text:
Advanced Interactive Media/How We Did This: A Brief History for One Interactive Project

The Advanced Interactive Media class of Spring 2008 laid out the foundations of a proposed immersive interactive project for the Elsing geological museum. This will detail that process and the reasoning for the decisions made:

- Why did we focus on the age-group that we did?

> The Elsing Museum’s primary clientele are 3rd graders (ten to twelve years old), so that naturally became this project’s primary focus. However for the website, it was decided to focus on the older age-group that would encompass the teachers and parents, since they were considered more likely to be the ones to consult that as part of their research and decision whether to include the museum as part of their children’s education.

- Colour Palette / Design layouts / Branding / Look & Feel:

> The final colour pallette was chosen due to its earth-tone like look that reflects the project’s subject, namely a geological museum. Nonetheless, that was also balanced by a desire to keep some element of saturated, advanced colours that more naturally appeal to children. In addition, while the somewhat more dull look of the pallette may not be so appealing to children of the focus age-group, we think the media interactivity offered by the Surround-Screen Video (hereafter: SSV) elements, such as the Wii-controllers, the chroma-keying and the touchscreen experience, as well as the use of high-energy, child-friendly fonts (developed in LiveType) will more than compensate.

> Similarly, there was an understood need to balance what could otherwise be opposing requirements in the look and feel to the VR environments, which would naturally be set within the context of a cave (designed in Lightwave). Normally, the interior of a VR cave might not be considered particularly appealing to children looking for an interactive experience, but the museum’s own artifacts come to the rescue in this with their amazing variety in colours and even picturesque, natural designs. Combining the two, with specimens interspersed throughout the VR cave, will more than resolve any balancing issues.

> These first two items gave rise to a natural branding of the museum, and therefore this project, by showcasing the most aesthetic of the museum’s artifacts on an interactive lazy-susan display. This would be used on the website to give interested parties a taste of what’s available without the complexities of the more extensive VR cave environment.

- Scripting and Games:

> The tone of any story to accompany the VR environments was naturally set from the beginning by the clientele, as well as any proposed length, taking into their attention spans. Initially, it was thought that competition would be a worthwhile interactive experience, with the children split into two or more groups which would then have a certain length of time to complete the challenges/games put before them. Eventually, this was relinquished in favour of merely running the games as separate entities which would nonetheless still fit into the context of the story intro presented at the beginning of the SSV experience.

> The story itself was developed along the ideas of a villain who steals and/or breaks certain artifacts leaving the children, directly as well as through their VR alter-egos to solve the clues and repair the damage, etc. It is a standard format that easily lends itself to all kinds of permutations but is also simple enough to be a format that children are familiar with.

> Both the story and the games were kept simple enough so that they would not become time consuming, but still retained the potential of fully engaging the children in the VR experience with the Wii-controllers. The story was consequently created with characters that the children could relate to, and the games broken down into two quickly-understood, self-explanatory categories:

1) The jigsaw puzzle

2) Identify the piece/item that doesn’t belong, and

Breakdown of the project : > Phase 1: Colour palette, look & feel of layout, fonts, livetype templates, motion backgrounds, basic navigation mind map & buttons > Phase 2: Detailed navigation including all globals, plus applying branding to all screen formats > Phase 3: To create at least one working prototype for every format (VR, touch screen, video   projection, brochure, touchscreen, etc.) and finally the client demonstration

Creation of the 3 teams : > Design : o Development of the Colour Palette, Design layouts, Branding and overall Look & Feel that will be the confines within which the other teams will develop their products. o Creation of PR materials such as non-standard brochures, etc. that can be disseminated to   schools in the effort to bring the SSV experience to those places. > Interactive : o Creation of the game concept, including the Wii-controller interface, and attendant story script and characters. o To create the navigational user-interface and apply it to the touchscreen and website templates produced by the Design Team, primarily using Director software. > AV : o Naturally, the filming and recording of elements such as the real-life, curator character as well the lazy-susan display. o Creation of a mini-documentary about the museum, done in coordination with the curator. This documentary would then be featured on the website and be part of the promotional material for the museum and the SSV.

In keeping with the developing media world that is the focus of this class, communication was conducted primarily through Facebook (http://www.facebook.com/group.php?gid=2248278720)and a job/task list in Google docs (http://docs.google.com/#all)

As the weeks in this Spring 2008 semester began to pass by, a better understanding was gained about how far this current class could be expected to go in completing the concepts being developed. As a consequence, the last two weeks were focused on developing prototypes of all the material and concepts that will serve as the foundation for the completed project. Those prototypes were then presented to the curator and Grants coordinator who will then take those prototypes and use them in grant applications for the funds necessary to continue. Further development and the final completion will now be passed to future Ai Classes!


text:
Bartending/Equipment

Without adequate equipment a bar cannot serve drinks efficiently, to a good quality, or perhaps at all. Bar setups range wildly in each establishment depending an array of conditions such as the bar's angle, size, stock, staff, menu, glassware, equipment, appliances, and even what time of day it is; so no two bar setups are ever the same. However the main dogma behind the format of every setup is usually about optimizing the work space and ensuring good speed and quality of service. This section will take you step by step through the standard, intermediate and advanced equipment bartenders require to meet demands depending on their nature.

Below is general overview of the equipment that must be made to bartenders depending on their establishment type in ascending order.

Beer & Lager Centralised Establishments
Many central European establishments such as Beer Halls and Taphouses are bar the norm devoted to the mass consumption of beer, where variety is not necessarily the focus as much as large quantities of a handful of quality products. It is not uncommon for such establishments to be tied to a single brewery thus restricting their products to one source. Understandably, bartenders will be required to carry a few sturdy items to cope with this unilateral demand; namely a Bottle Opener.

Restaurants
Primarily considering the focus of the establishment on food, it is understandable that the bar be limited to convenient beverages so as to keep up with the service of food. However this is not always the case. Many modern restaurants have taken to shifting their unique selling point to their bars as opposed to their cuisine. In which cases a bartender would require much more than the following equipment examples. It is usually the standard for a restaurant bar to have a selection of spirits, liqueurs and wines to accompany the diners' eating experience; thus standard equipment such as a Jigger and a Waiter's Friend would be required for an efficient service.

Pubs & Bars
The most common drinking establishment type worldwide must be sure to equip it's bartenders well for a variety of reasons. Generic bars must be prepared to meet demands of seasonal events as well as economize in times of little business to be successful. These bar setups can vary depending on the time of day, week and month of the year. With the emphasis being primarily on alcohol, bars offer a wider variety of beverages to the public which in turn may require a wider variety of glassware. Ice Wells are imperative in bars and pubs today as approximately 50% of their available beverages are served with ice.

Clubs & Night Clubs
Clubs tend to have a broader selection of spirits available to the public which allows for an exponential increase in the number of drinks available. Again this may not always be the case. Massive demand through the late hours can also lead to a requirement for more efficient waste disposal and glass washing systems in order to maintain good service. Depending on the nature of the establishment, cocktails may or may not be available to order. However the addition of cocktails to any establishment's menu introduces whole sets of equipment to bartenders which must be used and maintained respectively.

Vodka, Rum, Cocktail and Specialty Bars
In the modern day, the public has somewhat lost it's taste for stronger spirits in favor of other beverages which lead to a decline in clientele to such establishments as Vodka and Rum bars in the western world. To re-excite the interest of the public many of these spirit bars have now turned to incorporating cocktails into their menus using the already stocked quality ingredients. With an infinite list of possible combinations of ingredients in most cocktail bars, there is a requirement for much more advanced equipment to either fulfill or complete a recipe. Shakers, Strainers, Bar Spoons, Bar Knives, Muddlers and Juicers are all examples of additional equipment used in these establishments. Certain recipes require certain equipment by the requirement of copyright law (see Dark And Stormy). Each new cocktail could require a new glass or a new piece of equipment not before used in the establishment and thus bartending staff must be adequately trained prior to their debut.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Bartending/Equipment


text:
Erlang Programming/Using lists

=Using lists=

foreach
3> lists:foreach( fun(X)->X*X end, [1,2,3]). >

produces no output because the purpose of foreach is to generate side-effects. However,

4> lists:foreach( fun(X)->io:format("~w ",[X]) end, [1,2,3,4]). 1 2 3 4

does work, because io:format is a side effect function.

sequence of numbers
lists:seq(1,100) is like range(1,101) in python.

5> lists:seq(1,10). [1,2,3,4,5,6,7,8,9,10]

sort
lists:sort( A ) is what you think.

6> lists:sort([1,3,2,6,5,4]). [1,2,3,4,5,6] 7> lists:sort([a,d,b,c]). [a,b,c,d] 8> lists:sort([f,e,a,"d","c",{b}]). [a,e,f,{b},"c","d"]

title: Erlang Programming/Using lists


text:
Erlang Programming/Macros

=Macros=

-define(LIKERT_SCALE, lists:seq(1, 5)). A = ?LIKERT_SCALE.

The code makes A = [1,2,3,4,5].

Some handy predefined macros include:

?MODULE (module name) ?LINE   (line number) ?FILE   (filename as a string)

% =========================================== % Example code % =========================================== -module(test_macros). -define(LIKERT_SCALE, lists:seq(1, 5)). -compile(export_all). start -> io:format("likert scale: ~w \n", [?LIKERT_SCALE]), io:format("module name:  ~w \n", [?MODULE]), io:format("line number:  ~w \n", [?LINE]), io:format("filename:     ~s \n", [?FILE]), ok.

% =========================================== % Example output % =========================================== %  % 6> c(test_macros). % {ok,test_macros} % 7> test_macros:start. % likert scale: [1,2,3,4,5] % module name:  test_macros % line number:  10 % filename:     ./test_macros.erl % ok

title: Erlang Programming/Macros


text:
Erlang Programming/Using regexp

= Regular Expressions =

2> re:run("hello world","w.+d"). {match,[{6,5}]}

The regular expression, "w.+d" matches the string, "hello world" at location 6 for 5 chars.

9> re:replace("10203040","([2-4]0)+","01",[{return,list}]). "1001"

Makes a substitution, replacing "203040" with "01".

title: Erlang Programming/Using regexp


text:
User:Nick

Nick on English Wikipedia. Nick on Commons. NotASpy on Flickr I'm an administrator at English Wikipedia and Wikimedia Commons. I'm also an OTRS volunteer. Don't hesitate to contact me if you require assistance

title: User:Nick


text:
User:Lupus~enwikibooks

Hi!. I don't use to be here, so in order to contact with me you can send an email. Thanks!

title: User:Lupus~enwikibooks


text:
C++ Programming/Code/Standard C Library/Functions/atan2

atan2
The atan2 function computes the arc tangent of y/x, using the signs of the arguments to compute the quadrant of the return value.

 * Related topics: acos - asin - atan - cos - cosh - sin - sinh - tan - tanh


text:
C++ Programming/Code/Standard C Library/Functions/cos

cos
The cos function returns the cosine of arg, where arg is expressed in radians. The return value of cos is in the range [-1,1]. If arg is infinite, cos will return NAN and raise a floating-point exception.

 * Related topics: acos - asin - atan - atan2 - cosh - sin - sinh - tan - tanh


text:
C++ Programming/Code/Standard C Library/Functions/sin

sin
If you don't want cmath you can write sin function it is;

using namespace std;
 * 1) include

double sin(double x) //sin function {return x-((x*x*x)/6.)+((x*x*x*x*x)/120.);}

int main { double a;   cin>>a; cout<<"sin("<<a<<")="<<sin(a*(3.14159/180.))<<endl; return 0;}

The function sin returns the sine of arg, where arg is given in radians. The return value of sin will be in the range [-1,1]. If arg is infinite, sin will return NAN and raise a floating-point exception.

 * Related topics: acos - asin - atan - atan2 - cos - cosh - sinh - tan - tanh


text:
Aros/Developer/NightlyConfig

=Creating a Nightly Build=

Prerequisites

 * SSH Account for the AROS Project on sourceforge.net
 * Space to store all the required source for your chosen packages,
 * Space to store the generated binaries and log files.
 * A valid svn account for https://svn.aros.org/ with read-access.
 * An email account subscribed to the AROS dev mailing list.

Terms Used

 * :                           The location the nightly scripts are stored in
 * :        A valid SSH account on sourceforge with access to AROS.
 * :             The local user account you will build binaries with.
 * :      An email account with access to the AROS Dev mailing list.
 * This will be used to send the log to the mailing list.

 * :

Nightly Package Names
The build script uses package names to identify what to build for a given server. the packages tell the build script which files in /pkg/ to run, which in turn creates the chosen binaries.

Valid Package Names
Files which are common to all platform generally have a name that is just their target, with all as the build target. Currently defined ones are

Common Package Names
sources.svn.all changelogs.svn.all documentation.svn.all

Arch Specific Package Names
sdk.svn.pc-i386 sdk.svn.pc-x86_64 contrib.svn.pc-i386 contrib.svn.pc-x86_64 bootiso-grub.svn.pc-i386 bootiso-grub2.svn.pc-i386 bootiso-grub2.svn.pc-x86_64 bootfloppy.svn.pc-i386 hosted.svn.linux-i386 hosted.svn.linux-x86_64 hosted.svn.linux-ppc

Disallowed Names
bootiso                           generic routine used by the arch specific nightly targets hosted sdk

Prepairing the system
Firstly you will want to set up a user account on your system that will be used to perform the build. create a user

adduser

The first step is to obtain the nightly build scripts. cd to the location you will perform the build in and checkout the scripts as follows -:

cd /home/ mkdir -p MyNightlyBuildDir cd MyNightlyBuildDir svn checkout https://svn.aros.org/svn/aros/trunk/AROS/scripts/nightly

you will have a new subdirectory called nightly that contains all the scripts used to build the nightly binaries. from here on in we will refer to this directory as .

now cd into that directory and make a directory tree to hold the AROS source

cd  mkdir -p Source mkdir -p Source/Contrib mkdir -p Source/Documentation mkdir -p Source/Necessary

Finaly create an initial cache of the svn files used to build AROS

IMPORTANT: You must set the correct LOCALE for your system at this point!!

type the following -:

cd Source export LANG=en_US.ISO8859-1

then checkout all the source repositories needed to perform your nightly build packages.

ie...

svn checkout https://svn.aros.org/svn/aros/trunk/AROS cd Contrib svn checkout https://svn.aros.org/svn/aros/trunk/contrib cd ../Necessary svn checkout https://svn.aros.org/svn/aros/trunk/contrib/necessary

You may omit complete contrib if you don't want to bundle it within your archive. Just leave the corresponding directory empty.

Prepairing your servers config file
Create a plain textfile in the cfg directory called  Open it in a text editor and add the packages you'd like to build.

For example .. CFG_PACKAGES=(sources.svn.all documentation.svn.all sdk.svn.pc-x86_64 bootiso-grub2.svn.pc-x86_64)

To enable sending an email report of the build results to the dev mailing list add the following lines ..

CFG_MAIL_ENABLE=yes CFG_MAIL_CC=

And to enable uploading the binaries so they may be listed on AROS.org add ..

CFG_UPLOAD_ENABLE=yes CFG_DS_USER= CFG_BACKUP_ENABLE=no

Performing your first run
The first time you check the system is operational ... perform a manual dry run by doing the following

cd ; ./build

And watch as it displays all your build output.

If everything went well and you have enabled the email logging in your server config, you should see a confirmation email on the AROS developer mailing list confirming the build completed successfully. If you have also enabled uploading the files, and have a valid AROS SSH account on sourceforge, the binaries should be uploaded into the temporary storage location ready to be moved into the live folders on the AROS.org download pages

Setting it up to run automatically
You will need to create a cron job for the build step we performed doing the dry run. You will want to add a command that is performed once every 24hrs, which runs the build script with your server's configuration and uploads the created archives to AROS.org.

0 2 * * * ( cd  && ./build <Your Server's Config-Name> ) > /dev/null

This will make the build machine attempt to build and upload AROS at 2am. All output will be directed to NULL (except of course the generated logfiles)

Adding the binaries to the AROS.org site
If you are simply providing binaries for existing AROS targets, the previous steps should be enough to add the files to the AROS.org site. If however you wish to provide new flavours of AROS then you will need to update the AROS.org site with information about them. Read on...

The scripts and page definitions to list the binaries on AROS.org are maintained in svn at https://svn.aros.org/svn/aros/trunk/documentation

To edit the files you'll need to checkout the repository and customise to suit.

listing new downloads
If you are providing binaries for AROS targets that don't currently exist, you will need to update the files which list them on AROS.org.

Edit the db/download-descriptions/<Language> files to list descriptions of the packages you build. make sure and follow the naming convention used by the existing packages.

Corrupt localised filenames in the binaries
This usually means the source files used have utf8 filenames in the source repository caches resulting from the initial checkout being done when the system locale was set to utf8.

to fix it change directory into each source directory in turn and delete all files and directories EXCEPT the top level .svn directory.

now type ..

export LANG=en_US.ISO8859-1

.. and finally do an "svn update" in each directory.

Maintaining the Main Nightly Build Machine
The main nightly build machine has the additional responsibility of running scripts to move the uploaded binaries to the live AROS.org download pages and clean up the upload directory. Other build machines don't need to run these scripts.

Step1
Moving the uploaded binaries to the live AROS.org download pages.

You will need to create a cron job that runs at regular intervals during the day, currently every hour. This job however isn't performed locally, and will require an AROS SSH account on sourceforge. Its goal is to move the archives uploaded by the various build machines into the correct folder on the AROS.org site so that they become available for download. It also deletes any old versions of these files so that only the most up-to-date version is available (there isn't enough space to keep multiple versions of each file).

00 * * * * ssh <Sourceforge SSH Account Name> 'bash -l -c aros/scripts/nightly-update' > /dev/null

This will make the server transfer any files present in the upload dirextory to the AROS.org site once an hour. Again all output is sent to NULL.

Step2
This step removes incomplete or broken nightly build files from the upload directory.

51 7 * * * ssh <Sourceforge SSH Account Name> 'bash -l -c aros/scripts/nightly-clean' > /dev/null

This will clean up the uploads directory on the AROS.org server at 7:51am each day.

title: Aros/Developer/NightlyConfig


text:
User talk:Epicurean~enwikibooks

Welcome message
Welcome, Epicurean!

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally. Webaware talk 05:15, 20 April 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Epicurean. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Epicurean~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:08, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 04:56, 19 April 2015 (UTC)

title: User talk:Epicurean~enwikibooks


text:
User talk:Gaspah

Welcome message
Welcome to Wikibooks, Gaspah!

title: User talk:Gaspah


text: User talk:202.125.156.122

title: User talk:202.125.156.122


text:
User talk:Jmroczek~enwikibooks

Welcome message
Welcome to Wikibooks, Jmroczek!

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Jmroczek. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Jmroczek~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:20, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:05, 19 April 2015 (UTC)

title: User talk:Jmroczek~enwikibooks


text:
IB Chemistry/Extended Essay

=Suggestions=

Here is a list of suggestions for Science Investigations. Please add/amend/extend the list as you see fit. This is essentially a brainstorm...

Non-coalescing drops
Some industries will mix liquids by cascading one liquid into another, assuming mixing will be complete. These drops may prevent complete mixing and need to be investigated.

Recent research has shed light on this phenomenon: The fast-moving shiny drops that skid across the surface of soapy water. Detergents play a crucial role – but how does the concentration of detergent affect the drops? What about ‘non-ionic’ detergents? What about a surface of anionic detergents and drops of cationic ones?

Challenges: Making drop-producing apparatus, Researching the chemistry of detergents, Assessing the numbers of drops

Thiosulphate/acid reaction (‘disappearing cross’)
Chemical engineers wishing to check if turbidity is a useful surrogate measurement for whether a reaction has reached completion.

How does the reaction respond to changes in acid concentration (it is first order for thiosulphate concentration but varies from first to zero order with acid).

How can we assess the amount of oil contamination on beaches?
Various groups will want to assess oil contamination, and a simple quick method would be appreciated.

The method used solvent extraction. Sand samples were laced with oil and the efficiency of recovery was measured. The sensitivity of the method was also estimated.

Synthesis and isolation of propanoic acid from propan-1-ol.
Chemical suppliers who want to investigate production of this chemical.

Trialling different methods.

Making biodiesel from waste vegetable oil.
Motoring enthusiasts who want to make their own environmentally-friendly and cheap fuel.

By using methanol and sodium hydroxide it is possible to make diesel fuel from old chip fat.

Analysing the gases remaining after burning a candle in a gas jar.
Motor vehicle engineers who want to check the composition of exhaust gases.

CO2 is easy; it is readily absorbed with NaOH.

O2 is harder; an oxygen probe is very useful but otherwise oxygen could be absorbed by, for example, the slow reaction of steel wool.

CO is challenging; it will be absorbed by CuCl solution but it is in very low concentrations.

Extraction of caffeine from tea.
Suppliers or independent analysts want to assess the caffeine content of tea products. Chemical suppliers want to provide pure caffeine to clients.

This method usually involves solvent extraction with a chlorinated alkane (e.g. dichloromethane or 1,1,1,-trichloroethane). With care it is possible to recycle the solvent and avoid health and environmental damage.

Other suggestions

 * Investigation of weak acid strengths/synthesis of new acids

 * Rate expression for tin/iodine reaction – kinetics of a reaction

 * Nobili’s rings – strange colours seen during electrolysis.

 * A transient red colour – the chemistry of iron/sulphate complexes.

 * Liesegang rings – stripey effects in gels.

 * Polarimetry to investigate concentrations of sugars. Suppliers of sugar syrups who want to assess their products.

 * Flame photometry to determine concentrations of aqueous ions. Analysis of mineral water samples to check manufacturers' claims.

 * Is the level of chloramines found in swimming pool water affected by temperature? Swimming pool operators concerned about eye irritation amongst their customers.

 * Effectiveness of two major washing powders on five common household stains. Consumer testing organisation who wants to test commercial claims.

 * Factors affecting the decomposition of biodegradable plastics. Environmental agencies testing claims of environmental-friendliness from plastics manufacturers.

 * Studies of infrared spectroscopy: nature and applications. Applications include breath-testing for alcohol and testing vehicle exhaust emissions.

 * NaCl and potential alternatives to NaCl as a roadway de-icer. Highways agency investigating de-icers which cause less environmental damage - but at what cost?

 * Synthesis and analysis of hexane-3,4-diol.

 * Calcium carbonate content of seashells.

 * Characterisation of New Polyethylenes Prepared by Single-Site Catalysis.

 * An Investigation Into The Uses Of Environmentally Friendly Additives/Alternatives To Petrol.

 * Protein Purification.

 * A study into the colour of ketchup.

 * Comparison By Spectral Characterisation Of Myoglobin In Beef, Pork and Turkey.

 * Spectrophotometric Determination of Trace Amounts of Iron in Plant Samples.

 * The effect of disulphide reduction on the structure, and activity of Yeast Alcohol Dehydrogenase.

 * Investigation into the effect of pH on the mechanism of glycation of phenylalanine as a model of the glycation of the B chain of insulin.

 * Analysis of Vitamin C in Orange Juice through a Chemiluminescent Clock Reaction.

 * Investigating the Effect of Different Halide Ions on the Stern Volmer Constant.

 * Developing and Validating an Assay for Tissue Thiols and the Analysis of the Effects of Aging and Taurine on Tissue Thiols.

 * Relation between the Slow Relaxation Time and Foaming of Selected ICI Surfactants.

 * Purification of ibuprofen from commercial tablets.

 * Building a breathalyser.g

 * Titration of various commercial vinegar samples.

 * Titration of vitamin C content of commercial fruit juices.

 * Investigation in the synthesis of acetylsalicylic acid (aspirin).

title: IB Chemistry/Extended Essay


text:
Irish/Unit 3

Unit 3 of WikiBooks Irish consists of a series of detailed lessons on specialized subjects, intended for the advanced learner. These lessons will contain detailed and comprehensive discussions of the vocabulary, terminology, and, where appropriate, grammar and syntax related to a specific subject.

Unlike the numbered lessons of Unit 1 and Unit 2, these lessons are purely subject-based and may be explored in any order. They are listed below by subject matter.

These lessons address real world subjects, like clothing, cooking, plumbing, or sport, not general language topics like pronouns or plurals. Comprehensive overviews of language can be found instead in the Irish reference pages.

General
Verbs

Home
House • Kitchen • Living room • Bedroom • Bathroom • Garden

Body
Body parts • Clothes

Food
Fruit • Meals • Vegetables

Geography

 * Tíortha an Domhain (Countries of the World)

Nature
Animals

Work
Professions

Society
School subjects • Insults

title: Irish/Unit 3


text:
User talk:Wuyqpy

1000 最常用单音节拼音单词定型(第1版)

词频 单字词 拼音 491 阿 ah 700 啊 a 111 爱 ai 217 安 an 690 案 am 866 按 an4 883 暗 anm 653 凹 ao 379 巴 ba 70 把 ba3 401 爸 ba4 405 吧 bab 596 八 ba1 222 白 bai 402 百 bai3 947 败 bai4 413 半 ban 431 办 bam 555 般 ban1 634 班 banm 669 板 ban3 948 伴 ban4 707 帮 bang 247 报 bao 356 保 bao3 467 包 bao1 747 抱 bao 748 宝 bao3 128 被 bei 390 北 bei3 568 备 bei4 662 背 bbei 695 贝 bei 809 悲 beib 151 本 ben 183 比 bi 341 必 bi4 563 笔 bi3 848 毕 bib 185 便 bian 273 边 bian1 283 变 biam 961 遍 bian4 981 编 bian1 205 表 biao 686 标 biau 178 别 bie 177 并 bimg 301 病 bing 731 兵 bing1 850 冰 bimg1 644 波 bo 716 伯 bo2 835 博 buo 6 不 bu 189 部 bbu 367 布 bu4 407 步 bub 184 才 cai 845 采 cai3 876 彩 caai 952 财 cai2 970 菜 cai4 595 参 cam 890 餐 can 660 草 cao 326 曾 ceng 897 层 cemg 586 查 cha 645 察 chha 818 差 cha1 371 产 chan 106 长 chang 136 常 chamg 245 场 chang3 713 唱 chung 666 朝 chao 839 超 chao1 167 车 che 566 沉 chen 743 陈 chem 66 成 cheng 448 称 cheng1 461 城 chemg 527 程 cheng2 873 承 chhemg 281 吃 chi 479 持 chi2 82 从 chong 688 充 chomg 790 冲 chung 34 出 chu 238 处 chu4 597 除 chu2 608 初 chu 354 传 chuan 523 穿 chuan1 701 船 chuam 651 创 chuong 765 床 chuang 823 窗 chuamg 560 存 chun 629 春 chum 124 次 ci4 134 此 cii 815 词 ci 813 村 cun 593 错 cuo 18 大 da 180 打 da3 303 达 dah 372 答 da2 209 代 daib 270 带 dai 520 待 daid 814 戴 daia 983 袋 daic 89 但 dan 504 单 dam 804 担 dan1 833 弹 dan4 64 当 dang 784 党 damg 16 到 dao 75 道 daod 535 倒 ddao 636 导 dao3 902 岛 daoo 1 的 de 36 得 de 206 德 de2 193 等 deng 769 登 demg 877 灯 deng1 19 地 di 73 第 di 589 底 di3 624 帝 di4 656 弟 dik 718 低 di1 147 点 dian3 197 电 diam 657 店 dian 982 典 diam3 606 掉 diao 150 定 ding 858 顶 ding3 100 动 dong 235 东 dung 906 懂 domg 705 斗 dou 59 都 du1 266 度 du4 357 读 du 493 独 du2 459 断 duan 648 段 duam 819 短 duan 990 端 duan1 50 对 dui 391 队 duie 637 顿 dun 52 多 duo 910 恶 e 831 恩 en 45 而 er 107 儿 era 146 尔 er3 353 二 eer 817 耳 err 63 发 fa 117 法 faa 339 反 fan 625 饭 fanp 764 范 fam4 867 犯 fan4 886 翻 fan1 976 凡 fan2 101 方 fang 250 放 famg 361 房 fung 317 非 fei 328 飞 fui 534 费 fei4 110 分 fen 760 份 fem 274 风 feng 667 封 femg 710 佛 fo 658 否 fou 186 夫 fu 241 父 fup 460 福 fu2 485 妇 ffu 494 复 fu4 591 富 fut 837 府 fuu 840 负 fud 898 付 fug 963 副 fuk 342 该 gai 553 改 gai3 962 盖 gai4 140 感 gaan 331 干 gan 787 赶 gan3 810 敢 gam 539 刚 gang 129 高 gao 255 告 gau 1000 搞 gao3 12 个 ge 294 格 geh 335 各 ged 542 歌 gef 679 哥 gea 908 革 ge2 115 给 gei 425 根 gen 522 跟 gen 233 更 geng 152 公 gong 173 工 gomg 514 共 gong4 518 功 gong1 929 宫 gung1 933 供 gung 621 够 gou 904 狗 gou3 434 古 gu3 462 故 gu4 724 顾 gu 742 姑 gud 871 骨 gub 953 孤 guf 755 怪 guai 249 关 guam 381 管 guam 395 观 guan 519 官 guan1 673 馆 guan3 221 光 guang 581 广 guamg 771 贵 gui 822 规 gui1 973 归 gui1 26 国 guo 46 过 guo4 159 果 guo3 156 孩 hai2 202 海 hai 659 害 hai4 708 汉 han 937 喊 ham 985 含 han2 60 好 hao 528 号 hau 846 毫 hao2 31 和 he 265 何 hea 338 合 hek 677 河 heh 796 喝 he1 456 黑 hei 114 很 hen 400 红 hong 38 后 hou 261 候 heu 393 乎 hu 680 呼 hu 692 护 hu4 792 胡 hu2 930 忽 hu1 135 话 hua 229 花 hua1 311 化 hhua 355 画 hua4 435 华 hua2 971 划 hua4 578 怀 huai 870 坏 huai4 71 还 huan 333 欢 huam 881 环 huan2 884 换 huan4 548 黄 huang 647 皇 huang 41 会 hui 108 回 hui2 950 挥 hui1 386 婚 hun 138 活 huo2 223 或 hue 366 火 huo 641 获 huo4 957 伙 huo3 85 己 ji 149 几 jii 172 机 ji1 237 记 ji4 336 即 jid 359 及 ji2 420 极 jih 437 计 ji4 486 纪 jig 487 基 jit 612 际 jif 620 激 jib 630 集 jip 649 急 jil 722 击 jix 753 级 jik 785 继 jiq 793 吉 jis 851 既 jiz 885 技 jji 959 迹 jif 987 济 jic 42 家 jia 196 加 jiak 536 价 jia4 600 假 jia3 830 架 jiad 118 间 jiam 141 见 jian 471 建 jiank 751 简 jian3 779 渐 jiand 795 坚 jian1 945 健 jiam4 161 将 jiang 426 讲 jiang3 928 奖 jiamg 938 降 jiang4 966 江 jiamg1 227 教 jiao4 276 叫 jiau 344 交 jiao 550 脚 jiao3 570 校 jiao 670 角 jiau3 776 较 jiaw 199 界 jie 230 结 jie2 254 接 jie1 275 解 jie3 453 节 jiea 687 姐 jiel 738 街 jiek 958 杰 jied 994 介 jie4 133 进 jin4 278 金 jin 306 今 jinn 369 近 jjin 397 尽 jim 468 紧 jim3 511 仅 jin3 955 禁 jim4 105 经 jing 374 精 jimg 464 惊 jinng 490 静 jing4 500 竟 jimg4 607 京 jing 668 睛 jing 683 境 jingt 694 警 jingl 854 景 jimgl 29 就 jiu 392 久 jeu 502 酒 jiu 521 究 jiu1 799 救 jiu4 827 旧 jeu4 849 九 jiu3 510 举 ju3 516 句 ju4 529 居 ju1 538 具 juk 554 据 jud 699 剧 jut 715 局 ju2 878 巨 ju 268 觉 jue 352 决 juek 473 绝 juee 260 军 jun 506 卡 ka 61 看 kan 79 开 kai 932 康 kang 530 考 kao 811 靠 kao4 43 可 ke 208 克 ke4 329 科 ke1 450 客 kep 457 刻 kee 935 课 ked 711 肯 ken 312 空 kong 956 恐 komg 214 口 kou 365 苦 ku 736 哭 ku1 279 快 kuai 549 块 kuai 671 况 kuang 940 困 kun 234 拉 la 15 来 lai 508 兰 lan 974 浪 lang 112 老 lao 739 劳 lao2 4 了 le 231 乐 le4 841 勒 lee 382 类 leik 725 泪 lei 836 雷 lei2 212 棱 lemg 693 冷 leng 37 里 li 144 力 li4 145 理 lii 201 利 lli 286 立 lit 327 离 li2 449 丽 lip 463 历 lic 545 礼 li3 616 李 li 774 例 lik 377 连 lian 465 脸 liam 484 联 lian 864 练 lian4 911 恋 liand 123 两 liang 364 量 liamg 599 亮 liang4 909 良 liang2 579 料 liao 590 列 lie 741 烈 liek 346 林 lin 802 临 lim 403 令 ling 452 领 ling3 470 另 limg 552 灵 ling2 332 流 liu 411 留 lieu 643 六 liu4 941 刘 liu 770 楼 lou 269 路 lu 733 露 luf 750 鲁 lud 781 录 lu4 853 乱 luan 343 论 lun 691 伦 lum 320 罗 luo 396 落 luo4 891 洛 lue 737 旅 lv3 758 律 lv 220 马 ma 282 妈 ma1 322 吗 mah 900 玛 mak 488 买 mai 794 卖 mmai 291 满 man2 709 慢 man 675 忙 mang 515 毛 mao 65 没 mei 80 美 mei3 176 每 mei3 960 妹 mei4 14 们 men 200 门 men 592 梦 meng 988 蒙 memg 447 米 mi 587 密 mi4 767 秘 mid 825 迷 mi2 86 面 mian 880 免 miam 915 妙 miao 262 民 min 116 名 ming 142 明 mimg 228 命 ming4 57 么 me 571 默 mo 744 莫 muo 946 模 mo2 565 某 mou 240 母 mu3 298 目 muu 632 木 mu 882 姆 mu3 30 那 na 436 拿 na2 533 哪 na3 617 纳 na4 914 娜 na 978 奶 nai 225 难 nanh 272 男 nan 451 南 nank 540 脑 nao 300 呢 ne 258 内 nei 49 能 neng 23 你 ni 444 尼 ni2 27 年 nian 509 念 niam 575 娘 niang 936 鸟 niao 438 您 nin 735 牛 niu 862 农 nong 921 弄 nomg 924 诺 nuo 88 女 nv 896 欧 ou 513 怕 pa 661 排 pai2 859 派 pai4 868 拍 pai 895 判 pan 965 牌 pai2 729 旁 pang 598 跑 pao 384 朋 peng 532 皮 pi 901 批 pi1 319 片 pian 754 票 piao 368 品 pin 246 平 ping 556 破 po 635 普 pu 62 起 qi 102 期 qi1 132 其 qi2 175 气 qi4 380 奇 qid 443 妻 qie 654 器 qif 732 七 qik 761 汽 qqi 989 棋 qih 91 前 qian 321 钱 qien 480 千 qian 432 强 qiang 954 枪 qiamg 315 切 qie 385 且 qie3 131 亲 qin 879 琴 qim 84 情 qimg 297 轻 qing1 309 清 qimg1 347 请 qing 389 青 qingc 308 求 qeu 351 球 qiu 968 秋 qiu1 48 去 qu 429 取 qu3 627 趣 qqu 652 区 que 672 曲 qul 168 全 quan 749 权 quam 195 却 que 501 确 queb 893 缺 que1 646 群 qun 53 然 ran 239 让 rang 414 热 re 7 人 ren 253 认 ren4 277 任 renk 889 忍 ren3 559 仍 reng 95 日 ri 419 容 rong 797 肉 rou 68 如 ru 256 入 ru4 681 若 ruo 969 萨 sa 585 赛 sai 892 塞 sait 171 三 san 824 散 san4 232 色 se 714 沙 shae 757 杀 sha 267 山 shan 806 善 sham 13 上 shang 476 伤 shamg 544 商 shang1 207 少 shao 995 烧 shau 497 社 shee 602 设 she 482 谁 shei 126 身 shen 236 神 shem 304 深 shenc 564 甚 shen4 816 森 shenm 25 生 sheng 166 声 shemg 728 圣 sheng4 759 胜 shengc 980 升 shengd 986 盛 sheng4 3 是 shi 17 时 shi 69 事 shhi 121 世 shi4 122 什 shib 125 使 shic 163 十 shid 164 实 shi2 285 师 shif 290 失 shi1 295 士 shih 302 始 shik 375 视 shil 406 识 ship 412 市 shir 428 式 shis 433 石 shit 440 似 shiz 477 示 shix 481 史 shiq 492 诗 shiy 524 室 shiic 558 食 shis 577 试 shig 828 适 shi4 927 势 shill 96 手 shou 203 受 shou4 423 收 shou1 424 首 shoud 949 守 sheu 972 授 shhou 188 书 shu 288 数 shub 410 术 shuc 474 树 shud 723 速 shuf 923 熟 shu2 498 算 shuan 442 双 shuang 190 水 shui 263 岁 shhui 421 随 shui2 605 虽 shui1 638 睡 shui4 967 顺 shun 24 说 shuo 119 斯 sii 216 死 si 289 四 ssi 314 思 sik 378 司 sil 782 丝 sis 415 送 song 801 松 somg 446 诉 su 583 苏 su1 778 属 su3 860 素 su4 887 束 suk 93 所 suo 698 索 suoo 9 他 ta 33 她 tae 103 它 tac 181 太 tai 387 台 tai2 752 态 taik 418 谈 tan 916 探 tan4 964 坦 tan3 730 堂 tang 931 套 tao 153 特 te 194 体 ti 307 提 tih 340 题 tid 47 天 tian 299 条 tiao 580 调 tiaoh 640 跳 tiao4 834 铁 tie 204 听 ting 611 停 timg 650 庭 tingb 94 同 tong 284 通 tong1 458 统 tong3 495 痛 tong4 857 童 tomg 104 头 tou 832 投 touh 531 突 tu1 537 图 tu2 573 土 tu 727 团 tuan 682 推 tui 919 退 tui4 991 腿 tui3 786 托 tuo 861 脱 tuoo 155 外 wai 271 万 wan 305 完 wan2 349 晚 wan3 719 玩 wanr 226 望 wamg 264 往 wang3 310 王 wang 626 忘 wung 943 亡 wung2 20 为 wei 109 位 weib 383 未 weic 472 维 weid 551 味 weie 574 微 weif 689 围 weih 807 卫 weik 912 委 weil 143 问 wen 148 文 wen2 604 温 wen1 712 闻 wem2 5 我 wo 83 无 wu 174 物 wu4 376 五 wu3 572 务 wuk 721 屋 wuf 766 舞 wud 768 午 wul 907 武 wull 996 误 wup 158 西 xi 325 喜 xii 408 希 xik 475 系 xib 517 息 xic 546 细 xid 569 习 xip 772 吸 xif 805 戏 xxi 844 洗 xil 855 席 xih 44 下 xia 939 夏 xia 90 现 xian 213 先 xiam 584 显 xiand 603 线 xianf 702 险 xiian 951 鲜 xiann 76 想 xiang 157 相 xiumg 165 向 xiung 191 像 xiamg 287 象 xiamg 543 响 xiang3 610 香 xiamg1 829 乡 xiang1 903 项 xiang4 51 小 xiao 257 笑 xiao 496 消 xiao1 92 些 xie1 218 写 xie 567 血 xieb 655 谢 xie4 944 鞋 xie2 55 心 xin 169 信 xin4 182 新 xind 98 行 xing 219 性 xing4 373 星 xingd 398 形 xingc 416 兴 xingk 507 幸 xxing 740 型 xingh 843 醒 xing3 979 雄 xiong 800 修 xiu 905 休 xiu 211 许 xu 503 需 xuu 576 须 xuc 678 续 xu4 466 选 xuan 925 宣 xuan1 56 学 xue 697 雪 xue3 756 寻 xun 409 亚 ya4 783 牙 ya 838 压 ya1 917 呀 yah 192 眼 yan3 330 言 yan 422 演 yen 618 验 yan4 622 严 yan2 633 研 yand 703 烟 yam 74 样 yang 609 养 yang3 614 阳 yang2 763 洋 yamg 32 要 yao 808 药 yaoc 920 摇 yao2 35 也 ye 313 业 yye 388 夜 ye4 791 叶 yep 865 野 yee 2 一 yi 40 以 yii 99 意 yyi 130 已 yib 348 医 yic 454 衣 yif 499 义 yid 525 易 yig 588 议 yik 642 艺 yil 704 依 yir 746 异 yis 798 遗 yiz 812 伊 yiq 863 疑 yi 894 忆 yi4 993 释 yit 113 因 yim 296 音 yin 557 引 yin3 601 印 yin4 926 银 yin 242 应 ying 259 英 yimg 399 影 ying3 918 营 ying2 72 用 yong 541 永 yomg 913 拥 yung 10 有 you 97 又 yiu 215 由 yeu 248 友 yiu3 526 游 you2 762 右 yeu4 826 油 yiu2 934 优 you1 54 于 yu 137 与 yuh 323 语 yu3 674 育 yu4 684 遇 yuz 685 雨 yul 745 鱼 yur 872 余 yuc 977 预 yuk 210 员 yuen 280 原 yuem 293 远 yuam 324 元 yuan 363 院 yuand 478 愿 yuan4 734 园 yuan2 162 月 yue 334 约 yue1 394 越 yuec 358 运 yun 821 云 yum 842 杂 za 8 在 zai 187 再 zai 803 藏 zang 875 臓 zamg 362 早 zao 360 则 ze 773 责 zee 316 怎 zen 888 增 zeng 292 战 zhan 455 站 zham 639 展 zhan3 252 张 zhang 631 丈 zhang4 788 章 zhamg 318 找 zhao 417 造 zhaoz 430 照 zhaok 992 招 zhaoo 11 这 zhe 28 着 zhed 127 者 zhee 179 真 zhen 856 珍 zhen 899 阵 zhem 139 正 zheng 427 整 zheng3 469 争 zhemg 623 证 zhemg4 869 征 zheng1 58 之 zhi 67 只 zhii 120 知 zhib 243 直 zhic 251 至 zhid 337 指 zhif 350 制 zzhi 489 志 zhih 505 治 zhik 613 致 zhhi 615 纸 zhig 628 支 zhil 663 止 zhip 676 质 zhis 706 值 zhir 777 职 zhio 789 智 zhix 874 置 zhiz 22 中 zhong 78 种 zhong4 81 总 zhong3 170 重 zhung1 345 终 zhung4 512 钟 zhong 561 众 zhongk 404 周 zhou 665 州 zheu 726 洲 zheu 198 主 zhu 224 住 zhub 562 注 zhud 619 助 zhug 664 组 zhuc 696 著 zhhu 717 族 zhuu 445 转 zhuan 547 专 zhuam 439 装 zhuang 852 状 zhuamg 87 最 zhui 775 追 zhui1 483 准 zhun 922 桌 zhuo 21 子 zi3 39 自 zzi 244 字 zi 720 资 zi1 160 走 zou 441 足 zu 820 祖 zu3 847 嘴 zui 942 罪 zui4 77 作 zuo 154 做 zuo 370 坐 zuoc 594 座 zuod 780 左 zuo3Kib

title: User talk:Wuyqpy


text:
Cookbook:Galaktoboureko (Greek Semolina Custard Pastry)

| Greece | Dessert

The galaktoboureko (Γαλακτομπούρεκο) is a traditional Greek dessert. It is made with phyllo, also known as filo or fillo dough, which is filled with a custard of semolina flavored with citron. It is served with a sugar syrup.

Pastry

 * 150 g white granulated sugar
 * 6 eggs
 * 150 g semolina
 * 1 liter milk
 * 2 vanilla beans
 * ½ ea. lemon, juiced
 * 1 pinch of salt
 * 450 g phyllo
 * Melted butter

Syrup

 * 125 ml water
 * 150 g white granulated sugar
 * 2 Tbsp lemon juice

Pastry

 * 1) Combine sugar, eggs, and semolina in a saucepan. Mix in milk and vanilla beans. Stir in lemon juice.
 * 2) Bring the mixture to a boil, stirring with a fork the whole time. When it starts boiling, reduce heat to low and keep stirring for another 10 minutes. Remove mixture from the heat, and let cool.
 * 3) Brush some melted butter over each sheet of phyllo dough.
 * 4) Butter a pan, and place half the phyllo in it sheet by sheet. Pour the cooled semolina custard over the pastry, and cover with the remaining buttered phyllo.
 * 5) Sprinkle pastry with a bit of water.
 * 6) Bake pastry at 356°F / 180°C / Gas mark 4 for 1 hour. Let cool.

Syrup

 * 1) Combine water and sugar in a saucepan.
 * 2) Bring to boil, and simmer for about 10 minutes.
 * 3) Let syrup cool.
 * 4) Stir in the lemon juice once cooled in order to preserve the vitamin C.
 * 5) Brush the finished pastry with syrup, and serve.


text:
Strength of Materials/Introductory Concepts

Stress
Stress is defined on the average as the force divided by the area of the body over which the force acts. More precisely, we can talk about a stress at a point, or simply a stress, in the limiting case where the area (an infinitesimal area dA)   approaches zero. The normal stress occurs due to the infinitesimal force normal to the infinitesimal area, while shear stresses occur due to the infinitesimal force in the plane of the infinitesimal area. (Note: strictly speaking, stress does not exist. Elongation ($$ \epsilon$$) and stiffness (E) do.  Stress is a convenient mathematical construct that allows easy manipulation of formulae)

Consider a force, dF, acting on an elemental area, dA, at any arbitrary angle. τxx = dFx/dA is the normal stress, also denoted as σx. The shear stresses are given by τxy = dFy/dA and τxz = dFz/dA, where dFy and dFz are the y and z components of force dF respectively. These shear stresses are in the plane y-z.

Now, instead of an infinitesimal area, consider an infinitesimal volume at the point in question. Let this volume be a parallelepiped with the sides dx, dy, and dz. In this case there are, in general, nine non-zero stresses. They form a stress tensor represented by the following 3x3 matrix.

$$ \left(      \begin{matrix}        \sigma_x & \tau_{xy} & \tau_{xz}\\        \tau_{yx} & \sigma_y & \tau_{yz}\\        \tau_{zx} & \tau_{zy} & \sigma_z      \end{matrix}  \right) $$

Accordingly, in the plane stress case (taking, for instance, the plane x-y) we will have a four component stress tensor. Further, in view of the shear stress symmetry (τxy = τyx = τ), the above 3x3 matrix reduces to the 2x2 symmetric matrix representing a symmetric plane stress tensor.

$$ \left(      \begin{matrix}        \sigma_x & \tau \\        \tau & \sigma_y      \end{matrix}  \right) $$

The shear stress symmetry holds in the three-dimensional case. This makes the above 3x3 matrix--and hence the corresponding stress tensor--symmetric: only three of its six shear stress components are independent ones. Consequently, the stress tensor has in general six different components: three normal stresses and three shear stresses. As already shown above, in the two-dimensional case the stress tensor has only three different components: two normal stresses and one shear stress.

Suppose the body is in equilibrium and is acted upon by the forces fx and fy (per unit volume) in the x and y directions respectively. It then can be shown that the stress equilibrium equations, in Cartesian coordinates, has the form:

$$ \frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau}{\partial y} = f_x $$

$$ \frac{\partial \tau}{\partial x} + \frac{\partial \sigma_y}{\partial y}  = f_y $$

Thus, in the plane stress case, we have two equations involving three unknowns.

Now turn to an example of the simplest one-dimensional loading of a body. Consider a rod, pulled at each end, along the longitudinal axis.

If we take a section perpendicular to the axis, it is easy to see σx = P/A where P is the load and A is the cross sectional area of the section.

Now consider the case where the cross section is angled θ with respect to the longitudinal axis. In this case, we can resolve the force P along and perpendicular to the section. The force components are, P cosθ and P sinθ respectively. The area of the cross section is now A / cosθ. The normal stress is σθ = P cosθ/(A/ cosθ) = (P/A) cos2θ. Similarly, the shear stress is τθ = (P/A) sinθ cosθ.

Strain
The strain in a material is defined as the ratio of change in dimension to the original dimension, and is a dimensionless quantity. It is denoted by the symbol ε, where ε = (L − L0)/L0. Here L0 is called the gauge length.

The strain defined above is called the engineering strain or the nominal strain, and is different from natural strain or true strain, which is defined as

$$ \epsilon_n = \int_{L_0}^L \frac{dL'}{L'} = \ln \left(   \frac{L}{L_0}  \right) = \ln (1 + \epsilon) $$

Note that the engineering strain and natural strain are the same for small values of ε, and for the most part for the kinds of loads and displacements in this book, the differences are not relevant. Also, it is not uncommon to use the symbol e to represent engineering strain and the symbol ε to represent true strain.

In the infinitesimal case, we have, strain εx = du/dx, where du is the change in length for the segment dx. Thus, the total change in length is given by Δ = ∫0L du = ∫0L εx dx.

Hooke's Law
Hooke's law states that the stress is linearly related to strain for some materials. This is an empirical law by Robert Hooke, who observed this behavior in springs. Thus, for 1 dimension we have,

$$ \sigma = E\epsilon $$

where E is the constant of proportionality called Young's modulus. Note that we have considered this value of E to be the same in all directions. Materials with whose properties don't have a directional variation are called isotropic materials. Materials which have different properties in different directions are called anisotropic. The most common cause of anisotropy is the crystalline nature of materials. However, most common structural materials do not have the same crystal orientation over large ranges. Another cause for anisotropy is the kind of processing done on a material. Some processes like drawing tend to create stresses in a particular direction.

Applying Hooke's law to the definition of length change, we have,

$$ \Delta = \int_0^L\frac{\sigma}{E}dx = \int_0^L\frac{P}{AE}dx $$

Or,

$$ \Delta = \frac{PL}{AE} $$

From mechanics we know that a spring has linear variation of extension with force, the constant of proportionality usually denoted by k. Thus, the spring constant for a beam under axial load is AE/L. This can be extended to components of different shapes, so that a structure is an assembly of springs arranged in a complicated manner.

Poisson's Ratio
The longitudinal strain is linearly related to lateral strain in many materials. The ratio of the lateral to the longitudinal strain is called Poisson's ratio and is denoted by ν.

$$ \nu = -\frac{\epsilon_{lat}}{\epsilon_{long}} $$

The value of Poisson's ratio varies from material to material, and is about 0.30 for most common metals.

Recently,the discovery of re-entrant materials has excited much interest. These are unique honeycombs and foams which exhibit negative Poisson's ratios. The allowable range of Poisson's ratio in three dimensional isotropic solids is from -1 to 1/2 (though some polymer foams have been found to exhibit a value of 1.0).

Temperature Effects
Materials change length due to changes in temperature. To quantify this value, the length and temperature must be referenced to a determined temperature. The rate of change is quantified by the coefficient of linear thermal expansion, denoted by variables, CoF  &  α, By definition of strain, we have, ε = α ΔT, where ΔT is the change in temperature (compared to a given reference point). If we have a structural member between rigid supports, a strain of ΔT develops within the member. There is also an associated stress, which is referred to as a thermal stress, however no stress is created in a member whose length changes due to a temperature change but is unconstrained. Structural members whose cross-section is composed of materials with differing CoFs, present a much more complex problem.

Energy Stored due to Deformation

When a rod is compressed, the work done on it is stored as energy. In the above figure, the energy stored is shown as the shaded area. Now, we know that the energy stored in a spring is 1/2 k x2, for a spring constant of k and extension x. For a rod, whose constant is AE/L and extension is PL/AE, the energy stored is 1/2 P2L / AE. From the above equation, we see that it is better to use long bolts rather than short ones as they will have lower peak load P for the same diameter and material, without doing any dynamical analysis.

Elastic and Plastic Deformation
So far, we have considered members which deform elastically when a load is applied. Some of them obey Hooke's law, so that the relationship between stress and strain is linear. Elastic deformation refers to the ability of the material to regain its original shape after the external load is removed. However, we know that for large loads, the material deformation is permanent, and this is called plastic deformation. Metals which can undergo plastic deformation are called ductile. Materials which cannot undergo plastic deformation are called brittle.

Materials for which the elastic region is very small are called plastic materials. The above image shows the idealization for such a material. The material flows when a certain stress state is reached.

Materials which are elastic for large deformations are called elastic materials. The above image shows an ideal elastic material--in this case, it also obey's Hooke's law, and is called a linearly elastic material. Note that a substance can be elastic and still not obey Hooke's law (non-linearly elastic).

Most metals undergo elastic deformation for small loads and deform plastically for larger loads. There is certain amount of increase in stress when the material deforms plastically for most metals. However, the elastic plastic approximation shown above is good enough simple analysis.

Note that material which exhibit such behavior will show what is known as elastic recovery, where the sample will show some tendency to the earlier state. In the above image, the amount of recovery is Δ after the load has been removed. Note that there is a residual deformation in the material in the unloaded state.

Some materials show increased stress during plastic flow, with a phenomenon called strain hardening. In this case they can be approximated with a stress strain curve with two slopes, but this kind of analysis is rarely used in practice as most structural deformation is in the elastic zone, and plastic flow will most likely be considered a failure mode.

Shear Strain and Modulus

Earlier we introduced stress-strain relationship for materials under an axial load, where we found that the stress σ was proportional to the strain ε. If we plot the shear stress τ against the shear strain γ, we get a straight line for material which obey Hooke's law. Thus, Hooke's law can be extended to shear stress using the following equation: τ = Gγ, where G is the shear modulus, and γ is the shear strain.


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Talk:Strength of Materials/Introductory Concepts

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Strength of Materials/Unsorted topics

Compatibility Conditions
In an Engineering Mechanics course, we use force equilibrium to solve for forces in individual members. However, there were cases where it was not possible and such problems were considered statically indeterminate problems. They require some constitutive equations to solve completely, and Hooke's law is such an equation.

For instance, consider a column fixed at both ends, and a load P applied at the center. Now, we know that reaction forces R1 and R2 act at either ends. Further, using force equilibrium, we have R1 + R2 = P. Here we have two unknowns, R1 and R2, but only one equation. We can apply Hooke's law, with displacements Δ1 and Δ2, where Δ1 = R1L/2AE and Δ2 = R2L/2AE. The second equation is Δ1 = Δ2, since the supports are fixed. Thus, we can solve the problem.

Finite Element Method
Consider a rod with different cross sections Ai and lengths Li, so that each section has an equivalent spring constant of AiE/Li. If the forces at each node (where the area changes) are Fi, then we can write equations in terms of displacements of each node and the spring constant of each section for the force Fi. If the beam is fixed at both ends, then the forces can only act on the other nodes.

In the general case, if you have loads being applied to connected thin rods, you can use this method to solve for forces and deformations. The whole structure can be discretized, and the nodes numbered so that we get the matrix [kij].

This procedure of breaking down a complicated problem into a large number of simple ones is called the Finite Element Method.

The individual pieces are called elements, and in the above case we have used the line element. For a general 2-D problem, we would need to use triangular elements at least. The matrix for a typical problem will contain hundreds of such elements and the size of the matrix be same as the number of elements. However, as most of the members are zero, the matrix is called a sparse matrix, and it can be solved much faster than the typical matrix you would encounter, say, in a mathematics course.

Saint-Venant's Principle

So far we have been dealing with point loads acting at the ends of beams. We have assumed that the stress is the same as σav = P/A. However, this is not true close to the point of application of the load. Saint-Venant's principle (by the French elastician Saint-Venant) states that the stresses remote from the point of application of the load are not affected by the precise behaviour of the structure close to the point of application of the load. In the drawing shown, this means that although the stress field at A-A might be hard to calculate, at B-B the stress can be approximated as P/A. S-V's principle is consistent with many years of experiment and analysis, but has not been proven.

Stress Concentration Factors
So far we have assumed that the stress at all points in a cross section is the same, and by and large that is correct. However, there is an increase in stress near the point loads and discontinuities. The peak loads are a function of the average stress calculated by methods stated previously. The correct peak loads can only be obtained by methods like finite element analysis. However, charts have been published for several typical scenarios. As mentioned earlier, the peak stresses σp is a function of the average stress σav. Now, we have σp = K×σav, where K is called the stress concentration factor and can be looked up from tables for common scenarios. The study of stress concentration factors is also a design guideline, and indicates the types of features that you should avoid in a real world situation. For instance, the stress concentration factor for a fillet radius increases rapidly beyond r < 0.1 d, so you should specify large fillet radii for your shafts or eliminate the change in diameter.

The above schematic shows the possible values of the stress concentration ratios K for a plate with a hole in it. The stress at the narrowest section is calculated based on d = D − 2r, where r is the radius of the hole and D is the width of the slab. As can be seen, the stress concentration is not significant till about r/d = 0.3.

Note that local stresses might not be as high as that implied by the stress concentration factors due to plastic flow of the material, since the stress concentration analysis is done for elastic flow.

Generalized Hooke's Law
Poisson's ratio shows the amount of change in dimension that occurs in another direction when a certain change occurs in one direction. The strain in each direction is the superposition of all the strains--the direct strains and the ones due to Poisson's ratio. We can therefore state the generalized Hooke's law as follows:

 * $$ \epsilon_x = {\sigma_x \over E} - {\nu\sigma_y \over E} - {\nu\sigma_z \over E}$$

 * $$ \epsilon_y = {\nu\sigma_x \over E} - {\sigma_y \over E} - {\nu\sigma_z \over E}$$

 * $$ \epsilon_z = {\nu\sigma_x \over E} - {\nu\sigma_y \over E} - {\sigma_z \over E}$$

 * $$ \gamma_{xy} = {\tau_{xy} \over G} $$

 * $$ \gamma_{yz} = {\tau_{yz} \over G} $$

 * $$ \gamma_{zx} = {\tau_{zx} \over G} $$

Note that these equations apply only to isotropic materials (steel, aluminum, copper, etc.) whose elastic properties are identical in all directions.

Relationships Between Moduli
Consider an element which has stresses σ1 and σ2 acting on it along the x and y directions respectively. Now consider further that σ1 = σ and σ2 = −σ.

The shear modulus can be expressed in terms of the Young's modulus:

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

Consider a body undergoing compression, so that the strains are εx, εy, and εz respectively. The fractional change in volume is thus εx + εy + εz. Applying the generalized version of Hooke's law, we have, the bulk modulus k is given by

$$ k = \frac{E}{3 \left(   1 - 2\nu    \right)} $$

Pressure Vessels
Vessels, tanks, and pipelines that carry, store or receive fluids, at a pressure different than its surroundings, are called pressure vessels. Pressure vessels usually are thin walled structures containing fluids at high pressure. The design decision is to find the value for the thickness t of the pressure vessel for a given maximum pressure p. Since the thickness is much smaller than the radius of the vessel r, the inner and outer radius are taken to be equal. For the longitudinal direction, we have the force due to pressure πr2p. This is balanced by the stress developed in the walls, acting on the circumference of the circle 2πrtσ. Thus the stress in the longitudinal direction is pr/2t. For the tangential direction, the pressure force is π2rl and the stress force is 2tlσ. The stress is pr/t.

Brittle materials
When for a particular property a material shows a large scatter in measured data (e.g. strength measurement results of brittle materials, or of threads with important variations in thickness), a characteristic value for, or the weakest link of such a property can be obtained by analyzing the measured data on the basis of three parameter Weibull statistics.

title: Strength of Materials/Unsorted topics


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Talk:Strength of Materials/Unsorted topics

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Strength of Materials/Torsion

Torsion acting on a long bar tends to twist it in the direction of the torque. The analysis is performed on a section perpendicular to the axis, and the sum of the internal resisting torque is set equal to the external torque acting on the system. The following assumptions will be made for the bars studied in this chapter:

 * 1) Plane sections remain plane after the torque is applied.
 * 2) The shear strain γ varies linearly in the radial direction.
 * 3) The material is linearly elastic, so that Hooke's law applies.

Torsion Formula
We want to find the maximum shear stress τmax which occurs in a circular shaft of radius c due to the application of a torque T. Using the assumptions above, we have, at any point r inside the shaft, the shear stress is τr = r/c τmax.

∫τrdA r = T

∫ r2/c τmax dA = T

τmax/c∫r2 dA = T

Now, we know,

J = ∫ r2 dA

is the polar moment of inertia of the cross sectional area..

Thus, the maximum shear stress

τmax = Tc/J

The above equation is called the torsion formula.

Now, for a solid circular shaft, we have,

J = π/32(d)4

Further, for any point at distance r from the center of the shaft, we have, the shear stress τ is given by

τ = Tr/J

We only consider the torsional loading of simple circular shafts in this analysis, ie cylinders or non-eccentric tubes without splits. Circular shafts are most commonly used as torque carrying members in machinery with rotating parts (like drive shafts of motors). This is fortuitous, as the analysis of non circular members under torsion is not simple to perform analytically.

Angle of Twist

The above image shows the twist of a shaft acted upon by a torque T at one end. We know that the shear angle γ is given by

γ = τ/G

For a shaft of radius c, we have

φ c = γ L

where L is the length of the shaft. Now, τ is given by

τ = Tc/J

so that

φ = TL/GJ

The angle of twist φ for a circular shaft acted upon by a torque Tx at a point x along its axis is given by:

$$ \phi = \phi_B - \phi_A = \int_A^B\frac{T_x}{J_xG}dx $$

where Jx is the moment at section x. Note that you can use the torsion formula for shafts with slowly varying area as long as they are circular.

title: Strength of Materials/Torsion


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Strength of Materials/Loading of Beams

One of the areas where solid mechanics as discussed in this book is most effective is in the case beam loading. The loads on a beam can be point loads, distributed loads, or varying loads. There can also be point moments on the beam. The beam itself is supported at one or more points. The conditions at the support depend on the kind of support used. If the support is a roller, it can only have a reaction perpendicular to the motion of the roller. If the support is a pin, it cannot carry a moment. If the support is fixed, then it can have a reaction in any direction and support a moment as well.

In the above image, a simple beam is loaded at the center by a load P. It has a pinned contact at one end, and a rolling contact at the other end.

The above image shows an ideal moment acting at the center of a beam. An ideal moment is one which is not associated with a force.

The general method is analyzing beam problems is to find the loads, reactions, and moments, and come up with the values for the loads and moments at each section. This, in general will be a piecewise function of the distance along the beam.

For loads, we set up axes and this decides the sign of the force. For moments, we set the convention that the clockwise moment is positive.

Examples

1. Consider the case of the beam discussed earlier shown above.

The beam has reactions R1 and R2 acting on each of the supports.

R1 + R2 = P

Applying symmetry, we have,

R1 = R2

So that,

R1 = R2 = P/2

The above image shows the shear diagram for this problem. Note that the positive and negative directions are conventions, but it is important to choose one direction for positive shear and stick to it. As can be seen, the shear value changes at the point of application of the load.

The above image shows the moment diagram for the beam. The moment varies linearly from the supports to the middle. (The value at the middle should be PL/4). Note that the moment is maximum at the middle and zero at the ends, so that the effect of the moment is maximum at the center. In later chapters we will see that the stresses and strains due to moments are the most important ones for beams.

Exercises

1. In the above beam, find the reactions in the supports and the shear force at a position x. Also, find the moment at that point.

2. The above beam shows loading by two separate point loads. Find the shear and moment at location x from one end of the beam.

3. Find moment at each point along the beam with a point load and moment acting at two different points. How can a point moment occur in practice?

Calculus for Solving Beam Problems
Methods of calculus can be used to deal with continuous load functions. However these methods can be extended to point loads and moments by the use of the Dirac delta function.

The above shows a beam with uniform load per unit length w. Such loads are used to model the self weight of the beam where it acts uniformly throughout its length. This can also be used for the loading of, say, a bridge due to all the vehicles on it. The individual point loads acting through the tires can be modeled as a continuous load if the number of vehicles is large.

Consider a beam with load per unit length q(x) for any point x along its length. Consider an elemental length dx of the beam. Applying force equilibrium for this segment, we have,

V + dV = q dx + V

where V is the shear at the point x. Thus, we have the relation

dV/dx = q

Applying moment equilibrium to the same segment, we have,

M + dM − V dx − M − q dx dx/2 = 0

Neglecting the second order terms in dx, we have,

dM/dx = V

Or,

d2M/dx2 = q

The above differential equations can be integrated with appropriate boundary conditions to get the shear and moment at each point.

Examples

Consider a beam of length L supported at its ends by two pins, with a uniform load per unit length of w.

The total load on the beam is thus wL. Thus, the reaction at each support is wL/2.

We have, the general relation for shear

$$ V = \int_0^x q dx + C_1 $$

The shear at origin is just the reaction at that point (=wL/2). If we take the vertical direction as the positive direction, we have, the shear at the origin is wL/2. Also, q = −w. Thus, we have, V = −wx + wL/2 for any point x.

The Moment at the centre should be wL^2/4

Similarly, we have the moment M

$$ M = \int_0^x V dx + C_2 $$

It is easy to see that the moment at the origin is zero. Thus, we have M = w x (L − x)/2.

Calculus for Point Loads
The above works well for well for continuous, smooth load functions. But in real world situations, we have to deal with point loads and moments. Thus we use the Dirac functions for a point load P at location a as

$$ P\left\langle x - a \right\rangle^{-1} $$

and for a point moment M at location a as

$$ M  \left\langle x - a \right\rangle^{-2} $$

Now the previously stated equations can be used under the rules for the Dirac function. The function itself is defined as

$$ \left\langle x - a \right\rangle^n = \left\{ \begin{matrix} 0 & x < a\\ (x - a)^n & x > a \end{matrix} \right. $$

Consider a beam of length L with supports at both ends and a point load P at the center. From the Dirac definition, we have,

$$ q =  -P\left\langle x - L/2 \right\rangle^{-1} $$

The shear at one end is just the reaction P/2. The shear at any point is given by

$$ V = \int_0^x q dx = \int_0^x -P\left\langle x - L/2 \right\rangle^{-1} dx = -P \left\langle x - L/2 \right\rangle^{0} + P/2 $$

The moment at one end is zero. The moment at any point is given by

$$ M = \int_0^x V dx = -P \left\langle x - L/2 \right\rangle + \frac{Px}{2} $$

Applying the definition of the Dirac function, we have, the shear

$$ V = \left\{ \begin{matrix} P/2 & x < L/2\\ -P/2 & x > L/2 \end{matrix} \right. $$

$$ M  = \left\{ \begin{matrix} \frac{Px}{2} & x < L/2\\ \frac{P(L - x)}{2} & x > L/2 \end{matrix} \right. $$

Note that the above is a special case of the general method called Laplace Transforms.

Exercises

1. Find the shear and moment in the above beam using Dirac methods.

2. The beam shown above has two loads which can be modeled as shown. Find the shear and moment at any point along the beam.

3. Consider the beam shown above with an overhang. Find the shear and moment at points along the axis.

Stresses and Strains in Beams
Given the loads and moments at each cross section, we can calculate the stress and strain at each location.

Consider the bending of a slender beam (one for which the cross section is much smaller than the length). A moment acting on the beam causes a deformation called flexure. Let the radius of the osculating circle of the beam be ρ. Consider an elemental length ds in the neutral plane (for which the deformation is zero). This element subtends an angle θ at the center of curvature, so that

ds/dθ = ρ

If we move a distance y along the radius, we have the length of the arc subtended would be (ρ − y) dθ. Thus the elemental extension would be y dθ. For a small enough curvature, we have the distance along the neutral plane would be the same as the initial undeformed length, dx. This gives us the axial strain at any point x as

εx = y/ρ

Also, using the Hooke's law, we have,

σx = E εx = Ey/ρ

Now, we know that a pure moment M acts on the beam, so that there is no axial force. Or,

∑ Fx = 0

∫ σx dA = 0

∫ y dA = 0

The above equation gives us the location of the neutral plane.

Further, applying the moment conservation part of the equilibrium relations, we have,

M − ∫ σx dA y = 0

Now, we know,

σx = Ey/ρ

and

I = ∫ y2 dA

where I is the moment of inertia, so that

M = E/ρ I

Or

E/ρ = M/I

Thus, we have the expression for the stress due to a pure moment as

σx = My/I

Suppose the maximum value of y is c (the distance from the neutral plane). Then, we have,

σmax = Mc/I

The quantity I/c is called the section modulus of a beam and is denoted by S.

σmax = M/S

Beams with Arbitrary Moments
So far we have considered beams which were acted upon by moments in the Mz. Suppose there is another moment in the My direction. Note that a moment in the Mx direction would simply be a torsional effect as the axis is in the x direction. Now, it is easy to see that the combination of moments is, in fact, equivalent to a moment acting on a beam of arbitrary cross section. It is fairly straightforward to show that, in this case, the stress at any point (y, z) on the face of the section for the beam is a function of Iz, Iy, and Iyz.

Shear in Beams
Earlier, we saw the methods to calculate the shear forces on a beam. Now we can analyze the stresses due to shear forces, like we did for stresses due to bending moments. We showed that the shear force V is given by V = dM/dx, where M is the moment acting at the point x.

title: Strength of Materials/Loading of Beams


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Strength of Materials/General State of Stress

Principal Stresses
The general state of stress can be represented by a symmetric 3 x 3 matrix.

It is always possible to choose a coordinate system such that all shear stresses are zero. The 3 x 3 matrix is then diagonalised, with the three principal stresses on the diagonal, and all other components equal to zero. The three principal stresses are conventionally labelled σ1, σ2 and σ3. σ1 is the maximum (most tensile) principal stress, σ3 is the minimum (most compressive) principal stress, and σ2 is the intermediate principal stress.

Stress Invariants
$$ I_{1} = \sigma_{3} + \sigma_{2} + \sigma_{1} $$

$$ I_{2} = \sigma_{1} \sigma_{2} + \sigma_{2} \sigma_{3} + \sigma_{3} \sigma_{1} $$

$$ I_{3} = \sigma_{1} \sigma_{2} \sigma_{3} $$

Hydrostatic Stress
The so-called hydrostatic stress, σh, is given by:

$$ \sigma_{h} = \left. \frac {\sigma_{1} + \sigma_{2} + \sigma_{3}} {3} \right. $$

Deviatoric Stresses
normal stress - pressure where pressure = average stress

Mohr's Circle
Consider the two dimensional stress condition where the stresses are σx, σy, and τxy. We have, for another set of orthogonal axes  x'-y'  at angle θ with x-y, the stresses are

$$ \sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau\sin 2\theta $$

$$ \tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta $$

From the above equations, we can see that for any stress states given by σx, σy, and τxy, we can find a value of θ such that the value of σx' is maximum. This value is called the principal stress σ1 (for maximum) or σ2 (for minimum).

The principal stresses are given by

$$ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left(   \frac{\sigma_x - \sigma_y}{2}    \right)^2 + \tau_{xy}^2} $$

and the maximum shear stress is given by

τmax = (σ1 − σ2)/2

From the definitions for σx and τx'y, we have

$$ \left(    \sigma_{x'} - \frac{\sigma_x + \sigma_y}{2}  \right)^2 + \tau_{x'y'}^2 = \left(   \frac{\sigma_x - \sigma_y}{2}  \right)^2 + \tau_{xy}^2 $$

In the σ-τ graph, this is a circle with center on the x axis, and the distance of the center from the origin is given by (σx + σy)/2 and a radius given by

$$ R = \sqrt{ \left(   \frac{\sigma_x - \sigma_y}{2}    \right)^2 + \tau_{xy}^2} $$

This circle is known as Mohr's circle, and is useful for visualizing the stress state at a point.

The above figure shows Mohr's circle for a stress state (σ, τ). The center and radius of the circle are obtained from equations stated above. The other stress σy can be read off by the point diametrically opposite the (σ, τ) point. The stress at any plane can be found using simple geometrical constructs.

Mohr's Circle for Common Cases

The above shear stress in this case is σ1/2.

A liquid is one which by definition cannot take a shear. Thus the Mohr-circle diagram is just a point. (The dot should be on the negative side of the axis as the liquids can resist only compression)

For pure shear, the Mohr circle is centered at the origin.

Failure Criteria
In the case of isotropic materials, the state of stress at any point of the body is completely defined by the triad of the principal stresses. Now that we are able to transform stresses to get the principal stresses, we can use these stresses to consider some of the criteria (theories) postulated for the failure of materials in two- and three-axial states of stress usually based on experiments on yielding and fracture of materials in the uniaxial state of stress. According to such experiments, the kind of failure depends on the type of material. Failure of ductile materials (most of metals) occurs when the elastic limit is reached and yielding commences. Failure of non-ductile materials (e.g., cast iron, concrete) occurs by brittle fracture.

Maximum Shear Stress Criterion
For ductile materials, one failure theory is that of maximum shear stress. We know that the maximum shear stress is given by τmax = (σ1 − σ2)/2. The yield stress, σy can be determined by uniaxial tensile tests. Thus, if the maximum shear stress theory is valid, failure occurs when the maximum shear stress reaches σy/2.

In the above image, the material will fail if the stress state is outside the shaded region.

Maximum Distortion Energy Criterion

If the maximum shear stress theory is valid, failure occurs when the maximum shear stress reaches $${\frac {\sqrt{2} \sigma_y} {3}}$$.

Failure of Materials
The actual failure mode of each material is unique, though certain criteria can be applied to classes of materials. mode of failure. fatigue failure,


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Strength of Materials/Introduction

Solid Mechanics is the study of load carrying members in terms of forces, deformations, and stability. There are three length scales: Atomic scale, Micro scale, Continuum scale. In this book, we take the continuum mechanics approach. We will consider the material to be homogeneous and isotropic (unless specified). Thus the material properties are even when we consider infinitesimal areas and volumes. The alternative approach is to build up material properties from basic equations relating atomic forces and interactions, and extending it to larger sets of such entities (e.g., ab-initio, molecular dynamics). Other approaches include micro-scale theories like dislocations to explain behavior of materials, especially for plasticity. However, the approach in this book is applicable to a large variety of practical problems, where we deal with structural members, usually under elastic deformation.

The images on the right show a body which is acted upon by external forces $$ \mathbf{F}_1\,$$, $$ \mathbf{F}_2\,$$, etc. The supports provide reaction forces ($$ \mathbf{R}_1\,$$, $$ \mathbf{R}_2\,$$) so that the body is stationary.

In addition to external forces, which act at points on the surface, we can also have forces which are body forces ($$ \mathbf{b}\,$$), i.e., they act on each point in the body. Forces acting at a distance (Eg: Gravity, Magnetic) can be examples of body forces.

Consider a body which has several external forces acting on it. If a body is in static equilibrium, then every section of the body should also be in static equilibrium. Now, consider an elemental area dA in that section. Let the force acting on the elemental area so that the whole body is in equilibrium be dF. We can resolve this force in three mutually perpendicular directions. Let the component of the force normal to the surface be dFz. Let the components in the plane of dA be dFx and dFy respectively. Then for the body to be in static equilibrium, we should have:

$$\sum {\mathbf{dF}_i} = 0, i=1,2,3$$

There are three other conditions concerned with moments for completely defining static equilibrium. We will look in to it in the later chapters.

title: Strength of Materials/Introduction


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IB History of the Americas/Chapter 11

The Great Depression in the Americas (Topic 11)
Background to the Great Depression
 * The US’s production in the late 20s were sky-rocketing, stimulated by the automobile, radio, film and other industries. There was a new plateau of prosperity, but few saw what was coming.
 * Prices on the stock market were rising at incredible rates.
 * The crash was partially triggered by the fact that the British had raised interest rates to try and bring back capital lured abroad by American investors.
 * Foreign investors and local speculators decided to sell their “insecure” stocks which led up to a selling frenzy.
 * On October 29th 1929, 16,410,030 shares of stocks were sold; the day was also known as the Black Tuesday.
 * As the supply of stocks in the market increase the price decreases. With low prices for stocks the demand for them decreases and there is less activity in the stock market. People with stocks lose out as the value for their stocks decrease and there is no demand.
 * By the end of 1929 stockholders had lost $40 billion
 * The stock market crash triggered a world-wide depression and by the end of the 30s more than 4 million workers in the US were jobless.
 * Over five thousand banks collapsed in the first three years of the depression and took with them the life savings of tens of thousands of people.

Causes of the Depression
 * The reason for the depression was overproduction in the agricultural and industrial sector.
 * There was too much production and not enough consumption.
 * More money was going into the hands of the wealthy that invested and increased productions, but not enough to the workers in terms of salaries. There was a lack of purchasing power.
 * Another reason was credit through installment; these methods caused many consumers to spend more than they could afford thus bringing them into bankruptcy.
 * The rise in unemployment due to new labor-saving technologies also played a part in the depression.
 * Financial problems abroad also helped to trigger the depression. There was the lack of international trade due to problems facing European nations.
 * Europe had not recovered fully from WW1 and the financial collapse aided to the depression.
 * The Hawley-Smoot Tariff of 1930 was designed as a protective tariff that placed a 38.5% duty on nonfree goods; it was designed to protect the local agricultural industry. The tariff also helped in the decline of international trade.
 * Hoover’s Fight against the Depression
 * Initially began with “Laissez-Faire” policies but moved to more intervention type ones towards the end of his presidency.
 * He fought against schemes that he saw as “socialistic”.
 * Hoover sought to use public works to get the economy running again. He was granted $2.25 billion from Congress for the projects.
 * Hoover Dam (1930 – 1936)
 * In 1932 the Reconstruction Finance Corporation (RFC) was started. It was a government lending bank that provided indirect relief by helping insurance companies, banks, agricultural organizations, railroads and governments. “Pump-Priming”
 * Congress passed the Norris-La Guardia Anti-Injunction Act in 1932 to give more rights for labor. Under the act there would be no “anti-union” acts and unions had the right to do boycotts, strikes and peaceful picketing.

FDR and the New Deal
 * FDR’s New Deal programs aimed at three R’s – Relief, Recovery and Reform.
 * Congress gave FDR extraordinary blank-check powers for his New Deal program.
 * The Emergency Banking Relief Act of 1933 gave FDR the power to regulate bank transactions and foreign exchange and also to reopen solvent banks.
 * He used the radio for his “fireside chats” in which he persuaded the 35 million Americans that it was safe to put their money in reopened banks. People had then started to deposit and banks were opening up once again.
 * The Glass-Steagall Banking Reform Act created the Federal Deposit Insurance Corporation which insured deposits of up to $5000, this further increased the confidence in depositing money and ended several bank failures.
 * The gold standard is a monetary system in which the normal economic unit of account is a fixed weight of gold coin. Under the gold standard, banknote issuers can redeem notes, upon demand, in that amount of gold coin.
 * At the time the gold reserve was declining and to prevent any sort of chaos he forced all private holdings of gold to be surrendered to the Treasury in exchange for paper notes. The value of the currency would become unmanageably high since the price of gold was increasing due to its growing scarcity. He then took the nation off the gold standards to maintain a more “managed economy”.
 * With the new “managed currency”, that was a result of the US leaving the Gold Standard, was aimed to bring about inflation in the economy. Since the price of the dollar was not allowed to increase with the price of gold the prices of goods in the market rose but not the price of dollar; thus resulting in inflation where the price of goods increase. FDR believed that the increased levels of inflation would stimulate new production.
 * However, he brought the country back later to Gold Standards for purposes of international trade only.

title: IB History of the Americas/Chapter 11


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