viga antes de la deformación, permanece plana pero no necesariamente ortogonal al eje del elemento después de la deformación. Esto supone la presencia de un estado de tensiones cortantes en la sección de la viga ().
Esta última suposición es la principal diferencia entre la teoría de vigas de Euler-Bernoulli y Timoshenko, que representa una mejor aproximación de la deformación de la sección transversal en vigas de gran peralte.

Campo de Desplazamientos
Teniendo en cuenta las suposiciones anteriores podemos definir el campo de desplazamientos. Así, por suposición número tres tenemos que:

En donde  representa la pendiente del eje de la viga y  representa una rotación adicional debida a la distorsión de la sección transversal.
Por lo tanto,

Campo de Deformaciones
A partir del campo de desplazamientos, las relaciones deformación-desplazamiento para el caso de deformaciones infinitesimales proporcionan las deformaciones normales y angulares del siguiente modo:

Deformaciones normales:

Deformaciones angulares:

Se observa que la hipótesis cinemática principal de la Teoría de Vigas de Timoshenko (suposición básica n°4) introduce una deformación angular , que expresada en valor absoluto es igual a la rotación  y que, dado que es independiente de las coordenadas , está uniformemente distribuida en la sección transversal de la viga. Esta es la principal diferencia con la teoría de vigas de Euler-Bernoulli, cuyas hipótesis cinemáticas conducen a .
Como se verá más adelante, en virtud de las ecuaciones de Lamé para el esfuerzo cortante, la presencia de la deformación angular  en la teoría de Timoshenko supone la existencia de tensiones tangenciales a la sección transversal de la viga, uniformemente distribuidas en ella.

Campo de Esfuerzos
Teniendo en cuenta que esta teoría supone el Coeficiente de Poisson es despreciable para el cálculo de los esfuerzos normales pero no para los esfuerzos tangenciales, podemos expresar el campo de esfuerzos con la ayuda de las ecuaciones de Lamé como sigue:

Y que los esfuerzos tangenciales se escriben como:

En donde tenemos que  es el Módulo de Young,  es el Módulo Cortante o Módulo de Cizalladura,  es uno de los Parámetros de Lamé y  es la dilatación cúbica, que se calcula como .

Fuerzas Internas (momento flector y fuerza cortante)
Podemos definir el momento flector  actuante sobre el elemento, como la sumatoria de una par de fuerzas que actúan sobre la sección transversal, representadas por los esfuerzos normales actuando sobre el área de la sección transversal de la viga. Podemos expresar esto matemáticamente como:

Del mismo modo podemos definir la fuerza cortante  actuante en la viga, como la sumatoria de las fuerzas actuantes sobre la sección transversal de la viga, es decir, los esfuerzos cortantes que actúan sobre el área de la sección transversal del elemento. Matemáticamente se expresa como:

Reemplazando en estas expresiones el valor de los esfuerzos normales () y los esfuerzos cortantes () obtenidos en el apartado anterior, tenemos que:

En donde  es la deformación por flexión del elemento. Es de interés apreciar que  y  dependen únicamente del eje coordenado .

La expresión para los esfuerzos normales (), nos indica que la variación a través del espesor del elemento es lineal, la cual podemos considerar exacta, de acuerdo con la teoría clásica de vigas. En contraparte, la expresión para los esfuerzos tangenciales (), nos muestra que su distribución en la sección transversal es uniforme (independiente de las coordenadas  y ). Este resultado puede considerarse como una "mejora" en comparación con la teoría de Euler-Bernoulli, que predice  y  incluso si la fuerza cortante  es no nula.

Aun así, el resultado de la teoría de Timoshenko de  uniformemente distribuido en la sección transversal tampoco puede considerarse como exacto, ya que está en clara contradicción, por ejemplo, con la distribución parabólica de  presente en los elementos de sección rectangular  que se deduce en forma exacta utilizando la fórmula de Collignon. Con el fin de contrarrestar este problema, se adopta un coeficiente de forma o distorsión , que busca hacer coincidir el trabajo de deformación constante con el exacto. De este modo podemos expresar el esfuerzo tangencial () y la fuerza cortante  como:

Donde  se denomina área reducida de la sección transversal. El coeficiente de forma  depende de la forma de la sección transversal y del coeficiente de Poisson y está en el rango . La determinación precisa de este coeficiente no es sencilla en el caso general y varios autores han proporcionado estimaciones para distintas secciones transversales; por ejemplo, Cowper (1966) proporcionó expresiones para  para secciones de tipo rectangular, circular y tubo de pared delgada, entre otras.

La tabla siguiente muestra el factor  de corrección de cortante para vigas con materiales elásticos isotrópicos, según Cowper, para las secciones más comunes.

donde   es el coeficiente de Poisson.

Deformada de la viga en el caso cuasiestático

Interesa a continuación encontrar el campo de desplazamientos transversales , que nos da la forma de la viga deformada en términos de las acciones externas, que pueden ser fuerzas perpendiculares al eje de la viga (puntuales o distribuidas), así como pares concentrados. Eventualmente, podría ser necesario también obtener el giro de la sección transversal . Comenzaremos planteando las ecuaciones de equilibrio para un elemento diferencial de viga como el que se muestra en la figura, en la que además se explicita la convención de signos que se utilizará en este desarrollo.

Del equilibrio de fuerzas en dirección  para el elemento diferencial, se obtiene:

Del equilibrio de momentos, se obtiene:

Por otra parte, combinando la expresión para el momento flector  obtenida en el apartado anterior, con la expresión para  proporcionada más arriba (en el apartado sobre Campo de Desplazamientos) y utilizando , se obtiene:

 donde, de lo visto en el apartado anterior,

Alternativa 1

Si se combina las dos expresiones anteriores y se considera, además, el equilibrio de fuerzas verticales, se obtiene una ecuación diferencial de segundo orden en incógnita , que puede resolverse por doble integración si se conoce dos condiciones de contorno y si se conoce la distribución del momento flector . (Se observa que la fuerza distribuida  usualmente es un dato del problema). Esta ecuación es:

Alternativa 2

Si se deriva con respecto a  la ecuación de equilibrio de momentos y se considera la relación  obtenida en el apartado sobre Fuerzas Internas, se obtiene:
 donde :
Si se combina estas dos expresiones considerando nuevamente la relación entre  y el cortante  y asumiendo sección transversal constante, se obtiene una ecuación diferencial en la que aparece la derivada de la fuerza cortante  con respecto a . Esta derivada se puede eliminar usando el equilibrio de fuerzas expresado más arriba, para obtener una ecuación diferencial de cuarto orden en incógnita  que podrá resolverse en términos de la carga distribuida , siempre que se conozca cuatro condiciones de contorno. Esta ecuación es:

Alternativa 3

En forma alternativa, es posible encontrar la deformada de la viga resolviendo en forma escalonada dos ecuaciones diferenciales de primer orden: la primera de ellas es simplemente la relación  ya mencionada, mientras que la segunda proviene de reescribir la expresión para  dada en el apartado sobre Campo de Desplazamientos, teniendo en consideración (1),  la relación ; (2), la relación entre  y el cortante  y (3), la ecuación de equilibrio de momentos.
Las dos ecuaciones diferenciales mencionadas son las siguientes:

La primera ecuación puede resolverse para  en términos del momento flector si se conoce una condición de contorno para el giro de la sección transversal de la viga. Una vez hallado  podrá resolverse la segunda ecuación para  en términos del cortante si se conoce una condición de contorno para la deformada.

Bibliografía

Referencias

Mecánica de medios continuos
Análisis estructural


text:
Peru's left-wing president tried to dissolve Congress and was impeached and arrested instead
Carlos Garcia Granthon/Fotoholica Press/LightRocket via Getty Images
Pedro Castillo, the left-wing president of Peru, tried to dissolve Congress and was instead impeached and arrested by police.
Vice President Dina Boluarte was sworn into the presidency on Wednesday and became the first female president of the Latin American nation.
Castillo announced in an address to his nation that he was dissolving the Congress ahead of a potential vote to remove him from office. He was reportedly under investigation for alleged corruption over accusations that he was personally profiting from public works. Castillo denied the allegations and said he would rule by decree until another Congress could be elected and another constitution drawn up.
Instead, the members of the Peruvian Congress voted to impeach Castillo and remove him from office.
Boluarte criticized his attempt to dissolve Congress in a statement on Twitter, saying, "A coup that aggravates the political and institutional crisis that Peruvian society will have to overcome with strict adherence to the law."
Boluarte was sworn in as the sixth Peruvian president in five years.
The U.S. government opposed Castillo's order and called for the democratic process to continue in Peru under the current constitution.
"We will continue to stand against and to categorically reject any acts that contradict Peru's constitution, any act that undermines democracy in that country," said spokesperson for the U.S. State Department Ned Price.
The left-wing leader had narrowly risen to power in a presidential vote. He previously worked as a schoolteacher and a union leader and ran under the platform of rewriting the constitution and the redistribution of wealth through state power.
Castillo's status is unknown.
Here's more about Peru's political turmoil:
Peru's President Ousted After Saying He Will Seize Powerwww.youtube.com
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Woman traveled 3,000 miles to meet online boyfriend, then is gruesomely murdered, decapitated, and has her organs harvested ›
Damar Hamlin speaks publicly for first time since on-field collapse: 'With God's guidance, I will continue to do wonderful and great things'
Ex-judge in Las Vegas previously under ethics investigation dead of apparent suicide: 'They have terrorized and lied about me'


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